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PHILOSOPHIC PRACTICAL MATHEMATICS
DESIGNED FOR THE USE OF
ACCOUNTANTS, MERCHANTS, BUSINESS MEN,
PRIVATE LEARNERS, HIGH GRADE COMMERCIAL COLLEGES
AND NORMAL SCHOOLS.
THE PHILOSOPHIC SYSTEM
Is Presented and Applied to the Solution of COMMERCIAL, FINANCIAL, MECHANICAL and
PROFESSIONAL CALCULATIONS and PROBLEMS, covering
the various fields of Business Life.
{
IT SOLVES AND DISCUSSES HUNDREDS OF PRACTICAL QUESTIONS-COMPLEX
AND COMPOUND–NEVER BEFORE PRESENTED IN ANY
SIMILAR TREATISE.
It Sparkles with New Thought, Advanced Methods, and Higher Work. It presents the Rarest
Gems of the Science of Practical Mathematics, and it Teaches that a NEW TRUTH
is better than an OLD ERROR, however ancient or renowned.
By GEO. souT.E.
Tractical and Consulting JAccountant, Lecturer on Commercial Sciences, Tresident of Soule Commercial
College and Literary Institute, Fellow of the Institute of Accounts of JNew York, &Member of the
JAssociated JAccountants of JNew Orleans, and Director in and President of various Corporations.
JAuthor of Soule’s “Intermediate Philosophic Arithmetic,” “Contractions in Ö(umbers,”
“Thilosophic Drill Problems,” “Analytic and Philosophic, Commercial and Exchange Calculator,”
“&Manual of JAuditing,” and “JNew Science and Practice of JAccounts.”
FIFTH EDITION, REWISED 1905.
PUBLISHED BY THE AUTHOR.
NEW ORLEANS. *
ENTERED ACCORDING TO ACT OF CONGRESS IN THE YEARS 1872-1895-1898-1901-1905
By GEO. Soulé.
IN THE OFFICE OF THE LIBRARIAN OF CONGRESS, AT WASHINGTON
ALL RIGHTS RESERVED,
COMPOSITION ELECTROTYPED
BY BY
MALUS & HOFELINE. T. A. SLATTERY.
PRINTED AND BOUND
BY
L. GRAHAM & SON, LTD.
PREEYACE.
Twenty-four centuries ago the renowned Grecian mathematician, Pythagoras, an–
nounced to the world his first great discovery in Mathematics, and since that eventful time,
the science of Mathematics has been by master minds enriched with new discoveries until,
as a science, it stands the equal of any, and is as unerring in its results as the laws which
govern the planets in their revolutions through the realms of space. But, while as a science,
Mathematics stands without a peer, and in general usefulness, without a superior, the first,
and by far the most generally used subdivision of the science, Arithmetic, is uniformly pre-
sented in the text books of the day, by methods which are general'y admitted to be more or
less defective in logical reasons for the operations, and in clear elucidations of the subjects.
In addition to these defects, the various treatises on the subject, almost without an
exception, are deficient in practical problems, and are by far too brief and of too low a
grade to meet the commercial and business demands of this age of progress.
With a view, therefore, to remedy in a measure the imperfections named, the author
presents a thousand page work on Philosophic Practical Mathematics, replete with all known
abbreviated methods and with practical problems covering the various fields of trade,
finance, mechanics and business all of which are fully elucidated by operations, philosophic
reasoning and pertinent discussions. Thus the work supplies a long-felt want, and is of
great value as a book of reference and authority for all classes of business men and
practical book-keepers. For private learners, teachers, commercial colleges and literary
institutions, it is believed to have no equal and to possess the crown of superiority.
The Science of Numbers, or Practical Arithemetic, is one of the queenly products of
the human mind. With man, it has come down the evolving centuries, compounding itself
with his thought and occupation, and unfolding its beauties as the human race advanced
on the planes of civilization. Receiving new inspirations and contributions from profound
thinkers as the ages passed by, it has attained to the dignity of a queenly science before
whose throne the logic of mind and the industries of the world worship. The erudite
Hindoo and the learned Egyptian communed at its infant shrine. Before it Thales and
Pythagoras reasoned with uncovered heads; at its feet Plato laid the genius of his great
mind; and Archimedes and Aristotle employed their eloquence and philosophy in unfold-
ing its mysteries. Newton and La Place adorned it with modern thought, and Pestalozzi
leavened it with the great principle of analytic reason.
But while the mathematicians of ancient fame were enthusiastic regarding numbers,
and spent much time in fanciful contemplation and speculation upon their properties, they
did but little for the advancement of practical arithmetic. Their numerical recreation and
labors were largely with magic squares and circles, while their philosophic talents were
bestowed largely upon the noble science of geometry. This science was so highly esteemed
by Plato that over the door of his lecture hall he wrote: “Let no one enter here who is
ignorant of geometry.”
Thus, until the time of Recorde, I500 to 1545; Ramus, I517 to 1572; Stevinus, I550
to I585; Newton, I642 to 1727; Leibnitz, I646 to 1716; and later Pestalozzi, 1746 to 1827,
arithmetic did not receive the attention due to its merits, and which modern business
methods now demand.
The Pestalozzian process of analytic reasoning has been, during the past seventy
years, utilized by more than a hundred authors. Colburn, Emerson, Hobart, Stewart,
McCormick, Ray, Greenleaf, Dodd, Perkins, Brooks, Davies, Robinson, Thompson and all
authors since 1810 have used in various modified forms, more or less, the analytic methods
of Pestalozzi, and thus they have moved on with the evolving world and performed a
measure of efficient missionary work in the pagan realms of practical mathematics.
The analytic method of Pestalozzi is the foundation of the golden Philosophic Sys—
tem presented in this and the author's other works on the Science of Numbers.
In the higher evolution of this system, the author of this work employs and combines
the three great processes of acquiring knowledge and eliciting truth, viz.: comparison,
analysis and synthesis. These three processes constitute the trinity of mathematics, and by
an ingenious use of them, the author is believed to have advanced the science to loftier
planes and to more rational methods than were ever before achieved. In no ancient or
modern work on numbers have comparison, analysis and Synthesis been woven into such an
extended chain of logical and philosophic reasoning as is herein presented.
| 4:4556
IV PREFACE.
... By the Philosophic System, all the difficult scientific statements of true proportion are
avoided, all the objections raised against the purely analytic method are obviated, the some—
what entangling cause and effect method is surpassed, and that monstrosity of a process
which “considers whether the answer is to be greater or less than the third term " is
anathematized and consigned to the shades of the dark ages of arithmetic.
By the Philosophic System, all the arbitrary rules that overload the faculty of
memory and prevent the expansion of the higher faculties of causality and comparison are
abandoned, and the reasoning organs of the mind are brought into action, thereby capaci-
tating the learner not only to produce the results of problems, but to observe fine distinctions,
reason logically, and deduce correctly. Thus like geometry, it serves to enlarge the pow-
ers of the mind and to qualify for high planes of usefulness, not only in the fields of mathe-
matics, but in all the vocations of life.
& The Philosophic System is believed to be the most valuable improvement yet made to
impart a thorough knowledge of the principles of numbers and to capacitate the student
to utilize the same in the practical affairs of business life. But notwithstanding , its
superiority and the fact that its advocates include many of the most profound mathematical
minds, yet, like every other improvement or discovery in education, commerce, art, or science,
it has some opponents and is regarded with indifference by those who are satisfied with the
non-progressive and non-reasoning methods of past ages.
For forty-eight years the author has labored with tongue and pen in the develop-
ment of the Philosophic System of Arithmetic, and has tested its superior merits in the
school room and lecture hall with over 18,000 students; and from a full knowledge of
its advantages, he conscientiously assures his co-laborers in the mathematical field of edu–
cation, that a more thorough knowledge of the science of numbers can be imparted, and in
far less time, by this system than by the usual methods and systems of work.
© It stands without a peer in the annals of mathematics, and it is believed that the day
is not far distant when its banner will wave in triumph from every spire, pinnacle and
dome of the Temple of Practical Mathematics.
. The method of disseminating or imparting arithmetical knowledge, that is in almost
universal use by the great army of authors and teachers, is the rule method, or in other
words, the non-reasoning method. By this method the principles involved are generally
lost sight of, and results without reasons are sought for. Being fully convinced of this
fact and knowing that the best results are to be attained only by thoroughly understanding
the principles of the science, the author has in this treatise labored most earnestly to
dethrone the venerable absurdities that have, as he believes, obscured the science of num-
bers for ages.
The entire work sparkles with the rarest gems of the science of numbers and
teaches that a new truth is better than an old error, and that facts and reasons are better
than fallacious theories, however ancient or renowned.
To obviate a formal introduction to the work, the subject matter usually classed
under that head has been presented in remarks and discussions throughout the work, in
connection with the various topics treated. By reason of this arrangement, the remarks
and discussions have been made more pertinent, and will be read at the time when the
learner most needs them.
The special features of the work and its many points of superior merit are so num-
erous and the number of subjects treated is so great as to preclude the propriety of giving
them space in the preface. . A cursory look through the alphabetical index following the
preface, will give full information thereof.
In the preparation of the work, the author has been guided by an experience of
forty-eight years as a teacher and lecturer on the science of numbers and as a practical
accountant and commercial lawyer in daily professional intercourse with all classes of busi-
ness men; and many of the intricate computations that have been . Submitted to him for
adjustment are herein presented and fully elucidated. . These problems and the solution
thereof constitute a special and most valuable feature of the work.
In concluding the preface, the author desires to extend his thanks to:
I. Miss Carrie McGuigin, his faithful amanuensis, for six years of efficient service
in the preparation of the copy. s
2. Mr. J. M. Buchee, thirty-five years his sincere friend and co-worker in the cause .
of practical education, for his cheerful assistance rendered in proof-reading and in work-
ing problems. To him is to be credited the preparation of Involution and Evolution, Square
and "Cube Root, Arithmetical and Geometrical Progressions and Annuities. His discus.
sions and criticisms on various matters in connection with the work have been valuable and
will be kindly remembered. r
To his sons, Albert Lee Soulé, Edward Everett Soulé and Frank Soulé, teachers
in the Soulé College, for assistance in proof-reading and in working problems. To A. L.
Soulé, is to be credited the work of preparing Continued Fractions, Savings Banks, Per—
mutations, Combinations. Probabilities and Chances.
PREFACE. V
In presenting this work to the public, the great tribunal that is to pass judgment
upon it, the author, , with the deepest convictions of the superior merit and value of the
philosophic system, indulges the hope that it may prove acceptable and he asks, that in
behalf of justice, in the interest of progress and the science of numbers, a decision may
not be rendered until a thorough examination shall have been made and its merits contrasted
with those of other systems. &
GEO. SOULE.
PREFACE TO EDITION OF 1905.
* A A
~–wr-w
In 1872 the Author published a work entitled Soulé's Analytic and Philosophic Com—
mercial and Exchange Calculator, which comprised about one-half the matter contained in
the present work. Five years’ labor was devoted to this treatise and it was received with
So much favor by Business Men, Accountants and Learners, that the Author was encour—
aged to enlarge the work and to include, elucidate and discuss many practical subjects and
problems from various fields of business mathemathics, that were not presented in the first
work. The new and enlarged work was given to the public in 1895 under the name of
Soulé's Philosophic Practical Mathematics. This work was revised and a new edition pub—
lished in 1898. In extent of subjects, in material, in original and in philosophic solutions
and discussions of problems, covering the whole field of Practical Mathematics, this work
is believed to be far in advance of any similar publication. In 1901 another edition was
published with several new problems.
The present edition, 1905, has been revised and enriched with the presentation and
solution of many new practical problems drawn from different lines of trade, finance,
mechanics, agriculture, general business, etc. Large numbers of these problems have been
presented to the Author, for solution, by accountants from all sections of the country, and,
as they are more or less complex coming from all kinds of business, they are gems of
varied hue in the crown of Practical Mathematics.
The brilliant conquests of commerce, the amazing magnitude of finance and exchange
and the higher demands of business, have produced a need for more contracted and rapid
systems for ordinary computations, and for a higher knowledge of the more intricate and
complex practical mathematical problems that are daily occurring in business affairs. To
meet this demand this work has been prepared and revised from time to time. It is now
tendered to the public in its improved and complete form, with the confident hope that
it will prove valuable to those for whom it has been written and published.
GEO. SOULF
New Orleans, January 2, 1905.

ALFF/AEET/CAL //VDEX.
A.
PAGES
Abbreviations, list of, ................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . e e e o e e e e e º e º e º e & Q & Q & P 38–39
Abstract number, def, of,................ . . . . . . . . . . . . . . . . . . . . . . . . . . . .................. to e º 'º º 26
, reasons for multiplying, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
, reasons for dividing, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200
Account sales, def of, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 480
, equation of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 643
Accounts current and interest accounts, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 653–665
showing interest method, product method, and equation method. g
Accounts current and interest accounts, in English money, ....... . . . . . . . . . . . . . . . . . . . . . . . . . 507, 661
, With brokers, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 705
, in German money, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 664
Actuaries’ table for life insurance, ......................................................... 536–539
4ddition, civil service method of, .................................................... • * * * * * * * * 46
def of . . . . . . . . . . . . . . . . . . . . . . . * * * * * * * * * * e s s e e s e e s e e s e º e º e s e e s e s e e s s e e s e e s a c e s a e º 'º' 40
grouping the figures in, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
horizontal, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
of decimals, ......... * . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223
of denominate numbers, ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . tº ſº º e º ſº e º e º ſº ºn tº gº © tº & © tº 267–269
of dollars and cents, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
of fractions, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175–180
of metric numbers, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . e ſº tº e º e º ºs º ºs º 936
of several columns at once, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
• Principle of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
Proof of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
, tables, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . e = s. sº tº e s e º ºs e º 'º e º ſº º e º e º e º ſº e º 'º e º 'º e º 'º 41, 42
Vicenary system of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
Adjustment of interest on partners’ accounts, .............................................. 845
“ the calendar, history of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237
“ losses in insurance, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512, 523, 525
“ insurance under the # loss clause, ........................................... 524
4 ( & & & 4 “ # Value clause, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525
& 4 “ “ “ 75% co-insurance clause,.............................. 524
& & & 4 & 4 “ average clause, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524
& 4 4 & marine loss, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524
( & & & mdse. loss, full insurance, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525–526
Admission of new partners, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 858
Ad-Valorem duty, def. of, . . . . . . . . . . . . . . . . . . . . . . . . . . . e e º e e s is e e s e e º sº e e s a e s is is e a e e s tº e º ºs e º sº * * * 500
Aliquots, table of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94, 139
Alligation, alternate and medial by several methods, . . . . . . . . . . . . . . . . . . . . . . . . . . © tº gº & s tº tº e º ºs e & 789–803
American experience table in insurance, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . .536–538, 541
Angles, def of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337
Angular measure, . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 252
Annual interest, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 607
ALPHABETICAL INDEX, VII
PAGES.
Annuities, ....... * @ e º 'º 6 e e º e º e & o e º e º & G & o e - to e º e º e e e º e º e e º e © e º e º 6 & e s is © & © e > * > 0 ºr a e s a e e s e s e e º e e º e 920–928
Apothecaries fluid measure, .................. & © tº c & © e º e º e o e * * * @ 9 • e º e e s e o e º e e'• * - - - - - - - - - 251
Weight, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250
Arabic notation, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ....... 30
Arbitration of exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 736–743, 774
Area, def of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338
Argentine republic exchange, . . . . . . . . ...................................................... 771
Arithmetic, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ....... 27
, philsophic system of, .......... • * * * - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - . . . . . . 26
Arithmetical complement, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
supplement, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
units of order,. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
Progression, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 901–908
Articles of agreement, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 806
Artificiers' work in surfaces,............................................................... 360
Assayers' weight, ..., . . . . . . . . . to º e º 'º e g º º & © e º e º 'º e º e º 'º e s e e º e e s e e s e s - e. e. e. e. e. e. e. e s e e o e s e e o e º e e s e e e 251
Assessment roll, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 489, 677
Assets, def of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . © e º 0 & 9 e º 'º e º 'º e º 3 s • e s e e s e e º ºs e s e < e e s e s sº e e º e e s e e 487
Assignee, def of, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . tº gº e º e º e º e º 'º - e. e º e º e º e º e s e º e • e s tº e º e º is º e º e 487
Assignment, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . & © tº e º 'º e º e o dº º e s e e º ºs e º º v c e e º e s e e s e º e º e s a s 487
Austrian exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . © e o 'º e e º e e e s e e º e e e s e e º ºs e e e s e e e º e s e o a e e º e s e e 767
Average adjuster in insurance, ... . . . . . . . . . . . . . . . . . . . . . © e o e º e º 'º e e g º e º e º e s e º e º a c e s e e e o e s e o e 527
“Average Clause” in insurance, . . . . . . . . . . . . . . . . . . . . . . '..................................... 512
Average of cotton, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 868
of payments or accounts, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .627—653
, simple and compound partnership, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 861
storage, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 778
Averaging sales in commission, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 867
Axiom, def of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
E
Balance of trade, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 711
Bankers’ interest on daily balances, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . - - - - - - - - - - - - - - - - 665
method of interest, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 571
and merchants' discount, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 576
Bankruptcy, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 487–489
Base, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434
Basis of life insurance, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 536
Belgium exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 764
Benevolent insurance associations, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 535
Bill for storage, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 784
Bills, short sight, bankers, commercial, bill of lading, a 1 bills, prime bills, . . . . . . . . . . . . . . . . 716
and invoices, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283–296
of exchange, forms and defs. of, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 713
Binary scale, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
“Black” method of interest, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574
Boards, lumber, timber and logs, measured, . . . . . . . . . . . . . . . . . . . . . . . . . . cº-oº º e s e º 'º - e & a tº e º e º e º e e 384, 388
Bonded Warehouses, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501
Bonds and stocks fully treated (see stocks and bonds), . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 676–710
Books and paper table, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A. e. e. e. 254
VIII ALPHABETICAL INDEX.
PAGES.
Porrowers and lenders, .................................................................... 546
Borrowing method in subtraction, ......................................................... 66
Bottomry and respondentia,............................................................... 512
Brazilian exchange, ............................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 768
invoice of coffee, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505–506
Brokerage and commission, . . . . . . . . . . . . . ................ . . . . . . . . . . . . . . . . . . ................. 480–487
on “cash notes”. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 584
Budget, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 489
Buying and selling exchange, ...... © tº e º 'º º e º e s e º 'º e º 'º e s e º e º 'º e º e s e º e s a s e e s a e s e s e e s s a s a e s is a e e s e 717
Buying exchange on own account, ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 717
& 4 through agent, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 718
é & & & “ and broker, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 718
Building and loan associations, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 977–1004
Building and loan associations, history and definitions; kinds of members; issuing of
shares; premiums or discount and net loans; dividends and withdrawals; four leading plans—the
Terminating, Serial, Permanent, and Dayton plans, defined and discussed; gross, met, installment,
and interest in advance plans of charging interest and premiums on loans discussed; State and
United States leagues; National building associations; plans of loaning money; plans of with-
drawal; plans of distribution of profits; problem by the gross plan, worked in full; different
methods of some leading New Orleans associations.
Problems worked and explained to illustrate the following important points:
1. To find the rate per cent of interest received by a non-borrowing member on maturing
shares.
2. To find the rate per cent of interest paid by a borrower, on the net plan.
3. To find the gain per cent realized by building and loan associations on investment of dues.
at simple interest.
4. To find the per cent realized by building and loan associations on investment of dues at
simple interest weekly.
5. To find average annual gain per cent realized on weekly payments at simple interest, and
annual amounts to be put at compound interest to close of period.
6. To find total cost of a loan to a borrower at 6 per cent simple interest, Gross plan, with
monthly payments of $1 per share.
7. To find actual premium charged on loans and refunded on their payment.
8. To find average annual gain per cent realized on monthly payments for each year, and
the annual amounts put at compound interest to close of period. Four solutions given, embracing
the compound interest table, annuity table, and logarithms. r
9. To find the length of time required by a purchaser to pay for a piece of property by
monthly payments including the interest. Also required to find the number of payments to be
made, total amount of interest, average per cent paid, and gain made by the company.
10. To find the face of mortgage notes, payable in successive months, to pay for a certain
loan.
11. Miscellaneous problems.
ALPHABETICAL INDEX. IX
C:
PAGES
Calendar, adjustment of, ............. gº tº tº e º 'º ſº tº e º 'º - e º & Cº tº gº tº e º 'º º º tº dº º ſº tº e º e o 'º e º e º º e º º º e e e g º ºs e e º e 237
def of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239
*money, ............................................................................ 234
*tion, .............................................................................. 161–164
9°pital stock dividend,.................................................................... 678
Carlisle table in life insurance, .......................................... • - - - - - - - - - - - - - - - - - 536–539
9*Peting floors, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368
Cash balances, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 653
dividend, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 678
storage, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 777
“Cash notes,” with commission and brokerage, ............................................ 582–589
and exchange combined, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 732
Certificates of stock,...................................................................... 676
Chain measure, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . tº e e º 'º e º e º ei e º gº 244.
Character and face of foreign bills,..... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 715
Check figure, use of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . -- . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
Chilian exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 771
Chinese exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 774
Circle, def of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338
Circular measure, with deſs......................................... tº º º tº e º 'º tº e º 'º º º ſº e º e º 'º º © Cº tº 252
Clandestine stock, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . tº º ſº e º ſº e º 'º e 679
Classification of fractions,. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . tº º tº e º e º e º e ºs 165.
Clearance, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501
Cloth measure, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243
“Co-insurance clause,”. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513
Cold storage, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 777
Colombian exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 771
Collateral note with margin, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 958
Combinations and permutations and chance, ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 943, 944, 945
Commercial instruments of Writing, forms of, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 551–556
Commission, averaging sales in, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 867
on “cash notes,”. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 584
and brokerage fully treated, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 480–487
Common fractions, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
Companies, kinds of insurance, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 511
Comparative weights, measures and values, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . © e s e e º e s a e 254
Comparison, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . & © e o e s & e e e e 26
of simple, annual and compound interest, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 615.
of the foot of different nations, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253
of thermometers complete, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274–277
Complement of number, def. of, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
( & “ multiply by, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
Complex fraction, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
Composite factor, def. of, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
number, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
Compound denominate numbers complete, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229–278
equation, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 627
fraction, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
interest, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 609, 917
interest tables, . . . . . . . . . . . . • . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 560, 561, 576
Check figure proof • e º e º e º ºs e º e º e º 'º - e º e º e º ºs º gº tº gº tº e º 'º º e o e º c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1005-1008
X ALPHABETICAL INDEX.
PAGES
Compound number, . . . . . . . . . .… tº e º s e º e 26
Proportion, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312
Computing interest, universal formula for, ...................................... A c s e s e e s e s a 561
99ncrete number, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
99ne, def of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370
Conjoined proportion, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312, 333
Connecticut system of partial payments, .................................................. 618
Consignee, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 480
Consignment, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 480
Consignor, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 480
Consols. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 680
Continued fractions,. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 949
Contractions in division, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138–142
in multiplication, ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88–121
in multiplication of fractions,................................................ 190–197
in practical problems, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96–101, 278–283
in subtracting dollars and cents, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
Contracting methods of vicenary addition, ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
Conventional interest, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 547
Co-operative insurance associations, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 535
Corporation, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 676
Cost of imported goods, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 752
Costa Rican exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 771
Cotton average, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 868
Coupon bonds, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * 679
Country clause in insurance policy, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514
Credit storage, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 777
Creditor, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 487
Cross multiplication, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
Cuban exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 771
Cube, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370
root, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 891
Cubic measure, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247
Currency, gold, silver, and uncurrent money, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493–500
Customhouse business, . . . . . . . . . ‘. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * - - - - - - - - - - - - - - - - - - - - - - - 500–508
Customs, def. of, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 500
in interest and discount, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 556
Cylinder, def. of, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370
mensuration of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374
I O.
Daily balances, interest on, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 665
Danish exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 773
Days of grace, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549
Debentnre insurance or investment, def. of, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 535
stock, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 677
Debtor, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 487
Decimal fractions, def. of, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165, 216
fractions treated in full, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216–229
scale changed to other scales, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28, 29
ALPHABETICAL INDEX. XI
PAGES.
Definition, * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * s a e e s a e s e e s s e a s • * * * * * * * * * * * * * * * * * * * * a e º e º e s e < * * * * * 25
Demonstration, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • * * 26
Denominate number, def. of, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26, 229
numbers treated in full, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .229–278
Penominator, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
Derivation of names of months, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240
of year, month, week, day, etc. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239
Determining gain or loss of business, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * * * * * * * * * * * * * * * * 809
Diagonal, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • * * * * * * * * * * * * * * * * * * * * * 337
rods, use of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a s a s tº e º e º & © tº e º e º e e 413
Diagram of United States lands divided into townships, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t 246
Diamond Weight, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251
Difference or remainder, def. of, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64, 434
between true discount and bank discount, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604
K
of latitude, . . . . . . . . . . . . . . e e s tº e e s e e - © º º is e º e º e º 'º - - e º s e a tº 4 s - º º e º s a e < * * * * * * * * * * * * * * * 301
of longitude, . . . . . . . . . • e º 'º e º e º e º e º e º e e s e s e s • * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 302
Discharge, def. Of, . . . . . . . . © º e º e e & e º 'º - e º e g º ºs e e º a tº w e s = e º is a n e º 'º e º is a e º 'º e º 'º e s ∈ e º 'º e o e º s e º 'º e º 'º & 487
Discount day, . . . . . . . . . . © o º 'o e º tº º te e º 'º - © tº e º is ſº tº e º 'º - © e º a s a tº tº º w tº ºr s a s - - - - e º e º e - - Q - e º e = ** - - - - - - 551
on time drafts, . . . . . . . . . . . © e º 'º e & © e º 'º • * - e. e. e º tº e º 'º - * * * * s e s - a - e. e. g. a s e º 'º - - - a a s e º s - - - - - 732
ID1scounting notes, . . . . . . . • * * * e o s * @ e º 'º tº s e º ºs e e º e º e º e e s e s is a e e s e s e e s a e º 'º e a s s at e º e º 4 º e s e º e < e < * * * 576
notes bearing interest, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 580
Dissolution of partnerships, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • e e s e - e. e º a s e a e e º e s e e 806
Dividends, defs. Of, . . . . . . - e = e s e e g º m - - - e, e º e º e - - - - e º e s tº a s - - - - e º s a s a - - - e. e. e - - - - - - . . . . 123, 487, 678
J”visibility of numbers, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * * * * * * * * * * * 147
* . Vision, Whole numbers, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122–143
, fractions, . . . . . . * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * is s a s e - 4 a º 197—213
, contractions in, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138–142
def, of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
, logic of . . . . . . . . * * * * * s tº e º & w is e is a s s s e º ſº º e º e > * * * * * * * * * * * * * * * * * * * e s a s a m + e º ºs e e s p → * ~ * 122
of decimals, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226
of denominate numbers, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273
of metric numbers, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 939
principles of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
proof of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . tº @ tº e º e º e º e º - - & © e º e º 'º - - e º 'º 123
Divisors, deſs. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123, 147
Dollars and cents, addition of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
{ { “ subtraction of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • * * g e º º ſº º ºſ e º & tº gº tº sº e e º 'º e º 'º 67, 68
Doyle's rule in lumber measure, . . . . . . • * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * e º e º e º a s e º 'º e º e s - e. e. e. e. 389
Draft, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . © e º ºs s > e - e º 'º e 502
Drawback, def. of, . . . . . . tº e º e s sº tº e º e < * * * * * * * * * * * * * * * * * * * * * * * * * * * * s s e º e º ºs e e s = e º 'º e º e º a s a s e e º s e 502
Drawing notes, points in, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . © tº ſº e º 'º º e º 'º e º 'º e 559
Drill exercises, addition, . . . . . . . . . . . tº ~ * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * e o e e º a 43, 44, 45, 48, 50, 55, 58
( & mensuration of solids, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * * * * * * * * * * * * * 4.17, 418
problems in contracted methods, . . . . . . . . . . • * s s tº is e º a e < * s e e s e e º e < e < e < * tº e º º e º dº º ºs e º e º tº e º º 115
& 4 “ simultaneous multiplication, . . . . . . . . . . . . . . . . . . . . . . . . . tº gº tº e º 'º - tº º tº gº tº e º 'º - e º e 90
Dry Imeasure, - - - - - e a e s e e º e º 'º e º e º e º e º e º ºs e e º e º e º ºs e e s is a e s e º e s e s - sº e s s s a e e s a e * * * * * * * * * * * * * * * * * * * 248
Due date in equation, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 627
Duodecimal scale, tº s e e is * * * * * * * * * * * * * * * * * * * * * 0 e o e º e º is e e º e º e º e º e s e e e º e o e º e s e º e o e g c e s e e o e e s e s 27
XII ALPHABETICAL INDEX,
PAGES.
Plevens, properties of ....................................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150–158
*lipse, def of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339
Plucidation of vicenary addition,.................................... . . . . . . . . . . . . . . . . . . . . . 56
*gineers' measure, . . . . . . . . . . . . ............................. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244
*nglish consols, . . . . . . . . . . . . . . . .................. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 680
exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 740–745.
life insurance table No. 3, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 536, 537, 538
money, freight in . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7:53
“ interest on, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 569
“ reduced to decimal of a pound, ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 450
“ table, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 680
System of numeration, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
Equador exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 771
Equal annual payments of principal and interest, ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 588
Equation method of account current and interest accounts, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 653, 656
Equation of accounts or payments, ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 627-653-1009-1012
Treating equation of sums due at different times, with different credits, from earliest
and latest dates; time counted forward and backward; simple and compound equations;
equation of account sales and on joint account; equation of notes bearing interest; rent
Inotes; notes maturing at different times, etc.
Equilateral triangle, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338
Equivalency of numbers, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311
Equivalent single discount of a series, ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459
Even numbers, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25,
Evil effects of usury laws,. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 545.
Evolution, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 885–900
Exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 710–777
Treating of balance of trade; origin and importance of exchange; bills of exchange;
letters of credit; travelers’ checks; character of foreign bills; par of exchange; rate of
exchange; buying, selling, and investing on account, through agent, and through agent
and broker; time exchange; exchange and cash notes combined; discount on time drafts;
arbitration of exchange; history of English monetary pound; cost of imported goods;
freight in English money; filling orders of English merchants, exchange between London
and other countries; arbitration of foreign exchange; unification of the money of the
world; exchange with Argentine Republic, Austria, Belgium, Brazil, China, Cuba, Colombia,
Costa Rica, Chili, Denmark, England, Equador, Finland, France, Germany, Greece,
Guatemala, Honduras, Italy, Japan, Mexico, Nicaragua, Norway, Peru, Portugal, Russia,
Salvador, Spain, Sweden and Switzerland.
Excise duties, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502
Exercises in addition, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43, 44, 45, 48, 50, 55, 58
in notation and numeration, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33, 34
Expectancy of life table, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 537
JF
Face and character of foreign bills, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 744
of note, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . s º e g º e º g º a tº a gº tº e V & e e º 'º t e º ºs e º e g º º º ſº tº º 549
Factoring, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157
Factors, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75, 84, 146.
146.
, prime and composite, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
ALPHABETICAL INDEX. XIII
PAGES. PAGES.
Fictitious dividend, . . . . . . . . . • * * * * * * * * g ∈ 678 | Fractions, decimal, . . . . . . . . . . . . . . . . . . . . .216–229
Figures, def. of, . . . . . . . . . . . . . . . . . . . . . . . . 27 , def of . . . . . . . . . . . . . . tº gº tº e º & a tº 164
, local value of, . . . . . . * * * * * * * * * * * e 27 , division of, . . . . . . . . . . gº tº º & & tº ſº tº 197—213
, order of . . . . . . . tº e < * * * * * * * * * * * * * 29 , G. C. D. of, . . . . . . . . . . tº t e º e º e & 213
, simple value of, . . . . . . . . . . . . . . . . 27 , L. C. M. of . . . . . . . . . . . . . . . . . . 214
Finland exchange, . . . . . . . . . . . . . . . . . . . . . . 764 , multiplication of, . . . . . . . . . . . . 183–197
Fire losses adjusted, . . . . . . . . . . . . . . . . . . . . 525 , Teduction of, . . . . . . . * @ e º ſº tº º gº a 167–175
Floating policy in insurance, . . . . . . . . . . . . 514 , Subtraction of, . . . . . . . . . . . . . . 180–183
Focal date in equations, . . . . . . . . . . . . . . . . 627 | Framing, building, laying floors, wain-
l'Orced cash dividend. . . . . . . . . . . . . e e º 'º e 678 Scoting, etc., . . . . . . . . . tº e º º gº tº dº e º & ºt 361
Foreign coins, table of, . . . . . . . . . . . . . . . . . 743 | Free list, def. of, . . . . . . . . . . . . . . § º g º º G & © tº 501
Forfeited stock, . . . . . . . . . . . . . . . . . . . . . . . . 677 | Freight in English money, . . . . . . . . . . . . . . 753
Formation of partnership, . . . . . . . . . . . . . . 805 | French currency, interest on, . . . . . . . . . . . 570
Formula, def. of, . . . . . . * * * * * * * * * * * * * * * * * * 26 exchange, . . . . . . . . . . . . . . . . . . . . . . 759–764
Formulae for complement and supple- money table, . . . . . . . . . . . . . . . . . . . 235
ment numbers, . . . . . . . . . . . . . * * * * * * 114 Tentes, . . . . . . . . . . . . . * * * * * * * * * * * * 680
Fractions, addition of, . . . . . . & sº e º ºs e g º ºs e º s 175–180 system of numeration, . . . . . . . . . . 30
, continued, . . . . . . . . . . . . . & © tº º ºs ºs 949 | Frustum, def. of, . . . . . . * * * * * * * * * * * * * * * * * 370
Gauging, . . . . . . . . . . . . . . . . . tº gº tº dº e º $ tº * * * g e º a tº e º ºs º ºs º e º is $ sº * * * * * * * 413
General and particular average, (liscussion and problems, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 527
Geographical mile, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * e s tº e º e s ∈ s e e s s is s e s m = < e < n e a s e e s ∈ e º e a e 24.2
Geometrical progression. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 908-920
German currency, interest on . . . . . . . . . . . . . . . . . . . . . . * * * * * * * * a s = * * * * * * * * * * * * * * * * * * * * * * * * * g e 571–664
exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . is tº e º $ tº gº tº gº 764-767
money table. . . . . . . . . . . . . * * * * * * * * * * * * * * s a 4 s e º e a tº a º e & e º e º $ sº º e < * * * * * e s tº * * * * * * * 236
Glazing and painting, . . . . . . . . . . . . . . * * * * * * * * * * * * * * * * * * * * * * * * * * * * is e e º is e s e s is a tº e a º e º ºs e e s ∈ s 366
Gold, silver, currency and uncurrent money, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . tº e º ºr e º 'º gº 493-500
Goods, cost of importing, . . . . . . . . . . * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * = 752
, insuring. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * * * * * * * * 526
Greatest common divisor, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158
of fractions, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213
German exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 764
Gross and Inet earnings, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * * * * * * * * * * * * * * * * * * * * * 677,678
Weight, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * • * * * * * * * * * * * 502
Guarantee, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * * * * * * * * * * * * * * * * * * * 480
Guaranteed stock, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * * * * * * * * * * * * * 677
Guatemala exchange. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * * * * * * * * * * * * * * * * * * * 771
Hamburg exchange on other cities, . . . . . . * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * s e º a s tº e © tº º 766
History of English monetary pound. . . . . . . . . . . . . . * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * g e 740
of interest, . . . . . . . . . . . • * * * * * * * * * s • * * * * * * * * * * * * * * * * * * * * e º e º s e s e s e i s s sº e s = e º e < * * * * * 543
of the year and adjustment of the calendar, . . . . . . . . . . . * & © e º 8 s tº g º e º a e º is $ $ s tº e º & 237
Homestead associations, to find rate of interest made by, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 592-595
See Building and Loan Associations, for full and extended treatment of methods in
use by Homestead Associations.
Honduras exchange, . . . . . . . . . . . . . . . . . . • * * * * * * * s is is e º e º a tº e s a s = n < n < e a e e s is e a e º e a e s e e tº º e º is a s e e is 771
Horizontal addition, . . . . . . . . . . . . . . . . . . tº gº tº e º tº € $ tº º e s tº e º 'º º e * * * * * * * * * * * * * * * s & e º e s tº a º e º 3 e º 'º e º sº. 49
68
subtraction, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * e º e e º 'º e * e e º tº º ſº tº tº tº e º º ºs
XIV ALPHABETICAL INDEX.
PAGES.
Importance of exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... 712.
of insurance, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ; . . . . . . . . . . . . . . . . . . . . . . . . . . 514
of life insurance, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . be e s & a s 532
of multiplication, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
Imported goods, cost of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 752
Improper fractions, def. of, . . . . . . . . . . . . ... • e º e < * * * * * * * * * * * * * * * * * * * * * * * * * * s • * * * * * * * * * * * * * * * * * * 165
Inclined plane, use of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 427
Indorsements on notes and bills, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 550
Installment, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 677
dividend, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 678
Installments, payments by, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 616–627
Insurance, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 508–543
defs. of property, fire, river, marine, transit, live stock, personal, life, health, accident,
fidelity and co-operative; policies and rates defined; kinds of insurance companies;
adjustment of losses; bottomry and respondentia ; average and co-insurance clauses;
lightning clause, # loss, # value, etc.; unearned premiums; general and particular
average; life insurance fully treated.
Integer, def. of, ... . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * * * * * * * * * * * * * * * is e º e º e º s tº e º is a e s is º e º is is as 146
Interest, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 543–616,
full definitions; history of usury laws; simple interest; annual, semi-annual, and
quarterly interest; compound interest; table of interest laws; promissory notes and
negotiable paper; interest divisors; interest on United States bonds; interest on
English money; interest for 360 and 365 days to a year; interest on French and
German currency; different methods of calculating interest; discounting notes in
full ; “cash notes”; renewing notes; notes bearing interest; interest on installment
plan ; true discount; maturity values.
Interest divisors, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... • . . . . . . . . . . . . 559
method of account current and interest accounts, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 653
on daily balances, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 665
term in Savings banks, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 669
Internal revenue, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502
Interval between dates, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296
Intrinsic value of stock, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 677
Investing in exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 721
in exchange through agent and broker, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 722, 724
See also Homesteads and Building Associations for extended treatment.
Investment and loan companies, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 675
Invoice or manifest, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 500
Invoices and bills, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283–296
Involution, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 883
Isosceles triangle, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338
Italian exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 764
J
Japanese exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 772
Jettison, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 528
Jºº.
Key figure, use of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
a t t e º s & a s a º ºs e º & tº 242
Knot, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
ALPHABETICAL INDEX.
PAGES.
Land measure. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * * > 0 s e e e s e o e s s e s e s e e s a e s s a e s e e s e e s • * * * * 272
Latitude, difference of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . º, º e º 'º 301
Leakage and breakage, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . e e s e e e s a e e s e º e s e º e º e º e = * * * * * * 502
Leap years, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • e o e s e e s e e º e º e s a e e º e º e º e º e s e e < * * * * * * * * * * 236
Least common multiple, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I59
common multiple of fractions, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214
Legal interest, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 547, 548
Length of day or night, to find, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297
Letters of credit, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 714.
Levers, kinds and application of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419
Liabilities, def of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 487
Licenses and taxes, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 489–493
Life insurance, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * & © tº e s e º e º ſº e º e º 'º & º 'º - * * * * * * tº e º e º 'º e º a sº e º e e 532–543
defs. of and basis of, with tables. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 536–541
Lightning clause in insurance, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513.
Like fractions, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175
numbers, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
Linear measure table, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242–243
Lines, defs, of . . . . . . . . . . . . . . . . . . . . . . . . . . . • * * * * * * * * * * * * * * * * * * * * * * * * * * * e a e < e < e < e < e º e s e e 336
Liquid measure. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247
Liquidation value of stocks, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 677
List prices and trade discounts, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457–466
Loan and investment companies, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 675
Local value of figures, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
Logic of division, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
Logs, lumber, etc., measured, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384
London exchange on New York, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 757
& 4 on Calcutta, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 758
& 4 on Paris, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 757
{ { on Shanghai, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 758
Long division, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
OD, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502
Longitude, difference of, . . . . . . . . . . . . . . . . . . . . . . . . . . . . • * > * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 302
and time, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303
Loss or gain of a day in traveling, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306, 307
Losses in insurance adjusted, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512, 523, 525
Lumber, logs, timber and boards measured, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384
Lumber and board measure, Doyle & Scribner Rules. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388
Manifest or invoice, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 500
Manufacturing on shares and tollage, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 785–789
Margin on collateral notes, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 958
to carry stock, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 704
Mariners’ measure, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242
Market value of Stocks and bonds, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . s s e s a s • * * * * * * * * is a e s - e. e. 677
Marking goods, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . - - - - * * * * * - - - 451–457
Maturity of note, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • * * 549
values in Series of notes or payments, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 595
Measures, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229
of time, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236
tables, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231
Measures of extension, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241
of Weight, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249
Measuring oil tanks, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . s tº a º is s a 3 & 4 ~ * * * * * * * * * * * 398
Mechanical powers, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * * * * * * * * * * * 419–434
Mechanics’ work in surfaces, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . - - - - 360
Medial proportion, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333
{ { or alligation, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 789
Mensuration of surfaces and solids with diagrams and cuts, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336–419
MENSURATION –LINEAR MEASURE.
To find diameter and circumference of circle, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340
hypotenuse, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341
base or altitude, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341
XVI ALPHABETICAL INDEX.
PAGES.
To find diameter or circumference from area, . . . . . . . . . . . . . . . ................................ 341
side of Square equal to given circle, ...... & e º e º e º e º e º ºs e º e º e º e º ºs e º 'º º e º 'º - e º ºs e e º ºn tº e * @ e > 342
“ of inscribed square, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342
height of tree, or to a certain point, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343
length of part of tree broken at top, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344
distance between two points when one is accessible, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345
distance between two inaccessible objects,. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345
MENSURATION.—SURFACES.
To find area of parallelograms, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346
of trapezoid, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347
of trapezium, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347
of triangle, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 348
of circle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349
of circular ring, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349
of ellipse, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 350
of irregular polygon, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351
convex surface of prism or cylinder, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351
area of segment of a circle, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352, 355, 356
chord of segment of a circle, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356
radius of inscribed circle, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357
Paving yards and walks, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360
Framing, building, laying floors, wainscoting, etc., . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361
Plumbers' work, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363
Slating and shingling roofs, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364
Plasterers' work, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365
Painting and glazing, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366
Carpeting floors, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368
MENSURATION.—SOLIDS.
To find solidity of a prism or cube, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371
surface of a prism or cube, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371
solidity of square or rectangular pyramid, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371
surface of Square or rectangular pyramid, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372
solidity of hexagonal or other pyramid, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372
surface of octagonal or other pyramid, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372
solidity of frustum of square pyramid, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373
surface of frustum of Square pyramid, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373
solidity of prisms, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374
solid contents of cylinders, . . . . . . . . . . . . . * * * * * * * & & ſº tº gº tº £ tº $ ſº º ºs & & ſº tº º e º e s s e º 'º a gº e º gº tº º e º 'º ºf 374
surface contents of cylinders, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374
solid contents of cone, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375
surface of cone, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375
solid contents of frustum of cone, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375
surface contents of frnstum of cone, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377
solidity of a sphere, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377
surface of a sphere, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377
solidity of hemisphere, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378
& 4 of segment of sphere and cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378-398.
& 4 of prolate spheroid, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379
surface of prolate spheroid, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * 379
solidity of oblate spheroid, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * 37
surface of oblate Spheroid, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380
solidity of semi prolate spheroid, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380
{ % of segment of prolate spheroid, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * 381
4 & of segment of oblate spheroid, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . gº. 381
‘‘ of middle frustum of prolate Spheroid, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383
“ of middle frustum of elliptic spindle, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383
4 & of casks or barrels, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383
4 & of cylindrical ring, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384
4 : of Wedge. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384
& 4 of round timber or logs, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385
To reduce round timber to square. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386
‘‘ find solidity of four sided timber, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387
“ “ solidity of four sided timber that tapers regularly, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387
“ “ length of timber to obtain given Solidity, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388
** reduce round timber to board measure, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388–390
** find board feet in planks, girders, joists, etc., . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . sº-º:
* * * * * * * * * * * * tº º º & 272
ALPHABETICAL INDEX. XVII
PAGES.
MENSURATION.—PRACTICAL PROBLEMS.
To find cubic feet and inches in solids, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .................. 393
cost freight per cubic feet, . . . . . . . . . . . . . . . . . . ....................................... 394
bushels in boxes, bins, etc., . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394
cubic yards in levees or excavations, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395
gallons in vessels, cisterns, etc., . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396
“ in oil tanks, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398
“ in horizontal cylinder, partly filled, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398–400
barrels in a cistern, . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 400
loads of earth to fill lots, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401
Squares of earth in grading, etc., . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402
cords of Wood in a rank, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * * * * * * * * * * * * * * * 402
Inumber of cars to carry Specified amount, . . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * * * * 403
& 4 of bricks in walls and foundations, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403
4 & perches in a stone wall, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * - - - - 404
cubic contents of various vessels and solids, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405
diameter of cylindrical vessels to contain specified gallons, . . . . . . . . . . . . . . . . . . . . . . . 408
To 8*8° coal, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409
To find tons of hay in stacks, mows, loads and windrows, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110, 411
“ of hay in circular stacks, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410
“ of hay in rectangular, stacks, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411
approximate bushels in cribs, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412
bushels of grain in regular heaps, . . . . . . . . . . . . . . . . . y - e s • * * * * * * * * * * * * * * * * * * * * * * * * * * * * 413
To gauge barrels, casks, etc.,. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413–416
To find tonnage of vessels, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416
Merchants and bankers’ discount, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * - e. 576
Merchants’ system of partial payments, ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • - - - - - - - - - - 617
Metric system of weights and measures, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 928–942
Mexican exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 771
Minuend, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
Mint Weight, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249
Miscellaneous problems of all kinds, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 951
Mixed numbers, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • * * * * * * * * 165
Monetary units of different ages and nations, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . - * * * * 230
Mortality table, . . . . . . . . . . . . e e o e º 'º e s e e e º e e s is e e s a • * * * * * * * * * * * * * * * * * * e e s s e < * * * * - - s e = . .538, 541
Mortgage, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . e e < * * * * * * * * * 680
Multiples, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . - 147
Multiplicand, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
Multiplier. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
Multiplication, . . . . . . . . . . . e e s e º e º e º e s a e º e º e º ºs e e s e a s a s e s = e s e e s = e s - e. e. e. e s e s - e e s e s e e s - e. e. e. e. e. e. e. e. 75–122
, by aliquots, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • e º sº e º sº. 4 e º a sº e • & © tº tº gº tº sº 94–101
, by factors, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
, contractions in, . . . . . . . . • * * * * * * * * * c e s e e s tº e º e a e e s a tº e º e s e e º 'º - e. e. e. e. e. g. e. e. e. e. n e s a e s - 88–121
, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75, 183
, directions for, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
importance of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
of abstract numbers, with reasons, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8()
of decimals, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a s a 5 - - a tº s tº e º O - © tº e º 'º - 225
of denominate numbers, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271
of dollars and cents, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .... • 84
of metric numbers, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 938
, philosophic system of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
, simultaneous or cross, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
table to 25 × 25, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
With naughts on right, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3
Nautical mile, def. of, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 242
Negotiable paper, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * - - 549
New Ham Sphire system of partial payments, . . . . . . . . . a e s a s s e s - e. e. e s a - - - e º e < * * * * * * * * * 618
Nicaraguan exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • , s , = e s a • * * * * * * * * * * * * * * * * * * * * * * 771
Nines, property of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113, 150
Non-participating policies in insurance, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 511
Norwegian exchange. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . - * s - - - - - e º º is - - - - - e º 'º - - - 773
Notation, Arabic. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * * * - - 30
, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * - - - & Gº tº e - 30
XVIII ALPHABETICAL INDEX,
PAGES.
Netation, exercises in. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33, 34
of decimals, . . . . . . . . . . . . . . . . . . . • * * * * * * * * > * > * > * * * * tº s º & e º º te e º ºs e g º º e º ºs e tº e º e ºs e º e e s is a e 216
, TOIll all. . . . . . . . . © tº 6 tº e º 'º e s e - e º us tº - • 6 e s tº e - © tº e e - is º ºs e º 'º tº tº 4 - - - 8 & 9 & 3 tº g º & © e º a tº - e g º e - ... ... 30, 31
Notes, discounting, . . . . . . . . . . . & © tº gº tº 3 º' tº e º e º 'º e º 'º º * @ e º 'º e º e e º e © tº e º e o e º ſº tº e º 'º e s tº e º 'º a s º e º e º e º e º e º 'º 576
, equation of interest bearing, . . . . . . . . . . . . e e e º 'º - © e º 'º e s to e º 'º tº tº ſº e º 'º e º 'º - e s e s tº e º e º º • e º e s a sº 649
& “ of maturing at different times, etc., . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 651
{ { of Tent, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 648
, interest bearing, to collect and discount. . . . . . . . . . . . . . . . . . . . . . . . . . . . . is e - a s gº - - - e s - - 557, 580
, partial payments on, . . . . . . . . . . . . . * * * * * * * * * * * * * * * e a e e s a e a e s = s. s = < * * * * * * * * * * * * * * * * * * * 624
points in drawing, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a * * * * * * * * * * * * * * * * * * 559
, promissory, . . . . . . . . . . . . . . . . • * 0 e º e º 'º - e º 'º - - e s tº a tº º 6 e º 'º - e s tº e º 4 & a tº 4 - e. e º 'º - e º 'º e - © e º e º - s e º e - 549
, with collaterals on margin, . . . . . . . . . . . . . . . . . . © e º e o e º e º e º e s ∈ e º sº a ºn s a a t t t w e º s ºn tº t e o 'º º º 958
Numbers, defs. Of all kinds, . . . . . . . . . . . . . . . . . . . . . . . . • e s e º e º 'º e º e a s e e º a e s ºf s - s • * * * * * * * * * * * 25, 26, 146
including abstract, concrete, odd and even, perfect and imperfect, prime and composite,
like and unlike, complement and supplement, etc.
Numeration, def. Of, . . . . . . . . tº º e º e º 'º - - - tº e º 'º e º e s - - e º is º - e º & e º 'º e © & a - 4 º' tº e º 'º º e • * * * * * * * * * * e s e e º a 30
, English system of, . . . . . . . . . . © 3 & 4 º º o e º e º e e s sº e º 6 e º e º e º e s a e s & e º e s - e º e a º 'º tº € 8 & a 31
, exercises in, . . . . . . . . . . . . . . . . tº e º e º e º e º e º 'º - e º e - a tº e º 'º - • e s - tº s - e. e. e. * * * * * * * * * * * * * 33, 34
, French system of, . . . . . . . . . . . . . . . . • e e s e º e s a • * * * * * * * * * * - • * * * g e º e º 'º & w tº a 30
of decimals, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219
Numerator, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . - e. e - - - - - - - - - - - - 165
Numerical equation, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . e - - - e < * - e. e. e - - 40
Oblate spheroid, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * * * * * * * * * * * * * * * * * * * * * 370
& 4 mensuration of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379, 380
Odd numbers, def. of, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . - - - e º - - - e. e - - - e. e. e - - - iº & - - 25
Oil tank measurements, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398
Old French and Spanish lineasures, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253
Open policy, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • * * * * * * * s e º e e º 'º w s & © - * * * * * * 510
Operation, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . e - - - ºr e - - * * * * * * ~ * * * - - 26
Order of figures. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
Origin of exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • 711
Painting and glazing, . . . . . . * * * * * * * * tº e º is e º e º e º 'º º ºs e º a º e º e º 'º º is e º 'º - e º s º e < e e * - © s tº e º a tº e tº & B tº ſº tº ſº a w 366
Par of exchange, . . . . . . * * * * * * * * * * s e a e º e s e tº e e º e - © tº e º 'º tº º s tº e º e º a tº dº e º e º 'º - a s & © to e º e º 'º - e º 'º tº º e º e 4 o e 716
of Stock, . . . . . . . . . . . . . . . . • * * * * * * * * * * * * e s e e º e s is a • * * * * * * * * * * * * * * * * * * * * * * * * * * * e e a e e s e s - e. 677
Parallelogram, def. Of, . . . . . . . * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 338
to find area of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346
Parenthesis and Vinculum, use of, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
Paris exchange on other cities, . . . . . . . . . . • * * * * * * * is a e < * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 764
Partial payments, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • e s e º e s e º 'º e s a s e e s s a s e o e e s s e s e º e s e s e e s s , 616–627
Five systems defined; problems worked in full.
Particular and general average, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • * * * * * * * * * * * * * * 527
Parties to notes, bills and drafts, . . . . . . tº e - © e º 'º e º ºs e º 'º - e º e s - - e. e. e. e. • * g e º - - - tº e º 'º e º e < e - e. e. e. * e º e e s - 549, 554
Partitive proportion, . . . . . . . . . © e º ºs s tº º ſº º º ºs e º e º 'º º ºs º º º © º e º e º & © tº dº tº tº º e º e º e º & tº e º e • * * * * * * * * * * * * • * 332
Paving yards and walks, . . . . . . . . * @ & E * : * ~ 3 - e. e. e. e. e. e ſº e º O & e º e º & º e s - a e º e tº - e º 'º e º o * * * * * * * * * > e s sº e - e. 360
Percentage, . . . . . . . . . . . . . . . . . . . . tº tº º º & © w is e - * g e º - - tº º e s - © tº e º e º s º e º 'º - a s e - e º a tº e º ºs e e º 'º' tº e & s • * * e º ºn a 434–480
definitions; tables of aliquots; to find percentage, base, rate per cent, amount or
difference in all combinations; per cent on English money; marking goods; impor-
ting English goods; trade discounts and list prices; series of discounts reduced to one
equivalent; discounts on bills; tables of trade discounts; value of sugar cane per
ton; quantitative chemical analysis; to find sucrose multiplier; practical questions.
Permutations and combinations and chance, . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * * * * * * * * * * * * * * * * * * 943
Peruvian exchange, . . . . . . . . . . . . . . . . . . . . . * * * * * * * * * * * * * * e s - e. e. e. e. e. e. e. e s = e º e s = • * * * * * * * * * * * * * * * * * 771
Philosophic solution, def, of, . . . . . . . . . . . . . . . . . . . . . . . . • e s tº a s e s e º 'º e º 'º e - e. e º e º e a s e º e º e s e e e e º 'º e e 26
system, def. of . . . . . . . . . . . . . . . . . & © tº 0 ° tº º & © tº e º & - - e º e º - 4 º' s e < * * * * * * tº e º 'º e º e e º 'º - e. e. e. e. e. 26
‘‘ of contracting interest, . . . . . . . . e - e s e - e. e s s a s e e s - 4 e 4 e º e º 'º - e s s e º 'º e º ºs e - 566
& 4 of multiplication, . . . . . . . tº º 0 e º a 9 • * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 77
( & of proportion, . . . . . . . . . . . . . . . . . . • e e s e º a s e e s e a s a • * * * * * * * * * * * * * * * * * * * * * * 315
Phrases and words used by bankers and brokers, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 682
Plane figure, def. of . . . . . . . . . . . . . . . . . . . . . • * g º e º 'º a ſº e g º a e g º e s a s a 4 s a tº a s a s is e º e s s a e º e º e º º e º e º e - 337
Plasterers’ work, . . . . . . . . • * * * * * * * * * * ~ * * * ~ * c e e e e º e s e e e a e s e e e º a 4 • e s e s a s a s = e º s s e s a e s a e s a e e s - 365
Plumbers’ work, . . . . . . . . * * * * * * * * * * * * * * * * * * * g e º s e e s e º e s e a º e º e º s - s = - * g is - - - - - - - - - * * - - - - - - - 363
Points in drawing notes, . . . . . . . . . . . . . . . . . . . . . © tº t e º e tº º e º ſº tº e e e s a e º me tº e - sº e e º tº © tº e º is - © e g tº tº º ºr º e º 'º 559
ALPHABETICAL INDEX. XIX
PAGES.
Policies, insurance, defs, of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • * * * * * * * c tº e º & 509, 510, 533
Polygon, def. of, . . . . . . . . * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 337
Ports of entry, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501
Portuguese exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 770
Powers of numbers, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
Practical operations with aliquots and other numbers, . . . . . . . . . . . . . . . . . . . . . . . . . 96–101, and 278–283.
problems in mensuration of surfaces and solids, . . . . . . . . . . . . . . . . . . . . . . 357–360 and 393–417
{ { Ill percentage, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 470–479
& 4 in stocks and bonds, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 698.
Preferred creditors, def. of, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 487
stock, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 677
Premises of problems, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
Price of stocks from English and French quotations, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 708, 709
Prime factor, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
number, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • a • * g º e a 9 - e º s e s tº e s a e º 'º - tº e º ºs e s tº e º 'º. 146
Principles of decimals, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . s e s - e. e. e. e. e. e s m = * * * * * * * * * * * * * * 220
of fractions, . . . . . . . . . . . . . . . . . . • * * * g e º º a s a e s a 4 e º e º a s e º 'º - - - - e º e º a tº e º 'º º 166
Problem, def. of, . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * * * * * > * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 26
• PſelhlSeS OT, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26.
, solution of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
Product, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
method of account current and interest accounts, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 653
Profit and loss, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . e is a e s º e s is e º e º & © 2 450
Progression, arithmetical, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 901–908
, geometrical, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • * * * * * * * * * * * * 908–920
Prolate spheroid, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • * e s tº e º e º 'o 370
* { mensuration of, . . . . . . . . . . . . . . . * * * - - - - - e. • a • * * * * * * * * * * * * * * * 379
Promissory notes and negotiable paper, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549–554
Proof of addition, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47, 152, 1005, 1008
of division, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123, 154
of multiplication, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75, 154
of subtraction, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64, 153
of transfers and postings, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156
Proper fraction, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
Properties of nines and elevens, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150–157
Proportion, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310–336
definitions and terms; general principles; relationship and equivalency of numbers;
Solutions by proportion, analysis, and the philosophic systems compared; partitive
proportion; medial proportion; conjoined proportion; reciprocal proportion.
Proportional division, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 871–876.
Pulleys, kinds and use of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424–427
Purchasing French exchange from distant cities, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 762
Pyramid, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • * * * * * * * * * * * * * * * * * * * * * * * * * * * * 37().
PARTNERSHIP-SETTLEMENTS.
All problems worked in full.
Partnership, definition of, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 803
, sharing gains and losses of, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 803
, commercial
, ordinary |
º!. | In Louisiana, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 803–804
, in commendam |
, limited
, trading or commercial \
, non-trading |
3 ºr ~~~ |
yº In common law States, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 804–805.
, particular |
, limited
, formation of three methods, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 805
, dissolution of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 806
Partners, nominal or ostensible, silent, secret, sleeping or dormant partners and sub-partners, 805
Partnership, articles of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 806–807
, settlements, discussion of, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 808
Resources, assets or effects, defined. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 808
Liabilities, defined, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • * * * * * * * * * * * * * * * s tº e º e º e º e 808.
XX ALPHABETICAL INDEX.
PAGES.
Investment, net investment, average investment, defined, , , , . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 809
Capital, net capital, insolvency, net gain, and net loss, defined, . . . . . . . . . . . . . . . . . . . . . . . . . . . . 809
Two methods of determining gain or loss, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 809–810
Oral exercises, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 810
Settlement, involving net investments, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 811
y * { “ Withdrawals, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 811
% * { & & { { and net investments, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 811
from statement of resources and liabilities, and gains and losses, by two methods,
With Variations, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 812
made in practical accounting form, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 813
from trial balance, by resource and liability method, . . . . . . . . . . . . . . . . . . . . . . . . . . 814
& 4 { { by loss and gain method, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 815
between physicians, in practical form, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 815–816
where no contract was made regarding gains and losses, . . . . . . . . . . . . . . . . . . . . . . 817–818
by note, from stated conditions, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 818
from trial balance, with conditions, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 819
between partners, with salary allowances, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 819
with steamboat owners, with conditions, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 819–820
to adjust current accounts with the entries, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 820–821
* { cost of driving logs, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 821–822
between manufacturer and planter (on shares), . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 822
{ { merchant and planter, to gin, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 822–82
* * son and daughter, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 823
• { children, from a will, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 823
to find gain from resources and liabilities, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 823
{ { net capital of partners at closing, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 824
from unequal investments and unequal shares in gains and losses, . . . . . . . . . . . . . . 824–825
to find net capital at commencing, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 825
* * original investments, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 825–826
net insolvency at commencing and capital or insolvency at closing, . . . . . 826
k . net capital, from irregularly kept books, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 826–827
* { respective net gains, with salaries allowed, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 827
& 4 gain and capital from trial balance, when goods and books are burned, . 828
{ { value of merchandise from trial balance, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 829
{ { met investment after reinvestment of profits, . . . . . . . . . . . . . . . . . . . . . . . . . . . 829–830
& 4 amount of cash stolen by dishonest book-keeper, . . . . . . . . . . . . . . . . . . . . . . . 830
to adjust accounts of partners, without entries in books, . . . . . . . . . . . . . . . . . . . . . . . 830
• { debit balances of partners, with unequal interests in gains and losses, 830
{ { cost of cotton factory between contractors, . . . . . . . . . . . . . . . . . . . . . . . . . . 831
between cattle dealers, from stated conditions, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 831–832
to adjust gain proportionately, with allowance for reserve fund, . . . . . . . . . . . . . . 832–833
between principal and agent, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 833–83.1
to find cash balance, net gain and present worth of a branch store—three solu-
tions, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 834–837
to find gain and capital of proprietor of branch store; also condition of mana-
ger's account. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 838
with agent of branch house, to determine cash on hand, net gain and net worth, .838–839
With branch store—two solutions, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 840–841
between temperance town and druggist agent, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 842—843
of interests in proportion to amounts invested, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 843
to find net gain of a manufacturing agency, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84.4
( & net Sales, list price, and cost, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 845
to adjust 1nterest on partners’ accounts, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 845–846
to find partners' balances, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 846–818
between book-keeper and firm, with entry, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 848
& 4 partners, with salary allowance, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 848
to adjust interest balances, with entries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 849
& f & 4 in case of dissolution, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 849
between mechanics, for lost time, with entry, . . . . . . . . . . . . . . . . . . . . • * * * * * * * * * * * 850
to adjust gain on admission of new partner, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 850
{ { anticipated loss, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * = e is e e s s • * * * ‘. . . . . 850
( & commission, charges and interest, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 851
( & coin and currency accounts, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 851
to reduce coin balances to currency, and vice versa, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 851—852
between partners of a note, considered doubtful, with entries, . . . . . . . . . . . . . . . . . 852–853
of partners' balances after division of resources, with entry, . . . . . . . . . . . . . . . . . . ;
o
changing single entry to double entry, using single entry ledger, . . . . . . . . . . . . . .
ALPHABETICAL INDEX. XXI
PAGES.
Settlement to open new books from irregular set, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 855
between partners from given resources and liabilities, .........................855–856
e & 4 “ , one of whom withdraws,....................................856-858
Admission of a new partner into a commercial firm, (four cases), . . . . . . . . . . . . . . . . . . . . . . . . . . .858–860
Partnership and partnership average, ...... * * * * * * * * * g e º e - • e º e º 'º e º e º a e e º e e e º e º e º 'º e & © e º e º ºs e º e 861
Simple partnership average problems, solved, ...................... • * r * e s a s e º e s e e º e º sº e s e - 861–863
Compound “ & 4 { { { { * * * * * * is a e e º e a e s a s e tº ſº tº a 6 - - e º is e º e . . . . . . . . . . . . .863–867
Averaging Sales on commission, ....................... * * * * * * * * * * * * * * * * * * * * © º ºr e º 'º e º ſº tº e º tº e 867
Cotton average problems, solved, . . . . . . . . . . . . . . . . * * * * * * * * * * * * * * e s a e e º e s a a e s a © e º a º ºs e e s is e º 'º a 868
Division and proportional division, . . . . . . . . . . . . . . . . . . . . . . . . * * * * * * * * * * * * * e s a e º e º e º a © tº e º 'º tº e º & 871
Settlements, with Special conditions, . . . . . . . . . . . * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 872
Division of proceeds of partial interest sold, in order to equalize remaining investments, ... .872–873
Division of proceeds of a dinner in proportion to the quantities furnished. . . . . . . . . . e e º s º e - 873
To find average rate per cent of insurance on cotton for different times and rates, . . . . . . . . . . 873
Settlement with the different grades of cotton, . . . . . . . . . . . . . . . . . . . . tº e º 'º e º is is e e º e º e º º is a s & tº e º e 874
Settling estates, . . . . . . . * * * * * * * * * * * * * * * * * * * * * * * * * * * * e e s e e o e s - e º e s e e tº e º e º e º 'º e g º e º e º 'º º ºs e º 'º º sº º º 874
Division of remainder of an estate in reciprocal proportion to ages, . . . . . . . . . . . . . . . . . . . . . . . .874–875
Division between two surviving heirs, in certain proportions, . . . . . . . . . . . . . . . . . . . • e s e e º e º e º 4 875
Division of an estate, dependent upon return of certain heirs, . . . . . e - e º s tº e º e tº e º e º e º e e e s & 6 tº 876
To find amount of an estate, when shares are given, . . . . . . . . . . . . . . º e < e < e < * * * * * • * g º e º is tº s e s 6 tº 876
Proportional division, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * * * * * * * * • ‘º e º e º e º e º 'º º . . . . . . . . . . .876–877
The land problem, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 * * * * * * * * * * * * * * * c e is e e s & e º 'º tº º e º 'º e º 'º & 878
The ditch problem, . . . . . . . . . . . . . . . . . . . . to e º e º e º 'º - e º e g º ºs e º 8 tº º º * * * * * * * * * * * * * * * * e º e º e º e º e tº e º 'º º º 878
Division of work and funds, . . . . . . . . . . . . . . . . . e - e º e s e e o e • * * * * * * * * * * * * * * * * e s e e s e e s e º e s a e e s e -
To find number of stolen articles, when shares are given, . . . . . . . . tº e º e º 'º - e. g. g º ºs e º 'º - - - - . . . . . . . .879–880
Division of land in shares, . . . . . • * * * e º e º e º e º e º e º 'º e º e º e e tº e e º e º 'º e * 9 & & © tº tº 6 tº tº e º 'º e º 'º - º ºr tº º º tº º tº e º ſº. 880
To find interest in mining company, . . . . . . . . . . . . . . * * * * * * * * * * * * * * tº e º e º e º e º te e º e º 'º e ‘º e º 'º º e s e º sº º 880
with portion of gain to be credited to
Division of gain dependent upon fixed investments,
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .881-882
TěSèI'We fund, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
C.
Quantitative chemical analysis, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Quantity, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . © º e º e s ∈ e s a e - © tº e º e º 'º e º &
* 607
• * * * * c e e º e º 'º - e < * * * * * * * * * * * * * * * * * * * * * * * * * s e tº º º e e
Quarterly interest, . . . . . . . . . . . . . . .
Quarternary scale reduced to decimal, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . e - © e º e º a º e º 'º - 4 e º & e º 'º º 28
Quinary scale reduced to decimal, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . - - - © tº e º g º º e e º º te tº © e º e e 28
© º e tº e º ſº º ſº 123
Quotient, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . tº ſº e º ſº º tº e º e tº e º 'º a tº e
J R
Radix, def of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . tº e
e e e º e e º e 434
Rate, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • e º ſº e º 'º -
of commission and brokerage, . . . . . tº e s e º e º a tº e º e º 'º e º e º 'o • * * * c e a e º e • * e s e e s e º e - tº º e º e e e s e e 480
of exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . & e º e º 'º e º sº e º e s s a tº e © tº º º & © tº e e ... 716, 727, 729, 744
of insurance premium, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . e - © tº tº º gº e º 'º e © tº o e º ſº tº º e º 'º - . . . . .510, 515, 534
of interest in sale by installment plan, . . . . . . . . . . . . . . • ‘º e º e º e s e º sº e < e < * * * * * e º ºs e º e º e º º e º e 592
of storage, . . . . . . • * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * e < * * * * * tº e º e º is e º 4 s e s a e & © - s e º 'º e º e 777
Ratio, defs, of . . . . . . . . . . . . . . . . . . . . . . . . . . . . • e e º 'º - e. e. e. e. © e º e º 0 & © tº e º s a a s e º a • * * * * * * * * * * * * * * * 308
, principles of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * tº a º e º 'º - e s tº e tº dº tº e > © e º º s º º & © tº 3 º' tº a tº g º g g tº e 309
of diameter and circumference, . . . . . . . . . . . . tº tº ſº e º e & © e º 'º e º ºs e e s º º e º e º º * is e º 'º º tº ºn e º a tº e º is e - 338
of sphere and cube, . . . . . . . . . . . . . . . . © e º ºs e s - e. * * * * * * * * * * * * * * * * * * * * * * * e º e º e tº e º 'º e º sº tº e e s = 377
of square and circle, . . . . . . . . . . . . . . . . . . . © e º e º 'º e º 'º e º 'º e º e º tº e g g g tº e º a • * * * * a s e s - e. e. e. e. e. e. e. g. a e 33
Receipts, forms of, . . . . . . . . . . . . e & © tº e º e e º o e e º e º 'º * - - - - e º e º 'º - tº - - - 6. © e. e. e º s - - - - e s e º e º e s - - - - tº º e e º e 551
, storage or Warehouse, . . . . . . . . . . . . . . . . . . . . . . . . . © e º s a e º e s e < e < e < e e s a s e = * * * * * * * * * * * * 777
Reciprocal of a number, . . . . . . . . . . . . . . . . . . . . . . • * e º e º & tº e º & e • * * * * * * * * * * * * * * * * * * * * * * * * * * * * * e 147
of a fraction, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
proportion, . . . . . . . . . . . . . . • * * * * * * * c e s = e º e º e e s • * * * * * * * * * * * * * * * * * * * * * * * * * 6 g º a º is s tº e 334
Rectangle, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ºn tº e s s a s 337
Reduce shillings and pence to decimal of a pound, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 450
Reduction of decimals, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * c e s - e. e. e. e. a = e º º e s s e a s = * * * * * * * * * * * * 220–223
of denominate numbers. . . . . . . . e e s e º a s a s a s s e º e º 'º - e < * s e º a e < * * * * * * * * * * * * * * * * * * * * * 256–267
of fractions, . . . . . . . . . . . . . . . . & e e º ºs e º 's e º e º 'º e º 'º º tº a º * @ e º 'º e tº e º 'º e º 'º - e º ºs e º 'º e º 'º - © 4 & 8 g º e & 167–175
of metric numbers . . . . . . • * * * * e s a e e < * * * * * * * * * * * * * * * * * * * • e º e º 'º e º s e e º e º ºs tº s º dº º is tº e a 935
of “ ( 4 to American equivalents and vice versa, . . . . . . . . . . . . • * * * * * * * * * 940
Registered bonds, . . . . . . . . . . . . e e º e s - e. e s e e º f * * * * * * * * * * * * * * * * - e s e º e g is e º e º 'º e º s e º e < e s - © e º e º e º a tº 680
Reinsurance, . . . . . . . . . . . . . . . . . • * * * * * * * * * is e s a s = e º e º e º 'º e s is a • * * * * * * g e º 'º e º e º e º 'º e a tº e º e º 'º e º a tº e º e º 511
Relationship of numbers, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . © e º a tº &
XXII AL PHABETICAL INDEX,
PAGES.
Remarks on measurement of segments, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382
Rene Wing notes, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589
Rent notes, equation of, . . . . . . . . . tº tº 4 & © - tº & e º 'º & * & ſº & e º 'º e º 'º e º e º e g * * * * * * * * * * * * * * * * * © tº - tº e º 'º º e º s e a 648
Repetends, def. of, . . . . . . . . . . . . . . . . . . . • * tº e º 'º e g tº e º e º e a * * * * * * * * * * * * * * * * * * * is º a s s s a º a s e s a e s = n e s 217
Reserve fund, . . . . . . . . . . . . . . . . . . . . . . © & © tº ſº e º ºs * * * * * * * * * * * * * * * * * * * * * * * * * * * * = e s a s • e a e º a a s = e a e s , 678
Resources, def. of, . . . . . . . . . . . . * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * s s a s a s is a e e s = e s s a e s e a < * * * * * * * 487
Respondentia and bottomry, . . . . . . . . * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 512
Return premiums, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 510
Rhombus, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . tº 4 s tº e º 'º - tº * * * * * * 338
Right angled triangle, def. of, . . . . . . . . . . . . . . . . . . . . . * * * * * * * * * * * * * * * * * * * * * * * s e a ºn a a e s = < e < e e s s 337
Risks, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * * e º e º tº º a ºn tº e º 'º e & 4 & 6 tº 6 e 51()
Rods, diagonal and ullage, use of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413
Roman system of notation, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30, 33
Bussian exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 771
Sales on joint account, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 662
Salvadorean exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 771
Salvage, def. of . . . . . . . . . . . . . . . e & tº e º e º 'º - e. e. e. e. e. e. e º 'º e º e º 'º e º e
Savings banks and savings bank accounts, fully illustrated, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 669–675
Scale beam, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • * tº e a tº e º a s 420
of numbers, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . tº e e º e e * tº e º 'º - tº e º 'º e º e º e e 27
Scaline triangle, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . tº gº tº 9 tº a 6 - - - - - © e º 'º is is e 338
Schedule, def. of . . . . . . . . . . . . . . * * * * * * * * * * * * * e º e º º e º e º 'º e º e s e e s e º e s e e º e s e e e s e e s s a e e e - © e º t e º e e 487
Science, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . tº º & & & e º e tº © e º 'º - e º º ſº tº e º e º e 25
Screw, use of the, . . . . . . . . . . . . . . . tº & © º 'º - e º tº tº º te e º tº e º º º & © & Cº tº ºn tº tº e º & © tº a º º e º 'º e º e º e º 'º e º e º & tº º ſº tº e º e s 430–434
Scrip insurance, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . tº e º 'º & © tº e º e & e º 'º © tº º e º 'º e º ºs is tº gº tº e º 'º º e º s tº tº e º e 5 12
Selling exchange on OWn account, . . . . . . . . . . . . . . . . . * * * * * * * * * * * * * * * * * * * * * * * * * g e º e - e. * * * * * 719
& 4 through agent, . . . . . . . tº º t e de e º e º e s e - - - - - e º ſº tº e º e - © tº e º a tº e a tº e s e s a 4 - - - - - * * * * * * * 719
& 4 { { “ and broker, . . . . . . . . - - - - - tº e º e º 'º e º 'º - e. e. e. e. tº e s e s e e < * * * * * * * * * * * * 720
Semi-annual interest, . . . . . . . . . . * * * * * * * * * * * * * e = c is e e º e - e º e º e º e º e º e = e s º s e e s e e s e º 'º - - - - a e s ∈ e e a e 607
Set of foreign B. E. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . tº e º e º 'º tº dº º tº e º º º & º 'º e º sº s ºr e º 'º - tº e º e º e s s sº e 713
Settling estates, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * * * * * * * * e º 'º is 874
Settling with insurance companies, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525
Shares, manufacturing on, . . . . . . . te e º 'º e º 'º e a s e º e s is tº e tº e º e º ºs e º e º 'º e º e e º e - * * * * * * * * * * * * * c e s is tº e. 785–789
Sharing gains and losses, . . . . . . . . © e º 'º e º 'º & e º e º ſº e º & © - - - - e º e º e º e º e - - - - e. e. e. e. e. e. e. e. e. a 4 - * * * * * * * & e - 861
Shingling and slating roofs, . . . . . . . . • e º ºs e º e º e s e º 'º e > * > * * * * * e º e º e º e º e - e. e. e. e. e. e. e. e. e. e. e. e. e. * * * * * e s s e 364
Shoemakers’ measure, . . . . . . . . . . . . . . . • * > * * * * * * * * e e e s • * * * * * * * * * * * * * tº e º 'º - - - - - © tº e &
Short cuts in division, . . . . . . . . . . . . . . . • * c e º 'º & ſº gº tº e º 'º - © º 'º e º 'º º tº $ tº dº e º 'º - e º 'º e º is a e º is tº is a 4 - © tº º tº s º e . 138–142
“ in multiplication, . . . . . . . . . . . . . . . . . . . . . . . . . . . • * * * * * * * * * * * * * * * * * * * * * * * * * * * * . . . . S8–121
‘‘ in & 4 of fractions, . . . . . . . . . . . . . . . . . . . . e tº & © º e º sº º e s ∈ & e s a s a tº 4 - e º s is e e s ∈ s 190–197
division, . . . . . . . . . . . . . . . . • e s - c e ºs tº e º 'º e s tº e e tº e º e º 'º tº e e º e º e º 'º - e tº © e º 'º e - - - - - - a s * * * * 128
Signs and symbols, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . & e º e º e s - a tº e º ſº s e s m tº ºt & e º sº - - - tº e e s tº e < 36–39
Silver, gold, currency and uncurrent money, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193—500
Simple equation, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . e is e o e º 'º - - - a s a e e s tº a e - a - - e º 'º e º e º sº e 627
fraction, . . . . . . * * * * * * * * * * * * * * * * * * e e s e º e a tº e e s a 4 - e. e e g º e s r. º. º ºs s ºn e º e s e s a e s ∈ a • * • * * * * * * * 165
value of figures, . . . . . . . . . . . . . tº º e º 'º º e º ºs e º º º ºs e - © e º g º e s e º 'º - 6 * * * * * * * * * * * * * * * * * * * * * * * * * * 27
Simultaneous multiplication, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . - - - - - tº e º 'º º º 88
Single discount from a series, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459
Slating and shingling roofs, . . . . . . . . . . . . . . . . . . . . . . . . . . . e e e s is a e s - e º e e s a s s s s e º a • * * * * * * * * * * * 364
Smuggling, def. of . . . . . . . . . . . . . . . . . . * * * * * * * * * * * * * * * * * * * * * a e º e º e - • * e s e e g = e s - - - e º ºs e º e º º º 501
Solar days and years, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . tº e º e º e º 'º e º 'º - * = c → - - - - - • tº e º t t e s is 236
Solid, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • e º e º e s - © tº e º º e º ºs e º e º 0. e - e º s s tº s & s is 370
Inleasure, . . . . . . . • * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * & a e º º e º e 4 & 3 - a tº tº a sº * 2.47
Solution of problems, def. of . . . . . . . . . . . . . . . . * @ e º e º 'º e º e º s a e s is a s s a • * * * * g e º e º e º 'º e º a e s e º e e s a s is 26
philosophic, def. of, . . . . . . . . . . . . . . . • * * * * * * * * * * * * * * * * * * * * * * tº gº gº e º e s tº º e º º s a tº º º & 26
statement, def. of, . . . . . . . . . . . . . . . . . . . . • * * * * * * * g º e s e e a e s ∈ e e s e º e s e s is a m e º sº a • * * * * * * * * 26
Spanish exchange. . . . . . • e e º e º e e s tº 0 - - - - - - e. e º ºs tº e & e º 'º - - - - e e s tº a e g º g c → • - e. e. a g º e e e s a 4 - - - - - - * * * * * * * 764
Special laws and business customs in interest and discount, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 556
names given to partners, . . . . . • * * * * * * * * * * e e s e e o e & © e º ſº & e e º 'o e • e e s s e e s s = * * * * * * * * * * * * * * * 805
Specific duty, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . © º e º e º 'º s e s ∈ e. e. e. e s = n < e < * * * * 500
Sphere, def. of . . . . . . . . © º & © - e º º * * * * * * * * * * * e e tº tº º tº a tº º & * * * * * * * * * e e s m e º e a tº a s a e e º ºs e º 'º e º 'º a º º º s e tº 370
, ratio of, to cube, . . . . . . . . . . . . . . . . . . © & © tº e º 'º e º 'º e º e º e º 'º e s e º e - e º 'º e º $ tº a “* * * * * * * * * * e e s & 377
Square, def. of . . . . . . . . . . . . . . . . . . . . . . - - - - - e º s e e e • * * * * * * * * is e s is a e º a e º e o e a s e s e s a e < * * * * * * * * * * * * 337
, ratio of, to circle, . . . . . . . . . . . • * * > 0 tº e e s m e º e - e - e. e. e. e. a e s a e e e s e e s e e s e e s is e º 'º - e. e. e. e. e. e s is a tº e 338
square measure. . . . . . . . . . . . . . . - * * * * * * * * * * * * * * * * * * * * * e s - e º a tº e - e - e º e º ºs e º 'º - - - - - e. e. e. e º e º º 243
e º s 885
tº ſº tº gº tº tº dº e º & ſº tº e º e º ſº tº e º tº e º 'º º tº e º 'º e º is e e 512
I’OOt. • * * * * * e e s - a e º 'º º e º 'º e º e o ºs e e º 'º e º e g º e º ºs e s is e e º e e s - e s e e s • * * * * * * * * * * * * * * * * * * * * * * * * *
ALPHABETICAL INDEX. XXIII
PAGES
Squaring numbers, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . { } & © tº $ $ ſº tº & © tº e s & ſº tº g g g g g & 118
Statement line, . . . . . . . . . . . . * * * * * * * * * * * * * * * * * * * * * * * * * * s e º e o e s a s g g g º e º a s s * tº $ tº a tº ſº gº º g g tº e º gº tº e & 161, 184
solution, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . e & a 2 tº e s e º is e º s º tº & # tº tº gº tº dº º tº $ tº º tº # 6 º' tº ę & © e & 26
Steelyard, def. of, . . . . . * * * * * * * * * > tº e º 'º e º 'º e s e º e º ºs e e ſº * * * * * * * * * * * * * * * * * * * * * tº º sº e º e º is ſº e º s is e º 'º e 421
Stocks and bonds, . . . . . . . . . . . • * * * * * * * * * * * * * * * * * * * * * a se e is tº e º 'º º e s tº dº º ºs e s tº 8 º' & © e º is tº º is & & G e º e . . . . . 676–710
definitions of all kinds; dividends; words and phrases of brokers; to buy, sell, and
invest in stocks and bonds; to buy, sell, and invest through a broker; to find gain
per cent on ; price of stocks; amount to be invested in ; par value of; price to realize
certain rate; value at maturity; present worth of; value of bonds under varying
conditions; value of stocks after dividends; watering stocks; interest accounts with
brokers; margin to carry stock; stock exchange; United States stocks and bonds in
Europe; operations in French and English quotations; practical problems in all
phases of stocks and bonds.
Stock dividend, . . . . . . . . . . . . . . & e º e º a g º e º º is º º ſº tº tº e º 'º º e º º ſº tº º ºs º 'º e º ſº tº e º 'º e º e º 'º e º tº º tº e º e º a tº e © tº e g g is e. 678
exchange, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . © e º e e ſº tº e º & 681
Storage, . . . . . . . . . & º º 'º C & ſº & © tº e º 'º º e º º te e º 'º * † tº º ſº tº º º gº tº ſº º º ſº tº º tº gº º º tº º tº g º ſº º . . . . . . 777–785
defs. and kinds of; storage per month; for average period; for average period with
deliveries; storage when received and delivered on different dates; storage on grain.
Subtraction, . . . . . . tº e º ſº tº º & tº dº e º º © tº e º º ſº e º & © tº $ tº e º ſº ſº tº tº tº dº ſº e s tº tº e º G & º e º ºs e g º sº º e s tº º tº e º e º e g g = * * * * g º & 64–7:5
, borrowing method improper, . . . . . . . . . . tº ºs º º t e º º & © tº ſº. tº g º ºs tº e º 'º $ tº s is tº t e s tº e g & tº £ tº º e º 'º 66
, by addition, . . . . . . . . . . . . . . . . . . . e e º te e º ºs e e º e e º 'º is tº a º ºs tº tº gº g º ºn tº e s tº $ tº dº º tº º sº º e º sº º & e g º & º 66
, by commencing on the left, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
, by complement of ten, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . tº y º e º a 2 & e s tº gº tº g tº gº gº 69, 70
horizontal, . . . . . . . tº e º sº e º 'º tº e º 'º & s is º º e º 'º a tº e º 'º e º dº º e s tº e e º º tº e º tº tº & E tº dº e º ºs e e g º s & © e º 'º e ge 68
of decimals, . . . . . . . . . . . . * e e º 'º e º e º 'º e º 'º º e º e º e e e e g g º e e º ſº e º º º e s tº gº º tº gº tº g tº e º º sº e º º ºs º e ºs 224
of denominate numbers, . . . . . . . ë º e º º º q & & & © tº a tº e º tº e º $ tº & º ºs e º is º gº tº e º 'º e e º e e s tº e º ºs & e & e 269
of dollars and cents, . . . . . . . . . . . . . . . . . . . . . . . . . . . . tº g g tº 9 & © tº e e g º º tº $ tº gº & & 9 º' g º gº tº a tº t. tº 67
§ of fractions, . . . . . . . . . . . . . . . . . . . * & e º ºs e º e º e º 'º a º e º e º 'º e < * * * * * * * * * * * * * * * * * * * * * * * * * * 180–183
of metric numbers, . . . . . . . . . . tº dº º e º ſº º ſº º e º º º e º e & tº dº ſº º tº ſº. © º º tº $ tº º sº a tº 8 tº $ a tº a tº e s & © tº tº ſº & º 937
, proof of, . . . . . * @ e º ſº e º e º e e º 'º e • * e º 'º e º e < * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 64
table, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
Subtrahend, . . . . . . . e e º sº e º se º 'º e º ºs e e º e s e s e º ºs e e s a e e s a e º e a s e e s = e e s tº e e s s = * * * * * * * * * * * * * * * * * * * * * * * 64
Sucrose multiplier, . . . . . . . . . se g is e º e º ºs e º e º ºs e e s e º e s a e < e < e < e < e s = e s a e s sº e s e º e º 'º º º is a s n e s a s ∈ s & s & e º º 467
Sugar cane, to find value perton, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 466
Supplement of number, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
of number, multiply by, . . . . . º, e = e º ºs e º e s e º is a s e º e e s s e º e º º e º e s a # & © e º s a sº s º º tº e º e º ºs & 108
Surfaces, def. of, . . . . . . . . e e º e e g g g g º e e º a e s e a º ºs e e a s e s e s e s = e e s e e s e º e e s a s a sº e º sº * * * * * * * * * * * * * * * * * 337
Surface measure, . . . . . . . . . . . . . . . tº e º e s e e s tº e a tº º e s tº e s tº $ tº e e º e º e s e º sº e s e a tº e s e º ºs e º e s e e s s e º e º e º ºs e s e 243
Surveyors’ measure, . . . . . . . . . . . . * * * * * * * * * a s e s e º e s = e tº e s tº º º e º e º e º e º 'º a ſº * * © it tº ſº tº $ tº e º t e tº tº $ tº $ tº e º ſº s 244
Sweden exchange, . . . . . . . . . . e e º e s s a e º e a e s e a e s e e s is e e s e e s e º 'º e º 'º s º e º 'º º º * g º e g º e º º $ $ tº $ tº $ tº e º ſº tº º ſº & 773
Swiss exchange, . . . . . . . . . . . . . . . . • * g e º 'º e º e s is s e º e º e º e º & © e º e g * * * * * * * * * * * * * tº e º º ºs e º e º 'º º s º e s a s is e 764
Symbols and signs, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . tº e º e º is e º e e is tº e º e º us tº t tº e º s tº $ tº e s s & 36–39
Synopsis for review, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .35, 63, 74, 121, 142, 149, 160, 163, 166
System, check or key figure, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . * e g g s e º e º e º ſº e s tº e & & º e tº $ w e 155
of addition, Vicenary, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . tº dº º ºs e e º 'º e a tº e º ºs tº º e º ºs e e s & e º º 56
of multiplication, philosophic. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . & e g º ºs e e º e e s a s tº e º e e 77
of notation, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
of numeration, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . tº e º e g º º ºs e º E tº e º & e º ſº º 'º & a tº € e 30, 31
of numbers, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . © tº $ tº e º 'º & e º e º s = e s e e s is a tº * * * * * 27
Table of addition, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . º, e º is e º 'º c e s e º ºs º º e º $ tº e & e e 41, 42
of aliquots, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . e & s e º e e s = e tº tº g º e º ºs e & e º ſº e º e º is s 94, 139, 436
of annuities, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . g e s s e e s is is e e s is a e s s e º e º e º e s s º ºr a º & © tº 925, 926, 927
of areas of segments of a circle, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353, 354
of casting unearned premiums, . . . . . e e = s e º e e º s e º e º s is e º s a e s & e º is s = e º 'º e s a e s = s a tº ſº tº ſº a g º 'º gº 522
of compound denominate numbers, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .231-256
of compound 111terest, . . . . . . . . . s e e s s e s o e e º e s e e s e e º e s e s s e s s a s e º e º e s = e º e s s & e s e e º e = * * * 611, 612
of decimals, . . . . . . . . . . . . . . . . . . . . . . . . . * @ º & & & © e & • e s = e < e < e < e < e e s e e s a • * * * * * * * * * * * * * * * * * * 218
of equivalents of metric and American weights and measures, . . . . . . . . . . . . . . . . . . . . . . . Q33
of foreign coins, . . . . . . . . . . . . . . . . . • e o e s e e s e e º e s e s e e s s a e º 'º & e s s a e s is e º e s e e s s e e s a e º 'º * * * > * 743
of interest divisors, . . . . . . . . . . . . . . . . . . . . . . . . . . . * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * tº tº e º 'º º 560, 561
of Interest laws of states and territories, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 548
of life insurance, . . . . . . . . tº e e is e º ºs e º s • e o e s e e o e s a • * * * * * * * * * * * * * * * * * * * * * * * * * * * * 537, 538, 539, 541
of marking goods. . . . . . . . . . . . s e e s e e s e s is a e s e e s = * * * * * * * • * g e º e º e e s a e s e s e º 'º e º e º e s e e e s tº e s is 454
of metric systems, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .931, 932, 933
XXIV
INDEX.
ALPHABETICAL
Table of multiplication, . . . . . .
of multiples in mensuration,
of permutations,
of principal to double itself,
of regular polygons,
of roman characters, . . . . . . . . . . . . . . . . . . . . . .
of subtraction,
of time, between two dates,
of trade discount. . . . . . . . . . . . . . . . . . . . . . . . . .
Tare, def. of,
Tariff, def. of,
Taxes and licenses,
Ternary scale,
Theorem, def. of,
Thermometers, comparison of,
Thirty-six per cent method of interest, . . . . . . . . . .
Timber, logs, etc., measured,
Time exchange,
& 4 , discount on,
measure table,
Storage,
table, between two dates, . . . . . . . . . . . . . . . . . .
Tollage,
Tonnage duty, def. of, . . . . . . . . . . . . . . . . . . . . . . . . . . .
of vessels, to find,
Trade discounts and list prices,
Trapezium, def. of,
mensuration of,
Trapezoid, def. of,
Imensuration of,
Travelers’ checks,
Triangles, def. of,
Troy weight,
True discount,
remainders, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
PAGES.
Ullage rods, use of, . . . . . . . . . . . . . . . . . . . . . 413
Uncurrent money, currency, gold and
silver, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493–500
Unearned premiums, table for, . . . . . . . . . . 522
Unification of the money of the world, .. 776
Unit, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . . 25, 164
, fractional, . . . . . . . . . . . . . . . . . . . . . . . . 25, 164
of a number, . . . . . . . . . . . . . . . . . . . . . . . 25
United States bonds outstanding, . . . . . . . 680
& 4 coin table, . . . . . . . . . . . . . . . 232
& 4 money table. . . . . . . . . & º ºs 231.
4 & government lands, . . . . . . . . 245
& 4 system of partial pay-
ments, . . . . . . . . . . . . . . . . 616
Universal formula for computing interest, 561
Unlike fractions, . . . . . . . . . . . . . . . . . . . . . . . 175
numbers. . . . . . . . . . . . . . . . . . . . . . . . 25
Use of compound interest table, . . . . . . . . 613
Usury laws, evil effects of . . . . . . . . . . . . . . 545
Value, def. of . . . . . . . . . . . . . . . . . . . . . . . . . . 229
of English money in U. S. . . . . . . . 235
of figures, . . . . . . . . . . . . . . . . . . . . . . . 27
of gold and silver, . . . . . . . . . . . . . .233, 234
tº º ſº e º 'º º e º e º s a s e e º e =
& e is a s e s is e e º 'º e ºs
* * * * * * * * * e º ºs e a e º s a t e e º e º s &
* * * * * * * s & e º s e º º
& 8 & # * * * * * * * * * * * * g g g g º ºs
tº º e º 'º e º º e º e º e s tº º e º is s e e s º e º s e
* e s 3 & s º ºs e e º s tº $ is
* * * * * * * * * * * * * * * * * & & e e
* * * * * > e º 'º tº º is is a e e º is º º e º e is e < * * * * g e º sº e e
* * * * * * * * * * * * * * * * is a s tº e º e s e º a s e s e s e e e
& © e º & gº e º ſº ſº tº e º & 9 º' tº 9 & © & © s e º e º s s e is sº tº gº & e g g g = e º 'º gº e º ſº a s & e º & tº º º
& © & # tº ſº e º e º & e º 'º e º ºs e g º 'º e º 'º º e º 'º e s s e º & e
* * * * * * * * * * e º 'º s e º 'º e º 'º $ tº $ sº tº e s e º e º e º a s is a º e
e e º 'º e º E & E; e º º e º s º e º e o 'º & “ e
tº ſº e º e g º e g º ºs e º 'º a dº e s s
tº tº 4 tº & e º 'º e º & e º e º 'º e s e g
* * * * * * * e º ºs e e º ºs e e º º e e º 'º e º e º e = e º s e g º e s e as a sº tº w tº e º e º e º e º 'º e s e s is
* * * * * * * * * * * * * e º s º e = * * * * * * * * * * * * * * * *
tº e º º º ºs e º 'º e º e s tº $ tº $ tº º º te s
tº gº e º e g º dº e º e º e º e º ºs e g º ºs e º ºs e e º ſº e s e º e º e s 6 s e s s e º sº tº $ & & sº tº º is © tº * * * *
tº dº ſº tº e º ſº e º 'º º ºs e º ſe e º e s ſº e º e º e e e g º º
tº e º ºs e º e º is is e º s is is a us tº e º ſº º 'º e g is e s is e º e º is e e s s = e º e º 'º
tº e º 'º e º tº º e º O & e º & e º 'º e º e º is tº s = e s tº gº º
ê º $ ſº tº ſº gº tº º $ tº e º e º is gº tº t e º &
* @ Tº e º 'º ſº º ºs º gº & © tº e º e s tº e º 'º g is s ſº º is tº e e
tº e º ºs º ºs e º e º e º e º e º e s e e s sº e s e º a s a s e º e s e e s e s a e s e s e. g. a e s a s a wº e º º sº e < * * * * * *
tº ºp º ºs º º º Gº tº º ºs º 'º e º e º e de e º 'º e s s as e s tº e º e º e s tº
• e º e º e e º ºs e º e º is is e º e s e e e s m e º s e < * * * * * * * * * * * * * * * * * * * g e e s = e º e s e s tº e º sº * * * * * * * * * * * * *
• * g e º & © e º e s e º a tº gº º is a e º e & © tº e º ºs º is tº º is 6 & © & © tº tº tº tº
* * * e e º e º e e º e s tº e s is s = e e º e e s ∈ n e º e º e s e º is e s e s a e
is e s is e s e e e º e º e º e s e a tº e º ºs º is e g º e s p & & e º tº * * * * * * *
tº e s tº s e e º e º e s tº e s tº t is g a £ tº º e º ºs e º & & © tº e º g tº e º ºs e º 'º
e º s & e º e º ſº e s tº a tº º s e º e s m & e º sº e º e º º e º e º e s tº e º e º 'º'
tº e º e s tº s º is e º º e º gº tº $ is tº t tº $ ſº º e s & 6 & 8 tº e º e º ſº º sº tº e º &
• * * * * * g e e º 'º g g g º e e g º e º sº gº w tº e s a s e º ſº e º ºr e º e º 'º e º
• * * * * * * * * * * * * * * * e < e º e º e º ºs s = e º e s s e º e s º is e e º &
* tº & e º e s º is t w e s tº tº s = e s tº e º ſº tº & tº º & º º $ tº tº ſº º # * * * s tº tº
* * * is a tº e g º gº e º e º e s e < * g º e º ºs º a ſº e s is tº e º ºs e & © & © tº & B
* * * * * g g g g º º is º ºs e º gº tº gº & tº gº tº sº º ſº is g g & G - tº gº tº # 8 º 'º & ſº tº
* * * g º e s is º is a º º a 2 m º º e º e º s & e & 6 º e º ſº is º º ºs e º tº º e &
& a g º g g g g g g g g g g & & 4 & e º º & tº tº $ tº º º
e = 8 s g º e º e º i e º º gº tº & s º º is tº $ 4 & 8 tº º ſº & a tº 8 & 8
* e e s tº e s e s a & E 9 & e < * * * * * * * * * * * * * * * * * * * * * * * * *
s = e s e e s e e s sº º ºs e e s w e º e s & e e s m e º 'º is 4 e = * * * * * *
* * * * * * * * * * * * * * e s e s a e º e º ºs e s e ºs e º & tº e s e º 'º e s e º ºs
tº s = e º is a e g g g e s sº a e º e < e < e º ºs & & & & & a tº * * * * * * * * * *
* e º e º 'º & e s sº º e g º e g º e º ſe a e g º e s s e e s s tº tº * * * * * * * * *
* * s s is s is º gº a g º is a g º e º e º e º a s s a w e º 'º e º sº e º ºs e º 'º & &
* e º e º a tº e º e º e s ∈ is & e º is tº a tº º
sº e g º ºs e º e e s & = < * * * s e s e º e s is s a tº e º is tº e º ºs º & © e º a º
s & e e s e º ºs e º g e º e º a s e e s is e º is e e º e s tº e º e º ſº we tº e & © º ºs
e e e s a s is a e e s a s a sº e s s e º a s a s is e º ºs e e is e º 'º º º e º 'º e º 'º
* * * * * * * * tº G & © e s º e #
e e s s a e s e e s e º a e s e s e º e e s e s is e s e s e e s ∈ e e s e º e s a “
e = e g º e s s a e g º sº e s a e s = ºr = g is a e e s ∈ s & & © tº e º 'º e º is a e
• e. e. g. a s a s = e s = e e s is s a º e º sº a tº e º 'º e s is s & e º ºs e º 'º e º 'º ºf
tº gº tº º ſº
e e º e = e s = e s e e s e s a s e e s is e e º 'o e º e º 'º - * * * * * * * * * * *
e e a tº e s e e º e s e º ºs s e º e s tº gº º e º º & © º º º tº e º e < * * * * * * *
Value of insurance, & © tº
of notes bearing interest and mat-
uring in consecutive months,
of stock after declaring dividends, .
of stock having specified time
to run,
of United States stocks and bonds
in London,
of watered stock,
Valued policy, def. of,
Various methods of interest,
Vermont system of partial payments, . . .
Vicenary scale changed to decimal,
system of addition eludicated,
Vinculum, use of, . . . . . . . . e e º 'º -
Wr
Warehouses, grain and bonded,
Watered stock,
Wedge, use of,
cubical contents of . . . . . . . . . . . . .
Weight, def. of, . . . . . . . . . . . . . . . . . . . . . . . . &
of coin,
Wheel and axle, use of,
Wine measure, . . . . . . . . .
Words and phrases used by brokers and
bankers, ..
* g e º e º e s e º e º º e
& e g º a dº e º 'º is e e s ∈ tº e º is
tº gº e g º ºs e º 'º º & tº $ tº tº º
e e º e e º sº e º a e º 'º º 'º e º 'º
tº e º g º e º e º ºs
e tº e º e º e g o ºs e º sº e º e s is tº e º º ºs e º e
• e a e s , s s e º e s e s is e s is © & 8 e º 'º'
& sº e º ºs e º 'º e º e º ºs e º º sº
502
416
457–466
338.
347
338
347
56–58
134
P- ºr a y
{ { {
678
429.
384
229
233
422
247
• * e o e o e o e < * * * * * * * * * * * * * * * * * *
682,
Sl's Pilsit Pºdiul Millimits.
*1|efinitions.
º =N
1. A Definition is the meaning or import of a word or words expressed by
Other words.
2. Science is classified knowledge.
3. Art Is the practical application of the principles of science, according to
prescribed methods.
4. Quantity is anything that can be increased or diminished.
5. A Unit is a single thing of whatsoever denomination or nature, as one
Orange, one pound, etc.
6. A Number is a unit or a collection of like units.
7. Like Numbers are those which express units of the same kind. Thus:
Five apples and six apples, seven pounds and nine pounds, are like numbers.
8. Unlike Numbers are those which express units of different kinds. Thus:
Four pounds, seven hours, ten peaches, are unlike numbers.
9. Odd and Even Numbers. An ODD NUMBER is one that cannot be
divided by 2 without a remainder, as 1, 3, 5, 7, 35, 479, etc. An EVEN NUMBER is
one that can be divided by 2 without a remainder, as 2, 6, 18, 54, 536, etc.
10. The Unit of a Number is one of the collection of units forming that
number. Thus the unit of twelve hats is one hat; of five is one ; of four pounds,
one pound.
11. The Unit is the universal basis of numbers and the foundation of
arithmetic. From unity arise two distinct classes of number: 1. Integers, 2. Frac-
tions. The first class, Integers, has its origin in the multiplication of the unit:
and the second class, Fractions, results from the division of the unit. The first is
synthetical, the second is analytical.
12. A Fractional Unit is one of the equal parts into which any integral
unit is divided. If the integral unit is divided into two equal parts, each is called
a half; if into three, each is called a third; if into four, each is called a fourth ;
and so on according to the number of parts into which the integral unit is divided.
See Fractions, page 164.
13. Numbers are expressed by words, figures, or letters.


(25)
26 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. 4×
14. An Abstract Number is one in which the kind of unit or quantity is
not designated. Thus: Three, four, five, etc.
15. A Denominate or Concrete Number is one in which the kind of unit
is designated. Thus: Two pounds, five yards, nine dollars, etc
16. A Compound Number is a denominate number expressed in two or
more denominations. Thus: Five years, four months, and eight days; two miles,
five furlongs, and ten rods; two yards, two feet, and five inches.
17. The Arithmetical Units of Orders are 1, 10, 100, 1000, 10000, etc.
18. An Arithmetical Complement of a Number is the difference between
the number and a unit of the next higher order. Thus: 3 is the arithmetical com-
plement of 7; 26 is the arithmetical complement of 74; 19 is the arithmetical com-
plement of 981.
19. An Arithmetical Supplement of a Number is the difference between
the number and a unit of the same order. Thus: 7 is the arithmetical supplement
of 17; 12 is the arithmetical supplement of 112.
20. A Problem is a question proposed or given for solution, or it is a ques-
tion requiring some unknown result from stated conditions.
21. The Premises or Conditions of a problem are the known facts and
truths, therein given as a basis or predicate, for the solution.
22. A Solution of a problem is the process of determining the required
result or demonstration.
23. A Philosophic Solution is, in this work, a full numerical statement
showing, step by step, how the result of a problem is obtained, with a logical reason
for each conclusion reached in the solution. This process of solution and reasoning
from the premises and conditions of all problems constitutes the Philosophic
System of Arithmetic. *
24. An Axiom is a self-evident truth. Aacioms are the laws which govern
the logical mind in processes of reasoning. -
25. A Theorem is a statement, the truth or falsity of which is to be deter-
mined by reasoning.
26. Demonstration is a process of reasoning by which the truth of a
theorem is proved.
27. Comparison is considering the relations between numbers; the rela-
tions between the premises and conditions of problems and the required result.
NOTE. – Comparison is the true method of investigation—it is the royal road to the fields of
unknown truth and knowledge.
The comparison of two numbers as to their unit value, gives ratio; the comparison of two
ºratios gives proportion ; and the comparison of several numbers which differ by a common ratio con-
stitutes progression.
28. A Solution Statement, or an Operation, is a statement of the figures
employed in solving a problem.
29. A Formula is the expression, by symbols, of general principles appli-
cable to the operations of particular problems.
30. Philosophy is the knowledge of phenomena as explained by, and resolved
into, causes and reasons, powers and laws.
3r DEFINITIONS. 27
31. Arithmetic is the science of numbers; or to define it more extendedly,
it is that branch of mathematics which treats of the principles, properties and rela-
tions of numbers when expressed by the aid of figures, either singly or combined.
These principles and relations of numbers, combined with the facts relating to
problems, are applied, by the reasoning powers of man, to the solution of all numer-
ical problems of business affairs and of practical life.
32. Figures.—Figures in arithmetic, are characters used to represent num-
bers. The ten Arabic figures which we use, are
Naught or Cipher. One Two Three Four Five Six Seven Eight Nine
O 1 2 3 4 5 6 7 8 9
By properly combining these ten figures, all possible numbers may be repre-
sented.
The 1, 2, 3, 4, 5, 6, 7, 8, and 9 are sometimes called digits. They are also
called the significant figures, because each signifies a number when alone.
The naught (0) is so called, because by itself it does not signify or express
any number. It expresses number only when used in connection with other figures.
33. Walue of Figures.—The value of a figure is its power to express
quantity.
FIGURES have two values: 1. A simple value. 2. A local value.
34. The Simple Walue of a figure is the quantity expressed when standing
alone, or in the unit's place. Thus, when we write 3, independently of other figures,
or in the unit column of numbers, it has only a simple value expressing 3 units.
35. The Local Walue of a figure is the quantity expressed when standing
to the left of other figures. Thus, 3 in 34 expresses a quantity of thirty units; and
3 in 345, expresses three hundred units.
The local value depends upon the scale or system of numbers employed, and
the location in the Scale.
36. The Radix is the number of units of one order which it takes to make
one of the next higher order; thus the radix of the common system is 10; of the
quinary system (not in use), it is 5.
37. A Scale or System of Numbers, in Arithmetic, is a succession of
units, increasing and decreasing according to some established custom in the opera-
tions of numbers. Thus there is the binary, the trinary, etc., etc., and the decimal,
the duodecimal, etc., etc.
38. In these four named different scales, the value of four ones (1111) would
be as follows: In the binary scale or system, the first 1 on the right is 1; the second
1 is 2; the third 1 is 4; and the fourth 1 is 8; making 15 altogether.
39. In the trinary system or scale, the first 1 on the right is 1; the second
1 is 3; the third 1 is 9; and the fourth 1 is 27; making 40 altogether.
40. In the decimal scale, the first 1 is 1; the second 1 is 10; the third 1 is
100; and the fourth 1 is 1000; making in all 1111.
41. In the duodecimal scale, the first 1 is 1; the second 1 is 12; the third 1
is 144; and the fourth 1 is 1728; making in all 1885.
From the foregoing, we see that the system of scale derives its name from
28 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. M.
the Radia or Ratio of value given to each succeeding figure from the right toward
the left.
42. The Decimal Scale or System is one in which the Radix or law of
increase and decrease is always ten. This system is in general use, and derives its
name from the Tatin word decem, which means ten.
In any scale, the number of figures or characters required, including the 0, is
the same as the radix.
TO CHANGE ONE SCALE TO ANOTHER.
43. The following elucidations will show the manner of reducing one scale to
the notation of another:
1. Reduce 2436 of the Decimal Scale to the notation of the Quinary Scale.
OPERATION.
2436 – 5 = 487 fives and 1 unit over.
487 fives – 5 = 97 fives of fives or 25's and 2 fives over.
97.25's -i- 5 = 19 fives of fives of fives or 125's and 225's over.
19 125's -- 5 = 3 625's and 4 125's over.
Hence 2436 of the Decimal Scale = 34221 of the Quinary Scale.
There is no conventional method of reading numbers written in other scales
than the decimal. This number 34221, Quinary Scale, would be read, 3.625's, 4 125's,
225's, 25's and 1 unit.
2. Reduce 34221 of the Quinary Scale to the Decimal Scale.
FIRST OPERATION. SECOND OPERATION
- BY SIMPLE REDUCTION.
1. Hait of the Quinary Scale = 1 unit of the Decimal Scale. 625's 125's 25's 5’s units.
C 5 S & 4 & & & 4 & 4 - 10 units & 4 & 4 & & 3 4 2 2 1
2 25’s & & & 4 { { & 4 - 50 4 & & 4 & 4 & &
4 125's & & • { & 4 & 4 - 500 & 4 & 4 & & & & 5
: 5 & 4 & & ( & 4. - & 4 & 4 4 & 4-
3 625's * =1875 “ 19 125's
Hence 34221 of the Quin- 5
ary Scale - - - =2436 units of the Decimal Scale. ºm-
- 97 25's
5
487 5's
5
2436 units.
3. Reduce 2031 of the Quaternary Scale to the notation of the Decimal
Scale.
FIRST OPERATION. SECOND OPERATION
BY SIMPLE REDUCTION.
1 unit of the 4 scale = 1 unit of the 10 scale. 64's 16’s 4's units.
3 fours ** * * 4 “ — 12 units “ 10 “ 2 0 3 1
0 sixteens ** * * 4 ** = 0 units “ 10 “ 4
2 sixty-fours “ “ 4 “ =128 units “ 10 “ -
Hence 2031 “ “ 4 ** =141 units ** 10 “ 8 16's.
4
35 4’s
4.
141 units.
.# DEFINITIONS. 29
4. Reduce 141 of the Decimal Scale to the Quaternary Scale.
OPERATION.
141 – 4 = 35 4's and 1 unit over.
35 4’s — 4 = 8 16's and 3 4's over.
8 16’s — 4 = 2 64’s and 0 16's over.
Hence 141 of the Decimal Scale = 2031 of the Quaternary Scale.
This 2031 of the Quaternary Scale is read, 2 64's, 0 16's, 3 4's and 1 unit.
5. Reduce 4893 of the Decimal Scale to the notation of the Vicenary or
Vigesimal Scale and reverse the work.
Before this reduction can be performed, we must adopt 20 characters to repre-
sent 20 figures in the Vicenary Scale. To do this in the plainest manner, let us use
the 10 figures of the Decimal Scale and the first 10 letters of the English alphabet,
for the required additional numbers.
10 11 12 13 14 15 16 17 18 19
Thus, 1 2 3 4 5 6 78 9 a b c d e f g h i j 0
Here we have 20 characters, each, except the naught, representing a single
number or integer in the Vicenary Scale. The numerical power of the letters is
shown by the small figures placed directly above them.
Having the required figures, we make the reduction by the following
OPERATION.
4893 – 20 = 244 20's and 13 (d) units over.
244 20's — 20 = 12 (c) 400's and 4 20's over.
Hence 4893, Decimal Scale, – 12 400's 4 20's and 13 units. Or, expressed in the figures
adopted as above, it equals c 4 d.
REVERSE OPERATION.
(12) (13)
c 4 (l
20
244 20's or, thus: d 2– 13 units.
20 4 20’s – 80 “
c 400's == 4800 “
4893 units. *====º
4893 “
44. Order of Figures.—The successive places occupied by figures are
called orders. Thus in the Decimal System, a figure in the first place is called a
figure of the first order, or of the order of units ; a figure in the second place is a
figure of the second order, or of the order of tens ; in the third place, of the third
order, or of the order of hundreds, and so on; each figure next to the left belonging
to a distinct order, the unit of which is tenfold the size or value of a unit of the
order of the figure on its right.
3O SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS *
45. From the above, we see 1st: That ten units of any order in a number,
in the Decimal System, make one unit of the next higher order.
2d. That moving a figure one place to the left, increases its representative
value tenfold.
3d. That moving a figure one place to the right, decreases its representative
value tenfold.
46. Notation is a method of writing numbers. There are two Systems, the
Arabic and the Roman.
47. By the Arabic notation, numbers are expressed or Written by figures.
This system is in general use and is so called because it was introduced into Europe
by the Arabians, in the 10th century.
48. By the Roman notation, numbers are expressed or written by letters.
This system is now used chiefly to number chapters and divisions of books. It is
so called because it was used by the ancient Romans.
49. Numeration is naming the places which figures Occupy.
50. Reading Numbers is expressing their value orally.
51. There are two systems of numeration, or reading numbers, the French
and the English.
The French system is the one in general use in the United States and on
the Continent of Europe.
The English system is that generally used in England and in English Provinces.
one septillions, three hundred thirty-eight sextillions, one hundred seventy-
nine quintillions, six hundred four quadrillions, nine hundred thirty-two
trillions, four hundred eighty-seven billions, two hundred sixty-three millions,
one hundred ninety-six thousand, five hundred sixty-eight.
JD --> h = <!
º P .3 : E
E SP ă ă ş
§ 3. P . .
spºt: * > # # 3
• *ſ ad CO 3 : ‘f
©
: cº CD 2- - O .S. g g r: § 3 ;
e • F-4 º C C := E = 2 o, tº at: P = 3
& 5 SP E F 2 + 2 + 3 + 2* .3 C C 2 tº # E =
E. : z = z = # 5 # E 5 § 3 E . E . = . 5 tº 3 E =
E 5, & # 5 § 3 ; 3 =# #5 E # = z = z = É # & 5
* pºss * ºt p- • *=d
3 = + c = 0: # 02::= Gº & E H 5 ºn 5 = 5 F : # g’.
ſº to ~ - E + E * : * = 2 = 3; 3 = F = < * E = 3 3 #
= < * > . . ; ; 2 º 3 × 2 = 3 S ; # 3 F 2: 3: ; E 2 # = 3
e L' C/2 . ſº "25 -
: § – 3 C : 302 = 3 dº #3 & 3.33.33E- 23 ° .32 , 3 F : 3 § 3 ;
º Q # = 5 ; * = # = 3; a + š + F : *- : * * * * * * : *- : ; # # *
- ºn CŞ # 53 : 3 E = 3 - 3 3 E : 3 °F. : 3 5 : 3 + 3 & F : 5 ; ; s o E =
c Gº Gº Tº J. F. "E. J. F. Tº A F ºf . º 'º A rººt /, … 'E 2, o Tº c r- : "E 2, 2 3 + 35
C/2 = 2 E = 2 + = 2 +5 = 2 = = 2'3 = 2 = E Z = E Z = E Z E E 2:5 .c. : :
E - 3 ... E + 3 E → * E * : E = E E * : E F = E = E E F = E = 3 E = E 5, & 3.
5 §2 3 * = 3; 3 = 5 & E 35 E. E. 35 = P 35 ; E 3: E → E + 35 = P 35 F # ſº
5 § E3 = 5.5H5% = 5% = 5 &#5 ºr 55FF as #355,5555 ; ; ;
* .E & 8
#: @## 2 8 4, 5 6 1, 3 3 8, 179,604,932, 48 7, 2 G 3, 19 6, 5 6 8. § =
&O CO. \__N/~/ \-e-Na-> \->N/~/ \->N/~ \->~~~~~\_ºvº-Z \->-SA-Z\->TV-Z \->~~ § 3 ;
:= -s. 3 Period Period of Period of Period of Period of Period Period Period Period of Period 5 £ :
a $2 : O ep- Sex- uin- uad- of of Of Thou- of g = 3
2. 5 : Octillions tillions. tillions. tillions. rillions. Trillions. Billions. Millions. sands. Units. 2 aſ .:
ſº sº. § 3 ;
: H 5 The number is ‘read two hundred eighty-four octillions, five hundred sixty- 3 3 &
c5 ºr =
QD a $3 'S
&Q Q
##. 3 ##
; : 5
3. C. §
* T â,
# 3 &
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3rd E.
# 3
§
i
* DEFINITIONS. 3 I
Q Ú) 9 * g :
# E É #####
+ 3 .S. e wº :- -> *** = .
a.º. *- JO g: r CD GD # a 5
S2 cº F = C 3 C . ...+,+,+:
tº º £ 3 := cº • * +: #####
F. cº - . & E := aff c º T= 2, 3 #5
cº QD (10 × 2– H - • F- 2/2 m{ ºš'g'53
: â, a 5 & 5 H.3 à = = 3 ###
3. # £5 # 2 o *:::= ** = * = #####
ă ă ă ă ă ă ă ă ă c E Ž e ####g
E- L; - s CAE e CAE CO UD 9
3 ####### #: ; ; # 2, #: 2, # #####
* • * e -á r- Q
; : , ; ; ; EF 3 ; ; 5 § 3 ; # 3 ; ; ; ; ; . # 4 g #5
:= 3 (2 + Žiž Tº E 5 3.2.É3 2 to .3 & É,2: .3 Éſ. * ####
# * * = E * : E 3 - E = 2; 5 § H = 5 Tº F = . 5"; . Tº P iſ o
: É's # 3 #5, # 5 #### 5 #3; # 5 #3; # = É ă ăgă
CD . º e p- • *t Úº
s sº 's 3 × 3 × 33 3:35 s : 335 s : ; # = 3, #####
= ## = #4255.25:42.É .2.É.33: 2.É3.25 Zä33 ###
§ 3. * : *- : * – F : *- : * * * * *- : 5 - 2 : * : * ~ Z * * = 3 Q = * : 3
5: ; Pºo : 5 ; ; 3 F : 3 ; ; 3 = % cº º c - 3 × = : 3 ; ;- 3 + c : ;
(2 32 15 × 2 = 2.5 + ... à 5 × 3.5 ° à 5 ° 55 ° à 5 & 3 + ... à 5 , º, .33%
> P - ric o E = a TE = to E'E 2: E to # = 0.3 E 2 : E a 3 E. z ř; E ºf 3 ####3
2 º' 6' 35 B # 3 F # & F # 5 E É E E # 35 = # E = É 3 E É H = É 3 E É E 5: * g :
- Z # = AE = E.: E.B.: # = 3.5 E3 = E 5: 5.35 E.; F = <= 5.3 F #####
# 3 #; F ============================F ####
; : # 2 s.45 ° tº 3 s.1799 o 49 32.4 S T 2 & 3, 1965 6 sº:
Ǻ 5 §§ \ -N- /N A *—y- a -N- ~~ —’á5 3 ºf
3 * C § Perºof Quad- Period of Trillions. Period of Billions. Period of Millions. Period of Units. #####
I’ll]OIl S. +º
c + 2 f 33 . as a
E- .F. 3 The number is read two hundred eighty-four thousand five hundred sixty- a #35 ââ
2 : one quadrillions, three hundred thirty-eight thousand one hundred seventy- #3; #:
3 : -> - e !", * e º * º 3 2 5.35 %
5 §2 So nine trillions, six hundred four thousand nine hundred thirty-two billions, ... à .3?
ce B.: e º * * * * * #3: ... à
S㺠E Z four hundred eighty-seven thousand two hundred sixty-three millions, one º żºłż ż
Öſ) dº e - t º - q := 8 &
ep rº hundred ninety-six thousand five hundred sixty-eight. ###3's
THE ROMAN SYSTEM OF NOTATION.
54. In the Roman system of notation the letter I represents one ; V, five :
X, ten ; L, fifty; C, one hundred ; D, five hundred, and M, one thousand.
The Intermediate and succeeding numbers are expressed according to the
following principles:
1st. Every time a letter is repeated, its value is repeated; thus II represents
two ; XX represents twenty; CCCC represents four hundred.
2d. When a letter is placed after one of greater value, the sum of their
values is the number expressed. Thus, VI expresses sia: ; XVII expresses seven-
teen.
3d. When a letter is placed before one of greater value the difference of their
values is the number expressed., Thus, IV expresses four ; XL expresses forty.
4th. When a letter is placed between two letters of a greater value, it is
combined with the one following it. Thus, XIX expresses nineteen ; CXLVI
expresses one hundred forty-sia.
5th. A dash or bar placed over a letter multiplies its value by 1000. Thus,
V expresses 5000; X expresses 10,000; L expresses 50,000; C expresses 100,000;
D expresses 500,000; M expresses 1,000,000.
6th. In like manner, a double or a treble dash placed over a letter multiplies
its value by 1000 two times for the double dash, and three times for the treble dash.
Thus, Cexpresses 100,000,000; Mexpresses 1,000,000,000; Cexpresses 100,000,000,000;
M expresses 1,000,000,000,000.
32 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
TABLE OF ROMAN CHARACTERS.
I OD18. XXV twenty-five.
II tWO. XXVI twenty-six.
III three. XXVII twenty-seven.
IV four. XXVIII twenty-eight.
V five. XXIX twenty-nine.
VI six. XXX thirty.
VII SeWOIA, XL forty.
VIII eight. L fifty.
IX nine. LX sixty.
X ten. LXX seventy,
XI eleven. LXXX eighty.
XII twelve. XC ninety.
XIII thirteen: C one hundred.
XIV fourteen. CC two hundred.
XV fifteen. CCC three hundred.
XVI sixteen. CCCC four hundred.
XVII seventeen. D five hundred.
XVIII eighteen. DC Six hundred.
XIX nineteen. DCC Seven hundred.
XX twenty, DCCC eight hundred.
XXI twenty-one. DCCCC s nine hundred.
XXII twenty-two. M one thousand.
XXIII twenty-three. MM two thousand.
XXIV twenty-four. MDCCCXCIV=1894.
MDCLXVII = 1000667.
TMCCCCXXVMDXLII = 1426542.
MMMMDCCXXMCLMCXIV = 1003721151114.
NOTE.4–It may be proper to remark that the methods used by the ancient Romans to express
large numbers were varied and somewhat different from the method now used, and as here explained.
Because of the complex system of the Roman numerals and the great difficulty in using them, in the
study of arithmetic and in business calculations, the Romans themselves used a mechanical contri-
vance called an Abacus. This Abacus consisted of a table or board with compartments or grooves
representing a different value to be given to the counters or pebbles placed therein. In later years,
the Abacus was made with wires on which were placed movable counters, each wire representing a
unit of a different value. And in this form, the Abacus is still used by some Europeans.
The Roman notation is chiefly used in numbering chapters and sections in books, marking
dials of clocks and watches, and in writing physicians' prescriptions.
-º-º-º-º-º: 2
& Q) C) C) Q) 3
ºotation and Numeration,
— OR —
WRITING AND READING NUMBERS.
55. To Write, or Notate Numbers, begin at the left and write the figures
of each period in their proper places, filling the vacant orders, if any, with naughts.
56. To Read or Numerate Numbers, begin at the right and point off the
Inumber into periods of three figures each. Then commence at the left and read in
Succession each period with its name.
57. To Werify the Notation, or Writing, numerate the number and see
if it agrees with the number given.
EXERCISES IN NOTATION AND NUMERATION.
58. Write seven ; nine; ten; fifteen; thirty-two; eighty-nine; ninety-nine;
One hundred nine.
Write six hundred twenty-two; twenty; one hundred twenty-two; thirty-
SOVCI).
Write four; forty; forty-four; four hundred; four hundred four; four hun-
dred forty-four.
Write two; ten; two hundred eight; two hundred twelve; three hundred;
four hundred four; five hundred fifty-four; six hundred; seven hundred; eight
hundred; nine hundred; ten hundred ten.
Write one thousand one; two thousand two hundred one; three thousand
thirty-one; forty-four hundred forty; five thousand five hundred five; sixty hun-
dred sixty; eight thousand nine hundred eighty-eight; twelve thousand two;
twenty-three thousand forty-five; thirty thousand fifty-four.
59. Read the following numbers:
307 13:28 34546 142462 815621.11 220 8123 19703 770077 40404040
170 2813 10010 555055 27624555 202 7890 88888 809010 3333333333333
200 3218 40.101 (306060 11111110 322 9087 90909 100107 1201050607082
60. Write the following numbers:
1. Write 1 unit; 1 ten; 1 hundred; 1 unit and 1 ten; 1 hundred, 1 ten and
1 unit.
2. Write 1 unit, 2 tens, and three hundreds; 4 units, 5 tens, and 6 hundreds;
7 hundreds, 8 tens, and 9 units.
3. Write 0 units, 1 ten, and 0 hundreds; 4 units, 0 tens, 0 hundreds and 4
thousands. º
(33)
34 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. +&
4. Three hundred forty-one thousand, twenty-two.
5. Sixty-five million, one hundred thirty-two thousand, three hundred eighty-
SeVéll.
6. Twelve billion, sixteen million, forty-three thousand, one hundred eleven.
7. Nine hundred thousand, three hundred fifty.
8. Six million, one hundred sixty-nine thousand, four hundred thirty-seven.
9. Twenty-two billion, one hundred three million, five hundred seventy-six
thousand, one hundred two. t
10. One hundred two trillion, one hundred twenty-five million, four hundred
three.
11. Write 208 million, 18 thousand, 1 unit.
12. Write 10 billion, 8 million, 103 thousand, eleven.
13. Write 200 sextillion, 1 quintillion, 100 quadrillion, 10 trillion, 111 billion,
1 million, 10 thousand, 10 units.
14. Write 97 million, 14; 5 thousand, 5.
15. Write eleven thousand, 11 hundred, 11; 16 thousand, 16 hundred, 16.
16. Five billion, fourteen million, nine.
17. Two hundred one million, twenty thousand.
18. Seventy-seven trillion, seventy-six.
19. Eight thousand, nine hundred, ninety.
20. Four hundred twenty million, one.
21. Thirty thousand million, thirty.
22. Three hundred eleven Octillion.
23. Four hundred two vigintillion.
24. Fill all the orders of figures with 9's from the order of units to the order
of octillions, both inclusive, point off, and read the same according to the French
and the English systems of numeration.
61. Write in figures the following numbers, and numerate them according
to the English system of numeration:
1. Twelve hundred forty-one million, one hundred twenty-three thousand
five hundred fourteen.
2. Four hundred nineteen thousand one hundred fifty-two million, twenty-one
thousand forty-Seven.
3. Fifty-three billion, two hundred twelve thousand twenty-six million,
seventy-five thousand three hundred eighty-four. Ans. 53,212026,075384.
4. 1342 trillion, 11122 billion, 14 million, 19 units.
pºse
5. 1 quadrillion, 1 trillion, 1 million, 1.
62. Read the following numbers:
XXXIV. LI. C. CCC. DC. MMDCLI.
XXXV. LXI. CI. CD. DLXIX. MMMXC.
|XXXVI. LXIX. CX. CDXIV. M. MMCCX.
XLIX. XC. CL. D. MC. MMCMMDX.
Write, in the Roman System of Notation, the following numbers:
9, 12, 14, 37, 49, 83, 108,519, 1519, 14704, 88976, 13140363, 1001001001.
SYNOPSIS FOR REVIEW.
1.
:
i
19.
20.
21.
22.
23.
24.
25.
26.
27.
11
SYNOPSIS FOR REVIEW.
Cº-º-
63. Define the following words and phrases:
A Definition.
Science.
Art.
Quantity.
A Unit.
A Number.
Like Numbers.
Unlike Numbers.
Odd and Even Numbers.
The Unit of a Number.
The Unit.
A Fractional Unit.
Numbers.
An Abstract Number.
A Denominate or Concrete Number.
A Compound Number.
The Arithmetical Units of Orders.
An Arithmetical Complement of a
Number.
An Arithmetical Supplement of a
Number.
A Problem.
The Premises or Conditions.
A Solution.
A Philosophic Solution.
An Axiom.
A Theorem.
Demonstration.
Comparison.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
41.
42.
43.
44.
45.
46.
47.
48.
49.
50.
51.
52
- e
53.
54.
A Solution Statement, or an Opera-
tion.
A Formula.
Philosophy.
Arithmetic.
Figures.
Value of Figures.
The Simple Value.
The Local Value.
The Radix. f
A Scale or System of Numbers.
The Binary Scale.
The Trinary Scale.
The Decimal Scale.
The Duodecimal Scale.
The Decimal Scale or System.
To Change one Scale to Another.
Order of Figures.
Decimal System of Figures.
Notation.
Arabic Notation.
Roman Notation.
Numeration.
Reading Numbers.
Two Systems of Numeration.
French System of Numeration.
English System of Numeration.
The Roman System of Notation.


gºigns and Syrrhboſs.
signs and s>yrnees:
_-. *—ºº--A
64. Signs and Symbols are used to abridge arithmetical operations. They
also indicate some relationship existing among numbers, and what operation is to
be performed.
The signs in general use in Arithmetic are as follows:
+, −, X, +, = , (), or
, , :, ; :, ..., 16°, v', ', 19, 29, 3°, ,-
65. The perpendicular or Greek Cross, (+) is the sign of Addition ; it is
called plus, and is read plus, or and. It means more, and indicates that the num-
bers between which it is placed are to be added. Thus, 7 + 9 is read 7 plus 9, and
means that 7 and 9 are to be added. When used after a number, thus, 5 +, which
is read 5 plus, it means 5 and a small excess.
66. The horizontal line (–) is the sign of Subtraction, and is called minus.
It means less, and indicates, when placed between two numbers, that the one that
follows it is to be taken from the one before it. Thus, 7–3 equals 4.
67. The oblique or Saint Andrew's Cross, (x) is the sign of Multiplication.
It is read multiply by, or times. It indicates that the numbers between which it is
placed are to be multiplied together. Thus, 7 × 5 is read, 7 multiplied by 5, or 5
times 7.
68. The horizontal line with a point above and a point below it, (--) is the
sign of Division. It is read divided by. It indicates, when placed between two
numbers, that the one before it is to be measured or divided by the one after it.
Thus 12-i-4 equals 3.
69. The parallel horizontal lines (=) are the sign of Equality. It is read
equals or is equal to. It indicates that the quantities between which it is placed are
equal. Thus, 5+3=12—4. A statement of this kind is called an equation, because
the quantity of 5+3 is equal to 12–4.
70. The () or is the sign of Aggregation. The first is the Parenthesis,
the second is the Vinculum. They are both used for the same purpose. They
indicate that the numbers within the parenthesis or below the vinculum, are to be
considered as one quantity. Thus, 16–(5+4)=7, or 16—5+4=7. See page 134.
71. The single point (...) is the Decimal sign. It indicates that the num-
bers which follow it are tenths, hundredths, etc. Thus, .5, .05 are read 5 tenths, 5
hundredths. Af
72. The (: ) is the sign of Ratio. It is read, is to or the ratio of.
73. The (::) is the sign of Proportion. It is read, as, or equal. Thus, 3: 6
:: 5: 10 is read 3 is to 6 as 5 is to 10; i.e. 3 is such a part of 6 as 5 is a part of ten.
(36)
Yºk SIGNS AND SYMBOLS. 37
74. The (16*) is the sign of Involution. The small figure to the right and the
top of the number indicates the power to which the number is to be raised. Thus,
8* indicates that 8 is to be raised to the second power or taken as a factor twice.
Thus, 8×8=64. 8* indicates that the third power of 8 is required. Thus,
8 × 8 × 8–512.
75. The radical sign (V) is the sign of Evolution. It indicates that some
root of a number is to be found. Thus, V64 indicates that the square root of 64 is
to be found. V512 indicates that the cube root of 512 is required. V4096 indicates
that the fourth root of 4096 is to be extracted. The small figure within the branches
of the radical sign is called the index, and indicates what root is required.
Other signs of evolution, now often used, are as follows:
Thus, 256* indicates that the square root is to be extracted.
625* indicates that the cube root is to be extracted.
1463 indicates that the square root of the cube of the number is required.
76. The (...) is the sign of Deduction. It is read, therefore, hence, or con-
8èquently.
77. The Interrogation (?) is the sign meaning what, or how many. It signi-
fies that the answer to the question asked is to be found.
78. The expressions 19, 20, 30, denote first, second, third, etc.
79. The sign of the comma (,) is the sign of Numeration. It is used to
separate large numbers into periods, to facilitate the reading of them.
SIGNS PAND DABBREVIDTIONS,
_*.
-r
Aº-
wº-
80. The following are the principal signs and abbreviations in general use
among merchants and business men, and are used in this work:
Abbreviation, contraction of a word or phrase by
omitting letters or substituting certain charac-
ters.
Ø, at, to, per, for, or for
each.
% or acct., account.
A. D., the year of our
Lord.
Agt., Agent.
. M., Forenoon.
amt., amount.
ans., answer.
Apr., April.
ass’t'd, assorted.
Aug., August.
bal., balance.
bbl., or brl., barrel.
bgs, bags.
bdls., bundles.
bls., bales.
B. L., Bill of Lading.
bkts., baskets.
bot. bought.
B. M. the gay world.
B. Pay., Bills Payable.
B. Rec., Bills Receiv-
able.
To O., buyer's option.
|bgue., barque.
br., brig.
bu., bushel.
bXs., boxes.
c. or ſº, cents.
Cap., Capital.
chts., chests.
cks., checks, casks.
Co., Company.
C. O. D., Cash on De-
livery.
C. B., Cash Book.
com., commission.
const., consignment.
Cr., Creditor.
C.S., Cases.
cwt., hundred weight.
(l. pence.
D. or d., or dol., or $,
dollar.
do., or “the same.
dft., draft.
NotE.—The origin of the dollar sign, $, has been variously accounted for:
some to be an abbreviation of the letters U. S., the initials for United States.
Dec., December.
doz., dozen.
Dr., Debtor.
ds., days.
dwt. or pwt., penny-
weight.
ea., each.
E. E., Errors excepted.
E. & O. E., Errors and
omissions excepted.
emb’d., embroidered.
Eng., English.
En Ville, or E. W., in
the town or city.
ergo, therefore.
Et als., and others.
Ex Rel., on the relation
or information of the
aggrieved party.
ex., example.
exch., exchange.
exp., expenses.
f., farthing.
fav., favor.
Feb., February.
fig'd, figured.
f. o. b., free on board.
fol., folio.
fr., franc.
ft., feet or foot.
fwd. or for’d, forward.
frt., freight.
gal., gallon.
gr., grain, grOSS.
hf., #. §
hf chts., half chests.
hhd., hogshead.
in., inches.
I. B., Invoice book.
i. e., that is.
ins., insurance.
inst., present month.
int., interest.
inv., invoice, inventory
I. O. U., I owe you.
Jan., January.
Ibs., pounds.
Led., Ledger.
L. F., Ledger folio.
M., Noon.
m/a., months after date.
Mar., March.
mdse., merchandise.
MS., manuscript.
mo., month.
No., number.
Nov., November.
N. B., note well.
N. P., Notary Public.
Oct., October.
O. I. B., Outward In-
voice Book.
O. K., all right.
Oz., ounce.
p., Dage.
p])., pages.
pay’t., payment.
pd., paid.
pkgS., packages.
pcs. Or ps., pieces.
pr., pair.
per or #3, by.
plts., plates.
pwt. or dwt.,
weight.
P. M., afternoon.
P. O. Post Office
P. P. C., to take leave.
Prox., Next month.
P. S. Postscript.
pun., puncheon.
pts., pints.
qr., quarter.
qts., quarts.
R. R., Railroad.
rec'd, received.
rec’t., receipt.
R. S. V. P., answer if
you please.
s. or / shilling.
penny-
S. B. Sales Book.
Sept., September.
Schr., Schooner.
Sh., Ship.
Shpt., Shipment.
Sunds., Sundries.
S. o., seller's option.
Sr., Senior,
Str., Steamer.
Supt., Superintendent.
T. O., Turn. Over.
tres., tierces.
ult., last month.
U. S., United States.
ves., vessel.
W. I., West Indies.
wt., weight.
yd., yard.
yr., year.
# or L., pound.
%, per cent.
, number.
+, sign of addition.
—, sign of subtraction.
×, sign of multiplica-
tion.
--, sign of division.
=, sign of equality.
1", one and one-fourth.
1*, one and one-half.
1°, one and three-
fourths.
Mk., Marks, the Ger-
man monetary unit.
V, Check mark; cor-
rect; approved.
GB or $, Cifrao, used
to separate the mil-
reis from the reis in
Brazil money.
17 doz. $, ºr, $1%, $º = 17 doz., 4 of which are Ø
$10 per doz., 6 @ $12 and 7 (2 $15.
8 doz. 3 (a) 5 ſ : a 4/6, = 2 doz. No. 4 @ 5 shillings
per doz., and 6 doz. No. 5 (2 4 shillings six-
pence per dozen.
It is thought by
But it was in use long
before the adoption of the Federal currency, and it is probably a modified figure 8, denoting a piece
of eight, i. e., eight reals—an old Spani
The character £, pound sterling,
the initial letter of the Latin word Librae, pound.
sh coin, the value of which was a dollar. .
is but a capital L with a line drawn across it, and represents
(38)
X- SIGNS AND ABBREVIATIONS. 39
The sign ſ, ºr Ş. is believed to have been originally a capital S (written thus, / ), the initial
letter for the word shilling.
The sign fö., pound weight, is formed from the first and third letters of the Latin word Libra,
and connected with a horizontal line.
The sign (3) is a modification of the Latin word ad, which means at or to.
The sign ºl, fºr penny, one-twelfth of an English shilling, is a coin corresponding in value to
the gºd of the old Saxon pound, which was the same in value as the Roman demarius, the go of the
Libra. Hence used to denote penny.
The character 36 is a modification of +, the sign of division. Thus eight per cent. may be
expressed by Iño, or 8 - 100, or by omitting the denominator and writing it thus, 8-i-, or in rapid
writing thus, 8 °ſ, or as we now write it, thus 8%.
The character #3 is another form of p, the initial letter of the Latin word Per.
The abbreviation Q. K. is used by business men and telegraph operators, and signifies “all
right,” or “inspected and approved.”
The abbreviation Fr. is a contraction of the word France, which is derived from the German
Word Franke, a name of the German tribe who, in the fifth century, overran and conquered Gaul,
and established the Kingdom of France.
S. L. & C. means Shippers Load and Count. These letters are used on Railroad Bills of
Lading and restrict the responsibility of the Railroad Company regarding the correctness of the
articles shipped.
C. I. F. or Cif. These letters are used by buyers when making offers for goods, and mean
Cost, Insurance and Freight, and all charges to the wharf of the buyer. g
Shippers’ Order Notify. When these words are indorsed on a Bill of Lading, they mean
that the consignee cannot take possession of the goods until he has paid the Draft drawn on him
by the shipper and received the Bill of Lading for the goods. The Draft and Bill of Lading are
sent by the shipper to a Bank which notifies the consignee that they hold a Draft on him, which
by paying he will be given the Bill of Lading for the goods.

SSG) @2%)
* *-*-es---
3.3
Fº */
:------------Rs
81. Addition—Increasing—is the process of uniting two or more numbers
of the same name or kind, into one equivalent number.
82. The number obtained by this process is called the sum or amount.
83. The Sign of Addition is a perpendicular cross, +, called plus; it
means more; thus, 7 -i- 9 is read, 7 plus 9, and indicates that 7 and 9 are to be
added. When used after a number, thus, 5 +, which is read 5 plus, it means 5 and
a Small excess.
84. The Sign of Equality is =. It is read equals or is equal to, and denotes
that the numbers between which it is placed, are equal to each other; thus, 7 H-9
= 16, means that 7 and 9 added, are equal to 16. The expression is read, 7 plus 9
equals 16.
85. A Numerical Equation is an equality between two numerical expres-
sions, which, though differing in form from each other, are equivalent. Each expres-
Sion is called a term of the equation. Thus, 5 + 8 = 13 is a numerical equation in
which the 5 + 8 is called the first member of the equation and 13 the second
member, and both are called the terms of the equation.
86. Principle of Addition. Numbers of the same kind, order, or character
only, can be added. Thus we cannot add 2 apples and 3 oranges; or 5 pounds of
sugar and 6 boxes of peaches; or 6 units and 5 hundreds; or ; and #, etc. We
can only add apples to apples, Oranges to oranges, sugar to Sugar, peaches to
peaches, units to ultiis, hundreds to hundreds, halves to halves, fourths to fourths,
etc. We can collect together things of different kinds, apples, peaches, oranges, etc.,
but by collecting them together we do not increase the number or sum of either,
and hence there is no addition.
Addition is the basis of all numerical operations, and being so constantly
used in all departments of business life, therefore it is essentially necessary that the
accountant and business man should be both accurate and rapid in the various
methods of adding numbers. By due attention to the principles and combinations
of numbers, as herein presented, aided by practice, the learner can, in a few weeks,
acquire a degree of proficiency that will place him in the front rank of experts, and
doubly capacitate him for service to himself or to others for whom he may labor.
To assist in disciplining the mind and to point out the best and most rapid
systems of adding columns, we present the following: -
(40)
ADDITION TABLES.
4I
87.
5
7
apply accurately, the result of two or more figures.
ADDITION TABLES.
NOTE.—In learning these tables and in handling all numbers, all intermediate words and
thoughts that occur between the numbers to be combined, should be omitted. Thus, instead of saying
or thinking that 2 and 2 are 4, 3 and 5 are 8, etc., say or think 4; 8; etc.
*|1
;
;
}.
.
}.
i.
.
ee}#.
i
;
:
i
;
* "a,Tole
INTo- L-
Explanation.—In this table, we
show 20 different combinations of
the 9 significant figures, to pro-
duce results from 1 to 9. It may
be said that three 1’s make 3,
three 2's make 6, etc., and that
they are regular combinations;
but we see by the table that two
1’s are 2, and that two 2's are 4,
etc. Hence, though the table
does not contain all the possible
combinations, it does contain all
that are essential and of value in
this connection.
10
11
13
14
16
17
1
8
* "a Tole
*
:
$.
3.
3.
º
|
.
;
3.
;
INTc. 22.
Eacplanation.—In this table,
we show the 25 different combi-
nations of the 9 significant fig-
ures, the sum of which equals
ten or more. To attain rapidity
in adding, it is absolutely neces-
sary that the learner should be
so familiar with these combina-
tions that he can instantly See
the result without adding, i. e.
he must know the result by the
combination, just as he knows
the value of 4 or 5, by the com-
bination of lines forming the
figure, or as he knows the pro-
nunciation of a word without,
spelling it.
The rapid increasing and decreasing operations in the Science of numbers,
depend largely upon the capacity of the calculator to apprehend instantly and to
is to aid the learner in acquiring the desired rapidity.
And the object of these tables
88. ADDITION AND SUBTRACTION TABLE.
* Talole INTc- 3-
1 & 3–9 1 & 3–8 || 1 & 3–7 || 1 & 3–6 || 1 & 3–5 & 3–4 || 1 & 2–3 || 1 & 3–2
2. “ “ 9 2 4 4% 8 2 * { 4 7 2 : {{ 6 2 “ {{ 5 2 44 4 4 2 “ 44 3
3 “ “ 9 || 3 “ “ 8 || 3 “ “ 7 || 3 “ “ 6 || 3 “ “ 5 || 3 “ “ 4
4 (4 “ 9 4 44 “ 8 || 4 44 4 7 || 4 44 (4 6 || 4 44 (4 5
5 (€ 4 9 || 5 4 “ 8 || 5 4 “ 7 || 5 4: “ 6
6 44 4 9 || 6 4' 48 || 6 4' 4 7
7 * 4 9 || 7 44 (+ 8
8 4 “ 9
the difference between each is to be supplied by the learner,
Explanation.—In this table, we present the 36 combinations of the significant figures, in which
This is a very important table for
rapid work in subtraction, by the addition method, and should receive careful attention.
42 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS.
89. ADDITION AND SUBTRACTION TABLE.
"T"a Tole INTo- 4:-
1 & 3–100 26 & 3–100 51 & 3–100 76 & 2–100
2 * 44 100 27 44 44 100 52 “ “ 100 77 44 44 100
3 ** {{ 100 28 4 44 100 53 44 44 100 78 “ “ 100
4 “ 44 100 29 4 44 100 54 ($ 44 100 79 (4 tº 100
5 “ 44 100 30 46 (£ 100 55 “ 44 100 80 ($ 44 100
6 4' 44 100 31 4 44 100 56 4' 44 100 81 tº 44 100
7 “ (: 100 32 “ “ 100 57 “ “ 100 82 “ “ 100
8 “ (: 100 33 44 4 100 58 44 44 100 83 44 44 100
9 ** {{ 100 34 “ 4 100 59 “ (: 100 84 44 44 100
10 * $ 44 100 35 “ 4 100 60 “ “ 100 85 ($ 44 100
11 ** {{ 100 36 44 °4 100 61 (< 44 100 86 4' 44 100
12 4° 44 100 37 44 44 100 62 “ (: 100 87 “ “ 100
13 ** 44 100 38 44 44 100 63 44 44 100 88 “ 44 100
14 “ “ 100 39 4 44 100 64 “ “ 100 89 “ 44 100
1 5 “ 44 100 40 “ 44 100 65 “ 44 100 90 46 “ 100
16 “ “ 100 41 4 44 100 66 ($ 44 100 91 4 44 100
1 7 “ 44 100 42 4° 44 100 67 tº 4 100 92 “ “ 100
18 “ 44 100 43 44 44 100 68 44 44 100 93 44 44 100
19 “ 44 100 44 44 44 100 69 “ 44 100 94 44 44 100
20 ($ 44 100 45 % ($ 100 70 (£ 44 100 95 44 44 100
21 “ (4 100 46 (£ 44 100 71 44 44 100 96 “ (: 10()
22 “ “ 100 47 44 44 100 72 4° 44 100 97 (£ 44 100
23 & 4 100 48 (£ 44 100 73 44 44 100 98 “ 44 100
24. 44 44 100 49 44 44 100 74 44 44 100 99 44 44 100
| 25 “ (: 100 50 44 44 100 75 (£ 44 100
Ea:planation.—We present this table to aid the learner in instantly seeing the difference
between 100 and any number from 1 to 99. It is of special value in addition and subtraction, and
all who expect to become rapid Calculators must be proficient in this character of work.
These tables constitute the alphabet of numbers, and render obsolete the dis-
gusting and mind weakening practice of counting fingers, birds
ponds, apples on trees, or rabbits in the yard, etc., which is so often seen in primary
arithmetics.
on limbs, ducks in

:
:
&
i
i
90. With one look at each of the following groups of numbers, name the
result and repeat the operation until you can call the amounts at sight.
4
t;
:
i
i
i
i
4-:
|
:
!
9.
*
7
#
:
i
*
-;
f
3 : 1 2 4 1 6 3 7 1 8 3
4 3 3 2 5 2 6 2 2 1 8
4 5 6 7 3 4 1 4 2 9 3
4 3 2 5 9 5 6 3 5 0 2
8 9 8 9 8 9 8 8 7 9 5 6
8 6 6 4 4. 2 2 9 8 7 9 7
3 8 7 6 8 5 5 6 9 4 7 5
9 3 3 4 4 7 5 6 1 7 7 8
3 4 5 1 6 9 8 2 3 6 5
5 2 2 3 3 2 1 5 7 4 7
7 9 8 7 8 2 4 6 8 9 3
7 1 8 4 6 5 4 9 7 9 7
4 8 2 7 7 8 5 6 8 8 5
5 8 9 9 8 9 8 8 5 6 9
DRILL EXERCISE NO. 2.
6 7 8 9 4 7 2 4. 1 7 8
6 7 5 8 3 2 9 8 0 2 4 6
3 2 8 9 7 3 7 9 1 2 3 4
3 7 4 4 5 6 8 4 1 3 9 3
9 3 1 5 0 9 8 7 6 5 4 3
9 5 5 4 4 3 3 2 2 1 1
6 7 8 9 0 3 1 8 2 4 ()
6 0 9 6 8 2 5 2 7 5 8
— — — — — — — — — — —
(43)
44
SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. 4%
DRILL EXERCISE No. 2.—Continued.
9 0 8 7 5 3 4 2 1 8 6 4 9 4 1 7 2 8 9
l 2 3 4 5 6 7 8 9 0 9 1 8 2 7 3 7 4 5
8 1 2 7 0 4 7 8 9 9 8 8 3 2 5 1 9 0 1
2 7 8 1 9 2 6 7 4 5 1 2 3 4 5 6 7 8 9
7 1. 8 2 0 3 7 4 6 5 0 9 8 7 5 6 4 3 2
1 2 3 4 5 6 7 8 9 0 8 4 7 2 9 1 4 8 2
9 0 9 8 8 7 7 6 6 5 5 5 4 3 3 2 1 0 6
5 6 9 8 5 4 8 2 9 3 8 4 1 5 2 6 9 5 2 8
2 2 9 9 8 8 7 7 6 1 0 6 8 2 4 9 0 3 1 0.
1 2 3 4 5 6 7 8 9 0 9 8 7 6 5 4 3 2 0 1
3 4 7 5 6 8 2 1 9 3 0 9 6 7 4 3 2 8 1 9
1 2 3 4 5 6 7 8 9 3 8 4 9 5 2 1 6 8
9 8 7 6 5 3 2 1 7 0 4 9 1 2 8 9 0 3
3 5 8 9 1 0 3 8 2 6 4 9 0 7 8 9 1 8
4 5 4 5 4. 5 4 5 4 5 4 5 4 5 4 5 4 5
9 8 7 6 5 4 3 2 1 8 5 6 4 3 8 1 6 3 1
4 2 5 9 6 0 2 5 6 8 9 1 3 9 0 2 5 9 7
3 5 7 9 1 8 3 8 6 0 4 8 2 8 6 3 9 1 7
4 1 9 2 8 3 8 4 7 5 6 0 9 1 8 3 6 2 5
8 1 1 2 6 3 5 4 8 9 1 9 2 8 3 7 4 6 5.
8 9 0 6 3 9 2 8 1 0 6 2 4 5 9 6 8 2 9
6 7 8 9 0 1 2 3 4 5 6 7 8 9 3 8 5 9 3
2 1 4, 9 8 1 4 0 3 6, 8 9 1 6 3 8 2 6 9
3 2 8 7 8 9 7 8 3 9 3 9 8 5 2 8 2 7 7
4 5 6 7 4 8 8 9 4 7 8 9 8 8 9 3 2 1 5
7 6 5 8 3 7 9 7 5 8 4 8 7 2 7 5 2 6 5
1 2 3 4 5 6 7 8 9 () 1 2 3 4 5 6 7 8 9
DRILL EXERCISE NO. 3.
91. 1. Write all the combinations of two figures that make 2, 3, 4, 5, 6, 7,
8 and 9.
2. Write all the combinations of two figures, that make 10, 11, 12, 13, 14, 15,
16, 17 and 18.
Repeat the following exercises several times, naming the results as rapidly as
you can, and continue the exercise for several consecutive days:
3. Commence with 1 and orally add thereto 2, and continue to add 2 to the
successively occurring sums, until you produce 101. Thus, 3, 5, 7, 9, 11, 13, etc.
DRILL EXERCISE.
45
DRILL EXERCISE No. 3.-Continued.
4. Commence with 1 and in like manner add 3 until you produce 100. Thus,
4, 7, 10, 13, etc.
5. Commence with 1 and in like manner add 4 until you produce 101. Thus,
5, 9, 13, 17, etc.
6. Commence with 1 and in like manner add 5 until you produce 101.
7.
:
11. Orally add by 2’s until you produce 20.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
{{
{{
{{
4%
{{
{{
{{
6é
& 4
{ % 3'S
{{ 4’s
{{ 5’S
4% 6'S
4% 77S
{{ 8’S
44 9'S
** 107S
46 11?S
<< 127S
{{ 137S
{{ 14?S
{{ 15'S
4%
{{
{{
4%
4%
{{
{{
4% 30.
44 40.
{{ 50.
44 60.
44 70.
& 4 80.
{{ 90.
{{ 100.
4 110.
44 120.
{{ 130.
** 140.
4, 150
Commence with 1 and in like manner add 6 until you produce 61.
Commence with 1 and in like manner add 7 until you produce 71.
Commence with 1 and in like manner add 8 until you produce 81.
Commence with 1 and in like manner add 9 until you produce 91.
25. Commence at 1 and orally add 3 and 5, alternately, until you produce 100.
26. Commence at 1 and orally add 4 and 7, alternately, until you produce 100.
30. Add by 2's, 3's and 4's alternately from 0 to 99.
92. Add the following problems orally:
5-H6+8=?
7+4+9=}
6-H7+5=}
9-H8+8=?
4-H 7–H9–?
8-H 9-H6=?
14+ 7+ 3=}
21–H 0–1– 2=?
7+12+ 0=?
26-1- 9-H12=?
8-H23–H 5=}
10–1–15–H16=}
22–H 8+ 1=?
12+ 9-H 7=?
4+ 9-H 8=?
20–H40–H 3=?
22+14+ 4 =?
17+19.--18=?
What is the sum of 8 peaches and 10 Oranges }
What is the amount of four 0's plus three 0's?
Add twelve thousand, twelve hundred and twelve.
What is the sum of twenty blackbirds, twenty sick birds and twenty-six red birds?
24+16+ 5+ 8=?
7+ 9-1-30+ 5=?
18-H 12–1–15+ 6=?
43+ 7+14+ 9=?
72-H21+ 9-H 14=?
69-H15+12+36=?
46 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. Aſ.
WBITTEN PROBLEMS IN ADDITION.
93. Add the following numbers, 6376, 564, 309, 485, and 5092:
OPERATION.
:
:
|
3
()
9
485
5092
Sum, 12,826
Carrying figure8, 132
Ea:planation.—In all addition problems, we first write the numbers
so that units of the same order stand in the same column, i. e. whits in the
units’, or first column; tens in the tens', or second column; hundreds
in the hundreds', or third column, and so on through the numbers. We
then begin at the units’ or first column and add the columns separately.
In adding the first column, we commence with the 2 and 5, and name only
the successive results; thus, 7, 16, 20, 26, which is 2 tens and 6 units; the
6 we write in the first place, or column of units, and place the 2 tens
which is to be carried to the column of tens, directly below the 6 in a
small figure. Then adding the 2 tens to the tens' column, we say, 11, 19,
25, 32—which is 3 hundreds and 2 tens; the 2 tens we write in the column
of tens, and place the 3 hundreds which is to be carried to the hundreds'
column, directly under it. Then adding the 3 hundreds to the hundreds'
column, we say, 7, 10, 15, 18, which is I thousand and 8 hundreds ; the 8
hundreds we write in the hundreds' column, and the carrying figure, 1
thousand, directly under. Then adding the 1 thousand to the fourth, or
thousands' column, we say, 6, 12, which is 1 ten thousand and 2 thousands, and this being the last col-
umn to add, we write the figures in their respective columns and produce 12826 as the 8wm of all
the numbers.
NOTE.--When adding, set the result in pencil figures, being careful to place the carrying fig-
ure or figures directly beneath the answer figure of each column added, as shown in the preceding
problem.
Different methods of using the carrying figure, elucidated by the following
examples:
THIRD METHOD.
FIRST-REGULAR METHOD. SECOND METEIOD.
$ 5462.48 $ 5462.48 37 $ 5462.48
848.19 848.19 31 848.19
761.78 761.78 25 761.78
81484.52 81484.52 33 81484.52
7948.93 7948.93 40 7948.93
852.57 852.57 13 852.57
smºmºmºsºm-mººms *=ºmºsºmºmºsºms 8 **E=
$97358.47 .37 OI’ smm mºsº, sºmmºns 8
1432,33 3.1 $97358,47 13
* > 25. 40.
33 33
40 25
13 3.1
S .37
$97358.47, $97358.47
Explanation.—By the first or Regular Method, which is preferable for accountants, the carry-
ing figure is written in a small figure under the resultant figure of the column added.
By the second method, we commence on the right and add each column independently of the
other columns, and set the results under the numbers added, or to the right of them, as is shown in
the operation; then these several results are added, and produce the sum or amount.
By the third method, we commence on the left and add each column independently of the
other columns, and set the results as shown in the operation; then we add these several results and
produce the sum or amount.
It will be observed that in the second and third methods the units, tens, hundreds, thousands,
etc., are written in their respective columns.
NotE.-In the second and third methods, the results of the separate columns may be written
to the right or left of the numbers added, or on a separate piece of paper, and the final result
thereof written beneath the numbers added.
NOTE 2. The second method, shown above, is sometimes called the Civil Service Method,
because it is used by many of the Civil Service Clerks.
x- GENERAL DIRECTIONS FOR ADDITION. 47
GENERAL DIRECTIONS FOR ADDITION.
94. From the foregoing elucidations, we derive the following general direc-
tions for addition :
1. Write the numbers so that units of the same order stand in the same column.
See explanation, page 46.
2. Begin at the units’, or first column on the right, and add each column sepa-
rately, writing the unit result under the column added, and carrying the tens, if any,
to the tens’ column. At the last column, write the full sum of the column. See explan-
ation, page 46.
3. To add horizontally, the numbers are not written in columns of like orders;
and the result, or sum, is written to the right of the numbers.
PROOE OF ADDITION.
95. The best proof of the correctness of addition is for the calculator to be
proficient in his work, and then re-add the columns in the reverse direction. Cast.
ing out the 9's, as is sometimes done, is not positive proof of correctness, as the
figures may be transposed, or replaced by wrong ones having the same sum, or
mistakes may be made that balance each other, and the work may appear to prove
right when it is wrong; though it will never prove wrong when it is right; hence
many accountants in verifying their calculations, prefer to repeat, or go over the
addition in a reverse direction.
When a column is added in the reverse direction and the result disagrees
with the first addition, then omit the bottom figure of the column when adding it up
and the top figure of the column when adding it down, until all the other figures of
the column have been added, and then add the omitted figures. By this means, the
combination in which the error has been made is changed, and the error, which
might be repeated indefinitely, is avoided.
See page 152 for a system of proof of addition, subtraction, multiplication,
and division, and also of the proof of the correctness of the transfer of numbers.
WRITTEN PROBLEMS.
96. Add the following numbers, and verify the work by reverse addition:
359 Ea:planation.—Commence at the top of the right hand column and add, naming
; results only, thus: 13, 20, 22, 30, 37, 42,49; then add the 4 tens to the second column
892 and proceed in the same manner, 9, 10, 13, 22, 28, 32, 34, etc.
768 In adding, the thought of words and the repetition of results should be
; banished from the mind. Never allow yourself to spell your way or crawl up or
587 down a column, as is too often done by many teachers and their pupils. Thus: 9
--> and 4 are 13, 13 and 7 are 20, 20 and 2 are 22, 22 and 8 are 30, 30 and 7 are 37, 37 and
5429 5 are 42, 42 and 7 are 49.
48
soule's PHILOSOPHIC PRACTICAL MATHEMATICs.
(2)
47
72
12
45
68
21
33
64
25
35
(3)
63
28
34
98
92
32
99
55
67
23
97.
WRITTEN PROBLEMS.–Continued.
(4)
124
690
658
67
623
489
109
327
432
399
ADDITION
This is a very rapid system of adding and consists in grouping two or
. (5)
845
324
97
8
326
72
216
123
78
666
(6)
4072
1867
504
1090
68
507
92
417.3
8293
8902
(7)
6913
2456
4379
308
6547
7085
2123
3458
6565
4567
(8)
540314
123456
342189
67310
102938
125439
334455
121212
304958
123456
BY GROUPING THE FIGURES.
(9)
687.659
789021
90876 .
427893
608091
623784
989790
602548
553241
789021
more figures in a column and naming the result of the group, in the same manner
as we group and name the result of several letters in words, when we read. The
following problems will illustrate this method:
11;
31}
37;
º
;
69
#18
22
8
6
34
4
41
9
61
6
ſi
9
1
|
Explanation.—Commence at the top of the first column and add thus: 13, 22,
34, 41, 49, 61. Then add the 6 tens or carrying figure to the second column and
continue as before, 17, 31, 37, 49, 61, 69.
NOTE.--To attain a high degree of skill in this method, the student
should know perfectly the alphabet of numbers, as given on page 41.
3
DRILL EXERCISE.
98. In the second, third and fourth examples following, the columns are
grouped, as indicated by the braces, as they Would naturally be grouped mentally,
in practical addition :
|}
(
(
jºr HORIZONTAL ADDITION. 49
In adding long columns, the rapidity may be often increased by grouping and
adding such numbers as make 10 or 20, independently of other groupings.
Where the same figure occurs more than once, multiply it instead of adding.
Thus, if the figure 7 occurs 4 times, multiply it by 4 and use the product instead of
adding.
Where figures occur in regular order and the number is odd, thus, 5, 6, 7, or
19, 18, 17, 16, 15, then as many times the middle number as there are numbers in
regular order, will be their sum.
When the number in regular order is even, thus, 2, 3, 4, 5, or 17, 16, 15, 14,
13, 12, then one-half as many times the sum of the extremes as there are numbers
in regular order, will be their sum.
Accountants generally add down the column so that the sum, when found, is
at the foot of the column where it is to be written, and to prove the result, they
re-add the column upward.
EHORIZONTAL ADDITION.
99. This method of addition consists in adding numbers from right to left, or
from left to right, when they are written in horizontal lines or when they are not
written in vertical columns of like orders.
The method is very valuable in business and is in general use by the most
efficient clerks and accountants.
NOTE 1.--When using this method great care is required to add, only FIGURES OF LIKE
ORDER.
NOTE 2.—The method of adding two figures of the numbers at a time may be used when
adding horizontally.
Add horizontally the following numbers:
(1) 456, 248, 3947, 1084, 2681, 89020 = 97436.
NOTE.--When adding horizontally, add toward the side at which the total result is to be
written. Thus in the above we commence on the left and say 6, 14, 21, 25, 26; then (carrying the 2
to the 5) 7, 11, 15, 23, 31, 33; then (carrying the 3) 7, 9, 18, 24; then (carrying the 2) 5, 6, 8, 17;
then (carrying the 1) 9, which completes the result.
\
Add the following numbers horizontally:
(2) 245, 362,2581, 409, 687, 5498, 64, 791062 =
(3) 79, 508, 309, 84, 506, 1742, 18, 164307, 4235, 78215 =
(4) 516, 4027, 684, 97, 68, 1493, 54, 352085, 781, 5638, 721409 =
Add the following numbers horizontally and vertically:
(5) 4345 1542 5136 6362 3809 3908 2621 # $: ; ; ; ;
2214 214 136 1043 2132 1025 47 # # 3&#: ;
3625 2153 214 256 3429 836 215 # #35:5: ;
5625 2101 9876 2123 1215 3724. 215.2 § 3; ::::::
1212 3.120 215.2 1312 21.21 3002 82 # $ $35: ;
1198 28 73 2006 7849 2150 6 $ $: ; ; ; ; ;
4621 982 4215 2982 1215 1098 4635 # * : * : *: ;
5731 4842 1563 2842 1563 4910 3125 #$: ; ; ; ; ;
#::::::::::A; ####:}; ***** | ***** ## 8:3# $$$$$$$. :*:::::::::::$ *:::::::: ;
5o 'SOULE's PHILOSOPHIC PRACTICAL, MATHEMATICS. *
DRILL EXERCISES.
100. What is the sum of each of the following Columns of numbers?
(1) (2) (3) (4) (5) (6)
4304 780 890 777 9040 984.75
291 1261 706 888 1288 8697
643 537 73 999 9907 46789
98 309 4009 666 6543 57698
1400 6987 8888 645 2018 94.576
2099 7286 5679 388 5281 49567
8877 1896 2834. 777 6745 84685
101. Add the following numbers horizontally:
(1) 248, 3936, 409, 1278, 97, 563, 9210, 45682 =
(2) 335, 1468, 87, 911, 1809, 19068, 54, 17401 =
(3) 72, 13615, 41848, 1905, 8, 9763, 691, 8625 =
102. Drill daily on the following columns of numbers with and without
grouping, until you acquire great rapidity:
(1) (2) (3) (4) (5) (6) (7) (8) (9) (10)
818 412 582 328 809 98.1 864 677 595 849
390 297 578 346 523 350 363 305 249 283
970 318 757 386 605 269 629 420 463 327
276 824 420 672 848 789 145 982 830 651
752 932 731 793 945 696 174 217 221 543
843 373 542 864 397 136 144 326 232 502
865 576 853 965 684 169 176 111 151 113
129 876 684 448 976 295 767 871 387 4.38
768 444 743 404 666 468 644 512 516 455
904 102 915 151 217 687 747 814 247 328
972 814 686 148 879 825 156 376 331 633
114 331 637 263 516 951 106 468 281 624
346 554 917 295 259 784 872 189 828 581
545 161 650 161 896 122 694 177 986 491
622 197 411 461 864 440 788 885 817 888
749 490 237 874 565 450 866 264 918 992
717 876 349 898 150 414 944 294 289 202
222 902 489 769 514 654 922 896 259 548
234 396 698 243 446 789 116 597 381 365
166 484 228 174 576 458 911 814 329 208
365 235 433 952 489 747 866 277 678 662
272 386 949 683 394 636 179 476 640 764
729 624 687 574 407 241 129 716 821 287
955 897 762 956 812 477 659 802 457 848
177 477 849 658 798 681 778 584 587 255
tºm- tº mºmmasºme - - tº- - -** *- -*m-ſº
12978
TWOTE.-By prefixing 1 to the second line from the top, and affixing the unit figure of the third
line to the second line, the footing or sum of each problem is shown.
Yºr ADDITION OF FRACTIONAL NUMBERS 5 I
TO ADD THE FRACTIONAL NUMBERS, 4 or g.
103. In business, the measurements of some lines of goods are recorded to
the accuracy of 4 and # of the unit of measurement, and such parts or fractions are
Written in Small figures to the right and at the top of the whole number thus, 41*,
39°, 40° would be read, in cases where the small figures represent FOURTHs,
FORTY-ONE AND ONE-FOURTH, THIRTY-NINE AND THREE-FourTHs, and FORTY AND
TWO-FOURTHS. In case the small figures indicated EIGHTHs, the numbers would
be read 41 and ONE-EIGHTH, 39 and THREE-EIGHTHs, and 40 and Two-EIGHTHS.
NOTE:-This work is presented in advance of fractions, on account of its importance in busi-
ness. It is believed that it can be readily understood by the practical sense of students who have
not yet studied fractions.
Add the following numbers vertically and horizontally:
41*, 39°, 40°, 418, 391, 388, 38°, 40°, 418, 398.
OPERATION
wºux. OPERATION HORIZONTALLY.
398
402 41*, 39°, 40°, 418, 39%, 38°, 38°, 40°, 41*, 39* = 4013.
413
391
388 Explanation.—Commencing at the top of the vertical column or at the
; left of the horizontal line and adding the small figures indicating FOURTHS, we
4.18 obtain 23 FourTHs. Then, since 4 fourths equal on E, in 23 fourths we have 5
393 and 3 fourths. Then adding the 5 to the other whole numbers, we produce 401
401 # and 3 fourths.
NotE.—In business, where there are many bales, bags, gallons, yards, pounds, etc., to add,
it is usual to set the numbers to be added in columns of 10, as shown in the following examples:
What is the total weight of the following 78 bales of cotton ?
(Add the footings of the columns horizontally).
(1) (2) (3) (4) (5) (6) (7) (8)
425 465 525 546 499 500 458 519
465 519 423 424 426 427 428 425
501 502 503 501 504 506 507 508
509 510 511 512 513 514 515 516
430 431 423 433 444 445 . 446 447
517 518 519 520 523 524 525 526
448 449 450 451 452 453 454 455
527 546 537 536 534 532 529 544
528 529 530 531 532 533 534
546 545 543 540 538 458 455
* * * * + 3& M. Jº 3: 39 ºf jº. # * * * * * * * % # # * + 3 + +& * * * * – ##### lbs.
NoTE.—When weighing bales of cotton, it is not the custom to consider the fractions of pounds.
Find the total weight of the following 96 bags of coffee :
(Add the footings of the columns horizontally).
(1) (2) (3) (4) (5) (6) (7) (8) (9) (10)
160 161 162 163 164 165 163 160 162 161
161 ,165 164 160 163 162 161 165 163 163
165 164 163 162 161 160 161 162 161 162
164 163 162 161 160 165 164 163 161 162
163 162 161 160 165 164 163 162 162 160
162 161 160 165 164 163 162 160 160 161
165 160 161 162 163 164 165 162 162
160 161 162 1:63 164 165 160 164 161
165 164 163 162 161 160 165 161 160
160 161 162 163 164 165 163 160 163
* =º tº-º — = ***** lbs.
52 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS.
Weekly Report of Mail Matter Received at a City Post Office.
CLASS OF MAIL. Mos. TUEs. | WED. | THURs. FRI. SAT. | TOTAL.
-- Qrdinary Letters . . . . . . . . . . . . . . . . . . . . . 29246 || 28561 29203 || 27693 27367 39985 | *****
Registered Letters. . . . . . . . . . . . . . . . . . . . 1657 1823 1654 1236 1342 394 | *****
Postal Cards. . . . . . . . . . . . . . . . . . . . . . . . . . 5129 4214 3942 4870 5986 4652 | * * * * *
Book Packets . . . . . . . . . . . . . . . . . . . . . . . 1844 1297 1329 1656 894. 773 | * * * * *
Farcels * * * * * * * * * * * * * * * * * s • * * * * * * * * a e s a 950 720 604 898 471 504 | *****
Newspapers . . . . . . . . . . . . . . . . . . . . . . . . . . 41695 || 42322 || 44.275 || 46889 573.18 40207 | *****
Total * * * * * * * * * * * * * * * * e = * * * * * * * * * * * * * * + 2 + # # tº * * * * + ºf # * * % ºf 34, 3- 36. * * * * * # * * * $6.
Weekly Report of a Circulating Library.
CLASS OF BOOKS. MON. | TUES. WED. THURS. FRI. SAT. ToTAL.
Fiction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4761 3857 1935 4560 2976 4085 | *****
History. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1322 1421 1202 2121 1322 1382 | *****
Biography. . . . . . . . . . . . . . . . . . . . . . . . . . . . 1215 1110 1313 1214 1124 1174 *****
Science . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2424 1324 1387 1216 1317 1566 | *****
Poetry. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1316 1218 1400 878 420 817 | *****
Religion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198 408 108 135 308 297 | *****
Total . . . . . . . . . . . . . . . . . . . . . . . . . . # 4t 36 & # * * * + 3 + ºf * ºf 34 × * * * * 2::::kx sk::::::::::
|ADDITION OF DOLLARS AND CENTS.
104. Dollar and Cent Signs.—The dollar sign is $, and the cent sign is 2.
When the dollar sign is placed before numbers, they are read as dollars. Thus, $45
is read 45 dollars. When the cent sign is placed after numbers, they are read as
cents. Thus, 14g is read 14 cents. When dollars and cents are written together,
the cents are separated from the dollars by a decimal point (...) and the sign of
cents is omitted. Thus, $16.45 is read 16 dollars and 45 cents.
Since there are 100 cents in 1 dollar, cents always occupy two places and only
two, in connection with dollars. When the number of cents is less than 10, a naught
must be used to fill the tens’ column, or the first place at the right of the point.
Thus, 8 dollars and 5 cents are written $8,05.
When cents only are written, they are expressed as follows: 25 cents or 25g
or $.25. -
When writing numbers representing dollars and cents for the purpose of
addition, they must be set so that dollars will be under dollars, and cents under
cents, in the regular order of units, tens, hundreds, etc., and the decimal points (...)
that separate dollars and cents must be in a vertical line.
The dollar sign ($) and the point (...) should never be omitted when writing
dollars and cents, except when writing in books, or on paper, which contains dollar
and cent columns.
105. The United States Monetary Units are as follows:
10 mills (m) = 1 cent, 2. 10 dimes = 1 dollar, $.
10 cents = 1 dime, d. 10 dollars = 1 eagle, E.
* : PROBLEMS. 53
PROBLEMS.
(1) (2) (3) (4) (5) (6)
106. Add $321. $521.16 $ 9.45 $ 431. $194.15 $684,22
640.80 83.25 80. 124. 8.05 708.10
9.13 19.30 17. 381. 73.75 276.14
75.20 S. .65 569. 6.13 582.98
100.05 4,07 6.10 827. .95 928.35
$1146.18 $635.78 $113.20 $2332- $283.03 $3179.79
7. Add $108, $97.16, $84.12, $ .75, $8, $6.40, 25g, $18. Ans. $322.68.
8. Add $580.10, $671.23, $794.98, $88, 45%, 5%, $3.10. Ans. $2137.91.
9. Add $999.99, $888.88, $777,77, $666.66, $555,55, $444.44, $333.33, $222.22,
$111.11, and 12. Ans. $4999.96.
10. Add $987.65, $876.54, $765.43, $654.32, $543.21, $123.45, $234,56,
$345.67, $456.78, $567.89, $678.90, and $789. Ans. $7023.40.
11. James has $420; Conrad has $130 more than James; and Henry has as
much as James and Conrad together. What sum have the three ? Ans. $1940.
12. A lady paid $42.75 for a dress, $80 dollars for a shawl, $21 for a bonnet,
and $8.75 for a pair of shoes. What was the total cost 3 Ans. $152.50.
13. Add the following columns vertically and horizontally, and verify the work
by adding the results:
Totals.
$468|24 || $306|50 || $ 48|21 || $516|18 || $233|40 || $108|96 || $410 |25 || $337|68
123 25 789 |06 437 |52 213|78 478 || 09 890 43 289 |67 869 || 30
371 || 60 930 || 48 270 |78 418 55 407 || 54 173| 26 157 |87 687 || 23
987|25 270 45 103 |48 682 |76 823 47 527 | 28 193 |16 184 || 50
818|25 390 || 31 970 |87 276|43 752 |34 843| 57 865 |12 976 | 89
972|11 434 || 65 545 |62 274|90 717|30 222 |34. 166||36 527 | 72
729 |55 217|41 297 ||31 318|82 249 || 32 576 87 554 |43 161 || 19
235 |45 490 || 78 684 || 44 582|57 757 |42 543 | 85 684 |68 560 || 43
328|80 809 | 66 981 |32 346|24 523 |45 350 | 67 386 |45 605 || 78
848|77 789 |55 169 |45 123|45 234 56 456 || 78 890 |71 256 |87
ADDITION OF SEVERAL COLUMNS IN ONE OPERATION.
107. In many cases, the addition of several columns at one operation will
greatly expedite the accountants' work; but in long columns of solid figures, the
addition by grouping, as above explained, is by far the most rapid and practical.
The following will illustrate the primary work of adding several columns at
ODCO :
e d :
1 º Explanation.—Commence at the top and proceed by adding to the upper
71 number 47, first the tens of the next number below, then the units, and in like
62 manner continue till all are added, thus: 47 -- 10 – 57 –– 8 = 65 + 70 = 135 +
. 1 = 136 -- 60 = 196 -- 2 = 198 + 90 = 288 + 4 = 292 + 20 = 312 + 6 = 318,
the sum of the two columns.
318
54 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. º:
By naming results only, which in practice should be done, we have 47, 57, 65,
135, 136, 196, 198, 288, 292, 312, 318.
2. Add :
; Explanation.—Commencing at the top, we have :
1587 . 1 3691 2 5691 3 5791 4 5796
4210 5 6796 6 7296 7 7376 8 7383
11593 9 11383 10 11583 ** 11593, the answer.
These operations show the basis of adding several columns simultaneously,
and though the method is too laborious to be of material advantage in long columns,
it should be practiced until the mind can easily and readily retain large and vary-
ing numerical results.
In practice, we shorten the work very much by combining and adding whole
amounts at once. Thus, in the second problem, by combining, as we would naturally
do in practice, we would have in the two upper numbers 5796, then adding the
hundreds' and thousands' figures (15) of the third number, we have 7296; then add
the units' and tens’ figures of the third number, we have 7383, to which add the
fourth number at once, and we have 11593, the correct result.
To show clearly how to contract, and to practicalize the above system, we
present the following combinations, which should be carefully studied.
ADD THE FOLLOWING NUMBERS :
(3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14)
23 52 84 68 98 88 97 96 95 89 99 89
16 36 95 72 59 46 89 99 88 87 98 98
39 88 179 140 157 134 186 195 183 176 197 187
Explanation.—In the third, fourth and fifth problems, the sum of the units' figures being less
than 10, the whole sum is instantly seen.
In problems 6, 7, 8, 9, 10, 11, 12, 13 and 14, we instantly see that the sum of the units' figures is 10,
or an excess of 10, and hence we know that the sum of the tens is to be increased by 1, and without
thinking what the excess is, we write the result of the tens’ figures, which, according to the combi-
nations, shown in tables 1 and 2, we know without adding, and while writing this result, we give an
Instantaneous thought to the exact sum of the units' figures.
ADD THE FOLLOWING NUMBERS :
(15) (16) (17) (18) (19) (20) (21) (22) (23) (24) (25) (26)
74 86 88 68 126 342 485 1627 178 526 395 2653
98 95 97 91 75 92 304 211 201 308 296 1210
172 1st 185 159 201 434 739 138s 379 s54 691 3863
Explanation.—In these problems, from 15 to 20 inclusive, we show how to facilitate and expe-
dite the work when one number approximates 100. Thus, in problem 15, we first mentally add 100
to the 74 which makes 174, and then, because the 100 added is 2 more than the 98, which should
have been added, we mentally deduct 2 from 174, and write 172, the correct sum. In problems 16, 17,
18, 19, and 20, we mentally add to each upper number, 100, and then from the several sums
thus produced, we mentally deduct, respectively, 5, 3, 9, 25, and 8. In problems 21 and 22, we first
mentally add, respectively, 300 and 200, and then to the respective sums thus obtained, we mentally
add 4 and 11. The other numbers are added in like manner.
NOTE.-Combinations similar to the foregoing problems, 3 to 26, must be practiced until the
student can easily and rapidly perform the work, otherwise, proficiency in addition and multipli-
cation cannot be attained.
Rapid work in simultaneous multiplication requires the instantaneous sum of two pairs of
numbers, and hence, again, the importance of proficiency in this work.
NOTE.—In using numbers mentally or orally, the calculator should always think or speak the
result in pairs, thus, 2476 and 158, should be read mentally or orally, 24, 76, and 1, 58.
DRILL PROBLEMS.
55
DRILL PROBLEMS.
108. Add the following columns at one operation, and drill on the Same, at
different sittings or lessons, until you achieve rapidity and accuracy:
87 146 184 64 85 240 264 429 793 843 798
96 95 72 98 132 193 187 367 241 427 524
64 27 92 58 42 321 284 316 475 762
27 81 36 21 88 406 343 509 666 533
18 45 47 76 98 463 220 704 988 487
74 47 362 432 567 4807 463 65342 123456
21 88 406 234 405 621 247 37286 906342
36 92 29 444 219 1282 5002 22406 421362
48 81 391 843 924 1096 1603 1321 187367
19 67 132 777 633 2761 4120 74350 8899
27 41 254 96 111 7932 2031 64275 364819
86 78 345 123 456 101 501 11111 300405
32 72 789 202 404 8763 1004 24359 87.1460
59 33 876 463 99 5832 803 606 463220 -
63 12 101 508 399 1847 160 10009 768
58 92 999 387 555 1234 4512 56789 102938
ADDITION BY ONE OF TEIE PROPERTIES OF 9
109. 1. What is the sum of 5 lines of numbers, the first being 467 ? Ans. 2465°
OPERATION: 467 1st line.
382 26 %
815 3d 44
617 difference between 26 line and 9.
184 {{ {{ 3d line and 9.
2465 Ans.
Explanation.-In all problems of this character, with any number of odd lines, the answer
may be produced as soon as the FIRST line of numbers is given, by writing the first line minus the num-
ber of 9's to be produced, and then prefixing the number of 98 to this result. Then, when the other
lines are set, to insure the result, every two lines must make 9, or, in other words, one-half of the
other lines must be such numbers as will, when added to the remaining half alternately, produce 9
in each column of the two lines added.
In this problem, we subtract 2 from the first line 467, and write 465, to which we prefix the 2
and thus produce the answer 2465. It should be borne in mind that 5 lines give two 9's, 7 lines three
9's, 9 lines four 9’s, etc.
2. What is the sum of 9 lines of numbers, the first being 10644? Ans. 410640.
OPERATION: 10644. 1st line.
23456 2.d 4%
13482 3d 44
96.780 4th {{
25621 5th €4
76543 difference between 26 line and 9.
865.17 44 {{ 3d “ and 9.
03219 4% {{ 4th “ and 9.
74378 4% 4% 5th 4 and 9.
What is the sum of 7 lines of numbers, the first being 86021? Ans. 386018.
56 soul E's PHILOSOPHIC PRACTICAL MATHEMATICs. A.
TEIE VICENARY SYSTEM OF ADDITION
110. The Vicenary System combines the grouping method of adding, with
a method of recording the 20's on the joints of the fingers as fast as they are pro-
duced by the operation of adding.
NotE.—This system was elucidated and published by the author in 1873.
All expert calculators admit that the most rapid system of adding long col-
umns of numbers is by grouping two, three or more numbers and mentally naming
the result of the group.
Btut such a system requires a constant strain on the mind to retain the large
and varying results and to make the several additions of the different groups, that
with many it is wholly impracticable. To avoid this mental strain, several different
methods of operation have been suggested by different calculators, but all have had
some demerit that has rendered them more or less objectionable. With a view,
therefore, to obviate the objections and difficulties of the usual methods and render
the work simple, comprehensible and practical, we present a new system, which,
from the character of the work, we name the Vicenary System. The leading pecu-
liarity of this system consists in recording on the different joints of the fingers and
thumb of the left hand, the 20's as soon as they are produced, and thus freeing the
mind from all effort to retain large results, or of making the rather difficult addi-
tions resulting from the several groupings.
THE WICENARY SYSTEM ELUCIDATED.
111. We have before stated that the leading
characteristic or peculiarity of the Vicenary Sys.
tem of addition consists in recording the 20's as
Soon as produced, on the joints of the fingers of
the left hand; and in order to render this part of
the operation clear, we present a cut of a hand
with the different joints of the fingers numbered,
so as to represent the values that we record on them in adding. The manner of
making the record of the 20's on these joints is, by simply placing the end of the
thumb on the finger, or the end of the first finger on the thumb, at such joint as
represents the value to be recorded. The manner of making the record, and the
value of the several joints, should be well understood before the operation of adding
is commenced. The addition tables, on preceding pages, should also be well under-
stood.
In making the addition, we first mentally group enough figures to produce a
result not less than 10, and then to this result add the result of enough other fig-
ures to produce a second result not less than 20; then having 20 or an excess of 20,
we record the 20 on the hand, by placing the end of the thumb on the first joint of
the first finger which represents the value of 20, then we add the excess, if any, to the
next group, and continue to produce and record the 20's until the column is
added. If it is desired, enough figures to produce a result in excess of 20, may be
grouped at once, and the record made and the eaccess carried on as above eaplained.

THE VICENARY SYSTEM OF ADDITION, 57
112. To better elucidate the vicenary system, we present the following
problems and explanations:
97
76
89
64
38
36
92
74
53
45
27
84
65
36
87.
1045
(1) Add the following numbers:
Explanation.—Commencing at the bottom of
the units' column, we first mentally see 13, (7,6),
to which we add 9 (5,4), which makes 22; we then
record the 20 by placing the end of the thumb on
the first joint of the first finger. Next, we see 14
iTººl (7, 5, and 2 in excess of 20), to which we add 9
|| || º i_-_º (3, 4, 2), and obtain 23; then we record the 20 by
*/ lº, / / ! ; ; ; 3. passing the end of the thumb to the end of the
W *. f. Agi #_i_3
º
A
*
§
§
|g
|
gets
fº
t>
gº
#
;
second finger. We then see 17 (6, 8, and 3 in ex-
cess of 20), to which we add 4 and obtain, 21;
then we record the 20 by passing the end of the
thumb to the end of the third finger. Next we see 16 (9, 6, and 1 in excess of 20), tº
which we add 9, (7,2), and obtain 25, the 20 of which we record by passing the end of
the thumb to the first joint of the fourth finger, and the 5 we write in the units' place
of the answer. We now have the first column added, and by inspecting the position of
the thumb and fingers of the left hand, we find 80 recorded, which, with the 5, the last
excess of 20, make 85, the sum of the column. Then, to add the second, or tens' column,
we first mentally see 16 (8 and the 8 tens from the units' column) plus 9 (3, 6) = 25; wº
then record the 20 on the first joint of the first finger, as above directed. Next we see 15
(8, 2, and the 5 in excess of 20), plus 9 (4, 5), - 24, then recording the 20, by moving,the
end of the thumb to the first joint of the sécond finger, we next see 20 (7, 9, and the 4 in
excess), which we record by passing the end of the thumb to the first joint of the third
finger." Then we see 12 (3, 3, 6), plus 8 = 20, this we record by placing the end of the
thumb on the first joint of the fourth finger. Then we see 16 (7,9), plus 8 = 24; the 20 We record
by placing the end of the first finger on the first joint of the thumb, and as the addition of the col-
umn is now completed, we inspect the position of the fingers and thumb of the recording hand and
from the position last above named, find the record to be 100, which, with the 4 last excess, makes
104, the result of the column. This 104 we write in the total result, and obtain 1045, as the Bum of
all the numbers. -
5
!
|
1
6
2
4.
01.61.|
||;
2
7
2
1
2
8
ſ
3
(3) Add the following numbers:
2
1.
Explanation.—Having made the explanation of the preceding example very
full, we will therefore, in this, omit some of the details. To make the operation
clearer, we have linked together, with brackets, the numbers used to produce the
various intermediate results. Commencing at the bottom of the column, we pro-
ceed thus: 11, 14, 25; then recording the 20 and adding the 5 excess to the next
2
1
group, we have 15, 7, 22; then, recording the 20 and adding the 2 excess we have
15, 13, 28; then, recording the 20 and adding the 8 excess to the next group, we
have 23; then, recording the 20 and adding the 3 excess to the next group, we
have 18, 10, 28; then, recording the 20 and adding the 8 excess to the next group,
we have 21; then, recording the 20 and adding the 1 excess to the next group, we
23 have 10, 11, 21; then recording the 20 and adding the 1 excess to the next group,
28 we have 16, 11, 27; then recording the 20 and adding the 7 excess to the next group,
we have 17, 7, 24; then recording the 20 and adding the 4 excess to the next group,
we have 21; then recording the 20 and adding the 1 excess to the next group,
*sº
2
1
6
we have 16. This completes the addition, and by inspecting the recording hand, we
25 find the end of the first finger on the second joint of the thumb, which shows that
200 has been recorded, which, with the last obtained 16 makes 216, the total sum
of the column.

58 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
DRILL PROBLEMS,
Add, separately, the following columns of numbers:
(3) (4) (5)
}}º 1 s 6 4. 11
8 8 8
3 ) 7 ) 24 : 13 :
#: 4 8
7 5 g
; : }} 7 #} º
'7 ) 23 7 3 2 ) 4
§§ § #}
3 ) 11 s 28 3 ) 1 9 5
: | : {. 9
2 2 9
9 ) 17 9 ) 7 9
§ { | 9
7
3
7
3
9
1.
Explanation.—In the third example, we added in the same manner as in the preceding example.
In the fourth and fifth examples, in stating the results of the grouped figures, and also in the
result in excess of 20, we have set only the excess figure, or figures, which, in practice, are the only
figures that should be mentally noticed. In the fourth example, we grouped differently from the
bracketing of the third example in order to show that no particular form of grouping is necessary,
and that the more figures grouped, the more rapidly the addition is made.
In the fifth example, we find five figures alike, and hence, by multiplying, we instantly see 45;
we then record the 40, and continue in the usual manner, until we come to three figures that are
alike; we then multiply them, and to the product, add the 3 excess, and 4, the last figure in the
column, and produce 11 in excess of 20.
CONTRACTING METEIODS.
113. To be observed when adding by the Vicenary System:
1. In mentally naming the results of the various groups, the unit figure only
should be named; thus, in the addition of the fourth example, instead of mentally
naming or thinking 17, 11, etc., name or think only 7, 1, etc. Remember, in prac-
tice, these results are never set on paper. It is here done to elucidate the work.
2. In adding the two results of the grouped figures to produce an excess of
20, add only the unit figures, and only mentally name the excess of 20.
These points were elucidated in adding the fourth and fifth examples.
3. Whenever the same figure occurs connectedly several times, the sum should
be obtained by multiplying instead of adding. This principle was elucidated in
example 5.
DRILL PROBLEMS,
59
Add the following columns of numbers by the Vicenary Method until you
acquire rapidity and accuracy:
(6) (7)
63
25
23
93
83
72
43
72
11
63
15
63
00
92
31
41
57
23
35
21
00
51
24
33
49
75
17
35
23
87
15
23
95
64
53
68
72
19
47
x==
eighty-seven thousand five hundred sixty-two.
2. Find the sum of 4888765,
4100000808707 and 222222333333444444.
68
(8)
24
63
75
82
29
76
84
57
38'
92
14
68
74
99
76
57
81
92
25
63
48
78
97
62
54
38
91
56
72
85
39
24
76
58
49
62
14
17
27
19
(
9
)
*
(
1
0
)
*-
(11)
49
83
27
51
43
02
13
38
55
28
33
24
81
91
88
92
02
48
65
08
62
64
87
48
55
82
78
57
20
31
42
53
84
43
90
18
22
37
40
50
(12)
39
77
62
35
56
50
46
34
40
19
58
49
68
62
93
82
55
23
21
30
74
28
84
25
62
49
89
98
28
33
49
87
62
49
77
83
92
11
13
14
(
)
(14)
7563
28325
461523
6393
21783
151672
3243
72
MISCELLANEOUS PROBLEMS IN ADDITION.
114. 1. Add seven million four thousand ninety-six, and three hundred
Ans. 7391658.
92238, 1600084, 8888888, 99999999999,
(15)
1963
2345
1215
1872
7312
910
2311
617
99
313
4632
516
3313
88
200
3915
617
3129
28
319
4615
313
3239
272
99
333
8917
222
6717
516
817
999
1322
400
620
189
4002
841
264.
974.
311
3263
70015
8063
1000
56792
107331
2441
3457
67423
87635
297521
80000
6751
324
233
49
275
3617
81535
71.23
60737
3015
40123
891.395
206
1482
5014
382
1571
932.176
35489
Ans. 2222265.33349723125.
6O soul.E's PHILOSOPHIC PRACTICAL MATHEMATICs. 4×
3. A merchant bought at one time 768 barrels flour for $4032; at another,
273 barrels for $1774.50; and at another, 410 barrels for $2921.25. How many bar.
rels did he buy and what was the total cost? Ans. 1451 bbls. $8727.75 cost.
4. From New Orleans to MacDonoughville is 1 mile; thence to Algiers,1; thence
to Old Spanish Fort St. Leon, 16; thence to Poverty Point, 18; thence to Point
Celeste, 7; thence to Pointe-à-la-Hache, 3; thence to Sixty Mile Point, 15; thence to
Quarantine, 9; thence to Bolivar Point,3; thence to Forts St. Phlip and Jackson, 2;
thence to The Jump, 10; thence to Head of Passes, 11; thence to Pilot Town, 10;
thence to Port Eads, 1. How many miles from New Orleans to Port Eads?
Ans. 107 miles.
5. From New Orleans to Algiers Depot is 1 mile; thence to Gretna, 3; thence to
Jefferson, 9; thence to St. Charles, 6; thence to Boutte, 6; thence to Bayou des
Alemendes, 8; thence to Raceland,8; thence to Ewing's, 6; thence to Lafourche, 6;
thence to Terrebonne, 3; thence to Chucahoula, 6; thence to Tigerville, 5; thence to
L'Ourse, 4; thence to Bayou Boeuf, 3; thence to Ramos, 3, thence to Morgan City,
4; thence to Galveston, 240. How many miles to Galveston 3 Ans. 321 miles.
6. During the fiscal year ending September 1, 1893, the sugar production?
in the different sugar producing parishes of Louisiana, Was as follows:
PARISHES IN POUNDS. | PARISHES. IN POUNDS. PARISHES. IN POUNDS.
Ascension . . . . . . . . . , 39,754,703 || Lafourche. . . . . . . . . . 39,875,463 | St. John the Baptist 21,196,526
Assumption . . . . . . . . 49,939,541 Orleans . . . . . . . . . . . . 2,626,000 || St. Landry......... 4,282,245
Avoyelles. . . . . . . . . . . 2,574,891 |Plaquemines. . . . . . . . 18,943,010 | St. Martin. . . . . . . . . . 7,024,291
East Baton Rouge. . 4,324,213 || Pointe Coupee. . . . . . 7,058,314 || St. Mary ..... tº e o 'º e e 79,641,726
Iberia. . . . . . . . . . . . . . 18,247,813 || Rapides . . . . . . . . . . . . 4,567,404 || Terrebonne. . . . . . . . . 68,456,522
Iberville . . . . . . . . . . . 38,755,524 || St. Bernard. . . . . . . . . 2,666,908 || Vermilion... . . . . . . . 3,452,639
Jefferson . . . . . . . . . . . 6,661,494 || St. Charles. . . . . . . . . 13,693,265 | West Baton Rouge... 17,424,725
Lafayette. . . . . . . . . . . 323,438 St. James . . . . . . . . . . 30,554,329 || Other small parishes 2,640,000
How mamy pounds were produced in the State during the fiscal year.
Ans. 484,684,984 pounds.
7. During the fiscal year ending September 1, 1893, the different COuntries
of the world produced the following quantities of Sugar :
COUCJINT* tº R.I.E.H.S.
CANE SUGAR. IN TONS. CANE SUGAR. IN TONS. BEET SUGAR. IN TONS.
Cuba. . . . . . . . . . . . . . . . . 850,000 || Java. . . . . . . . . . . . . . . . . 430,000 || Germany . . . . . . . . . . . . 1,225,000
Porto Rico. . . . . . . . . . . 70,000 || British India......... 50,000 || Austria-Hungary..... 780,000
Trinidad. . . . . . . . . . . . . 50,000 | Brazil. . . . . . . . . . . . . . . . 200,000 || France. . . . . . . . . . . . . . . 590,000
Barbados . . . . . . . . . . . . 70,000 | Natal . . . . . . . . . . . . . . . . ,000 || Russia.... . . . . . . . . . . . . 460,000
Jamaica. . . . . . . . . . . . . . 28,000 | Fiji. . . . . . . . . . . . . . . . . . 23,000 || Belgium . . . . . . . . . . . . . 180,000
Antigua and St. Kitts, 30,000 || Phillipine Tslands. ... 225,000 | Holland . . . . . . . . . . . . . 70,000
Martinique. . . . . . . . . . . 30,000 || Manila, Cebu, Hoilo... 250,000 | United States, short
Guadeloupe... . . . . . . . . 50,000 | United States. . . . . . . . 246,094 tons . . . . . . . . . . . . . . . 13,284
Lesser Antilles. . . . . . . 27,000 | Peru. . . . . . . . . . . . . . . . 40,000 | Other countries . . . . . . 97,000
Demerara... . . . . . . . . . . 110,000 | Egypt . . . . . . . . . . . . . . . 65,000
Reunion . . . . . . . . . . . . . 35,000 | Sandwich Islands . . . . 125,000
Mauritius . . . . . . . . . . . . 75,000
How many tons each, of cane and beet sugar, were produced in the world
during the fiscal year, and how many tons of both kinds 3
Ans. 3,087,094 tons of cane sugar,
3,415,284 tons of beet sugar,
and 6,502,378 tons of both.
* MISCELLANEOUS PROBLEMS IN ADDITION. 61
8. From New Orleans to the mouth of Red River, is 210 miles; thence to
Black River, 40; thence to Alexandria, 110; thence to Grand Ecore, 120; thence
to Grand Bayou, 95; thence to New Hope, 60; thence to Waterloo, 30 ; thence to
Shreveport, 35. How many miles to Shreveport by river ? Ans. 700, miles.
9. From New Orleans to Carrollton, is 7 miles; thence to Donaldsonville,
71; thence to Plaquemines, 32; thence to Baton Rouge, 20; thence to Port Hudson,
23; thence to Bayou Sara, 12; thence to mouth of Red River, 40; thence to Nat-
chez, 72; thence to Rodney, 45; thence to Grand Gulf, 18; thence to Vicksburg,
61; thence to the Louisiana Line, 97; thence to Helena, 230; thence to Columbus,
329; thence to Cairo, 20; thence to Cape Girardeau, 50; thence to St. Louis, 151.
How many miles to St. Louis by river. Ans. 1278 miles.
10. According to the most reliable information, the following figures show the
area, in Square miles, and the population of the different countries of the world:
Square Miles. Population. Square Miles. Fopulation.
North America. . . . . . . . . 9,349,741 88,005,695 || Africa. . . . . . . . . . . . . . . . . . 11,514,985 168,497,091
South America ........ 6,887,794 33,565,882 || Australasia and Poly-
Europe. . . . . . . . . . . . . . . . . 3,942,530 360,580,788 nesia. . . . . . . . . . . . . . . . . 3,456,103 5,683,968
Asia, including Malaysia 16,956,284 823,155,251 | South Polar Regions ... 253,678 . . . . . . . . .
What is the entire area and the whole population of the world?
Ans. 52,361,115 square miles area, 1,479,488,675 population. .
11. Bought at one time 343 yards of calico and 132 yards of silk; at another,
104 yards of calico and 24 yards of silk; at another, 96 yards of calico and 48 yards
of silk. How many yards of each kind did I buy? Ans, calico, 543; silk, 204.
12. Newman, Soulé, and Lynch, form a copartnership; Newman invests
$5400, Soulé, $5000, and Lynch, $1000 more than both Newman and Soulé. What
is the capital of the firm 3
13. The cotton crop of the Southern States, from 1885 to Sept. 1, 1893, was
as follows: º dº
1885-6. 6,550,215 bales. . . . . . . . . . . . . . New Orleans received 1,764,013 bales.
1886-7. 6,513,263 “ . . . . . . . . . . . . . . 1,762,798 “
1887-8. 7,017,707 “ ..... .* * * * * * * * * 1,780,375 “
1888-9. 6,935,082 “ . . . . . . . . . . . . . . 1,697,376 “
1889-0. 7,311,322 “ . . . . . . . . . . . . . . 1,973,571 “
1890-1. 8,652,597 “ .............. 2,077,744 “
1891-2. 9,035,379 “ .............. 2,503,251 “
1892-3. 6,700,365 “ . . . . . . . . . . . . . . 1.602,079 “
How many bales were produced in the eight years, and how many of them
did New Orleans receive? Ans. 58,715,930 bales. N. O. rec'd 15,161,207 bales.
14. The Mayor of the City of New Orleans, receives a yearly salary of
$3500; the treasurer, $3500; the Commissioner of Public Works, $3500; the
Comptroller, $3500; the Commissioner of Police and of Public Buildings, $3000;
the City Attorney, $3500; the City Surveyor, $2500; the City Superintendent of
Public Schools, $3000; the City Superintendent of Fire Alarm Telegraph, $1800.
What are the total salaries of these officers ? Ans. $27800.
15. The Governor of Louisiana, receives a salary of $4000 per annum; the
62
soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. X.
Lieutenant Governor, $8 per day during the 60 days' session of the Legislature;
the Secretary of State receives $1800 per annum; the Auditor of Accounts, $2500;
the State Treasurer, $2000; the Attorney General, $3000; the five Justices of the
Supreme Court, $5000 each; the two Judges of Criminal Court in N. O., $4000
each; the two Judges of the Court of Appeals in N. O., $4000 each; the five
Judges of the Civil District Court, Parish of Orleans, $4000 each; the State Super-
intendent of Education, $2000. What is the total salary of all these officers, includ-
ing the per diem of the Lieutenant Governor ? Ans. $76780.
16. A father gave to his son seven thousand eight hundred dollars; to his
daughter, nineteen hundred and fifty dollars more than he gave to his son; and to
his wife, three thousand five hundred more than he gave to both the son and the
daughter. What sum did he give away ? Ans. $38600.
17. The length of the Mississippi River is 4200 miles; of the Nile, 4000;
Amazon, 3750; Yenisei, 3400; Obi, 3000; Yang-tse-Kiang, 3320; Niger, 3000;
Lena, 2700; Amoor, 2650; Volga, 2000; Ganges, 1600; Brahmapootra, 2300; La
Plata, 2300; Mackenzie, 2300; St. Lawrence, 2000; Saskatchewan, 1900; Orinoco,
1550; Columbia, 1020; Colorado, 600; Yukon, 1600; Red River, 1500. What is the
combined length of all? AnS. 50690.
18. The following statement compiled from the New Orleans Cotton
Exchange Report of September 1st, 1893, shows the consumption of cotton by the
Southern States, from Sept. 1, 1892, to Sept. 1, 1893: *

No. of Mills | Looms Spindles Bales Pounds Average
ST A T E S . in in in Weight
Operation. Operation. Operation. Consumed. Consumed. Per Bale.
Alabama. . . . . . . . . . . . . . . . . . . . . . . 22 2,875 159,912 50,222 24,449,891 486.83
Arkansas. . . . . . . . . . . . . . . . . . . . . . . 1 108 3,000 765 382,500 500.00
Georgia. . . . . . . . . . . . . . . . . . . . . . . . . 57 11,243 || 511,336 | 184,403 || 86,931,608 || 471.45
Kentucky . . . . . . . . . . . . . . . . . . . . . . 6 688 50,543 || 17,004 || 8,339,968 || 490.47
Louisiana . . . . . . . . . . . . . . . . . . . . . . 4 1,518 52,800 || 14,408 || 6,862,896 || 476.32
Mississippi. . . . . . . . . . . . . . . . . . . . . 7 1,812 53,692 | 15,955 7,479,433 468.78
Missouri. . . . . . . . . . . . . . . . . . . . . . . 1 240 10,500 1,931 942,115 487.89
North Carolina. . . . . . . . . . . . . . . . . 122 11,108 || 524,770 | 181,996 || 82,913,899 || 455.50
South Carolina. . . . . . . . . . . . . . . . . 53 13,537 525,457 203,533 94,013,304 461.90
Tennessee . . . . . . . . . . . . . . . . . . . . . 26 2,621 128,215 33,443 16,239,159 485.58
Texas . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1,362 44,436 12,396 6,478,066 522.59
Virginia. . . . . . . . . . . . . . . . . . . . . . . . 12 3,109 106,486 27,792 12,961,202 466.36
How many each, of mills, looms and spindles were in operation, and how
many each, of bales and pounds of cotton were consumed during the year by the
twelve states? Ans. 318 mills, 50,221 looms, 2,171,147 spindles.
743,848 bales, 347,994,041 pounds.
19. A man made a will and bequeathed $4000 to each of three sons; and to
each of two daughters, $500 more than to each son; and to his wife, $1000 more than
to a son and a daughter together; and the remainder of his estate, which was $1000
more than he bequeathed to his wife and children, he left to benevolent institutions.
What was the sum received by the wife and by each daughter; what was the
sum given to benevolent institutions, and what was the amount of the whole estate?
Ans. Wife, $9500; each daughter, $4500; Benevolent
Institutions, $31500; whole estate $62000.
64.
65.
66.
67.
68.
69.
70.
71.
72.
73.
74.
75.
76.
77.
78.
79.
SYNOPSIS FOR REVIEW,
117. Define the following words and phrases:
Signs and Symbols. What are they
used for ?
Sign of Addition.
Of Subtraction.
Of Multiplication.
Of Division.
Of Equality.
The Parenthesis and Winculum.
Decimal Point.
Sign of Ratio.
Sign of Proportion.
Sign of Involution.
Sign of Evolution.
Sign of Deduction.
Interrogation Point.
The Signs for First, Second, etc.
Sign of the Comma.
80.
81.
82.
85.
86.
Signs and Abbreviations in Busi-
Il CSS,
Addition.
Sum, or Amount.
Numerical Equation.
Principle of Addition.
87, 88, and 89. The Alphabet of Num-
96.
99.
103.
104.
105.
107.
109.
bers.
Proof of Addition.
Horizontal Addition.
Adding Fractional Numbers, 4 or $.
Adding Dollars and Cents.
U. S. Monetary Units.
Addition of Several Columns in-
One Operation.
Addition by one of the Proper-
ties of 9.

(63)
*.
subtraction.
º---------------------N
(DECREASI NG).
*—e-º-Lº-
118. Subtraction is the process of finding the difference between two num.
bers of the same kind.
119. The result obtained by subtraction is called the Difference, or
Remainder.
120. The greater number is called the Minuend, which means a number to
be decreased. -
121. The lesser number is called the Subtrahend, which means the number
to be subtracted.
122. The Sign of Subtraction is a horizontal line —. It is read minus,
and means less.
When this sign is placed between two numbers, it indicates that the number
after it is to be subtracted from the number before it. Thus, 8–3 is read, 8 minus
3 = 5, or 8 less 3 = 5.
123. The Principle governing all problems in subtraction is, that like num-
bers and units of the same order only, can be subtracted, the one from the other.
124. To Prove the operation of subtraction, add the remainder to the sub-
subtrahend; if the sum is equal to the minuend, the work is correct.
125. Subtraction is the reverse of addition, and by it we find what number
added to the lesser of two numbers will produce the greater.
126. SUBTRACTION TABLE.
2 — 2 = 0 || 3 — 3 = 0 || 4 — 4 = 0 || 5 – 5 = 0 || 6 – 6 = 0 || 7 – 7 = 0 || 8 — 8 = 0 || 9 — 9 = 0
3 — 2 = 1 || 4 — 3 = 1 || 5 — 4 = 1 || 6 — 5 = 1 || 7 – 6 = 1 || 8 — 7 = 1 || 9 — 8 = 1 || 10 — 9 = 1
4 – 2 = 2 || 5 — 3 = 2 || 6 — 4 = 2 || 7 — 5 = 2 | 8 — 6 = 2 | 9 – 7 = 2 | 10 — 8 = 2 | 11 — 9 = 2
5 — 2 = 3 || 6 — 3 = 3 || 7 — 4 = 3 || 8 — 5 = 3 || 9 – 6 = 3 || 10 – 7 = 3 | 11 — 8 = 3 | 12 — 9 = 3
6 — 2 = 4 || 7 — 3 = 4 || 8 — 4 = 4 || 9 — 5 = 4 || 10 – 6 = 4 || 11 — 7 = 4 || 12 – 8 = 4 || 13 — 9 = 4
7 — 2 = 5 || 8 — 3 = 5 9 — 4 = 5 | 10 — 5 = 5 | 11 – 6 = 5 | 12 — 7 = 5 || 13 — 8 = 5 || 14 — 9 = 5
8 — 2 = 6 || 9 — 3 = 6 || 10 — 4 = 6 || 11 — 5 = 6 | 12 – 6 = 6 || 13 — 7 = 6 || 14 – 8 = 6 | 15 – 9 = 6
9 — 2 = 7 || 10 — 3 = 7 || 11 — 4 = 7 | 12 — 5 = 7 || 13 — 6 = 7 || 14 – 7 = 7 || 15 – 8 = 7 | 16 — 9 = 7
10 — 2 = 8 || 11 — 3 = 8 || 12 — 4 = 8 || 13 — 5 = 8 || 14 – 6 = 8 || 15 — 7 = 8 || 16 — 8 = 8 || 17 — 9 = 8
11 — 2 = 9 || 12 – 3 = 9 || 13 — 4 = 9 || 14 — 5 = 9 || 15 – 6 = 9 | 16 – 7 = 9 || 17 – 8 = 9 | 18 – 9 = 9
127. ORAL EXERCISES.
1. Commence at 50 and orally count to 0 by continually subtracting 1, thus:
49, 48, 47, 46, 45, etc.
2. Commence at 50 and orally count to 0 by continually subtracting 2, thus:
48, 46, 44, 42, etc. :
3. Commence at 51 and orally count to 0 by successively subtracting 3, thus:
47, 44, 41, 38, etc.
(64)
Y ORAL EXERCISES. 65
4. In like manner, commence at 50 and subtract respectively 4, 5, 6, 7, 8,
9, 10, until you produce 1, thus: 46, 41, 35, 28, etc.
5. Commence at 50 and subtract alternately 2 and 5 until you produce 1,
thus: 48, 43, 41, 36, etc.
6. Commence at 50 and subtract alternately 8 and 3 until you produce 6,
thus: 42, 39, 31, etc.
7. 60 — 2 + 12 + 6 — 2 — 2 — 3 + 5 – 8 + 11 – 6 + 9 — 3 equals &
8. 75 + 5 — 20 — 20 — 5 + 8 + 7 – 9 + 6 – 5 + 4 — 3 + 2 – 45 = }
128. To subtract one number from another, when any figure of the subtrahend
is less than the corresponding figure of the minuend.
1. From 897 Subtract 641.
OPERATION.
897 641 Explanation.—First write the numbers with the lesser wºnder or
641 897 over the greater, so that units of the same order stand in the same
column. Then commence with the units’ figure and Subtract each
-º- o-e order separately, thus: 1 from 7 leaves 6; 4 from 9 leaves 5; 6 from 8
256 256 leaves 2. By this work we obtain the difference, or remainder, 256.
Subtract the following :
278 843 384 978 425 9876 6203
499 521 762 655 679 3456 7021
--as -- -º- - smºs--
129. Demonstration—to prove that the difference between two numbers is
the same as the difference between the two numbers when equally increased.
OPERATION.
6 6 + 3 = 9 5 5 + 2 = 7 23 23 + 10 = 33
4 4 + 3 = 7 2 2 + 2 = 4 11 11 —H 10 = 21
– The difference is — T The difference is T - The difference i ºmme
2 the same. 2 3 the same. 3 12 * . IS 12
Ea:planation.—Here we see that the difference between 6 and 4 is 2, and that the difference
between 6 and 4 equally increased by 3, is also 2. The operations with 5 and 2, and 23 and 11 show
similar results. Hence the law that the difference between two numbers is the same as the difference
between the two numbers when equally increased.
The application of this numerical law is shown in the following problem, and
it governs all operations in subtraction, where the subtrahend figure of any order
exceeds the minuend figure of the same order.
130. To subtract one number from another, when any figure of the subtrahend
is greater than the corresponding figure of the minuend.
1. From 4173 Subtract 2346.
FIRST OPERATION.
#### ####
ÉÉÉÉ ÉÉÉÉ
Minuend - - - 417 3 Subtrahend - - - 2346
subtrahend - - 2346 ° Minuend - - - - 4173
Difference - - 1827 1827
Explanation.—Having written the numbers with the lesser under or over the greater, so that
units of the same order stand in the same column, we commence at the right hand to perform the
operation. º ©
We first observe that 6 units cannot be taken from 3 units. We therefore, according to the
66 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS.
foregoing numerical law, mentally increase the 3 units by 10 units, making 13 units; from this, we
subtract the 6 units and set the remainder, 7 units, in the line of difference. Then, as we added 10
units to the minuend, we now. to compensate therefor, mentally add 1 ten, the equivalent of 10 units
}. the tens’ figure of the subtrahend, and say 5 from 7 leaves 2, which we write in the line of dif-
€I'ellC6.
We next observe that 3 hundreds cannot be taken from 1 hundred ; and, therefore, for reasons
above given, we mentally add 10 hundreds to the 1 hundred, making 11 hundreds, and then say 3 from
11 leaves 8. Then having added 10 hundreds to the hundreds' figure of the minuend, we now men-
tally add 1 thousand, the equivalent of the 10 hundreds, to the thousands' figure of the subtrahend,
and say 3 from 4 leaves 1. This completes the operation and gives 1827 as the difference between
the two numbers.
131. TELE BORFOWING METELOD IMPROPER.
The foregoing is the only rational and true method of subtraction, and it
should be universally substituted for the absurd and unmathematical “borrowing ”
method, which is given by nearly all the authors of arithmetics now before the
public. In the language of Shakespeare: “Neither a borrower nor a lender be.”
Subtract the following:
7843 582 6213 98761 1234 1407.28 34007
5907 427 7302 5496 87.601 60893 65100
132. TO SUBTRACT BY ADDITION.
1. From 4173 Subtract 2346.
OPERATION,
4173 2346 Explanation.—We will here perform the operation by addition,
2346 ° 4173 which consists simply in adding to the subtrahend such a number as
will make it equal to the minuend. Thus, commencing with the
units' figure of the subtrahend, or smaller number, we say, 6 and 7
1827 1827 make 13; and write the 7 in the units' place of the difference; then
carrying 1, we Say 5 and 2 make 7, and write the 2 in the tens’ column of the difference; then we
say, 3 and 8 make 11, and write the 8 in the third column, or hundreds' place of the difference; then
carrying 1, we say 3 and 1 make 4, and Write the 1 in the fourth place of the difference. This com-
pletes the operation.
2. From 73245 Subtract 1228.
OPERATION BY SUBTRACTION.
*:::: Explanation.—Here we say 8 from 15 leaves 7; 3 from 4
leaves 1; 2 from 2 leaves 0; 1 from 3 leaves 2; 0 from 7
72017 leaves 7.
OPERATION BY ADDITION.
73245
1228 Ea:planation.—Here we say 8 and 7 make 15; 3 and 1
make 4; 2 and 0 make 2; 1 and 2 make 3; 0 and 7 make 7.
72017
3. From 56802 Subtract 50531.
OPERATION BY SUBTRACTION.
56802
50531 Explanation.—Here we say 1 from 2, 1; 3 from 10, 7; 6
* from 8, 2; 0 from 6, 6; 5 from 5,0; which being the last
6271 figure on the left, has no value, and hence is not written.
OPERATION BY ADDITION.
56802
50531 Ea:planation.—Here we say 1 and 1 = 2; 3 and 7 = 10;
tº-ºº-e 6 and 2 = 8; 0 and 6 = 6; 5 and 0 = 5. The naught is not
6271 written for the reason given in the first solution.
Y GENERAL DIRECTIONS FOR SUBTRACTION. 67
Subtract the following by both methods:
428 284 2087 4892 78091 5647832
537 192 5301 1309 60742 4909271
GENERAL DIRECTIONS FOR SUBTRACTION.
133. From the foregoing elucidations, we derive the following general
directions for subtraction:
1. Write the numbers so that units of the same order stand in the same column.
2. Begin at the units' figure and take successively each figure of the subtrahend
from the figure of the corresponding order of the minuend.
3. When a figure of the subtrahend is greater than the figure of the same order
in the minuend, add 10 to the minuend figure, perform the subtraction, and then add 1
—the equivalent of the 10—to the meat subtrahend figure.
4. To subtract by the addition process, proceed as elucidated in problem 1
above.
PROBLEMS.
134. Write the following groups of numbers as they are here written, and
Subtract the lesser from the greater of each group :
(1) (2) (3) (4) (5) (6) (7)
467 1807 3842 607 3001 6879 12004
342 4251 1291 8013 1009 9640 17862
Find the difference between the following numbers:
8. 624 and 507. Ans. 117. 12. 754 and 621. Ans. 133.
9. 1148 and 5162. Ans. 4014. | T3. 41074089 and 1875429.
AnS. 39198660.
10. 7013 and 904. * * 14. 9876543210 and 1234567sgo
11. 237 and 894. Ans. 657. Ans. 864.1975320.
TO SUBTRACT DOLLARS AND CENTS.
135. What is the difference between $483 and $51.65% Ans. $431.35.
OPERATION.
$483.00 Explanation.—In all problems of this kind, we first write the
51.65 numbers in the same manner as when adding dollars and cents, with
dollars under dollars, and cents under cents, so that units of the same
$431.35 order stand in the same column, and the points in a vertical line.
When there are no cents in the minuend, we fill the place of cents with
naughtS.
The operation of subtraction is performed with dollars and cents, the same as
with other numbers.
What is the difference between the numbers in each of the following
groups:
(1) (2) (3) (4) (5) (6) (7)
$16.25 $8.00 $.75 $41.04 $10.50 $1.93 $408.76
9.38 3.75 .50 6.61 4.78 .47 291.85
*=mº-ºmmº
68
MATHEMATICS. *
SOULE's PHILOSOPHIC PRACTICAL
(8)
$681.85
90.38
(14)
$576.00
132.85
(9) (10) (11)
$127.05 $248.00 $49.11
105.50 181.15 9.89
(15) (16) (17)
$87.45 $ 482.68 $14.00
19.38 2246.10 68.25
(12) (13)
$8529.09 $560784290.15
2798.17 91842531.83
(18) (19)
$279107.16 $4764000.24
91020.48 7150831.50
BUSINESS CONTRACTIONS IN SUBTRACTING DOLLARS AND CENTS.
136. What is the “Change” (difference) to be given to the purchasers in
the following transactions?
NOTE.-Make the subtractions mentally, using pen or pencil, only to write the difference.
Repeat the work until you acquire rapidity and accuracy.
Amount of Bills Amount of Bills Amount of Bills
Purchase. Presented. Purchase. Presented. Purchase. Presented.
$ 1.45 $ 3.00 Ans. $ $ 2.75 $ 5.00 Ans $ $ 4.65 $ 10.00 Ans. $
.65 2.00 ($ $ 1.63 5.00 % $ 43.40 60.00 ($ $
4.27 10.00 ($ $ 23.42 50.00 “ $ 72.20 100.00 ($ $
1.85 5.00 ($ $ 3.85 10.00 (€ $ 1.15 5.00 % $
12.20 15.00 ($ $ 38.25 50.00 “ $ 67.75 100.00 € $
16.63 20.00 “ $ 14.10 20.00 (€ $ 25.10 40.00 % $
7.18 10.00 “ $ 6.38 10.00 (€ $
Explanation.—In all cases like the above, first add (mentally) to the amount of purchase
enough cents to make even dollars; then make the (mental) subtraction, and to the difference add
(mentally) as many cents as you added to the purchase. Thus, in the first example, 55c. added to
the $1.45, makes $2.00; and $2.00 from $3.00, leaves $1.00; and 55c. added to $1.00, makes $1.55 change.
With a little practice, these mental additions and subtractions may be performed instantly
and without any mental labor. See page 42, for table to qualify to see instantly the difference
between any number and 100.
NOTE.-In business practice, the book-keepers, salesmen and “change ’’ clerks, generally
count out to the purchaser, when the amount of purchase embraces dollars and cents, sufficient
change to make with the purchase even dollars, and then count even dollars to the amount of
money presented by the purchaser. Thus, if the purchase were $2.85 and the money presented was
$5.00, the party making change would count out 15 cents, thus making $3.00, and then count out
$2.00 more, thus making, with the purchase, $5.00.
EIORIZONTAL SUBTRACTION.
137. Accountants and business men frequently find it convenient to sub-
tract horizontally, i.e., when the minuend and subtrahend are on the same line.
The process or operation in this case, is the same as explained in the General Direc-
tions, except that neither number is written under the other.
Find the difference, by horizontal subtraction, between the following
numbers:
(1) (2)
84 – 52 = 32. 143 – 62 = 81.
(5) (6)
87 — 39 = ? 643 – 214 = ?
(4)
(3) 4
721 – 95 = 626. 62 — 18 = ?
- (7) (8)
$486.48 – $91.35 = ? $327.15 – $168.08 = ?
(11)
$2476.40 and $6871.55 = }
(9) (10)
$386.14 — $169.87 = ? $480.23 and $291.45 = ?
㺠(13)
$84936.14 and $57170.28 = ? $4238.64 and $56789.34 = ?
+x SUBTRACTION BY THE METHOD OF COMMENCING ON THE LEFT 69
SUBTRACTION BY THE METHOD OF COMMENCING ON THE LEFT.
138. In the operations of addition and subtraction, we generally commence
to add and to subtract on the right, but we have shown in adding several columns
at once, that we may commence to add on the left; and in like manner, we may
commence on the left in the operations of subtraction, and perform the work towards
the right with the same rapidity and ease as in the ordinary way of work.
It is often convenient, and sometimes necessary, to know almost instantly the
difference between the hundredth, thousandth or millionth figures independent of
the difference of the lesser orders of figures, and by this new method of work, this
important result is obtained.
The following example and the solution of the same will render the operation
clear:
From 43827 subtract 17495.
OPERATION. Explanation.—We commence on the left at the 1, and because the adjoining
43827 subtrahend figure (7) is greater than the minuend figure of the same order, we
17495 add 1 to the first subtrahend figure and say 2 from 4 leaves 2, then because. We
added 1 to the subtrahend figure of the fifth order, we now add its equiva-
* lent, 10, to the minuend figure of the next lower order, and say 7 from 13 leaves 6;
26332 then passing to the next lower order, for the reason that the subtrahend figure
of the next lower order is greater than the minuend figure of the same order, we
add 1 to the 4 and say 5 from 8 leaves 3; then, because we added 1 to the subtrahend figure of
the third order, we now add its equivalent, 10, to the minuend figure of the second order, and say
9 from 12 leaves 3; then 5 from 7 leaves 2, and thus we have the correct remainder.
Working by this method, what is the difference between 389210 and 190768?
Ans. 198442.
2487 and 568; 4784 and 4819; 9621 and 6876; 534672 and 270819.
SUBTRACTION BY THE COMPLEMENT OF 10.
139. In many lines of business, banking especially, and in the operation of
closing accounts, it is often convenient, and a saving of time, to be able to write the
difference between the sums of several unadded debit and credit numbers, without
first finding the amount of the respective debit and credit numbers of which the
difference or balance is desired.
1. What is the difference or balance of the following account?
Ans. 6790.
I)R. C.R. Explanation.—Here we add the units' figures of the smaller, or the
*==mºsºs credit side, and obtain 22; this we subtract from the next higher number
8472 || 3876 of tens (30) and obtain 8, which we add to the units' figures of the debit.
2168 || 4589 side and produce 40. The 0 we write as the first figure of the differ-
ence. Here we observe that there were three tens in the credit side and
4607 || 764 four in the debit, and hence we have 1 ten to add to the tens of the debit, or to
596 || 1483 subtract from the tens of the credit. We will subtract it from the credit.
1659 We now add the second column and obtain, minus the 1 ten, 28; this taken
6790 Bal from 30, leaves 2, which, added to the second column of the debit, gives 29;
* the 9 we write as the second figure of the balance. Here we observe that
fººms. there is one more ten in the credit side than in the debit. This 1 ten we add
to the third column of the credit and obtain 25, and this taken from 30
leaves 5, which we add to the third column of the debit and produce 27; the 7 is written as the third
figure of the balance. We now observe that there is an excess of one ten in the credit, which We
add to the fourth column of credit and obtain 9, this taken from 10 leaves 1, which is added to the
fourth column of debit and produces 16; the 6 is written as the fourth figure of the difference; and
as the tens are now equal in the debit and credit numbers, the difference or balance is complete.
7 O SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs.
2. What is the balance of the following account ?
DR. CR. DIRECTIONS: First. 7 -- 3 +8 = 18; 20 — 18 = 2; 2 + 1 = 3, the first
*-*mºs figure of the balance — 2 tens excess on the debit.
948 || 4351 Second, 2+2+9+ 4 = 17; 20–17 = 3. ... 3 + 5=8, the second fig-
ure of the balance — 2 tens excess on the debit.
793 Third, 2-H4+ 7+ 9 = 22. 30–22 = 8; 8+3= 11. This gives 1 as
427 the third figure of the balance, and as there were 3 tens in the debit and
1 ten in the credit, there are two tens excess in the debit, which, sub-
tracted from the fourth figure of the credit, gives 2 as the fourth and final
B alance, 2183 figure of the balance. ?
3. A depositor has a credit balance of $7206. He draws the following
checks: $527, $1318, $98, and $1642. What is the credit balance? Ans. $3621.
IXIRECTIONs. First. Add horizontally the units’ figures of the checks = 25; 30–25 = 5.
Then 5 +6= 11. (2 tens excess on the checks).
Šćcond. Add, horizontally the tens' figures of the checks and the 2 tens excess, = 18, 20–18
= 2. Then 2+0=2, the second figure of the balance. (2 tens excess in the check numbers). In
the same manner, proceed with the remaining orders.
4. What is the debit balance of the following account”
DR. CR. Directions. First 10–6–4 ; 4 + 2 + 5 = 11.
1375 || 356 Second. 10–5 = 5; 5 + 9 +7= 21.
8692 Third. (3–1 =2) 10–2 = 8; 8 + 3 + 6 = 17.
ºmºsºm- | 9711 Balance. Fourth, 8-1-9.
5. Find the difference between the following numbers by the same process
38071 and 934068.
Ea:planation. 9 and 8 = 17; 3 and 6 = 9; 9 and 0 = 9; 1 and 4 = 5; 6 and 3 =9; 9–1 = 8.
6. Find the balance of the following accounts:
IDR. . CR. • DR. C.R. DR. CR.
$ 482.45 $345.36 571 | 6416 4123.14 5421.45
1243.91 || 92.85 899 342.37
562.87 223 1691.52
7. Find the balance of the following accounts:
Rohlman, Levy & Hiller.
1891 1891
Jan. 1 To Merchandise. . . . . . . . 4 || 184|50 Jan. 15|By Cash. . . . . . . . . . . . . . . 21 || 7500
Feb. 18 || “. “. . . . . . . . 10 || 75|65 Feb. 21 | “ “ . . . . . . . . . . . . . . . 28 || 225/55
March 29 || “ “. . . . . . . . 46 || 212||13 April 17 | “ “ . . . . . . . . . . . . . . . 40 || 65|25
April 4 || “. “ . . . . . . . 54 || 478|14 May 1 | Balance. . . . . . . . . . . . . .
Bolton, Ricks & Schwamacher.

1894 1894
Aug. 3 |S. B. . . . . . . . . . . . . . . . . . . 14 ||1468|15 July 6 II. B. . . . . . . . . . . . . . . . . . . 31 ||2476/58
* | 17 S. B. . . . . . . . . . . . . . . . . . . 46 || 527|28 Aug. 10 |C. B. . . . . . . . . . . . . . . . . . 23 ||3823|47
Sept. 4 loſ. . . . . . . . & s e e s e e s - e e s • 74 || 193|16 & 4 24 J.. . . . . . . . . . . . . . . . . . . . 35 ||1068|75
Oct. 14 IS. B. . . . . . . . . . . . . . . . . . . 98 ||2809|75 Oct. 5 C. B. . . . . . . . . . . . . . . . . . . 67 || 483|22
Nov. 18 S. B. . . . . . . . . . . . . . . . . . . 116 || 79.4|88
Dec. 1 | Balance . . . . . . . . . . . . .
z-
NOTE.-If it is desired to balance and foot the account, as in the work of closing Ledgers,
first add the greater side of the account, and set the amount. Then add the lesser side and find the
difference by the method shown in problem 2, above.
This method of subtraction should be practiced by all students who aspire to be accountants,
until they acquire accuracy and rapidity. Some bankers will not employ book-keepers who cannot
subtract in this manner.
+). MISCELLANEOUS PROBLEMS IN SUBTRACTION. 71
8. A depositor has a credit balance of $4208.10, he deposits $684.50 and
draws checks for the following amounts: $528.00, $476.40, $78.00, $287.44. What
is his credit balance 3 * Ans. $3522.76.
9. My cash balance in bank was $1846.15; I drew out $920.00, then depos.
ited in currency, $1200.00, and in checks, $1719.25; I then drew out $408.50. How
much is my present cash balance in bank 2 Ans. $3436.90.
MISOELLANEOUS PEROBLEMS IN SUBTRACTION.
140. 1. Bought a lot of flour for $2225, and sold the same for $2800.
What was the gain } Ans. $575.
2. It is 700 miles to Shreveport and 321 miles to Galveston. How much
farther is it to Shreveport than to Galveston ? Ans. 379 miles.
3. The ant has fifty eyes, and the dragon fly 12000. How many more has
the dragon fly than the ant? Ans. 11950 eyes.
4. The Old Testament, King James' edition, contains 39 Books, 929 chapters,
23214 verses, 592439 words, and 2738100 letters. The New Testament contains 27
Books, 260 chapters, 7950 verses, 182253 words, and 933380 letters. How many
more of each, Books, chapters, verses, words and letters, does the Old Testament
Contain than the New
Ans. 12 Books, 669 chapters, 15264 verses, 410186 words, and 1804720 letters.
5. Mercury, the smallest planet of the Solar system, is 36,000,000 miles dis-
tant from the Sun. Neptune, the most distant, but not the largest planet, is
2,746,000,000 miles distant from the Sun. How much farther from the Sun is Nep-
tune than Mercury Ż Ans. 2,710,000,000 miles.
6. Physicists have determined that to produce the color, dark red, 395,000,-
000,000,000 ethereal waves strike the eye per second ; and to produce violet,
760,000,000,000,000 ethereal waves strike the eye per second. How many more
waves per second are required to produce violet than dark red 3
Ans. 365,000,000,000,000.
7. Sound travels through the air at the rate of 1118 feet per second, and a
bullet fired from a rifle, travels 1750 feet per second. How much faster does the
ball travel than sound 3 Ans. 632 feet per Second.
8. Physiologists have determined, with the aid of the microscope, that the
lungs of a man contain not less than 600,000,000 air cells; they have also deter-
mined that a single drop of human blood contains more than 4,000,000,000 of
corpuscles. How many more corpuscles in one drop of blood than air cells in the
lungs? Ans. 3,400,000,000.
9. General George Washington was born in 1732, and died in 1799; General
Robt. E. Lee was born in 1807, and died in 1870. How much older was General
Washington than General Lee, when he died ? Ans. 4 years.
10. What is the difference between 23222 and 11 thousand 11 hundred 11 ?
Ans. 11111.
72 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS.
11. What number must be added to 68741 to make a million ?
Ans. 93.1259.
12. Hubbard has $500 which is $150 more than I, and I have $75 more than
Reiffer. How much has Keiffer, and how much have I? Ans. Keiffer has $275.
I have $350.
13. There are two parties who owe me $8000, and one of them owes $4250.
The other wishes to pay me $1700 on account. How much will he then owe ?
Ans. $2050.
14. A speculator bought a lot of apples for $215, and sold them at such a
price, that if he had gotten $22.50 more, he would have gained as much as they
cost him. How much did he sell them for 3 Ans. $407.50.
15. I gave a $20 bill for a purchase of $8.15. How much change should I
receive 3 Ans. $11.85
16. The common house fly makes 330 beats per second with his wings, and
the honey bee makes 190 beats per second. How many more beats will the fly make
in One minute than the bee ? AnS. 8400.
NOTE.-There are sixty seconds in a minute.
17. From New Orleans to Vicksburg is 401 miles, and to Natchez, 277 miles.
How far is it from Natchez to Vicksburg 3 Ans. 124 miles.
18. What is the difference between one million, seventeen thousand seven,
and one thousand, sixteen hundred sixteen 3 Ans. 1014391.
19. The sum of two numbers is 1463, and one of the numbers is 628. What
is the other ? Ans. 835.
20. The velocity of our earth on its yearly voyage through space, around
the sun, is 99733 feet per second; the velocity of a 12-pound cannon ball, fired
from a gun, with an average charge of powder, is 1734 feet per second. How many
feet farther does the earth travel, in each second, than a cannon ball ?
Ans. 97999 feet, or 18 miles and 2959 feet.
21. What number is that, to which, if 17821 be added, the sum will be
3790.7% AnS. 20086.
22. At an election, the defeated candidate received 23742 votes; but had he
received 5112 votes more from new voters, he would have been elected by 1000
majority. How many votes did the elected candidate receive? Ans. 27854.
23. How many years have elapsed since the birth of the following named
persons: Zoroaster, according to Aristotle, was born B. C. 5429; Abraham, B. C.
2000; Menes, B. C. 2000; Moses, B. C. 1570; Solomon, B. C. 1033; Homer, B. C.
1000; Lycurgus, B. C. 850; Thales, B. C. 640; Solon, B. C. 638; Pythagoras, B. C.
600; Confucius, B. C. 551; Buddha, or Guatama, B. C. 500; Sophocles, B. C. 495;
Socrates, B. C. 470; Hyppocrates, B.C. 460; Plato, B.C. 429; Aristotle, B.C. 384;
Demosthenes, B.C. 382; Praxiteles, B. C. 360; Alexander, B. C. 356; Euclid, B. C.
323; Cicero, B. C. 106; Seneca, B. C. 5; Plutarch, A. D. 50; Justinian, A. D. 483;
Bacon, A. D. 1561.
24. A father divided his plantation, consisting of 4500 acres, among his
* MISCELLANEOUS PROBLEMS IN SUBTRACTION. 73
five sons: Albert, Edward, William, Frank, and Robert. To Albert, he gave 800
acres; to Edward, he gave 150 acres more than he gave Albert; to William, he gave
100 acres less than he gave Edward; to Frank, he gave as much as he gave
Edward; and the remainder he gave to Robert. How many acres did Robert
receive 3 Ans. 950 acres.
25. The Equatorial diameter of the earth is 7925.65 miles, and the Polar
diameter is 7899.17 miles. How much greater is the Equatorial diameter than the
Polar 3 Ans. 26.48 miles.
26. Find the difference between 1001734 and MMDCCXLIV. Ans. 10.
27. A carriage, a horse, 4 mules, and 25 sheep, are worth $1425. The car-
riage is worth $450, the horse, $150, mules, $175 a piece. What are the sheep
worth 3 Ans. $125.
28. The cash on hand in the morning was $278.50. Received during the
day from sales, $482.10, and from other sources, $675. There is on hand at the
close of the day, $330.50. How much was paid out during the day ?
Ans. $1105.10.
29. The cash balance in bank, in the morning, was $2860.15. Deposited
during the day, $472. Drew checks in favor of A. $95, of B. $106, and of C. $400.
There remains on hand undeposited, $150. What is the balance in bank and on
hand, at the close of the day's business? Ans. $288.1.15.
30. December 1, 1893, a merchant had cash in his safe, $2340, and on deposit
in bank, $6800. During the month he received $4520, and paid out $5350. What
is the balance in Safe and bank & Ans. $8310.
31. January 1, 1894, a retailer deposited $35 change in his cash drawer, in
the morning, as per ticket: paid out during the day, $320; received from J. Jones,
during the day, $410; from S. Smith, $80. At night there is $328 in the drawer.
What have the cash sales been 3 Ans. $123.
32. There is $22.40 change in the drawer in the morning, as per ticket:
during the day, cash was received from A. $60, from B. $40.50, from C. $121, and from
bills receivable, $600. There was paid out for expenses, $34.50, and to D. $28.
There was $996.15 on hand in the drawer at night. What were the cash sales?
Ans. $214.75.
33. A merchant purchased merchandise to the amount of $65,421.10, and
sold merchandise to the amount of $42,412.90. He then took an account of stock
and found that he had $31,018.40 on hand. How much did he gain 3
* Ans. $8010,20.
74.
SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. º:
118.
119.
120.
121.
122.
123.
124.
129.
syNOPSIS FOR REVIEw.
Subtraction.
Difference, or Remainder.
Minuend.
Subtrahend.
Sign of Subtraction.
Principle of Subtraction.
To Prove Subtraction.
What is the Numerical Law re-
garding the Difference between
TWO Numbers ? *
The Borrowing Method Impro-
º:
O
131.
per.
132.
133.
135.
136.
137.
138.
139.
141. Define the following words and phrases:
To Subtract by Addition.
General Directions for Subtrac.
tion.
To subtract Dollars and Cents.
Business Contractions in Sub-
tracting Dollars and Cents.
To Subtract Horizontally.
Subtraction by the Method of
Commencing on the Left.
Subtraction by the Complement
of 10.

sultiplication.
-------------------------Rs
( INCREASING ).
142. Multiplication is the process of increasing one of two numbers as
many times as there are unitſ, in the other. Or, differently explained, it is the pro-
cess of finding the product of two numbers.
143. The number to be multiplied is called the Multiplicand.
144. The number which shows how many times the multiplicand is to be
increased, or repeated, is called the Multiplier.
145. The result obtained by multiplying is called the Product.
146. The multiplicand and multiplier are called Factors. The meaning of
the word factor is maker or producer.
147. The Sign of Multiplication is an oblique cross, x. It shows that the
numbers, between which it is placed, are to be multiplied together, and is read
multiplied by, or times. Thus, 8 × 3, is read 8 multiplied by 3, or 3 times 8.
Changing the order of the factors does not change the product or result.
Thus, 8 x 3 may be read 3 times 8, or 8 times 3.
148. To Prove the operations of multiplication, repeat the work, or multiply
the multiplier by the multiplicand. If the result is the same as the first, the work
is probably correct.
NoTE.—See page 145 for préof of multiplication by casting out 9's and 11's.
PRINCIPLES OF MOILTIPLICATHON.
149. 1. In all multiplication operations, the product is the same, in name or
kind, as the multiplicand. 2. In all cases, the multiplier must be regarded as an
abstract number. Two denominate numbers cannot be multiplied together as denom-
inate numbers. Thus, we cannot multiply 5 apples by 5 apples, 10 cents by 10
cents. 12 yards by 8 pounds, or 6 boxes by $2. All such questions are absurd and
inSolvable.
To illustrate this principle, we present the following problem:
1. What will 8 pounds of rice cost at 11 cents a pound 7
OPERATION. Explanation.—Here we have 11c., the price of 1 pound, to be increased
- * * > 8 times, and our reasoning is, since 1 pound costs 11c.18 pounds will cost 8
Multiplicand, 112 times as much, or 8 times 11c., which is 88c. To use the 8 as a denominate
Multiplier, 8 number and say 8 pounds times 11c., would show a deficiency of head
sense on the part of the calculator. It is true that in this problem the
Product 88 multiplicand and multiplier are both denominate numbers; but since the
roduct, g multiplier shows the number of times the multiplicand is to be increased
or taken, it must always be considered and used as an abstract number. ... Thus, in this example, we
do not multiply 11c. by 8 pounds, but we take or increase 11c. 8 times, for the reason that 8 pounds
will cost 8 times as much as 1 pound. e º
To our mind, nothing in connection with the Science of numbers is more absurd or meaning-
less than to speak of multiplying 11c. by 8 pounds, or $2, by 5 gallons, or 50c., by 500. To propose
to multiply 4 oranges by 5 peaches would be equally good sense. We 'know what is meant y tak-
ing or increasing lic. or 4 oranges a certain number of times; but we cannot understand or com-
prehend the taking of anything so many cent times, gallon times, or peach times.

(75)
76 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS.
150. MTULTIPLICATION TAET,E.

*
3)
10
15
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
20| 21
40 42
60 63
80| 84
100,105
12012;
140,147
160|168
180|189
210
231
252
273|2
294
315
336
357
378
399
420
441
16
32
48
64
80
96
112
128
144
160
176
192
208
224
240
256
272
288
304
320
336
18
36
54
72
90
108
126
144
162
180
198
216
234.
252
270
288
306
324
ºAL9
tº tºº ºded
360
378
4
8
12
16
20
24
28
32
36
40
44
48
52
56
60
64
68
72
76
80|100
84|105
9
18
27
36
45
54
63
72
81
90
99
108
117
126
135
144
153
162
171
180
Tsº
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
160
170
180
190
200
º
12
24
36
48
60
72
84
96
108
120
132
144
156
168
180
192
204
216
228
240
252
15
30
45
60
75
90
105
120
135
150
165
180
195
210
225
240
255
270
285
300
315
:
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
91
104
117
130
143
156
169
182
195
208
221
234
247
260
2732
42
48
54
60
66
72
78
84
90
96
102
108
1
2
299
322|33
345
368
391
414
437
460
104
112
120
128
136
144
114|133|152
120 140|160
Tºmºrºs
1
3
1
4
105
112
119
126
:
:
2
0
168
37
2
2
44
46
48
50
:
72
75
88||110
92||115
96|120
132|154
138|16.1
144|168
100|125
150|175
176
184
192
198
207
216
200
220
230
240
225
250
264
276
288
286
299
330
345
312
300
325
360
375
352
368
384
400
396
414
432
450
456
475
462
460i483
480504
500|525
|2
5
We here present the Multiplication table of Pythagoras, which extends to 25
times 25. If the commercial student excuses himself from learning the Whole
table, he should at least learn from 1 to 20 times 10, or that part bounded by the
heavy lines.
This portion of the table is far more valuable than the 12 times 12 table which
is given in school arithmetics, and should be thoroughly learned.
When learning this table, or when using it in the solution of problems, be
careful not to use any intermediate words orally or mentally, but simply note the
figures and pronounce the product at once; thus, do not say or think 7 times 3 are
21; 14 times 8 are 1.12; 17 times 9 are 153, etc.; merely look at the numbers to be
multiplied and say, or think, 21, 112, 153, etc. In reading, we do not stop to spell
orally or mentally the words that compose the sentences; from the combination of
the letters we see what the words are, without looking specially at each individual
letter; and to read or operate with rapidity in the combination of numbers, we
must omit all Superfluous talk or thought.
EXAMPLES, WITH PHILOSOPHIC SOLUTIONS. 77
THE IMPORTANCE OF MULTIPLICATION.
151. Next to a thorough knowledge of addition stands multiplication, as an
important acquisition to be possessed by business aspirants. All who are experts
in these two subjects of practical arithmetic, combined with a neat and rapid hand-
Writing, hold passports to well paying positions. No young man or lady who is
ambitious to achieve success and honor in the fields of business can afford to be non-
efficient in these topics. Such proficiency is worth more than unnumbered testi-
monials of competency from honored merchants, noted bankers, railroad kings,
political leaders, or popular governors,
THE PEIILOSOPHIC OR LOGICAL SYSTEM OF MULTIPLICATION.
152. At this point of our work, we introduce, and throughout the book
shall continue to use the logic of numbers—the Philosophic System of Solving
problems. (See page 26 Article 23).
By this system, the reasoning faculties of the mind are brought into action,
invigorated, strengthened, and capacitated to see fine distinctions, to consider Con-
ditions, to investigate facts, to reason logically, and to deduce correct conclusions,
from not only the premises and conditions of problems, but upon all matters and
questions that the changing affairs of this world's life may present for consider-
ation.
EXAMPLES, WITH PHILOSOPHIC SOLUTIONS.
153. 1. One orange costs 5 cents. What will 6 oranges cost 3
Ans. 30 cents.
SOLUTION STATEMENT.
5%
6 Reason.—One orange costs 5 cents. Since 1 orange costs 5 cents,
--> 6 oranges will cost 6 times as much, which is 30 cents.
30g Ans.
2. If one hat costs $2, what will 4 hats cost? Ans. $8.
SOLUTION STATEMENT.
$2
4 Reason.—One hat costs $2. Since 1 hat costs $2, 4 hats will cost
--> 4 times as much, which is $8.
$8 Ans.
3. At 8 cents per yard, what will 5 yards cost 7 Ans. 402.
SOLUTION STATEMENT.
8g
5 Reason.—One yard costs 8 cents. Since 1 yard costs 8 cents, 5
ºmº yards will cost 5 times as much, which is 40 cents.
40% Ans.
(Continued.)
78 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. *
THE REASON, WHY, AND WHEREFORE, CONTINUED.
Question.—How do you know, that if 1 yard costs 8 cents, 5 yards will cost
5 times as much 3
Answer.—By the exercise of my judgment—by the use of the reasoning
faculties of the mind.
Question.—What do you mean, in this connection, by judgment?
Answer.—The conclusion arrived at by the operations of the mind, after duly
considering the premise, the facts, and the conditions of the problem.
Question.—What do you mean by premise or premises 2
Answer.—The proposition, declaration, truth, or fact, which is asserted as
the basis or predicate of a question. In this problem, the premise is, one yard costs
8 cents.
Question.—Why will 5 yards cost 5 times as much as 1 yard 3
Answer.—Because 5 is five times as much as 1.
Question.—What kind of reasoning is the foregoing?
Answer.—Analogical and axiomatical. Analogical, because there is analogy,
relationship or likeness existing between the cost of 1 yard and the cost of 5 yards.
Axiomatical, because the premise and question considered, the conclusion is self.
evident.
Question.—What is reason ?
Answer.—The faculty or power of the human mind, by which truth is dis-
tinguished from falsehood, right from wrong, and by which correct conclusions are
reached by considering the logical relationship which exists between the premises,
the facts, and the conditions of particular statements and questions.
4. What will 4 books cost at 40 cents each 3
SOLUTION STATEMENT.
40g Questións.—1. How do you
4 Reason.—One book costs 40c. know that 4 º will gºt 4
times as much as 1 book # 2.
ºmºsºs Since one book costs 40 cents, 4 What do you mean by judgment?
$1.60 Ans. books will cost 4 times as much. 3. Why will 4 books cost i times
-- as much as 1 book?
NOTE.—The answers given to like questions in the preceding problems, are the proper answers
to these and similar questions.
5. At 18 cents per dozen, what will 5 dozen cost 3
SOLUTION STATEMENT.
18g Questions.—1. How do you
5 Reason.—One dozen costs 18c. 1-now that 5 º Y. *;
& times as much as OZell º
*º-> Since 1 dozen costs 18 cents, 5 Why will 5 dozen cost 5 times as
902 Ans. dozen will cost 5 times as much, much as 1 dozen # 3. What do
you mean by judgment 3
6. If 1 hat costs $4, what will 6 hats cost?
SOLUTION STATEMENT.
$ 4 º Questions.—1. How do you
6 Reason—One hat costs º º: know they will ? 2. Why will
1 hat costs $4, 6 hats will cost they # 3. What is judgment in
$24 Ans. times as much. this connection ?
+): EXAMPLES, WITH PHILOSOPHIC SOLUTIONS. 79
7. Flour is $7 per barrel, what will 20 barrels cost 3
SOLUTION STATEMENT. W.
Questions.—1. How do you
$ 7
Reason.—One barrel costs $7. e -
20 Since 1 barrel costs $7, 20 barrels know this? 2. Why will they
--sº will cost 20 times a h 3. What do you understand by
$140 Ans. S Illll CIſl. judgment in this connection?
8. Bought 12 pounds of sugar at 72 per pound. What was the cost of the
Whole %
SOLUTION STATEMENT.
72 Reason.—One pound costs 7c. º
12 Since 1 pound costs 7 cents, 12 Questions.—1. How do you
tº-º- pounds will cost 12 times as know this 2. Why will it?
84% Ans. much.
9. At $7 per cord, what will 123 cords cost 7
SOLUTION STATEMENTS Ea:planation.—In the second
statement, for convenience, the
lSt. 2d. multiplicand is used as the mul- e
$ 7 123 tiplier. Questions.—1. How do you
Reason--One cord costs $7. ſº * * * * *
123 7 Since 1 cord of wood costs $7, know this? 2. Why will it *
4- m-m- 123 cords will cost 123 times as
$861 Ans. $861 Ans. much.
10. If an employé receives $4 per day for services and he works 22 days,
how much money has he earned ?
SOLUTION STATEMENT. Reason.—One day's service is
22 worth $4. Since 1 day's service Questions.—1. How do you
is worth $4, 22 days' services are know this ; 2. Why will he
4 worth 22 times as much, or since 3. What d e
*- he receives $4 for 1 day’s work, * at do you mean by judg-
$88 Ans. for 22 days' work he will receive ment #
22 times as much.
11. There are 60 minutes in an hour. How many minutes are there in a
day of 24 hours? Ans. 1440 minutes.
SOLUTION STATEMENT. -
60 Reason.—In 1 hour are 60 min- Questions.—1. How do you
24 utes. Since there are 60 minutes know this What do you mean
- in 1 hour, in 24 hours there are by judgment #
1440 Ans. 24 times as many.
The reasoning of these problems and the answers to the questions following
the reasoning, should be repeated until the mind is fully capacitated to solve, in
like manner, all similar problems.
DRILL PROBLEMS.
154. Solve the following problems in like manner as the foregoing, and
write the reason for each :
1. What will 6 books cost, at 40g each Ans. $2.40.
2. At 152 per dozen, what will 5 dozen cost Ans. 752,
3. If 1 box costs $2, what will 24 boxes cost? Ans. $48.
8o soule's PHILosophic PRACTICAL MATHEMATICS. 4×
4. At $6 per cord, what will 41 cords of wood cost? Ans. $246.
5. Paid $4 per barrel for potatoes and bought 36 barrels. What did they
cost ºf Ans. $144.
6. What will 9 yards cost at 162 per yard 3 Ans. $1.44.
7. Flour is worth $6 per barrel. What are 25 barrels worth ? Ans. $150.
8. 12 inches make a foot. How many inches in 15 feet 3 Ans. 180 inches.
9. 4 quarts make a gallon. How many quarts in a barrel that holds 41
gallons ? Ans. 164 quarts.
10. If you buy 17 boxes of peaches @ $2 per box, what will they cost?
Ans. $34.
11. If you buy 9 pencils at 5 cents each, and hand to the seller 50g, how
much change should you receive 3 Ans. 52.
12. A merchant bought 43 barrels of apples at $4 per barrel, and paid $120
on account. How much does he still owe ? Ans. $52.
TO MULTIPLY ABSTRACT NUMBERS AND GIVE REASONS THEREFOR.
155. 1. Multiply 7 by 6.
OPERATION.
7 Ea:planation and Reason.—Plato tells us that one is the basis of all
6 things; but whether this statement be true or false, 1 (one) is certainly
the basis of all numbers. Multiplication is the process of repeating one
* number as many times as there are units—ones—in another. Considering
42 Ans, these facts, we first multiply the 7 by 1, and in the product obtain a pre:
mise for our argument. Thus, axiomatically 1 time 7 is 7. Since 1 time 7
is 7, 6 times 7 is 6 times as many, which is 42. This is the long looked for reason for the multipli-
cation of abstract numbers.
Multiply the following numbers, and write the reason:
6 × 4; 8 × 5; 17 × 12; 23 × 16; 234 × 157; 341 × 256.
TO MULTIPLY, WHEN THE MULTIPLIER CONSISTS OF ONLY
ONE FIGURE.
156. 1. What is the product of 947 multiplied by 6%
OPERATION. Ea:planation.—In all problems of this kind, write the multiplier
under the units' figure of the multiplicand; then commence with
the units’ figure, and say, 6 times 7 are 42, which is 4 tens and 2 units;
# g the 2 units write in the units’ place of the product and retain in the
# §§ mind the 4 tens to add to the column of tens ; next say 6 times 4 tens
#5 are 24 tens plus the 4 tens retained in the mind, are 28 tens, which is 2
Multiplicand 94.7 hundreds and 8 tens; the 8 tens write in the tens’ column of the product,
Multipli 6 and retain in the mind the 2 hundreds to add to the column of hundreds.
Ipſler Then say 6 times 9 are 54 hundreds, plus 2 hundreds are 56 hundreds,
emºsºmeºmº which is 5 thousand and 6 hundreds, which write respectively in the
Product 5682 thousands' and hundreds' columns of the product. This completes the
operation and gives a product of 5682.
In practice, instead of saying 6 times 7 are 42, 6 times 4 are 24, etc., we should only name
the result of the combination. Thus, 42, 24, etc. In handling figures, we show.ld always pronounce
the result of the combinations without naming the figures that make the result, just as we pronounce words
without spelling or naming the letters that make the words.
* MULTIPLICATION ELUCIDATED. 8I
f
Perform the following multiplications:
• (2) (3) (4) (5) (6)
Multiplicand, 543 983 2769 76895 81.453
Multiplier, 7 8 5 9 6
Product, 3801 7864. 1384.5 692055 488718
7. At $85 each, what will 7 wagons cost 3 Ans. $595.
8. What will 8 lots of ground cost, at $1875 each 3 Ans. $15000.
9. At $6 per barrel, what will be the cost of 245 barrels of flour?
SOLUTION STATEMENT. Reason.—One barrel of flour costs $6. Since 1 barrel of flour
Multiplier, 245 costs $6, 245 barrels will cost 245 times as much. The $6 is the
Multiplicand, $ 6 real multiplicand, but in the operation we use it as the multiplier.
p 7 This is done for convenience, in all problems where the multipli-
cand is less than the multiplier. The result is the same which-
$1470 AllS. ever factor is used as a multiplier.
10. What will 42 dozen boxes cost, at $9 per dozen? Ans. $378.
11. At $7 a piece, what will 48 chairs cost? Ans. $336.
12. A clerk receives $75 per month. If he spends $40 per month, how much
can he save in one year, or 12 months? y Ans. $420.
13. Multiply 1, 2, 3, 4, 5, 6, 7, 0, 8, 9, and 10 together. Ans. 0.
14. Bought 44000 pounds of cotton at 8% per pound and sold it at 9% per
pound. What was the gain 3 Ans. $440.
15. What is the difference in the cost of 150 sheep at $4 a head, and 80 head
of cattle at $12 a head. Ans. $360.
TO MULTIPLY, WHEN THE MULTIPLIER CONSISTS OF MORE THAN
ONE FIGURE.
157. 1. What is the product of 397 multiplied by 653?
OPERATION.
F.3 g .
#
'E 2 #"> 23
& E #:
#####
Multiplicand, 397
Multiplier, 653
First partial product by 3 units, 1191 = 3 times the multiplicand.
Second “ 4% by 5 tens, 1985 = 50 “ {{ 4%
Third {{ é & by 6 hund's, 2.382 = 600 ($ {{ {{
Total product, 25924.1 = 653 {{ 44 & 4
Explanation.—In all problems of this kind, write the multiplier under the multiplicand, so
that units of the same order stand in the same column, and then multiply by one figure at a time.
82 - SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
First multiply by the units’ figure, then by the tens', hundreds', and so on, in regular order, through
the multiplier; then add the several partial products together and thus obtain the required product.
In this problem, first multiply by 3, the units’ figure, in the same manner as explained in the
first problem where there was but one figure in the multiplier and obtain 1191 as the first partial
product. Write this below the multiplier so that units of the same order stand in the same
column.
Next multiply by the 5 tens; say 5 times 7 are 35, which is 3 hundreds and 5 tens ; write the 5
tens in the tens’ column directly below the multiplying figure, and reserve in the mind the 3 hundreds
to add to the hundreds’ column. Then say 5 times 9 are 45 + 3 hundreds, which were reserved, mak-
ing 48 hundreds which is 4 thousands and 8 hundreds ; write the 8 hundreds in the column of hundreds,
and reserve the 4 thousands to add to the thousands’ column; then say 5 times 3 are 15 plus 4 thous-
º reserved, are 19 thousands which is 1 ten thousand and 9 thousands, which write in their respective
COLUIDOll].S.
Then, in like manner, multiply by the 6 hundreds in the multiplier, being careful to write the
first figure obtained (2) in the hundreds' column, directly under the 6 of the multiplier, and the
other figures in their respective columns, thousands, ten thousands, and hundred thousands ; then add
the partial products together, which gives 259241 as the whole product of 397 multiplied by 653,
NoTE.—In practice, remember to name or think only the results of the numerical combina-
tions, when adding or multiplying.
EXAMPLES.
2. Multiply 3426 by 457.
OPERATION.
* É
... F & .
gºš
# tº 3 ; ; • *
######
© sº; e. >|r. HP tº #5
Multiplicand, 3.426
Multiplier, 4 57
First partial product by 7 units, 23982 = 7 times the multiplicand.
Second “ “ by 5 tens, 1713 () — 50 % “ {{
Third 4 “ by 4 hund's, 13704 = 400 “ “ {{
Whole product, 1565 682 = 457 46 “ 4%
3. Multiply 28433 4. Multiply 989769 5. Multiply 21794
by 4172 by 248.193 by 2365
Multiply the following numbers:
6. 483 by 569 8. 581 by 76 10. 671 by 508 12. 1683 by 328
7. 924 by 237 9. 1847 by 84 11. 8765 by 2046 13. 2346 by 127
Operation of the
10th problem.
671
508 Ea:planation.—In all problems where there are naughts in
5368 the multiplier, multiply by the significant figures only, for the
3355 reason that the product of any number by 0 is
340868 Ans.
X- MULTIPLICATION. 83.
º
TO MULTIPLY, WHEN EITHER FACTOR IS 1 WITH NAUGHTS
a' ANNEXED, AS 10, 100, 1000, ETC.
EXAMPLES.
158. 1. Multiply 465 by 100.
OPERATION. Explanation.—In all problems of this kind, we annex as many naughts to
the multiplicand as there are naughts in the multiplier. In this problem, we
46500 annex two 0's and produce 46500, the answer. The basis or reason for this is
in accordance with the Principles of Notation, (Arts. 44 and 45) where it was
º shown that annexing one naught to a number removes or changes the unit of
such number into tens, the tens into hundreds, the hundreds into thousands, and so on, and hence mul-
tiplies by 10, 100, 1000, etc., for the higher powers of ten.
2. Multiply 500 by 10. Ans. 5000 || 4. Multiply 1850 by 1000. Ans. 1850000
3. Multiply 1000 by 100. Ans. 100000 || 5. Multiply 7020 by 10000. Ans. 70200000
TO MULTIPLY, WHEN EITHER THE MULTIPLICAND OR MULTIPLIER,
OR BOTH, HAVE NAUGHTS ON THE RIGHT.
159. Multiply 463 by 200.
OPERATION.
463 463 Explanation.—In all problems of this kind, we first multiply
2 Of 200 together the significant figures and to the product annex as many
º º, thºs are on the right of the multiplier or multiplicand,
*- or of both. In writing the numbers to be multiplied, place the
92600 Ans. 92600 Ans. right hand significant figures of the multiplicanã and muftiplier in
º e the same column, omitting the naughts or extending them to the
right. In this problem, we first use the multiplier 2 hundred as 2 units; hence the first partial pro-
duct, 926, was 100 times too small. We then, by annexing the two naughts, multiplied it by 100,
º ºned 92600 as the correct product. For further explanation and reason, see problem 1,
rt. 158.
2. Multiply 940 by 4700.
OPERATION.
94 940 Explanation.—94 multiplied by 47 equals 4418 as the
47 4700 first partial product; then, since the 94 was used as 94
-ºm- OI’ units instead of 94 tens, we annex one naught to 4418
658 658 making 44180; then, since the 47 was used as 47 units
376 376 instead of 47 hundreds, we annex two naughts to 44180,
**- e-mºmº-mº- making 4418000, the correct answer. See Problem 1. Art.
4418000 Ans. 4418000 Ans. 158.
3. Multiply 3400 by 26. Multiply 5020 by 420. 4. Multiply 82000 by 483.
OPERATION. OPERATION. OPERATION.
3400 5020 82000 r 483
26 420 483 O 82000
204 1004 246 966
68 2008 656 3864
*-ºs- - 328 4-
884.00 Ans. 2108.400 Ans. ——— 39606000 Ans.
39606000 Ans.
5. Multiply 842 by 600. 8. Multiply 23500 by 12030.
6. Multiply 1208 by 1020. 9. Multiply 1000 by 6208.
7. Multiply 9900 by 707. 10. Multiply 81000 by 90200.
84 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs.
TO MULTIPLY BY TEIE FACTORS OF A NUMBER.
160. Factors of a number are such numbers as will, when multiplied
together, produce the number. Thus, 6 and 6 are the factors of 36; 7 and 8 are
the factors of 56.
PRINCIPLE.
The product of any number of factors will be the same, in whatever order
they may be multiplied.
1. Multiply 2435 by 42.
OPERATION.
*: Explanation.—In all problems of this kind, separate the multiplier
into two or more factors; then multiply the multiplicand by one of
17045 the factors, the resulting product by another factor, and so on, until
6 all the factors have been used. The last product will be the correct
product.
102270 Ans.
Use the factors and multiply the following numbers: 2. 480 by 361.
3. 1756 by 125. 4. 3281 by 128. 5. 781 by 63. 6. 3140 by 36. 7. 588 by 81.
8. 473 by 432. 9. 5107 by 336.
To MULTIPLY, WHEN THE MULTIPLICAND OR MULTIPLIER CON-
TAINS DOLLARS AND CENTS.
161. 1. Multiply $342.15 by 6.
OPERATION.
$342.15 Eacplanation—In all problems of this kind, multiply in
the regular manner, then prefix the dollar sign ($) and place
the point (. ) two places from the right. The answer is then
Product, $2052.90 in dollars and cents.
2. What will 1682 pounds of sugar cost, at 9% per pound 7 Ans. $151.38.
3. A merchant's monthly expenses are $1342.75. What are they for 12
months & Ans $16113.00.
4. It costs a family $2.30 a day for marketing. What will be the expense
for 30 days? Ans. $69.00.
5. What will 37 boxes of oranges cost, at $3.75 per box 3 Ans. $138.75.
6. At 16 cents per pound, what is the value of 23780 pounds of cotton ?
Ans. $3804.80.
7. If it costs $17500 to construct one mile of railroad, what will be the cost
to build 364 miles 3 Ans. $6370000.
8. What will 875 tons of railroad iron cost at $55 per ton ? Ans. $48125.
4× MISCELLANEO US PROBLEMS IN MULTIPLICATION. 85.
GENERAL DIRECTIONS FOR MULTIPLICATION.
162. From the foregoing elucidations, we derive the following general
directions for multiplication:
1. Write the multiplier under the multiplicand, so that units of the same order
stand in the same column, and draw a line beneath.
2. When the multiplier consists of one figure, begin at the units’ figure and
multiply each figure of the multiplicand by the multiplier. Write in the product line
the units of each result, and add the tens, if any, to the newt result.
3. When the multiplier consists of two or more figures, begin at the units’ figure
and multiply successively, each figure of the multiplicand by each figure of the multiplier,
placing the right hand figure of each partial product under that figure of the multiplier
which produced it.
4. Draw a line beneath the several partial products and add them together; the
sum will be the required product. If there are any decimals in the factors, point off
as many figures from the right of the product as there are places of decimals in the
multiplicand and multiplier.
PROOF.—1st. Carefully review the work. 2d. Multiply the multiplier by
the multiplicand; if the results are the same, the work is probably correct.
NOTE.-See proof of multiplication by casting out the 9's and 11's, page 154.
MISCELLANEOUS PROBLEMS IN MULTIPLICATION.
163. 1. What is the value of the following numerical expressions:
(43 —6)+(8 × 2). Ans. 53.
2. What is the product of 8 + 7, multiplied by 204—101? Ans. 1545.
3. What is the product of 16 + 18–10 by 12 × 23 Ans. 576.
4. What is the difference between 50—(5 × 4) and 25 + 4–83 Ans. 9.
5. Multiply 240 – 50+ 22 by 14 × 16 – 112–65. Ans. 29736.
6. Multiply 16 thousand 16 hundred 16, by 11 thousand 11 hundred forty
and 11. Ans. 214052016.
7. Multiply together the following numbers:
9 × 8 × 7 × 6 × 5 × 4 × 3 × 0 × 2 × 1. Ans. 0.
8. What will 6 dozen dozen boxes cost, at one half a dozen dozen cents
per box 3 Ans. $622.08.
9. The pulse of a healthy middle-aged person beats 72 times per minute,
how many times does it beat in 1440 minutes? * Ans. 103680.
10. Multiply one million twenty-six, by nineteen thousand seven hundred
ten. Ans. 19710512460.
11. One cubic foot contains 1728 cubic inches. How many cubic inches
in 324 cubic feet 3 Ans. 55.9872.
86 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
12. One square foot contains 144 square inches. How many square inches
in 95 square feet? Ans. 13680.
13. One gallon contains 231 cubic inches. How many cubic inches in a cis-
tern that holds 3500 gallons 2 AnS. 808500.
14. One bushel contains 2150.42 cubic inches. How many cubic inches in
20 bushels 3 Ans. 43008.40.
15. One mile contains 5280 feet. How many feet in 25 miles?
Ans. 132000.
16. The human heart beats 4200 times an hour. How many times does it
beat in 10 years, there being 24 hours in one day, and allowing 365 days in each
year 3 AnS. 3679.20000.
17. Sound travels 1118 feet per second. How far will it travel in 10 minutes,
there, being 60 seconds in a minute 3 Ains. 670800 feet.
18. A railroad train runs 25 miles an hour. How far will it go in 3 days,
allowing 3 hours for lost time in stoppages } Ans. 1725.
19. Light travels 192500 miles per second. How many miles will it travel
in 1 day, there being 24 hours in a day, 60 minutes in an hour, and 60 seconds in a
minute % Ans. 1663.2000000.
OPERATION INDICATED.
24 hours, x 60 minutes, x 60 seconds, x 192500 miles per second, = the answer.
20. If a person respires 17 times in a minute, how many times will he breathe
in a day? Ans. 24480.
21. At $17 per ounce, what is the value of 9 pounds of gold, there being 12
ounces in a pound, Troy or Mint weight? Ans. $1836.
22. What will 27893 pounds of tobacco cost, at 56 cents per pound 3
Ans. $15620.08.
23. What will 1870 acres of land cost, at $18 per acre 2 Ans. $33660.
24. The Senate and the House of Representatives of the State of Louisiana,
consist of 137 members, who receive $4 per day. The regular session continues
60 days. What is the yearly expense for the salaries of the State's law makers ?
Ans. $32880.
25. How many hours, minutes, and seconds in the month of July 2
NOTE.—See Problem. 19.
Ans. 744 hrs. 44640 min. 2678400 sec.
26. What will 1463 ounces of silver cost, at 87.2 per ounce?
Ans. $1272.81.
27. A contractor has 865 men employed at $1.50 per day. What are the
weekly wages of all, for 6 days’ labor? Ans. $7785.
28. What will it cost to build 37428 cubic yards of levee, at 23 cents per
cubic yard & Ans. $8608.44.
29. A steamboat arrives with 3840 bales of cotton, 1320 sacks of cotton Seed,
and 580 barrels of molasses. Her freight charges are $2 per bale for Cotton, 25% per
+& MISCELLANEOUS PROBLEMS IN MULTIPLICATION. 87
Sack for Cotton seed, and 50% per barrel for molasses. What is the amount of her
freight bill? Ans. $8300.
30. A drayman charges 75 cents a load, and he has hauled 63 loads. How
much is due him & Ans. $47.25.
31. What will it cost to slate the roof of a house containing 52 squares, at
$13.25 per square 2 Ans. $689.
32. The walks around a dwelling contain 129 square yards. What will it
cost to flag them with German flags, at $3.10 per square yard 2 Ans. $399.90.
33. What will it cost to pave a street containing 20000 square yards, with
stone, at $4.75 per square yard 2 Ans. $95000.
34. Bought 2180 barrels of coal at 48% per barrel. What was the cost?
t Ans. $1046.40.
35. Multiply 5 billion 16, by 5 million 1 thousand.
All S. 25005000080016000.
36. A hogshead of sugar contains 1085 pounds. How many pounds in 107
hogsheads of equal weight? Ans. 116095.
37. A planter produced 68 bales of cotton. If the average weight of the
bales was 460 pounds, and the cotton sold for 13 cents per pound, how much money
would it bring ? * Ans. $4066.40.
38. What will 3 cases, containing 2 dozen pairs each, of shoes cost, Ø $2.90
per pair 3 Ans. $208.80.
39. If it costs $1.50 a day to support one person, what will it cost to sup-
port a family of 13, for one year or 365 days? Ans. $7117.50.
40. A merchant sold three dozen dozen ladies’ hose at one-quarter of a dozen
dozen cents a pair. How much did he receive for them 2 Ans. $155.52.
41. The pressure of the atmosphere is 15 pounds on every Square inch of
surface. The exterior surface of a man of average size is about 2500 square inches.
How many pounds weight does he sustain } Ans. 37500 pounds.
42. How many dollars are 375 $10 gold pieces worth ? Ans. $3750.
43. What is the value of 2146 dimes 3 Ans. $214.60.
44. What is the value of 1010 quarter dollars? Ans. $252.50.
45. What is the value of 728 nickles 2 Ans. $36.40.
46. What is the value of 1612 half dollars? Ans. $806.00.
47. 4875 is the thirteenth part of a number. What is the number 3
Ans. 63375.
48. The sun is 1384500 times as large as the earth; the earth is 45 times as
large as the moon. How many times is the Sun as large as the moon?
Ans. 62302500.
49. A man's receipts are $1800 a year, and his disbursements are $1125 a.
year. How much are his net receipts in three years? Ans. $2025.
50. The population of a city is 250,000. If the average expense of each
inhabitant is $1 per day, what is the total yearly expense for an ordinary year?
Ans. $91,250,000.
51. A merchant sold 52 barrels of molasses, each containing 41 gallons, at
38g per gallon. How much did he receive for it? Ans. $810.16.
88 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS Yºr
52. Light travels 192500 miles a second, and it requires 100000 years to travel
to us from some of the fixed stars that are seen with the telescope. Allowing 365
days, 5 hours, 48 minutes, and 49 seconds to a year, and remembering that there
are 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute, how far
distant are such stars 3 AnS. 607470883250000000 miles.
SIMULTANEOUS, OR CROSS MULTIPLICATION.
164. This system of multiplication is of inestimable value. It is, so to
speak, the accountants' and calculators' magic wand, by which they may produce
results in multiplication operations with almost lightning rapidity.
No one can be proficient in the handling of numbers without a thorough
knowledge of this system of work in addition to the several other contracted
methods, which follow this work.
The operations of this system are based upon the following principles:
Hundredsxtens “ thousands.
Tens × hundreds 66 thousands.
10. Units X thousands “ thousands. J
11. Ten thousands x units produce ten *:::::::)
1. Units × units produce units.
2. Tens × units { % ten S.
3. Units X tens {{ tenS. }
4. Hundreds × units produce hundreds.
5. Tens × tens “ hundreds. |
6. Units x hundreds {{ hundreds.
7.
8.
9
Thousands × units “ thousands. l
12. Thousands X tens 44 ten thousands.
13. Hundreds × hundreds “ ten thousands. }
14. Tens × thousands 44 ten thousands.
15. Units x ten thousands 6% ten thousands.
16. Hundred thousands × units produce hundred thousands.)
17. Ten thousands X tens { % hundred thousands.
18. Thousands × hundreds {{ hundred thousands.
19. Hundreds x thousands “ hundred thousands. }
20. Tens × ten thousands {{ hundred thousands.
21. Units × hundred thousands 4% hundred thousands.
and so on for higher numbers. J
PROBLEMIS.
1. Multiply 54 by 37.
OPERATION. Explanation. First multiply together in the ordinary manner, the units' fig-
54 ures, thus, 7 times 4–28, and write the 8 in the first place of the product, and
37 - carry 2. Next multiply the tens' figure of the multiplicand by the units' figure
of the multiplier, thus, 7 times 5 (+2 to carry)=37, which retain in the mind and
1/ add thereto the product of the units’ figure of the multiplicand by the tens,
*º-º-º: figure of the multiplier, thus, 3 times 4 = 12, H-37 = 49. Write the 9 in the
1998 tens' place of the product and retain the 4 hundreds in the mind to be added to
the column of hundreds. The product of the multiplicand by the units' figure
of the multiplier, and also of the units' figure of the multiplicand by the tens' figure of the multi-
plier is now produced in the two figures (98) of the final product, and hence we have no further use
for the units’ figure of either factor, therefore check and pass the unit column and proceed to mul-
tiply the tens’ figures of the two factors, thus, 3 times 5 are 15, plus the 4 retained in the mind, makes
19, which write to the left of the two product figures first obtained, and complete the product.
SIMULTANEOUS OR CROSS MULTIPLICATION. 89
To elucidate the operation by figures only, we present the following;
OPERATION. *|†
54 Baplanation.— 5 ** >~~ g
3. 7 x 4 = 218 units.
*== 7 × 5 + 2 + 3 × 4 = 419 tens.
1998 3 × 5 + 4 = 119 hunds.
Multiply the following numbers:
(2) (3) (4) (5) (6) (7) (8) (9)
35 62 87 76 93 89 93 86
46 23 42 58 64 97 78 92
10. Multiply 5417 by 62. Ans. 335854.
OPERATION. Explanation.—In this, as in the above problems, first multiply the units
5417 figures and set the result in the first place of the product. Thus, 2 times 7
$) are 14; set the 4 and carrying the 1 proceed as follows: 2 times 1 are 2 plus
62 1 makes 3, to which we add 6 times 7 are 42, making 45; set the 5 and carry
V 4. Then 2 times 4 are 8 plus 4 makes 12, to which we add 6 times 1 are 6,
Tºº-º- making 18; set the 8 and carry the 1. Then 2 times 5 are 10 plus 1 makes 11,
335854 to which we add 6 times 4 are 24 making 35; set the 5 and carry the 3. Hav-
ing now as many figures in the partial product produced as we have in the multiplicand, for the
reason given in the first problem we check and pass the units’ figure of the multiplier, and with the
tens' figure multiply 6 times 5 are 30 plus the 3 to carry makes 33, which completes the product.
11. Multiply 62549 by 53.
Ea.planation.— Carrying and
product figures.
OPERATION. 3 × 9 = 27 or thus: 27; 12, 14, 45, 59;
62549 3 × 4 + 2 + 5 × 9 = 59 15, 20, 20, 40;
1/ 3×2 + 4 + 5 × 5 = 35 18, 21, 10, 31;
*=== 3 × 6 + 3 + 5 × 2 = 31 , (check the 3).
3315097 5 × 6 + 3 = 33 30, 33.
Multiply the following numbers:
(12) (13) (14) (15) (16) (17)
13502 3243 14907 897 52061 298.1453
43 27 64 48 83 39
18. Multiply 7524 by 346.
OPERATION. Ea:planation.—In the elucidation of this problem, the principles
7 524 \ used being the same as in the above examples, we will therefore con-
34 6 dense the explanation of the work.
6 times 4 = 24, set the 4 and carry the 2.
1/1/ th ; times 2 = 12 + 2 = 14 + 4 times 4 = 30, write the 0 and carry
2603304 6 6 times 5 = 30 +3 =33 + 4 times 2 = 41 + 3 times 4 = 53, write
the 3 and carry the 5.
6 times 7 = 42 +5 = 47 -- 4 times 5 = 67 + 3 times 2 = 73, write the 3 and carry the 7.
Having now as many figures in the partial and final product as we have figures in the multi-
plicand, we observe that the product of the multiplicand by the units' figure of the multiplier is
already produced in the final product, and hence we check and pass the units’ figure and commence
with the tens' figure of the multiplier; and as the product of the first three figures of the multiplicand
90 soule's PHILosophic PRACTICAL MATHEMATICs. {
has been taken by the tens' figure of the multiplier, we commence with the fourth or last figure of
the multiplicand. 4 times 7 = 28 + 7 = 35 + 3 times 5 = 50, set the 0 and carry the 5. *
Now, as the product of the multiplicand by the tens' figure of the multiplier is produced in
the final product, we check and pass the same and commence with the third or hundreds' figure of
the multiplier; and as the product of the first three figures of the multiplicand, by the third or
hundreds' figure of the multiplier has already been produced, we commence with the fourth or
thousands' figure of the multiplicand.
3 times 7 = 21 + 5 = 26, which we set and complete the product.
19. Multiply 74018 by 45602.
OPERATION. f
74 () 18 ing an
45 602 Elaplanation.— º: #.
1/1/v/v/ 2 × 8 = 16 or thus:
1. 8 = wº & A ºn a tº
3.37 53 68836 2 × 1 + 1 + 0 × 3 16; 2, 3; 48;
2× 0 + 0 × 1 + 6 × 8 = 48 8, 12, 6, 18, 40, 58;
2× 4 + 4 + 0 × 0 + 6 × 1 + 5 × 8 = 58 14, 19, 5, 24, 32, 56;
2 X7 -- 5 + 0 × 4 + 6 × 0 + 5 × 1 + 4 × 8 = 56 (Check 2) 5, 24, 29;
0 × 7 + 5 + 6 × 4 + 5 × 0 + 4 × 1 = 33 33; (check 0) 42, 45,
6×7 -- 3 + 5× 4 + 4 × 0 = 65 20, 65; (check 6) 35,
5 × 7 -- 6 + 4 x 4 = 57 41, 16, 57; (check 5)
4 × 7 -- 5 = 33 28, 33.
The multiplication sign X, in the above elucidation, should be read “TIMES.”
NoTE 1.-it will be observed that the general principles governing this system of multiplication
have been strictly used in the operations and explanations. Thus, in the last problem, units are multi-
plied by units, (8 by 2); tens by units, (1 by 2); units by tens, (8 by 0); hundreds by units, (0 by 2);
tens by tens, (1 by 0); units by hundreds, (8 by 6); thousands by units, (4 by 2); hundreds by tens,
(0 by Ö); tens by hundreds, (1 by 6); units by thousands, (8 by 5); ten thousands by units, (7 by 2);
thousands by tens, (4 by 0); hundreds by hundreds, (0 by 6); tens by thousands, (1 by 5); units
by º (8 by 4); ten thousands by tens, (7 by 0); thousands by hundreds, (4 by 6); hun-
dreds by thousands, (0 by 5); tens by ten thousands, (1 by 4); ten thousands by hundreds, (7 by 6);
thousands by thousands, (4 by 5); hundreds by ten thousands, (0 by 4); ten thousands by thous-
ands, (7 by 5); thousands by ten thousands, (4 by 4); ten thousands by ten thousands, (7 by 4).
NOTE. 2.--By these lengthy elucidations, the work seems more tedious than it really is in
practice. In performing the operation practically, we name results only, and thereby very much
lessen the labor

DRILL PROBLEMS, IN SIMULTANEOUS MULTIPLICATION.
165. Students should drill on the following problems, at different times
until they can call the intermediate results as rapidly as they can talk.
A few weeks of practice, an hour or two per day, will capacitate the learner
to see mentally the intermediate results faster than he can talk or write them.
1. Multiply 1234 by 37.
OPERATION.
1 2 3 4. *
3 7 Intermediate results.-28; 21, 23, 12, 35; 14, 17, 9, 26;
1/ 7, 9, 6, 15; (check 7) 3, 4.
456 58
SIMULTANEOUS OR CROSS MULTIPLICATION. 9 I
2. Multiply 438 by 395.
OPERATION. Intermediate results.-40; 15, 19, 72, 91; 20, 29, 27,
4.38 56, 24, 80; (check 5) 36, 44,
sº, 9, 53; (check 9) 12, 17.
- or thus: 40; 91; 80; (check 5).
1 73 0 1 0 53; (check 9) 17.
3. Multiply 3602 by 5184.
OPERATION. Intermediate results.-8; 16; 24, 25, 2, 27; 12, 14,
360 2 48, 62, 10, 72; (check 4) 24,
5 18.4 31, 6, 37; (check 8).3, 6, 30, 36;
1/1/1/ (check 1) 15, 18,
1867 2 768 or thus: 8; 16; 27; 72;
(check 4). 37; (check 8) 36;
(check 1) 18.
4. Multiply 90472 by 8706. &
OPERATION. Intermediate results.-12;. 42,43; 24, 28, 14, 42. —t –
9 () 4 7 2 49, 4, 53, 16, '69; 54, 60,
8 7 0 6 t 28, 88, 56, 144; (check 6).
1/1/1/ 14, 32, 46; (check 0) 63, 67;
(check 7) 72, 78;
78 7 6 49 2 3 2 or thus:... 12; 43; 42; 69;
%, 144; (check 6) 46;
(check 0) 67; (check 7) 78.
5. Multiply 503102 by 430240.
OPERATION. Intermediate results.-8; 4; 4; 12, 2, 14, 6, 20;
5 0.31 02 2, 6, 8, 8, 16; 20, 21, 3, 24;
430 24 0 (check 4) 10, 12, 9, 21, 4, 25;
1/VVV (check 2) 2, 12, 14; (check 0)
15, 16; (check 3) 20, 21;
21 64 54 6 04480 then annex the 0.
or thus: 8; 4; 4; 20 ; 16; 24; (check 4)
25; (check 2) 14; (check 0)
16; (check 3) 21; then annex
the 0.
§
$ore.—when there are naughts on the right of either or both of the factors, treat them as
explained in Problem 1, Article 159.
6. Multiply 4081700 by 53920.
OPERATION.
4081 700 Baplanation.—14; 66; 52; 115;
53% 48; (check 2) 80; (check 9) 20;
(check 3) 26; annex 3 naughts.
2 2008 52 640 00
92
SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs.
7. Multiply 240700 by 68100.
Multiply by the following problems:
(8) (9) (10) (11) (12) (13)
321 2334 4354 270018 554.422 9080706
34 136 809 30402 19345 1122033
10914
(14) (15) (16) (17) (18) (19)
456 4120 5841 16190 2736 88997
123 54 325 2305 2107 11238
56088
(20) (21) (22) (23) (24) (25)
6208 894 2382 6709 4087 19205
340 143 751 284. 2614 41003
21107.20
(26) (27) (28) (29) (30) (31)
14901 789 999 666 5566 5070608
782 567 888 777 4488 107024
11652582
32. Multiply 222333444555 by 219543620324 OPERATION.
- 222333444555
21954.3620324
assilssºgliooasſºszo Ans.
A DIFFERENT METHOD OF SIMULTANEOUS MULTIPLICATION.
166. Multiply 673 by 452. AnS. 304196.
OPERATION. © Ea:planation.— 2 × 3 = 6
"ş, 2 × 7 – 3 x 5 = 29
2 × 6 + 2 carried -- 7 × 5 + 3 × 4 = 61
304196 5 × 6 + 6 carried -- 4 × 7 = 64
4 × 6 + 6 carried = 30
It will be observed that the problem is to repeat 673,452 times; this has been
done in the operation, though by reason of reversing the multiplier it appears at
first sight that the multiplicand has been repeated only 254 times. By following
Yºr SIMULTANEOUS OR CROSS MULTIPLICATION. 93
the different steps of the solution with care, it is clear that the 673 has been repeated
2 times + 50 + 400 times, as the problem required.
This method of simultaneous multiplication is preferred by some calculators
to the method first presented. We however much prefer the first method, for the
following reasons:
1. There are no less figures to make and no less thinking to do in this
method than there is in the first method.
2. By reversing the multiplier errors are much more liable to occur, and
increased work is required in rewriting the multiplier.
3. When making extensions in bills or invoices, the reversing of the multi-
plier would add mental labor to the operation, lead to confusion, and increase the
liability to make errors.
We present this method more to show its deficiencies, when compared with
the regular simultaneous multiplication first presented, than for any merit it pos-
$62SSéS.
2. Multiply 6758 by 927. Ans. 6264666.
OPERATION BY THE ABOVE METHOD.
67.58 Eacplanation.— 7 x 8 = 56
729 7 × 5 + 5 + 2 × 8 = 56
sºmeºmºmºmºmºmºmºsºme 7 x 7 -- 5 + 2 × 5 + 9 × 8 = 136
6264666 7 × 6 + 13 + 2 × 7 -- 9 × 5 = 114
2 × 6 + 11 + 9 × 7 = 86
9 × 6 + 8 = 62
TO MULTIPLY WHEN THE MULTIPLIER CONSISTS OF TWO FIGURES.
167. 1. Multiply 5234 by 23.
orrºs. Explanation.—First. 4 × 23 = 92; set the 2 and carry 9.
5234 Second. 3 × 23 = 69 + 9 = 78; set the 8 and carry 7.
23 Third. 2 × 23 = 46 -- 7 = 53; set the 3 and carry 5.
Fourth. 5 × 23 = 115 –– 5 = 120.
120382 +
2. Multiply 6845 by 34.
OPERATION. Explanation.—
6845 5 × 34 = 170.
34 4 × 34 = 136 -– 17 = 153.
gºmºmºmºmºsº 8 × 34 = 272 + 15 = 287.
232730 6 × 34 = 204 -- 28 = 232.
Multiply the following problems in the same manner:
3 4 * (6) (7) (8) (9)
§ 13; ački 63024 725 309 8523
42 26 54 63 37 86 47
amm amº- mº
amme sº-s m-- *-*
94. SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. A.
CONTRACTIONS WHEN TEIE MULTIPLIERS ARE CONVENIENT ALL-
QUOTS OF 10, 100, or 1000, OR SOME MULTIPLE THEREOF.
NOTE.—Those who are unfamiliar with division and with fractions may pass such problems
as involve this knowledge. *
168. TABLE OF ALIQUOTS.
To multiply by 1% multiply by 10 and divide the product by 9
{ % 1+ {{ 10 {{ {{ {{ 8
{{ # {{ 10 {{ {{ {{ 7
{{ 13 { % 10 4. {{ {{ 6
{{ 24 4% 10 {{ 4% {{ 4
44 3# , 4% 10 {{ {{ 4% 3
44 6# {{ 100 {{ {{ {{ 16
{{ 8% {{ 100 {{ {{ {{ 12
4% 12; * { 100 {{ {{ {{ 8
{{ 14% {{ 100 44 44 {{ 7
{{ 16# {{ 100 {{ {{ {{ 6
{{ 18; {{ 300 44 {{ {{ 16
{{ 25 {{ 100 4% \ {{ {{ 4
{{ 31#. 4% 500 4% {{ {{ 16
{{ 33% 4. 100 {{ {{ “ 3
{{ 37; 44 300 {{ {{ 4 * 8
{{ 624 {{ 500 {{ {{ {{ 8
44 663 44 100 and subtract # of product; or
4% 200 and divide by 3.
{{ 75 “ 100 and subtract # of product.
{{ 83% {{ 500 and divide by 6, or x 1000 and +12.
{{ 87; & 4 700 and divide by 8, or
{{ 4% 100 and deduct # of product.
{{ 1124 4% 100 and add # of product.
{{ 125 44 1000 and divide by 8.
{{ 133; 4% 100 and add # of product.
{{ 1663 4% 1000 and divide by 6
4% 250 {{ 1000 {{ 4
{{ 33.3% 4% 1000 {{ 3
{{ 375 {{ 3000 { % 8
{{ 625 {{ 5000 {{ 8
{{ 833% 44 5000 {{ 6
{{ 875 {{ 7000 {{ 8
NOTE.-Learn this table perfectly.
The reason for the above methods of work will appear plain by a careful
examination of the following problems:
PROBLEMS.
169. 1. Multiply 289274 by 24.
OPERATION. -
4) 2892740 Ea:planation.—In answering the conditions of this simple problem,
which is to repeat the multiplicand 24 times, we first observe that 2% is + of
723185. An 10, or that 4 times 2} make 10, and hence to facilitate the work, we first
S. multiply by 10, which is done by annexing one naught. This gives us a
product of 2892740 which is four times too #". for the reason that 10 is 4 times 2}. To produce
the correct result, therefore, we divide by 4, which gives us 723185.
NotE.—In practice, we would simply divide by 4 and carry the division one place.
* CONTRACTIONS IN MULTIPLICATION BY ALIQUOTS. 95
2. Multiply 79862 by 124.
OPERATION.,
8) 7986200 Explanation.—The conditions of this problem require that the multi-
... plicand be taken or repeated 124 times; but in performing the operation,
998.275 Ans we first observe that 12} is # of 100, and hence to save time and figures, we
{ * first take or repeat the multiplicand 100 times, as indicated by the small
naughts, which gives a product as many times too great as 100 is times as great as 12%, which is
8 times. Therefore, we divide by 8 to produce the correct product.
NOTE.-In practice, the annexing of naughts should be made mentally, or, we would 8imply --
8 and extend the operation two places. 4.
3. Multiply 937104 by 25.
OPERATION.
4) 93710400 Explanation.—We first multiply by 100, which is done by annexing
gº two naughts to the multiplicand, and then divide by 4.
T e º & & * fº
23427600 Ans. º reason for this work is the same as that given in the preceding
NOTE.-In practice, divide by 4 and extend the operation two places.
4. Multiply 94.26 by 663.
OPERATION.
3) 94.2600 Explanation.—We first multiply by 100, and then deduct one-third of
314200 the product from itself; the remainder is the correct result or product.
The reason for this is, that as 663 is only # of 100, when we multiply by 100,
we produce a product #'too great, and hence by deducting # of the product,
6284.00 we have in the remainder the correct result.
Multiply the following numbers by the above methods:
(5) (6) (7) (8) (9) (10) (11)
32 36 96 168 40 124 238
6# 8% 16# 62% 37; 125 112
200 300 1600 10500 1500 15500 26775
(12) (13) (14) (15) (16) (17) (18)
817 640 264 640 800 696 1215
375 625 875 75 18; 833% 1334
Multiply 321, 192, 248, 576, 643, and 764,
by 24, 3}, 84, 124, 14%, 163, 18%, 25, 33%, 373, 62%, 75, 873, 1123, 125,
375, and 625.
TO MULTIPLY BY 73, 174, 224, 274, 324, 15, 35, 45, AND 55.
170. 1. To multiply by 73, multiply by 10, and deduct # of the product.
The reason of this is 73, being # of 10, multiplying by 10 gives a product # too large,
therefore, by deducting 4 of the product by 10, we obtain the correct result.
2. To multiply by 174, multiply by 20, and deduct 4 of the product. The
reason is 17; being # of 20, multiplying by 20, gives a product # too large.
Hence, by deducting # of the product by 20, we produce the correct result.
3. To multiply by 22%, multiply by 20 and add 4 of the product to itself.
96 soule's PHILOSOPHIC PRACTICAL MATHEMATICs. *
Considering the reasons given in the two preceding problems, this operation is clear,
from the fact that 224 is ; more than 20.
4. To multiply by 27%, multiply by 30, and deduct fºr of the product. This
operation is evident because 27# is +} of 30.
5. To multiply by 32% multiply by 30 and add #3 of the product.
NOTE: 1.-324 is T', more than 30.
NOTE 2.—There are two 24's in 5; three 24's in 74; four 24's in 10; six 24's in 15; eight 2% in 20;
nine 23's in 224; eleven 23's in 27+, and thirteen 23's in 32%.
6. To multiply by 15, multiply by 10 and add 4 of the product to itself, or
multiply by 30 and divide by 2.
7. To multiply by 35, 45 or 55, multiply respectively by 70, 90, and 110, and
divide the product by 2; or halve the multiplicand and double the multiplier.
NOTE.-This principle may be applied to 65, 75, 85, and 95, by parties who know the multi-
cation table to 20 times 10.
Multiply the following numbers:
(1) (2) (3) (4) (5)
247 214 586 637 4826
15 35 45 55 15
74.10 7490 26370 70070
3705 35035
PRACTICAL OPERATIONS WITH ALIQUOTS AND OTHER NUMBERS.
NotE.—Students who are not familiar with division and fractions, may omit such problems as
involve this knowledge in their solution.
171. 1. What will 462 pounds cost at 25¢.
Directions. Divide by 4, carrying the division two places for cents.
2. What will 840 yards cost at 1242 per yard?
Directions. Divide by 8, carrying the division two places for cents.
3. What will 563 bushels cost at $1.25 per bushel?
Directions. To the number of bushels add 4 of same, carrying the divis-
ion two places for cents. Or, divide by 8, carrying the division three places.
4. What cost 1267 gallons at $1.6247
Directions. To the number of gallons add 3 and # of same, carrying the
division two places for cents.
5. What cost 4102 bushels at $1.374?
Directions. To the number of bushels add + and # of same, carrying the
division two places for cents.
6. What cost 65 turkeys at $12.334 per dozen :
Directions. Multiply the number of turkeys by 12, add to the product
# the number of turkeys and divide the sum by 12, carrying the division two places
for cents.
PRACTICAL OPERATIONS WITH ALIQUOTS AND OTHER NUMBERS. 97
7. What cost 1183 pounds at $1.16%
Directions. Add to the price # and #5 of itself. Or, add to the price #
of itself plus # of the #.
8. What cost 375 boxes at $2.35%
Directions. Multiply the price by 3 and divide the product by 8, carry-
ing the division three places.
9. What cost 76 yards at $1.75%
Directions. Multiply the number of yards by 2 and deduct # of product.
NOTE. $1.75 is g of $2.
10. What cost 1428 pounds at 95%
Directions. From the number of pounds deduct #6, carrying the divis-
ion two places for cents.
11. What cost 50 barrels at $7.45?
Directions. Divide the price by 2, carrying the division 2 places for cents.
12. What cost 84% gallons at 75%
Directions. From the gallons, with the fraction expressed decimally,
deduct # of the same.
13. What cost 2347 pounds corn at 374.2 per bushel ?
Directions. Multiply the pounds by 3 and divide the product by 8, 7,
and 8, carrying the division two places.
14. What cost 87 dozen at $1.20 per dozen”
Directions. Add # of the number of dozens to the same, carrying the
division two places for cents. *
15. What cost 625 pounds at $1.68 per pound?
Directions. Multiply the price per pound by 5 and divide the product
by 8, carrying the division three places.
NOTE. 625 is # of 1000. •
16. What cost 83% dozen at $6.70 per dozen 7
Directions. Multiply the price by 5 and divide the product by 6, carry-
ing the division two places.
NotE. 83% is 3 of 100.
17. What cost 94 yards at 74% per yard?
Directions. Annex 0 to the yards and deduct # of the product from itself.
18. What cost 95 yards at 1742 per yard 3
Directions. Multiply the number of yards by 20 and deduct 4 of the
product from itself.
NOTE. 174 is g of 20.
19. What cost 187; gallons at $3.42 per gallon ?
Directions. Multiply the price by 200 and deduct # of the product from
itself. Or, remove the decimal point two places to the right of the price, and add g
of the amount to itself. Or, shorter still, multiply the price by 3 and divide the
product by 16, carrying the division three places.
NotE. 1874 is ſº of 1000.
20. What cost 250 barrels at $8.15 per barrel.
Directions. Divide the price by 4, carrying the division three places.
98 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. 4%
21. What cost 139 pounds at $1.16% per pound 3
Directions. To the number of pounds add # of the same, carrying the
division two places for cents.
22. What cost 6871 yards at 242 per yard
Directions. Divide the number of yards by 4, carrying the division one
place and point off two.
23. What cost 433 yards at $1.1633
Directions. To the yards, with the fraction expressed decimally, add #
of the result.
24. What cost 87.4 pounds at $2.65%
Directions. From the price per pound omit decimal point, deduct # of
same, carrying the division two places. Or, from the price deduct # and express
the fractional remainder decimally. Thus, 265 – 33} = 231% = $231,874.
25. What cost 235 pounds at $1.80?
Directions. Multiply the pounds by 2 and deduct #5 of the product, or # of
the multiplicand from the product.
26. What cost 196 pounds at $1.75%
Directions. Multiply the pounds by 2 and deduct # of the product from
itself. Or, to the number of pounds add 4 and # of the same.
27. What cost 164 yards at 55g 3
Directions. To # the number of yards extended two places, add ºr of
the #.
28. What cost 50 pounds at $1.29%
Directions. Divide the price by 2, carrying the division two places for
CentS. f
29. What cost 58 pounds at 97%
Directions. Annex 2 naughts to the 58 and deduct 3 times 58 from the
product.
30. What cost 8863 pounds of wheat at $1.15 per bushel ?
Directions. Multiply the pounds by 2 and from the product deduct ºr of
8863 × 1.15
60
NOTE. $1.15 - 60 pounds per bushel = 14c. per pound.
31. What cost 6420 pounds corn at 49% per bushel?
Directions. From the number of pounds deduct # and point off two fig-
6420 × 49
56
NOTE. # = {c. per pound.
32. What cost 582 gallons at $1.314 3
' Directions. To the number of gallons add + and ſº (4 of the 4) of the
the number of pounds. Or thus,
ures. Or thus,
gallons.
33. What cost 465 yards at $2.93%g 3
Directions. Multiply the number of yards by 3 and from the product
deduct # of the number of yards.
Yºr PRACTICAL OPERATIONS witH ALIQUOTS AND OTHER NUMBERS. 99
34. What cost 3840 pounds of hay at $183 per ton ?
Directions. sº or point off three figures in the pounds and mul-
tiply the pounds by # the price, or the price by 4 the pounds.
NOTE.-In this case, when the price is in dollars there will be three decimals in the answer.
When the price contains cents, there will be five decimals in the answer.
35. What cost 14675 pounds of oats at 482 per bushelf
º e 14 4
Directions. gº tº
by 2 and point off two places.
36. What cost 28760 pounds of barley at 52g per bushel?
28760 × 52
48
or multiply the pounds by 3, divide the product
Directions.
two places.
37. What cost 34 bushels and 24 pounds of corn meal at 75% per bushel?
Directions. To # of the bushels add # of 75,
thus, *so
or to the pounds add +3 of the same and point off
or thus, 34 at 75 = $25.50
$25.50 = cost of 34 bushels.
.375 = “ “ , { { or 22 lbs. 24×75= .41
35 = ** ** 2 lbs.
$25.91 practically.
$25.91 practically.
38. What cost 475 lbs. oats at 36% per bushel?
Directions. Since there are 32 lbs. in a bushel, mentally separate the
price 36% into 32%-i-4g. Point off two places in 475 and then add of it to itself.
Thus: 4.75
.593
$5.34%
39. What cost 1663 pounds of beans at $1.30 per bushel ?
Directions. lºgº or 1663 + 62 = 26# bus.
26 bus. $1.30 = $33.80 $26. = 26 bus. at $1.00
31 lbs. .65 7.80 – 26 “ .30
15% “ say .33 or thus, .65 = 31 lbs.
4; “ “ . 9 .33 = 15% “
. 9 = 4; “
$34.87 practically.
40. What cost 2781 pounds of Small hominy at 64 cents per bushel ?
© 1 & 2781 & º tº
Directions. tºº. Or, since there are 50 pounds in a bushel and since
the price is equal to 50 + 4 + +", of # of 50, therefore to the number of pounds, add
# and #5 of # of the pounds, and point off two places.
$34.87 practically.
Thus: }) 27.81
Yº) 5.562
2.2248
$35,5968 :
IOO SouLE's PHILOSOPHIC PRACTICAL MATHEMATICs. A.
41. What cost 473 bushels and 55 pounds of sweet potatoes at $1.40 per
bushel?
NOTE. 60 pounds make a bushel of sweet potatoes in New Orleans.
Directions. 473x $1.40 = $662.20 to which add # of $1.40 which is $1.28
(# or # of $1.40) = $1.40–12 (fly of 140) = $1.28 + $662.20 = $663.48. Ans.
or thus: To the bushels add # of themselves, and to this result
add +} of $1.40 thus:
473.
94.60
94.60
1.28
$663.48
42. What cost 53 tons, 1670 pounds of hay at $16.25 per ton?
43. What cost 25 sacks, 3750 pounds corn at 63% per bushel ?
Directions. Add to the weight # of itself.
NotE. 63 is # more than 56, the number of pounds in a bushel of corn.
44. What cost 25 sacks, 4375 pounds of bran at 874.2 per cwt. ?"
Directions. Subtract from the weight # of itself.
87
NotE. 873c. for 100 pounds is #= #c. per pound = #c. less than 1c. per pound.
45. What cost 30 quarter bales, 3214 pounds hay at $15 per ton?
Directions. Subtract from the weight # of itself.
NOTE. $15 per ton is #33 = #c. per pound.
46. What cost 176 sacks, 35220 pounds oats at 374.2 per bushel ?
Directions. Add to the pounds # of itself, # of the # and # of the #.
NotE. Since 32 pounds make a bushel, 37} is #, + of the #, and # of the + more than 32.
47. What cost 35220 pounds, 1100% bushels oats at 38% per bushel?
Directions. Add to the weight 4 of itself, and 4 of the #. Or, add to
the weight (38-32) = 62 for each bushel, and 42 (3%) for 20 pounds (3) 62 per bushel.
The above problems should be carefully noted by the student, as they contain
many points that will aid him in becoming a rapid calculator.
In all problems of multiplication, a quick, critical view should be given to the
factors, the price and the quantity, to see what advantages may be taken or what
short combinations may be made with them as aliquots, or as approximations to hun-
dreds or easy fractions of hundreds, or in consequence of their family properties
as tens, complements, supplements, etc. By so doing, the mind is made acute and
great facility is acquired in handling numbers. The table of aliquots should be
thoroughly understood.
There is almost no end to the various combinations that may be used in the
operations of multiplication, Particular attention is invited to the methods follow-
ing, wherein numbeºs are glassed into families and where each family possesses some
e
*q → Ç º
£ : © *e
Yºr CONTRACTIONS IN MULTIPLICATION. IOI
individuality of relation to the principles of numbers by which the operations of
‘multiplication may be contracted.
It is often convenient to consider the price as the quantity, and the quantity as
, the price, before making the multiplication, as indicated in the directions for work-
ing several of the above problems.
For further work of this character, see the miscellaneous problems following
Denominate Numbers.
TO MULTIPLY WEHEN TELE MULTIPLIEE IS A MIXED NUMBER, WEHOSE
FRACTIONAL PART IS ONE FRACTIONAL UNIT LESS THAN *
A WHOLE NUMBER.
172. 1. Multiply 243 by 113.
OPERATION.
243 Explanation.—Multiply by 12, and deduct from the product
11% + of the multiplicand. The reason for this is clear, for when we
2916 multiplied by 12 we repeated the multiplicand # time more than the
60; problem requires, and hence the deduction of one-fourth thereof gives
the correct product.
2855#.
PROBLEMIS.
Multiply the following problems in like manner:
(2) (3) (4) (5) (6) (7) (8)
358 149 87 246 • 892 675 486
7; 53 9% 14: 24; 68; 37+}
TO MULTIPLY BY ANY NUMBER BETWEEN EIGHTY-EIGHT AND ONE
HUNDRED, OR NINE HUNDRED AND EIGHTY-EIGHT AND
ONE TEIOUSAND.
173. 1. Multiply 216 by 98.
OPERATION. Explanation–In all problems of this kind, multiply by 100 and
0 0 then deduct as many times the multiplicand as the multiplier is lºë
216 than 100. In this problem, we first multiply the 216 by 100,
432 which is done by annexing two naughts; this gives us a product
2 times 216 too great, for the reason that 100 is two more than
21168 A. 98. Hence, to produce the correct result, we deduct 2 times
IlS. 216 which is 432 from the product by 100, and in the remainder
we have the correct product.
2. Multiply 4268 by 997.
OPERATION. Explanation.—In all problems of this kind, multiply by 2000
and then deduct as many times the multiplicand as the multiplier is
4268000 less than 1000. In this problem, we first multiply the .4268 by
12804 1000, which is done by annexing three naughts; this gives us a
mºmºmºmºmºse product 3 times 4268 too great, for the reason that 1000 is three
4255196 A. more than 997. Hence, to produce the correct result, we deduct
IlS. 3 times 4268, which is 12804, from the product by 1000, and in the
remainder we have the correct product.
I O2 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
Multiply the following numbers in the same manner:
3. 342 by 94 6. 528 by 93 9. 3284 by 995 12. 860 by 89
4, 684 ($ 88 7. 763 “ 91 10. 4009 {{ 989 13. 498 & 92
5. 507 4 91 8. 842 {{ 99 11. 3864 ($ 996 14. 6407 & 988
TO MULTIPLY BY ANY NUMBER BETWEEN ONE HUNDRED AND ONE
HUNDRED AND THIRTEEN, AND ONE THOUSAND AND ONE
TEIOUSAND AND THIRTEEN.
174. 1. Multiply 5239 by 107.
OPERATION. Explanation.—In all problems of this kind, first multiply by
523900 100, then add to the product thus obtained, as many times the
36673 multiplicand as the multiplier is greater than 100.
The reason for this is obvious, for the multiplicand is to be
560573 Ans increased 107 times, and by annexing two 0's we increase it 100
times, to which product we add the product of the multiplicand
by the 7, and thus repeat the multiplicand 107 times.
2. Multiply 2871 by 1005.
OPERATION. Explanation.-In all problems of this kind, first multiply by
28710 00 1000, then add to the product thus obtained, as many times the
14355 multiplicand as the multiplier is greater than 1000.
NOTE: 1.-In the foregoing problems, small naughts have
been annexed, in order to show more clearly the operations of
2885355 Ans. the work.
NOTE: 2. —The operations of the two preceding methods may be extended to the extent of the
students’ knowledge of the multiplication table.
Multiply the following numbers in the same manner:
3. 2564 by 103 6. 521 by 1001 9. 48071 by 1007
4. 809 44 106 7. 1685 4 1006 10. 2910 & 111
5. 4712 € $ 112 8. 2364 ($ 1012 11. 17036 44 1008
TO MULTIPLY BY ANY NUMBER BETWEEN TWELVE AND TWENTY.
175. 1. Multiply 564 by 13. *
OPERATION.
1692 Daplanation.-In all problems of this kind, multipl
3.
º OI’ * OT º; 13 by the unit figure of the multiplier and write the .
ºmºmºmºmº gº cºmme over or under the multiplicand, one place to the right, and
7332 1692 7332 add same to the multiplicand.
*º-º-º-º-º: By this means, we save multiplying by the tens' figure
7332 of the multiplier.
NOTE.-This method of arranging the figures may also be used to advantage when the multi-
plier is any number similar to the following: 106 or 1004. In such cases, the products would be
placed two places to the right for hundreds, and three, for thousands.
¥ CONTRACTIONS IN MULTIPLICATION. IO3
Multiply the following numbers as above:
2. 681 by 14 5. 1269 by 16 8. 728 by 13 11. 286 by 19
3. 754 £6 17 6. 5028 4: 18 9. 1407 4 15 12. 1675 4 17
4. 425 ($ 106 7. 1232 ($ 108 10, 3416 44 1004 13. 4062 ($ 1005
TO MULTIPLY BY ANY NUMBER OF TWO FIGURES WHICH ENDS
IN ONE.
176. 1. Multiply 452 by 41.
OPERATION. te
1808 Explanation.—In all problems of this kind, multiply
452 or 452 or 452 × 41 by the tens' figure of the multiplier and write the product
41 41 1808 over or under the multiplicand one place to the left and
-ºsmºs-smºs add same to the multiplicand.
18532 1808 18532 By this means, we save multiplying by the units' figure
gº-º-º: of the multiplier.
18532
NOTE 1.--This method of arranging the figures may also be used to advantage when the multi-
plier is any number similar to the following: 601 or 4001. In such cases, the products would be
placed two places to the left for hundreds, and three, for thousands.
NOTE 2.-In all these contracted methods, the operation should be repeatcd several times on
each problem.
PROBLEMIS.
Multiply the following numbers as above:
2. 648 by 31 5. 4235 by 51 8, 2890 by 61 11. 9027 by 71
3. 872 ($ 41 6. 2806 ($ 81 9. 5075 (€ 91 12. 2364 “ 111
4, 425 ($ 501 7. 1242 {{ 601 10. 6322 4° 4001 13. 4327 4 7001
TO MULTIPLY BY THE FACTOR'S OF THE MULTIPLIER.
177. Multiply 8421 by 64.
OPERATION.
8421 Ea:planation.—Instead of multiplying by 64, we use the factors 8 and
64 - 8; first multiplying by 8 we produce a product of 67368, which we multi-
e- ply by 8, and thus produce the correct result. By this system of work, we
67368 save one line of figures and the addition of two lines.
S age 84 for further explanations.
538944 Ans. ee pag p
PROBLEMS.
Multiply the following numbers as above:
2. 2407 by 32 4. 796 by 63 6. 527 by 48 8, 2840 by 56
3. 682 4° 45 5, 1428 4 36 7. 290 46 72 9. 7065 (€ 49
IO4. soule's PHILOSOPHIC PRACTICAL MATHEMATICs. 4×
TO MULTIPLY ANY NUMBER BY ELEVEN OR ONE EIUNDRED AND
ELEVEN.
178. 1. Multiply 62 by 11.
OPERATION. Erplanation.—In multiplying any number by 11, it is evident that
we have but to add the number to its decuple, i. e. to ten times the
682 A number, hence in this and all operations of two figures we have but
IlS. to add the two figures together and place the sum between them.
When the sum of the two figures is in excess of 9, carry one to the
left hand figure. Thus multiply 87 by 11; 8 and 7 added make 15; place the 5 between the figures
and add the 1 to the 8 which gives a product of 957.
2. Multiply 3456 by 11.
OPERATION.
3456 Explanation.—In this problem, according to the principle stated
11 in the first problem, we add the number to its decuple, thus 0-H. 6 =
6; 6 + 5 = 11; 5 + 4 + 1 to carry = 10; 4 + 3 + 1 to carry = 8;
38016 then bring down the 3, and we have the correct product.
3. Multiply 3456 by 111.
OPERATION. Explanation.-In this problem, we are required to repeat 3456
100 times plus 10 times, plus 1 time. Hence, if we add 3456 to 10
3456 times 3456 and 100 times 3456 we shall have the correct result. This
111 we do by bringing down the first figure 6, and adding as follows:
6 plus 5 = 11; 6 plus 1 to carry, plus 5 plus 4 = 16; 5 plus 1 to
& carry, plus 4 plus 3 = 13; 4 plus 1 to carry, plus 3 = 8; and then
383616 bring down the 3. By performing these operations in the usual
& manner, the work will appear plain.
Multiply the following numbers as above:
4. 463 by 11 6. 245 by 111 8. 823 by 111 10. 2407 by 111
5. 386 ‘‘ 11 7. 741 & 11 9, 568 46 11 11. 14586 “ 1111
TO MULTIPLY BY ANY NUMBER ONE PART OF WHICH Is A FACTOR
OF ANOTHER PART.
179. 1. Multiply 3246 by 328.
OPERATION.
3246 Ea:planation.—The conditions of this problem are, that we take or
328 repeat the multiplicand 328 times. In performing the operation, we
tºº, first multiply by 8, which repeats it 8 times; we yet have it to repeat
25968 320 times, and as 320 is equal to 8 40 times, therefore, 40 times the pro-
1038720 duct by 8 (25968) is the same as 320 times the multiplicand; hence, to
eme shorten the operation, we add to the product by 8, 40 times itself.
1064688 The sum of these two products is the correct product.
2. Multiply 7251 by 618.
OPERATION. Ea:planation.—In this problem, we first multiply by the 6 hundreds
"; which repeats the 7251 600 times, and leaves it unrepeated 18 times.
1 To repeat it 18 times more, we observe that 18 is equal to 6, 3 times,
4350600 and having already repeated the 7251 600 times, we therefore repeat
130518 the one hundredth part of the product by the six hundreds 3 times,
which is equal to 18 times 7251, and add the same to the product by 6
hundreds. The result is the correct product.
4481118
* - CONTRACTIONS IN MULTIPLICATION. IO5
3. Multiply 536183 by 27945.
FIRST OPERATION.
536183 --
27945 Eacplanation.—In thi bl re Ob that the first two fi
a planation.—In this problem, we observe that the first two figures
48.2564700 45, are equal to the hundreds' figure 5 times, and that the thousands'
1447694.1000 and ten thousands' figures 27, are equal to the hundreds' figure 3 times.
24128235 We therefore multiply first by the hundreds' figure 9, which repeats the
14983 638935 multiplicand 900 times. We have yet to repeat it 27045 times, and
SECOND OPERATION. (using the 27000 first), since 27000 is equal to 9 3000 times, we therefore
Omitting the naughts. *P* the product by 9 3000 times, instead of the multiplicand 27000
536183 times, which gives us 14476941000; and since 45 is equal to 9 five times,
27945 we therefore repeat the one hundredth part of the product by the nine
*=ºmºeºsºmmºn hundreds 5 times, instead of repeating the multiplicand 45 times, which
1.# gives us 24128235. We then add the several products together, the sum
24128235 of which is the true product.
14983633935
TO MULTIPLY BY ANY NUMBER OF TWO FIGURES ENDING WITH
NINE.
180. 1. Multiply 563 by 39.
OPERATION.
563 563 x 40 Explanation.—In all problems of this kind, in-
39 22520 crease the multiplier by 1 and from the product sub-
OI’ e - © º e
tract the multiplicand. In this problem we multi-
22520 21957 Ans. ply by 40, which gives a product of 22520, from
21957 Ans. which we subtract 563 and obtain 21957, the answer.
PROBLEMIS.
Multiply the following numbers as above.
2. 726 by 29 4. 864 by 49 6. 627 by 59 8. 9081 by 69 .
3. 683 (4 79 5. 927 & 89 7. 1074 “ 19 9, 23.45 % 99
TO MULTIPLY ANY TWO NUMBERS OF TWO OR THREE FIGURES
EACH, WHEN THE TENS AND HUNDREDS ARE ALIKE
AND THE SUM OF THE UNITS IS TEN.
181. 1. Multiply 86 by 84.
OPERATION. Explanation.—In all problems of this kind, first multiply the units'
86 figures and write the whole result, then add one to the multiplier of
84 the tens, and multiply the other tens, or tens and hundreds, by it, and
prefix the result to the product of the units' figures. In this problem,
we first multiply 4 times 6 are 24, which we set. Then add 1 to the 8
7224 of the tens' column, making it 9; we multiply 9 times 8 are 72, which
is set to the left of the 24 and produces the correct product.
I oé SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. - *
2. Multiply 132 by 138.
OPERATION. Explanation.--We first multiply 8 times 2 are 16, which constitutes
132 the units’ and tens' figures of the product. We then add one to 13,
138 making it 14, and multiply 14 times 13 are 182, which completes the
product. Or, after adding the 1 to the 13, we may multiply 14 times 3
are 42; set the 2 and multiply 14 times 1 plus 4 to carry, are 18.
18216 NoTE 1.--When the multiplication of the units' figures does not
give a product of two figures, the tens’ place must be filled with a 0.
NoTE 2.-The reason for this work is that since the sum of the units’ figures is 10, and the tens'
figures are alike, it is plain that the sum of the products of the tens by the units and the units by
the tens, is equal to 10 times the tens’ figure, and since its first figure occupying the second place
is naught it does not change the original product of the units, and since the second or carrying
figure of the units by the tens is always the same as the tens of the factors, we therefore add 1 to
either factor and multiply by the other for the final product.
Multiply the following numbers as above:
(3) (4) (5) (6) (7) (8) (9) (10)
38 61 223 149 326 417 311 208
32 69 227 141 324. 413 319 202
1216 4209 50621 21009 105624. 172221
(11) (12) (13) (14) (15) (16) (17) (18) (19)
29 47 81 91 126 206 164 102 113
21 43 89 99 124 204 166 108 117
essm -sºme mºm a-m mºm- ºmmº- mºmmam -m-m-m-, *-
TO MULTIPLY TWO NUMBERS OF TWO FIGURES EACH WHEN THE
TENS7 FIGURES ADD TEN AND THE UNITS? FIGURES ARE ALIKE.
182. 1. Multiply 47 by 67.
OPERATION. Ea:planation.—In all problems of this kind, first multiply the units’
47 figures and set the whole result in the product line, then multiply the
67 tens' figures and add to their product one of the units' figures, and
prefix the result to the product of the units' figures. In problem 1, we
first multiply 7 times 7 = 49; then 6 times 4 = 24, to which we add 7
3149 Ans. and obtain 31. Problems 2, 3, 4, and 5, were similarly worked.
NOTE-When the product of the units' figures is but one figure, the tens' place in the product
must be filled with a 0.
Multiply the following numbers as above:
(2) (3) (4) (5) (6) (7) (8) (9) (10) (11)
84 73 96 41 78 24 98 46 42 77
24 33 16 61 38 84 18 66 62 37
*-* *
" - - - - -
2016 2409 1536 2501
M. CONTRACTIONS IN MULTIPLICATION. Io?
TO MULTIPLY TWO NUMBERS WHEN THE UNITS? FIGURES ARE
FIVES AND THE SUM OF THE TENS, FIGURES IS MORE
OR LESS THAN TEN.
183. 1. Multiply 65 × 85.
65 OPERATION. Explanation. First, 5 × 5 = 25 units.
85 Second, 8 × 6 = 48 hundreds, to which add
— 5 × 5 = 25. 8t. × 6t.= 48h.-- # of 4 of the sum of the tens' figures, 8 + 6 - 2 = 7,
5525 (8+6) = 7h.= 55h. or 5525. = 55 hundreds. -
2. Multiply 45 × 35,
45 OPERATION. Explanation. First, 5 × 5 = 25 units.
35 Second, 3 × 4 = 12 hundreds, to which add
— 5 x 5 = 25. 3t. × 4t. = 12h. x 4 + of the sum of the tens' figures, 3 + 4 + 2 = 3
1575 of (3 + 4).3%h.-- 12h. = 1550 + hundreds 50 units, = 1550 + the 25 units first
the 25 = 1575. obtained = 1575 product.
45 SECOND OPERATION. Explanation.—In this solution, since the sum
35 of the tens’ figures is odd, we first find the pro-
duct of the tens' figures to which we add the pro-
3 x 4 = 12h.-- # of (3 + 4) = 3h., duct of the units figures and thus produce the
* 1575 º final product. This method may be used in all
50 units = 1550 + 5 × 5,- 1575. cases, if-desired.
PROBLEMIS.
Multiply the following problems as above :
(3) (4) (5) (6) (7) (8) (9) (10) (11)
85 75 45 105 165 115 175 225 185
25 65 95 85 95 115 15 35 145
a- - - - - - - - -
TO MULTIPLY TWO NUMBERS OF TWO OR THREE FIGURES EACH,
WHEN THE HUNDREDS AND TENS, OR THE UNITS OR
TENS FIGURES ONLY, ARE ALIKE.
184. Thus, multiply the following numbers:
(1) . (2) (3) (4)
54 87 124 116
34 82 123 146
1836 Ans. 7134 15252 16936 Ans.
Ea:planation.—In all problems of this kind, first multiply the unit figures and write the unit
result in the product line, then add the column of units or tens, that is not alike, or both, where
three figures are used, and with the sum multiply one of the numbers in the column, where they are
I O3 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. Yºr
alike, and add to it the carrying figure; then multiply the figures in the tens or tens and hundreds,
and add the carrying figure; the result will be the correct product.
In problem 1, we first multiply 4 times 4 = 16; the 6 we set and carry 1; then 3 plus 5 = 8,
and 8 times 4 = 32 plus 1 to carry make 33; the 3 we set and carry 3; then 3 times 5 = 15 plus 3
to carry make 18.
In problem 2, we first multiply 2 times 7 = 14; then 7 plus 2 = 9 and 9 times 8 =72 plus 1 to
carry = 73; then 8 times 8 = 64 plus 7 to carry =71. Problems 3 and 4 are similarly worked.
PROBLEMIS.
Multiply the following numbers as above:
(5) (6) (7) (8) (9) (10) (11) (12) (13)
73 65 106 45 123 89 154 93 146
23 64 107 85 113 85 154 97 106
sºm-º dº tºº ºmºmº ºmºmºmºm sºmmºn sºmmºn sº- tºmsºmºsº
TO MULTIPLY BY THE COMPLEMENTS AND SUPPLEMENTS OF
NUMBERS.
[See page 26 for a definition of Complements and Supplements of Numbers].
185. The following operations involving the complements and supplements of
numbers are deduced from Algebraic principles, and while these principles are easily
applied to Arithmetical Computations, the reasons therefor cannot be fully com-
prehended without a knowledge of Algebra, and hence are not given.
TO MULTIPLY TWO NUMBERS, BOTH OF WHICH ARE CONVEN-
IENT NUMBERS UNDER ONE HUNDRED, ONE THOUSAND, ETC.
186. 1. Multiply 97 by 94.
OPERATION.
97 (3 Complement of 97) ..º.º. all problems of this kind, 1st multiply
6 C l t of 94 the complements of the two numbers and write the result in
94 (6 Complement o ) the product line. 2. Subtract the complement of either
&=º number from the other number and prefix the remainder to
91.18 the product of the complements. Or, add the multiplicand
and multiplier, rejecting 1 at the left hand of the sum.
NOTE. 1.--When the product of the complements does not contain as many figures as there
are naughts in the order, 100, 1000, etc., to which the numbers belong, prefix naughts to supply the
deficiency. See Problems 2 and 3.
NOTE 2.—When the number of figures in the product of the complements exceeds the naughts
in the order, 100 and 1000, etc., to which the numbers belong, add the excess, or left hand figure
to the result obtained by subtracting the complement of either number from the other number.
See Problems 4 and 5.
* CONTRACTIONS IN MULTIPLICATION. IO9
2. Multiply 98 by 97. 3. Multiply 991 by 989.
OPERATION. Eacplanation. 1st. 3 times OPERATION. Ea:planation. 1st.
98 (2 Comp.) 2=6, which we write in the 991 (9 Comp. of 991.) 11 times 9 = 99,
97 (3 Comp.) product line and prefix a 989 (11 Comp. of 989.) which is written
Inaught as directed in Note 1. in the product line
2d. 3 from 98 = 95, which and a naught pre-
pletes the product. - Note 1.
2d. 9 from 989 = 980, which is prefixed to
the 099 and completes the product.
5. Multiply 978 by 940.
-ms-ºn-
4. Multiply 89 by 88.
OPERATION. Ea:planation. 1st. 12 times OPERATION. Explanation. 1st. 60
89 (11 Comp.) 11 =132, 32 of which is writ- 978 (22 Comp.) times *; = 1320, 320 of
88 (12 Comp.) ten in the product line, as di- 940 (60 Comp.) which is written in the
rººd * . º 77-H1 product line as directed in
e OIO &j = - Note 2.
7832 Ans. , 78, which is prefixed to the 919320 Ans. 2d. 60 from 978 =918-1-
32 and completes the product. 1 =919, which is prefixed to the 320
- and completes the product.
PROBLEMS.
Multiply the following numbers as above:
(6) (7) (8) (9) (10) (11) (12) (13) (14) (15)
94 92 88 82 99 73 97 74 85 99
93 93 88 91 96 98 62 85 91 99
(16) (17) (18) (19) (20) (21) (22) (23)
996 988 991 969 950 940 965 999
994 997 985 985 988 930 987 999
990024 9.85036 976135
TO MULTIPLY TWO NUMBERS, WHOSE COMBINATION IS SUCH THAT THE PRODUCT
OF THE COMPLEMENT OF THE UNITS, FIGURE OF THE MULTIPLIER, BY THE
TENS FIGURE OF THE MULTIPLICAND IS THE SAME AS THE PRODUCT
OF THE TENS” FIGURE OF THE MULTIPLEER BY THE UNITS’
FIGURE OF THE MULTIPLICAND.
187. 1. Multiply 48 by 34.
OPERATION. Explanation.—In this problem, the complement of the unit figure of
48 the multiplier is 6, and 6 times the tens' figure of the multiplicand is 24;
34 which is the same as the product of the tens' figure of the multiplier, and
the unit figure of the multiplicand. In all cases of this kind, we multiply
gººm-º: the units' figures and write the result in the product line. Then we add 1
1632 Ans. to the tens’ figure of the multiplier, and with the sum multiply the
tens' figure, or tens and hundreds of the multiplicand and prefix the result
to the product of the units' figures. In this problem, we say 4 times 8 are 32; 4 times 4 are 16.
NotE.—Any multiple of 11, may be multiplied as above when the figures of the multiplier
add 10.
I IO SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. M.
PROBLEMS.
2. Multiply the following numbers as above:
(2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13)
63 32 88 146 128 42 96 147 96 84 44 55
29 94. 73 77 94 86 38 124 66 67 73 82
TO MULTIPLY TWO NUMBERs, BOTH OF WHICH ARE CONVEN-
IENT NUMBERS UNDER ANY NUMBER OF EIUNDREDS.
188. 1. Multiply 497 by 294.
OPERATION. Explanation. — In this problem, we first multiply
3 C I t; Of 9 the complements of the multiplicand and the
30 500) 497 & Complement of 97 multiplier and set the result 18, in the product line.
9 U (300) 294 6 Complement of 94. (Where this result is but one figure, the tens' place is
- filled with a 0. When the result is more than two figures,
39 the hundreds' figure is added to the hundreds' column of
the product). 2. We mentally add the complement fig-
146118 ures to their respective numbers, and thus produce 500
and 300 as shown in the parentheses on the left. 3. We
multiply the 5 of the 500, by the complement of the multiplier, and the 3 of the 300, by the com-
plement of the multiplicand and add the products, which gives 39, as shown by the number within
the brace on the left of the operation. 4. We subtract the 39 from 100 and thus obtain
61 which is prefixed to the 18 in the product line. 5. We multiply 5 and 3, the significant figures
of 500 and 300, and thus obtain 15, from which we subtract 1, the number of hundreds used when
the 39 was subtracted, and prefix the remainder 14, to the 6118, which completes the product. All
similar numbers of hundreds and thousands may be multiplied in a similar manner.
NOTE: 1.—The figures shown within the brace and parentheses should be mentally made.
NOTE: 2.-When the sum of the products of the increased hundreds' figures by the comple-
ments, in that part of the operation where the 39 was produced, exceeds 100, then subtract the sum
from 200, and from the product of the increased significant figures of the multiplicand and multi-
plier, deduct 2.
PROBLEMS.
2. Multiply 595 by 393. 3. Multiply 698 by 497.
OPERATION. Elaplanation by figures. OPERATION.
(6) 595 (5) 7 times 5 = 35. (7) 698 (2)
(4) 393 (7) 7 times 6 + 5 times 4 = 62. (5) 497 (3)
62 from 100 = 38.
233835 Ans. 4 times 6 — 1 = 23. 346906 Ans.
Multiply the following numbers as above:
(4) (5) (6) (7) (8) (9) (10) (11)
(9) 891 (9) 291 892 991 394 791 392 488
(6) 592 (8) 295 596 892 592 495 788 692
(12) (13) (14) (15) (16) (17) (18) (19)
493 692 790 392 488 598 691 894
395 597 691 593 489 496 394 389
Yºr CONTRACTIONS IN MULTIPLICATION. I I I
TO MULTIPLY TWO NUMBERS, BOTH OF WHICH ARE CONVEN-
IENT NUMBERS AROVE, ANY NUMBER OF HUNDREDS.
189. 1. Multiply 704 by 305.
OPERATION. Explanation.—In this problem, we first multiply the supplement figures
4 and 5, and set the result 20 in the product line. 2. We multiply the
704. hundreds' figures of each number by the supplement figure of the other,
305 and add the two products thus, 5 times 7 = 35; 4 times 3 = 12; 35 and 12
= 47; this 47 we prefix to the 20 in the product line. 3. We multiply the
2 hundreds' figures and obtain 21, which we write in the product and com-
14,720 plete the operation. All similar numbers of hundreds and thousands may
be similarly worked.
NOTE. 1.--When the product of the units’ or supplement figures do not give two figures, fill
the tens' place with a 0. See Problem 2. When this product exceeds 2 figures, add the hundreds”
figure to the hundreds' column of the product. See Problem 3.
NOTE: 2. —When the sum of the products of the hundreds' figures by the units' figures is not
two figures, fill the thousands' place in the product with a 0. See Problem 4. When this sum
exceeds 2 figures, add the ten thousands to the ten thousands' column of the product. See Problem 5.
PROBLEMS.
(2) (3) (4) (5) (6) (7) (8)
301 312 501 912 813 612 1206
202 209 301 811 206 509 911
60802 6520s 150801 739632 - -
(9) (10) (11) (12) (13) (14) (15)
406 612 308 810 511 203 1412
507 109 702 305 406 202 1211
TO MULTIPLY TWO NUMBERS, BOTH OF WHICH ARE CONVEN-
IENT NUMBERS OVER ONE HUNDRED, ONE
THOUSAND. ETC.
190. 1. Multiply 108 by 104.
OPERATION. Explanation—In all problems of this kind, first write the product
of the supplement figures in the product line. 2. Cancel, mentally,
I08 (8 Supplement.) one of the left hand figures in one of the numbers, add the remain-
104 (4 Supplement.) ing figures and the carrying figure, if any, and prefix the sum to the
product of the supplement figures. Thus, 4 × 8 = 32; then cancel
1123 1 in the 108 and add 8 to 104 = 112, which is prefixed to the 32 and
232 Ans. completes the product.
Or, first cancel the left hand figure in one of the numbers, then add
the remaining figures and annex thereto the product of the supplements of the two numbers.
NOTE 1.-In case of hundreds, when the product of the supplement numbers does not contain
two figures, fill the tens' place with a 0.
In case of thousands, when the product or the supplement numbers does not contain three
figures, fill the hundreds' place or the tens' and hundreds' places with 0's.
NoTE 2.—In case of hundreds, when the product of the supplement numbers contains more
*:::::::: figures, add the hundreds to the hundreds' column, when adding the multiplicand and
multiplier.
In case of thousands when the product of the supplement numbers contains more than three
#; add the thousands to the thousands' column or place, when adding the multiplicand and
multiplier.
II 2 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. X-
2. Multiply 1014 by 1009.
OPERATION. Explanation.—Here we first multiply the supplement numbers and write
the result in the product line. 2. Cancel, mentally, the left hand figure
1014 in one of the numbers, add the remaining figures, and the carrying fig-
1009 ure, if any, and prefix the sum to the product by the supplement figures.
Thus, in this problem, 9 × 14 = 126; then cancel the left hand 1 in the
1023126 Ans 1014, add 14 to 1009, and prefix the sum, 1023, to the product by the Sup-
g plement numbers.
NOTE: 1.—When the product of the supplement numbers does not contain three figures, fill the
hundreds' place, or the hundreds' and tens', with 0s. See Problem 4.
NoTE 2. When the product of the supplement numbers contains more than three figures, add
the thousands to the thousands' column. For the higher numbers, as 10000, etc., the same principle
maintains. See Problem 5.
PROBLEMIS.
Multiply the following numbers as above:
(3) (4) (5) (6) (7) (8)
103 1004. 1036 116 121 1342
102 1001 1040 111 104 1015
10506 1005004 1077.440 12876 12584 1862130
(9) (10) (11) (12) (13) (14) (15)
103 114 109 1008 1024 1135 1037
108 112 112 1007 1016 1010 1040
TO MULTIPLY TWO NUMBERS, ONE OF WHICH IS A CONVENIENT
NUMBER OVER, AND THE OTHER A CONVENIENT NUMBER
UNDER ONE HUNDRED, ONE THOUSAND, ETC.
191. 1. Multiply 105 by 93.
OPERATION. º . problems § this º first multiply the
Complement and Supplement numbers and subtract the produ
105 (5 Supplement.) from 100, or 1000, . as the numbers multiplied tºº. to †.
93 (7 Complement.) order of hundreds or thousands, and write the remainder in the pro.
duct life, as the first two figures of the product. 2. Cancel the
9765 left hand 1 in the multiplicand, add the remaining figures of the
Same, from which sum subtract. 1, or the number of hundreds used
in subtracting the product of the complement and supplement fig-
ures, and prefix the remainder to the product first obtained. Thus, in this problem, we say, 7x5 =
35; 35 from 100 =63, which is written in the product line. Then cancel, mentally, the left hand 1,
we add 93 and 5 = 98, less 1, for the hundred used in subtracting the 35 = 97, which is prefixed to
the 65, and completes the product.
NoTE 1. . When the remainder, resulting from the subtraction of the products of the comple-
ment and supplement numbers does not contain two figures, then fill the tens' place with a 0. See
Problem 2.
NOTE 2.--When the product by the complement and supplement figures exceeds 100, or 1000,
then subtract the same from 200 or 300, or 2000 or 3000, according as the numbers multiplied are of
the hundreds' or thousands' orders, and then, from the sum of the added figures of the multiplicand
and multiplier, subtract as many as you used hundreds or thousands. See Problems 2 and 3.
CONTRACTIONS IN MULTIPLICATION. II 3
Multiply the following numbers as above :
(2) (3) MENTAL OPERATION.—6 × 223 =
112 MENTAL OPERATION.—11X12 = 132; 1223 1338; 2000–1338 = 662. Then 4+3
89 200–132=68. Then 9-1-2 = 11; 11–2 994. Tłº, º º
e Il "TV E 12.
— = 9. Then 8 + 1 = 9. Left hand 1 is *mºnºsmº tº Left hand figure of iºnisand
9968 canceled. 1215662 is canceled.
(4) (5) (6) (7) (8) (9) (10) (11)
116 1012 133 1023 125 123 1013 1123
97 988 85 997 96 92 996 988
11252 999856 11305 1019931 12000 11316 1008948 1109524
(12) (13) (14) (15) (16) (17) (18) (19) (20)
108 1012 1104 124 111 109 1018 1201. 1302
95 990 988 96 88 89 940 996 975
*E* **E= *==* *E-º-º: sº *E---> º-º-º-º-º: *==sº * tº-º-º-º-º:
A PECULLAR PROPERTY OF THE NUMBER NINE.
192. If we multiply the 9 figures in their order: 1, 2, 3, 4, 5, 6, 7, 8, 9, by 9, or
any multiple of 9, not exceeding 9 times 9, the product will be in like figures,
except the tens' place, which will be a 0. The significant figure of the product will
be the quotient of the multiplier divided by 9. Thus, to multiply by 9, will give a
product of 1s; 18, which is two times 9, will give a product of 2s, and so on with
the other multiples of 9, up to 81.
OPERATIONS.
(1) (2) (3) (4)
123456789 123456789 123456789 123456789
9 27 45 72
1111111101 3333333303 5555555.505 88888.88808
If we omit the 8 in the multiplicand the product figures will all be the same.
There are many peculiar properties belonging to the figure nine, by reason of
its being 1 less than the radix of our system of notation, but being mostly of no
practical value, we cannot give them space. There is, however, one property of the
nine, that may often be used with advantage by accountants, in the detection of
errors in posting or transferring accounts, and we will state it. It is as follows:
That the difference between any number, and the figures composing the num-
ber reversed in any manner, is a multiple of 9. Thus, the difference between 6871,
and any transposition or reversing possible to be made with the same figures, will
be a multiple of 9.
EXAMPLES.
6871 6871 6871 6871 6871 6871
7861 1786 6718 6781 8671. 8167
9) 990 9) 5085 9) 153 9) 90 9) 1800 9) 1296
110 565 17 10 200 144
NotE.—See Expert Accounting, in Soulé's New Science and Practice of Accounts, for an
extended application of the Properties of 9's and 11's, to the finding of Errors in Accounts. Also, see
the Properties of 9's and 11's following Division in this book.
II.4.
SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
ALGEBRAIC FORMUL.AE FROM WPHICH ARE DEDUCED THE OPERA-
TIONS OF MULTIPLYING COMPLEMENT AND SUPPLEMENT
NUMBERS.
193. (See Problem, page 105). (See Problem, page 106).
86 = 90 — 4 47 = 50 — 3
84 = 90 – 6 67 = 50 + 17
7224 90 × 90 – 4 × 90 3149 50 × 50 — 3 × 50
— 6 × 90 + 4 × 6 + 17 × 50 — 3 × 17
8100 — 10 × 90 –H 24 = 2500 + 14 × 50 — 51 =
8100–900 + 24 = 7224 2500 + 700 — 51 = 3149
(See Problem, page 107). (See Problem, page 108).
65 = 60 + 5 97 = 100 — 3
85 = 80 -H 5 94 = 100 – 6
5525 60 × 80′ + 5 × 80 9118 100 × 100 — 3 × 100
+ 5 × 60 + 5 × 5
4800 + 400 + 300 + 25 = 5525
(See Problem, page 109).
48 = 50 — 2 -
34 = 50 — 16
1632 50 × 50 — 2 × 50
— 16 × 50 + 2 × 16
2500 — 18 × 50 + 32 =
2500–900 + 32 = 1632
(See Problem, page 111).
704 = 700 + 4
305 = 300 + 5
214720 700 x 300 + 4 × 300
+ 5 × 700 + 4 × 5
210000+1200+3500–H20=214.720
(See Problem, page 112).
105 = 100 + 5
93 = 100 – 7
9765 100 x 100 + 5 × 100
— 7 × 100 — 5 × 7
10000–2 × 100 — 35 =
10000 — 200 — 35 = 9765
— 6 × 100 + 3 × 6
10000 – 9 × 100 + 18 =
10000–900 + 18 =. 9118
(See Problem, page 110).
497 = 500 — 3
294 = 300 – 6
146118 500 × 300 — 3 × 300
— 6 × 500–H3× 6
150000–900–3000+18=146118
(See Problem, page 111).
108 = 100 + 8
104 = 100 + 4
11232 100 × 100 + 8 × 100
+ 4 × 100 + 8 × 4
10000+ 800 + 400+ 32=11232
Tº2 = 40 – 8
43 = 40 + 3
1376 40 × 40 – 8 × 40
+ 3 × 40 – 8 × 3
1600 — 5 × 40 – 24 =
1600 — 200 –24 = 1376
CONTRACTIONS IN
115
MULTIPLICATION.
&
MISCELLANEOUS DRILL PROBLEMIS IN CONTRACTED METHODS.
194. The student who aspires to rapidity, accuracy, and superior merit as a
business calculator, should practice a few minutes on one of the following exercises,
each day, before entering upon the regular work of arithmetic.
This exercise is the secret road that leads to honor and fame as an expert
calculator.
NOTE.—Students who do not understand division and fractions, may omit such problems as
require such knowledge.
Add at one
Operation.
36
42
57
28
• 81
79
14
68
94.
86
23
*=
346
481
209
526
140
37
34
93
92
EXERCISE No. 1.
MULTIPLY SIMULTANEOUSLY.
46 62 58 - 2416 124 342
26 34 47 23 237 276
MULTIPLY BY ALIQUOTS.
63 432 56 344 1368 4026
25 33%. 12; 62% 125 375
COMPLEMENTS. SUPPLEMENTS.
98 91 991 108 104. 162
97 88 988 105 102 107
COMPLEMENTS AND SUPPLEMENTS.
126 146 1012 497 594 806
92 77 988 294 395 307
TENS’ FAMILY.
61 124 323 87 73 84 109
69 126 327 27 33 24 101
ADD HORIZONTALLY.
4708, 290, 728, 49, 864, 71403.
Add b
Vicenary
3204
2063
72
1.
1012
1011
2161
1024
3248
721
1809
109
93
44
46
714.
512
68
48
the
ystem.
58
34
68
17
45
78
12
34
56
90
13
51
42
67
19
78
55
89
76
90
68
34
79
23
31
84
45
54
55
66
77
89
I I 6 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
㺠195. DRILL ExeRcise No. 2.
327 SIMULTANEOUS MULTIPLICATION.
114
208 83 68 89 298 689 548 768 987 4067
75 79 97 76 89 98 287 578 876 352
320 * = e {-º-º: &_-ºl Gº-º-º: *= *=== tºº-ºº: emº
612
184 ALIQUOTS.
º 1462 at 12; g. 480 at 374 g. 654 at 87.4%.
544 at 18%. 326 at 16%. 562 at 83.4%.
* 116 at 25%. 183 at 334%. 275 at $1.25
COMPLEMENTS.
2362 87 85 78 82 . 99 988 985 976
89 91 88 89 99 989 975 974.
1427 °
5134 * > — —— . * tº
409 SUPPLEMENTS. COMPLEMENTS & SUPPLEMENTS.
1292 104. 123 1014 1126 || 137 1045 148
102 116 1009 1012 || 88 986 95
36021 196. DRI LL EXERCISE No. 3.
12003 MISCELLANEOUS CONTRACTING METHODS.
10400 96 78 593 412 42 68 246 6843 463
5026 94. 38 491 418 29 36 127 729 1624
10820 * = *s * º mºº smºs ºsmºs immºn $º *=ts *=s===
72035 se
7604 58924. 3456 432 268 2695 7321
964 36945 111 17% 75 1008 996
Subtract by addition: 4867 | 248 3842 | 163 460 | 6892
1261 963 49 588 | 4087
527 1125 | 1.407 123 || 9364
65 OPERATION. 45 OPERATION.
85 35 75 95
5 x 5 = 25. 8t. x 6t.= 48h.-H. — 5 × 5 = 25. 3t. × 4t.= 12h.-- 55 25
5525 - # of (8+6)= 7h.= 55h. or 1575 # of (3 + 4) = 34h.-- 12h.- || – —
55.25. 1550 – the 25 = 1575.
63 32 88 146 42
94 73 77 86
CONTRACTIONS IN MULTIPLICATION. I 17
197. DRI LL EXERCISE No. 4.
1. 62 × 25 6. 974 × 125 11. 328 × 375
2. 87 × 124 7. 65 × 224 12. 261 × 750
3. 140 × 18; 8, 184 × 1874 13. 414 × 35
4. 365 x 33% 9. 212 × 75 14. 807 x 84
5. 740 x 62% 10. 514 × 83% 15. 694 × 16#
198. DRILL EXERCISE No. 5.
NEW ORLEANs, Apr. 1, 1894.
NEUGASS & THOMPSON,
Bot. Of DELERNO & GRANTER.
H
{ 5234 lbs. Butter sº sº a 25 g
136; “ La. Pecans - tº-º * “ 124 g
56 doz. C. Eggs - $º “ 18; g
249 boxes Lobsters * tº “ 374g.
140 jars Pickles º “ 624%
48 lbs. Rice & º wº “ 64%.
83 bush. W. Corn e * “ 87 g
64 “ W. Oats - * * *º “ 44 g
92 bags Salt - sº tº & “ 93 g
84 lbs. B. Tea º tº- “ 97 g
108 gals. R. Whiskey tº sº 4 $1.06
143 doz. Pine Apples - sº * : {{ 1.15
119 lbs. Y. H. Tea, gº “ 96 g
324 boxes Shrimp - me ſº * “ 35 g
240 gals. Fire Proof Coal Oil 4- “ 15 g.
164 lbs. Lard sº ſº tº º “ 7.4%
286 “ Green Peas sº ſº 3- “ 24%
320 “ N. Y. C. Cheese º tº- “ 174g
112 gals. La. Molasses - - - “ 22; ſ!
124 “ B. Whiskey º 4 $1.25
72 bbls. La. Oranges gº {{ 3.75
61 “ N. Potatoes º * {{ 2.50
16; lbs. C. Soap - e- gº = “ 84 g
9+ “ Persian Dates - - - “ 63 g
1 bbl. 50g Sugar, 241-24-217 lbs. “ 54 g
43 lbs. Lard Oil tº gº tº “ 58 g
45 “ Elgin C. Butter R-8 º “ 35 g
34 bush. W. Oats tºº gº sº 4 48 g
425 bbls. Flour gº -> gº 4 $8.75 562070
NOTE 1.—All extensions of the above bill should be made mentally.
NotE 2.—See pages 96 to 101 for other Practical Drill Problems.
58, 68, 69, 70, 90, 91, and 95, for Rapid Methods of Handling Numbers.
Also, see pages 53, 54, 57,
II 8 sou LE's PHILOSOPHIC PRACTICAL MATHEMATICS. º:
MULTIPLICATION BY SQUARING NUMBERS.
199. TABLE OF SQUARES AND CUBES.
| -
Numb. Square. Cube. Numb. Square. Cube. Numb. Square. Cube. Numb. Square. Cube.
1 l 1 26 676 17576 T51 2601 |132651. 76 5776 438.976
2 4 8 27 729 || 19683 52 2704 ||140608 77 5929 || 456533
3 9 27 28 784 || 21952 53 2809 ||148877| 78 || 6084 || 474552
4 16 64 29 841 || 24389 54 2916 |157464 79 6241 || 493039
5 25 125 30 900 27000 55 3025 166375|| 80 || 6400 || 512000
6 36 216 || 31 961 29791 56 3136 |175616 81 6561 || 531441
7 49 343 || 32 || 1024 || 32768 57 || 3249 |185193, 82 6724 || 55.1368
8 64 512 33 1089 || 35937 58 || 3364 |195112| 83 | 6889 || 571787
9 81 729 || 34 || 1156 || 39304 59 || 3481 (205379| 84 || 7056 || 592.704


10 || 100 || 1000 || 35 | 1225 || 42875 60 3600 |216000 85 | 7225 | 614125
11 | 121 | 1331 36 | 1296 || 46656 61 || 3721 |226981| 86 || 7396 || 636056
- 12 || 144 1728 || 37 || 1369 || 50653 62 3844 |238328 87 7569 || 658503
13 | 169 || 2197 || 38 || 1444 || 54872 63 || 3969 |250047| 88 || 7744 | 681472
14 || 196 || 2744 || 39 1521 593.19 64 | 4096 |262144|| 89 || 7921 || 704969
15 225 | 3375 | 40 | 1600 | 64000 65 4225 274625 90 8100 | 729000
16 || 256 | 4096 || 41 || 1681 | 68921 66 4356 |287496. 91 8281 || 753571
17 | 289 || 4913 || 42 || 1764 || 74088 67 || 4489 |300763| 92 | 8464 || 778688
18 324 5832 || 43 | 1849 || 79507 68 4624, 314432| 93 || 8649 || 804357
19 || 361 | 6859 || 44 1936 || 85184 69 || 4761 |328509| 94 || 8836 || 830584
20 | 400 8000 || 45 | 2025 | 91125 70 4900 |343000 95 || 9025 || 857.375
21 || 441 9261 || 46 || 2116 || 97336 71 || 5041 |357911| 96 || 92.16 || 88.4736
22 || 484 |10648 || 47 2209 103823 72 || 5184 |373248 97 || 9409 || 91.2673
23 || 529 |12167 || 48 || 2304 |110592 73 || 5329 |389017| 98 || 9604 || 94.1.192
24 || 576 |13824 || 49 || 2401 117649 74 5476 |405224| 99 || 9801 || 97,0299
25 | 625 15625 | 50 || 2500 125000 75 5625 42.1875|| 100 10000 |1000000
SQUARING NUMBERS.
200. The usual methods of Squaring numbers on the basis of complements and
supplements, are attended with so many variations that they are of very little
practical value, especially to those who understand the system of simultaneous
multiplication, and hence, we shall present but little work of this kind.
TO SQUARE NUMBERS BETWEEN TWENTY-FIVE AND FIFTY.
201. 1. What is the square of 26 %
OPERATION. Explanation.-In all problems of this kind, mentally subtract
26–25 = 1 25 from the number to be squared, and the remainder from 25; then
25 — 1 = 24 square this second remainder and add to its hundreds' figure the first
242 = 576 remainder.
p- * To be proficient in this work, the squares of numbers as high as
576–H199–676 Ans. 25, must be thoroughly iearned from the table. g
X- CONTRACTIONS IN MULTIPLICATION. I IQ
TO SQUARE NUMBERS BETWEEN FIFTY AND SEVENTY-FIVE.
202. 2. What is the square of 63?
OPERATION.
63 — 25 = 38 Explanation.--In all problems of this kind, mentally subtract,
63 – 50 = 13 first 25, and then 50 from the given number; then square the second
132 = 169 remainder and add to its hundreds' figure the first remainder.
169-H3899 = 3969 Ans.
TO SQUARE NUMBERS BETWEEN SEVENTY-FIVE AND ONE HUN-
DIRED.
203. What is the square of 89?
OPERATION. Explanation.—In all problems of this kind, subtract
89 – 75 = 14 75 from the number, and the remainder from 25; then
25 — 14 = 11 square the second remainder and add to its hundreds' fig-
112 121 ure the difference between the number and the second
121+(8999–1199)= 7921 Ans. remainder.
PROBLEMIS.
What is the square of 28, 42, 51, 67, 78, and 93%
To MULTIPLY Two NUMBERS ON THE PRINCIPLE THAT THE SQUARE
OF THE MEAN OF TWO NUMBERS MINUS THE SQUARE
OF ONE–HALF OF THE DIFFERENCE, IS EQUAL
TO THE PRODUCT OF THE TWO NUMBERS.
204. Multiply 15 by 19.
OPERATION.
15 + 19 = 34 -- 2 = 17* = 289 – 2* = (4) 285 Ans.
Ea:planation.—From the square of the mean, subtract the square of half their difference.
1. Multiply 68 by 72. -
OPERATION.
(70 — 2) × (70 + 2) = 70° — 2* = 4900 – 4 = 4896 Ans.
I 20 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs.
2. Multiply 81 by 79.
OPERATION.
80°–1 = 6400–1 = 6399 Ans.
3. Multiply 92 by 108.
OPERATION.
100° — 8° = 10000 — 64 = 9936 Ans.
Multiply the following numbers as above:
4, 28 by 34 6. 67 by 69 8. 57 by 63
5. 34 ($ 32 7. 73 44 85 9. 74 44 86
TO MULTIPLY TWO NUMBERS, THE DIFFERENCE OF WHICH IS ONE,
THREE, FIVE, OR ANY SMALL NUMBER.
205. 1. Multiply 21 by 22.
OPERATION.
21* = 441 + 21 = 462 Ans.
Ea:planation.—In all problems of this kind, square the lesser number and add to the product as
many times the lesser number as it is less than the greater. The reason is that the lesser number
should be repeated as many times as there are units in the greater number; but in squaring the lesser
Inumber, you fail to repeat it as many times as there are units difference between the numbers.
2. Multiply 87 by 89. º
OPERATION.
87° = 7569 + (2 × 87) = 174 = 7743 Ans.
3. Multiply 236 by 241.
OPERATION.
236* = 55696 -- (5 x 236) = 1180 = 56876 Ans.
4. Multiply 96 by 92. Ans. 8832.
Multiply 354 by 361. Ans. 127794.
6. Multiply 32 by 43.
SYNOPSIS FOR REVIEW OF MULTIPLICATION.
A Lº - a
v- w wr
Note:-The numbers before the words and phrases refer to the articles treating the subjects
Ilºſſle (1.
206. Define the following words and phrases:
cand or Multiplier contains
Dollars and Cents.
142. Multiplication. 162. General Directions for Multiplica-
143. Multiplicand. tion.
144. Multiplier. 163. Miscellaneous Problems in Multi-
145. Product. plication.
146. Factors. 164. Simultaneous or Cross Multiplica-
147. The Sign of Multiplication. tion. - r - |
148. Proof of Multiplication. 165. Drill Problems in Simultaneous
149. Principles of Multiplication. Multiplication.
150. Multiplication Table. 166. A Different Method of Simulta-
151. The Importance of Multiplication. neous Multiplication.
152. The Philosophic or Logical System 167. To Multiply when the Multiplier
of Multiplication. consists of Two Figures.
153. Examples with Philosophic Solu- 168. Table of Aliquots.
tions. 169. Problems.
*154. Drill Problems. 170. To Multiply by 74, 174, 224, 274,
155. To Multiply Abstract Numbers 324, 15, 35, 45 and 55.
and give reasons therefor. 171. Practical Operations with Ali-
156. To Multiply, when the Multiplier Quots and other numbers.
consists of only one figure. 172. To Multiply, when the Multiplier
157. To Multiply, when the Multiplier * is a mixed number whose frac-
consists of more than one fig- tional part is one fractional
ll.T0. unit, less than a whole number.
158. To Multiply, when either factor is 173-93. Miscellaneous Contractions in
1 with naughts annexed, as 10, Multiplication.
100, 1000, etc. 193. Algebraic Formulae, for Comple-
159. To Multiply, when either the Mul- ments and Supplements.
tiplicand or Multiplier, or both, 194-98. Drill Problems in Contracted
have naughts on the right. Methods in Addition and Multi-
160. To Multiply by the Factors of a plication.
Number. 199. Table of Squares and Cubes.
161. To Multiply, when the Multipli- 200-5. Squaring Numbers.
(121)
IVISION.
a-a-a-a-a-a-a-a-a-a-/----------El
=N
**** ( DECREASI NG).
*—iº—dº-
-sº-sºr—wº-
207. Division is the process of finding how many times one mumber is equal
to another, which is used as a unit of measure. Or, differently defined, it is the
process of finding one of the factors of a given product, when the other factor
is known.
208. Erom the first and the better definition, we see that division is a process
of measuring some numbers by other numbers. And that it is not the process of
finding “how many times one number is contained in another,” as is taught by
nearly all the authors of Arithmetics. One number cannot go into another, how-
ever Small the one, or large the other; and the question of division of numbers
does not Warrant a definition so unmathematical, indefinite and illogical.
THE LOGIC OF DIVISION.
209. The following questions and operations will elucidate the true meaning
of division and its philosophy:
1. You have $6, and I have $2. How many times is your money equal to
mine *
Is it not clear, by the terms of the question, that your sum of money is to be
compared with, and measured by, my $2? And the contracted thought to do this
is, $6 is equal to $2, 3 times.
The full logical thought, recognizing 1, or unity, as the basis of all numerical
computations, would be as follows: $6 is equal to $1, 6 times. Since $6 is equal to
$1, 6 times, it is equal to $2, one-half the number of times, which is 3.
2. Again—you have 8 yards of cloth, and I have 4 yards. How many times
is your quantity of cloth equal to mine?
Here it is plain that my 4 yards is made, by the terms of the question, the
unit of measure; and your 8 yards is the thing to be measured. And the contracted
thought to do the work would be, 8 yards is equal to 4 yards, 2 times.
The full logical reason is as follows: 8 yards are equal to 1 yard 8 times.
Since 8 yards are equal to 1 yard, 8 times, they are equal to 4 yards instead of 1,
the fourth part of the number of times, which is 2.
3. Divide 10 by 5.
In this problem, the real question is, how many times is 10 equal to 5, not how
many times 5 can go into 10, or how many times is 5 contained in 10.
5 is the unit of measure, and 10 is the number to be measured or divided.
jº
The contracted reasoning is, 10 is equal to 5, 2 times; not 5 is contained or
º (122)
* * DIVISION. I 23
goes into 10, 2 times. The full logical reason is as follows: 10 is equal to 1, 10 times.
Since 10 is equal to 1, 10 times, it is equal to 5, instead of 1, the fifth part of the
number of times, which is 2.
This process of reasoning is applicable to every possible question of division,
and it is the only logical reasoning that can be given for abstract numbers. Yet
strange to say, it has escaped the attention or the approbation of all other authors
of Arithmetics. With due modesty, we claim some merit for its first introduction,
and earnestly commend it to the thoughtful consideration of authors, of students,
and of the public.
210. The Dividend is the number to be measured, or the number to be divided.
211. The Divisor is the number used as the unit of measure or the number by
which to divide.
212. The Quotient is the result of the division, and shows how many times the
dividend is equal to the divisor.
213. The Remainder is the number left after dividing dividends which are
not multiples of the divisor, or which are not equal to the divisor an exact number
of times. It must always be less than the divisor.
214. The Sign of Division is a horizontal line with a point above and below,
thus -i-. It is read divided by, or is equal to ; and it indicates that the number
before it, is to be divided by the number after it: thus 25 -- 5, is read 25 divided by
5, or 25 is equal to 5.
215. Division is also indicated by a horizontal line, a vertical line, or a curved
line, when placed between the dividend and the divisor. Thus, the following state-
Inents :
36 9 || 36 9) 36 36(9–
9
are all read 36 divided by 9, or how many times is 36 equal to 9, or 36 is equal to 9
how many times 3
PRINCIPLES OF DIVISION.
216. 1. When the divisor and dividend are both denominate or both abstract
numbers, the quotient will be an abstract number.
2. When the divisor is an abstract number and the dividend a denominate
Inumber, the quotient will be a denominate number.
3. When there is a remainder, it is a part of the dividend, and is therefore
the same in name or kind.
4. Multiplying the dividend or dividing the divisor, multiplies the quotient.
5. Dividing the dividend or multiplying the divisor, divides the quotient.
6. Multiplying or dividing both the divisor and dividend by the same num-
ber, does not change the quotient.
217. Proof.-Multiply the quotient by the divisor, and to the product add the
remainder, if any. If the result is equal to the dividend, the work is correct. See
Proof of Division by the Properties of 9's and 11's, at the close of this subject.
I 24 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. • {
218. The operation of division may be performed, either by addition or sub-
traction. Thus, in the following problem:
How many times is 43 equal to 12% Ans. 3 times and 7 remainder.
OPERATION BY ADDITION. OPERATION BY SUBTRACTION.
43 is equal to 12, 1 time. 43
43 “ “ “ 12, 2 times. 43 is equal to 12, 1 time.
24 31
43 “ “ “12, 3 times. 43 “ “ “ 12, 2 times.
36 19
and 7 remainder. 43 “ “ “ 12, 3 times.
43 and 7 remainder.
ORAL EXERCISES.
219. 1. How many times is 0 equal to 1 ? or 0 + 1 = ? no times.
2. 4% {{ 1. “ 0% Or 1 - 0 = }
Ans. An infinite number of times.
3. How many times is 1 equal to 13 or 1+1= ? | 12. How many times is 35 equal to 7? or 35— 7= ?
4. & 4 é & 2 & 4 1 ? or 2–1= ? | 13. 4 & & 4 56 ** 83 or 56— 8– ?
5 ( & & 4 3 ( & 1? Or 3–1=3 || 14. & 4 & 4 63 ‘‘ 93 or 63-- 9= ?
6. & 4 & & 4 “ 2% or 4–2–3 || 15. 4 & 4 & 72 ** 93 or 72– 9–?
7. & 4 & 4 8 ‘‘ 23 or 8–2 – ? | 16. & & 4 & 80 ‘‘ 10? or 80–10= ?
8. { { & 4 9 ** 33 or 9–3–? | 17. & 4 & 4 88 ‘‘ 11? or 88–11= ?
9. { { & 4 12 “ 4 3 or 12–4–? | 18. & 4 & 4 96 “ 12% or 96–12– ?
( & & & 2 & & º —5– *
#. ( & & 4 ; & & # : *::=; 19. §–? 4)42=? 9)45=? 77-7=? 12 | 84= ?
20. How many times is 36 equal to 3, to 4, to 6, to 9, to 12, to 36%
21. How many times is 42 equal to 2, to 6, to 7, to 42?
14 21 28 64. 8
22. – =? – =1 − =1 − =? 24(?-, 32(ºr 7) 56 =? 9 || 72=?
2 7 4. 8 — ? ~ :
FRACTIONAL NUMBERS.
220. When we divide a unit or a number of units of any kind into equal parts,
these parts are sometimes called fractions. The name of the equal parts varies
according to the number of parts into which the thing or number is divided.
When the unit or number is divided into two equal parts, 1 of the parts is
called one-half, and is written thus, #. If divided into 4 equal parts, 1 of the parts
is called one-fourth, and is written thus, 4; 3 of the parts are called three-fourths,
and are written thus, #.
In like manner, we obtain fifths, sixths, sevenths, eighths, twelfths, sia teenths,
twenty-firsts, etc.
* DIVISION. I 25
In Writing fractional numbers in figures, we place the number which shows the
name of the parts below a horizontal line as a divisor, and the number which shows
how many parts are taken, or used, above the line as a dividend.
The following examples will fully elucidate this work:
Two-thirds, 3. Five-eighths, #. Seven-twelfths, ".
Three-fourths, #. Seven-ninths, #. Nine-tenths, #.
Fifteen-sixteenths, #. Eleven-eightieths, #.
How do you find #, #, #, #, etc., of any number 2
How do you find 3, #, #, g, etc., of any number 2
What is 4 of 4% What is # of 15% What is 3 of 9% What is # of 15 +
“ (: ; “ 6 “ (: ), “ 18% “ (: ; “ 12? “ (: ; “ 633
“ “ # 4 83 “ “ , « 28 “ “g “ 40% “ “ H. “ 80?
221. ORAL AND WRITTEN PROBLEMs.
SOLUTION STATEMENT. *
1. Five pounds cost 40%. g Reason—Five pounds cost
40 40c. Since 5 pounds cost 40
5
What was the cost of 1 Cent8, 1 pound will cost the 5th
part, which is 8 cents.
Q - -
pound 4 82 Ans.
REASONS, WHY, AND WHEREFORE, CONTINUED.
Question 1. How do you know that if 5 pounds cost 40 cents, 1 pound will
cost the 5th part 2
Answer. By the exercise of my judgment—by the use of the reasoning
faculties of my mind.
Question 2. What do you mean in this connection by judgment 3
Answer. The conclusion arrived at by the operations of the mind, after hav.
ing duly considered the premise, the facts, and the conditions of the problem.
Question 3. What do you mean by the premise or premises 2
Answer. The proposition, declaration, truth or fact which is stated, or
asserted, as a basis, or predicate, of a question. In this problem the premise is,
5 pounds cost 40 cents. .
Question 4. Why will 1 pound cost one-fifth part as much as 5 pounds 3
Answer. Because 1 is the fifth part of five.
Question 5. What kind of reasoning is the foregoing 3
Answer. (See answer to the same question on page 78).
Question 6. What do you mean by reason 3
Answer. (See answer to the same question on page 78).
I 26 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS.
*
SOLUTION STATEMENT.
Reason.—Twelve yards cost
g
2. If 12 yards cost 60%. 60 60c. Since 12 yards cost 60c.,
What will 1 yard cost 3 12 1 yard will cost the 12th part,
which is 5 cents.
| 52 Ans.
Questions. 1. How do you know this? 2. Why will it? 3. What do you
mean by judgment 3
3. Nine barrels of flour cost $72. What was the cost of 1 barrel ?
& Ans. $8.
NoTE.—Work and reason as in above problems.
SOLUTION STATEMENT. -
4. Paid $18 for 6 days' $ . Reason.—Six days’ labor cost
- 7
labor. What was the rate | 18 $18. Since 6 days’ labor cost
6 $18, 1 day’s labor will cost the
Q
per day ? 6th part, which is $3.
$3 Ans.
Questions. 1. How do you know this? 2. Why will it?
SOLUTION STATEMENT. Reason.—The freight on 17
5. The freight on 17 bales $ bales was $51. Since the freight
of cotton was $51. What 17 51 on 17 bales, was $51, on 1 bale
was the freight per bale 3 - ºm- it would be the 17th part,
$3 Ans. which is $3.
Questions. 1. How do you know this? 2. Why will it? 3. What do you
mean by judgment 3
SOLUTION STATEMENT. Reason.—Eight persons are to
6. If you divide 2 dozen g e
Oranges equally among 8 *2, receive 24Oranges. Since 8 per-
| sons are to receive 24 oranges,
persons, how many oranges s e- 1 Jerson will receive the 8th
will you give each 3 | 3 oran ges Ans. part, which is 3 oranges.
Questions. 1. How do you know this? 2. Why?
SOLUTION STATEMENT. Reason.—In 12 hours 360 miles
7. A rail © IMI. are run. Since 360 miles are run
e road train runs 360 in 12 hours, in one hour the 12th
360 miles in 12 hours. * 12 part of the distance would be
* Tun, which is 30 miles. Or, since
What is the speed per hour?
12. hours' running gives 360
| 3 * º miles, 1 hour's running will give
0 miles Ans the 12th part. g Willg
Questions. 1. How do you know this? 2. Why?
DIVISION. 127
1st SOLUTION STATEMENT.
ſh. Reason.—Seven cents buy 1
p • 1. pound of rice. Since 7 cents
8. Rice is 7 cents per 7 || 35 will buy 1 pound, 1 cent will
* buy the 7th part, and 35 cents
5 lbs. Ans. will buy 35 times as much. Or,
can you buy for 35 cents? 2d solution statDMENT. since 7 cents buy 1 pound, for 35
cents you can buy as many
7 || 35 pounds as 35 cents are equal to
7 cents.
pound. How many pounds
*º-
5 lbs. Ans.
Questions. 1. How do you know this? 2. Why? 3. What do you mean
by judgment? º
NOTE--Work and reason the following problems in the same manner.
9. At 9 cents per pound, how many pounds can be bought for 45 cents?
Ans. 5 pounds.
10. Flour is worth $8 per barrel. How many barrels can be purchased for $56%
Ans. 7 barrels.
11. For 8.95, how many papers can you buy at 5 cents a paper? Ans. 19.
12. At $3 a piece, how many chairs can be bought for $36? Ans. 12 chairs.
13. If the printer charges $2.50 to set one page of this book, how many pages
can be set for $75 % Ans. 30 pages.
14. The dividend is 42 and the divisor, 7. What is the quotient? Ans. 6.
15. The divisor is 8, the quotient 3, and the remainder 2. What was the
dividend ? Ans. 26.
->
TO DIVIDE WEHEN THE DIVISOR DOES NOT EXCEED TWELVE.
222. 1. Divide 3648 by 5.
OPERATION. h Ea:planation.—In all problems of this kind, we write
tº º t - º the numbers as shown in the operation, and then begin on
Divisor 5) 3648 dividend. the left of the dividend to divide, we begin on the left
* > in order to carry the remainder, if any, of the higher order
Quotient 729, and 3 rem. of units to the next lower order. In this problem, we first
take the 3 thousands, and by inspection, we see it is not
equal to 5; we therefore unite it with the 6 hundreds, making 36 hundreds, which, by trial multipli-
cation and subtraction, mentally performed, we find is equal to 5, 7 hundreds times and 1 remainder;
the 7 we write in the hundreds’ column of the quotient line, directly under the 6, the last figure of
the dividend used. Then, to the 1 remainder, we mentally annex the 4 tens, making 14 tens, as the
second partial dividend, and which, by mental multiplication and subtraction, we find it equal to 5,
2 tens times and 4 remainder; the 2 we write in the tens’ column of the quotient line, and to the 4 we
mentally annex the units' figure of the dividend, making 48 units as the third and last partial diyi-
dend; this we find, by mental multiplication and subtraction, to be equal to 5, 9 times and 3 remain-
der.
This remainder is usually expressed fractionally by writing it over the divisor; thus # which
expresses the part of a unit of times that the remainder is equal to the divisor.
I 28 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. X-
SHORT DIVISION.
223. Operations in division according to the foregoing method, are called short
division, because the multiplication and subtraction work, in finding the remainder
of the partial dividends, were mentally performed.
2. How many times is 846 equal to 6%
OPERATION. º'. * preceding problem, we gave a full and
; rr; c. awº ivi o explicit, explanation for each step of the operation. In practice
Divisor 6) 846 dividend much of the explanation therein given is omitted, and the work
e -*- performed thus: Commencing with the left hand figure we say 8
Quotient 141 is equal to 6, 1 time and 2 remainder; 24 is equal to 6, 4 times;
tº 6 is equal to 6, 1 time.
PROBLEMS.
Work the following indicated divisions:
(3) (4) (5) (6) (7) (8)
7) 847 8) 12327 9) 1085 11) 2386 2344 7 | 93020
121 1540% 120; 2164? 3
(9) (10) (11) (12) (13) (14)
8) 1471 4) 11899 9) 81018 12) 10824 21345-1-8 9) 76451
Divide the following:
15. 9872 by 4 17. 10286 by 6 19. 1691 by 9 21. 10008 by 9
16, 1483 “ 7 18, 4.8710 ($ 7 20. 4.1070 € $ 8 22. 199999 44 8
23. What is + of $ 528? 25. What are ; of $ 448%
24. “ are # of $1005? 26. “ “ g of $6448?
27. How many apples can be bought for $2.25, at 5 cents a piece 2
Ans. 45 apples.
28. At 15 cents per pound, how many pounds can you buy for $3.15%
Ans. 21 pounds.
29. Paid $90 for 10 volumes of Chambers' Cyclopedia. What was the price
of one volume 3 Ans. $9.
30. If 8 men are to receive $5791 in equal parts, what will be each man's
share 2 Ans. $7.23%
31. The dividend is 63, and the quotient is 9. What is the divisor 3
Ans. 7.
32. The quotient is 36, and the divisor is 6. What is the dividend ?
Ans. 216.
33. The dividend is 72, and the divisor is 4. What is the quotient 2
- Ans. 18.
34. The quotient is 15, the divisor is 3, and the remainder is 2. What is the
dividend ? Ans. 47.
35. The divisor is 24, the quotient is 6. What is the dividend? Ans. 144.
36. How many pounds of cotton, at 11 cents a pound, will be required to pay
for 33 pounds of sugar (a) 8 cents per pound 3 Ans. 24.
+): DIVISION. I 29
TO DIVIDE WHEN THE DIVISOR EXCEEDS TWELVE.
224. 1. Divide 7387 by 36.
OPERATION. th †W. first write º ii. shown i.
tº ge ivri e e Operation, and commence to divide as explained in the
Divisor, Dividend, Quotient. first written example. But as the divisor is ; large to be
36) 7387 ( 205; conveniently used mentally, we therefore write the operation
72 of multiplying the divisor by the quotient figures, and sub-
* tracting the successive products from the several partial
dividends. In performing the division, we first see by
187 comparison, that 7 thousands are not equal to 36, and hence
180 there will be no thousands in the quotient, We then annex
to the 7 thousands the 3 hundreds, making 73 hundreds as the
7 ind first partial dividend; this is equal to 36, 2 times, and a re-
I’6100.3.IIlCl62I’. mainder; we write the 2 in the hundreds' column of the
-- uotient, multiply the divisor by it, write the product under,
and subtract the same from the 73 hundreds of the dividend. This work gives us 1 hundred remain-
der, to which we annex the 8tems, making 18 tens as the second partial dividend; this partial dividend
not being equal to 36, we write 0 (no tens) in the tens’ column of the quotient, and annex to the 18
tems the 7 whits, making 187 units as the third and last partial dividend. This is equal to 36, 5 times,
and a remainder; we write the 5 in the quotient, multiply and subtract as we did with the first
obtained figure of the quotient, and thus produce 7 remainder, which we write over the divisor as
explained in short division.
LONG DIVISION.
Operations in division, according to the above method, are called long division,
for the reason that the multiplication and subtraction work in finding the remain-
ders of the partial dividends is written. See Contractions in Division.
.GENERAL DIRECTIONS FOR DIVISION.
225. From the foregoing elucidations, we derive the following general directions
for the operations of division. For the process of reasoning given in connection With
the operations or solution statements, we refer to pages 122 to 129.
1. Draw a vertical or curved line, and write the dividend on the right, and the
divisor on the left. *
2. Take the least number of the left hand figures of the dividend that are equal
to or greater than the divisor, find how many times the same is equal to the divisor, and
write the result in the quotient line.
3. Multiply the divisor by this quotient figure, subtract the product from the
partial dividend used, and to the remainder annew the succeeding figure of the dividend
and divide as before.
4. Proceed in like manner until all the figures of the dividend have been used.
5. When the partial dividend is not equal to the divisor, write a maught in the
quotient, annea, the succeeding figure of the dividend to the partial dividend, and pro-
ceed as before.
6. If there is a remainder after the last division, write it in the quotient, draw
a line below it, and write the divisor underneath, as a part of the quotient.
I 30 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. 4×
PROOF. Multiply the quotient by the divisor and to the product add the
remainder, if any. If the result is equal to the dividend, the work is correct.
NOTE: 1. -The product referred to in No. 3, must never be greater than the partial dividend
from which it is to be subtracted; if it is larger, the quotient figure is too large, and must be
diminished.
NOTE: 2-The remainder, after each subtraction referred to in No. 3, must always be less than
the divisor; if it is not, the last quotient figure is too small and must be increased.
NOTE: 3.-The order of each quotient figure is the same as the lowest order in the partial
dividends from which it was obtained.
PROBLEMI.
How many times is 66804 equal to 53° Ans. 1260}#.
OPERATION.
Divisor, Dividend, Quotient, Proof.
53) 66804 (1260% 1260 quotient.
53 53 divisor.
138 3780
106 6300
º-s, 24 remainder.
320 -emmºmºm.
3.18 66804 dividend.
24 remainder.
PROBLEMIS.
Divide the following numbers:
1. 784 by 82. Ans. 9;;. 2. 91070 by 8761. Ans. 10####.
3. 107941 by 396. Ans. 4. 7167901 by 11267. Ans.
5. 37021 by 2002. Ans. 6. 8888888 by 332311. Ans
7. Divide $6805 equally among 5 men. What will be the share of each *
Ans. $1361.
8. What is the sixty-fourth part of $44800 % Q Ans. $700.
9. 145 men picked 1305000 pounds of cotton. Suppose each picked an equal
quantity, how much did each man pick 3 Ans. 9000 pounds.
10. A father gave his 7 Sons a Christmas present of $353.50 to be shared
equally among them. What was each one's share ? Ans. $50.50.
TO DIVIDE WHEN TEIERE ARE NAUGHTS ON THE RIGHT OF THE
DIVISOR.
226. 1. Divide 2843 by 200. Ans. 14.5%;.
OPERATION. Explanation.—Since by our scale of numbers they increase from
2 00)28 43 right to left in a ten-fold ratio, and decrease from left to right in a cor-
responding manner, it is clear that the removal of any order of figures
from left to right diminishes its value ten times for each place of
14 and 43 rem. removal. And as previously shown on page 83, that the annexing of
naughts multiplies numbers, by removing them to places of higher
value, so in like manner, cutting figures off from the right of a number removes the remaining orders
to the right, and hence decreases them ten-fold for every figure cut off. Hence to cut off one figure is
dividing by 10; to cut off two figures divides by 100; to cut off three figures divides by 1000; and
SO OI).
Considering these principles, in all cases of this kind we cut off the naughts from the right
of the divisor and the same number of figures from the right of the dividend; and then divide
the remaining figures of the dividend by the remaining figures of the divisor. When there is a
remainder, annex the figures cut off, and we obtain the true remainder.
DIVISION. I 3 I
3.
l
2. Divide 87931 by 1000. Ans. 87%
OPERATION.
1|000)87931
Quotient 87 and 931 remainder.
0
O
3. Divide 178 by 10. 4. Divide 6581 by 300.
OPERATION. OPERATION.
1|0,178 3|006581
Quotient 17, and 8 remainder. Quotient 21, and 281 remainder.
Ans. 17%. Ans. 21}}}.
5. Divide 71468071 by 341000.
OPERATION.
341 | 000)71468||071(209344%; Ans.
682
323s
3069
199
6. Divide 88.97600 by 8100. Ans. 1098;#.
7. Divide 1000000 by 10000. Ans. 100.
8. Divide 99999 by 9000. ” Ans. 11%.
9. Divide 33440 by 270. Ans. 123}#}.
10. Divide 140817 by 6800. Ans. 20áš.
TO DIVIDE BY THE FACTORS OF A NUMBER.
227. 1. Divide 936 by 24. Ea:planation.—In all problems where the divisor is a com-
OPERATION. posite number, we may divide by the factors and thus
4)936 shorten the operation, In this example the factors are 4 and
-º-º: 6, and we first divide by 4 which gives a quotient 6 times
6)234 too large, for the reason that 4 is but of 24, the true divisor.
--> We therefore divide this quotient by 6 and obtain the true
39 quotient.
2. Divide 588 by 28. The factors are 4 and 7. Ans. 21.
3. Divide 6976 by 32. The factors are 4 and 8. Ans. 218.
4. Divide 2583 by 63. The factors are 7 and 9. Ans. 41.
5. Divide 10206 by 81. The factors are 9 and 9. Ans. 126.
6. Divide 11984 by 56. The factors are 8 and 7. Ans. 214.
I 32 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS.
PROBLEMS INVOLVING THE ENGLISH MONEY OF ACCOUNT.
228. 1. What will 13840 pounds of cotton cost, at 8 pence a pound.
Ans. £461. 6s. 8d.
OPERATION. Explanation.—In this problem, the price is given in one
of the subdivisions of the English monetary unit, and hence
13840 we must know what that unit and its subdivisions are, before
8d. we can solve the problem. The English monetary unit is
gº-me the pound 8terling, yº,jº divided into 20 shillings; each
shilling is divided into 12 pennies, and each penny into 4
12) 110720d. farthings. With this knowledge of English money, we can
work all problems of the above character. In this example,
29) 922% 8d. * we first multiply the price of one pound by the number of
pounds, and thus produce the value of the whole in pence.
fº 461 6S. Then to reduce the pence to shillings, we divide them by 12,
and obtain 9226 shillings and a remainder of 8, which being
a part of the dividend, is therefore 8d. Then, to reduce the shillings to pounds, we divide them by
20, and obtain 461 pounds and a remainder of 6, which being a part of the second dividend, is there-
fore 6s. In the English monetary system the following abbreviations are used: £, represents
pounds; s. represents shillings; d. represents pence, and f. represents farthings.
2. What is the value of 483 yards 3. What will 241 boxes cheese cost,
of cloth, at 16 shillings per yard 3 at £3 per box”
Ans. £386. 8s. Ans. £723.
OPERATION. OPERATION.
483 - 241
16 3
20) 7723 shillings. 36 723
# 386 8S.
4. Sold 486 yards of calico at 5 pence a yard. What did it amount to ?
* Ans. £10 2s. 6d.
5. Bought 38495 pounds of good middling cotton at 7 pence per pound.
E[ow much did it cost 3 Ans. £1122 15s. 5d.
6. What is the value of 850 barrels of flour at 34 shillings a barrel ?
Ans. £1445.
7. How much will 1812 tons of iron cost, at £52 4s. per ton ?
Ans. £94586 8S.
8. Bought 38421 pounds of cotton at 9 pence per pound. What did it cost 3
Ans. £1440 15S. 9d.
Yºr DIVISION. I33
TO FIND THE TRUE REMAINDERS WHEN DIVIDING BY THE FAC-
TOR'S OF A NUMBER.
229. Divide 1607 by 72, using the factors 3, 4, and 6, and find the true remainder.
Ans. 22 quotient, and 23 remainder.
FIRST OPERATION. Explanation.—In this problem, using the factors 3, 4, and 6, we
3) 1607 obtain for remainders 2, 3, and 1.
* The first remainder 2, is clearly units of the given dividend,
ſº o and hence a part of the true remainder.
4) 535 2, 1st remainder. The second remainder 3, being fourths of the second dividend
535, which are reciprocal thirds of the given dividend, it is hence
6) 133 3, 2d remainder. # of the reciprocal of + of # = 9, of the given, dividend and true
7 remainder.
ºp The third remainder 1, being sixths of the third dividend, 133,
22 1, 3d remainder, which are reciprocal twelfths of the given dividend, it is
hence # of the reciprocal of # of + of # = 12, of the given divi-
dend and true remainder. Therefore 2, the first remainder, plus 9, the unit value of the second
remainder plus 12, the unit value of the third remainder = 23, the true remainder.
SECOND OPERATION. TEIIRD OPERATION.
3 | 1607 3 | 1607
4 || 535 = 2, 1st remainder
4 | 535 = 2 = 3 of 1. 6 || 133 = 3, 2d remainder.
6 || 133 = 3 = } + 3 = + + 4 = +}. 22 = 1, 3d remainder.
tº-º-º-º: First remainder - - E. tº- di - 2
2 3 e Plus 2d remainder 3, Xthe preceding divisor 3=9
mºmes - 2 ll. * -ºº º - ------- e
22 = 1 = }} + +} = }} + 6 = ##, Plus 3d remainder i. all the preceding divisors
or 23 the true remainder. 4 and 3- - - - - - - 12
which added, gives the true remainder - 23
Eplanation.—The second operation involves fractions, and hence not regularly in order at this
F. In this operation each subsequent remainder is reduced to the fractional unit of the preced-
ing fraction.
g By the third operation, we see that the true remainder may be obtained without considering
the reciprocal relationship of the quotients and divisors.
The general method shown in the third operation is as follows: Add to the first remainder, the
product of the other remainders by all the divisors preceding the one which produced it.
2. Divide 7851 by 64, using the fac- 3. Divide 17803 by 96, using the fac-
tors 8 and 8. a tors 2, 3, 4, and 4. Ans. 185%.
Ans. 122 quotient, 43 remainder. OPERATION.
OPERATION. Ea:planation.— 2)17803 Elap lanation.
8)7851 Here the 1st re- *-
mainder is 3, to 3)8901–1 1st remainder - - º 1
º which we add
8)981—3, 1st remainder. the product of -
the 2d remain- 4)2967–0 2d remainder, 0x2 = - - 0
122–5, 2d remainder. Herămultipºl
- by the preced- 4)741–3 3d remainder, 3x3x2 = - 18
ing divisor 8, equals 40, making 43, the true re- --
mainder. 185–1 4th remainder 1X4×3×2 = 24
True remainder, - - 43
I 34. soul E's PHILOSOPHIC PRACTICAL MATHEMATICS.
Divide 27865 by the factors of 81. Ans. 344sºr or 344 and 1 remainder,
Divide 101041 by the factors of 84. Ans. 1202%
Divide 899 by the factors of 108. Ans. 81% s.
If $4691 are divided equally among 35 men, what will each one receive?
Ans. $134;.
:
TEIE PARENTEHESIS AND WIN CULUM.
230. The parenthesis, (), or vinculum, -, denotes that the numbers within the
parenthesis or below the vinculum are to be considered as one quantity. Thus, as
shown on page 36, 12 — (4-H3) = 5, or 12 – 4 + 3 = 5.
In all operations involving the signs of +, −, x, and -- the following prin-
ciples govern :
1. The signs +, and — affect only the number or expression which immedi-
ately follows either of them.
2. The signs x and -- indicate an operation to be performed with the
numbers or expressions between which either may be placed, and such operation
must be performed before that of any + or — which may immediately precede
or follow. r"
3. When the parenthesis or vinculum is used, the operations indicated thereby
must be performed or simplified into one expression before connecting them with
any other numbers in the problem. This is done for the reason that all numbers
included within the parenthesis or the vinculum are considered as one number.
4. In the operations of all problems involving these principles, commence
with the (), or , if any, then with the expressions indicated by the x and --
signs, and lastly with the numbers connected with the signs of + and —, which
should be considered from left to right.
5. In cases where there are double parentheses, the operation indicated by
the interior parenthesis must be performed first.
PROBLEMIS.
1. Simplify the following expression:
16 + 24 + 8 + 4 — 3 × 5. Ans. 27.
2. What is the value of the following numerical expression ?
(740 + 63 — 200) -- 450 — 325 + 76 x (65 – 35 + 10) *
OPERATION.
740 + 63 = 803 – 200 = 603; then 450 — 325 = 125 + 76 = 201; then 603
-- 201 = 3; then 65 — 35 = 30 + 10 = 40; then 3 × 40 = 120 Answer.
3. Simplify the following expression:
(740 + 63 — 200) + (450 — 325 + 76 ) × (65 – 35 + 10)?
† DIVISION. 135
OPERATION.
740 + 63 = 803 – 200 = 603; then 450 — 325 = 125 + 76 = 201; then 603
+ 201 = 804; then 65 – 35 = 30 + 10 = 40; then 804 x 40 = 32160 Ans.
4. Simplify the following expression:
(740 + 63 – 200) + 450 — 325 + 76 x (65 – 35 + 10).
OPERATION.
450 — 325 = 125 + 76 = 201; then 65 — 35 = 30 + 10 = 40; then 201 x
40 = 8040; then 740 + 63 = 803 – 200 = 603; then 603 + 8040 = 8643 Ans.
5. What is the value of the following expression ?
(420 – 60) — 576 — 490 + 312 – 10 + 16 × 5 + 8 + 6 – 4. Ans. 584.
6. What is the value of 9500 + 6200 – 9000 -- 3400 + 2100 — 5490 × (6500
– 6400 + 200) * Ans. 15697.
7. What is the value of 2765–1624 × (320 + 96) -- (3645 – 3376).
Ans. 1764}{}}.
8. What is the value of 47 –H 1561 -- 7 -- 142 × 20 + 522 -i- 87 –H 207 --
23 + (342 — 150) × 62. Ans. 15029.
231. MISOELLANEOUS PROBLEMS IN DIVISION.
1. One of the factors of 10800 is 225. What is the other ? Ans. 48.
What number multiplied by 137, will give 959137 for the product?
Ans. 7001.
3. Multiplying 372 by an unknown number gives 44640. What is the num-
ber 3 Ans. 120.
4. What is the quotient of 9126 divided by 9% AnS. 1014.
5. Divide four million eight thousand sixteen by MMDCXLIV.
Ans. 1515}}#}.
6. The great church of St. Peter, in Rome, will accommodate 57000 people.
The largest church in New Orleans will accommodate 2500. How many times is the
capacity of the church of St. Peter equal to that of the largest church in New
Orleans ? Ans. 22#.
7. What number is that to which, if sixteen be added, the sum multiplied
by 8, and 13 subtracted from the product, the remainder will be 3393 Ans. 28.
OPERATION.
339 + 13 = 352 -- 8 = 44 — 16 = 28 Ans.
The student will write the full reasoning.
8. There is a number from which, if you subtract 55, and divide the remain-
der by 12, your quotient will be 36. What is that number 3 Ans. 487.
I 36 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. X.
OPERATION.
The student will write the full reasoning.
9. A merchant owes a debt of $1875, which he agreed to pay by weekly
installments of $25. He has made 55 payments. How many more payments has he
to make 3 Ans. 20.
10. The velocity of the earth on its yearly voyage around the sun is 99733
feet per second. The velocity of a cannon ball fired from a gun with an average
charge of powder is about 1750 feet a second. How many times as fast as the veloc-
ity of a cannon ball, is the velocity of our earth ? Ans. 56}#}.
11. About one-siath of a man's weight is blood. How many pounds of blood
in a man whose weight is 168 pounds 3 Ans. 28 pounds.
12. A merchant bought 350 barrels of flour at $6 per barrel, and sold it at
$7.50 per barrel. The gain he gave, in equal parts, to four worthy boys, to aid
them in obtaining an education. What was the cost and the selling price of the
flour, and how much money did each boy receive 3 Ans. $2100 cost,
$2625 selling price, $131.25 each boy received.
13. The air which surrounds our earth, and of which we each inhale about
600 gallons every hour, is composed of four parts of nitrogen, and one part of oxy-
gen. How many gallons of each are there in a room 22 feet long, 21 feet wide, and
10 feet high, and which contains 34560 gallons of air 3
Ans. 6912 oxygen, 27648 nitrogen.
14. An acre contains 160 Square rods. How many acres in a plantation con-
taining 123200 square rods 3 Ans. 770 acres.
15. A boy sold 50 oranges at 52 each, and thereby gained $1.50. At what
rate did he buy the Oranges } Ans. 2% a piece.
16. How many times 136 will produce 1768 % Ans. 13.
17. Divide the product of 750 and 875, by their difference. Ans. 5250.
18. The diameter of the earth at the equator is 7925 miles. How long would
it take a locomotive to travel that distance, at the rate of 25 miles an hour *
Ans. 317 hours = 13 days, 5 hours.
19. It is estimated that, by reason of intemperance, the United States loses
annually $98400000. How many school houses costing $5000 each, and how many
libraries costing $3000 could be established with this amount of money 2
Ans. 12300 of each.
20. The first Atlantic telegraph cable, as originally made, cost $1258250.
Ten miles of deep sea cable were made at a cost of $1450 per mile, and 25 miles of
shore ends were made at a cost $1250 per mile. The remainder cost $485 per mile.
How many miles of cable were made 3 Ans. 2535 miles.
^. 21. The circumference of our earth at the equator is 24899 miles, and the
mean diameter of the earth is 79.12 miles. How many times is the circumference as
great as the mean diameter ? Ans. 3.}}#} times.
22. A grocer wishes to put 3335 pounds of sugar in three kinds of boxes,
containing respectively, 20, 50, and 75 pounds, and use the same number of boxes
of each kind or size. How many boxes will he require? Ans. 23 of each size.
4× DIVISION. I37
23. The capacity of steam engines is measured by horse power; and 1 horse
power is a force that will raise 33000 pounds, one foot in one minute. How much
horse power has a steam engine that possesses a capacity of 1188000 pounds 2
Ans. 36.
24. The average weight of a man is 150 pounds, and about # of this weight
is blood. Allowing that the heart throws out two ounces of blood at each pulsation,
that it beats 72 times a minute, and that 16 ounces make a pound, how long will it
take the heart to circulate all the blood in the body? Ans. 2+}# minutes.
25. Ten freedmen agreed to pick 20000 pounds of cotton and receive for
their labor # of the cotton picked. After they had picked 7000 pounds, four freed-
men quit, leaving the other six to finish the work. How much cotton is each entitled
to When the work is finished ? Ans. 140 pounds each for those who left, and
573% each for those who remained.
26. The Old and New Testaments contain 1189 chapters and 31164 verses.
If a person were to read 12 chapters every Sunday, how many Sundays would it
require to read the contents of the whole Bible? If he were to read 85 verses a
day, how many days would it require to read the contents of the whole book?
Ans. 99 tº Sundays; 366 # days.
27. A merchant bought 800 gallons of molasses at 652, and sold # of it at
72% a gallon. From the profit he bought his children a set of Cutter's Anatomical
and Physiological Charts, and had $8.20 left. What did the charts cost 3
Ans. $19.80.
28. A man produces, by breathing, at least six gallons of carbonic acid gas
every minute; a single burning gas jet, ten gallons; an ordinary stove, sixty gal-
lons. How many gallons of carbonic acid gas will two heated stoves, fifty burning
gas jets, and an audience of 1000 people, produce in three hours, and how many
times would the quantity fill a room 100 feet long, 50 feet wide, and 30 feet high 3
Ans. 1191600 gallons. 1% times.
29. A merchant pays $180 for 12 dozen shirts. At what price per shirt must
he sell them to gain $723 Ans. $1.75.
30. A tailor paid $117.60 for a piece of cloth, the price was $2.80 a yard.
How many yards did it contain } Ans. 42.
31. Paid $4569 for paving 1523 square yards. What was the price per square
yard? Ans. $3.00.
138 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. {
CONTRACTIONS IN DIVISION.
232. 1. Divide 9746521 by 634.
OPERATION.
ū34)9746521(15373 ºr Ans.
3.06. Ea:planation.—The operation of this problem is abridged
2365 by performing mentally, the subtraction of the product of
1632 the divisor by the quotient figure, instead of placing it
-º-º-ººs under the dividend and then subtracting, as is ordinarily
1941 done.
39
Or, to arrange the figures differently, which we much prefer, as follows:
3241 -
4369 In this problem, the subtraction is performed mentally,
0634. as in the example above, but the remainders are placed in a
634)9746521 º * * * sº tº t o
- vertical position over the dividend, so as to avoid repeating
15373 ºr Ans. the figures of the dividend.
TO DIVIDE BY THE FACTORS OF THE DIVISOR.
233. Divide 8316 by 36.
OPERATION.
36) 8316 Ea:planation.—Here we first divide by 6, which gives
*m- e us a quotient 6 times too large, because 6 is but # of 36
1386 guotient by 6 © º 5 y
Q. y the true divisor; and hence, to obtain the correct result,
231 Ans. we divide the first quotient by 6.
* DIVISION. I 39
CONTRACTIONS.
234. Peculiar to problems when the divisor is an aliquot part of 10, 100, or 1000,
or some convenient multiple of 10, 100, or 1000. Thus, to divide by
1% multiply the dividend by 9 and divide the product by 10
1+ {{ {{ {{ 8 {{ 4. {{ 10
1#. 4. $é {{ 7 44 4% « 10
1#. {{ {{ {{ 6 {{ {{ {{ 10
2} {{ \ {{ {{ 4 {{ 4% {{ 10
3} {{ {{ {{ 3 {{ 4% {{ 10
6# {{ {{ {{ 16 4. {{ {{ 100
8% {{ {{ {{ 12 {{ {{ {{ 100
12#. {{ {{ {{ 8 {{ * {{ {{ 100
14% {{ {{ 44 7 {{ {{ {{ 100
16# & 4 & 4 & 4 6 {{ {{ {{ 100
18; {{ {{ {{ 16 {{ 4. 44 300
22} & 4 44 & 4 4 {{ {{ . {{ 90
25 44. {{ {{ 4 {4 {{ {{ 100
31#. {{ & 4 {{ 16 {{ {{ - {{ 500
33% 44 {{ {{ 3 {{ {{ {{ 100
37; {{ * .. {{ 8 {{ {{ {{ 300
624 {{ {{ {{ 8 & 4 {{ {{ 500
66# {{ {{ 46 3 {{ {{ {{ 200
75 4% {{ {{ 4 44 {{ {{ 300
83; {{ {{ {{ 6 { % { % { % 500
87; {{ { % {{ 8 {{ 44 44 700
88; {{ {{ {{ 9 44 {{ 44 800
125 {{ & 4 44 8 {{ {{ 44 1000
133} { % 44 44 3 & 4 & 4 44 400
1663 {{ {{ { % 6 44 {{ & 4 1000
225 {{ { { {{ 4 44 {{ 4. 900
250 {{ 4% {{ 4 4% 4. {{ 1000
333; {{ {{ 44 3 {{ {{ 44 1000
375 {{ {{ {{ 8 {{ 44 {{ 3000
625 44 4% 4% 8 {{ {4 {{ 5000
833% {{ 4% {{ 6 {{ {{ 44 5000
875 {{ 4% 4% 8 {{ {{ {{ 7000
15 {{ 4% 4% 2 {{ {{ {{ 30
35 4% 44 4% 2 {{ 44 {{ 70
45 {{ 44 {{ 2 {{ {{ 44 90
The reasons for these contractions are based upon the fact that in all division
operations the result or quotient is not changed by multiplying the divisor and
the dividend by the same number. .*
I4O SOULE's PHILOSOPHEC PRACTICAL MATHEMATICs.
To elucidate this work, we present the following problems: *.
3. Divide 7242 by 24.
OPERATION.
2} | 7242 Explanation.—By inspection, we observe that 4 times 2} are 10; we
4 4 therefore, to save time and labor, first multiply it by 4, to produce a
**- new and more convenient divisor; and in order not to effect a change
10 28 • or error, in the quotient, we also multiply the dividend by 4 for a new
96.8 Ans. dividend, and then divide in the regular manner which, in this case, is
done by simply pointing off one figure.
4. Divide 179801 by 334.
OPERATION. Eacplanation.—-In this problem, we also see by inspection that 3
33 #) 179801 times the divisor makes 100, hence for the reasons given above, we mul-
tiply the dividend by 3, and divide the product by 100. In practice, we
-* *me do not set the figure (3 in this example) by which we multiply the
5394.03 Ans. dividend.
5. Divide 12358 by 373.
OPERATION.
373) 12358 Explanation.—We here observe, by inspection, that the divisor 37}
-*mºms is § of a hundred, and hence, if we multiply it by 8, we will have for
39 Ø ) 98864 a new divisor 300; therefore, for reasons given above, we multiply the
divisor and the dividend by 8, and then divide in the usual manner.
329}# Ans. “ y &,
6. Divide 97450 by 75.
OPERATION.
75) 97450 tº - tº -
Explanation.—By inspection, we here observe that 4 times the
399) 38.98% divisor makes 300; we therefore multiply the divisor and dividend by 4
and with their products proceed to divide.
1299; Ans.
7. Divide 4307491 by 125.
OPERATION.
125) 4307491 Ea:planation.—In this problem, we observe that
8 times the divisor makes 1000; we therefore multi-
1000) 34459928 = 34459 +}} Ans. ply the dividend by 8 and divide the product by 1000.
PECULLAR CONTRACTIONS.
235. In the preceding contractions the work was based upon the principle that
multiplying the divisor and dividend by the same number does not effect a change
in the quotient.
In the following contractions the work is based upon the principle that, to add
to, or subtract from both divisor and quotient, the same aliquot parts, will effect no
change in the result.
¥ DIVISION. I41
By the application of these principles, we can often increase or diminish
either divisor or dividend, or both, by aliquot parts, and thus obtain a more con-
venient divisor.
8. Divide 426 by 74.
OPERATION. Ea:planation.—In this problem, we first divide by 10, and then
73) 42.6 add # of the quotient to itself. This we do because by inspection,
14.2 we observe that 73 is # of 10, or, that it is # of itself less than 10;
hence it follows that if we divide by 10, our quotient will be # of
4 itself too small, and to obtain the correct quotient we must add #
56.8 = 56% Ans. of the quotient by 10 to itself.
NOTE.-To add to, or deduct from, the dividend before dividing, will produce the same result
as adding to, or deducting from, the quotient.
9. Divide 2548230 by 24.
OPERATION.
# of 24 254823%
is 6 — Explanation.—In this problem, we first divide by 30 and then add
39 | #, + of the quotient to itself. The reason for the work is the same as
*--> that given in the above problem.
106.1764 Ans.
10. Divide 8316 by 36.
FIRST OPERATION. SECOND OPERATION. Explanation.—In the first operation, we de-
1. g duct from the dividend and divisor # of them-
6 Of 36 # 1 36 8316 selves, which decreases each in like proportion,
IS 6 = 6 - 6 and then divide.
*- : - 39 277 ºr In the second operation we divide by 30,
39 || 6939 46 ºr which gives a quotient # too large, because
the 30 with which we divide is # º *
36. the true divisor : hence we deduct # of the
231 Ans. 231 Ans. quotient by 30 from itself, and obtain in the re-
mainder the true result.
NoTE.—In nearly all contractions like the above, it is easier and shorter to divide by the faç-
tors of the divisor. Hence we give but few problems, merely to show the application of the princi-
ple.
PROBLEMS.
Divide the following numbers:
1. 4897.20 by 54. 2. 8725 by 35. 3. 26748 by 48. 4, 8316 by 36.

SYNOPSIS FOR REVIEW OF DIVISION.
named.
207.
209.
210.
211.
212.
213.
214.
215.
216.
217.
218.
219.
220.
221.
-º-
-Q.
©
NOTE.—The numbers before the words and phrases refer to the articles treating the subjects
How to Divide when the Divisor
does not exceed Twelve.
Short Division.
To Divide when the Divisor ex-
ceeds Twelve.
Give the General Directions for
Long Division and all the De-
tails for the Full Operation.
To Divide when there are Naughts
on the Right of the Divisor.
To Divide by the Factors of the
Divisor.
Operation in English Money.
236. Define the following words and phrases:
Division. 222,
The Logic of Division.
The Dividend. 223.
The Divisor. 224.
The Quotient.
The Remainder. 225.
The Sign of Division.
Other Signs of Division.
Principles of Division. 226.
Proof of Division.
How many ways may Division be 227.
performed.
Oral Exercises. 228.
Fractional Numbers. 229.
The Philosophy of Division.
To Find-the True Remainder.

→3' T
TO
*.
(142)
MISCELLANEO US PROBLEMS
INVOLVING THE PRINCIPLES OF ADDITION, SUBTRACTION, MULTIPLICATION,
AND DIVISION.
a .44 º A-La
y—w Yºrv -w
237. 1. The subtrahend is 216, and the remainder is 184. What is the min-
uend ? Ans. 400.
2. A grocer paid $350 for some tea and some coffee, and for the tea he paid
$50 more than for the coffee. What did he pay for each 3
Ans. tea, $200; coffee, $150.
3. The sum of two numbers is 787, and one of them is 298. What is the
other ? Ans. 489.
4. The difference between two numbers is 97, and the lesser number is 103.
What is the greater ? Ans. 200.
5. What number multiplied by 4 will give the same product as 16 multiplied
by 127 Ans. 48.
6. If the speed of the steamer J. M. White is 15 miles per hour in still
water, and the velocity of the river is 3 miles per hour, how far will she run up the
river in 4 hours? How many miles down the river in 4 hours ? How far if she runs
in still water 4 hours ? Ans. 48 miles up; 72 miles down;
60 miles in still water.
7. If two men start from the same point and travel in opposite directions,
one at the rate of 20 miles per day and the other at the rate of 25 miles per day,
how many days will they travel before they are 495 miles apart 3 Ans. 11.
8. The greater of two numbers is 897 and their difference 598; what is the
lesser ? Ans. 299.
9. A newsboy sold 20 papers at 5% each, and with the money bought oranges
at 4% each. How many oranges did he get? Ans. 25.
10. A boy sold 5 chickens at 25% a piece, and 8 ducks at 50% each. He
received in payment 3 pigeons at 30% each, and the balance in money. How much
money did he receive 3 Ans. $4.35.
11. The divisor is 187 and the quotient 101, what is the dividend?
Ans. 18887.
NoTE.–In division operations, the worker must remember, 1. That the divisor, while it is
the unit of measure, is, also, one of the factors of the dividend. 2. That the quotient is the other
factor of the dividend. 3. That the dividend is the product of the divisor and the quotient, plus
the remainder when there is one, and hence, dividing the dividend by either factor will give the
other, and the remainder if any.
(143)
I44 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. X-
12. The product of two numbers is 42230 and one of them is 205; what is the
other ? Ans. 206.
13. A grocer bought 28 barrels of apples at $4 per barrel; at what price
must he sell them per barrel to realize a gain of $42? Ans. $54.
14. The dividend is 5484888 and the quotient 81864, what is the divisor 3
Ans. 67.
15. What number divided by 33 will give the quotient 30449, and remain-
der 18% Ans. 1004835.
16. J. J. Foley bought a barrel of sirop de batterie containing 43 gallons at
95% per gallon; 4 gallons having leaked out, he sold the remainder at $1.05 a gallon.
How much did he gain by the transaction ? Ans. $.10.
17. R. S. Soulé bought 354 barrels of flour for $2478, and sold the same at
$7.50 per barrel. How much did he gain 3 Ans. $177.
18. The capital stock of a manufactory is $100000, which is divided into
200 shares. What are 5 shares worth ? Ans. $2500.
19. R. D. T. Sherwood sold to H. Ries 25 barrels of apples at $4 per barrel,
and 124 barrels of potatoes at $3.25 per barrel. He received in payment 1
hogshead of Sugar containing 1143 pounds at 82, and the remainder in money.
How much money did he receive 3 Ans. $411.56.
20. A speculator bought 528 cords of wood at $6.50 per cord. He re-corded
the wood so that it measured 579 cords, which he sold at $6.75 a cord. How much
did he gain } Ans. $476.25.
21. The dividend is 7359, the quotient 198, and the remainder 33; what is
the divisor Ž Ans. 37.
22. A. and B. Were partners in business, equal in gains and losses; A.
managed the business, for which he was to receive from the gains $1200. The
gains of the business were $18524 and the losses $7108. What amount of money
was due each on settlement 3 Ans. A. $6308, B. $5108.
23. A fruit dealer bought 250 boxes of peaches for $375; he sold 94 boxes at
$2 per box, and 147 boxes at $3 per box. The remainder of the peaches spoiled in
his store; his expenses in buying and selling were $45. How many boxes were
spoiled, and how much did he gain? Ans. 9 boxes; $209 gain.
24. A cistern holding 1500 gallons, has two pipes, one of which will fill it in
10 hours, and the other empty it in 30 hours. If both pipes are left open, how long
will the cistern be in filling 3 Ans. 15 hours.
25. A cistern holding 3000 gallons, has three pipes; by one 30 gallons per
minute, and by another 20 gallons per minute, are conducted into it, and by the
third it may be emptied when full in 24 hours. If all three pipes are left open, in
what time will the cistern be filled? Ans. 100 minutes
NOTE.-There are 60 minutes in an hour.
Yºr MISCELLANEOUS PROBLEMS. I45
26. Edward has $165.00. One-fifth of the amount is in 50-cent pieces, the
remainder is in 25-cent, 10-cent, 5-cent, 3-cent, and 1-cent pieces; allowing that he
has an equal number of each coin, how many has he 3
Ans. 66 50-cent pieces, and 300 of each of the other coins.
27. A merchant had 200 bbls. of potatoes for which he was offered $4.00 per
barrel, which price would have given him a gain of $20. Having declined this
offer, on a declining market, he finally sold at a loss of $20. How much per
barrel did his potatoes cost, and how much per barrel did he sell them for?
Ans, cost $3.90, and sold for $3.80 per bbl.
28. A government ration consists of 1 lb., 4 oz. of beef, and 1 lb., 6 oz. of
flour; if beef costs 1032 a lb. and flour 3g2 a lb., what will be the cost of 10840
government rations, and how many lbs. of, beef and flour will there be? -
Ans. $1983.38, cost; 14905 lbs. of flour; 13550 lbs. of beef.
NOTE.—There are 16 ounces in a pound.
OPERATION INDICATED.
1 lb., 4 oz. = 14 lbs. beef (a) 10% =
1 * 6 “ = 1; “ flour à 3%g =
Cost of one ration - - - - = $
1+ x 10840 = lbs. beef.
13 × 10840 = “ flour.
29. A speculator bought an invoice of flour and pork for $297; the flour cost
$7 a barrel, and the pork $12; there were 3 times as many barrels of flour as of pork.
How many were there of each 3 Ans. 27 bbls. flour, 9 bbls. pork.

£roperties of Numbers.
D E FINITION S. *
238. The Properties of numbers are those qualities which belong to them.
239. An Integer is a whole number; as 1, 5, 6, 18, etc. Whole numbers
are divided into two classes, Prime and Composite.
240. A Prime Number is one that can be divided, without a remainder,
only by itself and 1; as 1, 2, 3, 5, 7, 11, 13, 17, etc.
241. A Composite Number is one that can be divided, without a remainder,
by some other whole number than itself and 1; as 4, 9, 12, 15, 24, etc.
All Composite Numbers are the product of two or more other numbers.
Numbers are prime to each other when they have no common factor that
will divide each without a remainder; as 6, 13, 29, etc.
242. An Even number is one that can be divided by 2, without a remain-
der; as 4, 8, 12, 56, etc.
243. An Odd number is one that cannot be divided by 2, without a remain-
der; as 1, 7, 19, 45, 133, etc.
244. The Factors of a number are the numbers which multiplied together
will produce it. Thus, 4 and 4, or 2 and 8, are factors of 16; 2, 3, and 4, or 2, 2, 2,
and 3 are factors of 24.
Every factor of a number is a divisor of it.
*NOTE.-For other Definitions of Numbers, see pages 25 to 31.
245. A Prime Factor of a number is a prime number that will divide it
without a remainder. Thus, 1, 2, 3, and 5, are the prime factors of 30.
246. A Composite Factor of a number is a composite number that will
divide it without a remainder. Thus, 6 and 8, are composite factors of 48.
247. A Perfect Number is one which is equal to the sum of all its divisors,
except itself; as 6 = 1 + 2 + 3.
248. An Imperfect Number is one, the sum of whose divisors is more or less
than itself. Thus, 14 is greater than 1 + 2 + 7, its divisors; and 18 is less than
1 + 2 + 3 + 6 + 9, its divisors. This last is called an abundant number.
(146)
Yºk PROPERTIES OF NUMBERS. I47
249. An Aliquot part of a number is such a part as will divide it without a
remainder. Thus, 1, 2, 3, 4, 6, and 8, are aliquot parts of 24. All aliquots are
factors of the number.
250. The Reciprocal of a number is the quotient of 1 divided by the
number. Thus, the reciprocal of 8 is 1 + 8 = }; and the reciprocal of # is 1 + 4 = 4.
251. Powers of a number are as follows: The first power is the number
expressed by itself; as 5 is the first power of 5. The second power is the product
arising by using the number as a factor twice; as 5 x 5 = 25, which is the second
power of 5. The third power is the product obtained by using the number as a
factor three times; as 5 × 5 × 5 = 125, which is the third power of 5. And in like
manner higher powers of numbers are obtained. *
252. A Multiple of a number is one or more times the number; or it is any
product, dividend, or number of which a given number is a factor, or which is
exactly divisible by a given number; as 25 is a multiple of 5; 24 is the multiple of 2, 3,
4, 6, 8, and 12. A number may have an indefinite number of multiples; as 2 will
divide 4, 6, 8, 10, 12, 14, etc., indefinitely.
253. A Common Multiple of two or more given numbers is a number
divisible by each of them without a remainder. Thus, 24 is a common multiple of
1, 2, 3, 4, 6, 12, and 24.
254. The Least Common Multiple of two or more given numbers is the
least number that is divisible by each of them without a remainder. Thus, 12 is the
least common multiple of 1, 2, 3, 4, 6, and 12.
NOTE.-Since every number is divisible by 1 and itself, the factors 1 and the given number
aré not usually given when naming the multiples. We shall not hereafter name them as multiples.
DIVISIBILITY OF NUMBERS.
255. A Divisor or measure of a number, is any number that will divide it
without a remainder. Thus, 4 is a divisor, or measure, of 12, and 5 is a divisor, or
measure, of 20. g’
One number is said to be Divisible by another when there is no remainder
after dividing.
256. A Common Divisor of two or more numbers is a number that will
divide each of them without a remainder. Thus, 2 is a common divisor of 12, 18,
and 24.
257. The Greatest Common Divisor of two or more given numbers is the
greatest number that will divide each of them without a remainder. Thus, 6 is the
greatest common divisor of 12, 18, and 24. -
I 48 soule's PHILOSOPHIC PRACTICAL MATHEMATICS. 2}
258. Every number is divisible by 2, whose unit figure is divisible by 2.
Thus, 34, 176, 790 are each divisible by 2. *
259. Every number is divisible by 4 when its units' and tens' figures are
divisible by 4. Thus, 156, 264, 34512, 56.1308, are each divisible by 4.
260. All numbers are divisible by 3, the sum of whose figures are divisible
by 3. Thus, 114, 225, 4101, are each divisible by 3.
261. All numbers ending in 0 or 5 are divisible by 5. Thus, 10, 15, and 35,
are each divisible by 5.
262. All numbers whose unit figure is divisible by 2, and the sum of whose
figures is divisible by 3, are divisible by 6. Thus, 36, 102, 678, 15936, are each
divisible by 6.
263. Every number is divisible by 8, when the units', tens', and hundreds'
figures are divisible by 8. Thus, 3824, 12512, 190720, are each divisible by 8.
264. All numbers are divisible by 9, the sum of whose figures are divisible
by 9. Thus, 441, 3456, 123453, are each divisible by 9.
265. All numbers ending in naught are divisible by 10. Thus, 20, 380,
11750, are each divisible by 10.
266. A number is divisible by 12 when the sum of its figures is divisible by
3, and the units and tens are divisible by 4.
267. The combination of numbers that are divisible by 7, and 11, are so
Varied and intricate that they cannot be conveniently classified or formulated under
any brief general law, and being also of very little practical value, they are not
presented.

SYNOPSIS FOR REVIEW.
Inamed.
238.
239.
240.
241.
242.
243.
244.
245.
246.
247.
248.
249.
º
NOTE.-The numbers before the words and phrases refer to the articles treating the subjects
268. Define the following words and phrases:
The Properties of Numbers.
An Integer.
A Prime Number.
A Composite Number.
An Even Number.
An Odd Number.
Factors of a Number.
A Prime Factor.
A Composite Factor.
A Perfect Number.
An Imperfect Number.
An Aliquot.
252.
253.
254.
255.
256.
257.
358.
259.
250.
251.
The Reciprocal of a Number.
The Powers of a Number; 1st.
2d, 3d, etc.
The Multiple of a Number.
A Multiple.
A Common Multiple.
Least Common Multiple.
A Divisor. -
A Common Divisor.
Greatest Common Divisor.
What numbers are divisible by 2,
3, 4, 5, 6, 8, 9, 10, 12.

(149)
*roperties of Nines and Elevens.
===N
*_* a Lºh-ººmm
269. The number nine has been called the “Magical Number.” It pos-
sesses a great many peculiar properties, some of which are more curious than
practical, and will be but briefly treated here. Among the Ancients, the special
properties of 9, and of some other figures, were used in their magic squares and
circles, and in their numerical combinations and puzzles, to such an extent as to
induce the belief, among the ignorant, that they possessed supernatural power, or
that numbers possessed material virtues.
On page 55, we presented a very peculiar property of 9, applied to the opera-
tion of addition.
A curious principle of 9 is shown in the following: Take any number of
two places of different figures, transpose the figures, and take the difference between
the two numbers, and name one of the figures of the difference, and the other figure
is known. Thus, take 73, transpose the figures, and we have 37; then 73–37 = 36
the difference or remainder, in which we see that the sum of 3 and 6 = 9. In all
similar cases, the sum of the remainder will be 9, and hence when one of the figures is
known the other may be found by subtracting the one known from 9.
Another interesting puzzle based upon the same principle is to write any
number of three or more figures, divide by 9, and give the remainder; then
erase any one of the figures in the number and divide by 9, and give the remainder.
The figure erased will be the difference between the remainders, when the second re-
mainder is less than the first ; but when the second remainder is greater than the
first, the figure erased will be the difference of the remainders subtracted from 9.
Thus, 4381 divided by 9, gives 7 remainder. Then cancelling 3 and dividing
481 by 9, gives 4 remainder, and 7–4 = 3 the figure erased.
Again, taking 4381 -- 9 gives 7 remainder. Then erasing 8 and dividing 431
by 9, gives 8 remainder. Then 8–7 = 1 and 9 — 1 = 8, the figure erased.
Again, let the reader take any number and divide it by 9, and give the
remainder. Then multiply the number by any number given, and divide the prod-
uct by 9, and the remainder may be told before the work is performed.
Thus, 428 –- 9 gives 5 remainder. Now multiply the 428 by 4 and divide
the product by 9, and the remainder will be 2. To determine this remainder, we
multiply the first remainder by the number used to multiply the first number
and divide the product by 9; the remainder thus obtained will be the same as the
remainder in the second division.
But, as our purpose is now not to present a discussion of these questions
except to apply one of the properties of 9 and 11 to the proof of addition, sub-
traction, multiplication, and division, we will not occupy space to discuss the
plailosophy or reason of these puzzling results.
(150)
* PROPERTIES OF NINES AND ELEVENS. I5 I
The basis or radix of our numerical system being 10, equal to 9 + 1, if we
multiply the radix and its equivalent by the same digit, thus 10 x 4 = 40, and
9 + 1 × 4 = 36+ 4 = 40, we find the product of 9 alone to lack just the multiplier
of being equal to the product of 10, and as this will hold good of any multiple of 10,
we see that if we divide any decuple as 40, 60, 80, etc., by 9, the remainder is
always the same as the number of tens divided, viz: 4, 6, 8, etc., and if we multiply
these numbers by 10, or any power of 10, as 400, 6000, etc., the remainders will
still be the same digits 4, 6, etc. Therefore we may conclude that any number as
467853 can be resolved into as many multiples of 10 as it contains digits, thus:
400000 + 60000 + 7000+ 800 + 50 + 3, and since the excess of 9's in each of these
round numbers is the same as its significant figure, it is plain the excess of 9's in
the entire number 467853 is equal to the excess of 9's in the sum of its digits, thus:
4 + 6 + 7 + 8 + 5 + 3 = 33; divide this by 9 gives quotient 3, and a remainder or
excess of 6. In finding the excess of 9's, it is best to reject the 9's as soon as its sum
occurs in the additions, thus: 4 + 6 = 10, dropping 9 = 1 + 7 + 8 = 16, dropping
9 = 7 + 5 = 12, dropping 9 = 3 + 3 = 6 excess.
This property is peculiar to 9 by reason of its being 1 less than the radix of
Inotation. This property belongs also to any number that will give a remainder 1
when dividing the radix by it, as 3, and 3 and 9 are the only numbers that will
divide 10, with 1 for a remainder; if the basis of our system were 9, then 8, 4, and 2,
would possess these properties.
The number 11 has several special properties, one of which is similar to that
of 9 and results from the fact that 11 is 1 more than the radix 10. The usual
method of finding the excess of 11's is somewhat different, viz.: by adding the alter-
nate figures; thus, to cast the 11's out of 92257458, we first find the sum of the
figures in the odd places, beginning at the right hand, thus: 8 + 4 + 5 +2 = 19,
casting out the 11's gives us an excess of 8; we then add the figures in the even
places, thus: 5 + 7 -H 2 + 9 = 23, giving an excess of 1; subtract this last excess
from the first, 8–1 = 7, the excess required; if the first excess was the smaller, we
would increase it by 11, then subtract the second. The excess of 11’s may be cast
out at the partial additions, as for 9’s.
These two properties of 9 and 11 can be used in proving addition, subtrac-
tion, multiplication and division, depending on this principle, that any number
divided by 9 or 11 leaves the same remainder as the excess obtained by casting out
its 9's Or 11's.
These methods of proof, however, are not absolutely infallible, as will be
shown on the second page following. By reason, therefore, of the liability to error,
the great majority of business men, in verifying their calculations, prefer to repeat
or go over, in a reversed direction, the first operation.
I 52
*
SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs.
PROOF OF ADDITION BY CASTING OUT THE NINES AND ELEVENS.
270. 1. Add the following numbers and prove the work by casting out the
9's and 11’s.
OPERATION AND PROOF BY CASTING
OUT THE 9’s.
Excess of 9’s.
365592 3
286548 6
64320 6
94094 8
= 23 = 5 excess of 9’s
in the numbers.
Sum 810554–5 excess of 9's in the sum.
NOTE.-Instead of casting the 9's out of each
number separately, they may be cast out of all
the numbers continuously. Thus we may say
3 + 6 = 9, which is dropped. Then 5-H 5 = 10
— 9 = 1 (omitting the 9) + 2 = 3
= 13 — 9 = 4 + 6 = 10 — 9 = 1
10 — 9 = 1 + 8 = 9 — 9 = 0, 6
1 + 3 = 4 + 2 = 6 (omitting 9)
= 1, (omitting 9) + 4 = 5 excess
numbers.
+
+
of
OPERATION AND PROOF BY CASTING
OUT THE 11’s.
Excess of 11's.
365592 7
286548 #)
64320 3
94094 0 y
= 19 = 8 excess of 11’s
in the numbers.
Sum 810554–8 excess of 11’s in the sum.
NOTE.-Instead of casting the 11's out of each
number separately, they may be cast out of all
the numbers collectively. Thus we may add the
figures in the odd places of all the numbers,
making 56 = 1 excess of 11; then add the figures
in the even places of all the numbers, making
48 = 4 excess of 11's ; then subtract the last ex-
cess from the first increased by 11, thus 1 + 11 =
12 – 4 = 8 excess of 11’s in all the numbers.
Explanation.—To prove problems of addition by casting out 9's or 11's, first find the excess of
the 9's or 11's in all the numbers; then find the excess of 9’s or 11's in the sum of the numbers; if
the two excesses are the same the work is correct, as far as this proof can determine.
THE CHECK OR KEY FIGURE.
271. The excess of 9’s or 11's is called the check or key figure by account-
ants, and is used not only to prove the operations of addition, subtraction, multipli-
cation and division, but to prove other operations of accounting.
CONTRACTED METHODS OF FINDING THE CHECK OR KEY FIGURE.
272. 1st CONTRACTED METHOD,
Add the following numbers and prove
by the check figure system:
365592
286548
64320
94094 = 104 = 5 check figure of all the
sºmmemºmºsºms numbers.
810554 = 23 = 5 check figure of the Sum.
Explanation.—By the first contracted method,
add the figures in all the numbers as units = 104;
then add these figures as units = 5 the check fig-
ure of all the numbers. Then add the figures in
the sum as units = 23, which add as units = 5 the
check figure of the sum.
The addition of the several sums is to be con-
tinued as long as any succeeding sum has more
than one figure.
The check figure of all the nunabers, and the
check figure of the sum being equal, proves the
work as far as the method can.
|
2d CONTRACTED METHOD.
Add the following numbers and prove by
the check figure system:
365592 = 30 = 3
286548 = 33 = 6
64320 = 15
94094 = 26
*
*
Gº
ſº-
;
— 23 = 5 check figure.
810554 = 23 = 5 check figure.
Explanation—By the second contracted method,
add the figures of each number as units and then
add the figures of the result, when it has more
than one figure. The results are the check fig-
ures of each number. Then add the check figures
and the sum of the same, and thus produce 5 as
the check figure of all the numbers. Then add
the figures of the sum of all the numbers as '
units, and also add this sum which gives 5, the
check figure.
The first of these contracted methods of finding the check figure, or the earcess of
9's, is new, and is now published for the first time.
Y PROOF BY CASTING OUT THE 9's OR II's. 153
ERRORS NOT DETECTED BY CASTING OUT THE NINES, OR BY THE
NINE CHECK FIGURE METEIOD.
273. 1. When figures have been transposed. 2. When the value of one
figure is as much too great as that of another is too small. 3. When a figure or
figures have been omitted, and other figures having the same sum have been sub-
stituted. 4. When 9 or 9’s have been omitted.
ERRORS NOT DETECTED BY CASTING OUT THE ELEVENS, OR BY
TEIE ELEVEN OEIECK FIGURE METEIG) D.
274. When alternate figures, i. e. figures in the odd, or figures in the even
places, have been transposed, thus, 234567 = 3 excess; then 234765 = 3 excess; or
236547 = 3 excess.
2. When two figures of the same unit value in the odd and even places have
been transposed for two other contiguous figures, whose value is the same but differ-
ent from the value of the other two, thus, 335644 = 1 excess; then 445633 = 1
excess, or 564433 = 1 excess.
3. When figures of the same unit value in the odd and even places have
been omitted, thus, 45567 = 5 excess; then 467 = 5 excess. e
By the foregoing, we see that the method of proof by casting out the 11's,
is more reliable than the method by casting out the 9’s.
PROOF OF SUBTRACTION BY CASTING OUT THE NINES OR ELEVENS.
275. From 1748272 take 1451604, and prove the work by casting out the 9's
and 11's.
OPERATION AND PROOF BY CASTING OPERATION AND PROOF BY CASTING:
OUT THE 9’s. OUT THE 11's.
Excess of 9's. Excess of 11's.
Minuend, 1748272 4 1 = difference 1748272 = 9
Subtrahend, 1451604 3 } of excess. 1451604 0 | 9=difference of excess.
Remainder, 296668 = 1 excess. 296668 = 9 excess in remainder.
Explanation.—-To prove problems in subtraction by casting out the 9's or 11's, first find the
excess of 9’s or 11's in the minuend and subtrahend, and take the difference. Then find the excess
of 9’s or 11's in the remainder; if the excess in the remainder is the same as the difference of the
excesses of the minuend and subtrahend, the work is correct, as far as this proof can determine.
I54
SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. XF
PROOF
OF MULTIPLICATION BY CASTING OUT THE NINES OR
ELEVENS.
276.
and 11's. ,
OPERATION AND PROOF BY CASTING
OUT THE 9’s.
Excess of 9's.
Multiplicand, 350412 6 ) 6 × 3–18–
Multiplier, 29595 3 } 0 excess.
Broduct, 10370443140=0 excess.
Multiply 350412 by 29595, and prove the work by casting out the 9's
OPERATION AND PROOF BY CASTING
OUT THE 11’S.
Excess of 11's.
350412 7 7 JR_35_o a
10370443140 = 2 exceSS.
Explanation.—To prove multiplication by casting out the 9’s or 11's, first find the excess of 9’s
or 11's in the multiplicand and multiplier, and then find the excess of the product of these two
eXCeSSéS.
Then find the excess in the product; if the excess in the product equals the excess of the
product of the excesses of the factors, the work is correct, as far as this proof can determine,
PROOE OF DIVISION BY CASTING OUT THE NINES OR ELEVENS.
277. Divide 65358547823 by 2789, and prove the work by the excess of 9's
• and 11's.
OPERATION AND PROOF
Ol]T THE 9’s.
Dividend, 65358547823 = 2 excess of 9’s.
BY CASTING
Divisor, 2789 = 8 excess of 9’s.
Quotient, 23434.402 = 4 excess of 9’s.
Itemainder, 645 = 6 excess of 9’s.
8 × 4 = 32 = 5 excess of 9’s.
5 + 6 = 11 = 2 excess of 9’s.
OPERATION AND PROOF BY CASTING
OUT THE 11’s.
Dividend, 65358547823 = 8 excess of 11's.
Divisor, 2789 = 6 excess of 11’s.
Quotient, 23434.402 = 2 excess of 11's.
Remainder, 645 = 7 excess of 11’s.
6 × 2 = 12 = 1 excess of 11’s.
1 + 7 = 8 = 8 excess of 11’s.
Explanation.—To prove division by casting out 9's or 11's, first find the excess of 9’s or 11's in
the dividend, divisor, quotient, and remainder; then multiply together the excesses in the divisor
and quotient, and find the excess of 9’s or 11's of the product; then, to this excess, add the excess
of the remainder and find the excess of 9's or 11's in this sum; if the excess is the same as the
excess in the dividend, the work is correct, as far as this method of proof can determine.
EXCESS OF ELEVENS, OR THE CHECK OR KEY FIGURE.
278. The excess of 11's in a number may be found in various ways. We
here indicate three of the methods:
First method. By dividing the number by 11.
Second method. By casting the 11’s out of the sum of the figures in the odd
places, then casting the 11's out of the sum of the figures in the even places, and
7 : PROOF BY CASTING OUT THE 9's OR II's. I55
subtracting the excess of 11's in the even places from the excess of 11’s in the odd
places. In case the first excess is smaller than the second, increase it by 11 and
then subtract.
Third method. From the sum of the figures in the odd places, subtract the
sum of the figures in the even places. In case the first sum is less than the second,
add 11 or some multiple of 11 to the first sum and then subtract; or when the first
sum is less than the second, subtract 11 or some multiple of 11 from the second suin,
and then take the remainder from the first sum. t
TEIE CEHECK OR KEY FIGURE SYSTEM.
279. The excess of 11’s is used largely by accountants in locating errors,
proving transfers, etc., and in books of accounts as above stated, is known as the 11
Check or Key Figure System.
EXERCISES IN FINDING THE CEIECK OR REY FIGURE.
280. Find the check or key figures of the following numbers:
1. 64752. Sum of figures in odd places, 15. Sum of figures in even
places, 9. 9 from 15 = 6, check figure.
NOTE.-See three methods above for finding check figure.
2. 134091. Sum of figures in odd places, 4. Sum of figures in even places,
14. 14 from (4 + 11) = 1, check figure. Or, (14 – 11) =
3 from 4 = 1 check figure.
3. 47. Mentally think 4 from 7, 3, check figure.
4. 52. Mentally think 5 from 13, 8, check figure.
5. 368. Mentally think 6 from 11, 5, check figure.
6. 193. Mentally think 9 from 15, 6, check figure.
7. 507. Mentally think 0 from 12, 1 from 2, 1, check figure, or mentally
think 11 from 12, 1, check figure.
8. 20908. Mentally think 0 from 19, 1 from 9, 8, check figure, or mentally
think 11 from 19, 8, check figure.
9. 50. Mentally think 5 from 11, 6, check figure.
10. 7060. Mentally think 13 from 22, 9, check figure, or mentally think 2
from 11, 9, check figure.
11. 34567. Mentally think 7, 12, 15, and 6, 10, 5, check figure.
12. $ 7428.95 Mentally think 5, 13, 17, and 9, 11, 18, 7 from 17, 10, check figure.
13. $20000.00 Mentally think 0, 0, 0, 2, 0, 0, 0, 2, check figure, or mentally
think 0 from 2, 2, check figure.
14. $52978.10 Mentally think 0, 8, 17, 22; and 1, 8, 10, from 22, 12; 1 from 2,
1, check figure, or mentally think 0, 8, 17, 22 and 1, 8, 10,
from 11, 1 check figure.
15. $ 4872.27 16. 25607.14 17. 134560.05
18. 289176.40 19. 10806.70 20. 100000.00
21. 100708,05 22. 50000.20 23. 899887,69
I56 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. º:
APPLICATION OF THE ELEVEN CHECK FIGURE METHOD OF
LOCATING ERRORS AND OF PROVING TRANSFERS
AND POSTING.S.
281. Post or transfer the following sales book items:
LEDGEER ACCOUNTS.
SALES BOOK. JONES. & SMITH.
tº C ºffſ - * - - * - -
##|g 7| || 487|59 8| ||16541
§ 3. '
Jones bot. Mdse. 7 487.59
Smith ** ** 6|| ||16541120
BROWN. WOOD.
Brown. “ “ 7| || 90.7815 — - ,- -
Wood “ “ 10| ||10862|05 º 9087|51 | 0|| ||10862 d |
Ea:planation.—Before posting the item of Jones' purchase, find and write the check figure to
the left of the folio column in the Sales Book, or whatever book is used as a posting medium.
Then when the item has been posted, find and write the check figure of the posted item to the left
of the folio column in the Ledger, or whatever book the item was transferred to. If the check fig-
ures are the same, the posting is correct. If they are not the same, there is an error, and it has
been located in this posting.
In the first posting above, the check figures are the same (7’s) and the work is correct.
In the second item (Smith's purchase) the check figure in the Sales Book is 6, and in the Led-
ger account it is 8. This shows an error and locates it in this posting.
In the third item (Brown’s purchase) the check figure in the Sales Book is 7, and in the Led-
ger account it is 8, hence an error in this posting. *
The check figure in the Sales Book of the fourth item is 10, while it is 0 in the Ledger posting.
This difference in the check figures shows an error, and locates it in the item posted.
In the above manner, all postings and transfers may be proved.
For further work on the application of the properties of 9 and 11, to the
detection of errors in trial balances, transfers, postings, etc., see Soulé's New Science
and Practice of Accounts.
-->


| | f O
actoring.
282. Factoring consists in separating or resolving a composite number into
its factors.
The operation is performed by division.
NotE.-Before entering upon this work of factoring, and the two subjects following, the learner
should thoroughly understand what are prime and composite numbers, factors and multiples of num-
bers, and the divisibility of numbers as shown on pages 146 to 150.
283.
OPERATION.
2 5460
2|2730
1365
5
3| 273
7
91
13
WRITTEN EXERCISES.
What are the prime factors, or divisors, of 5460?
Ea:planation.—In all problems of this kind, we first divide the
given number by any prime factor, and the successive quotients by
prime factors, or divisors, until we obtain a quotient that is a prime
number. In this problem, our last quotient is 13, which not being
divisible, is a prime number. Hence the divisors 2, 2, 5, 3, 7, and the
quotient 13, are all prime factors, or divisors, of 5460.
PROBLEMS.
284. Find the prime factors of the following numbers:
1. 84
2. 376
3. 864
4. 6105 7. 25600 10, 32460
5, 1683 8. 10376 11. 13532
6. 3560 9. 71460 12, 96033
285. What are the common prime factors of 28, 64, and 72%
OPERATION.
2 ) 28 64. 72
2 || 14 32 36
7 16 18
Explanation.—In all problems of this kind, we divide the given
numbers by any common prime factor of all the numbers, and the
quotients thus obtained are divided in the same manner, till they
have no common factor or divisor. The several divisors will be the
common prime factors of the numbers.
NOTE.—A number that is a factor, or divisor, of two or more numbers is called a common
factor of these numbers.
PROBLEMS,
286. Find the prime factors common to the following numbers:
1. 18, 24, and 36 4.
2. 54, 72, and 84
3. 506, 436, and 308
44 and 280 7.
5. 148, 256, and 320
6. 325, 635, and 550
28 and 64
8. 112 and 250
9. 526, 400, and 780
(157)
I 58 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. X-
$º.
GREATEST COMMON DIVISOR.
287. For the definition of a divisor, a common divisor, and the greatest common
divisor, see page 147.
G. C. D. is the abbreviation for the greatest common divisor.
1. What is the greatest common divisor of 42, 56, and 210
OPERATION.
2 |42, 56, 210 Explanation.—In all problems of this kind, we first divide by
any prime factor that will divide all the numbers; then we divide in
7| 21, 28, 105 like manner the successive quotients thus obtained, until we obtain
| *E===mEºmºmºmºmº quotients that have no common factor, or are prime to each other;
3 4 15 then we multiply all the divisors together, and in the product we have
the greatest common divisor.
2× 7 =14 Ans.
NOTE 1.--When there is no number greater than 1, that will divide all the numbers without
a remainder, then 1 is the greatest common divisor.
NOTE: 2.—When there are two large numbers, the operation may be more easily performed
by first dividing the larger number by the smaller, and if there is a remainder divide the preceding
divisor by it, and thus continue until there is no remainder.
NOTE: 3.—When there are more than two numbers, proceed as with two, and then, with the
atest common divisor of the two and one of the other numbers, and thus continue until through
with all the numbers. The last divisor will be the greatest common divisor.
Problems 2 and 3 elucidate this operation:
2. What is the greatest 3. What is the greatest
common divisor of 88 and common divisor of 195,
24 % Ans. 8. 285, and 315? Ans. 15.
OPERATION. OPERATION.
24)88(3 285)315(1 15)195(13
72 285 15
1624(1 30)285(9 45
16 270 45
8)16(2 & 15)30(2
16 30 f
GENERAL DIRECTIONS FOR FINDING THE GREATEST COMMON
DIVISOR.
288. From the foregoing elucidations, we derive the following general direc-
tions for finding the G. C. D.
1. Write the numbers on a horizontal line and divide by any prime number that
will divide all without a remainder, and write the quotients in a line below.
2. Continue this process of dividing the successive quotients, until quotients
are obtained which have no common factor, or divisor.
FACTORING, I59.
3. Multiply together all the divisors, and their product will be the G. C. D.
NOTE.-When there are two or more large numbers, it is often more convenient to work by
successive divisions, as elucidated in problems 2 and 3.
PROBLEMIS.
289. What is the greatest common divisor of the following numbers?
4. Of 441 and 567 ? Ans. 63.
5. Of 90, 315, and 810? Ans. 45.
6. Of 654, 216, and 108? Ans. 6.
7.
Robinson has 25, and Blaise, 45 dimes. How shall they arrange them in
packages, so that each will have the same number in each package 3
Ans. 5 in each package.
8. A planter has 697 bushels of corn and 204 bushels of rough rice, which
he wishes to put into the least number of bins containing the same number of
bushels, without mixing the two kinds. How many bushels must each bin hold?
Ans. 17 bushels.
9. A commission merchant has 2490 bushels of wheat, 1886 bushels of corn,
and 8438 bushels of oats, which he wishes to ship in the least number of sacks of
equal size, that will exactly hold either kind of grain. How many sacks will he
require? Ans. 6407.
OPERATION INDICATED.
Find the greatest common divisor as above, (it is 2).
2)2490 1886 8438
1245-H 943-H4219–6407 Ans.
LEAST COMMON MULTIPLE.
290. For the definition of a multiple, a common multiple, and the least common
multiple, see page 147.
L. C. M. is the abbreviation for the least common multiple.
1. What is the least common multiple of 5, 6, 8, 21, 28?
OPERATION.
2) 5, 6, 8, 21, 28 Explanation.—In all problems of this kind, we first
arrange the numbers on a horizontal line, and then divide by
2) 5 3 4 21 14 the smallest prime number that will divide two or more
3) 5. 3 2 21 7 without a remainder, and write the quotients and undivided
4./
numbers in a line below; this process of dividing we cont
7) 5 1 2 7 7 tinue until there are no two numbers that can be divided by
the same number without a remainder; then we multiply
the divisors and the numbers in the last line together, and
2 × 2 × 3 × 7 × 5 × 2–840 Ans. the product is the least common multiple.
5 1 2 1 1
NoTE.—When there is any number of the problem that will divide any of the others without
a remainder, it may be cancelled before commencing to divide.
I6O SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS.
GENERAL DIRECTIONS FOR FINDING THE LEAST COMMON
MULTIPLE.
291. From the foregoing elucidations, we derive the following general direc-
tions for finding the L. C. M.
1. Write the numbers in a line and then divide by the smallest prime number
that will divide two or more without a remainder, and write the quotients and undi-
wided numbers in a line below.
2. Continue this process of dividing, until there are no two numbers, with
common factors, or that can be divided by any number greater than 1.
3. Find the continued product of the divisors and the numbers in the last line,
and it will be the L. C. M.
g PROBLEMIS.
2. What is the least common multiple of 4, 9, 12, 15, and 24? Ans. 360.
What is the least common multiple of the following numbers:
3. Of 8, 4, 9, and 30? Ans. 360.
4. Of 50, 27, 3, 45, and 63? Ans. 9450.
5. Of 21, 36, 11, and 22% Ans. 2772.
6. Of 800, 600, 10, 40, and 12% Ans. 2400.
7. Of 8, 18, 20, and 70 ? Ans. 2520.
8. A drayman has 2 drays and 2 floats. On 1 dray he can haul 9 barrels of
flour, and on the other 12 barrels; on 1 float he can haul 18 barrels, and on the
other 21 barrels. What is the least number of barrels that will make full loads for
either of the drays or the floats? Ans. 252.
9. A fruit dealer desires to invest an equal amount of money in oranges,
peaches, and grapes, and to expend as small a sum as possible. The price of
oranges is $2.40 per box; peaches, $1.60; and grapes for a medium article, 90g.,
and for first quality, $1.20; of these two qualities the fruit dealer took the cheaper.
How much more money did he invest than he would have done had he taken the
grapes at $1.20 per box 3 Ans. $28,80.
PARTIAL OPERATION.
L. C. M. of $2.40, $1.60, .90 = $14.40.
$14.40. × 3 (kinds of fruit) = $43.20 spent by purchasing grapes a 90g.
L. C. M. of $2.40, $1.60, $1.20 = $4.80.
$4.80. × 3 = $14.40, what he would have spent by taking grapes (a) $1.20.
$43.20–$14.40 = $28.80, Ans.
MSYNOPSIS FOR RE VIE W OF FA CTORING!
NOTE.-The numbers before the words and phrases, refer to the articles treating the subjects
named.
292. Define the following words and phrases:
282. What is Factoring 3
285. A Common Factor.
288. General Directions for Finding Greatest Common Divisor.
291. General Directions for Finding Least Common Multiple.
293. Cancellation is the process of shortening the operations of division, or
of the indicated result of multiplication and division operations combined, by
rejecting equal factors from both dividend and divisor, or from both increasing and
decreasing numbers, of an indisated result.
The operation is performed by drawing a line across each factor cancelled, or
cut out; thus, 3, 7, 25.
294. The Principles of Cancellation, are : º
1. Rejecting, or cancelling a factor from any number, is in effect dividing the
number by that factor.
2. Rejecting, or cancelling equal factors from both dividend and divisor, or
from both increasing and decreasing numbers in an indicated result, does not
change the quotient or result.
NoTE.—A good understanding of factors, multiples, and the divisibility of numbers, will
greatly facilitate the student in the operations of cancellation.
STATEMENT LINE.
295. A statement line, as used in this work, consists of a vertical line upon
whose right and left sides are placed respectively the increasing and decreasing
numbers of the problem. This statement is made to facilitate the operation, as will
be shown in the problems following.
In the solution of all practical problems, when writing the numbers on the
statement line, whether increasing or decreasing, a reason must be mentally given
for each number written, as is elucidated in the 12th Problem of Article 297, follow-
ing. See pages 77 to 81, and 125 to 128, for a full elucidation of the reasoning given
in the solution of problems, and of the philosophic system presented in this work.
PROBLEMIS.
Divide 7 × 3 × 4 by 7 × 4.
Operation by Cancellation. Explanation.—In all problems where we have both mul:
tiplication and division operations to perform, we use a vertical
ë or perpendicular fine, which we call the statement line. This
43 line is used to facilitate the work by separating the dividends
4. and the divisors, or the increasing and the decreasing numbers.
The dividends, or the increasing numbers, are always placed
* upon the right hand side of the line, and the divisors or
3, Alls. decreasing numbers, are always placed upon the left hand side.
In this example, having written the numbers that constitute the dividend and the divisor,
respectively upon the right and left hand sides of the statement line, we cut out, or cancel, the equal
factors 7's and 4’s in the numbers constituting the dividend and the divisor, and thus obtain 3 as
the answer to the problem.
To perform the work without the aid of cancellation, we would be obliged to make the
following figures: 7 × 3 = 21, which x 4 = 84, the dividend; then 7 × 4 = 28, the divisor; then
28)84(3, Ans.
84
amº

(161)
I62 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. x-
2. Multiply 25, 48, and 88 together, and divide the product by the product
of 10, 36, and 8.
Operation by Cancellation. Explanation.—In this example, we write the numbers
3 1% 25 5 2 on the line as above directed and then cancel 10 and 25 by
º ; # - 5; then 36 and 48 by 12; then 8 and 88 by 8; then 4 and 2 by
2. This is all that can be cancelled, and we then multiply
110 together 5, 2, and 11, divide the product by 3, and thus
36; A. obtain the true result, 36%.
Il S.
7
3. Divide the product of 32 × 3 by 8 × 9 × 16.
Operation by Cancellation. Explanation.—Having written the numbers on the
§ |32 4 statement line, we first cancel 8 and 32 by 8; then 9 and 3
3 g 3 by 3; then 16 and 4 by 4. Now having no more numbers on
4 I6 the increasing side of the line to cancel, we multiply
- together the remaining numbers on the decreasing side of
12| 1 = T's, Ans. the line and thus produce the correct result, Tºg.
NOTE.—In all cases where, after cancelling, no factor appears on either side of the statement
line, the factor 1 is always understood as being there. Its non-appearance is in consequence of
not having written it when we cancelled a number by itself, as in the following problem.
4. Multiply 14, 5, and 3 together, and divide the product by the product of
2, 15, and 7.
OPERATION.
2 | 1.4 7 1 2 || 14 7 1 Ea:planation.—In this problem, we first
3 15 # 13 Iš # 1 cancelled 2 and 14; then 5 and 15; then the 3
1. 1. and 3; and lastly the 7 and 7. The work might
7 | 3 OI’ 7 || 3 have been abridged by multiplying the 3 and 5
and then cancelling both 15's, and then multi-
1. Ans. 1. 1 = 1, Ans. plying the 7 and 2 and cancelling both 14's.
7
GENERAL DIRECTIONS FOR, CANCELLATION.
296. From the foregoing elucidations, we derive the following general direc-
tions for cancellation :
1. Cancel all the factors common to both dividend and divisor, or to the increas-
ing and decreasing numbers. *-
2. Divide the product of the remaining factors of the dividend by the product
of the remaining factors of the divisor.
NOTE.--When a factor cancelled is equal to the number itself, the unit 1 remains, since a
number divided by itself gives 1 as a quotient. If the 1 is in the dividend it must be retained ; if in
the divisor, it may be disregarded, since dividing by 1 does not change the quotient.
297. Cancel and work the following line statements:
(1) (2) (3) (4) (5)
2 || 6 12 || 45 31 || 124 9 || 76 10 | 16
3 || 9 9 || 56 5 || 17 20 | 91 135 | 45
18 || 54 7 | 84 51 || 10 70 || 140 24 || 60
27 16 || 4 4 25 *-
1, Ans. 70, Ans. #, Ans. 1921%, Ans. 14, Ans.
jºr CANCELLATION. I63
6. Divide the product of 6, 7, 12, and 22 by the product of 11, 3, 14, and 8.
Ans. 3.
7. What is the quotient of 28 × 65 × 7 × 78 + 56 x 130 x 42 × 13%
Ans. #
8. Multiply 21, 55, and 128 together, and divide the product by 14 x 25 x 64.
Ans. 63.
9. How many bushels of corn, at 70% a bushel, will pay for 140 gallons
molasses at 65 cents a gallon ? Ans. 130 bushels.
10. Bought 420 pounds of sugar at 6 cents a pound, and gave in payment
360 pounds of rice. What was the price of the rice per pound 7 Ans. 7 cents.
11. Sold to a drayman 64 bushels of oats at 75 cents a bushel, for which he is
to pay in drayage at 50 cents a load. How many loads must he haul ?
Ans. 96 loads.
12. Paid 65% for 5 yards of calico. What will 27 yards cost at the same rate?
Analytic Solution by Explanatºon.—In all practical problems of this kind,
Cancellation we give a reason for each step of the operation, and make
e the whole statement to indicate the final result without
§§ 13 performing any of the intermediate work. In this problem,
#|27 we place the 65c. on the increasing side of the statement line
as our premise, and reason thus: 5 yards cost 65c. Since 5
e yards cost 65c., 1 yard will cost # part of it, and 27 yards will
$3.51, Ans. cost 27 times as much as 1 yard. * j
13. A merchant sold 25 boxes of candles containing 36 pounds each at 162
per pound, and received in payment starch at 6 cents per pound. How many boxes,
each containing 30 pounds, did he receive? Ans. 80 boxes.
14. How many pounds of butter at 35% per pound, will pay for 245 pounds
of rice at 5 cents per pound 3 Ans. 35 pounds.
15. If 17 barrels of flour cost $110.50, what will 500 barrels cost at the same
rate? Ans. $3250.
16. Sold 340 boxes of soap, containing 45 pounds each, at 122 per pound,
and received in payment starch at 9% per pound. How many boxes, each contain-
ing 24 pounds, did I receive 3 Ans. 850 boxes.
a - -aº a
w ur-w
SYNOPSIS FOR RET/IE W OF CANCELLATION.
NoTE.—The numbers before the words and phrases, refer to the articles treating the subjects
named.
298. Define the following words and phrases?
293. Cancellation.
294. Principles of Cancellation.
295. Statement Line.
296. General Directions for the Operation.
E. ſo N O
fractions.
36 S. 3: tº s º ºs e - e º º ºs e º e º e º e Eºcºcº =N
—-tº-4(O)-4—
299. According to the general methods prescribed by the text books of Our
country, for the various operations of fractions, the subject is correctly considered
by both the teacher and the pupil to be the most difficult in the science of numbers.
But by our fully evolved, logical, and philosophic system of handling fractional
numbers, the subject is simplified, rationalized, and rendered pleasing to the student.
By our method of work, fully one-half of the time and figures required by the ordi-
nary methods is saved, and all the arbitrary and absurd rules which overload the
organ of memory and prevent the expansion of the higher faculties of causality and
comparison, are abandoned to the shades of the dead past, and entombed with the
ingenious minds which gave them birth.
By our system, all the reasoning faculties of the mind are brought into action
and exercised in a manner to give logical strength and acuteness to work, not only
in the fields of mathematics, but upon all the plains of life.
In behalf of truth, education, and humanity, we lament the non-progressive
methods of the general school and college arithmetics.
The present authors of school arithmetics, with but few exceptions, are timidly
following in the obscure paths of arithmetical Science which were marked out ages
ago, when the science was in its infancy—too cowardly, non-progressive, and con-
tracted in their views to seek and explore more direct and comprehensible routes to
the fountains of mathematical knowledge.
300. The Unit is the universal basis of numbers, and the foundation of arith-
metic. From unity arise two distinct classes of numbers. 1. Integers. 2. Frac-
tions. The first class, integers, has its origin in the multiplication of the unit;
and the second class, fractions, results from the division of the unit. The first is
synthetical, the second is analytical.
I) EFINITIONS.
301. A Fraction is one or more of the equal parts of a unit, or of a collection
of units. Or more briefly, it is a part of anything, or a numerical expression of
a part of a unit; thus, 1 half and 3 fourths are fractions.
302. A Fractional Unit is one of the equal parts into which any integral
unit is divided. If the integral unit is divided into two equal parts, each is called
a half; if into three, each is called a third ; if into four, each is called a fourth ;
and so on, according to the number of parts into which the integral unit is divided.
303. Fractions are divided into two kinds, Common, or Wulgar, and Decimal
Fractions.
(164)
Y FRACTIONS. I65
304. Common Fractions are expressed by two numbers, one written above
the other, with a horizontal line between them. The number below the line is called
the Denominator, and the number above the line is called the Numerator. Thus,
# (one-half), # (three-fourths), # (five-sixths), (seven-eighths), and +} (thirteen-seven-
teenths), are common fractions, the denominators of which are respectively 2, 4,
6, 8, and 17. The numerator and denominator together, are called the terms of the
fraction.
305. The Denominator of a fraction shows the number of equal parts into
which the unit is divided. Thus in the fraction #, the 8 is the denominator and
shows that the unit is divided into 8 equal parts called eighths.
306. The Numerator of a fraction shows the number of equal parts taken to
form the fraction.
Thus in §, the numerator is 5 and shows that 5 of the 8 equal parts are taken,
or expressed, by the fraction.
All fractions arise from division and are expressions of unexecuted division
in which the numerator is the dividend, the denominator the divisor, and the fraction
itself the quotient.
307. Decimal Fractions are those in which the denominators are not generally
expressed, but are always 10, or some power of ten; thus, .5, .75, .821, read respec-
tively five tenths, seventy-five hundredths and eight hundred twenty-one thousandths,
are decimal fractions. To write these fractions as common fractions, they would be
written thus, #3, #3, and #5.
The point (...) placed before the 5, 7, and 8, in the above decimally expressed
fractions, is called the decimal point, and is used to abbreviate the work.
CLASSIFICATION OF FERACTIONS.
308. For convenience, fractions are classed under the following heads: Proper
Fractions, Improper Fractions, Simple Fractions, Mia!ed Numbers, Compound Frac-
tions, and Complea, Fractions.
309. A Proper Fraction is one in which the numerator is less than the
denominator; as 3, #, g. *
310. An Improper Fraction is one in which the numerator is equal to or
greater than the denominator; as, #, 3, #, and *.
311. A Simple Fraction is one in which both terms are whole numbers, and
may be either a proper or an improper fraction; as, #, #, ++, or *.
312. A Mixed Number is a number composed of a whole number and a
fraction; as, 24, 5}, and 21+%.
313. A Compound Fraction is a fractional part of a fraction or mixed
number; as, # of 3, 4 of +3 of 12#.
I 66 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. X}
314. A Complex Fraction is one that has one or more of its terms fractional;
aS,
# 6# 3 # 6#
– of —, — of — of —
# 5% # 1#. 8
315. The Reciprocal of a Fraction is the result of 1 divided by the fraction.
Thus, the reciprocal of # is 1 -- # = # = 14.
316. The Walue of a Fraction is the result of its numerator divided by its
denominator. Thus, # = 4, * = 23.
GENERAL PRINCIPLES OF FERACTIONS.
317. 1. Multiplying the numerator, or dividing the denominator, multiplies
the fraction.
2. Dividing the numerator, or multiplying the denominator, divides the
fraction.
3. Multiplying or dividing both numerator and denominator by the same
number does not change the value of the fraction.
SYNOPSIS FOR RE VIEW OF FIRACTIONS.
ºm-E-3
NOTE.-The numbers before the words and phrases, refer to the articles treating the subjects
named.
318. Define the following words and phrases:
301. A Fraction. 310. An Improper Fraction,
302. A Fractional Unit. * 311. A Simple Fraction.
303. How Fractions are Divided. 312. A Mixed Number.
304. Common Fractions. 313. A Compound Fraction.
305. Denominator. 314. A Complex Fraction.
306. Numerator. 315. The Reciprocal of a Fraction.
307. Decimal Fractions. 316. Value of a Fraction.
308. Classification of Fractions. 317. General Principles of Fractions.
309. A Proper Fraction.

iteduction of Fractions.
(3) º *H*=N
319. Reduction of Fractions is the process of changing their form without
altering their value.
320. A fraction is reduced to Higher Terms when the numerator and the
denominator are expressed in larger numbers. Thus, # = }, or $, or ; ; # = #, or
-Ér, Or ##, etc. † wº-
321. A fraction is reduced to Lower Terms when the numerator and the
denominator are expressed in Smaller numbers. Thus, # = }, or 3; # = }, or #.
322. A fraction is reduced to its Lowest Terms when its numerator and its
denominator are prime to each other, or have no common divisor, Thus, #, #, and
*#, are in their lowest terms.
323. Whole Numbers may be reduced to fractions having any desired denom-
inator.
Whole line. Half lines.
Third lines. Fourth lines.
ORAL EXERCISES.
324. 1. If a line, an Orange, an apple, or a unit of any kind is divided into
two equal parts, what is each part called? Ans. #.
2. If divided into three equal parts, what is each part called ? Ans. #.
3. If divided into four equal parts, what is each part called? Ans. 4.
- 4. When divided into four equal parts, what are three of those parts called ?
Ans. #.
5. How would you get # of an apple?
Ans. Divide it into 4 equal parts and take 3 of the parts.
6. When any number or thing is divided into five equal parts, what is one
of those parts called? AnS. #.
7. What are two, three, and four of the parts called respectively?
Ans. #, #, and #.
8. 1 unit, abstract, or denominate of any kind, equals how many halves?
thirds? fourths? fifths? sixths? sevenths? eighths? ninths?
(167)
I 68 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. X-
ge 9. 2 = how many halves? thirds? fourths? fifths? sixths? sevenths? eighths?
*. 3 = how many halves? thirds? fourths? fifths? sixths? sevenths? eighths?
*. 4 = how many halves? thirds? fourths? fifths? sixths? sevenths? eighths?
ninths, 5 = how many halves? thirds? fourths? fifths? sixths? sevenths? eighths?
13. 6 = how many halves? thirds? fourths? fifths? sixths? sevenths? eighths g
What kind of numerical work is the above called ”
325. 4 = how many fourths? sixths? eighths? tenths? twelfths? fourteenths?
# = how many sixths? ninths? twelfths? fifteenths? eighteenths? twenty-firsts?
# = how many eighths? twelfths? sixteenths? twentieths? twenty-fourths?
# = how many tenths? fifteenths? twentieths? twenty-fifths? thirtieths?
# = how many sixteenths? twenty-fourths? thirty-seconds? fortieths? sixty-
fourths # & -
+ = how many thirty-seconds? forty-eighths? sixty-fourths? eightieths?
What kind of numerical work is the above called 3
326. Answer the following numerical questions:
# = | #iº | #E} | #: | # =}| #:
# = } | . . ; ; # =; #F# # =; # = }
12 – s 2 4 - 4 8 - - 6 4 - 9 6 - 7 5 O –
*i; } :=}| #E} | #E} | . E! | #:
What kind of numerical work is the above called ?
327. How many halves = a unit? How many eighths = a unit #
{{ “ thirds = {{ 44 “ ninths = {{
64 “ fourths = “ 4% “ tenths = **
44 “ fifthS = “ ‘‘ ‘‘ elevenths = %
{{ “ SixthS = “ 6% “ twelfthS = {{
{{ “ Sevenths = “ 64 “ thirteenths = “
1 unit equals how many eighths? § = ? 13 units equal how many thirds? # =?
1. {{ & 4 {{ twelfths? 43 = ? 13 “ 64 {{ sixths? *=?
3 units equal “ thirds? § = ? 13 “ 4% 44 eighths 3 4.g4–?
4. {{ {{ 4. fourths? 14 = ? 3+ “ {{ é- fourths # 14–?
5 {{ {{ {{ halves? 19 =? 2; “ { % {{ sixteenths?#=?
14 “ 44 & 4 halves? # =? 43 “ & 4 & 4 eighths? §§ – ?
14 “ {{ {{ fourths 7 # =?
328. What is the reciprocal of 1, of 2, of 3, of 4, of #, of 23
329. Analyse the fraction #.
Analysis.—# is a proper fraction, since the numerator is less than the denominator; 4 is the
denominator, and shows that the unit is divided into 4 equal parts; + is the fractional winit, since
it is ONE of the four equal parts into which the unit is divided; 3 is the numerator and shows that
three of these equal parts are taken; 3 and 4 are the terms of the fraction, and its value is less
than 1, or unity.
* - REDUCTION OF FRACTIONS. I 69
TO REDUCE FRACTIONS TO HIGHER TERMS.
330. 1. Change # to a fraction whose denominator is 64.
OPERATION.
64 - 16 = 4 Explanation.—In all problems of this kind, we first divide
the required denominator by the denominator of the given
13 × 4 = 52 fraction. Then with the quotient thus obtained, multiply both
º- — Ans. terms of the given fraction, and in their products we have the
16 × 4 = 64 required fraction.
GENERAL DIRECTION FOR REDUCING FRACTIONS FROM LOWER
TO HIGHER TERMS.
331. From the foregoing elucidations, we derive the following general direction
for reducing fractions from lower to higher terms:
Divide the required denominator by the denominator of the given fraction, and
multiply both terms of the fraction by the quotient.
PROBLEMIS.
332. Change ++ to a fraction whose denominator is 75
{{ # to {{ {{ {{ 176
4% # to & 4 4% 44 384
{{ # to & 4 4. 4é 2180
TO REDUCE FRACTIONS TO THEIR LOWEST TERMS.
333. Reduce # to its lowest terms.
FIRST OPERATION. Ea:planation.—In all problems of this kind, we divide
2)# = ##; 4) # = g Ans. both the numerator and the denominator by their common
factors. Or, as shown in the second operation, we may
SECOND OPERATION. produce the same result with less figures, by dividing
8)# F § Ans. both terms of the fraction by their greatest common divisor.
By this reduction we change the form of the fraction ##, but we do not alter,
or change, its value, for the fractional unit of the resulting fraction (g) is 8 times as
great, while the number taken is ; as great.
When the terms of the fraction have no common factor greater than 1, the
fraction is in its lowest terms and is called an irreducible fraction.
The object of reducing fractions to their lowest terms is to enable us to understand their
value more easily and readily. -
I7o SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS.
GENERAL DIRECTION FOR REDUCING FRACTIONS TO THEIR
LOWEST TERMS.
334. From the foregoing elucidations, we derive the following general direction
for reducing fractions to their lowest terms:
Cancel all the factors common to both the numerator and the denominator;
Or, divide both numerator and denominator by their greatest common divisor.
PROBLEMIS.
335. Reduce the following fractions to their lowest terms.
2. #4, ##, ##, ##, Ans. #, #, #, #. 7. Hººs, Ans. #.
3. ###, #3, ###, Ans. #3, #, #r. 8. #####, Ans. ####.
4. ####, AnS. #. 9. #####, AnS. #.
5. Hº, Ans. ###. 10. ####, Ans. #.
6. ###, Ans. #. 11. §§§, Ans. ##.
TO REDUCE WHOLE OR, MIXED NUMBERS TO IMPROPER
FRACTIONS.
336. 1. Reduce 53 to an improper fractions, or to thirds.
OPERATION.
5; Explanation.—In all problems of this kind, we reason
tº- thus: Since there are 3 thirds in every unit or whole
14. A number, in 5 units there are 5 times as many, which are
*ā- Ans. * + the # make *.
2. Reduce 9 to a fraction whose denominator is 6.
OPERATION.
9 × 6 = ** Ans.
GENERAL DIRECTION TO REDUCE WHOLE OR MIXED NUMBERS
TO IMPROPER FRACTIONS.
337. From the foregoing elucidations, we derive the following general direction
for reducing whole or mixed numbers to improper fractions:
Multiply the whole number by the required denominator and to the product add
the numerator of the fraction, and write the required denominator under the result.
* REDUCTION OF FRACTIONS. 171
PROBLEMS.
Reduce the following numerical expressions to improper fractions:
3. 84 Ans. *. 8. 714 Ans. 4+*.
4. 16# Ans. *. 9. 68% Ans. 44%.
5. 17; Ans. 4. 10. 21833 Ans. ***.
6. 323 Ans. *;1. 11. 23# Ans. ***.
7. 435% Ans. **iš. 12. 108Tºg Ans. 1-#4.
13. Reduce 14 to a fraction whose denominator is 9.
14. {{ 37 44 {{ 44 4% 24.
14. {{ 54; “ {{ & 4 {{ 16.
TO REDUCE IMPROPER FRACTIONS TO WHOLE OR MIXED NUMBERS.
338. Reduce 41 to a mixed number.
OPERATION. Ea:planation.—In all problems of this kind, we reason thus:
+ = 44 Ans. Since there are 4 fourths in 1 unit, or whole number, in 17 fourtha
OI’ there are as many units as 17 is equal to 4, which is 4 times with 1
17 -- 4 = 44 Ans. remainder, or altogether 4+ as the proper quotient, or answer.
*.
GENERAL DIRECTIONS TO REDUCE IMPROPER FRACTIONS TO
WEIOLE OR MIXED NUMBERS.
339. From the foregoing elucidations, we derive the following general direction
for reducing improper fractions to whole or mixed numbers :
Divide the Numerator by the IDenominator.
PROBLEMS.
340. Reduce the following improper fractions to whole or mixed numbers:
2. ** Ans. 4. 6. * Ans. 33.
3. * * Ans. 53. 7. # Ans. 5+.
4. 1 #4 Ans. 48. 8. 4.4 Ans. 9+.
5. 2.23. Ans. 18%. 9. *###9. Ans. 34&#.
l
2
172 soul.E's PHILOSOPHIC PRACTICAL MATHEMATICs. 4×
TO REDUCE COMPOUND FERACTIONS TO SIMPLE FRACTIONS.
341. 1. Reduce # of # of £ to a simple fraction.
OPERATION. Explanation.—In all problems of this kind, we multiply together
3 % = }}. A all the numerators for a new numerator and all the denominators
# × # × # = #& Ans. for a new denominator.
When a compound fraction contains whole or mixed numbers,
they must first be reduced to improper fractions.
When there are common factors in both terms of a compound fraction, they should be can-
celled before multiplying. By cancelling the common factors, the work is shortened and the result
* for the reason that dividing both terms of a fraction by the same Inumber does not alter
its value.
2. Reduce # of 73 of # of 4 of +43 to a simple fraction.
OPERATION.
3
3 I; 2 A 1 1
– × — × − x − x – = – Ans
§ 2 g 1 12 3
3 3
GENERAL DIRECTIONS TO REDUCE COMPOUND FRACTIONS TO
SIMPLE FRACTIONS.
342. From the foregoing elucidations, we derive the following general direction
for reducing compound fractions to simple fractions:
Cancel common factors if they occur in both terms of the fractions ; then multi-
ply the numerators together for the new numerator and the denominators together for
the new denominator of the fraction.
PROBLEMS.
343. Reduce the following compound fractions to simple ones:
1. Reduce # of ; of § to a simple fraction. Ans. Hºg.
2. Reduce # of 74 of # of 4 of Hº to a simple fraction. Ans. #.
3. Reduce # of 24 of # of # of 13 to a simple fraction. Ans. }}.
4. # of +% of Hºr. Ans. #. | 8. ºr of 96. Ans. 54.
5. # of § of #3. Ans. #s. 9. § of +3 of 17*. Ans. 6.
6. # of 3% of 4. Ans. 14. 10. § x #.x #. Ans. #.
7. # of 84. Ans. 24%. 11. 5 × 2 × #. Ans. ##.
TO REDUCE FRACTIONS OF DIFFERENT DENOMINATORS TO EQUIV-
ALENT FRACTIONS OF A COMMON DENOMINATOR, OR OF
TEIE LEAST COMMON DENOMINATOR.
344. A Common Denominator is a denominator common to two or more
fractions.
345. The Least Common Denominator of two or more fractions is the least
number divisible by each of the denominators.
* REDUCTION OF FRACTIONS, I73
846. A Common Denominator of two or more fractions is a common multiple
of their denominators; and the least common denominator of two or more fractions
is the least common multiple of their denominators, for the reason that all higher
terms of a fraction are multiples of its corresponding lower or lowest terms.
347. Reduce #, #, and #, to equivalent fractions having a common denominator.
OPERATION.
i; #, #, #,
3 × 4 × 8 = 96, common denominator.
# of = 32; hence ##, equivalent of #.
# of £96=72; hence #, equivalent of #.
§ of = 84; hence ##, equivalent of £,
Explanation.-In all problems of this kind, we obtain the common denominator by multiplying
together the denominators of all the fractions. Then to find the respective numerators we take
such a part of the common denominator as the respective fractions are parts of a unit, as shown in
the operation.
Or, divide the common denominator by the denominator of each fraction, and multiply the
quotient by its numerator.
GENERAL DIRECTIONS TO REDUCE FRACTIONS TO A COMMON
DENOMINATOR.
348. From the foregoing elucidations, we derive, the following general direc-
tions for reducing fractions to a common denominator:
1. Multiply together the denominators of all the fractions for a common
denominator.
2. Then to find the respective numerators, take such a part of the common
denominator as the respective fractions are parts of a unit. Or, divide the common
denominator by the denominator of each fraction, and multiply the quotient by its
77.0/merator.
PROBLEMIS.
349. Reduce the following fractions to equivalent fractions having a common
denominator:
1. #, #, and #. Ans. #, #, and #.
2, #3, #, and #. Ans. ###, ###. and #}.
3. *, and #. Ans. ###, and ###.
4. #, #3, #5, ##, and #. Ans. #####, #5, #####, #####, and #.
5. Hº, 3, 4, and 3%. Ans. ###, #1, ###, and ####.
I74 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. X-
350. Reduce 3, #, and g to equivalent fractions having the least common
denominator.
OPERATION.
2 || 3. 4. 8
2 || 3. 2. 4
3. 1. 2
º 2 × 2 × 3 × 2 = 24, least common denominator.
# of = 8, hence ºr is the eqtivalent of 3.
# of $ 24 = 18, hence # is the equivalent of #.
§ of = 21, hence # is the equivalent of #.
Eacplanation.—In all problems of this kind, we first find the least common multiple of the
denominators of all the fractions as explained in Article 290, page 159, which is the least common
denominator. Then, having the least common denominator, to find the respective numerators, we
take such a part of the ieast common denominator as the respective fractions are parts of a unit,
as shown in the operation. Or, divide the L. C. D. by the denominator of each fraction and multiply
the quotient by its numerator.
. Before finding the L. C. D., reduce mixed numbers to improper fractions, and the fractions to
their lowest terms.
GENERAL DIRECTIONS TO REDUCE FRACTIONS TO A LEAST
COMMON DENOMINATOR.
351. From the foregoing elucidations, we derive the following general direc-
tions for reducing fractions to the least common denominator.
1. Find the least common multiple of the denominators of all the given fractions.
2. Then to find the respective numerators, take such a part of the least common
denominator as the respective fractions are parts of a unit. Or, divide the L. C. D. by
the denominator of each fraction and multiply the quotient by its numerator.
NOTE.-Mixed numbers must be reduced to improper fractions, and the fractions to their
lowest terms before finding the least common denominator.
PROBLEMIS.
352. Reduce the following fractions to equivalent fractions having a least
common denominator.
*
1. 3, #, and #. Ans. ##, #}, and #.
2. Hºs, +, and #. AnS. ##, ##, and #.
3. #, ##, and #. Ans. #6, #3, and #.
4. I'd ##, ##, and #. Ans. #5, #3, #3, and #.
5. 54, #, 14%, and #. Ans. ***, #3, #, and 4g.
6. ##, 8, #}, and 1. e Ans. ##, *, #3, and #.
7. 34, #, #, and #. Ans. ##, #3, #, and 49.
8. Fºr, 3, 4, and #. Ans. ####, ####, #3, and +}#}.
e () () (e) (C) 2
GT
c-ā-ā-ā-āj-,-
Addition of Fractions.
===N
*—dº ſº-
-*ºr w -uu-
353. Addition of Fractions is the process of adding two or more fractional
numbers of the same kind, or of the same denomination.
354. Like Fractions are those which express like parts of like units or things.
Thus, # yard and # yard, also # and # are like fractions.
*
355. Unlike Fractions are those which express unlike parts of like units or
things, or parts of unlike units or things.
Thus, # of a pound and # of a pound are unlike parts of like units or things,
and # and # are unlike parts of unlike units.
In addition of whole numbers, we learned that we could not add apples to
oranges, pounds to boxes, or units to tens or hundreds ; that we could only add things
that were of the same unit kind.
This same principle maintains in the addition of fractional numbers. We
cannot add halves to thirds, fourths to fifths, etc. We can only add halves to
halves, fourths to fourths, etc.
ORAL EXERCISES.
356. 1. Add +, +, and #. Ans. 14.
SOLUTION.—Since the fractional parts are alike, we have only to add the numerators to obtain
the sum of the fractions. Thus, # -- 3 + # = } = 1%, or 14, Ans.
2. Paid $4 for a grammar and $; for an arithmetic. What did both cost 7
Ans. $1}.
SOLUTION.—Since the halves and fourths are unlike parts of the unit dollar, we cannot add
them in their present form. We must first reduce the 4 to the fractional unit of the fourths, and
by the exercise of our reason and knowledge of numbers, we see that 3 is equal to 3, and 3 added to
# equals #, and # equals $1+, Ans.
3. What is the sum of ; and g {
SOLUTION.—Since the fractional parts are unlike, we cannot add them in their present form.
We must first reduce the # to the fractional unit of eighths; and by the exercise of our reason and
knowledge of numbers, we see that # is equal to #, and # added to § equals º and *-equals 13, Ans.
4. What is the sum of #, g, ºr, and #2 Ans. 24;.
5. Add #, #3, ##, and ##, Ans. 2%.
Mentally add the following fractions:
# -- # = ? # -- # = ? +++3 = ? # + š ++* = ?
# -- # = ? § -- # = ? # +# = ? # + 3 ++’s = ?
# + # = ? # -- # = ? # + 4 + # = ? # + +º, +% = ?
# + # = ? # + 4 = ? # + # ++’s = ? # + 4 ++} = ?
# + # = ? # ++’s' = ?
(175)
176
soul E's PHILOSOPHIC PRACTICAL MATHEMATICS.
357. What is the sum of 3 and #3
OPERATION.
# #
8 9
17 – 1 -5
find the numerical fractional equivalent value of # and # in the unit of twelfths.
reason thus: # = +; and # = nº ; (Write the 8 under the # as a memorandum).
(write the 9 under the # as a memorandum). Tºg and g = +} = 1.1%,
2. What is the sum of ; and #%
3. What is the sum of # and + 3
4. Add {} and #;
358. Add 3, #, and #,
OPERATION.
2 7
# 3
;
16
MEMORANDUM
|UNITS. FRACTIONS.
i 3 ºf 1.
$º-º-º-ºms § 24 24
2#, Ans. 9
Add the following:
2, #, #, 3, and #,
3. I's, #, #,
4. #, #, and #,
359. Add #, #. #, #,
OPERATION.
1 2 3 4 5 7
amºmº ºms ºsmº ºsmºs smºs º-se
2 3 4 5 § 3
UNITS. MEMORANDUMI
IFRACTIONS,
1.
1. I I 3 ; 17
1. * *= *-* * *
tº 4 2 4 3 40
4}# Ans.
4 . 6 • 4
# and #; # and #,
Ans. 1-#.
Explanation.—Since the fractional units are unlike, and since
neither can be reduced to the unit value of the other, We, there-
fore, first reduce both to a common denominator, by multiplying
their denominators together, which gives 12, as the least common
denominator. Having the least common denominator, we next
To do this, we
+ = *; and # = |* ;
Ans.
PROBLEMIS.
Ans. 1:#.
AnS. 1+.
4 1
7 2 °
Ans. 234.
9 5 - 1 2 3 e
Ans. 'IOT 3 1#;
Ea:planation.—In all problems having more than two terms, we
select and add only two terms at a time. Here we select the #
and # and reason thus: # = #, which added to # = ** = 13. We
set the 1 in the column of units and the # in the column of mem-
morandum fractions, and cancel the terms added, } and #. We
now have the #, and # to add, and producing the least common
denominator, 24, we reason thus: # = # and # =#. (Write the
16 under the # as a memorandum). Then # = ºr and # = #.
(Write the 9 under the # as a memorandum). Then # and # =
# =1#a. We write the 1 in the column of units and the ºr in the
Column of memorandum fractions. We then cancel the ##, add the
Column of units and to the sum annex the remaining fraction and
obtain 23'ſ, the answer.
Ans.
Ans.
5
1 7
4}#.
#
, and g together,
Eacplanation.—Selecting the 4 and # to add first, we reason thus:
# = }, which added to # = # = 14. We write the 1 in the column
of units and the # in the column of memorandum fractions and can-
cel the # and #. Next we select the # and 3 and reason thus: # = },
and # = # which added to # = } = 13 = 13. We write the 1 in the
column of units and the # in the column of memorandum fractions,
and cancel the # and #. Then we select the + and the # (in the col-
umn of memorandum fractions) andreason thus: # = } which added
to # = #, We set the # in the column of memorandum fractions
and cancel the # and #. Next we select the # and the # and reason
thus: # = # and # = #, which added to g = i =1#. We write the
1 in the unit column and the # in the column of memorandum frac-
tions, and cancel the # and #. # and # remain to be added; and as
they are not of the same unit value and neither can be reduced to
the unit value of the other, we proceed to reduce both to equivalent fractions of the same unit, or to
the least common denominator, which is 40. Having the least common denominator, we then
reduce the # and # to their numerical equivalent fractional values in fortieths. To do this, we rea-
son thus: # = #, and # =#3; # = f; and # = #3, which added to # = # =1#. We writé the 1 in
the column of units and the #6 in the column of memorandum fractions and cancel the # and #.
* ADDITION OF FIRACTIONS. 177
Lastly we add the column of units and to the sum annex the remaining fraction, and thus produce
445 as the correct result.
NotE.—The foregoing problems illustrate the most easy, rapid, and rational system of adding
fractions known, and as fractions are so indispensable and of so frequent occurrence in practical
life, the principles involved in the system should be thoroughly understood.
In practical work, we would very much shorten the operation by adding several fractions at
once, and mentally performing the most, if not all of the reduction and addition work, without
stating the results. Thus, in the above problem, we would add the 4, #, and £ at once. We can
instantly see that their sum is 'º, or 2+, and without naming or setting the 24, we add to it mentally
the result of 3 and #, which we mentally see is #, or 13, making 33, which are the only figures we
write. Thus all the fractions, except # and #, are added at one mental operation. Then we men-
tally add the sum of # and # by the same process of reasoning as given in the illustration of the
above example, and obtain the correct result, 4+5.
3 4. 4, 17
Add #, 32 ºz. 2 T-
OPERATION.
3 4. A 17 Eacplanation.—Selecting the # and # to first add, we reason
smº sº-ºº ºne *-*. thus: The least common denominator is 20; # of 20 = # and #
4. # 9 24 = }}; then # of 20 = ºr and # = ##, which added to # = ## =
1}, which we set in their proper columns and cancel # and #.
We next select š and #4 and reason thus: The L. C. D. is 72;
UNITS. MEMORANDUM of 72 = *g and # = }}; then ºr of 72 = **, and #4 = }}; and
FRACTIONS. + ## = }} = 1%, which we write in the proper columns, and
cancel the # and the #4. We now have ## and ++ to add, and
1. II. If having obtained 360 as the least common denominator, we rea-
^^ T^ 3 #3, son thus:...g. = ##, and ## = ###; * = 3;u, and +} = #3; #
2 #5 3 29 72 3 6 O + $5 = ### which we write in the proper column and cance
2###, Ans. ## and ##. Then adding the units’ column and annexing to the
sum the final fraction, we produce the answer, 23#.
NOTE.—To contract the work of finding the least common denominator of two fractions, we
see, by inspection, what is the greatest common divisor of the denominators, and then divide either
denominator by it, and, with the quotient multiply the other denominator the product will be the
least common denominator.
2. Add + of ; of 24, 3, #, and # together.
OPERATION.
Statement showing the reduction of the fractions. Explanation.—Here we have compound frac-
3 e
1 2 g 3 5 9 tions and mixed numbers, and before adding,
* , 4 s g wzº we reduce the mixed numbers to improper frac-
* >
Statement showing the result of the reduction and tions, and the compound fractions to simple
the addition of the fractions ones. Then we add the # and #, which are
TNITS. MIEMORANDUMI FRACTIONS. equal to 1 and #; then the 3 and #, which are
3 3 # 9 I 23 equal to ##; then the ## and #, which are equal
1. A s 6 2; § 24 to 1 and ###. Then adding the whole numbers,
20 216 3 529 and annexing the fraction, we have 2#33 as the
correct result.
GENERAL DIRECTIONS FOR THE ADDITION OF FIRACTIONS.
360. From the foregoing elucidations, we derive the following general direc-
tions for the addition of fractions:
1. Select such two fractions as can be the most easily reduced to the same frao-
178 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS. X-
tional unit ; find the equivalent value of both the fractions in this fractional whit,
and add the numerators together; if the sum equals or eacceeds a unit write the unit
in the column of units, and the fractional remainder in the column of memorandum
fractions. If the sum is less than a unit, write the fraction in the column of memo-
ºrandum fractions. Then cancel the fractions added.
2. In like manner select, reduce, and add two more fractions, and thus proceed
wntil all are added.
3. When there are compound fractions, reduce them to simple ones before
adding. When there are mia!ed numbers write the whole numbers in the column of
wnits, cancel them from among the fractions, and then add the fractional numbers.
All fractional ea.pressions should be in their lowest terms before adding.
361. PROBLEMS IN ADDITION OF FERACTIONS.
1. Add #, 3, #, #, #, and Hº, Ans. 4%. 11. $, ºr, and #, Ans. 1%;"
2. Add #, #, and ##, Ans. 23%. 12. 34, 1}, 24, and 4*, Ans. 114.
3. Add #, #, #, and Hº, Ans. 243. 13. #3, #, ##, and ##, Ans. 2 #.
4. Add #, ##, ##, and ##, Ans. 34%, or 3%. 14. 24, 34, 44, and 5, Ans. 1513.
5. Add #, +%, #, and ##, Ans. 3.}}. 15. #, #, #, and #3, * Ans. 14%g.
6. # of # and 2% of § of #, Ans. #4. | 16. 73, 53, and 10}, Ans. 23}}.
7. Add 24, 6%, 5}, and 24, Ans. 17. 17. 44, ##, and ºr, Ans. 14%.
8. §, #, and #5, Ans. 21%. | 18. 14, 34%, 13, and #}, Ans. 21+3.
9. #, #, and #, Ans. 1-#. 19. 3, 1+3, 10%, and 5. Ans. 1834.
10. 11'5, 6%, 18}}, and 25%, Ans. 28%. 20. 1254, 327+, and 25+, Ans. 478+.
21. #3, #, 14%, ##, and ###, Ans. 3+}}.
22. What is the weight of 10 sacks of wheat which weigh respectively: 1544,
1494, 1603, 1573, 1524, 1413, 1633, 1584, 1394, and 1613 pounds? Ans. 15394 lbs.
23. Add + of ; of # and 24 of ºr of 1. Ans. 1 #.
24. Add # of ; of 4 and # of 4 of #. Ans. 23#.
25. How many yards in 8 bolts of domestic, measuring as follows: 40%, 394,
434, 42%, 433, 384; 39%, and 414 yards? Ans. 328}#.
26. 14 bags of coffee weigh as follows: 1624, 1633, 161+, 1644, 1654, 1643,
1653, 1623, 165*, 1643, 1653, 1654, 164++, and 1654 pounds. How many pounds
in all ? --- Ans. 2301.
27. A merchant bought 11534 pounds of rice for $924; 871; pounds of
sugar for $87%; 5803 pounds of coffee for $115g; 240+ pounds of cheese for $434;
and 408; pounds of Graham flour for $183. What was the total number of pounds,
and the total cost of all he purchased ? Ans. 32543 pounds; $357; cost.
28. Add #, #, 4, #, and 3% of # of 14. Ans. 5}}.
29. Add #, 3, #, +}, 1}, and 4 of 3. Ans. 5*.
* ADDITION OF FRACTIONS, I 79
ADDITION OF FEACTIONS BY A NEW METHOD,
362. The following new method of addition of fractions may be used to
advantage when the fractions are large, and the denominators are prime to each
other. *
The principles involved are novel and should serve to awaken increased
interest in the operation of numbers.
363. 1. Add the following fractions: #, #, and #.
OPERATION.
#, #, and #, added by the Explanation.—In the operation of problems by
usual method, gives 2:#. this method, we are governed by the following gen-
eral principles: 1. A proper fraction is less than
unity. 2. A complement of a fraction is the differ-
ence between the fraction and unity, and is found
by subtracting the numerator from the denominator.
OPERATION BY THE NEW METEIOD.
FRACTIONS. UNITS. COMPLEMENTS.
T. * = º — 4 e d
# = 1 9 3. The sum of several fractions is therefore equal
# = 1 — # to as many units as there are fractions, less the sum
4 I of the complements of the fractions.
# = 1 — # In this problem the sum of the units is 3, and
s=ºs amºe the sum of the complements is ###, and the differ-
3 — ### = 25%, Ans. ence is 2 ºr, the º: 3602
e 02 º
SECOND OPERATION BY THE NIEW METHOD,
FRACTIONS. UNITS. COMPLEMENTS.
# = 1 — #
# - 1 — ; ; 360 the C. D. x the 3 units = 1080 the sum of the
4. == 1. tº- l units in the C. D. unit.
5 5
3
Then # = 80 *
# of 360 = } = 135 K = 287 the sum of the complements.
5. J - 72 *.
1080 – 287 = 793 = the sum of the fractions. 793 -i- 360 = 2;, Ans.
NOTE.-In case some of the denominators are composite and not all prime to each other, the
L. C. D. may be found and the operation performed as above.
PROBLEMIS.
Add the following problems by the above method:
9 1 6 2 3 #5 8 1 O 7
2. Hºr, #, and ##. 4. ##, ##, 3, and ###.
1 2 1 1 3 1 _9.9 205.
3. ##, ##, #, and #}. 5. Hº, and ###.
subtraction Of Fractions.
=N
º-º-º-Lº-
364. Subtraction of Fractions is the process of finding the difference
between two fractional numbers of like units and of like parts.
365. The Principle is, that fractions can be subtracted only when they
express like parts of like units Or When they have a common denominator.
ORAL EXERCISES.
NOTE.-See Oral Exercises pages 167 and 168.
366. 1. One unit of any kind equals how many 3's?
# taken from # leaves how many 3's 2
One unit of any kind equals how many 3's 2
# from # leaves how many #’s?
# from 4 leaves how many #’s?
# from 4 leaves how many #’sº
# from 4 leaves how many 3's 2
i
367. Answer by mental work the following numerical questions:
#—# = ? * — ; = ? # — # = ? # -- #– # = ?
# — # = } # —# = ? 1 —# = ? # + 3–4 = ?
# —# = ? #—# = ? 34–13 = ? 24 — 3 + 4* = ?
PROBLEMIS.
1. What is the difference between +; and ++ 7 Ans. #.
OPERATION.
## Explanation.—In this example the denomina-
ll. OI’ 4-3 – 44 = + Ans.
** 1 5 T 1 5 57 tions being the same, we have but to take H.
#, Ans. from +3 to obtain the correct difference.
2. What is the difference between # and #3 Ans, go.
OPERATION.
Ea:planation —Here we see that the fractions are not of like
# = 15 or ; # units, and that neither can be reduced to the unit of the other,
# = 16 15 16 therefore, we must reduce the fractions to a common denomi-
nator. By inspection, and in accordance with the principles
—l. ºmº- 1 as explained in the first four problems of addition of fractions,
2 0 } Ans. 3 O2 Ans. We see that the least common denominator is 20; then, that #
are equal to #3, and that # are equal to #3, and that the differ-
ence is go.
** * *
(180)
¥ SUBTRACTION OF FKACTIONS. I8 I
3. From 28; take 74. Ans. 214.
OPERATION.
28; Ea:planation.—In this problem we first observe that the
7; # may be reduced to 8ths, and by the exercise of our
reason we see that it is equal to #, which taken from 3
amºs- leaves #; then we find the difference between the whole
21; Ans. numbers as in simple subtraction.
4. What is the difference between 37; and 12#3 - Ans. 24+}.
OPERATION.
72 99 Explanation.—By inspection, we here see that the frac-
373 = 27 tions belonging to the whole numbers are not of like units
12# = 56 or of the same denominator, and that neither can be
reduced to an equivalent fraction of the same unit as the
-> other, and, therefore, we must reduce both to fractions
24}#, Ans. of like whits or of a common denominator, before we can
subtract; and by multiplying together the denominators, we produce 72 as the least common
denominator, which, for convenience, we write below the fractions, and by the same reasoning as
given in addition of fractions, we see that § are equal to #3 and that § are equal to ##, which, for
convenience, we carry to the right of the respective fractions, and to economize time, we write only
the numerators. ... We now observe that the upper fraction, belonging to the greater number, is less
than the lower fraction, belonging to the lesser number. Therefore, before we can subtract the
fractions, we must add 1, reduced to 72ds, to ##, which gives us #3. We now subtract # from ## and
have a remainder of # as the fractional part of our answer. . We now add 1 to the subtrahend, to
compensate for the 1, previously added to the minuend, making it 13, which we subtract from 37
and have a remainder of 24, which we write below the line and thus complete the operation.
GENERAL DIRECTIONS FOR SUBTRACTING FRACTIONS.
368. From the foregoing elucidations, we derive the following general direc-
tions for subtracting fractions:
1. Reduce the given fractions to equivalent fractions of like fractional units, or
of the least common denominator; then take the difference of their numerators and
write the same over the common denominator.
2. When there are mia!ed numbers, subtract the fractional parts first and then
the whole numbers.
PROBLEMS IN SUBTRACTION OF FRACTIONS.
369. 1. What is the difference between +4 and #3 Ans. #.
2. What is the difference between # and #3 Ans. #.
3. What is the difference between 5% and 3; ? Ans. 23.
4. What is the difference between 7 and 3+6 & * Ans. 34%.
5. What is the difference between 23# and 14? Ans. 9%
What is the difference between the following numbers:
6. # and #, Ans. #. 9. § of # and # of #, Ans. #.
7. # and #, Ans. #6. 10. 84 and 3%, Ans. 4}}.
8. § and #, Ans. #'s. 11. 12; and 94, Ans. 2%.
I 82 soule's PHILOSOPHIC PRACTICAL MATHEMATICS. XF
12. Hºf and #, Ans. Tºg. 17. 25% and 945, Ans. 16#.
13. 4 and #, Ans. #. 18, 9 and 35%, Ans. 5}}.
14. # and #5, Ans. ##. 19. Hº, and 3Fs, Ans. 3 #.
15. 2; and 13, Ans. 1+. 20. 7+; and 4, Ans. 3+3.
16. 9; and 23, Ans. 6+5. 21. 31; and 173, Ans. 13%.
22. From 6% of + + 13; take # of # of 15—# of #. Ans. 13%.
OPERATION INDICATED.
* x * = 2; 2 + 13% = 15%; # × 3 × 15 = ** = 3; ; 3%– (#x #) = 2;.
Ans. 15%—2; = 13%.
23. From 84 + 63 — ºr take # of # of 3 of 13 + 2}. Ans. 10#3.
24. E. J. Jacquet had $38#. He gave $2} for a pair of Indian clubs, $5; for
books, $14 for a drawing board, and $3 for ink and pencils. How much had he left?
Ans. $283.
25. W. G. Crena had $74 and his friend gave him $4 more; R. C. Bush had
$16; and he spent $5g. How much more has R. C. Bush than W. G. Crena?
Ans. $24.
26. M. Gundersheimer bought 2 bags of coffee, each weighing 1634 pounds.
He sold 27+ pounds, 503 pounds, 87+ pounds, and 45% pounds. How many pounds
has he left? Ans. 1163 lbs.
27. S. Delerno bought 75% gallons of molasses. He used 44 gallons, lost by
leakage 23 gallons, and sold 22# gallons. How much has he left? Ans. 46% gallons.
28. What is the difference between a dozen times 6, plus 64, and 6 times a
dozen minus one dozen and a half dozen 3 • Ans. 244.
29. J. Astredo bought 6 chests of tea weighing 38%, 42%, 444, 413, 394, and
43% pounds. He sold 120g pounds and used 5% pounds. How many pounds has
he on hand? Ans. 123}.
30. J. Birba owned the steamer Isabel, and sold # of it. What is 4 of his
present interest ? AnS. #.
31. From the sum of 64 and 83, take the difference between 14; and 94 %
Ans. 10%.
32. What number is that to which if 16} be added, the sum will be 44; ?
Ans. 27%.
33. P. H. Weiss bought 3 of ; of a vessel and sold # of # of his share. How
much of the whole vessel has he left 2 AnS. #.
34. W. Van Benthuysen owned # of the steamer Natchez. He sold to
J. Maier + interest in the steamer, and to Leo. Winner, 4 of his remaining interest.
What is the present interest of each in the boat Ž
Ans. Van Benthuysen, §; Maier, #; and Winner, 3.
35. F. Brinker and Henry Pike, were each # owners of a broom and brush
factory. F. Brinker sold 4 of his interest to G. Wagner, and then # of his remaining
interest to Henry Pike, who subsequently sold # of # of his whole interest to
W. Lacoume. What is the present interest of each ownerº
Ans. F. Brinker, $; G. Wagner, #; H. Pike, # and Wm. Lacoume, #.
sultiplication of Fractions.
=N
Aº-sº
-*ºr v---wº-w
370. Multiplication of Fractions is the process of multiplying when one or
both of the factors contain fractional numbers.
`NotE.—See the Philosophy of Multiplication, pages 77 to 80.
In the multiplication of simple numbers, we saw that the result of multipli-
cation operations was increasing, but in the multiplication of fractions, when the
multiplier is less than a unit, the result is decreasing. This is evident from the fact
that multiplication is the process of repeating the multiplicand as many times as
there are units in the multiplier, and, therefore, when the multiplier is less than a
unit, the multiplicand will be repeated only a part of a time, or such a part of itself
as the multiplier is a part of a unit.
PRACTICAL DEFINITION OF MULTIPLICATION.
371. To elucidate the principles of the subject and render clear the reasoning,
We present our first questions in denominate numbers; and to aid still further in
comprehending the work, we give the following practical definition of multiplication.
372. Multiplication is that operation in the practical computation of
numbers, of finding the cost of either a part of one, or of many pounds, yards,
barrels, etc., when the cost of one pound, yard, barrel, etc., is given. On the prin-
ciple or fact embraced in this definition, we base our reasoning for the solution of
every practical question that can possibly be presented in multiplication, either of
Simple numbers or of fractions.
METHOD OF REASONING.
*
373. Considering the foregoing, we see that in all multiplication questions
of a practical nature, we must necessarily reason from one, or unity, to a part of one
or many. Thus, if 1 pound cost 50%, 4 of a pound will cost 4 part of it; and if 1
yard cost $2, 3 yards will cost three times as much, or 3 times $2.
In the Solution of questions in abstract numbers, we apply the same system
of reasoning without naming the factors, and thereby avoid all of the arbitrary
rules given in other arithmetics of the day.

(183)
184 soule's PHILOSOPHIC PRACTICAL MATHEMATICs. *
STATEMENT LINE.
374. In all problems where we have both multiplication and division work
to perform, we use a vertical line, to facilitate the operation.
The right hand side is the increasing or multiplication side, and the left hand
side is the decreasing or division side. At the top of the right hand side, we first
write the premise of problems, or the number to be multiplied or divided, or that
which the conditions of the question require the answer to be in. Then, continuing
with the operation, the increasing and decreasing numbers are written respectively
upon the increasing and decreasing sides. But be it remembered, we never place a
number on either side after the premise or nature of the answer is written, without
giving a reason therefor. In practice, the reason is instantly seen by the mind and
directs the placing of the figures. When a problem is fully stated on this line,
cancellation is applied and the operation is performed more readily than by any
other manner of statement.
NoTE.—See page 161 for a full elucidation of the Statement Line.
º ORAL EXERCISES.
375. 1. What will 6 pounds cost at 5% per pound 7 Ans. 30%.
SOLUTION.—According to Article 152, page 77, we reason as follows: 1 pound costs 5c. Since
1 pound costs 5c., 6 pounds will cost 6 times as much, which is 30c.
SOLUTION STATEMENT.
e g º
2. What will 6 pounds 2 | 11 Reason.—one pound costs 53c.
cost (a) 54% per pound 2 § 3 Since 1 pound COStS 5ie, or *c.,
* 6 pounds will cost 6 times as
Ans. 332. | 33g, Ans, much, which is 33c.
SOLUTION STATEMENT.
Reason.—One pound costs 5c.
3. What will 64 pounds 5 Since 1 pound costs 5c. 4 pound
cost at 52 per pound? 2 13 * will cost # as much, and 4% lbs.
Ans. 32.4% - will cost 13 times as much,
o º 32% Ans. which is 324c
SOLUTION STATEMENT.
Explanation and Ireason.—Here
4. One pound costs 54%. ; # we see by considering the ques
What Will 6 # pounds COSt tion, that 5% c. is the number to
*= - - sº be multiplied. Accordingly, we
at the same rate % 4 || 143 place the same at the top of the
º line; and to facilitate
the work, to free the operation of
º e 35% 2 Ans. jractions—we reduce #. 5}c. to
halves, making *c. the denominator of which we place on the degreasing side, and the numerator
on the increasing side, of the statement line. We then reason as follows: Since 1 pound costs 42-c.,
# of a pound Will cost 4 part of it, and as this conclusion is a decreasing one, we write the 2 on the
decreasing side; then, since 4 of a pound costs the result of the statement thus far made, *, 6%
reduced, will cost 13 times as much, which, because the conclusion is an increasing one, we writé on
the increasing side, and thus complete the reason and the statement, which worked out, gives the
answer, 35% c.
* MULTIPLICATION OF FRACTIONS, I85
SOLUTION STATEMENT.
g * Reason.—One pound costs 5%
5. One pound costs 54%. } § 4 or *c. Since 1 pound costs lºc.
What will 64 pounds cost -** -i tºº + pound will cost the 4th part
at the same rate % 3 || 100
and 34 pounds will cost 25 times
33.4%, Ans. as much.
THE REASON, WHY, AND wherEFORE, CONTINUED.
Question. How do you know that if 1 yard costs * cents, # of a yard will
cost the 4th part?
Answer. By the exercise of my judgment—by the use of the reasoning
faculties of the mind.
Question. What do you mean in this connection, by judgment 3
Answer. The conclusion arrived at by the operations of the mind after duly
considering the premise, the facts, and the conditions of the problem.
Question. What do you mean by premise or premises 3
Answer. The proposition, declaration, truth, or fact, which is asserted as
the basis or predicate of a question. In this problem the premise is, one pound costs
5% or * cents.
Question. Why will 64 pounds cost 64 times as much as 1 pound 3
Answer. Because 6+ is six and one-fourth times as much as 1,
Question. What kind of reasoning is the foregoing?
Answer. Analogical and axiomatical. It is Analogical, because there is
analogy, relationship, or likeness existing between the cost of 1 pound and the cost
of 64 pounds. It is Axiomatical, because the premise and question considered, the
conclusion is self-evident.
Question. What is reason 3
Answer. The faculty or power of the human mind by which truth is dis-
tinguished from falsehood, right from Wrong, and by which correct conclusions are
reached by considering the logical relationship which exists between the premises,
the facts, and the conditions of particular statements and questions.
SOLUTION STATEMENT.
- Reason.—One yard costs 163c.
6. At 1632 per yard, 3 || 50 Since 1 yard costs ºc., 4 of a
º 2 || 45 ard will cost the # part d
what will 224 yards cost 3 y part, an
# y * yards will cost 45 times as
$3.75, Ans. much.
SOLUTION STATEMENT.
g Ičeason.—One bushel costs 664c,
7. What will 4; bush- 2 | 133 Since 1 bushel costs 134c., 4 of a
els cost at 6632 per bush- 4 || 19 bushel will cost the 4 part, and
el ? wº-mº i =ºmºsºme * bushels will cost 19 times as
$3.15%, Ans. much.
I86 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. 4×
Question. How do you know this?
SOLUTION STATEMENT.
$ Reason.—One dozen costs $4.
8. Chickens are selling 4 . Since 1 doz.costs ºf dollars, , of a
at $3; per dozen. What 6 || 25 dozen will cost # part, and *
will 4+ dozen cost 3 gº | * dozen will cost 25 times as much.
$153, Ans.
Questions. 1. How do you know this? 2. What do you mean by judgment
SOLUTION STATEMENT.
Reason.—One gallon costs 42}c
9. What will 8 gallons 2 | 85 Since 1 gallon costs * cents, 8
cost at 424 cents per gal- 8 gallons will cost 8 times as
lon 3 | $3.40, Ans. much.
SOLUTION STATEMENT.
Reason.—One pound costs 12c.
12 Since 1 pound costs 12 cents, #
2 | 29 º of a pound will cost the # part,
* - sº and * pounds will cost 29 times
$1.74, Ans, as much.
SOLUTION STATEMENT. º
$ Reason.—One dozen costs $#.
11. What will # of a 4 || 3 * Since 1 dozen costs # of a dol-
g 3 || 2 lar, of a dozen will cost the
dozen cost at $# per dozº * ?
10. What will 144 lbs.
cost at 12 cents per pound?
gº tº sº # part, and 3 of a dozen will cost
$3, Ans. 2 times as much. .
SOLUTION STATEMENT.
Reason.—One day’s work is
12. A mechanic Works 4 13 worth $3+. Since 1 day’s work
153 days and receives $3} 2 || 31 is worth ** dollars, # of a day’s
per day. How much money s 403 work is worth # part, and **
is due him 3 days’ work is worth 31 times as
$503, Ans. much.
Questions.—1. How do you know this? 2. What do you mean by judgment 3
3. What is the use of the statement line? 4. What kind of numbers do you place
on the right hand side 2 5. What kind on the left # 6. What do you do before you
place a number on either side? Answer.—Give the reason for so doing. 7. What
do you mean by reason ? 8. What is Cancellation ? 9. Why do you use Cancel-
lation ?
GENERAL DIRECTIONS FOR MIUILTIPLYING FRACTIONS.
376. From the foregoing elucidations, we derive the following general
directions for multiplying fractions?
1. Write on the upper right hand side of the statement line the premise of the
problem ; or the number which is to be multiplied or divided ; or the number represen-
Yºr MULTIPLICATION OF FRACTIONS. 187
ting the answer. Then, reasoning from ONE, OR UNITY, To A PART OF ONE or to
many, write the other numbers upon the multiplication or the division side of the line,
according as the conclusion is increasing or decreasing.
2. Mia!ed numbers should be reduced to fractional eacpressions, and the reason
given for writing both the numerator and the denominator.
PROBLEMIS.
377. What will # of a yard cost, at # 2. What will 584 pounds cost, at 16#
of a dollar per yard 3 Ans. $4. per pound 3 Ans. $9.75.
OPERATION OPERATION
$ 2
4.
§ Write the reason.
.*
25
; # 39 Write the reason.
:
| $9.75, Ans.
| $#, Ans.
3. What will 3; dozen cost, at $3% per dozen ? Ans. $12}.
NOTE.—Make the statement and write the reason.
4. What will 53 bushels cost, at 154% per pint %
OPERATION.
Explanation.—Reducing and placing the 15% c. on the state-
2 31 ment line, we reason as follows: Since 1 pint costs *c., 2
pints or a quart will cost 2 times as much ; and if 1 quart,
2 costs the result of the statement now made, 8 quarts, or
* a peck, will cost 8 times as much; and if a. peck costs the
4 = 124 result of this statement, 4 pecks, or a bushel, will cost 4
8 || 43 times as much ; and if a bushel costs the result of this
- º # of a ºhel will cost the # part of it, and 4*
ºmº- Cost 43 times as much.
$53,32, Ans.
5. What will 50+ pounds of tea cost, at 1032 per ounce?
OPERATION.
Ea:planation.—Here we reduce and place the 10}c. on the
2 21 line, and reason thus: Since 1 ounce costs *c., 16 ounces or
2 1 pound will cost 16 times as much, and since 1 pound costs
16 the result of this statement, + part of a pound will cost +
4 || 201 part of it, and *%+ will cost 201 times as much. This com-
-ºms pletes the reasoning and statement, which worked, gives
$84,42, Ans. $84.42, answer.
Make the solution statement and write the reason for the following problems:
6. What will 24 pounds cost, at 94% per pound 3 Ans. $2.28.
7. What will 143 dozen cost, at $5 per dozen ? Ans. $73}.
- 8. What will 6 dozen and 7 chickens cost, at $4.87; per dozen 3
Ans. $32,093.
9, What will 42 pounds and 11 ounces of butter cost, at 2232 per pound?
Ans. $9.60+}.
SOLUTION STATEMENT.
Explanation and Reason.—As usual, we here place
2 || 45 the premise, the price of one pound, on the statement
6 | 683 line, and reason as follows: 1 pound, or 16 ounces,
cost #c. Since 1 pound costs *c., 1 ounce will cost,
the 16th part and 683 ounces (which is 42 pounds and
$9.60+}, Ans. 11 ounces) will cost 683 times as much.
1.
188 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. ¥
10. What will 193 pounds cost, at 1832 per pound? Ans. $3.60;#.
11. What will 253 yards cost, at 17% per yard 3 Ans. $4.55%.
12. What will 113 yards cost, at 1242 per yard? Ans. $1.42 #.
13. What will 213 yards cost, at 1642 per yard 2 Ans. $3.58%.
14. What will 14+ pounds cost, at 1242 per pound? Ans. $1.74%.
15. What will 31; pounds cost, at 11.3% per pound? Ans. $3.46%.
16. What will 13 pounds and 9 ounces of tea cost, at 87%g per pound?
Ans. $11.86%.
TO MULTIPLY ABSTRACT FRACTIONAL NUMBERS.
378. 1. Multiply 84 by 33.
OPERATION.
25 Ea:planation and Reason.—In this problem, both factors
4 Iš 5 are abstract numbers. Hence we cannot give the same
- analogical reasoning as we gave in the foregoing problems
125 º where the factors were denominate, or concrete numbers;
although were we to do so, the result, so far as the figures
are concerned, would be correct. We therefore reduce and
31%, Ans. place the 8%, the number to be multiplied, on the statement
li.e., and reason as follows: Axiomatically 1 time 8; or * is º. Since 1 time * is º. 4 time *
is + part of *, and * are 15 times as many.
Or, by analysis, thus: 1 time * is %, Since 1 time * is º, + time * is the # part = }},
and * are 15 times #3 = ** = 314, answer.
* NOTE.—It will be observed that we found our reasoning upon one, which is the basis of all
numbers, as explained in Article 155, page 80. Multiplication is the process of repeating one number
as many times as there are units—ones—in another. And it is self-evident that one time any number
is equal to the number. This self-evident conclusion is the premise for all questions in multiplica-
tion of abstract numbers.
2. Multiply § by 3, Ans. }.
SOLUTION STATEMENT.
4 || 3 Reason.—One time # is #. Since 1 time # is #, # time # is the
3 2 # part, and # times # is 2 times as many, which is #.
* - Or, by analysis, thus: Since 1 time # is #, time # is the #
#, Ans. part of # = 1%, and # times # is 2 times ºf = |*, or #, answer.
3. Multiply +} by $, and write the reason, Ans. Hºs.
SOLUTION STATEMENT.
4. Multiply 6 by É, and 6
: pº . 3}.
Write the reason. 8 || 5 Ans. 3
SOLUTION STATEMENT.
6 || 5
5. Multiply # by 9, and 9
write the reason. 7 * - Ans. 73.
|
6. 5; by ÉÉ, and write the reason. Ans. 24.
7. $4 by ##, and write the reason. AllS. #.
6
* MULTIPLICATION OF FRACTIONS, * 189
MISCELLANEOUS PROBLEMS IN MULTIPLICATION OF FERACTIONS.
379. 1. What will 16 yards cost, at 1432 per yard? Ans. $2.36.
2. What will 23# pounds cost, at 35% per pound? Ans, $8.31#.
3. What will # of a yard cost, at $3 per yard? Ans. $#.
4. What will 3 a yard cost, at $4 per yard? Ans. $4.
5. What will 84 pounds cost, at 74% per pound? Ans. 63%.
6. What will 103 pounds cost, at 94% per pound? Ans. 99 Hºg.
7. Multiply $4 by 12, Ans. 54.
8. Multiply 11 by Tºr, Ans. 1.
9. Multiply fºr by 11, e Ans. 1.
10. Multiply 13 by ºr, Ans. 94%.
11. Multiply 105 by gº, e * Ans. 12.
12. Multiply 136 by #, Ans. 44%.
13. Multiply 12 by 31#, Ans. 382.
14. Multiply # by #, AnS. ##.
15. Multiply 114 by 13, y Ans. 18%.
16. Find the value of #, of $, of #, of #, of 4, Ans. #.
17. What is the product of #, #, #, and #3 Ans, ºs.
18. What is the product of 13, #, 2, and 54% e Ans. 11}.
19. What is the product of # of 2+ by # of 74% Ans. 1 #.
20. What is the product of 124 multiplied by 54 times 63%. Ans. 4641%.
21. At +} of a dollar a pound, what will # of a pound of tea cost 3
Ans. *F of a dollar.
22. What will 5; dozen buttons cost, at gº of a dollar per dozen 3
ºs. Ans. # of a dollar.
23. What will 43 yards cost, at 442 per yard? Ans. 20%.
24. What will 143 dozen cost at $5 per dozen ? Ans. $734.
25. What will 154 pounds cost, at 1042 per pound 7 Ans. $1.62%.
26. What will 40g pounds cost, at 22#g per pound? Ans. $9.19%r.
27. What will 2812; gallons cost, at $4.50 per gal? Ans. $12656.25.
28. What cost 471; gallons, at $33 per gallon ? Ans. $1592:3.
29. Sold 9378524 pounds of cotton at 14% per pound. What did it amount to?
Ans. $135695.53%.
30. If a man earns $2; in a day, how much will he earn in 16; days?
Ans. $414.
31. A contractor pays $1+ per day for labor, and he has 370 men employed for
six days. How much money will it take to pay them 3 Ans. $2775.
32. P. Machray paid +% of a dollar for a book, and for paper # of the cost of
the book. How much did he pay for the paper? Ans. $3.
I90 soul.E's PHILOSOPHIC PRACTICAL MATHEMATICs. 4×
33. Distillers of the essence of rose have determined, by experience, that it
requires 48000 pounds of rose leaves to make or distill one pound of the ottar of
roses. How many pounds of rose leaves will it require to distill 50% pounds of the
Ottar of roses? - Ans. 2442000 pounds.
OPERATION INDICATED.
34. If one pound costs a
cent and a half, what will 2 3 e
254 pounds cost? 2 || 51 Write the reason.
Ans. 3842. | 38+2. wº
35. G. V. Hooper owned # of the steamer Katie, and sold # of his share to
Maschek. What part of the whole steamer did he sell? Ans. §.
36. E. Garner can work the problems in this book in 43 months. How many
months would it take him to work 3 of them? Ans. 3% months.
OPERATION.
MOS. e Reason.—He can work all the problems in
4 || 19 *months. Since he can York all the prob-
3 || 2 * r. l.9. = 34. tº lems in *months, to work"; of the problems
Or, "4 x # 3}, Ans he would require's part as many months, aná
*- to work #, he would require twice as many
3%, Ans. months. "
37. W. J. Kearney paid $# for 1 gallon of molasses. What is # of a gallon
worth at the same rate 3 Ans. $3.
38. What will 74 boxes of raisins cost, at $24 per box : Ans $16%.
39. On one occasion at the New Orleans Opera, 4 of the ladies and gentlemen
present were French; # of the remainder, American; # of the remainder, German;
and the others were of different nationalities. What part were Americans, what
part were Germans, and what part were of different nationalities?
Ans. # Americans, #3 Germans, and # of different nationalities.
40. C. Reynolds owned g of a plantation and sold # of his share to D. C.
Williams, who sold # of what he purchased to Frank Soulé, who sold # of what he
purchased to L. B. Keiffer. What is Keiffer's share in the plantation ? Ans. #'s.
41. W. Weiss owned # of 2000 acres of land and sold # of his share to H.
Marsden, who sold # of what he purchased to J. T. Finney. How many acres have
each 3 Ans. W. Weiss, 400; H. Marsden, 450; and J. T. Finney, 750 acres.
MENTAL CONTRACTIONS IN MULTIPLICATION OF FERACTIONS.
NoTE.—See pages 88 to 120, for Contracted Methods of Multiplication
380. In problems of multiplication of fractions in which the factors are small,
the operation may be performed mentally, or with but few memorandum figures,
without reducing or making the statement as above, and the result produced almost
instantly. -
1. Multiply 124 by 8+, Ans. 103}.
OPERATION.
12; Ea.planation. 8 × 12 = 96 + (4 of 12 = 3) => 99, -- (# of
8#. 8 = 4) = 103, which we set in the product line. Then, + of
# = }, which is annexed to the 103, and completes the an-
103}, Ans. SWeI’.
CONTRACTIONS IN MULTIPLICATION OF FRACTIONs. I9 I
2. What will 123 pounds cost, at 9% per pound 3 Ans. $1.23%.
OPERATION.
9;
12; Explanation. 12 × 9 = 108 + (# of 9 = 6) = 114, + (#
$1.23%, Ans.
3. Multiply 13; by 84,
OPERATION.
13; #
8%
1.16%, Ans.
4. Multiply 113 by 84,
OPERATION.
11; ;
8# 3
964, Ans.
Multiply 124 by 44, Ans.
Multiply 94 by 63, Ans.
Multiply 54 by 54, Ans.
Multiply 11% by 12 ºr, Ans.
i
Multiply 10% by 15, Ans.
of 12 = 9) = 123, -ī- (§ of # = |* = }) = $1.23%, answer.
Ans. 116%.
Explanation. 8 × 13 = 104, -i- (# of 13 = 6}) = 110, with
# reserved; + (# of 8 = 6) = 116, which is written in the prod-
uct line. Then # of # = # -- the # reserved = }, which is an-
nexed to the 116 and completes the answer.
Ans. 96%.
Explanation. 8 × 11 = 88, + (; of 11 = 5}) = 93, with 4
reserved; + (# of 8 = 2+) = 95, with # reserved; (then 4 of +
= }, + + reserved = # + # reserved = }=1}, which is added
to the 95 and completes the answer.
PROBLEMIS.
53%. , 10. Multiply 243 by 93, Ans. 2394.
613. 11. Multiply 37% by 25%, Ans. 948#.
160}#. 12. Multiply 714 by 524, Ans. 37274.
28%. 13. Multiply 3704 by 434, Ans. 160244.
142*. º
PRACTICAL APPROXIMATIVE CONTRACTIONS.
TO MULTIPLY ANY FRACTIONAL NUMBERS TO THE NIEAREST UNIT.
381. 1. Multiply 9% by 84.
OPERATION.
9% Explanation.—In this practical system of contraction, we
8; first multiply the whole numbers, and retain in our mind the
product, to which we mentally add, first the product, to the
nearest unit, of the multiplicand by the fraction of the multi-
77, Ans. plier, and then the product, to the nearest unit, of the multi-
plier by the fraction of the multiplicand, and set the result in
the product line, the same being the practical answer. -
In this example, we see that the product of 9 and 8 is 72; that the product of 9 by the + to
the nearest unit is 2; that the product of the 8 by the # to the nearest unit is 3, and that the sum
of the three products is 77.
2. Multiply 104 by 7%.
OPERATION. *
10} Laplanation.—Here we see that the product of the whole
7; numbers is 70; that of 10 to the nearest unit is 1, which added
to the 70 makes 71; then that the # of 7 to the nearest unit is 2,
*º which added to the 71 makes 73, the practical answer; 733, is
73, Ans. the exact result.
I92 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. †
3. What will 15; pounds cost, at 1342 per pound 3
OPERATION.
15; Earplanation. 15 × 13 = 195, -i- + of 15 to the nearest unit
13# is 4, - 199, -i- # of 13 to the nearest unit = 10, − 209, as the
$2.09, Ans. practical result. The exact result is $208}}.
4. What will 253 yards cost, at 1732 per yard 2
OPERATION.
25; Explanation. 25 × 17 = 425, -H # of 25 to the nearest unit
17; is 19 = 444; + 3 of 17 to the nearest unit = 11, = $4.55, the
practical answer. The exact result is $4.551.
$4.55, Ans.
NoTE.—In multiplying the 25 by #, we gained #c., and in multiplying the 17 by #, we lost #c.;
this lºc, excess of loss, and the loss of #c. by reason of not multiplying the fractions, account for
*gc. deficit in the answer. In practice, it is easy to see whether the loss or gain, by reason of the
contractions, exceeds a unit; and if so, to increase or decrease the result accordingly.
With large numbers, the regular system, as first presented, under the head of multiplication
of fractions, is preferable to this approximative method.
-
- PROBLEMS.
5. Multiply 94 by 64, Ans. 593.
6. What will 113 yards cost, at 12% per yard 3 Ans. $1.42%.
7. What will 213 yards cost, at 1642 per yard 3 Ans. $3.58g.
8. What will 83 pounds cost, at 162 per pound? Ans. $1.40.
382. TABLE OF THIRTY-SECONDS AND THEIR EQUIVALENT VALUES IN DECIMALS.
Fractions. Decimal IFractions. Decimal Fractions. Decimal Fractions. Decimal
Lquivalent. e Equivalent. IEquivalent. Equivalent.
# = .03125 ºr .28125 |}} = .53125 | # = .78125
# = Tº = .0625 # = ſº = .3125 # = # = .5625 }} = +3 == .8125
ºr - .09375 |}} = .34375 |}} = .59375 3% = .84375
32 = .15625 |}} = .40625 |3} = .65625 | }} = .90625
† = * = .1875 # = 15 = .4375 # = +} = .6875 % = }} = .9375
** = .21875 $3 = .46875 |3} = .71875 |3} = .96875
& = *; - + = .25 #}=#=#=#= .5 #=}}=}=#= .75 #=}{=}=#=}= 1
383. TABLE SHOWING THE PRICE OF A SINGLE ARTICLE WHEN THE PRICE PER
DOZEN IS GIVEN.
Per Doz. Per Piece. Per Doz. Per Piece. Per Doz. Per Piece. Per Doz. Per Piece.
25c. = 21; c. $3.75 = 31+c. $7.25 = 60% c. $10.75 = 89.3 c.
50c. = 4}c. 4.00 = 33}c. 7.50 – 62}c. 11.00 = 91}c.
75c. = 6+c. 4.25 = 351%, c. 7.75 = 64 ºc. 11.25 = 93#C.
$1.00 = 8#c. 4.50 = 37.4c. 8.00 = 66;c. 11.50 = 95;c.
1.25 = 10; c. 4.75 = 39%;c. 8.25 = 68}c. 11.75 = 97}}c.
1.50 = 12; c. 5.00 = 41}c. 8.50 = 70;c. 12.00 = $1.00
1. 75 = 141;c. 5.25 = 43#c. 8.75 = 72}}c. 12.25 = 1.02*, c.
2.00 = 16; c. 5.50 = 45;c. 9.00 = 750. 12.50 = 1.04% c.
2.25 = 18#c. 5.75 = 47}}c. 9.25 = 771, c. 12.75 -- 1.06+c.
2.50 = 20; c. 6.00 = 50c. 9.50 = 79;c. 13.00 = 1.08% c.
2.75 = 22+}c. 6.25 = 52; 2 c. 9.75 — 81+c. 13.25 = 1.10% c.
3.00 = 25c. 6.50 = 54;c. 10.00 = 83; c. 13.50 = 1.12% c.
3.25 = 27 ſºc. 6.75 = 56+c. 10.25 = 851% c. 13.75 = 1.14 'gc.
3.50 = 29;c. 7.00 = 584c. 10.50 = 87}c. 14.00 = 1.163c.
º: CONTRACTIONS IN MULTIPLICATION OF FRACTIONS. I93
PECULIAR CONTRACTIONS OF MULTIPLICATION OF FERACTIONS.
384. The preceding problems, and the explanations given, fully illustrate
the work, or the general principles of contraction of fractions, without regard to
any peculiar combination of numbers.
The principles upon which the work is based are so few and simple, the
operations So short and easily performed, and the practical advantages of the work
So great, that We specially commend it to the earnest attention of all classes of com-
mercial men and accountants.
In the following methods of contraction, the process depends somewhat on
the peculiar combination of numbers, and hence, although shorter than the first
contractions, is less valuable.
TO MULTIPLY NUMBERS OF TWO OR MORE FIGURES EACH, WHEN
ONE OR MORE OF TEIE RIGHT HAND FIGURES CONSIDERED
As TENS, HUNDREDs, ETC., MAY BE REDUCED TO
FRACTIONS OF HALVES, FOURTHS, OR EIGHTHS.
385. 1. Multiply 425 by 875.
OPERATION.
4+ hundreds. Explanation.—In this problem, we reduce the units' and
8; hundreds. tens' figures of both the multiplicand and multiplier to
fractions of + and # respectively and thus obtain 4+ and
*-mºmº-º-º- 8% to be multiplied together, which operation we perform
371875, Ans. according to the work elucidated in Problem 1, page 190,
and obtain a product of 37*. Then we reduce the frac-
tional part, º, of this product, to a whole number, and annex it to the integral part of the prod-
uct; 3 = 1875, which annexed to 37 gives 371875, the correct product.
This nº is reduced to a whole number for the reason that it represents ºr of 10000; and it
represents this for the reason that we first reduce four figures, two in each factor, to fractions.
NOTE.-See Table of Thirty-Seconds, on page 192, for the decimal equivalents of fractions
from ºf to ##.
2. Multiply 850 by 450.
OPERATION.
84 Explanation.—By reducing the units’ and tens' figures
4; of both numbers as explained in the preceding problem,
* we have 8+ and 44 to be multiplied together. The prod-
*-mºmºmº uct of 84 × 44 is 38+. This + represents, for reasons
382500, Ans. given in the preceding problem, 4 of 10000, and hence we
reduce it to a whole number and annex the same to the
38, 4 of 10000 = 2500, which annexed gives 382500, the correct product.
3. What will 2812; gallons cost, at $4.50 per gallon ?
OPERATION.
284 Explanation.—The reason for the work of this problem
4; is the same as elucidated in the two preceding, and hence
$12656.25, Ans. is omitted. t
I94. soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. jºr
4. What will 1650 bbls. pork cost, at $13.75 per bbl.
OPERATION,
1. Explanation.--In this problem, we use the units', tens’
16# and hundreds' figures of the price as #, and the units’ and
& tens’ figures of the barrels as #, and thus produce a
º-º-º-mºmºmºmº-º-º-º-º: product of 22#; the # we reduce and annex as above
$22687.50, Ans. directed and obtain $22687.50.
Multiply the following numbers:
5. 12250 by 4750. 7. 3456 by 2125.
6. 24250 by 8125. 8. 379 by 425.
OPERATIONS.
(5) (6) (7) sº
12+ 24+ 3456 7
4; 8} 2} 4+
58187500, Ans. 197031250, Ans. 6912 1516
432 94;
7344000, Ans. 161075, Ans.
9. Multiply 775 by 425; 1850 by 475; 3250 by 875; 14625 by 16125.
TO SQUARE NUMBERS THE SUM OF WHOSE FRACTIONS ADDS ONE.
386. 1. Multiply 94 by 94.
OPERATION.
9% Ea:planation.—In this problem we add 1 to 9, making it
94 10; then we multiply 10 times 9 are 90, and 3 of # is +,
and thus produce the correct result. We do this because
one half of 9 taken twice and added to its square, is the
904, Ans. same as to multiply the 9 by 1 more than itself. This
F. and process of work are applicable to the mul-
tiplying of any two like numbers whose fractions add unity or 1.
PROBLEMS.
2. What will 124 pounds cost, at 1242 per pound? Ans. $1.56}.
3. What will 64 pounds cost, at 6% per pound? Ans. 42*.
4. What will 83 pounds cost, at 85% per pound? Ans. 72%.
4. What will 193 pounds cost, at 193% per pound? Ans. $3.80%}.
6. What will 49 ºr pounds cost, at 49+2 per pound? Ans. $24.50%r.
TO MULTIPLY ANY TWO NUMBERS, THE DIFFERENCE OF WHICH
IS ONE, AND THE SUM OF WHOSE FRACTIONS IS ONE.
387. 1. Multiply 54 by 43.
OPERATION.
5} Explanation.—In all problems of this kind, we add 1 to
4; the larger number, and then multiply the sum by the
lesser number, and set the result in the product line; then
1 5 we square the fraction of the larger number and subtract
24}}, Ans. the result from 1, and annex the remainder to the whole
number in the product. In this example, we added 1 to
5, which made 6: this we multiplied by 4, and produced 24; we then squared #, which gave us 's;
We º ºcted the ſº from 1, and obtained #, which we annexed to the 24 and completed the
COTTOC tº TêSUllt.
* CONTRACTIONS IN MULTIPLICATION OF FRACTIONS. 195
PROBLEMS. **
2. What will 84 pounds cost, at 7:42 per pound 7 Ans. 63%.
3. What will 103 pounds cost, at 94.2 per pound? Ans. 994'sſº.
4. What will 193 pounds cost, at 1832 per pound? Ans. $3.60%.
TO MULTIPLY ANY TWO LIKE NUMBERS, WHOSE FRACTIONS HAVE
LIKE DENOMINATORS, OR, WHOSE DENOMINATORS MAY BE
EASILY REDUCED TO THE SAME DENOMINATION.
388. 1. Multiply 84 by 83.
OPERATION.
84 Explanation.—In all problems of this kind, we add to
8; the multiplicand the fraction of the multiplier, and then
multiply by the whole number of the multiplier and set
Lºs the result in the product line; then we multiply the frac-
72%, Ans. tions together and annex the result to the whole numbers,
& & & and obtain the answer. In this problem, we added # to
8}, which gave us 9, which we multiplied by 8 and obtained 72; we then multiplied the fractions
and obtained º. The reasons given in the first problem, under the head of contractions, cover
all these special cases, and hence we here only give the directions for performing the operation.
PROBLEMS.
2. What will 43 yards cost, at 442 per yard 3 Ans. 20%2.
3. What will 94 yards cost, at 9% per yard? Ans. 92%.
4. What will 124 yards cost, at 1242 per yard? Ans. $1,60}}.
TO MULTIPLY ANY TWO NUMBERS WHOSE FRACTIONS ARE ONE.
HALF, ONE THIRD, ONE FOURTH, ETC., OR WHOSE FRACTIONS
ARE OF THE SAME VALUE.
389. Multiply 7; by 54.
OPERATION.
7; Explanation.—Here we multiply the 7 by 5, and produce
5} 35, to which we mentally add # of the sum of 7 and 5,
and obtain 41, which we set in the product line; then we
sºmº multiply the fractions together and obtain +, which we
41%, Ans. annex to the 41 and obtain the correct result. The reason
* for adding 4, the sum of 7 and 5 is, because # of either
number added to # of the other is the same as 4 of their sum. Were the fractions #, +, +, etc., we
would add +, +, +, etc., of the sum.
I96 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
2. Multiply 83 by 43.
OPERATION.
8; Explanation.—In this problem, we first multiplied the 8
4; by 4, and obtained 32, which we retained in the mind;
then we added the 8 and 4, which gave us 12; we then
* = multiplied the 12 by #, and obtained 8, which we added
40%, Ans. to the 32, and set the result, 40, in the product line; then
we multiplied the fractions and annexed the product to
the product of the whole numbers, and thus obtained the correct result.
PROBLEMS.
3. What will 94 pounds cost, at 54% per pound? Ans. 49%.g.
4. What will 14+ pounds cost, at 1242 per pound? Ans. $1.74%.
5. What will 31; pounds cost, at 1142 per pound? Ans. $3.46%.
6. What will 93 pounds cost, at 1132 per pound? Ans. $1.14%.
7. What will 154 pounds cost, at 1042 per pound? Ans. $1.62%.
8. What will 405 pounds cost, at 22.Ég per pound? Ans. $9,194.
NoTE.—See pages 88 to 120 for other Contractions in Multiplication.
_º

à livision of Fractions.
=N
390. Division of Fractions is the process of dividing when the divisor or
dividend, or both, contain fractional numbers.
In the division of simple numbers, we saw that the result of division opera-
tions was decreasing; but in the division of fractions, when the divisor is less than
a unit, the result is increasing. This fact is plain, for the reason that the operation
of division is the process of finding how many times the dividend is equal to the
divisor, and, hence, when the divisor is less than 1, the dividend will be equal to the
divisor as many times itself as 1 is equal to the divisor. *…
QUESTIONS IN DIVISION.
391. In practical operations, we usually have the three following cases or
questions in division of fractional numbers:
1st. To find the cost of one pound, yard, or article of any kind, when we
have the cost of many pounds, yards, or articles of any kind given.
2d. To find the cost of one pound, yard, or article of any kind, when we
have the cost of a part of a pound, yard, or article of any kind given.
3d. To find the number of pounds, yards, or articles of any kind that can be
bought with a specified sum, when we have the price of one, or a part of one pound,
yard, or article of any kind given.
METHOD OF REASONING.
392. From these questions, we see that division is the converse of multiplica-
tion, and that from the nature of the question, we must reason from many to one or
from a part of one to one. Thus: 1st, if 5 pounds cost 50g, 1 pound will cost the #
part of it; in the 2d case, if # of a yard cost $2, # of a yard will cost the # part
of it, and #, or a whole yard, will cost 4 times as much; and in the third case, if *g
buy 1 yard, or any other thing, #2 will buy the # part of it, and #, or a whole cent,
will buy 2 times as much.
NotE.—See Introductory Remarks and Elucidations of Division on pages 122 and 123.
(197)
I 98 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. 4×
393. ORAL AND WRITTEN PROBLEMs.
5 e
1. If fºr cost x, what will 1 book cost
Answer. If 5 books cost x, 1 book will cost the 5th part of x.
-*- cost x. What will 1 hat cost?
9.
* hats
Answer. Since 8 hats cost X, 1 hat will cost the 8th part of x.
3. *... cost x. What will 1 pencil cost 3
pencils
4. —"— cost x. What will 1 orange cost 3
oranges
5. --— cost x. What will 1 chicken cost
* chickens
6. g of a yard cost x. What will # of a yard cost 7
7. If 3 of a yard cost 102, what will # of a yard cost 7
Analytic Solution.—If # of a yard cost 10 cts., # of a yard will cost the half part of 10 cts.,
which is 5 cts.
Question. How do you know this?
8. # of a pound cost 15 cts. What will # of a pound cost 3
9. § of a pound cost 20 cts. What is the cost of $ 3
10. # of a pound cost 12 cts. What did 1 pound cost?
Analytic Solution.—Since # of a pound cost 12 cts., 4 of a pound will cost # of 12 cts., which
is 4 cts., and #, or one pound, will cost 4 times as much, which is 16 cts.
11. § of a yard cost 40 cts. What will be the cost of 1 yard 3
12. § of dozen cost $16. What is the value of ; of a dozen 3
Analytic Solution.—Since # of a dozen cost $16, # of a dozen will cost 4 part, which is $8, and
# or a whole dozen will cost 3 times as much, which is $24; and since 1 dozen costs $24, # of a
dozen will cost + part, which is $6, and # of a dozen will cost 3 times as much, which is $18.
SOLUTION STATEMENT.
Reason or the Philosophic
$ Solution. 2+ or # yards cost
13. 24 yards cost $64. $6+, or $*#. Since # yards cost
: ; * dollars, # of º: will cost
º 1. the 5th part, and # or a whole
What will 3% yards cost 5 | 16 yard wifi cost 2 times as much;
and since 1 yard costs the result
of this statement, # of a yard
at the same rate? $8, Ans. will cost the 5th part, and ºl
will cost 16 times as much.
SOLUTION STATEMENT. -
Reason.—% pounds cost 14*c.
g
14. 84 pounds cost 62; 2 | 125 Since * pounds cost 135 cts., #
cents. What was the cost 25 | 3 of a pound will cost the 25th
1. º * - mºmme part, and #, or a whole pound,
of 1 pound | 74%, Ans. win cost 3 times as much.
Questions. 1. How do you know this? 2. What do you mean by judgment?
3. What kind of reasoning is this? 4. What is the premise in this problem?
5. What do you mean by premise?
* I)IVISION OF FRACTIONS,
I99
SOLUTION STATEMENT.
15. 33 yards cost 37; 2ſ 75
cents. What did 1 yard 15 || 4
cost 7 tºmºmºmº || s=s=mºmº
102, Ans.
SOLUTION STATEMENT.
$
16. A man receives $21; 2 || 43
for 64 days' services. What 25 4
s the rate per day ? tº tº ,
Wa, |) y $3.44, Ans.
Reason.—4 yds, cost #cts.
Since "# yards cost ºf cents, #
of a yard will cost the 15th
part, and # or a whole yard will
cost 4 times as much.
Reason.—% days' services cost
$#. Since he receives * dol-
iars for * days' services, for +
of a day's service he will receive
the 25th part, and for #, or a
whole day, 4 times as much.
Or, since ºf days' services are worth ** dollars, 4 of a day's service is worth the 25th part,
and 4 or a whole day's service is worth 4 times as much.
SOLUTION STATEMENT.
$
17. 83 dozen cost $52. 26 º
What will 1 dozen cost? sº
| $6, Ans.
SOLUTION STATEMDENT.
$
18, 9 sheep cost $33}. : 135
What will 1 sheep cost 7 &=º
$33, Ans.
SOLUTION STATEMENT.
$
19. If ; of a pound cost : :
$g, what will 1 pound cost? dº
$1%, Ans.
20. Bought 12; dozen for $124. What was the cost per dozen 3
SOLUTION STATEMENT.
s fb
21. At 7:4 cents per 1.
pound, how many pounds 1. ão
can be boughtfor83} cents? tº ºsmºmº
11% lbs., Ans.
22. Chickens cost $# a piece.
Reason.—% dozen cost $52.
Since ºf dozen cost $52, # of a
dozen will cost the 26th part,
and 3 or a whole dozen, will
cost 3 times as much.
Reason.—9 sheep cost $1.3%.
Since 9 sheep cost +3* dollars, 1
sheep will cost the 9th part.
Reason.—# of a ſb, cost $3.
Since # of a pound cost # of a
dollar, 4 of a pound will cost the
third part, and # or a whole
pound, will cost 4 times as much.
AnS. $1.
Reason.—ºftc. buy 1 pound.
Since ºf cents will buy 1 pound,
4 of a cent will buy the 15th
part, and # or a whole cent, will
buy 2 times as much; and since
1 cent will buy this result, # of
a cent will buy the third part
and *;0 cents will buy 250
times as much.
How many can be bought for $134 3
Ans. 18 chickens.
2OO soul E's PHILOSOPHIC PRACTICAL MATHEMATICS.
2.
REASONING FOR THE DIVISION OF ABSTRACT NUMBERS.
394. 1. Divide 6 by 2.
STATEMENT.
2 || 6 Remarks.-The real question in this problem is to find
*s how many times 6 is equal to 2. Or, in other words, we
3. A are required to measure 6 by the unit of measure, 2. The
, Ans. basis or unit of all numbers is 1; and hence in our rea-
soning for division of abstract numbers, we use 1 as our unit of first measure. The following is our
premise, reasoning, and conclusion: 6 is equal to 1, 6 times. Since 6 is equal to 1, 6 times, it is
equal to 2, instead of 1, # as many times, which is 3,
2. Divide g by #.
SOLUTION STATEMENT.
8 || 7 Explanation and Reason.—Here we are required to mea:
3 4 sure # by the tunit of measure, #. The basis or unit of all
numbers is 1. Hence, as explained on page 122, in our
“mºsºme ºmº reasoning for division of whole numbers, we use 1 as our
24 || 28 first unit of measure. The following is our premise, rea-
*mmº soning, and conclusion: # is equal to 1, § of a time. Since
1%, Ans # is equal to 1, § of a time, it is equal to #, instead of 1, 4
62 e times as many times; and to #, instead of #, the # part of
the number of times.
SOLUTION STATEMENT.
Reason.—A is equal to 1 *
e 4. 15 º Since + : equal to *
tº º times, it is equal to #, instea
3. Divide 33 by 23. 8 || 3 of 1, 3 'times as many times,
gºe and to #, instead of #, the eighth
1}}, Ans. part of the number of times.
SOLUTION STATEMENT.
Reason.—Writing 5 on the
statement line, we reason thus:
6 # 5 is equal to 1, 5 times. Since
4. Divide 5 by ºr. 5 is equal to 1, 5 times, it is
immºns mm sºmmºn equal to Tºr, 11 times as many
9%, Ans. times, and to ºr, the 6th part
of the number of times.
SOLUTION STATEMENT.
Reason.-In this example, we
15 2 write the dividend on the state-
tº º 2 8 ment line and reason thus: ºr
5. Divide # by 8. is = to 1, ºr of a time. Since
ºmmºns mºmmº *; is == to 1, 1% of a time, it is
go, Ans. = to 8, the 8th part of the
Inumber of times.
NOTE:-The solution of the five preceding problems elucidates the only correct reasoning for
dividing abstract fractional numbers. But for practical work we would not advise a change from
the reasoning given where the numbers are denominate.
6. Divide 223 by 54, Ans. 45%. 8. Divide 143 by 9, Ans. 13.
7. Divide 3 by 3, Ans. 44. 9. Divide 32 by 9, Ans. 33.
* DIVISION OF FRACTIONS. 2OI
GENERAL DIRECTIONS FOR DIVISION OF FBACTIONS.
395. From the foregoing elucidations, we derive the following general direc-
tions for division of fractions:
1. Write on the upper right hand side of the statement line the number which is
to be divided or measured. Then, reasoning from the basis of MANY to ONE, or from
a PART OF ONE to ONE, write the other numbers on the division or the multiplication side
of the statement line, according as the conclusion is decreasing or increasing.
2. Mia!ed numbers should be reduced to fractional eaſpressions, and the reason
given for writing both the numerator and the denominator.
MISCELLANEOUS ORAL EXERCISES.
396. 1. If 3 of a yard costs $24, what will 1 yard cost? Ans, $6.
2. If 3 of a yard costs $24, what will # of a yard cost? Ans. $#.
3. # of a number is 15. What is the number 3 Ans. 20.
4. If 3 of a number is 8, what is 13 times the number ? Ans. 21.
5. If 3 of a dozen cost $8, what will # of a dozen cost at the same rate 3
e Ans. $9.
6. What part of 4 is 33 AnS. #.
Analytic Solution—Here, by the terms of the question, we have 3 to divide or measure by 4”
and by the exercise of our reason, we proceed thus: Since 3 is equal to 1, 3 times, it is equal to 42
+ of 3 times, which is #. Or thus: Since 1 is # of 4, 3 is 3 times #, which is #.
7. What part of 5 is #3 AnS. #.
Analytic Solution.—Since # is equal to 1, 3 of a time, it is equal to 5 the # part of # of a time,
which is ſº.
8. What part of # is 7 ? Ans. 83.
Analytic Solution.—Since 7 is equal to 1, 7 times, it is equal to ł, 5 times 7 which is 35, and to
# instead of +, + part of 35, which is 8%.
9. What part of ; is ; ? Ans. 13%.
Analytic Solution.—Since # is equal to 1, § of a time, it is equal to 4, 8 times ; which is º. and
to § instead of , ; part of 49, which is #9, or 14%. *
10. What part of g is #3 Ans. #. 12. What part of 5 is # of 2? Ans. #.
11. What part of 34 is 24% Ans. Hºr. 13. What part of # is # of #3 Ans. #.
14. 9 is ; of what number ? Ans. 72.
Analytic Solution.—Since 9 is ; of a number, 3 or the whole number is 8 times 9, or 72.
15. 13 is + of what number 7 Ans, 91.
16. 21% is # of what number 3 Ans. 1064.
17. § is # of what number ? AnS. #.
18. 24 is # of how many times 3? Ans. 10.
Analytic Solution.—Since 24 is # of the number, # is + part of 24 which is 6, and # or the whole
number is 5 times 6, which is 30; and as 30 is equal to 3, 10 times, therefore, 24 is # of 10 times 3.
19. 32 is 4 of how many times 8? Ans. 7.
20. 28 is is of how many times 12? Ans. 5.
2O2 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS. X-
21. # of 48 is # of what number? Ans. 54.
Analytic Solution.—Since 48 is the whole of a number, 4 of the number is + part of 48, which
is 12, and # is 3 times 12, which is 36; and since 36 is # of an unknown number, # of it is # of 36,
which is 18, and 3 or the whole number is 3 times 18, which is 54.
22. § of 63 is ºr of what number 3 Ans. 154.
23. § of 3 of 64 is ºr of what number? Ans. 104.
24. # of # of 42 is # of what number ? Ans. 7%.
25. # of 32 is # of 4 times what number? * Ans. 9.
Analytic Solution.—Since 32 is the whole of a number, 3 of the number is + part of 32, which
is 8, and # is 3 times 8, which is 24; and since 24 is # of 4 times an unknown number, 3 of 4 times
the number is # of 24, which is 12, and 3 or the whole of 4 times the number, is 3 times 12, which is
36; and since 36 is 4 times the number, 4 of 36, which is 9, is the required number.
26. # of 40 is # of 7 times what number 3 Ans. 6.
27. # of 56 is # of 6 times what number 7 Ans. 12.
28. § of ; of 66 is 33 of 3 times what number? Ans. 3.
29. What is $ and # of a $, of 3 of 15? Ans. 5.
MISCELLANEOUS PROBLEMS IN DIVISION OF FEACTIONS.
397. 1. Bought 4 yards for $144. What was the cost per yard 7 Ans. $33.
2. Sold 84 pounds for $1.87. What was the price per pound?
Ans. 22 cents.
3. Paid 37; cents for 64 yards of calico. What was the price per yard?
Ans. 6 cents.
4. At $13 per gallon, how many gallons can be bought for $148.4%
Ans. 108 gallons.
5. Divide # by 2. Ans. #. 13. Divide 21 by #r. Ans. 33.
6. Divide ºr by 3. Ans. }. 14. Divide 105 by #. Ans. 119.
7. Divide +} by 5. Ans. #. 15. Divide +} by #. Ans. 4.
8. Divide # by 5. Ans. #. 16. Divide # by #. Ans. 12.
9. Divide 7+ by 9. Ans. #. 17. Divide 24 by 3. AnS. 3.
10. Divide 2 by # Ans. 23. 18. Divide ## by ++. Ans. 2*.
11. Divide 3 by ºr. Ans. 7. 19. Divide 34 by 24, Ans. 1%;
12. Divide 5 by +4. Ans. 5+. 20. Divide # of $4 by #. AnS. #.
21. If one pound of tea costs # of a dollar, how many pounds can be bought
for $25% Ans. 30 lbs.
22. 6 barrels of flour Were divided among some poor families in such a manner
that each received # of a barrel. How many families were there? Ans, 9 families.
OPERATION INDICATED.
1 = family, 6
i. Or, 2 || 3 Write the reason.
•= ± -º-º: 9 families.
9 families, Ans.
DIVISION OF FRACTIONS. 2O3
23. If a boy can earn ºr of a dollar in one day, how many days will it take
him to earn $21? Ans. 33 days.
24. Henry walked 25 miles, which was # of the distance Robert walked.
Płow many miles did Robert walk? Ans. 30 miles.
25. At the battle of Germantown, the British lost about 600 men; this was
# of the number lost by the Americans; and the number lost by the Americans
was # of the number they received as re-enforcements just before the battle. How
many men did the Americans lose, and how many did they receive as re-enforce-
ments? Ans. 1000 men lost, 2500 re-enforcements.
26. A man had his store insured for $9000, which was # of ºr of its value.
What was the store Worth 7 Ans. $12375.
27. Sulphur will fuse at 2320 Fahrenheit, which is 7+ times the temperature
required to melt ice. At what temperature will ice melt 3 Ans. 32°.
OPERATION INDICATED. 2329 -- 7# = 329.
28. A quantity of mercury weighed 320624 lbs., which is 134 times the weight
of an equal bulk of water. What would an equal bulk of water weigh 3
Ans. 2375 lbs.
29. A pound of water at 2120 F. was mixed with a pound of powdered ice at
32O. What was the temperature of the mixture? Ans. 1229.
OPERATION.
212 -- 32
## = 122°, Ans.
NOTE.-A body in cooling 19 gives out just as much heat as it takes to heat it 1°. In the above
mixture one pound gains exactly the temperature that the other loses. *
This will not be the case, however, when substances of different natures are mixed.
30. When the air was at the freezing point, a cannon 27613% feet distant from
New Orleans was discharged. 25% seconds elapsed after the discharge before the
sound reached New Orleans. How many feet per second did the Sound travel ?
Ans. 1090 feet.
31. Divide 287; by 5. Ans. 57+}.
ion without the Explanation.—We first divide the 287 by the process of
org. Line short division and obtain a quotient of 57, and a remain-
g der of 2; this remainder we reduce to a fraction whose
5)287 denominator is the same as that of the fraction to be
)287; divided, add it to this fraction, and then divide the sum
* -º by 5 and annex the result to the quotient 57. Thus 2 = }
57%, Ans. + # = **, and * +5 = ##.
32. Divide 1471 fºr by 9. Ans. 163+.
33. Divide 10443 by 12. Ans. 87 ſº.
34. E. T. Churchill divided 14+; dozen apples among 3 boys and 2 girls; he
gave to each girl twice as many as to each boy. How many did each boy and each
girl receive? Ans. 24; doz. each boy, 4% doz. each girl.
(Operation on other side).
2O4. SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS.
OPERATION INDICATED,
3 boys, each receives 1 apple, which makes 3 apples.
2 girls, each receives 2 apples, which makes 4 apples.
3 + 4 = 7, the sum of the proportion of the apples due the 3 boys and 2 girls.
Statement to obtain the amount Statement to obtain the amount
due each boy. due each girl.
IDOZ. DOZ.
12 || 175 12 || 175
7 || 3 7 || 4
3 2
35. Divide 1 by #. Ans. 5. 36. Divide # by 1. Ans. #.
37. If 44 pounds of coffee cost 90 cents, what will 223 pounds cost?
Ans. $4.55.
38. R. E. L. Flemming owns 3 of the capital stock of a factory valued at
$24000; he gives # of # to educational societies, and the remainder he divides
equally among his four children. How much does he give to educational Societies
and how much does each child receive?
Ans. $1500 to educational societies; $1875 each child receives.
39. S. J. Weis has 65+ yards of cloth, 2 yards wide. How many yards of
lining # of a yard wide will be required to line it? Ans. 1963 yards.
OPERATION INDICATED. (654 × 2) -- #.
40. Divide 18 oranges between A. and B. so that A. will have 4 more than B.
What number will each have 3 Ans. A. 10; B. 8.
OPERATION.
Statement showing Statement showing
what B. rec'd, what A. rec'd.
B. receives 1. 18 18
A. receives 14 9 || 4 9 || 4
ſº-º- * I ammº 4 5
A. and B. receive 2#. 8 tºmºg ºmºmº
10
41. Divide 18 oranges between A. and B. so that A. will have # less than B.
What number will each have # Ans. A. 7#; B. 10%.
42. A., B., and C. are to receive $26 in proportion to 3, #, and #. What will
each receive 3 Ans. A. $12; B. $8; C. $6.
43. Frank can work 100 problems in 4 hours, and Lillie can work the same
problems in 5 hours. How many hours will it require for both to work the prob-
lems? Ans. 2% hours.
SOLUTION.—Since Frank can work the problems in 4 hours, in 1 hour he can work # of them;
and since Lillie can work the problems in 5 hours, in 1 hour she can work # of them. Hence, ††:
= # of the problems, worked by both in 1 hour; and since ºr of the work required 1 hour, 3% of
the work will require # of an hour, and #3, or the whole work, will require 20 times as many hours,
or * hours, which is 23 hours. º
* DIVISION OF FRACTIONS. 2O5
44. A. and B. can do a piece of work in 14 days, A. can do # as much as B.
How many days will it take each to do it, working alone? Ans. ; days fºr B.
23 “ “ A.
OPERATION INDICATED.
1 equals the work done by B.
# “ {{ {{ ** A.
1#, or #, equals the work done by B. and A.
Hence, B. does # of the work in 1 day, and A. does # of the work in 1 day.
14 ds. -- # = 24; days, for B. to do the work alone. 14 ds. -- # = 323
days, for A. to do the work alone.
.45. Three persons, A., B. and C., do a piece of work; A. and B. together do
# of it, and B. and C. do # of it. What part of the work is done by B.? Ans. ##.
SoLUTION.—As A. and B. do of it, it is clear that C. does the remaining #; and as B. and C.
do ºr of it, it is clear that A. does the remaining Yºr. Then, as A. does ºr, and C. §, they, together,
do nºr-H # = ##; and if A. and C. do #5 of a piece of work done by A., B., and C., it is clear that B.
does the difference between #3 and ##, which is...}}.
46. If 6 oranges and 7 lemons cost 33%, and 12 oranges and 10 lemons cost
54%, what was the cost of 1 orange and 1 lemon each? Ans. 1 Orange, 22.
1 lemon, 32.
OPERATION.
6 oranges and 7 lemons cost 33%.
12 {{ 10 4% 542.
12 {{ 14 44 66 g.
12 “ 10 44 54%.
4 & 12.
122 + 4 = 3%, cost of 1 lemon.
32 x 7 (lemons) = 21%, cost of 7 lemons.
332 — 21% = 12¢, cost of 6 oranges.
12% -- 6 = 22, cost of 1 Orange.
Reason.-Since 6 oranges and 7 lemons cost 33 cents, twice as many, or 12, oranges and 14
lemons will cost twice as much which is 66 cents; and since, by the second condition of the prob-
lem, 12 oranges and 10 lemons cost 54 cents, the difference between 66 cents and 54 cents, which is
12 cents, will be the cost of the difference between (12 oranges and 14 lemons) and (12 oranges and
10 lemons) which is 4 lemons. And since 4 lemons cost 12 cents, one lemon will cost the fourth part
Which is 3 cents. And since 6 oranges and 7 lemons cost 33 cents, by subtracting the cost of 7
lemons which is 21 cents, we have I2 cents, the cost of 6 oranges. And since 6 oranges cost 12
cents, one orange will cost the sixth part which is 2 cents.
47. A miller invested $54 in grain of which #was barley at 624% per bushel;
# was wheat at $1.874 per bu.; and the balance oats (2 374% per bu. How many
bushels of grain did the miller buy 3 Ans. 40 bus.
OPERATION.
Bushels. BU.
* @ 13°C. = **c. = .18% proportionate cost of the barley. | 1
# (2) ###c.= *#5c.— 1.12% & 4 & 4 wheat. 1.35 54.00
To (a) ºc. = **c.— .03% & 4 4 & oats. | 40 bus. purchased, Ans.
$1.35 proportionate cost of 1 bu. of grain.
48. A. and B. can do a piece of work in 10 days; A. alone, can do it in 15
days. How many days will it take B. to do it? Ans. 30 days.
Wiscellaneous Problems,
=N
INVOLVING THE PRINCIPLES OF ADDITION, SUBTRACTION, MULTIPLICATION,
AND DIVISION OF FRACTIONS.
P-Jºyr-w
398. Find the difference between # and #; # and #; ; and #; 3% and 2%; 4%
and # of 3} + Ans. To last, 3.
2. Find the sum of # of ºr and # of #r. Ans. #.
3. To the quotient of 23 divided by 5}, add the quotient of 33 divided by ++.
Ans. 7#.
4. A number was divided by 3, and gave a quotient of 20. What was the
number 3 Ans. 15.
5. What number is that, which being multiplied by ºr, gives as a product ##
Ans. #.
OPERATION INDICATED. # -- + = #, Ans.
6. What number is that, from which, if you take # of itself, the remainder
Will be 12% & Ans. 30.
OPERATION INDICATED. 1 = } – } = }; if # = 12, # = 6, and # equals 30.
#
e
7. What number is that, to which, if you add # of itself, the sum will be 40?
Ans. 25.
OPERATION INDICATED. 1 = } + # = #; if # = 40, # = 5, and # = 25.
8. A. owns # of a store which is worth $25000, and sells # of his share.
What part does he still own, and what is it worth ? Ans. A. owns ºr, worth $2500.
9. Smith owns ºr of a cotton mill and sells #5 of his share to Jones for
$33000. What is the mill worth at that rate? Ans. $242000.
10. John has 5 cents, and James # of 8 cents. What part of James' money
is John’s 7 AnS. #.
OPERATION INDICATED. # of 82 = 22 and # = 62; then 5 -- 6 = #, Ans.
11. The sum of four fractions is 1}. Three of the fractions are #, #, and #.
What is the fourth” Ans. *r.
12. What number is that, to which if # of # of 14 be added, the sum will
be 1; ? Ans. 1.
13. Two boys bought a bushel of oranges, one paying 24 dollars and the
other 43 dollars. What part of it should each have 3 Ans. First, #; second, #.


(206)
jºr MISCELLANEOUS PROBLEMS IN FRACTIONS. 207
14. A farmer sold # of his mules on Monday; on Tuesday he bought # as
many as he sold, and then had 40. How many mules had he at first 3
Ans. 56 mules.
OPERATION INDICATED. # from # =
#.
# o
# =
0;
#
#
... } + # =
8
fi
7 *
7
, and #
;
#
15. A planter gave 50 bales of cotton at $50% per bale for flour at $7.4 per
barrel. How many barrels of flour did he receive 3 Ans. 334 bbls.
16. A merchant bought 20 barrels of flour at $6; per barrel. Having sold 5
barrels at $74 a barrel, and 6 barrels at $83 per barrel, at what price per barrel must
he sell the remainder to gain $10 on the whole 3 Ans. $64.
17. If it requires 34 yards of cloth to make a coat, 24 to make a pair of
pantaloons, and 13 to make a vest, how many suits may be made out of 1574 yards,
allowing # of a yard waste in cutting each suit 3 Ans. 21.
18. A. W. McLellan gave #, #, and # of his money to different benevolent
institutions, and had $1000 left. How much had he at first 3 Ans. $20000.
19. C. Manson owning ºr of a rice mill, sold # of his share for $8800. What
was the value of the mill? Ans. $24200.
20. A book-keeper worked 914 days, and after paying # of # of his earnings
for board and washing, had $438 remaining. How much money did he receive in
all, and how much per day ? Ans. $730 in all, $8 per day.
21. Prophet can do a piece of work in 6 days, and Fisher can do the same
work in 8 days. How many days will it take both together to do the work 3
Ans. 3.} days.
OPERATION INDICATED. # + 4 = #P. 1 day -- # = 3} days.
22. Myers, Levy, and Hoffman can do a piece of work in 10 days; Myers and
Levy can do it in 15 days. In what time can Hoffman do it, working alone?
Ans. 30 days.
23. A man died and left his wife $14400, which was # of # of his estate.
At her death she left # of her share to her daughter. How much money did the
daughter receive, and what part was it of her father's estate 3
* Aris. $12000, #4 of her father's estate.
24. A man engaging in trade lost # of the money he invested; he then
gained $1000, when he had $3800. What did he have at first, and what was his
loss % Ans. $4900 at first, $2100 loss.
oppEATION INDICATED. $3800–$1000 = $2800. $2800 -- # = $4900.
# of $4900 = $2100.
208 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICs. XF
25. A mule and a dray cost $240; the mule cost 13 times as much as the
dray. What did each cost? Ans. $90 dray, $150 mule.
OPERATION INDICATED. 1 (cost of dray)+ 1} (cost of mule)=2# = $240.
$ 240
240 8 || 3
8 || 3 3 || 5
| $90 cost of dray. | $150 cost of mule.
26. How many bushels of apples at $4 a bushel, will pay for ºr of a barrel
of Oranges at $63 a barrel? Ans. 73 bushels.
27. Sweeney paid # of his year's wages for board, 3 of the remainder for
clothes, and had $80 left. How many dollars did he receive for labor? Ans. $560.
28. Forcheimer lost 3 of his fish-line, and then added 254 feet, when it was
just # of its original length. What was its original length'? Ans. 204 feet.
OPERATION INDICATED. 1 = #–# = #. #–3 = } = 254 feet.
254 × 8 = 204 feet, Ans.
29. Purcell, having a certain number of cents, gave one-half of them and
half a cent over to one beggar; one-half of what he had remaining and half a cent
over to a second beggar; and to a third, one-half of what he then had and half a
cent over, and had left 3 cents. How many cents had he at first 3 Ans. 31 cents.
OPERATION INDICATED.
(3 + 42 over) × 2 = 72, had before making 3rd gift.
( 7 -- #2 over) × 2 = 15%, {{ << 2nd £6
(15 + #2 over) × 2 = 312, {{ 4% 1St 4
30. A housekeeper's weekly purchases from her butcher were as follows: on
Monday, 73 pounds of meat; on Tuesday, 64 pounds of meat and 1 duck; on
Wednesday, 84 pounds of meat; on Thursday, 63 pounds of meat; on Friday, 3
pounds of meat and 2 fish; and on Saturday, 93 pounds of meat, 1 chicken and
2 dozen crabs. The meat was 12#2 per pound; the fish were 75% each; the chicken
$1.40; the duck 802, and the crabs 40% per dozen. What sum of money was due
the butcher at the close of the week? Ans. $9.79%.
31. John lives with his parents, but works for Mr. Smith who pays him $210
per year. His parents board him, but he has his clothes to buy. He spends # of
his wages for cigars, # of the remainder for theater tickets, # of the remainder for
wine, and 4 of what he then has for novels. How much has he remaining at the
end of the year to pay for his clothes? Ans. $30.
32. Joseph worked on the same conditions as John, in the problem above.
He gave +4 of his wages to the cause of charity, ºr of the remainder for useful
books, # of the remainder for evening tuition, paid $100 for clothes, and deposited
the balance in the bank. How many dollars did he put in the bank? Ans. $50.
* MISCELLANEOUS PROBLEMS IN FRACTIONS. 2O9
33. A. G. Niehues and R. G. Jones have $1899, and Jones has 3} times as
much as Niehues. How much has each? Ans. Niehues $422, and Jones $1477.
NOTE.-For the operation, see problem 25.
34. J. C. Beals can solve 25 problems in 50 minutes and H. H. Barlow can
solve them in 30 minutes. In what time can both solve them? Ans. 18; minutes.
35. U. Burke purchased 200 barrels of flour for $1450, and sold # of it at a
profit of $4 per barrel, and the remainder at $74% per barrel. How much did he
gain } Ans. $67.50.
36. What is the numerical value of 44–3
2% + 1} AnS. 114.
37. M. Ernst bought 3841; pounds of cotton at 7; pence per pound. What
did it cost 3 Ans. £124, 0s, 113d.
38. R. J. Kennedy has 3 dozen oranges which he wishes to divide between
Miss Katie and Miss Tillie, so that Miss Katie will receive + more than Miss Tillie.
How many will each receive? Ans. Miss K. 20, and Miss T. 16.
OPERATION INDICATED
1 = Miss Tillie's proportional share. 36 36
1} = Miss Katie's proportional share. 4 9 || 4
# = sº ſº tº 4 || 5
2} = the sum of the proportional shares. 16 sº I -º-º-º-º:
20
39. A tree 110 feet high, had # of it broken off in a storm. How much of it
was left standing ? Ans. 44 feet.
40. What cost 223 pounds of coffee at 21% per pound 7 Ans. $4.94+3.
41. If 18; yards cost $3.37}, what will 34 yards cost 3 Ans. 63 cents.
42. Miss Cora has $600 of which she wishes to give to A. 3, B. 4, C. §, and
D. #. How much will each receive? Ans. A. $200, B. $150, C. $120, and D. $100.
43. Miss Mamie has $600 which she wishes to give to A., B., C., and D. in the
proportion of #, +, +, and #. How much will each receive?
Ans. A. $210+}, B. $157+%, C. $1264%, and D. $105*.
OPERATION INDICATED,
600
# + 4 + 4 + 4 = ##. 19 || 20 = $2104% A's share.
3
44. If a yard and a half cost a dollar and a half, what will twelve and a
half yards cost 3 Ans. $12#.
45. What part of 6 cents is # of 5 cents? Ans. #.
OPERATION INDICATED. (5 × #) + 6. Write the reason.
46. A planter remits his factor $500 to invest in rice and coffee, in equal sums.
He pays 94% per pound for rice, and 233% per pound for coffee. How many pounds.
of each did he purchase ? Ans. 27023; lbs. rice. 1069+ºf lbs. coffee.
47. If 3 of 6 be 3, what will # of 20 be? Ans. 73.
48. If 3 is the third of 6, what will the fourth of 20 be? Ans. 34.
2 IO SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. jºr
49. E. L. Hunt owned a quantity of rice, of which he sold # for $99.60. What
is § of the remainder worth at the same rate } Ans. $16.60.
50. M. Landman paid $60 for # of an acre of land. What is the value of §
of an acre ? Ans. $50.
51. J. J. Hauler bought 937852; pounds of cotton at 14% per pound. What
was the cost 3 …” Ans. $135695.53%.
52. E. T. Berwick invested # of his money in sugar, in rice, 3 in coffee, and
deposited in bank $2645. How much money had he at first? Ans. $63480.
53. L. Meyer spends # of his time in study, # in labor, $ in rest and recrea-
tion, and the remainder in sleep. How many of the 24 hours of a day does he
sleep 3 Ans. 7 hours.
54. An industrious young lady spends # of her time in the performance of
household affairs, # in reading good books, ſº in physical exercise in the open air
and Sunlight, $ in the practice of music, singing and parlor amusements, or Social
intercourse, 2 hours per day in eating, and the remainder of the day in sleeping.
How many hours per day does she devote to each 3
Ans. 6 hours to household affairs; 4 hours to reading; 2 hours to exercise;
3 hours to music, etc.; 2 hours to eating, and 7 hours to sleeping.
55. A loafer spends 4 hours per day sauntering on street corners, 3 hours
Smoking and drinking, # of the day in sleep, + of the day in drunkenness, +'s in
eating, Hº in quarreling, and the remainder of the day in gaming. How many hours
does he spend in gaming 3 Ans. 3 hours.
56. A fashionable young lady spends # of her time in dressing, painting, and
making her toilet, $ in reading novels and papers of senseless fiction, % in making
calls and gossiping, ºr in street promenading, ºr in criticising industrious young
men and speculating upon the qualities and fortune of an anticipated husband, ºr
in making remarks derogatory to the character of those who labor, while her own
mother is perhaps cooking or washing, ++, in entertaining young men, and the
remainder in eating and sleeping. How many hours does she devote to useful ser-
vice, and how many to eating and sleeping 3
Ans. 0 hours to useful service; 8 hours to eating and sleeping.
57. A man willed # of his property to his wife, # of the remainder to his
daughter, and the remainder to his son; the difference between his wife's and
daughter's share was $8000. How much did he give his son? Ans. $4800.
OPERATION INDICATED.
# = wife's interest; +* = daughter's interest; # = son’s interest; +”; – # =
# = $8000; then $8000 + + = $25600 the whole estate; # of which is $4800, Ans.
58. R. W. Tyler owned a # interest in a factory, and sold to C. Modinger 4 of
his interest for $15000. What interest does he still own, and how much is it worth
at the rate received for the part sold 2 Ans. He still owns 3, worth $15000.
59. A. and B. own a certain number of oranges; A. owns # and B. #. C. pays
A. 50 cents for # of his oranges, and B. 40 cents for # of his oranges. What part of
the whole number do A., B., and C. respectively own, and what is the remainder of
A. and B. Worth at the same rate? Ans. A. owns #. A’s share is worth $1.00.
B. owns #3. B's share is worth $1.60.
C. owns ºs.
Yºr MISCELLANEOUS PROBLEMS IN FRACTIONS. 2 II
60. J. Cassidy owned 4 of the steamer R. E. Lee. He sold to G. Buesing #
interest in the steamer for $20000; and to J. C. Beals + of his remaining interest
at the same rate. What did he receive for the last sale, and what is his remaining
interest in the boat? Ans. He received $30000; # remaining interest.
61. N. Puech and A. Palacio bought, on joint account, each 4, the New Orleans
Cotton Factory. N. Puech sold # of his interest to R. Krone, and subsequently 4 of
his remaining interest to A. Palacio, who subsequently sold # of ; of his whole'
interest to R. Lynd, for $7500. What is the factory worth at the same rate, and
what is each owner's interest ?
Ans. $32000 value of factory; Puech owns #;
Krone, #; Palacio, #; and Lynd, #.
62. A. has $100; he gives # of it for 5 barrels of flour, and # of the remainder
for 4 barrels of potatoes, and with the remainder he buys coffee at 20g per pound.
How much coffee did he buy? Ans. 200 lbs. Coffee.
63. A person owns ºr of a stock of goods; he sells # of his share for $5000,
and + of the remainder for $5000, and then the balance of his interest for $15000.
What part did he sell the last time, and what would the whole stock be worth at
that rate % Ans. # sold last. $27,777; value of stock.
OPERATIONS IN MIXED NUMBERS, COMPLEX, COMPOUND, AND
IMPROPER FRACTIONS, INVOLVING + , —, x AND + SIGNS,
AND ALSO THE PARENTHESIS OR VINCULUM.
NoTE.—For directions how to reduce or simplify fractional expressions indicated by the
+ , – , X and — signs, and by parentheses or vinculums, see Article 230, page 134.
- 7 6+
64. Multiply ſº by "#.
py; by si
OPERATION INDICATED.
3 22
11 || 2 3 22 4 || 25 3 || 4
4 || 25 OT 11 || 2 10 || 3 8 || 15
10 || 3 sº º ºs * º sº
* 1#. 1; 24, Ans.
Ans. 2;
s.sa., 7% 6#
65. Divide 5; by 3#
OPERATION INDICATED,
3 22
11 || 2 3 22 4 || 25 3 || 4
25 || 4 OI’ 11 || 2 10 || 3 15 | 8
3 || 10 # = ſº ſº-º-º-º: * = gº º gºmº,
sº-º 1; 1; #, Ans.
Ans. ##
2 I 2 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS.
66. Multiply # by # Ans. #. 67. Divide # by # Ans. 23#.
68. Reduce. * Of # of 8% -- 7# to a simple fraction. Ans. #.
69. What is the result of 3 + 3 x 3 + # -- # -- #3 Ans. 24%.
70. Find the value of 2; x +} + 7# — 63. Ans. 1%.
71. § -- # -- # – # × #} equals what number ? Ans. 11%.
72. # — ; x # -- # = what number ? Ans. };
73. What is the value of # + 3 — 3 × (## -- #3) -- #! Ans. #
74. Add + of 3, #, #, #, and 74—53. Ans. 44%.
75. What is the product of ; of 124 by § of 6%—#3 Ans. 34.3%.
76. What is the product of # of 13; +13 by # of (+}-#)? Ans. 4}}.
77. Divide the sum of 4, 3, #, and 3, by the difference between 3; and # of 3.
w Ans. 31's.
78. Multiply [60 divided by (# of 74 of 3)] by # of 7. Ans. 12.
NOTE.-When there are double parentheses, perform the operation indicatéd by the interior
Vºl. first. See page 134, Article 230, for full instructions regarding Parentheses and
1IlCl11UIIITS,
79 Simplif o # + # — # x + i + +, +- # 1 6.
º plify the fraction + 1. 2.4 - 2 1 7 Ans. 8+}.
# — # X ## -- # – ##
9 63 7 6 3 5
— —4 4 2. – 2. 2. – 5.2 - 5 — 1 1 1 – 81.
= #; # -- # = #; * —# = ##; # —## = } *** -- # = 84%, Ans
e e 8 .* 9
80. Divide (ś. of #-) by (#- of #) Ans. }
6+ T +} 4}
81. L. Kaiser bought # of ; of 284 barrels of apples, and sold to S. L.
Crawford # of 9 barrels for $204, which was $1.50 more than the same cost. What
was the cost of the whole, and how many barrels has he unsold 3
Ans. $394, cost; 7; barrels unsold.
82. What is the difference between # of # of eleven thousand eleven hundred
and eleven, and a third, plus one half a third of two thirds of 150$ 2
Ans. 7249-##.
gºats COIII]]|| Divis]ſ [f Fra(ti)|S.
*= tº
399. The Greatest Common Divisor of two or more fractions is the greatest
fraction that will divide each of them, without a remainder.
400. One fraction is divisible by another when the numerator of the divisor
is a factor of the numerator of the dividend, and the denominator of the divisor
is a multiple of the denominator of the dividend.
Thus +} is divisible by #; for +} = }}; and #} -- s = 4.
401. The Greatest Common Divisor of several fractions, is a fraction
whose numerator is the greatest common divisor of all the numerators, and whose
denominator is the least common multiple of all the denominators.
Thus the G. C. D. of # and ## is ###.
1. What is the greatest common divisor of 3, 4}, and #1 Ans. Töö.
Operation to find the G. C. D. Operation to find the L. C. M.
of the numerators : of the denominators :
3 || 3 9 18 2 |4 2 25
l 3 6 2 l 25
3 = G. C. D. 2 × 2 × 25 = 100 L. C. M.
GENERAL DIRECTION FOR FINDING THE GREATEST COMMON
DIVISOR OF FERACTIONS.
402. From the foregoing elucidations, we derive the following general direc-
tion for finding the greatest common divisor of fractions:
Find the G. C. D. of the numerators of all the fractions and write it over the
L. C. M. of their denominators.
NotE.—All the fractions must be in their simplest form before commencing the operation.
–2.
2. What is the greatest common divisor of #, #, #, and #3 Ans. Tåg.
3. What is the greatest common divisor of 8% and #3 Ans. Ps.
4. What is the greatest common divisor of 5, 3}, 63, and #53 AnS. #'s.
5. What is the greatest common divisor of #, #, #, #, and #3 AnS. gº.
6. A grocer has three kinds of molasses, which he wishes to ship in the least
number of full kegs. Of the 1st quality, he has 115% gallons; of the 2d quality,
1284 gallons; and of the 3d quality, 134% gallons. How many gallons will there be
in each keg, and how many kegs will be required?
Ans. 64% gals. in each keg ; 59 kegs required.
(213)
#as Umm Mlik ºf Fridiºs
-------------------------------------N
Aºa —-ºº- Æa
-* vºw ~gº w
403. The Least Common Multiple of two or more fractions is the least
number that each fraction will divide without a remainder.
NotE 1.-The G. C. D. of several fractions is always a fraction; but the L. C. M. of several
fractions may be a fraction or a whole number.
404. A fraction is a multiple of a given fraction when its numerator is a
multiple of the given numerator and its denominator is a divisor of the given
denominator.
Thus # is a multiple of #; for 4 is a multiple of 2, and 5 is a divisor of 15
Eſence # -- # = 6; or thus # = }#; and # -- # = 6.
405. A fraction is a common multiple of two or more given fractions when
its numerator is a common multiple of the numerators of the given fractions, and
its denominator is a common divisor of the denominators of the given fractions.
406. A fraction is the least common multiple of two or more given fractions.
when its numerator is the least common multiple of the given numerators, and its
denominator is the greatest common divisor of the given denominators.
1. What is the L. C. M. of 3, #2, and #3 Ans. 63.
Operation to find the L. C. M. Operation to find the G. C. D.
of the numerators : of the denominators :
2 2 5 4 3 || 3 12 15
1. 5 2 1 4 5
L. C. M. is, 2 × 5 × 2 = 20. G. C. D. is 3.
2
Hence the L. C. M. of the fractions is ºº = 63.
GENERAL DIRECTION FOR FINDING THE LEAST COMMON
MULTIPLE OF FEACTIONS.
407. From the foregoing elucidations, we derive the following general
direction for finding the least common multiple of fractions:
Find the least common multiple of the numerators and the greatest common
divisor of the denominators, and then divide the L. C. M. of the numerators by the
G. C. D. of the denominators.
NotE.—The fractions must be in their simplest form before commencing the operation.
(214)
Yºr LEAST COMMON MULTIPLE OF FRACTIONS. 2 I 5
2. What is the L. C. M. of #, #, and #g Ans. 4}.
3. What is the L. C. M. of 3, 3, #, #, and #3 Ans. 60.
4. What is the L. C. M. of 54, 74, #4, and # Ans. 44.
5. There is an island 15 miles in circuit, around which A. can travel in #
of a day, B. in § of a day, and a horse car in ºr of a day. Suppose all start
together from the same point to travel around it in the same direction, how long
must they travel before coming together again at the place of departure, and how
many miles will each have traveled ?
Ans. 104 days; A., 210 miles; B., 180 miles; Horse Car, 525 miles.
PARTIAL OPERATION.
2) 4 8 10 Denominators.
3) 3 7 3. Numerators. *-i-º-º-º-mºmºmºmº
- 2 the Greatest Common Divisor.
1 7 1
3 × 7 = 21 -- 2 = 104 days before they all meet; then the following propor-
tional statements give the miles traveled by each :
A. B. H. Car. or thus, 15 -- # x 104 = 210 miles
15 15 15 traveled by A.
3 || 4 7 || 8 3 || 10 15.-- g × 10% = 180 miles
2 21 2 21 2 || 21 traveled by B.
tº- 15 –- ºr × 10% = 525 miles
525 m., Ans, traveled by Horse Car.
210 m., Ans. 180 m., Ans.
6. What is the Smallest sum of money for which I could purchase a number
of bushels of oats, at $º a bushel; a number of bushels of corn, at $3 a bushel; a
number of bushels of rye, at $14 a bushel; or a number of bushels of wheat, at
$24 a bushel; and how many bushels of each could I purchase for that sum ?
Ans. $224; 72 bushels of oats; 36 bushels of corn; 15 bushels
of rye; 10 bushels of wheat.
PARTIAL OPERATION.
5) 5 5 3 9 Numerators.
2) 16 8 2 4 Denominators.
3) 1 1 3 9
*=º
1 1 1 3
2 the Greatest Common Divisor.
5 × 3 × 3 = 45 the least common multiple of the numerators, which divided
by 2, the greatest common divisor of the denominators, gives $224, the smallest sum ;
then 22; divided by $4%, $3, $#, and $# gives respectively the number of bushels
of each article represented by the different prices.
7. In December 1875, the Earth, Mars, and Saturn, were in conjunction.
The period of the revolution of Mars is 13 years, and of Saturn 294 years. When
will they be again in conjunction at the same point of their orbits?
Ans. In 1003 years.
Alecimal Fractions.
—sº a Aſſº–
-ºr-ºr-º-
408. A Decimal Fraction is one or more of the equal parts of a unit, which
is divided into tenths, hundredths, thousandths, etc., according to the decimal scale;
hence the denominator of decimal fractions is always 10 or some power of 10. The
word decimal is derived from the Latin word decem, which means ten.
409. The Decimal Point (. ) is used to distinguish decimals from whole
numbers. When there are mixed numbers, it also separates the whole numbers
from the decimals.
The following are decimal fractions: #3, Tºo, #3, and Tážg. They are here
written as common fractions, but generally the denominator of decimal fractions is
omitted and the value is indicated by Writing the decimal point before the numerator.
To write the above fractions according to the decimal notation, they would
be written thus:
fºr decimally expressed is .7 *śr decimally expressed is .137
ºr decimally expressed is .15 T###5 decimally expressed is .0423
410. Notation of Decimals. Whenever decimal fractions are expressed
decimally, the numerator must have as many decimal places as there are naughts
in the denominator. Thus, #6 = 4; Tº = .16; # = .1456. When the number
of naughts in the denominator is greater than the number of figures in the numer-
ator, naughts must be prefixed to the numerator until the number of places is
equal to the naughts in the denominator. Thus, rão = .04; tºo - .007; rºw
= .00125; etc.
411. A Pure or Simple Decimal consists of a decimal fraction, decimally
expressed or written. Thus .5, .42, .875, and .1256 are pure decimals, and are read
respectively 5 tenths; 42 hundredths; 875 thousandths; and 1256 ten thousandths.
412. A Mixed Decimal consists of a whole number and a decimal. Thus,
24.5 and 41.25 are mixed decimals. They are read respectively, 24 and 5 tenths; 41
and 25 hundredths.
NOTE.—A mixed decimal may be read as an improper fraction. Thus, 24.5 = 24%; = 2,4}.
413. A Complex Decimal consists of a decimal with a common fraction
annexed. Thus, .15% and .005% are complex decimals. They are read respectively,
15; hundredths; 5; thousandths.
414. A Terminate Decimal is one which ends, as # = .125. And an
interminate decimal is one that does not end, as # = .3333+.
NOTE.-Decimal fractions were invented by Regiomontanus, in 1464; but they were not in
general use until the latter part of the sixteenth century, when the first treatise on decimals was
written by Stevinus, in 1585.
•.
(216)
Yºr DECIMAL FRACTIONS. 217
f r
CIRCULATING DECIMALS.
415. A Circulating Decimal is one in which a figure or set of figures con-
stantly repeats itself. Thus, # = .3333+ , ºr = .181818--, + = .142857+.
416. A Repetend is the figure or set of figures which repeats itself, and it
is expressed by placing a dot over the figure repeated, thus: # = .3333+ = 3.
When the repetend consists of more than one figure the dot is placed over the first
and last figure, thus: ºr = 181818+ = i8; and + = 142857+ = i+2857.
417. A Pure Circulating Decimal is one which contains only the repetend,
as # = 6; } = .142857; # = 1.
418. A Mixed Circulating Decimal is one which contains other figures than
the repetend, as # = .16; ### = .647.
419. A Simple Repetend contains one figure, as .6.
420. A Compound Repetend contains two or more figures, as i8 and .342.
421. Similar Repetends are those which begin and end at the same decimal
place, as .536 and 427.
422. Dissimilar Repetends are those which begin and end at different decimal
places, as .205 and 3765.
423. A Perfect Repetend is one which contains as many decimal places,
less 1, as there are units in the denominator of the equivalent common fraction;
thus, 4 = i+2857.
424. Conterminous Repetends are those which end at the same decimal
place, as .50397 and 42618. }
425. Co-originous Repetends are those which begin at the same decimal
place, as .5 and 124.
Circulating decimals had their origin in reducing common fractions to deci-
mals.
The general law for the formation of repetends will be seen from the follow-
ing:
1. * = .1111+ = i. 5. # = .4444+ = 4.
2. ºs = .01010+ = .0i. 6. # = .2727-- = .27.
3. * = .001001-- = .00i. 7. § = 135135+ = .iáš.
4. nºw = .000100014--.000i. 8. §§§ = .17281728--- iT28.
2 I 8 soul E's PHILOSOPHIC PRACTICAL MATHEMATICs, X-
426. Decimal fractions, like whole numbers, decrease toward the right and
increase toward the left in a ten-fold ratio; and hence the prefixing of naughts
between the decimal figures and the decimal point, or the removal of the decimal
point towards the left diminishes their value ten-fold, or divides the decimal by ten
for each order or place removed. Thus: .5 = #; .05 = Hitſ; .005 = Tºro ; etc.
427. The removal of the decimal point to the right, increases the value
ten-fold or multiplies the decimal by ten for each place removed. Thus : .005 =
Tºo : .05 = Tâû ; .5 = *; ; etc.
428. Annexing naughts to decimals does not change their value, because
the significant figures are not thereby removed nearer to or farther from the decimal
point. Thus: .5 = #; also .50 = }}; ; .500 = ***, all of which are equal.
429. Decimal orders are also called decimal places, each order being counted
as one place. Thus, in .0043 there are four decimal places, although the 3 is of the
fifth decimal place from unity, the base of the system.
430. The following table will illustrate more fully the relation of whole
numbers and decimals, with their increasing and decreasing orders to the left and
right of the decimal point:
TABLE.
WHOLE NUMBERS. DECIMALS.
é
tº CD
# 5 . E &
C 32 zº r: .#
* gº tº F. -->
:= tº F: cº gº
*] F. : ſº rº º • C
C p F. s'E 3 # E
*:::= < 3 ‘E tº 3 F = E E
$2.5 ° E 2: . $ 35 ; E 2 = E
2 = 22 S- rº; tº P- := 3 £ l 2.3 te
ºr T 2 sº E ºf = . ; ; 3 = 3 E Tº
Gº SH 9 *- : SP º: 2, $2 & 23 SP E. F. Sº
3 & F : G 3.3 : E : º, $º ºt. # = }
5 : 3 E. × 2 = 2 & 5 # 5 3.5 ± 3: E
# * = É # 5 É # E & E E 5 # F# E É
E F = E = F E = P C F. E. F. F. H. E. E. E.
9 8 7 6 5 4 3 2 1 - 2 3 4 5 6 7 8 9
Orders of ascending scale. Orders of descending scale.
This number is read 987 million 654 thousand 321, and 23 million 456 thousand
789 hundred millionths.
431. In order to clearly understand decimals, we must bear in mind that
one is the basis of all numbers, integral and fractional, abstract and denominate,
and that all mathematical operations have this fundamental principle for their
origin, and every number is but a multiple, either ascending or descending of unity
OT O706.
432. The names of the decimal orders are derived from the names of the
orders of whole numbers. Thus the names of the orders in the ascending scale, are,
after units, tens, hundreds, etc., and the orders in the descending scale, are, after units,
tenths, hundredths, etc., the decimal orders being the reciprocal of the orders of whole
numbers equally distant with themselves from the units.
DECIMAL FRACTIONS. 2 I9
433. Numeration of Decimals. In reading decimal fractions, the entire
decimal is regarded as reduced to units of the lowest order expressed, and the name
of this order is given to the entire number of decimal units. Thus, .25 is read
twenty-five hundredths.
434. Before reading a decimal, we must determine 1st. How many units
are expressed. To do this, we numerate and read the significant figures of the
decimal as in whole numbers. 2nd. We must determine the name of the lowest
order in the decimal. To do this, we numerate the number decimally. Thus to
read .001073, we commence at the 3 and numerate to the 1 thousand, and thus find
that 1073 units are expressed; then we commence at the decimal point and numerate
decimally to the 3, and thus find that millionths is the lowest order; we then read
1073 millionths.
EXERCISES.
435. Read the following numbers:
1. 16.008 reads thus: Sixteen units and eight thousandths.
2. .94; reads thus: Ninety-four and three-eights hundredths.
3. 5067.4005 reads thus: 5067 units and 4005 ten-thousandths.
4. Write and read 197.8; 4.68907; .00073; 48.769146.
5. Write and read 2.491; 10.0101089167; 582.400410905.
6. Write and read 5841.291;; 8000.0000000217; 9876541.1000001.
436. Writing Decimals. In writing decimals, we write down the given
number as if it were a whole number; then, to facilitate the operation, we numerate
from right to left, beginning the numeration with tenths, and continue until we come
to the required place or order, always writing 0's to fill the places not occupied by
significant figures. Thus, to write 25 ten thousandths, we first write the 25; then
we begin at the right and numerate thus, tenths, hundredths, thousandths, ten-
thousandths; by this we find that four places are required and as there are but
two figures in the number, we prefix two 0's and obtain the correct result .0025.
1. Write 104 hundred thousandths. Ans. .00104.
Ea:planation.—According to the above directions, we write
> the 104 and then commence on the right and numerate thus:
OPERATION. - *>
00.104 Tenths, hundredths, thousandths, ten-thousandths, hundred-
thousandths. This numeration shows that five places are
required, and as we have but three, we therefore prefix two 0's.
EXERCISES.
437. 1. Write 10101 hundred billionths. Ans. .00000010101.
Write decimally, numerate, and read the following:
2. 31.4 millionths. 6. 1205 ten millionths. 10. Po.
3. 12 thousandths. 7. 897 hundred billionths. 11. ºr.
4. 107 billionths. 8. 1 Sextillionth. 12, Tºr.
5. 1 trillionth. 9. 21001 ten vigintillionths. 13 10423
te 100000
14. Taºğgo 16. 1000 dºwo 18. 1#.
15 748#. 100
d 1000000 || 17. Tºo 19. To Jºão
22O SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. Yºr
t
PRINCIPLES.
438. From the foregoing work we recapitulate the following principles: .
1. Decimals are governed by the same laws of notation as whole numbers;
hefice the value of any decimal figure depends upon the place it occupies.
2. Each removal of the decimal point one place to the right, is equivalent to
multiplying the decimal by 10.
3. Each removal of the decimal point one place to the left, is equivalent to
dividing the decimal by 10.
4. Annexing or rejecting naughts at the right of any decimal does not
change its value.
<>
<>
REDUCTION OF DECIMALS.
TO REDUCE DECIMAL FRACTIONS TO A COMMON DENOMINATOR.
439. 1. Reduce .7, .18, .2581, and .045, to a common denominator.
OPERATION.
.7000 Explanation.—To reduce decimals to a common denom-
º: inator, we have but to annex a sufficient number of 0’s to
.0450 give each decimal the same number of places.
TO REDUCE A DECIMAL TO A COMMON FRACTION. .
440. 1. Reduce .25 to a common fraction.
Ea:planation.—In all problems of this kind, we simply write
º the decimal as a common fraction and then’ reduce it to its
Töö = #, IlS. lowest terms.
2. Reduce .125 to a common fraction.
OPERATION. ** = #, Ans.
3. Reduce .593 to a common fraction.
FIRST OPERATION.
593 445. Explanation.—To reduce complex decimals
— = — and 44%-- #49 = 4% = 4%. Ans. * simple fractions, we first write the decimal
100 oo’ 8 T-T 8 8 O 0 - 3 22 as a common fraction; then we reduce both
8 go. the numerator and the denominator to, the
fractional unit of the denominator contained
in the numerator term of the fraction, and thus obtain a complex fraction, which we reduce to a
simple fraction.
* REDUCTION OF DECIMALS. 22 I
SECOND OPERATION. Explanation.—Here, when, reducing, the fraction to the
59; fractional unit of the denominator contained in the numera-
— = ### = 4%. Ans for term of the fraction, we shorten the work by omitting
8 0 O - 3 27 e the denominator (8) in both terms of the complex fraction,
100 and writing the result as a simple fraction. By this process,
we save the operation of division, the result of which is the
cancelling of the denominator in both terms of the complex fraction.
GENERAL DIRECTION FOR REDUCING DECIMAL FRACTIONS TO
COMMON FRACTIONS.
441. From the foregoing elucidations, we derive the following general direc-
tion for reducing decimal fractions to common fractions:
Write the decimal as a fraction, then reduce the fraction to its lowest terms.
Reduce the following decimals to common fractions:
4. .8, AnS. #. 10. .07, Ans. Tºo. 16. .484, Ans. #6.
5. .05, Ans. #5. 11. .005, Ans. #5. 17. .055#, Ans. #.
6. .25, Ans. 12. .1045, Ans. #6. 18. .008%, Ans.
7. .125, Ans. 13. .88, Ans. #. 19. .00054%; Ans.
8. .675, Ans. 14. .909, Ans. 20. .999, Ans.
9. .105, Ans. 15. .00025, Ans. 21. .4007 ºr, Ans. #####.
TO REDUCE COMMON FRACTIONS TO EQUIVALENT DECIMALS.
442. 1. Reduce # to a decimal.
OPERATION. Explanation.—To reduce common fractions to decimals, we an:
nex naughts to the numerator and divide by the denominator, and
8) 3,000 then point off as many spaces for decimals as there were 0's annexed.
When a remainder continues beyond four or six places, we discon-
tinue dividing and write the sign + to the right of the last figure
.375, Ans. obtained, which indicates that the quotient is not complete. The
annexing of 0's to the numerator is equivalent to multiplying it by
10 for each naught annexed, consequently the quotient obtained is as many times 10 too great as
there were 0's annexed; and hence the reason for pointing off as many places in the quotient as
there were 0’s annexed to the numerator.
f
2. Reduce # to an equivalent decimal.
OPERATION. 7) 5.000000
.714285-H, Ans.
222 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS.
3. Reduce ### to an equivalent decimal.
FIRST OPERATION. SECOND OPERATION. Explanation.—Here, in the
first operation, we annex six
725) 3.000000(4137-- 725) 3. (.004137-1-, Ans. 0's and obtain but 4 figures in
2900 .004137-1-, Ans. 30 tenths. the quotient. Therefore, in
*E=ºmºte * -º-; order to point off as many dec-
1000 300 hundredths. imal places as we annexed 0's,
725 . we prefix two 0's and thus ob-
3000 thousandths. tain the correct result. The
2750 2900 reason for this will appear clear
2175 if we consider each step of the
1000 ten thousandths. work as performed in the sec-
5750 725 ond operation. We are to di-
vide or measure 3 by 725, and
2750 hundred thousandths. we first see that 3 is not equal
2175 to 725 any whole or unit num-
ber of times; we, therefore,
5750 millionths. write the decimal point in the
5075 quotient, annex a 0 to the 3
units and thus reduce it to 30
675 remainder. tenths, which we also see is not
equal to 725 any tenth times,
and hence we write 0 in the tenths' place of the quotient; we then annex another 0 and thereby reduce
the 30 tenths to 300 hundredths, which we see is not equal to 725 any hundredths times, and hence we
write 0 in the hundredths' place of the quotient ; we then annex another 0 and thereby reduce the
300 hundredths to 3000 thousandths, which we see is equal to 725, 4 times, with a remainder. We
have now obtained the first significant figure of the decimal, and we continue the division in the
usual manner to the sixth decimal place and annex the + sign to indicate that there is still a remain-
der. The sign — is sometimes used to indicate that the last figure is too great. Thus, # = .1666-1-;
or, by abbreviating # = .167—.
4. Reduce 6; to a decimal.
FIRST OPERATION. SECOND OPERATION.
6# = ** and * = 4) 27.00 63 = 6 and #; and
* # = 4) 3.00
6.75, Ans.
.75 + 6 = 6.75, Ans.
GENERAL DIRECTION FOR REDUCING COMMON FRACTIONS TO
EQUIVALENT DECIMALS.
443. From the foregoing elucidations, we derive the following general direc-
tion for reducing common fractions to equivalent decimals:
Annea, naughts to the numerator and divide by the denominator. Then point
off, from the right of the quotient, as many decimal figures as there are naughts anneared.
Reduce the following fractions to equivalent decimals not exceeding 6 places:
5. #. Ans. .71875 9. §, Ans. .625
6. ###, Ans. .336 10. Ta, Ans. .076923-H
7. T#5, Ans. .032 | 11. 87%, Ans. .370625
8. § of +, Ans. .1071.42–H | 12. 47.18%, Ans. 47.1875
13. Reduce # to a complex decimal of 3 places.
OPERATION. &
3) 2.000
* .6663, Ans.
14. Reduce # to a complex decimal of 4 places. Ans. .4285#.
15. Reduce # to a complex decimal of 6 places. Ans. .2222223
* ADDITION OF DECIMALS, 223
ADDITION OF DECIMALS.
444. Addition of Decimals is finding the sum of two or more decimals.
Since decimals increase from right to left, and decrease from left to right in
a ten-fold ratio as do simple whole numbers, they may be added, subtracted, mul-
tiplied, and divided in the same manner.
1. Add. 785, .93, 166.8, 72.5487, and 4.17.
OPERATION.
.785 Explanation.—In all problems of this kind, we write the
.93 numbers so that units of the same order stand in the same
166.8 column, and the decimal points are in a vertical line; then we
e add as in simple whole numbers. When the addition is
72.5487 completed, we point off in the sum, from the right hand, as
4.17 many places for decimals as equal the greatest number of
decimal places in any of the numbers added. If there are
245.2337. A complex decimals they must be reduced to pure decimals, as
c , AmS. far, at least, as the decimal places extend in the other numbers.
GENERAL DIRECTIONS FOR ADDITION OF DECIMALS.
445. From the foregoing elucidations, we derive the following general direc-
tions for addition of decimals:
1. Write the numbers so that units or figures of the same order stand in the
same column. -
2. Then add as in simple numbers, and point off, in the sum, from the right, as
many decimal places as equal the greatest number of decimal places in any of the
ww.mbers added.
Add the following numbers:
2. 3.25, 42.348, 748.4, and 29.32. Ans. 823.318.
3. .0049, 47.0426, 37.041, and 360.0039. Ans. 444.0924.
4. 1121,6116, 61.87, 46.67, 165.13, and 676.167895. Ans. 2071.449495.
5. .8, .09, 34.275, 562,0785, and 1.01. Ans. 598.2535.
6. 81.61356, 6716.31, 413.1678956, 35.14671, 3.1671, and 31.4.6.
Ans. 7564,0052656.
7. 1.013, 240.064, 999.9, 80.6051, and .17. Ans. 1321.7576.
8. What is the sum of the following numbers: twenty-five, and seven
millionths; one hundred forty-five, and six hundred forty-three thousandths; one
hundred seventy-five, and eighty-nine hundredths; seventeen, and three hundred
forty-eight hundred thousandths. Ans. 363,536487.
9. A farmer sold at one time 3 tons and 75 hundredths of a ton of hay; at
another time, 11 tons and 7 tenths of a ton; and at a third time, 16 tons and 125
thousandths of a ton. How much did he sell in all? Ans. 31.575 tons.
224 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs.
SUBTRACTION OF DECIMALS.
446. Subtraction of Decimals is finding the difference between two decimals.
1. From 345.3046 subtract 92.1435847.
OPERATION.
345.3046 Ea:planation.—In all problems of this kind, we write the
92.1435847 numbers so that units of the same order stand in the same
column, and the decimal points are in a vertical line; then we
iº subtract as in simple whole numbers, and point off in the
253.1610153, Ans. difference, from the right hand, as many piaces for decimals
as equal the greatest number of decimal places in either the minuend or subtrahend.
When the decimal places in the subtrahend exceed those in the minuend, naughts are under-
stood to occupy the vacant places, and may be filled in if it is desired.
2. From 142.63 subtract 51.1111}.
OPERATION.
*::::::: Ea:planation.—In all problems of this kind, we reduce the
−: complex decimals to pure decimals of equal places and then
91.5555#, Ans. subtract as in subtraction of fractions.
GENERAL DIRECTIONS FOR SUBTRACTION OF DECIMALS.
447. From the foregoing elucidations, we derive the following general
directions for subtraction of decimals.
1. Write the numbers so that the units of the same order stand in the same
column.
2. Then subtract as in simple numbers, and write the decimal point as in addi-
tion of decimals.
NOTE 1.-If there are complex decimals of unequal places in either or both of the given dec-
imals, reduce them to pure decimals of equal places, and then subtract as in subtraction of fractions.
NOTE 2.—If there are not as many decimal places in the minuend as there are in the subtra-
hend, annex decimal naughts to it, until the decimal places are equal.
PROBLEMS.
(3) (4) (5)
81.04089 121.25 532.8
14,587 109.05438 9.00451681
66.45389, Ans. 12.19562, Ans. 523.79548319, Ans.
6. From 461.072 take 427.125, Ans. 33.947.
7. From 17.5 take 4.19, Ans. 13.31.
8. From 4000.0004 take 4.3, Ans. 3995.7004.
9. From three million take three millionths, AnS. 29.99999.999997.
10. From 11 take 1 and 9 trillionths, Ans. 9.999999999991.
11. From 24000 subtract 2.078, Ans. 23997.922.
12. From 886.333 subtract 98.5427, Ans. 787.7903.
* MULTIPLICATION OF DECIMALS. 225
MULTIPLICATION OF DECIMALS.
448. Multiplication of Decimals is finding the product, when either or
both of the factors contain decimals.
1. Multiply 26.58 by 4.3.
IFIRST OPERATION. Explanation.—In all problems of this kind, we multiply as in
whole numbers, and point off on the right of the product as many
26.58 places for decimals as there are decimal places in both the multipli-
4.3 cand and multiplier. The reason for thus pointing off the 3 deci-
mal places in this problem is obvious from the fact, that in the
multiplicand, we have 2 decimal places or hundredths, which we
7974 used as whole numbers, and thereby produced a product 100 times
10632 too great; and in the multiplier we have 1 decimal place or tenths
which we also used as a whole number, and thereby produced a
114.294. Ans product 10 times too great; and both together, give a product 1000
•- * *-2 © times too great; hence, to obtain the correct product, we divide by
1000, or point off 3 decimal places.
SECOND OPERATION.
100 2658 Explanation.—In this operation, we reduce the factors to com-
10 || 43 mon fractions, and then multiplying them together, we obtain a
*=sº product of ++$#4, which, written decimally, is 114.294. This pro-
114294 A cess shows, in another way, why we point off on the right of the
1000 g product as many places for decimals as there are decimal places in
or decimally written both factors.
114.294, Ans.
2. Multiply 4,024 by .0056.
OPERATION.
4.024 Explanation.—In all problems of this kind, where the num-
.0056 ber of figures in the product is not equal to the number of
decimal places in the two factors, we must prefix a sufficient
24144 number of 0's to supply the deficiency. In this example, we
20120 prefix one 0. The reason of this will appear evident by
e- working the example as a common fraction as shown in the
.0225344, Ans. second operation of the first problem.
GENERAL DIRECTIONS FOR MULTIPLICATION OF DECIMALS.
449. From the foregoing elucidations, we derive the following general direc-
tions for the multiplication of decimals:
1. Multiply as in whole numbers and from the right of the product point off as
many figures for decimals as there are decimal places in the multiplicand and multiplier.
2. If the product does not contain as many decimal places as both factors,
- Supply the deficiency by prefiasing naughts.
PROBLEMIS.
3. Multiply 27 by .9, Ans. 24.3.
4. Multiply .38 by 8, Ans. 3.04.
5. Multiply .75 by .42, Ans. .3150.
6. Multiply .006 by .0103, Ans. .0000618.
226 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS,
7. Multiply 9999. by .9999, Ans. 9998.0001.
8. Multiply 7.0007 by .0303, * Ans. .21212,121.
9. Multiply 340.012 by 61.23, AnS. 20818.93476.
10. Multiply .1234 by 1234. Ans. 152.2756.
11. Multiply 1500 by .00014, Ans. 21.
12. What is the product of one thousand twenty-five multiplied by three
hundred twenty-seven ten-thousandths? Ans. 33.5175.
13. What is the product of seventy-eight million two hundred five thousand
two, multiplied by fifty-three hundredths? Ans. 41448651.06.
14. Multiply one hundred fifty-three thousandths by one hundred twenty-
nine millionths. Ans. .000019737.
15. Multiply 1 thousand by 1 thousandth. Ans. 1.
16. Multiply 2 million by 2 billionths. Ans. .004.
17. What will 37.23 tons of hay cost at $20.75 per ton 2 Ans. $772.52-H..
18. What will 428,431 bushels cost at $1.125 per bushel? Ans. $481.98--.
TO MULTIPLY A DECIMAL OR MIXED NUMBER BY TEN, ONE
HUNDRED, ONE THOUSAND, ETC.
450. 1. Multiply 428.375 by 100.
OPERATION. Ea:planation.—In all problems where the multiplier is 10,
O'Q27 ſº 100, etc., we simply remove the decimal point as many places
42837.5, Ans. to the right as there are naughts in the multiplier, annexing
Inaughts if required, as shown in Articles 426 and 438.
2. Multiply 271.32 by 1000, Ans. 271320.
3. Multiply.756 by 100, Ans. 75.6.
4. Multiply .025 by 10, Ans. .25.
5. Multiply 61,052 by 10000, Ans. 610520.
A- -- * *
r −ww.r-w
DIVISION OF DECIMALS.
© 451. Division of Decimals is the process of finding the quotient when the
divisor or dividend, or both, contain decimals.
1. Divide 17.094 by 8.14.
FIRST OPERATION. º lanation.—In all problems of this kind, we divide as in whole
numbers, and then point off as many places for decimals from the
8.14) 17,094 (2.1, Ans. right of the quoiº dº jºinial places in the dividend exceed those in
16 28 the divisor, observing to supply any deficiency by prefixing naughts. In
this problem, the excess is one, and we therefore point off one dec-
814 ſº imal place in the quotient. The reason for thus pointing off is
814 obvious from the fact that in the dividend we had 3 decimals or
*QUSANDTHS, and in the divisor we had 2 decimals or hundredths,
Th Eºº ill al and thousandths divided by hundredths give tenths as a quotient.
§ Fºº, Y}* *ś2 *PPèar plain if we observe that the dividend is the product of the
divisor and quotient multiplied together, and hence we point off enough icº" places in the
quotient to make the number in the two factors equal to the number in the product or dividend,
according to the principles shown in the first problem of multiplication of decimals.
DIVISION OF DECIMALS. 227
SECOND OPERATION.
1000 || 17094 Ea:planation.--In this operation, we reduce the decimals to common
814 || 100 fractions and then proceed as in the division of mixed numbers.
The reduction of the dividend and divisor to common fractions,
2 and then the mixed numbers to improper fractions, is performed
.1, Ans. thus: The dividend 17,094 = 17+3+m = |}}}#; the divisor 8.14 = 8*śr
Decimally written 2.1 Ans. =###. This method also shows the reason for pointing off, and may
be used for all problems in decimal fractions.
PRINCIPLES OF DIVISION OF DECIMALS.
452. From the foregoing elucidations, we derive the following principles:
19. The dividend must contain at least as many decimal places as the
divisor; and when both contain the same, the quotient is a whole number.
29. The dividend is the product of the divisor and Quotient, and hence con-
tains as many decimal places as both the divisor and quotient,
3°. The quotient must contain as many decimal places as the number of
decimal places in the dividend exceeds the number in the divisor.
2. Divide 7898.56 by 2.4683.
OPERATION.
2.4683) 7898,5600 (3200, Ans. Ea:planation.—Here we have an excess of decimals in
74049 the divisor, and in all cases of this kind, we first make
49366 them equal by annexing naughts to the dividend, and
49366 the quotient will be a whole number. The reason for
annexing the naughts will appear more obvious by
00 solving the problem in the form of a common fraction.
3. Divide 7.0761 by 687.
OPERATION.
687) 7.0761 (103 or thus: aft;...º.º.º.
687 .0103, Ans. 10000 || 70761 none in the divisor; hence, according
*-*-* 687 to the foregoing instructions, we must
2061 *-i-º- point off 4 decimal places in the quo-
2061 | 0.103. Ans tient, and as there are but 3 figures in
ſº 7 • the quotient, we prefix 1 naught. In
4-mºmºs all problems of this kind, 0's are
prefixed to supply any deficiency of
figures that may occur.
4. Divide 47.789 by 39.27.
f OPERATION. Explanation.—In this problem we have a remainder,
39.27) 47.789 (1.2168+, Ans. after dividing the dividend, of 665; to this and the two
39.27 successive remainders, we annex 0's and continue the
-- division until we have produced 4 decimal places. The
8519 annexing of 0's reduces the successive remainders to the
7854 next lower order of tenths, and hence all quotient figures
**-* * produced by annexing 0's, are decimals. We therefore
6650 point off from the right of the quotient as many places for
3927 decimals as the number of decimals in the dividend exceeds
ºm- those of the divisor plus the number of 0's annexed.
27230 This is done in all division problems where 0's are an-
23562 nexed, and a sufficient number of 0's should be annexed
to produce 4 or 6 decimal places. When there is a remain-
36680 der after the last division, the plus (+) sign should be
3.1416 annexed to the answer to indicate that the quotient is
amº-mºº- incomplete.
228 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. 4%
*
GENERAL DIRECTIONS FOR DIVISION OF DECIMALs.
453. From the foregoing elucidations, we derive the following general
directions for dividing decimals:
1. Divide as in whole numbers and point off as many decimal places in the
quotient as those in the dividend eacceed those in the divisor.
2. When there is a remainder, annew naughts to the dividend and carry the
work as far as may be desired.
5. Divide .112233 by 12, 16. Divide 1.12233 by 12, 11. Divide.0004869 by .0396
OPERATION. OPERATION. FIRST OPERATION.
12) .112233 12) 1.12233 .0396).0004869(122--
396 =.0122--, Ans.
9352–H =.009352–H, Ans. 9352–H =.09352-H, Ans. | *=º
909
7. Divide 11.2233 by 12. |8. Divide 112.233 by 12. 792 -
OPERATION. OPERATION. •º. 1170
12) 11.2233 12) 112.233 792
.9352-H, Ans. 9.3527–H, Ans. 378
SECOND OPERATION.
9. Divide,0004869 by 396.10. Divide,0001809 by 3.96. Sººdºº-oº-, Ans.
OPERATION. OPERATION. 96
396).0004869(12–H 3.96).0004869(12-i- 909
396 =.0000012–1–, Ans. 396 = .00012-i-, Ans. 792
909 909 1170
792 792 Afte 792
117 117 378
12. Divide 67.8632 by 32.8. Ans. 2.069. 17. Divide 7,482 by .0006. Ans. 12470.
13. , Divide 983 by 6.6. Ans, 148.939-H | 18. Divide 1 by 999. Ans. .001001-H.
14. Divide 13192.2 by 1947. , Ans, 1260. | 19. Divide 84375 by 3.75. Ans. 22500.
15. Divide 67.56785 by .035. Ans. 1930.51. 20. Divide 1081 by 39.56. Ans. 27.3255+.
16. Divide 00125 by 5. Ans. .0025. 21. Divide 35.7 by 485. Ans. .0736+.
22. If rice costs 8.0775 per pound, how many pounds can be bought for
$40,64875? Ans. 524.5 pounds.
23. Sold 14.75 acres of land for $191,75. What was the price per acre 2
Ans. $13.
24. Divide four thousand three hundred twenty-two and four thousand five
hundred seventy-three ten-thousandths, by eight thousand and nine thousandths.
Ans. .5403+.
TO DIVIDE DECIMAL FRACTIONS BY TEN, ONE HUNDRED, on E
THOUSAND, ETC., ETC.
454. 1. Divide 48.76 by 10. *
OPERATION. Pºplanation.--In all problems of this kind, we simply remove
4,876. A the decimal point as many places to the left" as there are 0's in
.ö (0, Ans. the divisor. The reason for this was fully shown in Articles
© gº tº 426 and 438. When th g
in the dividend to allow this to be tº its ... be †: º, º; º ...” figures
2. Divide 875.25 by 100. Ans. 8.7525. 5. Divide 9.85 by 100. Ans. .0985.
3. Divide 5231 by 1000. Ans. .0005231. 6. Divide 025 by 200. Ans. .000125.
4. Divide 72 by 10000. Ans. .0072. 7. Divide 412.99 by 10. Ans. 41.299.
(ſºld Dºmint Nileſ.
—-mº-º-o-º:
DEFINITIONS.
455. A Denominate Number is a concrete number which expresses a
particular kind of unit or quantity, either simple or compound.
456. A Simple Denominate Number is a number which expresses a unit
or units or quantities of but one kind or denomination: as 5 dollars, 15 boxes, 8
pounds, 4 days, etc.
457, A Compound Denominate Number is a number which expresses units
or quantities of two or more denominations, under one kind of measure: as 8 dollars,
20 cents; 12 pounds, 10 ounces; 7 days, 18 hours, 21 minutes, etc.
In whole numbers and in decimals, the law of increase and decrease, between
units of lower and of higher orders, is by the uniform scale of 10; but in compound
numbers the scale varies according to the kind of measure employed.
MEASURES.
458. A Measure is a standard unit established by law or custom, by which
quantity, such as extent, dimension, capacity, amount, or value, is measured or
estimated.
There are eight kinds of measure: º
1st. Length. 2d. Surface, or area. 3d. Solidity, or capacity. 4th. Volume.
5th. Weight, or force of gravity. 6th. Time. 7th. Angles. 8th. Money, or Value.
WEIGHT.
459. Weight is that property of matter by virtue of which a body tends
toward the centre of the earth; and the resistance required to overcome this
centralizing pressure, or gravitating tendency of matter is named weight. Weight
varies according to the quantity of matter and its distance from the centre of the
earth.
VALUE.
460. Walue is the ratio or unit of measure of wealth existing between
different commodities with reference to an exchange. It is the sole condition of
Wealth and the universal name given to the inherent quality or power of ome thing to
command another in eachange. *
Briefly expressed, value is the worth of one thing as compared with some
other thing.
(229)
23O SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. x-
MONEY.
461. Money is stamped metal called coin, or printed bills or notes called
paper money, issued by the general government of States or Nations, and supplied
to the people to facilitate trade and commerce. It is the standard measure of the
Value of things and services, and hence, it is the universal medium of exchange
and settlement among all civilized peoples. See Domestic Exchange.
462. Currency (from curro, I run), is a term applied to the money of a
nation, whether it be coin or paper money.
463. Bullion is the name given to uncoined gold and silver, and includes
bars or ingots.
464. Legal Tender. By this term is meant money which is required by
law to be accepted in payment of debts.
465. Gold Coins are a legal tender for any amount, when not below the
Standard weight prescribed by law. When gold coins are reduced by natural
abrasion, below the standard weight, they are a legal tender at a valuation propor-
tional to their actual weight.
466. Standard Silver Dollars are by Act of Congress of 1878, a legal
tender for all debts, except in contracts where other money is specified.
467. United States Bank Notes are a legal tender for all debts except
duties on imports and the interest on the public debt.
468. Silver Certificates are a legal tender for all debts and are also
receivable for customs, taxes and all public dues. *
469. National Bank Notes are not a legal tender; but as they are secured
by a deposit with the Comptroller of Currency at Washington, and are redeemable
in lawful money by the National Banks, and the Treasurer of the United States,
they are perfectly good, and are accepted without question by individuals, firms,
corporations, States and the general government, in payment of all kinds of debts,
except interest on the public debt, throughout the United States of America.
NOTE.-Standard Silver Dollars may be deposited in amounts not less than $10 with the
Treasurer or any Assistant Treasurer of the United States, and Silver Certificates received
therefor.
470. MONETARY UNITS OF DIFFERENT AGES, NATIONS AND PEOPLES.
A great variety of articles and objects have been made to serve as money in different ages and
different countries. The money, or measure of value of the ancient Greeks was an ox. We learn
from Homer that the armor of Diomede cost nine oxen, and that of Glaucus, one hundred.
History tells us, also, that when Pythagoras made his first great discovery in Mathematics—
that the square described upon the hypotenuse of a right angled triangle, was equal to the sum of
the squares described upon the other two sides—he sacrificed 100 oxen, in honor of the discovery.
Evidence is abundant to show that, th the most distant regions and at different times, cattle
formed a currency for the early pastoral and agricultural nations. The ancient Germans used
cattle; and cattle and slaves, or “living money,” were in common use among the Anglo-Saxons.
* MONETARY UNITS OF DIFFERENT AGES, NATIONS AND PEOPLEs. 23 I
With the early Italians and Romans, sheep and oxen formed the oldest medium of exchange, ten
sheep equalling one ox. In the early English and Irish civilization, the sheep and ox were the
units of exchange. The Zulus and Kaffres estimate their wealth, to-day, in the unit of cattle, and
use them as money in all transactions which inyolve payment.
The Savage tribes of Australia use greenstone, serviceable for making hatchets, and red ochre,
Serviceable for painting their bodies, as a general medium of exchange or money.
The Indians of British Columbia, use for money haiqua shells; while the Indians of Alaska
use the skins of animals. The early Carthagenians and Spartans used for money, pieces of leather
and skins, marked with a government stamp. In Chinese Tartary, cubes of tea pressed together;
in Hindostan and the Western coast of Africa, shells called cowries, and in Abyssinia, blocks of
rock salt are to this day used as money. In Iceland, dried fish, and among the North American
Indians, the belt of wampum, or Indian beads made of shells, are used as money. The monetary
history of some of the American colonies, from 1620 to 1700, shows that almost every commodity pro-
duced or manufactured, was used as money.
The early settlers of Virginia made Tobacco a legal tender for the payment of debts; and
hundreds of Virginians bought their wives from English wife sellers with tobacco, paying from 120
to 150 pounds per wife. This was undoubtedly the best use ever made of body and brain injuring
tobacco.
Following these various rude forms of money, iron, lead, copper, tin, and later, silver and
gold, were introduced by the ancients, and copper, bronze, silver and gold, and paper bills are used
by most modern nations. Russia at one time used platinum for money, but found it unsuitable.
Nearly all modern civilized nations have adopted gold and silver as the general standards for
money. They are among the most imperishable of all substances, contain great value in small
volume, and are easily transported or secreted, which are important considerations in business, and
in case of unjust or insecure government.
Our own country, and many others, find it convenient to use copper, nickel and bronze for the
Subdivisions of the monetary unit in connection with gold, silver, and paper money.
MEASURE TABLES.
471. A Measure Table is a regularly arranged Statement showing how
many times a higher denomination of a system of measurement equals the next
lower denomination of the same system. Or, in other words, it shows how many
units of each lower denomination are required to equal one unit of the next higher
denomination.
472. TABLE OF UNITED STATES MONEY.
10 Mills (m) = 1 Cent, 2. E. $ d. Ct. II].
10 Cents = 1 Dime, d. | 1 = 10 = 100 = 1000 = 10000
T0 Dimes = 1 Dollar, $. 1 = 10 = 100 = 1000
10 Dollars = 1 Eagle,” E. 1 = 10 = 100
20 Dollars = Double Eagle. 1 = 10
NOTE 1.-The mill is not coined. It is used only in computations.
NOTE: 2.—The word dollar is derived from Thal, a dale or valley; or from Joachim’s Thaler,
first coined in the mines of the Bohemian Valley of Saint Joachim, from whence it was borrowed from
the Austrians by Spain, during her union with Austria, under the Empire of Charles the W., and
which was known as the old Spanish Carolus pillar dollar, which, for ages, had been used by
bankers as a standard of value for the English monetary pound, during its many changes.
The dime is from the French word disme meaning one-tenth ; the cent is from the Latin centum,
a hundred; the mill is from the Latin mille, a thousand; the eagle the $10 coin, is the name of the
national bird. See page 38, for the Origin of the Symbol of the Dollar ($).
232 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
473. United States Money is the legal monetary measure of value in the
United States of America. The unit of the measure is the gold and the silver dollar.
474. The U. S. monetary unit, the Dollar, was established by the Continental
Congress, August 8, 1786, with the proviso that it should be decimally divided,
Which gives great facility in commercial computations. .*
475. The coin of the United States consists of gold, silver, nickel, and
bronze. The following table shows the name, value, composition and weight of each
coin, as now issued (1894) by the Mints:
NOTE.-Paragraph 5 of Section 8, of Article I, of the Constitution of the United States, gives
Congress power “to coin money, regulate the value thereof, and of foreign coins, and to fix the
standard of weights and measures;” and Section 10 of Article I, says no State shall coin money,
emit Bills of Credit, or make any thing but gold or silver coin a tender in payment of debts.
By Act of Congress, of February 25th, 1862, the issue of United States Treasury Notes was
authorized and made “lawful money and a legal tender in payment of all debts, public and private,
§ sº United States, except duties on imports and interest on bonds and notes of the
Illt;0 3,150S,
476. TABLE OF UNITED STATES COIN.
COIN. VALUE. COMPOSITION. WEIGHT.
GOLD.
Dollar. . . . . . . . . . . . 100 cents. . . . . . . . . 90 parts gold, 10 parts alloy. . . . . . . . . . . . . 25.8 grains Troy.
Quarter Eagle. ...|2+ dollars. . . . . . . . . 90 “ “ , 10 “ “. . . . . . . . . . . . . . 64.5 “ & 4
Three Dollars....!3 dollars... . . . . . . . 90 “ , 10 “ “ . . . . . . . . . . . . . 77.4 “ § {
Half Eagle. . . . . . . 5 dollars. . . . . . . . . . 90 “ “ , 10 “ “. . . . . . . . . . . . . . 129 & 4
Eagle. . . . . . . . . . . . 10 dollars. . . . . . . . . 90 * * & 4 : 10 * * “. . . . . . . . . . . . . . 258 (« & 4
Double Eagle..... 20 dollars. . . . . . . . . 90 * * “ , 10 “ “. . . . . . . . . . . . . . 516 “ 4 &
SILVER. -
Dime . . . . . . . . . . . . . 10 cents. . . . . . . . . 90 parts silver, 10 parts alloy. . . . . . . . . . . . 38.58 grains Troy.
llar. . . . . 25 ts. . . . . . . . . & 4 << , 10 “ “. . . . . . . . . . . . . .4 & 4 & 4
$º: ; ::::::::::: ; ; ; ; ; ; ::::::::::: tº . .
Dollar. . . . . . . . . . . . 100 cents. . . . . . . . . 90 * * “ , 10 “ “. . . . . . . . . . . . . 412.5 & 4 & 4
NICKEL.
3-cent piece...... 3 cents. . . . . . . . . . . 75 parts copper, 25 parts nickel. . . . . . . . 30 grains Troy.
5-cent piece. . . . . . 5 cents. . . . . . . . . . . 75 “ “ , 25 “ “. . . . . . . . . . 77.16 & 4
IBRONZE,
One cent.......... 1 cent. . . . . . . . . . . . 95 parts copper, 5 parts tin and zinc.. . . . 48 grains Troy.
NoTE. 1.-From 1786 to 1834 the $10 gold piece weighed 270 grs., 4% pure, and all other
denominations were in like proportion.
From 1834 to the present time 1894, it has weighed 258 grains, ºr pure.
NotE 2.—The silver dollar weighed, from 1786 to 1837, 416 grains, # pure, and all other
denominations in silver were in like proportion. From 1837 to 1853, it weighed 412.5 grains, Tºo
pure, and the 50c, 25c, 10c, and 5c. pieces, in like proportion. In 1851, the 3c. piece was coined, #
pure. From 1853 to the present time, the Dollar has weighed 412.5 grains, but the subdivisions
thereof were reduced in 1853, to the weights shown in the above table.
477. Silver Coin of the denominations of 58%, 25¢, 10¢, and 5%, are all
made under the standard weight to prevent their being shipped abroad and are
legal tender only for amounts not exceeding $10.
478. The Nickel 5 and 3-cent pieces, and the bronze 1-cent, are legal tender
Only for amounts not exceeding 25 cents.
NoTE.—The 5c. nickel weighs 5 grams, or 77.16 grains, and is nearly 3% of a meter in diameter;
48.6, laid side by side, measure one meter. - ©
See Report of Director of the U. S. Mint of 1893, for all the laws relating to the coinage of U.
S. money, and the weights and purity of the various coins from the founding of the government to
July 1893.
COMPARATIVE VALUES OF GOLD AND SILVER, 233
COMPARATIVE VALUES OF GOLD AND SILVER.
As shown in the above table of coins, 25.8 grains of nine-tenths pure gold
coin and 412.5 grains of nine-tenths pure silver coin are each $1.00 United States
money. Hence the monetary ratio between gold and silver in the United States is
as 1 to 15.988. i. e. silver being 1, gold is 15.988.
In England this ratio is 1 to 14.287; in France it is 1 to 15.5; and in
Germany, 1 to 13.95.
ALLOWANCE MADE FOR DEVIATION IN THE WEIGHT OF COIN.
479. In the coinage of the United States money the following allowance is
made by law for a deviation in Weight:
# grain for the Double Eagles and Eagles;
# grain for all other gold pieces;
14 grains in all silver pieces;
3 grains in the nickel 5-cent piece, and 2 grains in the nickel 3-cent
piece, and bronze 1-cent piece.
480. The old silver half dime and 3-cent pieces, the bronze 2-cent pieces,
and the nickel 1-cent pieces are not now coined.
481. The Trade Dollar was coined for Asiatic commerce, and not for
currency. The weight is 420 grains, ſº pure.
482. The ALLOY of a coin is some harder metal mixed with the gold or
silver to harden it moderately and thus lessen the wear or abrasion. Gold and
silver, in a pure state, being very soft, would rapidly wear away, were they not
alloyed.
The alloy for American gold coin is composed of about ºf silver and ſº cop-
per. The difference in color of our gold coins is because of the different quantity
of silver in the alloy. -
The alloy for silver coin is pure copper. Coin thus alloyed is called standard.
WEIGEIT OF COIN.
483. $10000 gold = 258000 gr. = 44 lbs. 9 oz. 10 pwt. 0 gr. Troy.
$1000 standard silver dollars = 412500 gr. = 71 lbs. 7 oz. 7 pwt. 12 gr. Troy.
$1000000 gold weighs 53750 ounces Troy or 3685.71 avoirdupois pounds.
$1000000 standard silver dollars = 412500000 gr. Troy or = 58928.55+
pounds avoirdupois = 3.68+ car loads of 16000 pounds each.
$1000000 silver trade dollars weigh 87.5000 ounces Troy or 60000 pounds
avoirdupois. •
$1000000 silver, half and quarter dollars, 20-cent pieces and dimes,
weigh 803750 ounces Troy or 55114.28 avoirdupois pounds.
234. SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. Yºr
VALUE OF GOLD AND SILVER. e
484. The following values are based upon the present United States mone.
tary valuation (1894) of these metals:
1 ounce Troy of pure gold is worth $20.67–H
1 pennyweight Troy of pure gold is worth 1.03+
1 ounce Troy of pure silver is worth 1.29-H
1 pennyweight Troy of pure silver is worth .0645-H
CANADA MONEY.
º 485. Canada Money is the legal currency of the Dominion of Canada. It
consists of gold, silver, and bronze coin and of paper Imoney. s
486. The silver coins are the 50¢, 25¢, 102, and 52 pieces. The bronze
coin is the 12 piece. The gold coins in use are the sovereign and half sovereign.
ENGLISH MONEY.
487. English, or Sterling Money, is the legal currency of Great Britain.
NOTE.—Sterling is derived from Easterling, the name given to the early German traders
whose money was called Easterling money.
488. The Monetary Unit of Great Britain is the pound sterling which is a
gold coin weighing 123,274 grains, 4% pure. It is equivalent to $4.8665 U. S. money.
NoTE.—For exchange purposes between the United States and England, the pound sterling
is valued by bankers at $4.86% and the rate of exchange is quoted in dollars and cents, $4.864, more
or less, according as premium is charged or discount is allowed.
TABLE OF ENGLISH MONEY.
4 Farthings (far. or qr.) = 1 Penny d. 9. &
12 Pence = 1 Shilling S. 1 = 1; - 21 = 252 = 1008.
\ 1 Sovereign sov. 1 = 20 = 240 = 960.
20 Shillings F OI’ 1 = 12 = 48.
l 1 Pound £. 1 = 4.
21 Shillings = 1 Guinea. 8.
1 Crown = 5 Shillings C.
-
NotE.-1st. The Guinea has not been coined since 1816; the term is only used in trade. 2nd.
The symbols, £. s. d. qr. or far. are the initials of the Latin words, libra, solidus, denarius and
quadrans, signifying respectively, pound, shilling, penny, and quarter. The quarter means the
farthing which is derived from a modification of the word “four-things,” the old English penny
being marked with a cross so deeply made that it could be broken into two or four pieces, called,
respectively, half-penny and four-things.
489. The money of Great Britain consists of gold, silver, copper, and Bank
of England notes, or bills.
* VALUE OF FOREIGN MONIES. 235
VALUE OF ENGLISH MONEY IN THE U. S. MONETARY UNIT.
Present legal value. Value previous to 1834.
1 Pound or Sovereign = $4.8665 $4.44;
1 Shilling == .2433 .22%
1 Penny F. .02024- .01%
1 Crown F. 1.2166-1-
1 Guinea = 5.1098-1-
NOTE.-By Act 56, George III., which is still in force, one pound of standard Troy gold (## pure)
is coined into £46. 14s. 6d. The full weight of one gold pound or sovereign is therefore 123.274
grains of standard gold, or 113.001 grains of pure gold. But, allowing for the abrasion or wear,
a sovereign Weighing but 122.75 grains of standard gold, is in England a legal tender for the
payment of debts.
A pound of silver ## pure, is coined into 66 shillings. The full weight of the shilling is
therefore 87 ºr grains standard silver, or 80 ºr grains of pure silver.
A pound of copper is coined into 24 pennies.
A pound of bronze, 95% copper, 4% tin, and 1% zinc, is coined into 40 pennies or 80 half
pennies, or 160 farthings.
Bank of England Notes are a legal tender for any sum exceeding £5. Gold is a legal tender
for any amount; silver, not exceeding 40s.; and copper not exceeding 12d., when in pennies or half
pennies, and not exceeding 6d., when in farthings.
The gold coins are the sovereign and the half sovereign. The silver coins are the crown,
half crown, 4s., 2s., 1s., and 6d.
See English Exchange for the history of English money, and a full elucidation of the subject.
FRENCEI MONEY.
490. French Money is the legal currency of France. It is based on the
decimal system, and the Unit is the silver Franc, which equals 19.3 cents U. S.
Imoney.
NOTE.-In exchange transactions between the United States and France, the rate of exchange
is the variable number of francs and centimes allowed for $1. The basis for the rate is 5.20 francs
for $1. This rate is the par of ecclvange, and is quoted more or less as premium is declared or discount
is allowed.
TABLE OF FERENCEI MONEY.
fr. do. Ct. Iſl.
10 Millimes (m.)= 1 Centime Ct. --- 1 = 10 = 100 = 1000
10 Centimes = 1 Decime do. 1 = 10 = 100
10 Decimes = 1. Franc fr. 1 = 10
NOTE.-The Millime is not coined; the term simply means the tenth part of a centime.
491. The money of France consists of gold, silver, bronze, and National
Bank notes. All French coin is based upon the unit of weight—the gramme.
A kilogramme of standard gold, ºr pure, is coined into 155 Napoleons (20
franc pieces), or 3100 francs. The gold coins are the 100, 50, 20, 10, and 5 franc pieces.
A kilogramme of silver is coined into 200 francs.
The silver coins are the 5, 2, 1, 3, and 4 franc pieces.
The copper or bronze coins of the French are the 10, 5, 2, and 1 centimes,
which weigh respectively 10, 5, 2, and 1 grammes.
The Franc is used in Switzerland and Belgium, and under different names,
in Spain, Italy, Greece, and Venezuela.
NotE.—See the subject, Foreign Exchange, in this work, for a full elucidation of French
Exchange. For the history of French money, see the Metric System, in this work.
236 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. *
GERMAN MONEY.
492. German Money is the legal currency of the German Empire.
In 1871 the German Empire established a new and uniform system of money
of which the gold “Mark” (Reichsmark), is the Unit. The Mark is equal to 23.8
cents United States money. -
493. The coin of the Empire consists of gold, silver, and nickel. Paper
money is also used.
TABLE TOF GERMAN MONEY.
100 pfennige, marked Pf., make 1 mark, marked RM.
NOTE.-In exchange transactions with the German Empire, for convenience, bankers base
the rate of exchange upon the equivalent value of 4 marks expressed in dollars and cents.
The exchange par of 4 marks is 954 cents. The rate of exchange is 954, more or less, accor-
ding as premium is charged or discount is allowed.
See the subject, Foreign Exchange, in this work for further information of German money
and a full elucidation of German, English, French, Italian, Austrian, Belgian, Russian, Spanish,
Grecial), Portugese, Japanese, Chinese, and Brazilian, and other South American Exchange.
MEASURE OF TIME.
494. 1.-Time is a measured portion of duration.
495. 2.-The Unit of measure is the mean solar day.
496. 3.—A Year is the time of the revolution of the earth around the sun.
497. 4.—A Day is the time of the revolution of the earth on its axis.
498. 5.—The Solar Day is the interval of time between two successive
passages of the sun across the same meridian of any place, and they are of unequal
length on account of the unequal orbital motion of the earth and the obliquity of
the ecliptic.
499. 6.—The Mean Solar Day is the mean, or average length of all the
Solar days in the year. Its duration is twenty-four hours.
500. 7.—The Civil, or Legal Day used for ordinary purposes, and which
corresponds with the Mean Solar Day, commences at midnight and closes at the
Inext midnight. e
501. 8.-The Astronomical Day commences at noon, and closes at the
Inext in OOn.
502. 9.—The Solar Year is 365 days, 5 hours, 48 minutes, 49.62+ seconds.
503. 10.—The Common, or Civil Year consists of 365 days for 3 successive
years, every fourth year containing 366 days, one day being added for the excess of
the Solar Year over 365 days. This intercalary day is added to the month of
February, which then has 29 days, and the year is called bissextile or leap year.
504. 11.—Leap Year. To determine what years are leap years, the following
regulation has been adopted:
Every year that is divisible by 4 is a leap year, unless it ends with two
naughts, in which case it must be divisible by 400 to be a leap year. Thus, 1884,
1776, and 1600 were, and 2000, 2400, and 2800 will be leap years; but 1885, 1794, and
1800 were not, and 1900, 2100 and 2200 will not be leap years.
X- BRIEF HISTORY OF THE YEAR AND ADJUSTMENT OF CALENDAR. 237
505. A BRIEF EIISTORY OF THE YEAR AND OF THE ADJUSTMENT
OF THE CALENDAR.
The necessity of some division and measurement of time was early felt, and in all ages the
subject has received the attention of the most learned men.
The phases of changes of the moon supplied a natural and very obvious mode of dividing and
reckoning time, and hence the division into months of 29 or 30 days was, perhaps, the earliest and
most universal. But it would soon be observed that, for many purposes, the changes of the seasons
were more serviceable as marks of division; and thus arose the division into years, determined by
the motions of the sun. It was, however, soon discovered that the year, or larger division, did not
contain an exact number of the smaller divisions or months, and that an accommodation was neces-
sary; and various not very dissimilar expedients were employed for correcting errors that arose.
The ancient Egyptians had a year determined by the changes of the seasons, without refer-
ence to the changes of the moon, and containing 365 days, divided into 12 months of 30 days
each, with 5 supplementary days at the end of the year. The Jewish year consisted, in the earliest
periods, as it still does, of twelve lunar months, a thirteenth being from time to time introduced,
to accommodate it to the sun and seasons; this was also the case with the ancient Syrians, Mace-
donians, etc.
The Jewish months have alternately 29 and 30 days; and in a cycle of 19 years there are 7
years having the intercalary month ; some of these years having also one, and some two days more
than others have, so that the length of the year varies from 353 to 385 days.
The Greeks, in the most ancient periods, reckoned according to real lunar months, twelve
making a year; and about 594 B. C., Solon introduced in Athens the mode of reckoning alternately
30 and 29 days to the month, accommodating this civil year of 354 days to the solar year by the occa-
sional introduction of an intercalary month. A change was afterwards made, by which three times
in eight years a month of 30 days was intercalated, making the average length of the year 365+
days.
The divisions of calendars of time now in use in America, and throughout Europe, are the
same as those of the Romans. It is supposed that Romulus, the founder of Rome, first undertook
to divide the year in such a manner that certain epochs should return periodically after a revolu-
tion of the sun; but the knowledge of astronomy was not then sufficiently advanced to allow this
to be done with much precision. He placed the commencement of the year in spring, and divided
it into 10 months: March, April, May, June, Quintilis, Sextilis, September, October, November, and
December. March, May, Quintilis, and October, contained thirty-one days each; the other six
contained only thirty. The names Quintilis and Sextilis remained in the calendar till the end of
the republic, when they were changed into July and August; the former in honor of Julius Caesar,
and the latter, of Augustus. --
The year of Romulus contained only 304 days. Numa Pompilius, 672 B. C., added two
months: January to the beginning of the year, and February to the end. About the year 452 B.C.,
this arrangement was changed by the Decemvirs, who placed February after January; since that
time the order of the months has remained undisturbed. In Numa’s year, the months consisted of
29 and 30 days alternately, to correspond with the synodic revolution of the moon. The year
would therefore consist of 354 days; but one day was added to make the number odd, as being
more lucky. In order to produce a correspondence with the solar year, Numa ordered an intercalary
month to be inserted every second year, between the 23d and 24th of February, consisting alternately
of 22 and 23 days. Had this regulation been strictly adhered to, the mean length of the year would
have been 365+ days, and the months would have continued for a long time to correspond with the
$3,1116 SeaSOIlS.
But a discretionary power over the intercalary month was exercised by the pontiffs, who fre-
quently abused it for the purpose of hastening or retarding the days of the election of magistrates;
and the Roman calendar continued in a state of uncertainty and confusion till the time of Julius
Caesar, when the civil equinox differed from the astronomical by three months.
|Under the advice of the astronomer Sosigenes, Caesar, 46 B. C., abolished the lunar year, and
238 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. Y
regulated the civil year entirely by the sun. He decreed that the common year should consist of
365 days, but that every fourth year should contain 366. In distributing the days among the different
months, he ordered that the odd months, that is the first, third, fifth, seventh, ninth, and eleventh,
should contain each 31 days, and the other months 30; excepting February, which in common years
was to contain only 29 days, but every fourth year 30 days. This natural and convenient arrange-
ment was interrupted to gratify the frivolous vanity of Augustus, by giving August, the month
named aſter him, an equal number of days with July, which was named after the first Caes - c.
The intercalary day, which occurred every fourth year, was inserted between the 24th and
25th of February. According to the peculiar and awkward manner of reckoning adopted by the
Romans, the 24th of February was called the sixth before the calends of March, secto calendas, i. e.
six days before March 1st.
In the intercalary year, this day was repeated and called bis-secto calendas ; whence the term
bissextile, which means two days, both of which are six days before March 1st. The corresponding
English term leap year appears less correct, as it seems to imply that a day was leapt-over instead of
being thrust in. It may be remarked, that in the ecclesiastical calendar, the intercalary day is still
inserted between the 24th and 25th of February.
The Julian year consisted of 365+ days, and consequently differed in excess by 11 minutes
10.38 seconds, from the true solar year, which consists of 365 days, 5 hours, 48 minutes, 49.62+-seconds.
In consequence of this difference, the astronomical equinox in the course of a few centuries,
sensibly fell back towards the beginning of the year. In the time of Julius Caesar, it corres-
ponded to the 25th of March; in the sixteenth century, it had retrograded to the 11th.
The correction of this error was one of the purposes sought to be obtained by the reforma-
tion of the calendar effected by Pope Gregory XIII., in 1582. By suppressing 10 days in the
calendar, calling October 5th the 15th, Gregory restored the equinox to the 21st of March, the day
on which it fell at the time of the Council of Nice, in 325; the place of Easter and the other mov-
able.church feasts in the ecclesiastical calendar having been prescribed at that Council.
And in order that the same inconvenience might be prevented in future, he ordered the inter-
calation which took place every fourth year to be omitted in centennial years, excepting such as
could be divided by 400 without a remainder, as shown above in Article 504.
The Gregorian method of intercalation thus gives 97 intercalations in 400 years; consequently,
400 years contain: 400 × 365 + 97 = 146097 days, and therefore the length of one year is 365.2425
days. or 365 d. 5 h. 49 m. 12 sec., which exceeds the true solar year by 22.38 sec., an error which
amounts only to one day in 3866 years.
The Gregorian calendar was received immediately or shortly after its promulgation by all the
Roman Catholic countries of Europe.
The Protestant States of Germany and the kingdom of Denmark adhered to the Julian Calen-
dar till 1700; and in England and the English American colonies, the alteration was successfully
opposed by popular prejudice till 1752. In that year the Julian calendar, or old style, as it was
called, was formally abolished by the Act of Parliament, and the date used in all public transactions
rendered coincident with that followed in other European countries, by enacting that the day follow-
ing the 2d of September of the year 1752, should be called the 14th of that month. When the
alteration was made by Pope Gregory, it was only necessary to drop 10 days; the year 1700 having
intervened, which was a common year in the Gregorian, but a leap year in the Julian calendar, it
was now necessary to drop 11 days.
The year thus adjusted is called the new style; and the Julian year, before the adjustment, is
called the old style. The old style is still adhered to in Russia and the countries following the com-
munion of the Greek church; the difference of date in the present century amounts to twelve days.
Thus, when it is January 1st in Russia and Greece, it is January 13th in other countries.
The civil or legal year, in England and the English American colonies, formerly commenced
on the 25th of March, the day of the Annunciation, though the historical year began on the 1st of
January, the day of the Circumcision. By the Act of Parliament for the alteration of the style, in
1752, the beginning of the year was transferred to the 1st of January.
X- BRIEF HISTORY OF. THE YEAR AND ADJUSTMENT OF CALENDAR. 239
A CALENDAR.
506. A Calendar is a division of time into certain periods adopted to the
purposes of civil life, as hours, days, weeks, months, years, etc.
The names and orders of the months, and the number of days contained in
each, are now as follows:
Names. No. No. days. Names. No. No. days. Names. No. No. days.
January, 1st, 31 May, 5th, 31 September, 9th, 30
Rebruary, 2d, 28 June, 6th, 30 October, 10th, 31
March, 3d, 31 July, 7th, 31 November, 11th, 30
April, 4th, 30 August, 8th, 31 December, 12th, 31
The number of days in each, may be readily remembered by committing to
memory the following lines:
“Thirty days hath September,
April, June, and November;
And all the rest have thirty-one,
Excepting February alone;
To which we twenty-eight assign,
Till Leap Year gives it twenty-nine.”
507. The Seasons. March, April, and May, are called Spring. June, July,
and August, are called Summer. September, October, and November, are called
Autumn, and December, January, and February are called Winter.
508. TABLE OF TIME MEASURE.
60 Seconds (sec.) = 1 Minute... . . . . . min. yr. mos. wº. ds. g his. min. S60.
º = º º 1 = 12 = }; = 8760 = 525600 = 31536000
7 Days E i Wº......... wi. T † 366 = 8784 – 527040 = 31622400
365 Days = 1 Common year...yr. 1 = | = lº = lſº = *:::::
366 Days = 1 Leap Year. . . . . . yr. * 1 = 60 = 3600
12 Calendar months = 1 Civil Year. . . . . . yT. * i = 60
10 Years = 1 Decade. . . . . . . . . . d. -
10 Decades or 100 Years = 1 Century. . . . . . C.
509. DERIVATION OF THE YEAR, MONTH, WEEK, DAY, ETC.
The term Year is derived from the Saxon Gear, and is the time of the earth's revolution
around the Sun. *
The term Century is derived from the Latin Centuria, and means a collection of a hundred
things. *
The term Month is derived from the Saxon Monath, from Mona, the Moon, and means the tº
of one revolution of the moon around the earth. tº a º
Week. The origin of the term week, is not known. It is more ancient than the writings of
Moses, and without proper evidence it has been thought by some to have its origin in the phases
of the moon, or the number of planets known to the Chaldeans and Egyptians, or to the motion of
celestial bodies. Seven days have been used as a division or period of time by most eastern nations
from time immemorial. The Greeks, however, did not use it, and it was not introduced into Rome
until after the reign of Theodosius. g tº tº
The Egyptian week commenced with Saturday. When the Jews took their flight from Egypt,
from their hatred to their ancient oppressors, they made Saturday the last day of the week, and the
Christians have so continued it.
24O SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS, *
*
The term Day is derived from the Saxon Daeg, and means the time of the revolution of the
earth upon its axis.
The term Hour is derived from the Latin hora, and means a definite space of time fixed by
natural laws. >
The term Minute is from the Latin minutum, (very small). The sixtieth part of an hour of
time; also the sixtieth part of a degree of angular space.
Second, from the Latin secundus, (the next after the first). The sixtieth part of a minute of
time, or of a degree of angular space; that is, the second regular subdivision of the hour or of the
degree.
510. DERIVATION OF THE NAMES OF THE MONTHS.
January, was so named in honor of Janus, the Roman deity who presided over gates and doors,
and who was naturally presumed to have something to do with the opening of the year. This deity
was represented with two faces, one looking backward, the other forward, implying that he stood
between the old year and the new, with a regard for each.
February derives its name from the circumstance that during the month occurred the Roman
religious festival called the Lupercalia, and also Februalia, from februare, to purify.
March, the third month of the year, but originally the first, was named Martius, in honor of
Mars, the god of war and the reputed father of Romulus, the founder of Rome.
April. The Romans gave this month the name of Aprilis, from aperire, to open, because it
Was the season when the buds began to open and the earth to produce vegetation.
May. The common belief is that May was named in honor of Maia, the mother, by Jupiter,
Of the god Hermes, or Mercury; but the most reliable authority is that it was named in honor of
the majores, or maiores, the senate in the original constitution of Rome.
June. The name of this month has been claimed in honor of Juno; but better evidence
proves that it was dedicated a junioribus, that is, to the junior or inferior branch of the original
legislature of Rome. ' º
July was originally designated Quintilis, in reference to its fifth place in the calendar; but
having been the month in which Julius Caesar was born, after the death of this emperor, the
Romans changed the name to July, in honor of that great warrior.
August, in the old Roman calendar was called Sextilis, it being the sixth month in the ten
which constituted the year. But it was subsequently named August, in honor of the emperor
Augustus, and one day was taken from February and added to August to make 31, the same as July,
the month named in honor of Caesar.
September, October, November, and December, are named, respectively, from the Latin numerals,
Septem, Octo, Novem, and Decem, as when the year consisted of but ten months and began in March,
they were the seventh, eighth, ninth and tenth months, as their Latin names indicate.
511. DERIVATION OF TEIE DAYS OF THE WEEK.
Sunday is so called because it was anciently dedicated to the sun by the sun worshipers.
Romans, Saxons, etc. In the 4th century, A. D., it was made the Christian Sabbath by Constantine,
who left the Sun worshipers and became a Christian.
Monday, from the A. Saxon, monami-daeg, moon day; a day sacred to the moon as Sunday was
to the sun.
Tuesday derived from Tuisco, the Saxon god of war.
Wednesday derived from the Scandinavian deity, Woden, whose functions corresponded to those
of Mercury, in the Greek and Roman Mythology.
Thursday, named for the Scandinavian god Thor, the god of Thunder. Thor was equivalent
to Jupiter in the Greek and Roman Mythology.
Friday, so named in honor of Freya, a Saxon goddess, the wife of Woden or Oden.
MEASURES OF EXTENSION. 24. I
Saturday is so called from being dedicated by the Romans to Saturn, the god of Agriculture,
and who was the father of Jupiter, Neptune, and Pluto, and in whose honor were held the Saturn-
alian festivals.
Thus we see that the English names of the week are derived from the ancient Saxons, who
borrowed the week from some eastern nation, and substituted the names of their own divinities
for those of the gods of Greece and Rome.
512. THE FOLLOWING ARE THE NAMES OF THE WEEK IN LATIN, SAxon
AND ENGLISH:
~– LATIN.——s SAXON.—, ENGLISH.—N
Dies Solos, Dies Mercurii, Sun's day, Woden's day, º Sunday, Wednesday,
Dies Lunae, Dies Jovis, Moon's day, Thor's day, Monday, Thursday,
Dies Martis, Dies Veneris, Tiw’s day, Friga's day, Tuesday, Friday,
Dies Saturni. Seterne's day. Saturday.
MEASURES OF EXTENSION.
513. Extension is that property of matter by which it occupies space. It
may have one or more of the three dimensions—length, breadth, and thickness.
514. The Yard is the United States and English unit of measure of exten-
Sion, whether for a line, surface or solid. A yard is a rod or tape whose length is
equivalent to ###### of the length of a pendulum vibrating seconds of mean time
in the latitude of London in a vacuum at the level of the sea, when the temperature
is 620 Fahrenheit; or differently stated, a pendulum that vibrates seconds under
the conditions given, is divided into 391393 equal parts, and 360000 of these parts
are taken for a yard, i. e. the second’s pendulum is 39.1393 inches in length of which
the yard consists of 36 of the 39 inches.
NoTE 1.-The yard was thus determined in England, during the reign of William IV., and
has been adopted by, and is now in general use throughout, the United States. e
2.—By joint resolution of Congress, June 14th, 1836, the Secretary of State was directed to
furnish to the Governors of each State, for the use of the States, a complete set of all the weights
and measures adopted by the United States; this was done and the statutory standards of all the
States have been made conformable thereto, except the bushel of New York State, and the barrel
of Massachusetts.
3.—The metric system was legalized by the United States iu 1866, but its use was left
optional with the people, and is used only by certain professional and Scientific men
NOTE.-See Metric System, in this book.
515. A Line has only one dimension—length.
516. A Surface has two dimensions—length and breadth.
517. A Solid, or volume has three dimensions—length, breadth, and
thickness. -
LINE, OR LINEAR MEASURE.
518. Line, or Linear Measure is used to measure distances, or length,
in any direction. & -
One inch. Two inches.
Three inches.
Note.—For the operations of linear, surface, and solid measure, see “Mensuration of Surfaces
and Solids,” in this book.
242 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS.
TABLE
L. m. fur. rā. yd. . . ft. in.
12 Inches (in.) = 1 Foot. . . . . . . . . . . . . . . . ft. 1–3–24=960=5280=15840=190080
3 Feet = 1 Yard. . . . . . . . . . . . . . . . d. 1= 8=320=1760= 5280= 63360
5+ Yards, or 164 feet = 1 Rod, or Pole. . . . . . . . rd., or P. 1= 40– 220– 660– 7920
40 Rods = 1 Furlong. ... -- . . . . . . fur. (obsolete) 1= 5}= 16}= 198
8 Furlongs (320 rds.)= 1 Mile (Statute Mile). mi. 1= 3= 36 -
3 Miles = 1 League. . . . . . . . . . . . . L. 1= 12
NotE.—The units of length are nearly all derived from different parts of the human body,
and from other objects: Thus, the foot was originally the length of the king's foot, and consequently
varied as a king with a long or a short foot chanced to reign. The span was the space between the
end of the thumb and little finger, when both were extended.
A hand was the width of the hand across the palm. The cubit was the length of the arm
from the elbow to the end of the middle finger. The fathom was originally the distance between
the ends of the two middle fingers, when the arms and fingers were extended to their farthest limit.
Henry I., king of England, declared that the ell, or yard, should be the length of his own arm, from
the end of the longest finger to the middle of the breast, and that the other measures should be
. from this. An acre was originally as much land as could be ploughed in a day by a yoke
OI OX62E1,
Furlong, now obsolete, is from fur. meaning furrow; and long, i. e. the length of a furrow.
Rod (16+ ft.) came from a rod or stick used for measuring. An inch was, with Edward II., in 1324,
the length of 3 barley corns, sound and dry; and 12 of such inches was a foot.
See Mariners', Cloth, and Shoemakers’ Measures; also Table of Comparative Weights,
Measures and Values, and the old French and Spanish Measures, in this book.
MARINERS’ MEASURE.
519. Mariners’ Measure is used to measure distances at Sea and also to
measure the depths of seas.
TABLE.
6 Feet = 1 Fathom.
120 Fathoms = 1 Cable-length.
880 Fathoms, or 7; Cable-lengths = 5280 ft. = 1 mile.
520. A Minute of the earth's circumference is called a geographical, or
nautical mile, or knot, which is gº of #6 = grºwd of the circumference of the earth.
The circumference of the earth at the equator is 24899 miles, which, divided by
21600, gives 1.15273+ statute miles.
The length of a degree at the equator is 1.15273× 60 equals 69.1638 statute
miles. wº
SEHOEMAIKERS’ MEASURE.
521. Shoemakers’ Measure is used by shoemakers to measure the human
feet and in the manufacture of boots and shoes.
522. The Unit of Measure is # of an inch which is the same as the former
unit of 1 barley corn, when 3 barley corns made 1 inch.
No. 1, small size, is 4; inches; and every succeeding number increases #
of an inch to 13.
No. 1, large size, is 84+ inches; and every succeeding number increases #
of an inch to 15.
Y TABLES OF MEASURES, 243
523. MISCELLANEOUS UNITS of LINEAR MEASURE.
NOTE:-The barley corn was formerly a unit of measure in England; 3 well matured barley
corns placed together, lengthwise = 1 inch.
TABLE.
*; of an Inch = A Line (American). 9 Inches = A Span.
+% of an Inch = A Line (French). 3 Feet = A Pace.
4 Inches = A Hand. 24 Feet (28 in.) = A Military Pace.
3 Inches = A Palm. 18 Inches = A. Cubit.
NOTE:-The cubit of the Egyptians, about 3000 B.C. was 20.63 inches. The cubit of Moses
varied at different periods from 18 to 25.2 inches. Egypt, Chaldea and other nations had regular
though varying systems of weights and measures.
CLOTEI MEASURE.
524. Cloth Measure is used to measure all kinds of goods sold by the yard.
TABLE.
--" yd. qr, na. in.
2# Inches (in.) = 1 Nail - - na. 1 = 4 = 16 = 36
4 Nails (9 inches) = 1 Quarter - qr. 1 = 4 = 9
4 Quarters = 1 Yard - yd. 1 = 2+
This table formerly contained:
NoTE 1.-The Flemish ell, or yard, which equaled 3 quarters or 27 inches; the English ell, or
yard, which equaled 5 quarters or 45 inches; the French ell, or yard, which equaled 6 quarters or
54 inches.
NOTE 2.—All of the above units of measure are now out of use except the yard, which is
divided into halves, quarters, eighths, sixteenths, etc., in place of feet and inches. At the Custom-
house, the yard is decimally divided.
NOTE 3.—The nail of 2+ inches is derived from the six-penny nail of commerce, which is, or was
formerly, 2+ inches long.
The word penny when applied to nails refers to the size and is a corruption of pound. Thus
By a six-penny nail is meant a nail, 1000 of which will weigh 6 pounds. And by the 4, 8, 10 and
double 10-penny is meant that 1000 will weigh respectively 4, 8, 10 and 12 pounds.
525. SQUARE, OR SURFACE MEASURE.
3 Feet.
One
# Squar |e
Yard
3 ft. x 3 ft. = 9 sq. ft. = 1 sq. yd.
526. Square, or Surface Measure is used in computing surfaces or areas.
527. A Surface has length and breadth, but no measurable thickness.
528. The Area of a surface is the quantity of surface it contains, and is
expressed by the product of the length by the breadth.
529. A Square is a plane figure bounded by four equal sides, and having
four right angles.
244 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS.
530. TABLE.
144 (12 x 12) Square Inches (sq. in.) = 1 Square Foot - - sq. ft. *
9 (3 x 3) Square Feet = 1 Square Yard - sq. yd.
304 (54 × 53) Square Yards GEº { 1 Square Rod - - sq. rā.
or 2724 sq. ft. * --> Or Perch - - P.
160 Square Rods = 1 Acre - - - - A.
640 Acres = 1 Sq. Mile, or Section... sq. mi., or sec.
sq. mi. 3. sq. rà. Sq. yd. sq. ft. Sq. in.
1 = 640 = 102400 = 3097600 = 278784.00 = 4014.489600
1 = 160 = 4840 = 43560 = 6272640
1 = 30+ = 272} = 39.204
1 = 9 = 1296
1 = 144
NotE.—The term Perch is from the French perche, a pole or rod, used for measuring, and is a
surface equal to a square rod.
531. Architects, carpenters, slaters, and some other mechanics frequently
measure their work by the square, which is a space 10 ft. by 10 ft., equaling 100
Square feet.
532. A square foot, yard, or mile, is a square each side of which is one
foot, yard, or mile.
3 Feet. The number of small squares contained in any large square is equal to
the product of the number of units in one side, multiplied by the number
of units in the other side. Thus, in figure 2, each side of which is three
feet, there are 9 square feet.
The difference between 84ware feet and feet 8quare, 8quare miles and
miles square, etc., for any unit of measure, is not generally understood,
and because of its practical importance we solicit special attention to it.
By 3 yards square is meant a square figure, each side of which is 3 yards;
but by 3 square yards is meant 3 small Squares, each a yard long and a
Fig. 2. yard wide.

:
Figure 2 is 3 feet square, and contains 9 square feet.
The difference between 500 rods square and 500 square rods is 249500 square rods.
There is no difference between 1 yard square and 1 square yard, 1 foot square and 1 square
foot, etc., of any unit of measure, but increase the measure above 1 unit and there is a difference.
NOTE.-Formerly, but now obsolete, 40 square rods or perches = 1 rood. 4 roods = 1 acre.
CHAIN MEASURE, OR SURVEYORS AND ENGINEERS, MEASURE.
533. Chain Measure is used by surveyors and topographical engineers, in
measuring land, laying out roads, etc.
NoTE.—The chain here used is called Gunter's Chain and is named after Edmund Gunter, an
eminent English mathematician, born 1581.
534. TABLE.
7.92 Inches = 1 Link - - - 1.
25 Links F. 1 Rod, or Pole - rid., or P.
4 Rods, or te
66 feet” } – 1 Chain - - - ch.
80 Chains = 1 Mile - - - mi. (5280 ft).
º: UNITED STATES GOVERNMENT LAND, .. 245
Engineers commonly use another chain or tape line, which consists of 100
links, each 1 foot long.
mi. ch. rd. 1. in.
1 = 80 = 320 = 8000 = 63360
1 = 4 = 100 = 792
1 = 25 = 198
1 = 7.92
535. Surveyors’ Square Measure is used by surveyors' in measuring the
area or surface of land. The acre is the measuring unit for land.
TABLE.
625 Square Links (sq. 1.)
16 Square Rods
10 Square Chains
1 Square Rod, or Pole. sq. rd., or P.
1 Square Chain - - sq. ch.
1 Acre - - - - - A
:
"...º. Miles (6 1 Square Mile, or Sec. sq. mi., or Sec.
T13.1°C WL116S e
. es Squar§ } = 1 Township - : - Tp.
also 10000 Square Links, sq. L. = 1 Square Chain - Sq. Ch.
10 Square Chains = 1 Acre - - - - A.
Tp. sq. mi. A. Sq. ch. Sq. rd. Sq. 1.
1 = 36 = 23040 = 230400 = 3686400 = 2304000000
DNITED STATES GOVERNMENT LANDS.
536. In 1802, Col. Jared Mansfield, the Surveyor General of the North-
western territories, instituted a very convenient system of laying out government
lands. The system adopted is as follows:
FIRST.—A line is run North and South and is called the Principal Meridian,
P. M.
SECOND.—Other lines are then run 6 miles apart and parallel to the P. M.
THIRD.—A line is run on a parallel of latitude East and West which is called
the base line, B. L. ->
POURTH.—Other lines are then run 6 miles apart and parallel to the B. L.,
thus dividing the land into squares, 6 miles each way, called townships, each of
which contains 36 square miles.
A line of townships running North and South is called a Range which is
designated by its number East or West of the P. M.
Townships are also designated by their number North or South of the B. L.
Townships are also divided into square miles, called sections, which are
numbered from 1 to 36, beginning on the North-east corner of each township.
Each section is subdivided into half sections, quarter sections, half quarter
sections, and quarter quarter sections.
537. TABLE.
1 Township = 6 mi. Square = 36 Sq. mi. = 23040 A.
1 Section = 1 “ {{ = 1 {{ = 640 € $
1 Half Section = 1 “by 3 mi. = } “ = 320 4%
1 Quarter Section = 4 “ “ , “ = + “ = 160 “
1 Half Quarter Section = 4 “ “ + “ = } “ = 80 “
1 Quarter Quarter Section = # “ “ # “ = ** “ = 40 “
246
SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS, Yºr
The following diagrams will make clear the foregoing statements regarding
.." and sections, and the manner of designating the subdivisions of
SęCTIOILS :
538. DIAGRAM SEIOWING THE DIVISION OF THE GOVERNMENT
LANDS OF THE UNITED STATES INTO TOWNSEIIPS.
Figure 1.
N Ea:planation.—In fig-
fi ure 1, W. E. is the base
º line. N. S. is the prin-
cipal meridian. R. 1
E. is range 1, east ; R.
2 E. is range 2, east ;
2 3 R. 3 W. is range 3,
E west; T. 3 N. is town-
2 3 T. 1 S
ship 3, north; T. 1 S.
| is township 1, south ;
B. A. is township 2, north,
range 2, west; B. is
township 3, South,
range 2, east.
A SECTION ENLARGED,
539.
One Mile Square.
A TOWNSHIP,
Six Miles Square.
N N
6 5 4 3 2 1
N. : Sec.
7 8 9 10 11 12
320 Acres.
18 17 16 15 14 13
E 1 5: N. E.
19 20 2 22 23 24 of
W. 4. S. E. #
* 40 A.
30 29 28 27 26 25 S. W. + Sec. of
160 Acres. S. E. #
31 32 33 34 35 36 80 A.
S
The above section represents one of the 36 sections
of a township.
NoTE.—1. In some parts of Louisiana and Missouri, that were settled by the French, the old
French arpent is still used, instead of the acre, as the unit in measuring land... 2. An arpent (Paris)
contains 100 sq. rods, (Paris) or 36804.12+ sq. ft. (English), and is 191.1844 ft. (English) on each side.
3. An acre contains 160 sq. rods, or 43560 sq. ft., and is 208,7103-H ft. on each side. These figures
show that an arpent (Paris) is .8449–H per cent., a little more than } of an acre.
See Table of French and Spanish Measures, and Table of Comparative Weights, Measures, and
Values, in this book.

X- SOLID, OR CUBIC MEASURE. 247
SOLID, OR CUBIC MEASURE.
540. Solid or Cubic Measure is used in measuring the contents, or volume,
of Solids.
541. A Solid, or Body, has length, breadth, and thickness.
542. A Cube is a solid, bounded by six equal square sides, or faces; hence
its three dimensions are equal to each other.
543. The Contents, or Volume, of a regular body is expressed by the
product of its length, breadth, and thickness.
544. TABLE.
1728 Cubic Inches (cu. in.) = 1 Cubic Foot - - - - cu. ft.
27 Cubic Feet (3× 3x3) = 1 Cubic Yard - - - - cu. ya. ,
16 Cubic Feet = 1 Cord Foot - - - - cq. ft.
8 Cord Feet, or 128 cubicft. = 1 Cord of Wood - - - ca.
24; Cubic Feet, or 164 feet
long, 14 ft. high, and K = 1 Perch of Stone - - - Peh.
1 foot wide
40 Cu. Ft. in United States, and 42 Cu. Ft. in England, 1 Shipping Ton.
100 Cu. Ft., 1 Register Ton, used in estimating the tonnage of vessels.
cu. ya... cu. ft. cu. in. cd. c.d. ft. cu. ft. cu. in.
= 27 = 46656 1 = 8 = 128 = 221184
tri NOTE.-A cord of wood is so named from the fact that originally it was measured by a cord or
String.
545. A Square of earth is a cube 6× 6×6 = 216 cubic feet.
546. In civil engineering, the cubic yard is the unit for measuring excava-
tions, embankments, and levees. A cubic yard of earth is called a load.
547. In commerce, the cubic foot is often the unit for computing freight
charges.
LIQUID, OR WINE MEASURE.
548. Liquid or Wine Measure is used in measuring molasses, wine, oil,
etc., and in estimating the capacities of cisterns, reservoirs, etc.
549. The Unit of measure for liquids is the gallon, which contains 231
cubic inches.
550. The Gallon of the United States is the standard or Winchester wine
gallon of 231 cubic inches, and contains 8.3388822 avoirdupois lbs., or 58,372.1754
Troy grains of distilled water at 39.83° F., the barometer being at 30 inches. It is
equal to 3.785207 litres. The gallon of the State of New York is of the capacity of
8 lbs. of -pure water at its maximum density 39.2° F., or 221.184 cubic inches. It is
equal to 3.62436 litres.
551. TABLE.
4 Gills (gi.) = 1 Pint - - pt. gal. qt. pt. gi.
2 Pints = 1 Quart t. 1 = 4 = 8 = 32
4 Quarts = 1 Gallon - gal. = 231 cubic in 1 = 2 = 8
314 Gallons = 1 Barrel - bbl. - 1 = 4
2 Barrels, or 63 gallons 1 Hogshead hind.
NOTE.—1. In the old tables, 42 gallons = 1 tierce; 84 gallons = 1 puncheon; 2 hogsheads =
248 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS. X.
1 pipe or butt; and 2 pipes = 1 ton. But these measures are no longer used as units. They are
casks of variable capacity, and in commerce, are usually gauged separately and have their capaci-
ties in gallons marked on them.
2. The barrel is used as a unit of measure only in estimating the capacity of Wells, cisterns,
vats, etc.
'3. In Massachusetts, the barrel contains 32.28 gallons.
4. This standard unit for liquid measure is called the Winchester gallon, from the standard
having been kept at Winchester, England. It was called the wine measure, because wine was
measured by it, while beer was measured by a larger measure.
5. The gallon is from the Latin gelo, a vessel for holding wine. The quart is derived from
the Latin quartus, a fourth; the pint is derived from the Greek pinto, to drink; and the gill is from
the Latin gilla, a glass to drink from.
NOTE.—The units of the old beer and milk measure, now nearly obsolete, are :
2 pints = 1 quart; 4 quarts = 1 gallon = 282 cubic inches; 36 gallons = 1 barrel; 54 gallons
= 1 hogshead.
DRY MEASURE.
552. Dry Measure is used to measure grain, fruit, vegetables, etc.
553. The Unit of dry measure is the bushel, which contains 2150.42 (prac-
tically 2150.4) cubic inches.
NOTE.—This United States standard unit for dry measure is the old English Winchester
bushel, so called from the standard being kept at Winchester. It is a cylinder 18% inches in diame-
ter and 8 inches deep. It contains 543391.89 standard Troy grains, 2150.42 cubic inches, or 77.6274
ºis pounds of distilled water at the temperature of 39.83° Fahrenheit, and barometer
II].C. (162S,
TABLE. © *
2 Pints (pt.) = 1 Quart - - - - qt. bu. pk, qt. pt.
8 Quarts = 1 Peck - - - - pk. 1 = 4 = 32 = 64
4 Pecks = 1 Bushel - - - bu. 1 = 8 = 16
8 Bushels (480 pounds) = 1 Quarter - - - qr. 1 = 2
36 Bushels = 1 Chaldron - - ch
NotE.—1. The heaped bushel, the cone of which is 6 inches above the brim of the measure,
contains 2747.7 cubic inches.
2.—When corn is bought or sold in the shuck, 2% to 4 cubic feet as per agreement, are allowed
for a bushel, according to the quantity of shuck on the corn and the density of the corn in the
crib. If the corn is shucked or husked, 2 cubic feet are allowed for a bushel.
3.—In many sections of the country, 2 heaped bushels of corn “in the ear” are called 1 bu. of
shelled corn. In other places, 2 even bu. of “ears” are allowed for 1 bu. of shelled corn.
4.—In Maryland, 5 bushels of corn make a barrel of corn.
554. The dry gallon, or half peck, contains 268.8 cubic inches; and 10 dry
gallons, or 2688 cubic inches make a bushel of coal in Pennsylvania, Ohio, Kentucky,
and several other states, when gauging coal barges, or when measuring heaps
or piles of coal. -
The dry gallon of 268.8 cu. in. is also one-eighth of the American standard
bushel.
In Louisiana, the custom, when selling coal, is to allow 2.6 bushels of 2688 cu.
in. = 6988.8 (6988.865) cubic inches for a barrel. See page 409.
Article 3912, of Weights and Measures, Revised Statutes of Louisiana, says: “That Weights
and Measures are to be procured by the Governor, like those of the United States, and deposited in
the office of the Secretary of State.”
Article 3925, says: “There shall be in this State a dry measure to be known under the name
of barrel, which shall contain 34 bushels, according to American standard.”
MEASURES OF WEIGHT, 249
Article 3926, says: “Coal shall be sold by the barrel or bushel measure.” Accordingly, in
Louisiana, a barrel of coal is 34 times 2150.42 cubic inches, = 6988,865 cubic inches.
In England, the standard Imperial bushel contains 2218.192 cubic inches; the Imperial quar-
ter is 480 pounds; and the Imperial gallon, both dry and liquid measure, contains 277.274 cubic
inches, which is one-eighth of the English bushel.
The Imperial standard gallon, both for dry and liquid measure, contains 10 pounds avoirdu-
pois of distilled water, temperature 62° Farenheit, barometer 30 inches. In volume its capacity is
277.274 cubic inches.
See Mensuration of Solids for practical problems in measuring or gauging barges, barrels, etc.
MEASURES OF WEIGHT.
TROY, OR MIN T W EIGHT.
555. Troy Weight is used in weighing gold and silver, and in physical and
chemical experiments.
556. The Standard Unit of weight in the United States is the Troy pownd,
which contains 5760 grains.
TABLE.
º EQUIVALENTS.
lb. OZ. pwt. gT.
24 Grains (gr.) = 1 Pennyweight - pvt. or dwt. 1 = 12 = 240 = 576
20 Pennyweights = 1 Ounce - - - - - - oz. 1 = 20 = 480
12 Ounces == = 24
1 Pound - - - - - - lb. 1.
NOTE 1.-The Troy pound is identical with the Imperial Troy pound of England, and derives
its name from Troy Novant, the ancient name of London. Some derive its name from Troyes, a
town in France, where this weight was first used in Europe. It was brought from Cairo, Egypt, in
the 12th Century. It is equivalent to the weight of 22.79442 cubic inches of distilled water at its
maximum density, or of 22.8157 cubic inches 629 Fahrenheit, barometer at 30 inches in both cases.
2.—The term pound is derived from the Latin word pendo, meaning to bend or weigh. The
term ounce is from the Latin word wincia, meaning one-twelfth part of a pound. The pennyweight
was the weight of the old English silver penny, which was formerly used as a weight. The term
grain is derived from grains of wheat which were formerly used for weighing. At first 32 grains
taken from the middle of the ear or head, and well dried, was equal to a pennyweight. But subse-
quently the pennyweight was divided into 24 parts, each part being called a grain, though it is
heavier than a grain of wheat.
3.—The symbol lb. is from libra, the Latin word for pound; oz. is from the Spanish word onza,
meaning ounce; put. is a compound of p. meaning penny and wit, meaning weight; dwt. is from d.
in the Latin word denarius, which means penny or the 240th of a pound, and wi. in Weight; gr. is
the abbreviation of grains.
AVOIRDUPOIS, OR COMMERCIAL WEIGHT.
557. Avoirdupois or Commercial Weight, is used in weighing all coarse
articles: as groceries, cotton, iron, etc.
25o SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. X-
558. TABLE.
27# Grains = 1 Dram - - - - dr.
16 Drams = 1 Ounce - - - - oz.
16 Ounces = 1 Pound - - - - lb.
25 Pounds = 1 Quarter - - - - qr.
4 Quarters, or 100 pounds = 1 EIundredweight - cwt.
20 Hundredweight, or 2000 pounds = 1 Ton - - - - - T.
480 lºounds = 1 Imperial Quarter - Im. qr.
100 Pounds is also called 1 Cental - - - - c.
T. CWt. lb. OZ. dr.
1 = 20 = 2000 = 32000 = 512000
1 = 100 = 1600 = 25600
1 = 16 = 256 = 7000 gr.
1 = 16
NOTE.-1. The cwt. in England, and in some cases in the United States, is 112 pounds, or 4
quarters of 28 pounds. The ton English is 2240 pounds. This is called the long ton, and 2000
pounds, the short tom.
2.—The long ton is used in estimating duties at the U. S. Customhouse, and also at the mines
in weighing coal, ores, etc.
3.—The avoirdupois pound is equivalent to the weight of 27.7015 cubic inches of distilled
water at its maximum density, or of 27.7274 cubic inches at 62°Fahrenheit, barometer in both cases
at 30 inches. The avoirdupois pound of England and the United States is the same.
4.—1 ft. avoirdupois = }}#} of 1 fö. Troy, or ## of 1 lb. Troy. Or 144 f5. avoirdupois = 175
Ib. Troy. But the Troy ounce is heavier than the avoirdupois ounce. 175 Troy ounces = 192 avoir-
dupois ounces.
5.—The term avoirdupois, is derived from the French word avoir du pois, to have (a fixed or
standard) weight. The term ton is derived from the Saxon word tunne, a cask. The origin of the
pound, ounce and grains was given in the Troy table. The symbol cwt. is compounded from the
words, centum and weight. The term dram is from drachma, originally a small Greek weight; it is
now obsolete as the ounce, in commerce, is divided into halves and quarters.
See Miscellaneous Table for old English units of Stone, pig, fother, etc.
APOTEIECARIES” WEIGHT.
559. Apothecaries’ Weight is used by physicians and apothecaries in
weighing and compounding dry medicines. Medicines are bought and Sold by
avoirdupois weight.
TABLE.
lb. OZ. dr. SCI’. gr.
20 Grains (gr.) = 1 Scruple - scr. or 9 1 = 12 = 96 = 288 = 5760
3 Scruples = 1 Dram - dr. “ 3 1 = 8 – 24 = 480
8 Drams = 1. Ounce - oz. “ 3 1 = 3 = 60
12 Ounces = 1 Pound - Ib. 4 ſh 1 = 20
NOTE.-1. The grain, the ounce, and the pound of this weight are the same as those of
Troy weight, the ounce being differently divided.
2.--The terms pound, ounce, dram and grains were defined in the Troy and avoirdupois
tables. . The term scruple is from the Latin word scrupulus, a small stone. The symbols used in
this weight had their origin, according to Champollion, in the Egyptian hieroglyphics.
NOTE.-Physicians, when writing prescriptions use the Roman numerals in small letters,
prefixing the symbols and writing j for i when it terminates a number. Thus, 22 gr. is written gr.
xxij; 11 scruples, 9xj; 6 ounces, 3v.j. -
. B. is an abbreviation for recipe, or take; a., aa., for equal quantities; 88, for semi, or half as
3 viss., means 64 drams; gr. for grain; P. for particula, or little part; P. aeq. for equal parts; q.
p. quantum placet, as much as you please. -
Apothecaries and Physicians, of earlier times designed to conceal from others all knowledge
of the ingredients given as medicines, and hence the articles used as medicine were named in
Latin, indicated in Latin numerals, and arbitrary signs employed to express the quantity.
TABLES OF WEIGHTS. 25 I
560. Apothecaries' fluid measure is used in compounding liquid medicines.
ſº TABLE.
60 Minims (m.) = 1 Fluidrachm - - - - f 3
8 Fluidrachms = 1 Fluidounce - - - - f2
16 Fluidounces = 1 Pint - - - - - - O.
8 Pints = 1 Gallon - - - - Cong.
Cong. O. f3 f 3 Ill.
1 = 8 = 128 = 1024 = 61440 One M. = about 1 drop of water.
1 = 16 = 128 = 7680
1 = 8 = 480
1 = 60
NotE.—Minim (M.) is from the Latin minimus, the least; the minim is the smallest fluid
measure in use; one minim is equal to about a drop of water.
Fluidrachms and Fluidounces are formed by prefixing fluid to the terms drachms and ounces
of the apothecaries' weight.
O. is an abbreviation of octams, the Latin for one-eight; meaning one pint, or the eighth part
of a gallon.
Cong. is from congiarium, the Latin for gallon.
Drops, are indicated in physicians’ prescriptions by Gºtt, which is from the Latin guttae.
561. A single common teaspoonful, or 45 drops, makes about one fluidrachm;
4 teaspoonfuls make 1 tablespoonful, which is about half a fluidounce; a wine
glass equals about one and a half fluidounces; a common teacup holds about 4 fluid-
ounces; a pint of water weighs 1.0423-H pounds; practically 1 pound.
DIAMOND WEIGHT.
562. Diamond Weight is used in weighing diamonds and other precious
Stones.
TABLE.
16 Parts = 1 Carat Grain = .792 Troy grains.
4 Grains = 1 Carat = 3.168 “ {{
NotE.—The term carat is derived from Kuara, the name of a tree in Abyssinia, which bears
a bean which varies but little in weight from the time it is gathered. In early ages, it was used
as a weight for gold in Africa. In India, it was used as a weight for diamonds and gold.
ASSAYERS’ WEIGHT.
563. Assayers' Weight is used by assayers in determining the Quantity
of any particular metal in ores, or metallic compounds.
TABLE.
1 Carat grain 2 Pwts. and 12 grains, or 60 grains Troy.
1 Carat 10 Pwts. Troy.
24 Carats 1 Pound Troy.
NOTE.—1. This assay carat is entirely different from the carat in diamond weight.
2.—The term carat is also used by assayers to express the proportional purity, or the finenes;
of gold, each carat meaning the twenty-fourth part. Thus, gold 16 carats fine has 16 parts gold and
8 parts alloy.
252 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs.
CIRCULAR, OR ANGULAR MEASURE.
564. Circular, or Angular Measure is used in measuring angles, latitude
and longitude, the location of vessels at sea, of planets, stars, etc.
565. The Standard Unit for measuring angles is the degree, which varies
with the size of the circle.
566. A Degree is the angle measured by the arc of a #0 part of the circum:
ference of a circle.
567. A Circle is a plane figure bounded by a curved
line, every part of which is equally distant from a certain
point within called the center.

568. The Circumference of a circle is the curved A.
line by which it is bounded, as A. E. D. B. G. H. F.
569. The Radius of a circle is a line extending
from its center to any point in its circumference, as C. E. H
and C. D.
570. The Diameter of a circle is a straight line passing through its center
and terminating at each end in the circumference, as A. B.
571. An Arc of a circle is any portion of the circumference, as B. D.,
D. E., etc. - *.
572. A Chord of a circle is a straight line drawn within a circle and ter-
minating in the circumference, but not passing through the center, as F. G.
573. A Segment of a circle is any part cut off by a chord, as F. H. G.
574. A Sector of a circle is any part of a circle bounded by two radii and
the arc included between them; as the space between C. D., C. B., and B. D.
575. An Angle is the opening or space between two lines or surfaces which
meet in a common point, called the vertex. Thus, A. C. E., E. C. D., and D. C. B.
are angles, and C. is their vertex.
576. A Semi-Circumference is one-half of a circumference, or 1800.
577. A Quadrant is one fourth of a circumference, or 90°.
578. A Sextant is one-siath of a circumference, or 600.
579. A Sign is one-twelfth of a circumference, or 300.
TABLE.
60 Seconds (marked ”) = 1 Minute - '. C. S. O f f /
60 Minutes = 1 Degree - O. 1 = 12 = 360 = 21600 = 1296000
30 Degrees = 1 Sign - - S. 1 = 30 = 1800 = 108000
12 Signs, or 3600 = 1 Circle - - c. 1 = 60 = 3600
1 = 60
* MEASURES OF LENGTH, SURFACE AND SOLID.
253
THE OLD FRENCEL AND SPANISH MEASURES OF LENGTH, SURFACE,
AND SOLID.
580. Louisiana having been both a French and a Spanish province, the old
French and Spanish units of measure are often met with in private and public
records; and to aid in understanding such units, the following table is presented:
581.
OLD FRENCH SYSTEM.
1 Point
1 Line = 12 points
1 Inch = 12 lines
1 Foot = 12 inches
1 Ell = 43 inches 10 lines
1 Toise = 6 feet
1 Perch or Rod, (Paris) = 18 feet
1 Perch or Rod, (Royal) = 22 feet
TABLE.
ENGLISH AND AMERICAN MEASURE.
.0074 English inches.
.08884 English inches.
1.06577 English inches.
12.7892 English inches.
46.716 English inches.
76.7352 English inches.
19.1838 English feet.
23.447 English feet.
1 League (common)= 25 to a degree = 2280 toises = 14579.688 English feet = 2.761
miles.
1 League (post) = 2000 toises = 12789.2 English feet = 2,422 miles.
1 Fathom (brass) = 5 feet French = 63.946 English inches.
1 Cable length = 100 toises = 639.46 English feet = 106.58 English fathoms.
582. OLD SPANISH SYSTEM.
1 Foot = 11.1284 English inches.
1 Wara = 3 feet = 0.9274 English yard =
33.3864 English inches. .
1 Common League = 19800 Spanish feet.
1 Judicial League = 15000 Spanish feet.
583. OLD FRENCH SQUARE & CUBIC MEASURE.
1 Square inch = 1.13587 English Square inches.
*1 Arpent (Paris) = 100 square perches, =
36804,120336 sq. feet, English.
- 1 Arpent (Woodland) = 100 sq. perches (Royal)
= 54978.994576 sq. feet, English.
1 Cubic inch = 1.2106 cubic inches, English.
1 Cubic foot = 2091.85 cubic inches, English.
*Arpent is the old French name for acre.
584. COMPARISON OF THE FOOT OF DIFFERENT NATIONS, AS NOW
IN USE.
Russian foot (16 vershok)= 28 inches English or
U. S. measure.
Prussian foot (fuss)= 12.3567 in. English or U.
S. measure.
Bavarian foot (fuss)= 11.4204 in. English or U.
S. measure.
Hanoverian foot (fuss)= 11.4504 in. English or
U. S. measure.
Saxon foot (fuss)= 11.1494 in. English or U. S.
Idea,SULT6.
Austrian foot (fuss)= 12.4455 in. English or U.
S. measure.
Spanish foot (pié)= 11.1288 in. English or U. S.
IOlea SULT6.
French foot, old, (pied de roi)= 12.7892 in. Eng-
lish or U. S. measure.
NOTE.—France and Spain now use the metric system.
Egyptian foot (cubit)= 21.3, 22.667, 25.13, and
26.5.
Persian foot (gaz)= 25 common, 37.5 Royal, in.
English or U. S. measure.
Turkish foot (pic)=26.8 or 27.06 in. English or
U. S. measure.
Bohemian foot (fuss)= 11.88 in. English or U. S.
Illea,SUIT6,
Bremerian foot (fuss)= 11.386 in. English or U.
S. measure.
Danish foot (fod)= 12.3576 in. English or U. S.
IOlea,SULT6: .
Swedish and Norwegian foot (fot)= 11.6904 in.
English or U. S. measure. ... •
NOTE.—Sweden and Norway now use the metric system.
Hamburgian foot (fuss)= 11.2896 in. English or
U. S. measure.
Japanese foot (shaku) = 11.948 in. English or
U. S. measure.
Mexican vara or yard = 32.97 in. English or
U. S. measure.
Portugese foot (palmo or span)= 8.6568 in, Eng-
lish or U. S. measure.
Arabian foot (guz)=25 in.; (covid) 19 in. English
or U. S. measure.
254
SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. º:
585. TABLE OF COMPARATIVE WEIGHTS, MEASURES, AND VALUES
Avoirdupois. Troy. Apothecaries.
7000 gr. = 1 lb. 5760 gr. = 1 lb. 5760 gr. = 1 lb.
1 lb. = 1+* lbs. – 1+ºr lbs.
or 144 lbs. F 175 lbs. - 175 lbs.
1 Oz. - Hº Oz. - ## Oz.
Or 192 OZ. F 175 Oz. - 175 oz.
A Wine Gallon = 231 cubic inches. | Solidity of Cube = 1, Solidity of Sphere
The Old Beer Gallon = 282 ( & & 4 diameter of which = one side of Cube = .5236
A Dry Gallon = 268.8 & 4 { { 1 oz. pure gold. . . . . . . . . . . . . . . . . ... = $20.67–H
An Imperial Gallon = 277.274 “ 4 & 1 oz. pure silver. . . . . . . . . . . . . . . . . . = $ 1.29–H
U. S. Bushel = 2150.42 cubic inches. | 1 p wit. pure gold.................. = 1.03
U. S. Bushel heaped = 2688. & 4 é & 1 pvt. pure Slver................. F .0645.
An English Bushel 2218.192 * * & 4
Diameter of Circle = 1, Circumference = 3.1416
Area of a Square = 1, Area of Circle
diameter of which = one side of the
Square = .7854
1 pint of water weighs 1.0431 lbs.
1 Gallon. “ & 4 8.338882 lbs.
1 Cubic foot of water weighs 62.425 lbs. at 39.20 F.
The Common year has 365 days.
The Leap year has 366 days.
The Solar year has 365 days, 5hrs., 48 min., 49.7 sec.
A Yard. . . . . . . . . . . . = 36 in. 1 Milreis of Brazil... = 54.6 cts. I 1 Drachma of
A Vara of Spain... = 33.3864 “ | 1 Rouble of Russia Greece. . . . . . . . . = 19.3 cts.
A Vara. Of Mexico.. = 32.97 “ gold . . . . . . . . . . . = 77.2 ** | 1 Peso of Cuba. . . . = 92.2 “
A Meter. . . . . . . . . . . = 39.37 “ | 1 Yen of Japan gold = 99.7 “ | 1 Peso, or dollar of
$1. . . . . . . . . . . . . . . . . . =100 cts. 1 Lira of Italy..... = 19.3 ** Mexico Silver... = 57.7 “
1 Franc of France. . =19.3 “ 1 Peseta of Spain... = 19.3 “ | 1 Piaster of Egypt. = 4.9 “
1 Mark of Germany. =23.8 “ | 1 Crown of Sweden. = 26.8 “ | 1 Piaster of Turkey = 4.4 “
1 £ of England..... =$4.8665 “ | 1 Rupee of India 1 Sol of Peru silver. = 53.1 ‘‘
1 Crown of Austria.=20.3 & 4 silver. . . . . . . . . . = 25.2 * *
NOTE.-These monetary values are in terms of United States gold, October 1st, 1893.
NoTE--See Table of Values of Foreign Coins.
A Statute mile = 5280 ft.
A Geographical, or Nautical mile, or knot, - 6086.41 ft.
A Statute mile being 1, a Geographical mile is 1.15273–H
586. An acre contains 160 sq. rds., or 43560 sq. ft., and is 208,7103-1- ft. on
each side.
An arpent (Paris) contains 100 sq. rds. (Paris), or 36804,120336 sq. ft. (English),
and is 191.1844 ft. (English) on each side.
An arpent (Woodland) contains 100 sq. rās, (Royal), or 54978.994576 sq. ft.
English, and is 234.476 English ft. on each side.
NOTE.-See Metric System for other comparisons.
BOOKS AND PAPER.
SIZE OF PAPER.
587.
Inches.
Flat Letter . . . . . . . . . . . . . . . . . 10 × 16
Law Blank, or Small Cap ... 13 × 16
Flat Cap . . . . . . . . . . . . . . . . . . . . 14 × 17
Crown. . . . . . . . . . . . . . . . . . . . ... 15×19
Demy . . . . . . . . . . . . . . . . . . . . . . . 16 × 21
Folio Post. . . . . . . . . . . . . . . . . . . 17X22
Check Folio . . . . . . . . . . . . . . . . 17X24
Medium—Writing. . . . . . . . . . . 18×23
Medium. . . . . . . . . . . . . . . . . . . . . 18×24
Inches.
Extra Size Folio. . . . . . . . . . . . . . . . 19 × 23
Medium—Writing and Printing. 19×24
Royal—Printing. . . . . . . . . . . . . . . . 20X24
Medium–Printing. . . . . . . . . . . . . 20X25
Double Cap . . . . . . . . . . . . . . . . . . . . 17 × 28
Super Royal—Writing.......... 20X28
Card Board & Sup. Royal, Print’g 22×28
Imperial—Writing . . . . . . . . . . . . . 22X30
Imperial—Printing. . . . . . . . . . . . . 22 × 32
NOTE.-Printing paper is made of many sizes, according to the requirements of the printer.
In book printing 24 × 38 inches, called double medium, is used very largely.
WEIGHT OF GRAIN AND PRODUCE PER BUSHEL.
255
588. In copying legal papers, recording deeds, etc., clerks are usually paid
by the folio.
Thus: 100 words make 1 folio in New York, and New Orleans; 72
words make 1 folio in Com. Law; 90 words make 1 folio in Chancery. In printing
books 240 impressions, or 120 sheets printed on both sides, make 1 token.
A sheet (medium) folded in 2 leaves is called folio.
& & ( & & & 4
& 4 { { & & 8
& 4 & & & 4 12
& 4 & 4 & 4 16
& 4 4 & & & 18
& 4 & 4 & 4 24
& 4 & & & 4 32
24 Sheets = 1 Quire.
480 Sheets = 20 Quires = 1 Ream.
2 Reams = 1 Bundle; 5 Bundles = 1 Bale.
12 Units = 1. Dozen.
144 Units = 12 Dozen = 1 Gross.
12 Gross = 1 Great Gross. º
20 Units = 1 Score.
56 lbs. = 1 Firken of Butter.
100 lbs. = 1 Quintal of Dried Fish,
196 lbs. = 1 Barrel of Flour.
200 lbs. = 1 Barrel of Flour in California.
200 lbs. = 1 Barrel of Beef, Pork, or Fish.
280 lbs. = 1 Barrel of Salt.
100 lbs. = 1 Cask of Raisins.
14 lbs. Iron or Lead = 1 Stone.
A Stone of Fish or Meat = 8 lbs.
& &
4 &
“ quarto or 4to.
octavo or 8vo.
16mo.
18mo.
24mo.
32mo.
A Stone of Glass
A Seam of Glass
21; Stones
8 Pigs
7 lbs. Wool
2 Cloves
2 Stones
6}. Tods
2 Weys
12 Sacks
A Truss of Hay
A Truss of Straw
256 Pounds of Soap
25 Pounds of Powder
12 Barrels of Wheat
duodecimo or 12mo,
5 lbs.
24 Stones.
1 Pig.
1 Fother.
1 Clove.
1 Stone.
1. Tod.
1 Wey.
1 Sack.
1 Last.
56 lbs.
36 lbs.
1 Barrel.
1 Keg.
7 English Quarters.
589. *WEIGHT OF GRAIN AND PRODUCE PER BUSHEL,
AS USED IN NEW ORLEANS WHEN THERE IS NO AGREEMENT TO THE CONTRARY.
Wheat. . . . . . . . . . . . . bush. 60 lbs. Clover Seed. . . . . ... bush. 60 lbs. | Dried Peaches. . . . . bahs. 33 lbs.
Corn. . . . . . . . . . . . . . . “ 56 “ | Timothy Seed...... “ 45 ‘‘ | Dried Apples. . . . . . ‘‘ 24
Rye. . . . . . . . . . . . . . . “ 56 “ | Barley Malt..... ... “ 34 “ | Onions. . . . . . © e º e o º “ 56 “
Oats. . . . . . . . . . . . . . . “ 32 “ | Peas, Split......... “ 60 “ | Coarse Salt... . . . . . “ 50 * *
Barley. . . . . . . . . . . . . “ 48 “ Small Hominy..... “ 50 “ | Fine Salt. . . . . . . . . . “ 50 “
Irish Potatoes... . . . “ 60 “ | Flaxseed. . . . . . . . ... ** 56 ‘‘ | Stone Coal. . . . . . . . ** 80 * *
Sweet Potatoes. ... “ 60 ‘‘ | Hempseed. . . . . . . . . “ 44 “ | Corn Meal......... “ 44 “ .
Beans. . . . . . . . . . . ... 62 “ | Buckwheat. . . . . . . . “ 52 “ Plastering Hair.... “ 7 “
Bran. . . . . . . . . . . . . . . “ 24 “ | Castor Beans....... “ 46 “ | Blue Grass Seed. ... “ 10 “

tidim ºf Dºminate Numbers.
-------------------------------------N
*—aſh- A
NOTE.—For the Metric System of Weights and Measures, see Appendix.
590. Reduction is the operation of changing an expression in one or more
denominations to an equivalent expression in some other denomination or denomin-
ations of the same kind of measurement: Thus, $10 may be changed to its equiva-
lent 1000 cents; 36 inches may be changed to its equivalent 3 feet, and 5 lbs. 8 oz.
avoirdupois may be changed to its equivalent 54 lbs., or 88 ounces.
Reduction is of two kinds, descending and ascending.
591. Reduction Descending is changing the forms of denominate quan-
tities from a higher to a lower order of units, or denomination; as in changing or
reducing dollars to cents, pounds to grains, feet to inches, etc.
592. Reduction Ascending is the converse of reduction descending, and
hence it is the changing of the form of denominate quantities from a lower to a
higher denomination; as cents to dollars, grains to pounds, inches to feet, etc.
REDUCTION DESCENDING.
TO REDUCE A SIMPLE DENOMINATE NUMBER TO A LOWER DENOMINATION, (IN THE
SAME SYSTEM OF MEASURE).
593. 1. Reduce 6 feet to inches.
OPERATION.
6 feet. Ea:planation and Reason.—By considering the conditions
12 inches. of the problem, we see that we are required to reduce 6
— –— feet to inches. i. e. to find the equivalent of 6 feet in the
| 72 inches, Ans. 2 Q.
unit inches. Before we can perform the operation, we
ft. º, must know the units of the lower denominations from
6. 0 feet to inches. And by referring to the Table of Linear
12 Measure, we find that 1 foot = 12 inches. This gives the
6 premise for the solution and from it we reason as follows:
*E*- Since 1 foot is - to 12 inches, 6 feet are = to 6 times as
72 inches, Ans. many, which is 72 inches, the answer.
2. Reduce 3 bushels to pints.
FIRST OPERATION.
bu. pks. Qts. pts. Ea:planation and Reason.—In this problem, we are re-
3 0 0 quired to reduce bushels to pints; but before we can
===s** sºma ººm. perform the operation we must know either the different
4 8 2 units of the lower denominations from bushels to pints,
or the equivalent of one bushel in pints. By reference to
3 12 96 the Table of Dry Measure, we see that 1 bushel = 4 pecks;
ºme ººm- 1 peck = 8 quarts; and 1 quart = 2 pints. These equivae
12 96 192 pts., Ans. lents furnish our premises, and from them we develop the
7 Solution.
We first write the problem as shown in the operation, filling all vacant denominations, from
(256)
Yºr REDUCTION DESCENDING. 257
bushels to the denomination required, pints, with naughts; then below each denomination we draw
a line and write thereunder the number of units of each order which make one of the next higher
order. Having thus stated the problem, we reason as follows: Since 1 bushel is = to 4 pecks, 3
bushels are = to 3 times as many, which is 12 pecks; then since 1 peck is = to 8 quarts, 12 pecks
are = to 12 times as many, which is 96 quarts; then since 1 quart is = to 2 pints, 96 quarts are =
to 96 times as many, which is 192 pints, the answer.
i.e. If we work from the basis of the number of pints in a bushel, we would reason thus; 1 bushel
†gº º: pints: Since 1 bushel is - to 64 pints, 3 bushels are = to 3 times as many, which is
pln US.
SECOND OPERATION.
3 bu. 3 bu. Explanation.—In this operation, we reason as follows:
4 tº: In one bushel there are 4 pecks, and in 3 bushels there
12 pks. or 2 pts. are three times as many pecks, which is 12; then in 1
eck there are 8 quarts, and in 12 pecks, there are 12
192 pts, Ans. " quarts, 9
96 QtS. times as many, which is 96; then in 1 quart there are 2
2 pints and in 96 quarts there are 96 times as many, which
192 pts., Ans. is 192 pints.
3. Reduce $7 to mills. Ans. 7000 m.
4. Reduce 3 shillings to farthings. Ans. 144 far.
5. Reduce 4 pounds Troy to pennyweights. Ans. 960 pvt.
6. Reduce 5 bushels to quarts. Ans. 160 qts.
7. Reduce 3 days to minutes. Ans. 4320 m.
8. Reduce 20 to ". AnS. 7200 ".
* 9. Reduce 2 square feet to square inches. Ans. 288 sq. in.
10
. Reduce 16 pounds avoirdupois to grains. Ans. 112000 grains.
TO REDUCE A COMPOUND DENOMINATE NUMBER TO A LOWER DENOMINATION, (IN
THE SAME SYSTEM OF MEASURE).
594. 1. Reduce 3 bu., 2 pks., and 1 pint, to pints.
OPERATION.
bu. pks. Qts. pts. Ea.planation and Reason.—Here we are required to deter-
3 2 () 1. mine the number of pints in the whole expression. We
º t_º ºmº first state the problem as shown in the operation, filling
4 8 2 the vacant units or denominations in the scale with a
naught and writing under each term the number of units
3 14. 112 of that order which make one of the next higher order.
* *mºmºsºme mºº-ºº: Having thus stated the problem, * º * *: :
Since 1 bushel = 4 pecks, 3 bushels are = to 3 times as
14 112 225 pts., Ans. many which is 12 º + the 2. pecks = 14 pecks; then
since 1 peck = 8 quarts, 14 pecks are = to 14 times as
many, which is 112 quarts; then since 1 quart = 2 pints, 112 quarts are = to 112 times as many,
which is 224 pints –H 1 pint = 225 pints, answer. º
In all problems of like character to the three preceding, the form of operation and the pro-
cess of reasoning here given should be used. -
2. Reduce 3 £, 2s. 8d. 3 far, to farthings. Ans. 3011 far.
3. Reduce 6 lbs. 12 oz. 13 drS. to drams. - Ans. 1741 drs.
4. Reduce 5 lbs. 2 oz. 12 grs. to grains. -- Ans. 29772 grs.
258 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS, Af.
5. Reduce 3 rods 20 links 6 inches to inches. Ans. 758.4 or 758% in.
6. Reduce 3 f.oz. 4 f.dr.s. 20 m. to minims. Ans. 1700 m.
7. Reduce 3 cable-lengths 5 fathoms to feet. Ans. 2190 ft.
8. Reduce 3 rds. 12 ft. 6 in. to inches. Ans. 744 in.
9. Reduce 6 hbds. 42 gals. 3 qts. 3 gills to gills. Ans. 13467 gills.
- TO REDUCE A FRACTIONAL DENOMINATE NUMBER TO A LOWER DENOMINATION,
(IN THE SAME SYSTEM OF MEASURE).
595. Reduce # of a gallon to gills.
OPERATION.
8 || 3 gallons. Explanation and Reason.—In this problem, we are re.
4 quarts. quired to find the equivalent of § of a gallon, in the unit
2 pint of gills. By referring to the Table of Wine Measure, we
pints. see that 4 gills = 1 pint ; 2 pints = 1 quart ; 4 quarts =
4 gills. 1 gallon. Then with this data as our premises, we write
º-e mº- the # of a gallon on the statement line and reason as
12 gills, Ans. follows: Since 1 gallon = 4 quarts, there are 4 times as
many quarts as gallons; then since 1 quart = 2 pints,
ſº e there are two times as many pints as quarts; then since 1
pint = 4 gills, there are 4 times as many gills as pints. This completes the reasoning, and by
Working out the statement, we obtain 12 gills as the equivalent of # of a gallon.
2. Reduce # of a dollar to cents. Ans. 60 cts.
3. Reduce # of a pound (£) to farthings. Ans. 800 far.
4. Reduce ºf of a Troy pound to grains. Ans. 1728 grs.
5. Reduce fºr of a yard to inches. Ans. 13+ in.
6. Reduce # of a bushel to pints. Ans. 27# pts.
7. Reduce # of a week to seconds. Ans. 403200 sec.
8. Reduce # of a franc to millimes. Ans. 285% m.
9. Reduce # of a mark to pfenniges. Ans. 75 pf.
10. Reduce # of an acre to sq. in. Ans. 5488560 sq. in.
§
<A →
tillſtill Of DelDillie Fra(ii)S.
596. A Denominate Fraction is a fraction whose integral unit is a denom.
inate number.
Thus, # of a day, .9 of a mile, are denominate fractions.
TO REDUCE DENOMINATE FRACTIONS, OR FRACTIONAL DENOMINATE NUMBERS TO
LOWER DENOMINATIONS, OR TO COMPOUND DENOMINATE NUMBERS.
597. 1. Reduce # of a bushel to lower denominations.
OPERATION.
bu. pks. qts. ptS. Explanation and Reason.—According to the conditions of
0 0 this problem, we are to find not the equivalent of # of a
bushel in pecks, quarts, or pints, but we are to determine
what compound denominate number composed of the
denominations of pecks, quarts, and pints, will be equiva-
lent to # of a bushel. In the operation, we first write
the # of a bushel and fill each lower denomination with
a naught, below which we draw a line and underneath
24 Write the number of units of each order which make one
of the next higher order. We then draw a vertical line
am-º. sº-º-º-º: to the left of the 4 pecks and reason as follows: Since 1
13 4; 1; bushel = 4 pecks, # of a bushel = the # part, and # bushels
1 plc 4. T; pt = 2 times as many, which is, as shown by the operation,
DK. Qt.S. # pts. 13 pecks. The 1 peck is now written below a long hori-
zontal line which we call the answer line, and is the first denomination in the number sought.
Then we draw a vertical line to the left of the 8 quarts and reason thus: Since 1 peck = 8
quarts, # of a peck = the # part, and # = 3 times as many, which work gives 4} quarts.
The 4 quarts are written below the answer line and is the second denomination of the com-
pound number required by the terms of the problem. Lastly, drawing a vertical line to the left
of the 2 pints, we reason thus: Since 1 quart = 2 pints, # of a quart = } part and # = 4 times as
many, which is 13 pints. This being the last denomination in the Dry Measure Table of Measure-
ment, we write it in full below the answer line and thus complete the solution—having obtained
1 plc., 4 qts., and 13 pts., as the equivalent compound denominate value of ; of a bushel.
tº -º#
O
3.
5
i
5
:
2. Reduce £ to a compound denominate number. Ans. 6s. 8d.
3. Reduce # yard to a compound denominate number. Ans. 1 ft. 23 in.
4. Reduce # lbs. Troy to a compound denominate number.
* Ans. 5 oz. 2 p wit. 204 grs.
5. Reduce # mile to a compound denominate number.
Ans. 6 fur. 26 rds. 3 yds. 2 ft.
6. Reduce g degree to a compound denominate number. Ans. 52', 30”.
7. Reduce # Cong. to a compound denominate number.
Ans. 4 O. 12 fl. Oz. 6 fl. dr. 24 m.
8. Reduce # h ha. wine to equivalent integers in lower denominations.
Ans. 33 gals. 3 qtS. 1 pt. 1 # gills.
9. Reduce £ºs to its value in integers of lower denominations.
Ans. 6s. 4d. 3% far.

(259)
26O SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
TO REDUCE DENOMINATE DECIMAL FRACTIONS, OR DECIMAL DENOMINATE
NUMBERS, TO LOWER DENOMINATIONS (IN THE SAME SYSTEM OF MEASURE).
598. 1. Reduce .875 of a bushel to pints.
OPERATION.
bu. pks. qts. pts.
.875 0 0 0 Explanation and Reason.—This problem is very similar
4. s 2 to the first and second problems in denominate numbers,
.875 3.5 28 page 256. We first write the problem as shown in the
4-º operation, filling all vacant denominations from bushels
3.500 28.0 56 pts., Ans. to pints with naughts; then, below each lower denomi-
Or, nation we draw a line and write thereunder the number
sº of units of each order which make one of the next higher
tºº-ºº-ºo order. Then, from these premises we reason as follows:
3.500 Since 1 bushel = 4 pecks, .875 of a bushel = .875 times as
8 many, which is 3.5 pecks; then, since 1 peck = 8 quarts,
28.0 3.5 pecks = 3.5 times as many, which is 28 quarts; then,
2 since 1 quart = 2 pints, 28 quarts = 28 times as many,
which is 56 pints, answer.
56 pints., Ans.
2. Reduce 0.755 of a gallon to gills. Ans. 24.16, or 24; gills.
3. Reduce 0.375 of a rod to inches. Ans. 74.25, or 744 in.
4. Reduce 0.25 of an acre to sq. ft. - Ans. 10890 sq. ft.
5. Reduce 0.42 of a ton to drams. w Ans. 215040 drs.
6. Reduce 0.3 of a yd. to nails. Ans. 4; nails.
7. Reduce 0.16 of a chain to inches. Ans. 126.72 in.
TO REDUCE A DENOMINATE DECIMAL FRACTION, OR A DECIMAL DENOMINATE
NUMBER, TO LOWER COMPOUND DENOMINATE NUMBERS.
599. Reduce .374 lbs. apothecaries' weight, to a compound denominate number.
OPERATION,
lbs. OZ. dr. SC. gr. Explanation and Reason.—This prob-
.374 () 0 0 0 lem is very similar to the elucidated
smm mºa ºsmºsa problem on page 259. In the opera-
tion, we first write the .374 of a pound
12 8 3 20 and fill the place of each lower de-
.374 .488 .904 .712 nomination with a naught; then draw-
ing a line below each naught, we
4.488 3.904 2.712 14,240 write the number of units of each
order of apothecaries' weight, which
4 oz. 3 dr. 2 Sc. 14.24 grs., Ans. it takes to make one of the next higher
order. Having thus stated the prob-
lem, we reason as follows: Since 1
pound = 12 ounces, .374 of a pound = .374 times as many, which is 4.488 ounces. We now draw
the answer line and write the 4 ounces below it. Then, since 1 ounce = 8 drams, .488 of an ounce
= .488 times as many, which is 3.904 drams. The 3 drams we write below the answer line, and
• continue thus: Since one dram = 3 scruples, .904 of a dram = .904 times as many, which is 2.712
scruples. The 2 scruples we write below the answer line, and continue thus: Since 1 scruple
= 20 grains, .712 of a scruple = .712 times as many, which is 14.24 grains.
This being the lowest denomination, it is written below the answer line and completes the
solution.
REDUCTION OF DENOMINATE FRACTIONS. & 261
Reduce 0.27 of a bi, to a compound denominate number. º
Ans. 1 plc. 0 qts. 1.28 pts.
Reduce 0.7 of a lb. Troy to a compound denominate number.
* Ans. 8 oz. 8 pWt.
Reduce 0.35 of a mi. to a compound denominate number.
Ans. 2 fur. 32 rds.
Reduce 0.32 of a day to a compound denominate number.
Ans. 7 hrs. 40 min. 48 sec.
Reduce 0.3 of a cu. yd. to a compound denominate number.
Ans. 8 cu. ft. 172.8 cu. in.
Reduce 0.65 of a gal. to a compound denominate number.
Ans. 2 qts. 1 pt. 0.8 gi.
What is the value of .6875 of a circle. Ans. 8 S. 70 30'.
What is the value of .7525 of a day. Ans. 18 h. 3 min. 36 Sec.

eduction Ascending.
TO REDUCE A SIMPLE DENOMINATE NUMBER TO A COMPOUND DENOMINATE
NUMBER OF HIGHER DENOMINATIONS.
600. 1. Reduce 942 gills wine measure to a compound denominate number.
OPERATION.
4 942 gills
2 || 235 pints + 2 gi. remainder.
4 || 117 quarts + 1 pt. remainder.
* 29 gallons + 1 qt. remainder.
29 gal. 1 qt. 1 pt. 2 gi. Ans.
Explanation and Reason.—In this problem, we are
to determine the equivalent of 942 gills in higher
denominations. According to the Table of Wine
Measure, 4 gills = 1 pint; 2 pints = 1 quart, and 4
quarts = 1 gallon. Being in possession of these
equivalent denominations, we first reduce the gills
to the next higher denomination, pints; then. We
reduce the pints to the next higher denomination,
quarts; and thus continue to reduce each lower
denomination to the next higher, until the highest
denomination in the Table of Wine Measure is
reached, or until the last quotient is less than the
next higher denomination. Our reasoning is as follows: Since 4 gills = 1 pint, there are + as
many pints as gills, which is 235 pints and 2 gills remainder, as shown in the operation.
Then, since 2 pints = 1 quart, there are 4 as many quarts as pints, which is 117 quarts + 1
pint remainder. Then, since 4 quarts = 1 gallon, there are + as many gallons as quarts, which is
29 gallons and 1 quart remainder. This completes the reasoning and gives 29 gal., 1 qt., 1 pt., and
2 gills as the equivalent of 942 gills.
2. Reduce 3721 pints to a compound denominate number, (dry measure).
Ans. 58 bu. 0 p.k. 4 qt. 1 pt.
3. Reduce 1391 inches to a compound denominate number, (linear measure).
Ans. 6 rās. 5 yds.1 ft. 11 in.
4. Reduce 2756 grains, Troy, to a compound denominate number.
Ans. 5 oz. 14 pwt. 20 gr.
5. Reduce 3457 drams, avoirdupois, to a compound denominate number.
Ans. 13 lbs. 8 oz. 1 dr.
6. Reduce 17353 farthings to a compound denominate number.
Ans. £18 1s. 6d. 1 far.
7. Reduce 34567 inches, chain measure, to a compound denominate number.
Ans. 43 ch. 64 links 4.12 inches.
NoTE.—In chain measure, the chain consists of 100 links.
8. Reduce 4315100 seconds to weeks.
Ans. 7 W. 0 d. 22 h. 38 m. 20 Sec.
9. Teduce a million mills to dollars. Ans. $1000.
10. What is the value of 360502 inches in higher denominations?
Ans. 5 m. 5 fur. 20 rods 3 y. 2 ft. 10 in.

(262)
Y
REDUCTION ASCENDING. 263
TO REDUCE A SIMPLE OR A COMPOUND DENOMINATE NUMBER TO A DENOMINATE
FRACTION OF A HIGHER UNIT.
601. 1. Reduce 3 pints to the fraction of a bushel.
i
the division side of the statement line.
OPERATION.
3 pints.
the value of 3 pints in the fraction of a bushel.
2.
13.
14.
15.
16.
17.
Reduce 8s. 4d. to the fraction of a pound sterling.
OPERATION INDICATED.
8s. 4d. = 100d. #}} = £º, Ans. Or, 100d.
12
20
###, Ans.
Reduce 5 gills to the fraction of a gallon. Ans. # of a gal.
Reduce 72 drams to the fraction of a ton. Ans. ###55 of a ton.
Reduce 16” to the fraction of a degree. Ans. ### of a degree.
Reduce 2 hours and 20 minutes to the fraction of a day.
Ans. # of a day.
Reduce 3 minims to the fraction of a pint. Ans. ##5 of a pint.
Reduce 25 inches to the fraction of a yard. Ans. # of a yard.
Reduce # of # of # of an inch to the fraction of a yard. Ans. ºf yard.
Reduce # of an hour to the fraction of a week. Ans. Tºg W.
Beduce 3 R. 25 P. to the fraction of an acre. Ans. #} ac.
Reduce 3 oz. 4 pwt. 16 grs. to the fraction of a pound Troy.
OPERATION INDICATED.
3 oz. 4 pwt. 16 grs. = 1552 grs.
24 || 1552
; or thus, #### grs. = % of a pound, Ans.
#5 of a pound, Ans.
Reduce 1 cu. ft. 200 cu. in. to the fraction of a cu. ya.
Ans. #4 of a cu. yd.
Reduce 3' 5" to the fraction of a degree. Ans. #5 of a deg.
Reduce 5 cable-lengths 3 fathoms to the fraction of a mile.
Ans. ### of a mi.
Reduce 6 rds. 4 yds. 2 ft. 9 in. to the fraction of a furlong.
Ans. Hºls of a fur.
Reduce 50 min. 30 Sec. to the fraction of an hour. Ans. 444 of an hr.
1 2 O
Explanation and Reason.—Since we are to find the equivalent
of 3 pints in the fraction of a bushel, we must first know
either the scale of units from pints to bushels, or the number
of pints in a bushel. By referring to the Table of Dry Measure,
we find that 2 pints = 1 quart; 8 quarts = 1 peck; and 4 pecks
– = 1 bushel. With this knowledge, we place the 3 pints on the
#r bushel, Ans. right of the statement line and reason as follows: Since 2
6 4. 2 pints = 1 quart, there are # as many quarts as pints, which we
indicate by writing the 2 on the division side of the statement
line. This gives us the value of 3 pints in the fraction of quarts. Then, since 8 quarts = 1 peck,
there will be # as many pecks as quarts, which we indicate by writing the 8 on the division side of
the statement line. This gives us the value of 3 pints in the fraction of pecks. Then, since 4
pecks = 1 bushel, there will be + as many bushels as pecks, which we indicate by writing the 4 on
This gives us what the condition of the problem required,
264 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
TO REDUCE A SIMPLE OR A COMPOUND DENOMINATE NUMBER TO A DENOMINATE
DECIMAL OF A HIGHER UNIT,
602. 1. Reduce 10 inches to the decimal of a yard.
OPERATION
12 | 10 inches. Explanation and Reason.—This problem is closely
3 .83% = decimal of a foot. relatod to the one under Art. 601, and the operation
* and reasoning are somewhat similar. We first
.27% decimal of a yard. write the 10 inches on the statement line, and
, or thus: considering that 12 inches = 1 foot, and 3 feet = 1
10 inches. N e º
12 yard, we reason thus: Since 12 inches = 1 foot,
3 there are "g as many feet as inches, which is, as
* | *=== shown by the operation, .83% of a foot. Then, since
.27% of a yd., Ans. 3 feet = 1 yard, there are # as many yards as feet,
or thus:
which is .27% of a yard, answer.
#} of a yd. = .27% of a yd.
2. Reduce 3 quarts, 0 pints, and 1 gill to the decimal of a gallon.
OPERATION. *
4 | 1. gill.
— Ea:planation and Reason.—For convenience, we
2 0.25 pints. here write the different denominations in column
4 || 3.125 quarts. on the statement line, and remembering that 4 gills
* -º-e = 1 pint, 2 pints = 1 quart, and 4 quarts = 1 gallon,
.78125 of a gallon. we reason as in the above problem, and divide the
3 qts. 0 pt: *i- 25 gills. i. by 4. which gives & decimal of .25 of a pint.
25 gills. is decimal is written to the right of the 0 pints.
4 Then we divide the .25 of a pint by 2 and produce
2 .125 of a quart, which is written to the right of the
4 3 quarts. The 3.125 quarts are then divided by 4,
º 7s 25 of a gal, Ans. and the required decimal of .78125 of a gallon is
or thus:" obtained.
#} of a gal. = .78125 of a gal., Ans.
3. Reduce 3 gills to the decimal of a gallon. Ans. .09375 of a gal.
4. Reduce 2 ounces and 5 grains, Troy, to the decimal of a pound.
Ans. .1675% of a lb.
5. Reduce 3’ to the decimal of a degree. Ans. .0008% of a deg.
6. Reduce 7 seconds to the decimal of an hour. Ans. .0019; of an hr.
7. Reduce 108 square inches to the decimal of a sq. yj.
Ans. .08% of a sq. yd.
. Reduce 12 drams to the decimal of a ton. Ans. .0000234375 of a ton.
8
9. Reduce 3 dimes, 4 cents, 2 mills, to the decimal of a dollar. Ans. $.342.
0. Reduce 3 oz. 2 drams to the decimal of a pound. Ans. .1953125 of a lb.
1. Reduce 3 qts. 1 pt. 2 gills to the decimal of a gallon.
Ans. .9375 of a gal.
12. Reduce 3 sq. ft. 72 sq. in. to the decimal of a sq. yd.
Ans. .38% of a sq. yd.
jºr REDUCTION ASCENDING, 265
13. Reduce 3 quarters, 2 nails to the decimal of a yd. Ans. .875 of a yd.
14. Reduce 6 minutes, 40 seconds to the decimal of an hour.
Ans. .11% of an hr.
15. Reduce 3 days 15 hours 45 minutes 48 seconds to the decimal of a week.
Ans, .5224 + week.
16. Reduce 240 A. 2 R. 25 P. to the decimal of a square mile.
Ans. .3760 + sq. m.
TO REDUCE A DENOMINATE FRACTION TO A FRACTION OF A HIGHER DENOM.
INATION.
603. 1. Reduce # of an ounce avoirdupois to the fraction of a ton.
OPERATION,
5 4 5 || 4 Explanation and Rea-
16 16 * 8on.—This problem is
25 also very similar to the
2000 one under Article 601,
4. or thus: and requires about the
20 roºga of a ton, Ans. Same process of reason-
tº ſº-º-º-º-º-º-º-º-e Ing.
We first write the #
zoºs of a ton, Ans. * of an ounce on the state-
ment line, and then re-
duce it to the next higher denomination, pounds, by dividing it by 16; and thus by successively
dividing by that number of the lower order which makes one of the next higher, we reduce the # of
an ounce to the fraction of a ton, the denomination required.
The reasoning for the work is as follows: Since 16 ounces = 1 pound, there are Y's as many
pounds as ounces; then, since 25 pounds = 1 quarter, there are ºf as many quarters as pounds;
then, since 4 quarters = a hundredweight, there are + as many hundredweight as quarters; then,
since 20 hundredweight = 1 ton, there are go as many tons as hundredweight.
If it is desired, after dividing by 16, the reasoning may be given as follows: Since 2000
pounds = 1 ton, there are gºatſ as many tons as pounds.
2. Reduce # of a gill to the fraction of a quart. Ans. As of a quart,
3. Reduce # of a millime to the fraction of a franc. Ans. stºrs of a franc.
4. Reduce # of a fluid dram to the fraction of a gallon.
Ans. ##H of a gal.
5. Reduce # of a pennyweight to the fraction of a pound Troy.
- Ans. #5 of a pound.
6. Reduce # of a square rod to the fraction of a square mile.
Ans. TTào's of a SQ. mi.
7. Reduce # of an inch to the fraction of a quarter, (cloth measure).
Ans. # of a qr.
TO REDUCE A DENOMINATE FRACTION TO A DENOMINATE DECIMAL OF A HIGHER
DENOMINATION.
604. Reduce # of a penny to the decimal of a pound sterling.
OPERATION Explanation and Reason.—This problem is very
4 || 3 similar to the one under Article 602. Remembering
12 the Table for English Money, we write the #d. on the
statement line and reason thus: Since 12 pence = 1
20 Shilling, there are tº as many shillings as pence,
tºmºsºmºmºmºmºmºmºmºsas which we indicate by writing the 12 on the division
+}r = .003125 of a £. side of the statement line. Then, since 20 shillings
3 2 O = 1 fº, there are ºn as many pounds as shillings,
which is indicated by writing the 20 on the division
side of the statement line. The result of the work thus far gives the fractional equivalent of ºd.
in the unit of pounds, which is #5:6. This we reduce to a decimal in the usual way and obtain
.003125 of a £, answer.
266 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs jºr
Reduce # of a pint to the decimal of a bushel. Ans. .01302084 of a bu.
Reduce ºr of an inch to the decimal of a rod. Alls. .0035; of a rod.
Reduce # of a grain to the decimal of a pound Troy.
Ans. .00007445hr of a lb.
Reduce # of a square inch to the decimal of a sq. yd.
Ans. .0006% of a sq. yd.
Reduce # of a second to the decimal of an hour.
Ans. .00026 ºr of an hour.
7. Reduce # of a " to the decimal of a degree. Ans. .0001} of a degree.
:
TO REDUCE A DECIMAL DENOMINATE NUMBER TO A DECIMAL OF A HIGHER
DENOMINATION.
605. 1. Reduce .35 of a pint to the decimal of a gallon.
OPERATION.
2 | .35 of a pint. .35 Explanation and Reason.—
2 Here º #. have a problem
wº very much like the one under
4 | .175 of a quart. or thus: 4 Article 602. We are to find the
*memº. mºmºmºmºmº sºmmemºmº equivalent of .35 of a pint in
.04375 of a gallon. .04375 of a gallon. the decimal of a gallon. We
first write the .35 of a pint on
the statement line, and reason as follows: Since 2 pints = 1 quart, there are 3 as many quarts as
pints, which is, as shown by the operation, .175 of a quart. Then, since 4 quarts = 1 gallón, there
are + as many gallons as quarts, which is .04375 of a gallon.
In the second statement, the reasoning is the same as in the first, but the opération of division
is not performed in detail, with each separate divisor.
2. Reduce .75 of a grain to the decimal of a pound Troy.
Ans. .0001302+3 of a lb.
3. Reduce .96 of a minim to the decimal of a pint. Ans. .000125 of a pint.
4. Reduce .16 of a rod to the decimal of a mile. Ans. .0005 of a mi.
5. Reduce 36 of a foot to the decimal of a cable-length.
Ans. .0005 of a C.-l.
6. Reduce .012 of a rod to the decimal of a league. Ans. .0000125 of a L.
7. Reduce .072 of a farthing to the decimal of a £. Ans. .000075 of a £.
TO REDUCE A DECIMAL DENOMINATE NUMBER TO A FRACTION OF A HIGHER
DENOMINATION.
606. 1. Reduce .35 of a second to the fraction of an hour.
OPERATION.
100 | 35 of a second. Explanation and Reason.—In this problem, we wish
60 to determine the equivalent, in the fraction of an
60 hour, of .35 of a second expressed as a common frac-
tion. We therefore write the .35 of a second on the
statement line as a common fraction, ºr and then
###55 of an hour, Ans. reason as follows: Since 60 seconds = 1 minute,
there are tº as many minutes as seconds; then, since
60 minutes = 1 hour, there are gº as many hours as minutes. This statement, when worked, gives
7xãwo of an hour, answer.
2. Reduce.36 of a dram to the fraction of a pound. Ans. ##5 of a lb.
Reduce .75 of a pint to the fraction of a barrel. Ans. ### of a bbl.
Reduce .124 of a farthing to the fraction of a £. Ans. grãºod of a £.
Reduce .27 of a square inch to the fraction of a rod.
Ans. Trºws of a rod.
Reduce .0156 of an inch to the fraction of a mile.
Ans. Fasº's of a mi.
Reduce .324 of a grain to the fraction of a Troy ounce.
Ans. Tºro of an oz.
What would .7875 of a hogshead of wine cost at 5 cents per gill ?
Ans. $79,38.
.
• * * * * * * * * * * e s ſº tº * * * > * * * * * * * * * * * *
607.
A—a 4- A ºr
r–Wurr-w
Addition of Compound Denominate Numbers is the process of
uniting two or more compound denominate numbers into one equivalent number.
The process of adding is the same as adding simpſe numbers, except that the
scale of increase and decrease, by passing from one denomination to another,
varies with every system of measurement and with almost every denomination of
each system.
We shall, therefore, not discuss the subject at length, but shall proceed to
illustrate the process by which results in compound addition are determined.
1. Add 5 bu. 3 plºs. 7 qts. 1 pt., 8 bu. 3 qts. 1 pt., and 3 bu. 3 p.ks. 1 pint.
OPERATION.
bu. pks. Qts. pt.
5 3 7 1
8 O 3 1.
3 3 () 1
17 3 3 1, Ans.
3 pts. -- 2, No. of pts. in a
Qt., = 1 qt, and 1 pt.
11 qts. + 8, No. of qts. in a
pk., = 1 p.k. and 3 qts.
7 pks. -- 4, No. of pks. in a
bu., =1 bu. and 3 pks.
Explanation.—In all problems of this kind, we first
write the numbers so that units of the same denomina-
tion stand in the same column, and begin with the lowest
denomination to add.
Accordingly, we here first add the pints and find the
sum to be 3 pts., which we divide by 2, (since 2 pts. = 1
qt.) and obtain 1 qt. and 1 pt. remainder. The 1 pt. we
write under the column of pts., and carry or add the 1
qt. to the column of qts., which added gives 11 qts. and
which we divide by 8, (since 8 qts. = 1 pk.) and obtain 1
pk. and 3 qts. remainder. The 3 qts. we write under the
column of qts., and add the 1 plc. to the column of plºs.,
which added gives a sum of 7 plºs. This we divide by 4
(since 4 plºs. = 1 bu.) and obtain 1 bu, and 3 p.ks. The 3
pks. we write under the column of plºs. and add the 1 bu.
to the column of bushels, which gives a sum of 17 bu.,
which we write under the column of bushels.
This completes the operation and produces 17 bu., 3
pks., 3 qts., 1 pt. as the complete result.
2. Add 1536. 10s. 9d., 8:6. 9s. 7d., 14%. 12s. 10d., and 13. 18s. 6d.
*
Ans. 273E. 11s. 8d.
3. Add 12 yds. 3 qrs. 2 na. 2 in., 5 yds. 2 qrs. 1 na. 1; in., 2 q1's. 1. na.
13 in., 8 yds. 2 in.
Ans. 27 yds. 0 qrs. 3 na. 4 in.
NOTE.-In working the above problem, the student must remember, when dividing the .
inches, that whenever the dividend is fourths, the remainder is also fourths. The remainder is
always like the dividend.
4. Add 10 lbs. 13. 63.0 9. 10 grs, 103.23. 19. 16 grs, and 13 lbs. 23.
7 3. 3 B. 12 grs.
Ans. 24 lbs. 33. 0 3. 2 9. 18 grs.
5. Add 3 yds. 2 ft., 9 in., 4 rās. 2 yds. 1 ft. 11 in., and 5 rids. 4 yds. 2 ft. 8 in.
Ans. 11 rds. 0 yd. 1 ft. 4 in.
6. Add 21 gals. 3 qts. 1 pt. 3 gi., 32 gals. 1 qt. 0 pt. 2 gi., and 47 gals. 2 qts.
1 pt. 1 gi.
Ans. 101 gals. 3 qtS. 1 pt. 2 gi.
7. What is the sum of the following quantities, Troy weight? 25 lbs. 10 oz.
15 pwt. 18 grs.; 42 lbs. 9 oz. 20 grs.; 26 lbs. 4 oz. 7 pwt.; 11 oz. 19 pwt. 23 grs.; 100
lbs. 6 oz. 14 pvt. 16 grs.
Ans. 196 lbs. 6 oz. 18 pwt. 5 grs.
(267)
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608. Subtraction of Compound Denominate Numbers is the process of
decreasing one compound denominate number by another of the same system of
ImeaSurement.
As in addition, the scale of increase and decrease varies, otherwise the work
is the same as in subtraction of simple numbers.
1. From 1836. 4s. 7d. 3f. subtract 11:6. 9s. 11d. 2f.
OPERATION.
:8. S. d. f. Ea:planation.-In all problems of this kind, we first
18 4 7 3 write the numbers so that units of the same denomina-
11 9 11 2 tion stand in the same column, and begin with the
lowest denomination to subtract. Accordingly, we here
commence with the farthings and say 2 far. from 3 far.
6 : 14 8 1 Ans. leaves 1 far. which we write under the column of far-
things. We now come to the column of pence and
observe that 11d. cannot be taken from 7d., because the 11d. is the greater number; we therefore,
according to the law that the difference between two numberg is the same as the difference between the two
mumbers when equally increased, as demonstrated in Article 129, add 12d. to the 7d., making 19d.
From this we subtract the 11d. and have 8d. remainder, which we write under the column of pence.
Then, as we added 12d. to the minuend, we, now, to compensate therefor, according to the above
law, add 1s., the equivalent of 12d., to the column of shillings in the subtrahend and thus we have
10s. to subtract from 4s. which we cannot do. Hence, for reasons above given, we add 20s. to the
4s, which makes 24s., from which we take 10s. and have 14s. remainder, which we write under the
column of shillings. We now add 1:6, the equivalent of 20s, to the 1136, making 12.É., which we
take from 1836 and have a remainder of 636, which we write under the column of pounds. This
completes the operation.
It will be observed that when we added the 12d. to the column of pence, and the 20s. to the
column of shillings, that, in each case, we added such a number of that order as made one of the
next higher order. This must always be done in simple numbers or in any of the systems of com-
pound numbers, when the subtrahend number or denomination exceeds the minuend number of like
denomination.
2. From 45 bbls. 19 gals. 2 qts. take 24 bbls. 24 gals. 3 qts.
OPERATION, gº
bbl. gal. Qt. Fæplanation.—We proceed in this as in the previous
45 19 2 example, to find the several denominate remainders;
24 24 3 but since one of them, the 25% gals., is partly fraction-
al, we reduce the # gallon to quarts, equal to 2 qts.,
20 25; 3 which we add to the 3 qts. in the remainder, making
2 5 qts., equal to 1 gallon and 1 quart; we set down the
1 qt, and add the 1 gal. to the 25 gals., giving 26 gals,
20 26 1 for the final result.
3. From 3 mi, 5 fur. 30 rds. take 1 mi. 7 fur. 32 rās.
Ans, 1 mi. 5 fur, 38 rids.
4. From 7 lbs. 3 oz. 12 pywt. 20 grs. take 3 lbs. 5 oz. 10 pwt. 15 grs.
- Ans. 3 lbs. 10 oz. 2 pwt. 5 grs.
(269)
27o SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs.
5. From 30O 25' 32” take 25o 34' 357/. Ans. 4o 50' 57".
6. From 4 lbs. 12 oz. 13 drs. take 2 lbs. 9 oz. 15 drs.
- Ans. 2 lbs. 2 oz. 14 drs.
7. From 13 hrs. 24 min. 7 sec. take 10 hrs. 30 min. 12 sec.
Ans. 2 hrs. 53 min. 55 sec.
8. From 3 sq. rds. 5 sq. ft. 95 sq. in. take 1 sq. rd. 10 sq. ft. 100 sq. in.
Ans. 1 sq. rd. 267 sq. ft. 31 sq. in.
SOLUTION.
sq. rd. sq. ft. Sq. in. Explanation.—By proceeding as in the explanations
3 5 above given, we obtain 1 sq. rā. 2664 sq. ft. 139 sq. in.
10 100 as a result; but it is not proper to have a fractional
expression in any but the lowest denomination of a
compound denominate number, so we proceed to reduce
266+ 139 # of a sq. ft. to sq. in.; # of 144 sq. in. = 36 sq. in. ;
# = 36 139 sq. in. -- 36 sq. in. = 175 sq. in., or 1 sq. ft. an
#: *; àY. #: 31 sq. lº º # ft. #.
SCI. Itſ, all Ol IIIl © all SWOT to O6 L SC. TC1, SC - IUe
267 31 31 º in. Q. Q1
9. Subtract the following:
A. R. P. sq. yd. sq. ft. sq. in.
40 3 22 30 4 10
16 3 30 35 5 100
23 3 31 24+ 7 54
# = 2+
# = 36
Ans. 23 3 31 25 0 90
10. From # of a hogshead, subtract 3 of a gallon.
Ans. 37 gal. 0 qt. 1 pt. 0# gi,
OPERATION INDICATED.
# of a hind. = 37 gal. 3 qt. 0 pt. 1; gi.
# of a gal. = 2 qt. 1 pt. 14 gi.
37 gal. 0 qt. 1 pt. 04; gi.
11. From 2% bu. Subtract # of a peck. Ans. 2 bu. 1 p.k. 4 qt. 143 pt.
12. From 555 A. 3 R. 30 P. take .335 of a square mile.
Ans. 341 A. 2 R. 6 P.
13. From 5 m. 4 fur. 24 r. take 2 m. 6 fur. 37 r. 3 yd. 2 ft. 9 in.
g Ans. 2 m. 5 fur. 26 rā. 14 yd. 0 ft. 3 in.
14. From the sum of # of 365+ days and # of 23 hours, take # of ºr of 154
minutes. Ans. 156 d. 14 h. 47 m. 25* Sec.
15. From the sum of # of 54 miles and .875 furlongs take # of 24 yards.
Ans. 2 m. 2 fur. 34 rā. 34 yd. 0 ft. 4; in.
609. Multiplication of Compound Denominate Numbers is the process
of determining the product of two numbers, when the number to be multiplied
is a compound denominate number.
Compound multiplication differs from multiplication of simple numbers, in
that when the product in any denomination equals or exceeds one of the next higher
denomination in the same system of measurement, said product must, by a process
of division, be reduced to that next higher denomination, before the number in it
can be multiplied.
1. Multiply 4 gals. 3 qts. 1 pt. 2 gi. by 9.
OPERATION.
gal. qts. pt. gi. Explanation.—In all problems of this kind, we write
4 3 1 2 the multiplier under the lowest denomination of the
9 multiplicand and multiply. . We first say 9 times 2 gi.
are 18 gi. which, as shown in the operation, equals 4
pts. and 2 gi. The 2 gi. we write under the gills and
44 1 1 2 reserve the 4 pts. to add to the product of pints. We
then say 9 times 1 pt. are 9 pts, and 4 pts. added make
18 gi. -- 4 = 4 pts. 2 gi. 13 pts. which reduced to the next higher denomination
13 ot . . 2 = 6 otS. 1 pt equals 6 qts. and 1 pt. The 1 pint we write under the
plS. + 2 = 0 qLS. L. pt. unit or denomination pints, and reserve the 6 qts, to
33 qts. + 4 = 8 gals. 1 qt. add to the product of quarts. Then we say 9 times 3
qts. are 27 qts, and 6 qts. added make 33 qts. equal to
8 gals. and 1 qt. The 1 qt. we write under the quarts, and reserve the 8 gals. to add to the product
of gallons. Lastly we say 9 times 4 gals. are 36 gals. plus the 8 gals. reserved, are 44 gals, which
WO .* under the denomination of gallons. This gives 44 gals. 1 qt. 1 pt. 2 gi. for the entire
product. &
NOTE. 1.-The multiplier must always be considered an abstract number, Art. 149, page 75.
f NOTE. 2.--When the multiplier is large, and a composite number, we may multiply by the
actors.
2. Multiply 7 Cong. 3 O. 15 fº by 6. Ans. 44 Cong. 7 O. 10 fº .
3. Multiply 27 cords 34 cu. ft. by 12. Ans. 327 cords 24 cu. ft.
4. Multiply 3 mi. 7 fur. 20 rds. by 7. Ans. 27 mi. 4 fur. 20 rds.
5. Multiply 7 hrs. 30 mi. 24 sec. by 5. Ans. 1 da. 13 hrs. 32 min.
6. Multiply 3 francs, 7 dec. 3 cen. 4 mil. by 5. Ans. 18 fr. 6 dec. 7 cen.
7. Multiply 43. 3s. 4d. by 12. - Ans. £50.
8. In 5 bbls. of pecans, each containing 2 bu. 3 p.k.s. 5 qts. 1 pt., how many
bushels? Ans. 14 bu. 2 pks. 3 qts. 1 pt.

(271)
272 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. ºr
9. Multiply 25 gallons, 3 quarts, 1 pint, by 3.5.
OPERATION.
25 gals. 3 qts. 1 pt. = º ptS. 2) 724.5 pt..
4 3.
*-* -*= 4) 362 qts. and .5 or + pt.
103 1035
2 621 90 gals. 2 qts.
-* Ans. 90 gals., 2 qts, # pt.
207 724.5 pt.
10. How many bushels in 12 barrels of grain, each barrel containing 4 bus.,
3p};S., 5 qts., 1 pt. 2 Ans. 59 bus, 2 qts.
11. How many gallons in 15 casks of wine, each cask containing 42 gallons,
2 quarts, 1 pint, and what will they cost at $2.48 per gallon 3
Ans. Contents: 639 gals., 1 qt., 1 pt. Cost: $1585.65.
12. A man has 7 lots of land, each containing 24 A., 3 R., 25 P. How many
acres has he altogether, and what are they worth at $15.65 per acre ;
Ans. 174 A., 1 R., 15 P. Cost: $2728.47#.
13. If a boy studies 5 hours 25 minutes each day, how many days of 24 hours
each will he study in 54 days 3 Ans. 12 ds., 4 hrs., 30 m.
14. What is the distance around an octagonal or eight sided field, each side
measuring 1 fur., 25 ras, 3 yds., 2 ft., 8 in...? . e
Ans. 1 mi., 5 fur., 5 rds., 3 yds., 1 ft., 10 in.
15. A man bought 25 pieces of broad, cloth, each containing 32 yds., 3 grs,
2 na. How many yards did he buy, and what did they cost at $5.84 per yard
Ans. 821 yds., 3 qrs., 2 na. Cost: $4799.75.

A 4 acres. x 28 compasses. B SPECIAL PROBLEM.
A farmer has a piece of land that he measures with
a pair of compasses whose points are 6 feet apart,
and finds the following dimensions of his land: A to
B 4 acres 28 compasses; A to C 4 acres 13 compasses;
C to D 2 acres 9 compasses. How many acres in the
plat of ground 3 Ans. 15 acres, 98 sq. rods,
16 Sq. yds., 8 sq. ft., 5.52 sq. in.
REMARKS.–The compasses for measuring land is a primative
instrument in the form of compasses, whose points are 6 feet
apart. 35 spans or units of the compasses (210 feet) is consid-
ered an acre for approximate work on a farm. This is incor-
rect, as an acre or square piece of ground containing one acre,
ſº º is 208,7103 ft. on each side. See Comparative Table on page
#54 of this book, also table on page 244. Seepage 253 for arpent, Whieh is the old French word
for acre.
SOLUTION.
1: "Eé"*}918 ft. A to c.
|
i
C 2 acres 9 com. D
- 2 & & e
#"Tºº $47 ft. c. to D x 918–495132 sq. ft. in the main body of land.
Then 4 × 210 = 840 ft.
*::"E}}#} 1008 feet A to B.
Less - - - 474 “ C to D.
Gives the base of *=º
triangle - - - - 534 “ × to B.
Then 534×918 A to C – 2 for the triangle = 245106 sq. feet.
Total sq. feet in the plat of land e- 680238 ‘‘ ‘ ‘
680238 – 43560 sq. feet in an acre, gives
15.616 -i- acres. .616x160 = 98.560 sq. rods.
.56×30+ = 16.94 sq. yds.
.94× 9 = 8.46 SQ. ft.
.46 × 12 = 5.52 sq. in. t
MoTE.—See page 254 for Paris and Woodland Arpents.
3) in ºf CII Dºmint Ninters
• * > e s tº e º 'º e º e º & e º - º 'º º============N
610. Division of Compound Denominate Numbers is the process of deter.
mining any required part of a compound denominate number.
It is the reverse of compound multiplication, and we first divide the highest
denomination in the number, and if any remainder occurs in the result, it must be
reduced to the next lower denomination before the number in that denomination is
divided.
1. Divide 32 lbs. 12 oz. 12 drs. by 5.
OPERATION.
lbs. OZ. dr. Explanation.-In all problems of this
5)32 12 12 kind, we write the quantity to be di-
vided in the order of its denominations
6 8 15%, Ans. and place the divisor on the left, as in
division ofº numbers.
-- h = ſº Having thus stated the problem, we
32 lbs. -- 5 = 6 lbs. and 2 lbs. remainder. . first divide the 32ibs. By 5 and obtain
* a quotient of 6 lbs. with 2 lbs. remain-
2 lbs. × 16 = 32 oz. -- 12 oz. = 44 oz. der. We write the 6 lbs. in the quo.
44 oz. -- 5 = 8 oz. and 4 OZ. remainder. tient line under the pounds, and re-
duce the 2 lbs. remainder to ounces =
32 oz. –H 12 oz. = 44 oz. which we divide
4 oz. × 16 = 64 drs. -- 12 drs. = 76. drs. by 5 and obtain 8 oz. with a remainder
76 drs. -- 5 = 15 drs. and 1 dr. remainder. of 4 oz. The 8 oz. we write in the
quotient line, under the ounces, and
1 dr. -- 5 = + dr. reduce the 4 oz. to drams, = 64 drs. --
12 drs. = 76 drs. which - 5 = 15 drs.
with 1 dr. remainder, or 15% drs. ; this
is written in the quotient line under the drams, and completes the operation.
NoTE.—When the divisor is large, and a composite number, we may divide by the factors.
2. A box contains 8 bu. 3 p.k.s. 5 qts. How many smaller boxes, each
holding 1 pk, 1 qt. 1 pt., can be filled from the larger box 3 Ans. 30 boxes.
OPERATION INDICATED.
8 bu. 3 pks. 5 qts. = 570 pts. Explanation.—In all problems of this kind,
1 plc. 1 qt. 1 pt. = 19 pts. we first reduce both dividend and divisor to the
same denomination, and then divide as in simple
^ 570 pts. -- 19 = 30 boxes, Ans. numbers.
3. Divide 37 mi. 3 fur. 4 rās. by 4. Ans. 9 mi. 2 fur. 31 ris.
Divide 60 24, 32” by 7. Ans. 54' 56”.
Divide 24 lbs. 3 oz. 12 pvt. 12 grs. by 12. Ans. 2 lbs. 0 oz. 6 pvt. 1 gr.
Divide 25 yds. 3 qrs. 2 nails by 13. Ans, 1 yd. 3 qrs. 3 nails, 14% in.
Divide 3 cu. yas. 19 cu. ft. 996 cu. in. by 12. Ans. 8 cu. ft. 659 cu. in.
:
(273)
274 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. Yºr
8. Divide 3 mi. 75 ch. 21 l. by 7. Ans. 45 ch. 3 1.
9. What is 3 of 2 m. 3 fur. 1 yd. 2 ft. Ans. 7 fur. 5 rds. 1 ft. 104 in.
10. If 300 families live in one township, and the land be equally divided, how
large a homestead will each have 3 Ans. 76 A. 3 R. 8 P.
11. How long will it take 4 men to excavate a cellar 16 feet square and 12
feet deep, if 1 man removes 3 cubic yards and 15 cubic feet in 1 day ? Ans. 8 days.
12. If a man travels 771 miles in 30 days, what distance does he pass over in
each day ? Ans. 25 mi. 5 fur. 24 rol.
13. If one spoon Weighs 3 oz. 12 p wt. 12 gr., how many dozen spoons can be
made from 38 lbs. 11 oz. 12 pywt. 12 gr. of silver ? Ans. 10; dozen.
14. If a man steps 2 ft. 94 in., how many steps will he make in going 334 miles?
Ans. 63360 steps.
15. If the distance around the earth were 25000 miles, and a man could travel
33% miles each day, how long would it take him to make the circuit of the globe,
allowing 365+ days to the year 7 Ans. 2 years 19 days 12 hours.
/To TN COMPARISON OF THERMOMETERS.
f. Ar c 611. There are three scales in use for registering temperature :
¥ so-ſoo 1st. The Fahrenheit (marked F.) in which the freezing point
200 H of water is marked 329, and the boiling point 2120. Zero is 32°
190 – Hoſſ" below freezing in this scale.
180H —lso NotE.—This F. scale is generally used in the United States for ordinary
170H || |60 R. household and business purposes. \ * º
160 — 170 2nd. The Centigrade (marked C.) in which the freezing point
isol—|| of water is marked 09, and the boiling point 100°.
14OH- Ti”9| |60 NOTE.—This C. scale is used in France and several countries in Europe,
130 H r and by scientists, generally.
isol |H|40H50 3rd. The Reaumur (marked R.) in which the freezing point of
to – |—|40 water is marked 0°, and the boiling point 80°. g
100 H|||—|30ſ NoTE.—This R. scale is used generally in Germany and Spain.
90 – —lsº From the above facts, we make the following statement of
80 H|| ||20 ratios by which one scale may be reduced to any other :
70 H| | | | |30 1st. Since the number of degrees between the freezing and
*Tl|lio boiling points in the F. scale is (2120–320 =) 180° and 100° in C.,
50 H ‘H10 therefore, (a) 10 F. = +}}o C. = }o C; and (b) 19 C. = }#9
| 39 2| F. = }o F.
- *:: 0–H–0– 2nd. Since the C. and R. scales both have the freezing point
20 H at 00 or zero, and since the boiling point is 100° C. and 80° R.,
"Hº"T "I therefore, (a) 10 C. = #9 = to R.; and (b) 19 R. = ** = }*C.,

3rd. Since the number of degrees between the freezing and
boiling points in the R. scales is 800 and in the F. scale 180°, (212°–32° = 180°),
therefore, (a) 10 R. = lºſo F. = ?? F; and (b) 19 F. = #o R. = #9 R.
A. COMPARISON OF THERMOMETERS. 275
PROBLEMIS.
1. Convert 600 F. to its equivalent in Centigrade degrees or Scale.
NOTE.-In changing the F. scale to C. or R. scale, or in changing the C. or R. scale to F., it
must be remembered that the number of degrees that expresses the temperature on a F. Scale
(which marks freezing at 32°) does not express the number of degrees above freezing point, as it.
does on the C. and R, scales. Thus, 609 F. is not 600 above freezing point, but 60° — 32° = 28°
above it. Hence to reduce a Fahrenheit reading to a Centigrade reading, first subtract 32° from
the F. reading or number, and then multiply the remainder by +93 = }; and to reduce F. to R. first
subtract 32° and multiply by ºr or $.
To reduce a C. to a F, reading or scale, first multiply the C. number by ## =
# and then add to the product 320 ; and to reduce R. to F. first multiply the R.
reading by *" or ; and add 32° to the product.
OPERATION.
280 × 100 280 × 5
600–320 = 280, above the freezing point of F. Then sº OI’ # =15;9 C. Ans.
2. Reduce 350 C. to the F. Scale.
OPERATION.
350 × 180 350 ×9
100- or T5
= 630 above freezing of F., or zero of C.
Then 630 + 32O = 950 F. Ans.
3. Reduce 600 F. to Reaumur scale.
OPERATION.
600 — 320 = 28, above freezing point of F. and above zero of R.
289 × 8 280 × 4
Then ** or º =12;O R. Ans.
4. Reduce 129 R. to F. Scale.
OPERATION.
120 ×180 120 ×9
80 4
= 270 above the freezing point of F. Then 279-H 32° = 59°F. Ans.
5. Reduce 50° C. to R. scale.
OPERATION.
500 ×80 500 × 4
TDD- or -5
6. Reduce 600 R. to C. scale.
OPERATION.
= 400 R. Ans.
600 ×100 T 600 × 5
—sº- O T- = 750 C. Ans.
276 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS.
TO EXPRESS TEMPERATURE WEIEN BELOW ZERO.
612. When the temperature is below 0 or zero, in any scale, it is indicated
by a minus sign placed before the number of degrees. Thus: — 100 F. indicates
10° below 0, or 420 below freezing point in F. scale.
— 15° C. indicates 150 below 0, or zero, according to the C. scale.
— 20° R. indicates 200 below 0, or zero, according to the R. scale.
7. Reduce — 16° C. to its equivalent, in the F. scale.
OPERATION.
—160 × # = — 28.80 F. Then, –28.80 + 320 = 3.20 F. Ans.
- NOTE.—In adding or subtracting the 329 where we have minus numbers, it should be done
algebraically. *
8. Reduce 229 F. to its equivalent on the C. scale.
OPERATION.
— 100 × 100 — 109 × 5
220–320 = — 100 ; X * or – 10° x # = — 5:9 C. Ans.
Taj- or 9
9. Change — 139 F. to the C. scale.
OPERATION.
— 130 — 320 = — 450 below freezing of F.; — 45° x 5 = — 250 C. Ans.
--~~
10. Change — 12° C. to the R. scale.
OPERATION.
— 129 ×80 &
*i;- or — 120x # = – 9.60 R. Ans. ->
11. Change — 20° R. to the F. scale.
OPERATION.
T*.* or—20° x * = — 45° F.; Then, – 45°–322 = –13°F. Ans.
12. Change — 200 R. to the C. scale.
OPERATION.
— 200 x + = — 250 C. Ans.
Reduce each of the following problems:
13. 66O F. to C. 17. — 32O F. to C.
14. 800 C. to F. 18. — 14O C. to F.
15. 212O F. to R. 19. — 18O R. to C.
16. 750 C. to R. 20. – 24O F. to R.
iscellaneous Problems,
==
IN DENOMINATE AND IN COMPOUND DENOMINATE NUMBERS.
=s*—dº--º-
-sur-Nºr-wº-
ſ
613. 1. Paid $2 for sawing 3 cas. 26 cu. ft. of wood. How much could be
sawed for $1, at the same rate 3 Ans. 1 ca. 77 cu. ft.
ed. cu. ft.
OPERATION INDICATED. 2 3 26
1 77 Ans.
2. Henry traded 4 rubber balls for 3 p.ks. 4 qts. 1 pt. of pecans. How
much did he get for each ball? Ans. 7 qts. + pt.
3. An English weaver sold 7% webs of cloth of an equal number of yards in
each web at 30:6. 8s. 3d. 2 far. a web. How much did he receive for all ?
OPERATION INDICATED.
£. S. d. far. £. S. d. far.
30 8 3 2 30 8 3 2
7; or, 7;
228 2 1; 3 74 times 30:6 =225 0 0 0
2 7# times 8s. = 60 0 0
74 times 3d. = 22 2
228 2 2 1 Ans. 74 times 2 far. = 15
Ans. 228 2 2 1
4. 38 lbs. 12 oz. 15 drs. of butter cost $74. How much can be bought for $1?
* Ans. 5 lbs. 4 oz. 10%; drs.
OPERATION INDICATED.
fb. OZ. dr. Explanation and Reason.—In this problem, we
38 12 15 -- 7# or *. have the quantity that $7+ = $* will buy, and we
reason as follows: Since $2.3 will buy 38 lbs. 12 oz.
15 drs., 4 of a dollar will buy the 22d part and 3,
or $1 will buy 3 times as much. In the operation
22)116 6 13 of all problems of this kind, it is best to first
multiply the dividend by the denominator of the
5 4 10##, Ans. divisor, and then divide the product by the numer-
ator of the divisor, as shown in the above partially
worked operation.
5. A farmer sowed 3 bu. 3 p.k.s. 6 qts. of oats on each of 3.5 acres. How
much oats did he sow on all ? Ans. 13 bu. 3 pks. 1 qt.
6. Divide 26 tons, 13 cwt. 3 qrs. 12 lbs. of coal equally among 12 families,
and find what each will have to pay at $12 per ton. Ans. $26.6935.
7. How many days at $1.25 per day will it take a man to earn 200 lbs. 12 oz.
of beef at 6 cents a pound 7 Ans. 9.6 + days.

(277)
278 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS Y
8. A merchant exchanged 3 yds. 2 qrs. of broad cloth worth $4 per yard,
for 26 gals. 3 qts. 1 pt. of molasses. What was the molasses worth per gallon ?
Ans. 52; a cents.
9. At a certain distance 20 lbs. of steam expanded the mercury in a Fahr-
enheit thermometer 3.5 degrees. How many lbs. at the same distance would be
required to expand it from zero to the freezing point? Ans. 1824 lbs.
NOTE.—There are 32° between zero and the freezing point, F.
10. From 17 lbs. 5 oz. take 2 lbs. 9 oz. Ans. 14 lbs. 12 oz.
11. One side of a square field is 1080 ft. long; how many rods of straight
fence will inclose it 2 Ans. 261 ra. 4yd. 1 ft. 6 in.
12. The wheels of a locomotive are 18 ft. 3 in. in circumference and make 44
revolutions a second; what time will be required to run 99 miles?
Ans. 1 h. 46 mi. 5 sec.
PRACTICAL PROBLEMIS.
614. N. B. See pages 96 to 101, of this book, for various contracted methods
of Working business problems containing the elements of per bushel, per hundred
(c), per thousand (m), per ton and per dozen, and also problems where one or both of
the factors are aliquots or multiples of 10, 100, or 1000.
13. What cost 1521 pounds OPERATION INDICATED. Reason.—One bushel costs
84c. Since 1 bu. or 56 lbs.
of corn at 842 per bushel, and 1521 - 56 costs 84c. 1 pound will cost
how many bushels are there? = 27 bu. 9 lbs. 56 || $4, the ºthº, an ºl
Ans. $22.814; 27 bu. 9 lbs. 1521 pºin cost is times
14. What cost 2842 bu. 16 opBRATION INDICATED. Reason.—One bushel costs
$1.22. Since 1 bushel or 60
lbs. of wheat at $1.22 per bushel? $ lbs. costs $1.221 pound will
Ans. $3467.56°. 60 | 1.22 cost the 60th part, and 170536
170536 pounds will cost 170536 times
as much.
SECOND OPERATION. THIRD OPERATION.
2842 @ $1.22 = $3467.24 - 2842 bu.. (2) $1 = $2842.
## = # (a) $1.22 = .32%; (20g) $1 = 568.40
(2) (2
*=mºmºrs mºms mºs @ (2¢) tº of $# 56.84
$3467.56; 16 lbs. (a)
* $3467.56%
See pages 96 to 101, for work similar to that of the third operation.
PRACTICAL PROBLEMS.
279
15. What cost 342506 lbs. Of Wheat
at $9.80 per Imperial quarter ?
Ans. $6992.831's.
16. Corn is 60¢ per bu. What is it
worth per cental * Ans. $1.07%.
17. Wheat is $2.90 per cental. What
is it worth per bushel? Ans. $1.74.
18. Cloth is $1.80 per yard. What is
it worth per metre ? Ans. $1.968–H.
19. Cloth is $3.937 per metre. What
is it worth per yard? Ans. $3.60.
20. Cloth cost in Mexico $2 per vara.
What is it worth per yard?
Ans. $2.181 + practically.
$2.183 + accurately.
NOTE. –A vara of Mexico is 32.97 in., prac-
tically, 33 Inches.
21. What cost 376 pounds of hay at
$15 per ton ? Ans. $2.82.
22. Cloth is worth $1.45 per yard.
What is 1t worth per Spanish vara ?
Ans. $1.342 + practically.
$1.344 + accurately.
NoTE.—A Spanish vara is 33.3864 inches, prac-
tically, 33% inches.
23. Sold 5294 pounds of hay at $23.75
per ton. How many tons were there, and
what was the value of it?
Ans. 2 tons, 1294 lbs. $62.86; value.
24. What cost 4 tons 1420 pounds of
hay, at $16.25 per ton ? Ans. $76.53%.
2
OPERATION INDICATED.
9.80
480 || 342506
OPERATION INDICATED.
g
60
56 || 100
OPERATION INDICATED.
g
2.90
100 60
OPERATION INDICATED.
1.80
36 || 39.37
OPERATION INDICATED.
3.937
39.37 || 36
INDICATED.
Accurately.
$
2.
32.97 36.00
OPERATION
Practically.
2.
33 || 36
OPERATION INDICATED.
$ $
15.
376
15.
Or, 2 | .376
2000
OPERATION
Practically.
$
INDICATED.
Accurately.
$
1.45 1.45
36
3
100 36.0000 || 33.3864
OPERATION INDICATED.
$ 5294 - 2000 = 2 t. 1294 lbs.
23.75 OI’ $23.75 × 2 = $
5.294 $23.75 × 1294 = $
* ===
9000 $62,863
INDICATED.
OPERATION
$
16.25
2 || 9.420
or, as in the preced-
ing problem.
28O soule's PHILOSOPHIC PRACTICAL MATHEMATICS. º:
25. What cost 1265 pounds of bran at
80% per cwt.% Ans. $10.12.
26. What is the value of 5790 hoop
poles at $18 per M. ? Ans. $104.22.
27. What is the value of 1364 pine
apples at $11.4 per C.? Ans. $156.86.
28. What is the cost of 31845 feet of
lumber at $22.25 per M. ?
Ans. $708.55#.
29. Bought 3 coops of chickens con-
taining 2 dozen and 7 chickens each, at
$4.35 per dozen. What did they cost?
Ans. $33,714.
30. Butter is worth 35g per pound.
How much can you buy for 102*
Ans. 44 ounces.
31. Sell 4; inches of silk at $2.75 per
yard, and state the amount.
Ans. $.343.
32. A lady wishes to buy 402 worth
of silk which is $3.00 per yard. How
much will you sell her ? Ans. 4% in.
33. A clerk commenced work on the
17th of January, and discontinued April
1st. He received $65 per month. How
much was due him counting January
17th, but not April 1stº Ans. $160.334.
OPERATION INDICATED.
g
30, or, 12.65
100 | 1265 “” 80g
OPERATION INDICATED.
$
18. OI’ 5.790
1000 || 5790 * 18
OPERATION INDICATED,
11:50 or, 13.64
100 || 1364 ° 11;
OPERATION INDICATED.
$
22.25 or, 31.84;
1000 || 31845 " ? 22+
OPERATION INDICATED.
$
4.35
12 || 93
OPERATION INDICATED.
OZ.
16
35 | 10
OPERATION INDICATED.
$
2.75
36
2 || 9
OPERATION INDICATED.
in.
36
300 | 40
OPERATION INDICATED.
$
65 $65
30 74 ° 244
NOTE.-In computing salaries and rents, all months are considered as containing 30 days.
34. What will 4 bu. 3 pks. 1 qt. 1 pt. cost at 10g per pint? Ans. $30.70.
35. What cost 4 bu. 3 p.ks. 1 qt. 1 pt. at $6 per bushel?
36. What cost 4 bu, 3 pks. 1 qt. 1 pt. at 25g per quart?
Ans. $28,784.
Ans. $38,373.
37. What cost 4 bu. 3 p.ks. 1 qt. 1 pt. at $1.75 per peck & Ans. $33.57+}.
38. A grocer has 8 jars of butter, each weighing 14 lbs. 7 oz. How many
pounds in all, and what is it worth at 32.4% per pound?
Ans. $37.533.
Aſ. PRACTICAL PROBLEMS. 28 I
39. Bought 3 bales of hay weighing as follows: (1) 421 pounds, (2) 394 pounds,
(3) 487 pounds, at $22.50 per ton. What did they cost 3 Ans. $14.64%.
40. What is the cost of 2417 cocoanuts at $8.25 per C. ? Ans. $199.404.
41. What is the value of 8750 shingles at $8.75 per M. ? Ans. $76,564.
42. What is the value of 11428 fence pickets at $9 per M. 3 Ans. $102.852.
43. Maple syrup is worth $1.92 a gallon. How much can be bought for 25% #
Ans. 4% gills, or 1 pt. # gi.
44. Tea is worth $.75 per pound. How much will you sell for 20%
- Ans. 44; Oz.
45. Bought 4692 pounds of barley at $.88 per bushel. How many bushels
were there and what was the cost 3 Ans. 97 bu. 36 lbs. Cost $86,02.
46. Bought 2765 pounds of oats at 76% per bushel. What was the cost, and
how many bushels were there ? Ans. $65.66% cost. 86 bu. 13 lbs.
47. What is the cost of 4878 pounds of wheat at $2.45 per cental?
Ans. $119,511.
48. What is the cost of 200 sacks of guano each weighing 162 pounds, at
$524 per ton ? Ans. $846.45.
49. A planter shipped 6 dozen dozen boxes of peaches to market, but being
delayed on the way # a dozen dozen boxes spoiled; the remainder were sold at 70
cents per box. What did they amount to ? Ans. $554.40.
50. A grocer bought a cask of rum containing 84 gals., at $1.25 per gal.; he
lost # of it by leakage; at what rate must he retail the remainder per quart to gain
$27 on his purchase ? Ans. 55%.
51. A box, 16 in. long, 12 in. wide, and 11.2 in. deep, contains 1 bushel; how
many bushels will a proportionally shaped box contain, each dimension of which is
as many feet as the first is inches, that is, 12 times as great Ans. 1728 bu.
52. A box 16 in. square and 8.4 in. deep, contains one bushel, how much will
a box contain, whose dimensions are g times as great 3
Ans. 0.669921875 bu., or 2 p.ks. 5 qts. . pt.
53. A can or vessel, 11 in. deep, 7 in. long, 3 in. wide, contains 1 gallon; how
many gallons will a similar vessel contain, each of whose dimensions is 10 times
as great 3 Ans. 1000 gals.
The last three questions give convenient dimensions for constructing the bushel
and gallon measures; and the given problems are easily solved by cubing the mul-
tiple of the dimensions; thus, the 3d problem is equal to 12°, the 4th to (g)”, and
the 5th to 108.
54. If cloth costs $6.40 per yard, and it takes 6 yds. 3 qrs. 3 na. to make a suit,
and a tailor charges $15 a suit for the making, what will be the entire cost of
making up 173 yds. 1 qr. 3 na., and how many suits can be made 3
Ans. $1485. 25 suits.
55. How many bottles, holding 1 qt. 1 gi., will it require to hold the contents
of 1 had. of wine 3 Ans. 224 bottles.
282 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS *
56. A hardware merchant received from Liverpool an invoice of iron weighing
2.T. 2 cwt. 3 q1s. 20 lbs., long ton weight; the invoice price was £12. 17s. 6d. per
ton; what did the whole cost in sterling money?
FIRST OPERATION.
2 T. 2 cwt. 3 q1s. 20 lbs. = 4808 lbs.,
and £12 17s. 6d. = 30906.
1 T. = 2240 lbs. ) 14856720d.
12) 6632}#d.
mºmºmºmºmº
20) 552s. 833d.
Ans. £27 12s. 833d.
SECOND OPERATION.
2 T. 2 cwt. 3 qrs. 20 lbs. = T. 2.1464 +
and £12 17s. 6d. = £ 12.875
£27,6349000
20
S. 12.6980
12
d. 8.376
Ans. £27 12s. 8.376d. -
Eacplanation.—In this and similar
questions, we first reduce the inte-
gers of the commodity to the lowest
denomination given, = 4808 lbs.,
and then reduce the price to its
lowest term or denomination. We
multiply these two amounts togeth-
er, giving 14856720d., as the cost of
4808 lbs. at 30906. per pound; but
since this last number is the cost of
one ton instead of one pound, and
as there are 2240 lbs. in one ton long
weight we must divide by 2240 to
reduce it to the rate per ton, which
gives us 6632}}d. as the total cost,
and this, when reduced, gives £27.
12s. 84%d. for the answer.
Ea:planation.—Here we reduce each
term to the decimal of its own
highest denomination, then multi-
ply them together, and reduce the
result according to the principles of
reduction of denominate decimals.
57. A cargo of castings was received from Wales, amounting to 225 T. 17
cwt. 2 qrs, long ton weight; the invoice price was £10. 12s. 6d. per ton; the freight
was 4S. per cwt. Considering the pound sterling to be equal to $4.84, what would
be the cost in Federal money, of the invoice and the freight, and also the entire
cost of the whole cargo? Ans. Cost of invoice, $11615.62%.
Cost of freight, $ 4372.94.
Cost of the whole, $15988.56*.
58. What will be the cost of 1450 tons, 2 cwt. 3 qrs. of coal, at $4 per long
ton ? Ans. $5800.55
OPERATION BY OUR NEW METHOD.
Ea:planation.—1 cwt. = 112 lbs. = % ton, =
T. CWT. QRS.
1450 2 3 112 – 2240 = .05 ton. 1 qr. = + of a cwt. or #
.05 .014 = .0125 of .05 of a ton, which is .05 - 4 = .0125 of a
.10 -H .0375 = . 1375 ton.
1450.1375 × $4 = $5800.55. Ans.
NotE,--In calculating the wholesale cost of coal, it is the custom to consider the pounds only to the nealºst quarter-
SECOND NEW METHOD.
Multiply total number of pounds by the price, divide by 4, 4 – = Ibs.
8 and 7, and point off 1 place in the dollars. Or make the 8 — = price per ton.
statement thus, and cancel: 7
Hills and Invoices.
* * * * * * * * * * * * * * * * * * * * * * * * sº e º e s tº e e Tº e º a º ºs e º a===N
-*.
wºr
* = -d-.
wr vºy
615. Bills, in a general sense, embrace all written statements of accounts and
many legal instruments of writing; but in a more common and limited sense they
are statements of goods sold or delivered, services rendered or work done, with the
price or value, quality or grade of each article or item. Bills or invoices of mer-
Chandise should state the place and date of each sale, the names of the buyer and
seller, the price, the extra charges or the discount to be allowed, the marks and
numbers on the goods, and the terms of the sale.
When goods are bought to be sold, or when bills are rendered to the jobber
or retailer, or consigned to an agent, the bill is then called an invoice.
It is the custom of accountants and merchants, when making bills, to com-
Imence the name of each article with a capital.
When a charge is made for the box, barrel, jar, etc., containing goods, it is
customary to write its price above and to the right of it, and add the same to the
cost of the goods it contains.
In making extensions, fractions of cents are not used in the product; when
they are # or more, they are counted cents, when they are less than $, they are not
counted.
NOTE.-Some houses count only every other #c. as a whole cent, thus equalizing the matter
with their customers.
In making the following bills, students should use pen and ink, and give
earnest attention to the proper form and spacing, to plain, neat and rapid penman-
ship of both words and figures; and above all, to the accuracy of extensions and
additions.
NOTE.--When notes or drafts are given in payment, the student should draw the same and
correctly mature the notes.
[No. 1. Groceries].
NEW ORLEANS, Jan'y 2, 1894.
H. A. & R. C. SPENCER,
Bot. of A. L. dº E. E. Soulé.
Terms—Cash.
1893
Dec. 16 || 2 bags Rio Coffee, 325 lbs. *º tº &= * * &- (a) $ 234c. || $ 76.38
50c.
1 bbl. Sugar, 234 lbs. 4- tº º sº * se tºº 4 & 9 C. 21|61
# Chest Black Tea, 35 lbs. - - - - - - - “ 874c. 30|63
1 blol. Rice, 243 – 16 = 227 tº sº sº sº tº * & 4 8 C. 18|16
40 gals. N. O. Molasses * * wº e sº *- $º * , 75 c. 3000
6 doz. Brooms gºe &º *º wº -ºs sº -: •º * * 4.15 24|90
3 bbls. XXX Family Flour sº tº sº - * tº “ 8.12% 24|38
25 lbs. Cream Crackers wº- * es sº * - * º & 4 16 c. 4|00
50 lbs. Graham do. * º wº & sº- sº * & 4 15 c. 7|50
20 lbs. W. Butter s tº & ºt sº º &E, sº 4-8 * ** 30 C. 6|00
Rec’d payment, $ 243 56
A. L. & E. E. SOULE
Per F. Richardson.

284 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. Yºr
... [No .2. Groceries].
NEW ORLEANs, Apr. 1, 1894.
Bot. of P. H. Thoele dé, Co.
MAYER & BARRIERE,
5233 lbs. Butter * * tºº &º º * > tºº cº (2) 25 c.
136% “ La. Pecans gº tºº tº dº tº tº º & 4 12; c.
56 doz. C. Eggs tº { } gº sº tº tº e gº “ 18; c.
249 boxes Lobsters tºº gºe * > gº sº gºe tº & & 37+c.
140 jars Pickles & e sº tº tº gº tº gºe “ 62.4c.
48 lbs. Rice dº * & e ; : tºº gº tº tºº ( & 6+c.
83 bu. W. Corn e tº- tºº * > tº º * : ** 87 c.
64 “ W. Oats £º tº tº * * * * > tº e ** 44 c.
92 bags Salt e- * * * * * = ** 93 c.
84 lbs. B. Tea gº tº ſº tº gº g- gº gº ** 97 C.
108 gals. R. Whiskey sº º tº ſº tº gº “ $1.06
143 doz. Pine Apples º wº gº tº gº Eº ** 1.15
119 lbs. Y. H. Tea gº * e º tº gº º tº e ** 96 0.
324 boxes Shrimp tº a tºº º tº º tº * ** 35 C.
240 gals. Fire Proof Coal Oil tº º º * > ** 15 C.
164 lbs. Lard &= - tº {- º º & * * tº & & 7% c.
286 ‘‘ Green Peas - - - tº tº tº wº & 4 2}c.
320 “ N. Y. C. Cheese - * > tº gº tº º “ 173c.
112 gals. La. Molasses * * * * * * “ 22}c.
124 “ B. Whiskey gº * > dº sº tº & Cº. “ $1.25
72 bbls. La. Oranges * * º º sº gº “, 3.75
61 “ N. Potatoes { } tºº º º tº tºº ** 2.50
* 16# lbs. C. Soap * > {º tº º º º sº & & 84c.
9+ “ Persian Dates tº gº tº º gº * sº & & 64 c.
1 bbl. 50c. Sugar, 241-24 = 217 lbs. {º} sº º & & 5}c.
43 lbs. Lard Oi sº tº * tº º tº * ** 58 C.
45 “ Elgin C. Butte * = tº gº sº ** 35 C.
34 bu. W. Oats sº *_º * : gº tº tº- * * “ 48 C.
425 bbls. Flour * * * * = as sº “ $8.75 562070
NotE.—All extensions of the above bill should be made mentally.
—-º-
[No. 3. Drugs.]
Henry L. Landfried, bought of R. S. Winn, of Bayou Tunica Post Office,
La., November 2, 1894, the following Drugs: 60 oz. Morphine at $3.15; 20 lbs.
Insect Powder at 35%; 28 lbs. Gum Camphor, at 552; 15 lbs. Gum Opium, at
602; 2 lbs. Blue Mass, English, at 902; 46 lbs. Potassium Bromide, at 60g; 8
lbs. Senna Leaves, at 552; 1 Carboy Nitric Acid, 74 pounds, at 11%. What was
the amount of his bill? (Make and receipt the bill). Ans. $262.34.
—-º-
[No. 4. Books].
NEw York, Dec. 8, 1894.
H. C. SPENCER & CO.,
Bot. Of B. D. Rowlee dº Co.
TERMS-Cash.

(2)
$
1
5
i
20 doz. Missionary Bibles, gº
108 “ small New Testaments,
65 “ Prayer Books, g->
65 “ Hymn Books, gº
3 “ Bible Dictionaries,
# “ Webster’s Dictionary,
& &
& 4
;
i
& &
& &
5
Rec’d payment, $ 953 25
B. D. ROWLEE & CO.,
Per E. Conrad.
BILLS AND INVOICES. 285
[No. 5. Tobacco].
NEw ORLEANs, Jan'y 31, 1894,
Bot. of L. L. Williams dº Co.
WM. MELCHERT & Co.,
TERMS-Cash.
321 lbs. Tobacco, Low Lugs, – as tº tº * $º sº. * - tº- (a) 6 c.
1140 * * & 4 Med. Lugs, - * sº sº $º gº tº- º “ 74c.
509 & 4 “ Low Leaf, - - sº tº º ºs gº º “ 9+c.
965 “ “ Med. Leaf, - - * * * * * * “ 11}c.
398 “ ‘‘ Good Leaf, - - - - - - - gº “ 13+c.
2416 44 “ Fine Leaf, - - - - - - - - “ 15 c.
713 “ & & Selections - tºs sº {-> gº § { tº sº “ 163c.
Rec’d payment, $ 798 49
L. L. WILLIAMS & CO.
—-0-
[No. 6. Sugar].
NEW ORLEANs, Dec. 17, 1894.
Bot. of C. J. Sinnott.
F. L. RICHARDSON, JR.,
TERMS-Dft. 30 days.
1420 lbs. Sugar, Common, sº s sº e * * º tº Ø 5}c,
1927 “ “ Good, agº, * tº- * sº * sº * * “ 7+c.
2810 “ “ Fair, †º gº sº. - - - - - “ 7;c.
902 & 4 “ Prime, gº *- cº-e tº- gº * 4- &=e *- “ 84c.
813 ºf “ Choice, - - - - - - - - - “ 9+c.
2741 * * “ Yellow Centrifugal, sº tº sº tº sº. º “ 104c
Rec’d payment by dft. (a) 30 days' sight, $ 878 78
C. J. SINNOTT.
-º-
[No. 7. Tobacco].
NEW ORLEANs, Dec. 23, 1894.
Bot. of Sadler dº Smith.
MONTGOMERY & TREPAGNIER,
TERMS-3 mos.
7 Gross Chewing Tobacco, - - - - - - - - a $13.
180 lbs. Smoking do., - - - - - - - - ** 1.40
6 M. Havana Cigars - tº e tº º tº- fº wº- *- º ** 70.
2 M. N. O. Manufacture do - tº me *-, sº * . & º “ 30. I
Rec’d payment, .#

—-3–
[No. 8. Pork and Beef 1.
Jacob Broders, bought of C. F. Sauter, of St. Louis, Mo., October 15,
1894, the following articles: 15 bbls. of Prime Mess Pork, $13.50; 10 bbls. Extra
Mess Beef, $8.75; 152 lbs. Bacon, Shoulders, at 74 g; 75 lbs. Long Clear Sides, 74%;
146 lbs. Sugar Cured Hams, at 1242; 78 lbs. Breakfast Bacon, canvassed, at 1142.
Make out a receipted bill. Total, $333.58.
286 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS. X-
[No. 9. Candles].
NEW ORLEANs, Feb. 4, 1894.
F. L. & W. P. RICHARDSON,
IBot. of P. W. Sherwood dº Co.
TERMS-1 mo.
30 box. Sperm Candles, 596 lbs. - - - sº ºn. - º @ .35}
24 do. Adam Extra Candles, 483 lbs. - - - - - sº “ .28
15 do. Silver Gloss Starch 360 lbs. - - 4- - - º “ .10%
Rec’d payment, $ 385 52

—-4–
[No. 10. Shoes].
NEW ORLEANs, August 8, 1894.
M. G. WUERPEL,
Bot. of Albert Bohnet.
TERMS-Cash.
6 dozen Men's Boston, 17 in. Kip Boots - Ø $50.00
4 “ Ladies' Calf, Pink Lined, Balmorals - & 4 25.00
5 “ Youths' Wax Brogans - - - sº & 4 15.50
12 “ Youths' Buff Congress, Balmorals - & 4 17.45
1 “ Misses’ Lasting Congress, No. 1, sewed & 4 23.75
2# “ Misses' Goad Polish No. 1, sewed &= & & 32.60 |
2 “ . Boys' Buff Balmorals, Cloth Top, tº & 4 25.75 i
35 “ Men's 17 in. French Top Calf Boots - & 4 35.90 l
6 “ Misses' Calf Balmorals No. 1, pegged - & 4 17.00 ||
7 “ Misses’ Lasting Congress, No. 2, sewed { { 19.00
20 “ Boys' Split Wax Brogans - * * { { 12.00
2573|65
—º-
[No. 11. Domestic].
Eugene Ellis, bought of Wm. Dirker, of New Orleans, La., November
4, 1894, the following Dry Goods: 3704 yds. of Toledo Plaids at 152; 42 yds. Cotton
Check at 12.4%; 81 yds. Cotton Duck at 1042; 90 yds. of Hickory Stripes at 1442 ;
67 yds. Tickings at 132; 86 yds. Lowell Sheetings, 10-4, bleached, 26%; 82 yds.
Sheeting, Pembroke 10-4, at 294 g; 384 yds. Cotton Drills, Pride of the South, at
7#2; 143 yds., Lonsdale Cambric at 124%; 45 yds. Wamsette at 14%. , Make and
receipt the bill. Total, $164,72.
—#-
[No. 12. Drugs].
NEw ORLEANs, August 7, 1894.
Bot. of Henry Palmer.
HENRY HIRSCH,
TERMS-Note 60 days.
55 lbs. Acetic Acid, No. 8., º, º º sm. - (2) 25c.
123 “ Carbolic Acid, Cryst. * - gº * & & 60c.
98 “ Sulphuric Acid, Carboy, per pound - - & & 6c.
67 “ Citric Acid - sº º sº sº me 4 & 53c.
34 oz. Antipyrine - - - - - - - “ $1.50
78 lbs. Aqua Ammonia - - - - º - 4 & 15c.
88 “ Bismuth * - - * - - - ** 3.50
44 “ Cream Tartar, pure - - - - - & 4 35c.
35 “ Castile Soap, white mº ºme º tº º & & 25c.
2 cans Glycerine, 50 pounds - - - tº- gº & & 35c.
24 lbs. Gum Arabic, French, powdered - º & 4 700.
67 “ Oil, Bergamot - ſº º º dº tº ** 3.00
45 “ Potassium Chlorate - - - - - & 4 35c. ſ
34 “ Potassium Iodide sº sº dº º ºs ** 1.50 | 4.
| 82584
BILLS AND INVOICES. 287
[No. 13. Butter and Cheese].
NEw ORLEANs, August 4, 1894.
J. A. SABATIER,
Bot. of W. E. Slaughter.
TERMS-Due bill 1 Imo.
167 lbs. Extra Fancy Creamery, Butter - - (2) 6}c.
112 “ Fancy Creamery { { . - tº- & 4 25 c.
75 “ Choice Creamery & 4 sº - sº- * { 20 c.
55 “ Prime Creamery & 4 & - 4- & 4 17 C.
85 “ Fancy Dairy & & º - * & 4 16 c.
62% “ Choice Dairy é & * - * & s 14 c.
111 “ Prime Dairy & 4 4-> - - { { 12 c.
48 “ Common & 4 tº- - * { { 8#c.
75 “ New York Fancy Cream Cheese - - & 4 12#c.
88% “ Western Chedders, Full Cream Fancy Cheese 11;c.
&

74 “ America Full Cream Fancy Cheese 12 c.
90 “ Twins Full Cream Fancy & & - - & 4 11 c.
75 “ Flats Full Cream Fancy § { - * & 4 10}c.
38 “ Best Skims & 4 - º- & 4 8 c.
92 “ Skims & & - * & 4 5}c.
157|40
-º-
[No. 14. Wood and Coal].
--- NEW ORLEANs, Jan'y 29, 1894.
H. J. CALVERT,
In acc’t with L. B. Keiffer.
Jan. | 1 || To old balance as per bill rendered, 91|10
6 | 12 cords Ash Wood - - * - - - - ſº - (a) $7.
6 || 4 cords Oak Wood - -> e- - - e- - - - ** 6.50
14 |50 bbls. Pittsburg Coal - - - º - - º - { { 60c.
Cr. $231|10
8 || By Cash - sº * * * * * * * * - “ $50
29 || “.6 days’ Labor, at $4, - tº cº- tº º sº tº - ** 24 74|00
Balance due Jan'y 29, 1894, $ 157 10
Settled by note at 60 days.
L. B. KEIFFER.
[No. 15. Poultry and Eggs].
NEW ORLEANs, August 9, 1894.
J. H. McNEELY, w
Bought of S. Delerno.
TERMS-Note 30 days.
6# doz. Western Grown Chickens, - - - - Ø $ 5.00
7# ‘‘ Louisiana Young Chickens, º - & & & 3.00
41’s “ Ducks, - gº tº - tº - - & & 2.55
6; “ Western Geese, g- º tº gº - - & 4 6.25
2# “ Western Young Turkeys tº ſº e º 'º' ** 15.00
1ł “... Western Old Turkeys, - - - - - ** 18.00
76 “ Texas Eggs, - - - - - - “ .18%
18863
288
SOULE's PHILoSOPHIC PRACTICAL MATHEMATICS.
D. S. DRENNAN,
[No. 16. Rice].
TERMS-Due Bill, 30 days.
Bought of A. J. Duclos.
NEW ORLEANs, August 13, 1894.
Aug. | 13 | 160 lbs. Rice, No. 2, tºº sº
136 & 4 Common - *
240 & 4 Ordinary - sº
110 & & |Fair - º &-
273 & 4 Good - gº * ->
140 * * * Prime Gº tº-
310 & 4 Choice * > e.
150 & & Head, & •º
285 & 4 Fancy *E= tº
25 bbls. of 162 lbs. Rough Rice
FRANK DUMAS,
& 4
& 4
& 4
& 4
é &
& 4
& 4
& 4
[No. 17. Plantation Sugar, Open Kettle].
Bought of H. P. Knobloch.
'84.50
189|22
NEW ORLEANs, August 15, 1894.
July | 10 |280 lbs. Choice Sugar - 4- tº º ºs wº - Ø 4}c.
197 ‘‘ Strict, Prime *- -- “º * * e- * ** 5 C.
230 ** Prime tº * gº {º tº- º *- º- “ 4}c.
254 ** Good Fair 4-3 tº º tº tº- s e-º < “ 4#c.
290 ** Good Prime * º tº $º & gº q-, “ 43c.
260 “ Common sº * 3-º º & gº & “ 43c.
425 “ Inferior º gº tº gº * º ſº gº “ 3}c.
85|40
-O-
[No. 18. Cotton].
NEW ORLEANs, August 9, 1894.
s. ForFT,
To J. J. Foley, Dr.
TERMS-Cash.
2 bales, 452,461 = 913 lbs. Low Ordinary Cotton - a 53e
1 “ 458 lbs. Ordinary Cotton sº º sº Ǻ & & 6 c.
3 “ 450,455,460 = 1365 lbs. Good Ordinary Cotton “ 6}c.
1 “ 456 lbs. Low Middling, Cotton - - - “ 64c.
1 “ 462 lbs. Middling Cotton tº me gº “ 74c.
1 ‘‘ 455 lbs. Good Middling Cotton - - - “ 73c.
2 “ 452,460 = 912 lbs. Middling Fair Cotton - “ 74c.
1 ‘‘ 456 lbs. Fair Cotton & sº sº gº º ** 8 c.

372.56
* BILLS AND INVOICES. 289
[No. 19. Refined Sugar]. ©
NEw ORLEANS, August 10, 1894.
EDWARD LeBLANC,
|Bot. Of E. N. Moore.
TERMS-60 days.
252 lbs. Cut Loaf Sugar - - - - - - (a) 6; c
176# “ Powdered Sugar - *- sº sº- - “ 64 c
240 “ Standard Granulated Sugar tº º - “ 6+ c
175 “ Off Granulated Sugar * * * * “ 63 c
252 “ Candy A. Sugar - - - - - - “ 6% c
144 “ Confectioners' A. Sugar sº me m ms “ 6, c
208 “ Extra C. Sugar * * tº sº m s “ 5+ c
197 “ C. Sugar. - & º --> gº tº- - ** 5 C
99.95
-º-
[No. 20. Lumber].
NEW ORLEANs, Jan'y 12, 1894.
A. & S. H. SOULE,
To Jas. Mc. Hugh. Dr.

To 4378 feet Common Boards - - - º - º (a) $21.00 per M.
“ 1760 feet Dressed Flooring - - - -- - ** 28.50 “
** 5125 Brick * * - - - - - º ** 14.25 + .
“ 9250 Cypress Shingles - - - - - - é & 6.50 **
“Cartage and Labor, 14|25
Received payment, $ 289 51
—-º-
[No. 21. Freight Bill].
NEW ORLEANs, Jan’y 3, 1894.
GEO. F. BARTLEY & CO.,
To Steamship Knickerbocker and Owners, Dr.
For Freight on 4393 cubic ft. (a) 25c. The same being contents of 8 boxes
measuring as follows:
Nos. 1, 2 & 3: 5 ft. 4 in. x 4 ft. 6 in. × 2 ft. 8 in.
“ 4, 5 & 6: 6 ft. 2 in. x 3 ft. 0 in. x 2 ft. 11 in.
“ 7 & 8: 12 ft. 3 in. × 2 ft. 4 in. X 1 ft. 6 in.
Received payment, $ 109 91
E
—-º-
[No. 22. Services].
A carpenter receives $3.25 per day of 9 hours. How much will he earn work-
ing each week day in the month of January, 1894, which commenced on Monday?
(Make the bill). Ans. $87.75.
—º-
[No. 23. Services].
A laborer receives 30g per hour, and he works 104 hours per day. How much
will he receive for 6 days work? (Make the bill). Ans. $18.90.
29O SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
º [No. 24. Services].
A clerk commenced work June 20th, and worked till September the 15th, at
$65 per month. He paid $15 per month for board and $3 per month for washing,
including June 20th, but not September 15th, how much money has he, not allowing
for other expenditures? (Make the bill). Ans. $133.163.
NOTE.-In calculating salaries and rents by the month, 30 days are allowed for a month; i. e.
the same salary or rent is paid for each month whether of 28, 29, 30 or 31 days.
—-3–
[No. 25. Rent and Services].
NEw ORLEANs, Jan'y 4, 1894.
J. T. O'QUINN,
* To W. Hermann, Dr.
For Rent of house No. 386 Dryades Street, from Oct. 7, 1894, to Jan. 1, 1895,
1st date included, 23# months at $35 cº- -> - - º- º
“ Services as collector from Sept. 19, 1894, to Jan. 4, 1895, both dates
included, 3}} months, at $75 tº- -- e sº * - - tº
Received payment, $ 363 00
v--—
[No. 26. Services].
NEw ORLEANs, Feb. 1, 1894.
MISSISSIPPI VALLEY TRANSPORTATION CO.,
To Buck dº Richardson, Dr.

For services rendered in cause No. 55472. “Steamer R. E. Lee, and Owners
vs. Miss. W. T. Co.”
Received payment,
—eſpe—
[No. 27. Services].
A mechanic receives $4 per day of 9 hours, and 60g an hour for “overtime.”
During the first three days of the week he worked 30 hours, and the last three
days of the week he commenced at 6:30 A. M. and worked till 10 P.M., less two
hours for dinner and supper. What are his wages for the week's work &
(Make the bill). - Ans. $33.90.
—cºo-
[No. 28. Carpenters' Time Sheet]. Rate
M. T. W. T. F. S. ds. p: day.
*:
R. Hirsh... . . . Foreman . . Oct. 22 10 || 10 || 10 || 10 || 15 10 || 6; $4| 00
F. Smith. . . . . Carpenter. “ 22 10 || 10 || 10 || 10 | 15 || 10 || 64 3| 50
O. Wood. . . . . do. ** 22 10 || 10 || 10 || 10 | X | 10 5 3| 00
P. Murphy ... do. ** 22 10 || 10 || 10 || 10 || 10 || 10 || 6 3| 00
H. Brown.... do. “ [22] 10 || 5 || 5 || 5 || 10 || 10 || 44 3| 00
J. Fitzpatrick do ** 26 × | X | X | X | 10 || 10 || 2 2 50
j. s." Smith |Apprentice “ 23 ió | 10 || 3 || 10 | 16 | 16 || 5 || 2 00
114|00
T- Ans.
In this time sheet, what is due to each mechanic, and what is the total
amount at the close of the week; 10 hours being a day’s work, and double price
being paid for overtime? Ans. $114. Total amount.
NotE.—The numbers written after the names represent the number of hours that each
worked on the different days of the week. The X signifies no hours.
* BILLS AND INVOICES. 29 I
[No. 29. Factory Time Sheet].
Rate
M. T. W. T. F. S. ds. per day.
J. C. Bell... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Oct. | 8 || 10 | 10 | 10 | 10 | 10 | 10 || 6 || 400
Thos. A. Clark. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . “ | 8 || 10 || 10 || 10 | X | 10 || 10 || 5 || 350
A. J. Duclos. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . “ | 8 || 5 || 10 || 10 || 10 || 10 || 10 || 54 || 350
T. J. Fatjo. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . “ | 8 || 10 | X | 10 || 10 || 10 || 10 || 5 || 350
H. A. Hornor. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ** | 8 || 10 10 || 10 || 10 || 10 || 10 6 3|50
H. C. Lowry. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ** | 8 || 10 || 10 || 10 || 10 || 10 || 10 || 6 2|00
In this time sheet, what is due each workman Saturday night, 10 hours being
a day's work, and what is the total amount?
FLYNN &
[No. 30. Paints, Oils, etc.].
KERWIN,
TERMS-30 days.
—eº-
Bought
Ans. To the last, $111.25.
NEw ORLEANs, Mar. 11, 1894.
of Donelon dº Haight.

2 cases, 12 gals. Green Paint
# “ 3 ** Vermilion
12 gals. Roofing Paints
9 “ Carriage “
6 “ Copper 4 &
50 lbs. White Lead
75 “ Putty tº-º:
12 gals. Coach Warnish
18 ‘‘ West. Linseed Oil
45 lbs. Eng. Vermilion
20 “ Vandyke Brown
[No. 31.
M. A. TERRY & CO.,
TERMS Cash.
Building Materials].
$1.80
2.40
163|01
NEw GRLEANs, Mar. 9, 1894.
Bought of H. M. Thompson dº Son.
750 ft. 14x3+ Clear Pine Flooring
1200 “ No. 1, Weatherboards
550 “ Ceiling tºº gº
3500 Laths tº º ſº s
45 bbls. Lime sºme tº-
18 “ Domestic Cement
800 Paving Brick tº-
25 bbls. Sand * sº
12 “ Small Shells -
2000 Am. Fire Brick º
750 ft. Sewer Pipe § -º
600 “ Chimney Flue Lining
Eº* >arssºesº† :tº-ºfºsº*º-
per M. $25.00
& 4
& 4
& 4
(2)
& 4
per M.
(2)
& 4
per M.
(2)
4 &
13.00
15.50
1.90
46126
292
soule's PHILOSOPHIC PRACTICAL MATHEMATICs.
E. WEIDIG,
[No. 32. Hardware].
NEw ORLEANs, Dec. 22, 1893.
bot. of Reymes dº Carmam.
TERMS-Cash. —Days—% Discount—Days—% Interest after Maturity and subject to sight Draft without notice.

18}
16
& 4
49+ doz. Claw Hammers -
Lathing Hatchets -
Chain Bolts - -
. Nails, ea. 6d.-8d.-10d.
. Measuring Tapes -
Marking Brushes -
Butcher Saws -
Gem Wrenches -
Chisels, ea. § and # in.
File Handles - -
Curry Combs - -
Gas Plyers - -
10"/14" Shelf Brackets
Ivory Rules - tº
Pointing Trowels -
Mortise Locks tº-
Rim & & -
Shutter Bolts ->
Padlocks - -
Drawer Pulls - -
Dashboard Lanterns
[No. 33.
McDANIEL & KELLER,
sº - - Gº- (2) $ 4.95
º - - º & & 8.49
$º - - tº & 4 1.75%
º - - º & & .04%
s - -> * > & 4 4.50
gº -- º sº & 4 .95
gº - e- º & 4 5.65
º - - tº º 46 2.35
tº - º tº º & 4 6.80
gº º tº- tº & 4 .88
º -> sº tº º & 4 1.76%
º - - tº & 4 8.46
º tº- º « » é & 4.98
gº e- sº tº º & & 10.20
- - º * > & 4 4.75
º º º gº & 4 3.30
* {-> Gº tº & 4 2.85
ge tº- -> & & 4 1.60
- -> - gº & 4 2.38
gº & º gº & 4 1.86%
tº e º tº ſº &é 4.25
Grain, Hay, etc.].
NEW ORLEANs, Jan'y 25, 1894.
Bot. of Redfield dº Halsnith.
TERMS-Note at 60 days.
2144 lbs. bu. Yellow Corn, tº º - ſº tº - @ $ 63c.
1242 “ — “ Texas Wheat, * - sº sº - “ 1.70
852 “ —— “ White Oats, tº sº - we * - & 4 56c.
792 “ —— “ Barley, - tº º - * * - & 4 83c.
1427 “ — Cwt. Bran, - sº m sº sº tº tº & & 75C.
3745 & 4 Tons Timothy Hay, * - sº ºne º “ 18.50 *
1701 * * Tons Clover Hay, * * - sº gº - ** 20.
Rec'd pay’t by note at 60 days, $ 150 27

REDFIELD & HALSMITH.
BILLS AND INVOICES.
293
[No. 34. Sugar—Centrifugals].
J. HEINTZELMAN,
Bot. Of Wm. O’Briem.
NEw ORLEANs, August 4, 1894.
TERMS-Cash.
244 lbs. Plantation Sugar gº. * sº-s sº (a) 5% c.
355 ‘‘ Granulated 4 & q= - sº sº sº ** 6 C.
197 ‘‘ Off Granulated “ +-º *. º sº wº “ 6+ c.
280 “ Choice White & & * * sº * dº * “ 64 c.
175 ‘‘ Off White { { tº-3 E-> tº- sº sº “ 6; c.
290 “ Gray Off White “ sº & &= sº tº “ 64 c.
178 “ Fancy Yellow & 4 sº * sº tº-: sº “ 6+ c.
230 “ Choice Yellow “ Gº tºº gº sº &- ** 6 C.
312 ‘‘ Prime Yellow 4 & º * sº * &- “ 5}{c.
230 ‘‘ Off Yellow & & gº {- # e. †º tº-e “ 5+ c.
234 “ Yellow Fair & 4 & e -> & tº sº “ 4; c.
226 ‘‘ Seconds & 4 &- &= dº * “ 4; c.
174|34
J. P. POWELL,
TERMS-1 mo.
[No. 35. Produce].
Bot.
NEw ORLEANs, August 7, 1894.
of W. E. Richards.
15
6
150
& 4
& 4
{
& &
( &
& 4
& 4
27 bbls. Neshamnock Potatoes
Onions tº- * *
6 Crates Cabbage . . .
12 Strings La. Garlic *-* - *-
30 bbls. Sweet Potatoes, Yams *
Western Cucumbers
40 doz. Egg Plants tº
140 lbs.
&
Choice Northern White Beans
Northern Red Kidney & 4
Northern White Kidney “
Green Peas 3-3 e *
Dried Peaches & sº
Evaporated Peaches *
i
:
i
.20
332
26
2.94 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS
[No. 36. Character, Virtue, Knowledge, etc.].
NEW ORLEANs, Oct. 1, 1894.
*
ETHICAL AMBITION,
Bot. of Soulé College.
TERMS-Fidelity to Duty.
1163 lbs. Energy and Perseverance tºº gº º (a) .87%
45 ‘‘ Self Reliance sº sº tº gº Q = º & 4 .95
100 “ Honesty sº * * * * * sº sº. “ $100.00
17 “ Business Sagacity * * * * * ** 1.15
150 “ Punctuality sº gº tº º gº *...* é & 1.16#
49 “ Cheerfulness es &=º tº º * * tº & & .81
56 ‘‘ Faithfulness & º * gº tº gº é & .94
73 boxes Accuracy tº sº * * = * i e & & .75
38 “ Thoughtfulness tº & ºn tº * > tº & & .66%
+ yd. Talkativeness *- tº cºs gº º * & 4 .01
25 yds. Economy e- * * is ºs sº * = ( & .83
2} “ Recreation tº * º º tº º & & .33%
50 bu. of Politeness ſº tº º gº tº tº gº & 4 2.00
75 ‘‘ ‘‘ Neatness and Order tº º º sº & 4 1.00
75 “ “ Obedience and Good Deportment º & 4 2.50
+ “ “ Attention to other Peoples' Business - & & 1.00
50 “ “ Attention to my own Business - º & & 5.00
500 bbls. Practical English Education * gº ‘‘ 4.00
500 “ Practical Business Education tº- sº & 4 4.00
500 facts of Physiology and Hygiene - º * † & 4 5.00
500 “ “ Phrenology and Sociology tº- tºº ** 5.00

[No. 37. Dry Goods].
NEw ORLEANs, Feb’y 14, 1894.
WEHRT & GEORGE,
Bot. Of Morris & Childress.
TERMS-30 days.

25 yds. Serge e ‘º - * * - * * (Ø .22}
37% “ Hopsacking - tº- wº - * - & 4 .42%
18% “ Diagonal Serge * * * * * & 4 .39
15} “ French Sateens º gº e- * º & 4 .27#
12 “ Brocaded Silk * * *-* e- *g: gº & 4 .62%
47 ‘‘ Colored Satins gº * s * * sº gº * { .43
22 “ Table Damask se *º- tºº tº- * & 4 .47%
310 “ Crash º tº * sº sº tº &= & 4 .06
82% “ Canton Flannel * - sº gº & 4 .09%
64 “ Storm Serge 4- * * - sº & e .35
9 “ Brocaded Cashmere sº º sº tº- & 4 .14+
53 “ Ladies’ Cloth 3- *- - & * sº { { .87% Lºw
87 “ Figured Satinet º * as * sº ( & .95
20 “ French Velours tº- *-> * sº “ 1.47%
35 “ Surah Silk e-º *- e- sº .75
33+ “ Scotch Plaids «º *- * - *º- * { { .65
28 “ Whipcords *º * -º *E º { % .33%
17+ “ Henriettas tº- sº sº sº & ( & .36
142} “ Silk Ribbon * wº s e-8 & 4 .12%
75 ‘‘ Satin * * * wº wº º * & f .16#
1# doz. Initial Linen Handkerchiefs * '' 4.75
| 382
jºr BILLS AND INVOICES. 295
[No. 38. Invoice of Goods Shipped].
NEW ORLEANs, Jan'y 16, 1894.
Invoice of Sundries, purchased by H. Strong & Co., and shipped per Steamer
La Belle, for account and risk of Chas. B. Thorn, Shreveport, La. º
87 bbls. Molasses, 3498 gallons * - sº - - - - Ø 60c.
20 hbds. Sugar, 23780 lbs. se smº sº º sm * * * “ 90.
10 bbls. Rice, 2150 lbs. - - - " - - - - - ** 50.
Charges. 4346.9
Drayage, º s w = - º - - - - * - Q- 1750
Insurance on $4800.40 sº g- º gº tº «º * > --> <--> Ø #% 30|00
Commission on $4364.00 sº .* * * * * ** º “ 24% 109|10
$ 4503 10
-Ö-
[No. 39. Slaters' Bill].
NEw ORLEANs, Jan'y 14, 1894.
C. H. REYNOLDs,
To H. Marsden, Dr.

36.04}; sqs. - - *- e- * (a) $14.50
For 239 ft. Guttering - - - º * > s & & .90
Received payment, $ 737 77
For slating a roof measuring 72 ft. 4 in. by 49 ft. 10 in., and containing
[No. 40. Cistern Makers’ Bill].
NEw ORLEANs, Nov. 4, 1894.
NEW ORLEANS, ST. LOUIS, AND CHICAGO R. R.,
- To W. L. dº H. Jackson, Dr.
For 150 Cisterns holding 766782.45 gals. (a 24c. per gal. The inside meas-
urement of each cistern is as follows: 11 ft. 3 in. perpendicular height,
lower base 9 ft. 2 in. in diameter and upper base 8 ft. 5 in. in diameter.
Received payment, § 19169 56
NotE.—For other bills involving discount, rebate, etc. and embracing different lines of busi-
ness, see other pages of this book as per index. o

Tº Filiit Indlf.Tille Between TWO DâtéS,
—N
616. 1. How many years, months, days, hours, and minutes from 4:20
o'clock P. M. June 10, 1889, to 9:15 o'clock A. M. August 14, 1894, not allowing for
Teap years, and counting 30 days to the month?
OPERATION. Eacplanation.—In all prob-
lems of this kind, we write
yr. IſlO. d. hr. min. the later date first. since it
y
1894 8 14 9 15, expresses the greater period
1889 6 10 16 20 of time, and the earlier date
º & e beneath. Then subtract as in
O yrs. 2 mos. 3 ds. 16 hrs. 55 min. Ans. compound denominate num-
bers.
NOTE.—The months are numbered from January, and the hours are counted from 12 o'clock
at night.
- SECOND OPERATION
The time from June 10, 1889, to June 10, 1894, - * - - = 5 yrs.
66 “ June 10, 1894, to Aug. 10, 1894, - - tº - = 2 mos.
{{ “ 4:20 P. M. Aug. 10, 1894, to 4:20 P. M. Aug. 13, 1894, - 3 ds.
44 “ 4:20 P. M. Aug. 13, 1894, to 8:20 A. M. Aug. 14, 1894, - 16 hrs.
& 4 “ 8:20 A. M. Aug. 14, 1894, to 9:15 A. M. Aug. 14, 1894, = 55 min.
NotE.—While both of the above operations are correct according to the conditions of the
problem, and conform to the usual method of finding the time between dates, neither operation
gives an accurate result, for the reason that some years and some months contain more days than
other years and other months. The only accurate way is when the time is less than one year, to
count the exact number of days in each month of intervening time, or refer to a time table. When
the time is more than one year, find the days for the months, as above, and allow 365 days for com-
mon, and 366 for leap years.
2. The Declaration of Independence was ratified July 4, 1776; the battle of
New Orleans was fought Jan'y 8, 1815. What is the time between these two dates,
by the usual method? Ans. 38 yrs. 6 mos. 4 ds.
3. Washington was born February 22, 1732. How old was he at the Declar-
ation of Independence, July 4, 1776? How long did he live after that event, his
death occurring December 14, 1799, and what was his age 3
w Ans. 44 yrs. 4 ms. 12 ds.
23 {{ 5 4% 10 { %
67 4 9 4, 22 (4
Jamestown, Va., was first settled May 23, 1607, and the Pilgrims landed
at Plymouth, December 22, 1620. What time intervened, by the usual method?
Ans. 13 yrs. 6 mos. 29 ds.
5. A note was dated January 10, 1892, and was made payable 2 years after
date. When did it become due; how many days did it run, by the usual method;
and how many days counting actual time; no allowance to be made for days of
grace # Ans. January 10, 1894, it matured;
720 ds. by the usual method;
731 ds. actual time.
(296)
TO FIND THE LENGTH OF THE DAY AND THE NIGHT, 297
6. A note is drawn Oct. 15, 1894, and made payable 3 months after date.
When does it mature, allowing 3 days of grace according to business custom, and
how many days does it run, actual time 3 Ans. Matures Jan. 18, 1895; runs 95 ds.
7. A note is drawn, Oct. 15, 1894, and made payable 90 days after date.
When does it mature, allowing the customary 3 days of grace, and how many days
does it run, actual time? Ans. Matures Jan'y 16, 1895; runs 93 days.
NOTE.-It is the custom, when maturing commercial instruments, to be governed strictly by
the terms expressed therein. When they are made payable in years or months, they are matured
in years or months, counting them as they run from the date of the instrument. When they are
made payable in days, they are matured in days, the actual number of days in each month of the
intervening time being counted.
The day that a note or other instrument is dated, is not counted as one of the days which it
has to run. The day it matures is however counted.
When discounting notes, bills of exchange, etc., the actual number of days which the instru-
ment has to run, is counted, to compute the interest, whether it is drawn in years, months or days.
See Banker's Discount in this book for extended work of maturing and discounting commer-
cial paper.
HOW TO FIND TEIE LENGTH OF THE DAY OR NIGEIT WHEN WE
HAVE THE TIME TEIE SUN RISES OR SETS.
617. 1. If the Sun sets at 7h. 5m. 10s., what is the length of the day and
the night, and at what time does the Sun rise ?
OPERATION.
Sun sets 7h. 5m. 10s. × 2 = 14. 10. 20.
add 12h.
Time after midnight that the Sun Set = 19h. 5m. 10s.
24h.
Sun rises 4h. 54m. 50s.
19h. 5m. 10s.
Length of day 14h. 10m. 20s.
w 24h.
Length of night 9h. 49m. 40s.
Explanation.--In all problems like this, first add 12h., to the time the Sun sets, which gives
the time past midnight that the Sun sets. Then subtract this from 24h. and in the remainder we
have the time the Sun rises. Then, since the sun rises 4h. 54m. 50s. A. M. and sets 19h. 5m.
10s. past midnight the difference will be the length of the day. Then the length of the day
subtracted from 24h. will give the length of the night.
2. If the Sun rises at 4h. 54m. 50s., what is the length of the night and
the day ?
OPERATION.
Sun rises 4h. 54m. 50s. × 2 = 9h. 49m. 40s. length of night.
24h.
14h. 10m. 20s. length of day.
Explanation.—In all problems of this kind, multiply the time the Sun rises by 2 and in the
product we have the length of the night; Then, to find the length of the day we subtract the
length of the night from 24h.
NotE.—The above methods of finding the length of the day and the night, and the rising and
the setting of the Sun are not absolutely correct, on account of the obliquity of the earth's orbit,
and other astronomical reasons.
3. When the Sun sets 6h. 34m. 15s., what is the length of the day and what
time does the Sun rise? Ans. Length of day, 13h. 8m. 30s.
| Sun rises 5h. 25m. 45s.
4. When the Sun rises at 5h. 14m. 15s. past 12 midnight, what is the length
of the night? Ans. 10h. 28 m. 30S.
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º: DIRECTIONS FOR THE OPERATION. 299
TO DETERMINE TEIE DAY OF THE WEEK ON WEHICH AN EVENT
DID OR MAY. OCCUR.
619. A convenient method of determining immediately what day of the
week any date transpired, or will transpire, from the commencement of the Christian
era, for the term of three thousand years.
620. The following table shows the ratio to be added for each month :
TABLE OF MONTHS.
January, ratio is - 3 || May, ratio is - - 4 || September, ratio is 1.
February, “ - - 6 June, * { * = sº | October, {{ - - 3
March, {{ e-e 6 July, {{ - - - 2 | November, “ - - 6
April, {{ º 2 || August, “ - - - 5 || December, “ - - 1
*
NotE.—In leap years, the ratio of January is 2 and the ratio of February is 5. There is no
change in the ratios of the other months.
621. The following table shows the ratios to be added for each century of
the Christian era :
TABLE OF CENTURIES.
200, 900, 1800, 2200, 3000, the ratio is tº
300, 1000, the ratio is - gº tº sº tº
400, 1100, 1900, 2300, 2700, the ratio is - - -
500, 1200, 1600, 2000, 2400, 2800, the ratio is -
600, 1300, the ratio is tº- s *- * *
700, 1400, 1700, 2100, 2500, 2900, the ratio is -
100, 800, 1500, the ratio is º e- sa º º
i
DIRECTIONS FOR THE OPERATION.
622. 1. Add to the given year, (omitting the century figures) one-fourth part
of itself, rejecting the fractions, if any.
2. To this sum add, 10, the day of the given month; 29, add the ratio of the
month, as per table of months; 30, add the ratio of the century as per table of centuries.
3. Divide this sum by 7. The remainder is the day of the week counting
Sunday as the first day, Monday as the second, etc.
3OO SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS. x:
PROBLEMIS.
623, 7, The battle of Shiloh was commenced on the 6th of April, 1862.
*What was the day of the week? Ans. Sunday.
OPERATION.
The given year, omitting the centuries, is - 62
One-fourth of same, rejecting fractions, is - 15
The day of the month was the * ºs - 6
The ratio of April is * wº tº- * * - 2
The ratio of 1800 is sº gº sº tºp - 0
This sum, divided by 7, gives * * - 7) 85
12 and 1 remainder tºº gº gº 12+1
The 1 remainder signifies the first day of the week—SUNDAY.
2. The Declaration of Independence was signed July 4, 1776. What was the
day of the week 3 Ans. Thursday.
OPERATION.
Given year is - - - - 76
One-fourth of same is - 19
Day of month is - - - 4
Ratio of July is - - - - 2
Ratio of 1700 is - - - - 2
Divide by 7) 103
*
14–H5 remainder—THURSDAY.
3. Gen. R. E. Lee was born June 19, 1807. What was the day of the
week # Ans. Friday.
4. Martin Luther was born Nov. 10, 1483. What was the day of the week?
Ans. Monday. "
5. What day of the week will January 1 occur in 2000? Ans. Saturday.
NOTE.-When there is no remainder after dividing by 7, the day of the week is Saturday.
6. A note is dated Wednesday, October 24th, 1894, payable in 120 days.
In what year and month, and on what day of the month and day of the week, will
it become due, allowing 3 days of grace? Ans. Sunday, Feb. 24, 1895.
7. On what day of the week were you born, and if you live to be 150 years
old, as we wish you may, on what day of the week will you die
ifference of Latitude.
624. Latitude is the distance in degrees, minutes, and seconds, of any
place on the globe, North or South of the equator.
Latitude is reckoned from the equator to each pole of the earth, and like arcs of all circles,
is measured in degrees, minutes and seconds, and can never be greater than a quadrant, or 90
degrees. The earth not being a perfect sphere, but oblate, or flattened at the poles, the degrees
vary slightly in length toward the poles.
The difference of latitude between two places is found by subtraction or addition, as in
compound denominate numbers.
1. The latitude of Washington City is 38° 53' 39" North, and that of New
Orleans is 29° 56' 59° North. Required the difference of latitude.
OPERATION.
Lat. of Washington = 38o 53° 39” N.
* “ New Orleans = 29O 56° 59'ſ N.
Dif. of latitude = 8o 56° 40” Ans.
Pºplanation.-Since both places are on the same side of the equator, that is in North lati-
tude, we subtract the lesser latitude from the greater. When the latitude of one place is North,
and that of the other South, of the equator, we add the two latitudes together, and the sum will
be their difference of latitude.
2. The latitude of Mobile is 30° 41' 26" North, and that of Quebec is 460
48' 17" North. What is their difference? AnS. 16O 6' 51".
3. Philadelphia is in latitude 39° 56' 53' North, and Rio de Janeiro 220 54,
24" South. What is their difference # AnS, 62o 51/17//.
4. New York is in latitude 40 deg. 42 min. North, and Cape Horn is in lati-
tude 55 deg. 58 min. South. What difference intervenes? Ans. 96 deg. 40 min.
5. Havana is in latitude 23 deg. 9 min. North, and San Francisco lies 14
deg. 39 min, further North. What is the latitude of San Franciscot
Ans. 37 deg. 48 min.

(301)
3]ifference of Longitude.
5N
aſſº. —-dºh- A
-ºr wºr---ºut-w
625. The Longitude of a place is its distance in degrees, minutes, and
seconds, East or West, from a given meridian.
*
A degree of longitude on the equator is 69.1638 statute miles, but since all meridians are
drawn through the poles and meet in a point, they gradually converge as we advance from the
equator, and vary in length with each degree of latitude, until they meet in the poles, and the
longitude becomes nothing.
626. A Meridian is an imaginary circular line on the surface of the earth,
passing through the poles and any given place.
The meridian from which longitude is reckoned is called the first meridian, or
standard meridian, and is marked 0°. All places East of the first or standard
meridian, within 1809, are in East longitude; and all places West of the first or
standard meridian, within 1800, are in West longitude.
The English reckon longitude from the meridian of Greenwich; the French from that of
Paris. ' The government of the United States usually reckons longitude from the English standard
meridian, Greenwich. In American maps, the meridian of Greenwich is printed at the top and the
meridian of Washington, the capital of the U. S., is printed at the bottom.
The difference of longitude between two places is found like the difference of latitude, by
subtraction or addition, as in compound denominate numbers.
1. The longitude of Galveston is 940 47' 26" West; of Liverpool, 39 4' 16"
West. What is their difference # Ans. 919 43' 10".
OPERATION. Ezplanation.—Since both places .
in West longitude, we subtract the
Long. of Galveston = 94° 47' 26". W. iesser from the greater. When one
“ “ Liverpool = 39 4' 16" W. place is in West and the other in
*-**-*. East longitude, we add the longi-
º e O Al Q/ A / tudes of the two places together, and
Dif, of longitude 91O 43/ 10 Ans. if the sum is more than iso degrees
we subtract it from 360 degrees, because the shortest distance between the places would then
be the other way around the world.
2. Cape Flattery, Wash., is in longitude 124° 44' 30" W. Hong Kong, Ch.,
is in longitude 1140 9' 32° East. What is their difference 7 Ans. 121° 5' 58".
OPERATION.
Long. of Cape Flattery = 1240 44' 30" W. 3600 01 0//
“ “ Hong Kong = 114O 9/ 32// E. 238o 54/ 2//
238o 54/ 2// 121O 5' 58//
3. The longitude of Calcutta is 88.0 20' 11" East, of Trieste, 13° 46' East.
"What is their difference # An S. 74O 34' 11".
4. The longitude of Havana is 820 21' 17" West, and of Antwerp 4o 24!
44” East. What is their difference # AImS. 86O 46' 1".
5. St. Petersburg is in longitude 300 19' 22" East. Cedar Keys is in longi-
tude 83C 1’ 57’’ West. Find their difference. Ans. 113C 21’ 19//.
(302)
a a Aſº A- a
—w wºr--ºr-
627. The circumference of the earth, being a great circle, is divided into
360 equal parts called Degrees of Longitude.
The earth revolves on its axis, from West to East, once in 24 hours—which
gives the sun the appearance of passing around the earth from East to West.
*.
Now, since the earth revolves once in 24 hours, all parts of its surface or
circumference—the 360° of longitude—pass under the sun during that space of
time. Hence, since 360° are passed under the sun in 24 hours, ºr part of 360°, or
15° of longitude are passed in 1 hour. Since 150 are passed in 1 hour or 60 minutes,
ºr part of 15°, or 15' of longitude are passed in 1 minute. Since 15' of longitude
are passed in 1 minute, or 60 seconds, ºr part of 15', or 15" of longitude are passed
in 1 second.
Then since 150 of longitude 1 hour of time,
1O 4 4. = + of 1 hour (60 min.) = 4 min. of time.
Then Since 15' “ { % = 1 minute of time,
1/ 4. 4% = #5 of 1 minute (60 sec.) = 4 sec. of time.
Then since 15// “ & 4 = 1 second of time,
1// 44 4% = +3 of a second of time;
Or, since 1 “ {{ '- 4 seconds of time,
1// 4. § { # of 4 sec. = #5, or # of a second of time.
Töö)
628. From the foregoing, we deduce the following:
COMPARATIVE TABLE OF LONGITUTE AND TIME.
360° long. make a difference of 24 hours of time.
15O 4. {{ & 4 {{ 1 { % 4%
15/ 4% {{ 64 { % 1. minute 46
15// {{ {{ {{ { % 1 second { %
1O “ {{ { % 46 4 minutes 46
1/ {{ 44 {{ & 4 4 seconds { %
1// { % {{ { % { % *; second { %
629. Longitude and Time give rise to two classes of problems, as follows:
1. To reduce time to longitude, or to find the difference of longitude between
two places, when the difference of time is known. *
2. To reduce longitude to time, or to find the difference of time between two
places, when the difference of longitude is known.
& (303)
3O4. SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. º:
PROBLEMS UNDER THE FIRST CLASS,
OR, TO FIND THE DIFFERENCE OF LONGITUDE BETWEEN TWO PLACES, WHEN THE
DIFFERENCE OF TIME IS KNOWN.
630. 1. Reduce 11 hrs. 20 min. 40 sec. of time to longitude.
OPERATION.
11 hrs. 20 min. 40 sec. Explanation.—Since according to the foregoing eluci-
15 dations, 1 hr. = 15° of longitude, 1 min. = 15' of longi-
tude, and 1 second = 15" of longitude there are 15 times
as many 9, ', and " of longitude as there are hrs., min.,
1700 10/ 0’’ Ans. and sec. of time. Hence in all problems of this kind, we
multiply the different units of time by 15, as in multipli-
cation of compound denominate numbers, and thus convert them into units of longitude.
2. The difference of time between two places is 3 hrs. 15 min. 24 sec. What
is their difference of longitude 3 - AnS. 480 51' 0/.
3. The difference of time between New York and Chicago is 54 min. 19 sec.
What is the difference of longitude? Ans. 130 34' 45".
4. When it is 11 o'clock 16 minutes 1 # seconds A. M. at Boston, it is 10
o'clock A. M. at New Orleans. Find the difference of longitude.
OPERATION INDICATED.
Time at Boston 11 hrs. 16 min. 11%; Sec.
‘‘ ‘‘ New Orleans 10 hrs. 0 min. 0 Sec.
Difference of time 1 hr. 16 min. 1 # Sec.
15
190 0/ 19''. AnS.
5. The longitude of Rome, Italy, is 12° 27' E.; the difference of time
between Rome and Mobile, Ala., is 6 hrs. 41 min. 57+; sec. What is the longitude
of Mobile, West? AnS. 889 2/ 28// W.
OPERATION INDICATED.
6 hrs. 41 min. 57+; Sec. = dif. of time.
15
100C 29/ 28' = dif. of longitude.
12O 27, 0” = longitude of Rome, E.
88O 2/ 28' = longitude of Mobile, W.
NOTE.-When the Dif. of Long. given is that of two places which are in opposite longi-
tudes, and the Long. of one of them is given to find that of the other, we subtract the longitude
of the given place from the Dif. of Long. and thereby find the longitude of the other.
This is done because the Dif. of Long. between the two places is equal to their sum (one
being E. and the other W.) and consequently, when the Long. of one is given, that of the other is
found by subtraction. Should the Dif. have been that of two places in the same kind of Long.,
we would then have added the difference of longitude, if the lesser longitude were given; and sub-
tracted, if the greater longitude were given, to find the longitude of the other.
* 6. When it is 12 M. in New York, it is 18 min. 2; sec. past 11 o'clock at
Cincinnati. What is their difference of longitude 3 AnS. 10O 29' 21".
OPERATION INDICATED.
12 hrs. 0 min. 0 sec.
11 hrs. 18 min. 2; Sec.
41 min. 573 sec. × 15 = 100 29' 21".
7. The difference of time between Jerusalem and Baltimore is 7 hr. 28 min.
36 sec.; the longitude of Baltimore is 76 deg. 37 min. West; what is the longitude
of Jerusalem, East 3 Ans. 35 deg. 32 min. E.
* LONGITUDE AND TIME. 305
PROBLEMS OF TEIE SECOND CLASS.
OR, TO FIND THE DIFFERENCE OF TIME BETWEEN TWO PLACES, WHEN THE DIF-
FERENCE OF LONGITUDE IS GIVEN.
631. 1. Reduce 300 42' 50" of longitude to time.
OPERATION, Explanation.—As shown by the foregoing elu-
15) 300 42/ 50// cidated table of longitude and time equivalents,
159 of longitude = 1 hr. of time, 15' of longi-
º tude = 1 min. of time, and 15" of longitude =
2 hrs. 2 min. 51# Sec. Ans. 1 sec. of time; therefore, there will be ºr as
many hours, minutes, and 8econds of time as there
are degree8, minutes, and seconds of longitude, and hence we divide the 9, ', and '', by 15, as in divis-
ion of compound denominate numbers, and thus reduce longitude to time.
2. The longitude of Paris, France, is 2° 20' 15" E., and of Berlin, Germany,
13O 23' 44' E. What is the difference of time 3 Ans. 44 min. 13++ Sec.
OPERATION.
13O 23/ 44” = longitude of Berlin, E.
2O 20/ 15'' = & 4 “ Paris, E.
15) 11o 3' 29' = difference of longitude.
44 min. 13+3 sec. = difference of time.
3. The longitude of Bombay, India, is 72° 54’ East; of St. Louis, Mo., 900
15' 15° West. What is the difference of time, and when it is 10 A. M. in St. Louis,
what is the time in Bombay ?
OPERATION.
72O 54." 0’’ = longitude of Bombay, E.
900 15' 15’’ = % “ St. Louis, W.
15) 1630 9. 15” = difference of longitude.
10 hrs. 52 min. 37 sec. = difference of time, or the length of time it is 10
o'clock in Bombay, before it is 10 A.M. in St. Louis.
10 hrs. = St. Louis time.
20 hrs. 52 min. 37 sec., or 8 hrs. 52 min. 37 sec. P. M. in Bombay.
4. The difference of longitude between St. Paul and Cincinnati is 100 35'
24”. What is the difference of time 3 Ans. 42 min. 21; sec.
5. The longitude of New Orleans is 90o 4' 9". The longitude of San Fran-
cisco is 1220 26’ 45". What is the difference in time, and when it is 12 M. in New
Orleans, what is the time in San Francisco 3
OPERATION INDICATED.
122O 26% 45'' = longitude of San Francisco.
900 4' 9’’ = {{ “ New Orleans.
15) 320 22, 36” = difference of longitude.
2 hrs. 9 min. 30% sec. = dif. of time.—1st. Ans.
12 hrs. 0 min. 0 sec. = New Orleans time.
2 hrs. 9 min. 30% sec. = dif, of time, or time before 12 M. in San Francisco.
9 hrs. 50 min. 29# sec. = 50 minutes 29; seconds past 9 A.M., San Francisco
time.—2d Ans.
3O6 soul E's PHILOSOPHIC PRACTICAL MATHEMATICs. *X
6. The longitude of Boston is 71°3' 30", and the longitude of Chicago is
87° 37' 45”. What is the time in Boston when it is 10 o'clock A. M. in Chicago
Ans. 11 o'clock 6 min. and 17 Sec. A. M.
OPERATION.
87O 37. 45'' = longitude of Chicago.
71O 3' 30’’ = * { “ Boston.
15) 160 34' 15° = difference of longitude.
1 hr. 6 min. 17 sec. =difference of time, or the length of time it is 10 A.
M. in Boston before it is 10 A.M. in Chicago.
10 hrs. = Chicago time added.
11 hrs. 6 min. 17 Sec. A. M. Ans.
7. In traveling from Washington, longitude 770 0 15° W., to New Orleans,
longitude 90o 4' 9° W., how much time will an exactly running watch appear to
gain } Ans. 52 min. 15; Sec.
8. The longitude of Greenwich is 0, of Astoria, Oregon, 123° 49' 42*. How
much earlier does the sun rise in Greenwich than in Astoria, Oreg. 3
Ans. 8 hrs. 15 min. 18; Sec.
9. When it is noon at Pekin, longitude 118 deg. East, what time is it at
Paris, longitude 2 deg. 20 min. -15 sec. East? Ans. 4h. 17 m. 21 Sec. A. M.
OPERATION INDICATED.
118.
2. 20. 15. 12.
15) 115. 39. º 7. 42. 39.
7. 42. an. 4. 17. 21.
10. When it is midnight at Halifax, longitude 63 deg. 36 min. West, what
time is it at Canton, 113 deg. 14 min. East longitude 3
Ans. 11 h. 47 m. 20 Sec. A. M.
THE LOSS OR GAIN OF A DAY BY TRAVELING AROUND THE WORLD.
632. Two very curious facts would be noticed, could we start on any given
meridian with the sun on a certain date and journey with it around the earth, so to
speak: The first fact to be noticed would be that in going around the earth from
Blast to West, and upon arriving at our original meridian, we would find that we
had lost one day's time, i. e. that we were one day behind the correct date. The
jºr LOSS OR GAIN OF A DAY BY TRAVELING AROUND THE WORLD 3O7
Second fact noticed would be that if we should reverse our first direction and go
around from West to East, we would gain one day's time, i. e. we would be one day
ahead of the correct date.
The meridian at Greenwich, Eng., is one of the standards of longitude for
the world. When it is midnight on Sunday there, it is Sunday from there to 180°
East, and Monday from Greenwich to 1800 West, or to the same meridian. Hence
Ship captains change their time at this 180th meridian. Thus in traveling from the
West towards the East, and upon crossing the 180th meridian on, say Sunday, the
time on shipboard would be changed to Monday. And in traveling from East to
West (in the opposite direction) and arriving at the 180th meridian, on Sunday, the
time would be changed to Saturday." These changes are always made by nautical
men in order that they may have the correct time of all localities East or West of
the 180th meridian.
There is also another line which may be named an International Date Lºve.
This line differs slightly from the 180th meridian. The direction of this line, as
indicated by Mr. Olney, the famous Scientist and Mathematician, is that it starts
from the Chatham Islands, latitude 440 S., longitude 177° W., and runs north and
west, to the East of New Zealand, New Guinea, and Borneo, thence through the
Philippines, approaches China near Canton, and goes to the northwest of the Japan
Isles, through Behring Sea, and terminates in the North Pole. The date on the
east of this line is generally one day behind what it is on the west of it.

—º *—
/~ “g / tº sº

ATIO.
633. Ratio, in the mathematical sense in which it is here used, means the
measure of the relation which one quantity bears to another of the same kind, as
expressed by the quotient of the first divided by the second; that is, it is the num-
ber of times that one quantity (the first) is equal to another, (the second) which is
used as a unit of measure. Thus, the ratio of 6 to 2, is 6-2, and is equal to 3.
If we ask what is the relation of 6 to 2, the correct answer would be 6 is 3 times
2. We thus see that the ratio three is the number which measures the relation of 6
compared with 2, and therefore, that ratio is not merely the relation of two similar
numbers, but the measure of this relation.
NOTE 1.-Mathematical writers are not agreed in their definitions of the ratio of one num-
ber or quantity to another. Many contend that it is the quotient of the second quantity divided
by the first. Thus, the ratio of 6 to 2 is, 2 – 6 is #-- *. This is the preferred definition of Webster
and of the Dictionary of Mathematics. But nearly all of the later writers as well as the German,
English and most of the French, prefer the definition first above given, and we are in full accord
with them.
The result of the two systems is the same, although differently obtained. By the first definition
We saw that 6 is equal to 2, 3 times; and by the second definition we see that 2 is equal to 6, # of a
time, which expresses precisely the same ratio between the numbers. The question is, which
number, the first or the second, shall be used as the unit of measure, or the divisor ? The prepon-
derance of authority is largely in favor of the second.
NOTE 2.-Numbers or magnitudes can have no ratio to each other unless they are of the
Same kind. Thus, there can be no ratio between 6 dollars and 2 yards, for 6 dollars are not equal
to 2 yards any number of times.
634. The Terms of a Ratio are the numbers compared. The first term of
a ratio is the ANTECEDENT, which means going before ; the second term is the CON-
SEQUENT, which means following.
635. The Sign of Ratio is the colon (: ), which is the sign of division, with
the horizontal line omitted. Thus, the ratio of 6 to 2, is written, 6: 2. Ratio is
also indicated by Writing the consequent under the antecedent in the form of a frac-
tion. Thus, the ratio of 6: 2 is often written g.
636. An Inverse Ratio is the quotient of the consequent divided by the
antecedent. Thus, the inverse ratio of 8:4, is 4.
637. The Walue of a ratio is the quotient of the antecedent divided by the
Consequent, and is always an abstract number.
638. A Simple Ratio is the ratio of two numbers, as 8:4.
639. A Compound Ratio is the ratio of the products of the corresponding
terms of two or more simple ratios, as follows:
§ {}= 6 × 8 : 2 × 4 ; or 3 × # is a compound ratio, - 6.
640. The Reciprocal of a ratio is the quotient of 1 divided by the ratio, or
it is the quotient of the consequent divided by the antecedent. Thus, the ratio of
6 to 2 is 6: 2 or #, and its reciprocal is, 1 + 3 = }, or $.
641. The Ratio of two fractions is obtained by reducing them to a common
denominator and then comparing their numerators. Thus, the ratio of #: # is the
Same as 5: 2.
642. The Ratio of Compound Denominate numbers is found by reducing
them to the same denomination and then making the comparison.
(308)
GENERAL PRINCIPLES. 3O9.
From the foregoing definitions and elucidations, the following formulas and
general principles are deduced:
FORMUL.A.S.
643. 1. The Ratio = the Antecedent -- Consequent.
2. The Consequent = Antecedent -- Ratio.
3. The Antecedent = Consequent X Ratio.
GENERAL PRINCIPLES.
644. 1. Multiplying the antecedent or dividing the consequent multiplies the ratio.
2. Dividing the antecedent or multiplying the consequent divides the ratio.
3. Multiplying or dividing both terms by the same number does not change the
value of the ratio.
645. These general principles may be formulated into one General Law, as
follows: Any change in the antecedent produces a LIKE change in the ratio ; but any
change in the consequent produces an OPPOSITE change in the ratio.
PROBLEMIS.
646. What is the ratio of the following numbers:
1. 10 to 5 3. 4 to 8 5. 54 to 8 7. 4; to #;
2. 18 to 6 4. 7 to 42 6. 11 to 60 8. ## to ##
9. What is the ratio of 5 gallons to 3 quarts 2 Ans. 63.
10. What is the ratio of 20g to $3? Ans. I's.
11. The antecedent is 12 and the consequent 4. What is the ratio ? Ans. 3.
12. The antecedent is 12 and the ratio is 2. What is the consequent ?
Ans. 6.
13. The consequent is 16 and the ratio is 8. What is the antecedent ?
Ans. 128.
14. The antecedent is 5% and the ratio is 4. What is the consequent ?
AnS. 14.
15. What is the reciprocal ratio of 16 to 48? Ans. 3.
16. The reciprocal of the ratio of two numbers is 3 and the antecedent is 16.
What is the consequent 3 Ans. 48.
17. What is the ratio of a pound Troy to a pound Avoirdupois? Ans. ##.
18. What is the ratio in grains of a pound of iron to a pound of gold 3
Ans. }}}.
19. The ratio of two numbers is 3, the antecedent is 15. What is the con-
sequent? Ans. 5.
20. The ratio of two numbers is 4 the consequent is 14. What is the ante-
cedent 3 Ans. 5%.
21. The inverse ratio of two numbers is 34 and the antecedent is 20. What
is the consequent? Ans. 70.
º 8 : 12
22. What is the value of { 10 : 5 } º Ans. 14. -
º 23:4% OPERATION INDICATED.
23. What is the value of { 3} : 1; } º 3 || 8
30 7
24. What is the difference between the direct 3 || 10
and the inverse ratio of 2; and 7# 3 Ans. 24. 8 || 5
Ans. 1#.
ºroportion.
})
/9
eº --~~~~~~~~~~~~~~~~~~i=
Ç' S, \
647. Proportion arises from the comparison of ratios. It is a comparison
of the results of two previous comparisons. Every proportion involves three com-
parisons: the first two were those which produced the ratios, and the third, that
Which compares or equates the ratios.
Proportion is the expression of the equality of equal ratios, or, it is the
comparison of two equal ratios. Thus, 6 : 2::15:5 is a proportion, and is read
6 is to 2 as 15 is to 5, or the ratios of 6 to 2 equals the ratio of 15 to 5.
Thus, the ratio of 6: 2 as 15:5 is a proportion, i. e. four quantities are in
proportion, when the first is the same multiple or part of the second, that the third
is of the fourth.
648. The Sign of Proportion is a double colon (::), or the sign of equality
(= ). Thus, the above proportion is expressed 6:2::15 : 5, or 6: 2 = 15 : 5. The
first is read, 6 is to 2 as 15 is to 5. The second is read, the ratio of 6 to 2 equals the
ratio of 15 to 5.
649. The Terms of a proportion are the numbers compared.
650. The Antecedents of a proportion are the first and third terms.
651. The Consequents are the second and fourth terms.
652. The Extremes are the first and fourth terms.
653. The Means are the second and third terms.
654. In the proportion, 3: 6::: 4:8, all the numbers are the proportionals ;
3 and 4 are the antecedents ; 6 and 8 are the consequents ; 3 and 8 are the eactremes;
and 6 and 4 are the means.
655. Three numbers are proportional, when the ratio of the first to the second
is equal to the ratio of the second to the third. Thus, 4, 8, and 16 are proportional,
since 4: 8 : : 8 : 16, each ratio being #.
In this kind of proportion, the second term is called a mean proportional
between the other two.
656. A Simple Proportion is an expression of the equality between two
equal ratios, as above elucidated.
657. A Compound Proportion is an expression of the equality between
two ratios, when one or both of which are compound.
658. There are several other classes or divisions of proportion; viz: Con-
joined, Partitive, Reciprocal, Medial, etc., but as we shall perform all kinds of
proportional problems by our philosophic system, independently of the various
terms and the ingenious classifications, or mechanical formulas used by most authors,
we will not now occupy space with them. As we present the problems, which are
generally classed under these various heads or divisions of proportion, We shall
then define each, with our philosophic Solution.
659. The following General Principles are derived from the preceding
elucidations:
1. In every proportion, the product of the eatremes is equal to the product of the means.
2. The product of the eatremes divided by either of the means, gives the other mean.
3. The product of the means divided by either eatreme, gives the other eatreme.
(310) -
ăţillinºiſ all Elliſilly Of NIIIlberS.
660. In defining ratio, we showed that there is a difference between the
relationship and the ratio of two numbers; we shall now show in the following
work, other differences which were not then indicated.
Relationship of numbers is a more general term with a wider significance
than ratio. And used as we shall employ it, it includes far more than the numerical
measure, ratio, of one number by another. g
Through the relationship and the equivalency of numbers, we shall, by
comparison, analysis, and synthesis, solve all forms of ratio and proportional
questions, simple, compound, reciprocal, etc., without regard to any of the preced-
ing mathematical conventional terms, ratios, proportions, general principles, etc.
Different kinds of relationship and equivalency of value may exist between
things of the same, and things of different kind. Thus, the relationship and
equivalency of monetary value, exist when 5 yards cost 50%, 1 hat cost $4, etc.; the
relationship and equivalency of numerical value exist when 8 = 5, 2 = 3, etc.; the
relationship and equivalency of exchange, or barter value exist when 10 pounds of
rice = 24 oranges, 6 days labor = 18 bushels of corn, etc.; the relationship and
equivalency of service or force and results exist when 30 men make 50 yards of
levee in 20 days; or when a steamboat carries 1000 bales of cotton 500 miles for
$2000, etc.
And thus some kind of relationship and equivalency exists, like the law of
compensation in nature, in every conceivable statement of numbers.
The following problems and their solutions will more clearly show the rela-
tionship and equivalency of numbers, and at the same time elucidate the philosophic
system of work.
SIMPLE PROPORTION.
PROBLEMIS.
1. 9 pounds cost 45%. What will 3 pounds cost at the same rate 3
PROBLEMI CLASSIFIED. Explanation and IReason.—This is a problem in simple pro-
Ib portion, and by our system of reasoning, it is as easily
§ is solved as a question in division.
In all problems of all classes of proportion, there is one
3 number which is of the same name, nature, or denomination,
OPERATION. as the Inumber or answer required. This number we use as the
mature of the answer, and in the solution of problems, we first
5 write it at the top of the increasing side of the statement line.
4 In this problem, the nature of the answer is 45c. which
3 we Write on the statement line and reason as follows:
Since 9 pounds cost 45c. 1 pound will cost the 9th part,
152, Ans which we indicate by writing the 9 on the decreasing side
9 º of the statement line; then since 1 pound costs the result of
this statement, 3 pounds will cost 3 times as much, which
worked, gives, 15c. the cost of 3 pounds.
It will be observed that we classified the problem before stating the operation.
This the learner or calculator should always do, in order to see more clearly the
relationship and the equivalency of the numbers which are compared with the
nature of the answer.
C.
9
(311)
SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS.
COMPOUND PROPORTION.
2. Three boys earned $72 in 12 days. How many dollars can 2 boys earn in
15 days?
PROBLEMI CLASSIFIED.
Explanation and Reason.—This is a problem in compound
proportion, and by the philosophic system of work it is as
bºy. #. º easily solved as a problem in simple proportion. By inspec.
pº tion and reason, we see that $72 is the nature of the answer.
2 15 We therefore write the same on the iºgºs; º:
statement line and reason thus: 3 boys earned $72. Since
OPERATION. boys earned $72, 1 boy earned the 3d part, and 2 boys can
$ earn twice as much ; then, since $72 are earned in 12 days,
72 in 1 day the 12th part will be earned, and in 15 days, 15
3 || 2 times as much will be earned. This completes the reasoning.
12 | 15 The operation statement is worked out by cancellation, as
J explained in cancellation, and in other subjects treated.
&= gº-mº- It will be observed that in giving the reason for writing
$60, Ans. the terms, boys and days, on the statement line, we did not
state all of the conditions or relationships of the syllogistic
premises from which we deduced our conclusions of increase and decrease of the dollars. Thus,
per example, with the term boys, we said: Since 3 boys earned $72, 1 boy earned the 3d part, etc.
To consider the full conditions, or to state the full premises of this term, we would reason
thus: Since 3 boys earned $72 in 12 days, 1 boy earned the 3d part, etc.
To consider the full conditions of the term days, we would reason thus: Since $72 are
earned by 3 boys in 12 days, in 1 day the 12th part will be earned, etc.
To name all the conditions at each step of the analysis would require much time, and would
rather confuse than aid the learner. For these reasons, and because the analysis is clear, correct,
and logical, without naming all the conditions, we omit their useless repetition.
CONJOINED PROPORTION.
661. Conjoined proportion is a proportion in which there are several terms
of different kinds of which each preceding term or antecedent is of an equivalent
value to a succeeding term or consequent.
3. If 4 apples are worth 3 oranges, and 2 oranges are worth 100 grapes, and
200 grapes are worth 6 peaches, and 9 peaches are worth 30 cents, then how many
apples can we buy for 90 cents?
PROBLEMI CLASSIFIED.
Ap. Or.
4 - 3 Gr.
2 = 100 Pe.
9 = 30
90
OPERATION.
Explanation and Reason.—This is a problem in conjoined
proportion, and by the philosophic method of work it is
3 || 2 solved in the same manner as the two preceding problems.
100 | 200 Writing on the statement line the 4, which represents apples
and which is the nature of the answer, we reason thus:
6 || 9 Since 4 apples are worth 3 oranges, conversely 3 oranges are
30 || 90 worth 4 apples; and since 3 oranges are worth 4 apples, 1
orange is worth the # part, and 2 oranges are worth 2 times
as many; then since 2 oranges are worth 100 grapes, con-
versely 100 grapes are worth 2 oranges; and since 100 grapes
are worth two oranges, 1 grape is worth the Tło part, and
200 grapes are worth 200 times as many. Then since 200 grapes are worth 6 peaches, conversely
6 peaches are worth 200 grapes; and since 6 peaches are worth 200 grapes, 1 peach is worth the #
part, and 9 peaches are worth 9 times as many. Then since 9 peaches are worth, 30c., conversely
30c. are worth 9 peaches; and since 30c. are worth 9 peaches, 1 cent is worth the ºr part, and 90c.
are worth 90 times as many,
The above is the full reasoning for the problem, and it gives healthful exercise to the mind of
the student—exercise which sharpens and strengthens and logicalizes the mind for service not only
in the domain of mathematics, but upon all questions of life.
The reasoning may be abbreviated thus: Since 3 oranges = 4 apples, 1 orange = # part, and
APPLES.
24 apples, Ans.
* PROPORTION, 3 I 3
2 oranges = 2 times as many; then, since 100 grapes = 2 oranges, 1 grape = Tº part, and 200
grapes = 200 times as many; then, since 6 peaches = 200 grapes, 1 peach = }; part, and 9 peaches.
= 9 times as many; then, since 30c. = 9 peaches, 1c. = }; part, and 90c. = 90 times as many.
NOTE. 1.-The student should repeat the reasoning of the foregoing problems until it is
perfectly understood, before advancing to other problems.
NOTE. 2.-By this philosophic process of solution, we consider and compare with the nature
of the answer but one number at a time, and the conclusion of all comparisons either increases or
decreases the nature of the answer at the head of the statement line. Should the conclusion increase or
decrease anything else, it is wrong, and the reasoning must be revised.
NOTE. 3.-In all the reasoning for each number of every term, it will be observed that we
make 1, which is the basis of all numbers, the standard unit of comparison or measurement; and
that the reasoning is always from 1 to a collective number, or from a collective number to 1.
GENERAL DIRECTIONS FOR PROPORTION
662. From the foregoing problems, explanations, and reasoning, we deduce
the following general directions for all proportional work:
1. Classify the problem.
2. Write the nature of the answer at the top of the increasing side of the state-
nent line.
3. Commence with such number of any term in the classified statement, that is
the equivalent of the nature of the answer, and comparing one by one, each number of
all the terms with the nature of the answer, consider and determine the equivalency
or relationship easisting between them, and whether from the conditions and relationship
the nature of the answer is to be increased or decreased, and then write the number on
the increasing or decreasing side of the statement line, according to the conclusion
reached by the comparison, the relationship and the conditions of the case.
4. When each number of all the different terms of the classified statement has
been compared, considered, and written on the statement line, then work out the problem,
canceling as much as possible.
REMAIRKS.
663. Before presenting other problems, we desire to place before the student
a few more facts pertaining to proportion. We wish to compare true proportion,
the analytic system of proportion, and the philosophic system of proportion, and to
make a few pertinent remarks regarding the illogical and mechanical systems of
proportion taught by nearly all the text books now before the public.
The philosophic system of solving proportional problems is not proportion,
nor is it analysis. To elucidate this, let us take the following problem and Solve it,
19, by proportion; 29, by analysis; 3°, by the philosophic method:
5 yards of cloth cost $14. What cost 15 yards of cloth at the same rate 3
3I4. soul E's PHILOSOPHIC PRACTICAL MATHEMATICs. *
664. FIRST SOLUTION BY PROPORTION.
Cost of 15 yds. : $14: ; 15 yards: 5 yds. Explanation.—This is the state
14 × 15 ment by true proportion, and by
= $42, Ans. it we see that the cost of 15 yards
Cost of 15 yds. = bears th I $14, th
*> g ears the same relation to ©
Or, 15 : 5, 3 times; and 3 timeS $14 F. $42, Ans. cost of 5 yards, that 15 yards
bear to 5 yards.
By the arbitrary rules of nearly all the school arithmetics in the land, the
principles of proportion are ignored, and in place thereof are substituted illogical
and absurd mechanical operations by which answers are obtained.
More than one hundred authors say in effect: “Write that number which is
like the answer sought as the third term; then if the answer is to be greater than
the third, make the greater of the two remaining numbers the second term, and the
smaller the first term; ” etc. *,
This process will produce the answer, but it is illogical, arbitrary, and absurd.
It robs proportion of all claims to a scientific process, and the student who thus
learns to solve proportional problems knows no more of true proportion than the
Hottentot knows of the Christian religion.
In the language of Prof. Brooks, when thus treated: “The whole subject
becomes a piece of charlatanism, utterly devoid of all claims to science.”
If proportion is used at all, it should be used as a logical process of reasoning,
and the solution statements should be made by a truthful proportional comparison
of the elements of the problem.
For the higher mathematics, proportion is indispensable; but in arithmetic,
proportional problems are more easily solved by analysis than by proportion, and
still more easily by the philosophic method than by analysis.
665. SECOND SOLUTION BY ANALYSIS.
OPERATION.
$14 -- 5 = $24; Analysis.-5 yards cost $14. What will 15 yards cost? Since
5 yards cost $14, 1 yard will cost the # part of $14 = $2; Then
$24 × 15 = $42, Ans. *..."; $';*. º, Sºhº. $23 = $42.
By the analytic system no proportional statement is made, and no false claim
thereto is assumed. The process of reasoning is less Scientific than in proportion,
but much easier, and far more practical.
The only serious objection that can be urged against the analytic system is
the labor to produce each intermediate result of the alternate operations of division
and multiplication.
This objection is entirely obviated by the philosophic system, and all the merit
of the axiomatical reasoning is retained.
666. The analytic system originated with Pestalozzi, a Swiss teacher and
educational reformer, who was born in 1746, and who wrote and taught from 1781 to
1827. His first methods were quite crude, but yet they served as the germs from
which have grown a large part of all modern improved methods of teaching.
Yºk SOLUTION BY THE PHILOSOPHIC SYSTEM. 315
He aimed to take nature's method. He first taught his pupils to think, then
to write, and then to read. He claimed that “the first one to read must have Written
before he had anything to read.”
In arithmetic he commenced with the simplest problems and taught his pupils
to think, to reason, and to analyze every problem without regard to any rules.
In 1805, his disciple, Joseph Neef, came to this country and opened a school
on the Pestalozzian system near Philadelphia. But the people were not then pre-
pared for the improved methods, and in a few years the school was abandoned.
The spirit of the new system had however taken root in the minds of pro-
gressive teachers who were pleased to find something better adapted to the wants
of the educational world than the old arbitrary system; and from that time to the
present, the Pestalozzian system has been growing in popular favor, and it now con.
stitutes the basis of progressive educational systems throughout the civilized world.
For more than 60 years past, most of the authors of arithmetics have
utilized, to some extent, in various modified forms, the system of Pestalozzi. There
Was no system of mental arithmetic before his day. e
In 1820, Warren Colburn published what he called an “Introduction to
Arithmetic on the Inductive or Intellectual System.” In this work he presented
many of Pestalozzi's ideas and methods.
From Pestalozzi's Analytic System of Arithmetic has been evolved the
Philosophic System—a system without a peer in the annals of mathematics, and to
extend, to practicalize, to logicalize, and to perfect which, the author of this book
has labored with tongue and pen for more than a third of a century.
He has taught the system to over 9000 students and has by his lectures and
publications advanced the philosphic system to loftier plains and to more rational
and logical methods than were ever before achieved. In no ancient or modern work
on numbers, have comparison, analysis, and synthesis been woven into such a chain
of logical and philosophical reasoning as is presented in this and the author's other
works on numbers.
The day is not far distant, when the banner of the philosophic system will
wave in triumph from every spire, pinnacle, and dome of the Temple of Practical
Mathematics.
667. THIRD SOLUTION BY THE PEIILOSOPHIC SYSTEM.
OPERATION.
*, 4. Premise.—5 yards cost $14. Reason.—Since 5 yards cost $14, 1
# | 15 3 yard will cost the # part, and 15 yards will cost 15 times as
- $42, Ans much, which is $42.
7 e
By this philosophic system, all the difficult scientific statements of true pro-
portion and all the objections to the analytic method are obviated, and solutions
are performed with the least possible number of figures, and by the easiest possible
process of logical and axiomatical reasoning.
316 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. º:
*
668. Practical arithmetic has been sadly neglected by mathematicians,
ancient and modern. Pythagoras, Plato, Euclid, and Archimedes, of ancient fame,
Newton, La Place, Leibnitz, and La Grange, of modern renown, have enriched the
literature of arithmetic, but they did not give to it the penetrating thought which
they gave to the higher branches of mathematics. And our more recent authors of
note, Davies, Peck, Ray, Greenleef, Perkins, Dodd, Brooks, Robinson, Thompson,
Wentworth, Hill, and others, have adopted largely the arbitrary non-reasoning
methods of their predecessors. They have failed to keep pace with the progress of
an ingenious age.
The world of thought in other departments of applied science has moved
grandly forward and left the mathematicians of the 19th century contemplating
and approving many of the monstrous absurdities of the “dark ages” of arithmetic.
With the ancients, magic squares and the properties of numbers seem to have
been deemed more important than the science or the practical application of arith-
metic. With the moderns, mechanical methods to produce answers, without regard
to reason, science, or brevity, seem to be the leading characteristic. *
And thus arithmetic has been slighted. Only feeble efforts have been made
to unfold and evolve its logical relations as a science, and to elucidate the philo-
sophic principles which extend throughout all its parts.
The brain-entangling cause and effect method of Prof. Robinson and some
others, the complicated and confounding methods of Prof. Davies and a hundred
others, seem to us a disgrace to the science of numbers, a discredit to the intelli-
gence and genius of arithmeticians, and an insult to the common sense of mankind.
The damaging effect of the arbitrary rules of these authors, on the minds of
the rising generation, cannot be estimated. They fail to present the full meaning
of problems, prevent the exercise of the reasoning faculties of the student, and
produce a distaste for the science of numbers.
TO THE STUDENT.
The philosophic system, or the reasoning we employ in solving problems in
proportion, is applicable to the solution of all arithmetical questions, and hence
efficiency herein is a condition precedent to every other consideration. The reason-
ing combines that of multiplication and division of whole and fractional numbers,
in alternation, and there is no other exercise that can be given in numbers, that will
give so much strength, acuteness, flexibility and comprehensibility to the reasoning
organs of the brain. Thus preparing the mind not only to solve easily all practical
problems in the most rapid manner, but also capacitating the person for efficient
Service in all the positions and relationships of business life where conclusions must
be deduced from the facts or premises, or from the circumstances and conditions of
things relating to the subject under consideration. Therefore, we have presented a
large number of problems in proportion and given the full reasoning in the Solution
of many of them.
Some of the more intricate problems are presented and solved more as logical
drills for the mind than for the practical value of the problems. The student should
º:
PROBLEMS IN PROPORTION. 317
critically read and re-read the reasoning of such solutions and then Solve the prob-
lems and write the reasoning without reference to that in the book. By this means,
a high degree of capacity to reason, to conclude and to express thought with the
pen, logically and grammatically may be attained.
669. PROBLEMS IN PROPORTION.
CLASSIFICATION. Reason.—8 pounds cost 56c.
1. If 8 pounds of sugar Ig 㺠Since 8 pounds cost 56c., 1 pound
will cost the art, and 27
cost 562, what will 27 27 the part.
OPERATION. pounds will cost 27 times as
pounds cost at the same C. much.
# 7 NOTE.-The student should
rate? Ans. $1.89. 3 work this problem and write
| $1.89, Ans. the reasoning.
CLASSIFICATION. Reason.—8 pounds cost 56c.
lbs. C. Si 8 ds cost 56c. con-
2. If 8 1nce 8 poun ©
pounds of Sugar S 1; versely 56c. cost or bought 8
cost 562, how many poun pounds, and since 56c. bought
Ž, y pounds organos. 8 pounds, 1C. will buy the 'g
can be bought for $1.89% 7 ; 27 part, and 189C. will buy 189
| times as many.
Ans. 27 pounds. #5 139 Note—The student should
27 lb work this problem and write
7 lbs. Ans. the reasoning.
CLASSIFICATION. Reason.—$1.89 bought 27
$1.89 º; pounds. Since 189C. bought 27
3. If $1.89 will buy 27 56 pounds, 1c. will buy Tºg part,
OPERATION. and 56c. will buy 56 times as
pounds of Sugar, how many lbs many
pounds can be bought for 1% ; 8 . NOTE.—The student should
562 % A. 8 *- work this problem and write
g nS. 8 pounds. 8 lbs. Ans, the reasoning.
CLASSIFICATION. Reason. # of a pound cost
lbs $ $14. Since # of a ºn:
4. If # of a pound cost # 1; ; tºº : rº
e 23 will cost 4 times as much. Then
$1}, what will 23 pounds OPERATION. º 1 º fº. *i; * i.
of a pound will cost the # par
COSt 3 Ans. $54 2 $ 3 and * pounds will cost 21 times
tº Q as much.
3 || 4 We leave this operation un-
8 || 21 cancelled in order that the
- * statement may the better .
* the student in understanding
$54, Ans. &
the reasoning.
NotE.—The student should work the above problem, and should write the reasoning for it
and all other problems that are solved.
3.18 soul E's PHILOSOPHIC PRACTICAL MATHEMATICs. 4%
5. A man walks 320 miles in 25 days, when he walks 6 hours per day. How
many miles can he walk in 20 days by walking 8 hours per day? Ans. 341 miles.
6. § of a steamboat is valued at $35700. What is the value of the whole
boat at the same ratef Ans. $40800.
7. A merchant borrowed $4000 for 64 days at 6 pr. ct. He now desires to
compensate the party from whom he borrowed the money by loaning him $1500 at 8
Dr. Ct. How many days must he loan it? Ans. 128 days.
OPERATION INDICATED.
ds.
64 º
* NOTE.-The student should classify the problem and
1500 | 4000 l Still (16 Dºlj SILOUILOl C y P
8 || 6 write the reasoning.
tºms"º " sºmmemº
8. A merchant invested $15000 for 4 months and gained $800. How much
investment would be required to gain $1400 in 3 months, at the same rate 3
Ans. $35000.
9. A note broker loaned $5000 for 30 days at 8%. The party to whom he
loaned the money wishes to reciprocate the favor by loaning the note broker $2000
at 6%. How long a time should he lend it 3 Ans. 100 ds.
10. A farmer owned # of a farm and sold # of his share for $12400. What is
the value of the whole farm at the same rate 3 Ans. $49600.
11. Tea costs 80 cents per pound. How much can you buy for 152, at the
same rate % Ans. 3 ounces.
CLASSIFICATION: OPERATION INDICATED.
C. lb. OZ.
80 1 or 16 16 Oz. Write the
15 80 | 15 reasoning.
12. A baker charges 52 for a loaf of bread when he pays $7.25 per barrel for
flour. What ought he to charge when he pays $8.70 per barrel for flour? Ans. 62.
13. If a 52 loaf of bread weighs 15 ounces when wheat is worth $1.40 per
bushel, what ought to be the weight of a 1242 loaf when wheat is worth $1.00 per
|bushel? Ans. 524 ounces.
CLASSIFICATION. OPERATION.
C. OZ, $. OZ. The student should write the
5 15 1.40 15
12; 1.00 5 reasoning by which the figures
2 25 are placed on the statement
1.00 | 1.40 g #
line.
524 oz., Ans.
14. A baker charges ten cents for a loaf of bread weighing 8 oz., when he
pays $6.30 per barrel for flour. How many ounces should the same price loaf weigh
when he pays $7.20 per barrel? Ans. 7 oz.
15. If the fore-wheel of a carriage is 7 ft. 6 in. in circumference, and turns
round 616 times, how often will the hind wheel, which is 12 ft. 10 in. in circumfer-
ence, turn round in going the same distance? Ans. 360 times.
PROBLEMS IN PROPORTION. 3 I 9
OPERATION.
16. 13% gallons cost $4}. $
What will # of a gallon
cost at the same rate 3
Ans. $4.
NOTE.—The student should
classify the problem and write
the reasoning.
sº2
:
:
17. If it requires 12 yds. of cloth to make a dress when the cloth is 38 inches
wide, what must be the width of a piece 20 yds. long, to answer the same purpose?
Ans. 22# inches.
18. A grocer has coffee at 1242 per pound, and a coal dealer has coal at 40%
per barrel. If, in exchanging, the grocer puts his coffee at 14%, what should the
coal dealer charge for his coal? Ans. 442 per barrel.
CLASSIFICATION. OPERATION.
g g g
124 14 14
40 25 || 2 Write the
40 reasoning.
44%, Ans.
Or thus:
(LASSIFICATION. OPERATION.
2 gain.
2 || 3
142 — 12#2 = 14% gain 25 2
40g | 40
NOTE.—The student should write the rea- 4#2 gain.
soning for the operations. 40g value of coal.
44% price of coal.
19. A. has rice at 62 per pound, and B. has sugar at 72 per pound, which they
wish to barter. Before bartering it is agreed that B. shall value his sugar at 7:42,
and that A. shall advance his rice accordingly. What should be the exchange price
of A’s rice 2 Ans. 6%.
NOTE.-See Miscellaneous Problems in Percentage for full elucidation of this class of problems.
20. If, with $8.10, you can buy 9 yards of cloth, how many yards can you buy
for $5.40% Ans. 6 yds.
* 21. If 6 yds. cost $7.20, how much will 12 yds. cost 3 * Ans. $14.40.
22. If, for 12 cents, you can buy 3 of a yard, how many yards can you buy for
36 cents 3 Ans. 2 yds.
23. If ; of a ton of hay cost $15, how much will 3000 pounds cost
Ans. $30.
24. If 20 pounds of sugar will supply a family of six persons for 7 days, how
many pounds will be required to supply the same family 30 days? Ans. 85% lbs.
32O soulE's PHILOSOPHIC PRACTICAL MATHEMATICS. X-
25. If 3 men eat 3 dozen oysters in 3 minutes, how many men will it require
to eat 100 dozen oysters in 100 minutes?
CLASSIFICATION. OPERATION.
M. D. Oys. Min. Men.
3 3 3 3
100 100 3 || 100 Write the
100 || 3 reasoning.
3 men, Ans.
26. If 3 men eat 3 dozen oysters in 3 minutes, in how many minutes can 100
men eat 100 dozen oysters ? s
CLASSIFICATION. OPERATION.
M. D. Oys. Min. Min.
3 3 3 3
100 100 100 3 Write the
3 100 reasoning.
3, min. Ans.
27. If 3 men eat 3 dozen oysters in 3 minutes, how many dozen oysters can
100 men eat in 100 minutes?
CLASSIFICATION. OPERATION.
M. D. Oys. Min. D. Oys.
3 3 3 3
100 100 3 100 Write the
3 || 100 reasoning.
3333; doz. Oysters, Ans.
28. If 24 acres of land will produce 550 bushels of potatoes, how many
bushels will 8; acres produce at the same rate 3 Ans. 1925 bus.
29. If 13, hens lay 13, eggs in 14 days, how many eggs will 6 hens lay in 7
days? Proportional Answer, 28 eggs, Logical Answer, 24 eggs.
OPERATION FOR PROPORTIONAL ANSWER. Reason for the Logical Answer.—Since 1;
2 || 3 = eggs hens lay 13 eggs, in 13 days, the period
3 || 2 7 for laying is hence 36 hours. And as 7
T--> days = 168 hours there will be as many
6 Write the periods for laying as 168 is equal to 36,
3 || 2 reasoning. which is 168 - 36 = 4 periods and 24
7 hours over. Therefore the number of eggs
laid will equal the product of the 4 pe-
tºmº || “…sºmeºmºsºmºsºms riods by the 6 hens, which is 4 × 6 = 24
28 eggs, Ans. eggs answer.
NOTE.-Since 36 hours is the laying period, it is presumed in this solution that the hens do
not lay any eggs during the 24 hours.
30. If 14 cats catch 1; rats in 14 minutes, how many cats will it take to catch
200 rats in 50 minutes? Ans. 6 cats.
OPERATION.
Cats.
CLASSIFICATION. 2 || 3
C. R. M. 3 ão
1; 1; 1; 2 3
200 50 50
3. PROBLEMS IN PROPORTION. 32 I
31. A railroad contractor agreed to grade 450000 cubic yards in 90 days. He
employed on the work 560 men for 30 days; then, measuring the grading completed,
he finds that they have made 120000 cubic yards. How many more men must he
employ at the expiration of the 30 days, in order to complete the work in the con-
tract time 3 Ans. 210.
PROBLEM CLASSIFIED : OPERATION.
Yds. D. M. M.
450000 90 569 70
120000 30 560 2 60 || 30
--msºme *= 120999 || 3399%
330000 60 4. 11
770 men required.
560 men employed.
210 men needed.
Explanation.—In all problems of this character, we must first find the number of men that
will be required to perform the unfinished work in the unexpired time. We therefore place 560,
the number representing men, on the increasing side of our statement line; and then to find the
proportional numbers, we deduct from the whole work that was to be performed, and the time in
which it was to be finished, the work that has been completed and the time consumed in its com-
pletion, and in the remainder, as shown in the problem classified, we have the numbers
containing the direct relationship with the numbers representing the work performed and
the time consumed. Having these proportional numbers, we reason as follows: If it requires
560 men 30 days to grade 120000 yards, to do the work in 1 day instead of 30, it will
require 30 times as many men, and to do the work in 60 days instead of 1, it will require
the 60th part of the number of men; then, if it requires 560 men to grade 120000 yards, to grade 1
yard instead of 120000, it will require the 120000th part of the number of men, and to grade 330000
yards instead of 1, it will require 330000 times as many men. This completes the reasoning and
statement, the result of which is 770 men, as the number required to perform the unfinished work
in the unexpired time, and as 560 men are now employed, it is clear, by the exercise of our reason,
that as many additional men must be employed as 770 is greater than 560, which is 210.
32. If 125 men, in 45 days, working 8 hours per day, can build 15 walls, each
150 feet long, 3 feet thick, and 12 feet high, how many men will be required to build
21 walls, each 200 feet long, 5 feet thick, and 9 feet high, in 14 days, working 10
hours per day ? Ans. 750.
PROBLEM CLASSIFIED :
M. D. H. W. F. L. F. T. F. H. Student should make the
125 45 8 15 150 3 12 statement and write the
14 10 21 200 5 9 reasoning.
33. A merchant owning # of a steamboat, sold # of his share for $45000.
What was the steamer worth at that rate 3 Ans. $90000.
34. A besieged city, containing a population of 45000, has provisions for 4
weeks. How many citizens must be sent away, that the provisions may support the
remaining population 40 days 3 Ans. 13500.
35. If a man walks 50 yards in ; of a minute, how far will he walk in 16#
hours? Ans. 66400 yds.
36. Allowing the number of pulsations in a person to be 72 per minute, and
322 soul.E's PHILOSOPHIC PRACTICAL MATHEMATICs. Yºr
that the velocity of sound is 1118 feet per second, what will be the distance of the
electric fluid when 10 pulsations are counted between seeing the flash of lightning
and hearing the sound of the thunder ? Ans. 93163 ft.
PROBLEMI CLASSIFIED : STATEMENT :
P. M. Ft. Sec. Ft. P. Ft.
72 = 1 1118 or, 1 1118 72 1118
10 72 60
1 m. Or 60 Sec. 10 10
37. A builder contracts to erect a wall 140 feet long, 11 feet high, and 24 feet
thick, in 10 days. He employed 8 men who labored on the wall 6 days, during which
time they built the wall 6 feet high, the required length and thickness. How many
additional men must be employed to complete the work in the required time 3
Ans. 2.
38. In a piece of machinery there are two wheels running into each other:
one contains 42 cogs and the other 33 cogs; in how many revolutions of the larger
wheel will the smaller gain 10 revolutions Ans. 363.
PROBLEM CLASSIFIED : STATEMENT :
C. R. R.
42 10 1
33 9 || 33
*º-º-e 10
9 *
39. A contractor engaged to build 480000 cubic yards of levee in 90 days.
He employed 500 men who worked 60 days and built 300000 cubic yards. How
many additional men must be employed at the end of 60 days to complete the work
in the contract time 3 Ans. 100 additional men.
40. A passenger train leaves the depot and runs at the rate of 30 miles per
hour. 5 hours after the departure of said train, the fast mail train leaves and runs
at the rate of 40 miles an hour. How far will the fast mail train run before it over-
takes the passenger train } Ans. 600 miles.
OPERATION INDICATED.
30 x 5 = 150 miles = distance of passenger train when fast express starts.
40 miles, speed of fast express train.
30 miles, “ “ passenger train.
10 miles, gain in a run of 40 miles.
M. Fast Ex.
40
10
150 NOTE.—The student should write the
tº-º reasoning for the operation.
41. A thief steals an article, and in running off with it he takes steps 6 feet
long; after he has gone 150 feet, a police officer starts after him and takes steps 64
feet long; allowing that they both step equally fast, how far will the officer run
before he catches the thief ? Ans. 1950 feet.
PROBLEMS IN PROPORTION. 323.
42. The steamer Katie leaves the wharf at New Orleans, and runs at the
average speed of 15 miles an hour. When the Katie had run 25 miles, the steamer
R. E. Lee leaves the wharf and runs at the average speed of 18 miles an hour. How
far will the Lee run before she overtakes the Katie : Ans. 150 miles.
43. A man has six hands hired, four of them at 75g a day and two of them.
at $1.00 a day. He wants to advance them $6.00 on account, each one in the propor-
tion of his daily pay. How much will each receive?
C.
OPERATION. 600
4 men at 752 per day = $3.00 5 3
2 men at $1. per day = 2.00 &ºm º º
$3.60 + 4 = 902 to each of 4
$5.00 men at 75% per day.
C.
600 $ $
5 2 6 6
* | sºmºmºse or thus: 5 || 75 5 | 1.00
$2.40 - 2 = $1.20 to each of 2 *E= wºº ºmºsºme
men at $1. per day. 90g. $1.20
44. Mr. A. leaves New Orleans on a journey, and travels at the rate of 25
miles a day; 6 days after Mr. B. leaves New Orleans, and travels the same road at
the rate of 30 miles a day. How many days will it be before Mr. B. will overtake
Mr. A. and how far does each travel? Ans. 30 ds.
f A. travels 750 mi.
B. travels 900 mi.
45. The interest on $5000, for 135 days, at 8 per cent. is $150. What will be
the interest on $4500, for 96 days, at 6 pr. ct. ^ Ans. $72.
46. If a grocer uses a gallon measure that is ; a pint too small, what will be
the true measure for 250 gallons of the false? Ans. 2343.
OPERATION INDICATED.
1 gal. = 8 pts. – pt. = 73 pts. sold for 8 pts. = of 250 gals. =234 gals.
true measure. Or,
# pt. Short weight on each of 250 gals. = 125 pts. = 153 gals. short weight.
250 — 153 = 2343 gals. true measure.
47. A. paid B. $50 for goods, bought by the pound. When B. weighed the
goods, he used a false weight of 14 ounces. What were A's loss and B's gain, and
What was the true amount of goods received by A. ?
Ans. A. loses # of weight bought.
B. gains # of weight Sold.
A. received $43.75 true value of goods.
OPERATION.
A. lost 2 oz. on each 16 bought = # = } of the
weight bought. B. gained 2 oz. on each 14
A. having lost 2 oz. on each 16, he therefore received sold = # = } of weight
++ of $50 worth of goods = $43.75 = true value sold.
of goods received by A.
324 SouLE's PHILosophic PRACTICAL MATHEMATICS. Yºr
48. A grocer used a false weight of 15+ ounces, instead of 16 ounces, for a
pound; how many pounds did one of his customers lose when he bought 225 pounds
just weight? Ans. 11+ lbs.
OPERATION INDICATED. *
OZ.
OZ.
15} = false weight. 4
16 = true weight. 15#.
# Oz. loss on each pound of 15% oz.
Oz. loss on 154 oz.
3 =
16
225
177 ºr oz. = 11 lbs. 1 ºr oz. or 11:#f lbs.
49. In the above problem, what would the customer have lost had he bought
225 lbs. by the false weight? Ans. 10}} lbs.
OPERATION INDICATED.
OZ. -- OZ.
16 4
15+ 16
# oz. loss on a pound of 16 oz.
3 = Oz. loss on 16 oz. or
225 1 pound.
Or,
# oz. x 225 = 1685 oz. -- 16 = 10% lbs.
50. If a mixture of 80 gallons of brandy and water, consisting of ; of brandy
and + of water, be worth $400, what would be the value of 100 gallons of pure
brandy of the same quality? Ans. $666.663.
51. If a perpendicular post that is 6 feet high produces a shadow, on level
ground, of 8 feet, what is the height of a tree, the shadow of which at the same
time is 170 feet 3 Ans. 1274.
52. If $# will buy 12 peaches, how many peaches will $24 buy? Ans. 40.
53. If it requires 124 yards of cloth to make a suit, when the cloth is 14 yards
wide, how many yards will it require when it is 14 yards wide 3 Ans. 15.
PROBLEMI CLASSIFIED : STATEMENT :
Yds. YolS. W. Yds.
12% 1; 2 || 25
14 2 || 3
- 5 || 4
54. If a boy can run 320 rods in 84 minutes, how many minutes will it take
him to run 24 miles? Ans. 194 min.
55. 4 men can pick 1500 pounds of cotton in 24 days, how many men will it
require to pick 84 bales of 450 pounds each, in 14 days? Ans. 20 men.
56. If 5 men, in 64 days of 53 hours each, can build a wall 1494 feet long, 34
feet thick, and 5% feet high, how many men will it require to build 2 walls, each 903
feet long, 4 feet thick, and 54 feet high, in 104 days of 84 hours each 3 Ans. 3.
CLASSIFICATION.
M. W. D. H. F. L. F.T. F. H.
5 1 6; 5; 1494 # 5%
3
2 104 8% 90; 4 54
PROBLEMS IN PROPORTION. 325
OPERATION.
M. Reason.—Since it requires 5 men to build 1 wall, to build 2 walls it will re-
5 quire 2 times as many men; then since it requires 5 men when they work 6:
2 or * days, if they work but 3 day, it will require 13 times as many men, and
2 || 13 if they work # or 1 whole day, it will require # as many men; then since it
- requires the number of men shown by the result of this statement when they
21 || 2 work 1 day, to work # day, it will require 2 times as many men, and to work
5 28 * (10% reduced) days, it will require the 21st part of the number of men; then,
17 | 2 since it requires 5 men when they work 5% or * hours per day, if they work
448 || 3 but # of an hour it will require 28 times as many men, and if they work # or 1
O whole hour it will require + as many men; then, since it requires the number
3 || 272 of men shown by the result of this statement when they work 1 hour per day,
13 || 4 to work # an hour, it will require 2 times as many men, and to work 24 (8%
4 reduced) hours, it will require the 17th part of the number of men; then, since
16 || 3 it requires 5 men to build a wall #8 (149% reduced) feet long, to build it # of a
foot long will require the 448th part of the number of men, and to build it 3 or 1
4 21 whole foot long, will require 3 times as many men; then, since it requires the
number of men shown by the result of this statement to build it 1 foot long,
to build it of a foot long, will require # part of the number of men, and to build it 24% feet long
will require 272 times as many men; then, since it requires 5 men to build a wall 3+ or '#' feet
thick, to build it 4 of a foot thick will require the 13th part of the number of men, and to build it
# or 1 whole foot thick, will require 4 times as many men; then, since it requires the number of
men shown by the result of this statement to build it 1 foot thick, to build it 4 feet thick will
require 4 times as many men; then, since it requires 5 men to build a wall 5% or '#' feet high, to
Build it of a foot high will require the 16th part of the number of men, and to build it # or 1
whole foot high, will require 3 times as many men; then, since it requires the number of men.
shown by the result of this statement to build it 1 foot high, to build it 4 of a foot high will re-
quire + part of the number of men, and to build it ** feet high will require 21 times as many men.
This completes the reasoning and the statement.
OPERATION.
R * sº B º
57. º If 4 boys can work 60 prob º: MOTE.—The student should
lems in 3 hours, how many boys & wº
* * k 40 bl 60 | 40 classify the problem and write
will be required to wor proolems 8 || 3 the reasoning.
in 8 hours? Ans. 1 boy. º sº-º-º-º-º-º
OPERATION.
58. In what time can 50 men do "25 NOTE.-The student should
a piece of work that 2 men can do 50 || 2 classify the problem and write
in 25 days? Ans. 1 day. --- the reasoning.
59. If it costs $3750 to supply 250 laborers for 30 days, when they are fur-
nished with 28 ounces of provisions per day, what will it cost to supply 175 laborers
for 40 days, when they are furnished with 24 ounces of provisions per day ?
Ans. $3000.
60. If it requires 126 pieces of paper to cover the walls of a room when the
paper is 18 inches wide, how many pieces will it require when it is 21 inches wide?
Ans. 108.
61. If the freight on 200 bales of cotton for 450 miles is worth 2400 pounds of
sugar, when sugar is worth 1242 per lb., how many pounds of sugar would the
freight on 350 bales of cotton for 300 miles be worth, when the sugar is valued 83%
per lb 3 Ans. 4000.
PROBLEMI CLASSIFIED : STATEMENT :
B. M. P. C. f5.
200 450 2400 12; | 2400
350 300 8; 200 || 350 Write the
450 i 300 reasoning.
2 || 25
35 | 4
326 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. 4×
62. If 24 men, in 124 days of 94 hours each, dig a ditch, when the earth is
estimated at 4 degrees of hardness, 542; feet long, 4% feet wide, and 44 feet deep, in
how many days of 74 hours each, will 248 men dig a ditch, when the earth is esti-
mated at 74 degrees of hardness, 3873 feet long, 33 feet wide, and 13 feet deep,
'allowing that time is required in proportion to the degrees of hardness of the earth?
A Ans. 4% days.
63. How many planks 13 ft. 9 in. long and 3%in. thick are equivalent to
5000 planks, 93 ft. long and 23 in. thick? Ans. 2750 planks.
CLASSIFICATION. Reason.—Placing the 5000 planks on the statement line, we
P. Ft. L. In. T. reason as follows: Since there are 5000 planks ºf ft. long, if
5000 93 2# they were but # of a foot long there would be 77 times as
131°; 3# many, and if ; or a whole foot long the 8th part; then, since
f there is the result shown by this statement, when the planks
OPERATION. are 1 foot long, if they were but I's of a foot long there would
P. be 12 times as many, and if \# ft. long, there would be the
5000 165th part; then, since there is this indicated result when
8 || 77 the planks are + in. thick, if they were but 4 in. thick there
165 | 12 would be 11 times as many, and if 3 or 1 in. thick the 4th
4 || 11 part; then, since there is this result when they are 1 in. thick,
7 | 2 if they were 3 in. thick there would be 2 times as many, and
if in. thick the 7th part.
64. If it costs $59.75 to carpet a room 32 ft. by 12 ft. how much will it cost to
carpet a room 48 ft. by 16 ft. with the same kind of material % Ans. $119.50.
65. If 8 men in 6 days of 10 hours each cut 50 cords of wood, how many
cords will 15 men cut in 15 days of 8 hours each 3 Ans. 1874.
66. If 3 compositors, in 24 days of 94 hours each, set up 28} pages of type,
how many compositors will be required to set 680 pages in 44 days of 10% hours
each # Ans. 33; men.
PROBLEMI CLASSIFIED : STATEMENT :
C. D. H. P. º C.
3 2+ 94 28% -- 3
4}. 10% 680 4 || 9
21 5 Write the
2 | 19 reasoning.
76 7
85 3
680
67. If 53 yards cost $8.20, what will 224 yards cost? Ans. $36.
68. If + of a pound costs $4, what will 1++ pounds cost 3 Ans. $2.124.
69. If 10 men in 6 days earn $240, how much can 6 men earn in 10 days?
*. Ans. $240.
70. How many chickens at 70g a piece must be given for 2 barrels of flour
at $7 per barrel ? Ans. 20.
71. If it costs a family of 8 persons $4000 a year to live in New Orleans, what
will it cost a family of 9 persons to live in St. Louis in the same style for 10 months,
supposing the prices to be in St. Louis + less than in New Orleans?
Ans. $2812.50.
PROBLEMS IN PROPORTION. 3.27.
72. If # of a pint of wine cost 37.4%, what will # of a hogshead cost at the
same rate # Ans. $201,60.
PROBLEMI CLASSIFIED : STATEMENT :
P. C. C.
# 37; 2 || 75
3 || 4
H. 2 Write the
# 4 reasoning.
63
5 || 4
73. A garrison consisting of 4500 men, has provisions for 9 weeks, allowing
each man 24 ounces per day. How many reinforcements may be sent into the
garrison, so that the provisions will last but 5 weeks when the daily allowance is
reduced to 16 ounces 7 Ans. 7650.
74. If it requires 2 men # of a day to perform # of a piece of work, how many
days will it require 3 men to perform # of it? Ans. # of a day.
PROBLEMI CLASSIFIED : STATEMENT :
M. D. W. DS.
2 # # 4 || 3
3 # 3 | 2 Write the
2 || 3 reasoning.
9 || 8
75. If it requires 5 men # of a day to perform # of a piece of work, how long
will it require 3 men to perform # of it 3 Ans. 24 d.
76. If the freight on 85 bales of cotton for 300 miles is worth 1000 pounds of
sugar, when the sugar is worth 84% per pound, how many pounds of sugar must be
given for the freight on 110 bales of cotton for 200 miles, when the price of sugar is
6#2 per pound? Ans. 11734 pounds.
OPERATION INDICATED.
lbs.
1000
85 | 110 NoTE.—The student should classify the prob-
300 200 lem and write the reasoning.
2 17
25 || 4
77. If a loaf of bread worth 5% weighs 64 ounces, when wheat is worth $1.80
per bushel, what should be the weight of a 12#2 loaf, when wheat is worth $1.62%
per bushel? Ans. 18 oz.
78. If it requires 5 men # of § of a day to do # of § of 4 of a piece of work,
what time will it require 4 men to do # of # of the work? Ans. 1; d.
PROBLEM CLASSIFIED : STATEMENT : *
M. Ds. W. Ds.
5 # x 3 = + 3 × 3 × # = } 4 | 1
- 4 5 Write the
4 # × # = # 8 reasoning.
15 || 8
328 soule's PHILosophic PRACTICAL MATHEMATICs. *
79. If 12 pounds of rice are worth 9 pounds of sugar, how many pounds of
rice can be bartered for 217 pounds of sugar 3
Ans. 289; lbs.
80. If 25 oranges are worth 48 peaches, how many oranges can be bought
for 240 peaches 2
Ans. 125 Oranges.
81. If it requires 1800 brick, 8 inches long, and 4 inches wide, to pave a side-
walk 40 ft. long and 10 ft. wide, how many brick 10 inches long and 5 inches wide,
will be required to pave a yard 50 feet long and 30 feet wide Ans. 4320 brick.
82. If # of a gill of Wine costs 12#2, what will 42; gallons cost, at the same
rate #
Ans. $342.
83. If 4 boys can make 100 hats in 6 days of 10 hours each, how many hats
would 6 girls make in 10 days of 8 hours each, allowing that 4 girls can do as much
work as 5 boys in the same time
CLASSIFICATION,
Boys. Hats. Ds. Hr.
4 100 6 10
10 8
Girls.
6 = 7# boys.
FIRST OPERATION.
Hats.
100
4
2 | 15
6 || 10
10 || 8
250 hats, Ans.
SECOND OPERATION,
Hats.
100
5
6
10
8
10
:
250 hats, Ans.
Ans. 250 hats.
Eacplanation.—Since the working capacity of
each girl is greater than that of each boy, we
must first find the equivalent working capacity
of the 6 girls in the working capacity of boys.
And since 4 girls can do the work of 5 boys, 1
girl can do # part = 1+ or # times as much as 1
boy, and 6 girls can do 6 times as much, = # x
6 = ** = 73 boys; i. e. the work done by 6
girls is equal to the work done by 7# boys.
Explanation to the second operation.—Since 4
boys make 100 hats, 1 boy will make the + part,
and 5 boys or 4 girls will make 5 times as many.
And since 4 girls make the result of this state-
ment, 1 girl will make the + part, and 6 girls
will make 6 times as many. A ſhen, since 100 hats
are made in 6 days, in 1 day # part would be
made, and in 10 days 10 times as many. Then,
since 100 hats are made when the hatters work
10 hours a day, to work 1 hour a day they woul."
make T'ſ part, and 8 hours 8 times as many.
84. If 20 girls can make 200 shirts in 10 days, how many days would it re-
quire 15 boys to make 300 shirts, estimating that 5 girls can do as much work as 6.
boys in the same time?
PROBLEM CLASSIFIED :
G. B. S. D.
20 6 200 10
5 15 300
Ans. 24 ds.
OPERATION:
D.
10
5 20 Write the
15 | 6 reasoning.
200 300
PROBLEMS IN PROPORTION. 329
85. If it takes 7 men # of a day to do g of a piece of work, how long will it
require 5 men to do # of the work? w Ans. # day.
86. A garrison of 3000 men has provisions for 42 days, allowing each man 2
lbs. 4 oz. per day. How many reinforcements may be sent in, so that the provisions
Will last 10 weeks, when the allowance is reduced to 1 lb. per day ?
Ans. 1050 reinforcements.
87. If 200 pounds of first grade rice cost $12, what will be the cost of 600
pounds of second grade, estimating that 5 pounds of the first grade are equal to six
pounds of the second 7 Ans. $30.
CLASSIFICATION.
lbs.
200 1st grade $12.
600 2d grade,
X # =
500 1st grade.
FIRST OPERATION.
Explanation.—Since 1 pound of the second grade = } of the
first grade, 600 pounds of the second grade = 600 × # = 500
pounds of the first grade. Then, since 200 pounds cost $12,
12 1 pound will cost gºt part, and 500 pounds 500 times as much.
200 || 500
$30, Ans.
SECONID OPERATION. t
$ Ea:planation of the second operation.—Since 200 pounds of
12 first grade cost $12, 1 pound will cost gººd part, and 5 pounds
200 || 5 of first grade or 6 pounds of second grade will cost 5 times as
6 || 600 much. And since 6 pounds of second grade cost this result,
tºº is 1 pound of second grade will cost # part, and 600 pounds of
$30, Ans. second grade will cost 600 times as much.
88. If 45 pounds of coffee cost $9, what will be the cost of 90 pounds of a
different quality, estimating that 4 pounds of the first quality are equal to 5 pounds
of the second 3 Ans. $14.40.
89. If 5 oxen or 7 cows eat 3 ºr tons of hay in 87 days, in what time will 2 oxen
and 3 cows eat 6 ºr tons ? Ans. 210 days.
CLASSIFICATION. Explanation.—Since by the first condition of the problem,
Ox C T DS 5 oxen or 7 cows performed the eating, and by the second
5 7 3-4 87 condition 2 oxen and 3 cows performed the eating, it is there-
OI’ TT fore necessary to first find the equivalent eating capacity of
2 and 3 6-ºr 1 ox compared with that of 1 cow ; or the equivalent eating
capacity of 1 cow compared with that of 1 ox. Hence,
since 5 oxen eat as much as 7 cows, 1 ox will eat # part = 1%, or 4 times as much as 1 cow ; or since
7 cows eat as much as 5 oxen 1 cow will eat 4 part = } as much as 1 ox.
Having now the equivalent eating capacity of 1 ox and of 1 cow each measured by the
capacity of the other, we next find the eating capacity of the 2 oxen in that of cows; or the eating
capacity of 3 cows in that of oxen, and since the eating capacity of 1 ox is 1%, or # that of 1 cow,
the eating capacity of 2 oxen is 2 times as much = 3 × 2 = ** = 2; cows. This added to the 3
cows = 3 + 2} = 5% cows.
Or, since the eating capacity of 1 cow is # that of 1 ox, the eating capacity of 3 cows is 3
times as much = } x 3 = ** = 2+ oxen. This added to the 2 oxen = 2 + 2} = 4+ oxen.
Now, from these reductions of the capacity of oxen to cows, or cows to oxen, and the other
conditions of the problem, we may solve the problem from either of the following classifications:
33O SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs.
CLASSIFICATION.
With the eating capacity of oxen ex-
pressed in that of cows.
C. T. Ds.
7 3++ 87
5% #,
OPERATION.
Ds.
87
7
29 || 5
37 || 11
11 || 74
210 ds., Ans.
CLASSIFICATION.
With the eating capacity of the cows
reduced to that of oxen.
Ox. T. Ds.
5 3+*r 87
4} 6 ºr
OPERATION.
Ds.
87
5
29 7
37 || 11
11 || 74
210 ds., Ans.
90. If 4 men or 5 women can perform # of a piece of work in 27 days, how
long will it require 1 man and 1 woman, working together, to perform the whole
work?
CLASSIFICATION. 1st OPERATION.
Men. Wo. DS. W. Ds.
4 or 5 27 # 27
1 and 1 whole or 4
14 2#. # 9 || 5
3 4
80 ds., Ans.
Ans. 80 ds.
2d OPERATION.
Ds. Eacplanation.—Since
27 the work of 4 men is
5 = to that of 5 women,
9 4 the work of 1 man is
= to that of 14 women;
3 || 4 and the work of 1 man
tº-º-º: and 1 woman is = to
80 ds, Ans. 14 + 1 − 24 women's
? work.
Or, since the work of 5 women is = to that of 4 men, the work of 1 woman is = to that of #
of a man, or to # of the work of 1 man; and the work of 1 woman and 1 man is = to 3 + 1 = 14
men’s work.
From these reductions and the , conditions of the problem, the classification and operation
statements are made.
NotE-The student should write the reasoning for each operation.
This problem may also be solved as follows:
4 × 27 = 108 = days for 1 man to do # of the work.
5 × 27 = 135 = days for 1 woman to do # of the work.
O
l
3
º
:
#s = work done by 1 man in 1 day.
= work done by 1 woman in 1 day.
1. *E=g '- tº t
}rº + H+H = Ha = go which is
the work done by 1 man, and 1 woman in 1 day. Then, since tº of the # of work
is done in 1 day, to do #3 of the # work, will require 60 days. And since it
requires 60 days to do # of the work, to do + will require # part of the time, which
is 20 days; and to do # or the whole work will require 4 times as many days, which
is 80.
91. If 4 of a pound costs 50¢, what will 4 pounds and 11 ounces cost, at the
same rate 3
Ans. $9.374.
92. A grocer sold 40 gallons of a compound, consisting of 4 whiskey and 4
water, for $60. What would be the value of 50 gallons of pure whiskey at the same
rate %
Ans. $150.
34- PROBLEMS IN PROPORTION. 33 I
93. If 3 men and 5 women can do a piece of work in 14 days, 6 women and 8
boys can do the same work in 104 days, and 4 men and 7 boys can do it in 12# days.
How long will it take 2 men, 10 boys and 3 women to do it? Ans. 9+} ds.
OPERATION.
First Step.
Men. Women. Boys. Days. Work in 1 day,
3 + 5 E 14 = *r
6 + 8 = 10% = # Classification.
4 + 7 - 12; – #
M. W. M. W. Work.
(3 5 = +) × 6 = 18 + 30 = # = ºr Subtract upper Equa-
W. B. W. B. tion from lower.
(6 + 8 = #1) x 5 == 30 + 40 = }}
Second Step.
M. B.
— 18 + 40 = ºr Multiply by 2.
4 + 7 = ºr Multiply by 9.
M. B.
— 36 + 80 = ºr Add the two Equations.
36 + 63 = }}
B.
143 = ###
1 B. = ### Work in 1 day.
Third Step.
TM. B. M. W.
4 + 7 = #F = ºr 3 + 5 = 1°,
(1 = ###)x 7 (1 = Hºg) × 3
= 7 = #F Subtract. = 3 = +}g subtract.
M. 5 W. = +};
4. = #F © 1 W. = ++g Work in 1 day.
1 M. = ++g Work in 1 day.
Fourth Step.
M. W. B.
1. 1 1
Iłg + T+g + ###
M. W. B.
2 3 10
++, + ++, + ºr = Pºs = work in 1 day; 12138 + 1224 = 944 days, Ans.
3
32 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. -
PARTITIVE PROPORTION.
670. Partitive Proportion is a proportion where a number is divided into
parts which bear certain quantitative relations or ratio to each other.
OPERATION. * Ea:planation.—Since A. re-
1. tº e ſº ceives 3 to B's 4, they, together
Divide 35 oranges § # _A. § B. receive 3 + 3 + 7. "Hence, A.
between A. and B. in the + = { will receive 3 and B. # = % of
rati & A. B. the Oranges. And since + are
“. O of 3 to 4. How many 35 35 to receive 35 oranges, + will
Will each receive? 7 || 3 7 || 4 receive + part and 3, 3 times as
Ans. A. 15: B. 20. *= ºsmºs *= ºs many. This gives A's share,
y * Similar reasoning with the #
15, A. 20, B. gives B's share.
2. A father divided $66 as a christmas present among his six children in
proportion to their ages, which were as follows: Albert 16, Eddie 14, Mamie 12,
Frank 10, Robert 8, and Lillie 6. How many dollars did he give to each 3
Ans. Albert $16, Eddie $14, Mamie $12,
Frank $10, Robert $8, and Lillie $6.
3. Divide $2222 into 4 parts, which shall be to each other as 1, 2, 3, and 5.
How many dollars will there be in each part 3
Ans. 1st, $202; 2d, $404; 3d, $606; 4th, $1010.
4. A business man has $42000 invested in three kinds of stock—bank, insur-
ance, and manufacturing. He has three times as much bank stock as insurance
Stock, and 24 times as much manufacturing stock as bank and insurance stock
together. What is his investment in each kind of stock #
Ans. Insurance stock, $3000; bank stock, $9000;
* manufacturing stock, $30000.
Memorandum.—Insurance stock 1 + bank Stock 3 = 4 × 23 = 10 = manu-
facturing stock; then, 1 + 3 + 10 = 14 = $42000.
5. An Egyptian army commander had 1300 camels, which he distributed
among three separate general officers, in proportion to #, #, and #. How many did
each receive 7 Ans. The 1st, 600; the 2d, 400; and the 3d, 300.
6. A benevolent man has $600, of which he wishes to give to A. §, to B. #,
to C. #, and to D. #. What amount will each receive 3
Ans. A., $200 ; B., $150; C., $120; and D., $100.
7. Another benevolent man has $600 which he wishes to give to W., X, Y.,
and Z., in proportion to #, +, +, and #. What amount will each receive?
Ans. W., $210.53; X., $157,89; Y., $126.32; Z., $105.26.
NotE.—See Partnership Average for the solution of the two preceding problems.
8. James and Louis, together, have $124, and James has three times as much
as Louis. How many dollars has each 3 Ans. James $93., Louis $31.
Solution.—Let $1 represent what Louis has; then, since James has 3 times as many, he has
$3, and together they have $1 + $3 = $4; then, $124 divided in the ratio of 1 to 3, gives Louis $3).
and James $93.
PROBLEMS IN PROPORTION. 333
9. Henry and William, together, have $296, and three times Henry's share
equals 5 times William's share. How many dollars has each 3
Ans. Henry $185, William $111.
Solution.-Since 3 times Henry's share = 5 times William's share, 1 time Henry's share = }
of 5 times William's share, − 5 – 3 = 1; or § of William's share; and this added to William's
share, which is 1, = } + 3 = , = what both have, or $296. Then, $296 divided in the ratio of #
to 3 gives Henry $185 and William $111. *
Or, thus: Since 5 times William's share = 3 times Henry's share, 1 time William's share = }
of Henry's share which added to # (Henry's share in fifths) = $ the sum of both shares. Then
dividing $296 in the ratio of # to #, gives William $111 and Henry $185.
Or, thus: To avoid fractions, give Henry 5 shares and William 3, making 8 shares. Then
since 8 shares equal $296, 1 share equals the # part, which is $37. Then 5 times $37 equals $185
equals Henry's share; and 3 times $37 equals $111 equals William's share.
10. A. and B. have $1020. § of A's is equal to ; of B's. How much has
each 3 Ans. A. $540. B. $480.
Solution. # of A's = # of B's, then # of A's = } of B's, which is 3 of # = #, and 3 thirds or
the whole of A's = 3 times # or # of B's. Hence 3 + š, B's share = ** which is the sum of A's and
B’s shares, or $1020. And ºr of $1020 is $60; and 9 times $60 = $540, A's share; and 8 times $60 =
$480, B's share.
11. X. and Y. own a plantation containing 3500 acres, 4 of X’s interest is
equal to # of Y’s. How many acres has each 3 Ans. X. 1500; Y. 2000 acres.
12. A pomologist has an orchard containing 1744 trees, apple, peach and
pear; there are 3 times as many apple as peach trees, plus 150; and 4 times as
many peach as pear trees, minus 120. How many trees are there of each kind?
Ans. 122 pear, 368 peach, 1254 apple.
NOTE.-Problems of this character belong properly to Algebra; but as they are frequently
met with in arithmetics, we have given it place, and for those who have not studied Algebra, we
give a brief solution.
Solution.—Let pears = 1; then peaches = 4 – 120, (which is 4 times the pears — 120); then
apples = 12 — 360 -- 150, (which is 3 times the peaches + 150). Then, since the pears, peaches
and apples = 1744, it follows that 1 + (4 — 120) + (12 – 360 + 150) = 1744. Then, transposing
or classifying we have 17 = 1744 -- 360 + 120 – 150 = 2074. Then, 17 = 2074, 1 = 122 pear trees.
(122 × 4) — 120 = 368 peach trees. (368 × 3) + 150 = 1254 apple trees.
MEDIAL PROPORTION.
671. Medial Proportion is a proportion where two or more quantities of
different values are combined to find the mean or average value of the combination.
This division of proportion is generally, and will be in this book treated under the
subjects of Alligation Medial and Alligation Alternate.
CONJOINED PROPORTION.
672. For a definition of Conjoined Proportion, see page 312.
1. If 40 oranges are worth 60 apples, and 75 apples are worth 7 dozen
peaches, and 100 peaches are worth 1 box of grapes, and 3 boxes of grapes are
worth 40 pounds of pecans, how many pounds of pecans can be bought for 100
oranges? Ans. 22# lbs.
334 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
2. If 40 lbs. of sugar are worth 60 lbs. of rice, and 75 lbs. of rice are worth
500 oranges, and 150 oranges are worth 200 apples, and 250 apples are worth 16
doz. bananas, and 10 doz. bananas are worth 8 doz. peaches, and 16 doz. peaches
are worth 50 lbs. of pecans, and 40 lbs. of pecans are worth 8 boxes of grapes, and
3 boxes of grapes are worth 2 boxes of raisins, and 7 boxes of raisins are worth 2
bbls. of flour, and 3 bbls. of flour are worth 6 bbls. of potatoes, and 2 bbls. of
. potatoes are worth $7, then how many pounds of sugar can we buy for $40%
Ans. 70+'s lbs.
RECIPROCAL PROPORTION.
673. Reciprocal Proportion is a proportion where a number is divided into
parts which bear certain, reciprocal quantitative relations to each other.
1. Four citizens of a neighborhood, A., B., C., and D., agreed to build a
schoolhouse and to pay for the same in reciprocal proportion to the distance that
each resides from the selected location. A. resides 2 miles; B., 1 mile; C., # mile;
and D., 4 mile from the schoolhouse. The cost of the house was $1500. What does
each Owe ? Ans. A. $100; B. $200; C. $400; D. $800.
OPERATION.
A. 2 miles. The reciprocal of 2 is # Explanation.-In solving problems, of
B. 1 mile. 4% { % ** 1 is 1 this kind, we first find the reciprocal of
O il {{ { % { % is 2 each number which represents the dis-
. 3 mile. # is tance that the several parties named
D. # Imile. {{ & 4 “ # is 4 reside from the school, or other object,
for which the money is contributed. In
* * ... fºr this problem, the reciprocal numbers are
The sum of the reciprocals is 73 shown in the operation of the problem,
and were obtained in accordance with
& Article 250, page 147.
1500 These reciprocals show the respective
2 proportional parts that each party_is
liable for, and the sum of the same, (7%)
represents the whole cost of the school-
smm mºm-mº smºsºms ºmº * -ºº-º-º: house, $1500. Hence to determine each
$100 $200 $400 $800 party's liability, or the amount that each
has to pay, we write the $1500 on the
statement line, commence with A. and reason thus: Since "# (7%) parts are equal to, or have to
pay $1500, 4 part equals or will pay the T's part and #, or a whole part, twice as much; and since
i part equals or will pay this indicated result, 3 part (which is A's) is equal to, or wili pay, # as
much. This shows A's amount to be $100.
Similar reasoning is given for the statements of B., C., and D., which are exhibited in the
operation.
2. A father left an estate of $81600, to be divided among his 5 children in
reciprocal proportion to their ages, which are 3, 8, 10, 12, and 15 years. How many
dollars will each receive? Ans. 3 years, $38400. 10 years, $11520.
8 {{ 14400. 12 “ 9600.
15 ($ 7680.
A. B.
1500
2 15
500
500 1
15 2 15
2
º
1
º
4
PROBLEMS IN PROPORTION. 335
3. Suppose, in the above problem, the father had ordered that the money
should be divided in proportion to their ages, how much would each have received?
Ans. 3 years, $ 5100. 10 years, $17000.
8 * $13600. 12 “ $20400.
15 “ $25500.
4. A., B., C., and D., agree to contribute the necessary funds for the erec-
tion of a mill at a certain specified place, in reciprocal proportion to the distance
that they respectively reside from the location selected for the mill, on condition
that they should be joint owners in the mill in proportion to the amount contributed
by each in its erection. The distance that each resides from the Selected location
of the mill, is as follows: A. 4 of a mile, B. § of a mile, C. 1 mile, and D. 24 miles.
The mill cost $3500; what is the interest of each, A., B., C., and D., in the mill?
Ans. A. #; B. #; C. #; and D. #.

* * * * * * * * * * * * * * * * * > se e º ºs e º e s - e º 'º e º e s sº e º e e º e s e s tº tº e º "
674. To the Reader. Mensuration of surfaces and solids is, by most
authors, located in the latter part of arithmetics. But, because of the general
importance of the subject in business affairs, We present mensuration in advance
of square root and percentage and the several subjects involving per cent. The
subject is herein treated far more extendedly, practically and philosophically than
in any other treatise.
If the student has any special reason for acquiring a knowledge of percent-
age, interest, exchange, etc., before he learns mensuration, he may omit this work
until he has studied the percentage Subjects.
The few problems involving the principles of square root, may be omitted by
those unacquainted with that subject, or if preferred they may turn to the subject
of square root, and in a few minutes acquire the knowledge thereof necessary to
work the problems.
It is presumed that those who possess this book are already familiar with the
first principles of square root.
675. Mensuration is the process of finding the length of lines, the area
of surfaces, and the volume or solidity of solids. The principles that govern
the process of work are derived from Geometry, a very important and inter-
esting branch of mathematics, but which cannot be fully explained in a treatise of
this character. In order, however, to explain Some of the most simple and fre-
Quently occurring geometrical terms and figures, we present the following
GEOMETRICAL DEFINITIONS AND FIGURES.
676. A Point is that which has position with-
O out measurable length, width, or thickness.
677. A Line is that which has length without
measurable width or thickness.
678. A Straight Line is one that does not change
direction at any point.
_Ts 679. A. Curved Line is one that changes direction
at every point.
2Ts 2^ 680. A Crooked or Broken Line is one that
changes direction at some of its points.

(336)
*
MENSURATION. 337
681. A Surface is that which has length and width, without measurable
thickness.
682. A Plane Surface is a surface which lies even throughout its whole .
extent.
683. A Curved Surface is a surface that has length and width, without
measurable thickness, and like a curved line, changes its direction at every point.
684. A Wolume, Solid or Body, is that which has length, width and
thickness.
ANGLES.
685. An Angle is the 686. A Right Angle
Opening or divergence of is an angle formed by
tWO lines from a certain * One Straight line meeting
oint _f | \_ another, so as to make
p o equal angles.
687. An Acute Angle is | 688. An Obtuse Angle
A. less than a right angle. is greater than a right
angle.
FIGURES.
689. A Plane Figure in geometry is a portion of a plane bounded by lines,
straight or curved, or by both combined. When the lines which bound a figure are
straight, the space that they inclose is called a polygon, or rectilineal figure.
*— 690. A. Polygon is a plane
\ figure, or portion of a surface
bounded by straight lines.
691. A Square is an equi-
lateral rectangle.
B
F- 692. A Rectangle is a & E 693. A Right-An-
quadrilateral polygon which * ăgled Triangle is a tri-
has its opposite sides equal *:
and parallel, and all its an- # 5 angle that has one of
# its angles a right an-
gles right angles.
* gle.
#
A. Base,
In the right-angled triangle, A B C, the side A C, opposite the right angle
B, is called the hypotenuse; the side A B, the base, and the side B C, the perpendic-
wla,
Upward of 2400 years ago, Pythagoras, a celebrated Grecian geometer, discovered that the
square described on the hypotenuse of a right-angled triangle is equal to the sum of the squares described on
the other two sides. And on making this discovery, to express his joy and gratitude, we are told that
he sacrificed one hundred oxen to the Muses. Pythagoras also discovered that a circle is the great-
est of all figures of equal perimeter.
A 694. A Diagonal is a straight line connecting the vertices
of two opposite angles of a polygon. As the line A B, in this
| rectangle. The sum of all the lines bounding a polygon constitutes
- its perimeter.
B


338 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. 4:
695. A Parallelogram 696. A Rhombus is an
is a quadrilateral having equilateral oblique-angled
its opposite sides parallel. parallelogram.
ZLZ
697. A Trapezoid is a
quadrilateral having only
two of its opposite sides
' parallel.
698. A Trapezium is
a quadrilateral having no
two sides parallel.
A
701. A Sca-
lene Triangle
# is a triangle
f which has allits
--J sides unequal.
700. An ISOS-
celes Triangle
is one having
\ only two of its
lsides equal.
O OOO
Pentagon. Hexagon. Heptagon. Octagon. Nonagon. Decagon.
699. An Equi-
lateral Triangle
is one having its
three sides equal.
-
i
º
702. A Circle is a plane figure bounded by a regular curved
line, every point of which is equally distant from a point within
called the center.
For the definition of the Diameter, the Circumference, the Radius, a Chord, an
Arc, a Sector, an Angle, and a Segment of a Circle, see Circular Measure, page 252.
703. The Ratio between the diameter and the circumference of a circle
has been demonstrated in geometry to be as 1 to 3.1416; i. e., when the diameter of
a circle is 1, the circumference is 3.1416, practically.
NOTE.-This ratio cannot be exactly expressed in figures. The decimal has been carried by
mathematicians to two hundred and fifty places, and yet the exact ratio was not obtained. It is
impossible to square the circle.
Van Ceulen, a German mathematician, first extended the approximation to 36 places by
continually bisecting the arc of a circle. In honor of this achievement, the 36 numbers expressing
the ratio of the diameter of a circle to its circumference were engraved on his tomb-stone. The
following are the numbers: 3.1415926535897932.38462643383279502884,
704. The Area of a figure, or of a described object, is the measure of its
surface in some square unit, as the inch, the foot, the yard, the mile, etc.
100 705. The Ratio between the area of a circle and
of a square, one side of which is equal to the diameter
too of the circle, has been demonstrated in geometry to be
as 1 to .7854; i. e., when the area of a square is 1, the
area of the circle is .7854. Or, if we divide a square
surface into 10000 smaller squares and then cut off
Area 10000 Area 7854. the four corners so as to make a circle, we will thus
10000–2146=7854. cut away 2146 of the smaller squares and have
remaining in the circle 7854 of the smaller squares.
3.





* MENSURATION. 339
2 706. An Ellipse is a figure bounded by an oval curved line.

The transverse diameter, or azis of an ellipse is a line passing through its
& 4 center in the direction of its length, as the line A. B.
The conjugate diameter, or aſcis, is a line passing through the center of the
a' ellipse in the direction of its width, as the line E D. -
TABLE OF MULTIPLES.
707. The following table may be often used to advantage in the computation
of mechanical and geometrical problems: -
TABLE.
Diameter of a circle × 3.1416 = Circumference.
Radius of a circle × 6.28318.5 = Circumference.
Square of the radius of a circle × 3.1416 = Area.
Square of the diameter of a circle × 0.7854 = Area.
Square of the circumference of a circle × 0.07958 = Area.
Half the circumference of a circle × by half its diameter = Area.
Circumference of a circle × 0.159155 = Radius.
Square root of the area of a circle × 0.56419 = Radius.
Circumference of a circle × 0.31831 = Diameter.
Square root of the area of a circle × 1.12838 = Diameter.
Diameter of a circle × 0.866 = Side of inscribed equilateral triangle.
Diameter of a circle × 0.7071 = Side of an inscribed square.
Circumference of a circle × 0.2251 = Side of an inscribed square.
Circumference of a circle × 0.2821 = Side of an equal square.
Diameter of a circle × 0.8862 = Side of an equal square.
Base of a triangle X by 3 the altitude = Area.
Multiplying both diameters and .7854 together = Area of an ellipse.
Surface of a sphere × by # of its diameter = Solidity.
Circumference of a sphere x by its diameter = Surface.
Square of the diameter of a sphere × 3.1416 = Surface.
Square of the circumference of a sphere × 0.3183 = Surface.
Cube of the diameter of a sphere × 0.5236 = Solidity.
Cube of the radius of a sphere × 4.1888 = Solidity.
Cube of the circumference of a sphere × 0.016887 = Solidity.
Square root of the surface of a sphere × 0.56419 = Diameter.
Square root of the surface of a sphere × 1.7724.54 = Circumference.
Cube root of the solidity of a sphere × 1.2407 = Diameter.
Cube root of the solidity of a sphere × 3.8978 = Circumference.
Radius of a sphere × 1.1547 = Side of inscribed cube.
Square root of (; of the square of ) the diameter of a sphere = Side of inscribed cube.
Area of its base x by # of its altitude = Solidity of a cone or pyramid, whether round,
square, or triangular.
Area of one of its sides × 6 = Surface of a cube.
Altitude of trapezoid X # the sum of its parallel sides = Area.
34O SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS. º:
LINEAR MEMSURE,
-º-
708. In the mensuration of lines, surfaces, and solids, we have three kinds
of measure—line or linear, surface and solid.
The unit of measure for lines is an inch, a foot, a yard, a rod, etc., or some
subdivision thereof.
PROBLEMIS.
1. The diameter of a circle is 50 feet. What is the circumference?
Ans. 157.08 feet.

Explanation.—In all problems
oper ATION. of this kind, we multiply the
O diameter by 3.1416, which is the
50 x 3.1416 = 157.0800 ft., Ans. ratio between the diameter and
\ . the circumference.
NOTE.—In all problems in Linear, Surface, or Solid Measure, the student should draw on his
paper the outline of the figure to be measured, before performing the operation.
2. The circumference of a circle is 40 feet. What is the diameter ?
Ans.-12.73 + ft.
Explanation.—In all problems of this kind,
OPERATION. we divide the circumference by 3.1416, which is
the ratio between the diameter and the circum-
40.0000 -- 3.1416 = 12.73 + ft., Ans. ference of a circle.
3. What is the circumference of a circular garden, the diameter being 25
yards and 2 feet? Ans. 241.9032 ft.
4. What is the diameter of a circle whose circumference is 68 feet 9 inches f
Ans. 262.605 in.
5. The radius of a circle is 40 feet. What is the circumference #
Ans. 251.328 ft.
6. What is the perimeter of a nonagon, each side of which is 61 inches?
Ans. 549 inches = 45 ft. 9 in.
NOTE.-The perimeter of a body or figure is its outer boundary or the sum of all its sides.
7. What is the perimeter of a hexagon whose sides are each 4 ft. 3 in.?
Ans. 25 ft. 6 in.
8. The perimeter of a duodecagon is 27 feet. What is the length of one
side? Ans. 2+ ft. = 2 feet 3 inches.
º: LINEAR MEASURE. 34. I
TO FIND THE HYPOTENUSE OF A RIGHT-ANGLED TRIANGLE. WHEN
THE BASE AND THE ALTITUDE ARE GIVEN.
PROBLEMIS.
709. 1. The base of a right-angled triangle is 60 feet and the altitude 25
feet, what is the hypotenuse? Ans. 65 feet.
NOTE.-Students who have not studied square root may omit this and the two following
problems, until they have learned square root.
OPERATION.
60° = 3600 Explanation.—In all problems of this kind, we extract
= 2 *
25 wº-º 625 the square root of the sum of the squares of the base and
4225 altitude, and the result will be the hypotenuse.
V4225 = 65 feet, Ans.
2. If one side of a square park is one mile, what is the distance from the
center to either corner ? Ans. 3733.52 + feet.
TO FIND THE BASE OR THE ALTITUDE OF A RIGHT-ANGLED TRI-
ANGLE, WHEN THE HYPOTENUSE AND ONE OF THE
OTHER SIDES ARE GIVEN.
710. 1. The hypotenuse of a right-angled triangle is 50 feet and the base
40 feet, what is the altitude : .* Ans. 30 feet.
OPERATION.
50? – 2500
40? – 1600 Explanation.—In all problems of this kind, we extract
*mº the 8quare root of the difference between the squares of the
Difference, 900 hypotenuse and the given side, and the result will be the
required side.
V 900 = 30 feet, Ans.
2. The hypotenuse of a right-angled triangle is 410 feet, and the base is
90 feet. What is the perpendicular or altitude : Ans. 400 feet.
TO FIND THE DIAMETER OR THE CIRCUMFERENCE OF A CIRCLE,
WELEN TELE AREA IS GEVEN.
NOTE.—The area of a figure is the quantity of its surface expressed in units of square
measure, as an inch, foot, yard, acre, etc.
PROBLEMIS.
711. 1. The area of a circle is 47124 square feet, what is the diameter?
Ans. 244.94+ feet.
34.2 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. º:
Explanation.—In all problems of
this kind, we first divide the area by
OPERATION. .7854, and then extract the square root of
.7854 ) 47124.0000 ( 60000 the quotient. To find the circumference,
we would multiply the diameter by

/ V 60000 = 244.94 + ft., Ans. 3.1416, or divide the area by .07958,
y and extract the square root of the
quotient.
2. What is the diameter of a circle whose area is 5026.56 feet 3
Ans. 80 feet.
TO FIND THE SIDE OF A SQUARE THAT IS EQUAL IN AREA TO A
GIVEN CIRCLE.
PROBLEMS.
712. 1. What is the side of a square equal in area to a circle whose diame-
ter is 40 feet 3 Ans. 35.448 feet.
OPERATION.
.8862 Explanation.—In all problems of this kind, we multi-
40 diameter. ply the diameter by .8862, or where the circumference is
*mºnººmºsºsº given, multiply it by .2821; the result in each case is the
35.4480 feet, Ans. side of an equivalent square -
NOTE.--The decimal .8862, is the square root of .7854 and is the side of a square equal in
area to a circle whose diameter is 1. The decimal .2821 is the square root of .07958, and is the side
of a square equal in area to a circle whose circumference is 1
2. The circumference of a race track is one mile or 5280 feet. What is the
side of a square field of equal area 3 Ans. 1489.488 feet.
3. The diameter of a circle is 60 feet, what is the side of a square of equal
area 3 Ans. 53.172 feet.
TO FIND THE SIDE OF A SQUARE INSCRIBED IN A GIVEN CIRCLE.
PROBLEMIS.
713. 1. The diameter of a circle is 40 feet, what is the side of the inscribed
Square ? Ans. 28.284 feet.
OPERATION. Explanation.—In all problems of this
* .7071 kind, we multiply the diameter by .7071, or
40 where the circumference is given, multiply it
N 2% - by .2251; the result in each case is the side
28.2840 feet, Ans. of the inscribed square.

2. What is the width of one side of the largest square piece of timber that
can be sawed from a log 30 inches in diameter Ans. 21.213 inches.
2} LINEAR MEASURE, 343
TO FIND OR MEASURE THE HEIGHT OF A TREE, OR THE HEIGHT
TO A CERTAIN POINT OF TEIE SAME WITHOUT CUTTING
OR CLIMBING TEIE TREE.
PROBLEMIS.
714. The figure A B C D, represents a
tree. What is the distance from C to B 3
Ans. 36 feet.
NoTE.—The point C is placed 3 feet from the ground
to facilitate the observation by the eye.
Ea:planation.—First measure from the base of the tree
a distance of, say 10 feet, and at this point, vertically
erect a pole E F, that has feet and inches marked
thereon. Then, place a second pole GH, say 2 feet outside
of the first, and at a point I, 3 feet from the ground,
place the eye and prolong the sight by the side of the
first pole to the point B, on the tree and note at what
height the line of sight strikes the first pole. Then,
suppose in this example, that the line of sight touches
the pole at the height of 9 feet, we have the following
proportion of triangles:
As 2 feet is to 6 feet, so is 12 feet to the distance from
C to B, or, thus: If a triangle of 2 feet base gives 6
feet height, 1 ft. base will give # part, and a triangle of
12 ft. base, (the distance from C to I) will give 12 times
as many feet height.
STATEMENT.
6 = ft. high.
2 | 12
36 ft., Ans.
NOTE 1.-The 6 ft. height is found by subtracting 3 feet from 9; 9 being the pole mark of
the line of sight and 3 feet being the distance that the line C I is from the ground.
NoTE 2.—The difference in the diameter of the tree at the ground and at the point B, is not
considered in the problem. For practical work of this kind this element need not be considered.
This problem is of value to lumbermen, for by it they may determine the height of trees up
to a required diameter, or to a limb or crook before cutting the tree and in case the height is not
what is required, the tree need not be felled.
2. In the above problem, suppose the line of sight on the pole E F had
passed the figures or indicated a height of Say 10 ft. 5 in. ; What would have been
the height of the tree from the ground to B Ans. 474 feet.
STATEMENT.
Ft.
12 | 89 10 ft. 5 in. — 3 ft. = 7 ft. 5 in. = 7 a ft. = #ft.
2 12
443 ft. from C to B.
3 ft. = distance from C to the ground.
ºsmºm-
473 ft., Ans.

344 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
3. What is the height of a tree, from the ground to the top, when the pole E
F is placed 40 ft. from the tree, the pole H G is placed 14 feet beyond the pole
E F, and the line of sight strikes the pole E F at a height of 18 ft. 4% inches, and
the eye point being 3 feet above the ground? Ans. 4283 ft.
OPERATION. Ft. Ft.
18 ft. 4; in. — 3 ft. = 15 ft. 4; in. = 1844 in. = 334 in. 2 || 369 2 || 369
40 ft. -- 14 ft. = 414 ft. 12 Or 12 || 498
12 in. in 1 ft. 3 || 2 18
º 2 | 83
498 in. *ss= | flººmºsº
4253 ft.
3 ft.
4283 ft., Ans.
4. If in triangles like the above, the pole F E is 24 ft. from the tree,
the pole H G is 3 feet beyond pole F E, the eye point or line C I being 3 feet
from the ground, and the line of sight on pole FE crosses the figure 16 ft. 3 in.,
what is the length of the tree from C to Bº Ans. 1194 ft.
TO FIND THE LENGTH OF A PART OF A TREE BROKEN OFF AT
THE TOP WIEHEN ONE END OF THE BROKEN PART RESTS ON
THE STUMP AND THE OTEHER END IS ON TEIE GEROUND.
PROBLEM.
715. 1. A tree 100 feet high broke in a storm, and the top touched the
ground 40 feet from the foot of the tree. What was the length of the portion
broken off 3 Ans. 58 feet.

Explanation.—In all problems of this kind, we add the square of the base
to the square of the height of the tree, and divide the sum by twice the
height of the tree, the quotient will be the length broken off... Or, divide
the square of the base by twice the height of the tree, and to the quotient
add one-half the height of the tree.
OPERATION.
\ 402 = 1600 1600 + 10000 = 11600 -- 200 = 58 feet.
1002 = 10000
à
Y
LINEAR MEASURE. 345
TO FIND OR MEASURE THE DISTANCE BETWEEN TWO POINTS ONE
OF WHICH IS INACCESSIBLE.
A.
C
PROBLEMIS.
716. 1. What is the distance between A and B
measured from the point B without access to A *
Ans. 90 feet.
Explanation.—To determine this distance, we resort to the
principles of Geometry and proceed thus: First prolong an indef-
inite straight line drawn through the points A and B say to C.
Then from the point B measure at right angles to the line A B C,
any convenient distance, say 30 feet to the point D. Then project an
indefiniteline drawn from A through D making line A. D. Then from
point D and perpendicular to line A D project a line to the line A
C until it intersects the line A B C at the point, say O; then
measure the distance from B to O which, say is 10 feet. Then the
distance from A to B =#. Thus: 30 x 30 = 900 ft. 900 ft. – 10
= 90 feet.
This principle of Geometry may often be applied to advantage
by any one, to measure the distance across rivers or other inaccess-
ible places.
2. Considering the above diagram, what is the dis-
tance from A to B, when the distance from B to D is
50 feet and the distance from B to O is 6 ft. 4 in. :
Ans. 394++ ft.
3. What is the width of a river in paces and feet,
when the distance from B to D in a diagram like the
above, is 60 paces and the distance from B to O is
94 paces, supposing each pace to be 3 feet
Ans. 385% paces, 11574 feet.
TO FIND THE DISTANCE BETWEEN TWO INACCESSIBLE OBJECTS,
As Two TREES ON THE OPPOSITE SIDE OF A RIVER,
PROBLEMIS.
717. 1. What is the distance between A and B
in the diagram, which are located on the opposite side
of a river ? Ans. 700 feet.
Explanation.—In all problems of this kind, first find the dis-
tances AC and BC by the preceding proposition, which are, say,
500 and 600 feet. 2d. Divide these distances by 10 or 100, by point-
ing off one or two places and thus obtain proportional distances;
in this problem, we divide by 100 and thus obtain 6 and 5 as the
proportionals. 3rd. Lay off on the prolongation of AC and BC
their respective proportionals. 4th. Connect and measure ab,
which is the proportional for AB, and multiply the same by 100,
the product will equal the distance AB. In this problem, we
assume that ab is 7 feet, which multiplied by 100 equals 700 feet
the distance from A to B.

346 soul E's PHILosophic PRACTICAL MATHEMATICS. *
*
2. Standing on a plateau, I can see two mountain peaks to the westward. I
ascertain the distance to the one on the left to be 25 miles, and to the one on the
right to be 30 miles. Laying off the proportionals 25 feet and 30 feet, I find the
distance between them to be 40 feet. What is the distance between the peaks f
Ans. 40 miles.
MENSURATION OF SURFACES,
—-tº-
718. The unit of measure for surfaces is a square whose side is some fixed
length, such as an inch, foot, yard, rod, etc.
In the solution of problems, in mensuration of surfaces, much time may often
be saved by applying the following principle which has been demonstrated in
geometry:
That all similar surfaces or areas are to each other as the squares of their like
dimensions.
TO FIND THE AREA OF PARALLELOGRAMS, EITHER RECTANGLE,
REHOMBOLD, RHOMBUS, OR SQUARE.
PROBLEMS.
|| || Z 7
719. 1. What is the area of a rectangular garden 240 ft. long and 120 ft.
Wide 3 Ans. 28800 sq. ft.
OPERATION.
240 ft. long. Explanation.—In all squares or rectangular figures, we
120 ft. wide.
multiply the length by the width, in the same units of
*=º
28800 sq. ft., Ans. measure, and in the product we have the required area,
2. How many square yards are there in the surface of a wall 212 feet long
and 32 feet high 3 Ans. 753% sq. yards.
3. What is the area of a rectangular floor 20 feet long and 14 feet wide
Ans. 280 sq. feet.
* MENSURATION OF SURFACES. 347
f
4. How many Square feet in a right-angled triangle whose base is 12 feet, and
perpendicular height is 8 feet? Ans. 48 sq. ft.
OPERATION. Ea:planation.—In all problems of this kind,
12 ft. long. we multiply the length by the height which
8 ft. high gives the sq. ft. of a rectangle of equal
• Illgil. length and height. This result is then di-
*memºmºsºmamm- vided by 2, since a right-angled triangle is
96-1-2=48 SQ. ft., Ans. equal to but one-half of a rectangle of equal
length and height.
5. What is the area of a triangle whose base is 40 feet and altitude or per-
pendicular height 18 feet? Ans. 360 sq. feet.
TO FIND THE AREA OF A TRAPEZOID.
PROBLEMIS.
720. 1. How many square feet in the surface of a trapezoid whose parallel
sides are respectively 108 and 130, and the perpendicular distance between them
32 feet 7 Ans. 3808 sq. feet.
OPERATION. Explanation.—In all problems of
108 this kind, we multiply the sum of the
130 parallel sides by the perpendicular
238 × 32 = 7616 distance between them, and take # of
7616 -- 2 = 3808 sq. ft., Ans. the product.
2. What is the area of a trapezoid whose parallel sides are 30 and 40 feet
and whose perpendicular is 15 feet 3 Ans. 525 sq. feet.
TO FIND THE AREA OF A TRAPEZIUM.
PROBLEMIS.
721. 1. How many square feet in a trapezium whose diagonal is 42 feet,
and the perpendiculars 18 feet and 16 feet? Ans. 714 sq. feet.
OPERATION. Explanation.—In all
42 ft. length of diagonal. problems of this kind,
ë ; a 111. we multiply the diagonal
34 ft. Sum of the perpendiculars. by the sum of the per-
gº- pendiculars which in-
2)1428 product. tersect it from opposite
tº-º-º-º-e angles, and take # of the
714 sq. feet., Ans product for the area.
o • ? e
2. What is the area of a trapezium whose diagonal is 30 feet, and the
perpendiculars are respectively 7 feet 6 inches, and 5 feet 8 inches?
Ans. 1973 sq. feet.
NOTE.--To find the area of an irregular polygon, divide the figure into triangles and tra-
peziums, and find the area of each separately. The sum of these areas will be the area of the
whole polygon.

348 soule's PHILOSOPHIC PRACTICAL MATHEMATICS.
TO FIND TEIE AREA OF A TRIANGLE.
EXAMPLE.
722. What is the area of a triangle whose base is 40 feet and altitude
or perpendicular height 18 feet? * Ans. 360 sq. feet.
OPERATION. Explanation.—In all problems of this kind,
40 ft. length of base. we multiply the base by the altitude, and take
18 ft, altitude. one-half of the product, and the result will
2 ) 720 be the area.
360 sq. ft., Ans.
TO FIND THE AREA OF A SCALENE TRIANGLE, WHEN THE THREE
SIDES ARE GIVEN.
723. 1. How many square feet are there in a triangle whose sides are 30,
40 and 50 feet 1 Ans. 600 sq. feet.
OPERATION.
30 60 & 60 60
40 30 40 50
50 -> tº- º
-ºm 30 1st remainder. 20 26 remainder. 10 3d remainder.
2) 120
60
60 × 30 × 20 × 10 = 360000
V 360000 = 600 feet., Ans.
Ea:planation.—In all problems of this kind, we first add the three sides together and take
half their sum ; then, from this half sum we deduct each side separately; then multiply together
the half sum and each of the three remainders, and extract the square root of the product; the
result will be the required area.
2. Suppose the base of a triangle to be 200 feet and each of the other two
sides is 50 feet; what would be the area 3 AM.S. Nothing.
NOTE.-This problem has puzzled thousands of students. Does it mystify you ?
3. What is the area of an equilateral triangular figure each side of which is
40 feet, 7 Ans. 692.8-H ft.
OPERATION INDICATED. Explanation.—In all problems of this-kind,
square one side and multiply the product by
40 x 40 = 1600; .433. The .433 being the ratio between the
area of a square and the area of a triangle,
1600 × .433 = 692.8 + ft. = Area. the side of which is equal to one side of the
square. See Table on page 350.
To find the area of the face or base of different kinds of regular polygons,
multiply the Square of one side by the tabular area set opposite the polygon of the
same number of sides, as shown in the Table on page 350.
* MENSURATION OF SURFACES. 349
TO FIND THE AREA OF A CIRCLE, when THE DIAMETER OR
THE CIRCUMFERENCE IS GIVEN.
PROBLEMIS.
724. 1. What is the number of square yards in a circular piece of ground
which is 20 yards in diameter? Ans. 314.16 sq. yds.
OPERATION. Explanation.—In all prob-
lems º this kind, lº º:
e tiply the diameter by itse
20 yds. diameter = length. º§ shown in the operation,
20 yds. diameter = width. and then multiply this pro-
e-ºº-e duct by .7854, which has
400 = the sq. yds. in a square which is 20 yds., long º tºº.
and 20 yards wide. ratio between the area of
400 × .7854 = 314.1600 sq. yds., Ans. a square and the area of
a circle, the diameter of
which is equal to one side
of the square.
2. What is the area of a circle whose circumference is 157.08 feet?
Ans. 1963.5 sq. feet.
OPERATION INDICATED.
157.08 -- 3. 1416 = 50 ft. = diameter.
Then 50 × 50 = 2500; 2500 × .7854 = 1963.5 square feet area.
. Or, if it is preferred, the square of the circumference may be multiplied by
.0795S.
NOTE.-The decimal .7854 is + of 3.1416, and is the area of a circle whose diameter is 1; and
.07958 is + of (1 – 3.1416); or, .07958 is + of the diameter of a circle whose circumference is 1.
3. The diameter of a circular garden is 56.4188 + feet. How many square
feet does it contain 3 Ans. 2499.99 —H = practically 2500 sq. ft.
TO FIND THE AREA OF A CIRCULAR RING, OR THE AREA INCLUDED
BETWEEN THE CIRCUMFERENCES OF TWO CIRCLES
HAVING A COMMON CENTER.
PROBLEM.
725. What is the area included between two concentric circles, the diam-
eters of which are 15 and 8 feet? Ans. 126.4494 sq. feet.
OPERATION.
152 = 225 Explanation.—In all problems of
8* = 64 this kind, we multiply the difference
* =º between the squares of the diameters by
Difference, #, the decimal .7854, and the product will

be the area.
126.4494, Ans.
35o
SouLE's PHILOSOPHIC PRACTICAL MATHEMATICs. $ºr
726.
15 feet 3
PROBLEMI.
OPERATION.
20
15
300
.7854
235.6200 sq. feet, Ans.
TO FIND THE AREA OF AN ELLIPSE.
What is the area of an ellipse whose diameters or axes are 20 and
Ans. 235.62 sq. feet.
Explanation.—In all prob-
lems of this kind, we multiply
the two diameters together,
and their product by the deci-
mal .7854, and the result will
be the area.
For the operations to find the entire surface of cylindrical, triangular, octag-
onal, and other solids, see Mensuration of Solids.
TABLE.
727. The following Table gives the names of the ten regular polygons, and
shows their areas when the side of each is equal to 1. It also shows the length of
the radius of the inscribed circle.
No. of sides. Names. Areas. Radius of inscribed Circle.
3 Triangle, 0.4330127 0.2886751
4 Square, 1.0000000 0.5000000
5 Pentagon, 1.7204774 0.688.1910
6 Hexagon, 2.5980,762 0.8660254
7 Heptagon, 3.6339124 1.03826.17
8 Octagon, 4.8284271 1.2071068
9 Nonagon, 6.1818242 1.3737887
10 Decagon, 7.6942088 1.5388418
11 Undecagon, 9.3656404 1.2028437
12 Duodecagon, 11.1961524 1.8660254
To illustrate the use of this table, we present the following problems:
1. What is the area of an equilateral triangle, whose sides are 40 inches?
40 x 40 = 1600; 1600 × .4330127 = 692.8203200
OPERATION.
Ans. 692.82032 sq. in.
Explanation.—In all problems of this
kind, square one of the sides of the
polygon and multiply the product by
the tabular area number set opposite
the named polygon in the above table.
NOTE.-This operation is based upon the principle that the areas of similar polygons are to
cach other as the squares of their like sides.
*
2. What is the area of an octagonal polygon one side of which is 30 inches f
Ans. 4345.58439 sq. in.
Operation indicated. 30° x 4.8284271.
3. The side of a decagon is 20+ feet. What is its area 3
Ans. 3233.4912 -- sq. feet.
(204)” x 7.6942088. *
Operation indicated.

X- MENSURATION OF SURFACES. 35 I
TO FIND TEIE AREA OF AN IRREGULAR POLYGON OF FOUR OR
MORE SIDES.
728. 1. Divide the figure into triangles by diagonals connecting some one
angular point with each of the others. 2. Compute the area of each triangle. 3.
Add their respective areas and the sum will be the area of the polygon.
TO FIND THE CONVEX. SURFACE OF A PRISM OR A CYLINDER.
PROBLEMIS.
729. 1. What is the area of the convex surface of a
pentagonal prism whose altitude is 8 ft. and each side 4 ft. ?
Ans. 160 square feet.
sº
sº s
º - OPERATION.
|||{i : º º 4 ft. x 5 = 20 ft. perimeter, 20 ft. x 8 = 160 sq. ft.
i
i ††
º
d
ſ
Explanation.—In all problems of this kind, multiply the perimeter of
the base by the altitude.
|
2. What is the area of the convex surface of a trian-
ſº
•r
º
º
*-
| #|| gular prism whose altitude is 64 feet, and the sides of its base
i are 3, 5, and 7 feet respectively 3 Ans. 974 sq. ft.
| |
º | | | OPERATION.
| . |
##||
3 ft. -- 5 ft. -- 7 ft. = 15 ft. perimeter.
15 ft. × 64 ft. = 974 sq. ft. Ans.
*::
**esses
**-
--
ºs--~~~~
3. What is the area of the convex surface of a cylinder,
whose altitude is 4 feet 3 inches, and the circumference of its
| base is 5 ft. 6 in.
OPERATION.
4 ft. 3 in. = 51 in. ; 5 ft 6 in. = 66 in. ; 66 in. x 51 = 3366
sq. in. equals 23 sq. ft. 44 sq. in. convex surface.
. . . .
# , ; ; ;
NOTE.--To find the entire surface, add the area of the bases or ends to the convex surface.




























352 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. Yºr
TO FIND THE AREA OF A SEGMENT OF A CIRCLE WHEN THE
DIAMETER AND THE HEIGHT OR VIERSED SINE ARE GIVEN.
PROBLEM.
730. What is the area of a segment of a circle whose diameter is 72 inches
and the versed sine is 12 inches? Ans. 446,0313 + Sq. in., area.
OPERATION
BY THE VERSED SINE METHOD.
Versed sine 12 in. -- the diameter, 72 in. =
.1666; = quotient of versed sine.
Find the quotient of .166 in the column of versed sines, in the following
table; then take the area number which is in the next column to the right, and
multiply it by the square of the diameter, the result will be the area of the segment
A D B.
The tabular number is .08554 × 72* = 443,4393 + sq. in., area of segment.
As the quotient of the versed sine by the diameter was .166; instead of .106, the
area would be nearer accurate to use as a multiplier the number opposite .167 in
the table. This number is .08629; and this x 72° = 447,3273 + sq. in., area of
segment.
For greater accuracy we would proceed as follows:
.166 = .08554
.167 = .08629
Difference, .00075 × .63 = .00050 + .08554 = .08604, the sum by which the
square of the diameter is to be multiplied.
Then, .08604 × 72* = 446.0313 + sq. in., area of segment.
NOTE 1.--To find the area of a segment greater than a semi-circle, find the area of the lesser
segment and subtract the same from the area of the whole circle.
NOTE 2.--To find the area of a zone of a circle, find the areas of the two segments and
subtract the sum from the area of the whole circle.
NOTE 3.−To find the chord of a segment of a circle when the diameter and versed sine are
given, see Article 734, page 356, also Article 807, page 398.

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Yºr MENSURATION OF SURFACES. 355
TO FIND THE AREA OF A SEGMENT OF A CIRCLE WHEN THE
OHORD AND THE HEIGHT OR, VERSED SINE ARE GIVEN.
PROBLEMI.
732. What is the area of a segment, the chord of which is 9 ft. 3 in. and
the height or versed sine is 4 ft. ?
Ans. 4036.9828 + sq. in., area of the segment A. D. B. ,
OPERATION. C
ºt seºwº
A -11 | B A é. #. B
º :
D D
First Step to find the diameter of the circle of which the given segment is a part.
A B = 111 in. = whole chord. (55%)* -- 48 = 64.17 -- in. difference between
A P = 554 in. = # chord. (liameter and versed sine.
P D = 48 in. 64.17 – 48 = 112.17 in. diameter of circle.
112.17 - 2 = 56.085 in. radius A. C.
Second Step to find the quotient of the versed sine divided by the diameter.
48 in. versed sine -- the diameter, 112.17 = .4279, quotient.
The tabular number for .427 = .31996
{{ {{ 46 44 .428 = .32095
Difference is, .00099
Then, .00099 × .9, the difference between .4279 and .427, - .000891. Hence
.31996 -- .000891 = .320851, the sum by which the square of the diameter is to be
multiplied.
.320851 × 112.17% = 4036.9828 + sq. in., area of the segment A. D. B.
A SHORT APPROXIMATE METHOD TO FIND THE AREA OF A SEG-
MENT OF A CIRCLE, WITHOUT THE VERSED SINE TABLE.
PROBLEMI.
733. What is the area of a segment of a circle, the chord of the segment
being 24 feet and the height or versed sine of the segment 6 feet?
Ans. 1004 sq. feet. .
24 feet.
OPERATION. At B Explanation.—In all problems
of this kind, in which the seg-
4 x 24 × 6 = 576 ment is less than a semi-circle,
3 × 63 we proceed as follows: To 4
add, -ār- = 27 times the product of the chord
by the versed sine, add the quo-
tient of 3 times the cube of the
versed sine divided by the chord,
and then divide the sum by 6.
area, 100.5 sq. ft. 6 feet. Or, if preferred, we may perform
the operation as follows: To 4
divide by 6) 603
*=mºs,

356 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS. ºf
•e-
times the square of the chord, add 3 times the square of the versed sine; then multiply the sum
by the versed sine and divide the product by 6 times the chord. Thus:
24” x 4 = 2304
6° x 3 = 108 241.2 × 6 = 14472.
sºmº- 14472 -- (6 x 24) = 100%
2412 -
ADVANTAGE OF THIS METHOD,
The advantage of this method is that it can be used without the table of
versed sines. The result produced is slightly different from that produced by
the use of the versed sine tabular method as shown above, but for most practical
purposes, it may be used when the versed sine table is not at hand.
OPERATION OF THE ABOVE PROBLEM BY THE USE OF THE WERSED SINE
TABLE.
AT 24 feet. —B 24 ft. length of chord -- 2 = 12 ft., 4 of chord.
12* + 6 ft. versed sine = 24 ft., difference be-
tween the diameter and versed sine.
24 + 6 = 30 ft., diameter, 6 -- 30 = .2 quotient.
Tabular number for .2 is .111823.
D ..111823 x 30°= 100.6407 sq. ft., area of segment.

6 feet.
TO FIND THE CHORD OF A SEGMENT OF A CIRCLE WHEN THE
DLAMETER AND VERSED SINE ARE GIVEN.
PROBLEM.
734. What is the chord of a segment, of a circle, the diameter of which is
72 inches and the versed sine 12 inchest Ans. 53.664 in., length of chord.
OPERATION.
First Step.–Diameter 72 in. — versed sine 12 in. = 60 × versed sine, 12 in. = 720.
• Second Step.
V720 = 26.832 in. = # of chord.

4 2
4°) 320 = 53.664 in. chord of segment.
C 72" 276
528) 4400
4224
5363) 17600
A º D
SGL’ 16089
D 5366°) 151100
107324
Explanation.—In all operations of this kind, first subtract the versed sine from the diameter
and multiply the remainder by the versed sine; then extract the square root of the product and
multiply the quotient by 2; the result will be the length of the chord of the segment.
X- MENSURATION OF SURFACES, 357
TO FIND THE RADIUS OF THE INSCRIBED CIRCLE OF A REGULAR
POLYGON.
PROBLEMIS.
735. 1. What is the radius of an inscribed circle on a hexagon, whose
sides are each 50 inclies? Ans. 43.30127 radius of circle.
OPERATION. Explanation.—In all problems of this kind,
multiply one side of the polygon by the tabular
* f – radius number set opposite the named polygon
50 × .8660254 = 43.3012700 in the radius column in the table of areas, on
page 350.
2. The respective sides of a nonagon are 3 feet. What is the diameter of
the 1nscribed circle 3 Ans. 8.2424 + feet. Diameter of circle,
OPERATION.
3 × 1.3737387 = 4.1212161 = radius.
4.1212161 × 2 = 8.2424.322 diameter.
PRACTICA, PROBLEMS IN MENSURATION OF SURFACES.
PROBLEMS.
736. 1. How many square rods in a plat of ground, the length of which is
200 rods and the width 60% Ans. 12000 sq. rods.
2. How many square feet in a floor 16 feet 8 inches long, and 10 feet 3 inclies
wide # Ans. 1703 sq. feet.
OPERATION.
3 || 50 12 || 200
4 || 41 Or, 12 123
170% sq. feet., Ans. 1703 sq. feet., Ans.
3. How many square yards in a floor 60 feet long and 40 feet 10 inches wide #
Ans. 2723 sq. yards.
OPERATION. 60 × 40} = 2450 sq. ft.; 2450 -- 9 = 2723 sq. yds.
4. A water pipe is 50 feet 9 inches long, and its diameter is 30 inches. What
is its concave surface? Ans. 57397.032 sq. inches.
OPERATION.
30 x 3.1416 = 94.248 circumference, or linear width of pipe,
94.248 x 609 in., length = 57397.032 sq. in., Ans.
* 5. How many square yards of plastering in one wall of a house 32 feet 4
inches long and 15 feet 3 inches high 3 -- Ans. 54;s Sq. yds.

358 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. º:
6. How many square feet in the bottom of a cistern whose diameter is 11
feet 8 inches? Ans. 106.90.163 sq. ft.
7. What is the area of the base of a cylinder whose circumference is 62.832
inchest * Ans. 314.16 sq. in.
8. A lumberman has 25 boards, each 14 feet long and 15 inches wide. How
many square feet in all? Ans. 437.4 sq. ft.
9. Find the area in square feet of a plank 22 feet 3 inches long, 24 feet wide
at one end and 14 inches wide at the other ? Ans. 38.3 sq. ft.
10. Find the area of a piece of ground which is 406 feet long and is of the
following width at different points, equally distant from each other: at the wider
end, 210 feet; at the narrower end, 165 feet; near the wider end, 180 feet; near the
narrower end, 142 feet; in the middle, 300 feet? Ans. 821644 sq. ft.
NOTE.—Draw a diagram of the ground before working the problem.
OPERATION INDICATED
TO FIND THE AVERAGE WIDTH.
210 + 165 = 1874, + 180 + 142 + 300 = 2023 ft. average width.
*mºm,
2 4
11. A steamboat boiler is 45 feet long and 50 inches in diameter; it has 3
fluos, each 10 inches, in its exterior diameter; how many square inches does the
boiler contain on its entire interior surface # Ans. 88278.96 sq. in.
OPERATION
TO FIND THE CONCAVE SURFACE OF THE BOILE R.
50 × 3.1416 = 157.08 in., circumference.
540 in., length.
84823.20 sq. in., in concave surface.
3455.76 area of ends.
88278.96 sq. in., the entire interior surface of the boiler.
OPERATION OPERATION
TO FIND THE SURFACE OF THE 2 ENDS. TO FIND THE SURFACE OF THE 6 ENDS OF THE
3 FLUES.
50? – 2500
2500 × .7854 – 1963.50 103 = 100
2 100 × .7854 = 78.54
ºmsºmºsºmºsºms 6
Area of 2 ends, - - - - 3927.00 ºsmºmº
Area of the 6 ends of 3 flues, 471.24 || Area of the 6 ends, - - - 471.24
Area of 2 ends, - - - - 3455.76 &
12. If a man occupies a space 20 inches by 30 inches, how many men may
stand in a semicircular space which is 200 feet in diameter ? sº
Ans. 3769 men with 552 sq. in., unoccupied.
* * MENSURATION OF SURFACES. 359
13. A piece of board is 16 inches long, 9 inches wide. How can it be cut into
two pieces so that by placing them together in another form they will make a square
12 inches on each side 3

y— 4. Answer. 1. Commence
* | as shown in the first fig-
ure 4 inches from either
-4- 4. end, and saw down 3 in.
*: 2. Then at a right an-
4. 4 gle from this point, saw
3 3 toward the longer end
4. 4 inches. 3. Then at a
right angle from this
point, saw down 3 in-
ches. 4. At a right angle from this point saw 4 inches in the same direction as the
first 4 inches were sawed. 5. Lastly from this point and at right angles with the
last line, Saw 3 inches, through the board, and arrange the pieces as shown in the
second figure.
NOTE.-In order to saw the board as above described, it would be necessary to bore holes at
the four angles, where the changes of direction were made. These holes would be plugged when
the pieces were placed in the square form.
14. A circular piece of ground contains 10 acres; what is
º its diameter in rods, and feet?
Ans. 45.135 + diameter in rods. 744.7275 + feet.
\
OPERATION INDICATED.
10 × 160 = 1600 = the sq. rods in 10 acres; 1600 -i-.7854 = 2037.1785 =
the sq. rods in a square one side of which is equal to the diameter of the 10 acre
circle; VT2037.1785 = 45.135 + rods = one side of the square and the diameter of
the circle; 45.135 × 10% = 744,7275 + feet.

15. A gardener wishes to lay out a circular garden that
ſ will contain one acre of ground; how many feet long must be a
© rope that, fastened by one end to a stake in the center of the
\ garden, will enable him to describe the one acre ?
\ Ans. 117.75 + feet.
OPERATION INIDICATED.
1 acre contains 43560 sq. feet; 43560 -- .7854 = 55462.1848-|- ; v55462,1848-1-
=235.50 ft., diameter; 235.50+ -- 2 = 117.75 + ft. = radius or length of rope.
16. Cheops, the largest of the Egyptian pyramids, is square at its base and
measures 764 feet on each side; how many acres of ground does it cover?
Ans. 13 acres 63 sq. rās. 294 sq. yds. 1 sq. ft.
36o SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. jºr
OPERATION INDICATED.
764 x 764 = 583696 sq. ft.; 583696 sq. ft. -- 9 = 64855 sq. yds. and 1 sq. ft.;
64855 sq. yds. -- 30+ = 2143 sq. rods and 294 sq. yds.; 2143 sq. rods -- 160 = 13.A.
and 63 sq. rods.
17. A modern stock raiser has a cow that he wishes to feed upon 2500 sq. ft.
of pasture per day. If he ties the cow by the tail with a rope, and if the head and
body of the cow be 74 ft. long and the tail to the point where it is tied be 34 feet
long, what must be the length of the rope to give the cow the desired feet of past-
urage, no allowance being made for the curve in the line of measure, on account of
the height of the cow % Ans. 17.2094 + feet.
OPERATION INDICATED.
2500 + .7854 = 3183,0914-- ; V 3183,0914 = 56.4188 -- 2 = 28.2094 – (74 +
34) 11 = 17.2094.
ARTIFICERS OR MECHANICS' WORK, IN SURFACES,
-º-
PAVING YARDS AND WALKS.
737. 1. How many bricks will be required to pave a sidewalk 64 feet long
and 11 feet 8 inches wide, each brick being 8 inches long and 4 inches wide?
Ans. 3360 bricks.
OPERATION INDICATED.
Length of brick = 8 || 768 = length of sidewalk in inches.
Width “ “ = 4 || 140 = width of Sidewalk in inches.
| 3360 = number of bricks to pave the sidewalk.
Ea:planation.—In all problems of this kind, we first make the statement to ascertain the
number of square inches in the sidewalk, by multiplying the length by the width in the unit of
inches; and then we divide this result by the product of the length and the width of a brick.
2. In the above problem, how many square yards in the sidewalk, and what
would be the cost of the paving at 90% per square yard 3
Ans. 82% square yards
$74.66 -- cost.
OPERATION TO FIND SQUARE YARDS.
64
3 || 35
9
82## square yards at 90.2 = $74.663.
Y ARTIFICERs' OR MECHANICs’ work, IN SURFACES. 361
3. How many German flags, each 16 in. by 16 in., will it take to pave a
yard 45 feet square ? Ans. 11391's flags.
4. A yard is 24 feet 3 inches long by 11 feet 5 inches wide; how many bricks,
4 by 8 inches, will it take to pave it, no allowance to be made for the openings
between the bricks? Ans. 1245}} bricks.
5. A circular court is 75 feet in diameter; in the center there is a fountain
15 feet in diameter; how many tiles, each 6 inches square, will it require to tile the
court 7 Ans. 16964.64 tiles.
OPERATION.
75% = 5625 sq. of the diameter of court. Ea:planation.—We first find the dif-
152 = 2.25 4. ‘‘ ‘‘ 4% “ fountain. ference between the squares of the
diameters of the court and fountain;
we then multiply this difference by
5400 difference. .7854, and produce the number of
6 | .7854 square feet in the area of the ring; we
6 | 1.44 then reduce this to inches and divide
the same by the number of inches in a
tºº, wº. ººm-º. tile.
16964.64, Ans.
NotE.—In actual practice, by reason of the waste in making the circle, many more tiles
would be required than are produced by this operation.
6. A circular court is 30 feet in diameter. How many tiles, each 6 inches
square, will it require to cover the court, making no allowance for waste?
Ans. 2827.44 tiles.
7. What will it cost to pave a yard that is 140 ft. long, 34 feet 4 inches
wide, with shillinger pavement, at $3.25 per square yard 2 Ans. $1735.74%r.
8. A contract is made with a company to pave a street with rosetta pave-
ment and to curb the same with 4 x 18 inch granite. The price is fixed at $14 per
square for the pavement, and $1.10 per running foot for the curbing. What will be
the cost for paving one square or block which is 320 feet long, the street being 42
feet 6 inches wide 3 Ans. $1904 for paving. $704 for curbing.
9. How many sods, 12 by 16 inches will it require to sod a yard 40 ft. long
and 28 ft. 4 inches wide 3 Ans. 850 sods.
FRAMING, BUILDING, LAYING FLOORS, WAINSCOTING, ETC.
738. The work of carpenters and joiners is done by the lineal foot, the
Square foot, the square yard, or by the square of 100 square feet.
Framing the large timbers used in building, such as sills, posts, principal
rafters, etc., the framing of hips and valleys, and the work of chamfering posts and
girders, raising plates and bond timbers, bridging joists, making mouldings, archi-
traves, etc., are counted by the lineal foot. *
The charge for dressing posts and girders, making window shutters, doors,
wainscoting, shelving, ceilings, lattice-work, cornice, etc., is by the square foot;
sometimes some of this work is charged by the lineal foot.
362 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
;
The charge for framing joists, small rafters, studding, putting oil weather-
boarding, shingling, sheathing, laying floors, making partitions, board fences, etc., is
by the square yard or the square.
NOTE.-In measuring weather-boarding, all openings over 4 feet wide are to be deducted.
PROBLEMS.
1. A carpenter in repairing a house, framed 3 sills 8x6 inches, each 42 ft.
long, and 6 sills 8x8 inches, each 18 feet 6 inches long; what was the number of
lineal feet, and what was the cost at 7.2 per lineal foot ? Ans. $16.59.
OPERATION.
42 × 3 = 126
184 × 6 = 111 $23, feet at 7 g = $16.59.
2. A room is 22 feet 3 inches long, and 18 feet 6 inches wide; it has two
doors each 4 feet wide. What will be the cost to wainscot this room 3 feet 4 inches
high at 60% per square yard? Ans. $16.33%.
OPERATION.
224 × 2 = 44; 2 | 1.47 = distamee around room.
184 × 2 = 37 3 || 10 = height of wainscoting.
ºme smºs 9 = feet in Sq. ya.
81; sº
4 × 2 = 8 27# sq. yds. at 602. = $16.33%.
73%
3. What will be the cost to floor a room 34 feet 6 inches long and 22 ft. 4
inches wide, at $4 per square 3 Ans. $30.82.
4. What will it cost to make twelve 8 panel doors 84 ft. x 34 ft. 13 inches thick
at 25% per square foot of 1 inch thick? Ans. $127.50.
5. What will it cost to make 24 shutters each 22 inches wide 14 inches thick
and 74 feet long at 20g per square foot of 1 incli thick? Ans. $82.50.
6. How many squares in the floors of a house containing 8 rooms measuring
as follows: 2 rooms are 16 by 20 feet, 2 are 154 by 20 feet, 2 are 14 by 18 feet, and
2 are 12 feet 4 inches by 18 feet? Ans. 22#.
7. A sub-contractor puts up the frame work of a building at 90% per square.
The building is 85 feet long, 60 feet wide, and from the bottom of the sill to the top
of the rafter plate is 31 feet 8 inches. From the end of the rafters on one side,
measured over the peak or apex of the roof, to the end of the rafters on the oppo-
site side is 69 feet 2 inches. The studding in the partitions on the first story meas-
ure 270 feet long and 15 feet 6 inches high; and on the second story it is 310 feet
4 inches long and 14 feet 3 inches high. The sleepers in all the rooms on the lower
floor measure 84 feet 6 inches long by 59% feet wide; and the joists in all
the rooms on the second floor measure the same as the sleepers on the first floor.
The two gables, or the vertical triangular ends of the building from the eaves or
cornice to the top are 60 feet at the base, and as the rafters are + pitch, they have
an altitude of 15 feet. What is the cost of erecting the frame work 3
Ans. $311.37-H.
º: ARTIFICERS' OR MECHANICs' work, 1N surf Aces. 363
OPERATION INDICATED.
85 + 85 + 60 + 60 = 290 ft. the measurement around the building.
290 × 31# = 91834 sq. ft. in the 4 sides or walls.
85 × 69% = 5879% sq. ft. in the roof.
270 × 15% = 4.185 sq. ft. in the partitions on the first floor.
310# × 14+ = 44224 sq. ft. in the partitions on the second floor.
844 × 59% = 5013; sq. ft. in the sleepers on the first floor.
84 × 59% = 5013; sq. ft. in the joists on the second floor.
60 × 15 × 2 = 900 sq. ft. in the two gables.
2
34597-1's sq. ft. in the entire frame work.
34597-14 sq. ft. -- 100 = 345.97+); squares, at 90.2 = $311.373.
8. What is the cost to slate a roof 127 ft. 4 inches long and 56 feet wide,
at $11.50 per square, and what will be the weight of slate on the roof, allowing a
square of slate to weigh 600 pounds 3 Ans. $820.023.
42784 lbs. or, 21 tons and 784 lbs.
PLUMBERS’ WORK.
739. Plumbers’ work is done by the pound, hundred-weight, or by special
agreement.
The following Table shows the weight of a square foot of sheet lead, accord-
ing to its thickness; and also the common weight of a yard of leaden pipe, accord-
ing to the diameter of the bore :

Thickness of Lead. |Pounds to a square foot. || Bore of Lead Pipes. Pounds per yard.
Inch. Inch.
+% 5.899 0; 10
# 6.554 1 12
# 7.373 14 16
# 8.427 14 18
# 9.831 1; 21
# 11.797 2 24
1. What is the cost of 261 feet of leaden pipe, of 14 inch bore, at 11 cents
per pound, allowing as per the table that each yard weighs 18 pounds?
Ans. $172,26.
OPERATION INDICATED.
261 -- 3 = 87 yards. w 3 || 261
87 × 18 = 1566 lbs. Or, 18
1566 x 11 g = $172.26 11
| $172,26, Ans.
2. What is the weight of lead, ; of an inch thick, that covers a surface 32
feet long and 8 feet 6 inches wide, estimating the Weight at 73 pounds per square
foot ? º Ans. 2006 pounds.
364 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
SLATING AND SEIINGLING ROOFS.
740. 1. The roof of a building is 72 feet 6 inches long and measures 46
feet 9 inches from eave to eave. How many slates and how many squares of slating
are there in the roof, allowing a slate to cover a space 4; by 8 inches and not allow-
ing for the double course at the eaves? Ans. 13557; slates.
33}#} sq. of slating.
OPERATION
TO FIND THE NUMBER OF SLATES. Explanation.—In all problems of this
Width of slate = 8 || 870 in. = length of roof. kind. We first find the nºber of sºlºe
561 in. = width “ “ inches in the roof by multiplying togeth-
OOIL Ill. = W1 er the length and the width of the roof
Length of slate 9 || 2 in the unit of inches; and then divide
exposed on the – the same by the number of square inches
roof. 13557; slates, Ans. that each slate covers.
OPERATION
TO FIND THE NUMBER OF SQUARES OF SLATING. Explanation.—Here we first make the
12 || 870 statement to find the number of square
1. 561 feet and then divide by 100, which is the
number of square feet in a square, as per
33}#} Squares, Ans. Article 531, page 244.
2. How many shingles will it require to shingle a house that is 54 feet long
and 35 feet 10 inches from eave to eave, estimating that 5 inches of each shingle
will be laid to the weather, and allowing for the double course at the eaves on each
side 3 Ans. 14256 shingles.
NotE 1.-The unit width of a shingle is 4 inches.
NOTE 2.—Add 10 inches to the width for the double courses.
3. A flat roof is 208 feet 2 inches long and 28 feet 5 inches wide. How many
square yards of tin will be required to cover it, and what will be the cost at $1.15
per square yard 3 Ans. 657++3 sq. yds. $755.85-H cost.
4. A building is 64 feet long and 40 feet wide; the rafters have one-quarter
pitch, allowing a projection of 6 inches on each end of the roof, and 1 foot on each
side for the eaves, how many slates and how many squares of slating will it require
to slate the building, if the slates are 6 inches wide and 5 inches of their length
are exposed to the weather, no allowance to be made for the ridge of the roof or for
the double course at the eaves? Ans. 14577; slates.
30% ## Squares.
OPERATION

TO FIND THE LENGTH OF THE RAFTERS.
100
400
102
10 202
20 500
ARTIFICERS' AND MECHANICs' work, IN SURFACES. 365
V 500 = 22.36 + feet = length of rafter.
Practically, the length of rafter is ſº tºº 22 ft. 4; in.
Projection of rafter is es * * tº s sº 1 ft. 0 in.
Length of rafter, º as tº ſº is º 23 ft. 4; in.
2
Length of two opposite rafters, or width of roof, 46 ft. 83 in.
OPERATION OPERATION
TO FIND THE NUMBER OF SLATES. TO FIND THE NUMBER OF SQUARES.
64 ft. -- 6 in. -- 6 in. = 65 ft. = 780 in.
46 ft. 83 in. = 5603 inches. 65
* | 3 | 1682
780 12
3 | 1682 & 100
6 * * = * =
5 30+}# squares.
14577; slates, Ans.
NotE.—By one-quarter pitch is meant, in mechanical language, that the apex of the roof is
to be + the width of the building higher than the plane of the base of the rafters.
PLASTERERS’ WORK.
741. The work of plasterers is of two kinds, viz.: ceiling, which is plaster-
ing on laths, and rendering, which is plastering on Walls.
In measuring plasterers’ work, the square foot, square yard, or square is used.
The windows, doors, fire places, and base board are generally deducted. In
some states, one-half of all openings over two feet wide are deducted.
Plasterers have also adopted the following regulations regarding the measur-
ing of their work :
1. To measure all circular walls and ceilings, twice.
2. To charge extra for corners of chimneys and other external angles when
made with “gauged ” mortar.
3. To add one foot in length of the cornice for each mitre or “return.”
4. Stucco work, when more than 12 inches wide, is done by the square foot. -
PROBLEMIS.
1. How many square yards of plastering in the walls and ceiling of a room
344 feet long, 21 feet 2 inches wide, and 14 feet 10 inches high, deducting 6 windows,
7 by 4 feet, and 2 doors, 9 by 34 feet, and the base board 8 inches wide, and what
would it cost at 242 per sq. yard 1 Ans. 231; sq. yards.
• $55.50 practically.
366 sou LE's PHILOSOPHIC PRACTICAL MATHEMATICS. A.
OPERATION INDICATED.
34; + 34} + 21} + 21% = 111 ft. distance around the room.
14 ft. 10 in. height of room, - 8 in. width of base = 14 ft. 2 in. height
of room plastered.
1113 x 14% = 15773 sq. ft. in the walls of room.
34; x 21% = 730+ sq. ft. in the ceiling.
23074% sq. ft. in the walls and ceiling.
DEDUCTIONS.
9 ft. – 8 in. = 8 ft. length of doors, allowing for width of base board
not measured in the area of the Walls.
84 x 34 × 2 = 584 sq. ft. in the 2 doors.
7 × 4 × 6 = 168 sq. ft. in the 6 windows.
2263 sq. ft. deductions.
23073% – 2264 = 2081; sq. ft. 20813% + 9 = 231#, sq. yds. at 24 g =
$55.50, practically.
2. A room 20 feet 6 inches long, 17 feet 4 inches wide, has two chimneys
extending 14 inches from the wall; what would be the cost of a cornice 16 inches
wide around this room, allowing for-12 miters, 4 at the corners of the room and 8 at
the two chimneys, at 18% per square foot ? Ans. $22.16.
OPERATION.
20% + 20% + 17* + 17% = 75% ft. around the room, not counting chimneys.
14 × 4 = 56 inches = 43 ft. width of the chimney walls.
1 ft. allowed for each miter 12 ft.
*
92% = length of cornice.
924 × 1 = 1233 sq. feet, at 18% = $22.16.
3. What will it cost to plaster the walls and ceiling of a house of 10 rooms,
4 of which are 20x164 feet and 144 feet high; 3 are 18×16 feet and 13 feet high
and 3 are 164 × 12# feet and 13 feet high; allowance being made for 20 windows
74x34 feet, and 18 doors 94×34 feet, at 232 per sq. yard Ans. $275.42.
NOTE,--The measurements in this problem are exclusive of the base board.
4. What will it cost to plaster the walls and ceiling of 2 rooms at 374 cents
per square yard, each room 24 feet 3 inches long, 12 feet’9 inches wide, and 11 feet
6 inches high, deducting 5 windows, 6 by 3 feet 4 inches, and 2 doors, 8 feet 8 inches
by 3 feet 6 inches, in each room, and the base board 8 inches wide 3
Ans. $79,57+.
PAINTING AND GLAZING.
742. Painting is done either by the square yard, or the square, or by esti-
mate. Glazing is done by the light or pane.
Painters and glaziers have adopted the following rules for measuring their
Work:
1. In weather-boarding, measure not only the width of the board, but also
Y ARTIFICERS AND MECHANICs' work, IN SURFACEs. 367
the width of the exposed edge. For partition boards, when beaded, add to the width
of the boards ; inch for each bead.
*, 2. For paling or picket fences, measure the height, and add thereto the
width of each rail, which gives the width or height.
3. To fixid the length, add to the length of the fence twice the width of each
post, if rectangular, or one-half the circumference, if cylindrical.
4. In measuring the length and width of panel doors, shutters, etc., press
the measuring line into all the quirks and mouldings.
5. In measuring weather-boarding and wall painting, deduct half the open-
lingS. ºf
6. In measuring shingle or board roofs, measure the butts or edges; for
trellis or lattice work, double the measure, and for venetian shutters add one-half.
7. In measuring the width of pilasters, commence at the wall and measure
round to the bead at the end of the jamb casing; pressing the line into all the
quirks.
8. For plain rosettes in pilasters, add 1 foot to the length; for carved
rosettes, add 2 feet to the length.
9. In measuring window jambs, cornice, etc., press the line or tape into all
indentations and concave surfaces.
10. When the measurement is made by the square foot, it is customary to
include the whole sash, and circular or oval windows as square.
PROBLEMIS.
1. What will the glazing cost for a house containing 46 windows of 8 lights
each, and 7 windows of 18 lights, at 25g per light 3 Ans. $123.50
2. What will be the cost of the painting of 43 ft. high wainscoting of 4 inch
wide beaded boards, of a room 18 feet 6 inches by 13 feet 4 inches, at 242 per sq.
yard 7 Ans. $8,594.
OPERATION INDICATED
18} + 184 + 13% + 134 = 633 ft. measurement around the room.
633 = 764 in. -- 4 in., the width of a beaded board, = 191 beads, or half
inches, to add to the measurement around the room; 191 half inches
= 95% inches = 7# ft.
33 + 7# = 713 ft. length of wainscoting.
1; x 4 = 322# sq. ft. = 35+ sq. yds. at 242 = $8.59%.
3. What will it cost to paint both sides of a paling or picket fence 120 feet
long, 5 ft. high, the posts being 8 feet apart from center to center and 4× 6 in. With
the narrow side set to the fence; the rails being 2 × 4 inches with the wide Sidº
facing the fence, at 25% per sq. yard? Ans. $42.81% #.
OPERATION INDICATED.
5 ft. high -- 4 in. + 4 in. = 53 ft. = height of fence including
the width of the two rails.
16 posts 6 in. wide, 2 sides each = 1 ft. for each post, to be
added to the length of the fence.
120 + 16 = 136 ft. length of fence, one side.
136 × 2 = 272 ft. length of fence, both sides.
272 × 53 = 15414 sq. ft. 1541; -- 9 = 1714 sq. yds.
1713 ºr x 25% = $42,81% #.
368 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. ºr
4. How many panes of glass, 12 by 10 inches, in a box of 100 feet?
Ans. 120 panes.
OPERATION INIDICATED.
- Pane.
1 pane 10x12 = 120 inch. 1 sq. ft. = 144 sq in. 1. 12 || 144
144×100 = 14400 sq. in. in the box. 14400 or thus: 120 144 or, 10 || 100
-- 120 = 120 panes. 100
5. How many panes of glass in 44 boxes, each containing 50 sq. ft., the panes
being 22 × 24 inches, and what will be the cost at $7 per 100 sq. feet 3
Ans. 600 panes. $154.00 cost.
6. What will it cost to glaze 13 windows, 3 feet by 7 feet 4 in., and 22 win-
dows, 3 feet by 63 feet, at 25 cents per square foot ? Ans. $178.75.
CARDETING FLOORS.
PROBLEMS.
743. 1. How many yards of carpeting that is 27 inches wide, will be
required to cover the floor of a parlor that is 32 feet 4 inches long and 25 feet 6
inches wide, making no allowance for waste in matching or turning under ?
Ans. 122# yds.
FIRST OPERATION.
36 || 388 Explanation.—In this solution, we first find the number
27 || 306 of square inches in the floor, by multiplying together the
length and width in the unit of inches. We then divide by
the product of the length and width of a yard of the carpet,
122# yds., Ans. which is the number of square inches in one yard of it.
SECOND OPERATION INDICATED.
12 || 388 3 || 97 Explanation.—In this solution,
12 || 306 2 || 51 we first find the number of square
*
gards in the floor, which would be
9 Or, 9 the number of yards required if
27 | 36 27 36 the carpeting was 1 yard, or 36
- - inches wide. But since the carpet-
–4– -4- ing is not 1 yard wide, we multiply
122# yds., Ans. 122# yds., Ans. by 36, which gives the number of
* yards required if the carpeting was
but 1 inch wide, and we then divide by 27, the width of the carpeting, and produce the correct
result.
We reason as follows: Since it requires this expressed number of yards when it is 36 inches
wide, if it were but 1 inch wide, it would require 36 times as imany yards, and if 27 inches wide, the
27th part.
NOTE.-In this connection, we will remark that although the result of this problem is math-
ematically correct, it is not correct in a practical sense; i. e., in practice it would require more
yards of carpeting than the mathematical measurement of the room gives. This is occasioned by
the matching of the flowers or figures of the carpet, and by the unavoidable waste by reason of
the room being a little wider or longer than an even number of widths of the carpeting, which
necessitates the use of another width of carpet.
The increased number of yards over the mathematical result, by the foregoing causes, will
depend upon the size of the figure in the carpet, and whether the carpet is laid lengthwise or
crosswise of the room; and the flower or figure of the carpet, and the entrance to the room,
whether from the side or end, often determines which way the carpet shall be laid. . These are
important facts for parties who buy carpets on their own measurement of rooms, and if observed,
will often save expense and much vexation.
The two following problems elucidate these points and make clear the practical manner of
working problems of this kind :
Yºr ARTIFICERS AND MECHANICs' work, IN SURFACEs. 369
2. How many yards of carpet, 27 inches wide, will be required to cover a floor
21 feet 10 inches long and 18 feet 6 inches wide, laid lengthwise, and allowing 6
inches on each strip for matching? Ans, 67 yards.
OPERATION INDICATED.
18 ft. 6 in. = 222 inches. 222 inches -- 27 = 8*, practically, 9 widths of
Carpet, #} of one width will be waste, either cut off or folded under.
21 ft. 10 in. + 6 in., allowed for matching, - 22 ft. 4 in. = length of each
strip. 22 ft. 4 in. x 9 = 201 ft. -- 3 = 67 yds.
3. If in the above problem, the carpet had been laid crosswise with the same
allowance, how many yards would have been required ? Ans. 63.4 yards.
OPERATION INDICATED.
21 ft. 10 in. = 262 inches. 262 in. -- 27 = 94%, practically, 10 widths of
carpet. •
18 ft. 6 in. + 6 in. allowed for matching, - 19 ft. = length of each strip.
19 ft. x 10 = 190 + 3 = 634 yds.
4. How many yards of carpet 36 inches wide, and what will be the cost at
$1.35 per yard laid, to cover a floor 44 ft. 7 inches long, 31 ft. 6 in. wide, laid cross-
wise of the room, allowing 4 inches for matching Ans. 1594 yds.
$214.87; cost.
5. A parlor is 29 feet long by 22 feet wide. It is carpeted with 36 inch
carpet at $1.75 a yard, surrounded by a carpet border 18 inches wide, at $1.25 per
linear yard. If the border is laid without being mitered and the carpet is laid
crosswise of the room, and an allowance of 4 inches on each strip is made for
matching the carpet, what will be the total cost? Ans. $141.50.
OPERATION INDICATED.
29 ft. – 3 = 26 ft. -- 3 = 83, practically, 9 widths. 22 ft. – 3 = 19 ft. --
4 in. (matching) = 194 ft. length of strip. 9 × 19 = 174 ft. -- 3
= 58 yds. carpet.
58 × $1} = $101.50 cost of carpet. -
Then, 29 × 2 = 58 ft. length of carpet strips on two sides of room. 22 — 3
= 19 × 2 = 38 ft. length of carpet strips on two ends of room. 58 + 38 = 96 ft.
+ 3 = 32 yds. length of carpet strips.
32 × $1.25 = $40.00 + $101.50 = $141.50, Ans.
NoTE.—When the border is mitered, there will be as many yards of border as there are
yards distance in the four sides of the room.
37O SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
• MENSURATION OF SOLIDS,
744. The mensuration of solids is divided into two parts, viz.: 1st, the
mensuration of their solidities; and, 2d, the mensuration of their surfaces.
The Unit of Measure for solids is a cube whose side is some fixed length,
such as an inch, foot, yard, rod, etc.
The Area of a figure is the measure of its surface.
The Solidity or Contents of a solid is the number of cubes that it is equal
to, or contains; or its capacity expressed in cubic measure of any fixed unit.
In the solution of problems in mensuration of Solids, much time may often
be saved by applying the following principles which have been demonstrated in
geometry:
That all similar solids are to each other as the cubes of their like dimensions.
DEFINITIONS.
745. A Rectangular or
Quadrilateral Solid is a
solid which has length;
| width, and thickness, and
iſ is bounded by six sides or
| | | - Jaces.
747. A Cube is a Solid
whose sides or faces are
all equal Squares.
749. A Cylinder is a
solid having two faces or
bases, which are equal
parallel circles, and which
has an equal diameter in
any parallel plane between
them.
751. A Frustum of
a cone is the part which
# remains after the top is
cut off by a plane parallel
to the base.
746. A Pyramid is a
Solid whose base is any
kind of a polygon, and its
other faces, triangles unit-
ed at a common point
called the verteac.
748. A Frustum of a
Pyramid is the part
- which remains after the
# top is cut off by a plane
parallel to the base.
750. A Cone is a solid
having one face or base
which is a circle, and a
COnvex or curved surface
terminating in a point,
called the vertea.
752. A Sphere is a
solid bounded by a curved
surface, all the points of
which are equally distant
in called the center.
The Radius of a sphere is a line drawn from the center to any part of the circumference.
The axis, or diameter of a sphere is a line passing through the center and terminated by the cir-
cumference.
753. A Prolate Sphe-
roid is a solid, elongated
revolution of an ellipse
about its longer axis.
in the direction of a line
joining the poles; or, it is :
a solid generated by the W
754. An Oblate Sphe-
roid is a solid flattened
\ or compressed at the
A poles; or, it is a Solid gen-
7 erated by the revolution
of an ellipse about its
shorter axis.






X} MENSURATION OF SOLIDS, 37 I
TO FIND THE SOLIDITY OF A PRISM, PARALLELOPIPEDON, OR CUBE.
PROBLEMIS.
755. 1. A rectangular box is 4 feet long, 3 feet 4 inches wide, and 2 feet
high. How many solid, or cubic feet does it contain? Ans. 263 cu. ft.
OPERATION.
4 4 º ...
#= bical or rectangular solids
ſº * * 3 !” Or, 12 gº we multiply together thé
| | j length, width, and height,
ºff== — — gº tºº-º-º-º: in the same units of meas-
| | | ": 26; Cll. ft., 263 Cll, ft., ure, and in the product we
Ans. Ans. have the required solidity.
2. What are the solid contents of a prism, the sides of whose base are 3 and
4 feet, and whose altitude is 6 feet? • Ans. 72 cu. ft.
TO FIND THE SURFACE OF A PRISM, PARALLELOPIPEDON, OR OUBE.
PROBLEMIS.
756. 1. How many square feet of surface does the box in Problem No. 1
above, contain Ans. 56 sq. ft.
OPERATION. £
4 × 3 × 2 = 263 sq. ft. on two sides.
4 × 2 × 2 = 16 sq. ft. on the other two sides.
3% x 2 × 2 = 18% sq. ft. on the ends.
56 sq. ft. in the entire surface.
2. What is the surface of a rectangular prism, the sides of whose base are
3 and 4 feet, and whose altitude is 6 feet? Ans. 108 sq. feet.
TO FIND THE SOLIDITY OF A SQUARE OR RECTANGULAR PYRAMID.
PROBLEMI.
757. A square or rectangular pyramid is 8 feet high and each side of
the base is 4 feet 6 inches. How many cubic feet does it contain? Ans. 54 cu. ft.
OPERATION.
8 8 Fº * º:
2 4 of this kind, we multiply the
2 º r # # height by the area of the base,
† O * and then divide by 3, because a
ſº º º 3 3 8quare pyramid has been demon-
º
º f g º b
ź. |% * | *- gº- sº-º strated in geemetry to be # of a
4 º % 54 cu. ft., 54 cu. ft., rectangular solid of equal height
ſº Ans Ans and area of base.
NOTE.--To find the solidity or volume of a regular pyramid that is not square or rectangular,
multiply the square of one side by the tabular area set opposite the polygon of the same number
of sides as shown in the Table on page 350, which will give the area of the base. Then proceed as
in the above problem, by multiplying by the height and dividing by 3.





















372 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
TO FIND THE SURFACE OF A SQUARE OR RECTANGULAR PYRAMID.
PROBLEM.
758. How many square feet of surface in a rectangular pyramid whose
slant height is 8 feet and whose base is 44 feet? Ans. 924 sq. ft.
OPERATION.
4 × 8 × 4 = 144. 144 - 2 = 72 sq. ft. area of four sides.
4 × 44 = 20+ sq. ft. area of base.
72 + 20+ = 924 sq. ft. area, or surface of pyramid.
TO FIND THE SOLIDITY OF HEXAGONAL OR OTHER REGULAR
PYRAMIDS NOT SQUARE OR RECTANGULAR.
PROBLEMI.
759. What are the solid contents of a hexagonal pyramid whose altitude is
11 feet, and each side of whose base is 2 feet? Ans. 38.105.1176 cu. ft.
OPERATION
2* = 4 feet = square of one side. Explanation.—In all problems of
this character, we multiply the
4 × 2.5980762 = 10.3923048 sq. ft., area of base. square of one side by the tabular
* 11 altitude. area set opposite the polygon of
**msºmºmºmº the same number of sides, as shown
3) 114.3153528 in the Table, page 350, and this
*=====s**E=== product is multiplied by the alti-
38.105.1176 solid feet, Ans. tude and the result is divided by 3.
TO FIND THE ENTIRE SURFACE OF AN ocTAGONAL OR OTHER
REGULAR PYRAMID, NOT SQUARE OR RECTANGULAR.
PROBLEMIS.
760. 1. What is the entire surface of an octagonal pyramid whose slant
height is 11 feet and each side of whose base is 2 feet?
Ans. 107.3136 sq. ft. surface.
OPERATION.
2 × 8 = 16 ft., perimeter of the base.
16 × 5% (one half of slant height) = 88 sq. ft., convex surface.
2* = 4 ft., square of one side. 4 × 4,8284 = 19.3136 sq. ft., area of base.
88 + 19.3136 = 107.3136 sq. ft., entire surface.
Explanation.—In all problems of this character, we multiply the perimeter of the base by
half the slant height; the result is the convex surface. Then to find the area of the base, multiply
the square of the side by 4.8284, which is the tabular area set opposite the polygon of the same
number of sides, as shown in the Table on page 350.
2. What is the surface of a regular octagonal pyramid, whose slant height
is 12 feet, and each side of the base 5 feet 3 Ans. 360.7106775 sq. ft.
* - MENSURATION OF SOLIDS. 373
TO FIND THE SOLIDITY OF THE FRUSTUM OF A SQUARE PYRAMID.
PROBLEM.
761. A frustum of a square or four-sided pyramid is 8 feet high, lower base
7 feet, and upper base 6 feet. How many solid feet does it contain
Ans. 338; cu. ft.
OPERATION.
7* = 49 = area of the greater base.
6° = 36 = “ “ “ lesser {{
* Azº. 7 × 6 = 42 = geometrical mean proportional between the two bases.
º 3) 127
º *mºmºmºs
* 42} = average area of the frustum of the pyramid.
8 = height.
3383 cu. feet, Ans.
Explanation.—In all problems of this kind, to find the solidity of the frustum of a pyramid,
we add together the areas of the two bases, and a geometrical mean proportional between them,
then divide the sum by 3 and multiply the result by the altitude.
NOTE.—To find the solidity or volume of the frustum of a pyramid that is not square or
rectangular, we multiply the square of one side of each base, by the tabular area set opposite the
polygon of the same number of sides as shown in the Table on page 350, which will give the area
of the two bases. Then, we multiply the two bases together and extract the square root of the
product; then add the quotient and the areas of the two bases together and divide the sum by 3;
the result is the average area of the frustum of a pyramid which multiplied by the height, gives
the solidity of the frustum. i
TO FIND THE SURFACE OF THE FRUSTUM OF A SQUARE OR REC–
TANGULAR PYRAMID.
PROBLEMI.
762. How many square feet in the entire surface of a rectangular pyramid
whose slant height is 8 ft., lower base 7 ft., and upper base 6 feet?
Ans. 293 sq. ft. surface.
OPERATION.
7 x 4 = 28 ft., perimeter of lower base, 6 × 4 = 24 ft., perimeter of
upper base.
28 + 24 = 52 ft., sum of perimeters of bases. 52 x 8 = 416 - 2 = 208
ft., convex surface. º
7* = 49 sq. ft. area of lower base. 6* = 36 sq. ft. area of upper base.
208 + 49 + 36 = 293 sq. ft. entire surface.
Explanation.—In all problems of this kind, we first find the convex surface of the frustum by
multiplying the sum of the perimeters of the two bases by the slant height of the frustum and
dividing by 2.
Then, to find the entire surface of the frustum, add to the convex surface the areas of the
two bases.
NOTE.—When the frustum of the pyramid is not four sided, the area of the two bases is
found by multiplying the square of each side by the tabular area set opposite the polygon of the
same number of sides in the Table on page 350.


374 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. X-
TO FIND THE SOLIDITY OF PRISMS OR SOLIDS OF DIFFERENT SIDES.
PROBLEMIS.
763. 1. How many cubic inches in an octagonal prism or solid, one side of
whose base is 20 inches and whose length is 60 inches º Ans. 115882.2504 cu. in.
- OPERATION. Explanation.—In all problems of this
kind, first square one of the sides of
the polygon. 2. Multiply the prod-
202 = 400; 400 x 4.8284271 - uct by the tabular number set oppo-
1931.37084 square inches area of base. site the polygon of the same number
1931.37084 × 60 = 115882,2504 cu. in. of sides in the Table, page 350. 3.
Mºly the last product by the
ength.
2. The side of a pentagonal prism is 2 feet, its length is 30 feet. How many
cubic feet does it contain Ż Ans. 206,457288 cu. ft.
( ). PERATION INDICATED. 23 × 1.7204774 × 30.
TO FIND TEIE SOLID CONTENTS OF A CYLINDER.
PROBLEMIS.
764. 1. A cylinder is 8 feet high and 4 feet 6 inches in diameter. How
many cubic feet does it contain 3 Ans. 127.2348 cu. ft.
OPERATION.
===s 2 9 12 || 54 Explanation.—In all
fºr. 2 || 9 Or. 12 || 54 problems of this kind,
| 8 7 8 we multiply the square
| | pº of the diameter, by the
| #| || — .7854 .7854 height, which gives the
|# #. tºmº. solidity of a rectangular
§º 127.2348 cu. ft., 127.2348 cu. ft., solid, whose height and
Ans. Ans. width of sides are equal
to the height and diame-
ter of the cylinder. Then we multiply by .7854, the ratio between the area of a square and that
of a circle, whose diameter is equal to one side of the square.
NOTE.—The solidity of a cylinder is equal to the solidity of a square prism, whose sides and
altitude are the same as the diameter and altitude of the cylinder, multiplied by the decimal .7854.
2. What are the solid contents of a cylinder, whose altitude is 6 feet, and
diameter 4 feet? Ans. 75.3984 cu ft.
TO FIND TEIE SURFACE OF A CYLINDER.
PROBLEMIS.
765. 1. In the cylinder in Probem No. 1 above, how many square feet are
contained on its entire surface 3 Ans. 144.9063 sq. ft.
OPERATION. -
44 × 3.1416 = 14.1372 ft. circumference. Ea:planation.—In all prob-
14.1372 × 8 = 113.0976 sq. ft. area in convex surface. lems of this kind, we multi-
442 × .7854 = 15.90434 sq. ft. in lower base. ply the circumference by the
4}* x .7854 = 15.90434 sq. ft. in upper base. height, and to the product,
-ºº ºmis ºmºmºmº add the area of the two
144.9063 sq. ft. in surface of cylinder. ends or bases.

Yºr MENSURATION OF SOLIDS. 375
2. What is the surface of a cylinder 6 feet long and 4 feet in diameter ?
Ans. 100.5312 sq. feet.
TO FIND THE SOLID CONTENTS OF A CONE. º
PROBLEMIS.
766. 1. A cone is 8 feet high, and its base is 4 feet 6 inches in diameter.
How many solid feet does it contain 3 Ans. 42.4116 cu. ft.
OPERATION.
8 8 Explanation.—In all
2 9 12 || 54 problems of this kind,
2 | 9 OI’ 12 || 54 H. make the i. *
7 * 10n as 1n a cylinder an
| .7854 .7854 then divide by 3, because
W 3 3 the 8olidity of a come is #
-- tº-º tº- of #: whose º:
}\ | g t t
#º º 42.4116 cu. ft., 42.4116 cu. ft, ºr “”
Ans. Ans.
2. What is the solidity of a cone, the diameter of whose base is 5 feet, and
altitude 12 feet 3 Ans. 78.54 cubic feet.
TO FIND THE SURFACE OF A CONE.
PROBLEMIS.
767. 1. How many square feet on the entire surface of a cone whose Slant
height is 8 feet, and whose base is 4 ft. 6 inches in diameter ?
Ans. 71.4531 sq. ft. Surface.
OPERATION.
44 × 3.1416 = 14.1372 circumference of the base.
14.1372 x 8 = 113.0976 - 2 = 56.5488 sq. ft. convex surface.
4;” x .7854 = 14.9043 sq. ft. area of base.
56.5488 –H 14.9043 = 71.4531 entire surface.
Ea:planation.—In all problems of this kind, we multiply the circumference by the slant height,
and divide the product by 2; then, to the quotient thus obtained, add the area of the base, and the result
will be the entire surface. -
2. What is the surface of a cone, the diameter of whose base is 5 feet, and
slant height 12 feet? Ans. 113.883 sq. ft. Surface.
TO FIND THE SOLID CONTENTS OF A FRUSTUM OF A CONE.
PROBLEMI.
768. A frustum of a cone is 8 feet high, lower base 7 feet in diameter, and
upper base 6 feet in diameter. How many Solid feet does it contain 3
Ans. 265.9888 cu. ft.





376 soul E's PHILOSOPHIC PRACTICAL MATHEMATICs.
FIRST OPERATION.
square of greater diameter.
* = ** “ lesser & 4 3 || 127
= geometrical mean between the two .7854
diameters. 8
| 265.9888
42} = area of the base of a rectangular cu. ft.,
Solid, the sides of whose bases are Ans
equal to the average diameter of
the frustum of the cone.
Explanation.—In all problems of this kind, we first find, as is, shown in the operation, the
area of a rectangular solid, the sides of whose bases are equal to the average diameter of the
frustum of the cone; then we multiply this by .7854, to reduce the area of a square to the area of
a circle, and then multiply by the height.
NOTE.—In the above solution, we found the geometrical mean between the diameters of the
two bases by simply multiplying the two diameters together. This is by far the shortest and
easiest method of obtaining the geometrical mean between two numbers, and it is correct because,
by demonstration, it is proved that the product of any two numbers is equal to the square root of the
product of the 8quares of those two numbers.
The usual method of finding the geometrical mean between the diameters is
as follows: Square each diameter, multiply their products together, extract the
square root of the product, the result is the geometrical mean between the diameters
of the two bases.
OPERATION.
Thus: 72 = 49; 6* = 36; 49 × 36 = 1764.
V1764 = 42 the geometrical mean, the same as obtained in the solution
above, by simply multiplying the two diameters together.
NoTE.—The last method would be used when the areas of the bases and not the diameters
are given.
SECOND OPERATION BY THE PRISMOIDAL FORMUL.A.
7 × 7 = 49 = area of the lower base.
6 × 6 = 36 = area of the upper base.
2) 13 13
169 = 4 times the area of the middle Section
*E*==º between the two bases.
6)254
42} = average area of the frustum of a square
pyramid of the same dimensions as those
of this frustum of a cone.
42; x .7854 = 33.2486 = average area of the frustum of the cone.
33.2486 × 8 = 265.9888 = solid contents or cubic feet.
gº x 5 = 424 x 4
NOTE.-The above solution is in accordance with the prismoidal formula, by which the solid-
ity of cubes, rectangular solids, cones, cylinders, pyramids, frustums of cones or pyramids and
Several other forms of solids may be determined. *
The prismoidal formula is as follows: Add together the areas of the two ends or bases and
four times the middle section parallel to them. Then divide this sum by 6, and multiply the quo-
tient by the height, or depth.

* MENSURATION OF SOLIDS. 377
TO FIND THE ENTIRE SURFACE OF THE FRUSTUM OF A CONE.
PROBLEM.
769. In the above problem, what is the entire surface of the frustum?
Ans. 230.1222 sq. ft.
OPERATION
7 x 3.1416 = 21.9912 = circumference of greater base. Ea:planation.—In all prob-
6 × 3.1416 = 18.8496 = circumference of lesser base. lems of this kind, multiply
*mºsºsºme the bases by 3.1416 and the
40.8408 × 8 = 326.7264 sum of their products by the
326.7264 - 2 = 163.3632 sq. ft. convex surface. height; then divide this prod-
7° x .7854 = 38.4846 sq. ft. of larger base. uct by 2 which gives sq. ft.
6° x .7854 = 28.2744 sq. ft. of lesser base. in convex surface; to this
$ºmmºn e result, add the areas of the
230.1222 sq. ft. in entire surface. two bases.
TO FIND THE SOLIDITY OF A SPHERE.
PROBLEMIS.
770. 1. A sphere is 4 feet in diameter. How many cubic feet does it
contain 3 Ans. 33.5104 cu. ft.
º- OPERATION.
4°, or 4 × 4 × 4 = 64 cu. ft. in a cube which is 4 feet on each side.
64 × .5236 = 33.5104 cu. ft., Ans.
Explanation.—In all problems of this kind, we cube the diameter by multiplying it by itself
3 times, as shown in the operation, and then multiply this result by .5236, which is the ratio
between the solidity of a cube and that of a sphere, whose diameter is equal to one side of the
cube.
NOTE.--To find the side of a cube which may be cut from a given sphere, square the diam-
eter, divide by 3, and extract the square root of the quotient.
2. What are the solid contents of a sphere whose diameter is 12 feet?
Ans. 904.7808 cubic feet.
TO FIND THE SURFACE OF A SPHERE.
PROBLEMIS.
771. 1. In Problem No. 1 above, what is the surface of the sphere?
Ans. 50.2656 sq. ft.
OPERATION.
4 × 3.1416 = 12.5664 circumference. Explanation.—In all problems of this kind, wo
& ) ºf emºs PK48 & & multiply the circumference by the diameter,
12.5664 x 4 = 50.2656 sq. ft. surface. and the result is the surface.
2. What is the surface of a sphere whose diameter is 12 feet 3
& Ans. 452.3904 Sq. feet.

378 soule's PHILOSOPHIC PRACTICAL MATHEMATICS. A.
sº
t TO FIND THE SOLIDITY OF A HEMISPHERE.
PROBLEM.
772. How many cubic inches in a hemisphere whose diameter is 60 inches
and radius 30 inches 2 Ans. 56548.8 cu. in.
OPERATION INDICATED.
60
§ Explanation.—In all problems of this kind,
5236 multiply the square of the diameter by the
&-º-º-º-º-º: radius, and multiply the product by .5236.
TO FIND THE SOLIDITY OF A SEGMENT OF A SPHERE.
PROBLEMS,
773. 1. How many cubic inches in a spherical Segment which has a diam-
eter of 60 inches and a depth of 20 inches? Ans. 32463.2 cu. in.
OPERATION.
60 -- 2 = 30 in. radius.
69" 302 × 3 = 2700.
2700 + 20° = 3100 = three times the square of the radius
and the square of the depth.
3100 × 20 × .5236 = 32463.2 the cu. in. in the spherical
Segment.
§
Jºrplanation.—In all problems of this kind, to three times the square of the radius, we add
the square of the depth or height; then multiply this sum by the depth and the product by .5236.
NoTE.—When the segment is greater than a hemisphere, find the solidity of the lesser seg-
ment and subtract the same from the solidity of the entire sphere.
2. What is the solidity of a spherical segment, the height of which is 6 feet
and the diameter of the base 20 feet 7 Ans. 1055.5776 cu. ft.
NOTE 1.--To find the surface of a segment of a sphere, multiply the height by the circumfer-
ence of the segment, and add the product to the area of the base.
NOTE 2.—To find the solidity of a spherical zone, or the frustum of a sphere, add to the sum of
the squares of the radii of the two ends or bases, # of the square of the height of the zone or frus-
tum ; then multiply this sum by the height and this product by 1.5708.

* MENSURATION OF SOLIDS. 379
TO FIND THE SOLIDITY OF A PROLATE SPEIEROID,
PROBLEMI.
774. A prolate spheroid has a transverse, or longer diameter of 8 feet, and
a conjugate, or shorter diameter of 5 feet. How many cubic feet does it contain 3
Ans. 104.72 cu. ft.
OPERATION.
5 = diameter. 8
5 = diameter. Or, 5
.7854 = ratio of circle, etc. 5
8 = height or transverse diameter. .5236
3 || 2 = ratio between C. & P. S. *
| 104.72 cu. ft., Ans.
º º
| º 104.72 cu. ft., Ans.
Ea:planation.—In the first statement, we indicate the solution for a cylinder of equal height
and diameter as the prolate spheroid, and then multiply by #, because a prolate spheroid is equal,
to # of a cylinder of equal height and diameter.
In the second statement, we indicate the solution for a rectangular solid, by multiplying
together the three dimensions, and then multiply by .5236, which is the ratio between the solidity
of a cube and that of a sphere, the diameter of which is equal to one side of the cube.
TO FIND THE SURFACE OF A PEROLATE SPEIEROID.
PROBLEMI.
775. In the above problem, what is the surface? Ans. 104.7849 sq. ft.
OPERATION.
8* = 64. 5* = 25. 64 + 25 = 89 -- 2 = 444; V44} = 6.6708 × 3.1416
x 5 = 104.7849264 sq. ft.
Explanation.—In all problems of this kind, square the diameters, and multiply the square
root of half their sum by 3.1416, and this product by the conjugate or shorter diameter. The result
will be the surface.
TO FINID THE SOLIDITY OF AN OBLATE SPEIEROID.
PROBLEMI.
776. An oblate spheroid has a height or shorter diameter of 5 feet, and a
width or longer diameter of 8 feet. How many solid feet does it contain 3
Ans. 167.552 cu. ft.
OPERATION.
8 = diameter. 8
8 = diameter. Or, 8
.7854 = ratio of circle, etc. .5236.
5 = height or conjugate diameter. 5
3 || 2 = ratio between C. & O. S.
167552 cu. ft., Ans.
Explanation.—In the first statement, we indicate the solution for a cylinder of equal height










38o soulE's PHILosophic PRACTICAL MATHEMATICs. X\
and diameter as the oblate spheroid, and then multiply by #, since an oblate spheroid is equal to #
of a cylinder of equal height and diameter. --
For an explanation of the second statement, see the explanation in the problem under
Article 774.
TO FIND THE SURFACE OF AN OBLATE SPEIEROID.
PROBLEM.
777. In the above problem, how many square feet on the surface?
Ans. 167.6558+ sq. ft.
OPERATION.
5* = 25. 8* = 64. 25 + 64 = 89 + 2 = 44%. V 44; = 6.6708 × 3.1416
× 8 = 167.65588224 sq. ft.
Ea:planation.—In all problems of this kind, square the diameters and multiply the square
root of half the sum by 3.1416, and this product by the transverse or longer diameter. The result
will be the surface.
TO FIND THE SOLIDITY OF A SEMI-PROLATE SPEIEROID.
_-
PROBLEM.
778. How many cubic inches in one-half of a prolate spheroid, the conjugate
diameter A B, of which is 60 inches, and the semi-transverse diameter C D, being
40 inches f Ans. 75398.4 cu. in.
OPERATION INDICATED.
3. Or thus: 60° x 40 × .5236.
4
º NoTE.—See Article 774, page 379.
'D'
401?
TO FIND THE SOLIDITY OF A SEMI OBLATE SPHEROID.
PROBLEM.
779. How many cubic inches in one-half of an oblate spheroid, whose longer
diameter A B is 80 inches, and whose semi-conjugate diameter C D is 20 inéhes?
Ans. 67020.8 cu. in.

jºr MENSURATION OF SOLIDS. 381
OPERATION INDICATED.
Or thus: 80° x 20 × .5236.
r
|:
20
.5236 NOTE.-See Article 776, page 379.
20',
TO FIND THE SOLIDITY OF A SEGMENT OF A PROLATE SPHEROID.
PROBLEMI.
780. In a prolate spheroid, having a transverse diameter A B of 96 inches,
and a conjugate diameter C D of 60 inches, the height of the segment E B F is 20
inches. What is the solidity of the segment? Ans. 20289.5 cu. in.
96//
A OPERATION.
96 x 3 = 288; 20 × 2 = 40 ; 288 — 40 = 248; 248 x 20°
× .5236 = 51941.12 inches.
Then 962 : 60% : : 51941.12 : 20289.5.
Ea:planation.—In all problems of this kind, multiply the transverse
diameter by 3; then from this product, subtract two times the height
E F of the segment; then multiply the remainder by the square of the
height of the segment and this product by .5236; then multiply this
result by the square of the conjugate diameter and divide the product
E by the square of the transverse diameter.
2011
C D 60”
TO FIND THE SOLIDITY OF A SEGMENT OF AN OBLATE SPEIEROID.
PROBLEM.
781. In an oblate spheroid the transverse diameter A B is 96 inches, the
conjugate diameter C D is 60 inches, and the height of the segment E D F is 10
inches. What is the solidity or capacity of the Segment 7 Ans. 8770.3 cu. in.
60 t OPERATION.
96 × 3 = 288; 10 × 2 = 20; 288 – 20 = 268; 268
× 102 × .5236 = 14032.48.
Then 96 : 60 : : 14032.48 : 8770.3,
Earplanation.—In all problems of this kind, multiply the
transverse diameter by 3, and from the product subtract two
times the height of the segment; then multiply the remain-
der by the square of the height of the segment and this
product by .5236; then multiply this result by the conjugate
diameter and divide the product by the transverse diameter.
B 961,

382 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. jºr
REMARKS ON THE MEASUREMENT OF SEGMENTs.
The measurement of segments is attended with much difficulty in the prac-
tical affairs of life. As shown in Article 773, page 378, the measurement of a seg-
ment of a circle is a very simple operation. But, as shown in Article 780, page 381,
and Article 781, page 381, the measurements of the segments of prolate and oblate
spheroids are very difficult operations, and require as elements, in their solutions,
the transverse and the conjugate diameters of the spheroids. Hence the difficulty
of the operation in practice, where the diameter and the depth of the segment only
are known. The operations to measure the segments of spheres, of prolate, and of
oblate spheroids are, as shown in the above problems, very different, and unless we
know what kind of a segment we have to measure, we cannot perform the operation
correctly. And this knowledge cannot, in all cases, be determined from the diam-
eter and the depth of the segment. To illustrate this point, we will suppose that it
is required to find the capacity of a segment in the form of a sugar kettle, or a
boiler, whose diameter is 70 inches and depth is 25 inches.

2–GTS Before we can proceed with accuracy, we must
A g- }* X
—-1 70" know whether the segment is that of a sphere, or of
a prolate or of an oblate spheroid.
T) If we proceed to measure it as a segment of a
25'ſ sphere, the result of our operation will be greater
than if we had measured it as a segment of a prolate spheroid, and less than if
we had measured it as a segment of an oblate spheroid.
To have measured it as a segment of a prolate or of an oblate spheroid, we
must have known the transverse and conjugate diameters of the figure of which
the segment is a part. And this knowledge We could not have obtained from the
given diameter and depth of the segment.
From the above, we see the difficulty in practice of measuring segments of
spherical solids. The best that can be done in cases where the figure is not known,
and where the dimensions given are limited to the diameter and the depth of the
segment, is to measure it as a segment of a sphere, and then, if it is possible, to
judge from the form and dimensions to what class of segments it belongs; to allow
a small deduction in case it is a segment of a prolate spheroid, and to allow a small
increase in case it is a segment of an oblate spheroid.
NotE.—In many cases where the capacity of segments of an unknown class is required, the
most expeditious and accurate method of determining the matter, would be to discard mathema-
tics, and to measure the vessel by a gallon measure. sº
* MENSURATION OF solids. 383
TO FIND THE SOLIDITY OF THE MIDDLE FRUSTUM OF A PROLATE
SPEIEROID.
PROBLEMI.
782. The middle frustum of a prolate spheroid-i o, is 70 inches in length, 60
inches e d, conjugate diameter, and 50 inches at its ends, ef and g h. What is its
Solidity or capacity? Ans. 177762.2 cu. in.
OPERATION. k º all problems of this
ind, first square the conjugate diame-
60° × 2 = 7200 + 50° = 9700. Her and multiply the product by 2;
Fº then, to this product, add the square of
Then 9700 × 70 × .2618 = the diameter of one end; then multi-
177762.2 cubic inches. ply this sum by the length of the frus-
tum, and the product by .2618.
TO FIND THE SOLIDITY OF THE MIDIDLE FRUSTUM OR ZONE OF AN
ELLIPTIC SPINDLE.
--- -- PROBLEM.
783. In the frustum of an elliptic spindle, the length e f is 70 inches, the
greatest diameter a b is 60 inches, the least diameter 50 inches, and the middle
diameter gh, (equally distant from the middle and end) is 56 inches. What is the
solidity or capacity ? Ans. 170834.972 cu. in.
OPERATION. Explanation.—In all problems of this
* kind, add together the squares of the
2 2 — 2 3 c Q.
60° -- 50° = 6100 + (56 × 2) greatest and least diameters, and the
= 18644. square of double the medium diameter;
Then 18644 × 70 × .1309 = then multiply the sum by the length
170834,972. and the product by .1309.
TO GAUGE OR FIND THE SOLIDITY OF CASKS OR BARRELS.
PROBLEMI.
784. In a cask 70 inches long, 60 inches conjugate or bung diameter, and 50
inches head diameter, how many cubic inches 3 Ans. 1724.11.008 cu. in.
OPERATION.
60 — 50 = 10, difference be- Eacplanation.—In all problems of this
tween bung and head diame- kind, add to the lesser diameter .5.6
terS. y e
or .7 of the difference between the
10 × .6 = 6. -- 50 = 56 av-
erage diameter of cask.
Then 562 × .7854 × 70 = the staves may be slight, medium, or
17241.1.008 cubic inches. above medium.
NotE.—In the three preceding problems, the cask, the middle frustum of a prolate spheroid,
and the middle frustum of an elliptic spindle are very similar prisms, and to show the different
results by the different methods of work we gave the same dimensions to each prism, with the
additional diameter to the frustum of the elliptical spindle. The results are as follows:
1. By the method of measuring middle frustums of prolate spheroids 177762.2.
2. By the method of measuring middle frustum of elliptical spindle 170834.972.
3. By the method of measuring casks and barrels 17241.1.008 cubic inches.
diameters, according as the curve in



384 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs,
TO FIND THE SOLIDITY OF A CYLINDRICAL RING.
PROBLEM.
785. What is the solidity of a cylindrical ring that is 3 inches thick and
has an inner diameter of 15 inches? Ans. 399.7188 cubic inches.
OPERATION. Explanation.—In all problems of this
* kind, first add to the thickness of the ring
3 + 15 = 18 × 3° = 162 × the inner diameter, and then multiply
2.4674 = 399.7188 cubic inches. this sum by the square of the thickness,
==== and the product by 2.4674.
NoTE.—To find the convex surface of a cylindrical ring : First add to the thickness of the ring
the inner diameter, then multiply this sum by the thickness, and the product by 9.8696.
TO FIND THE CUBICAL CONTENTS OF A WEDGE.
PROBLEMIS.
786. 1. A wedge is 20 inches long, 6 inches wide and 4 inches thick. How
many cubic inches does it contain 3 Ans. 240 cubic inches.
FIRST OPERATION.
6 + (6 × 2) = 18; 18 × 20 x 4 = 1440; 1440 - 6 = 240 cu. in.
Explanation.—In all problems of this kind, add to the length of the edge or point, twice the
length of the back or head; then multiply this amount by the length and the thickness of the
wedge and divide the product by 6, the quotient will be the cubical contents.
NotE.—When a wedge is a true prism as the above, the solidity is equal to the product of its
three dimensions, divided by 2.
2. In the above problem, what would have been the number of cubic inches,
if the edge of the wedge had been 10 inches long Ans. 293; cubic inches.
-- OPERATION.
10 + (6 × 2) = 22 × 20 × 4 = 1760 - 6 = 2933 cubic inches.
MEASUREMENT OF LOGS, TIMBER, LUMBER, AND BOARDS.
787. There are various kinds of scales, and rules, and tables, that lumber
dealers have and often use in the measurement of lumber, by means of which very
little calculation is required. But as these scales and rules are readily understood
on inspection, and as the results obtained by their use are in many cases only
approximative, we therefore omit a description of them, and present accurate meth-
ods by means of figures.
In the measurement of round timber, different customs and methods prevail
in different localities, which we will point out in the problems presented a few pages
further on.
A Standard Saw Log is 12 feet long and 24 inches in diameter.

ROUND TIMBER. 385
ROUND TIMBER,
-º-
TO FIND THE SOLIDITY OR OUBIC FEET OF ROUND TIMBER OR LOGS.
PROBLEMS.
788. 1. The length of a log is 30 feet, and the average girt or circumfer-
ence is 7 feet; how many cubic feet does it contain? Ans. 117.6 cubic feet.
FIRST OPERATION.
7 Explanation.r-In all problems of this character, we
7 multiply the square of the average girt or circumfer-
30 ence by the length, and the product thus obtained by
8, and then divide by 100 or point off two figures in
100 || 8 this last product, and we have the solidity very
ºs- nearly. This method of work º: and º:
3. º than the usual methods, and while the result is a
1173 cubic feet, Ans. little in excess of the mathematical result based on
e º the presumption that the log, or tree, is a regular
º it is nearer the exact measure of the frustum of a cone than most of the usual methods
Of WOrk.
li º following statement gives the correct mathematical result considering the log as a
cylinder:
(7 x 7 x 7854 x 30) + (3.1416 x 3.1416) = 116.9786 + cu. ft.
2. What are the contents by the two above methods, of a log 62 feet 5
inches long, that girts 10 feet 9 inches at the larger end, and 3 feet 7 inches at the
Smaller end? Ans. 256.463 cu. ft. by the first method.
255.108 cu. ft. by the second method.
OPERATION BY THE FIRST METHOD,
Ft. In, 12 || 86
10 9 girt of larger end. 12 || 86
3 7 girt of smaller end. 12 || 749
*-*- 100 || 8
2) 14 4 -
& 256.463-H cu. ft., Ans.
7 2 approximate average girt or
circumference.
OPERATION BY THE SECOND METHOD,
12 || 86
3.1416
12 || 86
3.1416 | .7854
12 || 749
255.108-- cubic feet, Ans
3. A log is 50 feet long, 40 inches diameter at the larger end and 20 inches
diameter at the smaller end. How many cubic feet does it contain, measured by
386 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS Yºr
the first above method, and how many when measured accurately as the frustum of
a cone # Ans. 246.741 cu. ft. by first method.
254.527 cu. ft. accurate measurement.
FIRST METHOD.
40 x 3.1416 = 125.6640 circumference of larger end.
20 × 3.1416 = 62.8320 circumference of smaller end.
125.664 + 62.832 = 188.496 - 2 = 94.248 approximate average cir-
cumference or girt.
Then (94.248° x 50 x 8) + (144 x 100) = 246.741+ cubic feet.
OPERATION FOR, ACCURATE MEASUREMENT AS A FRUSTUM OF A CONE.
402 = 1600
202 = 400 × 2800 + 3 = 933% accurate area of a square one side of
40 × 20 = 800 which is equal to the average diameter of the log.
Then (9334 × 50 × .7854) -- 144 = 254,527+ cubic feet.
NoTE.—This is the only method that is mathematically correct; but it is not adopted by
lumber dealers when buying or selling logs. They make allowance for crooks, splits and other
defects, and hence adopt approximate methods. Logs are more generally bought by board meas-
ure, and the different methods adopted for board measure will be presented under the heading of
Lumber and Board Measure, two pages following.
TO REDUCE ROUND TIMBER TO SQUARE TIMBER.
PROBLEM.
789. How many solid feet of square timber are there in a stick of round
timber 28 feet long, 34 inches in diameter at the larger end, and 21 at the smaller
end ? Ans. 65+ cubic feet.
OPERATION. *
34 in., diameter of larger end. 3 || 55
21 4 { % “ Smaller “ ” 12 || 55
tº- 3 || 28
2) 55 12
*-
3) 27# in., approximate average or mean diameter.
9% = # deducted.
º-ºº-
18% in., average of each side of the square stick.
Explanation.—In working problems of this character, different methods are used by different
lumber dealers. In this solution we allowed # of the average diameter for the squaring of the
stick, and thus obtained 18% inches as the average measure of one side of the square stick. Then
having a square stick (18+ inches by 18% inches), we multiply the three dimensions together in the
unit of feet and obtain cubic feet.
This method of work is generally used, but where the timber is large, and especially where
there is but little taper to the stick, the deduction of # is found to be too great, hence some
dealers deduct but 4 of the average diameter instead of #.
To work the foregoing problem by deducting #, we produce the following
figures and result:
65;}} cu. ft. Ans.
4) 27# in., average diameter as above. | 8 | 165
63 in., - # deduction. 8 || 165
4- 12
203 in., average of each side of the square stick. 12 || 28
82}#} cu. ft., Ans.
4): ROUND TIMBER. 387
TO FIND THE CUBICAL OR SOLID CONTENTS OF SQUARED OR
FOUR-SIDED TIMEER.
PROBLEMI.
790. A piece of timber is 34 feet long, 16 inches wide, and 15 inches
thick; how many solid feet does it contain 3 Ans. 563 cubic feet.
OPERATION.
34 Explanation.—In all problems of this character, we
12 | 16 multiply the length, width and thickness together in
12 || 15 the same unit of measure, and the result is the cubical
contents.
563 cubic feet, Ans.
TO FIND THE CUBICAL CONTENTS OF SQUARED OR FOUR-SIDED
TIMBER THAT TAPERS REGULARLY.
PROBLEM.
791. The larger end of a rectangular stick of timber is 25 inches wide by
20 inches thick; the smaller 16 inches wide by 12 inches thick, and the length is 30
feet; how many cubic feet does it contain? Ans. 694; cubic feet.
FIRST OPERATION.
width. Thick-
* IleSS.
Larger end 25 × 20 = 500 sq. in. in larger end. 12 || 334
Smaller “ 16 × 12 = 192 sq. in. in smaller end. 12
*mmº tº 30
41 × 32 = 1312 = 4 times the middle sec- —
ºsmºmsºme tion between the ends.
6) 2004
69 is cubic feet, Ans.
334 sq. in., average area of the stick.
Ea:planation.—In all problems of this character, where the taper of the adjacent sides is not
in the same ratio, we first multiply together the width and thickness of the two ends, and then
add together the widths of the two ends and also the thickness, and multiply together their sums.
We then add the three products together and divide the sum by 6, which gives the average area
of the stick in square inches, which we reduce to the unit of feet, by dividing by 12 × 12, and
multiply by the length, and obtain cubic feet.
NOTE.—When the timber does not taper regularly, measure separately the parts that do,
and take the sum for the entire solidity.
SECOND OPERATION BY THE PRISMOIDAL FORMULA, ELUCIDATED ON PAGE 376.
25 × 20 = 500 sq. in. in larger end.
16 × 12 = 192 sq. in. in Smaller end.
2) 4.1 × 32
204 × 16 = 328 x 4 = 1312 = 4 times the middle section between the 2
-*- al’é2,S.
6 ) 2004
tº-º-
334 = Sq. in. average area of the stick.
Then 334 x 30 - 144 = 69 is cubic feet, Ans.
388 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
TO FIND THE LENGTH THAT MUST BE CUT OFF OF A SQUARE
PIECE OF TIMBER WHICH DOES NOT TAPER, TO OBTAIN
\ A GIVEN SOLIDITY. -
PROBLEMS.
792. 1. A piece of timber is 8 inches wide and 6 inches thick, how much
must be cut off to make 1 cubic foot ? Ans. 36 inches.
OPERATION.
1 cubic foot = 1728 cubic in. 8 || 1728
8 × 6 = 48 square inches. Or, 6
1728 -- 48 = 36 in., Ans. gº
36 in., Ans.
2. A piece of timber is 24 inches wide and 16 inches thick; what length is
required to make 4 cubic feet? Ans. 18 inches.
OPERATION.
24 || 4
16 || 1728
| 18 inches., Ans.
3. A stick of timber is 4 inches thick and 16 inches wide; how much must
be cut off to make 6 cubic feet 3 Ans. 134 feet.
©
º
LUMBER AND BOARD MEMSURE,
-Ö-
793. Lumber, as the term is here used, includes boards, planks, joists,
scantlings, and sawed timber of all kinds.
794. A Standard Board is one that is 12 feet long, 12 inches wide, and 1
inch thick. Hence 1 Board Foot is 12 inches long, 12 inches wide, and 1 inch
thick, or 1 foot long, 1 foot wide, and 1 inch thick, and contains 144 square or board
inches.
NOTE.- Since 1 board foot contains but 144 board inches, there are 12 times as many board
feet as cubic feet in lumber, timber, and logs. Hence to change board feet to cubic feet, divide by
12; and to change cubic feet to board feet, multiply by 12.
795. A Standard Saw Log is 12 feet long and 24 inches in diameter.
ROUND TIMEER, OR LOGS REDUCED TO BOARD MEASURE.
796. There are various methods of ascertaining approximately the number
Yºr LUMBER AND BOARD MEASURE. 389
of feet of square edged inch boards that can be cut from a given log; different
methods are adopted by different lumber dealers, which give approximate results
sufficiently accurate to answer practical purposes. *
The actual number of feet varies according as the logs vary in not being per-
fectly straight cylinders, and according to the manner of sawing them. If the log
is first sawed into inch boards, and each board is then separately edged, it will give
several feet more than if the log is first slabbed, and then sawed into inch boards.
PROBLEM.
797. How many feet can be sawed from a log 20 feet long, and 18 inches in
diameter at the smaller end ? Ans. 245 feet.
OPERATION BY THE DOYLE METHOD, OR, RULE.
18 in., diameter. d †† th. method, 4 inches are
e Or educted from the diameter in order to
4 in., allowed to square the log. reduce the log to a square. Then the square
º º º of the remainder will be the board feet of a
14 in., each side of the Square stick. log 16 feet long. Then since the yield of
14 logs is in the ratio of their lengths, we
20 therefore divide by 16 to obtain the board
feet in a log 1 foot long, and multiply by 20
to obtain the board feet in a log 20 feet long.
16
smºs-
245 sq. feet, Ans.
GENERAL DIRECTIONS FOR DOYLE'S RULE.
798. From the diameter of the smaller end deduct 4 inches; then square
the remainder and multiply the product by the length of the log given, and divide
this product by 16.
The above method is the basis of the table given in Scribner's Lumber and
Log Book, which is probably used more generally than all other books of this kind.
Though this method, from a mathematical point of view, is only approximate, it is
practical in its simplicity and is regarded with favor by lumber dealers.
This method favors the buyer in the case of small logs, and the seller in the
case of large logs. For logs 14 inches in diameter, this method gives the same
result as squaring the log and allowing # for saw kerf. For logs 35 inches in
diameter, it gives # of the solid contents; and for logs 50 inches in diameter this
method is equivalent to allowing nothing for slab and less than % for saw kerf.
In the combination of figures, to produce this method, it is estimated that
the saw cuts # of an inch each time it goes through the stick, and as the standard
board is 1 inch thick, it is clear that 4 of the stick is cut away by the saw, and
accordingly an allowance is made therefor.
The Doyle method is adopted by the New Orleans lumber dealers, with the
following regulations regarding t
CLASSIFICATION, INSPECTION AND DEDUCTION.
799. Logs are classified as cypress, pine and oak.
39C) SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. A.
DEDUCTIONS.
For the hollow butts of cypress, deduct the length of the hollow. For the red
heart of pine, and for all pecky logs, deduct 10 feet from the length of all such logs.
For crooks, knots and wind shakes, deductions are made as per agreement.
OTHER METEIODS.
800. Some lumber dealers adopt the following method: Subtract from
the diameter of the smaller end eacpressed in inches, # or 4 of itself, according as the
logs are small or large, and multiply the square of the remainder by the length in feet,
and divide the product by 12.
PROBLEMI.
How many board feet in a log 22 feet long, and 28 inches in diameter at the
smaller end, working first by the Doyle method, second by deducting #, and third
by deducting # of the diameter?
Ans. By the Doyle method, 792 board feet;
by deducting #, 638;# board feet;
by deducting #, 8084 board feet.
OPERATION. OPERATION. OPERATION.
By the Doyle method. When # of the diameter is When + of the diameter is
deducted. deducted.
28 28 28
4 9% 7
24 183 21
24 - 21
16 || 22 3 || 56 12 22
tº-º-º-e 3 || 56 +-------->
792 ft., Ans. 12 22 808; ft., Ans.
| 638% ft., Ans.
TO FIND THE BOARD OR SQUARE FEET IN PLANKS, GIRDERS,
SCANTLINGS, JOISTS, AND SQUARE TIMBER.
PROBLEMS.
801. 1. How many square feet of lumber in a board 16 feet 4 inches long
and 15 inches wide 3 Ans. 2013 Sq. ft.
OPERATION. * tº
3 49 12 196 Explanation.—In all problems of
4 || 5 or, 12 || 15 this kind, we multiply the length
º tº-º-º-º: and width together in the unit of
Practical Solution.—Since the board is 15 inches wide, it contains 14 square
feet for each foot of length. Hence 14 times 16# = 20+ square feet. Ans.
X} LUMBER AND BOARD MEASURE. 391
2. A board is 20 feet 6 inches long, 21 inches wide at one end and 15 inches
at the other end. How many square feet does it contain } Ans. 30% sq. ft.
OPERATION."
21 in., wider end. 2 || 41
15 in., narrower end. 12 | 18
2) 36 | 30% sq. ft., Ans.
18 in., average or mean width.
Practical Solution.—Since the average width of the board is 18 in. wide, it
contains 14 sq. ft. for each foot of length. Hence 14 times 20% = 30% sq. ft. Ans.
3. How many square feet in a plank 24 feet long, 22% inches wide, and 3
inches thick? Ans. 135 sq. ft.
OPERATION.
24 Explanation.—As the board foot is 1 inch in thickness,
2 || 45 it is clear that when the thickness exceeds 1 inch the
12 || 3 o measurement must be increased accordingly ; hence when-
ever the thickness exceeds 1 inch, we multiply by the
*º thickness. When the thickness is less than 1 inch, by
135 sq. ft., Ans. custom, no deduction is made, the measurement being in
that case the same as if the lumber was 1 inch thick.
Practical Solution.—Since the board is 224 in. wide, and 3 in. thick, it is
practically 67.4 sq. in. or 53+ sq. ft. for each foot of length. Hence 5}} times 24 =
135 sq. ft. Ans.
4. What is the number of board feet in 16 pieces of scantling each 20 feet
long, 4 inches wide, and 3 inches thick? Ans. 320 board feet.
Practical Solution.—4 in. wide and 3 in. thick, is equivalent to 12 in. wide or
1 sq. foot for each foot of scantling; and as there are 16 scantlings, hence 20
times 16 is 320 sq. ft. Ans.
5. How many board feet in 200 girders each 30 feet long, 15 inches wide,
and 2 inches thick. And what will they cost at $18 per M. ?
Ans. 15000 board feet; $270 cost.
Practical Solution.—15 in. wide, by 2 in. thick = 24 sq. ft. to the foot. Hence
24 times 30 = 75 × 200 = 15000 sq. ft.; and $18 × 15m = $270. Ans.
6. What will be the cost of 4 black walnut boards, each 10 feet 9 inches
long, 28% inches wide, and 1% inches thick, at $55 per M.3 Ans. $9.829##.
7. A piece of timber is 34 feet long, 16 inches wide, and 15 inches thick.
How many board feet does it contain 3 Ans. 680 sq. ft.
Practical Solution.— 16 in. = 1 ft. wide by 15 in. thick = 20 sq. ft. for each
foot of length. Hence 20 times 34 = 680 sq. ft. Ans.
8. A rectangular telegraph pole is 60 feet long, 16 inches square at the
larger end and 6 inches square at the Smaller end. How many cubic and how
many board feet does it contain } Ans. 53; cubic feet; 6463 board feet.
NOTE.-See Article 761, page 373.
392 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. *
9. A circular telegraph pole is 60 feet long, 16 inches in diameter at the
larger end, and 6 inches in diameter at the smaller end. How many cubic and how
many board feet does it contain, and what is the cost at 15% per cubic foot ?
Ans. 42.32434 cubic feet; 507.892 board feet; $6,34865 cost.
NotE.—See Article 768, page 375.
10. How many superficial or board feet in 40 scantlings, each 154 feet long, 4
inches wide, and 2 inches thick? Ans. 4134 feet.
SOLUTION STATEMENT.
1. 3. Practical Solution.—4 × 2 = 8 in. = 3 sq. ft. 154 ×
* | * 40 = 620 × # = 4134 sq. ft.
40 or thus:
- 153 x 40 = 620 – 4 of itself = (206§) 4134 sq. ft., Ans.
4135 sq. ft., Ans.
11. How many square feet in 62 joists, each 20 feet long, 14 inches wide, and
3 inches thick; and what will be the cost at $30 per thousand?
Ans. 4340 feet. $130.20, cost.
12. How many square feet (board measure) in a square piece of timber, 20
feet 8 inches long, 14 inches wide, and 9 inches thick? Ans. 217 feet.
13 What will be the cost of 846 feet of boards at $22.50 per M., and of 240
feet of flooring at $32.75 per M. 3 Ans. $26,894.
14. The width of a board is 16 inches, what must be its length that it may
contain 24 square feet? * Ans. 18 feet.
OPERATION.
24 sq. ft., contents of the board.
144 sq. in., in each square foot. 3456 sq. in. -- 16 in., the width, – 216 in. -- 12
*mme I = 18 feet, the length of the board.
3456 sq. in., contents of the board. -
15. A plank is 2 inches thick and 14 feet long, how wide must it be to contain
42 feet ſº Ans. 18 inches.
OPERATION.
42 sq. feet in the board.
144 sq. inches in each sq. foot. e
º 6048 - 168 = 36 inches, width of plank 1 inch
6048 sq. inches. thick.
14 feet, length of plank. 36 -- 2 = 18 inches, width of plank 2 inches
12 in. in 1 foot. thick.
168 inches, length of plank.
16. A board is 8 inches wide; what length will make 2 square feet?
w Ans. 3 feet.
OPERATION.
2 sq. feet.
144 Sq. inches in each sq. foot. 288 sq. in. -- 8 in., the width of the board, =
ºmºsºme 36 in. or 3 feet. Ans.
288 sq. in. in 2 sq. feet.
PRACTICAL PROBLEMS IN THE MENSURATION OF SOLIDS, 393
PRACTICAL PROBLEMS IN THE MENSURATION OF SOLIDS,
—-º-
TO FIND THE NUMBER OR OUBIC FEET AND INCHES IN BOXES
AND SOLIDS OF DIFFERENT FORMS.
PROBLEMIS.
802. 1. How many cubic feet in a box 6 feet long, 3 feet wide, and 2 feet
deep f Ans. 36 cubic feet.
OPERATION. Explanation.—In all problems of this character,
we have but to multiply the three dimensions
6 × 3 × 2 = 36 cu. ft., Ans. together in the unit of feet.
2. How many cubic feet in a box 4 feet 3 inches long, 3 feet wide, and 16
inches deep 3 Ans. 17 cubic feet,
OPERATION
12 || 51 4 || 17 Explanation.—Whenever any of
3 OT 3 the dimensions are given in inches,
12 | 16 ' '3 || 4 we either divide them by 12, in
order to reduce them to feet, or
cº-º ºmº gº sº else we use the inches as a fraction
17 cu. ft., Ans. 17 cu. ft., Ans. of a foot, as shown by the second
line statement.
OPERATION.
3. How many solid feet in 8 12 28
boxes, each 28 inches long, 10 inches }. 10
wide, and 6 inches deep 3 12 || 6
Ans. 7% solid feet. s
7
# solid feet, Ans.
4. How many cubic feet in a cylinder 6 feet long, and 3 feet 4 inches in
diameter ? Ans. 52.36 cubic feet.
5. How many cubic feet in a frustum of a cone, whose height is 6 feet,
diameter of the greater end is 4 feet and of the smaller end 3 feet?
Ans. 58.1196 cubic feet.
NOTE.-See Problem on page 376.
6. How many Solid feet in a sphere that is 30 inches in diameter 3
Ans. 8.18125 solid feet.
7. How many cubic inches in a cannon ball which is 13 inches in diameter 7
Ans. 1150.3492 cu. in.
8. Suppose an orange to be exactly spherical and 4 inches in diameter, how
394 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
many Oranges, 2 inches in diameter and spherical in form, will it take to equal the
larger orange? Ans. 8.
FIRST OPERATION. SECOND OPERATION.
4* = 64 cube of larger orange. | ; :
2* = 8 “ “ smaller 4 * *
64 + 8 = 8 oranges, Ans. 33.5104 -- 4.1888 = 8 oranges, Ans.
. . Explanation.—The first solution of this problem is based upon the geometrical fact that all
solids are to each other as the cubes of their like dimensions.
- In the second solution, we first find the solid contents of each, and then we find how many
times the contents of the larger orange is equal to the contents of the smaller.
9. If an apple, spherical in form, and 12 inches in circumference, sells for
10 cents, what is the proportional value of 3 apples, also spherical in form, that are
5 inches in circumference? Ans. 2 #2.
10. How many cubic inches in a semi-oblate spheroid which is 70 inches in
diameter, and 30 inches deep? Ans. 76969.2 cu. inches.
NOTE. – See Article 779.
11. An orange peddler sells two oranges which are each 3 inches in diameter,
or 3 oranges which are each 2 inches in diameter, for 52. Allowing the oranges to
be perfect spheres, which is the better purchase, and how many cubic inches of
orange would be gained ? Ans. The better purchase would be the
2 Oranges each 3 in. in diameter.
The gain would be 15.708 cu. in.
TO FIND THE COST OF FEEIGHT PER, CUBIC FOOT ON BOXES.
PROBLEMS.
803. 1. What will be the freight on a box 9 feet 3 inches long, 4 feet. 6
inches wide, 2 feet 10 inches deep, at 30 cents a cubic foot ? Ans. $35.38%.
2. What is the freight on 8 boxes, which are each 3 feet 3 inches long, 2 feet
8 inches wide, and 21 inches deep, at 202 per cu. foot ? Ans. $24,263.
TO FIND THE NUMBER OF BUSHELS IN VESSELS, BOXES, BINS, ETC.
PROBLEMS.
804. 1. How many bushels will a bin hold, that is 10 feet long, 8 feet 6
inches wide, and 5 feet 2 inches deep 3 Ans. 352.89-H bushels.
OPERATION.
#! Explanation.—In this problem, we multiply the three
215042 | 102 dimensions together, in the unit of inches, which gives
62 cubic inches, and then divide by 2150.42 the number of
100 cubic inches in a bushel. Iły reason of the decimals in
352.89-H. Ans the 2150.42 we also multiply by 100 to cancel the decimal.
NotE.—2150.4 cu. in. instead of 2150.42 cu. in. is often used, and gives a result sufficiently
accurate for practical purposes. Had we used 2150.4 the result would have been 352.9.
X- PRACTICAL PROBLEMS IN THE MENSURATION OF SOLIDS. 395
2. A wagon box is 12 feet long, 44 feet wide, and 15 inches high. How many
bushels will it hold when even full? Ans. 54.24+ bushels.
. NotE.-A very short way to find the number of bushels any box or vessel will hold is to
multiply the cubic feet it contains by 45 and divide the product by 56. This is correct because
the ratio between 2150.4 and 1728.0 is as 56 to 45.
3. A planter wishes to build a bin that will hold 100 bushels; it is to be 10
feet long and 44 feet high. What must be the width of the bin 3
Ans. 33.185-H inches.
OPERATION.
2150.42 = cu. in. in 1 bu. 120 in. = length.
100 = bushels. 54 in. = height.
215042.00 = cu. in. in 100 bu. 6480 sq. inches.
215042 + 6480 = 33.185+ inches, Ans.
4. How many bushels in a cylinder shaped box, whose height is 10 feet, and
diameter 10 feet 3 Ans. 631.125 bu.
1St OPERATION. 2d OPERATION.
120 10 Ea:planation.—In the second operation, we first
120 10 make the statement to find cubic feet in the
120 10 cylinder; then instead of multiplying by 1728
*f PF to reduce the cubic feet to inches, and dividing
2150.4 | .7854 .7854 by 2150.4 to reduce the cubic inches to bushels,
*m- I - 56 || 45 we multiply by 45 and divide by 56. These
- sºmeº- numbers are obtained by reducing # to its
lowest terms.
NotE.-In this problem, we use 2150.4 cu. in. as a bushel which is, for ordinary purposes,
sufficiently correct.
5. A sugar kettle in the form of a half of a prolate spheroid is 42 inches
deep and 50 inches in diameter. How many bushels, and how many gallons will it
hold # Ans. 25.566–H bushels. 238 gallons.
OPERATIONS INDICATED.
Bushels. Bushels. Gallons. Gallons.
42 42
50 50
42 50 42 50
| 50 or, .7854 50 Or, .7854
50 3 2 50 3 2
2150.42 .5236 2150.42 231 .5236 231
| 25.566 bu. 25.566 bu. 238 gals, 238 gals.
NoTE.—See Article 778.
TO FIND THE NUMBER OF CUBIC YARDS IN LEVEES OR EXCAVA—
TIONS.
PROBLEMS.
805. 1. How many cubic yards in a levee 80 rods long, 60 feet wide at the
base, 124 feet at the top, and 5 feet 4 inches average depth ?
Ans. 9451# cubic yards.
396 soule's PHILOSOPHIC PRACTICAL MATHEMATICs. *
OPERATION.
60 Width at the base. 80 Explanation.—Here
12; “ “ “ top. 2 || 33 we first multiply the
-º-e 4 || 145 three dimensions to-
2) 72} 12 || 64 gether in the unit of
27 feet, * then divide
* ~ * by 27, the number of
36+ average width. a- - mºs cubic feet in a cubic
945.13% cu. yas., Ans. yard.
2. A railroad contractor excavated a cut through a hill 840 feet long, 70
feet wide on top, and 30 feet wide on the bottom, average depth 50 feet. What
was the cost at 212 per cubic yard 7 Ans. $16333.33%.
3. A tunnel was excavated 1350 feet long with a cross section or area of
660 sq. feet. How many cubic yards, and what was the cost at 30¢ per cu. yard?
Ans. 33000 cu. yards. $9900, cost.
TO FIND THE NUMBER OF GALLONS IN VESSELS, CISTERNs, ETC.
PROBLEMIS.
806. 1. How many gallons will a box hold, that is 5 feet long, 2 feet 4 inches
wide, and 3 feet deep 3 Ans. 261.81–H gallons.
OPERATION.
60 Explannation,-In this problem, we multiply the three
231 ; dimensions together in the unit of inches, and divide by
231, the number of cubic inches in a gallon.
261.81-H gal., Ans.
2. How many gallons in a cylindrical cistern or tank 12 feet 4 inches high,
and 8 feet 4 inches in diameter ? Ans. 5032 gals.
1st Full Operation. 2d Contracted 3d Contracted 4th Contracted 5th Contracted
Operation. Operation. Operation. Operation.
148 148 3 || 37 3 || 37 3 ||37 feet.
100 100 3 || 25 3 || 25 3 || 25 “
100 { 100 25 25 3 || 25 “
231 | .7854 34 5.8752 5.875 8 || 47 “
5032.0000 gals. 5032 gals. 5032 gals. 5031.8287 5031,8287
Explanation to 18t Operation.—In all problems of this character, we multiply the height and
the square of the diameter together in the unit of inches, then the product thus produced by the
decimal .7854, and then divide by 231, the number of cubic inches in a gallon.
Explanation of 2d Operation.—Since.7854 is equal to 231, .0034 times, we may shorten the
work by simply multiplying by .0034, omitting the division by 231. In the operation we set 34
only, remembering there will be four decimal places in the answer.
Baplanation of the 3d Operation.—In this operation we use the three factoral dimensions in
the unit of feet, and multiply their product by 5.8752. This 5.8752 is produced thus: In a cubic
foot there are 1728 cu. inches; this we multiply by .7854, which is the area of a circle whose
diameter is equal to one side of a square, and then divide the product by 231 the number of cubic
inches in a gallon. Thus the 5.8752 is the number of gallons in a cubic foot reduced to a cylinder.
Explanation of the 4th and 5th Operations.—In the 4th operation we shorten the work by mul-
tiplying by 5.875 instead of 5.8752, and produce an answer nearly or practically correct. In the
5th operation we consider that as 5.875 gallons equals 5% gallons, and as 5% equals # gallons, we
hence, multiply by 47 and divide by 8.
NOTE.-The 4th and 5th methods are so nearly correct that they are used by many builders
and engineers. We prefer the 2d Contracted Operation, as it is exact and in the majority of cases
it is shorter than the other methods.
3. How many pints in a cylindrical vessel, whose height is 14 inehes and
diameter 12; inches 3 Ans. 59.5 pints.
#: PRACTICAL PROBLEMS IN THE MENSURATION OF SOLIDs. 397
OPERATION.
14
2 || 25 e
2 || 25 Ea:planation.—Here we first find the number of
34 gallons, the same as in the preceding problem, and
: then we multiply by 4, which gives us quarts, and
*- then by 2, which gives us pints.
59.5 pints, Ans.
4. How many gallons of water will a steamboat boiler hold, that is 40 feet
long, 5 feet 4 inches in diameter, allowing for 3 flues, each 15 inches in diameter
Ans. 5583.072 gals.
OPERATION. OPERATION.
TO FIND THE CONTENTS IN GALLONS OF THE | TO FIND THE DEDUCTIONS IN CONSEQUENCE
WHOLE BOILER. OF THE 3 FLUES.
40 40
12 12
64 15
64 15
34 34
-ms - sºmmº- sºmmº- 3
6684.672 sm - mammº mºnºmº
1101.600 1101.600
5583.072 gals., Ans.
5. How many gallons in a cistern which is in the form of a frustram of a
cone, whose height is 9 feet 6 inches, lower base 7 feet 2 inches, and upper base 6
feet 8 inches : Ans. 2671.3392 gals.
OPERATION.
862 – 7396 3 || 20676 Ea:planation.—In all prob"
80.2 = 6400 114 }..."; th; character, we
-> I'S nd the contents or
86 X 80 = 6880 34 - ::::::::::: the cistern i.
=sºmmºn wºmm mºm-º ºmme cubic inches, according to the
20676 2671.3392 gals., Ans. principles jº PP-
376 and 396.
NotE.—231 cancels .7854, giving a quotient of .0034.
6. How many quarts will a bucket hold, that is 10 inches deep, 12% inches
in diameter at the top, and 94 inches at the bottom ? Ans. 16.2038 quarts.
OPERATION,
124° = 156 + 16 || 5719
94* = 85% 3 || 10
124 × 94 = 115 § 34
4
357+'s º-
| | 16.2038 quarts, Ans.
7. How many pints in a conical cup, 4 inches in diameter at the top, and 6
inches deep 3 Ans. .8704 pint.
OPERATION.
; Explanation.—The contents of a cone being # of the
6 contents of a cylinder of equal height and diameter, we
* proceed the same as in the measurement of a cylinder, and
3 || 2 then divide by 3. We multiply by 4 and 2 to reduce gallons
mº mmmº to pints.
8704 pt., Ans
398 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. T.
8. How many gallons will a tub hold which is 34 inches upper diameter, 29
inches lower diameter, and 21 inches deep 3 Ans. 70.9954 gallons.
9. How many gallons in a boiler or kettle,
the lower part of which is in the form of a hemis-
phere, with a diameter of 60 inches and a depth
of 30 inches, and the upper part being in the form
of a frustum of a cone, whose lower base is 60
inches, upper base 70 inches and the depth or height
being 12 inches? Ans. 417.52 gallons.
OPERATION OPERATION
TO FIND THE CONTENTS OF THE TO FIND THE CONTENTS OF THE
HEMISPHERE PORTION. FRUSTUM PORTION.
60 60 60° = 3600 3 || 12700 3 || 12700
60 60 70? – 4900 12 12
231 | .5236 or thus : .00223 70 × 60 = 4200 231 .7854 or thus: .0034
30 30 $º ºsmºsº
tº ºmºmºmºsº 12700 — 172.72 gals.
172.72 gals.
244.8 gals. | 244.8 gals. 3
NOTE.—The decimal .00223 is the quotient of .5236 – 231; and the decimal .0034 is the
quotient of .7854 – 231.
Explanation.—See Article 772, page 378, and Article 768, page 375, for a full explanation for
finding the contents of hemispheres and frustums of cones. The reason for dividung by 231 is that
231 cubic inches constitute a gallon.
TO FIND THE NUMBER OF GALLONS IN A HORIZONTAL CYLINDER
WEHEN TEIE DIAMETER AND THE LENGTH OF THE CYLINDER
AND THE DEPTH OF THE LIQUID ARE GIVEN.
PROBLEMS.
807. 1. How many gallons in a cylinder 18 feet 4 inches long, 6 feet diam-
eter, the depth of the liquid being 12 inches Ans. 424.7917+ gals.
18° 4".
| 2.
!
i
s |
le i
e-T-7– : ~7
OPEIRATION.
First Step.–To find length of chord: Diameter 72 in. — 12 in. depth of liquid
= 60 × 12 in. depth of liquid = 720. V 720 = 26.832 + in. # of chord x 2 =
53.664-i- in. chord, or width of liquid surface.
NOTE.-See page 356, for the operation to extract the square root of 720.

* PRACTICAL PROBLEMS IN THE MENSURATION OF SOLIDS. 399
Second Step.–Depth of liquid: 12 in. -- the diameter, 72 in. = .16663 =
tabular number.
Third Step.–See Table, page 353, and find the quotient .166 in the column of
Versed sines. Then take the area of segment noted in the next column and multi-
ply it by the Square of the diameter, the result will be the area of the segment of
the cylinder. Thus:
The tabular area for .166 = .08554
The tabular area for .167 = .08629
Difference is .00075
.00075 × .63 or .66+ = .00050.
Then .166 = .08554
And .00063 = .00050
.08604 = the sum by which the square of the diameter is to
be multiplied.
Fourth Step.–.08604 × 72* = 446.0313 + sq. in. in area of segment.
Fifth Step.—To find gallons:
446.0313 x 220 in. length of cylinder.
231 cu. in. in...a gallon.
= 424.7917+ gallons.
NOTE: 1.--To find the area of a segment greater than a semi-circle, find the area of the lesser
segment, and subtract the same from the area of the whole circle.
NOTE 2.—To find the area of a zone of a circle, find the areas of the two segments and sub-
tract the sum from the area of the whole circle.
2. How many gallons in a tank of the following description and dimen-
sions: The upper portion is in the form a vertical rectangular prism, 13 feet 5 inches
deep, 11 feet 9 inches long, and 9 feet 3 inches wide. The bottom is a segment of a
horizontal cylinder, the chord being 9 feet 3 inches, and the height or versed sine
4 feet 3 Ans. 13372.404 gallons.
41 Feet 9. _56,ºrºC
- *. A *#. B
: :
: º OPERATION > -º
: D
First Step.—To find the gallons in the
º ... • *** cubical part of the tank:
.# +II" TB A. 161 in. deep.
A ſ Y. 141 in. long.
32; f{ 111 in. wide.
<!-f
161 x 14.1 × 111
Đ 231 cu. in. a gal.
= 10908.272 gallons.

4OO soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. jºr
Second Step.–To find the diameter of the circle of which the given segment
is a part :
A B = 111 in. length of chord. (55%)*-i-48 = 64.17+ in. = the difference be-
tween the diameter and the versed sine.
64.17+ 48 = 112.17 in diameter of circle.
112.17 - 2 = 56,085 in., the radius of the
circle A. C.
Third Step.–To find the quotient of the versed sine divided by the diameter:
48 - 112.17 = .4279 quotient.
A P = 554 in. # of chord.
P D = 48 in.
See Table on page 354, and find the quotient in the column of versed sines.
Then take the area number shown in the next column on the right and multiply it
by the square of the diameter. The result will be the area of the given segment.
The tabular area number for .427 = .31996
The tabular area number for .428 = .32095
The difference is .00099
Then, .00099 × .9, which is the difference between .4279 and .427, - .000891.
Hence .31996 -- .000891 = .320851 the sum by which the square of the diame-
ter is to be multiplied.
Thus, .320851 × 112.17% = 4036.9828+ sq. in. area of the segment A D B.
Fourth Step.–To find the gallons in the Segment :
4036.9828 × 141 in., length of segment in tank _
gº tº 2464.132 gallons.
231 cu. in. in a gallon
Add the gallons in the cubical part of tank, as above 10908.272
Total gallons in the tank - 13372.404 gals.
NotE.-See Haswell’s Engineers' and Mechanics' Pocket Book, for extended work on Men-
suration and on other subjects.
TO FIND TEIE NUMBER OF BARRELS IN A CISTERN.
PROBLEMI.
808. How many barrels will a quadrilateral cistern hold, whose height is 12
feet, and width of each side 5 feet 8 inches 3 Ans. 91 ºr barrels.
OPERATION.
144 Explanation.—In all problems of this character, we
231 | 68 multiply the three dimensions together in the unit of
68 inches, then divide by 231, which gives gallons; we
63 || 2 then divide by 31%, the number of gallons in a stand-
ard or conventional barrel, and thus obtain the con-
91+ºf bbls., Ans. tents of the cistern in the unit of barrels.
X- PRACTICAL PROBLEMS IN THE MENSURATION OF SOLIDS. 4O I
TO FIND THE NUMBER OF CART LOADS OF EARTH TO FILL LOTS.
PRORLEMS.

-º-º-º- -: 809. 1. A piece or lot of ground measures
_T) -ºl 63 feet 24 inches front by 120 feet deep. How
many square yards in the piece, and how
\
__* many cart loads of earth will it take to fill the
_.” pººr above stated lot 6 inches deep, the cart box
–– --"
- Cart Box. being 5 feet 4 inches long on the top, 4 feet
8 inches long on the bottom, 3 feet 6 inches wide, and 2 feet deep 3 How many
cubic yards of earth are used in the filling 3 &
Ans. 842% sq. yds.; 140}} cu. yds.; 108 ºr loads.
OPERATION OPERATION
TO FIND SQUARE YARDS. TO FIND CUBIC YARDS TO BE FILLED.
12 1758.5 = ft. front of lot." 12 || 758.5 = ft. front.
120 = ft. depth of lot. 120 = ft. depth or length. -
9 12 || 6 = ft. thickness to be filled.
27
842; Sq. yds., Ans. º-
= cu. ft. in a cu. yd.
-
-
140}} cu. yds.
Ö
OPERATION OPERATION
TO FIND CAPACITY OF THE CART BOX. TO FIND NUMBER OF LOADS.
2 || 7 = ft. wide. 140}} -- 1:# = 108 ºr loads.
5 = ft. average length (5'4" + 4'8")
27 || 2 = ft. deep. 2 or thus:
*-*-* mºmºmºsºs º 54 7585
1%; cu. yds. 35 27
| 108 ºr loads, Ans.
2. A yard is 144 feet long and 32 feet 4 inches and 7 lines wide, American
measure. It is desired to fill the yard 27 inches deep with earth. Allowing the
yard to be a perfect plane, how many cubic yards of earth will be required to fill
it; and at 70g per cubic yard, what will be the cost 3
Ans. 3884; cu. yds.; $272.003, cost.
PARTIAL OPERATION.
144
12 || 4663 (144 x 12) x 3884; x 27
12 Or, = cu. yds.
12 27 1728 × 27
27
4O2 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
TO FIND THE NUMBER OF SQUARES OF EARTH IN GRADINGS,
LEVEES, AND EXCAVATIONS. e
PROBLEM.
810. How many squares of earth are contained in a piece of grading or levee,
420 feet long, average width 66 feet and average depth 21 ft. 4 inches 3
Ans. 27.37% squares.
OPERATION.
420 Explanation.—As shown by the Table of Solid Meas-
, 66 ..ºf ure, page 247, a square of earth is a cube 6 × 6 × 6 =
2i. 64 216 cubic feet. Hence we multiply the three dimen-
* * sions of the grading together in the unit of feet and
2737; squares, Ans. divide by 216. - *.
TO FIND THE NUMBER OF CORDS OF WOOD IN RANKS.
PROBLEMS.
811. 1. ‘How many cords of wood in two ranks, each 44 feet long and 6
feet 3 inches high 3 Ans. 17% cords.
OPERATION.
44 Explanation.—In all problems of this character, it
12 || 75 is customary, in the absence of a special agreement
128 ; to the contrary, to estimate the length of the 8tick as 4
*E== feet, notwithstanding the fact that in practice it is
17 ºf cords, Ans. often much less than 4 feet.
NOTE.—In solving all problems of this character, we multiply together the three dimensions
in the unit of feet, and divide by 128, the number of cubic feet in a cord. In this problem we also
multiply by 2, for the reason that we had 2 ranks,
2. A rank of wood is 46 feet 4 inches long, 6 feet 6 inches high, and 3 feet
8 inches deep or length of stick. How many cords does it contain
Ans. 9; cords.
NoTE.-In commerce the length of stick, when less than 4 feet, is estimated as if it were 4
feet long, and the price is graded accordingly.
3. What will be the cost of two ranks of wood each rank 44 feet long and
6 feet 3 inches high, at $5 per cord 3 Ans. $85.93%.
4. How many cords of wood can be placed under a shed that is 40 feet long,
16 feet wide and 18 feet high 1 Ans. 90 cords.
X. PRACTICAL PROBLEMS IN THE MENSURATION OF SOLIDS. 4O3
TO FIND THE NUMBER OF CARS REQUIRED TO CARRY A. SPECIFIED
NUMBER OF BUSEHELS.
PROBLEMS.
812. 1. How many cars will be required to carry 15000 bushels of wheat,
each car carrying 20000 pounds? Ans. 45 cars.
g
OPERATIQN. INDICATED.
15000 × 60 = 900000; 900000 - 20000 = 45 cars.
2. If a freight car is 28 feet long and 8 feet wide, how deep must bulk corn
be, in order that the car may contain 22400 pounds, which is a large weight for a
single car? & **- Ans. 26.666-H inclues deep.
OPERATION INDICATED.
22400 - 56 = 400 bushels. 2150.42 × 400 = 860168 cubic inches.
28 × 12 x 8 × 12 = 32256 sq. inches. 860168 - 32256 = 26.666+ inches deep.
3. How many cars will be required to carry 120000 feet of lumber, allowing
6000 feet per car # Ans. 20 cars.
4. How many trains of 12 cars each, will be required to transport 600000
cubic yards of gravel, estimating that each cubic yard weighs 3200 pounds, and
that each car carries 18000 pounds & Ans. 8888; trains.
OPERATION INIDICATED.
600000
3200
18000 or thus: (600000 × 3200) + (18000 x 12).
12 es
*º
gººms-ºs-ºs-
TO FIND THE NUMBER OF BRICKS IN WALLS AND FOUNDATIONS
PROBLEMIS.
813. 1. How many bricks in a wall 230 feet long, 12 feet 9 inches high, and
21 inches thick, allowing 20 bricks to the solid foot ? Ans. 102637; bricks.
OPERATION.
230 r
4 || 51 Ea:planation.—Here we first find the number of cubic
12 : feet, and then multiply by 20, the number of bricks in
a cubic foot.
102637; bricks, Ans.
4O4.
SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. A.
2. What will it cost to build the foundation walls of a house of brick, the
house being 60 feet long and 28 feet wide, and the walls 6 feet high and 2 feet 6
inches thick, at $15 per thousand, allowing 18 bricks per solid foot ?
Ans. $672.30.
OPERATION.
60 × 2 = 120 ft. in the two sides. ‘.
28 × 2 = 56 ft. in the two ends.
176 ft. round the outside of all the walls.
24 x 4 = 10 ft. in 4 times the thickness of the walls, deducted for the
corners measured twice.
166 ft. actual length of walls.
-msmsº
166 = length of walls.
6 = height {{
2 || 5 = thickness “
18 = number of bricks in a cubic foot.
1000 || 15 = price per thousand.
$672.30 º
3. How many Milwaukee bricks which are 84 × 4 × 23 inches, will it take
to build a wall 60 feet long, 40 feet high, and 3 bricks thick, allowing # inch between
bricks for mortarº
FIRST OPERATION.
Ans. 45139%; bricks.
SECOND OPERATION.
# -- # = 8% in. length of brick and mortar. 60
# + 4 = 2; in. thickness of brick and mortar. 40
# × 23 = 22#} sq. in, surface of one brick and mortar. 12
44 + 22# = 6+}} bricks in one square foot of wall, 1 brick 12 º
thick. 8|99 = 3× 4
### × 3 = 18# bricks in 1 square foot 3 bricks thick. 35 |4
0 × 40 = 2400 x 18:### = 4513933, number of bricks in the 21||8
wall 3 bricks thick. 33 || 8
45139%; bricks.
TO FIND THE NUMBER OF PERCEIES IN A STONE WALL.
PROBLEMS.
814. 1. How many perches in a stone wall 60 feet long, 8 feet 6 inches high,
and 2 feet 8 inches thick?
Ans. 54}# perches.
OPERATION.
60
12 || 102 Explanation.—Here we multiply the three dimen-
O
; * sions together in the unit of feet, and then divide by
54% P., Ans.
24%, the number of cubic feet in a perch.
2. What will it cost to build a breakwater of stone, 220 feet long, 18 feet
thick, and 12 feet high at $1.90 per perch?
Ans. $3648.
PRACTICAL PROBLEMS IN THE MENSURATION OF SOLIDS. 405
3. How many perches of stone are contained in a wall 90 feet long, 12 feet
high, and 22 inches thick? Ans. 71% perches.
OPERATION. -
90 Explanation.—A perch is, as shown in the Table of
12 Cubic Measure on page 247, 244 cubic feet, But in
12 22 building stone walls, 23 cubic feet are allowed per
99 || 4. perch for mortar and filling. Hence, when built in a
wall, 22 cubic feet of stone make a perch. Then, as
sº ºf . *H # º 24%, ". therefore º the º
g er of perches in the whole wa of the same, and
sº, peº IIl .ll. & thus produce the perches of stone ff preferred, the
# = # or perches lin perches of stone may be obtained by multiplying the
†-Lºs Wall, whole number of perches by 22, and dividing the prod-
71} = perches of stone. uct by 24%.
NOTE.-It is estimated that three pecks of lime and 4 bushels of sand are used in a perch.
WIDTHS OF WALLS.
A wall 1 brick or layer thick is called, by brick masons, a 4% inch wall.
& & & & & 4 & & & & 4 & & 4 & & 4 & & 4 9 & & & 4
& & & 4 & 4 & 4 ( & & & & & & & & 4 & 4 13 ( & & 4
& 4 & & & & & & & & & & & & & & & & 4 18 4 & & 4
& & & & & & & & & & & & & & & 4 & 4 & 6 22 é & & 4
& & & & & & & 4 & & & & & & & 4 & 4 & 4 26 & 4 &
i
TO FIND THE NUMBER OF CUBIC YARDS, CUBIC FEET, CUBIC
INCHES, BUSHELS OR GALLONS, IN DIFFERENT KINDS
OF VESSELS AND SOLIDS.
PROBLEMIS.
815. 1. A rectangular box is 6 ft. 3 in. long, 3 feet wide, and 4 feet 6 inches
high or deep. How many cubic yards, how many cubic feet, and how many cubic
inches does it contain? Also how many bushels, and how many gallons will it hold?
Ans. 3; cu. yds.; 84; cu. ft.; 145800 cu. in. ; 67.8+bu.; 631.17–gals.
OPERATIONS INDICATED.
First. Second. , Third. Fourth. Fifth.
To find cu. yds. " To find cu. ft. To find cu. in. To find bul. To find gals.
4 || 25 75 75
3 4 || 25 75 36 36
2 9 3 36 * 54 54
27 2 || 9 54 2150.42 231
3# cu. yds. 84; cu. ft. 145800 cu. in. 67.8–H bu. 631.17— gals.
Explanation.—To find cubic yards, we multiply the three dimensions together in the unit of
feet, and then divide by 27, because 27 cubic feet make 1 cubic yard. .
To find cubic feet, we multiply together the three dimensions in the unit of feet.
To find cubic inches, we multiply together the three dimensions in the unit of inches.
To find bushels, we find the number of cubic inches and then divide by 2150.42, because
2150.42 cubic inches make a bushel.
To find gallons, we find the number of cubic inches and then divide by 231, because 231
cubic inches make a gallon.
2. A box is 5 feet 6 in. long, 3 feet 4 in. wide, and 2 ft. 8 in, deep. How
many cubic feet does it contain and how many gallóns will it hold?
Ans. 48# cu. feet; 365% gallons.
3. A cellar, in the form of an inverted frustum of a rectangular pyramid, is
40 ft. 4 inches long at the top and 30 ft. long at the bottom; it is 24 ft. wide at the top
4O6 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
and 18 feet 8 inches wide at the bottom; and it is 7 feet 6 inches deep. How many
of each, cubic yards, cubic feet, and cubic inches does it contain 3 Also, how many
bushels, and how many gallons will it hold? Ans. 2094%; cu. yds.; 5661; cu. ft.;
9782400 cu. in. ; 4549.06-i- bu.; 42348 ºr gals.
OPERATION BY THE PRISMOIDAL FORMUL.A.
40++, × 24 = 968 = area of the top of the cellar in sq. ft.
30 × 18+3 = 560 = “ “ “ bottom of the cellar in sq. ft.
2) 70-4; 42.É.
35% x 21.4% = 750% x 4 = 3000; = 4 times the middle section between the
gººmsº tWO area.S.
6) 4528;
754}} = average area of the cellar.
754}} x 7# (ft. deep) = 5661} cubic feet.
566.1% cu. ft. -- 27 = 20.94%; cu. yards.
5661; cu. ft. x 1728 (cu. in.) = 9782400 cubic inches.
9782400 cu. in. -- 2150.42 = 4549.06–H bu.
9782400 cu. in. -- 231 = 42348#r gals.
NotE 1.-If it is desired, the different dimensions in the above problem may all be reduced
to inches and the work performed in the same manner, thus avoiding much fractional work.
NOTE 2.—The above solution is in accordance with the prismoidal formula, by which the
Solidity of cubes, rectangular solids, cones, cylinders, pyramids, frustums of cones or pyramids
and several other forms of solids may be determined.
NOTE 3.−For an explanation of the Prismoidal Formula, see Article 768, pages 375 and 376.
4. How many cubic yards were excavated from a cellar 92 feet long and 50
feet wide at the top, 86 feet long and 44 feet wide at the bottom, and 8 feet 4 inches
deep 3 Ans. 1291}} cu. yds.
5. A cylinder is 7 feet 4 inches high and 3 feet 5 inches in diameter. How
many of each, cubic yards, cubic feet, and cubic inches does it contain? Also, how
many bushels and how many gallons will it hold, and allowing that à cubic foot of
water weighs 624 pounds, what would be the weight of water the cylinder could
contain } Ans. 2,4902-H cu. yds.; 67.2353–H cu. ft.; 116182.6512 cu. in. ;
54,028 — bu. ; 502.9552 gals.; 4202.2081 pounds.
OPERATIONS INDICATED.
Cu. Yols. Cu. Ft. ge Weight of Water.
22 3 || 22
12 || 41 3 || 22 12 || 41
12 || 41 12 || 41 12 || 41
.7854 12 41 .7854
27 .7854 2 | 125
2.4902-# gu. yds. 67.2353–H cu. ft. 4202.2081 pounds.
Cu. In. Bu. Gals.
88 88
41 41 41
41 41 41
.7854 † .7854 231
*mºmº || s=ºmº 2150.42 .7854
116182.6512 cu. in. *= | *=====
54.028— bu. 502.9552 gals.
NotE.—By inspection, we find that 231 always cancels .7854 and gives a quotient of .0034.
PRACTICAL PROBLEMS IN THE MENSURATION OF SOLIDs. 4O7
6. How many gallons in a cylinder that is 9 feet 3 inches high and 8 feet
4 inches in diameter ? Ans. 3774 gals.
7. A cone is 134 inches high and 45 inches in diameter at the base. How
many of each, cubic yards, cubic feet, and cubic inches does it contain 3 Also, if
it is a vessel, and these dimensions are inside measurements, how many bushels and
how many gallons will it hold? Ans. 1.5226+ cu. yds.; 41.1108— cu. ft.;
71039.43 cu. in. ; 33.035+ bu.; 307.53 gals.
OPERATIONS INDICATED.
Cu. Yds. Cu. Ft. Cu. In. Bu. Gals.
12 || 134 134
12 || 45 12 || 134 | 134 45 134
12 || 45 12 || 45 i 45 45 45
.7854 12 45 45 | 7854 45
3 .7854 .7854 3 3 | .7854
27 2150.42 231
1.5226–H | 41.1108– 71039.43 33,035–H 307.53
sº cu. yds. cu. ft. cu. in. bu. gals.
8. How many cubic inches in a cone that is 10 feet 2 inches high and 8 feet
3 inches in diameter at the base ? Ans. 313040.0196 cu. in.
9. A cistern in the form of a frustum of a cone is 10 feet 2 inches high, 9
feet 6 inches diameter of lower base and 8 feet 6 inches diameter of upper base.
How many bushels, and how many gallons will it hold, and allowing that a gallon
of water weighs 84 pounds, (8.3389 is the accurate weight of a gallon) what would
be the weight of the water the cistern could hold 2
Ans. 520.26+ bu. ; 4843.2048 gals.
40360.04 pounds.
OPERATIONS INDICATED.
Bushels. Gallons. sº
1142 = 12996 11676 11676
102* = 10404 122 122
114 × 102 = 11628 .7854 .7854
2150.42 | 231 |
3) 35028 * == i <===mºssºmsºmº
TTT676 520.26-H bu. | 4843.2048 gals.
4843.2048 x 84 = 40360.04 pounds.
10. A rectangular pyramid is 6 feet high, and has a base 4 feet 3 inches by 3
feet 9 inches. How many of each, cubic yds., cubic feet, and cubic inches does it
contain 3 And if hollow, how many bushels and how many gallons will it hold 3
Ans. 143 cu. yds.; 31% cu. ft.; 55080 cu. in. ; 25.61-- bu.; 238## gals.
OPERATIONS INDICATED.
Cu. Yas. Cu. Ft. Cu. In. Bu.
6
Gals.
72 72
4 || 17 6 72 51 | 51
4 || 15 4. 17 51 45 45
3 4 | 15 45 3 3 |
27 3 | 3 | 2150.42 231
1}} cu. yds. | 31% cu. ft. | 55080 cu. in. 25.61–H bu 238% gals.
$
408 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
TO FIND THE HEIGHT OR THE DIAMETER OF CYLINDRICAL YESSELS
TO CONTAIN A SPECIFIED NUMBER OF GALLONS.
PROBLEMS.
816. 1. The diameter of a cylindrical cistern is 6 feet 8 inches. What must
be its height to hold 3000 gallons? Ans. 137.8676 in. height.
ôPERATION INDICATED. .
3000 gallons.
231 cu. in. in 1 gallon.
693000 cu. in. in a cistern to hold 3000 gallons.
80 in. diameter of cistern.
80 46 46 wº
6400 = square of diameter.
.7854 = ratio of the area of a square to the area of a circle.
5026.5600 = area of a section of the cistern.
693000 ---5026.56 = 137.8676 inches height.
80 3000
80 231
Or thus: .7854
wº
137.8676 in. = 11 feet, 5.8676 inches.
2. The height of a cylindrical cistern is 7 feet 6 inches. What must be the
diameter so that it will hold 2000 gallons? Ans. 80.84 inches diameter.
º "OPERATION INDICATED,
90 | 2000
.7854 || 231
6535,9477
*/ 6535,9477 = 80.84 + inches., Ans.
wo §
PRACTICAL PROBLEMS IN THE MENSURATION OF SOLIDS. 4O9
GAUGING COAL.
817. A barge of coal measures as follows: Whole length from a to b 113
6", width of boat 25' 10", depth 8, 9.5". The length of bulkhead or rake No. 1 ls
79" and the depth is 92". The length of bulkhead No. 2 is 72" and its depth is
102". How many of each, bushels and barrels, does it contain?
Ans. 24096,1008 bushels. 9267,731 barrels.
No. 1. - –4– No. 2.
- b

(Lines showing side of boat).
OPERATION
To FIND THE CUBIC INCHES IN THE BULKHEADS.
No. 1, 79” length × 92” depth = 7268 sq. in.
No. 2, 72” length × 102” depth = 7344 sq. in.
Total sq. in. in the area of length and depth of the two bulkheads 14612
14612 × 310” width = 45297.20 cu. in. in the bulkheads.
tº
GENERAL OPERATION,
173' 6" = length of boat.
1/ = allowance for rail.
172' 6" = length including bulkheads.
12" 7" = length of bulkheads, (79” + 72" = 12' 7").
159' 11" = length less bulkheads. &
12 = in. in 1 foot. *
1919" = length in inches of main boat.
310" = width of boat.
594890 = sq. in. area of main boat.
ió5.8 = depth of main boat.
62939362 = cu. in, in main boat.
45297.20 = cu. in. in bulkheads.
67469082 = cu. in. in main boat and bulkheads.
2698763 = 4% deductions, for gunnels cross-ties, pumps, the adjustment of the bulk-
tº-º-º-º: heads, etc.
64770319 = cu, in net in the whole boat.
64770319 cu. in. -- 2688 cu. in., the number of cubic inches allowed for a bushel of coal,
= 24096.1008 bushels, 24096.1008 - 2.6 bushels, the number of bushels allowed for a barrel of
coal, -= 9267.7310 barrels.
NOTE.-The depth of the bulkheads or rakes varies according to the construction of different boats,
REMARKS ON THE ABOVE SOLUTION.
The above solution is made in accordance with the custom of the official gaugers of coal for
Louisiana. The allowance of 2688 cu. in. for a bushel, and 2.6 bushels for a barrel, is practically
in accordance with the law of Louisiana, which says, that 3+ bushels of the United States standard
shall be a barrel of coal. Accordingly, 2150.42 × 33 = 6988.865 cu. in. constitute a barrel of coal.
See pages 248 and 249 of this book.
…”
(Continued.)
º tºº,
4 IO SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. Af
The above 64770319 cu. in. divided by 6988.865 cu. in., gives 9267.6448 barrels.
The result obtained by dividing the bushels by 2.6 was 9267.7310 barrels.
The difference is .0862
The excess by the 2.6 method in a quantity of 9267 barrels; is so insignificant that the
method may be regarded as practically correct.
ANOTHER METHOD,
The following method of measuring or gauging coal is used in Ohio and some other states:
1. Find the cubic inches in the barge exclusive of the rakes”, and including the gunnels,
timbers, etc. ſº
2. To this add the cubic inches in the rakes.
3. Then find the actual cubic inches in the timbers, pumps, etc., and deduct the same from
the total cubic inches in the barge.
4. Divide the remainder by 2688, the number of cubic inches in a bushel of coal, in Ohio,
Pennsylvania and Kentucky and several other states—but not in Louisiana.
f The following operation elucidates this method. Af
DIMENSIONS OF BARGE. *Rakes are the ends of the barge or boat,
Long. Wide. Deep. not perpendicular to the keel or bottom. The
Body 1350 298 90 6-7 = 36551828 Cubic inches. word bulkhead is often used in the sense of
Rakes 185 298 76# = 4231227 ‘‘ { { rake. It was so used in the preceding problem.
*E= ºs fThis operation statement is an actual meas-
40783055 “ ( { urement of a barge of coal by Mr. E. A. BURN-
Deductions for timbers, etc., 1129646 “ & 4 SIDE, Coal Measurer of Cincinnati, through
whose courtesy we present the method.
2688) 39653409 (14752 bushels
in the barge.
§
APPROXIMATE METHODS OF FINDING THE TONS OF HAY IN STACKS, MOWS,
LOADS AND WINDROWS.
818. The only accurate method of measuring hay is to weigh it. But as this cannot
always be done, approximate methods are sometimes used.
The observation and experience of farmers and dealers in hay have shown the following
approximate facts:
1. That hay, well settled in stacks or mows, will weigh a ton for every 15 cubic yards.
2. That hay, slightly settled in stacks, mows or loads, will weigh a ton for every 20 cubic
yards.
3 That good, average, meadow hay unsettled in windrows, will weigh a ton for every 25
cubic yards.
4. That hay, when baled, will weigh a ton for every 10 cubic yards.
NOTE.-There is a difference in the weight of equal bulks of different kinds of hay, new or old, clover, timothy or
other kinds, and hence some allowance should be made therefor.
TO FIND HOW MANY TONS OF HAY IN CIRCULAR STACKS.
819. Directions.—Multiply the square of the circumference in yards by 4 times the height
in yards, and divide by 1500; or point off two places and divide by 15.
PROBLEM.
How many tons in a circular stack whose circumference is 30 yards, and whose height is
10 yards ; & Ans. 24 tons.
OPERATION.
302 = 900 × (4 × 10) = 36000; 36000 – 1500 = 24 tons, Ans.
XF PRACTICAL PROBLEMS IN THE MENSURATION OF SOLIDS. T. 4 II
TO FIND ELOW MANY TONS OF HAY IN RECTANGULAR AND SQUARE
STACKS,
820. Directions.—Multiply the length in yards, by the width in yards; then
multiply this product by half the height in yards, and divide the last product by 15.
FROBLEMI.
How many tons in a square stack 40 yards long and 10 yards wide and 9
yards high 3 * Ans. 120 tons.
OPERATION.
40 × 10 = 400; 400 × 44 = 1800 - 15 = 120 tons, Ans.
TO FIND ELOW MANY TONS OF ELAY IN A MOW.
º 821. Directions.—Multiply the length, width and height in yards, and divide
the product by 15.
PROBLEMI.
How many tons in a well settled mow whose length is 60 feet, width 40 feet
and height or depth 30 feet 3 Ans. 1773 tons.
IFIRST OPERATION. SECOND OPERATION.
60 ft. = 20 yds.; 40 ft. = 134 yds.; 30 ft. = 10 yds. 3 || 60
20 × 134 × 10 = 2666; cu. yds. -- 15 = 177% 3 | 40
tons 3 || 30
tº 15
NOTE.-If the mow is very high, as it often is where horse
or steam hay forks are used, the hay will weigh more
per cubic yard and hence the divisor should be 13
or 14.
177% tons, Ans.
TO FIND THE TONS OF HAY IN LOADS.
822. Directions.—Multiply the length, width and height, in yards, and
divide by 20.
PROBLEM.
How many tons of hay in a load 18 feet long, 10 feet wide and 6 feet high 3
Ans. 2 tons.
FIRST OPERATION. SECOND OPERATION.
3 | 18 18 ft. -- 3 = 6 yds. ; 10 ft. -- 3 = 34 yds.; 6 ft.--
3 || 10 3 = 2 yds.
3 || 6 6 x 34 × 2 = 40 cu. yds. -- 20 = 2 tons.
20 or thus:
* tº 18 × 10 × 6 = 2 t
2 tons, Ans. 3-3-3-30 = 4 tons.
4. I2 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. A.
TO FIND THE TONS OF HAY IN A WINDROW.
823. Directions.—Multiply the length, width and height together, in yards
and divide by 25.
PROBLEMI.
How many tons of hay in a windrow 600 ft. long, 6 ft. wide and 5 ft. high
s Ans. 263 tons.
FIRST opBRATION. SECOND OPERATION.
;|º 600 ft. -- 3 = 200 yds.; 6 ft. -- 3 = 2 yds.; 5
3 || 5 ft. -- 3 = 1; yds.
25
200 × 2 × 13 = 666; cu. yds. -- 25 = 263 tons.
-*
263 tons, Ans.
TO FIND THE APPROXIMATE NUMBER OF BUSEHELS OF CORN IN
CRIBS.
IPROBLEMIS.
824. 1. How many bushels of shelled corn in a crib whose inside measure
is 10 feet 6 inches high, 12 feet long, 8 feet wide at the top, and 6 feet wide at the
bottom, allowing 2 cubic feet of corn in the ear to make a bushel? º
Ans. 441 bushels.
OPERATION.
12 ft., length.
7 ft., average width (8 + 6 + 2). or thus: 12 × 7 × 104
21 ft., height (104 ft).
cu. ft. allowed for a bushel. 2
= 441 bushels.
2
2
tº-ºº-ººp
441 bushels, Ans. *
NOTE.—It is found by experiment that from 2 to 24 cubic feet of corn on the cob is required
to give one bushel of shelled corn.
2. How many bushels of shelled corn in a crib of corn in the shuck, the crib
being 10 by 8 feet at the bottom, 15 by 12 feet at the top and 11 feet high, allowing
3# cu. ft. for a bushel ? Ans. 418 bushels.
OPERATION.
Bot. measure, 10 × 8 = 80 sq. ft. on the bottom. 126;
Top “ 15 × 12 = 180 sq. ft. on the top. 11
Sum of top -º- or thus: 34
and bottom, 25 × 20 = 500 sq. ft. in 4 times the middle -->
section between top
6) 760 and bottom.
-*-
418 bushels.
1263 sq. ft. average area of crib.
1263 × 11 ft. height -- 3% = 418 bushels.
This crib is in the form of a frustum of a pyramid.
NotE 1.-See Article 791, page 387, for an explanation of the solution of Frustums of
Pyramidal Solids. Also see page 373.
NOTE: 2. —It is found by experiment that from 3% to 33 cubic feet of corn in the shuck is
required to give one bushel of shelled corn.
PRACTICAL PROBLEMS IN THE MENSURATION OF SOLIDS. 4. I 3
3. A corn crib, filled with corn on the ear, is 20 × 8 feet at the top, and 16
× 6 feet at the bottom and 11 feet high. Allowing two even bushels of corn on the
ear to make one bushel of shelled corn, how many bushels of shelled corn are there
in the Crib 7 Ans. 6963 bushels.
NOTE 1.-The form of this crib is a frustum of a pyramid.
NOTE 2. —For accurate methods of finding the number of bushels or gallons in boxes, bins,
tanks, etc., see page-394.
NOTE: 3.--Most corn cribs have parallel and vertical sides and ends. In all such cases, mul-
tiply the length, width, and height together in the unit of feet, thus producing cubic feet. Then
divide by 2 or 2% as above explained.
*
MEASUREMENT OF GRAIN WHEN IN REGULAR HEAPS.
IPROTRLEMI.
sº
825. A farmer has a heap of grain in a conical form, the diameter of which
is 14 feet 4 inches and the depth 5 feet 3 inches; how many bushels does it contain 3
Ans. 226.906 bu.
OPERATION.
172 Explanation.—The first part of this operation is the
172 same as for a cylinder, and then we divide by 3 for the
2150.4 || 63 reason that the contents of a cone are # of a cylinder of
ſº pºw equal height and diameter. Had the grain been heaped
.7854 against the side of a wall or building, we would divide
3 the result of this operation by 2, because in such a case
the heap would be but # a cone. Had it been placed in a
226.906 b A right angled corner, we would divide the result of this
* * **, operation by 4, for the reason that the pile would then be
+ of a cone.
GAUGING BARRELS, CASKS, ETC.
826. Gauging is a method of finding the contents, or cubical capacities of
casks, barrels, etc. In practice, the measurement or gauging of casks and barrels
is generally performed by means of instruments known as diagonal and ullage or
“wanting” rods.
THE USE OF THE DIAGONAL AND UILLAGE IRODS.
827. To gauge barrels or casks with these rods, place the diagonal rod into
the bung and extend the pointed end to the lower side of each end of the barrel or
cask, and thus find the average or mean diagonal distance; then on the side of the
rod at the point of the mean diagonal distance, will be found the number repre-
senting the full capacity of the vessel in gallons. If the barrel or cask is not
full, then the ullage or “wanting” rod is inserted vertically into the bung, and
at the highest point where the liquid touches this rod will be found a number
4. I4. soulE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
representing the number of gallons “wanting” to make the barrel full. The
difference between the full capacity of the barrel, as shown by the first rod, and
the gallons “wanting” will be the contents of the barrel or cask.
The diagonal rod may be prepared by finding the bung diagonal of a one
gallon barrel of medium curve, which is 7.15+ inches, and then to find the diagonal
number for each successive number of gallons, multiply the 7.15+ inches by the
cube root of the successive numbers, 2, 3, 4, 5, etc., of gallons. Thus, to find the
diagonal number of inches for 8 gallons, we proceed as follows : V8–2; then 7.15+
× 2 = 14.3+ inches.
There is no way of finding the exact capacity of casks, barrels, etc., for the
reason that they are not constructed in exact conformity with any regular mathe-
matical figure. The curvature of the staves, for nearly all kinds of casks and
barrels, is slightly different, and hence all measurements of their contents are only
approximately correct. *
The capacity of a cask is equivalent to that of a cylinder having the same
length and a diameter equal to the mean diameter of the cask.
To find the contents of a cask or barrel without the aid of the diagonal rods,
we first measure the length and the head and bung, or center diameters,
inside measure. Then, having the two diameters, we approximate the mean
diameter according to the quantity of curve, as follows: When the curve is but
slight, we add 4 or .5 of the difference between the two diameters to the lesser dia-
meter; when the staves have a medium curve, we add .6, and when they are above
a medium curve, we add 65 or .7.
Having thus produced the average or mean diameter, we proceed the same
as in the measurement of a cylinder.
A barrel with slightly A barrel with medium A barrel with above
curved staves. curved staves. medium curved Staves.
FROBLEMS.
1. A barrel is 26 inches long, 17 inches in diameter at the head, and 20
inches in diameter at the bung or center. The staves have a medium curve. How
many gallons will it hold 3 Ans. 31.244-H gallons.
OPERATION.
20 in., center diameter. 18.8 18.8
17 in., head diameter. * 18.8 18.8
sº .7854 or thus: .0034.
3 in., difference. 231 || 26 26
.6 adjustment for curve in staves.
tº-º-º-e 31.244096 gals., Ans. 31.244096 gals.
1.8 = .6 of the difference.
17 head diameter added.
18.8 average or mean diameter.
NotE.—By the use of the gauging rods, a barrel of the above dimensions gauged 31-F gallons.

* PRACTICAL PROBLEMS IN THE MENSURATION OF SOLIDS. 415
2. The inside measurements of a certain barrel are as follows: Length, from
chime to chime, 30 inches; bung diameter, 22.8 inches; head diameter, 19.5 inches.
How many gallons will it contain by the mathematical method of gauging?
º Ans. 47.0618-- gallons.
OPERATION.
22.8' = bung diameter. 21.48 21.48
19.5” = head diameter. 21.48 | 21.48
º-msm-mºs 231 .7854 or thus: .0034
3.3" = difference, 30 30 °
.6 -º-º: º-
*== 47.0618-1- gals. 47.0618–H
1.98" = .6 of the difference. gallons.
19.5" = head diameter.
21.48’’ = average or mean diameter.
COMPARISON OF MEASUREMENTS.
828. A barrel of the above dimensions was gauged by the gauging rods
and gave a capacity of 47+ gallons.
The same barrel, when measured by a gallon measure, held 47 gallons and 1
gill.
When the same barrel was measured with a head diameter of 19.4 inches,
the other dimensions being the same as above, it showed a capacity of 46.8867–H
gallons.
1. A cask is 40 inches long, 31% inches head diameter, and 38% inches bung
diameter. What is its capacity in wine and beer gallons?
Ans. 181.93–H gallons wine. 149.02+ gallons beer.
OPERATION. *
38# in., bung diameter. 38.75
31; in., head diameter. 31.50
amºmºmº OI’
74 in., difference. 7 7.25
.7 .7
5.0% = .7 of the difference. 5.075
31.5 head diameter added. 31.5
36.575 average or mean diameter. 36.575
OPERATION OPERATION
TO FIND WINE GALLONS. TO FIND BEER GALLONS.
36.575 36.575
36.575 36.575
.7854 .7854
231 40 282 | 40
181.9313+ gals., Ans. 149.0288-|- gals., Ans.
416 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs, *
TONNAGE OF WESSELS.
829. The Tonnage of a vessel is the number of tons weight it will carry
with safety. º
The carrying capacity of vessels, that is, the amount of space available for
stowing away the cargo is reckoned in tons. . The adopted measurement is one
hundred cubic feet of space to a ton. This measurement does not include the deck
space, where part of the cargo is sometimes carried. The registered tonnage is
always less than the actual tonnage.
To determine accurately the tonnage of a vessel or steamboat, is a very
difficult problem. The methods in practical use do not give accurate results. The
best method probably would be to divide the vessel into several sections, and then,
by numerous measurements, find the average length, width and depth of each sec-
tion, and having by this means found the contents in cubic feet, the actual tonnage
is found by dividing the cubic feet by 40.
The following is the United States method of finding the tonnage of vessels:
Measure in feet, above the upper deck, the length of the vessel, from the
forepart of the main stem to the afterpart of the sternpost. Then measure the
breadth taken at the widest part above the main wale, on the outside, and the depth
from the under-side of the deck-plank to the ceiling in the hold. Then, from the
length take three-fifths of the breadth, and multiply the remainder by the breadth
and depth, and the product divided by 95 will give the tonnage of a single-decker.
If the vessel is double-decked, substitute half of the breadth for the depth,
and proceed as directed for single-decked vessels. *
The following is the carpenters method of measuring the tonnage of vessels:
Measure the length, breadth, and depth, all in feet. Divide the continued
product of the dimensions by 95, and the quotient will be the tonnage.
If the vessel is double-decked, half the breadth is taken as the depth.
NOTE.-For more extended work on the Tonnage of Vessels and Steamboats, see Haswell’s
Book for Engineers and Mechanics, or the Merchants' and Shipmasters' Manual, or the Cyclopedia
Britannica or the Century Dictionary.

DRILL EXERCISE IN MENSURATION OF SOLIDS.
417
DRILL EXERCISE IN MENSURATION OF SOLIDS,
—-4–
880. Statements showing the capacity in gallons of each of the following fig-
ures. The dimensions of each are given above, and the statement to obtain gallons
is given below each figure:
Height 6 feet.
IIeight 6 feet.
º 6 feet.
Height 6 feet.
º
i
; lij
º
§
|
|
|
§
|
º
Diameter 40 in. Diameter of base Upper Diam. 30 in. Each side of base
40 inches. Lower Diam. 40 in. 40 inches.
# Gals.
~ - 3 || 3700
302 = 900 231 |.7854
40? – 1600 72
40 × 30 = 1200 *m.
231 | .7854 3700
72 3
sº*i. - Tranºlam.
* All the sides I) ia meter eter 6 ft. Transversel)iam 6ft.
- 30 inches. g e d g * : A ſh &
Side of lower base 40 inches. 40 inches. Conjugate do 40 in Conjugate 40 in.
40 inches. V- #: Fº
iſſiº Gals.
72
302 = 900 72
40? – 1600 * 231 | .5236
40 × 30 = 1200 40
3700
3
Diameter 60 in. Depth 30 in. Diameter 60 in. Depth 40 in. l)iameter 60 in. Depth 20 in.
& O "
's
**
60 – 2 = 30
30° X 3 – 2700
Gals. 2700 + 20% = 3100
60 Gals.
231 || 60 60 Gals.
.5236 231 || 60 3100
30 .5236 231 20
mºmsºmºmºsºs 40 | .5236






















418 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS. A.
See Problem and Solution,
page 398.
Length 60 in.
#. Diam. 40 in.
IH Diam. 32 in.
A
M-
: D *
See Article 781.
40 – 32 = 8
& is: = #'s See Article 783. See Article 782. See Article 785.
37.62 × 60 × .0034 =
. . . . gallons.
NotE.—To find the capacity in bushels of the above figures, divide by 2150.4 or 2150.42,
instead of 231.
To find the number of cubic feet instead of gallons, divide by 12, 12, and 12 or 1728, instead
of 231.





4% MECHANICAL POWERS. 4 IQ
MECHANICAL POWERS,
—º-
831. The Mechanical Powers consist of six simple machines, or the
elements of machinery, by the scientific use of which we are enabled with a given
force and velocity, to accumulate a greater force with less velocity, or a less force
with a greater velocity; or in other words, to change either the direction or the inten-
sity of a force, or both direction and intensity. These Powers are called the Lever,
the Wheel and Aacle, the Pulley, the Inclined Plane, the Wedge, and the Screw.
To fully understand the nature of a machine, four things must be considered,
viz.: 1st, The force or power which acts on, or is applied to the machine; 2d, The
resistance or weight which is to be overcome or moved by the power; 3d, The center
of motion, called the Fulcrum, which is the point over which the power and resist-
ance act; and 4th, The respective velocities of the power and the resistance.
A machine is in equilibrium when the power and the resistance are equal;
in which case it is either at rest or in a state of uniform motion.
TEIE LEVER,
832. The Lever is a rod or bar of metal, wood, or other material, which is
capable of moving around the fulcrum.
Levers may be either straight or bent; simple or compound. They are
usually divided into three classes, according to the position of the fulcrum in rela-
tion to the weight and power.
FIGURE 1, is a lever of the first class, where the
fulcrum is between the weight and the power.
FIGURE 2, is a lever of the second class, where
the weight is between the fulcrum and the power.
FIGURE 3, is a lever of the third class, where the
power is between the fulcrum and the weight.


42O SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. X.
The Arms of a lever are the lines on each side of the fulcrum, at right
angles to the direction of the power and the weight.
In the three figures given above, the levers being horizontal and the forces
vertical, the arms of the lever are evidently, in each case, the portions into which
it is divided. If however, the lever is bent or is inclined to the direction of either,
4 B or both, of the forces, then the arms are the perpen-
2^ diculars between the fulcrum and the directions of the
X forces. Thus, in FIGURE 4, the power acting in the
\ direction B P, the momentum or force of the power is
\ not expressed by P × A F, but by P × B F. The
(º) P distance from the fulcrum is called the leverage.
When the forces act perpendicularly to the arms of a straight lever, an
equilibrium is produced when the weight multiplied by its distance from the ful-
crum is equal to the power multiplied by its distance from the fulcrum; or, in other
words, the weight is to the power, as the distance from the power to the fulcrum is
to the distance from the weight to the fulcrum.
#
Application of the Lever.—Machines and utensils in daily use furnish us
familiar examples of the three classes of levers:
Of the first class, we name the crowbar and
poker, when used to raise the load or weight on
their points; scissors, snuffers and pincers are
pairs of levers of this class. The point C, FIG-
URE 5, which connects them being the fulcrum.
The power is applied at the handles, and the resistance is the object between
the blades.
Another example of the bent lever
is seen in the ordinary truck, FIGURE
6, used for moving heavy goods a
short distance. In this machine, the
axis of the wheels, F, is the fulcrum,
against which the foot is placed, while
the weight, at R, is raised off the
ground by the hand, applied at P.
The Scale Beam, or balance, is one of the most useful applications of the
first class of levers. The beam is a lever poised at its centre on a knife-edge of
steel, a, FIG. 7. From its ends, A B, are suspended the scale pans, C and E. The
Center of gravity, m, is placed below the fulcrum, a, to secure a horizontal position
of the beam when in equilibrium. If it coincided with the fulcrum, the balance
Would rest equally well in all positions, and if it were above the 5ulcrum the beam
Would be upset by a slight disturbance.


¥ MECHANICAL POWERS. 42 I
The Steelyard is a lever of the first class,
with unequal arms. The mass, Q, to be weighed, is
attached to the short arm, A, FIGURE 8, and it is
counterpoised by a constant weight, G, shifted upon
the longer arm, marked with notches to indicate
pounds and ounces, until equilibrium is obtained.
It is evident that a pound weight at G will balance
as many pounds at Q as the distance G C is greater
than A. C.
Levers of the second class occur less frequent-
ly. A pair of nut-crackers with the fulcrum at
the joint C, FIGURE 9, is a double lever of this
class. An oar is another example; the water is
the fulcrum, the boat is the weight, and the hand
the power. A door moving on its hinges, and a wheelbarrow, are other examples
of levers of the second class.
In levers of the third class, the power, being nearer the fulcrum, is always
greater than the weight. On account of this mechanical disadvantage, it is used
only when considerable velocity is required, or the resistance is small.
The common fire-tongs, sugar-tongs, and 10
sheep-shears are double levers of this class.
The most striking illustrations of this class of
levers are seen in the animal kingdom. The
compact form and beautiful symmetry of ani-
mals depend on the fact that their limbs are
such levers. The socket of the bone, a, FIG-
URE 10, is the fulcrum ; a strong muscle, b c,
attached near the socket, is the power, and the
weight of the limb and whatever resistance, w, may oppose to motion, is the weight.
The fore-arm and hand are raised through a space of one foot, by the contraction
of a muscle applied near the elbow, moving through less than 1-12 that space. The
muscle, therefore, exerts 12 times the force with which the hand moves. The
muscular system is the exact inversion of the system of rigging a ship. The yards
are moved through small spaces with great force, by hauling in a great length of
rope with small force; but the limbs are moved through great spaces with compar-
atively little force, by the contraction of muscles through small spaces with very
great force.
EXAMPLES.
1. In a lever of the first kind, the fulcrum is placed 10 inches from the
weight, the long arm is 8 feet; if a power of 150 pounds be applied at the end of
the long arm, what weight will it balance at the end of the short armº
Ans. 1440 pounds.
OPERATION. Eaplanation.—Since the product of the
150 lbs., power. weight by the distance it is from the fulcrum,
96 in. ión g arm. 150 is equal to the product of the power by the
7 or, 10 | 96 distance it is from the fulcrum, we therefore
I fulcrum together, and divide the product by
1440 lbs., . . length of the short arm, and obtain the
1440 lbs., Ans. AnS. required weight.
*=ºmºmº
10) 14400 | multiply the power and its distance from the



422 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
2. In a lever of the second kind, if a weight of 1200 pounds is placed 2 feet
from the end resting upon the fulcrum, what power will it require to sustain this
weight, the length of the lever being 12 feet 8 inches 3 Ans. 1894%; lbs.
OPERATION.
1200 lbs., weight. 1200
24 in., distance of weight from fulcrum. Or, 152 24
28800 product, which -- 152 (inches length of lever) gives |
1894% lbs., the required power.
3. In a lever of the third kind, what power will it require to sustain a
weight of 80 pounds, the distance from the fulcrum to the power being 4 feet 4
inches, and from the fulcrum to the weight 6 feet? Ans. 110+} lbs.
OPERATION.
80 lbs., weight. 80
72 in., length of lever. Or, 52 | 72
5760 product, which -- 52 gives 110}} lbs., the required power. 110+} Ans.
4. If the orifice of the safety-valve of a steam boiler is 2 inches in diameter
and 5 inches from the fulcrum of the lever, how far from the orifice must, a weight
of 60 pounds be placed so as to just equal the force of the steam when at a press-
ure of 80 pounds to the square inch, making no allowance for the concave surface
of the boiler or weight of lever ? Ans. 15.944 in.
OPERATION.
2” x .7854 = 3.1416 sq. in. on surface of safety-valve.
80 lbs. pressure to the square inch.
251.3280 pressure on safety-valve.
5 in. distance from fulcrum to the orifice.
69 ) 1256.640ſ, product -- the weight gives
20.94.40 in. distance of weight from the fulcrum.
5 in. distance from fulcrum to orifice deducted.
15.944 in. distance of the weight from the orifice.
TEIE WEHEEL AND AXLE.
833. The common lever is chiefly employed to raise weights through small
Spaces, by a succession of short intermitting
efforts. After the weight has been raised, it
must be supported in its new position until the
lever can be again adjusted, to repeat the action.
The wheel and axle is a modification of the lever,
which corrects this defect; and, since it converts
the intermitting action of the lever into a con-
tinuous motion, it is sometimes called the perpet-
wal lever.

MECHANICAL POWERS. 423
This machine consists of a cylinder called the axle, turning on a centre, and
Connected with a wheel of much greater diameter. The power
is applied to the circumference of the wheel, and the weight is
attached to a rope wound around the axle in a contrary direc-
tion. Instead of the whole wheel, the power may be applied to
a handle named a winch, or to one or more spokes inserted in
the axle. When the axle is horizontal, the machine is called
a Windlass, FIGURE 11, when it is vertical it forms the capstan
FIGURE 12, used on shipboard, chiefly to raise the anchor.
The head of the capstan is pierced with holes, in each of which a lever can be
placed, so that many men can work at the same time, exerting a great force, as is
often necessary in raising an anchor, while the recoil of the weight is arrested by
a catch at the bottom.
The law of equilibrium is the same as in the lever.
Draw from the center, or fulcrum c, FIGURE 13, the
straight lines c b and c a, or c a', to the points on which
the weight and power act; a c b, or aſ c b is evidently a
lever of the first class, in which the short arm c b is the
radius of the axle, and c a, or c a', the long arm, is the
radius of the wheel. Hence,
P × a c = W X c b.
P: W = c b : a c.
Or,
That is to say, the wheel and aale are in equilibrium, when the power is to the
weight as the radius of the aarle is to the radius of the wheel.
In one revolution of the machine, the power moves through a space equal to
the circumference of the wheel, and the weight moves through a space equal to the
circumference of the axle; hence the power and weight are inversely as their
velocities, or the spaces they describe.
PROBLEMS,
1. What must be the radius of a wheel, or the length of a crank, by which
500 pounds suspended from an axle of 4 inches radius may be kept in balance by
a power of 40 pounds? Ans. 4 feet 2 inches.
OPERATION.
4 inches radius. 4
500 pounds weight. 40 || 500
Or, - tº
49 ) 2009 product -- 40 the power. 50 in. = 4 ft. 2 in., Ans.
50 in. = 4 ft. 2 in., radius of wheel or length of crank.
Explanation.—The operation of this work is based upon the principle above elucidated, that
the power is to the weight as the radius of the axle is to the radius of the wheel or the length of the crank.


424 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
OPERATION.
2. What must be the diameter of 28 in., length of crank.
an axle in order that a power of 150 150 pounds power.
pounds, applied at the end of a crank 28
inches long, measuring from the center
of the axle, will balance 800 pounds? 54 in., radius of axle.
Ans. 10% inches. 2
8% ) 429% product -- the weight.
104 in., diameter of axle.
3. If the anchor of a vessel weighs 3000 pounds, what force will be required
from each of 5 sailors to raise the same, working at the capstan with bars 54 feet
long, the diameter of the capstan being 16 inches, allowing that each bar enters
the capstan 4 inches 3 Ans. 684 pounds.
OPERATION.
5 ft. 6 in. length of bar.
8 in. radius of capstan.
6 ft. 2 in.
4 in. distance the bar enters the capstan.
5 ft. 10 in. = 70 in., length of bar from the center of capstan.
16 -- 2 = 8 in., radius of capstan.
3000 lbs., weight of anchor.
79 ) 2400% product -- the length of bar.
5) 3424 lbs., power required -- 5, the number of sailors.
684 lbs., force required of each sailor.
THE PULLEY.
834. Fixed Pulley.—The usual form of this machine is a 14
small wheel, turning on its axis, and having a groove on its
edge, to admit a flexible rope or chain. In the simple fixed
pulley, FIGURE 14, there is no mechanical advantage, except
that which may arise from changing the direction of the N
power. Whatever force is exerted at P, is transmitted, with- b
out increase or diminution (except from friction and the º
rigidity of the rope), to the resistance at the other end of the
cord. From the axis C, draw C a and C b, radii of the wheel,
at right angles to the direction of the forces; a C b represents
a lever of the first class, with equal arms; hence, in equilibri-
wm, the power and weight must be equal, and they describe equal
8pacé8.

Wy
* MECHANICAL POWERS. 425
Movable Pulley.—When the block or frame is not fixed, TN 15 X
the pulley is said to be movable. The weight is suspended D F
from the axis of the movable pulley, and the cord is fastened
at one end, and passing over a fixed pulley, is acted on by the
power at the other. In this arrangement, FIGURE 15, it is
plain that the weight is supported equally by the power and
the beam at D. For the pulley acts as a lever of the second
class, whose arms are to each other as 1 : 2; the fulcrum is at
b, b c is the leverage of the weight, and b a the leverage of
the power. The diameter b a is twice the radius b c, therefore
equilibrium will obtain when the power is equal to one-half of
the weight: i. e., b{−Pi—a
P: W = b c : b a = 1 : 2,
therefore,
W
P = +
To raise the weight one foot, each side of the cord must be shortened one
foot, and the power, consequently, passes over two feet. The space traversed by
the power is twice the space described by the weight.
Compound Pulleys.-Sometimes compound pulleys are used,
each consisting of a block which contains two or more single pulleys,
generally placed side by side, in separate mortises of the block. Such
an arrangement is shown in FIGURE 16. The weight is attached to
the movable block, and the fixed one only serves to give the power
the required direction. It is easily seen that the power required at
P is just the same as would be required at any point between A and
B. The weight is divided equally among the pulleys of the movable
block, and, of course, among the cords passing around them ; and as
the power required to sustain a given weight is diminished one-half
by a single movable pulley, it follows that such a system will be in
equilibrium when the power is equal to the weight divided by the number
of cords, or by twice the number of movable pulleys.
P: W = 1 : 2n ; or, P = ".
2n
In this system, as in the single movable pulley, the Space through which the
weight is raised, is as much less than the space through which the power descends,
as the weight is greater than the power. e
*—
P: W = velocity of weight : velocity of power.
If the power is moved through 6 feet, fig. 16, each division of the cord in
which the movable block hangs will be shortened one foot, and the weight raised
one foot.

426 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICs. X.
Another system of pulleys is represented in FIGURE 17.
In this arrangement each pulley hangs by a separate cord,
one end of which is attached to a fixed support, and the other
to the adjacent pulley. The effect of the power is rapidly
augmented, being doubled by each movable pulley added to
the system. The numbers placed near the cords show what
part of the weight is sustained by each, and by each pulley.
Such a system, however, is of little practical use, on account
of its limited range. In the common block system, fig. 16 (in
practice the pulleys or sheaves of each block are placed side
by side, to save room, here they are separated for sake of
clearness), the motion may continue until the movable block
touches the fixed one; but in this only till D and E come
together, at which time A will have been raised only g of
fhat distance.
P; W = 1 + 2*, or, P = }
PROBLEMIS.
1. What power will support a weight of 1200 pounds in two movable
pulleys? Ans. 300 pounds.
OPERATION.
1200 weight + 4, twice the number of movable pulleys, = 300 lbs., Ans.
2. In a compound pulley (see fig. 16), containing three single pulleys in the
movable and immovable blocks, what power will support 3000 pounds?
Ans. 500 pounds.
OPERATION.
3000 - 6 = 500 pounds, Ans.
3. In a block and tackle with 4 sheaves in the fixed block and 3 in the
movable block, what power will be required to support 5000 pounds 3
Ans. 714% pounds.
OPERATION.
5000 -- 7 = 7.14% pounds, Ans.
NotE.—The pulleys in a block and tackle are called sheaves.

‘ſk MECHANICAL POWERS. 427
THE INCLINED PLANE.
835. This mechanical power is commonly used whenever heavy loads,
especially such as may be rolled, are to be raised a moderate height. In this way
casks are moved in and out of cellars, and loaded upon carts. The common dray is
itself an incline plane (as is 18
Clearly seen by inspecting FIG-
URE 18). Suppose a cask weigh-
ing 500 lbs. is to be raised 4
feet by means of a plank 12 feet
long; it is plain, that while the
cask ascends only 4 feet, the Mºtſº
power must exert itself through Wºº
tº:-
12 feet, and hence, 12:4 = 500 : 1663, the force necessary to roll the cask.
In mechanics, the inclined plane is a hard, Smooth surface, inclined obliquely
to the resistance. The length of the plane is R S, figure 19, S T its height, and R.
T its base. The power may be applied,
a—In a direction parallel to the length;
b—Or parallel to the base.
c—Or in any other direction.
In each case the conditions of equilibrium may be derived from those of the
lever.
Application of the power parallel to the length of the inclined plane.—
When a body is placed upon an inclined plane, FIGURE 19, its weight, which is the
19
resistance to be overcome, acts in the direc-
tion of the force of gravity, namely in the
perpendicular b a. Let the power, P, act, by
means of the cord, in the direction a c par-
allel to the inclined plane R S, then from the
point a, draw a d at right angles with the in-
clined plane, and complete the parallelogram
a c b d. The force of gravity will be resolved
into two other forces; one represented by b
c causing pressure on the inclined plane; the
other, represented by c a, tending to cause
motion down the inclined plane. This latter force is to be balanced by the power
applied to move the body. The body will therefore be sustained when
P: W = a c : a b, P: W = S T : B S;
OT
and since the triangles a b c and R S T are similar, y P = W x ST
R.S.
This may be illustrated by an apparatus constructed like that shown in the
figure.





428 soul E's PHILOSOPHIC PRACTICAL MATHEMATICs. *
If the direction of the power is parallel to the inclined plane, equilibrium will
obtain when the power is to the weight as the height of the plane is to its length.
While the weight is raised through a space equal to the vertical height of the
plane, the power must move through a space equal to its length. If the length of
a plane is 10 feet, and its height 2 feet, P must move 10 feet while W is raised 2
feet; hence the power and weight are inversely as their velocities.
Application of the power parallel to the base of the inclined plane.—In
the second case, let the power act in the direction of a 20
P, FIGURE 20, parallel to B C, the base of the plane;
and draw the lines b a and b c perpendicular to the di-
rection of the power and weight; then a b c is a bent
lever, having its fulcrum at b, and equilibrium will take
place when
P: W = b c : a b, P: W = A B : B C;
OI’ &
the triangles a b c and A B O being similar, } P = W X A B
tºmº B C.
If the direction of the power is parallel to the base of the plane, equilibrium
will obtain when the power is to the weight as the height of the plane is to the length
of its base. k
In this case, the space described by the power is to the space described by
the weight as the base of the plane is to its height.
PROBLEMIS.
1. The length of an inclined plane is 10 feet and its height 3 feet, what
power applied parallel to the plane will be required to balance a weight of 500
pounds? Ans. 150 pounds.
OPERATION,
500 pounds, weight. 500
3 feet, height. Or, 10 || 3
1500 product -- the length of the plane, 10 ft., gives 150 | *=.
lbs., the power required.
- 2. The height of an inclined plane is 12 feet and the length 15 feet, what
weight applied parallel to the plane will support a power of 150 pounds?
Ans. 1873 pounds.
OPERATION.
150 pounds, power. - 150
15 feet, length of plane. or, 19 || 15
2250 product -- the height, 12 feet, gives 1874 lbs., the | 1874
*- -- **
required weight.

MECHANICAL POWERS. 4.29
3. The base of an inclined plane is 12 feet and the height 4 feet, what power
applied parallel to the base of the inclined plane will support a weight of 500
pounds? Ans. 1663 pounds.
OPERATION.
500 pounds, weight. 500
4 feet, height. Or, 12 | 4 or, 500 x #
2000 product -- the length of the base, 12 feet, 166;
gives 1663 lbs., the power required.
TEIE WEDGE.
836. Instead of lifting a load by moving it along an inclined plane, the
same result may be obtained by moving the plane under the load. When used in
this manner, the inclined plane is called a wedge. It is custom- 21
ary, however, to join two planes base to base. In FIGURE 21, A
B is called the back of the wedge, A C and B C its sides, and d
C its length. The power is applied to the back of the wedge, so
as to drive it between two bodies, and overcome their resistance.
The resistance may act at right angles to the length or to the
sides of the wedge. In the first case, it resembles an inclined
plane, when the power is parallel to the base; and hence the forces
will be in equilibrium when the power is to the resistance as the back of the wedge is
to its length. In the second case, it is similar to a plane when the power is parallel
to the length; and therefore in equilibrium when the power is to the resistance as the
back of the wedge is to its side. -
If the wedge is isosceles, the power is to the weight as half the back of the
wedge is to one of its sides.
The power is supposed to move through a space equal to the length of the
wedge, while the resistance yields to the extent of its breadth.
Application of the wedge.—As a mechanical power, the wedge is used only
where great force is to be exerted in a limited space. In oil-mills, the seeds from
which vegetable oils are obtained are sometimes compressed with enormous force
by means of a wedge. It is everywhere employed to split masses of stone and
timber. The edges of all cutting tools, as saws, knives, chisels, razors, shears,
etc., and the points of piercing instruments, as awls, nails, pins, needles, etc., are
modified wedges. Chisels intended to cut wood, have their edge at an angle of
about 30 deg.; for cutting brass, from 50 deg. to 60 deg.; and for iron, about 80
deg. to 90 deg. The softer or more yielding the substance to be divided the more
acute the wedge may be constructed. In general, tools which are urged by press-
ure, admit of being sharper than those which are driven by a blow.

43O SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. jºr
The theory of the wedge gives but very little aid in estimating its effects, as
it takes no account of friction, which so largely modifies the results, and the pro-
portion between pressure and a blow cannot be defined.
PROBLEMS.
1. The head of a wedge is 2 inches thick and the length of each inclined
side is 8 inches; what will be the measure of its effect if struck by a force equal to
200 pounds? Ans. 1600 pounds.
OPERATION.
200 pounds force.
8 inches, length of inclined side.
1600 product + 1, which is ; the thickness of the head of the wedge, gives
1600 pounds, the measure of the effect required.
2. The head of a wedge is 4 inches thick and the length of each inclined
side is 15 inches; what pressure, applied to the head of the wedge, is required to
drive it between two surfaces, the resisting force of which is equal to 300 pounds?
Ans. 40 pounds.
OPERATION.
300 pounds, resisting force.
2 inches, # of the thickness of the wedge.
600 product -- 15, the length of the inclined side, gives 40 pounds, the pressure
required.
TEIE SCREW.
837. The screw is a spiral thread, or groove, winding round a cylinder, so as
to cut all the lines drawn on its surface, parallel to its axis at the same angle.
The screw has the same relation to the ordinary inclined plane, that a spiral
staircase has to a straight one.
The thread of a screw projects from the surface of the cylinder, and is de-
signed to fit into a spiral groove, cut in the interior of a block called the nut; a
lever is also fixed in the head of the cylinder to which the power is applied. The
combination of these parts forms the mechanical power technically called the Screw.
In working the screw, the resistance acts on the inclined face of the thread,
and the power parallel to the base of the screw. This corresponds to the case in
which the direction of the power is parallel to the base of the inclined plane.
Yºr MECHANICAL POWERS. 43 I
Equilibrium will, therefore, take place when the power is to the resistance as the dis-
tance between the threads of the screw is to the circumference described by the power.
P; W = h : 2 R. and P = W x sº-
2 Idir
h being the distance between the threads, R the radius of the screw, or of the
length of the lever attached to it, and the ratio of the circumference of a circle
to its diameter.
During each revolution the power describes a circle, whose circumference
depends on the length of the lever, but the end of the screw advances only the
distance between two threads; thus in this, as in all cases of the use of machines,
What is gained in power is lost in velocity.
Mechanical efficiency and applications of the screw.—The mechanical
efficiency of the screw is augmented, either by increasing the length of the lever, or
by lessening the distance between the threads. If the 22
threads of a screw are 3 of an inch apart, and the power à
describes a circle of 5 feet (120 half-inches) circumfer-
ence, a power of 1 pound will balance a resistance of 120
pounds; if the threads are # inch apart, 1 pound will
balance 240 pounds, the efficiency being doubled.
Fine screws are therefore more powerful than coarse
ones. The applications of this most useful mechanical
power are too numerous to mention, but no more fre-
Quent or important example of its use can be named
than is seen in its use in presses of nearly all kinds. A
good illustration of which is seen in the copying press,
FIGURE 22. -
The endless screw is a contrivance by which a slow motion is imparted to a
wheel, as shown in FIGURE 23. The threads of the
screw act upon the cogs of the wheel, and serve to move
the weight Q, attached to the axis M. L. If we call the
radius of the circle described by the winch D B = r,
and let h = the distance between the threads of the
screw, we shall have the power of the screw
O
=#. Let R = MF, and R' = M L,
l,
and the power of the wheel and axle will = #.
Then W: P = 2ar x R : h x R';
2tr R
E P -- tº
w 4 × b. R/
Therefore the weight is to the power as the circumference of the circle
described by the winch D, multiplied by the radius of the wheel, is to the distance
between the threads of the screw multiplied by the radius of the axis.


4.32 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. A.
PROBLEMIS.
e 1. The distance between the threads of a screw is # of an inch, and the
circumference described by the handle, or lever, is 4 feet; what weight can be
raised by a power of 250 pounds, applied at the end of the lever?
Ans. 48000 pounds.
OPERATION.
4 feet circumference of handle.
12 inches in 1 foot.
As {{ {{ 4 4.
4 + “ “ 1 inch.
192 + “ “ 48 inches.
250 pounds power.
48000 “ Ans.
2. The distance between the pitch threads of a screw is # of an inch, and
the length of the crank or lever is 16 inches, what weight can be raised by a power
of 400 pounds applied at the extremity of the lever? Ans. 32.1699.84 pounds.
* OPERATION.
1. inches length of lever
3 1ā; inches diameter of the circle described by the lever.
.141
100.5312 inches circumference of the circle described by the lever,
8 * inches in 1 inch.
804.2496 inches in 100.5312 inches.
400 pounds power.
321699.8400 pounds, Ans.
3. The distance between the threads of a screw is # of an inch, the handle
or lever is 10 feet long, what power will be required to sustain a weight of 50000
pounds? Ans. 26.5257+ pounds.
OPERATION.
10 feet length of handle.
2
20
3.1416
50000
62.8320 10
12 2
Oſ. 3.1416
753.9840 12
5 – # in. 5 2
2) 3769.9200 = } in.
1884.96 = #in.
50000 -- 1884.96 = 26.5257+ pounds, Ans.
MECHANICAL POWERS. 433
In the foregoing work on Mechanical Powers, much of the work, especially
the introduction to the subjects treated, was taken from Silliman’s Principles of
Physics, a book of great worth, and to which we refer for a more comprehensive
discussion of the subject here briefly treated.
RRºº as se
a S$$$$$...s
§§ § Nº.
º
º
Ž
%
Ş N sº § § w V)2 2,…º : 22 2% º ye ę
§§ sº sº *ść ãº
ºğ SRS S$ º ſº º
838. In the above diagram of wheels, A is a pinion with 14 teeth and makes
50 revolutions per minute. B is a spur wheel with 42 teeth and is connected with
pinion A. C is a pinion with 28 teeth and is carried by the shaft of wheel B. D
is a spur wheel with 104 teeth and is engaged with pinion C. E is a pinion with
24 teeth and is carried by engaging wheel B. F is a spur wheel with 112 teeth
engaged with pinion E. A is the engine shaft pinion at 50 revolutions per minute
and is therefore the pinion motor. What is the cycle of the whole double train in
time, and how many revolutions will each wheel make during the cycle?
Ans. 10% min. cycle of the whole train.
A makes 546 revolutions.
B, C and E each make 182 revolutions.
D makes 49 revolutions.
F makes 39 revolutions.
OPERATION.
A makes 50 revolutions per minute.
B #4 × 50 = ** revolutions per minute.
NOTE.-C and E have the same time of revolution as B.
D # x * = ** revolutions per minute.
F #4, x * = ** é & {{ & 4
Since A makes 50 revolutions in one minute, it will make 1 revolution in ºr
minute; in like manner B, D and F will make one revolution in ºr, ſº and
minutes respectively.
The L. C. M. of sº, º, ºr and # = ** or 10# min., Ans, to 1st, question.
Since A makes 1 revolution in ºf min., in 10# min., it will make as many
revolutions as 10# is equal to gº which is 546 times; in like manner B (also C and E),
D and F will make respectively 182, 49 and 39 revolutions during the cycle.
NotE.—See Problem 5, page 215.






...”
839. Per Cent is derived from the Latin phrase per centum, and means
literally on or of the hundred. Or more briefly, the hundredth.
840. The Sign of Per Cent is %. Thus 5% is read 5 per cent.
841. Percentage is a term applied to all calculations in which 100 is used
as the unit of measure or the basis of comparison. It may be applied to all tran-
sactions of business, to all departments of science, and to every thing in nature.
It is in general use in almost every department of practical life.
Thus we say 20% gain or loss; 10% of the number or quantity; 25% better ,
or lower quality; 20 is 80% of 25; 30% increase or decrease of the products of
field, mine, or factory; the general won 75% of the battles fought and lost 45% of
his army; the illiteracy of the State is 10% of the population; the scholarship of
the student is 98%, etc. -
842. In the application or the practical computation of Percentage, the
following elements are considered, viz: the Base, the Rate, the Percentage, and the
A mount or Difference.
843. The Base is the number on which the per cent is calculated.
Thus, in the expression, “What is 5% of 70 ?” the base is 70.
844. The Rate is the number of hundredths to be taken.
Thus, in the question, “What is 5% of 70 ?” the rate is 5.
845. The Percentage is the result obtained by taking the given per cent of
the base.
Thus, in the statement, 5% of 70 is 3.50, the percentage is 3.50.
846. The Amount is the sum of the base and the percentage.
Thus, if the base is 70, and the percentage is 3.5, the amount is 70 + 3.5 =
73.5,
847. The Difference is the remainder obtained by subtracting the percen-
tage from the base.
Thus, if the base is 70 and the percentage is 3.5, the difference is 70 — 3.5
= 66.5.
NOTE.-The relationship between the above elements or terms is such, that when any two of
them are given, the others can be found.
In all problems of per cent, two of these elements must be stated.
848. Since per cent means on the hundred, it is clear that 1% of any number
or thing is the hundredth part of it; and if 1% is the hundredth part, then 2%, 3%,
10%, 25%, etc., are 2, 3, 10, 25, etc., times as many or as much.
Thus, 1% of $400 is $4, and 2% would be twice as much, which is $8; and

(434)
º: PERCENTAGE. 435
1% of 200 pounds would be 2 pounds, and 10% would be 10 times as much, which
is 20 pounds.
Per cent may be expressed as a common or a decimal fraction.
Thus, 2% = +$5, or .02; 3% = rig, or .03; 25% = **, or .25; 624% = ###
or .62%; 100% = }}}, or 1; 125% = ##, or 1.25; 250% = }}}, or 2.50.
7.
849. PER THOUSAND, PER MILLE OR PER M.
In some of the professions and in some lines of business, the unit of measure
or the basis of comparison is 1000. --
Thus, the statistics and the tables of mortality are given per M ; the rate of
exchange is often quoted per M.; the price of lumber is quoted per M feet; and the
price of many kinds of articles of commerce is quoted per M.
NOTE.-M, is the Latin numeral meaning 1000.
PROBLEMS IN PIER CENTAGE.
850. 1. What is 6% of 150 bales? Ans. 9 bales.
OPERATION.
1.50 Explanation.—In all per cent problems of this kind, we first find 1% by
6 pointing off 2 places, and then multiply by the given rate per cent. We
+-º-º-º-º: reason as follows: 19% of any thing is the hundredth part of it; hence 1%
9.00 Ans. of 150 bales is 1.50 bales, and 6% is 6 times as much, which is 9 bales.
2. What is 4% of $5103 Ans. $20.40.
OPERATION.
$5,10 Explanation.—First find 196 by dividing by 100, which is done by
4 pointing off 2 places, and then multiply by 4. We reason thus, 1% of
$510 is $5.10, and 4% is 4 times as much, which is $20.40.
In all percentage operations, we always find 1% first, and then the 96
$20.40 Ans. required; by this system, we will always have in the answer as many
decimal figures as we had in the multiplicand. We should never decimal
the rate per cent.
3. What is 40% of 375 oranges } Ans. 150 oranges.
OPERATION.
3.75
40
150,00 Ans.
4. A stock raiser had 860 head of cattle and they increased 80%. What
was the increase, and how many has he now
Ans. 688 increase. 1548 present stock.
5. A man has $4854.20, and gives to various charitable and educational
institutions 25%. How much has he left 3 Ans. $364065.
6. A factor sold 48524 pounds of cotton at 1242 per pound, and charged
24% commission. What was his commission ? Ans. $151,6375.
7. The population of a city, ten years ago, was 200000. During the past
ten years it has increased 12.4%. What is the present population ? Ans. 225000.
8. A planter owned 550 sheep, and in 1 year they increased 50%; how many
did he then have 3 Ans. 825.
436 SouLE's PHILosophic PRACTICAL MATHEMATICS. º:
9. What is 34% of $1720.15? Ans. $55,90-H.
NOTE.-In practice, whenever the decimal or fraction of a cent, in the answer, exceeds #, it
is counted as a whole cent, and when it is less than 3, it is not counted.
10. What is 124% of a cargo that weighs 1280 tons 16 cwt. 3 qrs. and 20
pounds 3 Ans. 160 T. 2 cwt. 0 qr, 11 lbs. 14 oz.
851. When the rate per cent, is an aliquot part of 100, the per cent may be
obtained by a simple division operation. Thus, to obtain 25% of 940, we divide by
4, for the reason that 25 on the hundred, or ºr, is #, and # of 940 is 235, the
a DSWeI’.
To aid the calculator in this contracted work, we present the following
852. T A P T , FX.
14% = ſº-º; and conversely gº- 14% |50 % = }= }; and conversely # =50 %
#%= ºri “ . . .';= 1..? |534%=###=#; “ . . . =534%
13% = ºats; * * *= 13% |55.9% = ºf “ “ #==55.7%
23% =#=#; “ “ ,- 25% |564%=#= %; “ “ , =564%
34% =#=#6; “ “ # = 3;% |584% =#= , ; “ “ , =584%
64% =+% =#; “ {{ * = 64% |624% =###=#; “ {{ # =62.4%
63% =#=#; “ “ #= 63% 65 % =# =#; “ “ #=65 %
84% = #;= #3; “ {{ # = 84% |663% =###= #; “ {{ # =663%
10 % = Pº, -ºo; “ “ Th; 10 % |68#% =###=++; “ “ +}=683%
124% =###= }; “ “ # =124% |73;%=#=}}; “ “ #=73;%
134% =#=#; “ “ #=134% |75 % =# = #; “ “ # =75 %
163% =#= }; “ “ # =163% |814% =###=#; “ “ ##=814%
18;% =#=#; “ {{ #=18#% |834% =###= #; “ {{ # =834%
20 % =#% = }; “ “ # =20 % |85 % =# = }}; “ “ #}=85 %
25 % = ** = }; “ “ # =25 % |874% =###= }; “ “ # =874%
263% =###=#; “ “ #=263% |913% =###=}}; “ “ +}=913%
31}% =###=#; “ “ #=314%|933% =###=#; “ “ #=934%
33%=#- #; ; ; # =#%|3%=#=#; . . #=::3%
35 % =# =#; “ “ ſº-35 % |95 % =#%=#; “ . . “ , ##=95 %
3.3%=#= #: .. # =#% 100% =}}}= the . º or the
40 % =+** = #; # = O Whole OI a Uning.
413% =###- #: {{ 44 *E* 150% =}#}= one and one-half times the
43#% =###– #; “ {4 *=433% whole number.
45 % = ** = }; “ {{ 3% =45 % |200% =#}}= twº ºnes the Whole num-
463% =###- #; “ 44 # =463% €I’.
ALIQUOT PARTS OF 1000.
624% = +3 1874% = # 416#% = } 625 % = 3 8334% = }
§% = +'s 250 % = + 4373% = # #% = } | 8663% = }}
83; ſº = # 2663% = + 4663% = # 6874% = }} | 875 % = }
125 % = # 312.4% - +3 5334% = # 7334% -> +} 916.3% = +.
1334% = |* | 3334% = } 5624% = # 750 % = } 933}7% = #
1663% = } 375 % = } 5833% = # | 812.4% = # | 937.3% = };
PERCENTAGE. 437
PROBLEMS.
1. What is 20% of $4264.20% Ans. $852.84.
1st OPERATION. 2d OPERATION.
$42.64.20 Explanation.—Since 20% is +
20 5 || $4264.20 of the amount, we therefore,
$ 852.84 Ans i. the º *divide
$ 852.8400 Ans. e g ºnus own ºne co-
2. What is 24% of $500 Ans. $12.50.
3. What is 10% of 450 pounds? Ans. 45 pounds.
4. What is 12.4% of 1600 chickens? Ans. 200 chickens.
5. What is 20% of 444 apples? Ans. 88% apples.
CLASSIFIED PROBLEMS.
853. To find the Percentage and the Amount or Difference, when the Base and
the Rate Per Cent are given.
Or, in Commercial Problems, to find the selling price, when the cost and the
gain or the loss per cent are given.
1. What is 20% of 5503 Ans. 110.
OPERATION.
5.50 Eacplanation.—In all problems of this kind,
20 or 5 || 550 for reasons above given, we first divide by
*=- 100, which is done by pointing off two places
110, Ans. and then multiply by the given rate. Or,
110.00 Ans. when the rate is an aliquot part of 100, we
take such a part of the number as the rate
per cent is part of 100.
2. What is 25% increase of 500 pounds, and what is the amount?
Ans. 125 percentage increase, and 625 amount.
OPERATION.
5.00 e 6
º Eacol tion.—In th robl we find the
25 Or 4 || 500 :planation n Unis problem, we IIn
percentage as above, and then add the same
125.00 125 = percentage. to the base.
125 + 500 = 625 pounds, amount.
854. From the foregoing elucidations we derive the following general direc-
tions for finding the Percentage and the Amount or Difference:
1. To find the Percentage, divide the base by 100, and then multiply by the
rate. Or, take such a part of the base as the rate is part of 100.
2. To find the Amount, add the percentage to the base.
3. To find the Difference, subtract the percentage from the base.
NOTE.--When the base is a compound number, reduce the whole number to the lowest named
denomination; or, reduce the lower denominations to a decimal of the highest.
438 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
3. What is 163% of 1272? Ans. 212.
4. What is 17% of 3850 hats? Ans. 654.50 hats.
5. What is 40% of $1611.15? Ans. $644.46.
6. Bought goods at 24g, and sold them at a profit of 25%. What was the
selling price? Ans. 302.
7. Goods cost $2.40 a yard, and sold at 20% loss. What was the selling
price 2 Ans. $1.92.
8. What is 52% of $514.63? Ans. $267.60–1–.
9. What is 624% of 3240 peaches? * Ans. 2025 peaches.
10. What is 163% of 600 books? Ans. 100 books.
11. A steamer running at a speed of 12 miles an hour increased her speed
12.4%. What was her speed after the increase? Ans. 13% miles per hour.
12. Bought 2462 bushels of corn and sold 333% of it. How many bushels
were left 3 Ans. 1641; bushels.
13. A capitalist had $85000 and invested 25% of it. How many dollars did
he invest ? Ans. $21250.
14. The product of a factory in 1894, was 94.400000 pounds, and in 1895, it
was 12.4% less. How many pounds was the deficiency? Ans. 11800000 pounds.
15. Which is greater, and what is the difference between 8% on 1200, and
7% on 1300% Ans. 1st, 8% on 1200; 2nd, 5.
16. A bankrupt merchant who owed $43000, paid 21%. How much did he pay?
Ans. $9030.
17. A man had $1400 and spent 10% of it, and then 20% of the remainder.
How much did he spend? Ans. $392.
18. Paid $8.50 per barrel for flour and sold it at a gain of 10%. What was
the selling price? Ans. $9,35.
19. Paid $15 for a coat and sold it at 20% gain. What was the profit?
Ans. $3.
20. Goods cost $22.50 per dozen, and sold at 334% gain. What was the gain
per dozen 3 Ans. $7.50,
21. A grocer bought 160 dozen eggs and sold 84% of them. How many did
he sell, and how many dozen are left? Ans. 13% doz. sold; 1463 doz. left.
22. A resides 14 miles from school; he walked 37.4% of the distance and
rode the remainder. Eſow many feet did he ride 3 AnS. 4950 feet.
855. To find the Rate Per Cent, when the Base and the Amount or Difference
are given, or when the Base and the Percentage are given.
Or in Commercial problems, to find the Rate Per Cent when the cost and the
selling price, or the Cost and the Gain or Loss are given.
* PERCENTAGE.
439
1. The base is 40 and the amount is 50. What was the rate per cent 3
OPERATION.
40 = base.
50 = amount, or the base }
+ the Percentage.
10 = Percentage, gain, increase.
100
40
| 25 = gain or increase on 100.
Ans. 25%.
Explanation.—In all problems of this kind, we
first find the percentage, increase, or decrease
the gain or loss, as shown in the operation.
Then by inspection and reason, we see that it
was the base 40 which produced the 10 percent-
age or gain. We then reason as follows: Since
40 gained 10, 1 will gain the 40th part, and 100
will gain 100 times as much, which is 25.
It may and should be here asked, where we
get the 100 and how do we know that the result
represents per cent ? We answer: The 100
comes from the problem itself. The question of the problem, “What was the rate per cent,” if
expressed as it is understood, would read:
What was the gain or increase on 100 % By this we see
where the 100 comes from, and as the reasoning showed that the result was a gain on the 100, it is
therefore clear that it is gain per cent.
2. The population of a city in 1885, was 68000; in 1886, it was 72760. What
was the rate per cent increase?
OPERATION.
68000 = base.
72760 = amount, base + P. or
increase.
4760 = Percentage = increase
or gain.
68000 || 100
| 7 = increase or gain on every
100, and hence is 7%.
Ans. 7%.
Explanation.—The reasoning for this problem
is the same as in the preceding one, and hence
is omitted.
3. A hat cost $3.50 and sold for $4.20. What was the rate per cent gain }
OPERATION.
$3.50 = cost = base.
4.20 = selling price = amount.
702 = gain = percentage.
350 100
20 = gain on every 1002 and hence
is rate % gain.
Ans. 20%.
Ezplanation.—As in the second problem above,
We then
Since $3.50 gain 70c., 10. will gain
the 350th part and 100c. will gain 100 times as
we first find the gain or percentage.
reason thus:
much, which is 20.
4. Goods cost $3.50 and sold for $2.80. What was the rate per cent loss?
OPERATION.
$3.50 = cost = base. .
2.80 = selling price = difference.
70% = loss = percentage.
350 || 100
20 = loss on every 100% and hence
is rate % loss.
Ans. 20%.
Explanation.—In this problem, we first find the
loss or percentage, and then reason as in the
above problem.
44O soul E's PHILOSOPHIC PRACTICAL MATHEMATICs.
*
5. Goods cost $22 and sold at a profit of $5.50. What was the per cent
gain 3 Ans. 25%.
OPERATION.
5.50 = gain = percentage. Explanation.—In this problem, we have the
22.00 || 100 gain or percentage given, and hence have but
wº-tº 4. 25 = gain = gain 76. to make the proportional statement, the reason-
ing for which is the same as above.
6. Yesterday the thermometer registered 76 degrees; to-day it stands 11
degrees lower. What is the per cent decrease in temperature? Ans. 14 °, 7%.
OPERATION. g
11 = percentage = decrease. Explanation.—The reasoning for the propor-
76 || 100 tional statement of this problem is the same in
the main as in problem 3, and is therefore
14; , 7% decrease in temperature. omitted.
856. To find what per cent one number is of another, or the Base and Percent.
age given to find the 76 or Rate.
1. What per cent of 15 is 12? Or, 12 is what per cent of 153 Ans. 80%.
OPERATION.
100% of 15 is 15. Eacplanation.—In this problem, 15 is the base or unit of
dº comparison; and to produce the required proportional Ill IIIl-.
15 100% bers to solve the question, we find 100% of it, and reason.
12 thus: Since 100% gives 15, conversely 15 required 100%;
hence 1 will require the 15th part and 12 will require 12 times as,
80% Ans. much, which is 80%.
OPERATION.
100% of 12 is 12.
2. What per cent of 12 is 15? %
or 15 is what per cent of 12? 100
Ans. 125%. 12 || 15
125% Ans.
857. To find what % one number is greater than another, or the Amount and
Base given to find the ſ, or Rate.
1. What per cent is 15 more than 12? Or, 3 is what % of 12? Ans. 25%.
OPERATION.
100% of 12 is 12. Ea:planation.—Here 12 is the base or unit of comparison;
and to produce the required proportional numbers to solve
f 0 the problem, we find 100% of 12 and reason thus: Since 100%
12 | 3 gives 12, conversely 12 requires 100%; and since 12 requires
| 100%, 1 will require the Pº part and 3 will require 3 times as,
25% Ans. much.
* . PERCENTAGE. 44. I
858. To find what % one number is less than another, or the Difference and
Base given to find the 7% or Rate. *.
* OPERATION.
100% of 15 is 15.
1. What per cent is 12 less than 15 + fº
or 3 is what % of 15? Ans. 20%. 15 || 3
20% Ans.
859. From the foregoing elucidations, we derive the following directions
for finding the rate per cent:
1. To find the Rate % when the Base and Amount, or the Base and the Differ-
ence, or the Cost and the Selling Price are given, first find the percentage—the increase
or decrease, the gain or loss—and then make the proportional statement shown in the
operations; or thus : The base or cost: the increase or decrease, the gain or the loss : :
100: the required rate per cent.
2. When the base and the percentage of increase or decrease, or the base and the
gain or loss, are given, then make the above proportional statement at once.
Or, if it is desired to ignore all processes of reasoning, thus : Multiply the
increase or decrease, the gain or loss, by 100, and divide by the base or cost.
PROBLEMIS.
1. The base is 1750; the percentage is 43.75. What is the rate %
tº Ans. 24%.
2. A yard of cloth cost 162 and sold for 182. What was the gain ſº
Ans. 12; 9%.
3. Paid $120 for a horse and sold it at a profit of $22. What was the
gain Ż, 3 Ans. 18; 7%.
4. In a population of 240000, there were 6214 deaths in 12 months. What
was the rate per cent 3 Ans. 2.58++%.
5. What per cent of 54 is 6? Ans. 11% 7%.
6. What per cent of $540 is $67.50% Ans. 12.4%.
7. $2.50 is what % of $60,40? º * Ans. 4 ſº, 7%.
8. 9 is what % of 216? Ans. 4% 7%.
9. An invoice of goods cost in New York $3840; the freight was $57.60.
What was the rate % Ans. 14%.
10. On a bill of $421.80, a discount of $21,09 was allowed. What was the 7%
discount # * Ans. 5%.
11. What % of a number is ; of it? Ans. 624%.
OPERATION.
Let 100 = the number. Then 3 of 100 is 62%, and 62% on the hundred is 624%.
12. What % of a number is 5% of 15% of it? Ans. #%, or .75%.
442 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. . . . .
13. What % of a number is 10% of 20% of 25% of it? Ans, sº, or 5%.
14, 15% of 60 is what % of 75% Ans. 12%.
15. 40 is what per cent of 83 Ans. 500%.
16. 6 is what % of 24? Ans. 25%.
17. A passenger train runs 25 miles an hour; an express train runs 35 miles
an hour. What % slower does the passenger train run, and what % faster does
the express train run, when compared with each other ?
- Ans. Pass. train runs 284% slower. Ex. train runs 40% faster.
18. Single tickets cost 25%; in packages of 10 they cost 20g each. What is
the 7% gain by buying by the package, and what is the ſº loss by buying by the
single ticket? Ans. 25% gain by buying by the package.
20% loss by {{ “ single ticket.
19. What % of a number is # of it? Ans. 68%.
20. What is 2% of 124% of 663% of 25% of a number 3 Ans. # 7%.
OPERATION.
100 assumed as the number. Then rºw x # × ### × #5 or rāg x * x
#o; or 5% × # × 3 × # = gºod.
Then ga'on of 100 is gº, and gº on the hundred is gº. 76.
21. In the above problem, what would be the percentage on $4800 3
Ans. $2.
0 Q
#
X
O
O
OPERATION.
gº, 7 of $4800 is $2 or thus: 25% of $4800 is $1200 : 663% of $1200 is $800 : 12.4%
of $800 is $100; 2% of $100 is $2, answer.
22. What is 48% of 163% of 37.4% of 624% of a number 3 Ans. 13%.
23. The cotton crop of the Southern States ending the fiscal year, Sept. 1,
1884, was 5713200 bales; and for the fiscal year ending Sept. 1, 1885, was 5655900
bales. 1. What % more was produced in 1884 than in 1885? 2. What % less in
1885 than in 1884 3. What % is 5713200 bales of 5655900 bales # 4. What % is
5655900 bales Of 5713200 % Ans. 1.0131-H ſº more was produced in 1884.
1.00296-1- ?, less “ {{ 1885.
101.01314-?, 5713200 is of 5655900.
98.997.05+%, 5655900 is of 5713200.
24. A man has due him $45, and compounds on receipt of $36. What % did
he lose ? Ans. 20%.
25. A broker bought bonds at 90% on the dollar and sold them at 95% on the
dollar. What % did he gain } Ans. 53%.
26. In a year of 365 days, 67 days are rainy. What % of the days are not
rainy ? Ans. 8144%.
27. According to the Carlisle mortality tables, 43 persons of every 5879 of 25
years of age die annually. What is the rate % of deaths? Ans. .731-H 7%.
860. To find the BASE when the Amount or Difference and the Rate per cent
Increase or Decrease are given.
Or, in Commercial Problems, to find the cost when the Selling price and the
Gain or Loss per cent are given.
º PERCENTAGE. 443
1. The manufactured value of goods is $2100, which is 20% more than the
Value of the raw material. What was the value of the raw material?
Ans. $1750.
FIRST OPERATION.
$100 = assumed base or value of raw Explanation.—By considering the problem, we
material. see that the $2100 is the amount of the lº Of
9ſ) — e the raw material and the 20% cost to manufac-
20 = 20% increase value. ture the goods, we also see"that the 30% was
calculated on the value of the raw material,
$120 = manufactured value. and not on the $2100, the value of the manufac-
tured goods. Hence there are no figures in the
problem upon whose face we can calculate the
4) 100 = assumed base. 20% cost to manufacture. We therefore, as
120 | 2100 shown in the operation, assume 100, as the base
*- ºmºmºm- or value of the raw material, and on this we
$1750 cost of raw material. calculate and add thereto the 20% cost to
manufacture. This gives an amount of $120, as
the manufactured value of goods from a base or
raw material value of $100,
Now with these values which contain the same ratio of base and amount, or of raw material
and of manufactured goods, that exists between the $2100 of manufactured goods and the required
value of the raw material from which it was produced, we make the proportional statement,
shown by the operation and obtain the required base or raw material value.
In making the solution statement, we place the $100 assumed base or value of raw material
on the line and reason thus: Since $120 amount or manufactured value at a gain of 20% require
$100 base or value of raw material, $1 will require the 120th part, and $2100 amount or manufac-
tured value will require 2100 times as much.
In assuming a number to represent cost, it is immaterial so far as correct results are
ooncerned, what number we assume; but we always select 100 for the reason that per cent being
on the hundred, our operation is facilitated to a greater extent by 100 than by any other number.
The foregoing reasoning and solution are applicable to all problems of like character,
regardless of the rate of gain or loss per cent. But when the gain or loss per cent is an aliquot
part of 100, the operation may be very much shortened by the following process of work and
reasoning :
SECONTD OPERATION.
6 ) $2100 Explanation.—In this solution, since the rate
Tº RTNT_ 1 l f % is an aliquot of 100, we reason thus: Since
350 = # of base or value of r. m. the amount, $2100, contains a gain of 20%, and
5 since 20% is equal to # of a thing, the $2100 is
-*msºmº- therefore the base or value of the raw material,
$1750 = base or value of r. m. plus + of the same, which makes #, and since
$2100, is ; of the base, # of the base is the #
part, and ; or the whole base is 5 times as much.
2. Sold goods for $40 and gained 25%. What was the cost of the goods?
Ans. $32.
FIRST OPERATION,
$100 = cost assumed. ^. Explanation.—Here, omitting the explanatory
25 = 25% gain added. remarks given in the preceding problem, We
º assume $100 as the base or cost and reason thus:
$125 = selling price. To gain 25% we must sell what cost $100 for
$ $125; then since $125 selling price at a gain of
100 25%, required $100 cost, $1 selling price will
125 | 40 require the 125th part, and $40 selling price will
* : *- require 40 times as much, which is $32
$32 cost, Ans.
444 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
SECOND OPERATION.
5) $40 = selling price. Explanation.—For this operation, We T63, SOIl
T8 = + of cost. thus : Since the $40 selling price contains a gain
4 of 25% and since 25% is 4 of the cost, the $40 is
hence # of the cost; and since $40 is # of the cost,
sº- # of the cost is the # part, which is $8; and # or
$32 cost, Ans. the whole cost is 4 times as much, which is $32.
3. Sold goods for $30 and lost 25%. What did the goods cost? Ans. $40.
IFIRST OPERATION.
$100 = cost assumed. Ea:planation.—Thereasoning for the proportion-
25 = 25% loss deducted. al statement is as follows: Since $75 selling
-msm price at a loss of 25% required $100 cost, $1
$75 = selling price. selling price will require the 75th part, and $30
$ selling price will require 30 times as much.
100
75 30 NOTE.—In practice the line statement is all
that needs to be made. The assumed base or cost
-s | º and the amount or selling price resulting there-
$40 cost, Ans. from, should be mentally performed.
SECOND OPERATION.
3) $30 = selling price. Explanation.—In this solution, we reason as
10 = # of cost follows: Since the $30 selling price is the cost
º º less 25%, it is therefore # of the cost; and since
# of the cost is $30, 4 is the # part which is $10,
tºº-ºº- and # or the whole cost is 4 times as much which
$40 cost, Ans. is $40.
861. From the foregoing elucidations, we derive the following general direc-
tions for finding the base or cost when the amount, or difference, or selling price,
and the rate per cent are given :
1. To find the Base or Cost, first assume 100 to represent base or cost; on it
calculate the given rate % to find the percentage; then either add it to or subtract it
from the base or cost according as the rate % is a gain or a loss, an increase or a
decrease, and thus produce an amount or difference, which has the same ratio of value
to the 100 assumed base or cost as the given. Amount or Difference has to the required
base or cost. Now, with this ratio of values and the given amount or difference, make
the proportional statement shown in the operation ; or thus : The produced amount or
difference : the assumed base or cost: : the given amount or difference : the required
base or cost.
2. If it is desired to ignore all processes of reasoning in the solution, the
following brief and arbitrary directions will solve this class of problems :
Multiply the given amount or difference or selling price by 100, and divide by 100
plus the gain ſº, or minus the loss %. *
PROBLEMS.
4. Sold goods for $20,624 and gained 25%. What did the goods cost?
| Ans. $16.50.
PERCENTAGE. 445
5. Sold a watch for $70 and lost 20%. What did it cost? Ans. $87.50.
6. The rent of a house is $67.50 per month, which is 124% advance on last
year's rent. What was the rent last year? Ans. $60.
7. The amount is 2663, the rate % is 334. What is the base? Ans. 200.
8. The difference is 400, the rate % is 20. What is the base? Ans. 500.
9. What number increased by 10% of itself is 1815? Ans. 1650.
10. $18.17 is 15% more than what sum ? Ans. $15,80.
11. $4.50 is 50% less than what sum ? Ans. $9.
12. A produce merchant sold corn at 6042 per bushel, which is 10% more
than he paid for it. What did he pay for it? & Ans. 55%.
13. A carpenter, after using 80% of his lumber had 500 feet on hand. How
many feet had he at first 3 Ans. 2500 feet.
14. A shoe factory manufactured 2000 pairs of shoes, which weighed 5500
pounds, not considering the thread and nails. There was a waste of 84% in
manufacturing. How many pounds of leather were used in making the shoes?
\ Ans. 6000 pounds.
15. Sold sugar at 84% per pound and gained 64 per cent. What did it cost?
Ans. 82.
16. My merchandise account is debited with $78500 and credited with
$72225. The gain / on sales has been 124. What was the cost of the sales, and
how much merchandise have I on hand 3 Ans. $64200, cost of sales.
$14300, mase. on hand.
17. A planter lost by a storm 20% of his grain and has 29120 bushels left.
How many bushels had he before the storm, and how many bushels did he lose ?
Ans. 36400 bu. before the storm. 7280 bu, lost.
18. Sold goods for $100 and gained 40%. What did they cost 3
Ans. $71.42%.
19. Sold goods for $175 and gained 150%. What did they cost?
Ans. $70.
20. What number increased by 25% of itself is $535? Ans. $428.
21. $72 is 20% more than what sum ? Ans. $60.
22. What number diminished by 25% of itself is 5103 Ans, 680.
23. A firm lost 14% of its capital by the failure of another house. The
remaining capital was $278864,46. What was the whole capital? Ans. $324261.
STATEMENT.
$
100 = capital.
86 || 278864.46.
24. A firm pays $9600 rent for its store, which is 20% more than it paid
last year. What was last year's rent? - Ans. $8000.
25. If 40 boxes of peaches be sold for $113.75, at a gain of 624%, what was
the cost and selling price per box : Ans. $1.75 cost. $2.84; selling price.
446 soulE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
862. To find the Base when the Rate per cent and percentage are given.
Or, in Commercial Problems, to find the cost when the Rate per cent and the
Gain or Loss are given.
1. The rate per cent is 8 and the percentage is 36. What is the base?
Ans. 450.
OPERATION.
100 = assumed base. fi Explanation.—In all º”. ‘ā, this kind, we
rst assume 100 base and find the percentage
8% calculated on assumed base. thereon at the given rate %. By this work we
** produce a base and a percentage containing the
8.00 = percentage on assumed base. same ratio that exists between the given per-
centage and the required base. Having these
100 = assumed base ratio numbers, we then make a proportional
8 | 36 † statement as shown in the operation.
In this problem we find the percentage to be
wºm-º. 8. Then placing the assumed base on the state-
450 base, Ans. ment line, we reason thus: Since 100 base at 8%
produced 8 percentage, conversely 8 percentage
required 100 base; and since 8 percentage requires 100 base, 1 percentage will require the 8th part,
and 36 percentage will require 36 times as much, which is 450.
NotE.—In practice, only that part of the operation shown by the statement line should be
made; the other work should be mentally performed.
2. 40 is 5% of what number 7 Ans. 800.
OPERATION INDICATED.
100
5 | 40
3. On a sale of a lot of goods, a loss of $17 was sustained. The % loss was
20. What was the cost 3 Ans. $85.
OPERATION.
100 = assumed cost.
20%
$20.00 = loss on assumed cost. Explanation.—In this problem, the reasoning is
the same as in the first problem, and hence is
100 omitted.
20 || 17 &
| $85 cost, Ans.
863. From the foregoing elucidations, we derive the following general direc-
tions for finding the base or cost when the rate per cent and the percentage or gain
or loss are given :
1. To find the Base or Cost, first assume 100 to represent base or cost, and then
Jind the percentage or gain or loss thereon at the given Rate %. Then, for reasons
given in the eaplanation of the first problem, make the proportional statement as shown
in the operation.
* PERCENTAGE. 447
- 2. If it is desired to ignore all reasoning in the solution, then multiply the
percentage, gain or loss, by 100, and divide by the given rate %.
PROBLEMIS.
4. The rate per cent is 10 and the percentage is 108. What is the base ?
Ans. 1080.
5. 3021 is 12.4% of what number 3 Ans. 24168.
6. Goods were sold at a profit of 25% and a gain of $5.60 was realized.
What was the cost 3 Ans. $22.40.
7. Sold a watch for $26.50 more than it cost and gained 50%. What did it
cost, and what was the selling price? Ans. $53 cost. $79.50 selling price,
8. City taxes are 2%, and a man paid $886.42. What was the assessed
value of his property ? - Ans. $44321.
9. Ełła is #76 of what number ? Ans. #.
10. 84 is 1% of what number ? Ans. 850.
11. State taxes are .6%, and a man paid $67.50. What is the assessed value
of his property ? Ans. $11250.
12. 5% of $180 is 12.4% of what sum ? Ans. $72.
13. A man lost $35.88, which was 8% of what he had at first. How much
did he have left 3 Ans. $412.62.
14. A planter sold 2430 barrels of molasses and had 25% of his yearly
product left. What was his yearly product? Ans. 3240 barrels.
864. To find the Gain or Loss per cent, when the Cost and the Selling price
are given. -
1. Goods cost $3 and sold for $4. What was the gain ſ, 3 Ans. 33.3%.
NOTE.-For the operation and reasoning for this class of problems, see Article 855.
2. Bought a horse for $180 and sold it for $160. What per cent was lost 3
Ans, 114%.
3. Sold flour which cost $7.50 at $8 per barrel. What was the gain per
cent 3 Ans. 63%.
4. Bought shirts at $16.50 per dozen and sold them at $1.75 a piece. What
% did I gain 3 Ans. 27# 7.
5. Bought 3 apples for 102 and sold them at 5% each. What % did I gain?
Ans. 50%.
6. Bought tea at $1.25 per Ib, and sold it at 74% per oz. What % did I lose ?
Ans. 4%.
865. To find the Cost, when the Gain per cent and the Selling price or the Gain
or Loss are given *
1. Sold goods for $15 and gained 25%. What did they cost? Ans. $12.
NOTE.—For the operation and reason for the solution of this class of problems, see Article
859.
448 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
2. Sold a vest for $2 more that it cost and gained 333%. What did it *
Ans. $6.
OPERATION INDICATED.
100
100 cost assumed, 33% ſº of 100 = 334; then 334 2
3. If 32 per pound is lost by selling coffee at a loss of 20%, what was the
cost 3 Ans. 15% per Ib.
4. Sold goods for $19.80 per dozen and lost 10%. What was the cost per
single article 3 Ans. $1.83%.
5. A loss of 12 per pound was sustained by selling at 7 & 76 loss. What
was the cost 3 Ans. 13%.
6. Sold corn at 564% per bushel and gained 124%. What did it cost 3
Ans. 502.
7. Sold gloves at a profit of 252 per pair, and gained 30%. What was the
cost per dozen : Ans. $10.
8. Sold butter at 352 a pound and lost 63%. What was the cost 3
& * Ans. 3742.
9. I gained $15 by selling a horse at 163% gain. What did the horse cost
me? Ans. $90.
10. A clothier sold a coat for $18 and thereby lost 11:#7% What did the coat
cost 3 Ans. $20.25.
11. A dealer asked $25 for a suit of clothes, but took off 10% to effect a sale,
and thereby gained 12.4%. What did the suit cost him 3 Ans. $20.
12. 144% was lost by selling silk after deducting 25% from $4 per yard
asking price. What was the cost per yard 3 Ans. $3.50.
ONE SELLING PRICE AND THE GAIN OR LOSS PER CENT OF THE SAME BEING GIVEN,
TO FIND WHAT IT SHOULD BE SOLD FOR IN ORDER TO GAIN OR LOSE A
DIFFERENT PER CENT.
866. 1. Sold merchandise for $196 and lost 30 per cent; what should it
have been sold for to gain 20 per cent 3 Ans. $336.
FIRST OPERATION.
$ C
109
79 196
$280 cost.
56 = 20% gain.
$336 Ans.
SECOND OPERATION.
S. P.
$100 cost assumed. $100 cost assumed. $120
20 = 20% gain added. 30 = 30% loss. 70 | 196
$120 selling price. $70 selling price. $336 Ans.
NotE.—In practice, the assuming of figures and the adding to and deducting from them
should always be mental work.
2. Sold goods for 18% and lost 10 per cent; what should they have been sold
for, to gain 10 per cent? Ans. 222.
PERCENTAGE. 44.9
3. Sold hats for $4.50 and gained 50 per cent; at what price should they
have been sold to gain 25 per cent? Ans. $3.75.
4. Sold boots at $50 per dozen and lost 163 per cent; at what price per pair
should they have been sold to gain 20 per cent? Ans. $6.
5. Sold goods for 21+2 and lost 15 per cent? for what should they have been
sold to lose 10 per cent ? Ans. 224%.
TWO SELLING PRICES, AND THE GAIN OR LOSS PER CENT OF THE FIRST BEING
GIVEN, TO FIND THE GAIN OR LOSS PER CENT OF THE SECOND.
867. 1. Sold goods at $12 and lost 25 per cent; what per cent would have
been gained or lost had they been sold for $18? Ans. 124% gain.
FIRST OPERATION. SECOND OPERATION.
$ C G. $
100 $2 13 3
75 12 16 || 100 2 12 75
$16 cost. 12#% gain. Ans. 225
18 selling price.
um==ºe 1123,
$2 gain. 100
124% Ans.
2. If 20 per cent is gained by selling goods at 242, what per cent would be
gained if the goods were sold for 30¢ 3 Ans. 50%.
3. Sold flour at $4.80 per barrel and lost 40 per cent, what per cent would
have been gained or lost if the flour had been sold at $7.80 per barrel?
Ans. 24%.
4. If sugar is sold at 1232 per pound, and a gain of 124 per cent realized,
what would be the gain per cent if the sugar was sold at 13+2%
Ans. 20.4% gain.
TO COMPUTE PER CENT ON ENGLISH MONEY.
868. 1. What is 3% of £ 647 18s. 5d.” Ans. £19. 8s. 9d.
FIRST OPERATION, SECOND OPERATION.
36 6.47.18.05 fº 6.47.18.05 Ea:planation.—In the first operation,
3 3 we divide the £, s. and d. each by
100 as shown by the decimal points
sº ===ºmºsºmºmºmºmºsºms and thus obtain 196. Then we mul-
£19.41.54.15 = 3%. fºl9.43.15.03 tiply by 3 and obtain 396. We then
20 20 reduce the hundredths of £. s. and d.
to shillings and pence, according to
8s.74 8S.75 the principles of denominate numbers,
S.'ſ S. as shown on page 260.
12 12 In the second operation, we found 1%
ºsmº º gºmºmºmºe as in the first operation, and then, when
9d.03 9d,03 multiplying by the required rate per
cent, we reduced mentally the .156 to
hundredths of shillings,” by dividing
by 12, and the .55s, to hundredths of pounds, by dividing by 20. Then we reduced the hundredths
of £. s. and d. to shillings and pence as in the first operation.
45O soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. *
TO REDUCE SHILLINGS AND PENCE TO THE DECIMAL OF A POUND,
AND TO REDUCE THE DECIMAL OF A POUND
TO SEIILLINGS AND PENCE.
869. 1. Reduce 12s. 6d. to the decimal of a pound.
SECOND OPERATION.
Multiply the shillings by .05 and the pence by .004% and
12s. 6d. = £0.625 add the results together. Thus:
12 12s. 6d. 12s. 6d.
em-º- .05 .004; or thus, 5 (hund’ths) 4% (thous'ths)
240) 150 (.625 º- tº-º-º: -b
.60--.025=3.625. 60 (hund'ſ hs)+25 (thous'ths)=36.625
NoTE.—The multipliers in the second operation are produced thus:
Since 20s. = £1., 1s. = £ºſ, or .05 of a £.
Since 240d. = £1, 1d. = £3+n, or .004% of a £.
In like manner the decimal of a farthing would be £540, or .001 ºr of a £.
2. Reduce .625 of a pound to shillings and pence.
Explanation.—In the 2d operation of the first prob-
lem, we showed how the multipliers, .05 and .004%,
OPERATION. were obtained and used to reduce shillings and pence
to the decimal of a pound. Now since the decimal
FIRST OPERATION.
Divide .62 N 12S was produced by these multiplers it is clear that by
by 5 - *** or thus, using them as divisors, we will reproduce the shillings
º 625 and pence. We first divide the first two figures of
Divide .025 o the decimal, .625 by .05 which gives 12s. and .02 over.
IV1C16) . } = 6d. We then divide the .025 by .004; and obtain 6 pence.
by 4% 12s. 6d. In practice, the operation is performed by dividing
the first two figures of the decimal by 5 and the
remainder, if any, and the third decimal figure by 4%.
PROBLEMIS.
3. Reduce 15s. 11d. to the decimal of a pound. Ans. £.796:
4. Reduce .679 of a pound to shillings and pence. Ans. 13s. 7d.
5. What is 3% of £647 18s. 5d. } Ans. £19 8s. 9d.
SOLUTION OF THE FIFTH PROBLEM.
FIRST OPERATION. SECOND OPERATION. * THIRD OPERATION.
£647. 18s. 5d. = £ 647.921 :6647 18s. 5d. f:647 18s. 5d.
r 3% 3% %
£19.43.763 #19.41.54.15 :E19.43.15 3
20 20 20
8S, .75260 8s. 74 8S.75
12 12 12
9d. .0312 9d. 03 9d. 03
6. What is 5% of £2614 1s. 2d. Ans. £130 14s. 1d.
7. What is 1% of £100 1s. 1d. Ans. £1 0s. 0.136.
8. What is 4 per cent. of 12s. 7d. } Ans. 3f.
9. What is 60 per cent. of £500 1s. 8d.: Ans. £300 1s. 0d.
NOTE—See pages 484, 569, 661, 744 to 758 for further work in English money.
PROFIT AND LOSS.
870. Profit and Loss are terms which express the gain or loss in business
transacted, and which involve the principles of percentage in all its various com-
binations, as applied under different subjects, such as Commission, Brokerage, etc.
The subject is of great practical value and should receive the thoughtful
attention of the business aspirant.
* PERCENTAGE. 451
Profit or Gain is the sum resulting from business or from business trans-
actions in excess of the Cost and Expense of the Business.
Loss is the sum spent or incurred to carry on business or to effect business
transactions, in excess of the receipts or income of the business.
MARKING GOODS.
871. To Mark Goods at a Given Gaim or Loss Per Cent.
1. Bought butter at 25¢ per pound. At what price must it be sold to gain
25 per cent ? Ans. 31.4%.
25
OPERATION. 64 = 25% gain.
25% = ** = *-i-º-º
31#2, Ans.
2. Bought cheese at 15% per pound. At what price must it be sold to lose
10 per cent 3 Ans. 1342.
15
OPERATION. 1.5 = 10% loss.
1.
T
13.5%, or 1342, Ans.
NOTE.—When the per cent. gain or loss is an aliquot part of 100, all we have to do, when
marking goods, is to add to or subtract from the cost, such a part of itself as the rate per cent, is
part of 100.
10% = }}
O
PRACTICAL OPERATIONS IN MARKING GOODS.
872. In marking goods, it is the custom of most merchants to use a private
mark, either of arbitrary characters or letters, instead of figures, to show the cost
and selling price, or, as some mark goods, to show the cost, the wholesale and
the retail price. If letters are used, any word or phrase that contains 10 different
letters may be selected.
Any of the following words or phrases may be adopted for a Marking word
or Key word :
Cash Profit. Regulation. Importance. |Black Horse. Hard Moneys.
Now be sharp. Washington. My Wife's Hat. Dont be lazy. Brick House.
Buy for Cash. Corn Basket. Purchasing. God help us. X. Lord Save it.
To illustrate the work, we will select the words “Cash Profit,” and with the
ten letters represent the ten figures as follows:
1 2 3 4. 5 6 7 8 9 0
C. A. S. H. P. R. O. F. I T
To avoid the repetition of any letter or character, an extra letter or character
called a “Repeater,” is used. By this means, the uninitiated are less likely to
discover the key word or mark. -
When the repeater is used, it should follow the letter or character which it
repeats.
452 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. * .
With the above key word, and L as a repeater, what is the mark for the cost
and selling price of goods that cost $18.60, on which 20% gain is required?
Ans. C F R T cost. A L S A selling price.
Instead of using one mark, as above, for cost and selling price, it is the
custom of many merchants to use two different marks, one for cost and one for
selling price.
To illustrate this work, we will use the above mark, Cash Profit, for the selling
price mark, and select the word Regulation for the cost price mark. Our marks
then stand as follows:
Oost Price Mark. Selling Price Mark.
1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0
R. E. G. U L A T I O N C A S H P R O F I T
With these key words, and B as a repeater, what is the mark for the cost and
selling price of goods that cost 50g per yard and that are held for sale at a gain of
10 per cent 3 Ans. L N cost. P B selling price.
NOTE.--To distinguish dollars from cents, some use the dollar or cent mark, some the decimal
point, and some make no designation of either in the mark, but consider the mark as all cents.
Instead of using words for marks, many merchants use characters. Any ten
characters may be selected to represent the ten figures. The following double
cross is frequently used, thus:
The double cross may be
1 || 2 | 3 iº used in many ways, thus:
With 0 as the
alsº tenth charac- 1 2 3 4 5 6 7 8 9 0 3|4|8 7 || 8 || 9
ter and x as L. L T, [] D T [..] [T O 5 || 2 || 7 6 || 1 || 2
7 || 8 || 9 a repeater.
8|9||1 5|4|3
With Buy for Cash, as a cost mark, using D as a repeater, and with the
characters shown in the first double cross above as the mark for the selling price,
mark the cost and selling price of the following articles, at the gain or loss per
cent. given :
F D H
Cost $ 4.40 (a) 10% loss, mark, tº
Cost $ 2.50 (a) 20% gain, mark, U_O H.: ;
{{ .18 (a) 25% gain, mark, “ 65.00 (a) 40% loss, mark,
“ 24.00 (a) 12:4% gain, mark, 46 .15 (a) 334% loss, mark,
Mark in letters from the Key of Buy for Cash, with D for a repeater, Cost
and Selling Price of the following:
f h
Cost $ 2.40 a 25% gain, Ans. #. Cost $ 3.00 (a) 10% loss.
{{ .16 (a) 12:4% gain, {{ 4.50 (a) 40% loss.
“ 15.00 (a) 33.4% gain, * {{ .10 @ 20% loss.
“ 16.50 (a) 30% gain, “ 75.00 (a) 5% loss.
Find the selling price of the following:
1. Of goods which cost 182 and sold at 334% gain. Ans. 242.
2. Of goods which cost $240 and sold at 40% gain. Ans. $336.
3. Of goods which cost $4.25 and sold at 163% gain. Ans. $4.95%.
PERCENTAGE. 453
4. Of goods which cost $14.50 and sold at 30% loss. Ans. $10.15.
5. A merchant paid $108 per dozen for coats. At what price per coat must
he sell them to gain 50% º Ans. $13.50.
6. Bought silk for $1.60 per yard. At what price must it be sold to gain
140% Ans. $3.84.
A PUZZLING PEROBLEM.
873. A nest of five tubs cost $3.00. What should be the selling price of
each tub, to gain 334% on cost 3
OPERATION BY AssumED RATIOS OF VALUE, TO OBTAIN ONE OF MANY
ANSWERS.
$3.00 cost + 334% = $4.00, selling price for the nest of 5 tubs. 22 + 32 +
5% + 82 + 12% = 30%, which is the sum of the assumed ratio of values. Then the
following proportional statements give the respective value of each tub:
First or smallest Tub. Second Tub. Third Tub. Fourth Tub. Fifth Tub.
2g 3 g 5 g 8 g 122
30 || 4.00 30 || 4.00 30 || 4.00 30 || 4.00 30 || 4.00
| 26%, Ans. 402, Ans. | 66%, Ans. | $1.063, Ans. $1.60, Ans.
Practically, where the nickel is the smallest coin used in trade, the prices
would be as follows:
1st tub, 25g ; 2d tub, 402; 3d. tub, 65%; 4th tub, $1.10; 5th tub, $1.60.
SECOND OPERATION BY PROPORTION, ACCORDING TO THE SERIAL NUMBERS
OF THE TUBS.
The following statements give the value of the respective tubs:
No. of smallest tub, 1
4 &
** 2d “ 2 First Or g *
4t tº 3d 44 “ 3 Smallest Tub. Second Tub. Third Tub. Fourth Tub. Fifth;Tub.
• * : * 4th “ “ 4 $ $ $ $ $
“ “ largest “ 5 4.00 4.00 4.00 | 4.00 4.00
15 || 1 15 || 2 15 || 3 15 || 4 15 || 5
Sum of serial Nos. 15 *===ºmes mºs
26}c. 53;c. 80c. 1.06;c. 1.33:#c
NOTE.-Other ratios of value may be assumed according to the judgment of the calculator.
Or, if it is deemed proper, the prices may be made proportional to the capacity or volume of the
respective tubs.
2. What must be the selling price per box of a nest of 7 boxes which cost
$15, the gain ſ, being 25?
IN THE FOLLOWING ARTICLES THE LETTERS INDICATE THE COST
PRICE.
874. Write the cost in figures, and write the selling price in letters, at a
gain of 20%, using Dont be lazy as the key, and any other letter as a repeater.
1. Hats, $dayc per doz. 4. Oranges, beg per doz. 7. Flour, $t.lb per bbl.
2. Gloves, $o.cb per pair. 5. Books, $n.yc each. 8. Coffee, odg per lb.
3. Cuffs, $o.by per doz. 6. Shirts, $d.ob each. 9. Sheeting, dm3% per yd.
Ans. to the last, 1342 cost, 16#2 selling price.
Ans, to the first, $18.00 cost. $od.ey selling price.
'454 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. Y
A RAPID PRACTICAL PROCESS OF FINDING THE SELLING PRICE
PER SINGLE ARTICLE, AT A CERTAIN PROFIT WHEN THE
COST PRICE IS PER DOZEN.
875. 1. Bought shoes at $7.50 per dozen; at what price per pair must I
sell them to gain 20% º Ans. 752.
In all questions of this kind, when the goods are bought by the dozen and
the gain per cent is 20, we simply divide by ten or remove the decimal point one
place to the left, and the result is the selling price per article. This is a very
important contraction for merchants who buy at auction, as by it they can instantly
determine at each bid what the selling price must be per article, in order to give a
fair profit.
The philosophy of this work is simple and will be easily comprehended by
the elucidation of the following problem:
If shirts cost $22,50 per dozen, what must they be sold for, a piece, to gain
20% * Ans. $2.25.
OPERATION. y
$ Explanation.—For reasons often repeated we assume $100
cost, and reason as follows: In order to gain 20% we must sell
10 22.50 what cost $100 for $120. This selling price we place on the
Ø º statement line, and continue the reason thus: If $100 cost
I? require $120 selling price, $1 cost will require the 100th part,
and 22% cost will require 224%; times as much. . Then as the
result of this statement will give the selling price of a dozen, and as we wish to obtain the selling
price for a single article, we therefore divide by 12. This completes the reasoning and statement,
and by the same we see why we remove the decimal point. First cancel the 0 on the 120 and one 0
on the 100; then cancel the 12 on each side of the line, and we have the cost price remaining on
the right and 10 on the left. . It is obvious that this will always be the case when the per cent
gain is 20, and hence the work of dividing the cost by 10, which is done by simply removing the
decimal point, to find the Selling price per article, is made clear and comprehensible.
876. T A TE T, TE
FOR DIFFERENT RATES PER CENT TO BE USED ACCORDING TO THE ABOVE
PRINCIPLES : &
To gain 20 % remove the decimal point one place to the left
{ {
{{ 25 % 6 & 4 { {{ and add # itself.
{{ 26 % { % & 4 {{ {{ {{ 3% 44
46 28 % {{ & 4 {{ {{ {{ i’s {{
éé 30 % { % & 4 {{ {{ {{ * “
{{ 32 % {{ {{ {{ {{ {{ +'s {{
{{ 334% { % {{ {{ {{ {{ . # “
& 4 35 % {{ {{ {{ {{ {{ # “
& 4 *37.4% {{ 4% {{ {{ {{ + “
{{ 40 % {{ {{ {{ {{ {{ # “
{{ 44 % {{ {{ 64 {{ {{ # “
{ % 50 % {{ {{ {{ {{ 46 # {{
§ { 60 % 4% 6% {{ {{ (4 # “
{ % 80 % {{ {{ 4% {{ {{ # “
{ % 12.4% 4% {{ 44 {{ and subtract Tº “
{{ 4% 4% & 44 1
... }, . . . . . . .
Žt 70 Tº 6
&é 10 % 46 {{ 6% {{ {{ * 44
*NOTE.-This is practically, not accurately correct.
According to the above what would be the price per article to gain 20% of
the following goods bought by the dozen:
Hats, $32, Gloves, $18.50. Shoes, $44.50. Coats, $150. Pants, $37.50.
Knives, $3.75. Chairs, $21,25.
* PERCENTAGE. 455
TO FIND THE PRICE AT WHICH GOODS MUST BE MARKED SO THAT A GIVEN PER
CENT OF GAIN OR LOSS MAY BE REALIZED AFTER ALLOWING FOR
STATED DISCOUNTS, BAD DEBTS, WASTE, COMMISSION, ETC.
877. 1. A merchant marks his goods at 40% gain, but supplies his whole-
Sale buyers at a discount of 20% from list price. What is his gain ſ, 3
Ans. 12% gain.
OPERATION.
$100 = COSt.
40 = gain Ż. $112 = selling price.
*mºmºmºsºmeº 100 = cost.
$140 = list price. sºmºsºms
28 = 20% discount. $ 12 = gain, and as this gain is
*=mºm on the hundred it is
$112 = selling price. therefore gain Ż.
2. If a merchant marks his goods at 25 per cent profit and effects sales at
20 per cent discount on retail price, does he gain or lose? Ans. Neither.
3. If a merchant marks his goods at 50 per cent profit and effects sales at
40 per cent discount on retail price, does he gain or lose, and if so how much 3
Ans. Loses 10%.
4. A merchant wishes to increase the price of an article which he now Sells
for $9, so that he can deduct 10% and still receive the present price; what must be
the advanced price? Ans. $10.
OPERATION.
$100 advanced price assumed. $
10% deducted. 100
90 9
$90 selling price. ****
$10 Ans.
5. A merchant wishes to mark his goods in such a manner that when he
sells at wholesale he may discount 20% from his retail price and still gain 10% on
cost. Accordingly, what must be the retail price of goods that cost $1.60 per
yard 3 Ans. $2.20.
OPERATION.
$1.60 = COSt. $
16 = 10% gain. 100 = assumed retail price.
80 | 1.76
$1.76 = wholesale price. *E==
$2.20 = retail price, Ans.
6. What must be the asking price of goods that cost 602 per pound, So that
10% may be deducted and a gain of 20% still be realized ? Ans. 802.
7. What must be the asking price of goods that cost $15, so that 40% may
be deducted and the goods sold at cost 3 Ans. $25.
8. Goods cost 45g per pound; what must they be marked so that by deduc-
ting 25% sales may be effected at a loss of 10% Ans. 54%.
456 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. ºr
9. Paid 182 per dozen for eggs. Allowing 5% for breakage and 10% for
uncollectable sales, how much per dozen must they be sold for to gain 14% on first
cost 3 Ans. 242.
OPERATION OPERATION OPERATION
to find selling price to allow 10% for un- to allow for 5% Or thus:
to gain 14% collectable sales. breakage.
18g.
14% 100 100 114
90 |, 2052 95 | .2280 90 || 100
.0252 fººm. , — 95 | 18
.18 .2280g. $.2400 Ans. ºsº
24g
.2052 Ans.
10. Goods cost $10. What must be the asking price so that 15% may be
deducted and a gain of 25% realized on first cost? Ans. $14.70+}.
11. Goods cost $12. What must be asked for them so that 20% may be
deducted from the asking price, 64% be allowed for waste, 10% allowed for “bad
debts,” and 25% gain made on first cost 3 Ans. $22#.
OPERATION.
$12 100 100 | 100 125
3 = 25% gain. 93% 15 90 16 80 17; Or, 93; 100
*ss — — — tº-ºº-ºº ºmº 90 100
$15 16 17; 22} 80 12
$ 223
12. Bought a cargo of coffee at 10g a pound. Paid 10% duty. What must
I ask for the coffee per pound, so that I may fall 10% on my asking price, allow
10% for loss by shrinkage of coffee, lose 10% of sales in bad debts, and still gain
10% on my investment (full cost)? Ans. 16###2.
TO MARK DIFFERENT LOTS OF GOODS IN AN INVOICE.
878. Bought in New York an invoice of goods amounting to $2400. Paid
freight and other charges to deliver them in New Orleans, $102. In the invoice
are 8 dozen hats which cost in New York $27 per dozen. 1. What was the ſº of
the charges # 2. What was the cost of 1 hat in New Orleans? 3. What is the
retail price per hat, from which 20% may be deducted and the hats sold at whole-
sale at a gain of 25% on New Orleans cost?
Ans. 44% charges.
$2.34+cost of 1 hat in New Orleans.
$3.66—retail price of 1 hat.
MARKING GOODS, BASED ON FACTORY COST AND ON SELLING
PRICE, ETC.
879. Goods cost at the factory, $1400. Boxing, $4. Drayage to dock, $9.50.
Freight to destination, $30. Drayage from wharf to store, $9. (1). What is the
PERCENTAGE. 457
floor cost of the goods? (2). What is the percentage of the charges? (3). What
price must the goods be marked to gain 20% on the floor cost. (4). Based on
floor cost, what must the goods be sold for, so that the gain will be 20% of the
selling price? (5). When the goods are marked to gain 20% on the sales, What
per cent is such gain on the factory cost? Ans. (1). $1452.50 (2). 33%
(3). $1743.00 (4). $1815,625 (5). 29.6875%
OPERATIONS,
(1). To find floor cost. (2). To find percentage of (3). To find gain of 20% on
charges. floor cost.
1400.
4. 52.50 $1452.50=floor cost.
9.50 * 1400 || 100. 290.50=20% gain.
30. *
9. 3.75 or 33% $1743.00=selling price to gain
wºmeºs 20% on floor cost.
$1452.50
(4). To find gain of 20% on (5) To find what the 20% gain on
the sales over floor cost. sales is on factory CoSt.
$ 100. $1815.625 = selling price.
80 || 1452.50 1400.000 = factory cost.
| sists,825= selling price to $ 415.625=gain—$1400–29.6875%.
gain 20% on sales.
Goods cost at the factory, $130. (1). Estimating that the expenses to place
the goods in the store of the purchaser and the discount allowed to customers
equal 12% of Factory cost, what must the goods be marked to gain, 10% on the
Store cost? (2). Based on floor cost, what must the goods be marked and sold for
so that the gain will be 10% of the selling price? Ans. (1). $160,16
(2). $161.77;
OPERATIONS.
FIRST. SECOND. |
$130. $100.
15.60=12%. $90 145.60
$145.60=floor cost. $161.77% =Selling price to
14.56=10% on floor cost. gain 10% on sales.
$160.16=price marked.
TO IMPORT AND MARK DIFFERENT LOTS OF ENGLISH GOODS.
879%. Imported from England an Invoice of goods amounting to £1450. 8s. 4d. Paid freight
$190.50. Insurance $132.15. In the invoice were 40 dozen pocket knives which cost £160, 2s. 8d.
The duty on the knives was 25% ad valorem. What was the invoice cost in United States Money,
and what must be the retail selling price per knife to allow 20% discount and to sell at wholesale
at a gain of 25% on the full importing cost? Ans. $3.28%.
OPERATION,
£1450-8-4=#1450.416--x4.8665=$7058.45 inv. cost of goods in United States Money.
£160-2–8–42160.133—H·x 4.8665=$779.29 { { ‘‘ knives § { é &
$190.50+$132.15=$322.65, Insurance and freight.
(322.65×100)—$7058.45=4.57+% charges on invoice.
$779.29 ×4.57% =$35.61 charges on knives.
$779.29x25% =$194.82 duties on knives.
$779.29–H35.61-I-S194.82=$1009.72 full cost of knives in New Orleans.
$1009.72–40–$25.24 cost of one dozen knives in New Orleans.
$25.24–12=$2.10 cost of one knife in New Orleans.
$2.10–1–25% =$2.62% wholesale price per knife in New Orleans.
($2.62}x100)—80=$3.28; retail price of one knife in New Orleans.
REMARK. —To find importing cost, each lot of goods in a foreign invoice must be treated in manner as above shown.
due care being given to the different rates of duty on the different lines and the grade or qualities of the goods.
458 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. jºr
TRADE DISCOUNTS AND LIST PRICES.
880. Trade Discounts are allowances or rebates made by merchants, manufacturers and other
classes of business men from their regular marked or list prices.
It is customary for many classes of business men to have fixed price lists of their goods
based on the highest market rate, and when the market declines, instead of changing the fixed
price list, they allow a discount or a series of discounts from the list prices. These discounts will
be shown in the problems following.
Merchants and Manufacturers also frequently allow certain discounts or rebates from the
retail price, proportioned to the amount of the bill sold.
Some lines of goods are sold at “TIME PRICEs,” conditioned on certain discounts if paid
prior to the time specified. These terms and conditions are often printed on the “Bill Heads,”
and read as follows: “Terms 4 months, or in 30 days less 5 per cent.”; or “Terms 90 days or 2 per
cent. discount in 30 days, or 3 per cent. discount in 10 days,” etc.
In many cases, where goods are sold on Time prices, the rate per cent. allowed is a matter to
The agreed upon by the parties, and is generally 5 per cent., 10 per cent., or 25 per cent., according
to the length of time that the bill is paid before due, the hazard of the credit, the profit realized
on the sale of the goods, the necessities of the parties for money, and the rate of interest at which
money can be obtained at the bank.
The discount is allowed on the face of the bill or invoice without regard to time, notwith-
standing that time and the rate of interest on money were considered in determining the rate of
discount.
With a view to give a better understanding of this system of discounting, we will state that
196 discount on bills maturing in 30 days is equal to 12 per cent. interest.
2% discount on bills maturing in 30 days is equal to 24 per cent. interest.
3% discount on bills maturing in 60 days is equal to 18 per cent. interest.
5% discount on bills maturing in 3 months is equal to 20 per cent. interest.
5% discount on bills maturing in 4 months is equal to 15 per cent. interest.
5% discount on bills maturing in 6 months is equal to 10 per cent. interest
5% discount on bills maturing in 12 months is equal to 5 per cent. interest.
When no rate of discount has been agreed upon, and the bills are paid before maturity,
business men usually allow and deduct from the face of the bill the current rate of interest for
the time the bill was paid before due.
PROBLEMS. g
1. A bill of goods at list price amounts to $462.50. A discount of 15 per cent. is allowed.
what is the net amount 3 Ans. $393.12%.
OPERATION,
$46,250 = 10% $462.50
23.125 = 5% 69.37%
$69.375 = 15% discount. $393.12% Answer. *-
NOTE.—To find 15 per cent. multiply by 10 and add # of the product.
2 An invoice of merchandise at regular list prices amounts to $3480. What is the net
amount, the series of trade discounts being 25 per cent., 10 per cent and 5 per cent f
* Ans. $2231.55.
OPERATION.
$3480 = list price.
870 = 25% discount. Explanation.—From the list price, we first deduct
the 25 per cent, discount, and from the successive
$2610 * h t deducted
261 = 10% discount, remainders the other per cents are “ e
The order in which the series of discounts is
$2349 deducted does not change the result.
117.45 = 5% discount.
$2231.55 = net amount.
In this manner, the discount may be found in any series,
* PERCENTAGE. 459
TO FIND THE SINGLE OR EQUIVALENT DISCOUNT, FROM A SERIES
OF DISCOUNTS.
881. 1. In the above problem, what single discount on the list price is
equivalent to the series of 25 per cent, 10 per cent. and 5 per cent. Ans, 35%%.
FIRST OPERATION.
$100 = list price. º Discounts. Explanation.- We first assume
25 = 25 per cent. discount 25 $100 as the list price, and then
-** *º- deduct therefrom the 25 per
75 cent. discount; then from the
remainder † º, 10 per
.50 = 1 ... di .50 cent. discount: then from the
7.5 0 per cent. discount 7.5 second remainder $67.50, deduct
re 5 per cent. discount, which
67.50 . 3, * F.” § 64%. .
º subtracted from the assume
3.3750 = 5 per cent. discount 3.375 $100 list price, gives $35% dis-
-*-*-* * e -*=s*- count, which being on the
$ 64.1250 = net price. 35.875% =354% hundred, is 35% per cent, dis-
- º count. Or, by adding the sever-
100. = list price, as above. al discounts as shown in the
- operation,...we also obtain 35g
35.875 per cent. = equivalent single discount. Per cent discount.
A series of any number of discounts, may be Worked in the same manner.
SECOND OPERATION.
100 — 25% = .75 Then 75 x 90 × 95 = .641250 of 100, or
100 – 10% = .90 & 64.125 per cent. and 100 – 64.125 = 35.875
100 — 5% = .95 ( per cent, the single equivalent discount.
A series of any number of discounts may be worked in the same manner.
PROBLEMIS.
2. The list price of a bill is $1781.40; the rates of discount are 33% per
cent., 20 per cent., 10 per cent. and 5 per cent. What is the net amount?
Ans. $812.3184.
3. A merchant sold a bill of goods amounting to $530 on which he allowed
40 per cent. and # discount. What was the net value of the bill? Ans. $265.
4. Which is the greater discount, and what is the difference on a sale of
goods amounting to $1420, 40 per cent., or 25 per cent. and 15 per cent.”
Ans. 40 pr. ct, is $53.25 greater.
5. What is the equivalent per cent. of #, 33% per cent, #, and 10 per cent.
discount ºf Ans. 76 per cent.
6. A manufacturer after allowing # and 20% discount, received for a lot of
goods $3360 What was the catalogue or list price? Ans. $6300.
460 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICs. &
[7. I
FRANK DUMAs,
NEW ORLEANs, August 12, 1895.
Bot. of C. McGuigin.
TERMS-Cash; 10% off.
462 lbs. Prime Coffee,
375 ‘‘ Good é &
180 * * Fair é &
216 “ Ordinary “
74 “ Common. “
Less 10% discount.
I 8. I
JONES & WEISS,
TERMS-Cash; 10 per cent. and 5 per cent discount.
dº (2) 21 c. 97.02
º & 4 22 c. 82|50
º & & 21 c. 37|80
gºe & 4 20; c. 44|28
tº { { 193C. 14|43
$248 43
NEw ORLEANs, Jan'y 21, 1895.
Bot. Of Stewart dº Henderson.
34 bbls. La. Oranges, lar., &
27 boxes Messina Lemons,
63 cases Malaga Grapes, * *
45 boxes California Pears,
5 mats Dates, 593 lbs., gº tº
Less 10% and 5% off.
[9. I
J. M. BUTCHEE,
TERMS-Cash; 12% per cent. and 10 per cent, discount.
# tº * gº (2) $5.75
& gº * , & 4 6.00
* > gº tº-e é & 1.75
gº gº * & & 4.50
sº sº gº ( & 73c.
714|73
NEW ORLEANs, Jan'y 31, 1895.
Bot. of J. B. Cundiff.
875 bbls. Nes. Potatoes,
440 “ P. B. Potatoes,
325 “ Perfection Flour,
1324 “ St. L. XX. Flour,
112 “ Family Clear Pork,
650 “ Prime Pork, wº
220 kegs Pig Feet, tº º
124 half bbls. F. M. Beef,
1872 lbs. Choice Ham, * *
289 “ B. Bacon, * e-
106 Pig Tongues, tº gº
I)is. 124% and 10%.
* * sº tº- (2) $ 4.25
tº e * & 4 3.87%
* sº *- & & 8.50
g- º sº & & 6.62%
tº # * sº & 4 17.50
sº sº tº & 4 13.75
gº sº tº & 4 7.50
* - sº wº- & 4 11.
* cº gº & & 14 c.
sº sº gº tº & & 94 c.
tº tº cº-e & 4 8 c.
3.116727
* PERCENTAGE. 461
10.
[ 1. NEW ORLEANS, Nov. 17, 1895.
GEO: B. BRACKETT & CO.,
Bot. of R. Spencer Soulé.
TERMS-Cash; 163 per cent. off.
1427 bu. No. 1 Winter Wheat, - - - - - - a $1.5
856 “ No. 2 Winter Wheat, * * tº - * * {{ 1.47
420 “ Ill. No. 1 White do. -> •º º * tº- - {{ 1.41
3145 “ W. Corn, - - - - - - - - - “ .70
1040 ** B. Oats. * - º º - º - - --> * { .55
6835|87
Less 163%. s = º s - * * * ->
-º-
[11. I
NEw ORLEANs, August 12, 1895.
E. L. CHAPPUIS,
Bot. Of W. Reed Carradine.
TERMS-4 months or 5 per cent. and 2% per cent. discount in 10 days.
Aug. 12 |42 bbls. Flour, “Winter Patents” - - - - (a) $5.25 220 |50
16 ‘‘ “ Extra Fancy º º - * - { { 5.00 80|00
27 “ Fancy - tº- - * * - & 4 4.75 128|25
60 * * ‘‘ Choice - º - - º - { { 4.50 270|00
36 “ “ Family - - -> - º - { { 4.75 171|00
70 “ Minnesota Patents - - º - { { 5.75 402|50
89 • { “ Minn. Bakers’ Grade - - Q- - & 4 5.14 457|46
79 “ Corn Meal, - - - * - - - sº - & 4 3.20 252|80
12 “ Grits, - sº - * ~ * - -- - & 4 3.10 37|20
10 “ Hominy, º - º - - es - 4 & 2.65 26|50
5 “ Rye Flour, - - º º - º - { { 3.90 19||50
9 “ Oat Meal, º-s, - - º - tº - & 4 5.75 51|75
211746
Discount 5% and 24%. - - - - -
NotE 1.-The above bill was paid 6 days after the purchase was made.
See page 457, for an explanation of Trade Discounts.
When Trade Discounts are not allowed and bills are paid before maturity, interest for the
unexpired time, usually allowed the payor.
NOTE: 2. —For other bills covering many lines of business, see pages 283 to 295. Also, see
pages 278 to 283, for Rapid Practical Methods of Bill Computations.
12. A dealer bought goods at 25% and 10% discount on list prices and sold
at the list price. What was his gain per cent? Ans. 48#.
OPERATION. Then the following
Let $100 = the ſist price. Statement :
25 $100 = sales at list price. $
-ms 67# = cost of list price. 2 | 65
$ 75 ºsmºsºm- 135 | 2
7; $ 32} = gain on cost. 100
$ 67; cost. 48%. Ans.
13. A merchant bought an invoice of goods Sept. 15, 1890, amounting to
$1200, on the following terms: “4 months or 5% discount in 30 days.” The bill
was paid Oct. 13, 1890. How much was paid f Ans. $1140.
462 soul.E's PHILOSOPHIC PRACTICAL MATHEMATICS. *
14. Goods are bought at 33.4% and 20% discount from list prices, and sold
at 30% and ſº off. What is the gain per cent? Ans. 184%.
OPERATION.
$100 = assumed as list price. 100 = assumed list price.
20 = 20% discount on purchase. 30 = 30% discount on sales.
80 70
26% = 334% discount. 7 = 10% discount on sales.
$ 53% = cost. $ 63 = selling price of $53; cost.
53% = cost.
$ 93 = gain on $53% cost.
Then if $534 gain $93, what will 100 gain : Ans. 184%.
15. What % is gained by selling hats at $3 each, that cost $42 per dozen
with 163% and 20% off? Ans. 284%.
16. What was the list price of goods that sold for $1.26 per gallon, with
50%, 10% and 12.4% discount Ans. $3.20.
17. A business man insured property to the amount of $65000 at 13%. He
was allowed a rebate of 20% for cash. What was the net premium paid?
Ans. $910.
18. A wholesale dealer sold a bill of goods amounting to $2000, on 6 months'
credit. A few days after the sale had been completed, the purchaser proposed to
pay the bill if the wholesale dealer would allow him 5 per cent discount. His prop-
osition was accepted and the bill paid. What amount of money was required ?
AnS. $1900.
* OPERATION.
$2000 face of bill.
100 = 5% discount.
$1900 amount required to pay the bill.
19. At the same time that the above transaction was made, the same whole-
Sale dealer sold to another merchant, on 6 months' credit, goods amounting to
$3000. The purchaser of this bill proposed to pay $1900, provided the wholesale
dealer would allow him 5 per cent discount on the amount of invoice paid, which
the wholesale dealer agreed to do, and the $1900 was paid. What amount does the
purchaser of the goods still owe ? Ans. $1000.
PERCENTAGE. 463
OPERATION.
$100 invoice assumed. Explanation.—The solution of this problem is
5 = 5% discount deducted. entirely different from the solution of the pre-
ceding one. The discount being allowed on the
* amount of the invoice that the money will pay,
$95 cash. and as the whole of the invoice is not paid, we
cannot therefore calculate on the $1900 cash or
on the face of the $3000 invoice. We are hence
100 obliged to assume some number to represent
pºse invoice, and upon which we may calculate and
95 | 1900 deduct the 5 per cent discount, and thus obtain
* ====== the necessary equivalent values with which to
2000 make a solution. As shown in the operation,
3000 We assume $100 to represent the invoice, and
then by deducting the 5 per cent discount we
Rºssºmº have in the remainder $95 as the cash equivalent
$1000 Ans. Aft of the $100 invoice; or as the amount of cash
required to pay $100 invoice. Having these
equivalent values, we place the $100 assumed invoice on our statement line, and reason thus: If
$95 cash will pay $100 invoice, $1 cash will pay the 95th part, and $1900 cash will pay 1900 times as
much. The result of which is $2000. Then having the amount of invoice that $1900 will pay or is
equal to, we deduct the same from the whole invoice, $3000, and have in the remainder $1000, the
balance that the purchaser still owes.
20. A country merchant purchased $4860 worth of goods, on 4 months' time.
He then proposed to pay cash if the city merchant would discount 5 per cent on
the face of the bill, which he agreed to do. What amount will pay the bill?
Ans. $4617.
21. Another country merchant purchased $84.50 worth of merchandise, on 4
months' time. He then proposed to pay $5000 cash on condition that he be allowed
5 per cent discount on the amount of bill that he pays, which proposition is
accepted. What amount does he still owe ? Ans. $3186.84.
22. A retailer bought from a wine merchant, on 3 months' credit, goods
amounting to $1500, and from a grocery merchant, on 3 months' credit, goods
amounting to $1800. But, without waiting for the maturity of the bills, he pays
the wine merchant cash in full less 5 per cent discount allowed, and to the grocery
merchant he pays $1000 cash, for which he is to be allowed 5 per cent discount on
the amount of invoice that it will pay. How much did he pay the Wine merchant?
And how much does he still owe the grocery merchant }
* º Ans. $1425, he paid the wine merchant.
$747.37, he owes the grocery merchant.
23. A debtor pays his merchant $500 in nickles at 74% discount. How much
credit does he receive 3 Ans. $462.50.
24. A debtor owes $555 and pays the same in nickles at 73% discount. What
amount of nickles will it take to pay the debt 3 Ans. $600.00.
464 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. ºr
882. TABLES OF TRADE DISCOUNTS.
TABLE No. 1.
tº e tº te 4. g tº -- ſº
gº ºt and : pret # prºt of or § prºct. º:
5 & & é & 3 & 4 7; & & & 4 92; & 4 & 4
10 & & é & 5 & 4 sº- 14% & & & 4 85} {{ * & &
10 “ “ 5 “ and 23 pr. ct. 16; “ “ 833; “ “
15 & & 6 & 5 & t *-ºs * 19+ & & & 4 80% & 4 & 4
15 & 4 “ 10 & 4 * sº
; !, 10 and 5 pr. Ct.
20 “ 4 & 1. & 4 . -
20 “ “ 10 “ and 5 pr. ct
25 “ “ 5 “ c. v. $
25 “ “ 10 “ <- *-
25 “ “ 10 “ and 5 pr. ct
25 “ “ 20 “ e. --.
25 º : 20 and 10 pr. ct.
; : : * : * 10 " and
30 “ ** 10 “ . :
23# & 4 & & 76# & 4 & 4
27# “ & 4 723; “ “
24 & & & 4 76 4 & & 4
28 { { ( & 72 & ( & &
31} & 4 { { 68% ( & & 4
28% & 4 & & 71+ 4 & & 4
32+ & 4 & 4 67; ( & & 4
35} & & & & 64% & ( & &
40 & 4 & 4 60 4 & & 4
46 & 4 é & 54 ... & 4 & 4
481.5 & 4 s & 51.1% & & &&.
33} { { { { 66} & 4 & &
37 & 4 ( & 63 & ( & &
44 & 4 & & 56 4 & 4 &
47% & 4 & 4 52} 4 & 4 &
58 & 4 & 4 42 & 4 & 4
62} & 4 & 4 37} 4 & 64
64TÉd “ & 4 35 ºr “ “
38+ & 4 ( & 61% & 4 & &
41% & 4 ( & 58% & ( & &
48 & 4 { { 52 { { & 4
53} & 4 & 4 46# ( & & 4
55% “ & 4 44%; “ “
43 & 4 & 4 57° 4 & 6&
46 & 4 & 4 54 4 * 4 &
49 & 4 & 4 51 & 4 & 4
52 {{ é & 48 & 6 & 4
58 & & & 4 42 & & & &
62} & 4 & 4 37% & 4 & 4
66% & 4 & 4 33} 4 * * *
69}} “ & 4 30; “ “
52} & 4 & 4 47% 4 & & 6
pr ct.
30 & 4 “ 20 4 & tº-
30 6 & “ 25 & 4 rºs
30 “ “ 25 “ and 20 pr. ct. tº
30 “ “ 25 “ “ 20 “ and 10 pr. ct, *
§ 25 º ** 20 tº “ 10 “ and 5 pr. ct.
; & 4 & 4 1. & 4 . . gº tºp
35 “ ** 20 tº º tº
35 “ “ 20 “ and 10 pr. ct.
; º º 20 . ** 10 “ and
: & & * & 1; « :
40 & 4 “ 15 & 4 º
40 $4 “ 20 é & t-
40 6 & ** 30 & 4 *
40 “ “ 30 “ and 10 pr. ct
40 & 4 << 30 “ ** 20 ...” º
§ : : sº . “ 20 “ and 10 pr. ct.
p
I’,
C
t
:
50 “ << 20 “ tºº sºs tº 60 4 & & 4 40 4 & 6 &
50 “ “ 30 “ 4-ºxº * -> tº 65 & 4 & 4 35 4 4 & 8
50 “ “ 30 “ and 20 pr. ct. gº 72 “ “ 28 tº “
74% { { 4 & 25} é & & &
67." & 4 & 4 36 & 4 & 4
68 & 4 & 4 32 & 4 & &
71} & 4 & 4 28} & 4 & 4
76 & 4 { { 24 4 & & 4
82 & 4 & 4 18 4 & & 4
85% & & & 4 14% & 4 é &
86 & £ & 4 14 & 4 & 4
90 & 4 { { 10 & 4 & &
93 & 4 & 4 7 & 4 & 4
§ “ “ gº tº * 20".” and 10 pr. ci.
60 “ << 10 & & sº sº º
60 “ << 20 “ sº sº
60 “ “ 20 “ and 10 pr. ct.
70 “ “ 20 “ * “º
70 & 4 § { 40 é & sº s
70 “ “ 40 “ and 20 pr. ct.
80 ** << 30 “ sº gº
80 4 & “ 50 ( & g-º tº
80 ** “ 50 “ and 30 pr. ct.
90 ** “ 5 “ ſº sº 90% { { é & 9} ( & & &
90 é & é & 10 4 { * - gºs 91 { { & 4 9 & & 4 &
APPLICATION OF THE ABOVE TABLE. ©
A merchant sold goods amounting to $321.80, and allowed 20%, 10% and
5% off. What was the net amount of the bill?
FIRST OPERATION. SECOND OPERATION.
$321.80
313% the equivalent of 20%,
— 10% and 5%. $321.80
$99.7580 = 31% of $321.80. 68%% = % on.
1.9308 = } of $321.80. ººmsºmºsºms
*==ºssº $218.8240 = 68% of $321.80.
$101.6888 = discount. i:373 - 6° 3321.80.
$220.11 = net amount. $220.11 = net amount.
ºr PERCENTAGE. 465
TABLE No. 2.
*
Mark your And your net Mark your And your net
goods And take off gain goods And take off gain
at a gain of will be at a gain of Will be
5 % 2 % 21% 96 50% 30 96 gain 5 %
10 “ 2} & 4 7+ 4 & 50 “ 25 & 4 12; “
12} & 4 3 { { 9} & 4 50 “ 20 & 4 20 & 4
16; “ 5 & 4 10; “ 50 “ 15 & 4 274 “
20 “ 10 & 4 8 “ 50 “ 10 & 4 35 “
20 * * 7} & 4 11 & 4 50 “ 5 & 4 42% & 4
20 * * 5 & 4 14 (4 50 “ 20 & 10 * * 8 * *
25 15 & 4 6+ “ 60 “ 33; & 4 6; “
25 10 & 4 124 “ 60 “ 30 & 4 12 * *
25 5 & 4 18#. { { 60 * * 25 & 4 20 é &
25 “ 10 & 5 “ 6; “ 60 “ 20 & 4 28 “
25 tº 20 & 4 0 & 4 60 * * 15 & 4 36 ( &
30 “ 20 & t 4 * * 60 “ 10 & 4 44 * *
30 & 4 15 & 4 10} {{ 60 & & 5 & 4 52 & 4
30 “ 10 & 4 17 t < 60 * * 25 & 10 * * 8 * *
30 “ 5 4 & 23# & & 70 “ 40 & 4 2 & 4
30 * * 15 & 5 “ 4# “ 70 “ 33% & 4 13+ “
33; “ 25 & 4 O & & 70 “ 30 & 4 19 & 4
33} & 4 20 & 4 6# & 4 70 4 & 25 & 4 27} & 4
334 “ 15 ( ; 134 “ 70 20 & 4 36 4 &
33; “ 10 & & 20 “ • 70 30 & 5 “ 13; “
334 “ 5 4 & 26; “ 80 ‘‘ 40 & 4 8 & !
33; “ 10 & 5 & 23 “ 11%; “ 80 * * 33; & 4 20 <<
40 * * 25 & 4 5 * * 80 “ 30 & & 26 tº
40 * * 20 & 4 12 & 4 80 * * 25 & 4 35 tº
40 “ 15 & 4 19 < 80 “ 30 & 10 * * 13.4 “
40 “ 10 { { 26 & 4 90 “ 40 é & 14 “
40 * * 5 {{ 33 & & 90 “ 33% ( & 26; “
40 * * 15 & 5 & 24 “ 10}#} “ 90 * * 30 & 4 33 44
50 * * 40 ** | lose 10 “ 90 * * 20 & 10 & 5 “ 29## “
50 “. 33# & 4 0 & 4
APPLICATION OF THE TABLE.
Goods cost $12.50 and were marked at 20% gain, from which a discount of
10% was allowed. What was the net gain per cent? Ans. 8% net gain.
FIRST OPERATION. SECOND OPERATION
Cost $12.50 $13.50 selling price. $ $100 cost assumed.
20% gain 2.50 12.50 cost price. 20% gain.
•=s* =mmºns 1. -º-
Marked price $15.00 $ 1.00 gain. 124 || 100 120 marked price.
10% discount 1.50 -*=- I ºmsm-me 12 = 10% discount.
- =m- 8% -*-*=
Selling price $13.50 $108 selling price.
100 COSt.
$ 8 gain on a hundred
and hence 8%.
466 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS.
TABLE NO. 3.

If you are allowed And sell at the list If you are allowed And sell at the list
a discount from rice a discount from price
the list price of your gain will be the list price of your gain will be
2} % 23# 9% 33} % 50 %
5 & 4 5+ 4 & 40 { { 66# & 4
5 & 24 & 4 774; “ 40 & 20 & 10 “ 131}} “
8% & 4 9 ºr & 4 50 { { 100 “
10 & 4 # & 4 60 { { 150 “
10 & 5 & 4 16+?? “ 66# { { 200 “
12+ & 4 14% “ 70 & & 233; “
15 & 4 17}} “ 75 & 4 300 “
16# & 4 20 & & 80 { { 400 & 4
20 & 4 25 “
20 & 10 & 5 “ 46 ºr “
25 & 4 33; “
30 & 4 42% & 4
APPLICATION OF THE TABLE.
The list price of goods is $500, from which 5% and 24% discount is allowed.
What is the net gain per cent 3 Ans. 7###%.
FIRST OPERATION. SECOND OPERATION.
$500 list price. $100 list price assumed.
25 = 5%. 363 x 100 5 = 5%. 8 || 59
— = 7###%. 741 || 8
$475 463; 95 100
11% = 24%. 23 = 24%. *-
7###%.
$463; cost. 923 = cost.
500
$73 - gain.
$ 36% gain.
883. TO FIND TEIE VALUE PER, TON OF SUGAR CANE.
NOTE.-For the leading facts regarding this new subject, we are indebted to Prof. W. C.
STUBBS, Director of the La. Sugar and State Experimental Station.
The method to determine the price to pay for Sugar Cane, varies somewhat
with different manufacturers of Sugar in the different sections of the sugar district
of Louisiana. The following Work and statements will make clear the various
methods used:
Cane is usually sold upon three factors, as follows: 1. The price of refined
sugar in the New Orleans Market. 2. The percentage of juice extracted from the
cane. 3. The percentage of Sucrose contained in the cane. The value of the cane
being a certain price for each of these elements or factors. The element of glucose
is not considered as a material factor in determining the price of cane.
The average chemical analysis of Louisiana sugar cane is about 90 per cent
juice and 10% fibre. The juice contains on an average about 12% of sucrose, 2%
of glucose, and 1 to 14% of solids not sugar.
NOTE.-Very few of the sugar mills in Louisiana extract more than an average of 72 to 80%
of juice from the cane.
A} PERCENTAGE. 467
QUANTITATIVE CHEMICAL ANALYSIS.
884. Quantitative Chemical Analysis consists in the determination and
Separation of the relative proportions of the component parts of which a substance
is composed.
The Several parts are expressed in per cents of the whole substance. In the
analysis of cane juice, molasses, etc., the component parts are, as above stated,
Sucrose, glucose and solids, that are not sugar.
The Co-efficient of Purity is the per cent that the sucrose is of the whole
Solids extracted from the cane. Thus, as in the above average in which the juice
contains 12% sucrose, 2% glucose and 14% of foreign solids, not sugar, the
co-efficient of purity is found as follows:
12 + 2 + 1} = 15%; then, 12 × 100 + 15% = 77.4%%.
The Glucose Ratio is the per cent that the glucose is of the sucrose, and is
found as follows, using the above figures: (2 × 100) -- 12 = 163%.
The per cent of the foreign element, not sugar, is found as follows: (14 × 100)
—H· 15; F 9:4%.
Fertilizers. The component parts generally considered in fertilizing sub-
stances are potash, nitrogen and phosphoric acid.
THE SUCROSE MULTIPLIER AND BIOW TO FIND IT.
885. A Sucrose Multiplier is a number used as a factor in determining
the price of sugar cane per ton, and is found as follows:
1. By estimating an average extraction of 80% of juice from the cane. 2.
By allowing the seller 50% of the juice. 3. By estimating an average of 10 or
10.4% sucrose in the juice.
FIRST OPERATION TO FIND A SUCROSE MULTIPLIER.
886. 1 ton of cane = 2000 lbs. 80% of which is 1600 lbs. of juice. } of
-1600 lbs. = 800 lbs. juice allowed the seller. (800 or 8 is the sucrose multiplier).
Hence, 800 x 104% x 42, the price of Prime Yellow Clarified Sugar in New
Orleans, gives $3.36, the price per ton of cane. These figures practicalized become
8 × 104 × 4 = $3.36, the price per ton of cane. (The two 0's in the 800 and the
% in the 10.4% being omitted).
SECOND OPERATION TO FIND A MULTIPLIER.
887. 1 ton = 2000 lbs., 80% = 1600 lbs. -- 2 = 800. 10% of 800 = 80
pounds sucrose or sugar. (80 is the sucrose multiplier).
468 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. º
Hence, 80 × 42, the price of Prime Yellow Clarified Sugar in New Orleans,
is $3.20, price per ton of cane.
Or thus, 10.4% of 800 = 84. Hence 84 x 42. = $3.36, the price per ton of
CaLle.
NotE.-According to the above method and principles, when the 9% of juice extracted is 70,
T2, 75,833, 85 or 90, the multipliers would be respectively, 7 or 70, 7.2 or 72, 7.5 or 75, 8.34 or 83;
8.5 or 85 and 9 or 90 according to the method used. In like manner, the sucrose multiplier would
be found for other per cents of juice extracted.
PROBLEMS.
888. 1. If the price of refined sugar in New Orleans is 4 cents per pound
and the percentage of sucrose in cane juice is 12 per cent, what would be the value
of cane per ton, at 8 cents per unit of sucrose and of the price of refined sugar 3
Ans. $3.84.
OPERATION. 82 x 12 x 4 = $3.84 per ton.
2. A purchaser of sugar cane says to a planter “I will give you 84 cents
per unit for each 1% of sucrose that your cane contains, multiplied by the price per
pound of refined sugar in New Orleans.” If the price of refined sugar in New
Orleans is 442 per pound, and the juice contains 13.4 per cent sucrose, what is the
value of the cane per ton ? Ans. $5.1255.
OPERATION. 2 9 w
13.4
84 × 13.4 × 44 = $5.125-1- or thus, 2 17
*-
NOTE.—The 84 cents per unit in this problem, equals 85% (8.5) of juice extraction as shown
by the above operations to find multipliers.
889. Another method of determining the price of sugar cane per ton.
In some of the sugar districts of Louisiana, the price or value of cane per
ton is determined by allowing 834 pounds of sugar per ton of cane, and to estimate
its value at the New Orleans market price for Prime Yellow Clarified Sugar.
PROBLEMIS.
890. 1. Allowing 834 pounds of sugar per ton of cane, what is cane worth
when Prime Yellow Clarified Sugar is quoted at 3:42 in New Orleans?
Ans. $2.653.
OPERATION. 16 51
3 250
3 * x 83% = $2,653 or, thus: *
| $2.65;
NOTE:-The 83% multiplier here used, is found as shown above in the “second operation to
find a multiplier,” substituting 834% for juice extracted in place of the 80 there used.
2. Using the figures of the above problem, and allowing that sugar cane in
Louisiana will produce 152 pounds per ton, what would be the gross profit to a mill
that purchased and manufactured 4162 tons of cane. No allowance to be made for
molasses or syrup that may be made, or the expenses to manufacture, or to market?
Ans. $9109.573.
NoTE.—150 to 160 pounds of sugar per ton of cane is the average for the cane of Louisiana.
* PERCENTAGE. 469
OPERATION TO FIND COST. OPERATION TO FIND SALES AND GAIN.
16 || 51 16 || 51
3 || 250 152 $20164.89
4.162 © 4.162 11055.31%
$11055.31#. $20164.89 $ 9109.57; gain.
f
NoTE.-Some purchasers of sugar cane pay per ton, 80 cents, more or less, for each cent o
the New Orleans price of Prime Yellow Clarified Sugar.
This 80 which is used as a multiplier of the New Orleans price of Prime Yellow Clarified
Sugar, is found as above shown in the second operation, Article 887, to find a multiplier allowing
80% of extraction of juice from the cane.
3. How many pounds of sucrose are there in 1840 pounds of cane juice
which contains 13.6% of sucrose? Ans. 250.24 pounds.
4. If the solids in the juice of cane are 154%, and the co-efficient of purity is
77%, how many pounds of sucrose are there in 660 gallons of juice, allowing 84
pounds to the gallon ? Ans. 660 pounds of sucrose.
FIRST OPERATION. SECOND OPERATION.
774; x 15.4% = 12% sucrose. 660 × 8% = 5500 pounds.
660 × 8% = 5500 pounds of juice. 5500 x 15% = 852.50 lbs. solids.
5500 × 12% = 660 pounds of sucrose. 852.50 × 774% = 660 lbs. sucrose.
5. In the above problem, if the glucose ratio was 163%, how many pounds
of glucose would there be? Ans. 110 pounds.
OPERATION.
163 × 12 = 2% glucose.
660 × 8% = 5500 pounds of juice.
5500 × 2% = 110 pounds of glucose.
6. How many pounds of sucrose are there in 64 tons of sugar cane, of which
864% is juice, and 14% of the juice is sucrose? Ans. 1513.75 pounds.
OPERATION.
2000 x 64 = 12500 lbs. × 864% = 10812.5 lbs. × 14% = 1513.75 lbs.
7. From 8 tons and 500 pounds of cane which was passed through a mill,
the bagasse weighed 3630 pounds. What was the per cent of juice extracted?
Ans. 78%.
OPERATIONS INDICATEI).
% %
100 16500 100
16500 || 3630 3630 16500 | 12870
tº- Or thus: * ==
12870 78%
iſ: } 78% extracted.
8. 12870 pounds of juice were extracted from a lot of cane; if the rate per
cent of extraction was 78%, how many tons of cane were there? Ans. 84 tons.
OPERATION INIDICATET).
100
78 || 12870
2000 or (12870 × 100) -- (78 x 2000).
47O SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. A.
9. In 53 tons of cotton seed fertilizing meal, there is 64% of nitrogen, 34%
of phosphoric acid and 1.9% of potash. How many pounds of each fertilizing
ingredient? Ans. 715 pounds of nitrogen.
343.75 pounds of phosphoric acid.
209 pounds of potash.
10. In 3400 pounds of bone meal, there are 1604 pounds of nitrogen and 790.8
pounds of phosphoric acid; what is the per cent of each 3
Ans. 4.72+ 7% nitrogen. 23.26+ 7% phosphoric acid.
MISCELLANEOUS PRACTICAL QUESTIONS IN PERCENTAGE.
891. 1. Bought 6 dozen shirts for $129.60; at what price per shirt must I
sell them to gain 50 per cent? Ans. $2.70.
2. Bought butter at 48% per pound; at what price per ounce must I sell it
to gain 663 per cent 2 Ans. 52.
3. Bought flour at $7.50 per barrel; at what price must it be sold to lose 10
per cent 7 Ans. $6.75.
4. Sold goods to the amount of $250 and gained 25 per cent; what did they
cost 3 Ans. $200.
5. Sold flour at $8.75 and lost 124 per cent; what did it cost? Ans. $10.
6. Bought gloves at $4 per dozen and sold them at 50¢ per pair; what per
cent did I gain } Ans. 50%.
7. My merchandise account is debited $145000 and credited $174000. There
is no merchandise on hand. What per cent did I gain on sales? Ans. 20%.
8. My merchandise account is debited $280000 and credited $314212.50. I
have on hand $44797.50. What is my gain, and what my gain per cent?
Ans. $79010 gain. 28%, 7% gain.
9. A manufacturer received a certain quantity of cotton to be manufactured
into cloth. He delivered 182400 yards, averaging 44 ounces to the yard. Allowing
that there was a waste of 10 per cent in manufacturing the cloth, how many pounds
of cotton did he receive 3 Ans. 57000 pounds.
10. I sell a certain quality and style of goods at $1.10 per yard. My rival in
trade sells the same quality and style at $1.25 per yard. What per cent more does
he charge for his goods than I? Ans. 13+%.
11. I sell flour at $8.50 per barrel. My rival sells the same kind at $9.
What per cent less do I sell than he 3 Ans. 53%.
12. The assessment of real and personal property of a city amounts to
$150,000,000. The various expenses of the municipal government are approximated
and estimated at $2,800,000. What must be the per cent tax assessment to raise
the amount 3 & Ans. 1.863%.
13. Sold coffee at 18% per pound and lost 10 per cent; what should I have
sold it for, to have gained 124 per eent? gº Ans. 22#2.
14. Sold butter at 35% per pound and gained 25 per cent; what per cent
would I have gained, had I sold it at 30% Ans. 74%.
º: PERCENTAGE. 47 I
15. Sold potatoes at $3.15 per barrel and lost 10 per cent; what per cent
would I have gained or lost had I sold them at $4.3 Ans. 143% gain.
16. Sold rice at 842 and lost 84 per cent; what per cent would I have gained
or lost, had I sold it at 92% Ans. 1% loss.
17. Whiskey which cost 90% per gallon is compounded with water in the
proportion of 1 gallon of water to 2 gallons of whiskey, and the mixture is sold for
85% per gallon. What % is gained? Ans. 413%.
18. A faithful accountant, who was receiving a compeusation of $720 per
year, had his salary increased 33.4%. How much does he now receive per month ?
Ans, $80.
19. A firm paid $300 per month rent and was raised 25%. What is their
Current rent 3 Ans. $375.
20. A clerk who received $125 per month had his salary reduced 20%. What
did he then receive? Ans. $100.
21. A man who was receiving $2.50 per day demanded an increase of 40%.
What would then be his daily pay? Ans. $3.50.
22. A merchant compounded, in equal quantities, sugar that cost 62 per
pound and sugar that cost 92 per pound. He then sold the mixture at 9% per
pound. What % did he gain? Ans. 20%.
OPERATION:
* g
1 pound at 62 = 62. 92 sales. 2 || 3 gain.
1 “ “ 92 = 92. 7.4% cost. 15 2
-ºmsºmºms -ms-smºs 100
2) 15g. 14% gain. smºs |
1 lb. of the com- 20% gain, Ans.
pound cost 7#2.
23. A merchant bought 5 bbls. flour for $40 and sold the same at a gain of
25%. He then bought 5 bbls. more for $40, and sold the same at a loss of 25%.
Did he gain or lose, and if so, how much? Ans. He neither gained nor lost.
OPERATION.
$40 cost 2 25% gain gives $50 selling price = $10 gain.
$40 sold Ø 25% loss gives $30 selling price = $10 loss.
24. A merchant sold 5 bbls. of inferior flour for $40 and gained 25%. He
then sold 5 bbls. of superior flour for $40 and lost 25%. Did he gain or lose in the
two transactions, and if so how much Ans. He lost $5}.
OPERATION.
$40 selling price @ 25% gain gives $32 cost = $ 8 gain.
$40 selling price @ 25% loss gives $53; cost = $134 loss.
Net loss $ 5;
25. A grocer bought 10 barrels of molasses each containing 40 gallons, at
50g per gallon. He then put the molasses in kegs holding 8 gallons each, and sold
the same as 10 gallon kegs, at 55% per gallon. What was his profit, and what %
did he gain? Ans. $75 profit; 37; % gain.
472 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS.
jºr
* 26. A fruit dealer sold 4 peaches for 52 and gained 564%. What % would
he have gained by selling 5 for 6 cents? Ans. 50%.
OPERATION.
5 -- 4 = 1.4% 100 cost. 6 -- 5 =1}g 5 2 156+ sales.
156} | 1.4 4 || 5 5 4
**- 1#2—#2 =#2 gain. | 100 Or 5 || 6
#2 cost. - * : -º-º-º:
50% gain. 150 sales,
50% gain.
27. A. and B. are two merchants; they desire to barter rice and sugar. A.
has rice, market value 6% per pound; and B. has sugar, market value 82 per
pound. At the time of the exchange or barter, A. suggests to B. that in order to
influence the market reports, he will place his rice at 72 per pound and that B.
shall advance his sugar accordingly. B. accepts the proposition. What should be
the exchange price of B's sugar 3 Ans. 9% per pound.
FIRST OPERATION.
2
72 = exchange value of 1 = gain. 82 = market value of
A’s rice. 6 || 100 B’s sugar.
6% = market value of - º- 1$2 = 163% gain.
ººmsºmº Sal ſle. 163% gain, &= -
12 = the increase on 94% = exchange value.
Sal (162.
SECOND OPERATION
Ž **
6% + 12 = 7 g = exchange 1 = gain on 62. 82 + 142 = 9.4% = exchange
value of A's rice. 6 || 8 value of B's sugar.
13% gain on 82.
28. Suppose in the above problem, that B. had proposed to reduce his sugar
12 per pound, and that A. should reduce his rice accordingly. What would be the
exchange price of A's rice? Ans. 542.
29. Suppose in the above problem that A. and B. had each raised 1g on the
market value of their rice and sugar, how much 7, would B. have lost, and what %
would A. have gained ? Ans. B. would have lost 34%.
A. “ “ gained 3}#}.
OPERATION TO FIND B's 'ſ, Loss.
g
94% = correct exchange value, as above. 3 || 1 = loss.
9 g = incorrect exchange value, as supposed. 28 3
100
#2 = loss by incorrect exchange value. -
34% loss. Ans.
PERCENTAGE. 473
OPERATION TO FIND A's 9, GAIN.
g
92 = B's selling price. 72 = A's incorrect selling 4 || 1 = gain.
8 || 6 price. 27 | 4
gººmsº 6#2 = A's correct selling 100
6% = price A, should price. emº
have sold for when #2 = A's gain by selling 3}}% gain, Ans.
B. sold for 99. at 72.
EXPLAN ATION FOIR, SECOND OPERATION.
82, B's market value -- 12 = 92, B's selling price; then since 8% sell for 92,
12 will sell for the # part, and 6¢, A's market value, will sell for 6 times as many,
which is 6% ; then since A. really sold for 7%, when he should have sold for 632, he
gained 72 — 632, - #2; and if 632 gain #2, 12 will gain the 63 or *# part, and
100% will gain 100 times as much, which is 34% or %.
30. Bought coffee at 152 per pound and sold it at 182; what would I gain by
buying $75 worth of coffee? Ans. $15.
31. Bought coffee at 15% per pound and sold it at 182; what did I gain by
selling $75 worth of coffee? Ans. $12.50.
32. A real estate speculator sold a house and lot for $7200 and gained 20
per cent. He then invested the $7200 in merchandise which was sold at a loss of
20 per cent. Was the result of the two transactions a gain or a loss, and if so how
much 3 Ans. $240 loss.
33. A grocer sold a barrel of flour for $8.75 and gained 25 per cent. Soon
after he sold to the amount of $637 and gained 40 per cent. What was the last sold
at per barrel, and how many barrels were there ? Ans. 65 bbls. at $9.80.
34. A general fought 32 battles and lost 4; what per cent did he lose?
Ans. 12.4%.
35. A merchant has due him $32 and compounds the claim on receipt of $4;
what per cent did he lose ? Ans. 874%.
36. The net assets of a bankrupt are $72000, which is 15 per cent of his
liabilities. What are his liabilities? Ans. $480000.
37. A person pays $30 per month for board, which is 20 per cent of his
monthly salary. What is his salary per month # Ans. $150.
38. Paid $6 for sawing 4 cords of wood, which was 163 per cent of its cost;
what did it cost per cord 3 Ans. $9.
$ cost.
STATEMENT. 100
50 || 3
4 || 6
39. A mother having a basket of oranges, took out 50 per cent for her
children; of these she gave 40 per cent to her son who gave 4 of his to his sister,
who thus received 3 oranges. How many were there in the basket at first 3
Ans, 30.
OPERATION.
Let 100% = the whole number of oranges. 50% = the number taken out.
474 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICs. &
40% of 50% = 20% = given to son. # of 20% = 10% given to sister. 10% = 3
oranges. Then 10% = 3, 100% = 30 oranges, answer.
Or thus:
100% = the whole number of oranges. 50% = } taken out. 40% = # of 3
= } given to son. # x + = Pºs = given to sister. Pö = 3, then }} = 30, the whole
number in the basket.
40. A merchant marks his goods 50 per cent above cost, but supplies his
Wholesale patrons at 25 per cent discount on retail price; what per cent gain does
he make 3 Ans 12#7%.
41. A merchant asked for a lot of goods 20 per cent more than they cost, but
they being a little soiled he closed the sale at 10 per cent less than his asking price,
and realized a profit of $112. What did his goods cost? What was his asking
price? And for how much did he sell them 3 Ans. $1400 cost.
$1680 asking price.
$1512 was the selling price.
$100 - $
20 | 100
* 8 112
$120
OPERATION. 12 $1400 cost.
*=== 280 = 20% gain.
$108
100 $1680 asking price.
t-ºm-º-º-º: 168 = 10% discount.
$8 gain.
$1512 selling price.
42. If the cost of purchasing and shipping goods from New York to New
Orleans is 8 per cent on New York cost, what per cent gain on the first cost must
the New Orleans merchant mark his goods in order to clear a profit of 334 per cent
On full cost $ What amount of goods must he purchase to make a net profit of
$35200 on New York cost? What would be his New Orleans profit?
Ans. He must mark his goods at 44% gain.
He must purchase $80000 worth.
His N. O. profit would be $28800.
OPERATION.
$100 cost.
8% freight. $ C $80000
| 199 25 8%
$108 total cost. 11 A4 lºgg 2200 *=s*
36 == 334%. — $ 6400
| $S{}900 Ans. 35200
$144 N. O. S. P. * = º
100 $28800 Ans.
$44 gain.
43. A merchant bought flour at $7.50 per barrel; what must he ask for it in
order that he may fall 10 per cent on his asking price and still gain 20 per cent on
the cost 3 s Ans. $10.
PERCENTAGE. 475
44. Sold a barrel of oranges for $6 and gained 20%. I then invested the $6
in merchandise which I sold at a loss of 20%. What was the gain or loss by the
two transactions? Ans. 20% loss.
45. How many apples must I buy so that after allowing 25% of them to be
eaten and 20% of the remainder to be given away, I may sell just 1 dozen :
Ans. 20 appleS.
OPERATION.
100 100 100
75 | 12 80 | 16 Or, 75 100
e- *E-º: 80 | 12
16 20 Ans.
46. A fruit dealer has pears worth 52 a piece, but will sell 6 for 25%. What
per cent would be gained by buying 6 for 25¢, and what per cent would be lost by
buying them separately at 52 each 3 Ans. 19, 20% gain; 29, 163% loss.
OPERATION,
6) 25 6 | 5 6 || 5
T4, 4}=25 | 6 5 | 100
5 100 &= assºmsºmºsº-ºº:
tºº *=se smºº 163% loss.
# 20 %gain.
47. The New Orleans Cotton Factory receives from a merchant 200 bales or
90000 lbs. of Muddling Cotton to manufacture into sheetings. The charge for man-
ufacturing is fixed at 3% per yard, and payment is to be made in Middling Cotton
at 12g per pound. After the factory had delivered to the merchant 128800 yards,
which weighs 8 ounces per yard, a settlement was made on the following conditions:
It was agreed to allow 123 per cent waste in the manufacture; the factory to retain
the cotton remaining on hand at 12%., and for the balance due to receive Middling
fair instead of Middling at 13%. How many pounds of cotton does the merchant
owe the factory? Ans. 14584* lbs.
OPERATION.
100 128800 yds.
874 64400 16 || 8
73600 lbs. Cotton. | 64400 lbs.
90000 {{ 4 * g
128800 yards.
16400 lbs. on hand. 3 gº
12g *=º
$3864.00
$1968.00 1968.00
$1896.00 - 13 g = 14584; lbs.
48. Bought coffee at 192 per pound; allowing that it will fall short 5 per
cent in weighing and that 8 per cent of the sales will be uncollectible, for how
much per pound must it be sold to clear 15 per cent profit on the cost 3
Ans. 252.
476 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. º
OPERATION.
192 cost. 100 100 115 100 | 115
285 = 15% gain. 95 | .2185 92 || 23 or, 95 || 100 or, 95 || 100
*º-º-º: *mºmº || º gº ºmºmºmºmºmº 92 || 19 92 || 100
.21852. 23%. | 252 Ans. | 19
49. A grocer bought at auction 200 baskets of champagne for $2800; what
must he ask for it per basket so that he may discount 20 per cent on his asking
price, allow 10 per cent of sales to be bad pay, 24 per cent commission for selling,
and still make a profit of 100 per cent on cost 3 Ans. $40.
FIRST OPERATION. SECOND OPERATION.
$ 1998
100 $ 7 2009
80 | 200 100 374 || 250 109 || 23%. 2
874 100 8% 200 2%
100 2800 ** 2000 *m-emsmº,
200 |250 $40 Ans.
*= 285}
$40 Ans.
50. A merchant sold a pair of shoes for $3.20 and lost 20 per cent; what
should he have asked for his shoes in order that he might have fallen 20 per cent
on his asking price and still have gained 20 per cent on cost 3 Ans. $6.
51. If it costs a merchant 5% over factory price, to put his goods in store,
and he sells at a profit of 10% on full or store cost, and pays a tax of 35% on
sales, how much goods must he sell per month, to equal a salary of $300 per month
And what % did he actually make on the goods, including tax as part of cost, in
addition to the 5%; and what amount must be invested to produce $300 per month?
Ans. 10.44225% net gain on store cost.
$3316.59-i- monthly sales.
9,938+ 9, gain including tax as part of cost.
$2872.94-H amt. to be invested to gain $300 per mo.
OPERATION.
$100 = assumed factory cost. Cost at factory, $100.
5 = 5% to place goods in store. Charges to put in Store, 5.
-º-º-º-º: Tax on sales, .05775
$105 = Store COSt. gºmºsºme
10.50 = 10% gain on store cost. Total cost, $105,05775
$115.50 = selling price.
.05775 = }o'ſ, tax on sales. 10.44225
tºº-ºº: 105.05775 100
$115.44225 = net receipts after paying tax.
105. = Store cost. 9.939-1-7% gain including
&== tax as part
$ 10.44225% net gain on store cost. of cost.
$
115,44225 100 COSt.
10.44225 || 300 115,44225 || 33.16.591.25
$33.16.59125 monthly sales.
| $2872,94+ amount to be in-
vested to gain
$300 per mo,
PERCENTAGE.
477
52. An importer sold # of a cargo at 224% gain; # of remainder at 163%
gain; # of remainder at 10% loss, and the balance at 40% loss. There was a profit
of $335.40 on all. What was the cost of the cargo?
Let $100
Then #
Then # of remainder
Then + of remainder
Then the remainder #
# = }
# =
OPERATION.
= the cost of cargo.
$334 at 224% gain
33% at 163% gain
8; at 10 % loss
25 at 40 % loss
–4–
12.
:
$100
Then $13;g — $10# = $2% net gain.
Then $2; : $100:: $335.40 :
$15093 cost of cargo,
:
or thus,
$ 73 gain.
5; gain.
# loss.
10 loss.
$13+g gain $10; loss
100
9 •
335.40
20
$15093,00
53. A live stock dealer sold two horses, each at the same price. On one
horse he gained 25 per cent, and on the other he lost 25 per cent. His net loss on
the two sales was $12.
OPERATION.—FIRST PART.
What was the cost of each horse ?
• Ans. $72 cost of 1st horse.
$120 cost of 2nd horse.
1. Assume that one horse was sold for $100 at a gain of 25 per cent.
2 { % {{
Then find the cost of each horse, and
{{ {{ { %
then proceed as shown in the operation.
Statement to find cost of first
horse sold a 25% gain.
100
100
$80, cost 1st horse.
125
Statement to find cost of the
two horses actually sold.
$
213;
13; 12
lsº
OPERATION.—SECOND PART.
Statement to find cost of second
horse sold Ø 25% loss.
100
75 100
| $1334, cost 2nd horse.
OPERATION —THIRD PAR.T.
Statement to find cost of first
horse actually sold
$
0
2134 || 192
S
$
2
| $7
$100 at a loss of 25 per cent.
the loss on the sale of both horses;
$ 80 cost first horse.
133; cost second horse
$213; cost of both horses
200 assumed selling price
of both horses,
$13; loss on sale of both
horses.
Statement to find cost of
second horse actually sold
$
1334
2134 || 192
| $120
478 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. º
- - SECOND OPERATION.
Let $1 = sale.
# or 25% gain on cost = } or 20% on sales.
# or 25% loss on cost = } or 334% on sales.
OPERATION FOR FIRST HORSE. OPERATION FOR SECOND HORSE.
$1 = selling price. $1 = selling price.
less # = 25% gain on cost. plus # = 25% loss on cost.
$# = cost of 1st horse. $14 = cost of 2nd horse.
# loss — # gain = # = $12.
$ then, $90 = sale of 1st horse.
12 — # = 18 = gain on 1st horse.
2 || 15 *=º
dº I am-ems $72 = cost of 1st horse.
| $90 = sale of each horse.
and,
$90 = Sale of 2nd horse.
+ # = 30 = loss on 2nd horse.
$120 = cost of 2nd horse.
54. A. and B. are both merchants and conducting the same character of
business. A. desires to exchange or barter $9384 worth of sugar for $9384 worth of
rice. The market price of the sugar is 11g, and the market price of the rice is 82.
B. agrees to exchange but suggests to A. that each increase #2 per pound with a
View to ingrease the market price for future sales; to this proposition A. cheerfully
assents, and they exchange at the increased price, A's sugar at 114% and B's rice at
842, to the amount of $9384 worth.
By reason of the #2 increase of price, what was their respective gain and loss
in dollars; their gain and loss in pounds; their relative gain and loss per cent;
their respective real gain and loss per cent, and their respective real investment;
and at what price should A. have valued his sugar in order not to have lost by the
#2 increase on B's rice # *
Ans. $144, A’s loss and B's gain.
1309 fºr lbs. sugar, or 1800 lbs. rice, A. loses and B. gains.
1}}} %, A's relative loss per cent and B's relative gain per cent.
1}}#%, A's real loss per cent.
1#%, B's real gain per cent.
$8976, A’s real investment.
$8832, B's real investment.
11++g, A. should have valued his sugar, in order not to have lost
by reason of the #2 increase on B's rice.
* , PERCENTAGE. 4.79
CONTRACTED SOLUTION.
$9384.00 - 11c. = 85309 ºr lbs. sugar, should exchange, A. sells and B. receives.
$9384.00 - 11}c. = 81600 lbs. sugar, did exchange, A. sells and B. receives.
~
3709 ºr lbs. sugar, A. gains and B. loses.
$9384.00 -- 8c. = 117300 lbs. rice, should exchange, A. receives and B. sells.
$9384.00 + 84c. = 110400 lbs. rice, did exchange, A. receives and B. sells.
6900 lbs. rice, A. loses and B. gains.
Then,
6900 lbs. rice at 8c. = $552 = A's loss and B's gain.
And,
3709 ºr lbs. sugar at 11c. = $408 = A's gain and B’s loss.
$144 = A's net loss and B's net gain.
Also,
#c. —– 11}c. = ºrc. = A's gain and B’s loss on 1 cent of Sugar.
And,
#c. - 84c. = Tºrc. = A's loss and B's gain on 1 cent of rice.
Then,
*fc. — ºc. = 33rc. = A's net loss and B's net gain on 1C. of barter.
Which equals 1}}}% = A's relative loss per cent and B's relative gain per cent.
The real gain and loss per cent is found by increasing the relative per cent in the same ratio
as the original prices, thus:
#TC. × * = +37c. = 1+}}% = A's real loss per cent.
#I c. X * = rºzc. = 13.3% = B's real gain per cent.
Also,
$9384 × #= $8976 = A's real investment.
And, &
$9384 x + = $8832 = B's real investment.
Then,
$144.00 – 11c. = 1309 ºr lbs. sugar, A's loss and B's gain.
And,
$144.00 – 8c. = 1800 lbs. rice, B's gain and A’s loss.
Finally,
B's ic. increase on 8c. = 64%; and 64 per cent increase on A's 11c. = +&c. which added to the
11c. gives 11+}c. as the price that A. should have valued his sugar in making the exchange.
Tºsº (g ---
sº e º e º ºs e º 'º e º a s e s = • * * * * ºr e º sº e º 'º - e º 'º e º e s tº s is e º e º a s e s is e º s = e º 'º e s a sº e º ºs e º 'º a s m e º e º e s ∈ e s is a se –4%
Çommission and Brokerage.
892. Commission is the sum charged by an agent, factor, correspondent,
broker or commission merchant, for buying or selling goods or other property, col-
lecting debts, making advances, or transacting other business of a similar nature.
The sum thus charged is generally calculated at a certain rate per cent on
the amount of purchase, sale, collection, advance or other business transacted.
The party who transacts business for others is called the Agent, Broker,
Collector, Solicitor, Factor or Commission Merchant, according to the nature
Of the business.
The party for whom the business is transacted is called the Principal.
NOTE 1.-Commission merchants or factors are usually placed in possession of the goods
bought or sold, and transact the business in their own names.
NOTE 2.—Brokers do not have possession of the goods or securities bought or sold and
transact the business in the name of those who employ them
893. A Consignment is a quantity of goods shipped or sent by one person
or firm to another, to be sold on commission for account of the shipper.
894. A Consignor is a person who ships goods to be sold on commission for
his account.
895. A Consignee is a person to whom goods are shipped to be sold for
account of another.
896. An Account Sales, is an itemized statement rendered by the Agent
or Consignee to the Shipper or Consignor, showing in detail the sales, the charges
of all kinds, and the net proceeds.
897. The Net Proceeds of a consignment is the amount of money due the
consignor, after deducting from the total sales the commission and all other charges.
898. An Invoice or Account of Purchase is an itemized statement rendered
by an agent to his principal showing in detail the cost of goods bought, per his
order, and all expenses or charges attending the purchase and shipment.
899. Guarantee is a promise or obligation by which the Agent becomes
surety for the payment of goods sold on credit, or for the performance of some duty
devolving upon others, or for the grade or quality of goods bought ; and for this
responsibility a percentum charge is made, which is called Guarantee.
THE RATE OF COMMISSION OR BROKERAGE TO CELARGE.
900. The rate per cent commission or brokerage charged for transacting
business for others is usually fixed by the Chamber of Commerce or Board of Trade
(480)
* - COMMISSION AND BROKERAGE. - 481
of the city where the person transacting the business resides, and it varies according
to the nature and extent of the business. A less per cent is generally charged by a
broker than by a commission merchant, because the work he has to perform requires
less labor and time, and his expenses for office rent, services, employees, etc., are
less.
The charges made by brokers, and the customs regulating the business of
brokers, are generally fixed by the Association of Brokers and will be fully
presented under the subject of Stocks and Bonds. We will here remark that it is
the custom of the New Orleans brokers to charge on the par value of Gold, Stocks
or Bonds, and not on the purchase or selling price, except as follows: on stocks
selling at and over $50 per share, 50g per share is charged; on stocks selling at
$25 and under $50 per share, 25g per share is charged; and for stocks selling below
$25 per share, a charge is made as per agreement.
901. The principles of percentage apply to Commission and Brokerage and
the questions are classed as follows:
TO FIND THE COMMISSION WEHTEN THE COST AND SELLING PERICE
AND TEIE PER CENT OF COMMISSION OF BROKERAGE
ARE GIVEN.
PIROELEMS.
1. An agent sold $2845 worth of goods and charged 2 per cent commission.
What was his commission ? Ans. $56.90.
OPERATION.
$28.45 Explanation.—As was explained in the first problem of
2 percentage, page 435, we first get 1 per cent by dividing by
100, whichi. done by placing the º point º º
tens and hundreds figures, then 2 per cent by multi-
$56.90 Ans. plying by 2. gures,
2. Sold 5 hlids., 5418 lbs. Sugar at 1242, and charged 24 per cent commission.
What is my commission and what is the net proceeds?
Ans. $16.93 commission. $660.32 net proceeds.
OPERATION.
# 541890 Explanation.—In this problem, we first find the value of
the sugar, and then the 2% per cent of the same. The
677.250 Val f multiplication of the 5418 lbs. by 12} is performed, for
$677. value or Sugar. reasons given on pages 94 and 436, by multiplying by 100, as
amsºmºsºmºsºmºmºs indicated by the two small naughts, and then dividing by
$ 16.93 = 24% commission. 8; or to express it differently, divide by 8 and carry, the
: º P. 1. º : º º 2% i.
performed by multiplying by 10, as indicated by the Sma
$660.32 net proceeds. :*.*,gººd then *ś by 4, as explained on pages
4 and 436.
3. A real estate agent sold a piece of real estate for $28465, and charged 1
per cent commission. What was his commission ? Ans. $284,65.
482 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
4. Collected $2842.80, and charged 14 per cent brokerage. What is my
brokerage f Ans. $35,534.
5. Bought 20 shares of $100 each of Germania Bank stock, at 15 per cent
premium, and paid brokerage 3 per cent. What was the brokerage and the cost of
the stock # Ans. $10 brokerage. $2310 cost of stock. .
NotE.—Brokerage is charged on par value.
6. Exchanged through a broker $1280 fractional currency for bills of a larger
denomination. The broker charged + per cent; what was the brokerage 3
Ans. $3.20.
7. Sold 208 shares of $100 each of National Bank stock at $924. Paid
brokerage 4 per cent. What are my net proceeds, and what is the brokerage?
Ans. $19084 net proceeds. $104 brokerage.
8. A commission merchant has 10 hlids., 12000 lbs. of sugar on commission
and sells it through a broker at 82 per pound. His commission is 24 per cent, and
the brokerage is 14 per cent. All other charges amount to $27. What are the net
proceeds, the commission and the brokerage?
Ans. $36 commission and brokerage. $897 net proceeds.
TO FIND THE AMOUNT TO BE INVESTED AND THE COMMISSION,
- WHEN BOTH ARE INCLUDED IN A GIVEN SUM.
PROBLEMIS.
902. 1. Received $10000 from a correspondent with instructions to invest,
the proceeds, after taking my commission out, in merchandise for his account. I
charge 2 per cent commission. What is my commission and what amount do I
invest ? Ans. $196,08 commission. $9803.92 invested.
OPERATION.
$100 assumed investment. Explanation.—We first observe that, as com-
2 = 2% commission mission is allowed on the amount invested only,
ſº-ſº e we cannot calculate on the sº for º:
© º would be getting commission on the amoun
$102 cash required to invest $100. invested jià. º due us. According-
ly, for reasons previously given in percentage,
$ Invested. pages 443 and 463, we assume $100 to represent
100 the amount invested, * OIl tº: * ...
calculate and add to the same the 2 per Çen
102 10000 ($9803.92 commission, which gives $102 as the cash equiva-
gº-me ºmº lent of $100 investment and the commission,
for making the same. With these figures, it is
clear that we can invest as many hundred
& dollars as $10000 is equal to $102; or philosophiº;
ally, placing the $100 assumed investment on the statement line, we reason thus: If $102 cash will
invest $100, $1 will invest the 102d part, and $10000 will invest 10000 times as much, which is
$9803.92. On this amount we are allowed commission which at 2 percent is, to the nearest unit ºf
à, º,08. The commission may also be found by deducting the investment from the whole
alloull º
* COMMISSION AND BROKERAGE. 483
2. A country merchant purchased $84.50 worth of merchandise, on 4 months’
time. He then proposes to pay $5000 cash on condition that he be allowed 5 per
cent discount on the amount of bill that he pays, which proposition is accepted.
What amount does he still owe ? Ans. $3186.84.
3. The net proceeds of a consignment due a consignor is $2511.25, which I
am instructed to invest, after deducting commission, in sugar at 10 cents per
pound. I charge 24 per cent commission. How many pounds of sugar can be
purchased ? Ans. 24500 lbs.
4. A wine dealer remitted to a merchandise broker in New York $1965.60 to
be invested in claret. The broker charges 24 per cent commission for investing
and 24 per cent for guaranteeing the quality, both of which are to be deducted from
the $1965,60. How much money was invested in wine, and how many boxes were
there, the price per box being $6.50? Ans. $1872 investment.
288 boxes.
TO FIND THE AMOUNT INVESTED, OR TOTAL SALES, WHEN THE
COMMISSION AND THE RATE PER CENT COMMISSION
ARE GIVEN.
FIROBLEMIS.
903. 1. An agent's commission for selling goods was $56,90, the rate per
cent was 2. What was the amount or gross sales Ans. $2845.
OPERATION.
sº * Sales. Bæplanation.—In all problems of this kind,
O
assume $100 and find the commission or broker-
$2.00 commission on $100. age thereon. Then with the relationship num-
bers thus obtained, make a proportional state-
$ ment as shown in the operation, reasoning thus:
2 * Since $2 commission requires $100 sales, $1
sºmeºmºsºmsºmºmºmº commission will require the 3 part, and $56,90
| $2845.00 Ans. will require 56.90 times as much.
2. A commission merchant received $27.88 commission for selling merchan-
dise, and guaranteeing the payment thereof. The rate of commission was 5 per
cent and the rate of the guarantee was 3 per cent. What was the amount sold 3
Ans. $348.50.
3. A broker received $71.25 for selling gold at # per cent brokerage. What
amount did he sell? Ans. $28500.
484 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. †
TO FIND THE RATE PER CENT COMMISSION OR BROKERAGE WEHEN
THE COMMISSION AND THE AMOUNT OF SALES, COLLECTIONS
OR INVESTMENTS ARE GIVEN.
PROBLEMIS.
An agent collected $3400 and his commission was $85; what was
Ans, 24%.
904. 1.
the rate of commission ?
FIRST OPERATION.
100% rate of commission assumed.
100% of $3400 is $3400.
Explanation.—In this solution, we assume 100
per cent and find 100 per cent of the amount.
Then with the relationship numbers thus obtain-
"100 ed, we make a proportional statement as shown
3400 | 85
–
in the operation, reasoning thus: Since $3400
require 100 per cent, $1 will require the 3400th
24% Ans.
SECOND OPERATION.
part, and $85 will require 85 times as much.
Ea:planation.—In this second solution, we
85 reason as follows: Since the commission on
3400 | 100 $3400 is $85, the commission on $1 would be the
-*- : -m-m- 3400th part, and on $100, it would be 100 times
$24 = 24%. as much.
2. A broker received $180.25 for collecting $7210. What per cent broker-
age did he charge 3 Ans. 23%.
3. A commission merchant bought per order, 150 barrels flour at $8 a
barrel. He paid drayage and freight $42. The whole invoice cost $1279.26. What
was his commission and rate per cent of commission?
Ans. $37.26 commission. 3% rate of commission.
4. An attorney collected $2210 and sent me $2099.50. What rate per cent
did he charge? Ans. 5%.
5. Paid $602.16 taxes on property assessed at $23160. What was the rate
per cent? Ans. 2.6%.
TO FIND THE COMMISSION OR BROKERAGE AT ANY RATE PER
CENT ON ENGLISH CURRENCY.
905.
FIRST OPERATION.
1. What is 4 per cent of £5860 16s. 7d. 3
SECOND OPERATION.
Ans. £234 8s. 8d.
THIRD OPERATION. -
£58,60 16s. 7d. 3658.60 16s. 7d. #5860 16s. 7d.
4% 4% 5 4%
#234.40 .64 .28 fº234.43 6 4 .80+ .029%
20 20 See page 450 for reducing
shillings and pence to the deci-
8.64S. 8.66s. mal of a pound.
12 12 #5860.829
4%
7.966. 7,966. #234.43316
20
8.66320S
12
7.9584d
Explanation.—In the first operation, we reasoned as follows: 1 per cent of £5860 16s. 7d. is
* COMMISSION AND BROKERAGE. 485
£58,60.16s. .07d., and 4 per cent is 4 times as much, which is in the first operation #234.40.64s.
and .28d. These hundredths of pounds, shillings and pence we then reduce to shillings and pence
according to the principles of denominate numbers, as shown in the operation.
g In the second operation, we first found 1 per cent in the same manner as in the first Opera-
tion, but when we multiplied by 4, we reduced the .28d. to hundredths of shillings by dividing by
12, which gave us .02s. and .04d. We then multiplied the .16s. by 4, and produced .64s., to which
We added the .02s, and obtained 66s, which we reduced to hundredths of pounds by dividing by
20, which gave us .03 pounds and .06s. We then reduced the .43 pounds .06 shillings and .04 pence
to shillings and pence, as explained in the first operation.
In the third operation, we first reduce the shillings and pence to the decimal of a pound and
. 4 per cent on the whole amount. We then reduce the decimal of the pound to shillings
and pence.
2. What is 34 per cent of £78 4s. 10d. 3 Ans. £2 14s. 9d.
3. What is # per cent of £200 19s. 3d. } Ans. 10s. 0d. 2f.
4. What is 24 per cent on £426. 11s. 7d. ? Ans. £1013S. 3d.
NOTE. 1.--To compute 23 per cent, first compute it at 5 per cent and then divide by 2. Or
º : º: º per cent and divide by 40, as 2% per cent is #5. Or find 10 per cent and divide by
, aS 2# is + of 10.
NOTE 2.—See Interest on English Money.
MISOELLANEOUS PROBLEMS.
906. 1. What is the commission for selling $1620.80 worth of goods at 24
per cent” Ans. $40.52.
2. A commission merchant invested $5050 for a planter, and charged 1 per
cent commission. What was his commission ? Ans. $50.50.
3. A merchant for whom I do business remits me $5050 with instructions to
invest the proceeds, after deducting 1 per cent commission, in certain goods. What
is my commission ? Ans. $50.
4. Bought $5000 gold at $1124, and sold the same at 154 per cent premium.
Paid + per cent brokerage for purchasing and # per gent for selling. How much
did I gain 3 Ans. $112.50 currency.
5. Bought $5000 of bank stock at $1124, and sold the same at 10+ per cent
premium. Paid # per cent brokerage for purchasing and # per cent for selling.
How much did I lose ? Ans. $137.50 currency.
6. Invested through a broker $10825 in U. S. Bonds at 108 per cent. Paid
+ per cent brokerage. What was the brokerage, and what the face of the bonds?
Ans. $25 brokerage. $10000 face of bonds.
OPERATION.
$100 par value of bonds assumed. $ bonds.
8 = 8% premium. 433 100
+ = brokerage. 4
# = +% | 10825
$108} = cost of $100 bonds.
$10000 bonds.
25 brokerage.
7. A broker collected $14812.75 and re-invested the proceeds in Securities at
par. He charged according to custom # per cent for collecting on the amount
collected, and 3 per cent for investing on the amount invested. What was his
brokerage 3 Ans, $147,39-H.
486 soul.E's PHILOSOPHIC PRACTICAL MATHEMATICS. *
8. The commission for selling books is 25 per cent and an agent received
$235. What was the amount of sales Ans. $940.
9. An auctioneer paid me $2954.80 net proceeds for goods sold. The charges
for selling were $28.20 and his commission 5 per cent. What was the amount of
sales? Ans. $3140.
10. A salesman sold $6837 of clothing upon which there is a profit of 33;
per cent, of which he received 10 per cent commission. What was his commission?
Ans. $227.90.
11. A salesman sold $4800 worth of goods and received $240, commission.
What was the rate of commission ? Ans. 5%.
12. A commission merchant received $2000 to be invested in supplies. He
charges 24 per cent commission for investing, 1 per cent is allowed for charges on
the supplies, and 2 per cent for insurance on cost plus 10 per cent of cost. What
Sum Was invested in supplies, and how much was his commission ?
Ans. $1891.30 invested in supplies. $47.76 — commission.
OPERATION.
$100 = 1st cost assumed. $101 = cost of supplies.
1 = 1% charges. 10 = % of cost of supplies.
101 = cost of supplies. $10.10 = 10% of cost.
$ 2} = % ..º. 101 = cost of supplies added.
* * - ºf - 11.10 = t -- 10%.
$2.525 = 23% commission. $111 § - *iº.
Aggregate cost of $100 supplies. $2.2220 = 2%. Insurance.
$100,000 = cost assumed. 100 = cost assumed.
1,000 = charges. 105.747 | 2000.000
2.525 = commission.
2.222 = insurance. | $1891.30 = invested in supplies.
— 1 0
$105.747 = cost to buy and 18.91 = 1% charges.
ship $100 supplies. $1910.21 = cost and charges of supplies.
2} = % commission.
RECAPITULATION.
$1891.30 = 1st cost of supplies. $47.755+ commission.
18.91 = 1% charges.
47.76 = 24% commission. oPERATION TO FIND INSURANCE.
42.03 = 2% insurance. $1910.21 = cost and charges of supplies.
191.02 = 10%.
$2101.23 = sum insured.
2 = 96 insurance,
$42.0246 = insurance.
13. A retailer bought from a wine merchant, on 3 months' credit, goods
amounting to $1500, and from a grocery merchant, on 3 months' credit, goods
amounting to $1800. But, without waiting for the maturity of the bills, he pays
the wine merchant cash in full less 5 per cent discount allowed, and to the grocery
merchant he pays $1000 cash, for which he is to be allowed 5 per cent discount on
the amount of invoice that it will pay. How much did he pay the wine merchantº
And how much does he still owe the grocery merchant?
Ans. $1425, he paid the wine merchant.
$747.37, he owes the grocery merchant.
14. A merchant received 180 bales of cotton which average 450 pounds to
the bale. The cotton was sold at 16#g per pound, and 24 per cent commission and
# per cent brokerage were charged for selling. Allowing 4 per cent loss on the
cotton received for sampling, classing and loose, and approximating the other
charges at $756.20, what amount of goods could be purchased with the net pro-
ceeds, after deducting 24 per cent commission on the face of the money invested?
Ans. $12006.83.
$2000.00 = sum received.
º ankruptcy.
& ---
=ss-
907. Bankruptcy is a failure in business and the condition of being insol-
Vent or unable to pay indebtedness.
e NOTE.-The word bankrupt is of Italian origin and is derived from the custom in many towns,
in the middle ages, of breaking the counter or bench (bancus), in the public exchange or market
place, that had been occupied by merchants who had subsequently failed.
908. A Bankrupt or Insolvent is a person or business man who breaks or
fails, or becomes unable to pay his debts in the ordinary or regular course of trade.
909. An Assignee is a person that is elected by the creditors of the bank-
rupt, or appointed by the court, to receive the assets of the bankrupt, pay the
expenses of the assignment, and use the proceeds of the property in paying the
Creditors of the bankrupt, as far as the money will allow, according to law and
equity.
910. An Assignment is the transfer of the property of a bankrupt to the
aSS1gnee.
911. A Schedule is a list of all the assets and liabilities of the bankrupt
containing the names and the place of business or residence of all his DEBTORS
and CREDITORS, and the amount due from or to each.
912. A Debtor is a person or firm who owes the bankrupt.
913. A Creditor is a person or firm whom the bankrupt OWES.
914. The Assets or Resources of a bankrupt are the entire property except
Such as may be exempt by law.
915. The Liabilities are what the bankrupt owes at the time of his failure.
916. Assets that can be converted into cash at their stated valuation are
called AVAILABLE AssETs; and those that cannot be immediately converted into
cash without a discount or rebate are called NOMINAL ASSETS.
917. Preferred Creditors are persons whom the law allows to collect their
claims in full, such as operatives, and employes to whom limited sums are DUE for
services rendered within a few months of the assignment. &
918. A Dividend is the sum or per cent paid to the creditors by the assignee
out of the assets of the bankrupt. Dividends of a part of the funds of a bankrupt
are often made before the final settlement of the estate.
919. As Discharge is an order or decree by a court of bankruptcy, evidenced
by a certificate therefrom, that the bankrupt has complied with all the requirements
of law governing cases of bankruptcy, and is therefore released or discharged from
all his debts.
(487)
488 SOULE's PHILOSOPHIC PRACTICAL MATHEMATIcs. &
PROBLEMIS.
1. The assets of a bankrupt amount to $203413.39. The liabilities amount
to $872546. Estimating that the expenses for settling the estate will amount to
$7385, and allowing the assignee 2 per cent on the full amount of assets, what per
cent will the bankrupt estate pay ? Ans. 22%.
OPERATION.
$2034.13.39 assets. Explanation.—From the total assets, we deduct
the expenses and assignee's charges, and thus
§ the net proceeds. We then observe that
72546, the liabilities, are to receive $191960.12,
$191960.12 net proceeds. and to find the rate per cent that $191960.12 is
of $872546 we place the net proceeds on the
statement line, and reason thus: If $872546 00
are to receive $191960.12, $1 will receive the
11453.27 expenses and assignee's com.
tº 191960.12 872546th part, and $100 will receive 100 times
872546.00 || 100 as much, which is 22 per cent. Or, we may
-* first assume and place on our statement line 100
22% Ans. per cent, and reason as in Articles 856 and 857.
2. A. and B. failed. Their assets were $6720, liabilities $30750, expenses
$250. The assignee charged 5 per cent on the assets. What was his commission
and what per cent did the creditors receive? Ans. $336 commission.
19.947+%.
3. In the above problem, what would have been the assignee's commission,
and the per cent the creditors would have received if the assignee had charged 5
per cent on the amount he paid to creditors and for expenses?
Ans. $320 commission. 20% received by creditors.
OPERATION.
100 $6400 6150
105 || 6720 250 expenses. 30750 || 100
$6400 paid out. $6150 paid to creditors. | 20% paid to
5% ,-- creditors.
$320.00 commission.
4. X. and Y. failed. The net proceeds, after deducting $440 expenses and
24 per cent commission of the assignee, were $7555. The liabilities were $47970.
What per cent will the creditors receive and what was the assignee's commission
On the total amount of resources? Ans. 15.74%. $205 commission.
OPERATION,
7.555 100 8200 100
440 97% 7995 - 24% commission. 47970 || 7555
$7995 | $8200 total assets. $205.00 commission. | 15.74%
5. A firm becomes bankrupt; its liabilities amount to $340180, and its
assets amount to $73216.58. Allowing the assignee 5 per cent commission on the
first $1000, 24 per cent on the next $5000, and 1 per cent on the remainder of the
assets, and estimating the other expenses at $2810, what rate per cent will the
estate pay, and how much will a creditor receive to whom $41319.65 are due 3
Ans. 20.4478 + 9%. $8448.95 +, the creditor receives.
6. A bankrupt estate paid a dividend of 184 per cent. The amount received
by a creditor, to whom 4 of the liabilities were due, was $7400. What were the
liabilities of the estate? Ans. $320000.
axes and Licenses.
----------------------------iss
aº Aºk--4---
-º-war -wº-
920. Taxes are sums of money assessed according to law on persons,
property, incomes, or products, for the use of the nation, State, county or parish,
incorporate city, or society.
921. A Poll Tax or Capitation Tax is a certain sum assessed on persons
Without regard to property. It is generally assessed by the State on the legal
Voters. &
922. A Property Tax or Direct Tax is a certain sum assessed on the
estimated value of real and personal property. This tax is usually a certain per
cent or a specified number of cents or mills on $100. In some cities and States the
rate per cent tax is limited. *
923. An Indirect or Excise Tax is a tax on articles of consumption in
their transfer from one person to another. It is also levied on licenses to pursue
certain trades and to deal in certain commodities.
924. Real Property or Real Estate consists of houses, lands and all
immovable property.
925. Personal Property consists of money, merchandise, stocks, mortgages,
notes receivable, furniture, live stock, etc., etc.
926. An Assessor is a person appointed to appraise the value of property
subject to taxation. &
927. A Collector or Receiver of Taxes is a public or government officer,
who has been elected or appointed, to collect or receive taxes.
928. An ASSessment Roll is a list or schedule containing the names of all
persons liable to taxation in the State, parish, or city for which the tax is assessed,
and the valuation of each person's taxable property. The assessment roll must be
prepared before making an assessment of taxes.
929. A License is a special tax or charge imposed by the national, State, or
municipal authorities, upon merchants, bankers, manufacturers, and nearly all
classes of business men, for the privilege of prosecuting their various occupations
and businesses within the limits of the authority imposing the tax and granting the
license. A proper and equitable assessment of this tax requires much wisdom and
experience, a faithful regard to duty, and unalloyed honesty.
930. U. S. Income Tax is at present, 1895, an assessment made by the
general government on all incomes exceeding $4000, with special exceptions.
NOTE.-The Income Tax has been declared unconstitutional by the U. S. Supreme Court.
931. The Budget is a financial statement of items showing the amount of
taxes to be raised. It is also a financial statement with business men, showing the
items of expenditure of the business for the current or succeeding fiscal year.
(489)
490 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
932. The method of assessing taxes, though in some respects slightly
different in some States and cities, is nevertheless virtually the same. --
In making an assessment of taxes the first thing in order, after the assess-
ment roll has been prepared, is to ascertain the amount of tax required to be raised,
to which amount must be added the expenses to collect the same, and the sum or
per centum allowed as being uncollectible. From the sum total of these three
amounts are to be deducted any special tax, the amount of poll tax, or the amount
of license tax, or all according as the assessment is one for a State, parish, or city.
Instead of increasing the amount of tax to be raised by the per centum
allowed for non-collectible tax, we may deduct the same per cent from the taxable
property or assessment roll, and use the remainder as the amount to calculate the
rate per cent upon.
Having these facts and figures, we then find the per cent of tax.
PROBLEMIS.
933. 1. The budget for the estimated expenditures and liabilities of a
State, for the coming year, amounts to $4500000. The inventory or assessment roll
amounts to $325000000. Allowing $600000 to be received for licenses and $50000
from polls, and estimating that 10 per cent of the assessment will be uncollectible,
and that 5 per cent will be required to pay the expenses for collection, what will be
the rate per cent of taxes on the assessment roll ? Ans. 1.3855 + 76.
OPERATION.
$4500000 amount of budget. Explanation.—From the amount of the budget,
650000 licenses and polls deducted. we first deduct the amount of the license and
&ºi=-º-º-º-º-mº-ºm-º. poll taxes, which leaves a balance of $3850000.
$3850000 amount of net tax to be raised. We then find the amount to be assessed in order
to allow 10 per cent for non-collectible tax.
To do this we first assume $100 to represent the
amount to be assessed, from which we deduct the
10 per cent allowed for non-collections, and
100 have in the remainder $90, the net collections as
90 || 3850000 the equivalent of $100 tax to be assessed. We
Tºº------ then reason thus: If $90 cash collections require
$4277777.77 total amount to be assessed. $100 tax to be assessed, $1 will require the 90th
Statement to find the amount to be assessed and part, and $3850000 will require 3850000 times as
allow 5 per cent commission for collecting. much. This gives us $4277777.77 as the amount
of tax to be assessed in order to allow 10 per cent
100 for non-collections and still realize $3850000
95 || 4277777.77 cash. We next find the amount to be assessed
tº in order to allow 5 per cent commission for
collection charges. To do this we again assume
$100 to represent the amount to be assessed, and
Statement to find the amount to be assessed to allow
10 per cent for non-collections.
$4502923.97 total amount to be assessed.
Statement to find the rate per cent tax. from this we deduct the 5 per cent commission,
$ and obtain $95 cash as the equivalent of $100
4502923.97 assessment ; then placing the $100 assumed
325000000 || 100 assessment on the statement line, we reason
thus: If $95 cash in order to pay 5 per cent
1.3855 -- % Ans. commission require $100 assessment, $1 cash
will require the 95th part, and $4277777.77 will
- require 4277777.77 times as much. This gives
us the amount of tax to be assessed, which is $4502923.97. Having now the total amount of tax to
be assessed, and the total amount of property as shown by the assessment roll on which the assess-
ment is to be made, we can find the rate per cent of tax. To do this we first observe that
$325000000 are to pay or produce $4502923.97, and to ascertain the rate per cent we place the
$4502923.97 on the statement line and reason thus: If $325000000 are to pay or produce $4502923.97,
$1 will pay or produce the 325000000th part, and $100, 100 times as much. The result of this state-
ament is the correct answer of the problem, carried to four decimals.
In practice 18 per cent or 11% per cent would probably be used.
º: TAXES AND LICENSES. 491
2. The taxable property of a village amounts to $924600. The tax to be
raised including collection charges amounts to $23492. Allowing that $5000 will be
received for licenses and special taxes, and making no allowance for non-collections,
what is the rate per cent, and how much will be the tax of Mr. A., whose real and
personal property amounts to $15380? Ans. 2%, rate of tax.
$307.60 Mr. A’s tax.
3. The budget for the estimated expenditures and liabilities of a city are as
follows:
EXPENIDITURES.
Salaries of city officers, clerks, collectors, etc., * º e $454,580
Wages of street, wharfage, drainage, etc., departments, º wº 424,000
Office expenses, º - -> tº º º º - * 29,900
Printing, º - º º º º - tº gº - gº 56,000
General service, - º * º - * - tºs 1,282,700
Contingent, - - º - -> º º e sº * - 50,000
Deficit of previous year, º -> tº º sº º * * 475,000
Total estimate of expenditures, - º tº º & $2,772,180
LIABILITIES,
Bonds maturing and interest on Same, tº -- º $1,808,508
7-#% certificates to redeem, - - º - - º 860,600
Special bonds, to be paid, <- - sº tº gº sº 70,000
Total amount of liabilities, - - *- - tº º $2,739,108
Total estimate of expenditures and liabilities, - - - $5,511,288
The assessment roll for real and personal property of the city, on which
the tax is to be assessed, amounts to $139,848,204. Allowing 10 per cent for non-
collections, and estimating that the receipts from new issues of bonds, from the
markets, wharves, licenses, and all other revenues of the city will equal the amount
of liabilities, what will be the rate per cent tax? Ans. 2.2025 + 7.
4. I am assessed as follows: Real estate $18600. Personal property $4000
Money at interest $2500. Poll tax $2. The rate per cent for city tax is 2 per cent
and for the State, 6 per mille; there is also a special city tax of 45 mills on every
hundred dollars. How much is my tax, for each, the city, State and special taxes,
and what was the total amount 3 Ans. $502.00 city.
11.29% special.
150.60 State.
2.00 poll.
$665.894 total amount.
492 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
5. A citizen paid $502.00 for City taxes, including $2 poll tax and the rate
per cent was 2. What was his property assessed for 3 Ans. $25000.
6. If a property holder pays $663 for city, State and special taxes, and the
rate for City is 24 per cent, for State, # per cent, and for special 65 mills on the
hundred dollars, what is the assessed value of his property? Ans. $20000.
7. A town desires to raise a tax of 24 mills on the dollar. There are 700
polls at $1, the personal and real property is assessed at $1,400,000. What is the
rate of tax and the amount of tax of a citizen whose property, personal and real,
is assessed at $11550 % Ans. .2% rate of tax.
$24.10 tax of citizen, including poll tax.
934. To facilitate the operation of apportioning taxes among a large number
of persons, the tax on $1, $2, $3, etc., at the ascertained rate per cent may be found
first and arranged as in the following table:
TABLE AT THREE AND ONE-HALF MILLS TO THE DOLLAR.
WALUE AMOUNT VALUE AMOUNT VALUE AMOUNT VALUE AMOUNT
OIF OF OF OF
PROPERTY. | OF TAX. PROPERTY. OF TAX. PROPERTY. OF TAX. | PROPERTY, OF TAX.
$ 1 .0035 $ 10 .035 $ 100 .35 $ 1000 $ 3.50
2 .007 20. .07 200 .70 2000 7.00
3 .01.05 30 .105 300 $1.05 3000 10.50
4 .014 40 .14 400 1.40 4000 14.00
5 .0175 50 .175 b00 1.75 5000 17.50
6 .021 60 .21 600 2.10 6000 21.00
7 .0245 70 .245 700 2.45 7000 24.50
8 .028 80 .28 800 2.80 8000 28.00
9 .0315 90 .315 900 3.15 9000 31.50
10 | .035 100 .35 1000 3.50 10000 35.00

gºld. (IRI). Silver and UICITeſt MOſley.
- - C - C - C --> --> C_-_C - - - - - - - - - - - - C - C - C - Cº - c s - e s - CDC C Dºº-Dºº Cºcº C. CTGº-Tºc Cº-Cºcº C. C. Cºcºcº =N
935. Premium is the per cent that an amount is increased.
936. Discount is the per cent that an amount is decreased,
937. Currency is some circulating medium representing money, used as a
Substitute for gold or the fixed measure of value.
938. Uncurrent Money is some circulating medium that is valued less
than gold, or less than the fixed measures of value. The coin of some foreign
countries is also classed as uncurrent money.
When one currency is at a certain per cent premium over another, the per
cent discount on the currency of lesser value is not the same as the per cent
premium on that of greater value; thus, when gold is 25 per cent premium, the
corresponding discount on currency is but 20 per cent; and when gold is 200 per
cent premium, $100 in currency is worth $33; in gold, or 663 per cent discount.
TO FIND THE CORRESPONDING DISCOUNT ON PAPER CURRENCY
OR ON SILVER WHEN WE HAVE THE PREMIUM ON GOLD.
1. If gold is 25 per cent premium, what is the corresponding discount
Ans. 20%.
939.
on currency?
FIRST SOLUTION.
OPERATION.
$100 gold assumed.
25 = 25% premium.
sºmsºmºmºs
$125 currency value of $100 gold.
Explanation.—Here we are constrained
to assume some number to Work from.
Accordingly, for reasons often given in
similar questions, we assume $100 gold
and to the same add the 25 per cent
premium, and by this work we obtain
$125 as the equivalent currency value of
$ Gold. • $100 gold, and having these equivalent
100 values we place the $100 gold on the
125 statement line, and reason thus: If $125
100 currency is worth $100 gold, $1 currency
&===s**
$80 gold value of $100 currency.
100
*==
$20 = 20% discount, Ans.
is worth the 125th part, and $100 currency
instead of $1 is worth 100 times as much.
The result of this reasoning and state-
ment gives us $80, as the gold value of
$100 currency, which deducted from $100
currency gives 20 per cent discount on
currency.

(493)
494 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. º
SECOND SOLUTION.
OPERATION.
$100 gold assumed. ſº. ñº º: of º:
R – Q K te SOIUltil On 18 the SãIn 6 a.S that given ln the
25 = 2 5% premium. first solution, so far as º: the
sºº-º-º-º: equivalent value in currency of $100 gold
$125 currency value of $100 gold. at 25 per cent premium. Then instead of
finding the value of $100 currency, we
% find the per cent at once by placing the
3% 25 per cent on the statement line, and
reasoning thus: If $125 currency give 25
per cent discount, $1 currency will give
100 the 125th part, and $100 will give 100
times as much. It will º tº:
tº 25 per cent premium on old is the
| 20% discount, Ans. tº: discount on *:::::::::
$125 currency, which the result shows to
be 20 per cent.
125
2. Gold is 12# per cent premium; what is the corresponding discount on
Silver ? Ans. 114%.
3. Gold is 150 per cent premium; what is the corresponding discount on
paper currency Ans. 60%.
4. Gold is 500 per cent premium; what is the corresponding discount on
paper currency 3 * Ans. 834%.
TO FIND TEIE CORRESPONDING PEREMIUM ON GOLD WHEN WE
HAVE TEIE DISCOUNT ON CURRENOY OR ON SILVER.
940. 1. If currency is 20 per cent discount, what is the corresponding
premium on gold 3 Ans. 25%.
FIRST SOLUTION.
OPERATION.
$100 currency assumed.
20 = 20% discount deducted. Ea:planation.—In the solution of this
problem, we first assume $100 currency
and then find the equivalent value of the
$80 value of $100 currency in gold. same at 20 per cent discount in gold.
Having these two values, we reason
$ Cur. º thus: If $80 gold is equal to $100 cur-
100 rency, $1 gold is equal to the 80th part,
80 and $100 gold is equal to 100 times as
much. The result of this reasoning gives
100 $125 currency as the value of $100 gold.
sº-º-º: º:
e IS, Vallle à, D. , we nav *
*i; value º gold in currency. cent premium on gold, which corresponds
0 deducted from same gives. to or is the equivalent of 20 per cent dis-
tº count on currency.
! $25 = 25% premium, Ans.
º: GOLD, CURRENCY, SILVER AND UNCURRENT MONEY. 495
SECOND SOLUTION.
OPERATION. ~
$100 currency assumed. Eacplanation.—In this solution, we first
20 = 20% discount deducted. find the equivalent value in gold of $100
currency at 20 per cent discount, in the
º same manner as in the first solution. We
80 Value of $100 currency in gold. then place the 20 per cent discount on the
$100 currency on our statement line, and
% reason thus: If $80 gold give 20 per cent
20 premium, $1 gold will give the 80th part,
and $100 gold will give 100 times as
80 much. It will be observed that 20 per
100 cent discount on $100 currency is the
corresponding premium on (its equiva-
lent) $80 gold, which the result shows to
tºmº sº.
25% premium, Ans. be 25 per cent.
2. The discount on silver is 40 per cent; what is the corresponding
premium on gold 3 Ans. 663%.
3. The discount on currency is 90 per cent; what is the corresponding
premium on gold 3 Ans. 900%.
4. If the discount on currency is 99.1%+ per cent, what is the corresponding
premium on gold? Ans. 10000%.
5. If the discount on currency is 100 per cent; what is the corresponding
premium on gold 3 Ans. Infinite.
TO FIND THE CORRESPONDING PREMIUM ON CURRENCY OR ON
SILVER, WHEN WE HAVE THE DISCOUNT ON UNCURRENT MONEY.
941. In comparing gold and currency, or currency and uncurrent money,
currency is by the custom of bankers and money dealers used as the standard unit
of value.
1. When the old paper money of the city of New Orleans was at 10 per
cent discount, what was the corresponding premium on currency 3
Ans. 114% premium.
OPERATION.
$100 currency or silver assumed.
10 = 10% discount.
$90 value of $100 uncurrent money. Ea:planation.—The reasoning for this
problem is the same as the second solution
# of the preceding question. Hence it is
90 omitted.
100
| 11}} premium, Ans.
2. When uncurrent money is 50 per cent discount, what is the corresponding
premium on currency or on silver? Ans. 100% premium.
3. The discount on uncurrent paper money is 5 per cent; what is the cor-
responding premium on current money? Ans. 5; 7%.
496 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. tº
TO FIND THE WALUE OF GOLD IN CURRENCY OR IN SILVER OR TO
FIND WHAT AMOUNT OF CURRENCY OR OF SILVER IT
WILL REQUIRE TO PAY A CERTAIN DEBT IN
GOLD, WHEN THE GOLD IS AT A CER—
TAIN PER CENT PREMIUM.
942. 1. I owe a gold debt of $2000. Gold is 15 per cent premium. What
amount of currency will it require to pay the debt 3 Ans. $2300.
OPERATION.
$2000 Explanation.—It is clear that as the premium is on
1.15 the gold, $100 gold is equal to $115 currency, and
hence it will require as many times $115 as $2000 is
- equal to $100, which is 20; or as is shown in the
$2300.00 Ans. operation, we multiply by the 2000 and divide by
100; or we may reason thus: When gold is 15 per
cent premium, $1 is equal to $1.15 currency, and $2000 is equal to 2000 times as much.
2. Gold is 124 per cent premium; what is $6840 worth in currency or in
Silver ? Ans. $7695.
3. Gold is 185 per cent premium; what amount of currency will it require
to pay a gold debt of $18500? Ans. $52725.
TO FIND THE WALUE OF GOLD IN CURRENCY OR IN SILVER WHEN
TEIE DISCOUNT ON CURRENCY OR ON SLLVER IS GIVEN.
943. 1. Currency is 25 per cent discount. What is $2000 gold worth in
currency? Ans. $26663.
OPERATION.
$ $100 currency assumed.
100 currency. 25 = 25% discount.
75 2000 -me
$75 = gold value of $100 currency.
| $26668 Ans.
2. Currency is 124 per cent discount; what is $3672 gold worth in currency?
Ans. $4196.57%.
TO EIND THE WALUE OF CIUFRIENCY OR OF SILVER IN GOLD WEHEN
TEIE DISCOUNT ON CURRENCY OR ON SILVER IS GIVEN.
944. 1. Currency is 25 per cent discount below gold; what is $2000 worth
in goldf Ans. $1500.
OPERATION.
25% of $2000 is $500, $2000 – $500 = $1500.
2. Silver is 10 per cent discount below gold; what is $1755 worth in gold 3
Ans. $1579.50.
* GOLD, CURRENCY, SILVER AND UNCURRENT MONEY. 497
TO FIND WHAT AMOUNT OF GOLD IT WILL REQUIRE TO PAY A
CERTAIN DEBT IN CURRENCY, OR IN SILVER WHEN THE
GOLD IS AT A CERTAIN PER CENT PREMIUM.
945. 1. I owe a currency debt of $2000. Gold is 15 per cent premium.
What amount of gold will it require to pay the debt? Ans. $1739.13%.
OPERATION.
$100 gold assumed. Ea:planation.—In the solution of this problem, we first
15 = 15% premium. assume $100 gold, and then add thereto 15 per cent of the
tº-me same and thus obtain $115 as the value in currency of $100
*** * old; or, in other words, we find by these equivalent values
$115 currency. that sióðgoid wiil pay $115 currency debt. with these
ratio figures we make the solution statement, and placing
$ Gold. the $100 gold on the line, we reason thus: If $115 currency
100 is equal to $100 gold, $1 currency is equal to the 115th part,
115 and $2000 currency is equal to 2000 times as much ; or thus:
If $100 gold will pay a currency debt of $115, then by
2000 transposition $1 currency debt will require the 115th part of
*=- $100 gold, and $2000 currency debt will require 2000 times as
$1739.13% Ans. much. The result is $1739.13; gold.
2. Gold is 124 per cent premium ; what amount will it require to pay a
silver debt of $584.15% * Ans. $519.244.
3. A merchant owes in currency a debt of $21294.10, which he pays in gold
at 1083. What amount of gold will pay the debt? Ans. $19580.7844.
4. A currency debt of $5200 is paid in gold at 160 per cent premium. How
much gold is required ? Ans. $2000.
MISCELLANEOUS PROBLEMS IN GOLD, CURRENCY, SILVER AND
DNCURRENT MONEY.
946. 1. Gold is 334 per cent premium; what is the corresponding discount
on currency ? Ans. 25%.
2. Currency is 334 per cent discount; what is the corresponding premium
on gold 3 Ans. 50%.
3. The discount on uncurrent paper money is 124 per cent; what is the
corresponding premium on currency 3 Ans. 144%.
4. Gold is 143 per cent premium; what is $16000 worth in currency?
Ans. $18340.
5. Gold is 143 per cent premium; what amount will it require to pay a
currency debt of $18340? Ans. $16000.
6. Gold is 50 per cent premium; how much silver can I buy with $1000
gold 3 Ans. $1500.
7. Gold is 50 per cent premium; how much gold can I buy with $1000
Silver? Ans. $666,663.
498 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
8. Uncurrent money is 10 per cent discount, and gold is 10 per cent premium.
What almount of gold can I buy for $12000 uncurrent money? Ans. $9818.18; 1.
OPERATION.
$12000 uncurrent money. $ 90
1200 = 10% discount. 100 100 100
110 || 10800 or thus: 110 | 12000
$10800 currency.
| $98.18.18 ºr $98.18.18%r
9. How much uncurrent money at 373 per cent discount can I buy with $50
currency * Ans. $80.
10. How much uncurrent money at 124 per cent discount can be bought for
$4500 gold at 94 per cent premium ? Ans. $5618.57%.
11. Uncurrent money is 5 per cent discount, and a young dealer supposing
that currency was the same rate per cent premium, sold $10000 currency at 5 per
cent premium. What was his loss, and his loss per cent?
Ans. $26.31##, uncurrent money loss. #3%, loss on current money.
OPERATION.
100 96 $10000
95 || 100 5 *%
*=º. 95 || 100 esmºmºmºsºmsºmºsºme e
$1054; value $100 current money. or, – * $26.314 loss in un-
100 5 §96 premium. Current,
5 % discount. money.
51*;96 premium on current money.
5 *% discount. e *5% --º
*96 loss on clu'rent money.
Or thus,
$10000 currency. 100
5 95 || 10000
$ 500.00 premium. $10526.31+} amount of uncurrent money he should have received.
10000.00 10500.00
$10500.00 uncurrent money 26.31+; loss in uncurrent money. i
received.
12. A broker purchased from one class of customers uncurrent money at 5%
discount, and paid current money; and from another class of customers he purchased
current money at 5% premium, and paid uncurrent money. During the day he
purchased $100000 of each, current and uncurrent money. What was his gain,
and gain 'ſ', on the purchase of uncurrent money? What was his loss, and loss %
on the purchase of current money 2 and how much did the parties lose, who sold the
current money, by reason of their ignorance of the correct principles of percentage?
Ans. 1. The broker gained $5000 in uncurrent money purchased, and gained 5; 96 on the
current money paid out.
NOTE.--When uncurrent money is 59% discount, current money is 54% 96 premium.
Ans. 2. The broker lost $5000 in uncurrent money paid for currency, and lost 4}{96 on the
uncurrent money paid out.
NotE.—When uncurrent money is 4}{96 discount, current money is 5% premium.
Ans. 3. The parties who sold the current money lost $263.15}} in uncurrent money, or $250 in
current money by reason of their ignorance of percentage calculations.
NotE.—These parties should have received $105000 uncurrent money for $997.50 current money,
cr $105263.15}} for $100000 current money.
*
GOLD, CURRENCY, SILVER AND UNCURRENT MONEY. 499.
13. A broker purchased $100000 of uncurrent money at 10 per cent discount,
for which he paid current money. Then with the $100000 uncurrent money he
bought Current money at 10 per cent premium. What were his net profits and his
gain per cent Ans. $909.09 ºr gain in current money.
185% gain on currency invested.
OPERATION.
$100000 uncurrent money CUR
e t . MONEY.
10000 = 10% discount deducted. ić
º UNCUR. MONEY. UNCUR. MONEY.
$90000 cost in current money. 110 || 100000
$ GAIN. $90909.09: current mone
e W.
11 || 10000.00 º
90000.00 || 100 . 90000.00 current investment.
$909,09 ºr gain on current investment.
193% gain, Ans.
14. The selling price of coffee is 20g when gold is 25 per cent premium;
What should it sell for when the premium on gold is 15 per cent? Ans. 18%.
OPERATION.
$ Ea:planation.—In this solu-
100 tion, we first find the gold
125 value of the 20c. currency,
20 which is 16c. The reasoning
º for this work was fully explain-
ed in Article 945, page 497 ;
then having the value when
16 old pric º ©
#, 3. # tº: en gold is at par gold is at par, to find the
5 o D º value in currency when gold is
— e e tº 15 per cent premium we simply
18% selling price when gold is 15% premium. add the 15 per cent, as shown
in the operation, and have in
the result the selling price
desired.
15. When gold is 10 per cent premium, goods sell for $2.75; what should
they sell for when gold is worth 132? Ans. $3.30.
16. If goods are worth $4.50 when gold is 150, what are they worth when
gold is at par 7 Ans. $3.
17. If goods are worth $4 in currency, what are they worth in uncurrent
money which is 10 per cent discount? Ans. $4.44%.
18. The gold price of goods is $2; what are they worth in uncurrent money,
7
the premium on gold being 124 per cent, and the discount on uncurrent money
being 124 per cent ? Ans. $2.57%.
FIRST OPERATION. SECOND OPERATION.
$ $
2.00 $2.00 100
874 Or, 25 87;
112} mºmºmº 2.25
$2.25
$2.57% Ans. $2,574, Ans.
gºustomhouse Business.
--------------------------------se
947. Custom Houses are buildings or offices erected or established by the
Imational government, where the collection of duties or customs on goods imported
is made, and where vessels are entered and cleared, etc.
948. Duties or Customs are taxes levied on certain imported goods for the
benefit of the general government and the protection of certain kinds of home
industry.
NOTE.-In some countries, customs or taxes are charged both on imported and exported
goods. The government of the United States charges customs only on goods imported.
There are two kinds of duties, specific and ad valorem.
949. Specific Duty is a tax levied on the weight or measure per ton, bale,
pound, gallon, or yard, without regard to its cost or value.
950. Ad Valorem Duty is a certain per cent levied or assessed on the
actual cost of the goods in the country from which they were imported, as shown
by the invoice or the appraised value in the absence of an invoice.
NOTE.—On certain kinds of goods both a specific and an ad valorem duty is assessed.
951. An Invoice or Manifest is an itemized account of the goods imported
or shipped to a purchaser, agent, factor, etc., with the actual cost or value. Regard-
ing foreign invoices, the law requires:
sº
1. That all invoices of imported merchandise shall be made out in the cur-
rency of the place or country from whence the importations shall be made, or if pur-
chased in the currency actually paid therefor, shall contain a correct description
of such merchandise, etc.
2. That all such invoices, shall at or before the shipment of the merchandise
be produced to the consul, vice-consul, or commercial agent of the United States of
the consular district in which the merchandise was manufactured or purchased as
the case may be, for export to the United States, and shall have indorsed thereon,
when so produced, a declaration signed by the purchaser, manufacturer, owner, or
agent, setting forth that the invoice is in all respects correct and true, and was
made at the place from which the merchandise is to be exported to the United
States; that it contains, if the merchandise is to be obtained by purchase, a true
and full statement of the time when, the place where, the person from whom the
same was purchased, and the actual cost thereof, and of all charges thereon, as
provided by this act; and that no discounts, bounties, or drawbacks are contained
in the invoice, but such as have been actually allowed thereon, etc.
(500)
º: CUSTOMHOUSE BUSINESS. 5ot
3. That no different invoice of the merchandise mentioned in the invoice so
produced has been or will be furnished to any one.
The most of invoices are made out in the weights and measures of the
country from which the goods are imported.
952. Tariff is a list of goods alphabetically arranged, with the rates of
duties and drawbacks, established by the laws of the United States.
953. The Free List comprises such goods as are imported free of duty.
954. Tonnage Duty is a tax levied per ton on vessels for the privilege
Of entering ports.
955. Ports of Entry are places where customhouses are established, and
it is lawful to receive goods into a country only at ports of entry.
NOTE: 1.—There are several kinds of entries, such as, entry of a vessel from a foreign port
With passengers, entry of merchandise for immediate use, bonded warehouse entry, re-warehouse
entry, withdrawal entry for consumption, or exportation, etc., etc.
NOTE_2.—The forms for the various entries are generally furnished by the customhouse or
the custom-house broker.
956. Bonded Warehouses are buildings in which imported goods on which
duties have not been paid are placed under bond to the general government by the
importer, until he may wish to put them on the market or export them.
f d NOTE 1.-Goods may be withdrawn from bonded warehouses for export, without the payment
of duties.
NOTE: 2.—Goods remaining in a bonded warehouse more than one year, are liable to an addi-
i. duty. And goods remaining over three years are considered as abandoned to the govern-
Iſle Ilú.
957. Smuggling is the act of bringing foreign goods into this country
without paying duties on them. This is done 1, By not entering them at the
Customhouse, and 2, By showing less than their real value in the invoice. It is a
crime for the prevention and punishment of which stringent laws have been enacted
and a strict supervision is exercised by the general government.
958. A Clearance is a certificate given by the collector of a port after the
requirements of law have been complied with, that the vessel has been properly
entered.
959. Duties are collected at the port of entry by a Customhouse officer
appointed by the United States government, under the title of collector of the port.
NOTE.—Duties are not computed on the fractions of a dollar; when less than 50c. they are
rejected; if 50c. or more, they are counted as a dollar.
The collector of the port appoints deputy collectors, appraisers, weighers,
gaugers, inspectors, etc., and supervises all entries and papers, estimates, duties, and
receives all moneys and securities.
960. A Naval Officer is appointed at the more important posts, whose duty
it is to receive copies of all manifests, to countersign all documents issued by the
collector, to certify his estimates and accounts, etc., etc.
961. The Surveyor superintends the employees of the collector, inspects
vessels and cargoes, and revises all entries and permits. He is personally respon-
sible to the collector of the port.
5O2 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
962. The Appraiser is an officer whose duty it is to examine all imported
goods and to determine their dutiable value so that ad valorem duty may be charged.
963. The Store-keeper is an officer who has charge of the warehouse.
964. A Customhouse Broker is a person who is thoroughly acquainted
with Customhouse laws, and acting under power of attorney, makes entries, secures
permits and transacts business for importers, for which service he makes a charge.
965. Internal Revenue is the revenue derived from the sale of public
lands and postage stamps, and from taxes on certain home manufactured articles,
such as distilled and malt liquors.
The laws prescribing the Internal Revenue are called ExCISE LAWs, in
contradistinction to the Tariff Laws, which prescribe Import Duties.
966. Excise Duties are taxes or licenses for the manufacture or sale of
certain articles consumed at home or in their transfer from one person to another.
967. A Drawback is money refunded for import duties which has been
previously paid. When goods on which duty has been paid are exported, the
amount of duty thus paid is refunded.
968. Allowances are deductions made in accordance with law before
estimating the duties.
969. Gross Weight is the whole weight of the goods, including the weight
of the hogshead, barrel, box, bag, etc., that contains them.
970. Net Weight is what remains after all allowances have been deducted.
971. Leakage and Breakage. By Act of Congress of 1894, all allowance
for leakage and breakage was repealed.
972. Draft is the allowance made for waste in weighing goods.
973. Tare is the allowance made for the weight of the box, crate, barrel, or
bags containing the goods.
The long ton of 2240 pounds and 112 lbs. to the cwt. is invariably employed
in the Customhouses of the United States.
The United States Revenue Laws have been often changed during the past
few years and are still agitating Congress and the people of the country more than
almost any other question of government. Tariff or Free Trade, High or Low
Tariff, are questions demanding the serious attention of our best law-makers,
statesmen, and financiers.
PROBLEMI.
974. 1. What is the specific duty on 150 casks of alchohol, each cask 60
gallons capacity, but which when gauged, are found to be 240 gallons short of full
capacity ? What is the duty at 15% per gallon ? Ans. $1314.
OPERATION.
150 × 60 = 9000 gallons, gross quantity. Explanation.—In , , all cases
* - * - in which the specific duty is
9000 gals. – 240 gals. Short measure = required, find the net quantity
8760 gals. net quantity. and compute the duty there-
8760 gals. × 152 duty = $1314 specific duty. * comp y
NotE 1.--To find the net quantity of liquids in casks, the law requires that they shall be
gauged by a Customhouse gauger.
NoTE 2.—To find the net quantity of liquids in bottles, the bottles must be counted and
-certified by a Customhouse appraiser.
* CUSTOMHOUSE BUSINESS.
503
TO FIND AD VALOREM DUTY.
PROBLEM.
975. 1. What is the ad valorem duty on an invoice of 12 dozen Geneva
watches, amounting to 21800 francs, if the duty is 25 per cent?
Ans. $1051.75.
OPERATION.
21800 × 19.3 g = $4207.40 Ea:planation.—In all cases in which ad valorem duty is
$4207 × 25% = $1051.75 required, find the net value and multiply the same by
the rate of duty,
NotE 1.-The 40c., in the net value, being less than 50c., are not used in the Customhouse,
when computing duty.
NOTE 2.--The intrinsic and Customhouse value of a franc is 19.3 cents. See Talble of Foreign
Money.
TO FIND THE RATE OF DUTY WEIEN TEIE NET VALUE AND AMOUNT
OF DUTY ARE GIVEN.
PROBLEMIS.
976. 1. The net value of an invoice is $1260 and the duty is $441. What
was the rate 3
OPERATIONS INDICATED.
100% rate.
1260 || 441 Or, 1260
Ans. 35%.
2. A merchant imported from London, England, a case of woolen goods.
The cost of the goods was £310 5s. 6d; commission 24 per cent. Freight £3 4s.
Consul fees 16s. 8d. The weight of the case was 640 pounds gross; allowing 10 per
cent for tare, what is the amount of duty, the specific being 50% per pound and the
ad valorem 35 per cent 3
OPERATION.
# S. d.
310 5 6 cost of goods.
7 15 2 = 24% commission on cost of goods.
3.18 0 8 total dutiable cost.
fº318,0333 x $4.8665 = $1547.71 = dutiable cost in dollars.
$1548 x 35% = $541.80 ad valorem duty.
640 lbs. – 64 lbs. tare = 576 lbs. weight of goods.
576 x 50g = $288 specific duty.
$541.80 + $288 = $829.80 total duty.
Ans. $829.80.
Explanation.—Commis-
sion is by law a part of
the dutiable cost, but
freight and Consul fees
are not. The intrinsic
and the Customhouse
value of the £ is $4.8665.
See Table of Foreign
Money
The 71 cents in the
$1547.71 being more than
50, $1 is added to the
dollars, making $1548
dutiable cost. To find
the commission on £. s. d. and to reduce £. s. and d. to the decimal of a £, see page 484.
NotE.—The Tariff Law of 1894, reduces the duty on woolen goods to 40 per cent on goods
valued at not over 50c. per pound, and to 50 per cent on goods valued at more than 50c. per pound.
504 soulE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
3. Imported from Antwerp 3600 meters of Brussels carpet, 27 inches wide at
4.50 francs per meter. The specific duty is 442 per sq. yd, and the ad valorem 35
an ºrº º P-y
per cent. Allowing 39.37 in. to a meter, what was the duty ?
Ans. $2393.66.
PARTIAL OPERATION.
3600m x 4.50 francs = 16200 francs.
16200 fr. x 19.3 = $3126.60 dutiable value.
$3127 × 35% = $1094.45 ad valorem duty.
3600m = 3937 yds. 3937 yds. 27 inches wide = 2952; sq. yds.
2952; sq. yds. × 442 = $1299.21 specific duty. 42.
$1094.45 + 1299.21 = $2393.66 total duty.
NOTE.—The specific duty on Brussels carpets was repealed by the Tariff Bill of 1894, and the
ad valorem duty increased to 40 per cent.
4. Imported from Geneva, Italy, 20 blocks of marble each block measuring
8x4x2 ft. The duty is 50g per cubic foot. The total invoice is 33700 lire. What
is the duty ? Ans. $640.
NOTE. –A lire is the same in value as a franc, 19.3 cents.
5. Imported from Hamburg, Germany, an invoice of toys amounting in
dutiable value to 12200 marks. What is the duty at 45 per cent on 6000 marks and
30 per cent on 6200 marks? Ans. $1085.28.
NotE.—The intrinsic and customhouse value of a mark is 23.8 cents. See Table of Foreign
Money.
6. The dutiable value of an invoice of leather from Vienna, Austria, is 5850
crowns. What is the duty at 20 per cent ad valorem Ans. $237.51.
NOTE.-The value of a crown is 20.3 cents.
7. What is the duty on 449650 pounds of R. R. iron at #g per pound?
Ans. $1573.77%.
8. Imported from Havana 60m cigars, weighing 900 lbs., total Invoice in
American money is $1500. Under the Revenue Law of 1894, the specific duty is $4
per lb. gross, and the ad valorem duty is 25 per cent. What is the duty, what was
the cost of the cigars per m, in Havana, and what was the importing cost, not
counting freight and insurance?
Ans. $3975.00 duty. $25 per m, Havana cost. $91.25 per m, importing cost.
CUSTOMHOUSE BUSINESS. 505
977. AN INVOICE OF COFFEE IMPORTED FROM RIO JANEIRO, BRAZIL.
NOTE —This involce is made in the monetary unit and in the unit of weight of the United
States, which has been the custom sunce March, 1890.
Invoice of 1997 bags of coffee shipped by HERNANDEZ BROTHERs & Co., on
board the British Str. Hassell, for New Orleans.
For account and risk of whom 1t may concern and consigned to order.

L 'if; 200 bags coffee wg. 26,200 lbs. (a) 1741c. p. lb., cost and freight, - - $ 4561|42
N 10/21 |295 4 & & b 38,645 ** 1686C, & $ 4 & 6 & & 8 tº tº 6515|55
O 21/27 | 153 { { & 6 20,043 ‘‘ 1631C. $ $ 4 ( & & { * * *s 3269|01
P 28/43 || 702 * { 4 4 91,962 “ 1594C, & 6 & 6 & 5 & 4 as sº 14658|74
R 44/49 || 192 * * 4 & 25,152 “ 1558C, 4 & 6 & 4 & { { tº gº 39.1868
S 50/54 214 4. & 4 28,034 ** 1533c. * { { { { { 4 4 tº º 4297|61
T 55/60 |241 & 6 & 4 31,571 ** 1496c. ( ; ; 4 t 4 4 & tº º 4723|02
Less freight to New Orleans, on 1997 bags (a 60c. and 5 per cent per bag. *::::: º
4,8388.170 Exchange, 4.85 - - - - - - - - slºoessly,
By our draft (a) 90 d/st on London, £8388 17.0 *R* ==me ammºm,
E, & O. E
RIO DE JANEIRO, 28 June, 1894.
HERNANDEZ BROTHERS & Co.
Verify all the computations in the above bill. See Index, English Exchange,
for the method of reducing the $40,685.92 to the English money of account.
For the entries resulting from the importation of goods from Foreign
countries, and from the remittance of exchange in payment therefor, See Soulé's
new Science and Practice of Accounts.
g
506
Yºr
SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS.
978. INVOICE OF COFFEE IMPORTED FROM RIO JANEIRO, BRAZIL.
NOTE.-This Invoice is made in the monetary unit, and in the unit of weights, of Brazil,
which were used in Brazil, up to March, 1890. *
Invoice of 4000 bags Coffee shipped by C. SINNOTT & CO.,
on board the British Steamer “Plato, "
for NEw ORLEANs,
Messrs. J. M. BUTCHEE & CO.,
*
for account and risk of
and consigned to
the order of Messrs. R. SOULE & CO.


B
F 400
B
G 600
B
EI 483
B
J 450
B
FC 522
B
IN 545
B
G 500
B
IP 500
4000 bags of 60 kilos = 16337.64 arrs. © 3 $840 p. arroba. e gº º º tº
Charges:
Duty on Customhouse valuation of 290 reis p. kilo - - - - - - -
on 240,000 kilos = Rs. 69: 600,000 a 11 per cent. - & * . gº s º &
Difference of 79 rs. per Kilo = Rs. 18,960,000 & 4 per cent tº dº tº &
Capatazias 60 rs. p. bag s sº tº dº sº º º ſº tº ſº tº *
Brokerage 50 rs. “ • * * * * * * * * * * * *
4000 bags a 700 rs p. bag - - - - - - - - - - - -
Mending bags, &e. - - - - - - - - - - - - -
Fire Insurance as gº tº gº e º º ºs e s = * * *
Consular Certificate sº ºn m = ± is ºn as as as ºn s =
Negro hire º m & tº e º ºs º º is * as tº º &
Commission for purchasing a 24 per cent.
Shipped against cable order dated 2d Feb., 1890, from New York.
= E. dº O. E.
Rio DE JANEIRO, 27 February, 1890.
O. SINNNOTT dº CO.
NotE 1.-A Kilo is the French unit of weight and equals 2.2046 Avoirdupois pounds.
NOTE: 2. —Tho Arroba is the Brazilian unit of weight and is equal to 32.38 Avoirdu
NOTE 3.- 14.69 Kilos are equal to ONE Arroba. Hence to reduce Kilos to Arrobas
*
|
7656.000
º
wº
wº
2800 º
000
840|710
62,736 w
12,558,000
Rs.
is pounds.
ivide the Kilos b
75,294|540
1,882|360
77,176,000
14.69.
Nº. 4.—See Index, Brazilian Exchange, for the Brazilian Money of Account and the manner of reducing Rs. to
3. S. and d.
979. STATEMENT OF FULL COST OF INVOICE OF COFFEE PER “PLATO.”
J. M. BUTCHEE & CO.,
In account with C. SINNOTT & CO.
1890
Feby.
ſ
!
|
27 || To Invoice Cost of 4000 bags of Coffee, per “PLATO,” to New Orleans,
‘‘ Propo. of cost of Cablegrams - * * g gº ºn * º s
‘‘ Stamps on drafts &
“Dill Brokerage. 3 16 per cent on Rs 77 : 426 $ 100 sº me &
º gº * e-
Rs.
By our draft a 90 d/st, on Sunnott & Soule, London. exchange 21;d. = £6976.8.3
Rs.
Cost on board, including freight of 40c. and 5 per cent per bag = 6.78c. per it.
E. dº O. E.
Rio DE JANEIRO, Feby. 27, 1890.
O. SINNOTT dº CO.
77 : 176|900
18||000
86|030
170
100
Rs.
77 : 426
26|100
smº-mº
77 : 426
Verify the calculations in the Invoice and in the Account Current and Interest Account
which follows.
INOTE: 1.- It is the custom to charge ºft on the amount invested plus the brokerage.
INoTE 2.-Sometimes the statement of the fi
cost is made at the end of the Invoice.
f
NotE.-This Account Current and Interest Account, is presented in advance of interest in order to show the whole work of import-
ing and the payment of the invoice in a connected form. See Index, Interest on English Money and Accounts Current and Interest
Account on English Money.
980. ACCOUNT CURRENT AND INTEREST ACCOUNT.
J. M. Butchee dº Co.'s Coffee per “Plato” to New Orleans, in Acct. Current and Interest Acct, a 5% with Sinnott & Soulé.
Dr. sº Cr.
1890 1890
Mar. 5|To Draft at 90 dB. - . 4.6976|| 8 3|June. 6| 3 || 2 |17| 4 Apr. ; 2. By Remittance (a 60 ds. ||3:2000 0|0|| June || 4 5||1| 7| 5
Apr. 16. “ ** & 4 4|| “. do. & & & & 2000, O O 6|| 3 || |16| 5
f"350 (a) 19s. 96 - 64|| 2 || 5 || Moh. 2774|| |13| 0 * 7 do. 4 & 6 & 3000 0|| 0 9| 0
“Bank Com. on £7040 : ** 13 ** Balance of acct.
(a) #96. - - - - 8|16| 0 Coffee per “Cheviot” 66314|10|| May 29|11|| |19|10
“Stamps - - - - 3||10| 0 Balance of Interest 6|| 8
“Interest, (balance) 6|| 8
** Postages - - - - 12| 6
** Balance Cash 9 June 609||19|| 0
1884 10 3|10| 4 7663/14 10 3|10| 4
- || – |-- - || – || - E. E. F. Flirt
LONDON, April 16, 1890.
SINNOTT (; SOULE).
REMARKS ON THE DEBIT SIDE.
1. The above is an Account Current and Interest Account by the interest method worked to the latest date in the account, which
is June 9th. See the subject of Accounts Current.
The first money column on the Debit side contains the items that J. M. Butchee & Co. owe Sinnott & Soulé.
3. The next column shows when they are due. . 4. The next column shows the number of intervening days between the date that
the items are due, and the date to which the account is made up. (June 9).
The last column shows the interest on each item for the time it was due, counting 365 days in a year.
6. The Marine Insurance is effected on £7350 which is an estimated value with a certain per cent added.
7. The 19 shillings per cent, insurance on the value is subject to a varying per cent. discount. The rate of insurance also varies.
8. The Bank commission of ¥ per cent is computed on the first cost £6976–8–3 plus the insurance £64–2–5.
9. The items of debit below the item of insurance were not due till the day the account was made up; and hence no interest is
computed thereon. &
10. The 19 shillings per cent means 19s. on every £100.
REMARKS ON THE CREDIT SIDE.
1. On this side the first money column shows the payments.
2. The next column shows the dates when the payments were due.
3. The next column shows the number of days intervening between the date that the items of payment are due, and the date to
which the account is made up. (June 9).
deb ta The last column shows the interest on each item of payment for the time it was due, and also the balance of interest, from the
ebit side.
NOTE.—This Acceount Current and Interest Account, and the two preceding Invoices are exact copies except, the names of the
original instruments in an importing house in New Orleans.
3.

ITSLI1"a ITC E.
=ss-
981. Insurance is a contract by which one party, in consideration of a
eertain sum called a premium, and which is proportioned to the risk involved, under
takes to compensate the other party for loss on a specified thing, from specified
causes, within a specified time.
982. Insurance is divided into two general classes: 1. Property Insurance.
2. Personal Insurance.
983. Property Insurance is security against loss by fire or damage by
transportation, or the injury or death of cattle, etc., and includes Fire Insurance,
River Insurance, Marine Insurance, Transit Insurance and Live Stock Insurance.
984. Fire Insurance is the insurance against loss or damage by fire of all
kinds of property on land, such as houses, merchandise, factories, etc.
985. River Insurance is the insurance of river steamers, barges, and
their cargoes.
986. Marine Insurance is the insurance of ocean vessels and their cargoes.
NOTE.—In river or marine insurance, when the steamer or vessel only is insured, nu is called
Bull Insurance; and when the cargo only is insured, it is called Cargo Insurance.
987. Transit Insurance is the insurance of goods or property during
transportation by land, or by both land and water.
988. Live Stock Insurance is the insurance of live stock, such as horses,
mules, etc., against accident or death within a specified time.
989. Personal Insurance includes Life Insurance, Health Insurance and
Accident Insurance.
990. Life Insurance is the insurance on the lives of persons whereby a
certain sum of money is to be paid to them when they attain a certain age, or at
their death, the amount is to be paid to their heirs, or to some one named in the
policy.
991. Health Insurance is the insurance by which the insured parties
receive a weekly allowance in case of sickness.
992. Accident Insurance is the insurance of persons against death or
certain injuries received under certain circumstances while at home, or when
traveling.
NOTE.-The following conditions are embraced in the policies of most Accident Insurance
Companies, giving such companies an unjust advantage over their patrons.
“The insurance under this Certificate shall not extend to or cover disappearances, or injuries
whether fatal, or disabling, of which there is no visible mark on the body of the insured; nor
(508)
{ INSURANCE. 5og
extend to, or cover ACCIDENTAL INJURIES or death, resulting from, or caused directly or indirectly,
wholly or in part, by hernia, fits, vertigo, somnambulism or disease in any form, gas or poison in
any form or manner, contact with Poisonous substancEs, surgical operations or medical treat-
ment, dueling, fighting or wrestling, war or riot, lifting or over exertion, suicide, felonious or
otherwise, sane or insane, sunstroke or freezing, riding or driving races, voluntary exposure to
unnecessary danger, nor extend to or cover intentional injuries inflicted by the insured or any other
person, or where the accident, injury or death happens while the insured is under the influence of
intoxicating drinks or narcotics, or in consequence thereof; or while or in consequence of, violating
the law or the rules of any company or corporation; or while employed in mining, blasting or
wrecking, or in the manufacture, sale or transportation of gun-powder, or any other explosive
compound, (unless insured to cover such occupation). Standing or riding upon the platforms of any
conveyance using steam or electricity as a motive power, or entering or attempting to leave such
conveyance while the same is in motion, or walking on the road-bed or bridge of any railway, are
hazards not counted or covered by this insurance, and no sum will be paid for injuries or death in
consequence of such exposure, or while thus exposed, happening to any person other than railroad
employes, who shall have been accepted under such occupation.”
993. There are also other forms or kinds of insurance under different names
according to the nature of the subject of insurance, or the objects or interests
insured: as Guarantee or Fidelity Insurance, Co-operative Insurance, or Benevolent
Insurance, Boiler Insurance, and Plate Glass Insurance.
994. Guarantee or Fidelity Insurance is the insurance against loss that
may occur through fraud or unworthy agents or employés who occupy positions of
trust. Guarantee or Fidelity Companies issue bonds of suretyship for a specified
amount to employers, guaranteeing the faithful conduct and the honesty of the
party employed for a given time, for which they charge the employé a given per
cent on the face of the bond.
Employés must possess good characters and good business capacity before
they can be thus guaranteed. -
Some guarantee companies also guarantee titles to lands.
995. Co-operative Insurance or Benevolent Insurance is that in which the
members of some Company, Association or Society, contribute weekly or monthly,
or at stated times, to a fund from which a specified sum is paid to the heirs of
deceased members, or to some person named in the Benefit Certificate or Policy of
the deceased member. The sum contributed depends upon the age of the person at
the time he becomes a member, and the amount of his benefit certificate or policy.
Of this class the Knights of Honor, the Legion of Honor, the Catholic Knights of
America and the Royal Arcanum, are leading examples.
996. The Insurer, generally a stock or mutual company doing business
under the title of an Insurance Company, is the party that contracts to indemnify
the other. The Insurer, after having signed the policy, is called the underwriter, so
called from the fact of having signed under the conditions of the policy.
997. The Insured or ASSured is the party that is protected by the under-
taking or contract of the insurer.
998. The Policy is the written or printed instrument containing the numer-
ous conditions of the contract between the insurer and the insured. In marine and
river insurance, there are two kinds of policies, valued and open.
999. A Walued Policy is one in which a value has been set upon the
property or interest insured, and inserted in the policy. A valued policy is often
called a special policy.
5 Io SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. Yºr
On country risks, most Insurance Companies limit the amount of insurance
to three-fourths or two-thirds of the appraised value of the property insured.
In many policies, especially on buildings, the Insurance Companies reserve
the right to repair, rebuild or replace the property destroyed or damaged with
other of like kind and quality, within a reasonable time.
1000. An Open Policy is one in which the amount of interest or property
is not fixed by the policy, but is left to be ascertained in case of loss or damage.
In OPEN POLICIES for River and Marine Insurance, additional insurance
may be entered, from time to time, as goods are shipped, either on the policy or in a
pass book, which is regarded as having the force of the policy.
Premiums on OPEN POLICIES are paid at the end of a voyage, or monthly, or
quarterly, or otherwise, as may be agreed upon.
1000%. Open Endorsement Policy is a policy issued to the assured in
blank, to cover on property held by the assured, as described in the several entries
to book attached, while in such places and insured for such periods of time, and
at such rates of premium, and for such amounts as shall be written. Hence, no
liability on the part of the Company exists, until application is accepted. These
entries are hence called endorsements.
1001. Perpetual Policies are sometimes issued, the rate being usually equal
to that of TEN annual premiums.
In PERPETUAL POLICIES, the premium is considered merely a deposit with
the Insurance Company; for at any time, at the instance of either party, the policy
may be cancelled, and 90 per cent of the premium or deposit must be returned to
the policy holder.
Policies are not transferable without the consent of the insurers.
1002. The Subject of the Insurance is the property itself, and the title
or interest that the assured has, is called his Insurable Interest.
1003. The premium is the price paid for the insurance.
1004. The Rate of Premium is either a certain per cent on the amount
insured, as # per cent or 13 per cent, or it is a specified number of cents on one
hundred dollars, as a rate of 65%, or $1.25, which means respectively, 652 and $1.25
on each $100 insured.
1005. The Rate of Premium depends in a great measure upon the time
for which the policy is issued and the degree of risk, which is so varied, that the
calculation of the rate becomes one of probabilities.
1006. Short Rates. Rates for less than a year are called SHORT RATEs,
and are proportionally higher than yearly rates.
1007. Return Premium. When a policy is taken for a year and is can-
celled at the request of the insured prior to the close of the year, a Return Premium
is paid to the party insured. In such cases, the company has the right to charge
Short Rates for the time the policy has been in force.
1008. A Risk is the probability that the subject of the insurance may be
lost or damaged, or it is the danger to which the thing insured is exposed.
Risks are denominated “hazardous,” “extra hazardous,” and “Specially
hazardous,” according to the probability of loss, which depends very much upon
the combustibility or perishable nature of the subject of insurance and its sur-
roundings. For example, a hardwarc store would be less liable to burn than a dry
goods store, and a store in which inflammable substances or explosive materials
* INSURANCE, 5 II
were kept would be more liable than either; and a hotel is more liable than a
private residence.
1009. Reinsurance is a contract which the first insurer enters into, in
order to relieve himself from some of the risks that he has undertaken, by dividing
them with other insurers.
1010. Cancellation of Policies. A policy is cancelled when the agreement
between the party insured and the insurers is annulled.
1011. Transferring Policies. A policy is transferred when the insured,
With the consent and acceptance of the insurers, conveys or transfers his rights
under the terms and conditions of the policy to some third person.
1012. An Insurance Agent is a person who is duly authorized to act for
an insurance company in soliciting business, contracting risks, collecting pre-
miums, and adjusting losses.
1013. An Insurance Broker is a person who effects insurance and negoti-
ates policies, for which service he is paid a commission or brokerage by the com-
pany taking the risk.
Brokers are regarded as agents of the parties insured, and not of the insur-
ance company.
1014. Insurance Companies are of several kinds according to the principle
or form of their organization: 1. Stock Companies. 2. Mutual Companies. 3.
Mixed or Stock and Mutual Companies.
1015. A Stock Insurance Company is one in which the Capital is divided
into shares, which are subscribed, paid for and owned by persons or firms called
stockholders. The stockholders share all the profits and are liable for the losses,
to the extent of the value of the shares owned.
1016. A Mutual Insurance Company is one in which the profits and losses
are shared by all who are insured, who are the policy holders.
In this form of company, there are no stockholders or capital stock. The
capital consists of unused premiums of the reserved earnings and of the invest-
ments of the company.
1017. A Mixed or Stock and Mutual Insurance Company is one which
combines the principles of both the Stock and Mutual Companies, i. e. they have a
Capital stock and shareholders, and divide a portion of the net profits among share-
holders and policy holders. Generally in mixed companies a limited dividend of 3
to 10 per cent is first paid to the stockholders, a portion of the profits is carried to
8wrplus or reserve fund, and the surplus profits, if any, are divided among the
participating policy holders.
1018. Non-participating policies are sometimes issued at reduced rates, by
mutual and by mixed companies, and such policies do not share in the losses or
gains of the company.
In Some mixed companies, the capital is unlimited, provided it is not less
than a certain sum; and in the organization of the company, it is represented by
deposit notes; and subsequently by the installments collected, premiums paid on
5I2 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
insurances, and by the sums earned in the business. Iu these companies, the
insured becomes, by the mere fact of insurance, a member of the company, and is
obliged, therefore, to observe its rules and recognize the authority of its agents.
1019. Insurance Scrip consists of the certificates given by mutual and
mixed companies to the insured, in lieu of cash dividends. They are equivalent to
receipts or certificates for money loaned. They generally bear interest, are nego-
tiable by delivery and transfer on the books of the company, and are payable at
Such time as the company may subsequently specify.
1020. The Surplus of an INSURANCE COMPANY is the excess of the assets
OVer the liabilities. Capital stock and unearned premiums are considered a part
of the liabilities when estimating the surplus or when determining gains and losses.
1021. Misrepresentation. Any material misrepresentation prejudicial to
the insurers, whether intentional or not, and whether made by the principal or his
agent, Vitiates the policy. ,
1022. Bottomry and Respondentia. Bottomry is a contract by which the
Ship owner, or the master on his behalf, pledges the keel, tackle, and apparel of his
ship as security for money which he borrows for the use of the ship, in contempla-
tion of a particular voyage or for a particular and fixed period of time. It is a
Condition of a bottomry contract that if the ship should be lost during the Voyage,
Or by any of the perils enumerated in the contract, the lender shall lose his money;
but if the ship arrives in safety, then he shall receive his principal and also the
interest agreed upon, however much it may exceed the usual or legal interest for the
use of money. When the money is loaned on other property exposed to maritime
risks, the contract is called Respondentia.
º 1023. Salvage is a compensation allowed parties for their voluntary services
in Saving a ship, steamboat, cargo, or other property from the perils of fire or water.
ADJUSTMENT OF LOSSES.
1024. In Marine Insurance the policies contain an “Average Clause" by
Which, “if the vessel is valued in the statement thereof, beyond the amount
insured, the insurers are to pay such part only of the sum insured (or of the partial
loss) as the amount insured bears to such valuation.” And in case of the cargo,
if that “is valued beyond the amount insured, the insurers are to pay such part
only of the sum insured (or of the partial loss) as the amount insured bears to the
full valuation.” The effect of this principle is that the insured is himself the insurer
for that part or portion of the property insured.
For example: If a vessel valued at $40000 were insured for # value, or
$30000, and were totally destroyed, the company or the insurers would pay the
$30000; but if the vessel were damaged to the amount of $10000, the company
Would pay only three-fourths of the partial loss, or $7500. Again, if a cargo valued
at $40000 were insured for $10000, and sustained a loss of $10000 or more, the
insurers would pay only $2500.
1925. In Fire Insurance, under the ordinary Fire Insurance Policy, which
does not contain the average or the 75 per cent clause, the company will pay the
full amount of the loss or damage, provided it does not exceed the sum covered by
insurance. Thus, if a house or stock of goods were insured for $10000 and a
partial loss of say $7000 be sustained, the company would pay the full loss or
$7000. But if the loss should reach $11000 or more, the company would pay only
$10000, the amount specified in the policy.
Sometimes FIRE INSURANCE POLICIEs contain the “Average Clause,” and
then the company issuing such policies would pay only such portion of the loss as the
amount insured bears to the total value of the property.
NotE.—The usual form of the “Average Clause,” to be used when policy covers any mer-
chandise in two or more locations, is as follows: *
“It is a part of the consideration of this policy and the basis upon which the rate of premium
#: INSURANCE. 5I3
is fixed, that the assured shall maintain insurance on each item of property insured by this policy
9qual to the actual cash value thereof, and that failing so to do, the assured shall be an insurer
to the extent of such deficit, and in that event shall bear his, her or their proportion of any loss.”
1026. Adjustments of Losses, under policies containing the “Average Clause,” would be
made as follows: Suppose property valued at $20000 was stored in two warehouses, and insured
for $15000, and that the goods in one warehouse were destroyed by fire making a partial loss of
say $8000, the company would pay $6000, i.e. such proportion of the loss as the amount insured
bears to the full value of the property.
1027. To Adjust the loss when several different insurance companies have risks on the
Same property, the loss is divided among all the companies in proportion to the amount of their
respective risks, whether all the companies be solvent or not. Thus, if goods valued at $25000 be
insured in Co. A, for $3000, in Co. B, for $5000, and in Co. C, for $10000, making $18000, and a loss
of $9000 should be sustained, Co.'s A, B, and C, would pay respectively, º, ø, and +3 of the loss.
1028. The 75 per cent Co-insurance Clause, which is now generally inserted in fire
policies, reads as follows: “It is a part of the consideration of this policy and the basis uponl
Which the rate of premium is fixed, that the assured shall at all times maintain a total insurance
on the property, insured by this policy, of not less than 75 per cent of the total cash value thereof
(as covered under the several items of this policy,) and that, failing so to do, the assured shal
become co-insurers to the extent of the deficiency, and in that event, shall bear their proportion
of any loss occurring under this policy.” *
1029. The Lightning Clause reads as follows: “This policy shall cover any direct loss or
damage caused by lightning, (meaning thereby the commonly accepted use of the term Lightning;
and in no case to include loss or damage by cyclone, tornado, or windstorm,) not exceeding the
sum insured, nor the interest of the insured in the property, and subject in all other respects to the
terms and conditions of this policy. Provided, however, if there shall be any other insurance on"
said property, this company shall be liable only pro rata with such other insurance, for any direct
loss by Lightning, whether such other insurance be against direct loss by lightning or not.”
1029a. Iron Safe Clause.—1st. The assured will take a complete itemized inventory of
stock on hand at least once in each calendar year, and unless such inventory has been taken within
twelve calendar months prior to the date of this policy, one shall be taken in detail within thirty
days of issuance of this policy, or this policy shall be null and void from such date.
2nd. The assured will keep a set of books, which shall clearly and plainly present a com-
plete record of business transacted, including all purchases, sales and shipments, both for cash
and credit, from date of inventory as provided for in first section of this clause and also from date
of last preceding inventory, if such has been taken, and during the continuance of this policy.
3rd. The assured will keep such books and inventory, and also the last preceding inventory, if
such has been taken, Securely locked in a fire-proof safe at night, and at all times when the build-
ing mentioned in this policy is not actually open for business; or failing in this, the assured will
keep such books and inventories in some secure place not exposed to a fire which would destroy the
aforesaid building; and unless such books and inventories are produced and delivered to this
company for examination after loss or damage by fire to the personal property insured hereunder,
this policy shall be null and void and no suit or action shall be maintained hereon. It is further
agreed that the receipt of such books and inventories and the examination of the same shall not
be an admission of any liability under the policy, nor a waiver of any defense to same.
1029b. The Foundation Clause. It is understood that the foundations and piling on
which the within described building rests, are not included in this Insurance, and in no event
shall they be taken as part of the valuation of the building for the purpose of arriving at the
assured’s contributory proportion in the application of the Co-Insurance Clause.
The foundations being exempted from the operation of the co-insurance clause of this policy
and not being covered by this insurance, it is hereby understood and agreed that in the event of loss,
if the insuring company should elect to rebuild, and it be found necessary on account of statutory
or other causes to replace the foundations in whole or in part, the cost of replacing or repairing
same will be at the expense of the assured.
5 I4. SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. º
1030. The Three-quarter Loss Clause or Country Clause inserted in the
country policies issued by many Insurance Companies, reads as follows:
“It is agreed and understood to be a condition of this insurance, that in case of any loss or
damage under this policy, this company shall not be liable for an amount greater than THREE-
FourTHS of said loss, not exceeding the sum herein insured, the other ONE-FourTH to be borne by
the assured, and in the event of other insurance on property covered under this policy, this
company shall not be liable for more than its proportion of three-fourths of such loss or damage.”
1030a. The Three-fourths Walue Clause is as follows:
“It is understood and agreed to be a condition of this insurance, that in the event of loss or
damage by fire to the property insured under this policy, this company shall not be liable for an
amount greater than three-fourths of the actual cash value of each item of property insured by
this policy (not exceeding the amount insured on each such item) at the time immediately preceding
such loss or damage; and in the event of additional insurance—if any is permitted hereon—then
this company shall be liable for its proportion only of three-fourths such cash value of each item
insured at the time of the fire, not exceeding the amount insured on each such item.”
1031. A Floating Policy sometimes termed a “ BLANKET POLICY,” covers
goods in a number of different warehouses, docks, etc., with the “Average Clause”
added.
IMPORTANCE OR VALUE OF INSURANCE.
1032. Insurance, considered in all of its various relations and different
applications, is one of the most important and valuable institutions of the age,
without which trade and commerce would languish, and the poor and medium
classes of society would be deprived of many blessings and opportunities of success,
which through it they now enjoy.
Considering the hazards which threaten property and the uncertainty of
human life, prudence, wisdom, regard for financial obligations, and affections for
family, demand that every man should insure both his property and his life.
For a business man not to insure his property is evidence of his incapacity,
and unless his goods or other property is fully paid for, it is proof of his Want of
fidelity to his creditors. The only exception to this is where the party has sufficient
capital, outside of his business, to enable him to continue business, pay his financial
obligations and to support those dependent upon him, in case of loss.
In regard to Life Insurance Companies, it must be said that while they are
doing a grandly humane work and merit the favorable consideration and patronage
of all men, some of the companies are characterized for extravagance beyond
prudence or reason; and some have violated the principles of ethics in their prom-
ises and statements of facilities, advantages and resources.
These facts make it necessary for the applicant for Life Insurance to investi-
gate and exercise judgment and discretion in selecting a company.
The various Social, Co-operative or Benevolent Insurance Associations with
their low rate of assessments, and their economical system of management, offer to
the poor man facilities for insuring, which as he loves his mother, his wife, his
children, or his fellow-man, he should accept.
These Co-operative Insurance Associations are the children of this progressive
age; they are supplementing the work of the regular Life Companies, and are
proving a blessing to mankind.
1033. Read the Policy of Insurance. To all who insure, whatever may be
the subject of insurance, or the character of the company in which you insure, we
say, read your policies and fully understand your rights and duties.
In case of loss of furniture, books, pictures, wearing apparel, jewelry, etc.,
the insured is required to give a full itemized statement of the same before the
Yºr INSURANCE. 5 I 5
insurance will be paid. The importance therefore of having a list or an inventory
of them safely preserved is easily seen.
The PARTY INSURED should also fully understand that if he insures his house
and then sustains a loss on his kitchen, or his furniture, he has no claims on the
insurers for loss or damage. His policy protects only such property as is specified
therein. He should also understand that where his house, when insured, is damaged,
and said damage paid for or repaired by the insurers, that his policy is reduced to
the extent of the damage sustained. A want of this knowledge frequently results
in unpleasant controversies between the insured and insurers.
The problems in insurance involve some of the principles of per cent, and
sometimes, especially in life insurance, the calculation of the average length of
human life, and the worth of money invested at various rates of interest, and in
many different Ways.
TO FIND THE RATE OF PREMIUM.
1034. The first thing to be determined by an insurance company after its
organization, is the rate per cent premium to charge. To determine this for river
insurance, independent of the charges and statistics of other companies, we present
the following figures:
From the statistics of river commerce, we will suppose that during the past
season there were 40 steamers plying between the ports of New Orleans and St.
Louis, the aggregate value of the same being $2,400,000, and which, in accordance
with the regulations of insurance companies in New Orleans, was insured for two-
thirds of its value, $1,600,000; that during the season
1 steamer was lost, the value of which was $45,000, 3 of which is - $30,000
And 1 was damaged to the amount of $18,000, § of which is - - - 12,000
Making a total loss of - ſº sº * gº $42,000
STATEMENT TO OBTAIN TEIE PER CENT LOSS.
$
42000
wo
100
23% loss, Ans.
By these figures, we see that to insure a steamer in this trade, the probabili-
ties of loss are 23 per cent, and hence the rate per cent of premium must be 23 per
cent, plus such increase as may be required to remunerate the insurer for his
trouble, cover expenses, and leave a fair margin for profit.
To show approximately the per cent to charge for fire risks, we present the
5 I6 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICs.
*
following figures, which are very nearly the actual figures of one of the New
Orleans Insurance Companies for one year's unusually disastrous business:
Amount of Fire Risks, $8,500,000
Losses on Fire Risks, $75,540
{{ “ Marine { { 2,800,000 {{ {{ Marine {{ 51,000
{{ {{ River 4 1,750,000 {{ {{ River 4 20,210
$13,050,000 $146,750
The expenses of the Company during the year were - $36,000
“ taxes {{ {{ * { {{ {{ 46 - 12,000
By these figures, we see that $8,500,000 have lost - tº º $75,540
To which we add # of the expense gº tº {º º sº 12,000
And #####, of the taxes - tº tº tº i tº 7,816
Total loss and expense on fire risks e © ę $95,356
Having the total losses of the fire risks, we then find the per cent loss by the
following statement
100%
$
95.356
8500000 || 95.356
8500000
100
1.1218 + 7% (nearly 13%).
1.1218 + 7%
Ans.
The result of this work shows that the probabilities of loss, including
expenses and taxes are 1.1218 + 7%, and hence the rate per cent of premium must
be 1.1218 + 7%, plus such increase as may be necessary to produce such profit as
equity would warrant and the wisdom of the Directors of the Company may deter-
mine. To fix the premium at 14 per cent, the Company would realize on the fire
risks, as shown by the above figures, a profit of $10,894.
In actual practice. Imany other small items of expenditure and allowance
might be made, and thus bring the result still nearer the exact premium to charge.
The supply of water and the efficiency of the fire department would also be con-
sidered. Our subject in presenting this work is more to show the general manner
of arriving at a very important result than the minute details of the operation,
which must necessarily differ in every case, and can only be fully understood as they
occur in the actual business of a company; in actual operations, the figures showing
the amount insured and the amount of loss sustained for several years would be
takeºustead ºf: the result of one year's business.
ºp &
gº •e." gº &
% INSURANCE. 517
TO FIND THE AMOUNT OF PREMIUM, OR COST OF INSURANCE,
WEHEN THE AMOUNT INSURED AND THE RATE OF
PREMIUM ARE GIVEN.
PROBLEMs.
1035. 1. A real estate owner insures for one year two houses, valued at
$7000 each, at 13 per cent. What is the premium ? Ans. $157.50.
2. What is the premium for an insurance of $120000 on a hotel, at $2.25 per
$100% Ans. $2700.
3. A ship was insured for $60000 at 14 per cent, and her cargo for $25000 at
$1.30 on $100. What was the cost of the insurance? Ans. $1225.
4. What is the premium pro-rata on a $3500 policy, dated Sept. 12, 1894, and
expiring Jan. 30, 1895, at the annual rate of 75% per $100% Ans. $10,064.
NOTE.-30 days to a month is allowed in this problem.
5. What is the premium on a $3500 policy dated as in the above problem at
the short rate of 95g per $100? Ans. $12.74;.
6. Insured my house for $8000, furniture for $4500, library for $1500, paint-
ings for $1000, piano for $500 and stable for $200. The policy costs $1.50; the rate
of premium is 1+ per cent, with 20 per cent discount or rebate; what is the cost of
the insurance and what is the actual rate of premium charged ?
Ans. $158.50 cost of insurance. 1% actual premium.
NOTE.-The law of Louisiana does not allow insurance companies to give a rebate. The net
rate of insurance must be stated.
OPERATION TO OBTAIN COST OPERATION TO OBTAIN THE ACTUAL
OF INSURANCE. RATE PER CENT.
$15700 sum insured. |14% premium = #% or thus:
14% rate of prem. |20% discount = # of # = # per cent, $ PREMIUM
GE º which deducted from # leaves 1% 157 o
$196.25 amt. of prem. | actual premium. 15799
39.25 = 20% dis. 109
$ 57 00 t or thus:
157. net prem.
1.50 policy. 14% = $1.25 1% Ans.
20% of which is .25, and which
$158.50 cost of ins. deducted leaves $1, which = 1%.
7. A merchant insured for one year his house for $12000, his furniture for
$5000, library for $3000, paintings for $2000, statuary for $2500, silver-plate for
$1500, piano for $500, mirrors for $2000, cisterns for $400, stables for $1000, fences
for $500, and wearing apparel for $3000. The policy cost $1.50; the rate of pre-
mium was 14 per cent, with 15 per cent rebate. What was the cost of the insur-
ance, and what was the actual per cent paid 3
Ans. $427.35, cost of the insurance. 14% 76, actually paid.
8. What will be the cost to insure a stock of extra extra hazardous goods
valued at $28500, at $3.25 on $100? Ans. $926,25.
5 I 8 soul.E's PHILOSOPHIC PRACTICAL MATHEMATICS. *
9. A real estate owner insures 2 dwellings valued respectively at $8500 and
$7000; and also the rent of same, which is $1800 for the $8500 dwelling and $1500
for the $7000 dwelling. The rate was 13 per cent. What did the insurance cost
him # Ans. $282.00.
10. The following entries are made on an “open policy” or in the pass book
used with an open policy: Make the extensions and find the total amount due.
DATES NAMES OF STEAMER IFROM TO SUBJECT AMT. INs. RATE PREM.
1895
Nov. 3 || Garland, New Orleans. Shreveport, Mdse. 4100|00 24
“ | 5 || Pargoud, 4 t é & Wicksburg, Mdse, 650000| #
“ | 8 || T. P. Leathers, { { & 4 St. Louis, Mdse. 980000. 1;
“ 14| Queen City, * { & 4 Cincinnati, Moise. 2300|00| 2
“ 27| City of New Orleans. & 4 & 4 Memphis, Mdse. 1685|00| 1
$ 37335
11. What will it cost to renew a policy of insurance on property valued at
$24200 at the rate g per cent premium ? Ans. $211.75.
12. An Insurance company unsured a block of buildings valued at $400000
at 2 per cent, allowing a rebate of 15 per cent. The company subsequently re-insured
$100000 at 13 per cent, $100000 at 14 per cent, and $100000 at 13 per cent. How
much premium did the company receive? How much did it pay for re-insurance,
and at what rate per cent were its receipts on the amount that it protects?
Ans. $6800, premium received. $4125, premium paid. 2% 7%.
PARTIAL OPERATION.
$400000 at 2% = $8000 – 15% = $6800. $100000 at 14% = $1125. $100000 at 14%
= $1250. $100000 at 13% = $1750. $1125 + $1250 + $1750 = $4125. $6800 –
$4125 = $2675. $400000 – $300000 = $100000. ($2675 × 100) + $100000 = 2.675%
= 2%%.
13. You insure your house for $9000; furniture for $4500; jewelry for $6000;
books for $1800; paintings for $2500; statuary for $1400; silver-plate for $1200;
piano for $400; cisterns and outhouses for $1500; wearing apparel for $2500. The
rate is 14 per cent with 15 per cent rebate. What is the cost of the Insurance, and
what, in case of a fire and a total loss, would be your various duties in order to col-
lect the insurance? Ans. To the first, $327.25.
Answer to the second : Your first duty would be to read your policy and act
in accordance with its conditions. 2. To make an itemized statement or bill of the
furniture, jewelry, books, paintings, statues, silver-plate and wearing apparel, and
present the same with your account to the insurance company.
14. If you keep an insurance on an extra hazardous stock of goods for 20
years at 5 per cent, and they then burn and you collect the full insurance, have you
gained or lost 3 Ans. Making no allowance for interest, you have lost premium
paid to the amount of the value of the goods.
X}. INSURANCE, 519
TO FIND WHAT AMOUNT TO INSURE SO AS TO COVER BOTH THE
PROPERTY INSURED AND PREMIUM PAID, WHEN WE
ELAVE THE AMOUNT OF PROPERTY AND THE
RATE PER CENT PREMIUM GIVEN.
PROBLEMIS.
1036. 1. I wish to insure a stock of goods valued at $44100, for such a sum
that in case of loss, I will receive the value of the merchandise and the premium
paid for insurance, or in other words I wish to insure the merchandise and the
premium paid for the insurance. The rate is 2 per cent. For what amount must I
insure, and what is the premium ? Ans. $45000, amount to insure.
$900, premium.
FIRST OPERATION.
$100 assumed. $ ſº first
= 2 * e 3.SSlllll C o represent
2 % premium. 98 100 the value of the goods,
plus the 2% premium for
$98 value of goods. 44.100 insuring the goods. We
then deduct therefrom the
$45000 amount to insure. 2 per cent premium and
2 º obtain $98, as the value
% premium. of the goods, which, with
e the premium paid for in-
$900.00 premium. Suring the goods required
an insurance of $100. We
then reason thus: Since $98 value of goods require an insurance of $100, $1 will require the 98th
part, and $44100 will require 44100 times as much, the result of which is $45000, We then calculate
the 2 per cent on the $45000, and find the premium to be $900.
SECOND OPERATION.
$44100 value of goods. $
2% premium. 100
98
$882.00 premium on goods. 882
*-*s
$900 premium on goods and premium.
44100 value of goods added.
$45000 amount to insure.
2. My house, situated in a country village, is valued at $12000, and the
furniture at $5000. I wish to insure # of the value of the house and furniture, so
that in case of loss I may receive the # value of both and the premium paid for the
insurance. The rate is 1% per cent; policy cost $1.50; the rebate or discount is 10
per cent. What amount of insurance must I effect and what will be the cost 3
Ans. $12954.03, insurance. $205.53, cost.
PARTIAL OPERATION.
$12000 + $5000 = $17000. # of $17000 = $12750. 13% = $1.75, 10% of $1.75 =
.175g. $1.75— .175 = $1.575, $100–$1,575 = $98.425. ($12750 × 100) + $98.425
= $12954.03 to be insured. 13% on $12954.03 = $226.70 - 10% rebate = $204,03
+ $1.50 = $205,53.
52O SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
3. A cargo of cotton is valued at $260000; for what sum must a policy be
taken to cover both cotton and premium, the rate being $1.624 on $100
Ans. $264294.79-H.
4. For what amount must I insure my dwelling, valued at $9600, so that in
case of loss I may receive both the value of the house and the premium paid The
rate of insurance is 1% per cent, rebate 20 per cent 7 Ans. $9736.30-H.
PARTIAL OPERATION.
7
1#% = 4. 20% = }, + x # = ſ. 3 — ſº, = ## = 13% net premium. $100–13%
= $983. ($9600 x 100) + 98% = $9736.30-H.
TO FIND THE AMOUNT INSURED, WHEN WE HAVE THE AMOUNT
OF PREMIUM PAID AND THE RATE PER CENT PREMIUM GIVEN.
PROBLEMS.
1037. 1. The cost of insuring a house was $114.00; the cost of policy was
$1.50; the rate of insurance was 14 per cent. What was the amount insured ?
Ans. $9000.
& Statement to find the sum insured. Explanation.—We first
OPERATIO N deduct the i. º :
* olicy from the total COS
$114.00 cost of insurance. 100 | jº, and have in
1.50 cost of policy. 5 || 4 the remainder $112.50 the
112.50 amount of premium which
is 1+ per cent of the sum
insured. The reasoning
$9000.00 Ans. for the statement to find
this sum is the same aS
given on page 446.
2. . A merchant paid $121.50 for insuring his store. The rate of premium
was $1.50 per $100; the rebate was 20 per cent; the policy cost $1.50. For what
$112.50 amount of premium.
amount did he insure his store? * Ans. $10000.
OPERATION.
$121.50 premium and policy. $1.50 rate of premium. 100
1.50 policy. .30 20% rebate. 1.20 | 120.00
$120.00 premium. $1.20 net rate of premium. $10000.
3. The premium for insuring # of a stock of goods at # per cent was $137.814.
What was the value of the goods? Ans. $24500.
4. The cost of insuring an invoice of merchandise was $180,104. The rate
of insurance was 1% per cent; the rebate was 10 per cent; the cost of the policy
$1.50. What was the amount of the invoice 3 Ans. $17640.
OPERATION INDICATED.
14% = 1.125 – 10% rebate. (.1125) = 1.0125%. $180.105 — $1.50 = $178.605.
(178.605 × 100) -- $1.0125 = $17640.
* INSURANCE. 52 I
*
TO FIND THE RATE PER CENT PREMIUM, when WB HAVE THE
AMOUNT INSURED AND THE AMOUNT OF PREMIUM GIVEN.
PROBLEMs.
1038. 1. The cost of insuring a stock of extra hazardous goods, valued at
$38400, was $1153.50 including the cost of policy, $1.50. What was the rate per
cent premium ? Ans. 3%.
OPERATION. Statement to find the rate per cent Explanation.—We first de-
premium. duct from the total cost, the
cost of the policy, and thus
º 0 obtain $1152, which is the
$1153.50 cost of ins. 100 amount of premium at a certain
1.50 cost of policy. 38400 || 1152 per cent on the value of the
goods. We then make the
g statement to find the rate per
$1152.00 amt. of prem. •º 3% Ans. cent. The reasoning for this
statement is the same as that
given on pages, 439 and 440.
ºm-sam-mm-
2. A merchant insures his goods for $12000. The insurers allow a rebate of
25 per cent on the premium, and charge $1.50 for the policy. The cost of insurance
and policy was $163.50. What was the rate of premium on $100?
Ans. $1.80, or 1.8%.
Statement to find the amount of Statement to find the rate per
- before the 25
OPERATION. Pººr cent of premium.
$163.50 cost of insurance. 100 216
1.50 cost of policy. 75 162 12000 || 100
$162.00 net prem. paid. | $216 amt. of prem. chg'd | slº, = 1.8%.
3. The premium paid for insuring a house valued at $9400, was $70.50.
What was the rate per cent ? Ans. #%.
4. I paid $193.50 for insuring my house for $12000. The policy cost $1.50.
The rebate, in lieu of participation in the profits of the company, was 20 per cent.
What was the rate per cent premium charged, and what was the actual rate paid?
Ans. 2% premium charged. 14% actual premium paid.
522 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. fºr
PRACTICAL LABOR SAVING SYSTEM OF CASTING UNEARNED
PREMIUMS.
1089. The following useful table is the result of the experience and genius
of Mr. H. CARPENTER, an expert insurance agent of New Orleans, through whose
kindness we are permitted to use it.
TABLE FOR CASTING THE UNEARNED PREMIUMS, THE YEAR COMMENCING ON THE
k FIRST OF JANUARY.
KTo find the unearned premium, multiply the premium charged by the amount opposite the date of
the policy).
DATE. JANUARY. DATE. FEBRUARY. DATE. MARCH. DATE. A PRIL.
1. .0027 1 .0876 1 .1643 1 .2492
2 .0055 2 .0904 2 .1671 2 .2519
3 .0082 3 .0931 - 3 .1698 3 .2547
4 .0109 4 .0958 4 .1725 4 .2574
5 .0137 5 .0986 5 .1753 5 .2601
6 .0164 6 .1013 6 .1780 6 .2629
7 .0192 7 . 1041 7 .1807 7 .2656
8 .02.19 8 ..1068 8 .1835 8 .2683
9 .0246 9 .1095 9 .1862 9 .2711
10 .0274 10 .1123 10 .1889 10 .2739
11 .0301 11 .1150 11 .1918 11 .2766
12 .0328 12 .1177 12 .1944 12 .2793
13 0356 13 .1205 13 .1972 13 .2820
14 .0383 14 .1232 14 .1999 14 .2848
15 .04.11 15 .1260 15 .2027 15 .2876
16 .0438 16 .1287 16 .2054 16 .2903
17 .0466 17 .1314 17 .2081 17 .2930
18 .0493 18 , 1342 18 .2109 18 .2958
19 .0520 19 .1369 19 .2136 19 .2985
20 .0548 20 .1397 20 .2164 20 .3013
21 .0575 21 .1424 21 .2191 21 .3040
22 .0602 22 .1451 22 .2218 22 .3067
23 .0630 23 . 1479 23 .2246 23 .3095
24 .0657 24 .1506 24 .2273 24 .3122
25 .0685 25 .1534 25 ,2300 25 .3150
26 .0712 26 . 1561 26 .2328 26 .3177
27 .0739 27 . 1588 27 .2355 27 3204
28 .0767 28 . 1616 28 .2383 28 .3232
29 ,0794. 29 . 1643 29 .2410 29 .3259
30 .0822 30 .2437 30 .3287
31 .0849 31 .2465

The unit of this table is based on the premium of $1. for 1 year, which divided by 365, the
number of days in a year, gives the premium on $1 for 1 day, which, carried to 4 places of decimals,
is .0027; this being multiplied by the number of the days of the year, gives the tabular multi-
pliers, which in the table presented is only extended to 4 months, or 119 days of the year. Having
these multipliers, by simply multiplying the amount of the premium by the multiplier in the table
corresponding with the date of the policy, we produce the unexpired premium. For example, the
premium on a policy issued for 1 year from March 17th, amounts to $80. To find the unearned
premium on the 31st day of December, we look for the multiplier opposite March the 17th, in the
20. * INSURANCE. 523
table, and with it multiply the $80 thus: $80 × .2081 = $16,6480, or practically $17; or in practice
the premium may be multiplied by the cents only increased by 1, when the decimal is in excess of
one-half, thus:
$80 × .21 = $16.80; or practically $17.
The above table was prepared for a company, whose fiscal year begins on the 1st of January;
but it can easily be adapted to the work of other offices whose fiscal year begins at a different
date, by substituting the first day of their business year for January 1st.
Want of space prevents the insertion of the table for the entire year, but the 4 months
presented is sufficient to demonstrate the simplicity, convenience and value of It.
ADJUSTMENT OF LOSSES.
1040. 1. Jones & Smith hold three policies on a stock of goods; one policy
in Co. A, for $5000, one in Co. B, for $8000, and one in Co. C, for $3000. A fire
occurs and damages the goods to the amount of $9860. What is due from each
Company ? Ans. $3081.25 from Co. A.
$4930.00 from Co. B.
$1848,75 from Co. C.
OPERATIONS INDICATED.
$5000 Co. A. CO. B. CO. C.
8000 9860 9860 9860
3000 16000 || 5000 16000 || 8000 16000 || 3000
$16000
2. A dwelling is insured in Company X, for $5500, and in company Y, for
$3500. The house is destroyed by fire, and the companies find that they can re-build
it for $7000; this amount, in accordance with the conditions of the policies, they
offer to the owner, who accepts it. What sum will each company pay?
Ans. Co. X, $4277.78. Co. Y, $2722.22.
3. J. M. Butchee & Co., carry policies on goods in 16 companies for $150000,
Company A carries a risk of $10000. The whole stock of goods is destroyed by
fire. Merchandise Account in the books of J. M. Butchee & Co. is debited with
$875000 and credited with $862500. By comparing the cost and sales of 100 arti-
cles of goods, it is determined and agreed between the insurers and insured that
15 per cent gain has been realized on sales, and a settlement is made on this basis.
Make the bill for the amount due from Company A.
OPERATION.
$ $875000 total cost.
115 100 750000 cost of Sales.
86.2500
$125000 cost of goods lost.
$750000 cost of sales.
(BILL.)
NEW ORLEANs, July 18, 1895.
To J. M. Butchee & Co., Dr.
For Yºº, of $125000, Mdse. destroyed by fire at No. 108 Magazine Street,
New Orleans, June 2, 1895. - - - - - - - - - -
INSURANCE CO. A.,
- - - $8333,33
Received payment,
524 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS.
Adjustment under the Average Clause.
1. A merchant has goods valued at $20000 in two warehouses, on which he has
a policy, with the “Average Clause,” for $8000. The goods in the warehouse in
which he had $12000 was damaged by fire to the amount of $6000. What is due by
the Insurance Company ? Ans. $2400.
OPERATION.
$12000 insured by owner. $6000
8000 insured by company. 20000 || 8000 or, 20000: 8000 :: 6000 : 2400
s:0000 total insurance. $2400
Adjustments under the Three-fourth Loss or Country Clause.
1. A country merchant who had a $10000 policy on his goods, with the three-
quarter loss clause, sustained a loss by fire amounting to $8000. How much do the
insurers owe the merchant 3 Ans. $6000.
NotE.—In policies containing the three-quarter loss clause, the insurers, in case of a total loss
pay only three-fourths of the value of the property destroyed, and if the loss is less than the sum
insured, the insurers pay only three-fourths of the loss.
2. If in Problem 1, with the three-quarter loss clause, how much would the
merchant have received: 1. If the loss had been $10000; 2. If the loss had been
$12000; 3. If the loss had been $16000 Ans. 1. $ 7500 on a $10000 loss.
2. $ 9000 on a $12000 loss.
3. $10000 on a $16000 loss.
REMARKs.—In case of a $10000 or a $12000 loss, as previously stated in note
to Problem 1, the underwriters are responsible for # of the loss, because it does not
exceed the face of the policy and because of the # condition clause of the policy.
In case of the $16000 loss, # of the same being $12000, the underwriters or
insurers carrying the $10000, are responsible for the full face of the policy, $10000.
NotE.-$10000 is # of $13333; ; hence, for a loss of $13333} or more, the insurers are respon-
sible for the face of the $10000 policy.
*
Marine Adjustments.
1. The owner of § of a vessel insured # of his interest at 24 per cent; the
sum paid for premium was $562.50. The vessel sustained a loss by fire of $5000.
What sum do the insurers owe him # Ans. $1406.25.
OPERATION INDICATED.
$100 22500 5000
2 3 || 4 80000 22500
562.50 *- |
a-º. -msm--assº- $30000 = # of value of vessel. $1406.25
$22500.00 = # of his # 3 || 8
$80000 = value of vessel.
NOTE.-In Marine Insurance, according to the “Average Clause,” the insurers pay for a
PARTIAL LOSS of property PARTIALLY insured, only such a proportion of the loss, as the sum insured
bears to the whole value of the property. *
2. A cargo of fruit valued at $8500, and insured for $8000 was damaged at
sea and sold at auction on arrival in port for $7500. What was paid by the insurers
on the adjustment of the loss? Ans. $941.17-H.
Adjustment under the Seventy-five per cent Co-insurance Clause.
1. A merchant effected $40000 insurance on property worth $60000. His
policies contain the 75 per cent co-insurance clause. A loss of $30000 was
¥ INSURANCE. 525
Sustained ; what sum will the merchant receive from the underwriters in a correct
adjustment of the loss, and what will be his loss as a co-insurer?
Ans. $26666; due from underwriters.
3333; merchant's loss as a co-insurer.
NOTE.—See Article 1028, page 513, for the 75 per cent Insurance Clause.
FIRST OPERATION.
$60000 × 75% = $45000 amount to be, or which is really insured. $45000 – 40000 = $5000 the
sum that the merchant is co-insurer. Thus making, under the 75% clause of the policy, $45000
insurance on the property
45000 $30000 Ea:planation.—The reasoning for the line statement is as
40000 follows: Since $45000 insurance lose $30000, $1 will lose the
| 45000dth part and 40000 (the sum insured by the underwriters)
$26666; Ans. will lose 40000 times as much.
$30000 — 26666} = $3333; merchant’s loss as co-insurer.
SECOND OPERATION.
$30000 $40000 × 663 = $26666}, Explanation.—Since $45000 lose $30000, $1
45000 || 100 which is the sum lost and will lose the 45000dth part, and $100 will
g--- to be paid by the insur- lose 100 times as much. The loss thus pro-
663% loss. ance companies. duced, being on $100, dollars, is hence the
loss per cent.
$5000 × 663% = $3333; which is the sum lost and to be borne by the merchant under the
75% co-insurance clause.
2. Suppose in Problem 1, that the loss had been $45000, how much would
the Insurance Companies owe ? Ans. $40000.
REMARKS.—When the loss is equal to or greater than the full amount insured
by the underwriters and owner as co-insurer, then the underwriters will owe the
face of the policies. se
3. Suppose in Problem 1, that a total loss of $50000 or $60000 had been
sustained, how much would the Insurance Companies owe ? Ans. $40000.
REMARKS.—See remark of Problem 2. In no case on fire insurance can the
underwriters owe to the insured more than the face of the policies.
Adjustment under the Three-fourths Value Clause.
Under this clause, which is inserted in some policies, the insured will receive the full
amount of his loss, provided it does not exceed three-fourths of the value of the property.
Thus: 1st. Suppose property valued at $1000, is insured for $750, and a loss of $600 is sustained,
the insured will receive $600 from the underwriters. 2d. Suppose property valued at $1000 is
insured for $750, and a loss of $850 or more is sustained, the insured will receive $750 from the
underwriters.
REMARKS.—This three-fourths value clause differs from the three-fourths loss or country clause
explained in Articles 1030 and 1040, pages 514 and 524, inasmuch as under this three-fourths value
clause the insured will, in case of a partial loss not exceeding three-fourths of the sound value of
the property insured, receive the full amount of loss. But under the three-fourths loss claw8e the
insured will receive only three-fourths of the amount of his loss whether it be more or less, not
exceeding the face of the policy. See Article 1030a. &
FINDING THE LOSS OF MERCHANDISE DESTROYED BY FIRE AND AIDJUSTING TH
LOSS WITH THE INSURANCE COMPANIES CARRYING THE RISKS.
PROBLEMS.
1041. To find the loss of Mdse. by fire is a far more difficult question than is supposed by
one man in a thousand.
To elucidate this matter, let us assume the following conditions:
1. That Jones & Smith carry $90,000 insurance on Mdse. in 12 Companies.
2. That their entire stock is destroyed by fire.
3. That their Double Entry Ledger, with all purchases and sales posted to date of fire,
shows Mdse. Account debited $486,500 and credited $467,450.
From these assumed facts, what has been the loss and what does Insurance Co. A OWe,
which carried a policy of $8000 on the stock? º
The first step in a case of this kind is to read your policies and notify the underwriters of
all the known facts of the fire. Then consult with them regarding your loss, exhibiting all books
and papers pertinent thereto. When the underwriters are satisfied that all is fair, then the
question of GAIN or Loss per cent on the sales of goods must be determined. And the solution to
526 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS. Yºr
this question is the golden key that unlocks the treasury of the underwriters carrying the risks.
No correct adjustment of loss can be made before this question is solved.
It is to the interest of Jones & Smith to show a large gain per cent on sales; and it is to
the interest of the Insurance Companies to show a small gain per cent on sales.
If an agreement cannot be reached by the two parties in interest, from the general accounts,
then an average gain per cent must be determined by listing the cost and sales of a number of
the leading lines of goods, thus:
Statement to find average gain Statement to find cost of sales of Mdse. based on
Cost. Sales. per cent. an average gain of 16.19 per cent.
$ 4. 4.80 4.31 100
.22 .30 116.19 || 467450.00
1.50 1.65 26.62 | 100
2.75 ( 3.00 * | $402315.17 cost of sales.
.15 .18 | 16.19 96 gain.
18. 21.00
Total cost of Mdse., $486500.00
$26,62 || $30.93 Cost of sales “ 402315.17
Net Gain, $4.31 Loss by fire, $ 84.184.83
In the above statement of cost and sales, to abridge the work, we limited the statement to
six items. In practice, FIFTY TO ONE HUNDRED should be used. In the selection of the items lies
the great difficulty. The underwriters insisting on items upon which a small gain was realized,
and Jones & Smith insisting on items upon which a large gain was realized. A nice discrimina-
tion is required in making this statement to render justice to both parties. The author hereof, in
behalf of his client, had nine sittings of two hours each, with Insurance Adjusters, in a vain
: make a statement of costs and sales, from which the average gain per cent should be
accepted.
Where the parties in interest fail to agree upon this point, arbitration of the question seems
to our mind to be the most ethical and reasonable.
f $8 The following statement shows the amount of loss due from the Insurance Co. having a risk
Of $8000.
Total amount insured, $90000. Total amount lost, $84.184.83. Hence the Co. owes sº
of $8000 = $7483.10.
INSURING GOODS.
1042. Business prudence demands that merchants and manufacturers should give careful
attention to their stock of goods and to the amount of insurance covering the same, adding or
cancelling policies so as to keep fully protected and losing nothing by over-insurance. A careful
record should be kept of the date the policies expire, and daily inspection should be made thereof.
The insured should also note with care all the special clauses and conditions imposed by his
policy. The Lightning, the Average, the Foundation and the 75% or more Co-Insurance clauses
should be thoroughly understood when the policy is written.
Points to be Specially Observed in the Mdse. Insurance.
When considering merchandise account and the statement of costs and sales, due regard
must be given to the following elements on purchases and sales: e
Sal (1). Trade Discounts. (2), Cash Discounts. (3). Rebates. (4). Time Purchases and Time
ableS.
The merchandise returned to sellers, and the merchandise re-purchased from purchasers, if
entered in merchandise account, must be eliminated therefrom. After considering all these ele-
ments, and finding the loss, then the question of depreciation and appreciation of the goods lost
is to be considered, in order to find the cash value of the goods, which the policy of insurance
COVOTS.
It is the custom of Insurance Companies, when adjusting fire losses, to allow #9% discount
for each 30 days, on time purchases and on time sales. This is equal to 6% interest.
VALUED POLICIES ON IMMOVABLE PROPERTY.
- The Valued Policy Insurance Law of Louisiana, enacted in 1900, requires that the insurance
on immovable property shall be the value assessed or appraised “by the insurer, or as by him per-
mitted to be assessed at the time of the issuance of the policy,” and that this value “shall be
conclusively taken to be the true value of the property at the time of the issuance of the policy,
and the true value of the property at the time of damage or destruction.” sº g
On this value, Insurers are obliged to adjust the loss. The law, however, gives the insurers
the right to restore the damage or to replace the property destroyed if they prefer to do so,
instead of making payment therefor. *
By this law the adjustments of loss are easily and expeditiously made, and the vexatious
contentions, so often made by the insurers under the former law, are obviated.
GENERAL AND PARTICULAR AVERAGE. 527
TAKING STOCK OR INVENTORYING GOODS.
When “taking stock,” or inventorying goods and other property preparatory to closing the
books and declaring the gains and losses at the close of the fiscal year, or in case of a corpora-
tion, at a dividend period, the current market value should be placed upon the goods and all other
property, stocks, bonds, etc.
NOTE.-In the case of stocks and bonds which have an unstable value and a wide margin of fluctuation, the average
Weekly or monthly value is sometimes placed upon them. Circumstances, judgment and ethics are the proper factors to
govern such cases.
In the case of merchandise, instead of listing every article at the current market value
With allowances for depreciation on account of damage, out of fashion or style, the inventory is
often taken on different lines of goods, at the orignal invoice price, and then a specified per cent.
reduction or addition is made thereto, according to the present market price of such lines of goods;
allowances being made also for damage, old stock, etc.
In some cases of unimproved real property, it is often difficult to determine its market value.
Whether it is worth the original cost, more or less, plus taxes, is a question to be decided by the
facts attending the case, and the judgment of the appraisers.
General and Particular Average.
1043. General Average is the proportional contribution by all parties con-
cerned in a vessel, cargo or freight, towards a loss or damage sustained or incurred
for the common safety or benefit.
1044. Particular Average is the contribution for loss or damage incurred or
sustained by or for one or more of the interests, vessel, cargo or freight; such loss
much be borne by the interest or interests which sustained the loss.
1045. . An Average Adjuster is one whose profession it is to average and adjust
losses in which general and particular average cases occur. He is required to be
familiar with insurance laws governing the subject, and to determine which loss
should be charged to General Average and which to Particular Average, and then
to apportion the loss and expense among the several contributory interests.
To determine, in all cases, what kind and character of losses and damages are
subject to general average and what to particular average is a very difficult matter
and often gives rise to prolonged litigation. Arnould's Maritime Law and Abbott,
on Shipping, give valuable information on this topic.
The following are some of the losses and damages subject to general average:
1st. The damage sustained by a ship purposely run ashore to prevent her foundering
at sea or driving on the rocks. 2d. All damage purposely done to the vessel or
cargo, to preserve the whole from impending danger, such as the cutting away of
masts, riggings, etc., when the ship is in distress. 3d. All charges incurred for the
general good, by entering a foreign port in distress. 4th. The expense for dis-
charging the cargo for the purpose of repairing the vessel, or for floating her when
she accidentally gets aground. 5th. The expense of pilotage when putting into a
port in distress. 6th. The expense of labor to pump the ship after having sprung a
leak. In some cases, seamen's Wages and the provisions of the ship's company are
subjects of general average.
The following are a few of the losses and damages subject to particular
average :
1st. Damage by fire from accidental causes, unattended by a sacrifice to
extinguish the fire; also the damage to goods by water thrown upon the burning
goods to extinguish the fire, which is regarded not a sacrifice of goods, but a real
advantage to them.
2d. The damage done to a vessel and cargo by being run foul of accidentally
and without fault on either side.
3d. The springing of masts, breaking of the upper works or timbers, parting
of cables, and all other loss and expense upon the ship occasioned by the perils,
insured against not purposely incurred, and not amounting to a total loss.
528 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
1046. Jettison. The goods thrown overboard, the cutting away of masts or
cables, and whatever the master of a vessel does in distress for the safety or
preservation of the whole, is called jettison.
In making an adjustment of general average, it is necessary to first take an
account of the various losses which are to be paid by contribution, and secondly, to
take another account of the value of all the articles that are to contribute. In the
list of contributory articles—vessel, cargo and freight—must be included the goods,
etc., thrown overboard; for otherwise the owners of these goods would receive full
Value for them and pay nothing towards the loss.
In estimating the amount of loss to the vessel for masts, rigging, etc., it is
the custom to deduct of the cost of the new articles; for being new they will be
of greater value than the articles lost. In estimating the value of the contributory
interests, the general custom is as follows: To value the ship at what it is worth at
the end of the voyage; to value the cargo at what it would have brought at its port
of destination, but it is sometimes valued at its invoice price at the port of lading;
and to value the freight at # the amount due at the time of the jettison or other
sacrifice. But in some ports the value of the freight is determined by deducting
from the gross amount the wages for seamen, pilotage, etc. In New York, # of the
freight is deducted.
As remarked in the introduction of this subject, it is so vast and complicated
that it cannot be satisfactorily treated in a work of this kind, and it is questionable
Whether or not it is proper to present it in works of this kind. It is a subject that
merits and has received the attention of the most profound jurists of both hemi-
spheres. We have briefly introduced it, in conformity to custom, and present the
following, to show the operation for a general and particular average computation :
PROBLEMIS.
1. The steamer City of Baltimore left New Orleans for Philadelphia with a
general cargo amounting to $150000. During the voyage, the vessel encountered
very severe weather, and goods to the amount of $4500 were thrown overboard. Masts
and rigging were cut away which cost $810 to replace; one anchor cost $250. The
expenses of the vessel in port while being repaired were $540. The charges for
protest and adjustment of average were $250. The gross amount of freight amounted
to $12420. The value of the vessel is $90000.
There were 300 hbds. of sugar on board, 158 of which were damaged by
water, according to the report of the surveyor, to the extent of 34 hbds., which, at
$100 amounts to $3400.
The goods and vessel were fully insured.
What is the general and particular average loss per cent., and the amount of
loss sustained by each contributing interest, estimating the cargo at the port of
lading invoice? Ans, 2.44884 + 7% general average loss.
114% particular average loss.
$2203.97 to be contributed by vessel.
$3673.26 “ {{ “ cargo.
$ 202.77 4. 4% “ freight.
$3400.00 ($ {{ “ sugar.
* GENERAL AND PARTICULAR AVERAGE, 529
OPERATION.
General Average Loss.
Masts, rigging, etc., - - - -- * - &n $810
Less # deducted because new * - sº - * 270 $ 540
Goods thrown overboard - *-> - º - s 4500
Anchor, cost of new - T - º - sº • Ge 250
Expenses in port º º º - tº- - es 540
Protest and average adjustmen º - º - ºs 250
Total amount of general average loss - wº $6080
Particular Average Loss.
34 hlids. sugar (a) $100 tº- - i- º sº tº- - tº - $3400
Contributory Interests to General Average Loss.
Value of steamer (when repaired) - sº as $ 90000
Cargo (invoice value) º tº - - sº - 150000
Freight - - º º º º - s - $12420
Less #, allowed for expenses - - s - 4140 8280
Total amount of contributory interests $248280
Contributory Interests to Particular Average Loss.
300 hbds. sugar (a) $100 - - ºn - gº *- sº * ºn $30000
Statement to find the General Average per Statement to find the Particular Average
cent Loss. per cent Loss.
$ LOSS. $ LOSS.
6080 3400
248.280 100 º 30000 || 100
2####% = 2.44884 + 7. 11; %
Statement showing the amount of loss that each contributing interest must pay.
Wessel, $ 90000 (a) 2.44884% tºº º $2203.97
Cargo, 150000 (a) 2.44884% - Ҽ - 3673.26
Freight, 8280 (a) 2.44884% - tº- - 202.77
Sugar, 30000 (a) 118% - º wº- 3400.00
REMARKS.–In actual practice, average adjusters individualize every con-
signee's invoice, and specify the loss on each article of goods. They also itemize
every expense incurred in the settlement of the various losses.
When the rate per cent loss contains an inconvenient fraction, as is the case
in the above example, it is extended to three, four or more places of decimals, and
a table prepared therefrom to facilitate the operation of multiplying the amounts of
the different invoices composing the cargo. If the contributing interests were few
in number, time might be saved by making proportional statements to find the loss
that each interest must Sustain.
53o soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. jºr
*
2. In the above problem, suppose that Jones & Co. had an invoice of general
merchandise valued at $2000, fully insured, which suffered no damage, what would
be the sum due by Jones & Co. or by their insurers? Ans. $48.97+.
OPERATION.
$2000 x 2.44884+% = $48.97+.
NoTE.—Since the cargo is one of the contributory interests to the general average, each
. of goods in the cargo (or his insurer) must pay on his invoice the general average per cent
Of 108S.
3. In the same problem, suppose that Smith & Co. had an invoice of sugar
valued at $1000, which suffered no loss, what would be the sum due by Smith & Co.
or by his insurers? * Ans. $113.33%.
OPERATION.
$1000 x 113% = $113.334.
NOTE:-The loss on sugar is a Particular Average Loss inasmuch as it was not incurred for the
common safety of the vessel, cargo and freight, therefore each owner of sugar (or his insurer) must
pay on his involce the particular average per cent loss. -
4. The ship Queen of the Ocean, on her voyage from Bombay to New Orleans,
was struck by a severe gale in the Gulf of Mexico, and her sails, rigging and upper
Works were torn and damaged to the amount of $1500. To relieve her sufferings,
goods were thrown overboard to the amount of $9400. And one mast was cut
away, the cost of which, to replace it, was $360. The cost of the surveys, protest
and adjustment of average amounted to $850.
The vessel, when repaired, was worth $50000. The value of the cargo in
New Orleans was $110000. The freight amounted to $9840. The cargo and vessel
were fully insured.
What was the general and particular average loss per cent sustained by each
of the contributory interests 7
Ans. 6+++4% or 6.2980+% general average loss.
2% particular average loss.
OPERATION.
General Average Loss.
Value of goods thrown overboard º ge º- º $ 9.400
Cost of mast - º - º - tº - $360
Less # for new - - -> > * º -> - 120 240
Cost of protest, surveys and adjustment of average 850
Total amount of general average loss - $10490
º: GENERAL AND PARTICULAR AWERAGE. 531
Particular Average Loss.
Cost to repair damage to sails, rigging and upper works - - $1500
Less # deducted for being new tº gº * = g- tº º 500
Net amount of particular average loss - tº s * > ſº $1000
Contributory Interests to General Average Loss. .
Value of vessel (after being repaired) tº ſº tº $ 50000
Value of cargo in New Orleans gº ºn tº gº 110000
Freight sº tº tº sº º gº gº e tºº $9840
Less #, allowed for expenses s ſº ſº º ſº 3280 6560
º Total amount of contributory interests - $166560
Contributory Interests to the Particular Average Loss.
Value of vessel tº is tº ſº º gºe * > E = $50000
Statement to find the general average loss per
cent that each interest must sustain,
$ LOSS
10490
166560 100
| 6}#}% Ans. or 6.2980-- 7%.
Statement to find the particular average
per cent loss that the insurers of
the vessel must sustain.
$
1000
50000 || 100
2% Ans.

ife Insurance.
A-/-/-/-/-/-/******A*/~~~~~~ES.
*RS-
1047. Life Insurance is a contract by which a life insurance company, in con-
sideration of certain payments made by the insured, agrees to pay a specified sum
of money to his heirs at his death, or to himself if living, at a specified age.
1048. Life Insurance Companies are organized on the plan of Stock Com-
panies, Mutual Companies, Mixed Companies, and on the Co-operative and Benevo-
lent Association plan.
NOTE.—See Fire Insurance for definition of each plan, page 511.
IMPORTANCE OF LIFE INSURANCE.
1049. The subject of Life Insurance is of vast importance to all classes of
people, the rich as well as the poor. For in the ever-evolving drama of life, fortunes
often disappear, and in the contests for wealth and fame small gains to the many
are the rule, while to some, cold faced poverty is the life and death companion.
Hence the rich should insure to guard against misfortune; those who are not rich
should insure so that in case of early death, their families and those dependent upon
them for support, would be protected from want and suffering; or in case of old age,
they would possess the means through an Endowment, or an Annuity Policy, or a
Trust Certificate, of self-support.
From the policy treasuries of the various forms of Life and Co-operative
Insurance, come millions of dollars annually to relieve the necessities of distressed
widows, of orphan children, and of decrepit people. And for this, thanks and
gratitude are due to the wisdom, to the humanity, and to the practical sense of those
who provided the insurance.
The world is most liberal and gracious in bestowing upon needy, afflicted, and
suffering humanity, words of sympathy, tears of pity, and prayers of supplication.
But these gifts, however bounteous, do not pay rent bills; they do not supply
covering for naked bodies; they do not provide sustenance for hungry stomachs, or
medicine for fevered brains. Therefore, by the light of these facts, and since by
law of nature all must live before they die, we conclude that it is the first and high-
est duty of man to provide for himself and those dependent upon him, and to assist
in the financial charities of the world. And all this we submit can be best accom-
plished through Life or Co-operative Insurance.
(532)
LIFE INSURANCE. 533
The Policies of Life Insurance vary in their conditions and are named as
follows:
1050. ORDINARY LIFE, LIMITED PAYMENT LIFE, TERM. ENDow MENT,
RESERVE ENDOWMENT, ANNUITY, TonTINE SAVINGs or INVESTMENT, Joint
LIFE POLICY, AND TRUST CERTIFICATE POLICY.
There are also some other special policies issued by some companies.
1051. An Ordinary Life Policy is one on which the premium is paid annually
during the life of the insured, and at his death, the amount of the policy is paid to
the party named therein.
1052. A Limited Payment Life Policy is one on which the premium is paid
annually for a certain number of years, specified at the time the policy is issued or
until the death of the insured, should that occur before the elapse of the time speci-
fied.
This class of policies is also issued on single payments, or on 5, 10 or more
annual payments. When issued on one payment, the insured receives annual cash
dividends from the company.
1053. A Term Policy is one payable at the death of the insured if it occur
during a given number of years, the payment of the annual premiums to continue
till the policy expires. gº
1054. An Endowment Policy is one that is paid to the insured at the end of
a specified number of years, or to his heirs in case he should die within the time.
The premium is payable annually, or in cash, in 5 or 10 annual payments.
1055. A Reserve Endowment Policy is one combining the Life and Endow-
ment Plans, the premium being the same as in a Life Policy, but the insurance
terminates at such a time as the insurer names, when an endowment will be paid
equal to the legal reserve of the policy.
1056. An Annuity Policy is one which secures to the insured the payment
of a specified sum annually, as long as he lives. The premium is paid in one cash
payment.
1057. A Tontine Savings or Investment Policy is one payable at the death
of the insured, but in which the excess of premiums received over the claims by
death and expenses, is divided at the end of a specified period of time, among the
survivors of those insured for that specified time, and at this time these survivors
may either accept the money as in an Endowment Policy, or change the form of
their policy as may be agreed.
1058. A Joint Life Policy is one payable at the death of the first, of two or
more persons, to the survivor. The premium is payable in the same manner as in
Life Policies. -
1059. A Trust Certificate Policy is a policy issued at a rate of premium much
lower than the ordinary policies, and by which the beneficiary named therein receives
a specified number, more or less, as may be desired, of annual payments after the
death of the insured.
534 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
For the heads of families, this is one of the best methods of insurance, as by
it the money can neither be lost nor squandered, and the family receives a certain
annual income to protect them from care, to educate the children, etc.
1060. A Non-forfeiting Policy is one which does not become void because of
the non-payment of premiums; provided that two or three annual payments have
been made thereon.
NOTE:-The limited forfeit law of life insurance of Massachusetts, makes the policy non-
forfeiting when two full annual payments have been made; the laws of the State of New York,
make the policy non-forfeiting when three full annual payments have been made.
1061. A Dividend in Life Insurance Companies is a share of the profits or
surplus paid to the stockholders in stock companies, to policy holders in mutual
Companies, and to stockholders and policy holders in mixed companies, after the
Iiabilities of the company are reserved.
1062. The Reserve Fund of a Life Insurance Company is a sum which,
invested at a given rate of interest, together with the maturing premiums on exist-
ing policies will produce a sufficient revenue to pay all obligations as they mature
Or arise.
1063. Surplus is the sum left, after providing for the liabilities, claims and
expenses of the company.
1064. Script is a Certificate entitling the policy holder to certain profits or
Surplus, when payable.
1065. Loading or Margin, is a percentage added to the net premium to defray
expenses and provide for an excess of mortality.
1066. The Reserve of Life Insurance Companies is that part of the premiums
of all policies, with the interest thereon, which is reserved or set aside as a fund for
the payment of policies when they become due.
The law of New York State requires Life Insurance Companies to reserve such a part of the
premiums as will, at 4 per cent interest, amount to a sufficient sum to pay all policies. The State
of Massachusetts specifies 4% per cent interest on the reserve.
1067. Walue of a Policy. “The net value of a policy is the difference
between the net single premium for the sum insured at the age of the policy holder
when the policy is valued, and the present value of all future net premiums calcula-
ted to be received on the life of the party insured. The gross value of a policy is
the difference between the net single premium, as given above, and the present
Value of all future gross premiums to be received on the policy.”
1068. The Surrendered Walue or Market Walue of a policy is commonly
determined by deducting from the reserve 25 to 50 per cent, as a surrender charge,
and to pay the remainder as an equitable surrender value.
Much discussion has followed this method of finding the surrendered value
of policies, to which we refer those interested to the “Principles and Practice of
Life Insurance.”
1069. The Rate of Premium is based mainly upon the probability of life as
shown in the following tables; but, specifically, it involves six elements: 1. The
reserve fund, as above defined. 2. The probable rate of interest on the reserve
* LIFE INSURANCE. 535
fund. 3. The cost of mortality, which is the estimated amount of each insurer's
share of the losses by death each year. 4. The loading, which is the part of the
premium required for expenses. 5. The age of the insured. 6. The period of insur-
ance and the kind of policy.
CO-OPERATIVE OR BENEVOLENT INSURANCE ASSOCIATIONS.
1070. Associations of this kind, of which there are several, as described on
page 509, differ materially in their general features and plan of work, from the
regular Life Insurance Companies. They also differ from one another in Some
respects, while on many points there is a close similarity.
The Knights of Honor, which is one of the leading Co-operative Associations
or Orders of this class, works on the following plan:
Members are admitted by vote into social societies called Lodges, and on the
payment of an assessment of from one to two dollars, according to age, receive a
Benefit Certificate for $2000, These certificates are the equivalent of the policies
in Life Insurance Companies and are payable at the death of the member to the
beneficiary named therein. The continuance of the Benefit Certificate is conditioned
on the prompt payment of all subsequent assessments upon the holder or member.
To replenish the fund for the payment of the Benefit Certificates, assessments
are made at different times on a graded plan according to the age of the members,
which vary in amount from $1 to $1.50. The sum thus raised is paid into the
General Fund called the Widows' and Orphans' Benefit Fund, from which is paid
the Benefit Certificates of all deceased members until the fund is reduced below
$2000. Then another assessment is made and paid, and thus the fund is maintained
and the certificates of deceased members are paid. gº
Very similar principles and methods govern nearly all the Co-operative Insur-
ance Associations, and thus they offer economical advantages to all who desire to
insure, but more especially do they commend themselves to persons of small or limi-
ted means.
By the co-operative plan of insurance, the assessments are collected and paid
over to the general fund by the officers of the societies or lodges, without charge or
commission, and the reserve and surplus, it is claimed, are in the pockets of the
members.
The Knights of Honor issue half rate Benefit Certificates for $1000. Some
of the other Co-operative Associations issue Benefit Certificates ranging from $500
to $5000. f
- The amount of the assessments is determined upon the expectancy of life,
the age of the member when he joins the order, and some other elements.
1071. Debenture. Insurance or Investment, is an investment evidenced by
a written instrument specifying the amount invested, the revenue to be derived
therefrom, the conditions upon which it is received, the manner of payment, etc.
While some Debenture Companies are conducted ethically, and are based upon
correct financial and mathematical principles to ensure their solvency, many are
founded upon a chimerical and delusive basis, and are conducted through fraud and
536 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. ºr
false pretense. Hence all who would patronize Debenture Companies should
critically investigate their Charter, the financial principles upon which they work,
the management, promises, etc. of the company, before they invest in Debenture
Bonds.
1072. From the Manual of the Life Association of America, and the Principles
and Practice of Life Insurance, we collate the following Statistics and the Tables of
Comparative Expectancy of Life and of Mortality.
BASIS OF LIFE INSURANCE.
1073. The duration of human life is governed by a law as regular in its opera-
tion as any other law. While the actual continuance of any individual life is wholly
uncertain, the average duration of a large number of lives can be accurately fore-
told.
Ten thousand persons of ordinary health, at the age of 35 years, will survive
31 years longer, on an average. Consequently, the expectation of life for a person
35 years of age is 31 years. sº
The basis on which the whole structure of life insurance rests is the law of
mortality among the human race. Strange as it may seem, war, pestilence, and
famine cause but slight variations in the operation of this law. In spite of these
destructive agencies, happening as it were by chance, in spite of all the unforeseen
accidents and catastrophies to which human life is exposed; in spite of the absolute
uncertainty of prolonged life in any individual case, yet the average duration of
existence among a large number of persons, say 10,000 or 100,000 is a mathematical
verity—a certainty as fixed and invariable as any other law of nature.
HOW HAS THIS LAW BEEN DISCOVERED AND WHAT ARE THE PROOFS OF IT?—
1074. Within the last 200 years, close observation and actual measurement of
the duration of life have been made among different classes of persons in different
countries, and under entirely different sets of circumstances. These tests, made
independently of one another, extended to embrace large numbers of persons—in
some instances, entire populations—so fully corroborate and verify one another, that
there is left no reason to doubt that the law of mortality is accurately determined,
and that it is in fact an absolute law of nature.
The separate and distinct observations commonly adduced as proof or
authority for this law are thirteen in number. The following are those in general
use in America and England: The American Experience, the Carlisle, the Farr or
English Life No. 3, the Actuaries or Combined Experience, and the Northampton.
1075. The American Eaſperience is made from actual experience of several
leading American Companies, extending over a period of 25 years.
1076. The Carlisle table is a record of the mortality of the town of Carlisle,
England, from 1778 to 1787.
1077. The Farr or English Life No. 3, is a record of mortality of all ages
throughout Great Britain, from 1838 to 1844.
1078. The Actuaries or Combined Experience is a record of mortality based on
the record of 62,537 assured lives in 17 Life Companies in England.
The Northampton Table was prepared by Dr. Price of Northampton, England.
2} LIEE INSURANCE. 537
TABLE NO. 1.
1079. C4OMPARATIVE EXPECTANCY OF LIFE.
A Table showing the expectancy of life in years and hundredths of a year, deduced from four
English and the American Experience Tables.
Farr or |Actuaries |American Farr or |Actuaries|American
Age. North- | Carlisle. English , or Com- || Experi- || Age. North- Carlisle, English , or Qom. Experi-
ampton. Life, İbined Ex-| ence. ampton. Life, [bined Ex- ence.
No. 3. pelience. No. 3. perience.
1. 32.74 44, 67 || 46.65 º e º e º e º º 53 | 16.54 18.97 17.67 | 18.16 | 18.79
2 37.79 || 47.55 || 48.83 tº e º e a e e Q 54 | 16,06 18.28 17.06 17.50 18.09
3 39.55 49.81 || 49.61 tº e º e tº e o º 55 15.58 17.58 16.45 16.86 17.40
4 | 40.58 || 50.76 || 49.81 & © Cº. e. tº e º 'º 56 | 15.10 16.89 15.86 16.22 16.72
5 40.84 || 51.24 || 49.71 tº e º ſo 57 || 14.63 | 16.21 15.26 15.59 || 16.05
6 || 41.07 || 51.16 || 49.39 G e º O 58 || 14.15 15.55 14.68 || 14.97 15.39
7 41.03 || 50.79 || 48.92 tº e º e © e º tº 59 || 13.68 14.92 14.10 14.37 14.74
8 || 40.79 || 50.24 || 48.37 tº e º e - e º & 60 | 13.21 14.34 13.53 13.77 || 14.09
9 40.39 49.57 || 47.74 e - - - tº e º º 61 | 12.74 13,82 12.96 13, 18 13.47
10 39,78 || 48.82 || 47.05 || 48.36 || 48.72 62 12.28 13.31 12.41 12.61 | 12.86
11 || 39.14 || 48.04 || 46.31 || 47.68 48.08 63 || 11.81 | 12.81 11.87 12.05 || 12.26
12 38.49 || 47.27 || 45.54 47.01 47.45 64 11.35 12.30 11.34 11.51 11.68
13 || 37.83 || 46.50 || 44.76 || 46.33 46.82 65 10.88 || 11.79 10.82 10.97 || 11.10
14 37.17 | 45.74 || 43.97 45.64 46. 16 66 || 10.42 11.27 10.32 10.46 10.54
15 || 36.51 || 44.99 || 43.18 || 44.96 || 45.50 67 9.95 || 10.75 9,83 9.96 || 10.00
16 || 35.85 || 44.27 || 42.40 || 44.27 44.85 68 9.50 | 10.23 9.36 9.47 9.48
17 | 35.20 || 43 57 || 41.64 || 43,58 44.19 69 9,05 9.70 8.90 9.00 8.98
18 || 34.58 || 42.87 || 40 90 || 42.88 43.53 70 8.60 9.15 8.45 8,54 8.48
19 33.99 || 42.16 || 40.17 || 42.19 42.87 71 8.17 8.65 8.03 8.10 8.00
20 33.43 || 41.46 || 39 48 || 41.49 42.20 72 7,74 8.16 T.62 7.67 7.54
21 32.90 | 40.75 || 38.80 40.79 41.53 73 7.32 7.72 7.22 7.26 7,10
22 32.39 || 40.03 || 38.13 40.09 40.85 74 6,92 7.33 6.85 6.86 6.68
23 31.87 || 39.31 || 37.46 || 39.39 40. 17 75 6.54 7.00 6.49 6 48 6.28
24 31.36 || 38.58 || 36.79 |- 38.68 39.49 76 6,18 6.99 6.15 6.11 5.88
25 30.85 || 37.86 || 36.12 37.98 38.81 77 5.83 6.40 5.82 5.76 5.48
26 30.33 || 37.13 ig5.44 || 37.27 38.11 78 5.48 6.11 5,51 5.42 5.10
27 | 29.82 36.40 || 34.77 || 36.56 37.43 79 5.11 5.80 5.2.1 5.09 4,74
28 29.30 || 35.68 || 34.10 35.86 36.73 80 4.75 5.51 4.93 4.78 4.38
29 28.79 34.99 || 33.43 35.15 36.03 1. 4.41 5.20 4.66 4.48 4.04
30 28.27 | 34.34 || 32.76 || 34.43 35.33 82 4.09 4.93 4.41 4.18 3.71
31 27.75 33.68 || 32.09 33.72 34.62 83 3.80 4.65 4.17 3.90 3.39
32 27.24 || 33.02 || 31.42 33.01 33.92 84 3.58 4.39 3.95 3.63 3.08
33 26.72 32 36 || 30.74 32.30 33.21 85 3.37 4.13 3.73 3.36 2.77
34 26.20 31.68 30.07 || 31.58 32.50 86 3.18 3.90 3.53 3.10 2.47
35 25.68 31 ()0 || 29.40 30.87 31.78 87 3.01 3.71 3.34 2.84 2.19
36 25, 16 || 30.32 28.73 30.15 31.07 88 2.86 3.60 3.16 2.59 1.91
37 24.64 || 29.63 || 28.06 || 29.44 30.35 89 2.66 3.47 3.00 2.35 1.66
38 24. 12 28 65 27 39 28.72 29.63 90 2.41 3.28 2.04 2.11 1.42
39 23.60 28.27 | 26.72 28.00 28.90 91 2.08 3.26 2.69 1.89 1.19
40 23.07 27.61 || 26.06 || 27.28 28.18 92 1.75 3.37 2.55 1.67 .98
41 22.56 26.97 25.39 26.56 27.45 93 1.37 3.48 2.41 1.47 .80
42 22.04 26.34 24.73 25.84 26.72 94. 1.05 3.53 2.29 1.28 .64
43 21 54 25 71 || 24.07 25.12 25.99 95 .75 3.53 2.17 1.12 .50
44 21.03 || 25.09 || 23.41 24.40 25.27 96 .50 3.46 2.06 .99 tº º º
45 20.52 24.45 22.76 || 23.69 24.54 97 tº e º 3.28 1.95 .89
46 20.02 23.81 22.11 22.97 23.81 98 • Q ſº 3.07 1.85 .75 o
47 19.51 23.17 | 21.46 22.27 23.08 99 2.77 1.76 .50
48 19.00 22.50 20.82 21.56 22.36 100 2.28 1,68 tº dº º o
49 18.49 || 21,81 20.17 | 20.87 21.63 101 1.79 - e º º e
50 17.99 21, 11 | 19.54 20.18 20.91 102 1.30 tº gº
51 17.50 | 20,39 || 18.90 19.50 | 20.20 103 | . . . . .83 & e º º tº $ tº e © e º ºs
52 17.02 || 19.68 | 18.28 | 18.82 | 19.49 104 | . . . . .50 © c e e e - © º O Gº e 9

3.
538 SOULES PHILOSOPHIC PRACTICAL MATHEMATICS. ºf
1080 [TABLE NO. 2].
• COM PA RATIVE MoRTALITY TABLE,-showing the number expected to die out of 1000
a-- persons entering each year, according to three English and the American Experience Tables.
Age. |American Experience. Carlisle. | Actuaries. English Life No. 3. Age.
10 7.49 f • * * * . e e º & tº e º e C -
15 7.63 ū e s tº © e º ºf © e º 'º tº &
20 7.80 tº º tº e tº G & e tº e º e & e
21 7.85 Q & © e. tº - e. e. tº e º 'º tº e
22 7. 90 tº e º e e e & e tº a º e e
23 7.95 tº a tº e
24 8.01 tº º º º tº e º e tº º 4 ºn © º
25 8.06 7, 31 7.77 9.20 25
26 8.13 7, 36 7.88 9.38 26
27 8.19 7.76 8.00 9.55 27
28 8.26 8.69 8. 13 9.74 28
29 8.34 9.82 8.27 9.93 29
30 8.42 10.10 8.42 10. 13 30
31 8.51 10.20 8.57 10.34 31
32 8.60 10. 13 8.74 10.56 32
33 8.71 10.05 8.91 10.80 33
34 8.83 10.15 9.09 11.05 34
35 8,94 10.25 9.28 11.33 35
36 9.08 10.55 9.48 11.62 36
37 9.23 10.85 9.68 11.94 37
38 9.40 11. 16 9.90 12.29 38
39 9.58 11.87 10. 13 12.65 39
4 9.79 13.05 10.36 13,06 40
41 10.00 13.77 10.61 13.48 41
42 10.25 14.37 10.89 13.94 42
43 10.51 14.58 11.25 14.44 43
44 10.81 14.79 11.69 14,97 44
45 11. 16 14.80 12. 21 15.54 45
46 11.56 14.81 12.83 16. 15 46
47 12.00 14, 60 13, 51 16.18 47
48 12.50 13.93 14.25 17.49 48
49 13. 10 13.68 15.06 18.23 49
50 13.78 13.41 15.93 19,02 50
51 14.54 14.29 16, 89 20.42 51
52 15 38 15. 20 17.94 21.45 52
53 16,33 16 14 19.09 22,51 53
54 17.39 16.89 20.31 23.64 54
55 18,57 17.92 21.66 24.85 55
56 19.88 19.00 23.12 26.17 56
57 21.33 20.89 24.67 27.63 57
58 23.10 24 20 26.38 29.25 58
59 24.72 28. 27 28. 24 31.05 59
60 26.69 33.48 30 .33 33.05 60
61 28 88 35, 78 32.61 35.29 61
62 31, 29 37.40 35. 12 37.77 62
63 33.94 38.25 37.83 40.53 63
64 36,87 39.77 40.82 43.60 64
65 40, 12 41 08 44.08 46.98 65
66 43.76 42.50 47.61 50.71 66
67 47.64 44.38 51.47 54.83 67
68 52.00 46.45 55.63 59. 33 68
69 56.76 49. 10 60.08 64.25 69
70 61.99 51.64 64.93 69.62 70
71 67.66 58.85 70.01 72.70 71
72 73.73 68. 12 75.80 78.54 72
75 94.36 | . . . . . . . . . . . . . . . . . . . .
80 144. 46 | . . . . . . . . . . . . . . . . . . . .
85 235. 55 . . . . . . . . . . . . . . . . . . . "
90 454. 55 | . . . . . . . . . . . . . . . . . . . .
94 857.14 ' . . . . . . . . . . . . . " ' ". . . . . .

The above table shows how many of 1000 persons will die in the year, for each age
The foregoing tables and the intºrest and annui
tions of Life Insurance Companies.
determined, but that of a multitude of lives is easily ascertained.
iven.
e principal data for the calcula:
one life can be
of the calcula-
tables, given farther on, furnish º
y life insurers or others, that the duration of an
It is not claimed -
By reason of this, the ultimate resul
tions of life insurers are always correct, for if one person lives longer than those insured with him, he helps those who die
early; and if he dies early, they help him. .
º LIFE INSURANCE. 539
TABLE NO. 3.
1081. SEHOWING TELE ANNUAL PREMIUM RATES
FOR AN INSURANCE OF $1000.
LIFE POLICIES.—PAYABLE AT DEATH ONLY. EnſlºwmºntPOliſles—Payallbasińlltalèſ, OI at DBālliſ VIII, ll
Payments to continue for
- * Twent
Age. Fift T t Age.
ge Life Ten Years º º Tºy Ten Years. º ºy *ś. ge
21 $18.60 $40.50 $30.80 $26.10 $105,40 $66,90 $48.20 $76.30 21
22 19 10 41, 20 31,30 26.60 105.50 67.00 48.30 76.50 22
23 19.50 41.90 31 90 27.00 105,70 67.10 48.40 76.60 23
24 20.00 42.70 32.50 27.60 105.80 67.20 48.60 76.70 24
25 20.50 43.50 33.10 28, 10 105,90 67,40 48.70 76.90 25
26 21 00 44.30 33 80 28 60 106.00 67.50 48.90 77.00 26
27 21.50 45.20 34 40 29.20 106.10 67,60 49.00 77.20 27
28 22.10 46, 10 35 10 29.80 106.30 67.80 49.20 77.40 28
29 22.70 47.00 35 90 30 50 106,40 68.00 49.40 77.60 29
30 23 30 48 00 36.60 31.10 106,60 68.20 49.60 77.80 30
31 24.00 49. 10 37 40 31.80 106.80 68,30 49.80 78.10 31
32 24,70 50.10 38,30 32.60 107,00 68 60 50.10 78.30 32
33 25.50 51.20 39.10 33.30 107.20 68 80 50.30 78.60 33
34 26.30 52.40 40.00 34.10 107.40 69,00 50.60 78.90 34
35 27, 10 53.60 41 00 35.00 107,60 69.30 50.90 79.20 35
36 28 00 54,80 42 ()0 35,80 107.80 69.60 51.30 79.60 36
36 29 00 56 20 43,00 36 80 108 10 69 90 51.70 80.00 37
38 30,00 57 50 44 10 37.70 108.4() 70.20 52.10 80.40 38
39 31, 10 59 00 - 45 30 38 80 108.70 70,60 52.50 80.90 39
40 32,20 60 40 46.50 39.80 109,10 71,00 53.00 81.40 40
41 33 40 62.00 47 70 41.00 109,40 71,50 53.60 81.90 41
42 34,70 63.60 49 00 42.20 109,80 72.00 54.20 82.60 42
43 36. 10 65.30 50.40 43.50 110 30 72,50 54.80 83.30 43
44 37.50 67.10 51.90 44.80 110,80 73 10 55.60 84.00 44
45 39.10 69.00 53 40 46.20 111.30 73.80 56.40 84.90 45
46 40,70 70.90 55 10 47.80 112.00 74.60 57.30 85.80 46
47 42.50 72.90 56 80 49.4() 112,60 75.40 58.30 86.80 47
48 44.40 75.10 58.60 51.10 113.40 76 30 59.40 88.00 48
49 46.40 77.30 60 50 52,90 114.20 77.30 60.70 89.20 49
50 48.50 || 79.60 62.50 54.80 115.10 78.40 62.00 90.60 50
51 50.80 82.10 64.60 56.90 || 116.10 79.70 63.50 92.10 51
52 53 30 84,60 66 90 59,10 117.20 81.00 65.20 93.70 52
53 55 90 87.30 69 20 61.40 118.40 82.50 67.00 95.50 53
54 58 70 90.10 71.80 63.90 119,80 84, 10 68.90 97.50 54
55 61, 60 93,00 74.40 66,60 121,20 85.90 71.10 99.60 55
56 64.80 96.10 77.30 69.50 | 122,80 87.90 73.50 101.90 56
57 68.20 99 30 80 30 72,60 124.60 90.10 76.10 104.40 57
58 71.80 102.70 83 50 75.80 126.50 92.50 78.90 107.10 58
59 75.70 106 30 86.90 79.40 | 128.70 95.10 82.10 110.10 59
60 79.90 110.10 90.60 83.20 131.00 98.00 85.50 113.20 60
61 84.30 114.10 94 50 87.30 133.60 101.20 ! . . . . . . . . . . . . . 61
62 89.10 118 30 98 70 91.70 136,40 104.70 | . . . . . . g 62
63 94.20 122.70 103 20 96.40 139,60 108.50 | . . . . . . . . . . . . . 63
64 99 60 127 50 108 00 101.50 143,00 112.70 e e º e º e e e º e º 'º 64
65 105.50 132.50 113.20 107.10 146,80 117.30 . . . . . . . . . . . . . 65

Not E 1.-The above table is taken by permission from THE MUTUAL LIFE INSURANCE COM-
PANY of New York.
NotE 2.-Policies which do not participate in the dividends of a company are issued at a
lower rate of premium, than given in the above table.
NOTE 3.-The above rates of premium are for annual payments. In some cases the payments
are made semi-annually or quarterly, when a small per cent is added to the annual payment before
dividing by 2 or 4.
54o SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
PROBLEMIS.
1. What is the annual premium for a man, aged 21 years, on a Life Policy
for $5000 % Ans. $93,00.
OPERATION.
sisº, Explanation.—The premium on $1000 at the age of 21 is, as
+- shown in the table, $18.60; hence, for $5000, it is 5 times
$93.00 Ans. as much.
2. What is the annual premium on a 10-year Endowment Policy for $5000,
the age being 21 years? Ans, $527.00.
OPERATION.
$105,40
5 Explanation.—Since the premium on $1000 for the con-
-º-mºmºmºsº- ditions named is $105.40, for $5000 it is 5 times as much.
$527.00
3. A man is 32 years old and takes out a 20-year Endowment Policy for
$25000. What is his annual premium ? Ans, $1252.50.
4. A person, 28 years old, took out a 15-year Endowment Policy for $2500.
What is the annual premium ? Ans. $169.50.
5. What is the annual premium on a Life Policy for $8500, payable in ten
annual payments, the insured being 44 years of age? Ans. $570.35.
6. A man, 21 years old. took out a 10-year Endowment Policy, and at its
maturity, the premiums paid amounted to $5270. What was the amount of the
policy 3 Ans. $5000.
OPERATION. Explanation.—Since the premium for 10 years was $5270,
for one year it is the ºr part, $527. Then since the annual
$5270 - 10 = $527 payment on a $1000 ten-year Endowment Policy is, per
table, $105.40 for a person aged 21 years, there were as many
527 -- 105.40 = 5 thousands thousands insured as $537 is equal to 'siojºſo, which is 5
5 × $1000 = $5000 times. Hence there was $5000 insurance.
7. Mr. X, aged 22 years, took out a Life Policy payable in 15 years. After
making eight annual payments amounting to $876.40, he died. What was the face
of his policy 3 Ans. $3500.
8. A man, 21 years old, paid $93.00 premium annually, on a policy of $5000.
What was the rate per cent? Ans. 1.86%.
OPERATION INDICATED.
93 100%
5000 || 100 or, 5000 || 93
9. The annual premium on a $25000 Endowment Policy, payable in twenty
years, for a person aged 32, is $1252.50. What is the rate per cent? Ans. 5.01%.
10. A man, 28 years old, paid an annual premium of 6.78 per cent on a 15-
year Endowment Policy of $2500. What was the annual premium, and how much
would he pay in all ? Ans. $169.50 annual premium.
$2542.50 total premium.
$2500 x 6.78% = $169.50.
OPERATION INDICATED. }; × 15 = $2542.50.
11. A man, 25 years old, took out a Life Policy for $100000 and paid the
regular annual premiums until his death, at the age of 69 years. How much did he
pay in premiums? AnS. $90200.
LIFE INSURANCE.
541
TABLE NO. 4.
d
1082. AMERICAN EXPERIENCE TABLE OF MORTALITY.
Wumber Expecation mber Number IExpectation
Age. Nº. * º Age. *: Dying. . Life.
10 100000 749 48.72 53 66797 1091 18.79
11 99251 746 48,08 54 65706 1143 18.09
12 98505 743 47.45 55 64563 1199 17.40
13 97.762 740 46.80 56 63364 1260 16.72
14 97022 737 46.16 57 62104 1325 16,05
15 96.285 735 45.50 58 60779 1394 15.39
16 95550 732 44.85 59 59385 1468 14.74
17 94818 729 44.19 60 57917 1546 14.10
18 94089 727 43.53 61 56371 1628 13,47
19 93362 725 42.87 62 54743 1713 12,86
20 92637 723 42.20 63 53030 1800 12.26
21 91914 722 41.53 64 51230 1889 11.67
22 91.192 721 40.85 65 49341 1980 11.10
23 , 90471 720 40.17 66 47361 2070 10.54
24 89751 719 39.49 67 45291 2158 10,00
25 89.032 718 38.81 68 43.133 2243 9.47
26 88314 718 38.12 69 40890 2321 8.97
27 87596 718 37.43 70 38569 2391 8.48
28 86878 718 36.73 71 36178 2448 8.00
29 86160 719 36.03 72 33730 2487 7.55
30 85441 720 35.33 73 31243 2505 7.11
31 84721 721 34.63 74 28738 2501 6.68
32 84000 723 33.92 75 26237 2476 6.27
33 83277 726 33.21 76 23761 2431 5.88
34 82551 729 32.50 77 21330 2369 5,49
35 81822 732 31.78 78 18961 2291 5,11
36 81090 737 31.07 79 16670 2196 4.74
37 80353 742 30.35 80 14474 2091 4,39
38 796.11 749 29.62 81 12383 1964 4,05
39 78862 756 28.90 82 104.19 1816 3.71
40 78106 765 28.18 83 8603 1648 3.39
41 7734.1 774 27.45 84 6955 1470 3.08
42 76567 785 26.72 85 5485 1292 2.77
43 75782 797 26.00 86 4193 1114 2.47
44 74985 812 25.27 87 3079 933 2.18
45 74.173 828 24.54 88 2146 744 1.91
46 73345 848 23.81 89 1402 555 1.66
47 72497 870 23.08 90 847 385 1.42
48 71627 896 22.36 91 462 246 1.19
49 70731 927 21.63 92 216 137 .98
50 69804. 962 20.91 93 79 58 .80
51 68842 1001 20.20 94 21 18 ,64
52 67841 1044 19.49 95 3 3 .50
NoTE 1.-This table is adopted by the State of New York and several States, in estimating
life endowments.
NOTE: 2.—Massachusetts has adopted the TABLEs prepared by DR. DIGGLEsworth, for
estimating life estates.
NOTE 3.-The CARLISLE TABLES prepared by MILNE are generally used in England.
542 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. *
THE PROBABILITY OF HUMAN LIFE.
1083. The probability of life is the likelihood which a person has, according to the table of
mortality, of living a certain number of years. The probability of death, is the likelihood a
person has of dying within a given period. As it is absolutely certain that a person will always
be dead or alive, the probability of death is found by subtracting the probability of life from
unity, and vice versa.
PROBLEM.
1. What is the probability that a person, aged 30 years, will live to be 50 years old, according
to the figures of the American Experience Table 3 Ans. .81698.
OPERATION.
According to the table, the number living, of the 100000, at the age of 30, is 85441; and the
number living, at the age of 50, is 69804; consequently, 15637 have died.
The probability of any member surviving for 20 years, will-be expressed by the fraction
of which 69804 is the numerator, and 85441 is the denominator; and that the probability of dying
will be expressed by the fraction in which 15637 is the numerator and 85441 the denominator.
Thus ##### = .81698, the probability of living.
And ##### = .18302, the probability of dying.
It will be observed that these amounts stand to each other as complement and supplement.
Hence, when we have one of the amounts we can find the other by subtracting it from the unit.
Thus 1 — .81698 = .18302; and 1 — .18302 = .81698.
NOTE.—As there are 85441 chances in all of the persons aged 30 attaining to 50 years, and as the probability is that
only 69804 will live, the contingency that each individual will live, is .81698, and hence the contingency of his dying
is .18302
2. A man 25 years old pays $140.50 annual premium on a 20 year policy of $5000. What
would this annual payment be worth to the company if continued for 20 years, and the company
put the payments at 4% compound interest ? Ans. $4183.82.
Operation by the Annuity Table, page 925.
$140.50 × 29.778079 = $4183,82.
3. Suppose in the above problem that the payment of $140.50 was continued for 20 years
and was annually invested at 4% compound interest, what would be the value of the investments
at the expectancy of the policy holder's life, based upon the American Experience Table, (38.81
years)?
OPERATION.
First step by the Annuity Table, page 925.
$140.50 × 29,778079 = $4183.82 = value of the annual payments or annuity for 20 years.
38.81 years — 20 years = 18.81 years, which is the time the value of the Annuity $4183.82,
was at compound interest at 4%.
Second step by C. I. Table, page 611.
$4183.82 × 2.02581652 = $8475.65 = compound amount for 18 years.
Int, on $8475.65 at 4%, .81 years 274.61
$8750.26 Ans.
For further information regarding Life Insurance, see “The Principles and Practice of Life
Insurance,” or consult the officers of some Life Insurance Company. For Co-operative Insurance,
consult some member of a Co-operative Association.
nterest.
-----------------N
ƺ º Aº.
* ur-er
1084. Interest is defined by many political economists to be the sum the
borrower pays for the use of money. But this definition is opposed by the more
acute and critical political economists, who define interest to be the share of the
product obtained by the owner of capital.
1085. The Rate of Interest is the per cent paid for the capital or money
borrowed, for a specified time.
It was formerly believed that the rate depended upon the abundance or scar-
city of capital in the money market at any given time. But this opinion has been
proven to be erroneous. The rate is now shown, by our best economists, to be
governed by the productiveness or profits of capital employed in the various lines
of business.
MILL says, after thoroughly arguing the case, “it is an error to imagine that
the rate of interest bears any necessary relation to the quantity or value of money
in circulation.”
RICARDO says, that “the rate of interest is not regulated by the rate at which
the bank will lend, but by the rate of profit which can be made by the employment
of capital, and which is totally independent of the quantity and of the value of
money.” -
NOTE.-The value of money is its purchasing power.
But independent of this general law based upon profits, the rate is sometimes
affected by the greater hazard that is incurred by loaning to less responsible parties
or on less certain security, and also by reason of the difficulties that sometimes
attend the transactions of loan, because of the legal limit placed upon the rate,
called Usury Laws.
Again, the rate is influenced by the duration of the loan. Money loaned for a
short period, commands a higher rate than money loaned for a long period. And
call loans bear a lower rate than loans made for a definite period of time.
HISTORY OF INTEREST.
1086. For upward of thirty centuries, the mind of man has been engrossed
with this subject, and yet no unanimity of opinion has been reached, as regards the
individual rights of men who loan and borrow money, to fix the rate of compensa-
tion to be paid to the owner or lender thereof. In the early ages, the charging and
receiving of compensation for the use of money was called usury, and was denounced
(543)
544 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. º:
as unjust. This mistaken and prejudicial view undoubtedly arose from a misunder-
standing or misinterpretation of some of the enactments of the Mosaic laws, which
read as follows: “Thou shalt not lend upon usury to thy brother; usury of money,
usury of victuals, usury of anything that is lent upon usury; unto a stranger thou
mayest lend upon usury; but unto thy brother thou shalt not lend upon usury.”
(Deuteronomy, XXIII, 19, 20.) And, “If thou lend money to any of my people that
is poor by thee, thou shalt not be to him a usurer, neither shalt thou lay upon him
usury.” (Exodus, XXII, 25.) And again, in speaking of a poor brother, “Thou
shalt not give him thy money upon usury, nor lend him thy victuals for increase.”
(Leviticus, XXV, 37.)
These laws are specific and direct in their application, and do not forbid the
charging of usury except to poor brothers. For ages, the followers of Moses and
the disciples of his teachings were the money lenders of Europe for usury; and by
reason of this, and the fact that they were as a class shrewd, trafficking, and a
money dealing people, is to be attributed the unjust prejudice, that in a measure to
this day exists against them and others who follow the profession of money lending.
ARISTOTLE, the Grecian philosopher, gravely contended that, as money could
not beget money it was barren, and usury should not be charged for its use.
In this regard, as in many others, the old philosopher did not philosophize
aright. He did not see that with money the borrower could add to his flocks, his
fields and his merchandise, and thus profit by the increase of all.
The Greeks were not, however, governed by the opinion of Aristotle, for in
his age, they paid an average rate of about 30 per cent on the capital employed in
the shipping trade, and about 12 per cent for capital employed in business on land.
The Roman Law allowed interest or usury, and during the days of the repub-
lic, the rate of interest or usury was very high. This resulted mainly from the fact
that the debtors, the plebeians, repeatedly threatened and sometimes made efforts
to deprive their creditors, generally the patricians, not only of the interest on their
capital, but also of the capital or principal itself.
But from the time of the 4th century, the writings of the Romish Church
Fathers and the Canon Law prohibit interest. This prohibition was based upon the
Mosaic law as above quoted and upon the Sophistry of Aristotle.
The council of Vienna, Austria, in 1311, declared any defense of interest to
be heresy.
In England, from the earliest period of history to the reign of Henry VIII,
interest was forbidden to all persons except to Jews and foreigners, and they were
often plundered to enrich the crown, on the miserable pretext of punishment, for
what the government called their “hellish extortions.”
But the opposition of the people to this ruinous interdiction against interest,
by the government, resulted in a statute passed in 1546, allowing interest to the
extent of 10 per cent per annum. Yet, in the reign of Edward VI, in 1552, the
government again prohibited the taking or charging of any interest, and passed an
INTEREST. 545
act declaring that, taking interest was “a vice most odious and detestable” and
“ contrary to the word of God.”
On the passage of this act, interest rose to 14 per cent and continued at this
rate until 1571, when an act was passed repealing the act of Edward VI, and re-en-
acting that of Henry VIII, allowing 10 per cent interest.
The word interest was now substituted for the word usury, and the meaning
of usury was changed to mean the rate per cent charged in excess of the legal rate.
By the middle of the 17th century, nearly all the governments of Europe
followed England in abolishing the Usury Laws, and establishing fixed rates. But
as the fixed rates were evaded just as the prohibition had been evaded before, and
as the legally fixed rates resulted in making the rate of interest higher than it
would have been without such laws, all European States have since abolished Usury
Laws.
In 1839, England abolished Usury Laws on commercial paper running less
than twelve months; in 1850, on all capital except loans on real estate; and in
1854, they were abolished altogether.
The United States stands almost alone in retaining Usury Laws. The legal
rate of interest in the National Banking Law is 7 per cent, in States where no rate
is fixed. In States where the State Laws fix the rate per cent, the National Banks
are to be governed accordingly.
See Table of Interest Laws of the different States and territories, page 548,
for the various rates per cent allowed, penalties for usury, and days of grace.
EVIL EFFECTS OF USURY LAWS.
1087. Usury Laws, like other laws fixing prices, are not only ineffectual, but
they prevent all persons, whose credit is not first class, from obtaining loans from
capitalists of the highest character, and force borrowers to secure loans from those
who are less scrupulous and who are also compelled to charge above the current
rate a premium sufficient to compensate the lender for the extra risk and odium
incurred in violating the USury Law.
Usury Laws undertake to prevent what is not an evil, and hence what should
not be interfered with ; consequently, they are productive of injury instead of bene-
fit. Who would pretend that it is just for the legislature to compel underwriters to
insure extra hazardous or specially hazardous properties, on the same terms or rate
per cent as they write simply hazardous risks. And yet such a law would be no
more absurd and unjust than the law of usury which requires that the rate per
cent of interest shall be the same on capital loaned to those whose Security or
character are widely different.
Again, it is impossible to make the legal rate a fair rate; for the market rate
is continually changing, and independent of the hazard of the risk, a rate that
would have been fair at one time would be too low at another time, and perhaps
extortionate at another time. A loan for a short period commands a higher rate
546 soul.E's PHILOSOPHIC PRACTICAL MATHEMATICS. Yºr
than for a long period; and a rate that is fair for a short period would be unfair for
a long period. The only way a fair rate can be determined is by competition in the
money market, when all the elements of supply and demand, of risk, of time, and
of profit, are duly considered by the borrower and the lender.
BORROWERS AND LENDERS.
1088. When the Usury Laws were passed, the poorer classes were the bor-
rowers and the rich were the lenders, and hence the Usury Laws, were enacted
mainly with a view to protect the poor, against high rates of interest.
But at the present time, the borrowers are largely the rich men, and the rich
corporations, and a large part of the lenders, are the poorer or medium classes.
Hence, considering the reason for the passage of the Usury Laws, it follows logic-
ally that they should be now abolished.
One evidence of this change of borrowers and lenders is the fact that in 1894,
three million depositors, mostly poor people, had nearly 1,300,000,000 dollars on
deposit in the savings banks. Another evidence is the fact that hundreds of mill-
ions of dollars of bonds of various kinds of corporations, of States, and of nations,
are held by the poorer classes.
Considering the Usury Laws in all their bearings, the conclusions are as
follows:
1. They are unwise and unjust. 2. They are practically inoperative. 3. They
are evidence of financial inexperience on the part of the law makers. 4. They are
violative of the rights of borrowers and lenders, and of the ethical principles of
business. 5. They should be repealed. -
That a law regulating interest in cases of judgments, successions, and trans-
actions in which no contracts of interest are made, is necessary, there is no question
of doubt. But further than this, the question of interest should be left to the judg-
ment of the contracting parties.
DEFINITIONS.
1089. Interest is the sum paid for the productiveness of capital.
INTEREST is calculated on a scale of a certain number of units on every $100
for 1 year, and hence it combines per cent and per annum, from the Latin per centum,
on a hundred, and per annum, for a year.
1090. The Principal is the sum of money or capital on which interest is
charged or paid.
1091. The Rate of Interest is the per cent paid on the principal, for its use,
for a specified time.
1092. Time is the period for which the principal is loaned or bears interest.
2} INTEREST. 547
1093. The Amount is the sum of the principal and the interest.
1094. The Proceeds is the difference between the principal and the interest.
The subject of interest is classified into several divisions, viz: Simple Interest,
Annual, Semi-Annual, and Quarterly Interest, and Compound Interest, which we
will present in the order here given.
1095. Simple Interest is the interest on the principal unincreased by inter-
est, however long overdue.
Annual, Semi-Annual and Quarterly Interest, is Simple Interest, and the interest
On the Simple Interest from the time it becomes due until paid. . e
NOTE.-Annual, Semi-Annual and Quarterly Interest is allowed by some States on notes and
contracts where the interest is Payable Annually, Semi-Annually or Quarterly.
Compound Interest is interest on both the principal and the interest, i. e. it is interest where
the principal is increased at the expiration of each period of interest payment by the interest on
the principal for such period, whether for 12, 6, 3, or 1 month; or for one day.
NOTE.—Compound Interest is allowed in some States, under certain conditions.
• 1096. Legal Interest is the rate per cent fixed by the law of each state, to
apply when no agreement is made. In Louisiana, it is 5 per cent.
1097. Conventional Interest is the rate per cent agreed upon by the parties
concerned. The law of many of the States places a limit to this interest. In
Louisiana, the limit is 8 per cent.
1098. Usury is a higher rate per cent than the law allows. The law of dif.
ferent States prescribes different penalties for usury. In Louisiana, the penalty is
the forfeiture of all interest.
The Principal, the Interest, the Rate, the Time, and the Amount or Proceeds,
constitute the five quantities involved in interest questions; and when any three of
these are given, the others may be found. Hence there are five classes of interest
questions.
548
SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS.
1099.
TABLE OF INTEREST LAWS.
The following table shows the Interest Laws of the different States and Territories compiled from
the State and Territorial Statutes of 1891.
Rate allowed
States and Territories. *...* º §* Penalties for Usury Gº; Iſºſ)
writing. |
PEIR CFNT: PER CENT |
ALABAMA, . . . . . . . . . . . . . . . . 8 8 Forfeiture of all the Interest. . . . . . . . . . Grace
ARIZONA, . . . . . . . . . . . . . . . 7 Any rate |None . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Sº,
ARKANSAS, . . . . . . . . . . . . . . . 6 10 Forfeiture of principal and interest. ... No statute.
*CALIFORNIA, . . . . . . . . . . . . 7 * | Any rate |None . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . No grace.
COLORADO, . . . . . . . . . . . . . . e 8 Any rate |None . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Grace.
CONNECTICUT, . . . . . . . . . . . . 6 6 None . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Grace.
DAKOTA, NORTH... . . . . . . . 7 12 Forfeiture of excess. . . . . . . . . . . . . . . . . . . Grace.
& 4 SOUTH . . . . . . . . . . 7 12 Forfeiture of all the interest. . . . . . . . . . Grace.
DELAWARE, . . . . . . . . . . . . . . 6 6 Forfeiture of principal and interest. ...|Grace,
DIST. OF COLUMBIA, . . . . . . 6 10 Forfeiture of all the interest. . . . . . . . . Grace.
FLORIDA, . . . . . . . . . . . . . . . . . 8 Any rate |None. . . . . . 6 tº e º is e a e º a tº e º 'º e º & e s tº e º e ..|No statute.
GEORGIA, . . . . . . . . . . . . . . . . 7 8 Forfeiture of excess of interest. . . . . . . . Grace.
flipAHO, . . . . . . . . . . . . . . . . . . 10 18 Forfeiture of thrice the excess of int.. No grace.
ILLINOIS,. . . . . . . . . . . . . . . . . 6 8 Forfeiture of all the interest. . . . . . . . . . Grace.
INDIANA, . . . . . . . . . . . . . . . . e 6 8 Forfeiture of excess of interest........ Grace.
INDIAN TERRITORY, ...... 6 6 Forfeiture of principal and interest....|Grace.
IOWA, . . . . . . . . . . . . . . . . . . . e 6 8 Forfeiture of 10 per cent on amount....|Grace.
KANSAS, . . . . . . . . . . . . . . . . . . 6 10 Forfeiture of excess of interest. . . . . . . . Grace.
KENTUCKY, . . . . . . . . . . . . . . 6 6 Forfeiture of excess. . . . . . . . . . . . . . . . . . . [Grace.
fLOUISLANA, . . . . . . . . . . . . . . 5 8 Forfeiture of all the interest. . . . . . . . . . Grace.
MAINE, . . . . . . . . . . . . . . . . . . . 6 Any rate |None . . . . . . . . . . . . . . . e - a e e e s e º e = * * * * * * * * Grace.
MARYLAND, . . . . . . . . . . . . . . 6 Forfeiture of excess of interest. . . . . . . . Grace.
MASSACHUSETTS, ... . . . . . . . 6 Any rate |None . . . . . . . . . . . ... . . . . . . . . . . . . . . . . . . . Grace.
MICHIGAN, . . . . . . . . . . . . . . . 6 10 Forfeiture of excess of interest. . . . . . . . Grace.
MINNESOTA, . . . . . . . . . . . . . . 7 10 Forfeiture of excess over 10 per cent...|Grace.
MISSISSIPPI, . . . . . . . . . . . . 6 10 Forfeiture of all the interest. . . . . . . . . . Grace.
MISSOURI, . . . . . . . . . . . . . º 6 10 Forfeiture of all the interest. . . . . . . . . . Grace.
MONTANA, . . . . . . . . . . . . . . . . 10 Any rate |None . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... No grace.
NEBRASKA, . . . . . . . . . . . . . . . 7 10 Forfeiture of interest and cost.........|Grace,
NEVADA, . . . . . . . . . . . . © e º O e 7 Any rate |None . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Grace.
NEW HAMPSHIRE, . . . . . . . . 6 6 Forfeiture of thrice the excess. . . . . . . . . Grace.
NEW JERSEY, . . . . . . . . . . . . 6 6 Forfeiture of all the interest. . . . . . . . . . Grace.
NEW MEXICO, . . . . . . . . . . º 6 12 $100 fine. . . . . . . . . . . . . . . . . . . . . . . . . . . ... Grace,
|NEW YORK, . . . . . . . . . . . . . 6 6 Forfeiture of principal and int. and
fine not exceeding $1000, or imprison-X |No grace.
ment not exceeding 6 months or both
NORTH CAROLINA, . . . . . . . . 6 8 Forfeiture of all the interest, etc. . . . . . Grace.
OHIO, . . . . . . . . . . . . . . . . . . 6 8 Forfeiture of excess above 6 per cent. . Grace.
OKLAHOMA TERRITORY,... 7 10 Forfeiture of all the interest. . . . . . . . [Grace.
OREGON, . . . . . . & © tº º e º 'º e º º º 8 10 Forfeiture of principal and interest....|Grace.
§PENNSYLVANIA, . . . . . . . . . 6 6 Forfeiture of excess of interest. . . . . ... Grace.
RHODE ISLAND, . . . . . . . . . . . 6 Any rate |None . . . . . . . . . . a e s m = e º e º e º e º e © e º e º e - © tº e Grace.
SOUTH CAROLINA, . . . . . . . . 7 8 Forfeiture of all the interest. . . . . . . . ... Grace.
TENNESSEE, . . . . . . . . . . . . . . 6 6 Forfeiture of excess of interest. . . . . . . . Grace.
TTEXAS, . . . . . . . . . . . . . . . . . 8 12 Forfeiture of all the interest. . . . . tº e º 'º º º Grace.
UTAH, . . . . . . . . . . . . . . . . . . . tº 8 Any rate |None . . . . . . . . . . . . . . . . . . . tº 6 & e º ſº tº * * . . . . . Grace.
VERMONT, ..... tº e s tº e a t e e º a 6 6 Forfeiture of excess of interest. . . . . ...|Grace.
VIRGINIA, . . . . . . . . . . . . . . . . 6 6 Forfeiture of excess over 6 per cent....|Grace.
WASHINGTON, . . . . . . . . . . . . 10 Any rate |None . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . [Grace.
WEST VIRGINIA, . . . . . . . . . . 6 6 Forfeiture of excess of interest. . . . . . . . Grace.
WISCONSIN, . . . . . . . . . . . . . . 7 10 Forfeiture of all the interest. . . . . . . . ... Grace.
WYOMING, ................ 12 Any rate |None . . . . . . . . . . . . . . . . . . . . e e e s e s a e º 'º º e º 'º Grace.
*In California, compound interest is allowed by written contract.
Written contract.
e collected provided it be added to the principal and by a new contract made
iIn Idaho, compound interestls allowed b
†In Louisiana interest upon interest may
a new debt. On Bottomry and Respondentia Loans, a hi
§
lected provided it be embodied in the instrument as discount.
|In New York, on demand loans of $5000 and upward, which are secured by collaterals, any rate of Interest may be
agreed upon in Writing,
In
ennsylvania, Commission Merchants may contract for 7 per cent.
In Texas, open accounts draw 8 per cent interest from the 1st day of January after the same are made.
gher rate than 8 may be collected. Also a higher rate may be col-
* PROMISSORY NOTES AND NEGOTIABLE PAPER. 549
PROMISSORY NOTES AND NEGOTIABLE PAPER
* Ah, a
~ wºr-w
1100. A Negotiable Promissory Note is an unconditional promise in writing,
to pay to the order of a certain party, or to bearer, at a certain time, a specified sum
of money.
The following is the usual form of a negotiable promissory note:

NEW ORLEANs, March 21, 1895,
Thirty days after date, for value received, I promise to pay to the order
of P. W. Sherwood, Five Hundred Dollars.
H. JONES.
Due April 20/23, 1895.
1101. Parties to a Note. The original parties to a note are those whose
Inames appear in the paper, when it is made.
Subsequent parties are those who become interested in the paper after it is
made. In the above note, H. Jones is the promissor, maker and payor. P. W. Sher-
wood is the promissee, payee, holder, and first indorser. The holder of a note is the
party who owns it. The first indorser of a note, is he to whom it is made payable,
whether his name appears first on the back of the note or not. Parties whose
names are written above that of the first indorser are sureties, and like the maker,
are responsible without protest.
If, in the above note, the words “the order of,” before the name of the payee, had been
omitted, it would have been unnegotiable.
No particular form of words is necessary to form a valid promissory note. The omission of
the words “value received" does not render the note invalid, but in the hands of the original
holder it would be necessary, if required by the maker, to prove that value had been received,
A promissory note bears no interest until maturity unless specified, and in either case unless
the per cent is specified, it will bear only the legal rate.
1102. Negotiable Paper is that which may be transferred from one owner to
another by assignment or indorsement.
There are several kinds of negotiable paper, namely: Promissory Notes, Bills of Exchange,
Due Bills, Bank Notes, Checks on Banks or Bankers, Coupon Bonds, Certificates of Deposit, Let-
ters of Credit, and Bills of Lading. -- e e -
Negotiable paper that does not specify the time of payment, is payable immediately, and if
the place of payment is not specified, it is payable wherever 1t is held at maturity. When the
plače of payment is specified, it must then be presented at such place the day that it legally
matures. The customs and laws of the place in which negotiable instruments are made payable,
always apply.
1103. The Face of a note, draft, etc., is the sum specified or named therein
and promised to be paid.
1104. The Maturity of a note is the day that the note becomes due.
1105. Days of Grace are days allowed for the payment of a note, draft, etc.,
after the expiration of the time specified in the instrument. By custom in most of
55o SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. * .
the States 3 days of grace are allowed on all notes, drafts, etc., that are not drawn
without grace, at sight or on demand.
NOTE.-In some States sight drafts are accepted, and bear three days of grace.
1106. Dishonoring a note is the failure to pay it when due.
1107. The Protesting of a note is a notice to the indorsers that it is due, that
payment has been demanded and refused, and that the holder looks to them for the
payment. The object of protesting a note is to Secure the responsibility of the
indorsers. Protests are generally made by a notary,
1108. Discount Day is the day that a note, draft, etc., is discounted. Many
bankers and business men when discounting notes, etc., charge interest for this day.
1109. Discounting Notes, etc., consists, according to the custom of bankers,
in calculating the interest on their face for the unexpired time including 3 days of
grace, and in some communities discount day, at some specified rate per cent, and
deducting the same from the face.
1110. The Proceeds or Cash Walue of Notes, etc., is what remains after
the interest or discount is deducted.
INDORSEMENTS ON PROMISSORY NOTES AND BILLS,
1111. An Indorsement is anything written on the back of a note or bill,
which pertains to the payment thereof.
1112. An Indorser is the party who writes his name on the back of a note
or bill. The first indorser on a note, is he to whom the note is made payable. The
first indorser on an accepted bill, is the Drawer.
- A note or bill is not negotiable unless it is indorsed by the payee, except in the case of a
note which is made payable to the bearer.
Indorsements vary in form as follows: 1. Blank. 2. In full. 3. Qualified. 4. Restrictive.
5. Conditional. 6. Protest and Notice of Protest Waived.-
The following elucidates the above indorsements, J. B. Anderson being the
indorser:
1. (Indorsement in Blank). (Second form of Qualified 5. (Conditional Indorse-
J. B. Anderson, Indorsement). ment).
Pay to the order of A. B.
2. (Indorsement in Full). J. B. Anderson, Ray to A. B., when he shall
Pay to the order of A. B. Agent. have completed the house
J. B. Anderson, he is now building for me.
4. (Restrictive Indorsement). J. B. Anderson.
3. (Qualified Indorsement). Pay to A. B. only.
Pay to the order of A. B., J. B. Anderson. 6. (Protest Waived).
without recourse to me. or thus:
J. B. Anderson. Pay to the Fourth National Protest and Notice of Pro-
or thus: Bank of St. Louis, for my test waived.
Without recourse. account. - J. B. Anderson.
J. B. Anderson. J. B. Anderson.
NoTE.—When indorsing notes, bills, or checks, write your name on the back in the same
way as it is written on the face. If your name is wrongly spelled on the face, then you should also
mis-spell it in the indorsement. Or you may both mis-spell it as it is on the face of the paper, and
then write it correctly.
#: MISCELLANEOUS COMMERCIAL INSTRUMENTS OF WEITING. 55I
DISCOUNTING NOTES AND ACCEPTANCES.
1113. It is the custom of the banks and the business men of New Orleans,
When discounting notes, to count the actual unexpired days, including one
DISCOUNT DAY; i. e., the day the note is discounted.
REMARKS. In 1904, the Legislature of Louisiana adopted the New York
Commercial Instrument Law, thus making nearly half the States of the Union
which have adopted this Law, up to December 1904.
This Law does not allow days of grace on Commercial paper. Hence when
maturing notes and bills, and when discounting or computing interest on them in
this book, and in states which have adopted this Law, consider the notes to be
drawn for 3 days more than the time specified therein, except such notes as may
be drawn without grace.
NOTE. —For a full discussion and elucidation of the foregoing indorsements of accommoda-
tion paper, acceptance or payment for honor, treatment of accommodation paper, renewals or
extension of notes, and an extended presentation of commercial instruments of writing, see Soulé's
New Science and Practice of Accounts, pages 25 to 39 and 199 to 201.
MISCELLANEOUS COMMERCIAL INSTRUMENTS OF WRITING,
-º-eº. Pe-º-
FIECEIPTS.
1114. 1. Receipt in full where payments have been previously made.
NEw ORLEANs, November 6, 1895.
Received from H. J. Calvert, Three Thousand Dollars, Two Thousand
having been previously paid, making in all, Five Thousand Dollars,
R. W. ABBOTT.
2. Receipt for Rent.

NEW ORLEANs, December 4, 1895.
Received from W. A. Beer, Fifty Dollars, for rent of house No. 457
Baronne Street, for November, 1895.
E. R. GURLEY.
3. Receipt through a Third Party.

NEW ORLEANs, January 3, 1895.
Received from R. B. Montgomery, through F. A. Golden, Twenty-three
Hundred, Forty-seven and 3% Dollars, in full of all demands to date.
GEO. B. MUSE.
NoTE.—The words in italics have no money value whatever. A receipt is evidence for the
amount named therein, and no words can make it worth any more. Hence, the words in italics
are useless, save as general information.
552 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs.
4. Receipt to apply as an Indorsement on a Note.

NEW ORLEANs, November 16, 1895.
Received from Geo. B. Brackett, Three Hundred Dollars, the same to
apply on his note of One Thousand Dollars.
T. C. HENRY.
NOTE.-Money paid on account of notes should be, when practicable, indorsed on the back
of the note. The above payment thus indorsed would read as follows: Received on the within note
Three Hundred Dollars.
NEW ORLEANs, November 16, 1895. 4. T. C. HENRY.
N O T E S .
—-º-
5. An Individual Note.

NEW ORLEANs, July 16, 1895.
Sixty days after date, for value received, I promise to pay to the order
of H. A. Spencer, One Thousand Six Hundred Dollars.
Due September 14/17, 1895.
R. P. HENRY.
6. A Joint Note.
NEW ORLEANs, July 16, 1895.
Two months after date, for value received, we promise to pay to the
order of S. S. Packard, Sixteen Thousand Dollars.
Due September 16/19, 1895. A. G. MAYLIE,
C. S. CLARKE.
N. B.-In this note, the joint makers are each liable for one-half of the sum named in the
Inote.
7. A Joint and Separate Note.
JNEw ORLEANs, February 4, 1895.
Six months after date, for value received, we or either of us promise to
pay to the order of H. B. Stevens & Co., Nine Thousand Two Hundred Dol-
lars, payable at the Canal Bank in New Orleans.
Due August 4f7, 1895. O. H. KARSTENDIEK,
A. V. DEGRUY,
H. C., YOUNG.
N. B.-In this note, in place of the words, we or either of us, the words, we jointly and
severally, may be used.
{E NOTES. 553
8. Collateral Note.
$5000. NEW ORLEANs, May 14, 1895.
Ninety days after date, I promise to pay to the order of J. G. Anderson,
Five Thousand Dollars, for value received, with interest at the rate of six per
cent per annum from date hereof until paid.
This note is secured by pledge of the securities mentioned on the reverse hereof, and in
Qase of its non-payment at maturity, or should the drawer hereof, when called on, refuse or fail to
keep the margin hereon good, the holder is hereby authorized to sell the said securities at public
or private i. Without recourse to legal proceedings, and to make any transfers that may be
required, applying proceeds of sale towards the payment of within note.
Due Aug. 12f15, 1895. L. W. WELLS.
NOTE.-On the back of this note must be itemized the securities pledged for its protection.
9. A Note with a certain Party as First Indorser.
If Henry Levy wishes to give his note to Louis Winn, and to give J. D.
Purcell, as first indorser, the note would be drawn as follows:

$1621. NEW ORLEANs, February 4, 1895.
Thirty days after date, for value received, I promise to pay to the order
of J. D. Purcell, One Thousand Six Hundred Twenty-one Dollars.
Due March 6/9, 1895.
HENRY LEVY.
This note must now be indorsed by J. D. PURCELL, and delivered to Louis WINN.
10. A Note with Surety.

NEW ORLEANs, March 21, 1895.
Four months after date, for value received, I promise to pay to the order
of N. E. Ball, Eight Thousand Dollars.
B. TURNER, JR.
I hereby, for value received, agree to become security on the above note.
C. W. BUTLER,
NOTE.-The contract of surety may be indorsed on the back of the note.
11. A Note with a Guarantor.
NEW ORLEANs, March 21, 1895.
Sixty days after date, for value received, I promise to pay to the order
of L. Powell, Twenty-four Hundred Five and ſº Dollars.
J. W. LUCE, JR.
For value received, I guarantee the payment of the above note at
maturity. F. P. BOLTON.
NotE.—The contract of guarantee may be indorsed on the back of the note with the word
“within” in place of the word “above.”
554 SOULE's PHILOSOPHIC PRACTICAL MATHEMATIcs.
13. 4 Note of a Married Woman in Louisiana.
NEW ORLEANs, March 21, 1895.
Ninety days after date, for value received, I promise to pay to the order
of J. H. McNeely, Forty-six Hundred Seventy-three and ºf Dollars. -
MRS, KATIE HOPKINS,
I hereby authorize my wife to sign the above note.
A. B. HOPKINS.
13. A Note signed with a cross by a party who cannot write.
NEw Orleans, May 23, 1895.
$ One year after date, for value received, I promise to pay to the order of
S. S. Levy, Twenty-five Hundred Dollars.
his
HENRY X SCHNEIDER,
mark
Witnesses: E. P. DELANNE.
H. C. AYCOCK.
BILLS OF EXCHANGE, DRAFTS AND CHECKS,
—- sº.--º-º-
-sºr ºv-wº-
1115. A draft, check or bill of exchange is a written order for the payment
of money unconditionally. The bill may be printed, except the signature of the
drawer—that must be written. t
14. A Time Bill or Draft, Time reckoned from Acceptance.
$56231.3% NEW ORLEANs, January 1, 1895.
Thirty days after sight, pay to the order of W. Wintz, Five Thousand
|Six Hundred Twenty-three and Yºo Dollars.
To A. Greenwood, : W. C. KEEVER.
New Orleans.
PARTIES TO BILLS AND DRAFTS.
There are three original parties to a bill of exchange, viz: the drawer,
drawee and payee. The subsequent parties to a bill are those to whom it may
be successively transferred. In the above bill, W. C. Keever is the drawer, A.
Greenwood is the drawee, and when he accepts the bill, he will become the acceptor,
maker, and payor, and the drawer then becomes first indorser. W. Wintz is the
º: BILLS OF EXCHANGE, DRAFTS AND CHECKS. 555
payee. Sight bills are paid at sight in Louisiana; but in some States, they are
accepted and bear three days of grace. Drafts and bills should be presented for
acceptance Without delay, and if acceptance is refused, they should be protested for
non-acceptance in order to hold the drawer.
Accepted bills bear grace, are protested for non-payment the same as notes,
and are subject to all the laws governing promissory negotiable notes.
15. A Sight Bill.

$1500, NEW ORLEANs, February 3, 1895.
At sight, pay to the order of C. B. Mödinger, Fifteen Hundred Dollars,
and charge the same to account of
To W. H. Reynolds, A. L. SOULE.
Alexandria, La.
16. Bank Check.

NEW ORLEANs, May 1, 1895.
SOULE COLLEGE BANK.
Pay to the order of H. C. Brown, Two Thousand Seventeen and 54% Dol-
lars. $2017 fºr.
J. J. BELL & CO.
17. Order, Payable in Merchandise.
NEW ORLEANs, April 11, 1895.

MR. J. VEGA.
Pay to Mr. J. N. Harrison or order, Forty-four and #6 Dollars in
merchandise, and charge the same to my account.
C. W. BUTLER,
18. Certificate of Deposit.

No. 400. SouLÉ CollFGE BANK,
NEw ORLEANs, July 5, 1895.
N. L. Barfield has deposited in this bank One Thousand Dollars, payable
to his order on return of this certificate properly indorsed.
$1000. M. GREEN,
Cashier.
19. Due Bill.

NEW ORLEANs, April 1, 1895.
Due to Wm. Hardie or order, on demand, One Thousand Seven Hundred
Dollars, with interest from date at 8 per cent.
GEO. C. FAHEY.
556
soul E's PHILOSOPHIC PRACTICAL MATHEMATICs.
20. Due Bill Payable in Merchandise.

NEw ORLEANs, March 16, 1895.
Due R. C. Spencer or order, Seventy-five Dollars, payable in merchan-
dise from my store, May 1, 1895.
C. W. GLYNN.
21. A Set of Domestic Bills.
(1). SouLí, CoLLEGE BANK,
NEW ORLEANs, January 3, 1895.
At sight of this First of Exchange, (second unpaid) pay to the order of
G. Loomis, Two Thousand Dollars, value received.
To the City National Bank, EDWARD E. SOULE.
of New York. Cashier.
1116.
(2). SOULF COLLEGE BANK,
NEW ORLEANs, January 3, 1895.
At sight of this Second of Exchange, (first unpaid) pay to the order of
G. Loomis, Two Thousand Dollars, value received.
To the City National Bank, EDWARD E. SOULE.
of New York. * Cashier.
-º-
SPECIAL LAWS AND BUSINESS CUSTOMS
TO BE OBSERVED IN INTEREST AND DISCOUNT CALCULATIONS.
A__* —a
w-uy-w
enactmentS.
The following are some of these laws and customs governing the most of the
mercantile communities of America:
2. To consider the year as consisting of 12 months of 30 days each, or 360
days, for a divisor in all interest or discount calculations.
NOTE.—The American, English, and French governments, and the State of New York con-
sider the year as consisting of 365 days, hence in calculating interest on United States, English or
French bonds, or on notes maturing and payable in New York, the divisor would be 365 instead of
360. See Interest Divisor, pages 560 and 561.
1. In performing the operations of interest calculations, we must
observe certain special laws, and also certain business customs which, by reason of
being in general use or of long standing, have become law without any legislative
Yºr SPECIAL LAWS AND BUSINESS CUSTOMS. 557
3. To allow 3 days of grace on all commercial paper, not payable on
demand, at sight, or without grace, except in States in which days of grace are
not allowed.
NOTE.-Also to allow three days of grace on sight drafts or bills of exchange in States in
which sight bills are accepted.
4. In maturing notes, drafts, etc., that are drawn in years or months, it is the
custom to count from the day of the month that the instrument is dated to the same
day of the month in which it matures. Thus a note dated July 15, 1895, and made
payable 3 months after date, matures October 15/18, 1895.
5. In maturing notes, drafts, etc., that are drawn in days, it is the custom to
mature them in days. Thus, if a note was dated July 15, 1895, the same as above,
but made payable in 90 days after date, it would mature October 13/16, 1895.
6. Notes, acceptances, etc., maturing on the 29th, 30th or 31st days of
months having less than 29, 30 or 31 days, are considered due on the last day of the
month, and allowing for grace are payable on the 3d of the following month.
Thus, a note dated December 28, 29, 30 or 31, 1895, and payable 2 months
after date, matures February 28th and March 3d, 1896.
7. Notes, drafts, etc., maturing on Sunday or a legal holiday, are payable in
Louisiana, the first business day following. In most of the States, commercial
paper maturing on Sunday or a legal holiday, is payable on the preceding business
day.
8. When discounting notes or drafts drawn in years or months, it is the
custom to count the actual days of unexpired time, including 3 days grace and also
discount day, in cities or places where interest is computed for the day the note is
discounted. Thus, a note dated July 15, 1895, and made payable 3 months after
date, would mature October 15/18, 1895, and if discounted the day it was drawn, the
discount time would be 96 days, including grace and discount day.
If this same note had been made payable 90 days after date, it would have
matured October 13/16, 1895, and if discounted the day it was drawn, the discount
time would have been 94 days, allowing grace and discount day.
And in like manner a note made payable in one year would be discounted for
369 days, if discounted the day it was drawn, including grace and discount day.
In discount computations, the time that has elapsed is deducted from the actual
time the note has to run.
NOTE 1.-In cities or places where no interest is charged for the discount day, but where 3
days of grace are allowed, the discount time on the above notes would be 1 day less on each, making
them respectively 95, 93 and 368 days.
NoTE 2.—In States where days of grace are not allowed, and the discount day is not included,
the discount time would be 4 days less on each of the above notes, making respectively 92, 90 and
365 days.
TO COLLECT INTEREST-BEARING NOTES.
1117. In collecting a note that is drawn in months or years and which bears
interest, it is customary to charge interest for the months at 30, or years at 360
days each, instead of the actual days.
558 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICs.
X.
If a note is dated March 10, 1895, and made payable 1 year after date with
interest, it would mature March 10/13, 1896, and if held by the payee till maturity,
he would collect the face of the note and the interest for one year or 360 or 363
days.
In case the note was not paid until the expiration of the 3 days of grace, then
the holder would be entitled, in equity at least, to interest for 3 days of grace,
making 363 days.
REMARKS.—The question of charging interest for the 3 days of grace, as here
stated, when the payor avails himself of the 3 days of grace, is not as far as we can
learn, covered by any statute or legal decision; and our financial men are not agreed
upon this point. Some argue that in States where the law allows the maker of
notes 3 days of grace, he should not be charged interest for those days. Others
argue that while the law allows 3 days of grace in which to pay the note or bill, it
does not stop the interest on those days any more than it does on any of the other
days the instrument has to run.
The equity of the question is clearly in favor of charging interest for the
days of grace, and we shall so present the matter in this work. It is an ethical
principle in business that value must be given for value received. That no person
shall use the property of another without rendering compensation therefor. Hence,
if the payor of a note retains the use of the payee's money for the 3 days of grace,
he, the payor, should pay for such use. And moreover, the 3 days of grace in States
allowing grace, is as much a part of the specified time a note has to run as the time
written in the body of the note.
ELUCIDATION OF THIS PRINCIPLE.
1118. A note is given March 26, 1895, for $5000, payable one year after date
with interest at 8 per cent.
What is the interest due in States allowing grace, if collected March 26 or 27
or 28 or 29, 1896? Ans. See statements following.
Statement to find inter- | Statement to find inter- I Statement to find inter- | Statement to find inter-
est due March 26. est due March 27. est due March 28. est due March 29.
$50.00 $50.00 $50.00 $50.00
360 8 8 360 | 8 8
| 360 360 361 362 360 || 363
$400.00 Ans. $401.11; Ans. $402.22; Ans. $403.33% Ans.
TO DISCOUNT NOTES BEARING INTEREST.
1119. In discounting a note drawn in months or years and which bears inter-
est, it is the custom to add to the face of the note the interest for the months at 30,
or years at 360 days each, plus three days of grace, which gives the maturity value
of the note, and then to discount the sum thus produced, also for the actual number
of days including 3 days grace and also 1 discount day, in places where the discount
day is included. See page 580, for the operations of discounting notes that bear
interest.
jºr SPECIAL LAWS AND BUSINESS CUSTOMS. 559
POINTS IN DRAWING NOTES.
1120. The following may be useful to those who have frequent occasion to
draw notes:
1. 33 days = 4 weeks and 5 days; 63 days = 9 weeks; 93 days = 13 weeks
and 2 days.
Therefore a note at 30 days will fall due five days later than the day of the
week on which it was given.
A note at 60 days will fall due on the same day of the week.
A note at 90 days will fall due two days later.
2. In drawing notes or accepting drafts, it should be borne in mind that .
often bills of different dates, having the same time to run, will mature on the same
day and thus occasion unexpected inconvenience. Thus a note drawn January 31st
for 3 months, and another drawn January 30th for 3 months, both mature April
30th and May 3d. Sometimes one day’s difference in the date or acceptance will
make 2, 3 or 4 days' difference in the maturing of bills having the same time to
run, thus:
Notes drawn or drafts accepted April the 30th at 3 months, mature July
30th and August 2d, but if dated or accepted May 1st, they will mature August
1/4, 2 days later. If a note was drawn February the 28th at 6 months, it would
mature August 28th and 31st; but if dated March 1st, it would mature September
1st and 4th, 4 days later. An observance of these principles by accountants and
business men will frequently save much inconvenience and sometimes result in
preserving the right of action against indorsers.
3. In discounting notes, the most of bankers reject the cents if less than
50, and if in excess of 50, they increase the sum by $1 and reject the cents.
In the operations of interest in this book, we will use the exact figures.
For Time Tables, see page 298 of this book.
Table of Interest Laws of the States and territories, as existing, in 1891, See
page 548.
INTEREST DIVISORS,
1121. An Interest Divisor is a number which used as a divisor of any prin-
cipal will give Interest in the quotient. The Interest Divisor differs according to
the rate per cent, and also according to the number of days, 360 or 365, used in a
year as a basis for interest computations,
1122. Interest, as shown in the preceding remarks and definitions, is a com-
pound of per cent, on a 100, and per annum, for a year, 360 days. Hence to obtain
1 per cent interest for 1 year, on any principal, we simply divide by 100. And to
obtain 1 per cent interest for 1 day, on any principal, we first divide by 100 to get
1 per cent interest for 1 year and then divide that quotient by 360 to get the interest
for one day. Or we may divide the principal at once by 36000 which is the product
of 100, the basis of per cent, and 360 days, the basis of a year.
* W.
56O SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. -
The quotient arising by this division, being interest, we therefore name the
36000 the Interest Divisor for 1 per cent.
NotE.—when 365 days are used as the basis of interest computations: then the 365 would be
used where 360 is used herein, and the 1 per cent Interest Divisor would be 36500.
Having thus produced the interest at 1 per cent for one year or one day, to
find the interest at any desired rate per cent and for any desired number of years
or days, we have but to multiply this interest by the desired rate per cent and the
desired number of years or days.
TABLE OF INTEREST DIVISORS.
1123. To aid in understanding the interest divisor and the use of the same,
we present the following table which gives the interest divisors at 1, 2, 3, 4, 44, 5,
6, 8, 9, 10, 12, 15, 18, 20, 24, 25, 30 and 36 per cent:
Interest Interest Interest
% D1v1sors. % Divisors. % Divisors.
36000 - 1 = 36000 36000 - 6 = 6000 36000 - 18 = 2000
36000 - 2 = 18000 36000 - 8 = 4500 36000 - 20 = 1800
36000 - 3 = 12000 36000 - 9 = 4000 36000 -- 24 = 1500
36000 - 4 = 9000 36000 - 10 = 3600 36000 -- 25 = 1440
36000 - 4} = 8000 36000 - 12 = 3000 36000 - 30 = 1200
36000 - 5 = 7200 36000 - 15 = 2400 36000 - 36 = 1000
When the rate of interest will not cancel the 36000 without a remainder, then
we proceed as shown in the statement for the second operation following:
NoTE.—In many computations, Equation of Accounts, Accounts Current and Interest
Accounts by the product or equation method, Interest on Daily Cash Balances, Cash Notes, True
Discount, etc., the Interest Divisor is of great service, and should be well understood by all
accountants.
BASIS OF ALL INTEREST COMPUTATIONS.
1124. The foregoing is the basis of all interest computations, and by working
in accordance there with we avoid all the arbitrary rules which confuse and confound
the millions.
To perform the operations of interest in detail, as above indicated, would
require considerable time and labor. Hence, with a view to economize both, and
still work from the foundation principles of interest, we combine reason and can-
cellation with the foregoing principles and evolve a brief, a simple, and a universal
formula, applicable to all interest computations.
Ǻ INTEREST EVOLUTION. 561
A UNIVERSAL FORMULA FOR COMPUTING INTEREST.
1125. The solution of the following problem shows the application of the
foregoing principles of interest, and the evolution in the operations by which the
brief, simple, and universal formula is obtained:
What is the interest on $72000 at 8 per cent for 11 days? Ans. $176.
First Operation, in detail.
36000) $72000 ($2 = interest at 1% for 1 day.
= 8%.
$16 = interest at 8% for 1 day.
11 = 11 days.
$176 = interest at 8% for 11 days.
Fourth Operation in
Interest Evolution.
Third Operation in
Interest Evolution.
Second Operation
in Interest Evolution.
$ $
72000 or, 72000 72000 720.00
100 || 8 36000 || 8 4500 | 11 45 | 11
360 | 11 11 -*
-wºme sºm-- *-* : *- | $176, Ans. $176.00 Ans.
$176, Ans. $176, Ans.
Explanation.—In the FIRST operation, we divided by 36000, the 1 per cent interest divisor and
obtained $2, the interest at 1 per cent for 1 day; this we multiplied by the rate, 8 per cent, and
obtained $16, the interest at 8 per cent for one day; this we multiplied by the days, 11, and
obtained $176, the interest at 8 per cent for 11 days.
In the SECOND operation, we indicated, on the statement line, the work of the first operation,
and then cancelled.
In the THIRD operation, we mentally divided the 36000 by the rate per cent, 8, and produced
4500, the 8 per cent interest divisor. By this mental cancellation, we very much shortened the
operation.
In the FOURTH operation, we first produce the 4500, the 8 per cent interest divisor, then cancel
the two 0's and in compensation therefor point off two places in the principal. By this mental
cancellation, we shorten the operation to the greatest practical limit, and present a universal
formula for interest computations, far superior to the arbitrary rules given in the arithmetics and
calculators now before the public.
TABLE OF CONTRACTED INTEREST DIVISORS.
1126. The following table shows the Contracted Interest Divisors, for the most
usual rates per cent :
TABLE OF CONTRACTED INTEREST DIVISORS,
ID % Interest % Interest % Interest
S. Divisors. D8. 0 Divisors. DS. Divisors.
360 - 1 = 360 360 - 5 = 72 360 - 15 = 24
360 - 2 = 180 360 - 6 = 60 360 - 18 = 20
360 + 2} = 144 360 - 8 = 45 360 - 20 = 18
360 - 3 = 120 360 - 9 = 40 360 - 24 = 15
360 - 4 = 90 360 - 10 = 36 360 - 30 = 12
360 + 4 = 80 360 - 12 = 30 360 - 36 = 10
NoTE.—When the rate per cent is not a factor of 360, such as 7 per cent, 11 per cent, etc.,
then 360 will be the Interest Divisor, and the rate per cent will be used as a multiplier as shown in
the first operation of the first problem following this table.
562 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. ºr
A PHILOSOPHIO METHOD OF USING THE FACTORS OF THE INTER-
EST DIVISOR,
1127. In the practical computation of interest, we prefer to use the factors of
the Interest Divisor (100 and 360, or 365) in a slightly different manner from that
shown by the four operations above, yet strictly in accordance with reason and logic.
bl In order to fully elucidate the work, we will re-state and re-work the above
problem.
What is the interest on $72000 at 8 per cent for 11 days? Ans. $176.
FIRST OPERATION. SECOND OPERATION. Explanation.—In the first opera-
tion, our statement i.to the
statement made in the Second Oper-
720.00 720.00 ation preceding, except that we
8 45 || 11 first divide the principal, $72000,
360 | 11 ſº-ſº by 100, by pointing off two places,
•º-ºmmºn $176.00, Ans and then by 360, instead of using
$176.00. A • * ~ * o the 36000 as a single divisor.
.UU, Ans. The reasoning for the work based
upon the foregoing elucidations, is
as follows: 1 per cent interest on any principal for 1 year is the Tºo part of it, which we produce
by pointing off two places. Then, at 8 per cent it is 8 times as much, which is indicated by
writing the 8 on the increasing side of the statement line. Then, since the interest, as indicated
by the statement is for 1 year, for one day it is the 360th part, which is indicated by writing the
360 on the decreasing side of the statement line; and for 11 days, it is 11 times as much as for 1
day, which is indicated by writing the 11 on the increasing side of the statement line.
NOTE.-The reasoning for the second operation, is as follows: The interest on $72000 for 1
year at 1 per cent is the 100th part, which is $720; and for 1 day at 8 per cent it is the 45th
part, and for 11 days it is 11 times as much as it is for 1 day.
This is a philosophical and logical method of work, step by step, and it is
universal in its application.
In the second operation, our statement is the same as that in the fourth oper-
ation preceding and constitutes the most valuable method possible, of computing
interest.
The reasoning is the same as in the first operation, except, instead of multi-
plying by 8 and dividing by 360, we mentally divide the 360 by 8, and then use the
quotient, 45, as a contracted interest divisor. In this manner, we contract the opera-
tion to the greatest practical limit and use reason and logic throughout the Solution.
PROBLEMS IN INTEREST
WORKED EY THE PHILOSOPHIC SYSTEM.
1128. The Principal, Rate Per Cent, and Time given to find the Interest, the
Amount, or the Proceeds :
FOR YEARS.
1. What is the interest on $560 at 8 per cent for 3 years? Ans. $134,40.
OPERATION.
$5.60 = 1 % of principal. Explanation.—Considering that interest invol-
8 = 8 % ves per cent and per annum, as elucidated in
O e the foregoing work, and in consonance with the
e foregoing logical method of operation, we here
$44.80 = int. for 1 year. reason as follows: The interest on $560 at 1 per
3 = years. cent for 1 year is the hundredth part, $5,60,
* . we express by pointing off two *...*;
— i and at 8 per cent it is 8 times as much, which is
$134.40 = int. for 3 yrs., Ans. $44.80; and for 3 years it is 3 times as much as
for 1 year, which is $134.40.
INTEREST.
563
FOR YEARS AND MONTHS.
2. What sum must be paid for the interest on $820 at 8 per cent for 1 year
and 9 months.
OPERATION.
$
:* = 1% of principal.
0 e
12 131 = months.
$114.80 = int. for 1 yr. 9 mos., Ans.
Ans. $114,80.
Ea:planation.—In this and in all problems
where there are months in the time, we reason
as follows: The interest on $820 at 1 per cent
for 1 year, or 12 months, is the hundredth part,
$8.20; and at 8 per cent it is 8 times as much;
and for i month instead of 13, it is the i3:
part; and for 21 months (1 year and 9 months
reduced to months) it is 21 times as much as it
is for 1 month.
FOR YEARS, MONTHS AND DAYS.
3. What is the interest on $1230.40 at 9 per cent for 2 years, 5 months, and
24 days? Ans. $274.99-H.
FIRST OPERATION. SECOND OPERATION.
$ f
12.3040 = 1% of principal. 12.3040
9 = 9%. 40 | 894
360 | 894 = days. – 1 ——
$274,9944
$274.9944 = int. for 2 yrs. 5 mos.
and 24 ds.
THIRD OPERATION.
$12.30.40 $
t 9 12.3040
*E=E--msmºs 360 9
$110.73.60 = interest for 1 year. 174:
2 •º-ºººººº-
*mºmº -mºsºs 53.5224
$221.47.20 = interest for 2 years.
53.52.24 = interest for 174 days.
$274.99.44
Explanation.—In this and in all problems where there are days in the time, in accordance
with the foregoing elucidated principles, we reason as follows in the first operation : The interest
on $1230.40 at 1 per cent for 1 year is the hundredth part, $12.3040§ and at 9 per cent it is 9 times as
much; and for 1 day, instead of 1 year, it is the 360th part; an
and 24 days reduced to days) it is 894 times as much.
for 894 days (2 years, 5 months
In the second operation, the same reasoning governed the statement, but instead of writing
the 9 per cent and the 360, we used the Contracted Interest Divisor, as elucidated in the foregoing
work.
GENERAL DIRECTIONS FOR, CALCULATING INTEREST.
1129. From the foregoing elucidations, we derive the following general direc-
tions for calculating interest:
1. For years, first find 1 per cent of the principal, by dividing by 100 (pointing
564 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICs. 2%
off two places); then multiply by the rate per cent, and this product by the number of
3years. See problem 1, page 562.
2. For months, write the principal on the statement line and find 1 per cent, by
dividing by 100, or pointing off two places; then indicate on the statement line the
division by 12, and the multiplication by the rate per cent and the number of months,
and work out the statement. See problem 2, page 563.
3. For days, write the principal on the statement line and find 1 per cent by
pointing off two places, or by indicating the division by 100; then indicate on the state-
ment line, the divison by 360, (or 365 when that is used) and the multiplication by the
ºrate per cent and the number of days, and work out the statement, cancelling as much
as possible. See problem 3, page 563.
NOTE.-Instead of indicating the division by 360 and the multiplication by the rate per
cent, We may simply divide by the quotient (the contracted Interest Divisor) of 360 divided by
the rate per cent, as explained on page 562,
4.
4. To find the amount, add the interest to the principal.
5. To find the proceeds, subtract the interest from the principal.
NoTE.—There are always as many decimals in the answer as there are on the statement line,
and no more.
PROBLEMIS.
What is the interest on $1350 at 8 per cent for 64 days? Ans. $19.20.
What is the interest on $550 at 7 per cent for 72 days? Ans, $7.70.
What is the interest on $727.20 at 5 per cent for 11 days? Ans. $1.111.
What is the interest on $7200 at 10 per cent for 121 days? Ans. $242.
10. What is the interest on $3155.16 at 6 per cent for 5 years?
Ans. $946,548.
11. What is the interest on $2344.80 at 8 per cent for 3 months?
Ans. $46.896.
12. What are the interest and the amount of $5000 at 8 per cent for 2 years,
5 months, and 15 days 3 * Ans. $983.33% interest; $5983.33% amount.
NOTE.--To obtain the amount, add the interest to the principal.
13. What are the interest and the proceeds of $45000 at 4 per cent for 8
months and 20 days 3 Ans. $1300 interest; $43700 proceeds.
NotE.—To obtain the proceeds, subtract the interest from the principal.
14. What is the interest on $1440 at 7 per cent for 123 days? Ans. $34,44.
15. What is the interest on $1711.90 for 34 days at 15 per cent 7
Ans. $24.25-H.
16. What is the interest on $3240 for 94 days at 9 per cent 3 Ans. $76.14.
17. What is the interest on $21636.72 for 63 days at 10 per cent?
Ans. $378.64-H.
18. What is the interest on $6840 for 5 years, 8 months, and 14 days at 43
per cent? Ans. $1756.17.
19. What is the amount of $2700 for 1 year, 2 months, and 5 days at 5 per
cent 3 Ans. $2859,374.
:
* INTEREST. 565
20. What are the proceeds of $5284.50 for 4 months, and 12 days at 7 per
cent # Ans. $5148.86+.
21. What is the interest on $1500000 for 3 days at # per cent per day?
Ans. $11250.00.
OPERATION INDICATED.
$ $15000.00
15000.00 #%
90 = 4% per day = 90% per year.
360 || 3 or thus: $ 3750.00 = interest for 1 day.
3 = days.
$11250.00, Ans. -ºm-
$11250.00 = interest for 3 days.
22. What is the interest on $30000 at 4 per cent per month for 28 days?
Ans. $1120.00.
, OPERATION INDICATED.
$ $
300.00 A 300.00
48 = 4% per month = 48% per year. OI’ 4
360 | 28 % 30 28
$1120.00, Ans. | $1120.00, Ans.
23. What is the interest on $260000 for 11 days at # per cent per day?
Ans. $10725.00.
24. What are the interest and the amount of $1864.500 from July 12 to Sep-
tember 13, at 44 per cent per annum, last day inclusive?
Ans. $14682.94 interest. $1879182.94 amount.
25. What is the interest on $2700 for 50 days at 2 per cent per month?
Ans. $90.
26. A note for $4100, having 80 days to run, is discounted at 14 per cent per
month. What are the proceeds? Ans. $3963.33.
27. What are the interest and proceeds of $10000 for 3600 days at 10 per
cent” Ans. $10000, interest. No proceeds.
28. The cash price of a lot of goods is $300. A buyer purchases the goods
on a credit of 90 days when money is worth 8 per cent. What ought to be the
credit price # Ans. $306.00,
OPERATION INDICATED.
* $
3.00
45 || 90
| $6.00 int.
29. The terms for an invoice of goods amounting to $2350 are “4 months, or
5 per cent discount for cash in 10 days.” The buyer paid cash in 10 days. What
did he gain or lose, money being worth 6 per cent? Ans. $76.57 gain.
OPERATION INDICATED.
$
2232.50
$2350 – 5% ($117.50) = $2232.50. 60 | 110 $117.50 – $40.93 = $76,57.
º- | $40,9291+.
566 soul E's PHILOSOPHIC PRACTICAL MATHEMATICs. *
THE PEIILOSOPHIC SYSTEM OF CONTRACTING INTEREST QUESTIONS.
1130. There are a great many methods of contracting interest calculations,
but the greater number of them are applicable only to special combinations of
numbers, while others require the memorizing of arbitrary rules and exceptions
instead of the exercise of the reasoning faculties. The philosophic system which
we here present is general in its application, and is based upon one beautiful system
of work by which we solve every interest problem, and give a reason for every figure
of the work without the aid of rules, no matter what may be the sum, the rate per
cent, or the time.
The following problems will elucidate this beautiful system of contraction:
PROBLEMS TO BE SOLVED MENTALLY WITHOUT THE AID OF PEN OR PENCIL.
. What is the interest on $1200 for 64 days at 5 per cent?
What is the interest on $1400 for 78 days at 6 per cent?
What is the interest on $1500 for 34 days at 8 per cent?
What is the interest on $2000 for 124 days at 9 per cent?
What is the interest on $1528.60 for 51 days at 12 per cent?
:
OPERATION OF
FIRST QUESTION.
5%
$
12.00
72 | 64.
$10.663, Ans.
FOURTH QUESTION.
9%
$
20.00
40 | 124
$62.00, Ans.
SECOND QUESTION.
6%
$
14.00
60 78
$18.20, Ans.
OPERATION OF
THIRD QUESTION.
8%
$
15.00
45 || 34
$11.33%, Ans.
FIFTH QUESTION.
12%
$
15.28.60
30 51
$25,9862, Ans.
In working out these statements, we make no figures but the result, or answer; thus in
working the first question we see that the 1200 and the 72 may be canceled by 12, and that 1200 is
equal to 12, 100 times, and that 72 is equal to 12, 6 times; we have therefore but to multiply the
64 by 100 and divide by 6; all of which is mentally performed.
In the second example, we cancel mentally, one 0 on the 1400 and the 0 on 60, then the 78 by
6, which gives us a quotient of 13, with which we multiply 140, and produce the answer.
In the third example, we cancel, mentally, the 45 and the 1500 by 15; we then annex the two
0s to the 34 and divide by 3; the result is the answer.
In the fourth example, we cancel a 0 on each side of our statement line, then the 200 or the
124 by 4, and multiply the two factors together, which gives us the correct result or answer.
In the fifth example, we cancel a 0 on each side of our statement line, then the 51 by 3, which
gives us a quotient of 17, with which we multiply the 15286, and produce the correct result.
º: INTEREST. 567
In all interest calculations, we must be careful to point off 1 per cent in
commencing the work, and in the final result point off the same number of figures.
We commend this system of interest calculations and this method of con-,
traction over all others, because of their simplicity and their easy application to all
problems in interest without referring to rules, and because, taking the problems as
they naturally occur in buisness, they are the shortest and by far the easiest.
... NOTE.-In the following problems, place the figures on the statement line and solvementally,
Writing no figures but the answer.
6. Find the interest on $1440 for 67 days at 24% Ans. $6.70.
7. 44 {{ “ 8400 for 28 days at 3% Ans. 19.60.
8. {{ {{ “ 1600 for 54 days at 34% Ans. 8.40.
9. { % { % “ 2817 for 48 days at 44% Ans. 16.90
10. { % {{ “ 2160 for 64 days at 5% Ans. 19.20.
11. { % { % “ 3000 for 134 months at 6% Ans. 202.50
12. € $ {{ “ 450 for 40 days at 7% Ans. 3.50.
13. 44 {{ “ 15000 for 33 days at 8% 3 Ans. 110.00.
14. é & 44 “ 2408 for 10 days at 9% Ans. 6.02.
15. { % {{ “ 1137.60 for 72 days at 10% Ans. 22.752.
16. {{ 46 “ 25000 for 18 days at 12% Ans. 150.00.
17. 4% 4% “ 728.27 for 96 days at 15% Ans. 29.1308.
PECULIAR INTEREST QUESTIONS.
1131. 1. What is the interest on 102 for 10 days at 10 per cent?
Ans. a's g º
STATEMENT. &
Explanation.—In problems in which there are not two
100 | 10 figures to represent dollars, instead of indicating the
10 divison by prefixing and pointing off two 0's we divide
360 | 10 the sum by 100 and place the result or quotient on the
*== statement line, and then proceed as usual. In this prob-
#2 Ans. lem, the interest on 10c. for 1 year at 1 per cent is Tºc.
2. What is the interest on $1.15 for 15 days at 6 per cent? Ans. ##2.
3. What is the interest on 12#2 at 44 per cent for 104 days? Ans. Hºoſe.
OPERATION INDICATED.
g g g
2 || 25 * 2 || 25 2 || 25
100 OT 36000 OI’ 8000
2 9 2 || 9 . . 2 21
360 º 2 || 21
2 || 21
4. What is the interest on 1632 at 5% per cent for 8 days, 6 hours, and 24
minutes? Ans. Hääog.
5. What is the interest on 1242 for 54 days at 64 per cent f Ans. ###2.
6. What is the interest on 642 for 10 days, 4 hours, and 25 minutes at 4;
per cent 3 Ans, a ####52.
568 soule's PHILOSOPHIC PRACTICAL MATHEMATICS. †
\)IFFERENCE OF INTEREST BY USDNG THREE HUNDEED AND SIXTY
AND THREE HUNDRED AND SIXTY-FIVE DAYS TO THE YEAR.
1132. Since there are 365 days in a common year, it follows that when 360
days are used as an interest divisor on the basis of a year, the interest is ## = #
greater than would be produced by using 365 as the interest divisor, and when 365
days are used as the basis of a year or the interest divisor, the result is sås = }; less
than would be produced by dividing by 360.
The only exception to the above is when the time is in even years; then
there is no difference.
The following problem will more clearly elucidate this difference:
1. What is the interest on $7300 for 81 days at 8 per cent when using 360
days for a year, and also when using 365 days for a year?
Ans. See operations following.
Operation for 360 days to the year. Operation for 365 days to the year.
\ 7.300 7300
360 | 8 365 || 8
81 81
$131.40 int. $129.60 int.
Operations to show that the interest for 360 days is y's more than the interest for 365 days, and that
the interest for 365 days is ºg less than for 360 days.
$131.40 - 73 = $1.80. $129.60 - 72 = $1.80.
$131.40 — $1.80 = $129.60. $129.60 + $1.80 = $131.40
INTEREST AT THEEE EIUNDRED AND SIXTY_FIVE DAYS TO TEIE YEAR.
1133. 1. What is the interest on $5000 for 93 days at 7 per cent, using 365
days for a divisor 3 Ans. $89.18.
OPERATION.
* 00 Explanation.—The reasoning in this problem is as
follows: The interest on $5000 for 1 year at 1 per cent is
365 % $50, and at 7 per cent 7 times as much; and for a day,
t_j instead of a year, it is the 365th part, and for 93 days it
$89.18 Ans is 93 times as much as for 1 day.
2. What is the interest on $2500 at 6 per cent for 146 days, counting 365
days as the interest year? Ans. $60.00.
3. What is the interest on $1111.11 at 11 per cent for 11 times 11 days,
counting 365 days to the year? Ans. $40.51-H.
INTEREST. 569
INTEREST ON UNITED STATES BONDS.
1134.
registed bonds from July 1, 1885, to August 1, 1890?
6 OI’
$3049.32 Ans.
1. What interest has accrued on $10000 United States 6 per cent
Ans. $3049,32.
OPERATION.
$10000 $
6 % 10000
smºsºmºsºmºmºsº 6
$600.00 365 30
5 years.
=-y- $49,32
$3000.00 int. for 5 years.
49.32 int. for 30 days.
*=mºmºs
$3049.32 Ans.
NOTE.-The United States Government computes interest on the basis of 365 days to a year
and for any number of days less than a year a proportional part is taken: Thus, for 19 days, º:
would be taken.
2. What interest has accrued on $250000 United States 44 per cent funded
loan from September 1st, 1882, to December 1st, 1890?
$
250000
2 || 9 OT
365 3010
smºmºmºse ºmºmºmº tº
$92773.97 Ans.
Ans. $92773.97 gold.
OPERATION.
$250000 $
4.4% 250000
gººmsºn s = 2 || 9
$11250.00 365 | 90
8 years. ammº ºmºmº-3
tº-º-º-º-º: sº $2773.97
$90000.00 int. for 8 yrs.
2773.97 int. for 90 dS.
$92773.97 Ans.
INTEREST ON ENGLISEI MONEY.
1135. 1.
uary 1, 1894, to July 1, 1895?
STATEMENT.
£
1000.00
3
365 || 546
tºº
fºº!487.67 Ans.
What interest has accrued on £100000 English consols from Jan-
Ans. £4487.67.
Ea:planation.—Consols are English bonds that bear 3
per cent interest. See Stocks and Bonds for full explan-
ation.
NotE.—365 days are used in Interest Computations in England.
57o soul E's PHILOSOPHIC PRACTICAL MATHEMATICs. Af
2. What is the interest on £1420 12s. 9d. for 63 days at 5 per cent?
Ans. £12.5s. 2d. +
OPERATION.
£ S. d. fº £12.26–H Explanation. — We
1420 12 9 1420.638 20 here first reduce the
5 4% 365 || 5 *mma--mº shillings and pence to
63 5.20S. the decimal of a pound,
.60 +.037} = £.638. 12 and then proceed in
£12.26030+ -: the usual manner of
2.40d. computing interest.
NotE.—For an explanation of the method of reducing shillings and pence to the decimal of
a pound, see Article 869, page 450.
3. What is the interest on £31994s. 43d., at 6 per cent for 63 days allowing
360 days to the year? Ans. £33 11s. 10d.
º OPERATION.
4s. 4d. £ £33.59179
5 4% 31.99.219 20
60 | 63 -º-º-º-º-
.20 + .018; = £.219. -º- 2 tº- 11. 83580S.
| £33.59179 12
10. 02960.
4. What is the interest on 14s. 7d., for 50 days at 8 per cent, allowing 365
days to the year? Ans. 1.9176d. practically 26.
5. What is the interest on £8 3s. 10d., for 34 days at 5 per cent, allowing 360
days to the year? Ans. 9.284d.; practically 9d.
6. What is the interest on £15940 12s. 6d. for 125 days at 8 per cent, allow.
ing, as is the English custom, 365 days to the year? Ans. £436 14s. 7d.
NotE.—See Index, English Exchange, for operations in English Exchange.
INTEREST ON FRENCEI CURRENCY.
1136. 1. What is the interest on fr. 55620.25, for 73 days at 4 per cent?
Ans. Fr. 444.96.
STATEMENT.
Fr.
556.20.25
4
365 | 73
| -*Rººmsºmº-
| Fr. 444,9620 Ans.
Nore...—365 days are used in interest computations in France.
2. What is the interest on fr. 4190.80 for 165 days at 3 per cent?
Ans. Fr. 56.83+.
INTEREST. 57.1
INTEREST ON GERMAN CURRENCY.
1137. 1. What is the interest on 6870.60 marks for 64 days at 44 per cent?
Ans. 54.21+ marks.
OPERATION INDICATED.
68.70.60
2 || 9
365 | 64
NOTE.—365 days are used in interest computations in Germany.
2. What is the interest on 128400 marks for 235 days at 5 per cent?
# Ans. 4133.42+ marks.
DIFFERENT METHODS OF COMPUTING INTEREST.
1138. We present the following different methods, not because they possess
any great merit, but that they may be contrasted with our philosophic system, and
also that they may be used for reference by business men who have previously
learned one of these methods, and who may not have the time or desire to learn a
new, though better method, {
BANKERS METHOD, OR THE SIXTY-DAY OR SIX PER CENT METHOD.
1139. The following problems will show the method of calculating interest
as used by many bankers.
NotE.—It is the uniform custom of bankers in their daily operations of banking, to calculate
interest from interest tables.
1. What is the interest on $2485 for 75 days at 6 per cent? Ans. $31,06.
OPERATION.
# of $24 85 = interest for 60 days. Explanation.—The basis of this work is that...?
= 6 21 = 46 {{ 15 ($ per cent for 60 days is equal to 1 per cent for 360
days, and º obtain ºp. cent for 60 #. We
&__ simply divide the principal, or sum on which we
$31 || 06 Ans. “ 44 75 4 .# io find the it; §1. this we do in this
case by drawing a vertical line through the sum,
so as to cut off the two right hand figures of the dollars. Then, as 75 days is 3 more than 60, we
add + of the interest for 60 days to itself, and thus produce $31.06, the correct interest. In case
we had but one figure representing dollars, then we would prefix a 0 before drawing the line, and
if no dollar figures, then two 0s would be prefixed.
If we wish the interest for a different per cent than 6, we would add to or subtract from 6
per cent interest such a part of itself as the rate per cent was more or less than 6.
NotE.—The interest for years and months is found by multiplying the interest for 60 days, or
two months, by one half the number of months.
tº gº
*
572 soul.E's PHILOSOPHIC PRACTICAL MATHEMATICS. *
2. What is the interest on $543 for 63 days at 5 per cent? Ans. $4.75.
OPERATION BY THE BANKERS' METHOD, OPERATION BY OUR
PHILOSOPHIC METEIOD,
gº of $5 || 43 = interest for 60 days at 6%
- 2715 = 4% “ 3 “ “ 6% $
— 5.43
# of $5 7015 44 “ 63 “ “ 6% 8 72 63 7
E 9502 {{ 4 63 44 4. 1% sºme ºsmºs
e-º-º- ºms | 3801
$4 || 7513 Ans. “ “ 63 “ “ 5% tº-ºº ºsmº
$4.75; Ans.
Ezplanation.—Here we deduct # of the interest at 6 per cent, and thus obtain in the remain-
der 5 per cent.
3. What is the interest on $1428 for 35 days at 8 per cent? Ans. $11.103.
OPERATION BY THE BANKERS’ METHOD. OPERATION BY OUR PHILOSOPHIC
SYSTEM OR METHOD.
# ) $14 || 28 interest for 60 days at 6%
# ) 7 || 14 46 “ 30 “ “ 6% t $
1 || 19 { % “ 5 “ “ 6% 14.28
*mº emº 9 4; 3# 7
# ) $8 || 33 {{ “ 35 “ “ 6% ammº
2 || 773 is # of 6% interest added. | 9996
$11 | 103 Ans. $11.10; Ans.
4. What is the interest of $1260 for 106 days at 6 per cent? Ans. $22.26.
Operation by the Bankers’ or 60-day or 6 per cent Method. Operation by our Philo-
} sophic Method.
$12.60 = interest for 60 d. Cut off two places to the right.
6.30 = interest for 30 d. 3 of $12.60 = $6.30 $ 21
2.10 = interest for 10 d. 3 of 6.30 = 2.10 #"
1.26 = interest for 6 d. } of 6.30 = 1.26 39
$22.26 = interest for 106 days. $22.26
5. What is the interest of $2340 for 3 months, 27 days at 7 per cent?
Ans. $53,234.
Operation by the Bankers’ or 60-day Method. Operation by our Philo-
gº sophic System.
$23.40 = interest for 60 days, or 2 months.
e º 9
11.70 = interest for 30 days, or 1 month. * 3
7.80 = interest for 20 days, # of 23.40 = $7.80 369 || 7
2.34 = interest for 6 days, # of 11.70 = 2.34 6 || 117
39 = interest for 1 day, # of 2.34 = 39
$45.63 = interest for given time at 6% * 819
7.61 = interest for given time at 1% # of $45.63 = $7.61 31941
$53.24 = interest for given time at 7%
NOTE.—Add one cent when the fraction is # or more. $53,23# Ans.
º: INTEREST. 573
6. What is the interest of $3600 for 5 months, 19 days, at 4 per cent?
Ans. $67.60.
Operation by the Bankers’ or 60 day Method. Operation by our Philo-
sophic System.
$36.00 = interest for 60 days, or 2 months.
18.00 = interest for 1 month.
$
| 36.9% 40
90.00 = interest for 5 months, $18.00 x 5 = $90.00 gſ | 169
9.00 = int. for 15 d. 15 = 3 of 60. 4 of $36.00 = $9.00 - |
1.80 = int. for 3 d. 3 = } of 15. # of 9.00 = 1.80
60 = 1nt. for 1 d. = # of 3. § of 1.80 = .60
$101.40 = interest for given time at 6%
33.80 = interest for given time at 2%
$67,60 Ans.
67.60 = interest for given time at 4%
Take # of $101.40 from itself.
7. Which of the above methods has the greater merit?
Ans. The Philosophic.
Taking all principals, all rates per cent and all periods of time, the Philo-
sophic System is far superior to all others.
TEIE THIRTY-SIX PER CENT METHOD OF COMPUTING INTEREST.
1140. This method is based upon the Interest Divisor for 36 per cent which is,
as shown on page 560, 1000. Hence, as elucidated on page 561, to find the interest
on any amount for 1 day at 36 per cent, divide by 1000; then compute it for the
required time and rate per cent.
PROBLEMI.
1. What is the interest on $2758.40 for 75 days at 8 per cent?
Ans. $45.97+.
OPERATION INDICATED.
t Explanation.-Since the 36 per cent Interest
2,75840 = interest for 1 day at 36%. Divisor is 1000, we divide the amount by 1000
36 || 8 and thus obtain $2.75840 interest for 1 day at 36
per cent. Then, as indicated in the statement,
75 we divide the interest for 1 day by 36, to obtain
-* | *-ºº-mºmº- - the interest at 1 per cent; this * We *:
- e c ply by 8 to obtain the interest at 8 per cent, an
$45,97334 Ans the result thus produced, by 75 to obtain the
Interest for 75 days.
Various methods are used by different parties to conclude the operation after
dividing by 1000. But as none of the methods are equal to the above for ease and
simplicity, we will not occupy space with them.
The 100 day method, of computing interest is also based upon the 36 per cent
Interest Divisor, and like the above method, is far inferior to the unrivaled Philo-
574 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. ºr
sophic System, as shown in previous problems, pages 572 and 573 and by the foi-
Iowing operation of the above problem.
OPERATION INDICATED BY THE PEIILOSOPHIC SYSTEMI.
$ $
27.5840 27.5840
45 || 75 t Or, 3 || 5
$45,9733; Ans. $45,9733; Ans.
THE S. L. BLACK METHOD OF CALCULATING INTEREST.
1141. The following method is generally known as the S. L. Black Method, it
having been taught by him some 40 years ago. It is one of the best of the rule
methods and yet it surrenders without a condition to our Philosophic System.
NoTE.—See the operations of the same problems by the two systems.
1. What is the interest on $2400 for 1 year, 4 months and 15 days, at 6 per
cent 3 Ans. $198.
OPERATION.
$24.00 16.5 Explanation.—In this problem, we first reduce
1200 the 1 year to months and add to it the 4 months,
thus making 16; we then divide the days by 3
gºººº-ºº and annex the quotient 5 to the 16 months. We
$198,000 Ans. then multiply the number thus obtained by such
* a part of the sum $2400 as the 6 per cent is part
of 12, which is #, and produce the answer in mills.
By this method of computing interest, when the rate per cent is less than 12,
reduce the time to months and annea, thereto # of the days, if any, and then multiply
the sum or principal by such a part of the time, or the time by such a part of the sum
or principal as the rate per cent is part of 12, and point off three, as the answer is in
mills, when there are days. When there are cents in the principal, point off five
places. When the rate is 12, multiply all of the time by all of the principal. When
the rate is more than 12, first find the interest at 12 per cent and then add such a
part of the interest to itself as the excess of rate is part of 12.
The basis of this system is that 1 per cent for one year is 12 per cent for one
month, and hence to find the interest for months at 12 per cent, we have but to
multiply the sum by the number of months, or the number of months by the sum,
and the answer is in cents.
INTEREST. e 575
2. What is the interest on $1230 for 48 days at 8 per cent? Asts, $13.12.
OPERATION. OPERATION
#) $1230 BY OUR PHILOSOPHIC SYSTEM.
TIO x 2 = 820 $
#) 48 = 16 # 4; | º ;
$13.120 Ans. $13.12 Ans.
Earplanation.—In this example we see that 8 per cent is # of 12, and that § of $1230 is $820,
which we multiply by 16, which is # of the days, and we have the answer in mills.
3. What is the interest on $6400 for 94 days at 9 per cent # Ans. $150,40.
OPERATION OPERATION
BY THE S. L. BLACK METHOD, IBY OUR PEIILOSOPHIC SYSTEM.
# of $6400 $
* *
= $4800 4} |94
31#. smºm-
-*mºsºmºmºsºmº $150.40 Ans.
148800
1600
*mm mºme
$150,400 Ans.
4. What is the interest on $1512 for 35 days at 10 per cent? Ans. $14.70.
OPERATION BY THE S. L. BLACK SYSTEMI, OPERATION
BY OUR PHILOSOPHIC SYSTEM.
252 50 4
# $1512 $1512 $
11
| 420 _* 36 gº 42
1260 $16,632
113 or, 1.008 $14.70 Ans.
13860 1 ºr gao
840 # ) sº interest at 12%.
$14,700 Ans. $14.70 interest at 10%.
By the contrasts here made, it is clear that our philosophic system possesses
advantages over the very best rule work known.
576 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
A NEW TABLE.
1142. The following table is used by many accountants, with advantage, in
interest computations.
6% per annum = }% of the face of a note or debt for 30 days,
44
8% “ “ = 3% “ {{ {{ {{ 4 4 30
9% “ “ = #% “ 4% {{ {{ 44 44 30 44
10% “ “ = #% “ {{ {{ {{ 44 4 30 4
12% “ “ = 1 % “ {{ {{ {{ £4 4 30 44
15% “ “ = 14% “ {{ {{ {{ 44 & 30 (£
18% “ “ = 14% “ 4% {{ {{ {{ {{ 30 4
20% “ “ = 13% “ {{ {{ {{ {{ {{ 30 4
24% “ “ = 2 % “ 4% {{ {4 4 4 30 4
The following problem will illustrate the method of work by the use of this
table:
PROBLEMS.
1. What is the interest on $843.60 for 15 days at 8 per cent? Ans. $2.81.
OPERATION. Explanation.—As 8 per cent per annum is equal to 3 per
cent of the face for 30 days, and as 15 days is 3 of 30, it is
plain that the interest is # of 3 per cent (= #) of the face
$8.43.60 of the $843.60, which is $2.81. Had the time been 18
e- days, we would have first found it for 15 days, then added
$2.81.20 Ans. # of the interest to itself, as 18 is # or 20 per cent more
than 15. Had it been for 20 days, we could have found it
for 15, and then added # of the interest to itself, or we could have found it for 30 days and then
deducted # of the interest, or we could have reasoned thus: 20 days being # of 30, and 8 per cent
per annum being 3 per cent of the face for 30 days, then # of # (= 3 of the face) is the interest.
By a little reflection, it is readily seen that all of these contracted methods are but the
canceled results of our philosophic system, and to the ready calculator they can often be used
mentally to great advantage.
2. What is the interest on $1624.50 for 22 days at 6 per cent?
3. What is the interest on $420.80 for 34 days at 9 per cent?
BANKERS’ - AND MERCHANTS’ DISCOUNT.
1143. In the preceding work, we have defined simple interest without making
mention of Bank Discount ; and as there has been and still is much discussion on
the subject of bank discount, we will therefore appropriate a small space to a con-
sideration of the subject before giving examples. Bank Discount, or the more gen-
eral term, Discount, is interest paid in advance. Hence, when bankers and business
men discount notes or debts, where the element of time is considered as well as the
rate per cent, they get the interest on the face of the note or debt for the actual
time, and at the specified rate per cent. The sum thus obtained, and called interest
or discount, is then deducted from the face of the note or debt discounted, and the
remainder or difference is the cash value or proceeds.
ºf BANKERs' DISCOUNT. 577
Many persons uninformed on the subject of business customs pertaining to
the Operations of interest, earnestly contend that bank discount is the same as
another division of interest called True Discount. True discount differs materially
from bank discount as we shall elucidate a few pages further on, and consists in find-
ing what sum of money loaned for a certain time at a certain rate per cent will amount
to a given sum ; or, to define it differently, it is the finding and deducting of such a
sum called true discount, from the face of a note or debt, as will leave a remainder
which, if the simple interest be added thereto for the same time and rate, will reproduce
the original sum.
The real difference between bank and true discount arises from the fact that in
bank discount we operate on the face of the note or debt, which, of course, includes
the interest charged; and in true discount we operate on the true present value of
the note or debt, and hence the difference amounts to the simple interest on the true
discount.
The term true discount does not imply that bank discount is false or untrue.
It is thus named to distinguish it from bank discount, and because the true discount
of a note or debt is just equal to the interest on the proceeds of the sum from
which the true discount was deducted.
1144. PROBLEMS IN BANKERS" AND MERCHANTS’ DISCOUNT.
NotE,--The following examples are all of a practical character, and each involves some point
of law or of business custom necessary to be known in dealing in notes and securities, or in per-
forming the daily discount transactions of the merchant and the banker.
TO MATURE AND DISCOUNT NOTES, WHEN DRAWN IN MONTHS AND WHEN DRAWN
IN DAYS.
NoTE.—Observe carefully the difference in the maturity and discount of the two following
notes:
$2540.80 * NEW ORLEANs, December 18, 1895.
1. Two months after date, for value received, I promise to pay to the order
of Frank Draxler, Two Thousand Five Hundred Forty and #6 Dollars.
A. D. HOFELINE.
When does this note mature? What are the proceeds, if discounted the day
it was drawn at 9 per cent ? * Ans. It matures February 18/21, 1896.
The net proceeds are $2498.88.
OPERATION.
Explanation.—In maturing this note in accordance with
40 #10s law, we count the months, but in discounting in accord-
* ºsmºº-ºº: ance with business custom in New Orleans, we count the
| $41.9232 discount. actual number of days in the two months and add thereto
$2498.88 proceeds. 3 days of grace and discount day.
578 soule's PHILOSOPHIC PRACTICAL MATHEMATICS. *
$2540.80 NEW ORLEANs, December 18, 1895.
2. Sixty days after date, for value received, I promise to pay to the order of
C. Reynolds, Two Thousand Five Hundred Forty and #, Dollars.
J. B. ANDERSON.
When does this note mature? What are the proceeds, if discounted the day
it was drawn, at 9 per cent? Ans. February 16/19, 1896, it matures,
$2500.15, proceeds.
OPERATION.
25.40.80 Explanation.—In thus problem, according to law, we
mature in days, and according to custom we discount in
40 || 64
| $40,6528 discount. days, counting grace and discount day.
*-*=
$2500.15 proceeds.
GENERAL DIRECTIONS FOR BANKERS’ AND MERCEHANTS’ DISCOUNT,
1145. From the foregoing elucidations, we derive the following general direc-
tions for Bankers' and Merchants' Discount:
1. Calculate the interest on the note at the given rate for the actual number of
days that the note has to run, plus three days of grace and discount day.
2. Subtract the interest thus found from the face of the note ; the remainder
will be the proceeds.
NOTE: 1. – When notes bear interest, find the amount or value of the same at maturity, and
calculate the discount on such maturity value.
NotE 2 — In many cities and States, interest is not charged for discount day.
Not E 3.-In all problems following, discount day is included unless otherwise stated,
PROBLEMS.
3. January 3, 1895, a note was drawn for $1200 and made payable one year
after date, without interest. When does this note mature? If discounted on the
day that it was drawn at 8 per cent, what was the discount and proceeds?
Ans. January 3/6, 1896, it matures, $98.40, discount. $1101.60, proceeds.
OPERATION.
Explanation.—This note being drawn in years, we therefore, in
12.00 10 conformity to custom, mature it in years, and when we discount
8 it, in conformity to a different business custom, we count the
3 36% 369 123 actual days of unexpired time, including 3 days grace and dis-
| —— count day. In making the statement, we use the same reasoning
| $98,40 as in the third problem of Interest.
$6231.50. NEW ORLEANs, November 1, 1895.
4. Ninety days after date, for value received, I promise to pay to the order of
R. Cole, Six Thousand Two Hundred Thirty-one and º, Dollars; payable at the
Germania National Bank, New Orleans. '
J. P. WILKINSON.
Yºr BANKERs' DISCOUNT. 579
When does this note become due What are the proceeds, if discounted
December 23, 1895, at 6 per cent? Ans. January 30/2 1896, it is due.
$6187.88, proceeds.
OPERATION.
$ Explanation.—Had this note been discounted
62.31.5% the day that it was drawn, it would have been
discounted for 94 days; but as it was not dis-
counted until December 23d, we see that 52 days
§ 360 || 42 7 have already expired, and as discount is only
*m-meems charged for the time that the note or debt has
$43,6205 discount. to run, we therefore deduct the 52 days from the
$6TS7.33 T d 94, and obtain 42 days unexpired time for which
ºn 7. net proceeds we discount the note.
5. A note for $7281.27, bearing date September 14, 1895, and payable 90 days
after date, was discounted at the Germania National Bank, September 28, 1895, at
8 per cent. What were the interest and net proceeds, and when does the note
mature ? Ans. $129.44, interest. t;
$7151.83, net proceeds.
December 13/16, 1895, it matures.
6. A draft, bearing date July 18, 1895, and accepted July the 21st, 1895, for
$10000, payable 30 days after sight, was discounted August 7th, at 8 per cent. What
were the net proceeds, and when does the draft mature?
Ans. $9962.22, net proceeds. August 20/23, 1895, it matures.
7. A note for $5800, bearing date February 12, 1895, and payable 6 months
after date, was discounted by a note broker March 1, 1895, at 12 per cent. What
was the maturity of the note? What was the discount, and what the cash value or
proceeds? Ans. August 12/15, 1895, it matured.
$324.80, interest.
$5475.20, cash value.
8. How much is the interest on $20000, for 64 days at 8 per cent?
Ans. $284.44.
9. A note for $75000 dated August 19, 1895, and payable 60 days after date,
was discounted August 20, 1895, at 5% per cent. What were the proceeds counting
days of grace and discount day? Ans. $74245.31.
10. In the above problem, what would have been the proceeds, allowing 365
days to the year, and not counting discount day ? Ans. $74.267.47.
OPERATION INDICATED.
$
750.00
365 || 62
4 23
732.53 discount.
11. A note for $385000, bearing date May 7, 1895, payable 15 days after
date, was discounted May 8, 1895, at 64 per cent. What was the discount, not
counting discount day ? º Ans. $1181.74.
58O - SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. ºr
TO COLLECT NOTES THAT BEAR INTEREST AND EVENTUAL INTEREST.
1146. 1. January 3, 1895, a note is drawn for $1200 and made payable one
year after date, with interest at 8 per cent. What is the interest and amount due
the holder at the maturity of the note. When does the note fall due #
Ans. $96, interest. $1296, amount. Due, January 3/6, 1896.
OPERATION IF PAID JANUARY 3, 1896. OPERATION IF PAID JANUARY 6, 1896.
$1200 face of note. $ 1200
8% 1200 45 || 363
$96.80 int. for 1 yr. and 3 ds.
or, 45 || 360 –
| $ 96 $1296.80 amount.
$96.00 interest for 1 year.
$1296 amount.
& NOTE.-Had the note been paid January 4, interest would have been charged for 361 days;
if paid January 5, interest would have been computed for 362 days, See pages 557 and 558, for an
explanation for thus counting the days.
2. I receive for the sale of real estate a note for $9000, bearing date January
5, 1895, payable two years after date with 6 per cent current, and 8 per cent event-
ual interest. The note was paid August 10, 1897; what was the amount received?
Ans. $10512.50.
NOTE.-Eventual Interest is the interest the note bears after maturity, in case it is not paid
at maturity.
OPERATION.
Face of note - tº * * gºs * : ge º * tº a t- $9000
Interest for 2 years and 3 days, at 6 per cent - & $1084.50
Interest on faee of note from Jan. 8, '97, to Aug. 10, '97,
at 8 per cent (214 days) wº * tº- 428.00 1512.50
Amount due August 10, 1897 sº m 'm as ºn tº $10512.50
3. Received a note dated October 2, 1895, for $6666.67, payable one year after
date, without grace with 6 per cent interest. When does it mature, and if I hold
the note until it matures what will be the interest and what the amount due 3
Ans. October 2d, 1896, it matures.
$400, interest.
$7066.67, amount due.
TO DISCOUNT NOTES THAT BEAR INTEREST.
$4500. *. NEW ORLEANs, June 4, 1895.
1147. 1. Four months after date, for value received, I promise to pay to the
order of O'Neil, Sullivan & Co., Four Thousand Five Hundred Dollars, with 6 per
cent interest.
G. PHARR.
º: BANKERs' DISCOUNT. 58 I -
When does this note mature? If discounted the day it was drawn by a note
broker at 8 per cent, what proceeds would the holder receive #
Ans. It matures October 4/7, 1895. The proceeds are $4463.67.
OPERATION OPERATION
To find the amount or value of the note To discount the maturity value of the note
at maturity. $ and find the proceeds. *
4500 4592.25
60 | 123 45 126
$92.25 = int. for 123 ds, at 6%. $ 128,5830 = interest or discount.
4500. = face of note added. 4592.25 = value of note at matu-
wºmºsºms mºs 3. tº-º-º-º: rity.
$4592.25 = amount or value of note $4463.67 = proceeds of note.
at maturity.
Explanation.—In this problem, we first find the amount or value of the note at maturity.
This we do by calculating and adding to the face of the note, the 6 per cent interest that it bears.
for 4 months and 3 days. This work gives us $4592.25 as the value of the note when it matures;
hence it is clear that this is the amount to be discounted. We then discount the $4592.25 according
to business custom for the actual unexplped time, including 3 days of grace and discount day, at
the specified 8 per cent.
$9730. NEW ORLEANs, November 24, 1895.
2. One year after date, for value received, I promise to pay to the order of
E. B. Cope, Nine Thousand Seven Hundred and Thirty Dollars, with interest at 8
per cent.
LOUIS BUSH.
What is the above note worth on the 1st June, 1896, if discounted at 12 per
cent # - Ans, $9884.
OPERATION OPERATION
To find the worth of the note at maturity. To discount or find the cash value of the note
June 1, 1896.
$ 9730 $10514.89
45 363 30 180
$ 784.89 = int. for 1 yr. and 3 ds. $ 630.8934 = interest or discount.
9730. = face of note. 10514.89 = principal or value of
*E=º ºsmºsºms ºr note at maturity.
$10514.89 = value of note at matu. $9884.00 = proceeds or value of
rity. note June 1, 1896. -
Ea.planation.—In this as in the preceding example, we first find the value of the note at
maturity by adding to the face of the same the 8 per cent interest that it bears for one year and 3
days; and then we discount the maturity value at 12 per cent for the unexpired time, counting
actual days, with grace and discount day. In finding the time we count thus: June 29, July 31,
August 31, September 30, October 31 and November 24 days, 3 days grace and discount day (June
the 1st) = 180 days.
582 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. º:
3. A mortgage note for $20000, dated December 16, 1895, payable 18 months
after date, with 8 per cent interest, was discounted January 6, 1896, at 7 per cent.
What was the maturity value and the proceeds?
Ans. $224.13.33 = maturity value. $20103.51 = proceeds.
NOTE.-The mortgage matured June 16119, 1897. Interest added at 8 per cent for 18 months
and 3 days (543 days) equals $2413.33. Maturity value discounted at 7 per cent, for the time
§ January 6, 1896, to June 16/19, 1897, including three days of grace and discount day, equals
ayS.
4. A note drawn April 2, 1895, at 3 months, for $1635.15, bearing 8 per cent
interest, was discounted June 1, 1895, at 8 per cent. When did the note mature?
W. at is it worth at maturity, and what was the cash value when discounted
Ans. July 2/5, 1895, it matures.
$1668.94, worth of note July 8, 1895.
$1655.96, cash value June 1, 1895.
NOTE.-Interest is added for 93 days. The maturity value is discounted for 35 days, which
includes discount day.
4. CASEI NOTES 77
OR DRAFTS WHICH, WIHEN DISCOUNTED, WILL PRODUCE A SPECIFIED SUM.
1148. 1. For what sum must a 60 day note be drawn, so that when discounted
at 8 per cent, the proceeds will be $8872? Ans. $9000.
FIRST OPERATION. Explanation.—As it is the custom of bankers to calcu-
º º *: on the . º notes, 1. ºf'; *
plain that if we were to a the simple interest of the
$4500 note assumed. $8872 to itself for the time and rate º, and draw the
64 interest for 64 days at 8%. note for the amount thus produced, it would not, when
discounted, produce the . sum for the reason that
the interest on this increased amount would be more than
$4436 p roceeds, Cash. the interest on the first sum. The deficit would be the
interest on the interest first obtained, plus the interest on
each succeeding sum of 1nterest, interminably. Conse-
4500 quently, to produce exact results, we cannot work on the
4436 | 8872 face of the sum that we desire to obtain for the note
* when discounted. We are therefore constrained to assume
some number to represent the face of the note to be drawn,
$9000, Ans. and to facilitate the operation we assume $4500 the 8 per
cent Interest Divisor, to represent the face of the note.
We assume the Interest Divisor for the reason that the interest thereon is always as many dollars
as there are days in the time; and knowing this we are saved the labor of computing the interest
on the sum assumed. Accordingly, in this problem, the time being 64 days, the interest or discount
on $4500 for 64 days at 8 per cent is $64, which subtracted from $4500, the assumed note, gives
$4436 proceeds. We now observe that as we discounted the $4500 for 64 days at 8 per cent, the
same ratio exists between the $4436 proceeds and the $4500 assumed note as exists between the
$8872 proceeds and the face of the note required to produce the same when discounted for the given
time and rate pr. ct, Hence we have but to find the proportional result of these two ratios. To do
this we place the $4500 assumed note on the statement line and reason thus: Since $4436 cash
require $4500 note, $1 cash will require the 4436th part and $8872 will require 8872 times as much.
SECOND OPERATION
By assuming $100 as the face of the note.
$100 note assumed. $100 $
45 | 64 1.42% 100
- ammºms *- 887.20 | 9
$1.42% interest. $98.57% proceeds. 8872.00
| $9000 = face of note.
* CASH NOTES. 583
GENERAL DIRECTIONS.
1149. From the foregoing elucidations, we derive the following general direc.
tions for finding the face of “Cash Notes:”
1. Assume as the face of the note the Interest Divisor for the rate per cent
given or $100, and find the proceeds of the same for the given time and rate per cent.
2. Then divide the assumed note multiplied by the required proceeds, by the
proceeds of the assumed note.
PROBLEMIS.
2. What must be the face of a note, so that, when discounted for 94 days at
12 per cent, it will produce $10000 proceeds? Ans. $10323.47.
OPERATION
By the use of the Interest Divisor,
$3000 note assumed. 3000 note assumed.
94 interest for 94 days at 12 per cent. 2906 || 10000
$2906 cash proceeds. $10323.47, Ans.
3. A creditor owed me a balance of $3212.65 and settled the same with his
“Cash Note,” payable 4 months after date. The note was dated June 17, 1895.
Allowing 8 per cent interest, what was the face of the note, and when does it
mature ? Ans. $3305.19-H face of note.
October 17/20, 1895, it matures.
4. What must be the face of a note to net or produce $1777.95, when dis-
counted at 7 per cent for 63 days? AnS. $1800.
OPERATION.
$
$36000 = note assumed. 36000
441 = interest for 63 days at 7%. 35559 || 1777.95
$1800.00, Ans.
$35559 = proceeds.
Explanation.—There being no 7 per cent Interest Divisor, we assume as the face of the note
the 1 per cent Interest Divisor and then multiply the interest at 1 per cent, which is $63 (as many
dollars as there are days) by 7, the rate per cent and thus obtain $441 interest. The statement is
then made as in the preceding examples.
NotE.—Whenever there is no Interest Divisor for the rate per cent given, the Interest Divisor
for 1 per cent should be assumed and the interest found thereon as above, or, if preferred, by
assuming and discounting a $100 note, as shown in the second operation, page 582.
5. A creditor owed me a balance of $6812.45 and settled the same with his
“Cash Note,” payable 4 months after date. The note was dated June 17, 1895,
allowing 8 per cent interest. What was the face of the note, and when does it
mature? Ans. $7008.69, face of note.
October 17/20, 1895, it matures.
584 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
6. A merchant wishes to obtain $7500 from bank, on a 30 day note, The
rate of discount is 8 per cent. What size note must he offer so that, when dis-
counted, he will receive the exact sum ? Ans. $7557.10.
7. A creditor owes $1250 and wishes to pay it with a draft at 45 days sight,
without grace or discount day. What must be the face of the draft, the rate of dis-
count being 8 per cent? Ans. $1262.63.
8. For what sum must a note be drawn for 45 days, without grace or dis.
count day, to settle a cash balance of $10862.50, interest at 10 per cent
- Ans. $11000.
4. CASEH NOTES.”
WITH INTEREST, COMMISSION, AND BROKERAGE COMBINED.
1150. 1. A customer desires to obtain from a bank $8000 on his 90 day note.
In conformity with bank custom, which prudence and safety demand, he is required
to have one or more indorsers on the note, and as his correspondent I indorse and
negotiate the note for him. The rate of bank discount is 8 per cent; I charge 24
per cent commission for indorsing, and # per cent brokerage for negotiating. What
must be the face of the note % - Ans. $8406.80.
OPERATION,
Face of note assumed, - gº - E- - tº- -> tº- $4500.
Interest on same for 94 days at 8 per cent - - $ 94.
Commission on same at 24 per cent tº - wº 112.50
Brokerage on same at # per cent - - - - 11.25 — 217.75
Cash value, or proceeds of the assumed note. - - - - $4282.25
$ Explanation.—In this solution, for reasons
given in the solution of the first problem of
4500 = note assumed. * Cash Notes,” we assume the 8 per cent Interest
4282.25 | 8000.00 Divisor as the face of the required note, and
*-*sºn &mmº-º from it we deduct the interest, commission and
$8406.80, Ans. brokerage, and thus produce the necessary
relationship numbers, as explained in the first
solution with which we make the proportional statement, the result of which gives the correct
3D SWOI’.
SECOND OPERATION.
By assuming $100 as the face of the note.
- * $
100 face of note assumed. 100
45 94 856.45 || 9
8000.00
$2.08; interest. smm.sºmem-mº
2.50 = 24% commission. $8406.80 Ans.
25 = #% brokerage.
$4.83; amount of deduction.
100. note assumed.
$95.16% proceeds.
* CASH NOTES., 585
To discount the above note, the following figures would be produced:
Face of note, * * * * * * * * - - $8406.80
IXEDUCTIONS.
Interest on face of note for 94 days at 8 per cent, - $175.61
Commission on facé of note at 24 per cent, - tº 210.17
Brokerage on face of note at # per cent, tº- tºs 21.02 —-- 406.80
Net proceeds, - sº * tº- - - º - - $8000.00
2. A merchant owes a cash balance of $2540.37, which he wishes to settle
by note at 120 days for such a sum as, when discounted at 8 per cent, and allowing
24 per cent commission for indorsing, will net the exact cash balance. What must
be the face of the note? Ans. $2681.29.
3. What must be the face of a note for 60 days to net $1720, discount at 5
per cent and 2% per cent commission for indorsing ? Ans. $1780.33-H.
4. A merchant owes a cash balance of $4000, which he wishes to settle by
note at 120 days for such a sum as, when discounted at 8 per cent, allowing 24 per
cent commission for indorsing, will net the exact cash balance. What must be the
face of the note % AnS, $4221,88.
CASH NOTE WITH COMMISSION ON FACE AND BROKERAGE ON THE
BALANCE,
1151. 1. A merchant owes his correspondent $3250 cash, for which it is
agreed that he shall give his note for such a sum as will net the amount due, after
allowing for the following deductions: Interest for 63 days at 10 per cent; commis’
sion on the face of the note at 24 per cent; and 3 per cent brokerage on the balance,
viz., on the remainder of the note, after the interest and commission shall have been
deducted. For what face must the note be drawn 3 Ans. $3411,31.
OPERATION,
$3600 face of note assumed.
$63 = interest. $
90 = commission. 3600
6859.53 || 2
$153 = interest and commission. 3250.00
3600 = note assumed. *ms
$3411.31 Ans.
$3447 = proceeds.
17.23% = }% brokerage on $3447
$3429.76%
NotE.—Observe that the interest and commission are required on the face of the note, and
brokerage on the balance, after deducting interest and commission.
-
586 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
SECOND OPERATION.
$ 100 note assumed. $
sº 199 25 4.25 int. and com. 100
A 36 £3 7 190.5425 | 2
º $95.75 proceeds. 3250.0000
$1.75 interest. .47873 = }% brok. on $95.75. *m-mºsºmeºm,
2.50 = 24% com. — $3411.31
$95.27124 net proceeds. AnS.
$4.25 = int. &com.
To discount the above note, the following figures would be produced:
Face of note, - - - - - - - - - - $3411.31
DEDUCTIONS.
Interest on face of note for 63 days at 10 per cent, - $59.70
Commission on face of note at 24 per cent, tº - 85.28 — 144.98
Balance, . . . . . . . . . . $3266.33
Brokerage on balance at 4 per cent, - ſº tº- º º ‘º 16.33
Net proceeds of note, - tº - tº- - gn $3250.00
CASH NOTE WITH BROKERAGE ON THE FACE AND COMMISSION ON
THE BALANCE.
1152. 1. For what sum must a note be drawn to produce $742.50, after the
following deductions are made: 184 days interest at 12 per cent, 4 per cent broker.
age on the face of the note, and 24 per cent commission on the balance #
Ans. $813.46+.
OPERATION
$3000 face of note assumed.
$ 184 = interest. $2808.50
7.50 = brokerage. 70.2125 = com. On $2808.50 $3000
2738.2875 || 742.5000
*-m-mºs mºmsm-
$ 191.50 = int. and brok. $2738,2875
3000.00 = note assumed. | $813.46
$2808.50 = proceeds.
NotE.—Observe that the interest and brokerage are required on the face of the note, while
the commission is required only on the remainder, after deducting Interest and brokerage.
SECOND OPERATION.
$ 100. = note assumed. $
100 face of note assumed. 6.38% = int. and brok. 100
30 | 184 - * 182.5525 2
$93.61663 = proceeds. 742.5000
| $6.13% interest. 2.3404% = .24% com. on * * * *
.25 = #% brokerage. *-* * $93.61663. $813.46+
$91.27624 net proceeds. Ans.
$6.38% amt. of deductions.
*. CASH NOTES. 587
CASEI NOTE WITH INTEREST ON THE FACE AND COMMISSION ON
THE REMAINDER.
1153. 1. A merchant wishes to draw a note for such a sum that, when
discounted for 64 days at 9 per cent, and 24 per cent commission on the remainder is
deducted, it will net $25000. What must be the face of the note 3
Ans. $26057.95.
OPERATION
$4000 = note assumed. $ 4000
64 = interest. 3837.60 25000.00
$26057.95 Ans.
$3936 = proceeds.
98.40 = commission.
$3837.60 = net proceeds.
NOTE. –Observe that the commission is required on the remainder only.
SECOND OPERATION.
$ $ N.
10% note assumed. 100
49 || 64 95.94 || 25000.00
$1.60 interest on $100 note assumed. * | $26057.95 Ans.
100.00 note assumed.
$98.40 proceeds of $100 note assumed.
2.46 = 24% commission on remainder or proceeds.
$95.94 net proceeds of the $100 note assumed.
In discounting the above note, the following figures would be produced:
Face of note, es wº º & * a {-, } * * tº wº $26057.95
Less interest for 64 days at 9 per cent, a t- * > º º 416.93
Remainder, - * * > • * {- «º tº gº * . ge $25641.02
Less commission at 24 per cent, tº tº º ſº gº º 641.02
Net proceeds, † º º º * , º gº tº º 4- $25000.00
CASEH NOTE WITH INTEREST ON TEIE FACE AND COMMISSION ON
TEIE NET SUM.
1154. 1. A merchant wishes to draw a note for such a sum that, when dis-
counted for 64 days at 9 per cent, and 23 per cent commission on the net sum is
deducted, it will produce $25000. What must be the face of the note?
Ans. $26041.67.
OPERATION.
$4000 = note assumed. Net sum to be produced, $25000 $ 4000
64 = interest. Com. on same at 24%, - 625 || 3936 || 25625
$3936 = proceeds. Amt. to be obtained from | $26041.67 Ans.
bank, - - - $25625
NotE.-Observe that the commission is required on the net sum only.
588 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. *
SECOND OPERATION.
$ Net sum to be pro-
10% note assumed. duced, - - - $25000
4|| || 64 Commission on same 100
*º- at 24 per cent - gº 625 | 98.40 || 25625.00
| $1.60 interest. -
100. face of assumed Amount to be obtain- $26041.67 Ans.
note. ed from bank, - - $25625
$98.40 proceeds.
In discounting the above note, the following figures would be produced:
Face of note, 3- - - - tº * = gº * - º tº $26041.67
CHARGES OR, DEDUCTIONS.
Interest on face of note for 64 days at 9 per cent, . º º 416.67
$25625.00
1024 || $100
25025
$25000 net proceeds required.
For further work of a kindred nature to the foregoing, see Domestic Exchange.
COMPLEX AND EXPERT INTEREST PROBLEMS.
1155. 1. A planter borrows $3040, and agrees to pay at the close of each
year $380 for ten years. What rate of interest does he pay? Ans. 24%.
OPERATION.
| $760
100
25% for 10 years.
$380 for 10 years = $3800 paid.
3040
$3800 – $3040 = $760 interest paid in 10 years. -º-º-º-
25 + 10 = 24% for one year.
EQUAL ANNUAL PAYMENTS OF PRINCIPAL AND INTEREST.
1156. 2. Bought a piece of land for $5000.00 agreeing to pay 8 per cent
interest, and to pay principal and interest in five equal annual payments, how
much is the annual payment 3 *. Ans. $1252.28.
OPERATION.
$5000 - $3.9927 = $1252.28.
Explanation.—By the conditions of the problem, we observe that the $5000 is the present
worth of an annuity, the time being 5 years and the rate per cent 8. Hence, as the present worth
of $1, multiplied by the annuity, would give the full present worth, it is clear that if we divide
jºr CASH NOTES. 589
the given present worth ($5000) by the present worth of $1, for the given time and rate per cent, the
quotient will be the required annuity.
By referring to the annuity table, we find the present value of $1 annuity per annum at
compound interest for 5 years at 8 per cent to be $3.9927.
Then by dividing $5000 by $3.9927, we obtain $1252.28 as the equal annual payment. The
reasoning is, since $3.9927 (present value of $1 annuity, for 5 years at 8 per cent) require $1
investment, then $1 present value, etc. will require the 3,9927th part of $1, and $5000 present value,
will require 5000 times as much.
To obtain the present value of $1 annuity, per annum, at compound interest
for 5 years at 8 per cent, without access to annuity tables, first find the compound
amount of $1 for 5 years at 8 per cent. Thus:
FIRST3 SECONIO 3
$100 Then add
1.08
-***m- $100
1st, $108.00 108
1.08 116.64
-**- 125.9712
2d, $116,6400 136.048896
1.08 = *-* **-
- $586.660096 = final value of $100, annuity for 5
3d, $125.971200 years at 8 per cent compound
1.08 interest. Divide by 100 =
-mºmsm--sºms $5.8666 -i-, final value of $1.
4th, $136.04889600
1.08
5th, $146,9328076800 = compound amount of $100,
which divided by $100
= $1.4693280768.
The $5.8666 -- 1.4693280768 = $3.9927 present value of $1 annuity per annum
at compound interest for 5 years at 8 per cent.
A DIFFICULT PRACTICAL QUESTION IN RENEWING NOTES.
1157. 1. Jones has a note for $4000 in bank which falls due August 19, 1895.
It is agreed that he shall pay the bank $2000 cash and a new 90 day note for what
may be due. The bank is to deduct the interest or discount on the new note at 8
per cent for 94 days, from the $2000 cash paid, apply the remainder on the $4000
due, and take the new note for the balance.
What is the interest on the new note, how much will the cash pay on the old
Inote, and what is the face of the new note %
Ans. $ 42.67 is the interest on the new note.
1957.33 is the amount paid on the old note.
2042.67 is the face of the new note.
OPERATION TO FIND THE FACE OF THE NEW NOTE,
For the explanation, see “Cash Notes,” page 582.
4500
4406 || 2000
$2042.67 face of new note.
By the conditions of the transaction and in accordance with bank custom, the
interest or discount on the renewal note must be paid in advance, and the Sum
required to pay it is to be taken out of the $2000 cash paid. By the above opera-
590 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. *
tion, which involves the principles of cash notes, we see that the face of the renewal
note is $2042.67; and consequently, the discount is $42.67. This $42.67 discount
subtracted from $2000 cash paid, leaves $1957.33 to be credited on the old note.
2. A man owes $700 now due, and it is agreed that he shall pay $400 cash
and give his note for 1 year, without grace or discount day, for the balance that
may be due; the discount on the note at 6 per cent to be paid in advance out of the
$400 cash paid. How much of the $400 cash is to be credited on the $700 and what
is the face of the renewal note % Ans. $380.57 cash to credit on $700 debt.
$319.43 face of note.
NOTE.—This note is discounted for 365 days.
TO FIND THE VALUE OF NOTES BEARING INTEREST AND MATUR-
ING IN SUCCESSIVE AND CONSECUTIVE MONTEIS.
1158. April 8, 1894, a party purchased from a real estate holder a house and
lot for $7500, on the following terms and conditions: $300 was paid in cash. 36
notes of $200 each bearing 7 per cent interest were given; three of the notes were
made payable in 3 months, July 8, 1894. The other 33 notes were made payable
monthly after July 8, 1894. After the three notes payable July 8, and 19 notes
payable monthly after July 8 had been paid, the holder sold the 14 remaining notes
at 7 per cent per annum discount. Counting 30 days to the month without grace or
discount day, what were the proceeds of the 14 notes, Bankers' Piº
* AnS. $3136.74.
SOLUTION.
$7500 cost of house and lot, less $300 paid in cash = $7200 which was paid
in 36 notes of $200 each. The 3 notes paid July 8, and the 19 notes paid monthly
in successive payments after July 8, make 22 notes which deducted from the 36
notes leave 14 notes on hand, 19 months after July 8, which is February 8, 1896.
Each of the 14 notes of $200 has accrued interest at 7 per cent from April
8, 1894, and each will continue to bear interest until paid. And as they fall due
respectively and successively monthly after February 8, 1896, the following state-
ment and operation shows the maturity value of the 14 notes, 23 to 36 inclusive:
QC
# F É Maturi # §
= | Face #| *ºrneº || Maturi Date *| | Dis.
's Nº. Date. Matures. 2 *: º Discounted. Matures. # É Count. Proceeds.
3. #| cent. 3| 3
# =|3
23| $200 April 8, 1894 Mar. 8, 1896 s $26.83; $226.83%|Feb. 8, 1896 Mar. 8, 1896, 1||7| $1.32 $225.51
24 “ 8, 1894 Apr. 8, 189624. 28.00 228.00 || “ 8, 1896 Apr. 8, 1896 2.7 2.66 225.34
25 “ 8, 1894 May 8, 1896.25, 29.16# 229.16; “ 8, 1896 May 8, 1896 3|7 4.01] 225.16
2 “ 8, 1894 June 8, 189626. 30.33: 230.33}| “ 8, 1896June 8, 1896 4|7 5.37| 224.96
27 “ 8, 1894 July 8, 189627, 31.50 231.50 “ 8, 1896July 8, 1896 5||7 6.75 224.75
28 “ 8, 1894 Aug. 8, 1896.28. 32.66; 232.66; “ 8, 1896 Aug. 8, 1896 6||7 || 8.14, 224.53
29 “ 8, 1894 Sep. 8, 1896.29, 33.83%. 233.83}| “ 8, 1896 Sep. 8, 1896 7|7 9.53 224.30
30 “ 8, 1894 Oct. 8, 1896.30 35.00 235.00 || “ 8, 1896 Oct. 8, 1896 8|7| 10.97 224.03
31 “ 8, 1894 Nov. 8, 1896.31] 36.16; 236.16; “ 8, 1896 Nov. 8, 1896 9| 7 | 12.40 223.77
32 “ 8, 1894 Dec. 8, 1896.32. 37.33; 237.33}| “ 8, 1896 Dec. 8, 1896,10|7| 13.84 223.49
33 “ 8, 1894 Jan. 8, 1897|33 38.50 || 238.50 “ 8, 1896Jan. 8, 1897|11||7| 15.30 223.20
34 “ 8, 1894 Feb. 8, 189734, 39.663] 239.663 “ 8, 1896 Feb. 8, 1897|12| 7 | 16.78 222.89
35 “ 8, 1894 Mar. 8, 1897|35 40.83; 240.83; “ 8, 1896 Mar. 8, 1897|13|7| 18.26 222.57
36 7| 19.76) 222.24
“ 8, 1894 Apr. 8, 1897|36 42.00 242.00 “ 8, * 8, 1897|14
2800 = face of the 14 notes. sists&#3281.8% $145.09 $3136.74

VALUE OF STOCK. 591
This statement.and operation show the following results: 1. The face of
the 14 notes $2800. 2. The maturity accrued interest on each and all the notes.
3. The maturity value of each and all the notes. 4. The discount on each and all
!. : 5. The net proceeds or present value, February 8, 1896, of each and all
iſle Il O'DeS.
TO FIND THE WALUE OF STOCK WHICH PAYS A SPECIFIED
MONTHLY DIVIDEND OR REVENUE, AND HAS A
SPECIFIED TIME TO RUN.
1159. 1. A party holds $2700 of stock, which pays $80 per month revenue,
(353% interest per annum). The stock has 27 years to run. What is the present
value of the stock the current rate of interest being 6%, allowing 6% interest on
all the monthly payments of the revenue or income 3 Ans. $12614,02
A SOLUTION.
OPERATION Face of Stock º - - - - - $2700.00
Receipts by monthly payments of $80 or
To find the sum of the terms of the monthly interest for 27 years at 353% 25920.00
the monthly payments. Interest on $80 monthly payments on
1 + 323 = 324 $2700, for 52326 terms of 1 mºnº, (1 20930.40
324 - 2 = 162 to 323 inclusive) = 43603 yrs, at 6% - 30.
162 × 323 = 52326 Value of stock at maturity - - $49550.40
VALUE OF $100 STOCK FOR 27 YEARS AT 6 PER CENT.
Face of Stock assumed. sº -> - tº-e - mº - - $100.00 100
Interest on $100 for 27 years at 6 per cent - s - - 162.00 392.82 49550.40
Interest on 50c., the monthly interest on $100 at 6 per cent,
for 52326 terms of 1 month = 4360+ years at 6 per cent, - 130.82 $12614.02 Ans.
$392.82
2. Suppose in the above problem, interest is allowed only on the annual pay-
ments of the monthly revenue instead of the monthly payments, what would be
the present value of the stock # Ans. $12575,34.
SOLUTION.
OPERATION Face of Stock for 27 vears at 3 § - - - $ 2700.00
find th f the t f Monthly interest for 27 years at 355 per cent
To ". sº º O or at $80 per month º - ºn tº - 25920.00
Interest on $960 yearly payments made by
1 to 26 monthly installments on $2700 for 351 terms
1 or, 27 of 1 year, (1 to 26) at 6 per cent, value of
2 ) 27 13 stock at maturity, - - - - - - 20217.60
13} 351 $48837.60
26
351
VALUE OF $100 FOR 27 YEARS AT 6 PER CENT.
Face of stock assumed, - - - - - - - - $100.00 100
Interest on $100 for 27 years at 6 per cent, - º - - 162.00 388.36 || 48837.60
Interest on $6, the yearly interest on $100 for 351 terms of
1 year at 6 per cent, am me an as ºn tº me - 126.36 $12575.34 Ans.
Value of $100, - - - - - - - - - - $388.36
NoTE.—See Stocks and Bonds in this work for extended operations of a similar character.
592 soule's PHILOSOPHIC PRACTICAL MATHEMATICS. fºr
TO FIND THE RATE OF INTEREST RECEIVED AND PAID IN THE
SALE OF PROPERTY ON THE INSTALLMENT PLAN.
1160. A Homestead Association sold to Levin Cooper Soulé, a house and lot
for $5000, on the following conditions of payment: Cash $1000; the remainder in
80 notes of $50 each, bearing 7 per cent interest and payable monthly, as the
months expire, in 80 consecutive months.
In this transaction, what is the rate per cent interest that the homestead
makes and what per cent interest does Levin Cooper Soulé pay on the $4000?
Ans. The Homestead makes and Levin Cooper Soulé loses,
3.543% + 3.453% = 7% + .5512+ 9% use of inter-
est on interest payments, = 7.5512+ 7.
SOLUTION.—First Step.
Since the 80 notes bear 7 per cent interest and are payable in consecutive
monthly periods as the months successively expire, the interest thereon is obtained
by finding the interest on $50 at 7 per cent for the whole number of months that all
the notes bear interest. This we find as follows: The notes mature successively in 1 to
80 months, thus forming an arithmetical series of 80 terms; hence the sum of the
first and last terms, multiplied by half the number of terms, gives the total number
of months that all the notes drew interest. Thus, 1 + 80 = 81 × 40 = 3240 months. .
FIRST OPERATION SECOND OPERATION
To find the interest. To find the interest.
$ Int, on first note for 1 mo. is - $ .29#
50 Int, on last note for 80 mos. is - 23.33%
12 7
3240 Int, on first and last notes is - $23.62%
| $ 945 interest on all the notes. $23,624 — 2 = $11.84% average interest
per month.
11.844 × 80 months = $945 interest on
all the notes.
Not E. –This $945 interest was received on the $4000 during the 80 months.
Second Step.
STATEMENT REMARK.—The Homestead Association
To find the rate per cent interest on the $4000. receives only 3.543% interest . On the
w ſº $4000 but it had the use of the monthly
12 | 80 266; | 945 payments of the monthly maturing notes
tºms — & ſº sº
| $266; 3.543% Ans. with the accrued interest, thus making
Lº wº an arithmetical series of 79 terms of note
See page 599, for an explanation of º
this kind of work. and interest payments.
YY RATE OF INTEREST RECEIVED AND PAID, 593
Third Step.
To determine the worth to the Homestead Association of the 79 note pay-
ments at 7 per cent interest, we proceed as follows: The 79 note payments form an
arithmetical series of 79 terms; therefore, as above explained, the sum of the first
and last terms, multiplied by half the number of terms, gives the total number of
months that the Homestead received interest on the $50 monthly note payments.
Thus, 79 + 1 = 80, the sum of the number of terms. 80 × 39% = 3160 months that
the Homestead had the use of $50.
NOTE.-39% is # of the number of terms.
The interest on $50 for 3160 months at 7 per cent is (******) $9213. This
sum being the interest worth of the 79 note payments, during the 80 months, it is
as justly to be credited to the receipts of the $4000 as is the $945 interest on the
monthly maturing notes. The rate per cent interest that the $921; is of $4000 for
the 80 months, worked as in the above problem to find the rate per cent $945 was of
$4000, is 3.453 per cent, which added to 3.543 per cent obtained above by the interest
received, = 7 per cent the Homestead gains on the $4000, exclusive of the Interest
on the interest of the 79 monthly payments of interest received.
Fourth Step.
To include the interest on the interest of the 79 successive monthly payments
of accrued interest as gain for the Homestead, we proceed as follows: The payment
of interest on the first $50 note for one month at 7 per cent was 29% cents which
was used by the Homestead for 79 months; the interest on the second note was 584
cents which was used by the Homestead for 78 months; thus the interest increases
in an arithmetical ratio of 29% cents, and the time decreases by an arithmetical
ratio of 1 month. The following statement shows in detail the interest on 79
accrued interest monthly payments:
.29% × 79 = 23.04% 6,413 × 58 = 372.16; 12.54 × 37 = 464.04% | 18.66% x 16 = 298.66%
.584 × 78 = 45.50 6.703 × 57 = 382.37} | 12.83; x 36 = 462.00 | 18.95% x 15 = 284.374
.874 × 77 = 67.37+ 7.00 × 56 = 392, 13.124 × 35 = 459.37} | 19.25 × 14 = 269.50
1.163 x 76 = 88.66; 7.29% x 55 == 401.04% | 13.413 × 34 = 456.16# | 19.54% x 13 = 254.04%
1.45; x 75 = 109.37% 7.584 × 54 = 409.50 13.703 × 33 = 452.37} | 19.83% x 12 = 238.00
1.75 × 74 = 129.50 7.873 × 53 = 417.374 14.00 × 32 = 448.00 20.12% x 11 = 221.37%
2.04% x 73 = 149.04% 8.163 x 52 = 424,66; 14.294 × 31 = 443.04% 20,413 × 10 = 204.16%
2.33} x 72 = 168. 8,453 x 51 = 431.374 || 14.58; x 30 = 437.50 | 20.703 × 9 = 186.37%
2.624 × 71 = 186.37% 8.75 × 50 = 437.50 | 14.874 × 29 = 431.37} | 21.00 × 8 = 168.00
2.91; x 70 = 204.16# 9.04% x 49 = 443.04% 15.163 x 28 = 424.66% 21.29% X 7 = 149,04%
3 203 × 69 = 221.374 9.33% x 48 = 448. 15.455 x 27 = 417.37} | 21.58+ x 6 = 129.50
3.50 × 68 = 238. 9.623 x 47 = 452,37} | 15.75 × 26 = 409.50 21.87% x 5 = 109.37}
3.79% × 67 = 254.04% 9.913 × 46 = 456.16# | 16.04% × 25 = 401.04% 22.163 x 4 = 88.66%
4.08% x 66 = 269.50 10.203 × 45 = 459,37} | 16.33+ x 24 = 392.00 22.45% x 3 = 67.37%
4.374 × 65 = 284.374 || 10.50 × 44 = 462.00 | 16.624 × 23 = 382.37% 22.75 × 2 = 45.50
4.663 × 64 = 298.66; 10.79% x 43 = 464.04% | 16.91; x 22 = 372.16# 23.04 × 1 = 23.04%
4.955 x 63 = 312.373 || 11.08; x 42 = 465.50 || 17.20; x 21 = 361.37% 23.33.4 × 0 = 00.00
5.25 × 62 = 325.50 11.374 × 41 = 466.374 17.50 × 20 = 350.00 ——
5.543 x 61 = 338.04% 11,663 × 40 = 466.66; 17.79; × 19 = 338.04% |Tot. products, $24885.00
5.83% x 60 = 350. 11.953 × 39 = 466.37} | 18.08% x 18 = 325.50
6.124 × 59 = 361.374 / 12.25 × 38 = 465.50 | 18.374 x 17 = 312.37%
By multiplying each interest payment by the time it has to run, we have in
the product the sum upon which to compute the interest for one month to equal the
interest on the 79 interest payments for the 79 different periods of time.
594 soule's PHILOSOPHIC PRACTICAL MATHEMATICs. *
REMARK.—See Article 1161, following this solution for an abbreviated method
of finding the interest that would accrue on the interest payments made on a
Specified number of equal installments payable at successive and consecutive
periods.
Fifth Step.
By the above statement we find there was $945 interest paid, thus: 29, 4-
23.33% = 23.62% × (80 + 2). We also find that the interest on the interest of the
79 accrued interest payments amounts to $145,164, which is equal to .5512+ per
cent on $4000 for 79 months, as per the following operations:
OPERATION OPERATION
To find the interest on $24885 To find the per cent that or thus:
for 1 month at 7 per cent. $145.16+ is of $4000 for
$ 79 months, $145.164 - 263; 1%
248.85 40.00 = .5512+ ſº. 263% 145.16#
12 || 7 12 || 79 100
*= | s= summ-ass- –
$145.164 interest. $263;
This .5512+ 76 added to the 7 per cent that the Homestead gains, as shown
in the first part of the Solution, gives 7.5512+ 7, that the Homestead realizes on
the $4000 during the 80 months, the use of money being worth 7 per cent.
Siarth Step.
The buyer pays 7 per cent interest on the notes as they mature. Thus he
pays 29% cents on the first note and $23.33% on last note, the sum of which is
$23,623. This $23.62% -- 2 = $11,814 average monthly interest. $11.814 × 80 mos.
= $945 interest paid by the buyer on the $4000. This is, as shown in the first part
of the work, second step, 3.543 per cent.
By the monthly payments the buyer loses the use of the 79 successive
monthly payments of $50 and the value of such use as shown above in the third
step, is $921; which = 3.453 per cent on the $4000. Then 3.543 + 3.453 = 7 per
cent interest that he pays. .*
The buyer also loses the use (not the payment) of the interest on the interest
of the 79 successive monthly payments, which, as shown in the fifth step, amounts
to $145.16%. This amount, as shown in the fifth step, is equal to .5512+ 7% on the
$4000 for 79 months. Hence the buyer loses, altogether, 3.543% + 3.455% +
.5512+ 7% = 7.5512+ 76 on the $4000 for 79 months, the use of money being worth
7 per cent. &
TO FIND THE AMOUNT OF INTEREST THAT WOULD ACCRUE ON THE INTERESTS
RECEIVED ON THE SERIAL OR PERIODICAL PAYMENTS OF A GIVEN AMOUNT
DIVIDED UP INTO A CERTAIN NUMBER OF EQUAL INSTALLMENTS, AND
PAYABLE AT THE END OF SUCCESSIVE CONSECUTIVE PERIODS,
1161. The method of finding the sum of the payments and their regular
interests having been given in full in the preceding problem, we present the fol.
Iowing shorter method of finding the amount of the interest on the Interest pay
4× INTEREST ON INTEREST PAYMENTS. 595.
ments, supposing each original interest payment to be invested at simple interest
from the time paid until the maturity of the last payment.
The first interest payment will bear interest for one month less than the
whole number of months; hence, if we let A represent the interest on the first pay-
ment for one month, and let M equal the remaining number of months, we will have
the following series and their summation to represent the successive interests, and
their total aggregate or final sum.
Diagram for the summation of the compounding of the successive increasing monthly inter.
ests, with the successive decreasing monthly periods,
INTEREST.S.
1st month = a, X m = am — 0
2d “ = 2a × m — 1 = 2am — 2a.
3d ‘‘ = 3a × m — 2 = 3am — 6a
4th “ = 4a × m — 3 = 4am — 12a
5th ** = 5a × m — 4 = 5am — 20a
6th ‘‘ = 6a × m — 5 = 6am — 30a
21am — 70a
Sum of interests = a (21m – 70)
The quantity (21m) is the sum of the arithmetical series from 1 to 6 = (21)
multiplied by the number of months m (6) = (21 × 6). And the minus quantity to
be subtracted (70), the sum of the last column, is the sum of a compound series or
rectangular pyramidal series, and is found for any term, (n) by cubing it, and sub-
tracting (n), and dividing the remainder by 3, hence formula (**) or as in above
diagram (6°– 6) = (70).
3
Or, it may be found by multiplying the number of terms (n) by one less, and
one more than itself as 5 × 6 × 7 and divide their product by 3 = 70, which sub-
tracted from (21 m) the remainder is the equivalent number of months to be multi-
plied by one month’s interest (a), which will give the amount on which to calculate
the interest for one month, at the given rate.
When a number of payments have been made to find the final Maturity Value
of the interest on the remaining payments, proceed as follows: First find the value
of the interests for the whole period by the method just given; then find the value
of that part of the series that has been paid, and subtract the last from the first.
To find the Present Value of the unmatured part of a series, find the final
value as above, and discount the amount by the customary methods. The usual
way would be by the principles of Bank Discount. A more ethical way would be to
add the accrued simple interest on the unmatured notes to their face value.
CONDENSED METHODS FOR FINDING THE MATURITY VALUE OF
TEIE INTEREST ON THE INTERESTS OF A SERIES OF
PERIODICAL NOTES OR PAYMENTS.
1162. First.—Find the simple interest on the first payment for one month, or
period, at the given rate per cent, and call it (a) or 1st result. Second.—Find the
sum of the arithmetical series from one to the number of periods (s) and multiply it
596 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. †
by the number of months m = (sm) or 2nd result. Third.—Find the sum of a com-
pound series by multiplying the number of terms (n) by one less and by one more
than itself, and divide the product by three, or by this formula (n°– n) = 3rd result c.
3
Fourth-Subtract the third result from the second result, and multiply the
remainder by (a) or first result, and the product is the sum on which to compute the
interest for one month to equal the interest on the interest for the 79 different
periods of time.
These four operations may be expressed by the following formula :
A ( SIOl _nº-n
3
This formula gives the number of dollars, on which to calculate the interest
for one month at the given rate; (m) represents the number of months the first
months' interest would run, and is therefore one less than the whole number of terms
(n) of the original series.
To apply these methods to the foregoing problem we proceed as follows:
FIRST STEP SECOND STEP THIRD STEP
To find the interest on $50 for To find the sum of the arith- || To find the sum of a compound
1 month at 7 per cent. metical series and the prod- Ser JéS
$ uct thereof by the months. |80 × 79 × 81 = 511920 –
50 1 + 80 = 81 × 40 = 3240 || 3 = 170640 = c or Third
12 || 7 E SUIT], result.
tºmº ºmºmºs 3240 × 79 = 255960 = S,
| $.29% int. = a or First m. or Second result.
result.
FOURTH STEP.
Result S. m. 255960 – result c 170640 = 85320 = d or Fourth result.
FIFTH STEP.
Result d 85320 × result a 29%g = $24885.00, which is the amount upon which
to compute the interest for one month, to equal the interest on the 79 interest pay-
ments for 79 different periods of time, as shown in the above detailed solution.
TRUE DISCOUNT,
1163. True Discount is such a deduction from the face of notes or debts, as
is equal to the simple interest on the remainder for the same time and rate for which
the deduction was made. For a statement of the difference between True and Bank
Discount, see our remarks on Interest and Bank Discount, page 576.
1164. The Present Worth of a note or debt is the sum that remains after
the true discount is deducted from the face; or it is the principal which, at the
specified rate of interest and time, would amount to the face of the note or debt
when it becomes due, For example, if we loan $500 for 1 year at 8 per cent, it will
amount to $540. Hence we see that the present worth of $540, due one year hence
at 8 per cent, is $500.
Yºr TRUE DISCOUNT. 597
The proceeds or cash value of this $540, by the bankers' method of discount
is but $496.80, which is $3.20 less than by the true discount system.
True discount is used but very little in business. Business men generally
use the Merchants' and Bankers' System of Discount.
The function or purpose of true discount is properly to determine what sums
are to be invested to produce a specified amount at a given rate of interest and for
a Stated period of time. And in this respect it is the same as interest where we
have the time, rate per cent, and interest or amount given to find the principal.
The following problems will fully illustrate the operations of True Discount:
PROBLEMS,
1. What is the present worth or cash value, and the true discount of a note
of $9000 for 94 days at 8 per cent?
Or thus: What sum loaned or placed at interest for 94 days at 8 per cent,
will produce $184.15 interest, and amount to $9000.
Ans. $8815.85 present worth. $184.15 true discount.
OPERATION.
$4500 present worth assumed. 4500 $9000 face of note.
94 interest. 4594 || 9000 8815.85 present worth.
$4594 amount 94 days hence. | $8815.85 present worth. $ 184.15 true discount.
Explanation.—By inspection and reason, we see that as true discount is the interest on the
PRESENT worth, we cannot therefore operate on the $9000, the face of the note, and hence we are
obliged to assume some number to represent present worth; and to facilitate the operation. We
assume $4500, the 8 per cent Interest Divisor to represent the present worth. As explained in Cash
Notes, page 582, we assume the Interest Divisor for the reason that the interest thereon is always.
as many dollars as there are days in the TIME. Knowing this we are saved the labor of computing
the interest on the sum assumed. Accordingly the interest on the $4500 for 94 days at 8 per cent 23.
$94; which added to the $4500 gives $4594, as the amount of $4500 assumed present worth due 94
days hence at 8 per cent. º
We now observe that as we have used the time and rate given in the problem, there is the
same relationship between the $4594 amount and the $4500 of assumed present worth, as there is
between the $9000 amount of note and the unknown present worth of the same ; and hence to find
the present worth of the $9000 amount, we have but to perform a proportional operatiºn as shown
by the second line statement in the solution. The reasoning for this statement is as follows: Since
$4594 amount require $4500 present worth, $1 will require the 4594th part, and $9000 Will require.
9000 times as much.
SECOND OPERATION
By assuming $100 as the present worth.
$100 $
45 || 94 100
cº- 918.80 || 9
| $2.08; = interest. 9000.00
100. = assumed present Worth. smm.sºme ==
--> $8815.85 present worth.
$102.08% = amount 94 days hence. 9000.00 amount of note.
smm mºm-
$184.15 true discount.
598 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs.
GENERAL DIRECTIONS.
1165. From the foregoing elucidations, we derive the following general direc-
tions for calculating True Discount:
1. Assume as the present worth, the Interest Divisor for the given rate per
cent and find the amount of the same for the given time and rate per cent.
2. Then divide the assumed present worth multiplied by the given amount, by
the amount of the assumed present worth.
NOTE.-$100 may be assumed as the present worth, instead of the interest divisor, if preferred.
2. What is the true discount on $3450 for 1 year and 8 months at 9 per cent,
Counting 30 days to the month, and not allowing grace? Ans. $450.
NOTE.-This problem may be stated as follows: *
3. What sum loaned for 1 year and 8 months, at 9 per cent will produce $450
interest and amount to $3450%
- OPERATION INDICATED.
$4000 $3450 $4000
4600 || 3450 3000 Or thus: 600 || 450
Cºmº. | -E-ºsmº *- smºº-º-
0.
$3000 = present worth. $ 450 = true discount. $3000
Not E. –1 year and 8 months = 20 months = 600 days, and hence the interest on the assumed
9 per cent interest divisor is $600,
4. What sum must be loaned for 2 years at 7 per cent to amount to $8550?
o Ans. $7500.
5. What is the present worth of a claim of $2468.45 due in 123 days, money
worth 13 per cent per month ? Ans. $2325.44.
6. What principal must be loaned for 183 days to amount to $3915.90, inter-
est at 6 per cent? Ans. $3800.
7. A merchant sold $5000 worth of goods on 4 months credit, and then
offered the purchaser 3 per cent discount for cash, which was accepted. What was
the discount, and how much did it exceed the true discount at 8 per cent?
Ans. $150 discount. $129.87 true discount. $20.13 excess.
NotE.—In transactions of this kind count 30 days to the month and allow no grace or dus-
&ount day.
8. Goods were bought to the amount of $4200 on 3 months' credit when the
current rate of interest was 44 per cent. How much cash could be paid for the
goods without loss to either party ? - Ans. $4153.28.
OPERATION INDICATED,
$8000
8090 || 4200
9. A owes B $8000 due in 6 months. B proposes to deduct 5 per cent from
the face of the bill for cash. A accepts the proposition and borrows the money at
6 per cent. How much does A gain or lose? Ans. $172.00 A gains.
OPERATION INDICATED.
$8000
400 = 5 per cent discount. Interest on $7600 for 6 months at 6 per cent is $228
º-º-mºmºmº- $400 — $228 = $172 gain.
$7600
TO FIND RATE PER CENT. 599
10. What is the present worth of $4391.68 due in 1 year and 4 months, money
Worth 5 per cent, and no allowance for days of grace or discount day?
Ans. $4117.20.
11. What is the true discount on $8765.25 at 9 per cent for 495 days?
Ans. $965.25.
PROBLEMS INVOLVING INTEREST OPERATIONS.
1166. To find the Rate per cent when the Principal, Time, and Interest or
Proceeds are given :
1. The interest on $9000 for 94 days is $188; what was the rate per cent?
Ans. 8%.
OPERATION OPERATION Explanation.—In all problems
To find the 1nterest on $9000 To find the rate per cent. ** d *: * º: P.
for the given time at 1 per assume some rate to work from,
- 1
cent assumed. 23.50 | ić 00 and as the ratio will be the same,
o º whatever rate per cent we assume,
-- therefore, to facilitate the work,
9000 8% Ans. we assume 1 per cent, then, in
360 94 order to obtain a sum of interest
*: the same º
- to the 1 per cent assumed, that
$23.50 interest. the $188 bears to the required
rate per cent, we calculate the
lnterest on the principal for the given time at 1 per cent. Accordingly we find, in this problem,
$23.50 interest, We then reason as follows: Since 1 per cent gives $23.50 interest, by transposition
$23.50 required 1 per cent; and since $23.50 interest required 1 per cent, 1 cent interest will require
the 2350th part and 18800c interest will require 18800 times as much.
–
º-m-m-m-
GENERAL DIRECTIONS.
1167. From the foregoing elucidations, we derive the following general direc-
tions for finding the rate per cent:
1. Assume 1 per cent and find the interest on the principal for the given time
at the 1 per cent.
2. Then divide the given interest by the 1 per cent interest on the principal for
the given time.
NOTE.- When the proceeds are given, first subtract the same from the principal or face of
note, and thus find the interest or discount
PROBLEMS.
2. The interest on $1000 for 288 days was $48. What was the rate per cent?
Ans. 6%.
3. A merchant loaned $1580 for 120 days and received for its use $52,663;
at what rate per cent did he loan it? Ans. 10%.
4. A note for $524.80 was discounted for 47 days and $518.6336 proceeds
received. At what rate per cent was it discounted ? Ans. 9%.
5. A received $3679.20 for his note of $4200 which had 186 days to run. At
what rate per cent was it discounted ? Ans. 24%.
6OO SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. ºr
6. A note for $7543.80 was discounted for 34 days and $7401.306 proceeds
received. At what rate per cent was it discounted ? Ans. 20%.
7. A merchant invested in goods $14000 and commenced the merchandising
business. He continued 8 months, and then from his books he found that his gains
on sales were $5215, and his store expenses of all kinds, $2705. What was the rate
per cent per annum gain; what was his gain per cent on investment; and how
much more did he gain by his business than if he had loaned his money at 10 per
cent interest and received $125 per month for his services?
Ans. 17}} % gain on investment.
26%; % gain per annum.
$576.66% gain on his business in exeess of what he would
have gained by loaning his money at 10 per
cent and serving others at $125 per month.
OPERATION INDICATED.
Gain on sales, * → * * $5215 $2510
Expense, sº tº * * * 2705 14000 || 100
Net gain, tº tº e º $2510 | 17++% gain on investment.
(1743% x 12) - 8 = 26% 7% gain per annum.
Interest on $14000, 8 mos. at 10% = $933,-- salary 8 mos. at $125 = $1000,
= $1933}. Net gain of business $2510 – $1933% = $5763.
TO FIND THE PRINCIPAL WHEN THE RATE PER CENT, TIME, AND
INTEREST OR AMOUNT OR PROCEEDS ARE GIVEN.
1168. 1. What principal loaned for 94 days at 8 per cent will produce $188
interest ? Ans. $9000.
FIRST OPERATION.
$4500 assumed principal.
94 | 188
| $9000 principal, Ans.
• SECOND OPERATION.
OPERATION OPERATION Explanation.-In all prob-
lems of this kind, we as-
To find the interest on the To find the principal. ..". º Interest º
incipal. or $100 as principal, an
assumed principal $ then calculate the interest
100 thereon for 1. #". rate
tº e and time. This is done in
100 assumed principal. 18.80 || 9 order to produce an interest
45 | 94 188.00 which bears the same re-
* &ºm=s=mº lationship to the assumed
2.083 i I’OSU, O principal that the given in-
$ 08; interest $9000 Ans terest bears to the required
principal that produced it.
In the first operation we assumed the 8 per cent Interest Divisor, $4500, as the principal; and
Seeing that the interest on $4500 for the given time and rate per cent is $94, we reason thus: Since
$4500 principal gives $94 interest, conversely $94 interest required $4500 principal. And since $94
Yºr INTEREST To FIND THE PRINCIPAL. 6OI
interest required $4500 principal $1 interest will require the 94th part, and $188interest will require
188 times as much.
In the second solution we found the interest on the assumed principal to be $2,083, Having
this interest, we make the proportional statement, reasoning as follows: Since $100 principal gives
$2,085 interest, conversely $2.08; interest required $100 principal. And sincel&#9c. ($2.08; reduced)
interest require $100 principal, &c. interest will require the 1880th part, and # or a whole cent will
require 9 times as much; and since 1c. interest requires the result of the statement now made,
18800c. ($188 reduced to cents) will require 18800 times as much. -
NoTE.—We much prefer the first operation, for the reason that the interest on the Interest
Divisor, for the time and at the rate percent, will always be the same as the number of days. And
hence, this being known, we save making one interest calculation in the operation.
2. What principal loaned for 121 days at 10 per cent will amount to $7442?
Ans. $7.200.
OPERATION.
$ Explanation.—When the amount is
3600 = assumed principal. given, instead of the interest, then
Amount = 3721 7442 = amount. add the interest on the assumed princi-
! — pal to the assumed principal and then
| $7200, Ans. make the proportional statement.
3. What principal discounted for 369 days at 8 per cent will give $1836
proceeds? Ans. $2000.
OPERATION.
$ ſºlº º º
== & © are given, instead of the interest, then
Proceeds = 4131 § - tº: principal. subtract the interest on the assumed
I’OCCCCHS = = proceeds. principal from the assumed principal
-º-, - and then make the proportional state-
$2000, Ans. ment.
GENERAL DIRECTIONS.
1169. From the foregoing elucidations, we derive the following general direc-
tions for finding the principal :
1. Assume $100 or the Interest Divisor as principal, and calculate thereon the
interest for the given time and rate per cent.
2. Then divide the assumed principal multiplied by the given interest, by the
interest on the assumed principal.
NOTE.—If the amount is given instead of the interest, then add the interest on the assumed
principal to the assumed principal, and divide the sum into the product of the assumed principal
and the given amount.
If the proceeds are given instead of the interest, then subtract the interest on the assumed
principal from the assumed principal, and divide the difference 1nto the product of the assumed
principal and the given proceeds.
PROBLEMS.
4. A banker loaned a sum of money for 369 days at 8 per cent and received
$164 interest. What was the sum loaned ? Ans. $2000.
5. A merchant loaned a certain sum for 1 year and 3 months at 5 per cent,
and received $750 interest. What was the principal loaned ? Ans. $12000.
6. What principal loaned for 90 days at 6 per cent will amount to $2030%
Ans. $2000.
6O2 soul.E's PHILOSOPHIC PRACTICAL MATHEMATICs. º:
7. Discounted a note for 183 days at 5 per cent and received $1637.30 pro-
ceeds. What was the face of the note? Ans. $1680.
8. What principal loaned for 64 days at 8 per cent will amount to $1369.20%
Ans. $1350.
9. A merchant discounted a note for 64 days at 5 per cent and received
$178.40 proceeds. What was the face of the note * Ans. $180.
10. A father desires to place his son in college for a four years' course of
study. The yearly expense for board, tuition, etc., is $400, and for his clothes,
etc., he allows him $150 per year. He now wishes to know what sum of money he
must place at interest at 5 per cent, so that the interest will just pay the yearly
expense of his son 3 Ans. $11000.
11. What principal loaned for 124 days at 7 per cent will produce $130.20
interest ? Ans. $5400.
OPERATION
y 36000 = 1% int. divisor as assumed prin. Explanation.—Since there is no
868.00 130.20 7 per cent Interest Divisor, we
assume as principal 36000, the 1
per cent Interest Divisor, and then
$5400, Ans. multiply the interest at 1 per cent,
which is equal in dollars to the
number of days, by 7, the rate per cent, and thus produce $868.00 interest. The solution statement
is then made as in the preceding problems.
12. What principal loaned for 72 days at 11 per cent will produce $33
interest ? Ans. $1500.
TO FIND THE TIME, WHEN THE PRINCIPAL, RATE PER CENT, AND
INTEREST, OR WHEN THE PRINCIPAL AND THE AMOUNT OR
PROCEEDS AND THE RATE PER CENT ARE GIVEN.
PROBLEMIS.
1170. 1. A note for $9000 was discounted at 8 per cent, and the interest
amounted to $188. For how many days was it discounted ? Ans. 94 days.
1 year or 360 days assumed.
OPERATION OPERATION
To find the interest on the $9000 at 8 per cent To find the number of days.
for 1 year assumed.
ds.
$90.00 principal. 360
8% interest. 720 | 188
$720.00 interest for 360 days. 94 days, Ans.
Explanation.—In all problems of this kind, we first assume 360 days, or 1 year. Then, in
order to obtain a sum of interest that bears the same relationship to the 360 days of assumed time
that the $188 given interest bears to the required time, we calculate the interest, on the principal
loaned for the assumed time at the given per cent. Accordingly, we produce, in this problem, $720
interest. We then reason as follows: Since 1 year's time, with the given principal and rate, giye
$720 interest, by transposition $720 interest required 1 year's or 360 days’ time; and since $720
interest required 360 days' time, $1 interest will require the 720th part, and $188 interest will require
188 times as many, which is 94 days. *
Nºrz-In case the interest is computed at 365 days to the year, then 365 would be the days
tº SSUlſſle Ol.
ºr INTEREST TO FIND THE TIME, 603
2. A merchant borrowed $1850 at 8 per cent, and paid $197,334 for its use.
How long did he have the money? Ans. 1 year and 4 months.
OPERATION INDICATED.
1 year or 360 days assumed.
$18.50 Y. DS.
8% | 1. 360
tº Eºmº 148 148 || 592
$148.00 interest. 3 592 Or, 3
480 ds. = 1 yr. 4 mos.
14 yrs. = 1 yr. 4 mos.
GENERAL DIRECTIONS.
1171. From the foregoing elucidations, we derive the following general direc-
tions for finding the time:
1. Assume 360 days, 1 year, and find the interest on the given principal at #3
given rate per cent and for the assumed time. J.-, ºr
2. Then divide the given interest multiplied by the assumed time, by the iº
on the principal for the rate per cent and assumed time.
NOTE: 1.--When the amount is given, first subtract the interest from the same to find the
principal.
NOTE 2.--When the proceeds are given, first add the interest to the proceeds to find the
principal.
NOTE 3.—In case the interest or dišcount has been computed at 365 days to the year, then
assume 365 days.
PROBLEMIS.
3. A note for $6000 was discounted at 5 per cent and the interest or discount
was $78.33%. For how many days was it discounted ? Ans. 94 days.
4. A merchant borrowed $2500 at 44 per cent, and paid $38.75 for its use.
How long did he have the money? Ans. 124 days.
5. Loaned a sum of money at 8 per cent until it amounted to $461.37. The
interest was $6.37. How long was it loaned ? Ans. 63 days.
6. A note was discounted at 10 per cent and $800 proceeds received. The
discount was $200. For what time was the note discounted ?
$ Ans. 720 days, or 2 years.
7. A broker loaned $50000 at 12 per cent, and received $550 interest for it.
How long did he lend it? Ans. 33 days.
8. How long will it take for $2000 to amount to $3000 at 10 per cent interest?
Ans. 5 years.
9. A note dated November 1, 1895, was discounted December 23, 1895, at 6
per cent. The face of the note was $6231.50 and the proceeds $6187.8795. Grace
and discount day were counted. For how many days was it discounted and how
long did it run after being discounted ?
Ans. 42 days discounted.
41 days to run after discounted not including discount day.
OPERATION INDICATED.
$6231.50 DS.
6 360
373.89 || 43.6205
$ 373,8900
42 days.
6O4 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
10. A note for $9730 bearing 8 per cent interest dated November 24, 1895,
payable in one year, was discounted at 12 per cent, the proceeds being $9883.9934.
What was the time and the date of discount, allowing grace and discount day?
Ans. 180 days time. June 1, 1896 date discounted.
OPERATION INDICATED.
$ 1 year, or 360 days assumed.
9730. $10514,8866-1-
45 || 363 12%
$ 784.8866–1– interest. $ 1261.786392 interest for 1 year.
9730.00
$10514.8866–1– maturity value.
9883.9934 proceeds.
$ 630,8932 discount. 180 = days the note was discounted
NOTE. —365 days —H 3 days of grace and 1 discount day = 369 days had the note been dis.
counted November 24th, 369 – 180 = 189 that the note was discounted after November 24th, which
is June 1, 1896. See Time Table, page 298
1172. T A. B. L TE
SHOWING IN HOW MANY YEARS A GIVEN PRINCIPAL WILL DOUBLE ITSELF.
ds.
1261.7863 | 360 = 1 year.
630.8932
e AT COMPOUND INTEREST. | COMPOUND INTEREST ON 1 DOLLAR FOR 100 YEARS.
IRATE At Simple – |
& Compounded
Interest. cººl yº, º Amount. || Years. | Per Cent. Accumulation.
96. YEARS. YEARS. YEARS. YEARS. $1 100 1 $ 2.75
1 100. 69.666 69,487 69.400 1 100 2 7.25
1} 66.66 46.556 46.382 46.298 1. 100 2} 11.75
2 50.00 35.004 34.830 34.743 1 100 3 19.25
2} 40.00 28.071 27.899 27.812 1 100 3# 31.25
3 33.33 23.450 23.278 23.191 1 100 4. 50.50
3# 28.57 20.150 19.977 | 19.890 1. 100 4% 81.25
4 25.00 17.673 17.502 17,415 1 100 5 131.50
4} 22.22 15,748 15.576 15.490 1 100 6 340.00
5 20.00 14.207 14.036 13.946 1 100 7 868.00
5% 18.18 12.946 12.775 12,686 1 100 8 2,203.00
6 16.67 11.896 11.725 11.639 1 100 9 5,543.00
6# 15.38 11.007 10.836 10,750 1 100 10 13,809.00
7 14.29 10.245 10.075 9,989 1. 100 12 84,675.00
73 13.33 9.585 9.914 9.328 1 100 15 1,174,405.00
8 12.50 9.006 8.837 8.751 1. 100 18 15,145,007.00
8} 11,76 8 497 8.346 8.241 1 100 24 2,551,799,404.09
9 11.11 8.043 7.874 7,788 1 100
9} 10.50 7.638 7.468 7.383
10 10.00 7.273 7.121 7,026
12 8.34 6.110
DIFFERENCE BETWEEN BANK DISCOUNT AND TRUE DISCOUNT,
1173. 1. What are the proceeds or cash value of $8000 for 64 days at 10 per
cent, by the bankers' system of discount, and what is the present worth or cash
value by the true discount system of work?
Ans. $7857.78, proceeds by bankers' system.
$7860.26, present work by true discount.
º: BANK AND TRUE D1scount CoMPARED. 605
OPERATIONS.
Banking System. True Discount System.
$8000 $3600
36 || 64 3664 || 8000
8000.
$ 142.22 bank discount.
$7857.78 proceeds,
$7860.26 present worth.
8000.00
$ 139.74 true discount.
2. What is the bankers' discount and the true discount of $15000 for 75 days
at 1 per cent per month, and what is the difference 3
Ans. $375.00, bank discount.
$365.85, true discount. $9.15, difference.
º 3. What is the difference between the interest on $5000 for 3 years 4 months
and 20 days (1220 days) at 6 per cent, and the true discount for the same time and
rate #
Ans. $171.79.
4. I hold a note of $10000, due in 10 years or 3600 days without grace. How
much will I receive for it if I discount it at 10 per cent bank discount, and how
much if I discount it at the same rate per cent by the true discount system?
Ans.
Nothing by bank discount.
$5000 by true discount.
TO FIND THE RATE PER CENT OF INTEREST PAID, OR THE RATE
PER CENT TRUE DISCOUNT, WHEN NOTES, ETC., ARE
DISCOUNTED BY BANK CUSTOM.
1174. 1.
A banker discounted two notes of $500 each, one for 34 days at 6
per cent, and the other for 240 days at 6 per cent. What was the rate per cent true
discount charged on each note 3
500 face of note.
60 || 34
$2.834 interest.
$497.163 proceeds.
$
5.00 face of note.
69 || 249 4
$20.00 interest.
$480. proceeds.
Ans. 64%
% on the 34 days' note.
64% on the 240 days' note.
OPERATION FOR 34 DAYS.
$
3 || 14.91.50
360 || 34
| .4695% int. on pro-
ceeds at 1%.
OPERATION FOR 240 DAYS.
4šſ 4
3 369 || 24% 80
$3.20 int. at 1% on
proceeds.
%
1
8 3.2% | 20.9% 50
64%
Ans.
6O6 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICs. º:
łºplanation.-In answering these questions we first find, as shown by the first statement, the
Proceeds of each note according to the bankers' system of discounting. Then, considering the
proceeds as principal, we have the question contained in Problem No. 1, on page 599, which is the
principal, time and interest given, to find the rate per cent. To answer this question we first find,
as shown in the second statement, the interest on the proceeds of the notes now used as principals,
for the respective time at 1 per cent, and then make the proportional statement to find the true
discount with this interest, the 1 per cent, and the interest obtained by discounting the notes
according to bank custom. The reasoning for the last two statements of each question was fully
explained on page 599, and hence we now omit it.
NOTE 1–By the result of this work we see that the rate per cent of bank discount is larger
than the same rate per cent of true discount, and what is still more important, we see that this
increased difference of rate increases in proportion to the time for which the note has to run. This
being the case on all notes of the same face and at the same rate, it seems natural to conclude that
bankers would prefer long time paper. But the advantages of short credits are generally con-
sidered to be more than equivalent for any excess that might be gained by extending the time.
The reason for this increase of rate per cent on long time notes or transactions is that, on them the
discount becomes larger, and hence the proceeds or principal to which it is referred becomes smaller.
NotE 2. —Instead of working on the face of the notes in the above questions, we could as
well have assumed $100 or $1000, or better still the Interest Divisor. By working from assumed
figures the operation is often simplified, and in questions where no principal or face of note is
given we are obliged to work from assumed numbers
2. What is the rate of interest, or rate per cent true discount on a 63 day
note at 8 per cent and 10 per cent ? Ans. 8.*.*, 7% true discount, at 8%.
10...'", 9% true discount, at 10%.
3. What is the rate of interest, or rate per cent true discount on a 360 day
note at 5, 6, 7, 8, 9, 10 and 12, per cent 3
Ans. 5*, 6}}, 7#3, 843, 934, 11%, 13 F1, per cent.
TO FIND THE RATE PER CENT OF TRUE DISCOUNT CORRESPONDING
TO A GIVEN RATE OF INTEREST.
1175. 1. If I wish to realize 10 per cent interest on 94 day notes, at what
rate per cent is the true discount 3 Ans. 9;###3%.
FIRST OPERATION.
$3600 assumed present worth or proceeds. $36.94
94 interest on same for 94 days. 360 | 94
0
Int. 1”
9.643; 94.00 interest.
93$$$$ 96 Ans.
Explanation.— In this solution we first assume $3600 the Interest Divisor at 10 per cent as the
present worth or proceeds, and then add thereto the interest for 94 days at 10 per cent. This
operation gives $3694 which is the principal at interest for 94 days at 10 per cent which produced
$94 interest. The next step is to compute the interest on the $3694 for 94 days at 1 per cent, which
is $9.6443. Having this interest, and the interest on $3600 for 94 days at 10 per cent, we then make
the third statement, reasoning thus: Since $9.643; interest require 1 per cent, $1 will require the .
$9,6438th part, and $94.00 interest will require 94.00 times as much.
SECOND OPERATION
By assuming $100 present worth or proceeds.
$3694 principal or sum loaned. | $ 9.64% interest at 1%.
0
%
1
162
23.5000
93.3%; 96 Ans.
$ $
100 present worth assumed. 9 | 92350
36 94 360 | 94 43,404
s
| Tlºw. 9
$2.61% interest. 26.79%2c. interest.
$100. present worth.
$102.61 principal.
* INTEREST. 6O7
2. What is the corresponding rate of true discount for a 63 day note at 8
per cent ? Ans. 74}} %.
IFIRST OPERATION. h
$4500 = assumed present worth or proceeds. $ %
63 = interest on same at 8% for 63 days. 45.63 1
ºs-sms me 360 | 63 7.98% | 63,00
$4563 = principal. * |
$7.98% interest. 733} % Ans.
SECOND OPERATION
By assuming $100 as the present worth or proceeds.
$ $ %
100 P. W. 1.01,40 1
45 || 63 360 | 63 .3549 || 2
* - 1.4000
$1.40 interest. ,1774; interest.
100. P. W. t
ºs-e-mas
$101.40 principal.
3. What is the true discount to correspond with 93 day notes at 12 per cent,
and with 360 day notes at 15 per cent and at 20 per cent 3
Ans. 11 º', 'ſ, on 93 day note at 12%.
13% 7% on 360 day note at 15%.
163% on 360 day note at 20%.
7#3} % true discount.
ANNUAL, SEMI-ANNUAL AND QUARTERLY INTEREST.
1176. Annual, Semi-Annual and Quarterly Interest is the regular simple
interest on notes or debts, and the interest on the unpaid simple interest from the
time it becomes due until paid.
Annual, Semi-Annual and Quarterly Interest is due after it has been accruing
for the time specified for the interest payments, except the interest on the interest
which is not due till final payment.
NOTE. –In some States, a neglect to collect the annual interest on notes drawn, with interest
payable annually, is considered as a waiver of the contract to pay annual Interest.
It is a principle of law that, in the absence of any agreement, all money or
claims bearing interest, or that is due and not paid, shall draw simple interest from
the time that they become due until paid. Hence, to entitle a party to annual, semi-
annual or quarterly interest, a special agreement to that effect is necessary. In
case of notes, this agreement is inserted and becomes one of the obligations of the
instrument. Thus, the following note contains the annual interest contract:
$6000. ST. LOUIs, September 1st, 1895.
Four years after date, for value received, we promise to pay to the order of
D. G. HESSEE, Six Thousand Dollars with interest at 8 per cent, payable annually.
MONTGOMERY & POTTS.
In this note the words in italics make it an annual interest note, by specifying
when the interest shall be paid ; and hence, if the interest is not paid at the end of
each year, it becomes by virtue of the terms of the instrument, and in accordance
with law and equity, a new debt, and will draw simple interest until paid.
6O8 soule's PHILOSOPHIC PRACTICAL MATHEMATICS.
The following work will show the operation of annual interest:
1. In the foregoing note nothing being paid until maturity, what will be the
amount then due 3 Ans. $8150.40.
OPERATION
- Explanation.—By the conditions
of the note we see that the inter-
est was payable at the end of
Face of note, $6000.00
(Interest on note for 1 year at 8%, $480.)
Interest on note for 4 years at 8%, 1920.00 ...h year, but as it was not paid
Interest on $480 for 3 years at 8%, $115.20 until the maturity of the note,
Interest on $480 for 2 years at 8%, 76.80 }. º,º: find pººl.
0 º Interest ODI tºle Ia C6. O €. In Ote
Interest on $480 for 1 year at 8%, 38,40 230.40 for the 4 years at 8 per cent. We
g then observe that the first year's
Total amount due at maturity, $8150.40 interest, $480, was due and not
paid for 3 years; that the second
year's interest, $480, was due and not paid for 2 years; and that the third year's interest, $480, was
due and not paid for 1 year; and hence, by the conditions of the contract, and in conformity to
law and equity, we find the interest on the three sums of annual interest for the respective time
that they were due and not paid. The last year's interest being paid when due, there is no interest
to charge thereon. We now add the total annual interest and the simple interest on the annual
interest for the time that it was due and not paid, to the face of the note, and thus obtain the
amount due at the maturity of the note.
A Quarterly Interest Note,
CHICAGO, August 10, 1895.
Two years after date, for value received, I promise to pay to the order of
QUIN & HILL Four Thousand Three Hundred Seventy-two Dollars, with 10 per
cent interest payable quarterly.
$4372.
J. L. CAILLOUET.
2. If nothing is paid on the above note until maturity, what amount will
then be due # Ans. $5322.91.
OPERATION
Face of note, tº- *- - L- es º * - gº tº gº * * gº gº ºn $4372.00
Interest on same for 2 years at 10 per cent, tº • * & gº sº sº tºº º ſº 874.40
(The interest on face of the note for 1 quarter is $109.30.)
Interest on 1st quarter's interest, $109.30, for 21 months, E. : gº tºº wº $19.13
Interest on 2d quarter's interest, 109.30, for 18 months, tº ſº tº tº-> 16.40
Interest on 3d quarter's interest, 109.30, for 15 months, tº gº º ºs 13.66 gº
Interest on 4th quarter's interest, 109.30, for 12 months, º tº º ſº tº 10.93
Interest on 5th quarter's interest, 109.30, for 9 months, tºº * . * > tº e 8.20
Interest on 6th quarter's interest, 109.30, for 6 months, tº- g- tº tº 5.46
Interest on 7th quarter’s interest, 109.30, for 3 months, tº º tº a gº sº tº 2.73
Interest on 8th quarter's interest, 109.30, for 0 months, º gº ºne º ſº-
Amount of interest on quarterly interest, * º sº tº * . º {- gº 76.51
Total amount due at maturity, tº- * . & gº {_º tº wº * * gº * , $5322.91
NotE.—Instead of finding the interest on each quarter's interest separately, we could have
added the months of time together, making 84, and then found the interest on 1 quarter's interest
for the whole time, 84 months or 7 years.
A Semi-Annual Interest Due Bill.
$660.
NATCHEz, October 15, 1895.
Due A. & M. Moses, for value received, Six Hundred and Sixty Dollars, with
interest at 6 per cent payable semi-annually.
FOWLER & DAVIS.
Yºr INTEREST. 609
3. Nothing being paid on the above due bill until the day of settlement, two
years 4 months and 20 days after date, what was then due Ans. $760,01.
OPERATION.
Face of due bill, * * * * * * * * * * * * * * *
Interest on same for 2 years, 4 months and 20 days at 6 per cent, * * * * *
(The interest on the face of due bill for 6 months is $19.80.)
Interest on 1st semi-annual interest, $19.80, for 22 months 20 days.
Interest on 2d semi-annual interest, 19.80, for 16 months 20 days.
Interest on 3d semi-annual interest, 19.80, for 10 months 20 days.
Interest on 4th semi-annual interest, 19.80, for 4 months 20 days.
$660.00
94.60
5,41
$760.01
Interest on $19.80 semi-annual interest for 54 months 20 days at 6 per cent = -
Total amount due on settlement, * $º - - gº * -> sº -> -
COMPOUND INTEREST.
1177. Compound interest is interest in which the principal is increased at the
expiration of each period of interest payment by the interest on the principal for
such period. The amount thus produced forming a new principal on which the
interest is calculated till the next period of interest payment, when the new princi.
pal is increased by the addition of the last interest, and so on from one period of
interest payment to another, until final settlement is made.
The periods of time for the interest payments may be annually, semi-annually,
quarterly, monthly, weekly, or daily.
There is no law forbidding the charging and receiving of compound interest,
provided the person to whom the money was loaned is willing to pay it, but it cannot
be legally collected unless there had been a previous agreement to that effect.
1. What is the compound interest of $9000 for 4 years at 8 per cent?
Ans. $3244.40.
FIRST OPERATION. SECOND OPERATION.
Principal, * * * * * $9000.00 | Principal for the 1st year, - $9000.00
Interest on principal for 1 year at 8%, 720.00 1.08
New principal for the 2d year, - $9720.00 | Principal for the 2d year, - $9720.00
Interest on same for 1 year at 8%, 777.60 1.08
New principal for the 3d year, º $10497,60 | Principal for the 3d year, - $10497.60
Interest on same for 1 year at 8%, 839.80 1.08
New principal for the 4th year, - $11337.40 | Principal for the 4th year, - $11337.4080
Interest on same for 1 year at 8%, 907.00 1.08
The amount for 4 years is - - $12244.40 || Amount at the end of the 4th year, $12244.400640
First principal deducted, - - 9000.00 | First principal deducted, - 9000.00
Compound interest for 4 years, - $3244.40 | Compound interest, - - - $3244.40
Explanation.—The operation of this problem is so clear that an explanation is almost unnec-
essary. We first find the interest on the principal for 1 year, and add it to the same, which pro-
duces a new principal for the second year's interest; and in like manner we continue till the end
of the 4 years; then we deduct the first principal from the amount and in the remainder have the
compound interest for 4 years at 8 per cent.
In the 2d operation, we first assume $1 and then find the amount of the same for 1 year at 8
6 IO soule's PHILOSOPHIC PRACTICAL MATHEMATICS. *
per cent, which is $1.08. Then, having the amount of $1, we find the amount of $9000 by multi-
plying the same by the $9000, and thus we continue to multiply the $1.08 by the successive amounts
produced until the close of the time. Then to find the interest we deduct the first principal from
the last amount.
Had we raised the ratio $1.08 to the fourth power and then multiplied by the $9000, we
would have produced the same result. Thus, $1.08×31.08×31.08×31.08×39000–$12244.40064000.
2. What is the compound interest of $500 for 1 year and 6 months at 6 per
cent payable quarterly? Ans. $46.72.
OPERATION.
Principal, - * - - - - $500.00 | Principal for 5th quarter brought for’d, $530.68
Interest on same for 3 months at 6%, 7.50 | Interest on same for 5th quarter, - 7.96
New principal for 2d quarter, - - $507.50 | New principal for 6th quarter, - tº- $538.64
Interest on same for 2d quarter, - 7.61 || Interest on same for 6th quarter, º 8.08
New principal for 3d quarter, - - $515.11 || The amount for 1 year and 6 months is $546.72
Interest on same for 3d quarter, - 7.73 First principal deducted, - - º 500.00
New principal for 4th quarter, - $522.84 || Compound int. for 1 year and 6 months, $46.72
Interest on same for 4th quarter, * - 7.84
New principal for 5th quarter, - - $530.68 |
3. What is the compound interest of $1000 for 2 years 3 months and 5 days
at 10 per cent, payable semi-annually 3 Ans. $247.58.
OPERATION.
Principal, -> * $1000.00 | Prin. for 4th 6 months brought for’d. $1157.62%
Int. on the same for 6 months at 10%, 50.00 || Interest on same for 6 months at 10%, 57.88%
New principal for the 2d 6 months, $1050.00 | New principal for 3 mos. and 5 days, $1215.50;
Interest on same for 6 months at 10%, 52.50 | Int. on same for 3 mos. and 5 ds. at 10%, 32.08
New principal for the 3d 6 months, $1102.50 || Amount for 2 years 3 months and 5 ds. $1247.58
Interest on same for 6 months at 10%, 55.12% | First principal deducted, - - 1000.00
New principal for the 4th 6 months, $1157.62} | Compound int. for 2 yrs. 3 mos. and 5 ds. $247.58
4. What will $1450 amount to at compound interest for 1 year 2 months and
12 days at 6 per cent compounded monthly? Ans. $1557.97.
In practice, the finding of compound interest is rendered more expeditious by
the use of the following Compound Interest Table:
NoTE.—To find the sum of the amounts of compound interest in the tables from 1 year to
any other year inclusive, without making the additions, take the compound interest for the next
year greater than the highest given, at the same per cent, and subtract from this the amount of 1
year at the head of the table in the same column; then remove the decimal point two places to the
right, and divide by the rate per cent; considered as an integer, the quotient will equal the sum
of all the amounts in the column from 1 year to the given year inclusive.
INTEREST,
6 II
1178. A M O U N T
OF $1, £1, 1 FR., AT COMPOUND INTEREST IN ANY NUMBER OF YEARs, NOT EXCEEDING FIFTY-FIVE.
Yrs. 1 per cent. 13 per cent. || 2 per cent. 24 per cent. || 3 per cent. |3} per cent. || 4 per cent.
1. 1.0100 1.0150 1.0200 0000 | 1.0250 0000 | 1,0300 0000 | 1,0350 0000 || 1,0400 0000
2 1,0201 1,0302 1,0404 OO00 | 1,0506 2500 | 1.0609 OOOO | 1,0712 2500 | 1,0816 0000
3 1.0303 1.0457 1,0612 0800 | 1.0768 9062 | 1,0927 2700 | 1,1087 1787 | 1.1248 6400
4 1,0406 1.0614 1,0824 3216 | 1.1038 1289 | 1.1255 0881 | 1.1475 2300 | 1.1698 5856
5 1.0510 1.0773 1.1040 8080 | 1.1314 0821 | 1.1592 74.07 | 1.1876 86.31 | 1,2166 5290
6 1.0615 1,0934 1.1261 6.242 | 1.1596 93.42 | 1.1940 5230 1.2292 5533 1,2653 1902
7 1,0721 1.1098 1.1486 8567 | 1.1886 8575 | 1.2298 7387 | 1.2722 7.926 | 1.3159 3178.
8 1.0829 1.1265 1.1716 5938 1,2184 0290 | 1.2667 7008 || 1.3168 09()4 || 1,3685 6905
9 1.0937 1.1434 1.1950 92.57 | 1,2488 6297 | 1.3047 7318 || 1.3628 9735 | 1.4233 1181
10 1.1046 1.1605 1.2189 94.42 | 1.2800 8454 | 1.3439 1638 | 1.4105 9876 | 1.4802 4428.
11 1.1157 1.1779 1.2433 7431 | 1.3120 8666 | 1.3842 3387 | 1.4599 6972 | 1.5394 5406
12 1.1268 1.1956 1.2682 4179 | 1.3448 8882 | 1.4257 6089 | 1.5110 6866 | 1.6010 3222
13 1.1381 1.2136 1.2936 O663 | 1.3785 1104 || 1.4685 3371 || 1 5639 5606 | 1.6650 7351
14 1.1495 1,2318 1.3194 7876 | 1.4129 7382 | 1.5135 8972 | 1.6.186 9452 | 1.7316 7645
15 1.1610 1.2502 1.3458 6834 | 1.4482 98.17 | 1.5579 6742 | 1.6753 4883 || 1,8009 4351
16 1.1726 1.2690 1.3727 8570 | 1,4845 0.562 | 1,6047 O644 | 1.7339 8601 | 1.8729 8125.
17 1.1843 1.2880 1,4002 4142 | 1.5216 1826 || 1 6528 4763 | 1.7946 7555 | 1.9479 00:50
18 1.1961 1.3073 1.4282 4625 | 1.5596 5872 | 1.7024 3306 | 1.8574 8920 | 2.0258 1652
19 1.2081 1.3270 1.4568 1117 | 1.5986 5019 | 1.7535 0605 || 1 9225 0132 2.1068 4918.
20 1.2202 1.3469 1.4859 4740 | 1.6386 1644 | 1.8061 1123 | 1.9897 8886 2.1911 2314
21 1.2324 1,3671 1.5156 6634 | 1.6795 8185 | 1.8602 94.57 2,0594 3147 2.2787 6807
22 1,2447 1.3876 1.5459 7967 | 1.7215 7140 | 1.9161 0341 2.1315 1158 2.3699 1879
23 1.2572 1.4084 1.5768 9926 | 1.7646 1068 || 1.9735 8651 2,2061 1448 2,4647 1555
24 1.2697 1.4295 1.6084 3725 | 1.8087 2595 || 2.0327 9411 2.2833 2849 2,5633 0417
25 1.2824 1.4509 1.6406 0599 || 1.8539 4410 || 2.0937 7793 2,8632 4498 || 2,6658 3633
26 1.2953 1.4727 1.6734 1811 | 1.9002 9270 2.1565 9127 | 2.4459 5856 2.7724 6979
27 .1.3082 1.4948 1.7068 8648 || 1.9478 0002 || 2,2212 8901 || 2.5315 6711 2,8833 6858
28 1.3213 1.5172 1.7410 2421 | 1.9964. 9502 || 2.2879 2768 2,6201 7.196 || 2,9987 0332.
29 1.3345 1.5400 1.7758 4469 || 2.0464 0739 || 2 3565 6551 2.71.18 7798 || 3.1186 5145.
30 1.3478 1,5631 1.81.13 6158 || 2.0975 6758 2.4272 6247 2,8067 93.70 || 3,2433 9751
31 1.3613 1.5865 1.8475 8882 2.1500 0677 || 2.5000 8035 | 2,9050 3148 3.3731 3341
32 1.3749 1.6103 1.8845 4059 || 2.2037 5694 2.5750 8276 3.0067 O759 || 3.5080 5875
33 1.3887 1.6345 1.9222 3140 2.2588 5086 2.6523 3524 || 3.1119 4235 | 3,6483 8110
34 1.4026 1.6590 1.9606 7603 || 2.3153 2213 2.7319 O530 3.2208 6033 || 3.7943 1634
35 1.4166 1.6839 1.9998 8955 2.3732 0519 || 2.8138 6245 || 3.3335 9045 3.9460 8899
36 1.4308 1.7091 2.0398 8734 2.4325 3532 2.8982 7833 || 3,4502 6611 || 4,1039 3255
37 1.4451 1.7348 2.0806 8509 || 2.4933 4870 || 2.9852 2668 || 3.5710 2543 4.2680 8986
38 1.4595 1.7608 2.1222 98.79 2.5556 8242 || 3.0747 8348 || 3 6960 1132 || 4,4388 1345
39 1.4741 1.7872 2.1647 4477 2.6195 7448 || 3.1670 2698 || 3.8253 7171 || 4,6163 6599
40 1.4889 1.8140 2.2080 3966 2.6850 6384 || 3.2620 3779 || 3.9592 5972 || 4,8010 2063.
41 1.5038 1.8412 2.2522 0046 || 2.7521 9043 3.3598 98.93 || 4,0978 3381 || 4,9930 6145.
42 1.5188 1,8688 2.2972 4447 2.8209 9520 | 3.4606 9589 || 4.2412 5799 || 5,1927 8391
43 1.5340 1.8969 2.3431 8936 2.8915 2008 || 3.5645 1677 4.3897 0202 || 5,4004: 9527
44 1.5493 1.9253 2.3900 5314 || 2.9638 0808 || 3.6714 5227 || 4.5433 4160 || 5.6165. 1508
45 1.5648 1,9542 2.4378 5421 || 3.0379 0328 || 3.7815 9584 || 4.7023 5855 || 5,8411 7568
46 1.5805 1.9835 2.4866 1129 || 3.1138 5086 || 3.8950 4372 || 4.8669 4110 || 6.0748 2271
47 1.5963 2.0133 2.5363 4351 || 3. 1916 9713 || 4.0118 9503 || 5.0372 8404 || 6.3178 1562
48 1.6122 2.0435 2.5870 7039 || 3.2714 8956 || 4.1322 5188 5.2135 8898 || 6,5705 2824
49 1.6283 2.0741 2.6388 1179 || 3 3532 7680 || 4.2562 1934 || 5.3960 6459 || 6.8333 4937
50 1.6446 2.1052 2.6915 8803 || 3.4371 0872 || 4.3839 0602 || 5.5849 2686 || 7.1066 8335
51 1. . . . . . . . . . . . . . . . . . . . . . . . . . 2.7454 1979 || 3.5230 3644 || 4.5154 2320 5.7803 9930 || 7.3909 5068
52 |. . . . . . . . . . . . . . . . . . . . tº e ºs 2.8003 2819 || 3.6111 1235 || 4.6508 8590 5.9827 1327 7.6865 8871
53 l. . . . . . . . . . . . . . e tº e º e e 2.8563 3475 || 3.7013 9016 || 4.7904 1247 6.1921 0824 || 7.9940 5226
54 |. . . . . . . . . . . . . . . . . . . tº º º G 2.9134 6144 || 3.7939 2491 || 4.9341 2485 6.4088 3202 || 8.3138 1435
55 . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9717 3067 || 3.8887 7303 || 5.0821 4859 6.6331 4114 || 8.6463 6692
Subtract $1 from the amount in this table to find the interest.
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JL N ſl O TWIV

X: COMPOUND INTEREST. 613
TO COMPUTE COMPOUND INTEREST BY THE USE OF THE COMIPOUND
INTEREST TABLE.
1180. To find the amount or interest when the principal, time and rate are
given :
1. What is the compound interest on $9000 for 4 years at 8 per cent?
Ans. $3244.40.
OPERATION. Ea:planation.—To find
the compound interest §
36(). sº the use of the compoun
$1.360489. = compound amount of $1 for 4 years. interest tables, first find
9000 = principal. in the table the compound
amount of $1 for the time.
$12244.40.1000 = compound amount of $9000 for 4 years. ººl rate, then aultiply
this amount by the prin-
9000. = principal. cipal, and from the prod-
e uct subtract the principal;
$ 3244.40 = compound interest. the remainder will be the
compound interest.
NOTE 1.-If the interest is compounded semi-annually, then find the compound amount on $1
for twise the number of years and for # the rate per cent. *
Thus, the compound interest on $1 for 8 years at 6 per cent payable semi-annually, is the same
as the compºund interest on $1 for 16 years at 3 per cent, payable annually.
NOTE: 2. —Iſ the interest is compounded quarterly, then find the compound amount on $1 for
4 times the number of years and for + the rate per cent.
Thus, the compound interest on $1 for 6 years at 8 per cent, payable quarterly, is the same as
the compound interest on $1 for 24 years at 2 per cent, payable annually.
º 3.—When compounded monthly, consider each month a year and divide the rate per
cent by 12.
NOTE 4.—When the amount is required for a number of years not contained in the table,
compute the same by multiplying together the amounts of any two number of years, whose sum
equals the number of years required. Thus, if the amount of 86 years at 4 per cent is required,
multiply together the compound amount of 46 and 40 years, or 36 and 50 years: For example, the
amount for 46 years is 6.07482–H and for 40 years it is 4.80102-H and the product is 29.1663323+
which is the amount of $1 for 86 years at 4 per cent.
NOTE: 5.-If partial payments are made on notes or debts bearing compound interest, the
compound amount of the principal must be first found, and the sum of the compound amounts of
the payments subtracted from it.
2. What is the compound interest on $800 for 6 years, 4 months at 44 per
cent 3 Ans. $257.43.
OPERATION.
$1.30226012 = compound amount of $1 for 6 years at 44 per cent.
800 = principal.
$1041.80809600 = compound amount of $800 for 6 years.
15,6275 = interest on $1041.81 for 4 months, at 44 per cent.
$1057.43 = compound amount on $800 for 6 years, 4 months at 44 per cent.
800. = principal.
Ǻmºsºmºsºmsºmºsº
$257.43 = compound interest on $800 for 6 years, 4 months at 44 per cent.
NoTE.—When the time includes months or days, the interest for such number of months
or days is computed on the compound amount for the even years, and added thereto, as shown
above.
614 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
3. What is the compound interest on $5000 for 12 years, 1 month, 12 days at
10 per cent compounded quarterly 3 Ans. $16548.29 compound amount.
$11548.29 compound interest.
OPERATION.
$3.27148956 = compound amount of $1 for 48 years at 24 per cent.
5000 = principal.
$16357,44780000 = compound amount of $5000 for 12 years at 10 per cent.
190.84 = interest on compound amount of $16357.45 at 10% for 42 ds.,
**smºm or for 168 ds, at 24%.
$16548.29 = compound amount.
5000.00 = principal deducted.
&
sºmºmºmºsºs º ºsmº
$11548.29 = compound interest.
4. What is the compound interest on $3000 for 20 years, 3 months, 16 days
at 3 per cent, payable semi-annually? Ans. $2490.07.
TO FIND THE PRINCIPAL WHEN THE COMPOUND AMOUNT OR INTER—
EST, THE TIME, AND THE RATE ARE GIVEN.
1181. 1. What principal at 8 per cent compound interest will amount to,
$12244.401, in 4 years? Ans. $9000.
OPERATION.
$12244.401 - $1.360489 = $9000.
Explanation.—The compound amount of $1 for 4 years at 8 per cent is $1.360489 and hence,
$12244.401 is the compound amount of as many dollars as it is equal to $1.360489, which 1s $9000.
Therefore, in all problems of this kind to find the principal, we divide the amount or interest
by the amount or interest of $1 for the given time and rate.
2. What principal at 4 per cent compound interest will produce $10000
interest in 9 years? ... Ans. $23623.25.
OPERATION INDICATED.
$10000 - .42331181.
NotE.—The $.42331181 is the compound interest on $1 for 9 years at 4 per cent. See table
and subtract $1 from compound amount.
TO FIND THE TIME, WHEN THE PRINCIPAL, THE RATE AND THE
COMPOUND AMOUNT OR INTEREST IS GIVEN.
1182. 1. In what time will $9000 amount to $12244.401 at 8 per cent?
* Ans. 4 years.
OPERATION. :
$12244.401-- $9000 = $1.360489 = compound amount of $1 which, in the table cor-
responds to 4 years.
Explanation.—Since $12244.401 is the compound amount of $9000 at 8 per cent in a certain
* COMPOUND INTEREST. 6 I5
time, the amount of $1 will be the 9000th part, which is $1.360489 and which we find in the table
under 8 per cent to correspond to 4 years.
Therefore, in all problems of this kind to find the time, we divide the amount by the given
principal and in the quotient we have the amount of $1 for the time. Then we find this amount in
the table under the given rate, and opposite thereto is the required time.
2. In what time will $2000 amount to $4584,0366 at 5 per cent compound
interest? Ans. 17 years.
TO FIND THE RATE WHEN THE PRINCIPAL, THE COMPOUND AMOUNT
OR INTEREST, AND THE TIME ARE GIVEN.
1183. 1. At what rate will $9000 amount to $12244.401 in 4 years?
Ans. 8%.
OPERATION.
$12244.401 - $9000 = $1.360489, which in the table corresponds to 8 per cent.
Explanation.—Since $9000 amounts to $12244.401 in 4 years at the required rate, $1 for the same
time and rate will amount to the 9000th part, which is, $1.360489, and which, in the table opposite
4 years corresponds to 8 per cent.
Therefore, in all problems of this kind to find the rate, we divide the amount by the given
principal and the quotient will be the amount of $1 for the given tiime at the required rate.
Then we refer to the table and find this amount opposite the given time, and the rate at the
top of the column is the required rate.
2. At what rate will $8000 amount to $22072.252 in 15 years? Ans. 7%.
NotE.—See the subject of Annuities for further work involving the principles of simple and
compound interest investments.
SIMPLE, ANNUAL, AND COMPOUND INTEREST COMPARED.
1184. What is the simple, the annual, and the compound interest on $5000, for
4 years at 8 per cent? Ans. $1600, simple interest.
$1792, annual interest.
$1802.45, compound interest.
OPERATION FOR ANNUAL INTEREST.
OPERATION
FOR SIMPLE INTEREST. Interest on the $5000 principal for 4 years at 8%, - - $1600
(The annual interest is $400 per year).
$50.00 Interest on $400 for 3 years,
8% Interest on 400 for 2 years,
Interest on 400 for 1 year, =
$400.00 -
4 yrs. the 1nt. on $400 for 6 years at 8 per cent, - º º -> 192
$1600.00 Ans. $1792
OPERATION FOR COMPOUND INTEREST BY THE COMPOUND INTEREST TABLE.
$1.36048896 = compound amount of $1 for 4 years at 8 per cent.
5000 = principal.
$6802.4480000 = compound amount of $5000 for 4 years at 8 per cent.
5000. = principal deducted.
$1802.45 = compound interest on $5000 for 4 years at 8 per cent.
6 I 6 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
By the foregoing work we see that the difference between simple, annual, and
compound interest in their effect, depends upon the time when interest money, if
due and not paid, begins to draw interest. As remarked in annual interest, a debt
begins to draw interest when due; but the time when a debt of interest becomes
due, is a question in some cases governed by business custom, and in other cases it
is a question to be decided by the parties interested. In bank discount, the interest
is payable in advance.
In simple interest, it is not considered due until the final payment of the
principal.
In annual interest, it is due after it has been accruing for the time specified
for interest payments, except the interest on interest which is not due till final pay-
ment.
Compound interest supposes all interest, whether upon principal or interest,
to be due at the end of such equal successive intervals of time as may be agreed
upon. r
When the interest is considered due the instant it has accrued and all inter-
est is made to draw interest, it is termed instantaneous compound interest.
PARTIAL PAYMENTS, OR PAYMENTS BY INSTALLMENTS.
1185. Partial payments, or payments by installments as the terms imply, are
the payments made at different times, of promissory notes, acceptances, bonds,
personal accounts, or other obligations.
At the time these partial payments are made, the creditor should and usually
does specify in writing, on the back of the note or other instrument, the sum paid
and the date of payment, and then signs his name.
It is a principle of law that a note does not draw interest until it is due,
unless the words “with interest” are contained therein. *.
There are several systems of computing interest upon notes, bonds, personal
accounts, and other obligations upon which partial payments have been made; and
as the different systems generally give different results, according as the time
intervening between the payments is more or less than a year, or the interest more
or less than the payment, it is therefore impossible in advance of knowing the exact
facts of each case, to say which system would be strictly just to all parties.
Among these various systems, the five following are mostly used: The United
States System ; the Merchants' System; the Vermont System ; the New Hampshire
System ; and the Connecticut System.
A These five systems we will define, and then illustrate by examples the United
States and the Merchants' Systems, which are in general use outside of the bound-
aries of the States of Vermont, Connecticut and New Hampshire.
THE UNITED STATES SYSTEM.
1186. The following is the decision of Chancellor Kent for computing interest
where partial payments have been made, and having been adopted by the Supreme
4× PARTIAL PAYMENTS, 617
Court of the United States, as also by several other States, is hence termed the
United States System.
I.-The rule for casting interest when partial payments have been made, is to apply the pay-
ment, in the first place, to the discharge of the interest due.
II.-If the payment exceeds the interest, the surplus goes towards discharging the principal,
and the subsequent interest is to be computed on the balance of principal remaining due.
III.—If the payment be less than the interest, the surplus of interest must not be taken to
augment the principal; but interest continues on the former principal until the period when the
payments taken together exceed the interest due, and then the surplus is to be applied towards
discharging the principal, and interest is to be computed on the balance as aforesaid.
[See Chancellor Kent, Johnson's Chancery Rep., Vol. 1, p. 17].
The practical result of this system is the loss of the use of money paid by the
person owing the debt when the same is less than the interest, from the time paid
until the sum of the payments exceeds the interest, and also the compounding of
the money in favor of the creditor as often as the amount paid exceeds the interest
then due, and hence two reasons for the debtor to defer payment as long as possible.
Thus, presuming a debtor owes a note for $10000 bearing 6 per cent interest, and
that he pays thereon $50 per month, at the close of the year he still owes the
$10000. Had he loaned the $50 monthly payments at 6 per cent he would have had
at the end of the year $616.50 to pay on his note and the accrued interest, $10600,
leaving a debt of $9983.50 instead of $10000.
TELE MERCEIANTS’ SYSTEM.
1187. The Merchants' System consists in computing the interest on the prin-
cipal, or original debt, from the time it became due until the close of the civil, or of
the fiscal year, and adding it to the principal; and then computing the interest also
upon the payments made, if any, during the year, from the time they were made to
the end of the civil, or fiscal year, and adding the same to the payments made; then
deducting the sum of the payments and interest from the amount of principal and
interest. The remainder will be a new principal. In this manner the operation is
continued from year to year, or from settlement to settlement, if made oftener than
once a year, until the final settlement.
The periods of time for which the interest is computed and the balance
brought down are termed rests.
Merchants frequently close their accounts every six months, and bankers every
quarter, in which case the directions given above for yearly settlements would be
made semi-annually and quarterly. The effect of this system is the same as the
account current and interest account with the creditors, made as often as the
accounts are balanced, and the practical result is the compounding of money for
both parties as often as settlements are made.
The Merchants' System is the only system that is applicable to book accounts
as well as notes, bonds, etc., and it is, we think, practically nearer true justice to
both parties, debtor and creditor, than the United States System.
1188. VERMONT SYSTEM IS
I. “When payments are made on notes, bills, or similar obligations, whether payable on demand or
6 I 8 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. *
at a specified time, with interest, such payments shall be applied: FIRST, TO LIQUIDATE THE INTEREST
that has accrued at the time of such payments, and SECONDLY, TO THE EXTINGUISHMENT OF THE
PRINCIPAL.
II. “ The annual interests that shall remain unpaid on notes, bills, or similar obligations, whether
payable on demand or at a specified time, “with interest annually,” shall be SUBJECT to simple INTEREST
from the time they become due to the time of final settlement.
III. “If payments have not been made in any year, reckoning from the time such annual interest
began to accrue, the amount of 8 uch payments at the end of the year, with interest thereon from the time
of payment, shall be applied: FIRST, to LIQUIDATE the SIMPLE INTEREST that has accrued from the
UNPAID ANNUAL INTERESTS; SECONDLY, to LIQUIDATE the ANNUAL INTERESTs that have become due ;
THIRDLY, to the ExTINGUISHMENT of the PRINCIPAL.”
^
TEIE NEW HAMPSHIRE SYSTEM.
1189. The New Hampshire System is essentially the same as the preceding
when partial payments are made on notes “with interest annually.” But “where
payments are made expressly on account of interest accruing but not then due,
they are applied when the interest falls due, without interest on such payments.”
THE CONNECTICUT SYSTEM.
1190. The Connecticut System, according to the Supreme Court of Connecti-
cut, is as follows:
Compute the interest on the principal to the time of the first payment; if that
be one year or more from the time the interest commenced, add it to the principal,
and deduct the payment from the sum total. If they be after payments made,
compute the interest on the balance due to the next payment, and then deduct the
payment as above; and in like manner from one payment to another, till all the
payments are absorbed, provided the time between one payment and another be one
year or more.
If any payments be made before one year's interest has accrued then compute
the interest on the principal sum due on the obligation for 1 year, add it to the
principal, and compute the interest on the sum paid from the time it was paid up to
the end of the year; add it to the sum paid, and deduct that sum from the principal
and interest added as above. s
If any payments be made of a less sum than the interest arising at the time
of such payments, no interest is to be computed, but only on the principal sum for
any period.
NoTE—If a year extends beyond the time of settlement, find the amount of the remaining
principal to the time of settlement; find also the amount of the payment or payments, if any,
from the time they were paid to the time of settlement, and subtract their sum from the amount
of the principal.
Yºr PARTIAL PAYMENTS, 619
1191. PROBLEMS WORKED BY THE UNITED STATES SYSTEM.”
1.
$9000. NEW ORLEANs, February 3d, 1894.
On demand, for value received, we promise to pay to the order of CUMMINGS
& CENAs, Nine Thousand Dollars with interest at six (6) per cent.
g MOYSE & WILSON.
The following payments were made and indorsed on the foregoing note:
September 10th, 1894, - $1500.00 March 5th, 1895, - $3953.75
December 18th, 1894, - 100.00 July 20th, 1895, - 2000.00
Counting 30 days to the month, what was the amount due April 1, 1896?
Ans. $2177.43.
OPERATION.
1894 * $ | c.
Feb. 3| To face of note tº tº * º tº gº &_º tº s * sº * * * * > 9000:00
Sept. 10 “interest on same from Feb. 3 to Sept. 10, 217 days at 6 per cent - º 325|50
é & ** | ** amount due sº gº gº ſº sº * * * sº º &º * * * & 9325 50
é & & 4 By cash, (1st payment), * * * * * * * * * 1500|00
& 4 é & ‘‘ balance due º tº- gº & # * . * = gº ge gº dº tº gº 1825 50
Dec. 18| “ Interest on balance due from Sep. 10 to Dec. 18, 98 days at 6% - $127.82
& 6 & 4 By cash, (2d payment), gº * * * tº sº tº gº & 100.00
is “| “ balance of interest - - - - - - - - - $27.82
95
March 5 “ int. on bal. due ($7825.50) from Dec. 18 to March 5, 77 days at 6% 100.43
which added to balance of Interest due makes *- sº — | 12825
& 6 & & & 4 amount due &E tº sº E: * - gº gº «-» º tº gº - sº 7953 75
& 4 & ‘ By cash, (3d payment), gº ºn tº as tº dº tº º tº 3953|75
&: $ 6 & 4 balance due sº tº sº * * * tººk gº gº tºº * * & º 4000 00
July 20 “interest on balance due flom March 5 to July 20, 135 days at 6% - cº 9000
& 4 & 4 & 4 amount due * {-º * ſº †º tº & ºn * * - gº Cº. ſº ºn º sº 4090 00
é & & 4 By cash, (4th payment), gº tº º tº ſº tº gº gº * & 200000
4 & & 4 ‘‘ balance due - *º-º tº tº ºf tº e * * gº † gº ‘. * > ( : 2000 00
1896
April 1| “interest on balance due from July 20 to April 1, 251 days at 6% - - 8743
& 4 “| “...amount due April 1, 1896 - tº tº ſº gº * tº tº tº 2177.43

Explanation.—In the operation of this problem we found that the interest due on the 18th of
December was greater than the payment then made; hence, according to the principles of this
system of work, we did not add the same to the principal. Had we done so the result would have
been compound interest on the excess of interest, $27.82, to the disadvantage of the debtor, and in
a sense that the law of this system does not allow. We accordingly deducted the payment from the
interest, and reserved the excess until the next payment, to which excess we then added the interest
on balance due from December 18 to March 5, 1895, and thus obtain $128.25 interest, which being
less than the payment then made, we added to the balance due September 10, 1894.
Instead of deducting the payment made from the interest, we could have set the whole
interest aside until the sum of the payments exceeded the sum of the interests, and then proceeded
in the regular manner by adding the sum of the interest to the amount due, and then deducting
the sum of the payments therefrom. But we think that the method shown in the operation is the
clearest, and hence prefer it. *
62O
soul E's PHILOSOPHIC PRACTICAL MATHEMATICS.
2. “
$5200.
A
MARSHALL, TEXAS, July 28, 1894.
PATTERSON & TURMAN.
t
October 1st, 1894, - $1200 August
December 8th, 1894, - 1000
June 25th, 1895, - 100
what was then due #
1st, 1895,
November 11th, 1895,
The following payments were made and indorsed on the above due bill:
$1500
900
Ans. $938.30.
Due, from date hereof, to WOMACK & WILLIAMS, or order, Five Thousand
Two Hundred Dollars, for value received, with interest at 10 per cent.
A final settlement was made December 15th, 1895. Counting actual time,
OPERATION.
T1394 |_$.1%
July 28. To amount due - --> * se tº- gº * --> sº ſº sº {-º sº- $º 5200.00
Oct. 1| “ interest on same from July 28 to Oct. 1st, 65 days at 10 per cent - sº 93|89.
& & & & {6 amount due ge tº ſº º * * º * - tº tº gº &º gº * 5293 89.
& t $ 6 By cash, (1st payment), gºs º gº ºne ºs tº sº, sº sº 1200|00
& 4 & 4 { { balance due ſº sº tº tº- gº * * wº e- # = gº sº * -º & 4093 89.
Dec. 8| “ interest on same from Oct. 1 to Dec. 8, 68 days at 10 per cent - sº 77|33.
& 6 & & ‘‘ amount due - tº * > 4 º' * - sº *E* $º tº sº sº sº º 4171 22
& 4 & 4 By cash, (2d payment), - - - - - - - - - - 100000
& 4 & 4 & 4 balance due - tº * > * - tº- gº & º sº & gºe tº wº 3171:22.
1895
June 25 “ int. on same from Dec. 8, '94, to June 25, '95, 199 days at 10% - $175.30
& & & 4 By cash, (3d payment), a. * * * * * * * * 100.00
& 4 “| “ balance of interest due - - - - - - - - - $75.30
Aug. 1| “ int. on bal. due ($3171.22) from June 25 to Aug. 1, 37 days at 10% 32.59
$ which added to balance of interest makes - tº gº tº gº * - 10789
& 4 ** | ** amount due * sº sº * gº tº 4 º' & ºt gº gº gº sº 3279|11
4 & & 4 By cash, (4th payment), º, sº as tº tº º sº gº º 1500|00.
“ ** | ** balance due - - - - - - - - - - - - - 1779|11
rov. 11| “ interest on same from Aug. 1 to Nov. 11, 102 days at 10 per cent - tº 50|41
& & & 4 & 4 amount due gº tº sº tº sº gº º sº * * tº º &º 4.- : 1829 52.
& 4 4 & By cash, (5th payment), tº º ºs º we dº º ºs ºne 90000
& 6 & & & 4 balance due sº tº- * sº gº tºº {º tº ſº &= ſº tº 929.52
Dec. 15| ‘‘ interest on same from Nov. 11 to Dec. 15, 34 days at 10 per cent - - 8|78.
& 4 “| “ amount due on final settlement December 15, 1895, gº ºf tº gº 938|30.
Explanation.—In the operation of this problem, we have counted actual time, which we,

regard as the better way at all times.
{ PARTIAL PAYMENTS,
62 I
1192. PROBLEMS WORKED BY THE MERCEIANTS’ SYSTEM.
$9000. NEW ORLEANs, February 3, 1894.
On demand, for value received, we promise to pay to the order of CUMMINGS
& CENAS, Nine Thousand Dollars with interest at 6 per cent.
MOYSE & WILSON.
The following payments were made and indorsed on the foregoing note:
September 10th, 1894, $1500.00 March 5th, 1895, - $3953.75
December 18th, 1894, 100.00 July 20th, 1895, - 2000.00
Counting 30 days to the month, and making an annual settlement on the first
of January, 1895, what was the amount due April 1st, 1896? Ans. $2.165.40.
OPERATION.
1894 $ | c.
Feb. 3| To face of note - tº- - - tº ſº - - sº - - s º 9000|00
1895 | -
Jan. 1) “interest on same from Feb. 3 to Jan 1, 328 days at 6 per cent & # - 492|00
é & “| “ amount due to date * - * tº- - - ºr - - tº º 9492|00
1894
sº 10 By cash, (1st payment), - - - - - - - $1500.00
1895
Jan. 1 “ interest on same from September 10 to January 1, 111
days at 6 per cent - - - - * - º 27.75
1894
º 18 By cash, (2d payment), tº E- - - - - = 100.00
5
Jan. 1 “ interest on same from December 18 to January 1, 13
days at 6 per cent - & º - tº - - .22
& 4 & & “ total credits to date - - - - - - - - - 1627|97
{ { ‘‘I To balance due this date - - s º º ºs e - • * º 7864|03
1896
Jan. 1| “ interest on same from Jan. 1 to Jan. 1, 360 days at 6 per cent - - 471184
& 4 ‘‘ ‘‘ amount due to date º cº-º º tº tº tº- & 4- tº º *- 8335|87
1895 ***
March 5 By cash, (3d payment), ge ºn tº sº tº - - $3953.75
1896
Jan. 1 “ interest on same from March 5 to January 1, 296 days
at 6 per cent - - - - - º - - 195.05
1895 -
July 20 “ cash, (4th payment), º tº - * = * - 2000,00
1896
Jan. 1 “ interest on same from July 20 to January 1, 161 days º
, at 6 per cent - - - - tº ºs º – 53.67
& 4 ( & ‘‘ total credits to date - - - - sº - - º 620247
& 4 ‘‘l To balance due this date - - º - - º º- - º 213340
April 1| “ interest on same from Jan. 1 to April 1, 90 days at 6% - *- sº 3200
§ { “| “ amount due on final settlement * tº- - tº tº- - - gº 2165|40

Explanation.—In the solution of this problem, as shown by the operation, we made the first
partial settlement at the close of the civil year, notwithstanding a year had not elapsed from the
date of the obligation. We did this because it is strictly in conformity with the merchant's
system of closing his accounts at the end of each civil or his fiscal year, and because we believe
that it is the only correct way of applying the merchants’ system of partial payments. Merchants
enerally not only desire, but by the municipal, State and general government laws are required to
#. their gains and losses, resources and liabilities on the first of January of each year. With-
out this information, correct returns for the assessment of various taxes could not be rendered, nor
622 soul E's PHILOSOPHIC PRACTICAL MATHEMATICs.
º
would their books show a correct condition of their business. Had we, as some authors direct,
computed the interest from February 3, 1894, to February 3, 1895, and then from February 3, 1895,
to February 3, 1896, and then from February 3, 1896, to April 1, 1896, the final result would have
been slightly different, and when the books were closed on the first of January of each year, the
true gains and resources of the business would not have been shown by the books.
In counting the time we used 30 days to the month in order to compare the result of the
#.'; with the result of the first and same problem worked by the United States System, on
page 619.
By the United States System the amount due on final settlement is - $2177.43
And by the Merchants' System as above, * * º º ºs º 2165.40
Excess produced by the United States System, - - - $12,03
2. *
$5200. MARSHALL, TEXAS, July 28, 1894.
Due from date hereof to Wom ACK & WILLIAMs, or order, Five Thousand
Two Hundred Dollars for value received, with interest at 10 per cent.
PATTERSON & TURMAN.
The following payments were made and indorsed on the above duº bill:
October 1st, 1894, - $1200 August 1st, 1895, - $1500
December 8th, 1894, - 1000 November 11th, 1895, - 900
June 25th, 1895, - 100 f
A final settlement was made December 15, 1895. Counting actual time, and
making a partial settlement at the close of the civil year 1894, what amount was
due on final settlement” Ans. $927.77.
OPERATION.
1894 - $ | c.
July 28. To face of due bill - - - º tº- <- ſº - tº 5200|00
1895
Jan. 1| “interest on same from July 28 to Jan. 1, 157 days at 10% * * 22678
4 & “| ‘‘ amount due this date tº- tº º º wº- - º tº- sº 5426||78
1894
Oct. 1 By cash, (1st payment), - - - - - - - $1200.00
& 4 é & “ interest on same from October 1 to January 1, 92 days
at 10 per cent - - - gº tº º tº- tº 30.67
Dec. 8 “ cash, (2d payment), tº ºs - is sº tº tº 1000.00
& 4 & 4 “ interest on same from December 8 to January 1, 24
days at 10 per cent - - - & cº- Q- i.e. 6.67
1895 -*º-
Jan. 1 “ total credits to date - - - - - - - - - 2237|34
& 4 “| To balance due to date - - tº ſº tº . gº tº- º tº gº 3189 4.
Dec 15| “ interest on same from January 1 to December 15, 348 days at 10% tº º wº
& 4 “| “ amount due this date - - - tº ºs - - * * º 349775
June 25 By cash, (3d payment), sº º º ºs º º ºs $100.00
€C. 15 “ interest on same from June 25 to December 15, 173 f
days at 10 per cent - - - gº - - º 4.81
Aug. 1 “ cash, (4th payment), - - - - - - - 1500.00
Dec. 15 “ interest on same from August 1 to December 15, 136
days at 10 per cent - - - - - tº- tº 56.67
NOV 11 “ cash, (5th payment), - º - º - e- tº 900.00
Dec 15 “ interest on same from November 11 to December 15, 34
days at 10 per cent - . - - - - - 8,50
6& & 4 “ total credits to date - - - - - * - º ºs 256998
&é “| To amount due on final settlement • m as mº' s m º ºs ºn

Explanation.—By comparing the result of this problem with the result of the second and
same problem worked by the United States System, on page 620, we have the following figures:
* PARTIAL PAYMENTS. 623
Amount due on final settlement by the United States System, - gº $938.30
Amount due as above by the Merchants' System, is e º º 927.77
Excess produced by the United States System, - - - $10.53
Where the payments are small, or the interval of time between them is more
than a year, the Merchants' System would produce an excess over the United States
System. *
1193. MISCELLANEOUS PROBLEMS
INVOLVING THE UNITED STATES AND MERCHANTS’ SYSTEMS OF WORK.
1.
$3740. VICKSBURG, May 1, 1894.
Two years after date, we promise to pay to the order of STEWART & BRADSTON,
Three Thousand Seven Hundred Forty dollars, with interest at eight (8) per cent,
value received.
BOWLES & STEWAR.T.
The following payments were made and indorsed on the foregoing note:
September 16th, 1894, - $500 October 28th, 1895, - - $1200
February 1st, 1895, - 75 April 1st, 1896, - - 300
May 4th, 1895, - 1500
Counting actual number of days, and working by the United States System,
what was the amount due at the maturity of the note, May 4th, 1896?
Ans. $560.68.
2.
$2000. o LITTLE ROCK, ARK., January 12, 1894.
Due SCULL & MCREY, Two Thousand Dollars, with interest at 8 per cent.
FOSTER & SULLIVAN.
The following payments were made and indorsed on the foregoing note:
November 1st, 1894, - $600 May 9th, 1895, . . $400
January 18th, 1895, - 750 December 1st, 1895, - - 200
Counting actual time, and working by the Merchants' System, what was due
on final settlement, April 10, 1896? Ans. $249.84.
3.
$6840.50.
On the 18th of October, 1894, a judgment was rendered by one of the District
Courts of New Orleans for Six Thousand Eight Hundred Forty and #, Dollars.
(The legal rate of interest in Louisiana is 5 per cent).
The following payments were made thereon:
January 1st, 1895, - - $1000 November 25th, 1895, - $1000
May 14th, 1895, - - 1000 March 10th, 1896, - 3000
A final settlement was made April 21, 1896. Counting actual time and
working by the United States System, what was then due Ans. $1225.24.
624 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
*
4.
$20000. WACO, TEXAS, January 1, 1894.
On demand, for value received, we promise to pay to the order of HIGHTOWER
& ROSE, Twenty Thousand Dollars, with 10 per cent interest.
POLLY, VINYARD & Co.
- Semi-annual payments of $3000 each were made on the above note from date
to January 1, 1897, at which time a final settlement was made. Counting actual
time, what was then due by the United States System, and also by the Merchants'
System of partial payments? Ans. $9466.19, by the United States System.
$9329.23, by the Merchants' System.
1194. PARTIAL PAYMENTS
ON NOTES DRAWN FOR A SPECIFIED TIME AND BEARING INTEREST AFTER
IMATURITY. *
$7000. SUMMIT, MISS., June 1, 1894.
One year after date, for value received, we promise to pay to the order of
CURTIs & NEWMAN, Seven Thousand Dollars, with 8 per cent interest after
maturity until paid. TYLER & MILLS.
The following payments were made and indorsed on this note:
December 1st, 1894, º - º - s $2000
April 10th, 1895, - - s - - 2500
October 17th, 1895, º - º - - 1500
A final settlement was made December 20, 1895. Counting actual time, what
was then due 3 Ans. $972.48.
OPERATION.
1894 $...]9.
June 1| Face of note due June 1/4, 1895 ºs - º - - - - - º 700000
Dec. 1 By cash, (1st payment), - - - - - - - - - - - $2000.00
“ interest on same from December 1 to June 4, 1895, 185
1895 days at 8 per cent - tº- sº - tº ºs - 82.22 |
April 10 ‘‘ cash, (2d payment), - tº º tº tº- - 2500.00 |
‘‘ interest on same from April 10 to June 4, 1895, 55 days
at 8 per cent - - - - - - - - - 30.56
“ total amount of credits before the maturity of the –
note June 4, 1895 - - - - - - - - - 461278
June 4| To balance due at the maturity of the note - - - - - , , - - 2387|22
“ interest on same from June 4, 1895, to Oct. 17, 1895, 135 days at 8% sº 71.62
Oct. 17| ‘‘ amount due this date es e º sº º sº me - º tº e 2458|84
& 4 & 4 By cash, (3d payment), - - - - - - - - - 150000
‘‘ balance due - - & tº- - º º - º * - º 958.84
Dec. 20 “interest on same from Oct. 17, to Dec. 20, 64 days at 8% gº tº ºs 13|64
& & “| “ amount due on final settlement - - - - - - * * * 97248

Explanation.—Where partial payments are made on notes of this character before they
mature, we must first find the balance due at maturity. To do this we deduct from the face of the
note the payments made before due, and the interest on such payments from the date of payment
till the maturity of the note. Having thus produced the balance due on the note or obligation at
the time of maturity, we then proceed with it, and the remaining payments if any, the same as in
the preceding work of partial payments, either by the United States or Merchants' System. . In
the operation of the above problem, we worked by the United States System, which is the system
in general use for transactions of this kind.
* PARTIAL PAYMENTS. 625
1195. PARTIAL PAYMENTS
ON NOTES, ETC., DRAWN FoR A SPECIFIED TIME, AND which BEAR ANNUAL,
SEMI-ANNUAL OR QUARTERLY INTEREST.
$9000. NEW ORLEANs, March 17, 1889.
Three years after date, for value received, I promise to pay to the order of
FRANCIS & BAKER, Nine Thousand Dollars with eight (8) per cent interest, pay-
able annually. E. W. JONES.
The following payments were made and indorsed on this note:
November 1st, 1889, - $1000 May 15th, 1891, - - $2000
July 10th, 1890, - 1500 January 4th, 1892, - - 1800
Counting 30 days to the month, what was due June 17, 1892, at which time a
final settlement was made 1 Ans. $4552.69, by 1st operation.
$4559,63, by 2d operation.
IFIRST OPERATION.

1889 $ | c.
March [17] To face of note tº sº dº sº sº º ºs º ºr ºt tº mº 900000
1890 “ interest on same from March 17, 1889, to March 17, 1890, 1 year at 8% - 72000
Mº 17| ‘‘ amount due this date sº- tº-e º- sº $º- tº- gºs s * - tº 97.2000
1889
Nov. 1. By cash, (1st payment), tº G- - tº- - * - $1000.00
“ interest on same from November 1, 1889, to March 17,
1890 1890, 4 months 16 days at 8 per cent - - - 30,22
March 17 “ total credits to date - - - - - - - - – 1030/22
i s 91 “| To balance due this date am s m sº em me as * * * 8689|78
March 17 “ interest on same from March 17, 1890, to March 17, 1891, 1 year at 8% - 695.18
& 4 “| “ amount due this date - - - - - - - - - - - 9384|96
1890
July 10 By cash, (2d payment), - - - - - - - $1500.00
“ interest on same from July 10, 1890, to March 17, 1891,
1891 8 months 7 days at 8 per cent - - - - 82.33
March 17 “ total credits to date - - - - - - - –— 1582/33
is 9 “| “ balance due this date sºme º sm em as sº * * as tº 7802|63
2
March 17 “ interest on same from March 17, 1891, to March 17, 1892, 1 year at 8% - 624.21
“ I”, “amount due this date - - - - - - - - - - - 8,2884
1891
May 15 By cash, (3d payment), - - - - - - - $2000.00
“ interest on same from May 15, 1891, to March 17, 1892,
1892 10 months 2 days at 8 per cent gº tº tº * 134.22
Jan. 4 “ cash, (4th payment), - - - - - - - - 1800.00
‘‘ interest on same from January 4, 1892, to March 17,
1892, 2 months 13 days at 8% mº as as ºn 29.20
March 17 ‘‘ total credits to date cº- ºr º º * - - —— 1396342
& 4 “| To balance due this date - - - - - - - - - - - 4463|42
June 17| “ interest on same from March 17, 1892, to June 17, 1892, 3 months at 8% - 89|27
& 4 “| “ balance due on final settlement * = e º sº a ºn * 4552|69
Explanation.—Transactions of this character have not occurred with sufficient frequency to
establish any real custom pertaining to the method of their solution; and hence, in the absence
of both custom and law, and in consequence of there being so many different systems of partial
626 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
payments, accountants and authors are divided in their opinions as to which of several methods
should be used in cases of this kind. For this reason we present two solutions, to the first of
which we give preference. The contract of annual interest specifies the period of time that the
same shall be paid, and therefore it seems reasonable that the settlements for the partial payments
should be made at the same time; and in accordance with this view, we have made them at the
close of each year from the date of the note. Had the interest been payable semi-annually or quar-
terly, we would have made the partial payments accordingly.
As shown in the operation, we first find the interest on the face of the note
for one year and add it to the same, and then from this amount we deduct the pay-
ment made during the year, increased by the interest on the payment from the date
that it was made until the end of the year; and thus we continue until the maturity
of the note, and then for the remainder of the time up to the final settlement, we
proceed in the same manner as in the Merchants' System of partial payments. In
transactions of this nature, where there are payments made subsequent to the
maturity of the note, they would be treated the same as in the Merchants' System.
In fact the only difference between the operations of this work and the
Merchants' System is the different time or rests at which the partial payment
Settlements are made.
By this system of work both parties receive yearly compound interest.
SECOND OPERATION.
1889 $ |º:
March 17| To face of note - º - - - º - - - º - - 900000
1892 •
June 17 ‘‘ int. on same from March 17, '89, to June 17, '92, 3 yrs. and 3 mos, at 8% 234000
& 4 “l ‘‘ amount of note at time of settlement mºs - - - º - - 11340 oo
{ % “| “ interest on $720 (the 1st annual interest) from March 17, 1890, to
June 17, 1892, 2 years and 3 months at 8% - sº - - $129.60
& 4 “| “ interest on $720 (the 2d annual interest) from March 17, 1891, to
June 17, 1892, 1 year and 3 months at 8% º as tº - 72.00
é & ‘‘ ‘‘ amount of interest on annual interest º - -> - º - - 201|60
& 4 “| “ total amount due this date -> º dº Q- - º º --> &- 11541.60
1889
Nov. 1 By cash, (1st payment), * - - - we ºw - $1000.00
‘‘ interest on same from Nov. 1, 1889, to June 17, 1892, 2
years 7 months and 16 days at 8% e- - tº- 210.22
1890
July 10 “ cash, (2d payment), <- - º- º g- -> º 1500.00
‘‘ interest on same from July 10, 1890, to June 17, 1892,
1 year 11 months and 7 days at 8% tº dº º 232,33
1891
May 15 “ cash, (3d payment), gº sº - tº- º tº- º 2000.00
“ interest on same from May 15, 1891, to June 17, 1892,
1 year 1 month and 2 days at 8% - - - - 174.22
1892
Jan. 4 “ cash, (4th payment), gº tº - * º * tº- * 1800.00
‘‘ interest on same from January 4, 1892, to June 17,
1892, 5 months and 13 days at 8% tº º tº º 65.20
June 17 “ total credits to date - - - - - - - - - 6981|97
& 4 “| To balance due on final settlement as as sº gº ºn tº º º 4559|63
Explanation.—In this solution we first find the interest on the face of the note from the day
of date to the day of final settlement, and then the interest on the annual Interest from the time it
became due until the final settlement. Then, from the total of these three amounts We deduct the
sum of the payments, increased by the interest on each from the time paid until the day of settle-
ment. The remainder is what is then due,
NotE.—In the foregoing operations of interest and partial payments, where the actual time
was counted, 29 days were reckoned for February in all the leap years that occurred.
† EQUATION OF ACCOUNTS. 627
EQUATION OR AVERAGE OF PAYMENTS AND ACCOUNTS.
1196. Equation, or Average of Payments and Accounts, is the process of
finding the mean or average time for the payment, in one amount, of several sums
of money, notes or debts due at different dates; or of the balance of accounts that
contain both debit and credit items.
The Equated Time is the date at which the various sums due at different
times may be paid in one amount without injustice to either party.
The Adjustment of Interest is the real object of equating or averaging
payments and accounts.
The principles upon which the process of the work is based are that the
interest of any sum paid any length of time before it is due is equitably balanced by
the interest of an equal sum paid an equal time after it is due. Hence in all equa-
tions or averages, the interest on the sums paid, before due by equation, must equal
the interest on the sums paid after they are due by equation.
The Assumed Due Date, or Focal Date is the date at which we assume all
items of debit or credit or both to be due, and from which we count the debit or
credit of time in the equation.
The Debit or Credit of Time is the time an item has to run from the assumed
due date until it is due.
The Earliest Date in an account or of the items to be equated, is used in
this work as the assumed due date or focal date.
NOTE.—Authors and accountants differ in regard to the date to assume as the due or focal
date. Some assume the earliest date, some the latest date, some the first of the month in which
the earliest date occurs and some assume a date prior or subsequent to the earliest date. The
result is the same whatever the date assumed.
To work from the earliest date each item of debit, except the first, is subject
to bank discount ; and to work from the latest date, each item, except the latest, is
entitled to interest. º
The subject is divided into simple and compound equation or average.
Simple Equation, or Average, is the process of finding the average time of
payment for debts or accounts, which contain debit or credit items only.
Compound Equation, or Average, is the process of finding the equated
average time of payment for the balance of accounts, which contain both debit and
credit items.
Equations with Special Conditions. For equations involving special con-
ditions, see pages 643 to 652 and 1009 to 1013 of this book.
There are numerous methods given by various authors for the equation and
averaging of accounts; but we believe that the introduction of all of them would
confuse and bewilder the student rather than elucidate and make clear the
operations and principles of the Work.
The methods we present are universal in their application, have received
the approval of practical and scientific accountants, and are short, concise and
require in their use the exercise of reason instead of the memorizing of rules.
628 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
PROBLEMS
TO FIND THE EQUIVALENT OR AVERAGE TIME FOR THE PAYMENT
OF SEVERAL SUMS OF MONEY DUE AT DIFFERENT TIMES,
OR WITH DIFFERENT TERMS OF CREDIT.
1197. 1. January 1, 1895, Z purchased from A goods amounting to $3000,
which he agrees to pay as follows: $800 in three months; $1200 in eight months,
and the remainder, $1000, in twelve months. What is the equated or average time
for the payment of the whole amount, so that neither party will gain or lose by the
use of money? Ans. 8 months.
OPERATION BY THE PRODUCT SYSTEM.
Z's Z’s Z’s
1895 Debit of Dolls. Credit of Months. Credit of Months on $1.
January 1, $800 X 3 = 2400
46 1, 1200 X 8 = 9600
{{ 1, 1000 X 12 = 12000
mº-º-º-º: *
Total debit of dollars $3000 Total credit of months on $1, 24000
24000 -- 3000 = 8 months.
Ans. 8 months equated time.
Explanation.-By the terms of the contract we see that Z is debited with $800, $1200 and
$1000 in money, and credited on the respective amounts with three, eight and twelve months in
time, from January 1, which is the due date of all the items; and since he is debited with various
sums of money and credited with various periods of time, it is clear, by the exercise of our reason,
that, in order to solve the question, we must first find Z's total credit of time on $1, from the due
date. To find his total credit of time on $1, we reason thus: Three months’ credit on $800 is equal
to 800 times three months' credit on $1, which is 2400 months; then eight months’ credit on $1200
is equal to 1200 times eight months' credit on $1; and then twelve months’ credit on $1000 is equal
to 1000 times 12 months' credit on $1; which added gives 24000 months’ credit on $1. Having this
total credit of time on $1, we then find the total amount of his debit in dollars by adding the
various sums together. We then reason thus: Since Z has a credit of 24000 months’ time on $1, on
$3000 he will have the 3000th part, which is eight months. Here it may be asked, why Z should
Inot have 3000 times as much credit on $3000 as on $1, instead of the 3000th part 3 To this question
we reply that Z's credit on $1, for 24000 months, is but an equivalent for the various terms of
credit due him by the conditions of the problem; and hence, if he is to receive a credit on the
whole amount, $3000, that he owes, instead of on $1. the credit allowed to produce an equivalent,
to the 24000 months on $1, or the various terms of credit, according to the first conditions, must
necessarily be in reciprocal proportion to the amount to which it is to apply.
To illustrate the correctness of the foregoing operation, by the principles of
interest, we present the following figures:
By the first conditions of the problem, Z was to pay $800 in three months,
$1200 in eight months and $1000 in twelve months; but as Z did not pay the $800
until the equated time, eight months, he therefore gained and A lost the interest of
the same for five months, which at, say, 8 per cent, is $26.67.
The $1200 Z was to have paid in eight months, and as the equated time is
eight months, there is neither gain nor loss to either Z or A on the same, by reason
of the equation.
* EQUATION OF ACCOUNTS. 629
The $1000 Z was to have paid in twelve months; but by reason of the equa-
tion, he paid it in eight months; and hence he lost and Agained the interest of the
same for four months, which, at the same rate as above, 8 per cent, is $26.67. By
this we see that A's gains and Z's losses, or Z's gains and A's losses are equal, and
that the interest on the sums paid before they are due by equation, is equal to the
interest on the sums paid after they are due by equation. This will always be the case,
no matter how many sums are paid before or after the equated time.
To elucidate the work of equation further, we present the following operation:
1895. Months Cr.
January 1, $800 3 at 12% $24 interest.
1200 8 at 12% E '96" ,
1000 12 at 12% = 120 “
$3000 $39 ) $249 interest.
8 months.
Ea:planation.-By this method we see that if Z had paid the various debits of dollars, at the
time of purchase, without waiting for the expiration of the various terms of credit, he would have
been entitled, as shown in the operation, at 12 per cent interest, to $240 interest deduction on the
$3000. But as he does not wish to pay the whole amount in advance of the whole credit, but at a
period of time, so that the interest on the sums paid before due will equal the interest on the sums
paid after they are due ; it is, therefore, by the exercise of our reason, plain that the time it will
require the $3000, at 12 per cent, to produce $240, will be the equated time, or the time to pay the
whole amount, so that Z's gains and A’s losses, or Z's losses and A’s gains, by interest, shall be
equal. To find this time, we compute the interest on the $3000, for one month at 12 per cent, which
gives us $30. We then reason thus: If $30 interest requires one month's time, $1 will require the
thirtieth part, and $240 will require 240 times as many months, which, worked, gives eight months,
the equated time. Had the terms of credit been days instead of months, we would then have
found the interest on the whole sum for one day.
The equated time will be the same regardless of the rate of interest used. We
used 12 per cent because, in the most of cases, it is easier, and requires less time to
calculate interest at 12 per cent than at any other rate.
By this method of work, interest tables can be used to advantage by those
who desire. The interest method, however, we do not regard with much favor. The
product method, or system, is better and more easily comprehended by learners.
NotE.-The accuracy of the result of the foregoing methods of work has been questioned by
several good authors, and much warm discussion has been indulged in on the subject; but as the
arguments advanced against the correctness of these methods are alike subtle and fallacious, we
shall, therefore, be very brief in considering them.
The following example and argument is generally used to elucidate the asserted incorrectness
of the foregoing methods of finding the equated time:
“If a man owes me $200, $100 of which is payable now, and $100 payable in two years, the
equated time is not one year. For in deferring the payment of the first $100 one year, he ought to
pay the $100, plus the interest, which, at 6 per cent, is $106; , but for the $100 which he pays one
year before it is due, he ought to pay the present worth of $100, which, at 6 per cent, is $94.33%;
and $106 -- 94.33}} = $200.33}}; whereas, by methods in general use, he only pays $200.”
By this argument it is contended that I lose $6, the interest on $100, for 2 years only, and
gain only $5.66, the true discount on $100 for 1 year, making a difference of nearly 34 cents.
But the argument is fallacious, and the error is in supposing that I lose $6, when really I
only lose the present worth of $6, for the $6 interest is not due until the close of two years, and
hence I should receive $100, plus the present worth of $6, due in one year, which is $5.66–H and
which added to the $100 due the first year = 105.66+ that I should receive; but as I only receive
$100, I therefore actually lose $5,66–H which is just equal to the gain on the $100 received for the
present worth ($94,33}}) of the $100 for one year before it was due.
63o SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. jºr
2. A owes Z $200, payable now; $500 payable in 4 months; $1000 payable
in 6 months; and $1400 payable in 9 months. What is the equated time for paying
all ? Ans. 6 months and 19 days.
OPERATION.
Debit of Dollars, Credit of Months. Credit of Months on $1.
1000 X 6 -: 60
1499 X 9 2– 126
Total debit of dollars $31 Total credit of months on $1 206
206 -- 31 = 63% months = 6 months 1944 days.
Explanation.—We here see that A owes various sums of money, and has due him various
terms of credit of time, and as shown in the first problem, we first find a total credit of time on $1.
The reasoning for the work is the same as given in the first problem, hence omitted. Before
making the multiplications, we cancel each sum of dollars by 100, and thus save figures. When
the fraction of a day in the answer, or equated time, is more than }, it is customary to count 1
Whole day; and, when less than 3 to omit it.
3. A merchant owes 4 notes, dated January 1, 1895. The first is for $500,
payable in 30 days; the second is for $800, payable in 40 days; the third is for
$1200, payable in 90 days; and the fourth is for $2000, payable in 120 days. What
is the equated time for the payment of all?
Ans. 88 days, which, added to Jan. 1, 1895, the date of the notes,
gives March 30, 1895, as the date to pay all at one time.
Dr. of Dolls. Cr. of Days. Cr. of days on $1.
$500 X 30 = 15000
800 , X 40 E 32000
1200 X 90 E 108000
2000 X 120 3– 240000
$4500 total debit of dollars. 395000 total credit of days on $1.
395000 - 4500 = 873 days.
4. A loaned Z $600 for 9 months, how long ought Z to loan. A $800 to equal
the favor? Ans. 6 months, 22 days.
OPERATION.
Dr. of Dolls. Cr. of Months, Months Cr, on $1. 8% 54%
$600 X 9 E 5400 63 months = 6 mos. 224 ds.
5. A bought of Z the following invoices:
January 2, - tº 9 tº gº tº. $400
66 9, - tº tº gº gº 210
{{ 15, - tº sº tº tº 350
{{ 21, - tº sº º tº t 800
46 30, - tº tº * > tº 1100 ~
What is the equated time for paying all? Ans. January 20.
º: EQUATION OF ACCOUNTS. 631
FIRST OPERATION WORKING FROM THE EARLIEST DATE.
Dr. Of Dolls, Cr. of days. Cr. of days on $1.
January 2, $400 X 0 F.
{ % 9, 210 X 7 = 1470
4 - 15, 350 X 13 – 4550
§ { 21, 800 X 19 F 15200
& 4 30, 1100 X 28 F 30800
$2.860 ) 52020 (18 days.
Earplanation.—In this problem the various debits of dollars were credited at different dates,
3. as there was no credit of time allowed, each item of debut is therefore due at the time of pur-
CIla.S6),
In this operation we assume the earliest date January 2, as the due date or focal date and
then proceed to find the time that each purchase or debut item is due after the assumed due date.
We reason thus: The $400 being due the day of purchase, it has 0 days credit of time. The $210
purchase on the 9th is really due that day; but presuming 1 t due on the earliest date, January 2, Z
therefore owes A time from January 2 to January 9, which is 7 days’ credit on the same. The $350
bought on the 15th is really due that day; but presuming it due on the earliest date, January 2, Z
owes A time from January 2 to January 15, - 13 days' credit on the same. The same reasoning
applied to the $800 and $1100, purchases of the 21st and 30th, gives respectively 19 and 28 days’ credit
due to A. Having thus made or presumed all the purchases due January 2, we proceed to find the
credit that Z owes on $1, and then on the whole amount in the same manner as explained in the
first problem of equations. The result of the equation is 18 days’ credit due A on the whole
amount, $2860, which, for the reason that we make all the debit items due January 2, must be
added to that date in order to give A the credit due to him; 18 days added to January 2 gives
January 20 as the equated time for paying the whole amount, $2860.
To more fully elucidate the work of equation, we present the following:
SECOND OPERATION WORKING FROM THE LATEST DATE,
IDr. Of Dolls. Dr. of Days. Dr. of days on $1.
January 2, $400 X 28 = 11200
{{ 9, 210 X 21 E 4.410
“ 15, 350 X 15 - 5250
“ 21, 800 X 9 = 7200
“ 30, 1100 X 0
$28.6% ) 2806% (93.g3 days, practically 10 days.
Explanation.—By working from the latest date, January 30, we see that as each debit item
was due the day that it was purchased, that the $400 bought January 2 was therefore due 28 days
prior to January 30; and hence A owes the time as well as the money. The $210 bought on the
prior to January 30; and hence A owes the time as well as the money. The $210 bought on the
9th was due 21 days; and again A owes Z the time and money; the $350 and $800, purchased respec-
tively January 15 and 21, are due respectively 15 and 9 days, for which A owes Z the time; the
$1100 bought on the 30th is due that day, and hence A owes no time thereon. Having now the
days’ time that A owes Z on the various items of dollars to find his debit of days on $1, and then
on the whole amount that he owes, we proceed to multiply and divide, as shown in the operation,
and for the reasons given in the first problem on equations.
The result of the work is 10 days' debut of time that A owes Z on the whole
amount of $2860; and to give Z the credit or advantage of this time, it is clear that
the 10 days must be counted backward from January 30, the day that all of the debit
items of dollars were due ; and by deducting the 10 days, from. January 30, we
obtain January the 20th as the equated time for the payment of the whole amount.
632 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS.
6. What is the equated time for the payment of the following account 3
Ans. October 5, 1895.
S. J. & M. M. KOHLMAN.
To LEVY & HILLER, Dr.
1895. g 541|85
May 1 To merchandise at 2 months - - - - -> - - - 3)
June 28 ‘‘ cash º º * - * - º tº- cº- - - - 210|30
August 16 “ merchandise at 3 months - - -> e- - - - sº § §
September 20 “ merchandise at 3 months - - e- º -> - - - 525 -
$207805
SOLUTIONS.
FIRST operaTION, Assuming The EARLIEST DATE, MAY 1, To WORK FROM,
Dr. of Dolls. Contract Cr. DS. Cr. Days assuming each Cr. of
purchase due May 1. Ds, on $1.
May 1 $541.85 61 + 00 = 61 33062
June 28, 210.30 00 + 58 = 58 12180
August 16, 800.00 92 + 107 - 199 15.20%
September 20, 525.90 91 + 142 - 233 122558
$2078.05 327000
327000 days credit on $1 - 2078 = 157 days equated time for the payment
of the whole amount, after May 1, 1895, which is October 5, 1895.
Ea:planation.—In this problem, like the preceding, the various debits of dollars were created
at different periods of time; but unlike the preceding problem, all the items, except the cash, have a
credit of time allowed the purchasers. The principles involved in the preceding problems are
alike applicable to the solution of this and all other questions in equations.
In order to find the total credit of time due the purchasers on $1, we first find the credit that
each debit of dollars is entitled to by the terms of the contract from the assumed due date, May
1, in the operation. The debit of $541.85, of May 1, has, by the conditions of the purchase, 2
months (61 days) credit.
The debit of $210.30, June 28, has no contract credit of time; but because we presume it due
May 1, it has from May 1 to June 28, 58 days credit from that presumed time.
The debit of $800, August 16, has 3 months, 92 days contract credit; and because we assumed
it due May 1, it has, from May 1 to August 16, 107 days additional credit, making 199 days credit.
The debit of $525.90, of September 20, has 3 months, 91 days contract credit; and because
we presume or assume it due May 1, it has, from May 1 to September 20, 142 days additional credit,
making 233 days credit from May 1. Having thus the credit of time for each item of debit of
dollars, we proceed to find the equated time, the same as in the preceding problems, with this
exception: In multiplying the first and last terms of days and dollars, we used, respectively, $542
and $526, instead of the exact amounts, $541.85 and $525.90; and in the second term we multiplied
by $210, instead of $210.30; and in the division, we omitted the cents in the diviso.
. In practice the gents need not and by accountants, are not used in equation operations of
multiplying and dividing. Whenever they exceed fifty the amount is increased by $1; and when-
ever they are less than fifty, they are omitted. By operating thus with the cents, time and labor
are saved without effecting any perceptible difference in the result.
The result of the operation is 157 days credit due the purchasers on the whole
amount, $2078.05, after May 1, the assumed date that each debut of dollars was
due. This added to May 1, gives October 5, 1895, as the equated time.
* EQUATION OF Accounts. 633
SECOND OPERATION, Assum ING THE LATEST DATE, SEPTEMBER 20, 1895, As THE
& DUE OR FOCAL DATE.
Dr. of Contract Dr. Cr, Dr. of Ds. Cr. of Ds.
Dolls. Cr. of Ds. of Ds. of Ds, on $1. on $1.
May 1, $541.85 61 *- 142 +E 81 43902
June 28, 210.30 00 *- 84 2– 84 17640
Aug. 16, 800.00 92 *- 35 57 45600
Sept. 20, 525.90 91 * 0 91 47.866
$2078.05 6151, 93466
Debit of days on $1 . º - - ºn tº º - - 61542
Excess of credit of days on $1 is º * = º - - - 31924.
31924 - $2078 = 15.4, days, practically 15 days. .
Explanation.—In this operation we assume the latest date in the account to work from and
reason thus: The $541.85 purchased May 1, not being settled for until September 20, the purchasers,
therefore, owe time on the same from May 1 to September 20, which is 142 days; this we set in the
column of debit days, and deduct therefrom the 61 days credit allowed by the terms of the contract,
which leaves an excess of 81 debit days.
Then the $210.30 of June 28, not being settled for until the assumed date, September 20, the
purchasers owe time on the same from June 28 to September 20, which is 84 days; this we set in
the column of debit days, and having no contract credit to deduct therefrom, it remains a net
debit of days. Then the $800 debit of August 16 not being settled for until September 20, the
purchasers owe time on the same from the date of purchase to the assumed date of settlement,
which is 35 days; this we also set in the column of debit days, and deducting the same from the
92 days credit allowed by the contract of purchase, we have an excess of 57 credit days, which we
extend to the column of credit days. Then the $525.90 debit of September 20 being by assumption
settled on that day, there is no debit of time, and hence the whole credit of days, 91, allowed by the
terms of the agreement, is extended to the column of credit days. Having now found the various
debits and credits of days on the different debits of dollars, we next find, by multiplication, the
debit and credit of days on $1, and then the difference, which is 31924 days credit on $1; this we
divide by the whole amount of debit dollars, and thus obtain 15 days, the equated time which, for
#. pºly given, is counted forward from the assumed date, and gives October 5, 1895, as
the final TeSult.
7. A owes Z $2000 due January 1, 1895; May 1, 1894, he pays Z $800; when
is the balance due # Ans, June 12, 1895,
OPERATION.
$2000
800 245 days credit on $800 (from May 1, 1894, to January 1, 1895,) = 196000
tº-º-º-º- days credit on $1.
$1200
1200 ) 196000
1634 days credit on $1200 after January 1, 1895.
Ea:planation.—When the debtor makes partial payments on the amount that he owes before
it is due, it is clear that he is entitled to interest on the money paid, or to a credit on the balance
after it becomes due, which is equal to the credit on the amount paid before it is due.
8. A is owing Z $2000, which is due by contract July 4, 1894. On the 14th
of February, 1894, A wishes to pay Z such a sum as will justly extend the payment
of the balance until October 7, 1894. What amount must he pay, and what amount
will remain unpaid” Ans. $808.51, to be paid.
$1191.49, to remain unpaid.
From Feb. 14, 1894, to July 4, 1894, are 140 days.
From July 4, 1894, to Oct. 7, 1894, are 95 “
235 ($
634
SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
Here we see that the time from the date of payment (Feb. 14) to maturity of
the whole sum (July 4) is greater than the time from the maturity to the date of
the extension, Oct. 7; and as the credit of days on $1 is in direct proportion to the
amount of dollars multiplied by the credit of time thereon, it is clear that we here
require a less sum to be paid before due than to remain unpaid after it is due, to
produce equivalent credits; hence, the amount to be paid before maturity and also
the amount to remain unpaid after maturity will be in reciprocal proportion to the
two periods of time, 140 and 95 days.
The following statements produce the desired reciprocal results:
OPERATION OPERATION
To find the amount to be paid before maturity. To find the amount to be paid after maturity.
$ $
2000 2000
235 | 95 235 | 140
$808.51 Ans.
$1191.49 Ans.
To show the whole work of reciprocal proportion involved in the foregoing
statement, we present the following:
— —l— — — 23 –
+ + 1.9 = Hº = Tº
1 - 95 = GF = Tºššū
—235–
1 3 3 00
1.
By this we see that the reciprocal of 140 is ++o, and of 95 it is ºs, which
added give rāšāo. With these reciprocal proportional numbers, we make the fol-
lowing solution statements:
OPERATION OPERATION
To find the amount to be paid before maturity, To find the amount to be paid after maturity.
$ $
2000 2000
235 | 13300 235 | 13300
13300 | 95 13300 || 140
$808.51 Ans.
$1191.49 Ans.
To show that the credit of time on $808.51 for 140 days, is equivalent to the
credit of time on $1191.49 for 95 days, we present the following:
Cr. of Ds. Cr. of Ds: on $1.
$808.51 X 140 = 113191+
1191.49 X 95 F. 113191+
$2000.00
NoTE.—To those accustomed to work on the interest system, we could say that, in the main,
the same process of reasoning as given above, would be used. Where we use the term eredit of
day8, the interest method of reasoning would require the term credit of interest.
9. Sold merchandise to LACY & RogERs, as follows:
1895. 1895.
Feb. 1, $500.00 at 60 days. May 21, $200.00 at 00 days.
March 18, 1019.68 “ 90 “ July 28, 400.00 “ 60 “
May 6, 704.25 4 30 % ſº
When does the account mature by equation? Ans. June 13, 1895.
2}. EQUATION OF ACCOUNTS. 635
OPERATION BY FIRST METHOD,
1895. Dr. of Dolls. Cr. of Ds. Cr. of Ds. on $1.
60
Feb. 1, $500.00 + 0 = 60 = 30000
March 18, 1019.68 90 + 45 = 135 = 137700
May 6, 704.25 30 + 94 - 124 = 87296
6é 21, 200.00 00 + 109 F 109 S- 21800
July 28, 400.00 60 + 177 F. 237 F. 94.800
$2823.93 f 371596
371596 days credit on $1 – $2824 = 132 days credit on $2823.93, which
added to February 1, the assumed maturity of each item of debit, gives June 13 as
the equated maturity of the account.
10. Suppose in the above problem that the purchasers, LACY & RogFRs,
had desired on August 1, to settle the account by a “cash” note, payable November
1/4, 1895, allowing interest at 8 per cent, for what face must the note be drawn?
Ans. $2916.91.
OPERATION INDICATED.
Interest on $2823.93 from June 13 to August 1, = 49 days at 8 per cent =
$30,75. $2823.93 + $30.75 = $2854.68, due August 1, and for which a cash note at
8 per cent is given payable November 1/4, -95 days, + discount day, = 96 days.
4500 × $2854.68 + (4500 – 96) = $2916.91, face of cash note.
C. B. LOTT & CO.
To CAHILL & LEIDENHEIMER Dr.
1896.
Jan. 20|To merchandise at 4 months gº cº sº {- º sº t- - wº- gº 1500|00
Feb. 15|| “. { { • 3 “ & tº- s £º sº sº &- s {- sº 840|75
March 9| “ $ 4 • * 4 44 * º tº * sº sº sº &= tº- s 200|28
April 1| “ acceptance “ 2 “ (with grace), º •- sº e & - 500|50
3041|53
11. May 1, LOTT & Co. settle by note drawn for such a time as will mature
at the equated date of the account. For how many days will the note be drawn
Ans. 23 days without grace, or 20 days with grace.
NotE 1.-In the operation, count actual days and allow for leap years.
NoTE 2.—The equated date is 125 days after January 20 = May 25, less 1 day for leap year
= May 24th.
12. Suppose, in the above problem, that LOTT & Co. had settled the account
May 1, by paying cash, what amount would they have paid, money worth 8 per cent?
Ans. $3025.98.
OPERATION INDICATED.
Since, as shown in the above problem, the equated date for the payment of
the account was May 24, therefore if the same is paid on May 1, the amount is to
be discounted for 23 days.
$3041.53 $3041.53 amount.
Thus: 45 || 23 15.55 discount.
$15.5455–H discount. $3025.98 net proceeds.
636 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. *
REMARKS.–It will be observed that in all of our work, where dates were
given, we have counted the actual days intervening between two dates. This we
believe to be the only correct system.
Many accountants however, to save a little time, count 30 days to the month,
by which the final result is often changed.
To find the time between dates and also in extending the equated time forward
or counting it backward from the date assumed, whether the earliest or latest, a
Time Table is often used to advantage. See Time Table on page 298.
COMPOUND EQUATIONS, OR AVERAGE OF ACCOUNTS.
1198. When the equated time is counted forward and when it is counted
backward. -
In all compound equations of payments or average of accounts, when the balance
of dollars and the balance of days are on the same side of the account (the debit or the
credit) the equated time is counted forward. When the balance of dollars and the
balance of days are on the opposite sides of the account, the equated time is counted
backward.
An account with the earliest date on the debit side and the final balance of
dollars and balance of days on the debit side, equated time is counted forward when the
earliest date is assumed, and backward when the latest date is assumed as the maturity
of the dollar items.
1. What is the equated time for the payment of the following account?
Ans. July 17, 1895.
I) r. PATTON, SWIFT & Co. Or.
1895. 1895.
Mar. 5. To mdse., at 60 days - $80000|April |10| By cash, - A- sº- $50000
April 22 “ “ 90 “ * 200000|May 28 “ “ - - - 1000|00
June | 1 | * * * * * * 30 ** tº- 1200,00|Aug. 9| “ “ - - - 800|00
“ [20] ‘ ‘ & 4 - 400|00 “ balance, * * * 2100.00
\ $4400|00 sºoooo
FIRST OPERATION
Assuming the earliest date in the account to work from.
1895. ; º; %" º | 1895. ºf "5" pººl.
Mar. 5, $800 60 + 0 = 60 48000 April 10, $500 36 18000
April 22, 2000 90 + 48 = 138 276000 || May 28, 1000 84 84000
June 1, 1200 30 + 88 = 118 141600 || Aug. 9, 800 157 125600
“ 20, 400 00 107 42800
$4400 508400 $2300 227600
2300 227600
Bal. of $ 2100 Bal. of days 280800

280800 + 2100 = 134 days credit due PATTON, Sw1FT & Co. on the balance
Yºr EQUATION OF ACCOUNTS. 637
$2100, and which, for reasons given in the first problem, is counted forward from
the assumed maturity of all the debit and also credit items of dollars, and gives
July 17, 1895, as the equated time for settling the balance of the account.
Explanation.—In this problem, PATTON Swift & Co. have both a debit and credit of dollars,
and also a debut and credit of days, but by our method of work, the solution is not in the least
complicated by reason of the credit of dollars and debit of days. The operation is necessarily
longer, but the two, and only two principles that we use the (debit of dollars and credit of days) are
as applicable to questions of this character as to those containing a debit of dollars and credit of
days only. In the main, the operation of this problem is the same as in the preceding examples.
We first assume the earliest date, March 5, as the due date of all the items of debit and credit of
dollars on both sides of the account, and then found, as is fully explained in the preceding work,
and adjusted the proper credit and debit of days on each item of debit and credit of dollars, accord-
ing to the conditions of the contract and the assumption of March 5, as the maturity of all the
dollar items of the account. We thus produce a net credit of 280800 days on $1, due to PATTON,
SWIFT & Co., which, as shown in the operation, is equal to 134 days on the balance, $2100 that they
OW €,
SECOND OPERATION
Assuming the latest date in the account to work from.
gº . Of Cr. of Cr. of Cr. of
1895. §§ Gººf H.” f pº ºl. 1895. ići. B. DS. ºil.
Mar. 5, $ 800 60 – 157 – 97 77600 Apr. 10, $ 500 121 60500
Apr. 22, 2000 90 – 109 = 19 38000 || May 28, 1000 73 73000
June 1, 1200 30 — 69 = 39 * 46800 || Aug. 9, 800 0
“ 20, 400 00 — 50 = 50 20000
$4400 182400 $2300 133500
2300 133500
Bal. of $ 2100 Bal. of days 48900
48900 - 2100 = 23 days that PATTON, SWIFT & Co. owe, August 9, on the
balance $2100, and to give the parties to whom this 23 days credit is due, the
advantage of it in the settlement of this account, it is clear by the exercise of our
reason that we must count backward from August 9, which gives us July 17, 1895,
as the equated time.
Having in the last problem and in preceding work on equations illustrated the
principles and operation of work by assuming either the earliest or latest date as
the maturity of all the debit and credit items of dollars, and preferring the method
of assuming the earliest date, to save space, we shall therefore give but one solution
of subsequent problems, and that by the earliest date method.
An account with the earliest date on the credit side, the balance of dollars and
days on the debit side, equated time counted forward from the earliest date,
2. What is the equated date of the balance of the following account 3
Ans. Jan. 6, 1896.
Dr. ... WEIL, HONOR & Fox. Cr,
1895. 1895.
Oct. 4|To Mdse., at 2 months || $250000|Sept. 8|By Cash, $ 80000
Nov. 1| “ Cash, s ge 300000||OCt. 20 % {{ †º 1500|00
Dec. 18 “ Acpt. at 1 month 100000|| 4 ||25 | {{ {{ sº 20000
NOV. 14 “ “ tº 5 70000

638 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. ºr
OPERATION.
Dr. of Cr. of * Cr. of Cr. of Dr. . of
1895. #ji. #." D.I. 1895. §. *.” p.º.
Oct. 4, $2500 61 26 = 87 217500 I Sept. 8, $ 800 0
Nov. 1, 3000 0 54 162000 || Oct. 20, 1500 42 63000
Dec. 18, 1000 34 101 = 135 135000 “ 25, 200 47 9400
Nov. 14, 700 67 46900
$6500 514500 $3200 119300
3200 119300
$3300 bal. of dolls. 395200 bal. of days.
395200 + 3300 = 11933, practically 120 days.
Ea.planation.—This 120 days credit of time on the balance of dollars, is due to WEIL, HoNor
& FOX; hence the time of payment is extended 120 days after the due date.
An account with the earliest date on the debit side, the balance of dollars and
days on the credit side, equated time counted forward from the earliest date.
3. When is the balance of the following account due 3
Ans. Feb. 21, 1895.
I}r. WOODS & WINN. Or.
1895. 1895.
Jan. 21|To Cash, - - sº $1420,00||Feb. 1|By Cash, - $500000
Feb. 1| “ Mdse., tºº - 180000|Mar. [10] {{ {{ - 700000
April 7| “ “ at 2 months, 950|40
OPERATION.

Assuming the earliest date as the maturity of the debit and credit items.
Dr. of Cr. of Cr. of ſº Cr. of Dr. of IDr. Of
1895. łºń. 3. Dººl. 1895. ji. is.” pººl.
Jan. 21, $1420.00 0 Feb. 1, $5000.00 11 = 55000
Feb. 1, 1800.00 0 11 19800 || Mar. 10, 7000.00 48 = 336000
Apr. 7, 950.40 61 76 -: 137 130150
$4170.40 149950 $12000.00 39.1000
4170.40 149950
Bal of dollars, sº,60 Bal. del, 21030
241050 days -- 7830 ($7829.60) gives 31 days.
Explanation.—This 31 days time WooDs & WINN owe to the parties with whom they have
this account, on the balance $7829.60, and as this balance of money is due to them, it is clear that,
in order to settle the 31 days time, the payment of the balance must be extended 31 days after
January 21, the assumed maturity of all the dollar items of the account, which gives February 21.
An account with the earliest date on the debit side, the balance of dollars on the
debit side and balance of days on the credit side, equated time counted backward from
the earliest date.
4. What is the equated time for the payment of the balance of the following
account Ans. Aug. 5, 1894.
* EQUATION OF ACCOUNTS, 639
Dr. SPENCER & FOLSOM. Or.
1895. 1895.
July |1|To Mdse., at 60 ds. $1200,00|July 22|By Cash, $ 50000
Aug. 10 “ {{ {{ 60 4 130000|Aug. 10 “ { % 60000
Sept. 25. “ {{ {{ 30 % 110000|| “ |10| “ Acpt. at 30 ds. 50000
Oct. 6| 44 {{ {{ 10 % 2000|00||OCt. 6| “ Cash, 1500|00
Dec. 1| “ Note at 90 ds. 200000
OPERATION.
Working from the earliest date.
Dr. Of Cr. Of Cr. of Cr. Of Dr. Of Dr. of
1895. #. Sº p. 1. 1895. ji. *.” Dººl.
July 1, $1200 60 0 = 60 72000 || July 22, $ 500 21 10500
Aug. 10, 1300 60 40 = 100 130000 || Aug. 10, 600 40 24000 -
Sept. 25, 1100 30 86 = 116 127600 “ 10, 500 33 40 = 73 36500
Oct. 6, 2000 10 97 = 107 214000 || Oct. 6, 1500 97 145500
Dec. 1, 2000 93 153 = 246 492000
$5600 543600 $5100 708500
5100 543600
$500 balance of dollars. Balance of days, 164900
164900 days divided by 500 = 330 days.
Explanation.—By the operation of this problem, we find that SPENCER & FOLSOM owe both
a balance of dollars and a balance of days, and hence, by the exercise of our reason, we see that
in order to settle or pay in kind the 330 days, the balance of dollars must be made due 330 days
prior to July 1, the assumed due or focal date of all the debit and credit items of dollars. 330 days
º backward from July 1 gives August 5, 1894, as the equated time for paying the balance of
Olla,I’S.
By the foregoing problems and illustrations, we see, for reasons given in the solutions, that,
when the balance of dollars and balance of days occur on the same side of the account, whether the debit
or the credit, the equated time is counted forward from the earliest date; and that, when the balance
of dollars and balance of days occur on opposite sides of the account, the equated time is counted
backward from the earliest date.
An account with the earliest date on credit side and balance of dollars on credit
si:3, and balance of days on debit side, equated time counted backward from the earliest
date.
5. What is the average date of the following account 3
Ans. April 10, 1895.
Dr. LACY & ROBINSON. Or.
1895. 1895.
July | 6|To Mdse., 90 days, $210000|June 14|By Cash, $5441|00
Aug. 30. “ {{ 90 ($ 50000|July 28 “ “ 300000
Sept. 18. “ (4 60 (£ 20000|Aug. 1| “. “ 300|00
Oct. 1} {{ (< 30 % 90000
64O soule's PHILosophic PRACTICAL MATHEMATICS. &
OPERATION.
Dr. of Cr. Cr. Of * Cr. of Dr. of Dr. of
1895, #. “Bº” pººl. 1895, Éli. “5,” Dººl.
July 6, $2100 90 22 == 112 235200 June 14, $5441 00
Aug. 30, 500 90 77 = 167 83500 || July 28, 3000 44 132000
Sept. 18, 200 60 96 = 156 31200 || Aug. 1, 300 48 14400
Oct. 1, 900 30 109 = 139 125100
$3700 475000 $8741 146400
146400 3700
Bal. of days, 328600 $5041 bal. of dolls.
328600 days -- 5041 = 65 days.
Explanation.—By the operation of this problem, we find that LACY & ROBINSON, have due
to them both a balance of days and a balance of dollars, and hence, in order to give them credit
for the 65 days due to them, we must pay the balance of money 65 days before the assumed date,
June 14, and consequently the time is here counted backward from June 14.
6. A owes B $1600, due in 6 months. A pays $800 in 2 months. What is
the equated date for paying the balance? Ans. 10 months.
FIRST OPERATION.
$1600 × 6 months = 9600 = months credit due A on $1.
800 × 2 months = 1600 = months debit due by A on $1.
800 8000 = months credit balance due A on $1.
8000 -- $800 = 10 = months credit due A on $800.
SECOND OPERATION.
Dr. of Dolls. Cr. of Mos. Cr. of Mos. on $1. Cr. of Dolls, Dr. of Mos. I)r. of Mos. on $1,
$1600 X 6 ~ 9600 $800 X 2 2- 1600
800 1600
$ 800 8000
Explanation.—In the operations of the above problem, we first observe that while A owes B
$1600 in money, B owes A 6 months time to pay the money. And 6 months credit on $1600 is equal
to 9600 months credit on $1. ... Then, as A paid $800 in 2 months B owes A for the $800, and A owes
B for the 2 months credit on $800, which equals 1600 months credit on $1. The balance of account
i. º A debit $800 and credit 8000 months on $1. 8000 - 800 = 10 months credit on the
ala,D C6 g
7. A owes B a note dated March 10, 1895, due in 183 days for $1200. June
14, 1895, A paid on account $300. What is the equated time for paying the balance,
counting the actual days in each month ? Ans. Oct. 8, 1895.
FIRST OPERATION.
$1200 x 183 = 219600 = days credit due A on $1.
300 × 96 = 28800 = days debit that A owes B on $1.
$ 900 190800 = balance of days credit due A on $1.
190800 - 900 = 212 days credit due A on $900. And 212 days after March
10, 1895, – Oct. 8, 1895. See Time Table.
ti Explanation.—From March 10 to June 14, is 96 days. See above problem for further explana-
1OLl,
º: EQUATION OF ACCOUNTS. 641
SECOND OPERATION.
Dr. of Dolls. Cr. of Ds. Cr. of DS. on $1. Cr. of Dolls, Dr. of Ds. Dr. of Ds. on $1.
Mar. 10, $1200 183 219600 June 14, $300 96 28800
300 28800
$ 900 190800
190800 - 900 = 212 days credit on $900 after March 10, 1895, as Oct. 8, 1895.
8. May 14, 1895, A owes B $800 due in 4 months and $500 due in 3 months.
A paid on account $300 in 1 month and $100 in 3 months. When is the balance
due by equation ? Ans. 4 months 17 days.
OPERATION INDICATED.
Dr. of Dolls. Cr. of Mos, on $1. Cr. of Dolls. Dr. of Mos, on $1,
$ 800 X 4 E. 3200 $300 X 1 F 300
500 X 3 - 1500 100 X 3 – 300
$1300 4700 $400 600
400 600
$ 900 4100
43 mos. = 4 mos. 16; d.s.
9. May 14, 1895, A owes B $800 due in 4 months, and $500 due in 3 months.
A paid on account $700 in 4 months and $500 in 6 months. When is the balance
due # Ans. June 14, 1894.
OPERATION INDICATED.
$ 800 X 4 - 3200 $700 X 4 - 2800
500 X 3 - 1500 500 X 6 ~ 3000
$1300 4700 $1200 5800
1200 4700
$ 100 1100
1100 + $100 = 11 months, i. e. the $100 balance was due 11 months before
the date of May 14, 1895, which is June 14, 1894. In other words A owes B interest
on $100 for 11 months. Hence the time is counted backward.
10. May 14, 1895, A owes B $800 due in 4 months and $500 due in 3 months.
A paid $900 in 4 months and $600 in 5 months. What is the equated date for the
settlement of the balance due A { Ans. Feb. 29, 1896.
OPERATION INDICATED.
$800 X 4 = 3200 $900 X 4 = 3600
500 X 3 - 1500 600 X 5 -: 3000
$1300 4700 $1500 6600
1300 4700
$ 200 1900
1900 + $200 = 94 months, after May 14, 1895, - Feb. 29, 1896.
Explanation.—The result of the operation shows that B owes A $200 and that A owes B 94
months credit on the same, from May 14, 1895. Hence while Bowes A the $200 he need not pay it
until 94 months after May 14, 1895. It must not be forgotten that all items of debit and credit
are assumed to be due on the earliest date.
642 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. *e
AN ACCOUNT EQUATED AND SETTLED BY NOTE.
1199. 1. The following bills were sold to TYLER & MILLs on a credit of 90
days each :
January 4, 1895, - $824.15 May 21, 1895, - - - $718.15
February 19, 1895, - 371.68 June 7, 1895, - - - 2120.80
April 1, 1895, - 1410.00
The following payments were made on the above purchases:
March 1, 1895, - - $600 cash. May 10, 1895, - $800 cash.
April 1, 1895, - - 350 “ June 8, 1895, - 1000 “
What is the equated time for paying the balance of the account?
Ans. Oct. 4, 1895.
OPERATION.
Dr. Of Cr of Cr of Cr. of Dr. of Dr. of
1895. #ji. “Bº” D.'...'. 1895. ji. "Bº" pººl.
Jan. 4, $824.15 90 O - 90 74160 || Mar. 1, $600 56 33600
Feb. 19, 371 68 90 46 + 136 50592 || April 1, 350 87 30450
April 1, 1410.00 90 87 + 177 249570 || May 10, 800 126 100800
May 21, 718.15 90 137 = 227 162986 || June 8, 1000 155 155000
June 7, 2120.80 90 154 = 244 517524
$5444.78 * 1054832 $2750 3.19850
2750.00 3.19850
$2694.78 bal. of dolls. 734982 bal. of days.
734982 days -- 2695 = 273 days, which counted forward from J anuary 4,
1895, gives October 4, 1895, as the equated time.
2. Suppose in the above problem that TYLER & MILLs had, on the 8th of
June, given their note due at the equated date for the balance due, for how many
days would the note have been drawn 3
Ans. 118 days without grace, or 115 days with grace.
NOTE:-The equated date being October 4, 1895, we simply find the intervening time between
June 8 and October 4, which is 118 days.
3. Suppose again, in problem No. 1, that TYLER & MILLs had settled June
8, the balance due in cash, allowing 8 per cent for the use of money and not count-
ing discount day, what sum would be required to pay the balance?
Ans. $2624.12; which is called the cash balance.
OPERATION.
$ Fº Eaplanation.—The equated maturity of the balance being Oct. 4,
2694.78 1895, it is clear that if TYLER & MILLs pay it June 8, they should
45 || 118 pay
be allowed the interest on the same for the intervening time, 118
| $70,66 days, which is, as shown by the opposite statement, $70.66.
EQUATION OF ACCOUNTS. 643
4. May 10, 1895, A owes Z $3000, due in 3 months. June 4, 1895, A paid Z
$2000 on account, when is the balance due # Ans. Dec. 22, 1895.
OPERATION.
1895. Ds. DS. on $1. 1895. DS. Ds. on $1.
May 10, $3000 92 = 276000 June 4, $2000 25 = 50000
2000 50000
$1000 ) 226000
226 ds.
EQUATION OF ACCOUNT SALES.
1200. An Account of Sales is a statement of the articles and price of
merchandise or other property sold, the charges incurred in effecting the sales, and
the net proceeds due the party or parties for whose account the goods were sold, and
to Whom the account sales is rendered.
The net proceeds are due as cash at the equated date of the different sales
and charges.
When equating account sales, it is usual to consider the freight and drayage
due on the receipt of the goods. The commission is often considered not due until
the last day of sale, but some accountants equate the time for the charge of
commission. In large sales, or where there is a long interval of time between sales,
the commission should be equated. The other charges are sometimes considered
due on the last day of sale and sometimes on the day that each item was paid, or
the average date.
In case of account sales for merchandise sold on joint account, the interest
of the consignee is sometimes considered due at the date of shipment and sometimes
at the date the goods were received. In justice to both parties, and in order to
preserve a uniformity in the accounts of each party interested in the goods, this
matter should be agreed upon at the time the parties contract for the joint account
operation. The consignee's gain or loss in joint account transactions is generally,
and should always be made due, the same as the commission, at the equated date
of the sales.
NOTE,--It is the custom of cotton factors in New Orleans, to consider the proceeds of cotton
sales due 7 days after the sale was made. This 7 days use of the proceeds is considered a just
compensation for the interest on the money advanced for charges, and saves the time and labor
of equating the account sales.
Flour and produce dealers consider the proceeds of goods sold on commission due ten days
after the sale was made. This allowance is made in place of equating the account sales, and is
considered just to both parties.
644 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS.
PROBLEMIS.
1. When are the net proceeds of the following consignment due?
Ans. Nov. 4, 1895.
No. 1820.
Sales by W. H. MoRGAN & Co. of 1200 bbls. flour, received Oct. 11, 1895, per
steamer Susie Silver, for account of SCHREIBER & EICKE, Belleville, Ill.

Oct. |12|Sold 100 bbls. Sucker City, cash, a $8 * * * * * * * 80000
& 4 15| “ 200 “ & 4 & 4 4 & Ø $74 sº sº &- wº tº- <- * 1575.00
“ “| “. 284 “ & 4 “ 20 ds., a $7; tº s tº m s tº º 2130|00
& 4 21| 44 8 * * & 4 ‘‘ cash, (a) $8+ tº ess º º º & &- 66100
“ |26 “ 200 “ & 4 “ 10 ds., a $7+ * dº •-º * tº º <-º 1500|00
** 29 “ 408 “ & & “ 15 ds., a $7; sº *- º wº sº wº *- 3060|00
1200 $9131|00
CHARGES.
‘Oct. 11|Freight, 40 cts. per bbl., $480; Charges per B. L., $240, - tº sº 72000
‘‘ 29|Labor, $12; Tarpaulins, $6; Inspection, $24, º g- * : tº tºº 4200
“ “|Drayage, $49; Storage, $54; Insurance, Fire $21.84—River $67.50, - 192|34
“ “[Skidding, $1.60; Cooperage — tº gº tº gº tº ſº sº 1|60
Nov. 2 Commission on sales at 24 per cent, - - - * † tº º sº 228/28|| 1184|22
Net proceeds to your credit due by equation Nov. 4, 1895, - - - $7946|78
E. & O. E.
NEW ORLEANs, November 1, 1895,
WM. H. MORGAN & CO.,
per F, HORN.
OPERATION.
1895 Ds Ds. 1895. Ds Ds. Ds.
‘Oct. 11, $720.00 0 Oct. 12, $800.00 0 1 800
“ 29, 235.94 18 4248 “ 15, 1575.00 0 4 6300
Nov. 2, 228.28 22 5016 “ 15, 2130.00 20 4 = 24 51120
“ 21, 66.00 0 10 660
“ 26, 1500.00 10 15 = 25 37500
“ 29, 3060.00 15 18 = 33 100.980
$1184.22 9264 $9131.00 197360
1184,22 9264
$7946.78 188096
188096 - 7947 = 24 days, which counted forward from October 11, gives
November 4, 1895, as the equated date.
Explanation.—In this equation we considered the freight and drayage due on the receipt of
of sale.
& 4
& 4
( &
& 6
9
15,
21,
26,
29,
OPERATION
Ds. Ds.
$ 800 0
1575 3 =
2130 20 + 3 = 23 =
66 9 =
1500 10 —- 14 = 24 =
3060 15 + 17 = 32 =
$9131
188229 -- $9131 = 21 days.
This equation gives us 21 days, which counted forward from October 12,
gives November 2, as the equated date of the sales.
4725
48990
594
36000
97920
188229
To find the equated date of the sales only, at which date the commission is due.
Oct. 12,
& 4 1
the goods, the commission on the average date of the sales, and the other charges on the last day
EQUATION OF ACCOUNTS. 645
EQUATE THE FOLLOWING ACCOUNT SALES ON JOINT ACCOUNT.
2. Sales of 1000 bbls. potatoes, received per steamer R. E. Lee, Nov. 2, and
sold on the joint account of the shippers, SPENCER & FREDERICK, and ourselves,
each 3, by MOSES, LEHMAN & Co.

1895 -
Nov. 3|Sold 300 bbls. Peach Blows, cash, (a) $3 º - - - solo
& 4 9| “ 450 “ Neshannock, 20 ds., a $3.20 - - * - 1440|00
“ [15] “ 60 “ 4 & cash, @ $3} º - * - 195|00
“ |28 “ 50 “ Pink Eyes, 10 ds.; (a) $3} * - sº - 175100
Dec. 4 “ 140 “ Peach Blows, 15 ds., (a) $3} - - º 490|00
1000 $3200.00.
CHARGES,
Nov. 2 Freight, 40 cts. per bbl., - - - - - - - - - 400|00
Drayage, * - - - tº- - º - wº º º - 50|00
Insurance, - - º - - -> tº - sº - º - | 21|50
Storage, $51.25; labor, $8.50, - * †- - wº - - - 59.75
Commission (a) 2% per cent, dº - - - gº tº ºn - 8000 611|25.
Net proceeds, - * - º - s - º - tº $2588.75
Cost of in voice, * - &= - º - sº - º 2000|00.
Net gain, - - - - - - - - - - - - $588.75
Our # net gain, - - - º - º - * * * - $294,37
Spencer & Frederick's # net gain, - tº - * - sº º $294.38
4 & $ & # invoice, tº tºp tº- tº - gº - 1000 00
& 4 ( & net proceeds, * - sº º ºs º- 1294.38
(exclusive of $ invoice previously credited,) due by equation Dec.
19, 1895, * * * ºs & & - *> - sº ->
E. & O. E.
NEW ORLEANs, December 5, 1895.
MOSES, LEHMAN & CO.,
Per LOUIS LEVY.
OPERATION.
1895. Ds. Ds. 1895. Ds. DS Ds.
Nov. 2, Fr’t & Dray. $450.00 0 Nov. 3, $900.00 1 900
Dec. 1, Ins., Sto., &c. 81.25 29 2349 ‘‘ 9, 1440.00 20 7 – 27 *38880.
Nov. 24, Commission '80.00 22 1760 “ 15, 195.00 0 13 = 13 2535
“ 24, Our # gain 2.94.37 22 6468 “ 28, 175.00 10 26 = 36 6300
“ 2, “ , inv. 1000.00 0 Dec. 4, 490.00 15 32 = 47 23030.
$1905.62 10577 $3200.00 71645
1905.62 10577
$1294.38 61068.
according to agreement.
account sales does not give the date of purchase.
61068 - $1294 = 47 days.
Explanation.—In this operation we have considered the freight and drayage due the day that
the produce was received. The other charges, except commission, we considered due December
1, as the greater portion of them must by custom have been paid then, if not paid when they were
respectively incurred. Strict justice would require the equated date of charges in all account
sales, but in practice it is not always done. The commission and gain are due on the average date
of the sales, and the 3 invoice the day that the produce was received, or the date of purchase
We have here considered the # invoice due the date received, as the
646 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. Yºr
OPERATION
To equate the sales only, at which date the commission and gain are due.
1895. Ds. Ds. Ds:
Nov. 3, $900 0
“ 9, 1440 20 + 6 = 26 = 37.440
“ 15, 195 0 -- 12 = 12 = 2340
“ 28 175 10 + 25 = 35 = 6125
Dec. 4, 490 15 + 31 = 46 = 22540
$ 68445
68445 -- $3200 = 21 days.
NoTE.-In the equation of account sales for joint account transactions, in which losses are
sustained, they are entered on the credit side of the account in the operation of equation.
ACCOUNT SALES.
1895
Jan. 3|Sold for cash, 500 bbls., Ex. Flour at $9. tº- - - - - $450000
& 4 8|Sold at 10 days, 100 bbls., Flour at $8.50, - •- - - - 850|00
$535000
CHARGES.
Jan. 2|Sundry charges, - - - - - - - - - - $290|50
& 4 8|Commission on sales at 2% per cent, as sº-, * > • * 133|75|| 424/25
Net proceeds, - - - * º wº sº - -- º * $4925||75
What is the equated date of the proceeds in the above account sales?
Ans. Jan. 6, 1895.
OPERATION.
Jan. 2, Charges, - - $291 0 = 0 || Jan. 3, Sales, * - $4500 1 = 4,500
‘‘ 8, Commission, - - 134 6 = 804 “ 8, “ sº - 850 16 = 13,600
$425 804| $5350 18,100
425 804
17296 -- $4925 = 4 days, after Jan. 2, = Jan. 6/95. $4925 17,296
SALES OF MIDSE. ON JOINT ACCOUNT.
Folio.
1895 COST AND CHARGES.
Jan. 4|Received, per Steamer Golden Rule, from W. B. Henry, to be sold
for his and our joint account, each #:
400 bbls. Mess Pork at $20, * - * *- - - $8000
4|Com. Sales Dr. to W. B. Henry, our # º * - - - 11 400000|| 4000|00
To Charges: &
4|Cash for Freight, - - - - - - - - $120 | C. 1
4| “ for Drayage, - - - - - - - -> 24 o
Insurance, Fire, g- - º - - - - --> 20
( & River, * - - - - - º tº- 180
Inspection, $12; Cooperage, $3, - * * * º 15 7 359|00|| 35900
Com. S. Dr.—Our 3 and charges, - s º º - - 11 || 4359|00
11 To Commission on Sales at 2% per cent, - tº - - - 11 226 25 226,25
“ W. B. Henry's # Inv. me * * * *-> * $4000
& 4 do # Net Gain, -> - - - - 232.38
11|W. B. Henry's Net Proceeds, - º -> º- - - - 11 4232|38|| 4232.38
Due by Equation February 7/95:
11|To P. and Loss, our # Net Gain, - º º º- - - 12 232|37 232;37
11|Com. Sales Dr. to close account, º º tº e dº tº- - 11 4691|00

905000
* SALES ON Joint Account. 647
SALES OF THE ABOVE MDSE.
Folio.
1895
Jan. 6W. J. Vizard, Cash.|C. 1
100 bbls. Mess Pork (damaged) a $15, - - - - - 150000
9|S. J. Blessing, Note 15 ds.
100 bbls. Mess Pork a $26, E* - * . gº sº º Q- 2/11 2600|00
11|Geo. W. Weingart, Jr., 10 ds.
200 bbls. Pork a $24.75, - tº º †- gº tº sº gº 8/11 495.000
905000
When are W. B. Henry's net proceeds due # Ans. Feb. 7, 1895.
OPERATION.
Jan. 4, § Inv., & Chgs., $4359 0 = 0 1 Jan. 6, Sales, $1500 2 F 3,000
“ 11, Com & Gain, 459 7 = 32.13 “ 9, “ 2600 23 —r 59,800
*=====s== “ 11, “ 4950 17 - 84,150
$4818 3213 sºme sº-mºme *= º ºsmºs
$9050 146,950
4818 3,213
143737–$4232=34 days, after Jan. 4, which is Feb. 7/95. $4232 143,737
SALES ON JOINT ACCOUNT.
Folio.
1895 COST AND CHARGES.
Jan. 20 Received, per Steamer J. M. White, from D. Hughes, to be sold on
the joint account of himself, J. W. Moses and ourselves each # :
500 bbls. XX Flour (a) $7.50 = $3750.
To which we have added from store:
100 bbls. Molasses (a) $18 = $1800 Cr. Mdse., - tº- •ºm t- 1
D. Hughes, Cr, for our # of his Inv., & e tº sº gº & 12|| 1250,00|| 1250.00
Com. Sales, Dr. for our # of our Inv., * * * * * V 60000|| 60000
D. Hughes, Dr., for # of our Inv., $600. tº º * * * *- 12
J. W. Moses, Dr., for # of our Inv., $600. º, tº sº º 12|| 2 h
Charges :
20|Cash for Freight, $80; Drayage, $30, * tº ºn age º C, 1|| 11000
Labor, $2; Inspection, $10, * * * * * * * 1200
Insurance, Fire, * tº- 3- º * > Gºe • * ge *E* 1800
do River, &= * $º sº * - sº wº sº º 7 7500|| 21500
Com. Sales, Dr. for our # cost and charges, - - - - 11|| 2065!00
29|Commission on sales a 2, per cent, sº sº sº sº we 11|T10555|| 10555
29|Cash refunded to W. J. Vizard— Amount allowed on damaged { C. 2 20000
Molasses, gº º wº- 4- <-> gºe º * Ge. * *
Charges :
29|Storage, $10; Cooperage, $4, - - - - - - - 7 14|00 14|00
D. Hughes, # of his Inv., i- se <--> & * * * $1250.00
£ 6 # of our Inv., - *- * - ſº &º sº 600.00
$1850.00
Less his # Net Loss, 4-6 iº ** º ſº * * tº 620.95
29|D. Hughes' Net Proceeds, Cr. ——|| 12|| 122905] 1229,05
Due by Equation Feb. 6/95:
29|J. W. Moses, of D. Hughes' Inv., - º tº sº $1250.00
do # of our Inv., sº tº º * * sº 600.00
$1850.00
Less his # Net Loss, - gº tº e tº tºº Gº 620.95
J. W. Moses, Net Proceeds, Cr. sº º ºs º º sº 12|| 1229.05|| 122905
Due by equation Feb. 6/95. * -º
29|Com. Sales, Dr., to close account, - - - tº dº ſº 11|| 257765
4842.65-


648 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
SALES OF THE ABOVE MDSE.
Follo.
1895 *
Jan. 29|W. J. Wizard, Cash.
500 bbls. XX Flour (a) $5, - s ‘Es & gº * - * -; 250000
100 bbls. Molasses, 3826 gallons, a 45c., *E. * * g- C. 2 || 1721||70|| 4221||70.
29|By Profit and Loss for our # Net Loss, - dº $º sº tº- 12/11 620.95.
4842|65,
When are the net proceeds of D. Hughes and J. W. Moses, due by equation
in the above account 3 Ans. Feb. 6/95.
OPERATION.
Jan. 20, # Inv. & Chgs, $2065 0 = 0 || Jan. 29, Sales, $4222 9 – 37,998
“ 29, Com. & Chgs., 320 9 – 2880 “ 29, § Loss, 621 9 = 5,589
$2385 2880 $4843 43,587
2385 2,880
$2458 )40,707(17 ds-
40707 - 2458 = 17 days, after Jan. 20/95, - Feb. 6/95.
EQUATION OF RENT NOTES.
1201. 1. January 1, 1895, A. leases of Z. a store for 1 year, at $500 per
month, amounting to $6000, on the following conditions: The rent to be paid
monthly in advance, as follows: The first month's rent is to be paid in cash, and
monthly maturity rent notes given for the remaining 11 months rent.
After the lease is signed, A. proposes to give one note for the whole rent
payable at such a time that neither party will gain nor lose by the change of terms;
to this proposition Z. agrees. What is the time for the payment of the new note?
Ans. June 16, 1895.
FIRST OPERATION.—BY Months WITHOUT AL- SECOND OPERATION.—BY ACTUAL DAYS AND AL-.
LOWING FOR GRACE. LOWING FOR 3 DAYS OF GRACE ON EACH NOTE.
Mos. Mos Cr. on $1. Ds. Ds. Cr. on $1.
$500 0 0 C $500 0 O
500 1 500 1 500 33 16500
500 2 1000 2 500 61 30500
500 3 1500 3 500 92 46000
500 4 2000 4 500 122 61000
500 5 2500 5 500 153 76500
500 6 3000 6 500 183 91500
500 7 3500 7 500 214 107000
500 8 4000 8 500 245 122:500
500 9 4500 9 500 275 137500
500 10 5000 10 500 306 153000
500 11 5500 11 500 336 168000
$6000 ) 33000 $6000 } 1010000
5% months. June 16, ’95. º 1683 days. June 18, '95.
Yºr EQUATION OF RENT NOTES. 649
2. Suppose in the above problem the lessee had desired to give, and the
lessor had consented to receive, two notes of equal amount, payable in periods of
time in the ratio of 1 and 4, what would have been the time of each note, allowing
three days of grace on each of the 11 notes?
OPERATION.
As shown in the second operation above, A., the lessee, has a credit of 1010000 days on $1, on
the presumption that the $6000 are due January 1, 1895. Now, by the terms of the problem, we must
find the respective credits on the two new notes so that the total credit on $1 shall be equivalent to
º days on the 11 notes which he first proposed to give. This we do by the following
Statements:
1st note, $3000 × 1 = 3000 I day. | 4 ds.
2d note, 3000 x 4 = 12000 15000 || 1010000 15000 || 1010000
15000 | 67% ds. = 2 mos. 7# ds, 269; d.s. = 8 mos. 29% ds.
PROOF.
1st note $3000 × 67; d.s. = 202000 ds. credit on $1.
2d note $3000 × 269; d.s. = 808000 ds. credit on $1.
1010000 total ds. credit on $1.
The aggregate days credit on $1 is the same as shown above on the eleven notes.
See pages 651 and 652 for full discussion of thus kind of work.
EQUATION OF ACCOUNTS AND NOTES BEARING INTEREST.
1202. 1. A. owes Z. $500 due March 8, 1895, and bearing interest at 8 per
cent; $1700, due May 17, 1895, and bearing interest at 6 per cent; and $350, due
June 1, 1895, and bearing interest at 8 per cent. What is the equated time for the
payment of the whole debt? Ans. May 3, 1895.
OPERATION.
1895 Ds.
Mar. 8, $ 500 8% F $ 4000 0
May 17, 1700 696 = 10200 70 714000
June 1, 350 8% = 2800 85 238000
$17000 ) 952000
952000 days -- 17000 = 56 days.
2. A. holds Z's note, due August 10, 1895, for $5000, bearing 8 per cent
interest from maturity, and Z. holds A's acceptance, due October 24, 1895, for
$14000, bearing interest at 5 per cent from maturity. What is the equated time
for paying the balance? Ans. Feb. 1, 1896.
OPERATION.
1895, DS, Ds. 1895. Ds. Ds.
Aug. 10, $5000 8% = $40000 0 0000 || Oct. 24, $14000 5% = sº 75 = 5250000
4000
$30000 ) 5250000
******mº
175 ds.
65o SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. †
5250000 days -- 30000 = 175 days, which counted forward from August 10,
1895, gives February 1, 1896, as the equated date for the payment of the balance.
By this equation, we see that the $5000 due August 10, 1895, was paid 175
days after it was due; and that the $14000, due October 24, 1895, was paid 100 days
after it was due.
Interest on $ 5000 at 8% for 175 days = $194,44
“ “ 14000 “ 5% tº 100 “ = 194,44
3. A. holds notes and acceptances against Z. as follows:
His note due Jan. 28, '95, for $2000, bearing 8% interest.
{{ 4% 4 Feb. 4, '95, & 4 4200, {{ 6% 46
“ acpt. “ Mar. 10, '95, “ 7000, “ 6% “
44 note “ “ 31, ’95, “ 2500, {{ 7% {{
Z. holds the following claims against A. :
His note due Feb. 14, '95, for $2400, bearing 5% interest.
“ acpt. “ Mar. 21, ’95, “ 3000, “ 8% “
“ note “ April 9, '95, “ 5000, without interest.
What is the equated time for paying the balance of the above account, allow-
ing 8 per cent interest on the balance? Ans. March 5, 1895.
OPERATION.
1895. 1895.
Jan. 28, $2000 8% = $16000 0 = Feb. 14, $2400 5% = $12000 17 = 204000
Feb. 4, 4200 6% = 25200 7 = 176400 || Mar. 21, 3000 8% = 24000 52 = 1248000
Mar. 10, 7000 6% = 42000 41 = 1722000 || April 9, 5000 0% = 00000 71 = 0000000
“ 31, 2500 796 = 17500 62 = 1085000
$15700 $100700 2983400 $10400 $36000 1452000
10400 1452000
$5300 a 8% = $42400 ) 1531400
1531400 + 42400 = 36 ºr days.
Explanation.—By the operation we obtain practically 36 days, which counted forward from
January 28, gives March 5, 1895, as the equated date for the payment of the balance, allowing 8 per
cent interest on same.
NotE.—The operation also shows a debit of 1531400 days on $1, at 1 per cent, and as the
§. of the balance of $5300 multiplied by the 8 per cent and the 363& days just equals 1531400
ays, it is clear that the equated time is March 5, 1895.
EQUATION OF NOTES, 65 I
EQUATION OF NOTES MATURING AT DIFFERENT DATES, AND THE
SUBSTITUTION OF OTHERS OF DIFFERENT AMOUNTS AND
MATURING AT EQUAL INTERVALS OF TIME.
1203. 1. A. owes Z. three notes bearing date, and drawn for the amount and
time respectively, as follows: August 10, 1895, $2000 payable 90 days after date;
September 12, 1895, $1200 payable 6 months after date; and November 25, 1895,
$4800 payable 1 year after date.
A. wishes to take up or redeem these three notes with four others of equal
amounts and maturing at equal intervals of time, so that there will be no loss or
gain to either party. What is the time of each new note *
OPERATION.
1895 Dr. Of Contract Total Cr. Cr, of Ds.
<- Dolls. Cr. of DS. of Ds. on $1.
Aug. 10, $2000 90 -- 0 = 90 180000
Sept. 12, 1200 182 + 33 = 215 258000
Nov. 25, 4800 366 -- 107 = 473 2270.400
4 ) $8000 2708400
$2000 each new note.
- Having now found the number of days credit due A. on the three old notes,
from August 10, 1895, the day that we assume them all due, we have next to find
the respective number of days credit on each of the 4 new notes in proportion to 1,
2, 3, and 4, as required by the conditions of the problem, so that the total credit of
the same shall be equivalent to the 2708400 days on the 3 old notes. We therefore,
in order to produce the necessary proportional numbers to make a solution state-
ment, proceed as follows:
Ds. Cr. assumed. Ds. Cr. on $1.
1st New Note $2000 1 2000
2d “ & 4 2000 2 4000
3d 44 & 4 2000 3 6000
4th ** & 4 2000 4 8000
20000
By this work we obtain 20000 days credit on the 4 new notes, and as the
assumed credit of 1, 2, 3, and 4 days gave this 20000 days, by transposition, we see
that the 20000 days credit required the 1, 2, 3, and 4 days time, and as we wish to
distribute 2708400 days credit on $1, on the time of the 4 new notes, the following
proportional statements will give the correct time of each note.
Statement for the 1st Note. Statement for the 2d Note. Statement for the 3d Note. Statement for the 4th Note.
E). DS. DS. DS.
1 2 3 4
2708400 20000 | 2708400 20000 || 2708400 20000 || 2708.400
135%; | 270}}}}} *406 fºr 541}#}}}
652 - soule's PHILOSOPHIC PRACTICAL MATHEMATICS. A.
The fractions of a day in the result cannot be avoided, and to use them as
equitably as the conditions of the problem will admit of, we omit the fractions in
each result and increase the whole numbers of the 2d and 4th results, 1 day each.
The reasoning for the first statement is as follows: If 20000 days credit on
$1 require 1 day's time, or a note of $2000 payable 1 day after sight, 1 day's credit
will require the 20000th part, and 2708400 days credit will require 2708400 times as
much. For each of the other statements the reasoning is the same, except that we
have respectively, 2, 3, and 4 days assumed time on the 2d, 3d, and 4th notes.
Instead of making these 4 statements, we could, in this problem, have multi-
plied the result of the first by the assumed numbers, 2, 3, and 4, and thus have
obtained the result. This method of work is applicable to all questions of this
character, regardless of the proportion of time that the new notes may be desired
to run.
To elucidate still further that the credit on the new notes is equivalent to the
credit on the old notes, we present the following:
--- DS. Cr. on $1.
1st New Note $2000 135 days 270000
2d 44 & 4 2000 271 “ 542000
3d 44 6 & 2000 406 “ 81.2000
4th 84 & 4 2000 542 “ 1084000
Total credit on $1, 2708000
4 & & 4 “ $1, of old notes, 2708400
Excess of credit on old notes, 400
NOTE.-The small excess of credit on the old notes results from the fact that we did not and
cannot use the exact time as shown in the four statements above. In the first and third notes the
small fraction of time was omitted. While in the second and fourth notes the fractions of time
being each more than one-half day, they were each counted as one day. The small excess thus
Fººd is only equal to one-fifth of a day on one of the new notes, which is too small a credit to
© U186
~##–Z
^2(5¢)\ \
º
Aunt Current and Interest AC(OInts, aid Cash BālāICES.
=N
Aº Aº-º.
w = w wr
1204. When it is desired to settle an account by cash, or to find the balance
that is due at any specified date, for the purpose of closing the books or determining
the full resources and liabilities of the firm, one of two methods, the interest or
the product, is generally used.
THE INTEREST METEHOD.
1205. The first and most usual method is to find the interest on each item of
debit and credit of dollars from the time it became due up to the date of Settle-
ment, or balancing of the accounts, and then charging or crediting the debtor with
the balance of interest, according as it is against him or in his favor.
TEIE PRODUCT METEIOD.
1206. The second, and next most usual method, is to find the balance of
interest due the debtor or creditor, by first finding the balance of the product of
the debit and credit of days on $1, the same as in equation of accounts when work-
ing from the latest date, and then dividing the same by the interest divisor, and thus
obtaining the interest, which is charged or credited to the debtor as in the first
method. º
In our problems and operations, we will illustrate both of these methods and
also a new method by which the operation is shortened and the accountant facili-
tated in the discharge of his duties.
THE EQUATION METEOD.
1207. This is a new method which we name the Equation Method, for the
reason that we assume all of the dollar items due on the earliest date in the account,
and proceed with all the debit and credit items of dollars, the same as we worked
equations when we assumed the earliest date to work from.
PROBLEMIS.
1. What is the cash balance of the following account, December 31, 1895,
interest at 8 per cent ?
Ans. $1531.42.
IDr. RENCH & WURZLOW. Or.
1895 1895
Feb. 4To Mdse. (a) 90 days, - - - $ 50000||Mar. [10|By Cash, - - - - - - - $ 30000
Mar. 28 “Cash, - - - - - - - 120000|April 25 “ Acpt. 2 60 days, - - - 500|00
May | 3| “ Mdse. a 30 days, - - - 70000||June 7| “ Cash, - - - - - - - 40000
Sept. 14 “ “ (a) 30 days, - - - 150000|||Oct. |19| “ Proceeds of flour due by
equation October 22, 1250|00

(653)
654 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. Yºr
FIRST OPERATION BY THE INTEREST METHOD,
RENCH & WURzLow in Account Current and Interest Account at 8% to December 31, 1895, with
MITCHELL & BRANDNER.
Dr. Cr.
Yºº ps. at Amt Yº..." |Ds.| Int. || Amt.
ji
1895. | 1895.
Feb. 4 To Mdse. a 90 ds. May 5240|| 2667|$50000|Mar. 10By Cash, Mar. 10296, 1973|$30000
Mar. 28 “Cash, Mar. 28278|| 74|13||120000|Apr. 25 “ Acpt.(a) 60.ds.|June 27|187| 2078||50000
May |3| “Mdse. a 30ds.|June 2212|| 3298|| 70000|June 7 “ Cash, ** | 7207 1840|| 40000
Sept]14 “ “ “30 “ Oct. 14 78| 2600||150000|Oct. 19 “ Proceeds of
flour, due by
15978 equation Oct. 22 Oct. 22 70|| 1945||1250.00
Dec. 31. “Bal. of int. 78.36|| 8142
78.36
Dec. 31|By bal. due to
date, 1531|42
398142 398142

Explanation.—In this solution we first find, after maturing the debit and credit items of dol-
lars, the number of days that each debit and credit item of dollars has been due prior to December
31, 1895, and then compute the interest on the same for the time at 8 per cent. We then take the
difference between the debit and credit of interest, and as the excess or balance is against Rench &
Wurzlow, we charge it to them, and then to find the cash balance we take the difference between
the debit and credit amounts of the account.
SECOND OPERATION BY THE PRODUCT METHOD,
RENCH & WURzlow in Account Current and Interest Account at 8% to December 31, 1895, with
MITCHELL & BRANDNER.
I}r. Or.
DR. DR. CR. CR.
|
Wºn DS. Prod'ct || Amt. Yººn LS Prod’ct || Amt.
1895. 1895.
Feb. 4 To Mdse. a 90ds.|May 5240||120000|50000|Mar. |10|By Cash, Mar. 10.296 88800|| 30000
Mar, 28 “Cash, Mar. 28278||333600||120000|Apr. 25 “ Acpt.(a 600s. June 27|187| 93500|50000
May 3 “Mdse. (a) 30 ds. June 2212||148400|| 70000|June 7 “ Cash, ** | 7207|| 82800|| 40000
Sept14 “ “ (a 30 “|Oct. 14 78||117000||150000|Oct. 19 “ Proceeds of
flour, due by
Dec. 31.No. ds. int. on $1, 719000 equation Oct. 22 Oct. 22 70|87500||125000
352600
tºº Dec. 31|No. ds. int. on $1, 352600
“ 31|To Bal. of days |366400 “ 31|By Balance due
interest on $1 to date, 1531|42
which divided
by the 8 per




cent int. div-
isor 4500 gives
$81.42 int. 8142 }
398142 398142
Explanation.—In this solution, we first mature all of the debit, and credit items of dollars,
and then find the number of days that each item of dollars on the debit and credit sides of the
account has been due prior to the day of settlement, December 31. Having these number of days,
we proceed thus: Taking the first transaction on the debit side, we see that the $500 was due May
5, 240 days before December 31, and hence Rench & Wurzlow owe interest on the same for 240 days,
or on $1 for 500 times 240 days, which is 120000 days. In this manner we continue with all of the
* ACCOUNTS CURRENT AND INTEREST ACCOUNTS. 655
debit items, and produce a debit of 719000 days interest on $1 against Rench & Wurzlow. We then
pass to the credit side, and in the first transaction we see that Rench & Wurzlow paid $300, 296
days before December 31, and hence they have interest due them thereon for 296 days, or on $1 for
300 times 296 days, which is 88800 days. In this manner we continue with all of the credit items
and produce a credit of 352600 days on $1 in favor of Rench & Wurzlow. We then take the differ-
ence between the debit and credit of days interest on $1, and find the balance to be 366400 days
interest that Rench and Wurzlow owe on $1. This 366400 we divide by the 8 per cent interest
divisor, and obtain $81.42 interest, which is charged against Rench & Wurzlow, and the account
balanced as in the first solution.
THIRD OPERATION BY THE EQUATION METHOD.
RENCH & WURZLOW in Account Current and Interest Account at 8% to December 31, 1895, with
MITCHELL & BRANDNER.
Dr. Or,
CR. DR. - I).R. - - CR,
DS. |Prod'ct || Amt. Ds. |Prod'ct Amt.
1895. * 1895. -
Feb. |4|To Mdse. a 90 ds. -- 0 90 45000|50000; Mar. |10|By Cash, 34 10200|| 30000
Mar. 28 “Cash, 52 62400||1200.00 Apr. 25 “ Acpt. (a) 60 ds. --80|143| 71500|50000
May 3| “ Mdse. a 30 ds. -- 88||118, 82600|| 70000|June || 7 “ Cash, 123| 49200|| 400|00
Sept 14 “ “ a 30 ds. --222,252.378000||150000; Oct. |19| “ Cash proceeds of
flour due by equa-
No. of ds. Cr. on $1, 568000||390000 tion Oct. 22, 260|325000||1250.00
No. of ds. Dr. on $1, 934400 $3900 -
2450 tºº-º-º: -
Bal. of Dr. ds. on $1, 366400 ** 455900||245000
which – the 8 per Dec. 31|By $1450 × 330 478500
cent interest divisor -
4500 gives balance No. of ds. Dr. on $1. 934400
of interest 8142||Dec. 31|By Balance due to date, 1531|42
- 3981|42 3981|42
|

Explanation.—In this solution, we first assume the earliest date as the maturity of all the
debit and credit items of dollars, and then allow an equivalent credit of days on the purchases,
and charge the corresponding debit of days on the payments, in the same manner as we worked
equation of accounts, when we assumed the earliest date as the maturity of all the dollar debit and
credit items. By this work we obtain a credit of 568000 days on $1, in favor of Rench & Wurzlow,
and a debit of 455900 days on $1 against Rench & Wurzlow. It should be borne in mind that the
debit and credit of days are shown on the sides of the account, reverse to the debit and credit of
dollars; i. e., the debit of days on the credit side of dollars, and vice versa.
These debits and credits of days on $1, represent the contract credit of days on the various
transactions, including the debit and credit of days obtained by assuming all the dollar items due
on the earliest date, February 4, and hence, in effect, justly make both the debit and credit of all
the dollar items due on the earliest date, February 4; and as we settle or close the account Decem-
Ber 31, it is clear that Rench & Wurzlow owe interest on $3900, from February 4, to December 31,
330 days, which is equal to 1287000 days on $1; and that they have due to them, by Mitchell &
Brandner, interest on the money paid, $2450, from February 4, to December 31, 330 days, which is
equal to 808500 days on $1, and when deducted from the 1287000 days that they owe, leaves a bal-
ance of 478500 days that they still owe on $1. º
It is obvious that instead of taking the difference between the debit and credit of days, we
could have added the respective numbers to the debit and credit of days first found. By this elu-
cidation, we see clearly how the balance of days is obtained. * ©
But in practice, and in the operation of this account, to save time and labor, instead of mul-
tiplying both the total debit and credit of dollars by the days intervening between the earliest date
and the date of settlement, we first find the difference or balance between the debit and credit of
dollars, and then multiply the same by the intervening time, and thus produce the correct balance
of debit or credit of days. In this account the difference is, $3900–$2450 = $1450, which multi-
plied by 330 days, the intervening time, gives the 478500 days debit on $1, which we add to the debit
of days obtained by the first work, and thus obtain a total debit of 934400 days interest on $1;
Having now the total days interest on $1 that Rench & Wurzlow owe to Mitchell & Brandner, and
that Mitchell & Brandner owe to Rench & Wurzlow, we take the difference and obtain a balance
of 366400 days interest on $1, that Rench & Wurzlow owe, which, we divide by the 8% interest
divisor, 4500, and obtain $81.42. This interest we charge to Rench & Wurzlow, and balance the
account as in the preceding solutions.
656 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
THE EQUATION METHOD
presents advantages which no other method can. It is shorter than most other
methods, and identical with equation of payments when the earliest date is assumed
as the maturity date of the dollar items.
By the Interest and Product Methods, the interest or product on any item of
dollars cannot be found until the day that the account is settled or closed, or the
date for closing determined, and by reason of this, accountants are pressed with
Work at stated periods of rendering accounts or of closing books, even when “mak-
ing out” individual accounts. But by the Equation Method the products may be
Computed on all of the debit and credit items of dollars at the pleasure of the book-
keeper, and when the account is settled or closed, one subtraction, one multiplica-
tion, and one division operation is performed, and the balance of interest is obtained.
By this method, should it be desired to equate the account after the interest
products are found, one subtraction and one division operation accomplishes the
object as shown by the following operation:
No. of days Cr. on $1 is 568000 Debit, $3900
No. of days Dr. on $1 is 455900 Credit, 2450
Credit balance, 112100 Debit balance, $1450
112100 - 1450 = 77; days, which added to Feb. 4, gives April 22, the equated date. Inter-
est on $1450 from April 22, to Dec. 31, = 253 days, is $81.52. This gives 10 cents excess of interest
over the former methods and results, from the fact that we did not use the ºth of a day in the
equated date, and hence the 253 days used in computing the interest were too many by 38th of a day.
2. What is the cash balance of the following account, September 1, 1895,
interest at 8 per cent * Ans. $14075.37.
I)r. ROBERSON & BRICE. -- Cr.
1895. 1895.
Mar. [21]To Mdse., - - - - 150000 Jan. 24|By Cash, - & tº º 500000
May 16 “ Cash, - - - - 210000 || Mar. 8 “‘ ‘‘ * - - - 750000
June || 4 || “. “ - se - - 50000 || Apr. 28 “ “ - - - - 1200|00
July 3| “ Mdse. (2 4 months - 3200|00 June | 1 | * * * * - º ºs º- 1400|00
“ [11] “ Cash, - gº - sº 400|00 July |10| “ Acpt. (a) 90 days, - - 8000|00
Aug. 20 “ Mdse. (a) 2 months, º 1800.00

FIRST OPERATION BY THE INTEREST METHOD.
ROBERSON & BRICE in Account Current and Interest Account at 8% to September 1, 1895, with
SIMION & EHREN.
IDr. Cr.
|
Yº..." |Ds.|| Int. || Amt. Y." |Ds. Int. || Amt.
|
1895. | |, |1895.
Mar. 21 To Mdse., Mar. 21164|| 54.67|150000|Jan. 24. By Cash, Jan. 24220,24444|| 500000
May 16 “ Cash, May 16108||5040 210000|Mar. || 8 || “. Mar. 8|17729500|| 750000
.June 4. ‘‘ ‘‘ June 4, 89| 989|| 50000|Apr. 28 “ “ Apr. 28.126 3360|| 1200.00
July 3, “ Mdse., a 4 mos. Nov. 3 63| 4480|| 320000|June 1| “ . “ June 1| 92. 28.62|| 140000
‘‘ [11] “ Cash, July 11, 52|| 462|| 40000|July 10 “ Acpt. (a) 90 Oct. 11 40 7111| 800000
Aug.20 “ Mdse., a 2 mos. Oct. 20 49| 1960|| 180000 days,
Dr. of red
interest $64.40 | Total credit
Cr. of red of interest, 601|66
interest 71.11 ! Total debit of
** Bal. Of | ! interest, 126.29
int. in red ——" 671 By bal. of cre-
| | dit of int. 4.7537
|12629
Sep. 1|To Balance due this
date, | 140753
| 235753 e 2357537

4% ACCOUNTS CURRENT AND INTEREST ACCOUNTS. 657
Explanation.—In this account, we have what is called “red interest,” by which is meant
interest which appears by reason of the method on either the debit or credit side of the account
when injustice it belongs and will be carried to the opposite side. , The transactions of July 3 and
August 20, on the debit side, and of July 10, on the credit side of the account, did not mature
until after the date of settlement, and hence it is clear that the interest on each, from the day of
settlement until the maturity of these transactions, belongs to the reverse side of the account, and
to facilitate the addition of the column without including the same, and also to facilitate transfer-
ring it to the side to which it belongs, it is generally entered in redink. In this account, not being
able to print with red ink, we have indicated the numbers that the accountant would have made
red, by printing them in italic figures.
Instead of thus indicating by red, the interest that appears on one side when it belongs to
the other, some accountants have a debit and credit column for interest on both the debit and credit
sides of the account. *
SECOND OPERATION BY THE PRODUCT METHOD.
ROBERSON & BRICE in Account Current and Interest Account at 8% to September 1, 1895, with
SIMON & EHREN.
I)r. Or.
DR. DR. CR. CR.
Yºn DS. Prod'ct || Amt. When Ds. |Prod'ct. || Amt.
1895. 1895. *
Mar. 21. To Midse., Mar. 21164|246000|| 150000|Jan. 24. By Cash, Jan. 242201.100000|| 500000
May 16 ‘‘ Cash, May 16108||226800|| 210000|Mar. 8 “ “ Mar. 81.77||1327500|| 750000
Junel 4 “ M. Mºunt 4| 89|| 44500|| 50000 #. * #. *:: ; #}}
July 3. “ Se.. (2) U1.D.6) UIIlê
y months, Nov. 3| 63|201600|| 320000|July 10 “ Acpt. (a 90
“ 11| “ Cash, July 11, 52| 20800|| 40000 days, Oct. 11 40|| 330000|| 800000
Aug.20 ** M dse. (a) 2
g months, Oct. 20 49|| 88200 1800.00 No. of ds. Cr.
‘‘ Balance of on $1, 2707500
prod. in red, 30200 No. of ds. Dr.
on $1, 568300
568300
Bal, of ds. Cr.
* on $1, 2139200
which divid-
ed by the 8%
Sep. 1|To Balance due int. ii. 456%
this date, 1407538 gives 475.38
2357538 |2357538

Explanation.—By the Product Method of work, we have “red products” instead of “red inter-
est,” which are treated in every respect the same as “red interest,” and for the same reasons as
given in the solution of this problem By the Interest method. By reason of the fractions of a cent
interest in the operation by the Interest Method, we obtain 1 cent more interest by this Method,
than by the Interest Method.
658 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS.
THIRD OFERATION BY THE EQUATION METHOD,
Roberson & BRICE in Account Current and Interest Account at 8% to September 1, 1895, with
SIMON & EHREN.
I)r. Cr.
CR. IDR. DR. CR.
Ds. |Product. || Amt. DS. |Product. || Amt.
1895. 1895. *=-º-º-º-
Mar. 21|To Mdse., 56 84000|| 150000|Jan. 24|By Cash, 0 500000
May 16 “ Cash, 112. 235200 || 2100|00 || Mar. 8 ‘‘ ‘‘ 43| 322500|| 750000
June || 4 | ** ** 131|| 65500|| 50000|Apr. 28 “ “ 94| 112800|| 1200.00
July 3| “ Mºise. a 4 months June | 1. ‘‘ ‘‘ 128, 179200|| 140000
(123 days), 283 905600|| 320000|July 10 “ Acpt. a 90 ds. (93
“ [11] “ Cash, 168| 67200|| 400|00 plus 167), 2602080000|| 800000
Aug. 20 “ Mdse. (a) 2 months
(61 days), 269| 484200|| 1800|00 26.94500||2310000
Sept. 1|By Bal. of days Cr.
1841700|| 9500|00 on $1, 2139200
Sept. 1|By $13600 mult. by 2992000 which divided by
220, the 8% int. div.
“ 1|No. of ds. Cr. on $1, 4833700 4500 gives 4.7538
4833700
“ | 1|To Bal. due this date, 1407538
- 2357538 2357538

Explanation.—By the Equation Method, we have no red interest or red products. The reason
ing for the work is the same as given in the first problem by the Equation Method, and hence
is omitted.
3. What is the cash balance of the following account, October 24, 1895,
interest at 6 per cent 3 Ans. $1704.35
I)r. A. DOERR. Or.
1895. º -
Jan. 1|To Mdse. (3) 60 days, - - $50000
Mar. 10 “ “ (no credit), - - 80000
Apr. 1| “ Cash, * º º- - 15000
Sept. 20 “ Mdse. (a) 60 days, - - 200|00

OPERATION BY THE EQUATION METHOD,
Dr. A. DoERR in Account Current and Int. Acct. at 6% to Oct. 24, 1895, with A. Rivier. Or.
CR. IDR. T)R. CR.
Ds. Prod'ct || Amt. DS. Prod'ct || Amt.
1895. º- 1895, smm-
Jan. | 1 |To Mdse. (a) 60 ds, 60 30000|50000|Oct. 24|By $1650 × 296, No. of
Mar. 10 “ “ (no credit), 68 54400|| 80000 ds. Cr. on $1, 488400
Apr. 1| “ Cash, 90 13500|| 15000 “ 24|By Bal. due this date, 1704|35
Sept. 20 “ Mdse. (a) 60 ds. 322 64400|| 20000
Oct. 24|No. of ds. Cr. on $1, 162300||1650|00
“ 24|To bal. (326100) of ds.
Dr. on $1, which—the
6% int. divisor gives 54|35
1704|35 1704|35

* ACCOUNTS CURRENT AND INTEREST ACCOUNTS. 659
4. What is the cash balance of the following account, December 31, 1895,
interest at 10 per cent? Ans. $5549.72.
I)r. C. ENDERLIN. Or.
- T1895.
Oct. 8|By Cash, - - - - $300000
Nov. 6| “ Dft, a 30 days, - - 50000
Dec. |10| “ Acpt. (3) 60 days, - 2000|00

OPERATION BY THE PRODUCT METHOD,
C. ENDERLIN in Account Current and Int. Acct. at 10% to Dec. 31, 1895, with HAGAN & HENDERSON.
Dr. Or.
DR. DR. CR. CR.
Yº..." |Ds. Prodetſ Amt. Y." |Ds. * Amt.
1895. 1895.
Dec. 31|To Bal. due this Oct. 8By Cash, Oct. 8 84|252000|300000
date, 554972 Nov. 6 ‘‘ Dft. (a) 30 ds.|Dec. 9 11000 º
1896.
pe 10 “ Acpt.a 60ds.|Feb. 11 42|| 84000|200000
No, of ds. Cr. on
$1. 263000
No. of ds. Dr. on
$1 (red), 84000
Bal. of ds. Cr.
on $1, 179000
which div, by
the 10% int.
div. 3600 gives 49.72
5549 H 5549.72


AN ACCOUNT WITH SPECIAL CONDITIONS OF DISCOUNT FOR CASH
PAYMENTS FROM WEHICH AN ACCOUNT CURRENT AND INTER-
EST ACCOUNT AND AN EQUATION OF ACCTS. ARE MADE.
1208. J. McDonald bought of John Gunn, on the following dates and con-
ditions:
Jan. 3, 1895. $237.65 on 90 days, or 3% discount for cash in 5 days.
“ 17 { { 122.50 On 60 :* 3% { { ( & ( & {{
“ 24, “ 324.00 on 66 “ 3% “ “ “ .
“ 31, “ isoo) on 30 “ 3% “ “ “ .
Feb. 10, “ 200.00 on 130 “ 1% “ “ “ “
& 4 15, & 4 100.00 On 120 & 4 3% & 4 ( & & 4 & 4
“ 24, “ 75.85 met { {
J. McDonald paid cash on account of said purchases as follows:
January 10, 1895, $150.00. January 21, 1895, $100.00 April 5, 1895, $300.00.
REQUIRED: 1. What is the balance due July 1, 1895, at 6 per cent interest?
66O soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. º:
2. If equated, what would be the equated date for paying the balance of the
account 3
REMARKS. It will be observed that there was a conditional discount privi-
lege on each bill purchased. But since the purchaser did not avail himself of said
privilege by paying cash within the specified time, the privilege became inoperative
and cannot be considered in the solution of the problem.
Special or adjunct conditions and terms of contracts, are not binding upon
either party thereto, unless strictly conformed to by both parties.
OPERATION BY THE PRODUCT SYSTEM TO FIND BALANCE DUE JULY 1, 1895.
J. McDONALD in Account Current and Interest Account with JoHN GUNN, interest at 6% to
July 1, 1895.
I}r. Or.
DR. DR. CR. CR.
Y.* Ds.|Prod'etl Amt. Whº Ds. |Prod'ot| Amt.
1895, º 1895. *
Jan. 390 ds, or 3% for Jan. 10|Cash, - - 172|| 25800|| 15000
cash in 5 ds. ||Apr. 3 89| 21182||23765 “ 21 “ * > tº 161|| 16100|| 10000
“ 1760 ds. or 3% for Apr. 5 “ tº use 87| 26100|| 30000
cash in 5 ds. |Mar. 18105||12915|| 12250
“ 2460 ds. or 2% for No. ds. Cr. on $1, 68000
cash in 5 ds. “ 25 98|| 31752|| 32400|July |1|By Balance due, 66524
“ .3130 ds, or 2% for
cash in 5 ds. ** 2121|| 18150|| 15000
Feb. 10120 ds. or 4% for
cash in 5 ds. ||June 10 21|| 4200|| 20000
“ 15120 ds. or 3% for
cash in 5 ds. ** 15, 16|| 1600|| 10000
“ 24|Net, Feb. 24|127| 9652|| 7585
No. ds. Dr. on $1, 99.451
68000
To Bal. ds. Dr.
on $1, 31451
which div. by
6% int. div.
6000 gives 524
121524 121524
OPERATION TO EQUATE THE ACCOUNT.
1895. Ds. Cr. Ds. Cr. on $1. 1895. Ds. Dr. Ds. Dr. on $1
Jan. 3, $237.65 90 - - 21420 Jan. 10, $150.00 7 - - - - - - 1050
17, 122.50 60 + 14 = 74. - - 9102 21, 100.00 18 - - - - - - 1800
24, 324.00 60 + 21 = 81 . - 26244 Apr. 5, 300.00 92 - - - - - - 27600
* 31, 150.00 30 + 28 = 58 - - 8700 &=ºmºmºssº
Feb. 10, 200.00 120 + 38 = 158 - - 31600 $550.00 30450
15, 100.00 120 + 43 = 163 - - 16300
24, 75.85 net 52 - 3952
$1210.00 117318 *...,
550.00 30450
$660.00 86868 – 660 = 131# ds. practically, 132 ds., after Jan, 3,




1895, - May 15, 1895, equated date.
NOTE.-Interest on $660, May 15, to July 1, 47 days, = $5.17. Interest shown in the Account
‘. i. Interest Account is $5.24. The difference results from the fraction of a day lost in the
equated date.
ACCOUNTS CURRENT AND INTEREST ACCOUNTS
66 I
of a convenient width to copy in the account current book.
In practice, instead of arranging the credit of the account to the right of the
debit side, it is frequently placed below it, in order that it may be made on paper
The following account current illustrates this form :
f
E. RICKER in Account Current and Int. Acct. at 8% to October 1, 1895, with T. H. BRODE & Co.
I) EBITS.
1895.
April
June
Oct.
1895.
July
Sept.
Oct.
25
10
17
1
To Mdse. (a) 30 days, º
“ Cash, tº- 4-
“ Mdse. (3) 60 days, s
“ Red interest from credit,
Total debit of interest,
“ credit of interest,
To Balance of debit interest,
1
CREDITS.
By Cash, - - -
“ Acpt. 2' 30 days, tº
By Balance due this date,
When due. Ds. Int. Amt.
April 24, 160|| 6400||180000
** |10| 174|| 46|40||1200.00
Aug. 16 46|| 2044||200000
467
13551
30|67
10484
5104|84
When due. Ds. Int. Amt.
July 1| 92|| 30|67||150000
Oct. 8 7. 4|67|300000
30|67
60484
5104|84
ACCOUNTS CURRENT AND INTEREST ACCOUNTS IN ENGLISH MONEY.
*209. 5. What is the cash balance of the following account, July 1, 1895,
intelest at 6 per eent?
Ans. £462 5s.14d.
I)r. R. G. MUSGROVE, JR. Cr.
3' |8. d. 36 s.ld.
1895. T |T|T | 1895. * I - I -m-
Jan. 15|To Cash, - tº- * 500 8 6 Feb. 10|By Cash, <- ſº º 1100|00|| 8
April 8 “ “ º tº sº 1000|14|| 3 || Mar. 5 ‘‘ ‘‘ sº em - 800|10| 4.
June 18 “Ex. 2 60 days, tº 100000| 0 |May 21 “ “ tº gº - 1050" 6||11

662 soule's PHILOSOPHIC PRACTICAL MATHEMATICS. †
OPERATION BY THE INTEREST METHOD.
R. G. MUSGROve, J.R., in Account Current and Int. Acct. at 6% to July 1, 1895, with A. CARON.
JDr. Cº.
* DS. Int. Amount. Yººn * Int. Amount.
1895. £ s. d. 11895. f º d.
Jan. 15To Cash, Jan. 15167||131864|| 500 8 6 ||Feb. 10|By Cash, Feb. 10,141||251700 ||110000| 8
Apr. 8 “ “ Apr. 8 84|14 02:100014| 3 |Mar. || 5 || “. Mar. 5118||15|1410 || 800|10| 4
June 18 “ Fx. at May 21 “ “ May 21 41||7|3| 64|1050, 611
60 ds. Aug. 20 50|| 8 6 8||100000 0
July |1|Am't of Cr.
27|18, 9 int. 4815 4+
& 4 Red Int. 8| 6 || 8
Am’t of Int. 57 2 #
27|18, 9
July 1|To Bal. due “ 1|Bal. of Cr.
this date, 47818, 5} int. 29, 33}
2980 1'24 2980. 12;



NOTE.—See page 507 for a similar account,
Dec. 1, 1895, A. and B. agreed to enter into a joint speculation, each # in the
purchase of 200 shares of factory stock. A. is to advance the margins and to settle
with the broker for the purchase of the stock, and to render an account to B.
Accordingly, Dec. 5, A. advanced to his broker, H. Lee, $2000, and Dec. 20
he advanced $1000 more.
Dec. 5, the broker purchased 200 shares of factory stock at 933, $18750. Com-
mission for purchasing at $ per cent, $25.
Dec. 22, the broker collected a 14 per cent dividend $300, and March 23, 1894,
he collected a second 14 per cent dividend $300.
May 6, the broker rendered his account to A. which A. settled on the 8th of
May, and the stock was delivered to him. June 26, A. received a $300 dividend on
the stock.
June 30, 1896, A. rendered his account to B.
Required: 1. The broker's account showing the balance due him May 6th
and 8th. 2. The account of A. showing balance due him by B. June 30, 1894.
SALES ON JOINT ACCOUNT.
663
OPERATION.
Two Hundred Shares Stock at 6 per cent per annum to May 6th, 1895.



Date. Amt. |Ds
1894. -
Dec. 5|To Cash, 200 shares
@ 93 18775|00||152
“ 5 “ Com. @ 4%
Less Cr. Int.
Bal. Dr. Int. 393|18
- 1916818
1895. [T *me I sº-
May | 6|To Balance due, 15568|18
Int, on same for
2 ds. (a) 6%, 5||19
May | 8 Amount due, 15573|37
Int. Date. Amt. |Ds.
1894. -
Dec. 5|Cash, 200000||152
475|63 || “. [20] ‘ ‘ 100000||137
“ 22|Dividend 13%, 30000||135
82|45 1895.
Mar. 123 & & 13%, 30000| 44
Total Cr. Int.
360000
May | 6|By Balance due, 15568|Z8
1916818
Int.
5067
2283
6|75
82|45
ACCOUNT OF A. WITH A. AND B.
A. and B. each # in account with A. from Dec. 5, 1894, to June 30, 1895, at 6 per cent
per annum.
Date. Amt. Ds. Int. Date. Amt. Ds. Int.
1894. 1895,
Dec. 5|To Cash—margin to June 26|By Cash, Div. 13%, 300|00| 4 20
broker, 200000|207|| 6900 “ |30| “ Balance, 18511||73
** 20 To Cash—margin to
broker, 100000||192|| 3200
1895.
May | 8|To Cash—paid broker
• as per his account||15573|37 || 53||137|56
Total Dr. Int. 238|56
Less Cr. Int. 20
Bal. Dr. Int. || 238|36 -
- 1881173 1881173
June 30 To Balance, 185I1||73 *-
REMARKS. -
A's # - - $9255.86
B's # (B. owes A.
June 30) º 9255.87
$13511.73


When B. pays A. $9.255.87
then A. & B. will each
own # of 200 shares of
stock as per account with
broker.
NotE.—In this account we change the locations of the columns for amount, days and
interest, and omit the “when due” column.
This is in accordance with the form in general use by Stock Brokers.
See “Stocks and Bonds” in this work for further accounts with brokers.
Geo. Soulé & Sons, of New Orleans La., have paid the following drafts, drawn on them by J. L. Pvorman &
Co., of Bremen, Germany, a correspondent house:
February 15, draft dated January 15th, for one month without grace for 1000 marks,
exchange 954. February
28,-draft dated January 30th, for one month without grace for 1500 marks, exchange 93%. March 14, draft dated
February 14th, for one month without grace, for 2000 marks, exchange 96%.
Geo. Soulé & Sons have drawn on their Bremen correspondent, the following drafts:
marks, one month without grace, exchange 95.
What is the balance of marks and dollars due to Geo. Soulé & Sons, June 30, 1895, interest at 6% 7
Ans. 2233.24 marks. $534,38.
February 1, draft for 1052.63 marks, one month without grace, exchange 95. March 2, draft for 1263.16
OPERATION
J. L. Poorman & Co. in Account Current and Interest Account at 6 per cent to June 30, 1895, with Geo. Soulé dº Sons.
Dr. Or.
1895. Yººn DS Interest. Amount. 1895. Wºm DS Interest. Amount,
M. $. Marks. || Dollars. M. $. Marks. ||Dollars.
Feb. 15To Cash, dft. Jan. Feb. 1|By Cash, dft. 1 mo.
15, ex 95+, Feb. 15||135|| 2250|| 536||100000|| 238||13 (no grace) ex. 95|Mar. 1||121| 2123| 504|105263|| 25000
“ 28 To Cash, dft. Jan. Mar. 2|By Cash, dft. 1 mo.
30, ex. 93%, “ 28||122|| 30|50|| 7.16||150000|| 35203 (no grace) ex. 95||Apr. 2|| 89|| 1874|| 4|45||1263.16|| 30000
Mar. 14|To Cash, dft. Feb.
14, ex. 96%, Mar. 14||108|| 3600|| 869|200000|| 48250 3997|| 949
June 30|By Balance, 2333,24|| 53.433
89.00|| 21.21 -* | *-
3997 949 4903|| 11|72
4549.03||1084.38 --> 4549,03||1084|38
1895. sm- I* * | * | * -* -ºs- F
June 30|To Balance, * 53438
NotE.—See Index, “German Exchange” for the reduction of marks to dollars.
REMARKs.—See Stocks and Bonds in this work for Accounts Current and Interest Accounts with Brokers.
3.




33alkers' Interest I. Daily Balances.
gº arriagra=== *N-
— — ——se
1210. In the foregoing Interest Accounts Current, we have illustrated the
different methods of finding the cash balance of accounts at stated periods, or at
Such irregular times as a settlement may be desired.
It is a general custom with most bankers to balance accounts with their
correspondents every month, or every 3, 6, or 12 months, allowing interest on all
credits, and charging interest on all debits, and bringing the balance of interest to
the debit or credit side of the account, as the case may be. The balance of interest
thus entered and brought down to the new account, will subsequently draw interest
the same as any other item in the account. The effect of thus closing and bringing
down the interest balances is the compounding of money as often as the accounts
are thus closed, as we fully explained in the Merchants' System of Partial Pay-
ments.
But in addition to the accounts current and interest accounts with their
correspondents, some bankers allow interest to their depositors on the daily bal-
ances of their respective accounts, and to elucidate the method of computing
interest on such balances, as well as the interest on the daily balances of accounts
with correspondents, we present the following
IPROBLEMS :
1. What is the balance due February 1, 1895, on the following account, at 5
per cent interest on daily balances, counting 365 days to the year?
ALBERT LEE & EDWARD E. SOULE.
LMr. Interest at 5%. Or.
1895. 1895.
Jan. 1|To Checks, sº es 30000|Jan. 1|By Cash, tº gº 1200.00
é & 2 & 4 & 4 º tº º 900 00 & 8 2 & & & 4 º sº 80000
& & 3| “. 4 & tº ºt 400|00 || “. 4| < * * tº ſº 102000
& 4 6|| “. { { tº tº 1100|00 || “. 5| “ tº = 75000
& 4 7 & 4 é & tºº s 800 00 & 4 6 { % 6 & tº tº- 50000
€ $ 9 & 4 4 & g- sº 500 00 & 4 8 & 4 ( & sº ſº 90000
** 14 “ & 4 gº tº 60000 || 44 9| 4: “ tº Ee 20000
** 16| 44 & 4 gº tº 35000 || 4 |11| < * * - - 100000
** [21] & 4 & 4 ſº tº 200000 || “ 15| < * * * ſº º 40000
“ 25 tº & 4 tº & 10000 || 4 |17| < * * tº sº. 65000
<< |28 “ ( & tº dº 21000 || 4 |18 | < ** * = 15000
“ (30) “ & 4 a & 75|00 || 4 |22 4° 44 tº sº 34000
Feb. 1| “ Balance, - - 290%|73 || “ 24, 44 44 tº sº 22000
“ 27| “ “ tº ſº. 210000
Feb. 1|Interest to date, - 773
10237.73 1023773


(665)
666
SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS.
FIRST OPERATION BY THE USUAL METHOD,
1211. The above account presents the figures as they stand on the Individual
Ledger of the bank, and from this ledger or the pass book the daily balances
are taken and arranged on a separate sheet of paper, in a statement form as fol-
lows:
EQUITY NATIONAL BANK, N. O., in account with A. L. & E. E. SOULF, interest at
I)R.
l 90000
2 80000
3 40000
4. 142000
5 217000
6 157000
7 77000
8 167000
9 137000
10 137000
11 237000
12 237000
13 2370:00
14 177000
15 217000
16 182000
17 2470 ſº
18 26.2000
19 262000
20 2.20%
21 62000
22 96000
23 96000
24 118090
25 108000
26 108000
27 3180.00
28 2970.00
29 2970:00
30 2895|00
31 2895.00
Total daily bal. || 5643000
5 per cent to February 1, 1895.
Explanation.—In finding the daily balance, we
observe that, on the 1st of January, A. L. & E. E.
Soulé deposited - - * $1200
and drew out - gº tº {º tº 4- ſº º 300
leaving a balance of - , ,- ū-º º * =e $ 900
The second day they deposited - *º- *- tº 800
which added to the balance of the first day gives $1700
of this amount they drew out {E_* ſº 900
leaving a balance of - - tº º º º $800
On the 3d, they made no deposit but drew out - 400
leaving on deposit s & * -> tº * = $400
On the 4th, they deposited º * ºn sº * = 1020
this producing a balance of Eº gº - - $1420
In this manner we continue through the
month and find a total daily balance of $56430.00
for 1 day against the bank and in their favor,
which divided by 7300, the 5% interest divisor
for 365 days to the year, gives $7.73 interest due
to the depositors.
Had the rate of interest been 4 per cent,
we would have first found the interest at 5 per
cent, and then reduced it # of itself. Had it been
3 per cent, we would have found the interest at 5 per cent, and then reduced it # of
itself. Had it been 6 per cent, or 7 per cent, we would have found it at 5 per cent,
and then increased it # for 6 per cent and # for 7 per cent.
Or to find the interest at any rate per cent, we multiply the balance of the
daily balances by the rate and divide by 36500 when 365 days are counted a year,
and by 36000 when 360 days are counted a year.
LEDGER ACCOUNTS. 667
SECOND OPERATION.—(LEDGER ACCOUNT).
ALBERT LEE & EDWARD E. SOULE.
Daily Balances.
— — |Ds.|Tot. Daily Bal. || Monthly Interest.
I)?'. Cr. Dr. Cr. |
30000|By Cash Dep. 1200.00 90000|| 1 90000
90000|| “ “ “ 80000 80000: 1 80000||
40000 400|00|| 1 40000
4 & 1 & 4 & 1020.00 142000|| 1 142000
& ( & & 44 75000 2170ſ)0|| 1 217000
110000|| “ “ “ 50000 1570|00|| 1 157000
80000 770|00|| 1 77000
4t tº 4 & 6 90000 1670|00|| 1 167000
50000|| “ “ “ 200|00 1370|00|| 2 27.4000
{ q & 4 & 4 100000 2370|00|| 3 71 1000
60000 1770|00|| 1 177000
* 4& 64 q6 40000 2170|00|| 1 217000
35000 1820 |00|| 1 182000
64 && 64 65000 247000]. 1 247000
4 & 5 & 66 15000 2620 00|| 3 7860.00
2000.00 620 00|| 1 62000
4 & 84 & 6 34000 960|00|| 2 1920.00
4 & 4 & 64 22000 118000|| 1 lº
10000 108000|2 2100.00
64 & 6 4 & 2100.00 318000|| 1 3.18.9%
21000 297.000|2 594000
7500 2895'00|| 2 579000
290273|| “ Interest, 773 |. div. b
I * -º-º- sº. 5643000 73
1023773 10237.73 |
Explanation.—We here present a new form of Individual Ledger in which we keep the
depositors’ accounts, and perform the operation to find the interest on the daily balances. Where
there are a large number of accounts on which interest is allowed on daily balances, this form of
Ledger may be used to advantage.
2. What is the balance of the following account on March 1, 1896, making
monthly settlements and allowing interest on the debit of the daily balances at 8 per
cent, and on the credit of the daily balances at 6 per cent, and counting 365 days
to the year?
Ans.
$17372,18.
NEW YORK CITY NATIONAL BANK.—Interest 8% on debits and 6% on credits.
1896. Dr. Cr. || Daily Balances. Tot. Daily Bal. | Monthly Int.
Jan. wº By Cash, wº sºlo solo
4 tº º 5 || 2250000
04 100000|| “. “ : 350000|| 6 2100000
6& | || || 300000 650000|7 4;º
** 28 750000|| “. “ 1140000 1040000|| 4 41600 à
IFeb. 104.1333| “ Interest, 13:33
*- * I - 00.11110000| 1820
2541338||____ 2541333 '-1} = −.
Feb |By Balance, 104.1333 104.1333|| 4 4165332
* * 400'00|| “ Cash, 160000 1161333|| 3 34839.99
• * 180000|| ". . 2100.00 1191333| 5 59566.65
* , 600000||... . Žº 61.1333| 4 24453.32
{ * 100000|| “. “ 800000 13113|33 || 6 7867998
4 i t & 4 & 380000 16913:33|| 6 10147998
* { 1700'00|| “. “ 210000 17313|33 || 1 1731333
Mar. 1 17372,18|| “ Interest. 58.85
I lºº- I sºme 357.85: 588s
2827218 2827213 | || |-
| | !
Explanation.—In this account we have interest on the debit of daily balances at 8 per cent,
and on the credit of daily balances at 6 per cent, and hence to find the interest on the monthly
footings of the daily balances, we multiply the same by the respective rates per cent, 8 and 6, and
divide by the one per cent, interest divisor 36500.

668 soule's PHILOSOPHIC PRACTICAL MATHEMATICS. º
3. What is the balance of the following account June 1, 1895, allowing
interest on daily balances at 6 per cent, and counting 360 days to the year?
& Ans. $7083.75.
THIRD NATIONAL BANK, BALTIMORE.
Daily Balances. D Tot. Daily Bal.
S.
1895. I}r. o Dr. Cr. D?'. Cr.
May $ $ 4 & 5000 aw
& 4 10000
4 &
6 &
4 & 7500
4 4 2000
1500
& 4 200000|| 5500
57.100
Ea.planation.—In this account we present another style of ruling for the Gene: al Ledger.
Where each debit and credit item is posted separately, this form of ledger may be used to advan-
tage. The interest being the same on both sides of the account, the difference of the total daily
balances is here divided by 6000, the 6 per cent. interest divisor for 360 days to the year.

sºils his misſing Bill Aºus
---------------------------------------------N
*—aº—º-
--sºr—w-ur-
1212. Savings Banks are Financial Institutions for the receipt on deposit of
Small sums of money for which a moderate rate of interest is allowed on stated
Conditions. *
A savings bank furnishes each depositor with a pass book, in which are
recorded all sums deposited and all sums withdrawn. This book contains the
Conditions on which deposits are received and interest is allowed and credited. It
also contains the interest credited to the depositor at the end of each interest term.
Such interest is entered in red ink.
... NOTE.-It is not the custom of savings banks to calculate interest on a fraction of a dollar,
but in finding the amount at the end of each interest term, fractions of dollars are included.
1213. The Deposits in Savings Banks are usually paid on demand, though
the charter of the bank and the conditions of deposit give the institution the right
to require 60 or 90 days notice.
1214. The Interest Term is the period of time at the close of which interest
is computed and credited as a deposit. It is thus compounded, if not withdrawn.
The Interest Term varies with different savings banks; with some it is 1
month, with others 3 and with others 6. Interest is usually declared and credited
June 30 and December 31. In some savings banks, deposits commence to draw
interest on the first of each quarter, January 1, April 1, July 1, and Oct. 1. In
others the money must be on deposit at least three months before it will be entitled
to receive any interest.
Some of the Savings Banks of New Orleans give interest on net deposits that
have been in the bank 3 months; others require deposits to have been in bank 4
months before giving interest thereon.
PROBLEMIS.
1. Allowing interest at 3 per cent per annum, and declaring interest semi-
annually, what was due on the following account, January 1, 1896?
NOTE.-Deposits to be in bank 3 months, before they bear interest.
Dr. Soulé College Savings Bank in account with C. McGuigin. Cr.
1895. .* 1895,
Jan, 1|To Balance, - - tº 421165 Feb. 10 By Check, wº tº º 90|30
“ 22 “ Cash, tº e * sº 5000 June 16. “ & 4 {-º tº º 651:5
Mar. [16] ‘‘ ‘‘ tº ſº tº 115|50 Nov. 1| ** &&. ſº tº 3880
May 10| “ “ tº º ºs 80.25
Aug. 14 “ “ tº tº tº 10000
Oct. I 11 tº “ & Eº º 7500

(669)
67o soul E's PHILOSOPHIC PRACTICAL MATHEMATICs. º:
FIRST OPERATION.
NOTE.-In this operation, we have counted actual time and given the depositor credit of 1c.
for every fraction of #c. or over every alternate time.



Date. Deposits. Checks. | Int. on. || From *. º Interest. | Tot. Int.
|
1895.
Jan. 1|To Balance, - - 421165
“ 22 “ Dep. gº - 5000 90.30 381,00|Jan. 1||June 30|| 181 575
Mar. [16] ‘‘ ‘‘ sº- - 11550 11500|Mar. 16|| “ 30|| 107 1100 6.75
May 10 “ “ º - 80.25 65.15 1500|May 10|| “ 30|| 52 O7
June 30 “ Interest, - - 6||75|| 51870
“ Balance, - -
67415|| 674|15
July | 1.To Balance, - -) 51870 518,00|July | 1||Dec. 31|| 184 794
Aug. 14 “ Dep. º - 10000 10000|Aug. 14|| “ 31|| 140 116|| 17
Oct. 1| 4 || “. * > - 7500 3880
Dec. 31 “ Interest, - tº- 9.17
“ Balance, - - 66.407
70287| 70287
1896.
Jan. 1|To Balance, - - 66407

NOTE.-The depositor is credited 1 cent for each 3 cent interest every alternate time.
SECOND OPERATION.
Operation reckoning in months of 30 days and counting odd days to the nearest convenient
fraction of a month.




Date. Deposits. Checks. || Int. on. || From TO Mos. || Interest. || Tot. Int.
|
1895.
Jan º 2. Tº Hºnºe, - -> º, 65 90.30|| 381|00|J 1|J 30|| 6 572
ep. - - à Ll. LIIlê
Mar. 16 ‘‘ ‘‘ º - 11550 11500|Mar. 16 “ 30|| 34 100 672
May 10 “ “ †- - 80.25 65.15 1500|May 10 “ 30|| 13 06
June 30 “ Interest, - -> 6.7%
“ Balance, - 31867
67412 67412
July |1|To Balance, - - 51867 T| 518.00% uly |1|Dec. 31|| 6 777
Aug. 14 “ Dep. º - 10000 10000|Aug. 14 “ 31|| 44 1||13 896
Oct. 1} < * * * tº - 7500 3880 sº-º-º-º:
Dec. 31 “ Interest, - - 896
“ Balance, - 665&
70263| 702 63
1896. IT |
Jan. 1|To Balance, - - 66383
Ea:planation.—Since the balance $421.65 was not in for 3 months before a check was drawn
which would decrease it, it is clear that interest will not be calculated on that amount. The check
of February 10th is to be deducted first from the last deposit preceding February 10th, (deposit of
January 22d, $50), and then the excess of check over this deposit from the balance of January 1st,
1895; $471.65 sum of deposits of January 1st and 22d, less $90 leaves $381, on which we calculate
interest for 6 months at 3% per annum, which is $5,72. Next calculate interest on $115 for 34 months
at 3% per annum, which is $1.00. Deposit of May 10th not being in 3 months, no interest is
aredited on same at June period. Total interest to be credited June 30th, 1895, is $6,72.
NOTE.-Balance the book and bring balance down.
ºr SAVINGS BANKS AND SAVINGS BANK ACCOUNTS, 671
To calculate interest to be credited December 31st, 1895,
Calculate interest on balance $518 for 6 months at 3% per annum = $7.77 and add to this sum
the interest on $15 (the excess of deposit of May 10th over check of June 16th), for 13 months = 6
“ents. The same result would be obtained by calculating interest on the balance $518.67 minus the
$15 excess of deposit of May 10th over check of June 16th, for 6 months = $7.55, and adding to this
Sum the interest on the $15 for a period of 73 months = 28 cents. Worked by either method the
interest amounts to $7.83. Then calculate interest on $100 for 44 months at 3% per annum, = $1.13.
Total amount interest December 31st, 1895, - $8.96. Balance of account J anuary 1st, 1896, = $663,83.
To find interest when book is balanced but once a year, say on January 1st.
First find interest on $381 (difference between the sum of the deposits of January 1st and
22d, and the check of February 10th) for 6 months = $5.72. Next get interest on $115 for 34 months
*=$1.00. Total amount of interest to be credited June 30th, 1895, = $6.72.
To find interest due on December 31st, 1895.
First calculate interest on $381 for 6 months = $5.72; next calculate interest on $115 for 6
months = $1.72; next calculate interest on $15 (difference between deposit of May 10th and check
of June 16), for 7# months = 29 cents; next calculate interest on $6 for 6 months = 9 cents; next
ealculate interest on $100 for 43 months = $1.12. Total interest to be credited on December 31st,
1895, = $8.94. Balance of account January 1st, 1896, = $663.87.
NOTE 1.-On general principles, interest would not be allowed on balance of deposit of
October 1st, ($75–38.80). According to circumstances, interest would or would not be allowed.
NoTE 2.-The second operation above is the method generally adopted by Savings Banks.
It is labor saving and on the general run of savings accounts, there is very little difference in the
final result.
NOTE 3.-Fractional parts of a dollar are not considered in calculating interest.
NotE 4. In subtracting checks from deposits there is no settled custom. Some banks sub-
tract checks from 1st deposit, other banks subtract checks from the deposit immediately preceding
date of check and again, other banks count back 3 or 4 months (according to the time the deposits
Imust be on hand before drawing interest) and subtract the check from the deposit nearest to that
date.
2. How much was due on the following account January 1, 1895, interest at
3 per cent per annum, declared semi-annually, and allowed after 3 months?
Ans, $707.37.
Dr. Soulé College Savings Bank in account with Katie Childress. Cr.
1894. 1895.
Dec. 3|To Cash, tº- tº * 20000 Jan. 18|By Check, - - * 5000
1895. May 4 “ “ * - * 7000
Jan. [15] ** ** - * - º 11080
Feb. 20) ** ** 4- tº- º 50|40
Apr. 28 “ “ tº- i- e 150|00
June |10| ** ** * * * 300|00
672 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS.
OPERATION.
IXate. Deposits. Checks. || Int. on. || From TO Mos. || Interest. || Tot. Int.
1894. - |
Dec. 3To Deposit, - - $20000 $20000||Dec. 3|June 30|| 61%; 345
1895.
Jan. [15] ** & 4 $º tºº 11080 5000 6000|Jan. 15 “ 30|| 5% 83
Feb. 20. “ & & tº tº 5040 5000||Feb. 20 “ 30|| 44 54 482
Apr. 28. “ & & * --> 15000 7000
June 10 ** & & as ſº 30000
“ 30 “ Interest, - - 4.8%
“ Balance, - - 696 o:
s31602 s1802
July 1|To Balance, - - $696.02 31600|July | 1 Dec. 31|| 6 474
Apr. 28 “ Deposit, - - 8000|Apr. 28 “ 31|| 8ºr 161 tº
June 10 ** é & dºe - 30000|June 10 “ 31|| 6; 500 1135
Dec. 31 “ Interest, - º 1135
‘‘ Balance, - - 70737
$707|37 || $70737
1896. Jºl
Jan. 1|To Balance, - sº




NotE.—This operation for calculating interest due on Dec. 31, 1895, applies to 2d part of
following explanation for finding interest due on December 31, 1895.
In this account no interest would be credited January 1, 1895.
To find interest due on June 30, 1895.
Find interest on $200 for 61%, months at 3 per cent per annum = $3.45. Then calculate inter-
est on excess of deposit of Jauuary 15, over check of January 18, - $60 for 54 months at 3 per cent
per annum = 83 cents. Then calculate interest on $50 deposit of Feb. 20, for 4 months at 3 per
cent per annum = 54 cents. Total interest to be credited on June 30, is $4.82. Balance of account
on July 1, 1895, is $696.02.
To find interest due Dec. 31, 1895.
1. Calculate interest on balance due for 6 months = $10.44 and add to it the interest on
excess of deposit of April 28, over check of May 4, $80 for 2+ months = 41 cents, also add interest
on $300 for 3 months = 50 cents = total interest $11,35.
2. The same result would be obtained by getting interest on $696 — $380 = $316 for 6 months
= $4.74 and adding to this interest the interest on $80 for 8 ºr months, = $1.61 plus the interest
on $300 for 63 months = 5.00 = total interest credited Dec. 31, 1895, – $11.35. The reason that
we subtract the $380 and not $450 from $696, is because the check of May 4, for $70 cancels that
much of the $150 deposit of April 28. Balance due Jan. 1, 1896, $707.37.
SECOND OPERATION.
Where the book is balanced only on Jan. 1, of each year.
Since the deposit of Dec. 3, was not in for three months previous to the crediting up of inter-
est, none will be credited on the account on Dec. 31, 1894.
To find interest due June 30, 1895, proceed as follows: First calculate interest on $200 for
61%; months = $3.45. Next calculate interest on $60 (the difference between the deposit of the 15th
January and the check of the 18th), for 5% months = 83 cents. Next calculate interest on $50 for
4+ months = 54 cents. Total interest to be credited June 30, 1895, − $482.
To find interest to be credited Dec. 31, 1895, first calculate interest on $200 for 6 months =
$3.00; next calculate interest on $60, difference between check of Jan. 18, and deposit of Jan. 15,
for 6 months = 90 cents; next calculate interest on $50 for 6 months = 75 cents; next calculate
interest on $80, (difference between deposit of April 28, and check of May 4), for 81's months =
$1.61; next calculate interest on $300 for 63 months = $5.00; next calculate interest on $4 for 6
months = 6 cents. Total amount of interest to be credited Dec. 31, 1895, = $11.32.
The difference, 3 cents between the two statements, arises from the fact that the odd cents
which were considered when the account is balanced on July 1, 1895, amounts to $2 and the inter-
est on the same for 6 months is 3 cents.
* SAVINGS BANKS AND SAVINGS BANK Accounts, 673
THIRD OPERATION.
Operation when the checks drawn are cancelled into the first deposit regardless of the fact that other deposits
have been made between the time of the first one and the first check drawn.
To calculate the interest due June 30, 1895, first calculate the interest due on $150 (difference
between first deposit and first check) from Dec. 3, to May 4, − 5 months = $1.88; next calculate
interest on $80, (difference between balance of $150 and check of May 4), to July 1, = 14% months
= 37 cents; next calculate interest on $110 for 54 months = $1.51; next calculate interest on $50 for
4% months = 54 cents. Total amount of interest to be credited on June 30, 1895, - $4,30.
To find interest to be credited Dec. 31, 1895, without book being balanced on July 1, 1895.
First calculate interest on $4 for 6 months = 6 cents; next calculate interest on $80 (differ-
ence between deposit of Dec. 3, and sum of checks of Jan. 18 and May 4), for 6 months = $1.20
next calculate interest on $110 for 6 months = $1.65; next calculate interest on $50 for 6 months =
75 cents; next calculate interest on $150 for 8 ºr months = $3.02; next calculate interest on $300
for 63 months = $5.00. Total amount of interest to be credited Dec. 31, 1895, - $11.68. Balance
due Jan. 1, 1896, = $707.18.
FOURTH OPERATION.
To find interest due Dec. 31, 1895, if book had been balanced on July 1, 1895. Balance at
that date by above third method would have been $695.50.
Interest on balance for 6 months = $10.43. Interest on $150 for 2 ºr months = 77c. Interest
on $300 for # months = 50 cents. Interest to be credited Dec. 31, 1895, - $11,70. Balance due
Jan. 1, 1896, = $707.20.
The difference between the two amounts due is the result of the odd cents adding $1.50 and
drawing interest for 6 months = 2 cents. -
3. Allowing interest at 3 per cent per annum and declaring interest semi-
annually what was due on the following account, Jan. 1, 1896.
NotE.—Deposits to be in bank 3 months before they bear interest.
Dr. Soulé College Savings Bank in account with F. E. Gillespie, Cr.
1895.
Mar. 16|To Cash, - - - - 500|00
Apr. 10 “ “ tº º cº- - 20000
June || 3 ** ** sº ſº G- º 150|00
Sept. [15] “ “ - º tº- Q- 240|00
OPERATION.
Date. Deposits. Checks. || Int, on. From TO Mos Interest. || Tot. Int.
1895. | —”
Mar. 16 To Deposit, - - $50000 $50000|Mar. 16 June 30|| 34 4|39 439
Apr. 10 “ & 4 - º 20000
June 3. “ 4 & tº tº 15000
“ 30 “ Interest, - tº 439
$85439
July | 1ſo Amt. brought down, $85439 §4%|July ||Dºº. #| || 7|56
Apr. 10 “ Deposit, - - 200% Apr. 10 “ 31|| 84 433
June || 3: “ 4 & tº gºe 15000|June 3 ** 31|| 6 ºr 259
Sep. 15. “ £6 tº tº 24000 24000|Sep. 15 “ 31 210
Dec. 31 “ Interest, - - 1658
- 1658
$1110 97
1896. *-
Jan. 1íTo Amt. brought down, sinº



674 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs.
*
EXPLANATION.
To find the interest to be credited June 30, 1895.
First calculate interest on $500 for 33 months at 3 per cent per annum = $4.39. The deposits
of April 10 and June 3, do not draw interest since they were not on deposit for 3 months. Interest
to be credited on June 30, = $4 39.
To find interest to be credited Dec. 31, 1895.
First calculate interest on total amount deposited and interest = $854.38 for 6 months at 3
per cent per annum = $12.81. To this amount add interest on $200 for 2% months = $1.33 and also
add interest on $150 for 95 months = 34 cents. Then add interest on $240 for 34 months = 2.10.
Total interest to be credited on Dec. 31, = 16.58.
The same result can be obtained by getting the interest on $854–$350 = $504 for 6 months =
$7.56. Then add to this amount interest on $200 for 83 months = $4.33, and interest on $150 for
646 months = $2.59. Then add interest on $240 for 33 months = $2.10. Total interest to be credited
Dec. 31, = $16.58. Amount due Jan. 1, 1896, = $1110.97.
#####

1215. To find the interest and the rate per cent of interest realized by a Loan
and Investment Co. on certain conditions.
1. A Loan and Investment Co. loans $3000 for 5 years, at 8 per cent per
annum. The interest ($1200) is added in advance and the whole amount $4200 is
made payable in 60 notes of $70 each payable monthly, without interest. Allowing
that the company may re-invest the monthly collections of the notes at 8 per cent
simple interest, what would be the total interest received by the Company, and what
would be the rate per cent per annum ?
Ans. $2026 total interest; 13% rate per cent per annum.
OPERATION.
Interest on $3000 for 5 years at 8 per cent is, -- - * * * - º - - $1200.00
& 6 “$70 each monthly payment at 8 per cent for 1770 months which is the sum
of the arithmetical series from 1 to 59 months, for the 60 notes, the last note's
interest not being invested is, - = - - * * - - tº ame - 826.00
Total interest received by the Company for 5 years, * * - - tº º - $2026.00
$2026 – 5 = $405.20 = average interest for 1 year.
($405.20 × 100) -- $3000.00 = 13% per cent rate per annum. m
NotE.—The sum of the arithmetical series is found by adding the first and last terms of pay;
ment and multiplying the sum by one-half the number by terms or notes. Thus, 0 + 59 = 59 × 30
- #70 § ºlºng half the sum of the extremes by the number of terms, thus: (0 + 59)
- 2 = X -: e
2. Suppose in problem 1, above, that the monthly note and interest collec-
tions were re-invested at simple interest as therein stated, what would have been
the respective interest gain for each year?
OPERATION.
Interest on $3000 for 1 year at 8%, $240.00
& 4 “ $70 at 8%, 66 mos. {sum of series 0 to 11), ($30.80) 30.80 $270.80 1st year.
“ “ $3000 for 1 year at 8%, 240.00
“ “ $70 at 8%, 210 mos. (sum of series 12 to 23) (98.00) 98.00 338.00 2d year.
“ “ $3000 for 1 year at 8%, 240.00
& 4 “ $70 at 8%, 354 mos. (sum of series 24 to 35) (165.20) 165.20 405.20 3d year.
“ “ $3000 for 1 year at 8%, º 240.00
“ “ $70 at 8%, 498 mos. (sum of series 36 to 47) (232.40) 232.40 472.40 4th year.
“ “ $3000 for 1 year at 8%, 240.00
“ “ $70 at 8%, 642 mos. (sum of series 48 to 59) (299.60) 299.60 539.60 5th year.
Total interest for 5 years, as in the first problem, ($826.00) $2026.00
3. Suppose in problem 1, above, that the monthly collection of interest on
the notes and the interest on such interest were used, at the close of each year, as
a part of the dividend fund, what would be the amount of interest earned yearly 3
OPERATION FOR THE FIRST YEAR. OPERATION FOR SECOND YEAR.
Interest on $3000, 1 year at 8 per cent, $240.00
Interest on $70 for 66 months at 8 per
cent, which is the sum of the arith-
metical series from 11 to 0, is 30.80
Total interest revenue, first year, $270.80
NotE.—The first year there were but 11 invest-
ments made during the 12 months, as the first
note was not collected till the close of the first
month.
Interest on $3000, 1 year at 8 per cent, $240.00
Interest on $70 for 78 months at 8 per
cent, which is the sum of the arith-
metical series from 12 to 1, is 36.40
Total interest revenue, second year, $276.40
NoTE.—The $70 collected on the first of the
first month of the second year was invested, thus
making 12 investments. The operation of the
three subsequent years will be the same as for
the second year.
REMARKs.—See pages 590 to 596, for the Gain and Gain Per Cent per Annum, for certain kinds
of Loans and Investments.
(675)
NOTE.-See Soulé's New Science and Practice of Accounts for 200 pages of Corporation
Accounting.
1216. Stocks are the transferable shares of CORPORATIONS-STOCK COMPANIES-
Specifying the amount of capital owned therein by the stock or shareholders.
1217. The Stock or the Capital Stock of a Stock Company is the capital
Contributed by stockholders and invested in the business.
1218. A Corporation or Stock Company is an Association of men created
and authorized by law to act as a single person, under a common name. The mem-
bers of a corporation succeed each other, so that the Association continues always
the same, notwithstanding the change of the individuals which compose it. Cor-
porations are in law artificial persons.
The capital of corporations is divided into shares of equal value, which are
transferable and may be bought and sold like any other property.
The management of corporations is generally vested in a Board of Directors
who are named in the Charter, or elected by the stockholders, and who select the
officers to conduct the business of the Company or Corporation.
1219. Certificates of Stock are instruments issued by a CORPORATION, certi-
fying under the signatures of the proper officers that the holder is the owner of a
specified number of shares of its CAPITAL STOCK. Thus, the following is a general
form of a
CERTIFICATE OF STOCK :
Shares.
#
ŽVo.
#
exº OF Louisiawa
SOULE COLLEGE BANK.
his i; to {ertifu, That
es e e s is a e º sº s the Proprietor of...........Shares of One Hundred Dollars
each, in the Capital Stock of S0ULE COLLEGE BANK, which is
# transferable on the books of the Bank by the said Stockholder,
i.
# or.… Attorney or Agent, on the surrender of this
# Certificate.
& Witness, the Seal of the Bank attested by the Signatures
§ of the President and Cashier thereof.
; New Orleans, 189......
# Cashier. President.

(676)
vſ. STOCKS AND BONDS. 677
1220. The Par or Nominal Walue of Stock is the sum named in the instru.
ment or certificate of stock.
When stock sells for more than the original or par value, it is said to be above
par or at a premium. When for less, it is below par or at a discount.
1221. The Market Walue of stocks or bonds is what they will sell for in open
market. In the case of stocks, this value depends upon the dividends they pay
compared with the current rate of interest in the money market; and in the case
of bonds, the market value depends upon the degree of confidence entertained by
capitalists of their being paid at maturity, and the rate of interest they bear
compared with the current rate of interest in the money market.
The success of incorporated companies has much to do with the market
value, and the sobriety, industry, fidelity, and capacity of the officers and directors
of companies have almost all to do with the success of the company.
1222. The Intrinsic or Liquidation Walue of Stock is the sum that each
share would be worth in case the CORPORATION went into liquidation, or it is the
quotient of the net resources after all liabilities are paid, by the number of shares.
1223. Preference or Preferred Stock is stock issued by some CORPORATIONS
which is entitled to an extra amount of interest or to a specified per cent dividend
out of the profits, before the common stock dividend is declared.
The circumstances which give rise to PREFERRED STOCK are varied and are
generally based upon equity. A corporation may be in great distress for money, and
to relieve the company, certain stockholders or other parties may advance the
required money for which preferred stock is issued to them. Again, in the re-organ-
ization of railroads or banks, preferred stocks may be issued to save the sacrifice
of the company's property, or to obtain the means to carry on the company’s busi-
IłęSS,
1224. Guaranteed Stock consists of shares on which a certain amount of
money or interest is due, payment of which is guaranteed, thus rendering the shares
especially valuable.
1225. Forfeited Stock is subscribed stock on which installments may or may
not have been paid, and on which the subscriber fails to pay installments when
called for. Such stock is forfeited to the company according to the terms of the
Charter.
1226. Debenture Stock is a writing acknowledging a debt. 2. A deed of
mortgage given by railway companies for borrowed money. 3. The bonds and
securities issued for borrowed money, by towns and corporations.
NOTE.—This term is in general use with English Financiers, but it is rarely used by American
Bankers or Security dealers.
1227. An Installment is a certain per centum of the capital stock which
subscribers are required to pay at a specified time. --
1228. An Assessment is a per centum tax levied and collected from the
stockholders to make up losses or deficiencies.
1229. Gross Earnings are the total receipts of the company.
678 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. Yºr
1230. Net Earnings or Net Profits are what remains after all expenses are
deducted.
1231. A Reserve Fund, or Contingent Fund, or Redemption Fund is a
certain portion of the yearly profits of the company, reserved or set aside for the
purpose of creating a fund to meet unforeseen emergencies or accidents.
DIVIDENDS.
1232. A Dividend is a pro rata division of the profits or a portion thereof,
among the stockholders of a corporation.
1233. A Cash Dividend is a certain amount of the profit of a company dis-
tributed in cash among the stockholders annually, semi-annually, or quarterly.
NOTE.-Dividends, when declared, are generally a certain per cent on the par value of the
shares, but sometimes they are a certain per cent on the paid-up stock.
1234. An Installment Dividend is a certain amount of the profits of a, com-
pany apportioned to the stockholders and applied to the payment of their unpaid
Subscriptions. An installment diyidend is in effect the paying of a cash dividend
and the collection of an installment of equal amount.
1235. A Stock Dividend is a certain amount of the profits of a company
apportioned to the stockholders and retained by the company, for which new paid-up
stock is issued to the stockholders. Or, in the absence of a profit, it may be addi-
tional stock, to be paid out of the capital or the future profits of the company, in
which latter case, it is equivalent to the operation of “watering” stock.
1236. A Forced Cash Dividend, or a Capital Stock Dividend, or a Fictitious
Dividend is, in the absence of a profit or reserve fund, a certain per cent of the
capital of the company distributed among the stockholders.
The method usually adopted is to place an inflated value on the real estate,
plant, bonds, personal accounts, or other resources of the corporation, or by charg-
ing expense and repairs to the property accounts, by which means false profits are
shown, and then Forced or Fictitious dividends are declared. The object of a forced
or fictitious dividend is deception, and the result dishonest. For by it, the public
are led to believe that the company is prosperous when it is not, and hence pay
more for the stock than it is worth.
NOTE.-In this connection it must not be forgotten that sometimes, from unusual causes,
companies do not make a yearly or semi-annual profit, and then they declare a dividend from a
portion of the previous gains of the company, which have been set aside under the head of RESERVE
FUND for that purpose. A dividend of this character is not to be classed as a forced or fictitious
dividend. In the early history of railroads and some other companies, forced or fictitious dividends
were of frequent occurrence.
1237. Watered Stock is that which has been increased above the authorized
capital by the issuance and distribution among the stockholders of new stock, for
which no payment is or will be made.
When these waſtERED SHAREs are made transferable, the act of watering is
violative of every principle of ethics, and should secure for the parties thereto the
judgment of a Criminal Court.
Yºr BONDS, 679
When the WATERED SHAREs are not made transferable, then the transaction
is one of deception, the motive for which will appear plain when we consider that
the charters of some companies require that their dividends shall not exceed a cer-
tain per cent on the Capital Stock, and that whenever the net profits of the company
are sufficient to produce a dividend in excess of the specified per cent, that the
company shall reduce its rates of tariff. Hence, to avoid this reduction of its rates,
new stock is issued and distributed pro rata to the stockholders, by which means
the capital stock is increased and the rate per cent dividend decreased and kept
Within the prescribed limits, and the company is thus enabled to continue its high
charges and heavy profits.
1238. Stock is often watered in the re-organization of a railroad or the con-
Solidation of different railroads, or the consolidation of other corporations, when
the property or plants are estimated at an increased valuation. If the property has
increased in value from any cause, then such watering might be honest, though it is
not the legal way to increase the capital stock of a corporation.
NOTE.-The watering of stock has been indulged in by some of the leading railway companies
of the country and by other corporations, to an alarming extent. Watered stock amount does not
represent any real investment, but only increased earnings, and hence, with the reduction of freight
and passenger transportation charges, which the competition of new roads now building and to be
built will certainly effect all watered stock, will eventually prove a loss to the holders. For, when
the incomes of the companies are reduced to their normal level by the means indicated, the entire
issue of stock will in like manner be reduced to the original value of the first issue.
We can, in a measure, forgive, if we cannot justify, the vendor of watered whiskey or rum,
on the moral plea of his having performed a philanthropic act by selling a less poisonous article
when watered than when pure, but no justification whatever can be offered in extenuation of the
sin and crime committed by Watering Stock.
If the capital of the company is not commensurate with the workings and demands of the
company, the capital stock can be increased in a legitimate and honest manner
1239. Clandestine Stock is first, a new edition of stock issued without
public notice and placed on the market to raise money to cover losses or expenses
of which the public have no knowledge.
NOTE.—Clandestine Stock has also been issued under the following circumstances:
The market value of stock is very high, and the company issues and distributes pro rata, at
par, to the stockholders of the company, a certain amount of new stock. This new stock is placed
on the market, and the full market value of the original stock is realized before the fact is known
that it is Clandestine.
BONDS.
1240. Bonds are written or printed obligations of Individuals, Corporations,
Cities, States or nations, to pay a specified sum of money at a certain time. BONDS
generally bear a fixed rate of interest payable, annually, semi-annually or quarterly,
and when issued by individuals or corporations for loans are Secured by mortgages
upon their property.
The security of the Bonds of States or nations, usually have no other security
than the confidence of the people. BONDS are issued in denominations varying
from ten dollars to fifty thousand dollars.
1241. Coupon Bonds are those having interest tickets or coupons attached,
68O SOULE's PHILOSOPHIC PRACTICAL, MATHEMATICS. Y
Which specify the interest due each year, half year, or quarter, and which are cut
off as they are paid, and held as receipts.
1242. Registered Bonds are bonds that are payable to the order of the holder
or owner, and consequently cannot be negotiated without assignment duly acknow.
ledged.
BONDS are known as First Mortgage, Second Mortgage, Debenture, etc.,
Consols, Income, Sinking Fund, Adjustment, etc., according to their priority of
lien, the class of property upon which they are secured or other characteristics.
BONDS are also named from the rate of interest they bear, or from the dates
at Which they are payable, or from both. Thus: Louisiana 4's, United States 4's of
1907, New Orleans City Premium Bonds, United States 3's.
1243. A Mortgage is a conditional conveyance or sale of real or personal
property as security for the payment of a debt or for the performance of a duty.
The written instrument is termed the MORTGAGE and is property or a resource
to the holder.
1244. Consols (from consolidated), are bonds issued in redemption of two or
more classes of bonds which are consolidated into the new bond, and generally at a
reduced rate of interest. *
ENGLISH OONSOLS.
1245. English Consols are 3 per cent English bonds. The term is abbrevia-
ted from the words consolidated annuities.
NOTE: 1.—These bonds were issued by virtue of an Act of Parliament to redeem several
separate issues of government bonds bearing various rates of interest, and with the condition that
while the interest was regularly paid the government could not be required to pay the principal.
They are hence practically perpetual annuities. In the act creating them, they are termed consoli-
dated annuities.
The quotations of these 3 per cent bonds or “Congols” indicate ordinarily the state of the
money market, as they form a large part of the English public debt.
NotE 2.-In England the term STOCK is limited to the Government Stocks and Annuities;
and the term SHARE is used for the Capital Stock of Corporations.
1246. French Rentes are the Government Stocks, which bear various rates
of interest.
NOTE.-In France the term RENTES has the same limitation as STOCKS in England.
DNITED STATES BONDS.
The following are the interest bearing bonds of the United States, outstanding
April 1, 1895, as per the statement of the United States Secretary of the Treasury:
1247. 1. Funded Loan of 1891. These are bonds which were authorized
by an Act of Congress of July 14, 1870, and Jan. 20, 1871, to refund the national
debt. They bear interest at the rate of 44 per cent, payable quarterly in gold.
They are redeemable at the option of the United States 15 years after Act of
Authorization. Amount outstanding April 1, 1895, was $25,364,500.
Yºr STOCK EXCHANGE. 68t
1248. 2. Funded Loan of 1907. These bonds were authorized by an Act
of Congress of July 11, 1870, and Jan. 20, 1871. They bear 4 per cent gold interest,
payable quarterly, and are redeemable July 1, 1907. Amount outstanding April
1, 1895, was $555,624,850.
1249. 3. Refunding Certificates. These certificates were authorized by an
Act of Congress of February 26, 1879. They bear 4 per cent interest, payable
quarterly. $54,710 are outstanding April 1, 1895. *
1250. 4. Loan of 1904. These bonds were authorized by Act of Congress
of Jan. 14, 1875, they bear 5 per cent interest, payable quarterly, and are redeemable
February 1, 1904. Amount outstanding April 1, 1895, $100,000,000. They were
placed on the market in 1894.
1251. 5. Loan of 1925. These bonds were also authorized by the Act of
Congress of Jan. 14, 1875. They bear 4 per cent interest, payable quarterly, and are
redeemable Feb. 1, 1925. Amount outstanding April 1, 1895, was $28,807,900.
1252. 6. U. S. Pacific R. R. Currency Sixes. These are 30 year 6 per
cent currency registered bonds, which the government issued to the various com-
panies chartered by Congress to build railways to and from the Pacific coast. They
were issued on the completion of 20 miles of road at the rate of $16000, $32000, and
$48000 per mile, according to the expense and difficulty of construction. The
interest is payable Jan. 1, and July 1. Amount outstanding April 1, 1895, was
$64,623,512.
In addition to the foregoing interest-bearing debt the United States had,
April 1, 1895, the following indebtedness:
1. Bonded debt on which interest had ceased since maturity, $1,770,250.26.
2. United States Notes, National Bank Notes and Fractional Currency,
$381,025,096.92.
3. Gold and Silver Certificates, and Certificates of Deposit, all of which is
offset by an equal amount in the Treasury, $567,944,442.00.
STOCK EXCHANGE.
1253. Stock Exchanges are Associations of Brokers, generally incorporated
bodies, whose business it is to buy and sell stocks, bonds, and other securities. The
regular commission for buying and selling securities at the New York Exchange is
# per cent on par value, except for mining stocks, on which contract rates are
usually charged.
The NEW ORLEANS STOCK EXCHANGE is an Association organized for the
purpose of buying and selling stocks, bonds, and other securities. The regulations
of the association are very stringent and specific, in order to prevent unworthy
persons from being admitted, and to insure fidelity and capacity in the performance
of contracts. To become a full member of the New Orleans Stock Exchange, the
applicant must be of good character, must be regularly engaged as exchange, stock,
coin, or bill broker, and own at least one share of the stock of the exchange. The
682 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICs. y
number of shares of the New Orleans Stock Exchange is 70, and the par of each is
$10. The present (1895,) market value per share is about $800.
The NEW ORLEANS STOCK EXCHANGE has an INSURANCE FUND which was
Created by an assessment of $15 per member. Upon the death of any member of the
Exchange, each surviving member is assessed $15, which is paid into the INSURANCE
FUND, and within 60 days after the proof of the death of any member the amount
of money thus realized is paid to the legal heirs of the deceased member.
The commission for buying or selling in the New Orleans Stock Exchange is
as follows: # per cent on the par of City.and State Bonds. On all other bonds
selling at and above par, 3 per cent on par value.
H On stock selling at and over $60 per share, 50 cents per share is charged. On
stocks selling at $25 and under $60 per share, 25 cents per share is charged.
On-stocks selling below $25 per share, a contract charge is made. On Scripts,
Warrants, etc., 3 per cent is charged on par value. ~.
.* In the sale of stocks, the following regulations are also observed by the New
Orleans Stock Exchange: º
1. On all contracts of time exceeding 3 days, one day's written notice must
be given before the stocks or securities can be delivered or demanded.
2. All sales not “cash’” are considered time. No purchase on sale at the
option of the buyer or seller for 3 days, shall bear interest.
3. All purchases and sales beyond 3 days shall be “flat” unless otherwise
expressed. *
4. On all time bargains, the rate of interest shall be 8 per cent, to be cal-
culated by days according to bank usage.
1254. DEFINITIONS OF WORDS AND PHRASES USED BY FINAN-
CIERS, BROKERS AND BANKERS.
A Trust is a corporation formed by large dealers in a certain article, whose object is to control
the market, thereby forcing prices above the level established by free competition, or reducing
them below the cost to produce the article in order to suppress opposition. Trusts sometimes
fairly and honorably reduce prices, by reason of an extensive trade, of buying material and
manufacturing articles in large quantities at a few centers of trade, etc.
A Syndicate is a combination of Brokers, Bankers or Capitalists, who undertake to place large
loans and transact other financial business.
Collaterals, are Securities, Bills Receivable, Stocks, Bonds, etc., which are pledged and placed in
the possession of an obligee, in addition to the principal security, to secure the payment of
money of the performance of an obligation entered into by the obligor.
Hypothecating Stocks and Bonds is depositing them as collateral security for money borrowed or
debts contracted.
4 Margin is a deposit made with a broker by a person who wishes to speculate in buying or selling
stocks or bonds, to enable the broker to carry the stocks or bonds and protect himself against
any loss that may occur by an unfavorable change in the market price.
The margin varies from 5 to 20 per cent of the value of the stocks or bonds.
NOTE.-In case the market moves so as to lessen the per cent margin and threaten loss to the
broker who is “carrying” the stock or bonds, the margin must be made good, on the notice of
the broker, by the deposit of an additional sum. If this is not done the broker is justified in
; the securities to protect himself against the loss of the money he has advanced to or for
is principal. *
Bºrge interest on all sums advanced for carrying stocks or bonds, allow interest on
margins, and charge the regular per cent brokerage for buying and for selling.
jºr DEFINITIONS OF WORDS AND PHRASES. 683
Difference is the variation of the price for which stock is contracted and the market rate the day
of delivery. In cases like this the broker against whom the variation exists frequently pays
the difference in money instead of supplying or receiving the stock.
Banker, the proprietor of a bank; one who deals in money, and negotiates bills of exchange.
Broker, one who receives and executes orders for other parties.
Factor or Commission Merchant, one who sells goods on commission.
Jobber or Merchant, one who deals in Stocks or Merchandise on his own account.
Stag or Outsider, one who does business outside of the exchange.
Bear is an operator who is “Short of Stocks,” or who wishes to purchase stocks; and hence one
who labors to decrease or depress the price that he may buy or “fill” at lower rates.
The term is also applied to all parties who wish to purchase goods. The name is derived
from the nature of the bear to pull things down with his paws.
Bull is an operator who is “long of stocks,” or who wishes to sell stocks; hence one who labors
to advance or inflate the prices.
The term is also applied to all parties who have goods to sell. The name is derived from the
nature or character of the bull to throw things up with his horns.
Future Sales or Seller's or Buyer's Option, the privilege given to either party of delivering the
Stock at any time within the blank days specified.
Cash Sales, to be delivered and paid for the same day before 2 P. M., if sold before 1 P. M., and
before 3 P. M., if made after 1 P. M.
A Regular Sale, payable the next day before 2 P. M.
Selling Short, selling what you have not in the hope of buying for less before called upon for a
delivery.
- Long of Stocks, buying and holding the same in the hope of a rise.
Short of Stocks means that you have sold that which you do not possess in the hope of buying at
lower prices and thus realizing a profit. *
A Wash Sale, a sham transaction between two brokers for the purpose of causing an advance or
decline in prices.
Wash sales are not allowed in well managed exchanges.
To make a Turn, sell for Cash and then buy it back on regular sale.
Milking or Lowering, lowering prices to induce holders to sell, and then the clique step in and buy.
Salting Down, saving what you have made.
Busy Bees, a clique or several parties working together with some special object in view to influence
prices.
To Fill or to Cover, buying Stock or Merchandise to fill contract of short sales.
To Unload or to Get Out, to sell Stock or Merchandise that is burdensome, either at a gain or a
loss.
Lame Duck, one who has lost heavily or has been badly crippled, but who has not failed.
Dead Duck, one who is unable to fulfill his contracts and fails.
Opening Day, the first day that the books are opened after a dividend has been declared.
Settling Day, a day that the things bought or sold are delivered or paid for, or a day that the con-
tract matured.
Contango, the postponement of settling day.
Point, one of any unit or measure of price, either increase or decrease.
Black List, a record kept of members of the exchange or customers of banks who violate their
contracts.
A Put Contract, a contract by which you have the right to put or sell to your broker, One Hundred
Shares of specified stock at a fixed price, during thirty days. For this contract you pay your
Broker $106.25.
A Call Contract, a contract by which you have the right to call or buy from your Broker, One
Hundred Shares of specified stock at a fixed price, during thirty days. For this contract you
pay your Broker $106.25.
A Spread is a contract by which you have the right to put to, or call from your Broker, One Hun-
dred Shares of specified stock at a fixed price, in thirty days. For this contract you will pay
your Broker $212.50.
NOTE.-A Straddle is where a dealer is “long’’ of one option and “short” of another. He
has thus straddled the market. Also where a dealer buys for future delivery is one market and
sells in another.
684 soule's PHILOSOPHIC PRACTICAL MATHEMATICs. *
*
S. 3, 10, 30, etc., mean that the Seller has three, ten, thirty, etc. days in which to deliver Stocks
or Merchandise bought. -
B. 3, 10, 30, 30, etc., mean that the Buyer has three, ten, twenty, thirty, etc. days in which to
receive Stocks or Merchandise bought.
- Eac. Dividend means that the sale is made with the dividend off.
C. means a cash transaction.
R. means a regular sale.
P. F. means preferred stock.
G. T. D. means guaranteed.
Ex. Coup. means without coupons or interest.
B. & Int, means that the buyer is to receive the interest on the stocks bought.
B. Flat means that the buyer is not to receive the interest.
b. c., between calls, means that the sale was made between the first and second call of the stock,
there being two or more calls each day.
Corner is an operation by several jobbers or operator.—“longs”—who form a clique to buy and
hold stocks or merchandise, thus producing a scarcity in the market which results in high
prices, and a loss to the “shorts” or others who are obliged to buy, and are hence cornered.
Corners, frequently by result disastrously to the cliques or “longs” who organize them as well as to
the “shorts” as they are left with a load that they cannot discharge except at a great loss.
Spread Eagle is the operation of a broker who sells a given quantity of stocks on time, say, 60, and
buys the same quantity at a lower price, say, 60. If both contracts run 60 days he will make
.# difference; but if he is compelled to deliver by buyer or seller before the full time he may
suffer loss.
Bucket or Curbstone Brokers are brokers who are not members of the stock exchange and though
some of them are men of honor, many are lame ducks and some are only speculators without
capital; they are not governed by as strict rules and do not abide by the strict letter of the
law as do members of the exchange. -
Rite Flying is a term applied to a certain class of transactions of an extra hazardous character,
and engaged in only by persons who are non-observing of the principles of ethics. The follow-
ing will serve to illustrate the transaction: A, is without capital; he solicits of B. the privilege
of drawing on him at 10, 30, or 60 days, with the assurance that before the maturity of the
draft he will place funds in his hands to meet the same. B. consents; A, draws, negotiates the
draft, and uses the proceeds. When this draft is about to mature, he draws and negotiates
another sufficiently larger than the first to net the face of the first. This process of drawing
and selling, he keeps up until by prosperous business he clears enough to cover the discounts
on his drafts, and to meet the last, one ; or, until his repeated losses and clandestine dealings
involve him in inextricable difficulties, and he becomes a disgraced and ruined man.
PROBLEMS IN STOCKS AND BONDS.
1255. To buy and sell Stocks and Bonds on our own account.
1. What will 50 shares of Germania National Bank stock cost at 15 per cent
premium ? Ans. $5750.
OPERATION.
$115 × 50 = $5750. Ans.
2. Bought 20 shares of National Bank stock at 95 per cent. What did it
cost 7 Ans. $1900.
3. Sold 40 shares of City Railroad stock at 1254. What did Ireceive for it?
Ans. $5020.
4. Bought 75 shares of Chattanooga Railroad stock at 90 per cent, and sold
the same at 5 per cent premium. How much did I gain 3 Ans. $1125.
4x PROBLEMS IN STOCKS AND BONDS. 685
1256. To invest in stocks and Bonds on our own account.
1. How many shares of $100 each of the City National Bank stock can I
buy for $2280, the stock being 14 per cent premium ? Ans. 20 shares.
FIRST OPERATION.
1 share = $100 par value. Ea:planation.—In this solution, we first find the
14 = 14% premium. par amount of stock that we can buy with the
* * money, and then the number of shares. In the
$114 cash value. operation we first added to the $100, par value
of 1 share, the 14 per cent premium, which gave
100 us $114 cost value; we then placed the $100 par
114 || 2280 value on our statement line, and reasoned thus:
| Since $114 cash will buy $100 of stock, $1 cash
$2000 = par of stock. will buy the 114th part, and $2280 will buy 2280
times as much. The result of which is $2000.
which divided by $100 = 20 shares. Ans. It is then clear that as 1 share is $100, there are
as #ºny shares as $2000 are equal to $100, which
IS 2#.U).
SECOND OPERATION.
SHARE. Explanation.—Here we place 1 share on our
| 1 statement line, and reason thus: Since $114
114 || 2:280 cash will buy 1 share, $1 will buy the 114th part,
*=-º-º-e and $2280 will buy 2280 times as much. The
20 shares. Ans. result of which is 20 shares.
2. Invested $15375 in City bonds, at 384 per cent discount. What amount
did I buy? Ans. $25000.
OPERATION INDICATED.
$100 – $384 = $614; $15375 – $61} = $25000. Ans. '
3. A merchant invested $3675 in Louisiana Frear Stone Manufacturing
Company stock, at 5 per cent premium, and sold the stock at $1144. What was his
gain } Ans. $319.374, gain.
4. Invested $10450 in Louisiana 4's at 1044. What is my annual income
from the same 3 Ans. $400.
1257. To buy and sell Stocks through a Broker.
1. Bought 25 shares of New Orleans Wrecking Company stock, at 95 per
cent. Paid New Orleans Board of Brokers rates of brokerage. What did it cost?
Ans. $2387.50.
OPERATION INDICATED.
25 shares at $100 = $2500 par; $2500 × 95% = $2375. $2375 + 4% brok. $12.50 = $2387.50. Ans.
2. Sold 100 shares of New Orleans 5 per cent Water Works bonds at 40 per
cent discount. Paid 3 per cent brokerage. What were the proceeds?
Ans. $5950.
3. Bought 80 shares of $50 each of the Mechanics' & Traders' Bank stock,
at 75 per cent, and sold the same at 10 per cent discount. Paid brokerage # per
cent for buying and 4 per cent for selling. What was my gain? Ans. $560.
686 soule's PHILOSOPHIC PRACTICAL MATHEMATICs. .*
1258. To invest in Stocks and Bonds through a Broker.
1. Invested $9056.25 in Louisiana Savings Bank stock, at a premium of 20+
per cent. Brokerage 3 per cent. How many shares did I buy 3 Ans. 75.
FIRST OPERATION.
1 share = $100 par value. Explanation.—We here first find the cost of
20+ = 20.4% premium. $100 par value at the rate of premium and bro-
# = }% brokerage. kerage given in the problem. This we do by
** == adding to the $100 par value the 20+ per cent
$120% cost of $100 par. premium and 4 per cent brokerage, which gives
$120%; with these ratio figures we make the
$ PAR VAL. proportional solution statement, as shown in
100 the operation, to find the par amount of Stock
483 || 4 bought. The reasoning for the statement is as
9056.25 follows: Since #4 (120% reduced) dollars cash
Eº-º-º-º: will buy $100 par value of stock, 4 of a dollar
100 ) $7500.00 amount of stock. cash will buy the 483d part, and # or a whole
-- - dollar will buy 4 times as much, and $9056.25
75 shares. Ans. cash will buy 9056.25 times as much. The result
of which is $7500. Then as 1 share is worth
$100, it is plain that we have as many shares as the $7500 is equal to $100, which is 75.
SECOND OPERATION,
SHARE, Ea:planation.—In this solution we work the
1 same as in the first to obtain the cost of $100
3 || 4 par value of stock; then, instead of finding theſ
| 9056.25 N amount of stock that we can buy, we find at
once the number of shares. This we do by
75 shares. Ans. placing 1 share, instead of the par value of the
same, on our statement line, and reasoning
thus: Since $##3 ($120% reduced) cash will buy 1 share, ($4 will buy the 483d part, and $3 will buy
4i. as many, and, $9056%; will buy 9056%; times as many as $1. The result of this gives us 75
Bºla,TOSs .*
NOTE.-We present the two operations beeause in some problems we wish to know the
number of shares, and in others we wish to know the par value or amount of stock.
2. Paid $1162.50 for City of New Orleans 4's at 33 per cent discount. Bro-
kerage 3 per cent. What is my semi-annual interest on them? Ans. $24.
OPERATION INDICATED.
33% discount. $ PAR. $1200
* #96 brok, deducted. | 100 4%
96% | 1162.50
3#96 net discount. * * == $48.00 int. for 1 year.
$100 par val. of stock. $1200 bonds bot.
$96; cost of $100 par.
$48 – 2 = $24 int. for 6 mos.
3. Invested $15912 in City of New Orleans 5 per cent bonds at 1034, bro-
kerage # per cent. What is the annual income 3 Ans. $765.00.
4. What amount must I invest in 34 per cent bonds which are selling at
104, brokerage 4 per cent, so that I may have an income of $525 per quarter f
Ans. $62700.00.
OPERATION INDICATED.
$525 x 4 = $2100. 3.} | * } = $60000 x 4% = $2700 + $60000 = $62700. Ans.
* PROBLEMS IN STOCKS AND BONJOS, 687
1259. The capital and rate per cent given, to find the dividend.
1. A merchant owns 200 shares of bank stock, the par value of which is
$50. How much will he receive when a dividend of 6 per cent is declared ?
Ans. $600.
OPERATION.
$50 × 200 = $10000 × 6% = $600 dividend.
2. What is the amount of a 6 per cent dividend on 120 shares of New Orleans
Gas Light Company stock? Ans. $720.
3. Invested $5747.50 in State National Bank stock, at 104. Paid brokerage
for buying 3 per cent. How many shares did I buy, and what is my annual income
therefrom when the dividends are 4 per cent semi-annually?
Ans. 55 shares. $440 annual income.
1260. The capital or investment and net gain given, to find the rate per cent dividend
or per cent gain.
1. The capital of a Bank is $1500000. The net profits for six months' busi-
ness are $192814.58. The Board of Directors resolve to pass $12814.58 to the credit
of “reserve fund,” and to distribute the remainder to the stockholders. What is
the rate per cent of the semi-annual dividend ? Ans. 12%.
OPERATION.
$192814.58 met profits. $
12814.58 passed to reserve fund. 180000
smºmºmºmºsºmº-ºº 1500000 || 100
$180000.00 profit for a dividend.
12% Ans.
2. I subscribed for 100 shares of bank stock. After paying three install-
ments of 25 per cent each, a dividend is declared. The capital stock of the Bank
is $300000; the net profits are $38460.50. Passing $8460.50 to reserve fund, what
is the rate per cent dividend and the rate per cent profit on my investment?
Ans. 10% dividend. 134% gain on investment.
3. An investor has $14700 in manufacturing stock, and receives $367.50 per
quarter, dividend. What is his annual per cent gain? Ans. 10%.
4. Invested $20000 and gained $1500. What was the gain per cent?
& Ans. 73%.
OPERATION INDICATED.
$
1500
20000 100
5. Bought 50 shares of stock at $90 per share, and sold the same at 12% per
cent discount. What per cent did I lose on the investment? Ans, 24% loss.
OPERATION INDICATED.
$
50 × $90 = $4500 cost. * 125 = loss.
50 × $874 = 4375 selling price. 4500 | 100
688 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. jºr
1261. The rate per cent dividend or interest, and the price of Stock given, to find
the rate per cent gain on investment.
1. Bought New Orleans City 6's at 25 per cent discount. What is my per
cent gain on investment? Ans. 8%.
OPERATION.
$100 par value. $ GAIN. Ea:planation.—As shown in the operation, $100 par value
25% discount. | 6 cost but $75, and hence the 6 per cent interest is really
sºmsºmºsºm- 75 | 100 gained on the $75 cash investment. In making the pro-
$75 cost. portional statement to find the gain per cent, we reason
8% Ans thus: Since $75 gain $6, $1 will gain the 75th part, and
$100, 100 times as much.
2. Bought 6 per cent bonds at 5 per cent premium. What is my per cent
gain on investment, making no allowance for the use of the semi-annual interest?
Ans. 5}%.
3. If I buy 50 shares of bank stock at 190, and pay 3 per cent brokerage,
and the bank declares a semi-annual dividend of 15 per cent, what is my per cent
gain? Ans. 154*7%.
OPERATION INDICATED.
$190 cost. 15% semi-annual dividend. $
#96 brok. on par value: 2 30
_*% p 381 || 2
$1904 total cost of stock. 30% yearly dividend. | 100
-*
4. Which is the better investment, 8's bought at a premium of 20 per cent,
or 6's at a discount of 25 per cent 3 Ans. 6's are 14% better.
5. A capitalist buys Louisiana 6's, having 10 years to run, at a discount of
15 per cent. What will be his rate per cent interest gain if he holds them until
maturity and then collects the face and interest ? Ans. 844%.
OPERATION.
$100 par of bond. $100 par of bond. $100 face of bond.
15 = 15% discount. 6% interest. 60 int. for 10 years.
$85 cost of bond. $6.00 interest for 1 year. $160 value at maturity.
10 years. dº. 85 cost of bond.
$60.00 interest for 10 years. $75 gain in 10 years.
We now have to find at what rate per cent interest $85, for 10 years, will
produce $75 interest.
OPERATION.
$85 9% 75
10 years. 1° 0 - 85 || 100
-ºs- 8.50 || 75.0 OF 10
$8.50 int. for 10 years at 1%. | j
x- 8+} % Ans. 8}# 96.
Ea:planation.—For the reasoning of this last work, see Interest Problem, on page 599.
6. Bought 5 per cent interest bearing bonds at 98 per cent. These bonds
30. PROBLEMS IN Stocks AND BONDs. 689
have 7 years and 6 months to run. What will be the gain per cent, the interest
being paid semi-annually, making no allowance for the use of the semi-annual
interest ? Ans. 5; % or 5.102+%.
7. In the above problem allowing 6 per cent simple interest on the annual
interest of the bonds, what would be the annual gain per cent made on the invest-
ment 3 Ans, 5}}%.
OPERATION INDICATED.
Interest on $5 for 12 mos. at 6% = 30c. $5 + 30c. = $5.30.
$
5.30,
98.00 100
53.3%
1262. The per cent that Stocks bear, and the per cent that is desired to be gained,
given, to find the price at which Stock must be bought in order to
wealize the desired gain per cent.
1. At what price must stock be bought which pays 6 per cent, in order that
the investment will pay 8 per cent? * Ans. $75 per share of $100.
FIRST OPERATION.
$100 assumed price. Eacplanation.—Here, having no value to work
8% gain. from, we assume $100, and then find the 8 per
cent gain thereon to be $8. We then see that
$8.00 amount of gain. $100 value of stock gives $8 gain, and by trans-
position that $8 gain requires $100 value of stock.
The question is now simply this: Since $8 gain
100 require $100 value, what value will $6 gain
8 || 6 require? The exact reasoning for the statement
*º is, since $8 gain require $100 value, $1 will
$75 Ans. ...he 8th part, and $6 will require 6 times
8.8 IIlllCIls 3
2. If I loan my money, I can get 10 per cent interest. At what discount
must I buy Louisiana Levee bonds, that bear 8 per cent, in order to obtain the same
per cent on the investment? Ans. 20% discount.
OPERATION INDICATED.
10 | *}=sso. 100–80 = 20.
3. Interest on money is 44 per cent. What price can I pay for stock that
pays 6 per cent dividends? Ans. $133}.
OPERATION INDICATED.
$
100
100
9 || 2 or, 4% | 6
| 6
4. Which is the best investment to make with $52000: To loan it at 12 per
cent interest on good collaterals; to invest it in stocks at 65 per cent that pay
690 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICs. &A
semi-annual dividends of 4 per cent ; or to buy merchandise which we can sell
during the year for $67380, with an attendant expense of $9400?
Ans. Stocks by a per cent over the 12 per cent loan, besides the use of
one semi-annual dividend for 6 months; and by ## per cent over
the merchandise investment, not making any allowance for the
use of the money collected prior to the close of the year.
OPERATION INDICATED.
65 | io } = 12#%. $67380 – $52000 = $15380 gain on sales; less $9400 expense. = ($5980 net gain
× 100) -- 52000 = 11339%. 12.4% — 12% = *%. 12#96 — 11}}% = #96.
1263. The rate per cent that Stocks bear, the cost of them, and the gain or income
desired to be realized from them given, to find the sum to be invested.
1. What sum must be invested in Louisiana 6's, at 75 per cent, to realize an
income of $3000 per year 2 Ans. $37500.
OPERATION.
$100 par of investment assumed.
6% rate of interest. *
00
$6.00 amount of income on $100 stock,
$50000 par amount of investment.
75% cost of $100 stock.
$37500.00 cost of $50000 stock or amt.
of investment.
Explanation.-Here we assume $100 as the par of the investment, and find the 6 per cent
income on the same to be $6. Then, with these ratio figures and the $3000 income, we make the
proportional solution statement by placing the $100 par of investment on the statement line, and
reason thus: If $6 income require $100 par of investment, $1 will require the 6th part, and $3000
income will require 3000 times as much. The result of this is $50000, par of the investment, which
multiplied by the price, 75 per cent, gives a cash investment of $37500.
2. How many shares of bank stock, of $50 each, must be bought at 10 per
cent premium in order that an income of $1500 may be realized, on the presumption
that the semi-annual dividends will be 6 per cent, and what will be the amount
invested ? Ans. 250 shares. $13750, amount invested,
OPERATION.
$ $12500 par of stock.
100 1250 10 per cent premium.
12 || 1500
$12500 par of stock -- $50 = 250 shares.
$13750 amount invested.
3. If 6 per cent bonds are selling at 5 per cent discount, what amount must
I invest in them in order to receive a yearly income of $1200? Ans. $19000.
*
PROBLEMS IN STOCKS AND BONDS. 691
}
1264. The cost of Stocks, and the premium or discount at which they were bought,
given, to find the par value.
1. Bought railroad stock at 70 per cent discount, and paid $7.50 per share.
What is the par value #
OPERATION.
$100 par value assumed.
70 = 70% discount.
$30 cost of $100 par.
Ans. $25.
Explanation.—Having no face of stock to work
on, we hence assume $100 as the par value on
which we calculate, and then deduct therefrom
the 70 per cent discount, and thus obtain $80 as
the cost of $100 stock at 70 per cent discount.
This gives us the necessary ratio figures to solve
the question; and in the line statement of the
$ Solution we place the $100 par value on the line,
| 100 and reason thus: Since $30 cost at 70 per cent
30 || 7.50 discount requires $100 par value, $1 will require
| *== the 30th part, and $7,50 will require 7,50times
$25.00 Ans. as much.
2. Sold bank stock for $55.50 and gained 11 per cent. What is the par of
the Stock # Ans. $50.
OPERATION.
$100 par assumed. $
11 = 11% premium. 100
E-º-º-º- 111 || 55.50
$111 cost. mºmsºme
| $50 Ans.
1265. The time that Bonds have to run and the interest they draw, given, to find the
price at which they must be purchased so that, when paid at maturity without
estimating the use of the annual interest, the amount received will
equal a certain rate per cent on the investment.
1. What must be paid for State bonds bearing 4 per cent interest and having
8 years to run, in order to make the investment equal to 5 per cent ? No allowance
to be made for interest on the interest of the bonds paid annually, semi-annually or
quarterly 3 Ans. $94,285-H.
OPERATION.
$100 face of bond. $100 investment. $
4% interest. 5% interest. 100
140 || 132
$4.00 interest for 1 year. $5.00 interest for 1 year.
8 years. 8 years.
$ 40.00 interest for 8 years.
$94,285-H investing price
of the bonds.
$32.00 interest for 8 years.
100.00 face of bond. 100.00 face of bond.
$132.00 = value of bond at $140.00 = value of investment
maturity. at 5% at the end
of 8 years.
Ea:planation.—As shown in the operation, we first find the value of a $100 bond and the value
of $100 investment each for the time at the respective rates per cent. Having these respective
values, we see that $100 at 4 per cent produces $132, and at 5 per cent, $140; and as we wish to realize
5 per cent on the investment, we place the $100 assumed value on the statement line, and with
the ratio numbers produced reason thus: Since $140 value in 8 years at 5 per cent requires $100
investment, $1 will require the 140th part, and $132 value will require 132 times as much, which
gives $94,285-i- investing price of the bonds.
NOTE.—This method does not take into consideration the interest on the interest received
annually on the bonds. To make such an allowance the price of the bonds would be a little higher
according as the payments of interest were made annually, semi-annually or quarterly. See the
problems following, for such allowance. Also see problem on page 591.
692 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
1266. The face of Bonds or Notes, the time to run, and the rate per cent of yearly,
half-yearly or quarterly interest that they bear, given, to find
the value at maturity.
1. What is the value at maturity of $1000 Louisiana 8's due in 5 years,
estimating simple interest on the annual interest at 8 per cent? Ans. $1464.
OPERATION.
Face of bond, - - sº - agº - tº - tº - º - - - - - $1000
Annual interest for 5 years at 8 per cent, • * * * * * * * * * 400
Interest on $80 (the yearly interest on the $1000) for, 4 + 3 + 2 + 1 = 10 years at 8%, - 64
Value at maturity, - - - - - - - - - - - - - $1464
2. What is the value at maturity of a $10000 bond due in 10 years with 5
per cent semi-annual interest, allowing simple interest on the semi-annual interest
at 5 per cent? : Ans. $16,187.50.
OPERATION. OPERATION.
To find the terms and years. | Face of bond, - - - - - - - - $10000.00
1. Semi-annual interest for 10 years at 5 per cent, º 5000.00
19 Interest on $250 (the semi-annual interest on $10000)
- for the 190 terms of 3 years (the series of 1 to
20 × 9} = 190 – 2 = 95 years. 19 inclusive) = 95 years at 5 per cent, tº cº- 1187.50
Value of bond at maturity, º - - $16,187.50
1267. The face of Bonds or Notes, the time to run, and the rate per cent of yearly,
half-yearly or quarterly interest that they bear, given, to find the present value or
worth of the same in order to realize from the interest or dividends and final
payment an equivalent to a given rate per cent per annum on the investment.
1. What is the present value or worth of a $10000 bond due in 10 years, with
5 per cent semi-annual interest, so that the investment may be equivalent to 10 per
cent per annum payable semi-annually, allowing simple interest on all the semi-
annual interests on the bond at 5 per cent? Ans. $6540.41.
OPERATION
To find the value of Bond at maturity.
Face of bond, -- * tº - ſº tº tº º ºs º- º *i; };
Semi-annual interest for 10 years at 5 per cent, - ſº - tº - - " - tº-
Interest on $250 (the semi-annual interest on $10000) for the 190 terms (the series of 1
to 19 inclusive) of 3 years = 95 years at 5 per cent, - - - , - º 1187.50
Value of bond at maturity, - * * s - sº * º - - $16187.50
OPERATION
To find the value of a $100 Bond for 10 years, at 10% semi-annual interest, the investment rate.
Present value assumed, - - - - - - - - - - - - - - $100.00
Semi-annual interest for 10 years at 10 per cent, • * * * * * * * 100.00
Interest on $5 (the semi-annual interest on the $100) for the 190 terms (the series of 1 to
19 inclusive) of 3 years = 95 years at 10 per cent, - - - - - - - 47.50
Value of $100 in 10 years at 10 per cent, semi-annual interest, - - - $247.50
4. PROBLEMS IN STOCKS AND BONDS. 693
OPERATION Explanation.—As shown in the operation, we
To find the present value of the maturity value first find the maturity value of the bond. Then,
of the Bond in order to produce the necessary relationship
o numbers, representing present and maturity
$ ASSUMED VAL. values according to the conditions of the prob-
100 lem, we assume $100 present value, to which we
247.50 | 16187.50 add the semi-annual interest and the simple
| *- interest on the semi-annual interest for the 10
$6540.41 years at 10 per cent, and thus obtain $247.50,
the maturity value of the $100 assumed value
on the same conditions as we did for the $10000 bond. We now see that as $100 present value
require $247.50 maturity value, by transposition $247.50 maturity value require $100 present value.
The reasoning for the line statement with these relationship figures is so simple and has been given
so often that we here omit it.
THE OPERATION REVERSED
To elucidate the problem still further, we present the figures produced by reversing the work.
Present cash value of bond, * * * * * * * * * * * * $6540.41
Semi-annual interest on same for 10 years at 10 per cent, - º - - - tº- 6540.40
Interest on $327.02 (the semi-annual interest on $6540.41) for 190 terms of 3 years = 95
years at 10 per cent, - - - - - - - - - * - ºn are 3106.69
-**
Value of bond at'maturity, - - - - - - - - - - $16187.50
2. What is the present value of a $10000 bond due in 10 years, with 5 per
cent semi-annual interest, in order that the investment may be equivalent to 10 per
cent per annum payable annually, allowing simple interest on the semi-annual
interest or dividend of the bond accruing yearly at 10 per cent per annum ?
Ans. $7040.82.
OPERATION OPERATION
To find the terms and years. To find the value of Bond at maturity.
9 Face of bond, Eº - - º sº. * - sº tº $10000.
1 Semi-annual interest on same for 10 years at 5 per cent, 5000
- Interest on $500 (the sum of two semi-annual payments
10 × 4} = 45. on the $10000) for 45 terms of years (the decreasing
series from 9 to 1 inclusive) at 10 per cent, - -> 2250,
Value of bond at maturity. - - - - $17250.
OPERATION
To find the value of a $100 Bond for 10 years, annual interest at 10 per cent.
Present value assumed, - º tº - - º - -> º tºº - - - - $100
Annual interest for 10 years at 10 per cent, * * * * * * * * * * 100
Interest on $10 (the sum of two semi-annual payments on the $100) for 45 terms of years
(the decreasing series from 9 to 1 inclusive) at 10 per cent, - - - - - - 45
Value of $100 in 10 years at 10 per cent annual interest, tº- º -> - - $245,
OPERATION
To find the present value of the maturity value of the Bond.
Ea:planation.—The operation of
this and the preceding problem is
100
245 | 17250 so explicit that a special explana-
$7040.82 Ans. tion is considered unnecessary.
THE OPERATION REVERSED.
Present value of bond, m as * * * * * * * * * * = $7040.82
Annual interest for 10 years at 10 per cent * > * 7040.82
Interest on $704.08 (the annual interest on
$7040.82) for 45 terms of years as above, at
10 per cent, º * * * * * * , s =
º * * 3168.36,
Value of bond at maturity, - - - - - - - - - - $17250.00
694 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
3. , What is the present value of a $10000 bond due in 10 years, with 5 per
cent Semi-annual interest, in order that the investment may be equivalent to 10 per
cent per annum payable semi-annually, allowing simple interest on all the semi-
annual interests of the bond at 10 per cent 3 Ans. $7020.20.
OPERATION
To find the value of the Bond at maturity.
Face of bond, - tº a G gº tº gº tº gº º sº. tº sº tº - - - $10000
Semi-annual interest for 10 years at 5 per cent, - - s s *- tº E- s ſº-º: 5000
Interest on $250 (the semi-annual interest on $10000) for the 190 terms of 3 years (the
decreasing series from 19 to 1 inclusive) = 95 years at 10 per cent, - tº {-} sº 2375
Value of bond at maturity, - - - - - - - - - - $17375
OPERATION
To find the value of a $100 Bond for 10 years, semi-annual interest at 10 per cent.
Present value assumed, - se tº º * -º sº ºs Exe sº * s wº- * * - $100.00
Semi-annual interest for 10 years at 10 per cent, eme s ∈ º ºs ºs º gº 100.00
Interest on $5 (the semi-annual interest on the $100) for the 190 terms of 3 years (the
decreasing series from 19 to 1 inclusive) = 95 years at 10 per cent, * > tº gº 47.50
Value of $100 in 10 years at 10 per cent semi-annual interest, - tº tº - $247.50
OPERATION
To find the present value of the maturity value of the Bond.
$
100
247.50 || 17375
$7020.20 + Ans.
THE OFIERATION REVERSED.
Present value of bond, - wº º gº- º * wº &= º * - tº tº tº º ſº $7020.20
Semi-annual interest on same for 10 years at 10 per cent, * > * > ºs im tº º gºe 7020.20
Interest on $351.01 (the semi-annual interest on $7020.20) for 190 terms of 4 years = 95
years at 10 per cent, - " - tº e º 4- * * * . * * se wº ſº, & ſº 3334.60
Value of bond at maturity, sº - - - - - - - - - $17375.00
4. What is the present value of a $10000 bond due in 10 years, with 5 per
cent semi-annual interest, in order that the investment may be equivalent to 10 per
cent per annum, allowing compound interest on the semi-annual payments of inter-
est at 10 per cent 3 Ans. $7004.52.
OPERATION
To find the worth of the Bond at maturity.
Face of Bond, * tºp wº tºº {-ºr sº es tºº & &º * gº º ge gº $10000.00
Semi-annual interest on same for 10 years at 5 per cent, tº- gº º sº gº tº 5000.00
Compound interest on $250 (the semi-annual interest on $10000) compounded annually
at 10 per cent for the sum of the 3 yearly decreasing series of terms from 9% to #
years inclusive, - * sº gº tº sº tº gº * gº tº E tº º 3167.93
Value of bond at maturity, - - - - - - - - - - $18167.93
OPERATION
To find the present value of the maturity value of the Bond.
$2.5937425 ) $18167.9300000 ($7004.52 Ans.
117325000
135753000
60658750
We here divide the maturity valne by the compound amount of $1 for 10 years at 10 per cent,
3} PROBLEMS IN STOCKS AND BONDs. 695
Whigh, for convenience we obtain from the compound interest table on page 612. The result or
quotient, $7004.52, is the present value of the bond.
Pºplanation.-In the first part of the operation, to find the compound interest on the $250 for
the sum of the decreasing series of terms of # years, from 9% to 3, at 10 per cent annual compound
interest, to economize time we used the compound interest tables; but as our tables are calculated
9nly for the integral or round years, we find it necessary to separate the 4 yearly decreasing series
into two yearly series, viz.: A yearly integral series from 9 to 1 inclusive, and a yearly fractional
. from 9% to 1% inclusive, and also the final 3 year. We then find in the table, in the 10 per cent
COIll Iſln.
The compound amount of the integral series from 1 to 9 years to be, tºº tºº tº º $14,9374246
From which we deduct the sum of 9 additions of $1 principal, sº gº tºº gº 9.
And obtain, compound interest, - - - - - - - - - - - $5.9374246
Then to find the amount of the fractional series from 9 to 13:
We multiply the amount of the full series $14.9374246 by $1.05;
The product of this multiplication is $15.68429583, from which we deduct $9, the
sum of 9 additions of $1 principal, and obtain, compound interest, sºme 6.684295.83
To these sums we add the compound interest on $1 for 3 a year at 10 per cent,
which is * * ºn tº- sº tº gº & e º dº *E º ºs º .05
And thus obtain the total compound interest, - - - - - - - - $12.67172043
This we multiply by $250, the semi-annual interest on the bond, - gº * - * = 250
And produce, as the total annual compound interest on all the semi-annual inter-
ests of the $10000 bond, gº sº tº º tº- gº tº s tº mº gº $3167.9301075
TO FIND THE VALUE OF BONDs UNDER WARYING CONDITIONs,
STATED IN THE PROBLEMS.
1268. 1. What price must be paid for a $1000 bond bearing 4 per cent inter-
est payable annually, having 8 years to run, so that the investment may be equal
to 5 per cent 3 Ans. $935.367.
REMARKS 1.-In questions of this kind, in which there are no conditions
regarding, simple, annual or compound interest, compound interest is computed
upon the annual payments of interest on the bond. (But not on the interest of the
annual interest that may be realized on the investment).
When such rate is not specified, the rate given in the bond is taken in the
solution of the question.
2. The value of a bond in the market as an investment at a given rate,
depends, other things being equal, upon the rate per cent interest at which the
re-investments may be made of the various accruing interest payments.
3. When a bond is bought at a premium, some part of each interest install-
ment received on the bond must be reinvested at compound interest, and so con-
tinued until the maturity of the bond. The accumulations of this interest investment
must amount exactly to the premium paid on the bond. For when the bond is paid,
the face only will be received.
4. When a bond is bought at a discount, by the same reasoning as above, the
interest installments received will be in excess compared with the actual amount
paid for the bond, and hence there will be at each interest paying period a certain
amount that may be expended and yet leave at the maturity of the bond the exact
amount invested therein.
NOTE.-For the solution of problems for bond investments, tables have been prepared by
696
soul E's PHILOSOPHIC PRACTICAL MATHEMATICs. *
different authors, and are in general use by bankers, brokers, and investors,
These tables are
prepared on different plains, according to certain conditions of investing and re-investing, and by
their use various investing results are shown without the tedious computations involving the prin-
ciples of compound interest.
In the first solution of the above problem, we will make use of the Compound Interest Table,
page 611, and in the second solution we will use the table showing the Present Worth of an
annuity, and the table showing Present Value of $1, when time and rate per cent are given.
Index, Present Worth).
EIRST SOLUTION.
In this solution, we consider the problem as
an annuity of $40 per annum for 8 years, with a
payment of $1000 at the expiration of that time.
And since the present worth of an annuity may
be found by dividing the payment by the com-
pound amount of $1 for the specified time and
at the given rate per cent, hence we will so
divide each of the 8 payments on the bond.
Annuity Compound Yalue of
Payments. Amount of $1. Annuity.
1st $40 — $1.05 = $38.09
2d 40 –– 1.1025 = 36.28
3d 40 — 1.1576 = 34.55
4th 40 — 1.2155 = 32.90
5th 40 — 1.2763 = 31.34–H
6th 40 — 1.34 - 29.85–H
7th 40 — 1.4071 = 28.42
8th 1040 – 1.4775 = 703.93–H
Market value of bond on condi- $935.36+
tions named in the problem.
IPROOF.
Cost of bond, - - - , , , - - $935.36
Interest on same, 1 year at 5% tº 46.77
$982.13
Less interest on bond, 1st year, gº 40.00
Balance, - - - - - - $942.13
Interest on same, 2d year at 5% sº 47.11
$989.24
Less interest on bond, 2d year, {--> 40.00
Balance, - - - - - gº tº gº $949.24
Interest on same, 3d year at 5% * = 47.46
$996.70
Less interest on bond, 3d year, gº 40.00
Balance, - - - - - - - - $956.70
Interest on same, 4 years at 5% wº 47.84
$1004.54
Less interest on bond, 4th year, gº 40.00
Balance forward, - - - $ 964.54
(See
SECOND SOLUTION.
In this solution, we also resolve the question
into an annuity as in the first solution. Then
since the present worth of an annuity may be
found by multiplying the amount of an annuity
by the present worth of an annuity of $1 for
the time specified and the rate per cent given,
we therefore multiply the $40 annuity by
$6.463213, which is the present worth of an
annuity of $1 for 8 years at 5 per cent, as shown
in the annuity tables, which see. Thus we obtain
$258.52 as the value of the $40 annuity.
Then to find the present worth of the $1000
bond, we multiply the same by $.676839, which
is the present value of $1 for 8 years at 5 per
cent as shown in the table of present Worth,
which see.
By this we obtain as the present
worth of the bond, - " - - $676.84
Value of the annuity as above gº- 258.52--
Value of bond and annuities, - $935.36+
Balance brought forward, - - $ 964.54
Interest on same, 5th year 5% sº 48.23.
$1012.77
Less interest on bond, 5th year, - 40.00.
Balance, - - - - - - $972.77
Interest on same, 6th year at 5% - 48.64
$1021.41
Less interest on bond, 6th year, * } 40.00.
Balance, - - - - - - - $ 981.41
Interest on same, 7th year at 5% - 49.07
$1030.48.
Less interest on bond, 7th year, tº 40.00
$ 990.48
Interest on same, 8th year at 5% - 49.52
$1040.00
Less interest on bond, 8th year, tº a 40.00
Balance, - - - - - - $1000.00
Less bond redeemed, º gº tº 1000.00,
2. What price should be paid for a $500 bond bearing 6 per cent interest,
due in 9 years, so that the investment will produce 8 per cent income, no allowance.
to be made for the interest on the annual interest payments?
Ans. $447.67+.
4× PROBLEMS IN STOCKS AND BONDS. 697
2’
OPERATION.
Face of bond, $500.00 | Investment of - - - $500.00 500
Int, for 9 years at 6% 270.00 | Int. on same for 9 yrs. at 8% "360.00 860 | 770
Maturity value, $770.00 | Value of investment, tº $860.00 | — —
$447.67+3 cash val.
of bond.
FIRST PROOF.
Cash value of bond, $447,67+3 447.67+3 || $322.323#
Maturity value of bond, 770.00 9 || 100
Gain on cost of bond in 9 years, $322.32%; | 8% gain on investment.
SECOND PROOF.
Interest on $447.67 for 9 years at 8 per cent, - - - - - - - - - $322.33
Cash value of bond, sº - sº gº - sº tº º * * tº-3 * = * . 447.67
Maturity value of bond, ſº tº- *º * - tº sº tº ge s tº 6 º' sº $7000
º 3. If in the above problem, interest at 6 per cent is allowed on the annual
interest payments, what would be the market price of the bond to make the invest-
ment equal to 8 per cent simple interest ? Ans. $485.34%.
oPERATION.
Face of bond, - - - - - $500.00 | Investment of, $500.00 500
Int. for 9 yrs. at 6%, - - - 270.00 | Int. 9 yrs. at 8%, 360.00 860 834.80
Int, on annual int. ($30) for 36 yrs. 64.80 mºmºs º ºsmºs
, -— Value of investment, $860,00 $485.34%; val.
Value of bond at maturity, - - $834.80 of bond.
FIRST PROOF.
Market value of bond as above, - - - - - - - - - - - - $485.35
Interest on same for 9 years at 8 per cent, - - - - - - - - - - 349.45
Maturity value of bond, - - - - - - - - - - - - - $834.80
SECOND PROOF,
Cash value of bond, $485.34%; 349.45%;
Maturity value of investment, 834.80 ** 100
Gain on cost of bond in 9 years, $349.45%;
8% gain.
4. If in problem 2 above, annual interest at 6 per cent is allowed on the $30
annual interest payments on the bond, and annual interest at 8 per cent is also
allowed on the $40 annual interest payments on the investment, what would be the
market price of the bond to make the investment equal to 8 per cent?
Ans. $438.27—.
OPERATION
Face of bond, - - - - - $500.00 | Investment of, - ºp as sº - $500.00
Interest for 9 years at 6 per cent, - 270.00 || Int. for 9 years at 8 per cent, - - 360.00
Int. on annual int. ($30) for 36 yrs. - 64.80 | Int. on annual int. ($40) for 36 yrs, at 8%. 115.20
Value of bond at maturity, - - $834.80 | Val. of investment at the close of 9 yrs. $975.20
$500 investment.
975.20 | 834.80
$428.02 value, or market price of bond on conditions named.
698 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
PROOF,
Market value of bond, - gº * s º º tº * tº a sº tºº tº º tº $428.02
Interest on same, for 9 years at 8 per cent, - tº gº º tº sº gº º ºs & 308.17
Interest on $34.24 (which is the interest on $428.02 for 1 year at 8 per cent) for 36 years
(the series of 1 to 8 inclusive) at 8 per cent, iº tº ſ ºf tº º e. tº & e tºº 98.61
Maturity value of bond, - - - - - - - - - - - - - $834.80
|PRACTICAL MISCELLANEOUS EXAMPLES IN STOCKS AND BONDS.
1269. 1. When the annual dividend of stock is 12 per cent, and the interest
of money is 10 per cent, at what premium ought the stock to sell? Ans. 20%.
OPERATION INDICATED,
00
2
lo
| $120 — 100 = 20% Ans.
2. What would be my income by investing $4100 in 6 per cent stock at 82
per cent 3 Ans. $300.
3. I bought stock at 6 per cent below par, and sold the same at 24 per cent
above par. Paid brokerage 4 per cent for buying and 4 per cent for selling. My
gain in the transaction was $750. What was the amount of my investment, includ-
ing brokerage for buying, and what was the brokerage for buying and selling?
Ans. $9450 investment. $100 brokerage.
OPERATION.
6 % gain on purchase. $
23% gain on sale. 100
15 2
84% total gain. 750
1 % = 3% brok. for buying and 3% for selling.
$10000 par of stock.
7+% net gain. 69% discount.
$600.00 discount.
$10000 — $600 = $9400 + $50 brokerage for buying = $9450 investment.
Brokerage on $10000 at #96 for buying and 3% for selling = $100.
4. A merchant sells 6 per cent bonds at 96, and invests the proceeds in 8's
at par. How much per cent does he increase his income? Ans. 28%.
OPERATION.
$100 696 bonds. $96 8% bonds. $
6 8 1.68
* =º tºº (6.00 | 100
$6.00 income on 6% bonds. $7.68 income on 8% bonds.
6.00 28% Ans.
$1.68 gain.
* PROBLEMS IN STOCKS AND BONDS. 699
5. A merchant invested $13725 in Louisiana 6's at 914; he held them until
they rose to 94, then sold out and bought New Orleans City 7 per cent bonds at
70%. How much did he increase his annual income? Ans. $500.
OPERATION,
$ $
| 100 | 100
183 || 2 $15000 141 || 2
13725 94 14100
| $15000 $14100.00 $20000
% 7 O
$900.00 $1400.00 – $900 = $500 Ans.
The free Banking Law of Louisiana requires that the stocks deposited with the Auditor of
Public Accounts as security for bank note circulation “shall always be, or be made equal to stock
of the State bearing not less that 6 per cent interest.” The Auditor is prohibited from receiving
“stock or bonds at a rate above their par value, or above their market value.”
6. What amount of circulating notes could a bank receive on stock that
produces but 5 per cent? Ans. 833% of the par value of the stock deposited.
OPERATION.
$
100
6 || 5
| $835 – 83% Ans.
7. Suppose in the above example that the stock or bonds deposited had been
7 per cent, and the market value 90 per cent, what amount of circulating notes
Would the bank have received ? Ans. 105% of par value of stock.
OPERATION.
$100 par value. $
90% 100
-sº 6 || 6.30
$90.00 market value. -*
7% | $105 = 105% Ans.
$6.30 interest.
8. Suppose again that the stock or bonds deposited had been 8 per cent, and
the market value 110, what amount of circulating notes would the bank have
received ? Ans. 1334% of par value of stock.
OPERATION.
$
100
6 || 8
| $133} = 1334% Ans.
9. In the last question, had the Auditor been allowed to receive bonds at
their market value, which would have been a very impolitic transaction, what
amount of circulating notes would the bank have received ? •
Ans. 1463% of par value of stock.
OPERATION.
$100 alſ.
10 = 10% premium. $
-* 100
$110 market value. 6 || 8.80
8% —
$8.80 interest $1463 = 1463%
7OO SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. X.
10. I invested $2598.75 in 4 per cent bonds at 74%, and after receiving a
year's interest sold the bonds. My whole gain, including the year's interest, was
$315; at what price did I sell the bonds? Ans. 794%.
OPERATION.
$ Whole gain, $315.
| 100 Interest, 140.
297 || 4 - --
2598.75 Gain on sale, $175.
Cost of bonds, 2598.75
$3500 par value of bonds bought.
4% Selling price of bonds, $2773.75
$140.00 interest.
$2773.75 –– $3500 = $793 = 7.94% selling price of $100 par of bonds.
11. A merchant owns $40000 of 6 per cent railroad first mortgage bonds,
now selling at 85 per cent. He sells and invests the proceeds in 6 per cent bonds,
at 20 per cent discount. What will be his increase of income per year?
Ans. $150.
TO FIND DIVIDENDS AND THE VALUE OF STOCK AFTER DECLAR-.
ING CASH INSTALLMENT, STOCK, AND FORCED DIVIDENDS.
1270. 1. A capitalist subscribed $25000 stock in a bank, the capital stock
of which is $1,000,000, but only 50 per cent paid in. A cash dividend of 10 per cent
on the par value is declared and paid. What amount of money does the capitalist
receive? What would be the market value of the 50 per cent paid stock to make-
the investment equal to 12 per cent interest?
Ans. $2500 dividend received. $834 market value of stock.
FIRST OPERATION. SECOND OPERATION.
$ $
100 50
1 year's int. 12 || 20 one year's div. 6 || 10
$1663 - 2 = $83} 83% market value.
2. Suppose in the above example, that an installment dividend of 10 per cent.
had been declared and credited to the stockholders on their unpaid stock, and that
the market value of the stock before the declaring of the dividend was $60, what.
would be the value of the stock after the dividend had been declared?
Ans. $72.
OPERATION.
|;
50 || 60
lsº am.
PROBLEMS IN STOCKS AND BONDS. 7or
3. Suppose again that the stock with 50 per cent paid, was worth in the
market 60 per cent of its par, and that a stock dividend of 10 per cent be declared
and delivered to the stockholders, what would then be the value of the stock #
Ans. 54.4%.
OPERATION.
$ $
60 120
55 50 110 || 100
OF
$54, Ans. y $109 ºr value of par - 2 = $54+r Ans.
4. Suppose again that the stock with 50 per cent paid, was worth in the
market 60 per cent of its par, and that a forced dividend of 10 per cent be declared
and paid to the stockholders, what would then be the value of the stock?
Ans. 54%.
OPERATION.
$
120
90
$108 value of $100 par - 2 = $54 = 54%.
$
60
50 || 45 Or, 100
| $54 = 54% Ans.
5. Sections 15 and 16 of the United States National Banking law require
that every banking association, after having complied with the provisions of the
Banking Act, preliminary to the commencement of banking business under its
provisions, shall transfer and deliver to the Treasurer of the United States any
United States bonds bearing interest, to an amount not less than one-third of the
capital stock paid in, and that the banking association making such deposit shall
be entitled to receive from the Controller of Currency circulating notes equal in
amount to ninety per cent of the current market value of the United States bonds
so transferred and delivered, but not exceeding the par value thereof, if bearing
interest at the rate of six per cent, or of equivalent United States bonds bearing a
less rate of interest.
In accordance with this law a bank deposited with the United States Treasurer
$800000 United States 6 per cent bonds, the current market value of which was
par. What amount of circulating notes can the bank receive from the United
States Controller of Currency Ans. $720000.
OPERATION.
$800000
0%
$720000.00 Ans.
6. Suppose in problem 5, that the current market value of the United States
6 per cent bonds deposited with the United States Treasurer, had been 110, what
amount of circulating notes would have been received? Ans. $792000.
OPERATION,
$110 market value. $ $800000
6% interest 90 $110 99%
6 || 6.60 or, 90 ºmmº-
$6.60 $792000.00 Ans.
$99.00 = 99% $99.00 = 99%
7oz soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. *
7. Suppose in problem 5 that the United States bonds deposited had been
United States 44 per cent, at a current market value of 114, what would have been
the amount of circulating notes received? Ans. $615600.
OPERATION.
$114 market value. $ $114 $800000
43% interest. 90 90 76.95
*==ºms -sºmimsº 6 || 5.13 Or, gºmºmºmºs ºmºmºmºms gººmsºmºmºmºmºmºmºmº
$5.13 interest. $102.60 $615600.0000
$76.95 4%
6) $461.70
$76.95
8. Suppose again in problem 5, that the United States bonds deposited had
been United States 5 per cent bonds, the current market value of which was 84 per
cent, what would have been the amount of circulating notes received ?
Ans. $504000.
OPERATION.
84 market value.
5% rate of interest.
$4.20 interest.
84
90 $ $800000
*== ! 90 63%
90 or, 7560 Or, | 84 gºmmºnsººmmismº
6 || 4.20 5 6 || 5 $504000 Ans.
$63.00 = 63% 6) 37800 $63.00 = 63%
$63.00 = 63%
9. Suppose again in problem 5, that the $800000 United States bonds depos-
ited had been United States 6 per cent bonds, the current market value of which
was $1134, what would have been the amount of circulating notes received ?
Ans. $800000.
OPERATION.
$1.13% market value. $ Ea:planation.—In this problem we find that
6 % 90 the 6 per cent on the market value of the
6 | 6.81 bonds is equal to 61% ºf per cent on the par, and
$6.81 interest. * hence, if 6 per cent par bonds give 90 per
$102.15 = 102*f;96 cent issue of circulating notes, 6 ºf per cent
will give 102.1% per cent issue; but in accor-
dance with the law governing the case, as shown in the above extract, the Controller is allowed to
issue only 100 per cent of circulating notes on the par of the bonds deposited.
10. A merchant subscribed $10000 in a banking incorporation, the paid up
capital of which was $400000. A cash dividend of 4 per cent on the par value of
the stock was declared. The paid up capital being 40 per cent of the capital stock,
what per cent gain dots he receive on his investment, and what is the capital of the
bank? Ans. 10% gain. $1,000,000 capital.
4 ºf PROBLEMS IN STOCKS AND BONDS. 7O3
TO CONSOLIDATE CORPORATIONS, WATER THE STOCK AND FIND
TEIE MARKET VALUE OF THE WATERED STOCK.
1271. 1. Two railroad companies, A. with a capital of $25,000,000, and X.
with a capital of $15,000,000 consolidated under the corporate name of A. & X. R.
R. Co. New stock was issued to redeem the old in the following manner:
(1). For every four shares of the stock of the A. road, five shares were issued.
(2). The stock of the X. road was watered 20 per cent.
What was the capital stock of the Consolidated Co., and how many shares of
watered stock were added? Ans. $49,250,000 capital stock of Consolidated Co,
92,500 shares watered stock.
OPERATION,
250000 = Shares of CO. A. 150000 = shares of Co. X.
62500 = shares of watered stock. 30000 = 20% watered stock.
312500 = shares issued. 180000 = shares issued.
180000
492500 = shares at $100 = $49250000 capital of Consolidated Co.
62500 + 30000 = 92500 shares watered stock.
2. In the above problem, suppose the market value of the stock of the A.
road to have been before the consolidation, 120, and that of the X. road to have been
85, what would a share of the new consolidated company be worth, in the market?
FIRST OPERATION.
1 share $100 of pure stock of Co. A. = 120 in the market.
1+ shares $125 of watered stock of Co. A. = 120 in the market.
Hence º = $96 the value of 1 share watered stock of Co. A.
1 share $100 of pure stock of Co. X. = 85 in the market.
1 share $120 of watered stock of Co. X. = 85 in the market.
Hence *** = $70% the value of 1 share watered stock of Co. X.
$96 -– $703 = $166; -- 2 = $831; the value of 1 share of consolidated stock,
SECOND OPERATION.
$25,000,000 = capital of Co. A. 120
6,250,000 = watered stock of Co. A. 31250000 || 25000000
$31,250,000 = watered capital of Co. A. $96 value of one share.
$15000000 = capital of Co. X. 85 6.
3000000 = watered stock of Co. X. 18000000 || 15000000
$18000000 = watered capital of Co. X. $70; value of one share.
$96 -- 703 = 1663 – 2 = $83F2 value of one share of consolidated stock.
3. The paid up capital stock of a newly incorporated company is $300000.
The business of the company gives evidence of very large profits, and as a conse-
quence the market value of the stock is high, say $220 per share, and the demand
greater than the supply. The stockholders therefore, in order to supply this demand
7O4 soule's PHILOSOPHIC PRACTICAL MATHEMATICS. *
and enrich themselves, clandestinely resolve to “water the stock” to the extent of
$200000. Accordingly they issue and divide pro rata among the stockholders
$200000 new stock, thus increasing the capital to $500000. What would be the
market value of the stock after it was thus “watered”? Ans. $132.
OPERATION.
$100 original stock worth $220, , $300000 original stock.
66; watered stock. 200000 watered stock.
$1663 present stock. $500000 present stock.
$ $
220 220
1663 100 or, 500000 || 300000
| $132 Ans. - $132 Ans.
4. The capital stock of a corporation consists of $600000 regular, $200000
preferred, and $200000 guaranteed stock. A semi-annual dividend of 3 per cent is
declared on the regular stock, 5 per cent on the preferred stock, and 6 per cent on
the guaranteed stock. $7562.10 were carried to reserve fund. What was the net
gain of the company during the six months business? Ans. $47562.10.
To FIND THE MARGIN TO CARRY STOCK.
1272. 1. A speculator deposited $7000 margin with a broker and instructed
him to purchase 500 shares of X. insurance stock and 150 shares of X. bank Stock.
The margin is to be kept good at 10 per cent. The broker purchased at once the 500
shares of insurance stock at 90, and the 150 shares of bank stock at 110. At the
close of six days, the insurance stock had depreciated 4 points and the bank stock
had advanced 2 points. What additional sum must be deposited with the broker
to maintain the margin at 10 per cent, allowing # per cent commission for buying, #
per cent for selling, and 6 per cent interest? AnS. $1579.66.
OPERATION.
Dr. S P E C U L A T O R. Cr.
To 500 shares at 90 - - - $45000.00 | By Int. on $7000 margin, 6 ds, at 6% 7.00
** 150 shares at 110 tº * - 16500.00 | “ Value of Stock this date:
** Commission on 650 500 shares at 86 º - - - $43000.00
shares (a)"# per cent, - - 162.50 | 150 shares at 112 - - - - 16800.00
“ Int, on $61662.50, 6 ds. at 6%, 61.66 *s-º
“ Commission for selling, - - 162.50 $59807.00
Total indebtedness to date, $61886.66
Credit as per opposite, - 59807.00
Loss to date, - - - $2079.66
650 shares at $100 par = $65000 at 10% = $6500, margin required.
Loss to date, 2079.66 to be made good.
Margin required, $8579.66
Margin deposited, 7000.00
Additional margin required, $1579.66
NOTE. 1.-In making statements of this kind, the commission for selling is considered due to
the broker on the date the statement is made. For in case the required additional margin was not
furnished to the broker, he would have the right to sell the stock at once and thus the commission
would be earned.
NOTE_2.—Margin is estimated on the par value, unless otherwise agreed. The margin is to
protect the broker in case the stocks should decline in value.
* PROBLEMS IN STOCKS AND BONDS. 7 of
ACCOUNTS CURRENT AND INTEREST ACCOUNTS WITH BROKERS.
1273. June 19, 1895, Frank Soulé placed $5000 as margin with H. Lee, his
broker, and instructed him to purchase the following stocks: 100 shares of Crescent
City IR. R. stock; 200 shares of Metropolitan Bank stock, and 50 shares of Sun
Mutual Insurance stock. June 21st the broker bought on account of the above
order, 70 shares Crescent City R. R. stock at 105; 115 shares Metropolitan Bank
Stock at 165 and 30 shares Sun Mutual Insurance stock at 1124.
June 24, he bought 30 shares of Crescent City R. R. stock at 104; ; 85 shares
of Metropolitan Bank stock at 1684 and 20 shares of Sun Mutual Insurance stock
at 111.
June 25, pursuant to instructions, the broker sold 175 shares of Metropolitan
Bank stock at 1703; and June 28, he sold 50 shares Crescent City R. R. stock at 1043.
July 3, a settlement was made with the broker. Make the bill of the broker
showing balance of account, allowing interest on margin, on commission and on
purchases and sales at 8 per cent, and # per cent broker's commission on both
purchase and sale of stocks.
OPERATION.
Dr. Frank Soulé, in account with H. Lee, to July 3/95 at 8 per cent. Cr.
Amount. || Ds. || Int, Amount. || Ds. Int.
1895 1895
June 21|70 sh. C. C. R. R. June |19|Cash deposited as
stk. (a) 105, 7350|00 margin, 5000|00
21|115 sh. Metropol- Int. on same, 14
itan Bk. stk. (a) ds. (a) 8%, 14 || 15|56
165, 1897500 25|175 sh. Metropol-
21|30 sh. Sun Insur- itan Bk. stk. (a)
ancestk. (a) 112}|| 337500 * 1704, 2.9881|25
21|Com. on 215sh. Ø Int. on same, 8
3. 96, 53|75 ds. (a) 8%, 8 || 53|12
21|Int, on $29753.75 28|50 Sh. C. C. R. R.
12 ds. (a) 8%, 12 79|34 stk. (a) 104%, 522500
24|30 sh. C. C. R. R. Int. On same, 5
Stk. (a) 104%, 3135|00 ds. (a) 8%, 5 5|81
24|85, sh. Metropol-
itan Bk. stk. (a) Tot. credit int. 74|49 74|49
1684, 14322.50 July 11|By Balance, 9459|36
24|20 sh. Sun Insur-
ance stk, a 111, 222000
24|Com on 135 sh. (a)
#%, 3375
24|Int. on $19711.25
9 ds. (a) 8%, * 9 39|42
25|Com. on 175 sh,
(a) +%, 43|75|| 8 08
28||Com. 50 sh. (a)
#%, 12|50|| 5 01
Tot. debit int. 11885 118|85
* 49640.10|| 49640|10




NOTE: 1.—Commission is due as soon as a purchase or sale is made, and will draw interest
from said time.
NoTE 2.—See Accounts Current and Interest Accounts, in this book, pages 653 to 661.
7O6 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. \ jºr
ACCOUNT CURRENT WITH BROKERS, WITH “SHORT SALES.”
1274. 1. May 13, L. C. Soulé, deposited with his broker, S. Simon, $10000 and
instructed him to sell short S. 60, on his account, 100 shares Citizen's Bank stock
and 300 shares Carrollton R. R. stock. May 15th, said stock was sold at the follow-
ing prices: Citizen's Bank stock at 85; Carrollton R. R. stock at 1094. July 10,
notice was given to the buyer and stock was bought to “cover” short sales at the
following prices: Citizen's Bank stock at 81; 200 shares Carrollton R. R. stock at
107, and 100 shares of same at 1054.
Make the account showing balance due July 11, allowing 8 per cent interest
on margin and on commission, but not on short sales, and compute the commission
on sales and purchases each at # per cent? Ans. $1245.92 bal. due L. C. Soulé.
Account which the student will verify:
Dr. L. C. Soulé, in account with S. Simon, to July 11/95, at 8 per cent. Cr.

Amount. || Ds. Int. Amount. || Ds. Int.
1895 1895
May 15|Com. on 400 sh. May 13|Cash deposit as
Ø 4%, 10000 margin, 1000000
Int. on $100. 57 15|100 sh. Citizen’s
ds. (a) 8%, 57 1|27 Bk. stk. (a) 85, 850000
July |10|100 sh. Citizen's 15|300 sh. Carrollton
Bk. stk. (a) 81, 810000 R. R. stk. (a)
10|200 sh. Carrollton 1094, 32850|00
R. R. Stºk. (a) 107|| 21400|00 July 11|Int, on $10000
10|100 sh. Carrollton margin 59 d.s.
R. R. stk. (a) 8%, 59 || 131|11
105+, 1052500
10|Com. on 400 sh. (a) 131|11 131|11
#%, 10000
Int. on $40125, 1
day @ 8%, 1 8|92
10|19 10|19
To Balance, 1245||92
41481|11 41481|11
NOTE 1.-Brokers do not allow interest on “short sales,” as there is no payment or receipt
of money.
NOTE. 2.-It is the custom of brokers to allow interest on cash sales, and to charge interest
on purchases and also on the commission for buying and selling regardless of the condition of sale.
REMARK.—See Accounts Current and Interest Accounts, pages 653 to 661.
2. August the 1st, 1895, a capitalist placed in the hands of Z., a broker,
$25000 for investment and speculation in buying and selling gold, stocks, and
bonds, for 60 days.
Z. immediately purchased 200 shares of Texas Railroad stock at 60, “b.
26,” no margin required, and sold “short” 400 shares of New Orleans, Mobile and
Chattanooga Railroad stock at 1024, “b. 10.” Five days after Z. called in the
Texas Railroad stock and sold it to A. at 61%; August 5, he bought $20000 of five
X- PROBLEMS IN STOCKS AND BONDS. 707
twenties at 1094; at the same time the party to whom the New Orleans, Mobile and
Chattanooga Railroad stock was sold, called it in; Z. paid the difference, the stock
being valued at 1023; August 24, Z. sold the five twenties at 110+ cash, and at the
Same time made a further sale of $15000, in the same kind of bonds, at 1104, “s.
20," which he purchases and delivers in eight days, at 109%; September 17, Z. sold
$20000 in gold at 1374, “s. 10, " which was not delivered at the expiration of the
ten days, but settled at 137; Z, then bought “flat” $25000 City 8's at 91, and sold
the same “flat” at 954.
The capitalist and broker now settled.
By the terms of the contract, Z. is to receive # per cent brokerage on the par
of actual purchases and sales, no interest is to be allowed. What amount of money
is due the capitalist? Ans. $26375.
OPERATION.
Capitalist in account with Z. Dr. Cr. .
By cash deposited, gº º gº º $º gº se wº #Et sº 4 * * $25000.00
To 200 shares Texas R. R. stock at 60, t- * --> tº- sº cº - $12000.00
“. .9% brokerage on same, tº tº- º gº tºº ſº sº º 25.00
By 400 *:::. O. M. & Chattanooga R. R. stock at 1024, short b. 10,
$4 tº
To #96 brokerage on same, tº e º sº. tº º sº $º º 50.00
By 200 shares Texas R. R. stock at 61%, * * gº tº gº tº º 12375.00
To #% brokerage on same, - &= gº gºe tº º tº-> *-*. * - 25.00
“ $20000 U. S. 5–20's at 1094, & gº tº tº sº eme sº mº 21825.00
“ #96 brokerage on same, - - tº as tº º gº º ºs 25.00
“ loss on 400 shares N. O. M. & Chattanooga R. R. stock, bought
at 102} and settled at 1024, $41150, se tº e º º ºs 150.00
By $20000 U. S. 5–20's at 110+, tº º sº º º & * > gº tº 22050.00
To #96 brokerage on same, tº sº tº º tº- tº º 25.00
By $15000 U. S. 5–20's at 110}, sº em º gº ºn a tº gº º 16518.75
To #96 brokerage on same, tº º tº ºne ſº & E º 18.75
“ $15000 U. S. 5–20's at 1094, - - - tº ſº tº gº tº 16481.25
“ #96 brokerage on same, - - - - - - - - - 18.75
By $20000 gold, s. 10, at 1374, $27500.
To #96 brokerage on same, * tº º ºs sº ºne º º 25.00
By gain on $20000 gold, bought at 137} and settled at 137 ($27.400), * 100.00
To $25000 City Warrants at 91, - tºº gº * cº gº gº tº 22750.00
“ #96 brokerage on same, - - - - - - - - - 31.25
By $25000 City Warrants at 954, - - - - is tº & ºne tº 238.12.50
To #96 brokerage on same, * * * > tº º sº me tº 31.25
Capitalist's total debits, tº tº dº dº- tºº cº * * dº $73481.25
Capitalist's total credits, sº º sº ºr me tº dº sº $99856.25
73481.25
Amount due the capitalist, - tºº tº Gº tº º & tº º $26375.00
UNITED STATES STOCKS AND BONDS IN EUROPE.
1275. The principal markets in Europe, for United States stocks and bonds,
are London, Paris, Hamburg, and Frankfort.
The European quotations are the prices of $100 bonds or stocks in U. S. gold,
but not at the U. S. valuation, as will be seen in the two following articles. The
accrued interest on the stock or bond at the date of purchase generally constitutes
\ wº
708 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. º
a part of its value. In some places however they are sold “flat” and then the
accrued interest must be added.
The London quotations of United States stocks and bonds are based upon an
assumed value of the pound sterling. This assumed value is $5, modified by the
rate of exchange. Hence to find the actual value of English quotations of Ameri-
can Securities, we must multiply the quoted price by the rate of exchange. Thus,
if Illinois Central R. R. stock is quoted in London at 103; and the rate of exchange
is 4.87; we find the actual value of the quotation by the following operation:
103.75 × 4.875
łº
3)
In Paris at the Bourse, the quotations are based upon the conventional value
of 5 francs for $1, while the actual value of a dollar in exchange transactions is
more than 5 francs. Hence to find the true American value of the Paris Bourse
quotations, we must deduct therefrom the exchange rate.
The German price of American Securities is the marks given for $100 par of
American stocks or bonds. The rate of German exchange is based, by American
bankers, upon the equivalent value of 4 marks expressed in dollars and cents. The
exchange par of 4 marks is 9542; the rate is 95% more or less according as premium
is eharged or discount allowed. See German Exchange.
To ascertain the value of bonds for remittance and sale in Europe, bankers
base their computations upon the buying rate of exchange.
To ascertain the net proceeds of American stocks or bonds sold in Europe,
allowances must be made for cost of remittance, insurance and commission; and if
bills of exchange are drawn against the net proceeds, allowance must also be made
for the selling rate of exchange.
TO FIND THE PRICE OF STOCKS IN THE UNITED STATES, WHEN THE
ENGLISH QUOTATION AND THE RATE OF EXCELANGE ARE GIVEN.
PROBLEMIS.
*
1276. 1. Chicago, Milwaukee and St. Paul Railway stock is quoted in London
at 63% and the rate of exchange is 4,864. What is the true value of the stock?
Ans. $62,03.
OPERATION INDICATED.
63.75 4 || 255 63.75 × 4.865
5 || 4.865 OT, 2 || 973 or, –
te 5 5
2. Louisiana 4's are quoted in London at 96%, exchange is 4,854. What is
the equivalent American quotation? Ans. $94,02.
4× PROBLEMS IN STOCKS AND BONDS. 7O9
TO FIND THE PRICE OF UNITED STATES SECURITIES IN LONDON,
WHEN THE AMERICAN QUOTATION AND THE RATE OF
EXCEIANGE ARE GIVEN.
PROBLEMIS.
1277. 1. New York Central and Hudson River Railway stock is quoted in
New York at 112}, exchange is 4.863. What is the equivalent price in London?
Ans. $115.20+.
OPERATION INI) ICATED,
112.125 112.125 × 5.00
4.86625 || 5.00 or,
4.86625
2. When the Southern Pacific 1st mortgage 6 per cent bonds are quoted in
New Orleans at 1214, exchange being 4.87%, what should be the equivalent London
quotation? Ans. $124,294-.
OPERATION INDICATED.
- || 121.25 121.25 × 5.00
4,8775 || 5.00 Or, --ms-
4,8775
OPERATIONS IN FRENCEL AND AMERICAN QUOTATIONS.
PROBLEMIS.
1278. 1. If the Paris Bourse quotations of United States 4 per cent bonds
are 984, exchange 5.15, what would be the equivalent value in the United States
market? Ans. $95.63+.
OPERATION.
The $984 Paris Bourse quotation or price being based upon the conventional
value of 5 francs for $1, which is too high a value, it must therefore be reduced in
the ratio of 5.15 francs (the exchange value of $1) for 5 francs.
Statement to thus reduce it.
5 × 98.50
1
5.15 5.00 or, –—
98.50 5.15
$95.631 U. S. market value of Bourse quotation.
2. If the Paris Bourse quotation of United States bonds is 1063, exchange
5.644, what would be the equivalent value in the United States market?
Ans. $94.552--.
OPERATION INDICATED.
$
1 106.75 × 5.00
11.29 2 Or,
5.00 5.645
4 || 427
$94,552+ U. S. market value.
See Foreign Exchange in this work for Extended Operations in Exchange.
ſº – ;
(Exchange.
•e--------------------RN
—-º'-- * *
w-wºr-w
1279. The term Exchange has various significations or meanings, according
to the Sense or manner in which it is used. The general acceptation of the term is
the act of giving or receiving one thing for another. It is also used to designate
the place where merchants, brokers, or other classes of business men assemble to
transact business at certain hours as the Cotton Exchange, the Produce Exchange,
etc. When used in this sense, the term is often contracted into “change”; as at
change or on change, meaning at the exchange.
EXCHANGE, when the term is used by commercial men and bankers in con-
nection with transactions of finance, signifies the method or system of paying debts
in distant cities by means of BILLS OF EXCHANGE, without the shipment or trans-
mission of money. In this sense we here use the term, and now treat the subject.
To elucidate the workings and to show the practical utility of exchange, let us suppose, 1.
That the sugar and cotton merchants of New Orleans ship their sugar and cotton to New York for
sale, and receive in payment currency or coin, which is forwarded to them by express. 2. Suppose
that the dry goods and grocery merchants of New York ship their goods to New Orleans for sale,
and receive in payment currency or coin, which is forwarded to them by express. Now it is clear
that if the sugar and cotton purchasers in New York had paid the grocery and dry goods merchants
there, and the grocery and dry goods purchasers in New Orleans had paid the sugar and cotton
merchants there, the whole business might have been settled without the expense and risk that
attend the shipping of currency or coin from one place to the other. This would be accomplished
by the sugar and cotton merchants drawing drafts or bills of exchange on the sugar and cotton
buyers in New York in favor of the grocery and dry goods buyers of New Orleans, who having
loought and paid for these drafts, could remit them to the grocery and dry goods sellers in New
York, in payment of their purchases. On presentation to the sugar and cotton buyers, these drafts
would be paid, and in this manner the debts due from New York to New Orleans and the debts due
from New Orleans to New York are both paid without the double shipment or transmission of
eurrency or coin from each place to the other.
It is evident that to make this system general, and to include all kinds of trade in various
places, it would frequently be very difficult for those having bills of exchange to sell, to find
purchasers, and viee versa. To obviate this difficulty, exchange brokers and exchange banks are
established. \
They buy bills with their own capital from those who have money to receive, and send them
forward for credit, and then sell their own bills drawn against this credit in such amounts as
purchasers may require. *
The exchange broker and exchange bank continue the purchase and sale of bills of exchange
in this manner, charging a commission only for their services, use of capital, and risk incurred in
buying bills that might not be honored when presented, until the demand to purchase bills becomes
greater than the supply of bills offered for sale or vice versa. Then they either sell their bills at a
premium or par, and buy at par or discount.
When the demand is in excess of the supply, they will charge not only a commission, but a.
premium sufficient to cover the expense of freight and insurance to ship the funds to meet their
(710)
* EXCHANGE. 7 II
bills at maturity, or to pay the interest on the amount of their over-drafts until the commerce of
the two places shall have adjusted the difference. They would also, in this case, buy at a premium.
When the supply is in excess of the demand, they will sell at a discount sufficient to cover the
expenses of shipping their funds from the place where the drafts are payable to their own place of
business.
The principal cause that produces the excess of supply or demand of bills of exchange, to
which we have here made mention, is the
BALANCE OF TRADE.
1280. Over-trading, or the balance of trade, is in itself a very important
subject, and is worthy of the profoundest consideration by business men.
In this connection we do not propose to discuss the subject, but for the infor-
mation of students we will briefly define it. By the “BALANCE OF TRADE'' is
meant the monetary excess of business transactions that one country or city has
over another with which she deals. For example, if the United States sells more
to England than England sells to the United States, then the balance of trade is in
favor of the United States. If the reverse is the case, then the balance of trade is
against the United States. Or, if New Orleans sells more to St. Louis than she
buys from St. Louis, then the balance of trade is in her favor; but if the reverse is
the case, then the balance of trade is in favor of St. Louis.
1281. ORIGIN OF EXCELANGE,
The origin of negotiable bills of exchange has been a subject of discussion on the part of
legal and commercial writers for ages, and it is yet undecided in what age and among what nation
they were first used. Some claim that they were known among the Romans and Grecians, but
there is no proof to substantiate the claim. It is granted however, on good authority, that such a
transaction as a request by A., in Rome, to B., in Alexandria, to pay to C. on his, A’s account, the
money that B. owes A., was of common occurrence; but a request of this character, and a com-
pliance there with, do not constitute a bill of exchange. ,
In 1272 bills of exchange were in use in Venice, but there is no evidence to show that they
were then negotiable. There are, however, records of them at an earlier date, and various theories
have been presented, on questionable evidence, to account for their origin. The French writers
ascribe their origin to the Jews in 1181, when oppressively and wickedly expelled from their
homes, in the kingdom of France, by Philip Augustus. But this date and circumstance, or cause of
origin, is questioned by other writers. Mr. Reddie, in his Historical View of the Law of Maritime
Commerce, says, “that if we inquire what cause has led to the invention, the true answer is, the
necessities of commerce; but that if we inquire who were the inventors, in what position and by
whom these necessities were most strongly felt, and what persons, experiencing the urgency of
these necessities in the most lively manner, produced the thing invented, it would be absurd to
call the extension of commerce the inventor, for this would be to confound the mover with the
agent. It is also highly probable that the Jews, being in these ages, as we have seen, the chief
money lenders, persecuted from mistaken religious views and on account of their alleged pecuniary
extortions, scattered over the European kingdoms, yet in a manner forced to keep up a pretty con-
stant communication with each other, clever and acute naturally, and comparatively skillful in
such business from having been trained to it for generations, were really the first inventors of bills
of exchange in a rude state.” But he concludes that there is no evidence that they made the
invention at the precise time and by reason of their expulsion from France.
712 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. º:
1282. THE IMPORTANCE OF EXCHANGE.
Money was the first great Invention by or through which commercial transactions were
extended and facilitated.
It answered, in the language of Mr. Parson, in his treatise on Notes and Bills, “all the
purposes for which it was wanted, for many ages. It satisfied all the requirements of social life,
and of commerce, through the early Eastern empires, and those of Greece and Rome. It is said,
but upon somewhat doubtful authority, that some kind of paper money was used in Tartary, or
China, or Japan, a thousand years ago; but nothing is known certainly about this. In Europe,
gold and silver money were the only circulating medium, and were sufficient; but five or six hun-
dred years ago, the discovery was made of a new circulating medium, bills of exchange, of which
it is the characteristic quality that it represents that which represents everything else.
“The use of money enlarged human intercourse, or so much of it as may be included in the
widest sense of the word commerce. It made interchanges possible and easy, which would other-
wise have been very difficult, if not impossible. We cannot imagine, for example, the whole
commerce of Greece and Rome, or a hundredth part of it, carried on by actual barter of commodi-
ties. Precisely in the same way, the invention and use of paper to represent money gave a new
enlargement to commercial intercourse, and greatly increased its facilities and its possibilities.
For we could not now suppose the commercial intercourse between America and Europe, for
example, to be carried on wholly by actual exchange of the precious metals, as must be the case
if bills and notes were abandoned, without a cost and hindrance which would be fatal to a very
large part of it. The invention and use of money conferred upon mankind the vast benefits which
have ever flowed therefrom, because money represented all other commodities, and for no other
reason whatever. He who had any superfluities on hand was no longer obliged to take the trouble
of storing them and the risk of their destruction, or to save them by exchanging them for the
superfluities of some accessible neighbor, whether these were precisely what he wanted or not.
For now he might sell his superfluities, and their value was then invested in something easily
preserved, and which could always be exchanged for the very article he wanted, as soon as he
found it within his reach. But after a time, this exchange was to be made in such quantities, and
at such distances, that it cost too much in time and trouble to be profitable; and here is a natural
limit to commerce by mere money, which seems to have been reached by the nations of Europe
some few centuries ago.
“Beyond this, therefore, it is plain that commerce could not have grown, unless new facili-
ties, by means of new instruments, had been provided for it; and this was done by the invention
and use of bills of exchange. Just as, by the help of money, a hundred oxen could be exchanged
for a hundred pieces of cloth, at distances which would have made the actual transfer of the oxen
and the cloth too onerous to be advantageous, so now commercial transactions which would have
required large bags or boxes of money to be sent back and forth at great cost, both of time and
money, and with much trouble and some hazard, can be carried into full effect, with equal prompti-
tude, safety, and facility, by exchanging small pieces of paper. And these two inventions, one
made at the beginning of human society and the other but a few centuries since, are useful for
precisely the same reason: Money represents all commodities, and 80 prevents the necessity of an actual
exchange of commodities; and bills and notes represent money, and so prevent the necessity of an actual
transfer of money.
“Moreover, as gold and silver were first used by weight, and it was a distinct though very
speedy improvement to coin them into money, so bills of exchange were first used only for the
benefit of a specified payee, but were soon perfected into the indispensable instrument of commerce
which they now are by being made negotiable. For this adds an entirely new element to their
utility. By means of indorsements, which may be extended indefinitely, negotiable paper not
merely makes money in one place or at one time, but becomes money in another place or at another
time, without actual transfer; and not merely makes credits the equivalent of money, but it
represents and carries with it the accumulated credit of all who become parties to the paper.”
* EXCHANGE. 7I 3
BILLS OF EXCHANGE.
1283. A Bill of Exchange is a written order for the payment of a specified
sum of money unconditionally. (The bill may be printed with the exception of the
signature of the drawer, which must be written).
Bills of exchange are divided into two classes, Domestic or Inland, and
Foreign.
A domestic or inland bill is one drawn and made payable in the same country.
A foreign bill is one drawn in one country and made payable in another.
NOTE.-It is generally understood that a bill drawn in any city and State of the United
States and made payable in any other city and State of the United States, is a domestic or inland
bill; but it has been decided by our courts, that a bill drawn in one State and made payable in
another, is a foreign bill, for the reason that each State is a sovereign government in many
particulars. But nothwithstandin g this legal decision, in commercial parlance, domestic exchange
means exchange on any of the States and territories of the United States.
...For the form of Bills of Exchange, Drafts, Notes, etc., and for an explanation of the parties
#.# ºrge, and for other forms of commercial instruments of writing, see pages 549 to
OI UIllS OOOK.
1284. SET OF FOREIGN BILLS OF EXCELANGE.

361800.9.7. NEW ORLEANs, April 22, 1895.
Sixty days after sight of this First of Exchange (second and third
unpaid), pay to the order of J. M. Butchee & Co., Eighteen Hundred Pounds,
Nine Shillings and Seven Pence, Sterling.
TO MESSRs. QUEALY & HARVEY,
Liverpool, Eng. MEAD & WELSH.

3.1800.9.7. NEw ORLEANs, April 22, 1895.
Sixty days after sight of this Second of Exchange (first and third
unpaid), pay to the order of J. M. Butchee & Co., Eighteen Hundred Pounds,
Nine Shillings and Seven Pence, Sterling.
To MEssRs. QUEALY & HARVEY,
Iliverpool, Eng. MEAID & WELSH.

39.1800.9.7. NEw ORLEANs, April 22, 1895.
Sixty days after sight of this Third of Exchange (first and second
unpaid), pay to the order of J. M. Butchee & Co., Eighteen Hundred Pounds,
Nine Shillings and Seven Pence, Sterling.
TO MESSRs. QUEALY & HARVEY,
Liverpool, Eng. MEAD & WELSH.
NOTE.-Exchange is drawn in sets in order to secure greater safety in transmission through
the mails. All the bills of a set are of the same tenor and date, and are sent by different mails
and when either is received and paid the other or others become void. In domestic exchange, it is
a recent custom of many banks to fill but one exchange, and if that should be lost, then to fill the
duplicate. In foreign exchange, many merchants mail by different steamers two of the set and
hold the third.
7 I4 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. . A.
LETTERS OF CREDIT.
. . 1285. A Letter of Credit is a written instrument drawn by a banker,
instructing any of several correspondent banks in different cities and countries, to
pay to the holder such sums of money as he may wish, not exceeding the amount
specified in the Letter of Credit, and promising to re-pay the sums so advaneed to
the person or persons advancin g them, or to honor his or their bills drawn for such
amounts as they may have individually furnished.
... . N9 tº 1.-Letters of Credit are drawn in different forms according to the ideas of the parties
issuing them, and the requirements of persons applying for them.
NOTE.2-A Letter of Credit differs from a bill of exchange in that it is payable at different
Places, at different times and in different sums. While a bill of exchange is payable at a certain
Place, at a certain time and for a specified amount.
LETTERS OF CREDIT are of great convenience to travelers by enabling them
to draw such sums at such times and at such places as they may wish.
To procure a Letter of Credit, the party is required to make a deposit of cash
or Securities with a banker who is prepared to issue them, for the amount of the
Letter of Credit.
An Interest Account is kept by the bank or banker issuing the Letter of Cre-
dit with the holder thereof, and at the close of the journey, an account is rendered
and settlement made, of the interest and of the unexpended balance, if any. The
interest allowed on the deposit varies from 13 to 3 per cent, and the interest
charged on the payments is the current rate.
When the holder of a Letter of Credit wishes to draw any money, he presents
the same to any banker named therein or on the list of the drawer's correspondents,
who will pay him the amount he wishes, in the currency of the country in which he
may be, at the regular rate of exchange then existing, and either take his receipt
or a draft on the drawing bank for the amount. This receipt or draft is remitted
to the drawing bank and is charged to the account of the payee or holder of the
Letter of Credit.
Foreign bankers who make payment on Letters of Credit, usually charge the
payees or holders a small commission, varying according to amount paid, from # to
1 per cent.
Form of a Letter of Credit.
No. 504. Office of HARNEY, CADY & Co., Bankers,
For £350. NEw York, July 1, 1895.
To our Correspondents :
GENTLEMEN.—We have opened a credit for Three Hundred Fifty Pounds
Sterling, in favor of Mr. A. L. Green, whose name appears below in his own
hand-writing, and we respectfully request that you honor his drafts on us, at
sight, for such funds as he may require, not exceeding the available extent of
this letter, indorsing all such drafts paid by you on this credit, which will
continue in force until January 1, 1896.
Very Respectfully,
2%ary / &4. & Gó.
(Signature of the holder).
CZ 2% 22%ae.
NOTE.—The purchaser of this Letter of Credit is supplied with a list of Harney, Cady &
Co's correspondents and their addresses, to whom he is authorized to present it. This list is often
printed on the Letter of Credit.
º: EXCHANGE. 715
AMERICAN EXPRESS COMPANY TRAVELERS’ CEHECKS.
1286. As an equivalent of the Bankers’ Letters of Credit, the American
Express Co. now issue travelers' checks, payable in various parts of nearly all the
countries of the world. Like Letters of Credit, these checks are a great service to
the traveling public. They may be used by different members of a family or party,
are of convenient form, are easily utilized as money, offer a safe and a cheap way
of carrying money when traveling in our own or in foreign countries, and they may
be purchased in hundreds of cities where Letters of Credit cannot be obtained.
Form of an Express Company Travelers’ Check.
WHEN COUNTERSIGNED BELOW 2-ºricanºpressºs
WITH THIS SIGNATURE. * º Chêque *E OOOOOO
789.
ſimtriran ºxpress (jumpani,
Will Daytſ the Orlºſ Of #20.00;
—s º
In United States and Canadal; ; ; ; GERMANY §. §ºhol and
º £TSTD Fr.) (;t," Mks. Pf. Imreſt. Kr.TOre|FIF. Tº G
7%pen?y Øollars. #|f|}|{2|}|º]};}} ; 39 49 || O2
COUNTERSIGNED : (sEE signatur E ABovE).
TREAS.
TEIE FACE AND CEIARACTER OF FOREIGN BILLS.
1287. 1. Foreign Bills of Exchange are, with rare exceptions, drawn in the
monetary unit or money of account of the country where they are made payable or
in which the drawee resides. See Monetary Units of different nations.
2. Foreign Bills of Exchange are accepted and bear grace whether drawn
at sight or on time.
Bills are sometimes drawn with “acceptance waived, ” in which case they are
not subject to protest for non-acceptance, but if not paid at maturity, would then be
protested for non-payment.
The most of our foreign exchange is effected through a few of the most
prominent mercantile and financial centers of Europe, such as London, Paris,
Hamburg, Amsterdam, Frankfort, Bremen, Antwerp, Vienna, Berlin, and Leipsic.
In mercantile parlance, we have Short Sight Bills, Long Bills, Bankers’ Bills,
Commercial Bills, Clear Bills, Bill of Lading Bills, or Documentary Exchange, A
1 Bills, and Prime Bills.


7 16 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
A Short Sight Bill is a bill drawn at 3 days sight. The most of English
bills, and also other foreign bills, are drawn at 60 days sight, and are called Long
JBills.
Bankers’ Bills are those sold by bankers,
Commercial Bills are those sold by merchants.
Clear Bills are those that have no collateral security, but are negotiated
Solely upon the responsibility of their makers and indorsers.
Bill of Lading Bills or Documentary Exchange are those secured or
protected by the assignment and transfer of the bill of lading, and policy of insur-
ance, of Some merchandise or commodity consigned for sale under order at the
maturity and for account of the bill.
Bills are sometimes secured by other collaterals than bills of lading and
policies of insurance.
A 1 Bills are either commercial or bankers' bills that are considered by
reason of the wealth, capacity and integrity of the makers and drawees, perfectly
good.
Prime Bills are generally regarded as synonymous with A 1 bills, but by
some they are considered a shade better. We, however, cannot conceive how they
can be better than A 1, when A 1’s are perfectly good. *
PAR OF EXCHANGE.
1288. The Par of Exchange is the established value of the monetary unit
of one country in that of another, and it is either intrinsic or commercial.
The Intrinsic Par of Exchange is the value of the monetary unit of one
country expressed in that of some other, and is determined by comparing the
weight and purity of the coins of different countries. Thus the intrinsic par with
England is now, 1895, $4.8665; with France it is $.193, and with Germany it
is $.238.
The Commercial Par of Exchange is the market value of the coin of one
country when bought or sold in another; and this value varies somewhat according
to the law of supply and demand.
See English, French, German, and other Foreign Exchange for the quotations
of Foreign Exchange.
THE COMMERCIAL RATE OR COURSE OF EXCELANGE.
1289. The Rate of Exchange is the variable price that is paid for bills.
There are three conditions on which bills of exchange can be bought and sold by
first parties; they are, par, premium, or discount.
To understand fully the reasons that combine to produce these three different
rates, requires careful study of the whole subject of exchange. For domestic
exchange, the following reasons are the principal ones governing the rate: The
“balance of trade” and the probability of its remaining for some time in favor of
* & EXCHANGE. g 717
or against the city where the bills are drawn; the opportunity or advantages pre-
sented to the seller of the exchange to use his money in his own city or where the
exchange is made payable; the rate of interest that the seller has to pay on his
over-drafts; and the expense of shipping money to meet his over-drafts, or of
shipping his surplus funds in another city to his own.
The rate of exchange was formerly based exclusively upon per cent. But
during the past few years many bankers throughout the United States base the rate
for domestic exchange upon the M. (one thousand.) We shall therefore use both
methods in this work.
BUSINESS CUSTOMS TO BE OBSERVED IN BUYING AND SELLING
EXCELANGE.
1290. When exchange is bought or sold through agents, it is the custom, in the
absence of a special contract, to allow the agent a commission upon the face of the
bill, plus the banker's premium, or minus the banker's discount. And when bought
through an agent and a broker, to allow both the commission and brokerage on the
face of the bill, plus the banker's premium, or minus his discount.
When money is invested at a premium or discount, it is the custom of the
banker, founded on equitable principles, to charge a premium or allow a discount
on the face of the bill that the money will buy, instead of calculating the same on
the amount invested.
PROBLEMS IN DOMESTIC OR INLAND EXCHANGE.
Problems in buying on our own account.
1291. 1. What cost a bill for $15000, bought at par 3 Ans. $15000.
2. What cost a bill for $15000, bought at 1 per cent premium ?
Ans. $15150.
3. What cost a bill for $15000, bought at 1 per cent discount
Ans. $14850.
4. What cost a bill for $15000, bought at $3.25 per M. premium ?
Ans. $15048.75.
OPERATION FOR THE 1 ST PROBLEM.
$15000 face and cost of bill.
OPERATION FOR THE 2D PROBLEM. OPERATION FOR THE 3D PROBLEM.
$15000 face of bill. $15000 face of bill.
150 = 1% premium. 150 = 1 % discount.
$15150 cost. $14850 cost.
OPERATION FOR THE 4TH PROBLEM.
$15000 $15000
3+ M. 3.25
= <=º or, – or, $15000 × 1003.25 = $15048.75000 cost of bill.
$45.000 $48,75000
3.75
$48,750 = premium.
718 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
Problems in buying through an Agent.
5. What cost a bill for $15000, bought at par, paying the agent 1 per cent
commission ? Ans. $15150.
6. What cost a bill for $15000, bought at 1 per cent premium, paying the
agent 1 per cent commission ? Ans. $15301.50.
7. What cost a bill for $15000, bought at 1 per cent discount, paying the
agent 1 per cent commission ? Ans. $14998.50.
8. What cost a bill for $15000, bought at $3.25 per M. discount, paying the
agent 1 per cent commission ? Ans. $15100.76
OPERATION FOR THE 5TH PROBLEM.
$15000 face of bill.
150 = 1% commission.
$15150 cost.
OPERATION FOR THE 6TH PROBLEM. OPERATION FOR THE 7TH PROBLEM.
$15000 face of bill. $15000 face of bill.
150 = 1 % premium. 150 = 1% discount.
$15150 face + premium. $14850 face — discount.
151.50 = 1 % commission. 148.50 = 1 % commission.
$15301.50 cost. $14998.50 cost.
OPERATION FOR THE 8TH PROBLEM.
$15000 face of bill $14951.25
48.75 = $3.25 per M. discount. 149.51 = 1 % commission.
$14951.25 face — discount. $15100.76.cost.
Explanation.—In solving questions of this character, in conformity to business custom, we
allow the agent a commission on the face of the bill, plus the banker's premium or minus the
banker's discount.
*-
Problems in buying through an Agent and a Broker.
9. What cost a bill for $15000, bought at par, paying the agent 1 per cent
eommission and the broker # per cent brokerage? Ans. $15225.
10. What cost a bill for $15000, bought at 1 per cent premium, paying the
agent 1 per cent commission and the broker # per cent brokerage?
Ans. $15377.25.
11. What cost a bill for $15000, bought at 1 per cent discount, paying the
agent 1 per cent commission and the broker # per cent brokerage?
Ans. $15072.75.
12. What cost a bill for $15000, bought at $3.25 per M. premium, paying the
agent 1 per cent commission and the broker # per cent brokerage 3
Ans. $15274.49.
º: EXCHANGE. 719
OPERATION FOR THE 9TH PROBLEM.
$15000 face of bill.
150 = 1 % commission.
75 = } % brokerage.
$15225 cost.
OPERATION FOR THE 10TH PROBLEM. OPERATION FOR THI, 11TH PROBLEM.
$15000 face of bill. $15000 face of bill.
150 = 1 % premium. 150 = 1 9% discount.
$15150 face -- premium. $14850 face — discount.
151.50 = 1 % commission.
75.75 = } % brokerage.
$15377.25 cost.
148 50 = 1 96 commission.
74.25 = # 96 brokerage
$15072.75 cost.
OPERATION FOR THE 12TH PROCLEM.
$15000 face of bill. $15048.75
48.75 = $3.25 per M. premium. 150.49 = 1 % commission.
sºmºsºsºm-ºsmº? 75.25 = } % brokerage.
$15048.75 = face -- premium. *E-º
$15274.49 = cost.
Explanation.—Where the purchase of exchange is made through both an agent and a broker,
or where a charge is made for commission and brokerage, in conformity to custom, in the absence
of a special agreement to the contrary, we allow both the commission and brokerage on the face
of the bill, plus the premium charged or minus the discount allowed.
Problems in selling on our own account.
13. What will be realized for $15000 sold at par’? Ans. $15000.
14. What will be realized for $15000 sold at 1 per cent premium ?
Ans. $15150.
15. What will be realized for $15000 sold at 1 per cent discount?
Ans. $14850.
16. What will be realized for $15000 sold at $3.25 per M. premium ?
Ans, $15048.75.
NOTE.—For the solution of the above four problems, see the solution of problems 1, 2, 3 and
4, page 717.
Problems in selling through an Agent
17. What will be realized for $15000 sold at par, paying 1 per cent commis-
sion ? Ans. $14850.
18. What will be realized for $15000 sold at 1 per cent premium, paying
1 per cent commission ? Ans. $14998.50.
19. What will be realized for $15000 sold at 1 per cent discount, paying
1 per cent commission ? Ans. $14701.50.
20. What will be realized for $15000 sold at $3.25 per M. discount, paying
1 per cent commission? Ans. $14801,74,
72O soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. jºr
OPERATION FOR THE 17TH PROBLEM.
$15000
150 = 1 % commission.
$14850 amount realized.
OPERATION FOR THE 18th PROBLEM. ſ OPERATION FOR THE 19TH PROBLEM.
$15000 $15000
150 = 1 % premium. 150 = 1% discount.
$15150 face + premium. $14850 face — discount.
151.50 = 1 % commission. 148.50 = 1 % commission.
$14998.50 amount realized. $14701.50 amount realized.
OPERATION FOR THE 201EI PROBLEM.
$15000 $14951.25
48.75 = $3.25 per M. discount. 149.51 = 1 % commission.
$14951.25 face — discount. $14801.74 = amount realized.
Problems in selling through an Agent and a Broker.
21. What will be realized for $15000 sold at par, paying 1 per cent commis-
sion and # per cent brokerage 3 Ans. $14775.
22. What will be realized for $15000 sold at 1 per cent premium, paying
1 per cent commission and 3 per cent brokerage 3 Ans. $14922.75.
23. What will be realized for $15000 sold at 1 per cent discount, paying
1 per cent commission and 4 per cent brokerage 3 Ans. $14627.25.
24. What will be realized for $15000 sold at $3.25 per M. premium, paying
1 per cent commission and # per cent brokerage 3 Ans, $14823.01.
OPERATION FOR THE 21ST PROBLEM.
$15000
1 % oommission $150 } 225
# 96 brokerage 75 240
Amount realized $14775
OPERATION FOR THE 22D PROBLEM. OPERATION FOR THE 23D PROBLEM.
$15000.00 $15000.00
1 % premium, º ge =-> * 150.00 || 1 % discount, g; s ſº gº 150.00
1 9 $ $15150.00 $148 $14850.00
9% commission, $151.50 = | 1 % commission, $148.50
# % brokerage, '**ś, } gº º 227.25 # % brokerage, ? 74.25 } wº º- 222.75
Amount realized, tº cº e- $14922.75 Amount realized, gº * tº $14627.25
OPERATION FOR THE 24TH PROBLEM.
$15000 1 % commission = - - - - $150.49
48.75 = $3.25 per M. premium. # 96 brokerage = tº sº tº as 75.25
$15048.75 $225.74
225.74 = commission and brokerage.
$14823.01 = amount realized.
NOTE.-In the foregoing 24 problems we have omitted extended explanations, believing that
the operations alone will render them easily understood. The student must be careful to observe
X- EXCHANGE. 72 I
that the agent and broker are allowed, when buying or selling exchange, a commission and brokerage
on the face of the bill plus the premium charged or minus the discount allowed.
We now approach exchange work that is generally considered very difficult, and although
by our philosophic system it is rendered comparatively easy and simple, yet it will require the
exercise of unimpaired brains and a remembrance of the business customs pertaining to exchange
operations, to fully comprehend it. We, therefore, here solicit the student’s earnest attention and
closest reasoning.
Problems in investing money in Earchange.
25. What amount of bill can be bought for $15000 at parº Ans. $15000.
26. What amount of bill can be bought for $15000 at 1 per cent premium?
Ans. $14851.49.
27. What amount of bill can be bought for $15000 at 1 per cent discount?
Ans. $15151.52.
28. What amount of bill can be bought for $15000 at $3.25 per M. premium ?
Ans, $14951.16.
OPERATION FOR THE 26TH PROBLEM. Eacplanation.—Since by business custom, and
in this case strict justice, the banker is allowed
$100 amount of bill assumed. a premium only on the face of the bill that the
1 = 1 % premium. money will buy, after paying the premium, it is
*mº clear that we cannot compute the premium on
$101 cash cost of $100 bill. the $15000, for were we to do so, we would be
Computing premium upon premium, which is
not allowed. To illustrate this we will make
100 the figures, thus: 1 per cent of $15000 is $150,
101 || 15000.00 ($14851.49 which deducted from the $15000 leaves $14850,
ºmºsºmeºmºsº as the face of the bill, and as 1 per cent of
$14850 is but $148.50, it is plain that it is incor-
rect to work on the face of the sum to be invest-
ed. Having, hence, no figures upon the face of which we can work, we are obliged to assume
some number to represent the amount or face of the bill, and upon which we may work according
to the conditions of the problem and produce relationship numbers of face and cost of bill, with
which and the $15000 we may make a proportional solution statement. We then assume $100 as
the face of the bill, and compute and add thereto the premium charged, and thus produce $101 as
the cash cost of the $100 bill. This gives us the required relationship numbers with which and
the $15000 we make the solution statement.
In making this statement we first observe that as a $100 bill cost $101 in cash, by transposition
$101 in cash will buy a $100 bill, and hence we can buy as many $100 bills as $15000 cost is equal to
$101 Cash; but to make the operation clear and philosophic we place the $100 assumed face on the
statement line, and reason thus: If $101, cash will buy a $100 bill, $1 in cash will buy the 101st
part, and $15000 in cash will buy 15000 times as much, which is $14851.49, answer.
NOTE.-In all work of this character, in actual practice, the assuming of the $100, and the
finding and adding thereto the premium, or the deducting therefrom the discount, should be per-
formed mentally.
The foregoing work gives the only exact method of solving problems of this character, but
some accountants solve them approximately by first deducting the premium from the amount to be
invested, then add to the remainder, the premium on the first premium, and thus continue the
alternation of subtracting and adding until the desired degree of accuracy is obtained. To
illustrate this method, we will solve the foregoing problem by it.
OPERATION.
$15000.00 sum to be invested.
150.00 = 1% premium deducted.
$14850.00
1.50 = 1 % on $150 premium added.
sºmmemºs as sºmeºmºs
$14851.50
.015 = 1 96 on $1.50 premium subtracted.
$14851.485 Ans.
In some cases, in which the sum to be invested and the rate per cent premium or discount is
even numbers, this method can be used to advantage, but generally the exact method is preferable.
722 soul.E's PHILOSOPHIC PRACTICAL MATHEMATICS. *
OPERATION FOR THE 27TH PROBLEM. Explanation.—Again having no number repre-
senting the face of the bill, on which we can
$100 face of bill assumed. work, we therefore assume $100 in order to work
1 = 1 % discount. upon and produce the necessary relationship
* Inumbers, as explained in the preceding problem,
$99 cash cost of $100 bill. to solve the question. On this assumed $100 we
compute and deduct therefrom the 1 per cent
$ discount, and thus obtain $99 as the cash equi-
100 valent, or as the cash cost of the $100 bill. We
99 || 15000 then observe by the exercise of our reason that
* as a $100 bill cost $99, by transposition $99 cash
$15151.52 will buy $100 bill, and hence we can buy as
many $100 bills of exchange as $15000 are equal
to 99. But working more philosophically, we place the $100, assumed face of bill, on our state-
ment line, and reason thus: If $99 cash will buy a $100 bill of exchange, $1 will buy the 99th part,
and $15000 will buy 15000 times as much, which is $15151.52, answer.
OPERATION FOR THE 28TH PROBLEM.
$1000 face of bill assumed. $1000 = face of bill assumed.
3.25 = $3.25 per M. premium. 3+ = $3.25 per M. premium.
$1003.25 = cost of $1000 bill. $1003}
$ Or, 1000
1000 4013 || 4
1003.25 | 15000.00 15000
|$14951.16 Ans. $14951.16 Ans.
Problems in investing through an Agent.
29. What amount of bill can be bought for $15000 at par, paying the agent
1 per cent commission? Ans. $14851.49.
30. What amount of bill can be bought for $15000 at 1 per cent premium,
paying the agent 1 per cent commission ? Ans. $14704.45.
31. What amount of bill can be bought for $15000 at 1 per cent discount,
paying the agent 1 per cent commission? Ans. $15001.50.
32. What amount of bill can be bought for $15000 at $3.25 per M. premium,
paying the agent 1 per cent commission? Ans. $14803.38.
OPERATION FOR THE 29TH PROBLEM. Explanation.—In this problem the agent is
entitled to a commission on the face of the bill;
and as there is no premium or discount to con-
100 sider, the operation and reasoning is the same
101 || 15000 ($14851.49) as in the 21st problem of investing, where the
& banker was entitled to a premium on the face
of the bill.
FIRST SOLUTION OF THE 30TH PROBLEM.
OPERATION OPERATION
To take out the Agent's commission. To take out the Banker's premium and find
face of bill.
$100 amount invested assumed. $100 face of bill assumed.
1 = 1 % commission. 1 = 1 % premium.
$101 cash required to invest $100. $101 cash cost of $100 bill.
* * * 100
101 || 15000 ($14851.49 invested). 101 || 14851.49 ( $14704.45) Ans.
$14851.49 $14,704.45
1% 1%
$148.51 commission. $147.04 premium.
Explanation.—As it is the business custom, in exchange transactions, to allow the agent a
4× EXCHANGE. 723
commission on the face of the bill or drafts, plus the premium charged or minus the discount allowed by
the banker or seller of the exchange, we ºtherefore first take the agent's commission out of the
amount to be invested. To do this we work and reason the same as we did in finding the face of
the bill in problem No. 26, under the head of Investing. The result of this statement is $14851.49
to be invested, and $148.51 is the agent's commission.
Then, having the amount to be invested, we invest the same, as shown by the second state
ment, with the banker at 1 per cent premium. The reasoning for this second statement is the same
as that given in problem No. 26, under the head of Investing.
NOTE.—By finding 1 per cent on the amount invested, or by finding the difference between
the $15000 and the amount invested, we have the agent's commission; and by finding 1 per cent on
the face of the bill, or by finding the difference between the $14851.49 and the face of the bill,
$14704.45, we have the banker's premium.
To add the agent's commission and the banker's premium together, and make but one pro-
portional statement thus,
100
102 || 15000
would give an incorrect result, for the reason that the agent is entitled to a commission on the face
of the bill plus the banker’s premium.
SECOND SOLUTION OF THE 30TH PROBLEM.
Instead of making two separate line statements, we can state it thus:
$
100
101 || 15000
101 || 100
10201 | 150000000 ($14704.45) Ans.
FIRST SOLUTION OF THE 31ST PROBLEM.
OPERATION OPERATION
To take out the Agent's commission. To find face of bill.
$100 amount invested assumed. $100 face of bill assumed.
1 = 1 % commission. 1 = 1 % discount.
$101 cash required to invest $100. $99 cash cost of $100 bill.
$
100 100
101 || 15000 ($14851.49) Ans. - 99 || 14851.49 ( $15001.50) Ans.
Explanation.—The operation of this problem is so nearly the same as that of the preceding
one, and the reasoning for the statements being precisely the same, we therefore omit repeating
Whenn.
SECOND SOLUTION OF THE 31ST PROBLEM.
OPERATION. Explanation.—To save time and figures we
$ prefer the one line statement in all cases where
100 it can be made. But in case the agent takes
101 || 15000 his commission out before going to the bank to
99 || 100 purchase exchange, as he would naturally do,
then separate Statements would necessarily have
99.99 150000000 ($15001.50) Ans. to be made.
OPERATION FOR THE 32D PROBLEM.
i 1000 100
00 1003.25 | 1.4851.49 Or, 101 || 15000.00
101 || 15000 *-*mm- 1003.25 | 1000
*- $14803.38 -**m-.
| $14851.49
724 * soul E's PHILOSOPHIC PRACTICAL MATHEMATICs.
w
Problems in investing through an Agent and a Broker.
33. What amount of bill can be bought for $15000 at par, paying the agent
1 per cent commission and the broker per cent brokerage? Ans. $14778.33.
34. What amount of bill can be bought for $15000 at 1 per cent premium,
paying the agent 1 per cent commission and the broker # per cent brokerage?
Ans. $14632.01.
35. What amount of bill can be bought for $15000 at 1 per cent discount,
paying the agent 1 per cent commission and the broker # per cent brokerage 3
Ans. $14927.61.
36. What amount of bill can be bought for $15000 at $3.25 per M. discount,
paying the agent 1 per cent commission and the broker # per cent brokerage?
Ans. $14826.52.
SOLUTION OF THE 33D PROBLEM.
OPERATION. Ea:planation.—In this, as in the above problems
in Investing, and for reasons therein given, we
$100.00 assumed as amount invested. assume and work upon $100. And as it is the
0 = 1 % commission. business custom to allow both the commission
and brokerage on the face of the bill plus the
t-mº banker's premium or minus his discount, we
$101.50 cash cost of $100 bill. compute as shown in the operation, and add to
- the $100 the commission and brokerage, and
thus obtain $101.50 as the cash cost, or cash
100 equivalent, of a $100 bill bought at 1 per cent
15000.00 ($14778.33) Ans. Commission and 3 per cent brokerage.
.50 = } % brokerage.
101.50
The reasoning for the line statement is the
same as given in preceding problems where
commission only was charged.
FIRST SOLUTION OF THE 34TH PROBLEM.
&
OPERATION OPERATION
To take out commission and brokerage. To take out premium and find face of bill.
$100.00 assumed as amount invested. * $100 assumed face of bill.
1.00 = 1 % commission. 1 = 1 % premium.
.50 = } % brokerage. *-
$101 cash cost of a $100 bill.
$101.50 cash required to invest $100.
100 100
101.50 | 15000.00 ($14778.33 Ans. 101 || 14778.33 ($14632.01 Ans.
The reasoning for these statements is the same as in the preceding problems and hence
is omitted.
PROOF BY REVERSING THE OPERATION.
- - - - - - - - - - - - $14632.01
Face of bill, * - -
Banker's premium at 1 % added, - º º tº - #º - - tº gº º 146.32
Face of bill plus premium, º - - - - - - - - - - $14778.33
Commission at 1 % (on $14778.33); - - * - - tº * -> dº ſº tº 147.78
* * * * * * * * * * * 73.89
Brokerage at # 96 (on $14778.33),
Total amount, - - - - - - - - - - - - - $15000.00
EXCHANGE.
725
SECOND SOLUTION OF THE 34TH PROBLEM.
$
100
15000.00
100
$14632.01 Ans.
101.50
101
$
100
2
15000
100
$14632.01 Ans.
203
101
FIRST SOLUTION OF THE 35TH PROBLEM.
OPERATION
To take out commission and brokerage.
$100.00 assumed as amount invested.
1.00 = 1 % commission.
.50 = } % brokerage.
$101.50 cash required to invest $100.
100
101.50 | 15000.00 ($14778.33 Ans.
OPERATION
To find face of bill.
$100 assumed face of bill.
1 = 1 % discount.
$99 cash cost of $100 bill.
100
99 || 14778.33 ( $14927.61 Ans.
SECOND SOLUTION OF THE 35TH PROBLEM.
100
15000.00
100
$14927.61 Ans.
101.50
99
$
100
2
15000
100
$14927.61 Ans.
203
99
PROOF BY REVERSING THE OPERATION.
Face of bill, tº º ſº -
Banker's discount at 1 % deducted,
Face of bill minus discount, * * tº-
Commission at 1 % (on $14778.33),
Brokerage at # 96 (on $14778.33),
Total amount, - - - -
OPERATION FOR THE 36TH PROBLEM.
$100 assumed.
1 = 1 % commission.
.50 = } % brokerage.
$101.50
$
100
101.50 | 15000.00
$14778.33
$1000 assumed face of bill.
$14927.61
149.28
$14778.33
147.78
73.89
$15000.00.
, 3.25 = $3.25 per M. discount.
$996.75 cost of $1000 bill.
$
1000
996.75 || 14778.33
| $14826.52 Ans.
726 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS.
MISCELLANEOUS PROBLEMS, INVOLVING THE FOREGOING PRIN-
CIPLES.
1292. 1. Find the cost of $2850 New York exchange at # per cent premium.
Ans. $2860.69.
2. Find the cost of $6540 exchange on Boston at $2.75 per M. premium.
Ans. $6557.99.
3. Buy $16250 exchange on San Francisco at $4.50 per M. premium and find
the cost. Ans. $163.23.13.
4. Buy $30400 exchange on New York at $1.12% per M. discount and find
the cost. Ans. $30365.80. .
5. What cost a bill for $22368.40 bought at 3 per cent discount ?
Ans. $22228.60.
6. Find the cost of $5862.70 exchange bought through an agent at $4.50 per
M. premium paying the agent # per cent commission. Ans. $5918.53.
7. Find the cost of $7250.80 exchange at # per cent premium paying an
agent 1 per cent commission and a broker # per cent brokerage. Ans. $7396.36.
8. Bought $24910.60 exchange at $3.70 per M. discount, and paid 4 per
cent commission, and # per cent brokerage. How much did I pay ?
Ans. $25004.57.
9. What cost a bill for $38215 bought at par paying an agent # per cent
commission and a broker 4 per cent brokerage 3 Ans. $38501.61.
10. Invested $8500 in exchange at $5 per M. premium. What was the face
-Of the bill ? Ans. $8457.71.
11. Invested $9000 in exchange at $4.50 per M. discount. What was the
face of the bill ? Ans. $9040.68.
12. Exchange is quoted at 24 per cent premium, how much can be bought
for $12500 % Ans. $12.195.12.
13. Exchange is quoted at 13 per cent discount, how much can be bought
for $46490% Ans. $47257.94.
14. If you pay 1 per cent commission and 4 per cent brokerage and buy
exchange at $1.25 per M. discount, how much exchange can you buy for $15000!
Ans. $14796.83.
15. Bought the following: Exchange $4700 at # per cent premium; $6250 at
# per cent premium; $5600 at # per cent discount; $800 at $1.10 per M. premium;
$16420.65 at $1 per M. discount, and through a broker $12286.45 at par; paid the
broker $5 per M. on the last exchange. How much exchange did I buy ald what
-did it cost 3 Ans. $46057.10, bot. $46.141.75, cost.
16. You owe a New York merchant, for goods bought of him $18400, which
you wish to remit in exchange. The rate is $2.50 per M. discount. What is the
cost Of the bill ? Ans. $18354.00
NOTE.—The student of book-keeping may make the Cash Book entry for both firms for this
transaction.
* EXCHANGE. 727
17. You owe a New York merchant, for goods sold for him, in your capacity
as a commission merchant, $18400, with which you wish to buy and remit exchange.
The rate is $2.50 per M. discount. What is the face of the bill? Ans. $18446.12.
NOTE.-The student of book-keeping may make the Cash Book entry for both firms for this
transaction.
ſº
18. Jones of New Orleans owes Smith of Chicago, a balance of $1018 for
Which he buys exchange at $1 per M. premium, and remits to Smith. What did
the bill cost 3 Ans. $1019.02.
... NOTE:-The student of book-keeping may make the Cash Book entries in both Jones and
Smith’s books, for this transaction.
19. A. of Galveston owes B. of Dallas, for net proceeds of cotton, $8232.38
which he invests in New York Exchange at $1 per M. premium. What is the face
of the bill? Ans, $8224.16.
NOTE.-The student of book-keeping may make the Cash Book entries in the books of A. and
B. for this transaction.
20. A commission merchant sold a consignment of goods for $28900. The
charges were $2140, commission 24 per cent. With the net proceeds he purchased
©xchange at 14 per cent premium. What was the face of the bill?
Ans. $25652.71.
NOTE 1.- The student of book-keeping may make the Cash Book entry for the purchase of
the bill by the consignee and for its collection by the consignor.
NoTE 2.-See Soulé's New Science and Practice of Accounts, pages 178, 179, 272 and 273,
edition of 1895 or 1896, for the proper entries for such transactions.
TO FIND THE RATE OF EXCELANGE.
1293. 1. Paid $2436 for a bill of $2400. What was the rate per cent
Ans. 13%.
OPERATION,
$2436 $
2400 36
tº ºsmºmes sº 2400 100
$36 = premium. tº-mº ºsmº
13% Ans.
2. The cost of exchange was $3484.60; the face of the bill was $3500. What
was the rate per M. and the rate per cent discount 3
Ans. $4.40 per M. or, .44% discount.
OPERATION.
$3500 15.40 15.40
2484.60 3500 || 1000 3500 || 100
* =
$15.40 = discount. | $4.40 per M. discount. .44% or Iº96.
728 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
3. Exchange cost $4050.15, face of bill $4000; paid brokerage for buying
# per cent. What was the rate per M. premium and the brokerage?
Ans. $5 per M. premium. $30.15 brokerage.
OPERATION.
$4020 $
100 4000 20
100.75 | 4050.15 * &=ºmº-ºº-ºº. 4000 || 1000
*==== &ºm==& $20 premium. mºsº
$4020.00 = face of bill -- premium. $5 per M. premium.
4050.15 = cost of bill.
$30.15 = brokerage.
4. A bill of exchange cost $6564.514, the face was $6500; paid commission
1 per cent and brokerage 3 per cent. What was the rate per cent discount, the
commission and the brokerage 3 * Ans. See operation.
OPERATION.
$100 $ $6500
1} 100 6467.50
1014 || 6564.51+
$32.50 = discount on $6500.
$1013
| $6467.50 = face of bill — ; 96 discount.
$6467.50 at 1 % = $64.67+ commission. 32.50
$6467.50 at # 9% = $32.33% brokerage. 6500.00 || 100
#96 discount.
TO FIND THE FACE OF BILLS WHEN THE COST AND RATE ARE
GIVEN.
1294. 1. The cost of a bill was $9024.75; the rate was $2.75 per M. What
was the face of the bill ? Ans. $9000.
OPERATION.
$
1000 assumed face of bill.
Cost of assumed face, 1002.75 |9024.75 cost of bill.
$9000 Ans. Face of bin.
2. Paid $3244.14 for a bill of exchange bought at $ per cent premium, bro.
kerage 4 per cent. What was the face? Ans. $3200.
OPERATION.
$ $
100 assumed face. 100 100
100.50 || 3244.14 807 || 8 100.50 | 3244.14
3228 or, 100
| $3228. = face + premium. msºmºsº sº, 100%
$3200 Ans.
$3200 Ans.
EXCHANGE. 729
3. The cost of a bill was $15377.25. Paid 1 per cent premium, 1 per cent
commission and 3 per cent brokerage. What was the face? Ans. $15000,
4. Invested through a broker $15000. Paid brokerage 1 per cent; face of
bill was $14803.38. What was the premium per M 7 Ans. $3.25.
OPERATION.
100 48.11
101 || 15000 14803.38 1000
| $14851.49 = face of bill -- banker's premium. | $3.25 per M. premium.
14803.38 = face of bill.
$48.11 = banker's premium.
NOTE.—In working the statement to find the $3.25 per M. premium, the last dividend which
gives the final figure 5 is a very little too small to produce even 5. This results mainly from the
fact that the divisor $14803.38 is slightly in excess of the true amount, which is $14803.379-H.
5. The face of a bill for a sum invested at 1 per cent discount, paying an
agent 1 per cent commission and a broker ; per cent brokerage, is $14927.61. What
was the amount invested? An S. $15000.
NotE.—See problem 35, page 725 for aid in the solution of this problem.
6. Invested in exchange at 14 per cent discount and received a bill for
$4100. What was the sum invested? Ans. $4038.50.
OPERATION.
$4100
61.50 = 1; 96 discount.
$4038.50 = sum invested.
TO FIND THE RATE OF EXCHANGE ON TIME BILLS.
1295. 1. What would be the equivalent rate per cent premium or discount
on a 30 day bill, money being worth 6 per cent and sight exchange on the same
place being 2 per cent discount? Ans. 2%% discount.
OPERATION.
$100 assumed to work on.
$100
$ 2%
100
60 || 33 $2.00 discount due the purchaser.
—— .55 interest ** **
55c. interest. —— .
| $2.55 sum of dis. and int. due the purchaser, and which
being on the $100, is hence equal to 2% 9% dis.
73o SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
\,
2. What would be the equivalent rate per cent premium or discount on 60
day bills, money worth 8 per cent and sight exchange 1 per cent premium ?
w Ans. #% discount.
OPERATION.
$100 assumed to work on.
$100
foo 1% premium due the banker.
1
45 || 63 $1.00 premium.
| $1.40 int. due the purchaser.
1.00 premium.
.40 excess of interest, which being on the $100 and in favor of the purchaser is equal
to # 96 discount.
3. What is the equivalent per cent premium or discount on a 15 day bill,
money being worth 4 per cent and sight exchange $ per cent premium ?
Ans. ##% premium.
OPERATION.
$100 assumed to work on.
* $100
$ #%
º 100
90 18 $# = 873 c. prem. in favor of the banker.
-msg. *- 200. interest.
| 200. interest in favor of the purchaser.
67#c. excess of premium.
Eacplanation.—By these statements, we see that there is an excess of the premium of 67% c., and
as it is on the $100, and in favor of the banker, it is therefore the per cent premium on 15 day
bills. By reduction we find it equal to 3% per cent.
4. Sight exchange is $4 per M. discount, and money is worth 5 per cent.
What is the rate of exchange per M. for 90 day bills allowing grace, but not discount
day & Ans, $16,913 per M. discount.
OPERATION.
$1000 assumed to work on. $
4 = $4 per M. discount. 1000
72 93
$ 4.000 = discount in favor of buyer.
12.91} = interest in favor of buyer. $12.913 interest in favor of buyer.
-
*~.
$16.913 interest and discount in favor of buyer, and which being on the $1000 is,
therefore, the rate of discount per M.
TIME EXCELANGE.
1296. 1. What is the cost of a $15000 bill drawn at 60 days sight, money
being worth 8 per cent and sight exchange being 1 per cent premium, allowing for
3 days of grace? Ans. $14940.
FIRST OPERATION. w
$15000 face of bill. $ Eacplanation.—As the check or bill has
150 = 1% premium for sight. 15000 63 days to run after presentation, it is
45 || 63 clear that we pay it 63 days before due,
$15150 cost of a sight bill. and hence we find and deduct from the
210 int. on $15000 for 63 days at 8%. $210 sight cost the banker's discount on the
face of the bill, $15000, for the 63 days
$14940 cash cost of bill. at i.per cent, and thus obtain the present
Cash cost. &
NOTE.-In practice, 10, 30, 60, and 90 day exchange is generally drawn at a certain per cent
premium or discount, which renders work of this character of but little use.
x- EXCHANGE. 73 I
SECOND OPERATION.
$100.00 assumed cost. $
1.00 = 1% premium on sight bill. 100
--- 45 || 63
$101.00 cost of sight bill. -º-
1.40 int. on $100 ass. for 63 ds. at 8%. | $1.40
$99.60 cost of $100, 60 day bill.
15000.
$14940.0000 cost of $15000, 60 day bill.
THIRD OPERATION.
Premium as above on $100 = $1
Interest as above on 100 = 1.40
.40c. discount on the $100 or # 96 discount.
$15000 $15000
.40 60
$60.0000 = discount. $14940 = cost of bill.
2. What is the cost of $8000 exchange on Baltimore, drawn at 30 days
sight, money being worth 6 per cent, sight exchange being 2 per cent discount?
Ans. $7796.
OPERATION.
$8000 face of bill. $
160 = 2 % discount. 8000
*-- 60 || 33
$7840 cash cost of a sight bill. -e
44 interest. $44.00 interest.
$7796 cash cost of $8000, 30 day bill.
3. A merchant owes a balance of $2511.25, which he wishes to pay with a
draft on St. Louis at 30 days sight. Allowing money to be worth 6 per cent, and
sight exchange or drafts to sell at 1 per cent premium, for what amount must he
draw his draft so that, when sold, it will net the amount that he owes?
Ans. $2500.
OPERATION.
$100.00 face of draft assumed. $ $
1.00 = 1 % premium. 100 100
60 33 100.45 || 2511.25
$101.00 cash value of $100 sight draft. ---
.55 int. on $100 for 33 days at 69%. 55c. interest. $2500 Ans.
$100.45 cash value of $100, 30 day draft.
To sell this draft the following figures are produced:
OPERATION.
$2500.00 face of draft. $
25.00 = 1 % wremium for sight drafts. 2500
---, 60 || 33
$2525.00 value of $2500 sight draft. *
13.75 interest. $13.75 interest.
$2511.25 cash value of 30 day draft.
732 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. 2.
f
EXCELANGE AND CASEI NOTES COMBINED.
1297. 1. I have a balance due S. S. Packard & Co., New York, of $6500. He
instructs me to purchase sight exehange and remit to him. Sight checks are 4 per
cent discount; I charge 4 per cent commission for investing, and give my note, with
collaterals, for such an amount as, when discounted at 8 per cent for 34 days, will
net the amount due the banker for the exchange. What is the commission, the
face of the note, and the face of the bill?
Ans. $32.34 commission. $6516.90 face of note. $6500.16 face of bill.
S O L U TI O N .
OPERATION
To take out # 96 commission.
$
| 100
201 || 2
6500
$6467.66
OPERATION
To invest $6467.66 at # 96 disc’t.
199 | 2
| 6467.66
| $6500.16
OPERATION
To find the face of the note.
$ assumed note.
4500
4466 | 6467.66
$6516.90 Ans.
2. Bought of the Merchants' Bank the following exchange: A check on
New York for $2500, at # per cent discount; a check on St. Louis for $2000, at #
per cent premium; and a gold check on Galveston for $5000, at # per cent discount,
gold being 10 per cent premium. For the cost of these three checks I gave $4000
currency and my cash note at 30 days, at 8 per cent, secured by collaterals. What
is the face of the note, allowing grace and discount day? Ans. $6016.71.
OPERATION INDICATED.
$2500 at # 96 discount = $18.75 = $2481.25 cost.
2000 at # 9% premium = 17.50 = 2017.50 cost.
4500
4466 || 5971.25
|sºn Ans.
Gold } º at # 9% discount = 27.50 = 5472.50 cost.
$99.71.25 = total cost.
4000.00 = cash paid.
$5971.25
DISCOUNT ON TIME DRAFTS.
1298. When time drafts or bills of exchange are negotiated before maturity,
they are subject to discount on the same conditions as promissory notes. In trans-
actions of this character, the premium or discount, and the interest are computed
on the face of the draft or bill.
y
IEXAMPLES.
1.
$8000. NEW ORLEANs, July 10, 1895.
Sixty days after sight, for value received, pay to the order of Geo. Fernandez,
Eight Thousand Dollars, and charge the same to our account.
To “BAAB & MASPERO, SCRIBNER & WAGNER.
ST. LOUIs, Mo.
EXCHANGE. 733
What are the net proceeds of this draft if discounted the same day that it
Was drawn at 8 per cent, the rate of exchange being 1 per cent premium ?
Ans. $7966.22.
OPERATION.
$8000.00 face of draft.
80.00 = 1 % premium.
$8080.00 value of draft.
113.78 interest.
$7966.22 net proceeds. Ans.
$
80.00
45 || 64
| $113.78 interest.
2. What is the value of a draft for $10000, payable 15 days after sight,
exchange # per cent discount, interest 8 per cent?
Ans. $9882.78.
º OPERATION.
$10000.00 face of draft.
75.00 = # 96 discount.
- *-
$9925.00 value of draft.
42.22 interest.
$9882.78 Ans.
$
100.00
45 19
$42.22 interest.
To buy, sell, and invest in eacchange and pay commission and brokerage on the face of
the bill only.
1299. 1. What will a bill for $15000 cost, bought at 1 per cent premium,
paying 1 per cent commission and # per cent brokerage on the face of the bill?
OPERATION.
Face of bill, ** - - - - $15000
1 % premium, - - - $150
196 commission, - - - 150
# 96 brokerage, wº- me - 75 37
ºm- 5
$15375
Cost of bill, - - * -
Ans. $15375.
Explanation.—As the commission and bro-
kerage are here, as well as the premium,
charged on the face of the bill, they could
have been added, making 24 per cent, and
the correct result obtained by adding to the
$15000 the amount of 24 per cent of itself.
2. What will be realized for a $15000 bill sold at 1 per cent premium, paying
1 per cent commission and 3 per cent brokerage on the face of the bill?
Ans. $14925.
OPERATION.
Face of bill, - - * -
1 % premium, - tº
Value of bill, º gº
1 % commission, tº º
# 96 brokerage, e -
- - - - $15000
- - - - 150
- - - - - $15150
- - $150
- sº 75
-mº 225
$14925
734 soule's PHILOSOPHIC PRACTICAL MATHEMATICS. º
3. What amount of bill can be bought for $15000 at 1 per cent premium,
paying 1 per cent commission and 3 per cent brokerage on the face of the "bill?
Ans. $14634.15.
OPERATION.
$100.00 assumed face of bill. $ Explanation.—The reasoning
1.00 = 1 % premium. 100 e º
1.00 = 1 $ P. mission. 102.50 | 15000.00 for the line statement is the
.50 = } % brokerage. --- same as given in the 26th prob-
--- 14634.15 Ans. º e
$102.50 cash cost of $100 bill. $ lem, and hence is omitted.
NOTE.-By the foregoing work, we have illustrated 39 different transactions in purchasing,
selling, and investing in domestic exchange according to the custom governing exchange transac-
tions, and also several different problems where, by special agreement, the commission and brokerage
are charged on the face of the bill.
This classification is entirely new, much more extended than ever before presented, and fully
illustrates work that has always confused students and the great majority of business men.
MISCELLANTEOUS PROBLEMS IN DOMESTIC EXCEIANGE.
1299. 1. What is the amount of the banker's premium on $25000. New
Orleans exchange at 3 per cent premium; what is the agent's commission at # per
cent, and what is the cost of the bill? Ans. $156.25, banker's premium.
$125.78, agent's commission.
$25282.03, cost of bill.
2. A merchant invested through his correspondent $7840.50 in exchange on
Philadelphia at # per cent discount. The correspondent charged for investing 1
per cent commission and paid a broker + per cent for transacting the business.
What was the face of the bill, what the banker's discount, what the correspondent's
commission, and what the broker's brokerage 7 Ans. $7802.22, face of bill.
$58.52, discount.
$77.44, commission.
$19.36, brokerage.
3. Bought exchange at 1 per cent discount and paid $2385.90; what was the
face of the bill ? Ans. $2410.
OPERATION INDICATED.
$100–$1 = 99. ($2385.90 × 100) – 99 = $2410.
4. A merchant in New Orleans owes a merchant in St. Louis $15000.
Exchange on St. Louis is 1 per cent premium, and exchange in St. Louis on New
Orleans is 1 per cent discount; making no allowance for the use of money or
telegraph expenses, which is it better for the New Orleans merchant to do, buy in
New Orleans at 1 per cent premium and remit, or to instruct the St. Louis merchant
to draw on him for such a sum as when sold at 1 per cent discount will net the
$15000 % Ans. $1.52 better that the New Orleans merchant buy and remit.
NotE.—See solution for the following problem.
EXCHANGE. 735
5. In the above problem, estimating that money is worth to the New Orleans
merchant 8 per cent, and that he instructs the St. Louis merchant by telegram
costing $2.40, to draw on him at sight for such an amount as when sold at 1 per
cent discount will net the $15000, and that it will require 4 days for the St. Louis
bill to reach the New Orleans merchant, what will be the gain to each merchant by
this mode of settling, over that of buying a bill in New Orleans at 1 per cent
premium, money being worth 6 per cent per annum in St. Louis, and 4 days time
required to transmit the exchange if bought in New Orleans?
Ans. $9.55 gain to the New Orleans merchant.
$10 gain to the St. Louis merchant.
SOLUTION.
OPERATION OPERATION
Showing the cost of $15000 exchange bought in Showing face of bill drawn in St. Louis and
New Orleans at 1 per cent premium. sold at 1 per cent discount to net $15000.
$15000 face of bill. $
150 = 1 % premium. 100
99 || 15000
$15150 cost. -----
$15151.52 face of bill.
Face of St. Louis bill, * - * º º - -> - º cº use º º $15151.52
Cost of New Orleans bill, - - - º - - º - -> º sº sº * 15150.00
Loss to the New Orleans merchant by paying the St. Louis bill, - - - - - $1.52
To which we add telegraph charges, se ºn sº me m sº s º º º 2.40
And produce a loss of - sº me as as amº m * ºn e º 'º º $3.92
Thus far we see that the New Orleans merchant has lost $3.92; but by reason of the St. Louis
merchant drawing on him, he has had the use of $15150, the cost of a $15000 bill in New Orleans,
for 4 days at 8 per cent; and the St. Louis merchant has had the use of $15000, the proceeds of the
St. Louis bill, for 4 days at 6 per cent.
OPERATION O PERATION
To find the worth of $15150 for 4 days at 8 %. To find the worth of $15000 for 4 days at 69%.
$ $
15150 15000
45 || 4 60 || 4.
$13.47 worth or int. - $10 worth or int.
By these iast two operations we see that the St. Louis merchant gains by the use of $15000,
for 4 days at 6 per cent, $10; and that the New Orleans merchant gained by the use of $15150, for
4 days at 8 per cent, - sº tº es tº as tº m - - - - - - $13.47
From which we deduct the losses, as shown in the first part of the operation, - - * * 3,92
And thus obtain a net gain of - - - - • * * * * * * * $9,55

Arbitration of Exchange.
===N
—w
1300. Arbitration of Exchange is the process of finding the proportional
cost of exchange between two cities or countries, by means of one or more interme-
diate exchanges.
The object of this system of exchange is to ascertain, by comparison of the
costs or rates of exchange, between several places, the most advantageous channel
through which to remit bills.
The subject is divided into simple and compound arbitration.
Simple Arbitration is the process of finding the cost or rate of exchange
between two cities through a third.
Compound Arbitration is the process of finding the cost or rate of exchange
between two cities through several others.
Exchange of different articles of merchandise, and different weights and
measures of different nations, may be arbitrated in the same manner as bills of
exchange and currencies.
PROTELEMIS.
TO FIND THE COST OF A BILL THAT WILL PAY A. CERTAIN SUM
IN ANOTEHER PLACE.
1301. 1. A merchant of New Orleans owes a balance of $5000 in New York,
which he wishes to pay. Direct exchange on New York is 1 per cent premium;
exchange at the time on St. Louis is 4 per cent discount, and St. Louis exchange on
New York is # per cent premium. Allowing # per cent brokerage for reinvesting
on the net investment or banker's cost of the, bill in St. Louis, which is the better
way for the New Orleans merchant to remit, and how much will he gain by remitting
the better way? Ans. Through St. Louis is the better way.
$37.63, gain.
OPERATION
For direct Exchange on New York.
$5000 face of bill.
50 = 1 % premium.
$5050 cost of bill.
(736)
+): ARBITRATION OF EXCHANGE. 737
FIRST OPERATION
For finding the cost of Exchange by remitting through St. Louis.
$
wº Bill on St. Louis, 100 | 99.50 cost of bill on St. Louis at # % discount.
Bill on New York, 100 | 100.50 cost of bill on New York at # 96 premium.
Face of bill on N. Y. 100 | 100.25 cost of bill on New York at # 96 brokerage.
5000 dollars invested.
$5012.37 Ans.
SECOND OPERATION.
Statement of costs.
$5000.00 face of bill.
25.00 = } % premium in St. Louis. Cost by direct exchange, - tº- $5050.00
Eºmºsºm-º-mºm. Cost through St. Louis, - - 5012.37
$5025.00 St. Louis cost. *===-
12.56 = + 96 brokerage in St. Louis. $37.63 gain.
$5037.56 cost including brokerage.
25.19 = } % discount in New Orleans on St. Louis bill.
$5012.37 cost of $5037.56 in New Orleans.
2. Suppose in the above problem, that the # per cent brokerage had been
charged on the face of the bill, instead of the net investment or banker's cost,
What would have been the cost of the exchange through St. Louis?
Ans. $5012.31.
FIRST OPERATION.
$
Bill on St. Louis, 100 99.50 cost of bill on St. Louis at # 96 discount.
Bill on N. Y. 100 || 100.75 cost of bill on N. Y. at # 96 premium and # 96 brokerage.
5000 dollars invested.
$5012.31 Ans.
SECOND OPERATION.
$5000.00 face of bill.
25.00 = } % premium.
12.50 = + 96 brokerage.
$5037.50 cost of bill in St. Louis including premium and brokerage.
25.19 = } % discount, N. O. exchange on St. Louis.
$5012.31 net cost in New Orleans.
3. Suppose again, that in the above problem brokerage had been charged on
the whole amount for investing, by the St. Louis agent, what would have been the
cost of the exchange through St. Louis? Ans. $5012.40.
FIRST OPERATION.
Bill on St. Louis, - sº - º -º 100 99.50 cost of bill in St. Louis at # 9% dis.
Bill on New York, - - º 100 || 100.50 cost of bill on N. Y. at # 96 premium.
Proceeds of am’t invested, brok. deducted, 99.75 | 100 amount invested in N. Y. exchange.
5000 dollars invested.
| $5012.40 Ans.
SECOND OPERATION.
$5000 face of bill.
25 = } 96 premium on New York.
*==--
$5025 St. Louis cost.
99.75 || 100.00
$5037.59 total cost including brokerage on same for investing.
9 = } % discount on St. Louis exchange in New Orleans.
$5012.40 net cost in New Orleans.
738 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. A.
TO FIND THE PROCEEDS OR FACE OF A BILL FOR A CERTAIN SUM
INVESTED.
1302. 1. A merchant of New Orleans wishes to invest and remit $5000 to a
New York correspondent. Exchange on New York is 1 per cent premium; exchange
on St. Louis, at the same time, is 4 per cent discount, and St. Louis exchange on
New York is # per cent premium, allowing # per cent brokerage on the met amount
invested by the agent, which is the better way to remit and how much will he gain
by remitting the better way?
Ans. Through St. Louis is the better way. $37.16 gain.
SOLUTIONS.
OPERATION,
For direct exchange on New York:
$
100 face of bill on New York.
Cost of bill 101 || 5000
- me summing
$4950.50 face of bill. Ans.
OPERATION.
For remitting through St. Louis:
$
Cost of bill - - 99.50 | 100.00 face of bill on St. Louis.
Cost of inv. - 100.25 | 100.00 amount invested by agent in N. Y. exchange.
Cost of bill on N. Y. 100.50 | 100.00 face of bill on New York.
5000 dollars invested.
| $4987.66 proceeds or face of bill.
4950.50 proceeds by direct exchange.
--sº
$37.16 gain as above.
2. Suppose that in the same problem the agent had charged # per cent
brokerage on the whole amount invested, what would have been the proceeds on the
face of the bill? Ans. $4987.62.
OPERATION,
$
Cost of bill on St. Louis, 99.50 | 100 face of bill on St. Louis.
Cost of investment, ,100. 99.75 amount invested by agent in N. Y. exchange.
Cost of bill on New York, 100.50 | 100.00 face of bill on New York.
5000 dollars invested.
| $4987.62 proceeds or face of bill.
3. Suppose again, in the above problem that the agent had charged + per
cent brokerage on the face of the bill, what would have been the proceeds or face
Of bill? Ans. $4987.72.
OPERATION.
$
Cost of bill on St. Louis, 99.50 100.00 face of bill on St. Louis.
Cost of N. Y. exchange, 100.75 | 100.00 amt. invested in N. Y. ex.
Invested at # 96 prem, and # 96 brok. | 5000 dollars invested.
| $4987.72 proceeds or face of bill.
* ARBITRATION OF EXCHANGE. 739
In the foregoing problems, we have elucidated the three different methods of
agents of charging commission or brokerage for reinvesting, when the cost of a
bill for a certain sum is required, and also when the proceeds or face of a bill for a
certain sum invested is required.
There is no statute law regulating the method of making this charge; and
as the business custom relating thereto is not uniform, it is, therefore, a matter of
Special agreement.
In the arbitration of exchange, brokerage or commission is charged at each
intermediate place through which it is remitted.
In the arbitration of foreign exchange, the commission or brokerage for
reinvesting is sometimes included in the rate of exchange; but when not thus
included, a separate charge is made.
FORMULA OF STATEMENTS.
1303. To elucidate the operation of charging commission or brokerage on the
net investment or face of bill, or whole amount for investment, when buying or
investing, we present the following formula of statements:
When buying a certain amount of exchange. When investing a certain sum in exchange.
To charge commission or brokerage on the
To charge commission or brokerage on the
met cost of bill, the statement is made thus:
net cost of bill, the statement is made thus:
100 || 100.50 100.50 | 100.
To charge commission or brokerage on the To charge commission or brokerage on the
jace of bill, the statement is made thus:
100 100.50 + banker's premium or
— banker's dis.
To charge commission or brokerage on the
whole amount, the statement is made thus:
99.50 | 100
face of bill, the statement is made thus:
100.50 –– 100.
the banker’s premium
or — banker’s dis.
To charge commission or brokerage on the
whole amount, the statement is made thus:
|
100 99.50.
Explanation.—In these statements, we have assumed 3 per cent as the commission or brokerage
eharged.
By Net Investment is meant the sum invested after the agent's brokerage
or commission has been deducted.
By Whole Amount of Investment is meant the sum invested, or to be
invested, before deducting the agent's brokerage or commission.
By Face of Bill is meant the amount of the bill that is obtained from the
banker, after paying the agent's commission or brokerage, and the banker's premium,
or allowing for the banker's discount.
74O SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. &
To find the equivalent rate of direct exchange to correspond with the arbi-
trated rate, we would omit from the statements the last amount on the increasing
side; or, we would divide the final result of the statement, or answer, by the last
amount; or, if the last amount is of a different unit, as it often is in foreign
exchange, we would divide by its equivalent, reduced to the same unit.
To elucidate this, we present the third example above; thus:
99.50 | 100.00
100.75 || 100.00
.9975-1-96 dis. Ans.
ENGLISH EXCHANGE.
1304. In the introduction of exchange we defined it to be, in a commercial
sense, a system of paying debts in distant places by means of bills of exchange,
without the transmission or shipment of money. By the term English Exchange,
we mean the system of paying debts in England and her dependent Provinces
without the transmission of money. **
1305. HISTORY OF THE ENGLISH MONETARY POUND.
The English monetary pound is of French origin. In the year A. D. 768, Charlemagne, or
Charles the great, ascended the throne of France, and during his reign he wielded the scepter with
such consummate skill and wisdom as to win the plaudits of the world and weave around his name
a brighter luster than was radiated from the jewels in his crown. In the year 800 he was crowned
by the Pope of Rome as “Charles Augustus, Emperor of the West,” and during the reign of this
great king and emperor, which continued till 816, he accomplished what had been undertaken by
his imperial predecessor, Augustus the First, the unification of the money of his empire. He
ordained that the French livre or peºind weight of silver should constitute the money unit of value
of the French kingdom; and this monetary pound of Charlemagne, reduced in weight, lived in
daily use through the vicissitudes of ten centuries in France, until 1792 when the French Nation
adopted the “metric system” of weights and measures, and the Franc, decimally divided, as the
monetary unit. It then disappeared from the money drawers of France.
In A. D. 1066, this Livre or pound was carried across the channel into England by William of
Normandy, and as conqueror he imposed it on the English people. The “Tower Pound” of A. D.
1066, the first coined in England, actually contained one pound weight of silver. But the word
has long since ceased to possess any truthful significance, for the English monetary pound has been
for five centuries, since the year 1200, dwindling in weight, until now the 20 shillings into which
it is divided weigh but 3%; ounces troy.
In 1620, when America was colonized, the people brought with them, or soon adopted, the
English pound as their unit of money. Some of the colonies still further reduced the weight of
º: ENGLISH EXCHANGE. 74. I
the English twenty shillings, and some coined smaller denominations, with sub-divisions in copper.
This they continued until 1660, when Charles the second reascended the throne of England, and
issued his royal order prohibiting any further coinage by the colonies. The pound then remained
the monetary unit until 1783 when England recognized the political independence of the United
States, and further, until the adoption of the present unit of money, the dollar. On the 6th of
July, 1785, the Continental Congress passed the memorable monetary ordinance reported by the
“grand Committee of thirteen,” by which they rejected the denominations of pounds, shillings,
pence and farthings, and established the dollar as the permanent monetary unit of the United
States, with the proviso, that it should be decimally divided. The precise weight, however, was
not fixed until August 8th, 1786. The weight and value of the American silver dollar was then
made to conform in value, or very nearly so, to the value of the old Spanish Carolus pillar dollar
which, by the usage of bankers for ages had been the standard of value for the monetary pound
during its many changes in weight,
As a matter of history, we would here remark that this Spanish dollar was not original with
Spain, she having borrowed it from Austria, during her union with Austria under the empire of
Charles the Fifth. The true ancestor of the American dollar is the “Joachim’s thaler," first
coined in the mines of the Bohemian valley of Saint Joachim (or James).
From 1786 to 1834, the commercial value of the English monetary pound, compared with the
United States dollar, was $4.44%, or £9 sterling was equal to $40, and vice versa. With this
equivalent of value between the monetary units of the two nations, we have but to multiply
pounds by 40 and divide the product by 9, to find the equivalent of pounds in dollars, and vice
wersa to find the equivalent of dollars in pounds.
Thus, to reduce £270 to dollars. And to reduce $1200 to pounds.
40 9
9 ) 10800 40 ) 10800
$1200 38.270
But, by Act of Congress, taking effect August 1st, 1834, the weight of the American coin was
reduced. The ten dollar gold eagle, which then weighed 270 grains troy, +} pure, was reduced in
weight 12 grains, and all other denominations of gold coin were reduced in like proportion. Con-
gress, also, in 1837, further reduced the coin by increasing the alloy from 's to hi. By reason of
this decrease of weight and increase of alloy, the relative value of pure gold in the United States
was enhanced, and the intrinsic value of the pound sterling, the English gold sovereign, was thus
enhanced or increased in like proportion. This increased value of the sovereign or pound sterling
is ascertained by calculating the actual number of grains of pure gold contained in the standard
coin of Great Britain, and estimating the same according to the United States valuation of gold.
The following work gives this increased value of the pound sterling, elucidates
the manner of obtaining it, and furnishes the data for the basis of the bankers' rate
of English exchange, from 1834 to 1874, which was nearly 94 per cent, more or less,
(according as premium was charged or discount allowed in selling the exchange)
premium on the old $4.44% value of £1:
742 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS.
ºr
From the original weight of the U. S. eagle ($10 gold piece) which was - £ a 270.0 grs.
We first deduct - - - - - - - - - * * * * * 12.0 grs.
And have in the remainder * * * * * * * * Eº as ºs 258.0 grs.
its present weight, ºr of which is pure gold.
From this we deduct To alloy, tº- tº º * sº º dº tºº sº * * tº- º 25.8 grs.
And have in the remainder *º- tº sº tº gº gº * ºn ge iº º 232.2 grs.
of pure gold.
By Act of Parliament, one pound weight of standard (+} pure) gold is coined into
£46 14s. 6d., which gives the sovereign, the monetary pound, a weight of - 123.2744 grs.
From which we deduct ºf alloy - - - - - - tº sº me tº as 10.2728 grs.
And have in the remainder, of pure gold, gº º ºs ems tº tº me tº 113.0016 grs.
Having now the weight of pure gold in the monetary unit of England and of
One of the denominations of the American monetary unit, to ascertain the value of
the English pound in the American unit of value, we have but to measure or value
its contents in pure gold by the United States valuation of gold.
OPERATION Explanation.—According to the foregoing work, we
To find the value of the English see that $10 American gold coin contains 232.2 grains
monetary pound <> of pure gold, and hence, by transposition, 232.2 grains
y p * . are worth $10; with these facts and premises we place
$ the $10 on the statement line, and reason as follows:
10 ' Since 232.2 grains of pure gold are legally worth $10,
232.2 | 113.0016 1 grain is worth the 232.2d part, and 113.0016 grains
-- “” (the weight of the sovereign in pure gold) are worth
113.0016 times as much, which worked, gives $4.8665.
| $4.8665
By deducting from this $4.8665 the old par value of the pound, $4,4444, we
have a gain of 42.21 cents, which gives nearly 94 per cent gain on the old par
value, or the value before the changes in the American coin in 1834 and 1837. Thus,
(.4221 × 100) -- 4.4444 = 9}}}}#.
This value, $4.8665, is, by Act of Congress of March 3d, 1873, which went
into effect January 1st, 1874, the present value of the pound sterling and the basis
of English exchange. By the same act, bankers and exchange dealers were
required to base their exchange transactions with England upon this value, and
allowing for premium, or discount, quote the rate of exchange in the U. S. dollar
and cent equivalent of the English monetary pound. In pursuance of this law,
bankers and exchange dealers, have adopted $4.86; as the par of exchange and
quote the rate $4.86% more or less according as a premium or discount is declared.
º
#
$ g ãº- t
* CŞ COUNTRY. STANDARD, Monetary Unit. HääE; COINS.
• s
* -T Argentine Republic......|Gold and silver.......|Peso ... . . . . . . . . . . . . . . . . . $0.96,5 Gº argentine ($4.82,4) and # argentine. Silver; peso and
SO 1VISIODS.
3 > * Gold: former system-4 florins ($1.92,9), 8 florins ($3.85,8),
Š- Austria-Hungary. . . . . . . . Gold. . . . . . . . . . . . . . . . . Crown . . . . . . . . . . . . . . . .20,3 ducat ($2.28,7) and 4 ducats ($9.15,8). Silver: 1 and 2 florins.
§ : g Gold: present system-200rowns ($4.05,2); 10 crowns ($2.02,6).
Š- P Belgium. . . . . . . . . . . . . . . . . . Gold and silver... . . . . Franc . . . . . . . . . . . . . . .19,3 |Gold: 10 and 20 francs. Silver: 5 francs.
Bolivia. . . . . . . . . . . . . . . . . . . Silver . . . . . . . . . . ... ...|Boliviano . . . . . . tº e º s .45,5 |Silver: boliviano and divisions.
§ * Brazil. . . . . . . . . . . . . . . . . Gold . . . . . . . . . . . . . . . . Milreis. . . . . . . . . . . . . . . . . .54,6 |Gold: 5, 10, and 20 milreis, Silver: }, 1, and 2 milreis.
§ es Ho British Possessions N. Gold ...... * e º e g º ſº gº © tº º Dollar . . . . . . . . . . . . . . 2 | 1.00
3. .S. A. * Newfound-
§ - | land).
$ ź Central Amer. States—
Š S (- Costa Rica..........
Q > Guatemala... . . .....
RS SD Honduras.......... } |Silver . . . . . ... . . . . . . . . [Peso . . . . . . tº e º ſº e s = e º e .45,5 |Silver: peso and divisions.
"S <d Nicaragua. . . . . . . . . .
S5 §, }={ Salvador. . . . . ... º º e * *.
°s tº chile............ ... .....|Gold and silver. . . . . . . Peso ... . . . . . . 6 s a .91,2 |Gold: escudo ($1.824) doubloon ($4.56,1), and condor ($9.12,3),
Q § º Silver: peso and divisions.
s & H Shanghia . .67,3
S S. º Haikwan .74,9
Sw SS 8 China. . . . . . . . . . . . . . .......Silver . . . . . . . . . . . . . . . Tael ..... º). 71.4 The Cash of China is, 1000dth part of the Tael.
QC 1011 US 111 , . . • * * *
rS S H UChefoo..... .70,4
*}se § ºr: Columbia. . . . . . . . . . . . . . . . Silver . . . . . . . . . . . . . . . . Peso ... . . . . . . . . . . . . . .45,5 Gold: condor ($9.64.7) and double-condor. Silver: peso.
SS tº P- Cuba....... tº ſº º e s s e º ºs º º ...'Gold and silver.......|Peso ... ... -----. . . . . .92.6 Gold: doubloon ($5.0.1,7). Silver: peso.
s’ s Denmark..... ............|Gºld..................|grown ... ............. .26,8 |Gold: 10 and 20 crowns.
; ; ; Ecuador..................Silver ................Sucre --- ... . . ...... .45,5 |Gold: condor ($9.64,7) and double-condor. Silver: sucre and
S $ T ge divisions.
S $ 3 |Egypt....................|Gold. . . . . . . . . . . . . . . . Pound (100 piasters) .. 494,3 |Gold : pound (100 piasters), 5, 10, 20, and 50 piasters. Silver:
S. § 1, 2, 5, 10, and 20 piasters.
sº SS * Finland..................'Gold................../Mark ... . . . . . .... . . .19,3 |Gold: 20 marks ($3.85,9), 10 marks ($1.93).
§ s C France...................|Gold and silver.......|Franc ....... ........ .19,3 |Gold: 5, 10, 20, 50, and 100 francs. Silver: 5 francs.
S ºff tº German Empire .........|Gold... . . . . . . .........[Mark. . . . . . . . . . . . . .23,8 |Gold: 5, 10, and 20 marks.
& Ş c |Great Britain............|Gold... . . . . . . . . . . . . . . Pound sterling ... ... 4.86.6% Gold: sovereign (pound sterling) and # sovereign.
§ E3 |Greece . . . . . .............'Gold and silver.......|Drachma. . . . . . . . . . . . .19.3 |Gold: 5, 10, 20.50, and 100 drachmas. Silver: 5 drachmas.
'S Ś Q |Haiti............ ... . |Gold and silver.......|Gourde ... tº $ tº dº º º .96, 5 |Silver: gourde.
jº So ſº IIndia ....................|Silver. . . . . ... • - - - - .... [Rupee ......... .21,6 Gold: mohur ($7.10,5). Silver: rupee and divisions.
s s É Italy............... .....|Gold and silver.......|Lira . . . . Gold...... : § : ; º º lire. Silver: 5 lire.
S Olºl • . . . . . . • * > * > * }old: 1, 2, 5, and 20 yen.
S $ P Japan. ... . . . . . . . . . . s & ſº gº Gold and Silver*...... Yen. . . . . . §::::::: .49,1 Silver: yen. y y
s S ºf Liberia. . . . .........'Gold... . . . . . . . . . dº e g º e e Dollar............ ..... 1.00
• S S ºr: Mexico...................Silver. . . . . . . . . . . ..... [Dollar . . . . . . . . . . . . . tº e º º .49,5 |Gold: dollar ($0.98,3), 2%, 5, 10, and 20 dollars. Silver: dollar
5 S E-i * (or peso) and divisions.
3 Netherlands..............'Gold and silver. . . . . . . Florin ... ...... s & .40,2 |Gold: 10 florins. Silver: #, 1, and 23 florins.
S P. Newfoundland...... .....'Gold..... © e º 'º e º e & ... ... [Dollar. . . . . . . . tº º e º s 1.01,4 |Gold: 2 dollars ($2.02,7).
'S P4 ||Norway.................. Gold... . . . . ...........ICrown . . . . . . . . . . . . . . tº e .26,8 |Gold: 10 and 20 crowns.
S p Peru . . . . . . . . . . . . . . . . . . . . Silver. . . . . . . . . . . . . . $9. , . . . . . . . . . . . . . . . . .45,5 |Silver: sol and divisions.
SY ºi Portugal . . . . . . . . . . tº e º e o 'º ..]Gold...... ... . . . . . . . . . [Milreis.... . Gold ‘’’’ 1%, §: 1, 2, 5, º: # l f($
es e º + O .77,2 |Gold: imperial ($7.71.8), and # imperial f($3.86).
's : Russia. . . . . . . . . . . . . . . . . . . . Silver f........ ....... [Ruble.. } Silver . . . . . .36,4 |Silver: }, +, and 1 ruble. p (
S p. Spain e & tº tº e º º tº G tº Gold and silver....... Peseta, g is is tº a • * * * 19,3 |Gold: 25 pesetas. Silver: 5 pesetas.
g s fº Sweden,...;. • * * * * * * * e a e s a Gold...... .... • . . . . ... ... Crown . . . . . . . . . . . . .26,8 |Gold: 10 and 20 crowns,
se º |Switzerland......... . . . . [Gold and silver. . . . . . . [Franc.. . . . . . . . * * .19,3 |Gold: 5, 10, 20, 50, and 100 francs. Silver: 5 francs.
Š PH Tripoli . . . . . . ..... . . . . . . . . [Silver. . . . . . . . ........|Mahbub of 20 piasters . .41,1
*S. Turkey ..................|Gold.......... . . . . . . . . .[iºster . . . . . . . . . . . . . . .04.4 |Gold: 25, 50, 100,250, and 500 piasters. *
CO |vºeia ... ............'Gold and silver. . . . . . . [Bolivar . . . . . . . . . . . . . .19,3 |Gold: 5, 10, 20, 50, and 100 bolivars. Silver ; 5 bolivars.



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744 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
THE RATE OF EXCELANGE.
NotE.—For the Face and Character of Foreign Bills, see Article 1287, page 715.
1307. The basis of English Exchange, as above stated, is $4.86% more or less,
according as premium or discount is declared.
By custom of exchange dealers, this rate is generally increased or decreased
by a varying scale of 3 cent, or 3 point, which is very nearly, and for practical
convenience, considered and quoted tº per cent. The exact per cent figures for the
#2 increase or decrease on the intrinsic value of the pound sterling are #### per
cent or decimally expressed.102743-- per cent.
In some cases, the rate varies by # and 3 cent in the quotations.
TO FIND THE RATE OF EXCELANGE WHEN A PREMIUM IS CELARGED.
1308. 1. Considering the balance of trade, the use of money, etc., the seller
of English Exchange wishes to charge a premium of 1 per cent. What would be
the rate of bills 3 Ans. $4.914.
OPERATION OPERATION
Based on the intrinsic value of the pound.
$4.8665 = value of pound sterling,
.0486–H = 1 % premium charged.
$4.9151 = (practically $4.914).
Based on the exchange value of the pound.
$4.865 = banker's basis of exchange.
.05 = prem. charged, allowing tº 96 to equal
-- #c. increase.
$4.915 = (practically $4.91%).
2. What is the rate of exchange when per cent discount is allowed ?
FIRST OPERATION.
By subtracting # per cent to the dollar and cent
value of £1.
$4.8665 = value of £1.
.0243–H = } % discount.
$4.8422 = (practically $4.84) Ans.
Ans. $4.84.
SECOND OPERATION.
By allowing To per cent to equal #c. decrease
on the banker's basis of exchange.
$4,865 = banker's basis of exchange.
.025 = discount.
$4,840 – Ans.
Ea:planation.—In the second operation since, I'd per cent equals 3 c. 4 per cent, which is ºr per
cent, equals Éc. which is 2}c. or 2.5 cents discount.
TO FIND THE PREMIUM CHARGED OR THE DISCOUNT ALLOWED
WHEN THE RATE OF EXCEIANGE IS GIVEN.
1309.
FIRST OPERATION.
$4.915
$4.8665
.0485
.0485
4.8665 100 . .
##### 9% practically 1 % Ans.
rate of exchange.
value of pound sterling.
= the premium.
1. The rate of exchange is $4.91%, what was the premium charged 3
Ans. 1%.
SECOND OPERATION. º
$4.915
$4.865
.050
.050
4.865 | 100
#}}} % practically 1%. Ans.
*. ENGLISH EXCHANGE. 745
2. The rate of exchange is $4.84, what was the discount allowed?
Ans. #%.
FIRST OPERATION. SECOND OPERATION.
$4.84 = rate of exchange. $4.84 = rate of exchange.
4.8665 = value Of £1. 4.865 = basis or par of exchange.
.0265 = discount. .025 = discount.
.0265 .025
4.8665 | 100 4.865 || 100
###33 (practically # 96). | #}} (practically # 9%).
THE RATE AND THE PAR OF EXCELANGE GIVEN TO FIND THE PER
CENT PREMIUM OR DISCOUNT DECLARED.
1310. 1. The rate of exchange is 4,873, the par of exchange is 4.86%. What
is the per cent premium charged? Ans. .2569% premium.
OPERATION.
C. C.
4.87; 4 || 5 4 || 5
4.86% 4.865 || 1000 or, 973 || 2
-*- -* 100
1}c. | .2569 % | ——
2. The rate of exchange is 4.80; the par is 4.864. What is the per cent
discount allowed ? Ans. 1.33607% discount.
PROBLEMS IN ENGLISEI EXCELANGE
TO REDUCE THE ENGLISH MONETARY UNIT TO THAT OF THE UNITED STATES.
1311. 1. What is the cost of a bill of exchange on London for £5000;
exchange 4.87; # Ans. $24375.
OPERATION,
*º, * Ea:planation.—Since the rate of exchange is the
exchange value of £1, it is clear that by multiplying
2435000
2500
324375,00 Ans.
the rate by the number of pounds we will obtain the
cost.
2. What will flá86 11s. 8d. cost in currency, exchange at 4.85% º
Ans. $7217,36.
746
SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS.
SOLUTION.
OPERATION
To reduce the 11s. 8d. to the decimal of a pound.
11 8
5 4;
5- 0.5 = 2.5881.
OPERATION
To reduce pounds to dollars.
361486.5833
4.85% rate of exchange.
7209929005
7432916
$7217.361921 Ans.
NOTE.-See Article 869, page 450, for an explanation for reducing shillings and pence to the
decimal of a pound.
3. What will £21468.3s. London exchange cost at 5.32?
OPERATION
To reduce the 3s. to the decimal of a pound.
3s. × .05 = .15 pounds.
or, 3s. – 20 = .15 pounds.
Ans. $114210.56.
OPERATION
To reduce the pounds to dollars.
4.21468.15"
5.32
$114210.5580
Ans.
NOTE:-In this and many of the problems in exchange operations, we use the multiplicand
as the multiplier.
This we do in conformity with custom and for practical convenience.
4. A merchant imports an invoice of goods from England, amounting to
£1723. 14s. 11d. What is it worth in United States money at the legal value of the
sovereign 3
FIRST SOLUTION.
OPERATION
To reduce 14s. 11d. to the decimal of a pound.
14 11
5 4;
.70 .0453
.70
# .745%
SECOND OPERATION.
3.1723.746
$4
Ans. $8388.61.
OPERATION
To reduce pounds to dollars.
4. 1723.746
$4.8665 legal value of £1.
s388.809090 Ans.
Explanation.—This opera-
tion is based upon the pre-
sumption that £1. = $4.86;
# of $6894.984 = amount considering 1 pound worth $4. $4.00 instead of $4.8665; hence the
*g of 1378.996 = } of $4 valuation of 1 pound, .80 slight excess, of 28-1-c. in the
114.916 = g of + of $4 valuation of 1 pound, 6# answer. It is a shorter (Jeth-
od than the first and is used
$8388.896 = amount considering £1 to be = to $4.86% by many accountants. As
shown in the operation we
first assume the pound sterling to be worth $4, then add to that value # of the $4 valuation, and
also Yºg of the # of the $4 value.
5. A merchant imported an invoice of English goods, amounting to £12400
2s. 8d. The importation duty is 40 per cent on the invoice price.
What is the cost
2. ENGLISH EXCHANGE, 747
of the invoice in United States money, including the duty but no other importing
charges? Ans. $84483.35.
OPERATION INDICATED.
:E12400.1333
4.8665
$60345.25 cost of invoice.
24138.10 = 40 % duty on cost.
*m.
$84483.35 importing cost.
6. By telegram, we learn that medium cotton is selling in Liverpool at 7%d.
Sterling exchange is 5.29. The freight on cotton to Liverpool in sailing bottoms
is *d. per pound.
From these figures, making no allowance for the insurance, interest on
money, and the charges to effect sales, what is the equivalent value of cotton of
like classification in U. S. money? Ans. Practically 1642.
OPERATION
74d. Liverpool price.
1%d. freight charges deducted.
7#d. = 7.3125d., which -- 240 = .030469 pounds.
.030469 × 5.29 the rate of exchange gives $.16118101, practically 164c. Ans.
** d. d. Ea:planation.—As many producers
7.3125 or thus: 16 117 of cotton unacquainted with the
240 || 5.29 240 5.29 customs and workings of commerce,
* -º *m- -º look upon and regard these figures
$.1611797 Ans. $.1611797 Ans. as the price of cotton in our own
market, we will remark that tº
closely approximate even, from data of this character, the price of cotton in our market, we must
allow for the insurance to ship cotton to Liverpool, the various charges to effect sales in Liverpool,
the interest on money, and in addition to this we must take into consideration the probability of
the increase or decrease in the Liverpool price of cotton, the rate of exchange, and lastly the
margin for profit to the buyer must be considered.
7. Good middling cotton is quoted in Liverpool at 54d. What is the equiva
lent price in United States money, the rate of exchange being 4.90 and no allow
ances made? Ans. $.10718--, practically 10+$2.
A NEW METHOD,
1312. The following new method may be often used with advantage when the
amount is in pence and the rate of exchange is near 4.86.
1. What is the equivalent in United States money of 8d., exchange 4.88%
Ans. $.16263, practically 1642.
OPERATION.
8d X 26. = 16c. Ea:planation.—By this method of work we first multiply
8d x *iſc. = 00.2 the pence by 24, c. which is the value of 1d. when
.2c. × 43 = 00063 exchange is 4.86; then to that part of the product obtain-
ed by multiplying by Zºo, we add to or subtract Tº part
$.1626; of the same for every #c. increase or decrease of the rate
of exchange from 4.86. In this example the rate of
exchange being 4.88, the difference between the same and 4.86 in half-cents is 4, hence we multiply
the .2c. by ſº.
748 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
2. What is the equivalent in United States gold of 93d., exchange 4.87; #
Ans. Practically 1942.
OPERATION OPERATION
By the new method. By the first method.
93d X 24"; c. = 19.2375c. d.
.2375c. × hºc. = 00.0593–1– 2 | 19
*E* sº 240 || 4,875
19.2968–H Ans.
.19296––c. Ans.
3. What is the equivalent in cents of 2s. 6d., exchange at 4.90?
Ans. 61#2.
OPERATION OPERATION
By the new method. By the first method.
2s. 6d. = 30d. × 245c. = 60.75c. d. *.
.75c. × Tº = .5 30
* * ===mme 240 || 4.90
61.25c. Ans. * - sºmºsºms
61}c. Ans.
PROBLEMIS.
1. What is the cost of a bill for £65241.3. 11, exchange $4.88%?
Ans. $318703.24.
2. What is the importing cost of an English invoice of goods amounting to
#365241 3. 11 ? Ans. $317496.28.
3. Exchange on Liverpool is quoted at 490, and wheat in Liverpool is quoted
at 42s. 8d. per quarter, what is the equivalent value per bushel in the United States
monetary unit? Ans. $1.303.
OPERATION INIDICATED.
512
240 || 4.90
480 || 60
NotE.—480 pounds is one quarter. See Table of Weights and Measures.
4. A merchant imported from J. Wood, London, goods amounting to £460
14. 10, on which there is a duty of 50 per cent. What is the cost in United States
money? Ans. $3363.30.
NOTE.—Commercial students should make the Journal and Cash Book entries for the above
transactions.
5. A merchant bought and remitted to J. Wood, London, a bill for £460
14s. 10d. when exchange was 4.88. What was the cost? Ans. $2248.42.
NoTE.—Commercial students should make the Cash Book entries for the above transactions.
TO REDUCE THE UNITED STATES MONETARY UNIT TO ENGLISH.
1313. 1. What amount of English exchange can I buy for $24375; the rate
of exchange being 4.87; # Ans. £5000.
OPERATION.
4.875) $24375.000 (£5000. Ans. Ea:planation.—Since the rate of exchange is the dollar and cent
24375 exchange equivalent of £1 Sterling, it is evident that we ma
gº sº --- buy as many pounds as $24375 is times equal to $4.87#. $
º: ENGLISH EXCHANGE. 749
{ 2. What amount of English exchange can be bought for $40250.65, the rate
of exchange being 5.32? Ans. £7565 18S. 3d.
OPERATION,
$40250.65 – 5.32 = 67565.9116.
20
S. 18. 2320
12
d. 2,784
3. What will be the face of a 60 day bill of exchange on Liverpool, that can
be bought for $7217.36, exchange at $4.853; ; Ans. £1486 11s. 8d.
4. A merchant owes a London correspondent $8119.53 currency, which he
desires to invest and remit. The rate of exchange is 4.853, and gold is 12# per
cent premium. What is the face of the bill? Ans. £148611s. 8d.
FIRST SOLUTION. SECOND SOLUTION.
$4.855 gold rate of exchange.
100 .606875 = 12} % premium gold.
112} 8119.530 *-*
4.855 5.461875 currency rate of exchange.
$8119,53 – 5,461875 = 361486.5829
361486,5829 20
| 20 * = -se--º
*-*s, *-*- S. 11.6580
S. 11.6580 12
12 *º-º-º-º:
*-*-* d. 7.896
d. 7.896 *
THIRD SOLUTION.
Operation to buy gold. 4.855) $7217.36 ( cuses,
O
100 S. 11.6580
112} | 8119,53 12
$7217.36 gold. d. 7.896
5. A Manchester merchant consigned to us an invoice of goods which, pursu-
ant to his order, were sold for his account. The net proceeds are $11240.20 which
we wish to remit; exchange is 5.41. What is the face of the bill?
Ans. £2077 13s.5d.
6. I have a balance of $4210.12 belonging to a Liverpool merchant, which I
am instructed to invest in 60 day English exchange and remit. The rate of
exchange is $4.84%; I charge 4 per cent commission for investing; what is my
commission and what is the face of the bill?
Ans. $20.95 commission. £865 1s. 8d.-- face of bill.
PER CENT AND INTEREST ON ENGLISEI MONEY.
1314. For the method of computing per cent on English money, see Article
868, page 449, and for the method of computing interest on English money see
Article 1135, page 569. For accounts current and interest accounts in English
money, see Article 1209, page 661.
75o soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. Yºr
TO FIND THE RATE OF EXCHANGE WHEN THE FACE OF THE BILL
AND THE COST ARE GIVEN.
1315. 1. The face of a bill on London was £5000, the cost was $24375.
What was the rate of exchange 3 Ans. $4.874.
OPERATION INDICATED.
5000) 24375 ($4.875 Ans.
2. The cost of £46014s. 10d. was $2248.42. What was the rate of exchange?
Ans. $4.88.
PRACTICAL MISOELLANEOUS PROBLEMS.
1316. 1. A Liverpool cotton factor sold 200 bales of cotton, weighing 90000
pounds, at 8:#d. per pound. He charges 24 per cent commission for selling. What
is his commission, and what are the net proceeds?
Ans. £820s. 73d. commission.
3:3199 4s. 4d. net proceeds.
FIRST OPERATION.
90000 weight. Commission reduced to £. s. and d.
8#d. price.
12 ) 19687.50d. commission as above.
810000 -
22500 20 ) 1640.7; d. = 1640s. 73d.
787500d. value in d. £82
23% rate of commission. ººm---sºmºmºs
f'82 0s. 7#d. commission.
19687.50d. commission in d.
787500.00 value as above.
12 ) 767812.50d. Y proceeds in d.
tºmsºmº. reduced to £.
20 ) 63984.44d. S. and d.
:63199.4s.
SECOND OPERATION.
90000 lbs. weight of cotton. #23281 5s.
8#d. price per pound. 24% rate of commission.
810000 3682. 0s. 7#d. commission.
22500 3281.5 0 sales of cotton.
12 ) 787500d. ) value in d. #3199. 4s. 4d. net proceeds.
dººm-ºn-me l reduced to
20 ) 65625s. } {... and s.
º--me
#3281.5s. |
NotE.—To compute 24 per cent commission, consider the whole amount 100 per cent and
º by 40, as 2% per cent is zºo of 100. Or first compute the commission at 5 per cent and divide
2. A merchant in New Orleans has a balance of $5000 currency due to his
Liverpool correspondent, and is instructed to invest the same in 60 day sterling
exchange and remit. The rate of exchange is $4.86; the New Orleans merchant
eharges 1 per cent commission for investing. What is the face of the bill?
Ans. £1018 12s. 5d.
ENGLISH
EXCHANGE.
751
FIRST OPERATION
To take out 1 per cent commission.
$100 investment assumed.
1 = 1 % commission.
$101 cash required to invest $100.
$
100
101 || 5000
$4950.50
SECOND OPERATION
To convert dollars to pounds.
4.86 || 4950.50
#21018.6214
20
12,4280
12
5,136
3. What is the amount of duty on the following invoice, the United States
tariff rates being 132 per pound on the hoop iron, and 10¢ per pound on the files
not over 10 inches long, and 30 per cent on the cost of same; and 62 on the files
over 10 inches long, and 30 per cent on the cost of same?
Messrs. RAND & BERCEGEAY.
Ans. $276.98.
BIRMINGHAM, 17 September, 1895.
Bot. of JNo. LUDLOW & SONs.


235 Bundles. £ | 8. d. || 3 || 8. d.
N T. C. Q. L.
R. B. 75 bundles Elephant Hoop Iron 14 by 18 1 19 2 0 -
40 4 * * * * * { * 1% by 17 1 0 0 0 - -
30 “ § { $ 4 e is 13 by 16 13 0 0 - -
40 ** & 4 $ 4 t 4 24 by 14 1 0 0 0 - -
N. O. 50 “ { % & 4 t is 3 by 13 1 5 0 0 - -
5 17 2 0 - -
At £715s. per ton, - - - - - - • * * 45 || 10 || 7 ||
Delivered free in Liverpool. * * * * * * * *
45 || 10 || 7
Cask 348.
º 48 doz. W. & S. Butchers hand saw Taper Files 3% in 3/4 - - 8
Weight. 60 “ 4 t * ! * { 4 in 3/9 - - 11 || 5
75 “ & 4 * { tº e 4% in 4/3 - - 15 18 || 9
C. Q. L. 6 * * { { mill & 10 in 11/9 - - 3 || 10 || 6
1 2 8 -º-
38 || 14 || 3
Discount, 35 per cent. * tº e º as * * * 13 || 10 || 11
Weight. 25 || 3 || 4
9 doz. W. & S. Butchers mill saw Taper Files 12 in 16/6 - - 7 || 8 || 6
C. Q. L. 6 4 * { % * { 14 in 23/ - 6 || 18
| 2 0 *== *-
14 || 6 || 6
Discount, 32% per cent, - - as º º * G - 4 || 13 || 1
9 || 13 || 5
Cask, 7/9; carriage to Liverpool, 5/3, º tº º º º Q_º - 13
35 | 9 || 9
RECAPITULATION.
Amount of 235 bundles, - - - - - º es tº - ºn tº 45 10 || 7
Amount of cask 348, - - - *- - - º - º º - 35 | 9
31|| 0 || 4
CHARGES.
I 2 9 1 1 4
Commission, 2% per cent on £45 10s. 7d. ; 3 per cent on £359s. 9d. - - 2 || 4 || 1
- 83 || 4 || 5’
Shipping, port charges and bill lading, - - - - - - - 1 || 0 || 4
Consul's certificate, - - - - - - - - - - - 11
Due in cash this day, * gº tº tº º 'º - 84 || 15 || 9
E. E. JNO. LUDLow & SONS.

752 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. ºr
OPERATION.
Weight of iron, 5 t. 17 c. 2 qr. 0 lb, = 13160 lbs. at 14c. * sº º tº ge ę tº $197.40
‘‘ ‘‘ files not over 10 in. long, 1 c. 2 qr. 8 lbs. = 176 lbs. at 10c. &= gº $17.60
“ “ files over 10 in. long, 1 c. 2 qr. 0 lb. = 168 lbs. at 6c. gº a * = 10.08
Cost of files, £359s. 9d. at $4.8665 = $172,70; 30% on $173. is - sº •º º 51.90
*=s 79.58
Amount of duty * = tº me º 'º me - - - - - $276.98
TO FIND COST OF DIFFERENT ARTICLES OF IMPORTED GOODS.
1817. 1. To ascertain the cost of the different articles of imported goods, the
amount of all the charges is first found, and then the invoice price of each article is
increased in the proportion that the charges bear to the invoice price of the goods.
When the cost is thus found the profits are then added.
In foreign invoices it is frequently convenient to first find the per cent that
the charges are of the invoice, and then increase the cost of each article according
to that per cent.
2. A merchant imported from Birmingham, England, an invoice of goods
amounting to £1723 14s. 11d. In the invoice were 12 dozen knives which cost £28
8s. 6d; allowing the duty to be 40 per cent and the freight to be $234.90, what was
the per cent of freight on invoice and duty, and what was the retail selling price
per knife to gain 25 per cent on the importation cost?
Ans. 2%. $1.71 retail selling price.
OPERATION.
3.1723.14. 11 = £1723,746 × $4.8665 $8388.61 OPERATION
40% duty on $8388.61 3355.44
tºmºmºmºsºmºmºmº To find the rate per cent of freight.
Cost of whole invoice, $11744.05
:628 8s. 6d. £28.425 × $4.8665 $138.33 234.90
40% duty on $138.33, 55.33+ 11744.05 || 100
Cost of 12 dozen knives, $193.66–– |2.00016% practically 2%.
2% freight on knives, 3.87–1–
N. O. cost of 12 doz. including frt. $197.53–H
25% gain on N. O. cost, 49.38–1–
Price of 12 doz. knives, 12 ) 246.92–
Price of 1 doz. knives, 12 ) 20.57+
Retail price of 1 knife.' $1.71–H Ans.
* NoTE 1.—See pages 456 and 457, for Practical Work, Marking Goods and Computing the Rate
Per Cent of Freight.
NOTE 2.-See Customhouse Business for Importing Goods, pages 503 to 508 for other problems
in importing goods.
XP' & a ENGLISH EXCHANGE. 753
COMPUTATIONS OF FEEIGHT IN ENGLISH MONTEY.
1818. 1. What is the freight on 540 bales cotton, weighing 243084 pounds, at
#d. per pound from New Orleans to Liverpool? Ans. £633 0s. 8d.
FIRST OPERATION. SECOND OPERATION.
243084 D.
#d. 8 5
12 || 243084
8 ) 1215420 20
12 ) 151927.5d. £633.0318
tºº 20
20 ) 12660 = 8d.
tº ºs. OS. 6260
:6633 0s. 12
7d. 512
2. What is the freight and primage on 68 bales cotton, weighing 31416
pounds, at ##d. per pound; primage 5 per cent Ans. £122 14s. 4d., freight.
£62s. 9d., primage.
FIRST OPERATION. SECOND OPERATION
3.1416 #122 14s. 4d. amount of freight. To find freight.
+3d. 5 % rate of primage.
tº-º-º-º-ºººººmsº ** = me D.
16 ) 471240 £6.13.11.10% amount of primage. 16 || 15
tº sº, 20 12 || 31416
12 ) 29452.5d. 20
* -º 2S.71 wºmmºnsººmsºmº
20 ) 2454 = 43d. 12 £iºns,
E122 = 14s. 8d.62+ ººmsºmºsºme
14.S. 376
NOTE.-Primage is a charge in addition to the freight; 12
originally it was a gratuity to the captain of the vessel for his * -º-º-º-º-º:
particular care of the goods. It now belongs to the owners or 4d. 512
freighters of the vessel, unless by agreement, it is allowed in whole
or in part, to the captain.
3. What will be realized in United States money from the freight and
primage of 21000 bushels of wheat from New Orleans to Liverpool; freight 123d.
per bushel, primage 5 per cent; exchange being 5.40? Ans. $6.201.56.
FIRST OPERATION.
12; d. freight. Explanation.—We here reduce and place the
#d. = 5 % primage on 12#d. rate of freight and primage 13; d. on our state-
- *-*E* ment line, and multiply the same by 21000, the
13; d. freight and primage per bushel. number of bushels on which freight and primage
are charged. . This gives us the amount of
D. freight and primage in pence, which we reduce
8 || 105 to pounds by dividing by 12 and 20, or 240.
12 || 21000 Then, as the result thus obtained is pounds, or
20 || 5.40 pounds and decimals of pounds, we reduce the
| * = same to United States money by multiplying by
$6201.56 Ans. the rate of exchange 5,40.
754 soulE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
SECOND OPERATION.
21000 bushels. £11488s. 9d. = £1148.4375+.
12}d. rate of freight. 5.40
262500d. freight. $6.201.56250 Ans.
13125d. = 5 % primage.
12 ) 275625d. freight and primage.
20 ) 22968 9d.
.61148 8s.
4. What is the freight and primage in United States money on 25000
bushels of corn from New Orleans to Liverpool, at 24s, per imperial quarter of 480
pounds; primage 5 per cent, and exchange 4.86? .* Ans. $17860.50.
FIRST OPERATION. SECOND OPERATION.
S. 25000 × 56 = 1400000 pounds.
24
480 || 56 1400000
25000 480 || 24
100 105 {-º-º-º:
20 || 4.86 20 || 70000S.
$17860.50 Ans. 423500 amount of freight.
175 = 5% primage on £3500.
£3675 freight & prim. in Liverpool.
4.86 rate of exchange.
$1786.50 Ans.
TO FIND THE PURCHASE PRICE OF ARTICLES TO FILL THE ORDERS
OF ENGLISH MERCHANTS, AT A SPECIFIED PRICE, FREE OF
ALL CHARGES WHEN DISCHARGED IN BRITISH PORTS.
1319. 1. A Liverpool correspondent instructs me to purchase and ship to
him 7200 centals of white wheat, on condition that the total cost, including all
expenses for purchasing, shipping, etc., shall not exceed 11s. 8d. per cental when
placed on the dock at Liverpool. The freight is 25s. per imperial quarter, shipping
expenses and commission for purchasing estimated at 9 per cent; exchange $4.90.
What is the maximum price to be paid per bushel? Ans. 87;ºg.
SOLUTION.
Statement to find the freight on 1 Cental.
S.
25
480 || 100
| 5s. 24d. freight on 1 cental.
11s. 8d. Liverpool price per cental.
5 2+ freight expenses.
6s. 53d. = 774d. = ***.d. to pay for wheat and remaining charges.
* ENGLISH EXCHANGE. 755.
• STATEMENT Explanation.—We first find, as shown in the
first statement, the freight on 100 pounds, and
To find the purchasing price of wheat. then deduct the same from the Liverpool limits
of price, and thus obtain 6s. 5%d. = 77.3d., with
which we are to buy the wheat and pay purchas-
2 | 155 ing and shipping charges. Then to find the
100 || 60 rice that this sum will pay per bushel, after
240 eing reduced to American money at the rate
4.90 of exchange given, and paying all charges, We
109 || 100 place the +3*.d. (774) on our statement line, and
-*- reason thus: If 100 pounds (1 cental) cost fººd,
87: §c. Ans. 1 pound will cost the 100th part, and 60 pounds.
or 1 bushel will cost 60 times as much. Then
to convert the pence to pounds, we divide by 240; to convert the English monetary unit to
American, we multiply by the rate of exchange 4.90, then to deduct the shipping expenses and our
commission, both being on the same value, we multiply by 100 and divide by 109. The final result
of the entire statement is the sum that we can pay for one bushel of wheat and comply with the
conditions of the problem.
2. A London correspondent advises the purchase and shipment of 5,000
bbls. of Extra Flour at 32s., 6d. per bbl. C. I. F. (That is with Cost, Insurance,
Freight and all other charges to deliver the same on the wharf at London, paid.)
The freight is 2 s. 8 d. per bbl.; primage, 5%; all other shipping expenses,
including Insurance, Freight, Drayage, Labor, etc., are estimated at 4% on first
cost; Commission for purchasing, 5%; Exchange, $4.85. What price can be paid
for the flour in the American market. Ans. $6.5954-H.
OPERATION.
2s. 8d. freight. 2s. 8d. freight. D. 32s. 6d. Liverpool price.
5% primage. 1} primage. sº }: 2 9% deducted.
am-mºº- -º 0 || 4.85 *-ºsmºsºm-
1.60 = 1; d. primage. 2s. 9%d. 104 || 100 29s. 8%d. = 356%d. = +**d.
105 || 100
$6.595-H Ans.
To reverse the question and operation, and find the London cost of flour
bought here at $6.5954+, with all the specified charges added, we produce the
following figures:
FIRST OPERATION.
6.5954 cost of 1 barrel in New Orleans.
100 || 104
100 || 105
4.85
£1,485 = £1 9s. 8.4d.
2 9.6 freight and primage.
£1 12s. 6d. Ilondon cost. Ans.
SECOND OPERATION.
$6.5954 purchasing price of flour in New Orleans.
.2638 = 4% shipping expenses on $6.5954.
$6.8592 cost of flour with 4% shipping charges added.
.3429 = 5% commission.
$7.2021 cost of flour in London in U. S. money, freight and primage unpaid.
$
4.85 | 7.2021
|ELºs = <1 ºr 84
20 2 9.6 freight and primage added.
9,700 £1 12s. 6d. London cost. Ans.
12
ºmº-º-
8d. 4
756 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
3. American tallow is quoted in London at 44s. 6d. per cwt. Allowing $124
per American short ton for freight, 5 per cent for primage, 3 per cent of prime cost
for shipping expenses, and 5 per cent of sales for London charges, what is the
maximum price to be paid in New Orleans when exchange is 4.80, in order to be
sold at the London quotation, and a gain of 10 per cent on prime cost realized?
Ans. $,0743599.
OPERATION.
44s. 6d. London selling price. 44s. 6d. London selling price.
5% rate of London charges. 2 216 London charges.
2s. 2,64, commission. 42s. 3 ºd. = 507 ºd. London net proceeds, per cwt.
Statement to find the net proceeds of 1 pound in U. S. money.
D.
5073
240 || 4.80
112
$.0905892–H net proceeds of 1 pound.
;
Operation to find the freight and primage on 1 pound.
Freight on 1 ton, - - - - - - - - - - - - - - - $12.50
Primage at 5% on $124, * sº tº * * * * tº º ºs º º ºs .625
Freight and primage on 2000 pounds, tº gº ºs - - - - - 2000 ) $13.125
Freight and primage on 1 pound, tº ſº * * * * ~ * = - $.0065625
Which we deduct from the London proceeds of 1 pound, tº º sº º tº sº .0905892
N. O. cost, including 3% shipping charges and 10% gain on prime cost, dº gº º $.0840267
Operation to deduct the 3 per cent shipping charges and 10 per cent gain.
| 100
113 | .0840267
$.0743599 Ans. N. O. prime cost, or the maximum N. O. purchasing price.
EXCHANGE BETWEEN LONDON AND WARIOUS OTHER COUNTRIES.
1320. London, England, has been called the great clearing house of the world.
The following table shows the rates of exchange of recent dates between
London and the cities and countries named in the list.
It should be remembered that these rates do not show the intrinsic value of
the different monetary units, and that they are subject to a limited increase or
decrease through the same causes that govern the rate of exchange in our own
country. They are presented as average rates, for general information concerning
exchange. 4.
#: PROBLEMS INVOLVING ENGLISH EXCHANGE. 756%
A New Orleans house received a cable message offering 8d. for 100 bales of
cotton, good middling, good color and staple, f. i c. and 6% tare in Manchester-
What would be the total New Orleans selling price and the price per pound, allow-
ing freight at 282 per 100 lbs., and insurance at $74, exchange at 4.86%
Ans. $7400.72 selling price. Price per lb. $.1480144.
FIRST OPERATION.
100 B/C = 50000 lbs.
Less 6% tare, 3000 “
mºm-ammº
47000 lbs. net (a) 8d.:-376000d. at 4.86 = $7614.00
Less freight on 50000 lbs. (a) 282 per C. lbs. 140.00
Operation to Reduce
pence to dollars, exchange 486 $7474.00
d. $7614.4-10% ($761.40) = $8375.40 the
376000 amount to insure.
240 || 4.86 #% insurance on $8375.40 = 73.28
| $7614.00 Selling price, - - º - - $7400,72
$7400.72 - 50000 lbs. = $.14801-1-
PROOF. 8d, a 486 = $162 = price of 1 lb.
8ól. .00972 = 6% tare.
240 || 486 * -
-Eºmº- || Pº-smº- .15228 = net price.
$.162 .0014657 = g- insurance.
Operation to find the amount .1508143 = price less insurance.
to insure. .0028 freight at 282 per C. pounds.
$ .15228 = net price. tº ºn e
.015228 = 10% increase. $.1480.143 net price of 1 pound.
E-smº -m- 50000 = lbs.
$ .167508 = amt. to insure (a) sº-º-º: • -
#9% = $,0014657. $7400,715 = purchasing price.
IPROOF.
SECOND OPERATION, * 6
The following inaccurate but approximate method is used .48 = 6%
by some cotton house accountants. 7.52
.0724 = ins.
100 B/C = 50000 lbs.
Less 6% = 3000 “ * 7.4476
*º- .14 = freight, Ex. 480.
47000 (a) 8d. = fºlò66–13–4 || –
Less freight @ 282 per 100 lbs., $140 = 29– 3–4 || 7.3076d.
(at Exchange, 480) (7.3076d. x 4.86)
£1537–10–0 || -- 240 =$.1479789 per lb.
Less gºſ, ins. on £1566 plus 10% 15— 1–5 × 50000 lbs... = $7398.95
Difference caused by
Selling price, - *- - tº- - £1522–8–7 the decimals, -> - 6
Exchange (a) 486 = - tº - - $7399.01 *m tºm-mm-
$7399.01
NOTE: 1.-The inaccuracy of this method consists in using 480 as the value of the pound
sterling, in calculating the freight and in omitting the shillings and pence on amount insured.
NOTE: 2. —The first operation is strictly accurate and much shorter and easier, as we work
in dollars and cents, instead of pounds, shillings and pence, after the first item.
--
2 O 's PHILO A A Q º
61 SOULES P SOPHIC PRACTICAL MATHEMATICS jºr
A firm in Manchester, Eng., instructs its correspondent in New Orleans to
buy 100 bales of good middling, good color and staple cotton on the following
conditions: 8d. per pound f. i. C. and 6% tare.
What price per pound may the New Orleans house pay, allowing $1.10 per
bale expenses in New Orleans, 50 cents per bale commission, 30 cents per 100 lbs.
freight, 1% insurance, estimating each bale to weigh 500
Ans. for 1st operation, $.1427.
{ % {{ 20. as tº .1448.
IFIRST OPERATION. SECOND OPERATION.
86. (a) Ex. 480 = .16 cents 8d. Ex. 48665 = .1622–1–
1% insurance = .0016 1% insurance = .0016
= .1584 = .1606
6% tare 9504 6% tare 96+
30g frt. per 100 lbs. .1489 30g frt, per 100 lbs. .1510
= .32 on 1 lb. 30 = .32 on 1 lb. 30
50g com. on 500 .1459 50¢ com. on 500 lbs. .1480
lbs. = .1% on 1 lb. 10 = .1% on 1 lb. 10
1102 exp. on 500 .1449 110g exp. on 500 lbs. .1470
lbs. =.222 on 1 lb. 22 = .22g on 1 lb. 22
N. O. buying price .1427 N. O. buying price .1448
NOTE: 1.-It is the custom of cotton buyers in problems of this kind to base their calcula-
tions on a 480 valuation of the pound sterliug, thus facilitating the operation to some extent.
NOTE: 2.—The difference between Exchange 480 and 48665 is .0665c. = 1.3854% or, practi.
cally 1396. Hence, by adding 1; 96 to the 480 Exchange price we produce the 48665 Exchange
price nearly.
NoTE 3.-By multiplying the .480 Exchange price by the current rate of exchange and
dividing the product by 480 we would produce the price at the current rate of exchange.
#: ENGLISH EXCHANGE. - 757
EXCELANGE BETWEEN LONDON AND WARIOUS OTHER COUNTRIES.
1820. London, England, has been called the great clearing house of the world.
The following table shows the rates of exchange of recent dates between
London and the cities and countries named in the list.
It should be remembered that these rates do not show the intrinsic value of
the different monetary units, and that they are subject to a limited increase or
decrease through the same causes that govern the rate of exchange in our own
Country. They are presented as average rates, for general information concerning
exchange.
Exchange on Paris and other cities in France 25.25, fr. and cts. for £1.
& 4 “ Amsterdam 4 & * { ‘‘ ‘‘ Holland 12.3, fl. and stiv. for £1.
4 & “ Brussels & ( & & “ “ Belgium 25.30, fr. and cts. for £1.
4 & ** Rome & 4 { { “ “ Italy (gold) 25.40, lire and cent per £1.
4 & ‘‘ Rome & & ( & “ “ Italy (paper) 28.10, lire and cent per £1.
& 4 ** Berlin 4 & { { “ “ Germany 20.55, mks. and pf. for £1.
& & ** Athens & 4 & 4 “ “ Greece 28.25, dr. and cts. for £1.
6 & ** Vienna, 4 & 2 & “ “ Austria 23.90, crowns.
& & ** Madrid & 4 é & “ “ Spain. 25.25, peseta and cts. for £1.
& 4 ** Lisbon “ “ “ “ Portugal 52d. for 1 Milreis.
4 & “ St. Petersburg “ “ “ “ Russia 30d. for 1 Rouble.
é & “ Hong Kong 4 * & 6 ** ** China. 46d. for 1 Dollar.
& & “Shanghai ** ** ** “ China. 62d. for 1 Tael.
4 & “ Calcutta, “ “ “ “ India 22d. for 1 Rupee.
4 & “ Rio de Janeiro ** ** “ “ S. America 26d for 1 Milreis of Brazil.
& 4 “Copenhagen ** ** ** ** Denmark 13d. for 1 Crown.
& 4 ** New York 4 & 4 & ** ** United States 49d. for 1 Dollar.
NOTE. 1.-The milries of Portugal is equal to 52}d. or $1.08 which is a much larger unit than
the milries of Brazil, which is equal to 54.6 cents. The table for Portugal money is as follows:
1 milreis = 2; crusados = 25 reales = 1000 reis.
I = 10 ** = 400 * *
1 é & = 40 4 &
NotE 2. —By the decree of October 19, 1868, the money of Spain was assimilated to that of
France, and the Peseta of reales was considered as equivalent to a franc and was taken as the
monetary unit. -
NotE 3.—The new monetary unit of Denmark is the Krona or Crown which equals 100 Ore.
1 Crown = 26.8 cents or 13+d.
NotE 4.—The money of account of the Netherlands (Holland) is the florin of 100 cents,
worth 40.2 cents United States gold. The florin was formerly divided into 20 stivers.
London. Eachange on New York. London. Eacchange on Paris.
1. What is the cost of a bill on New 2. What is the cost of a bill on Paris
York for $5000, exchange at 494d. 3 for fr. 10540.50. exchange 25.25%
Ans. £1026 0s. 10d. Ans. £417 8s. 11d.
OPERATION. OPERATION,
$500 face of bill. 25.25) 10540.50 (£417.4455
494d. exchange. 20
12) 23250d. * *
20) 20520 10d. 10d. 92
381026 08.
758 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. &
London. Eacchange on Calcutta.
3. What is the cost of a bill on Calcutta for 2000 company rupees 4 annas;
exchange 23d. } Ans. £19113s. 10d.
OPERATION. *
2000 Company rupees 4 annas = 2000.25 Company rupees.
23d.
4 Explanation.—The money of ae-
3, Illſlå.S = *====ºs
= .25 12 ) 46005.75d. count in Calcutta is the rupee, which
is divided into 16 annas, and each
20 ) 3833 9}d.
) *E=-º-º-º-º-º: # anna into 12 pice.
fº 191 13s.
London. Eachange on Shanghai.
4. What is the cost of a bill for 1250 taels 5 maces; exchange 584d.
Ans. £304 16s. 24d.
FIRST OPERATION, Ea:planation.-Accounts are kept in China in
tael, maces, candarines and cash. The table is
1250 taels 5 maces = 12505 maces = 125050 10 cash, 1 candarine, 10 candarines, 1 mace and
candarines = 1250500 cash. 10 maces, 1 tael. The tael of Shanghai is equiv-
alent to $1.14 while the tael of Haikwan or
1250500 customs tael, is equivalent to $1.27.
1000 || 584 C * º Il Oº gold and silver coins in
gamsºs ºssº ina. e only money made is the cash, and
12 ) 73154.250d. it is * coined %. ºft It is composed of 6
**E= parts of copper and 4 of lead. It is circular in
20 ) 6096. 2.250d. = 2+d. form, marked on one side only and has a square
amºsºmsºmº hole in the center, by which means they are
3304 16s. carried like beads on a string or wire. Foreign
coins are in general circulation in the ports
SECOND OPERATION. open to the commerce of the world and they
are considered as bullion rather than coin, and
1250 taels 5 maces = 1250.5 are usually taken by weight. , Silver ingots are
1250.5 × 58+ = 73154.25d. also used as money, being weighed by the unit
73154.25 reduced to £. s. and d. = £304 16s. tael. See Exchange on China.
t 2#d. as in the first operation.
UNITED STATES VALUE OF UNITED STATES STOCKS AND BONDs,
WHEN QUOTED IN LONDON, AND THE LONDON VALUE OF
SAME WEEN QUOTED IN THE UNETED STATES.
PROBLEMIS.
1321. 1. United States 4 per cent bonds are quoted in London at 1154; what
is the equivalent quotation in the United States, sterling exchange being 4.84?
*~ Ans. 112.26;.
OPERATION.
4.86 × 1153 = 56133 – 5 = 112.26%.
NOTE.-All American stocks and bonds are quoted in London on an assumed value of the
pound sterling. This assumed value is $5, modified by the rate of exchange. See Stocks and Bonds
in this book, pages 708 and 709.
2. The quotation of American stocks in New York is 934, exchange is 4.864.
What is the equivalent London quotation ? Ans. 95.8-1-.
OPERATION.
93.25 × $5 = 46625 - 4.86} = 95.8+.
3. Louisiana 4 per cent consols are quoted in New Orleans at 101, sterling
exchange is 4.85. What is the equivalent London quotation ? Ans. 104.12+.
FRENCH EXCHANGE. 7.59
FRENCEI EXCELANGE.
NOTE.-See French Monetary Table, page 235.
1322. The intrinsic value of the franc and the United States legal value for
Customhouse computations, is $.193 or 1932.
The rate or course of eachange on France is always the variable number of
francs, or francs and centimes, allowed for a dollar.
The basis for the computations of the rate of exchange is 5.20 francs for $1
gold, and it is quoted 5.20 more or less according as premium is charged or discount
allowed.
NOTE.—In English exchange the rate is the dollar and cent value of £1, according as
premium is charged or discount allowed. But in French exchange the rate is the franc and
centime value of $1. Hence in English exchange the higher the rate the dearer is exchange. But
in French exchange the higher the rate the cheaper is exchange, and the lower the rate the higher is
exchange.
TO FIND THE RATE OF EXCHANGE WEHEN A PEREMIUM IS OHARGED,
1323. 1. Taking into consideration the balance of trade, the worth of money,
etc., the banker or seller of French bills determines to charge say, 2 per cent pre-
mium. What would be the gold rate, and if gold is 15 per cent premium, what
would be the currency rate for French bills? Ans. Fr. 5.09%, gold rate.
Fr. 4.433+, currency rate.
OPERATION.
Fr. 5.20 = $1.00 par of exchange. Ea:planation.—By the operation, we see that
02 = 2 % premium added. the cost of fr, 5.20 exchange, at 2 per cent
$1.02 cost of 5.20 exchange. premium, is $1.02. The reasoning for the line
Fr. statement is as follows: If $1.02 buy fr. 5.20,
5.20 ic. will buy the 102d part, and $1, or 100c. will
1.02 1.00 Thuy 100 times as much, which is practically fr.
| Fr. 5.09%+ practically fr. 5.09%. 5.09%. *
To find the currency rate of exchange at 2 per cent premium, gold at 15 per
cent premium, we make the following statements:
FIRST STATEMENT. SECOND STATEMENT.
Fr. 5.20 = $1.00 par of exchange,
5.20 .02 = 2% premium.
1.02 | 1.00 -->
115 100 $1.02 gold cost of fr. 5.20.
wº-ºº-º- .153 = 15% premium on gold.
Fr. 4,433 Ans. -----
$1.173 currency cost of fr. 5.20 exchange.
Practically fr. 4.433.
Fr.
5.20
1.173 | 1.000
f Fr. 4,433 Ans.
760 soul E's PHILOSOPHIC PRACTICAL MATHEMATICs. . º
TO FIND THE RATE OF EXCHANGE WHEN A DISCOUNT IS ALLOWED.
1324. 1. What would be the gold rate if the banker or seller allowed 1 per cent
discount, and if gold was 124 per cent premium, what would be the currency rate?
Ans. Fr. 5.252, gold rate. Fr. 4.6689, currency rate.
OPERATION.
Fr. 5.20 = $1.00 par of exchange. Fr.
.01 = 1% discount. | 5.20
.99 il.00
Fr. 5.252+ Ans. Practically fr. 5.25%.
Operation to find the currency rate.
$.99 cost of fr. 5.20 exchange.
Fr. Fr. 5.20 = $1.00 Fr.
5.20 .01 5.20
.99 || 1.00 * *- 1.11375 | 1.00000
112} | 100 or, .99
*m-º-º-º-º: .12375 = 124% prem. Fr. 4.6689 Ans.
Fr. 4.6689 Ans.
$1.11375 cost of fr. 5.20.
TO FIND THE PREMIUM CHARGED OR THE DISCOUNT ALLOWED,
WHEN THE RATE OF EXCELANGE AND THE PAR ARE GIVEN.
1325. 1. The rate of exchange is 5.09%+, the par is 5.20, what was the rate
per cent premium charged? Ans. 2%.
OPERATION.
Fr. 5.20 par of $1. G. Or, with the G.
5.09%+ = rate of exchange deducted. 10}} fractions re- 51 || 5.20
•ºsmºs - 5.093+ | 100 duced thus: 26000 || 51
.10}} gain on fr. 5.09%+. | - 100
| 2% Ans.
2% Ans.
NotE.—It will be observed that the seller gains on his investment, of 5.09%+ francs and not
on the par of 5.20.
2. The currency rate of exchange is 4.433, the premium on gold is 15 per
cent, what was the rate per cent premium charged, or discount allowed by the seller
of the exchange? Ans. 2% premium.
OPERATION.
Fr. 4.433 currency rate. G.
s .665 = 15% premium on gold. ,102
-º-º: 5.098 || 100
Fr. 5.098 gold rate. -ºmsºme -
5.20 par of exchange. 2+ practically 2% premium.
.102 gain on fr. 5.098.
4k FRENCH EXCHANGE. 761
3. The rate of exchange is 5.25+, and the par 5.20, what was the rate of dis-
count allowed by the seller of the exchange? Ans. 1% discount.
OPERATION.
Fr. 5.20 par of $1. L.
5.25+ rate of exchange. 4 || 2
dºme 2101 || 4
100
| $38% praetically 1% discount.
NOTE.--It will be observed that the seller loses on his investment of 5.25+ francs and not on
the par of 5.20.
We have omitted the reasoning in the line statements of the foregoing prob-
lems, because so often given in previous operations.
Pages 437 to 448 give full explanations of similar percentage line statements.
.05% loss by the exchange.
TO CONVERT UNITED STATES MONEY INTO FRANCS,
1326. 1. What is the equivalent value in francs of $5000, exchange 5.15%
Ans. Fr. 25775.
OPERATION.
Fr. § exchange value of $1. Ea:planation.—As the rate of exchange is the
number of francs and centimes given for $1,
2575000 and as the rate is here fr. 5.15%, it is clear that
2500 for $5000 we can buy 5000 times the fr. 5,154;
&= - hence we have but to multiply the rate of
tºm-º. exchange by dollars to convert dollars to francs.
Fr. 25775.00 Ans.
2. What amount of exchange can I buy for $1410.60, exchange at 4.45}}
Ans. Fr. 6280.6965.
OPERATION.
$1410.60 × 4.45+ = fr. 6280.6965 Ans.
TO CONVERT FRENCEI MONEY INTO UNITED STATES DOLLARS.
1327. 1. What is the equivalent value in dollars of fr. 25775, exchange 5.15%
Ans. $5000.
OPRRATION. * Explanation.—As the fr. 5.15} is the exchange
equivalent of $1, it is clear that in fr. 25775 we
5.15} = 5.155 ) 25775,000 ($5000 Ans. have as many dollars as it is times equal to fr.
5.15}. Hence we divide the francs by the rate
of exchange.
2. What is the cost of fr. 50500.404, exchange fr. 4.05%
Ans. $12446.18.
OPERATION.
N
T.
2 || 101000.81 or, 4.0575 ) 50500.405 ( 12446.18 Ans.
1623 || 4
$12446.18 Ans.
762 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. *
TO INVEST IN FRANCS FOR A CORRESPONDENT AND CHARGE
COMMISSION.
1328. 1. Invested $10000 in French exchange at 5.20. My commission on the
amount invested is 3 per cent; what is the face of the bill bought?
Ans. Fr. 51741.24, face of bill.
OPERATION.
$
100
201 || 2
10000
5.20
Fr. 51741.24 Ans.
*
TO PUFCHASE FRENCEI EXCELANGE ON ORDERS FROM DISTANT
CITIES.
1329. 1. A banker in New York instructs his correspondent in New Orleans
to purchase fr. 200000. He limits the price, including all expenses, at 5.20. New
Orleans sight exchange on New York at the time is # per cent discount. The cor-
respondent who makes the purchase charges 4 per cent commission on prime cost.
What is the cost of the bill, what is the commission and discount, and what is the
maximum purchasing rate of exchange?
Ans. $38461.54, cost.
$189.93, commission.
$284.90, discount.
$5.26; maximum purchasing rate of exchange.
solution.
Operation to find the cost of the 200000 francs at 5.20, the limited price.
Fr. 200000.00 -- 5.20 = $38461.54 cost, including 4% commission and #96 discount.
Operation to deduct #96 commission and #96 discount, and thus produce the prime cost.
$100.00 prime cost assumed.
.5
0 = }% commission. 100
.75 = #96 discount. 101.25 || 38461.54
$101.25 = cost including the commission and discount. | $37986.71 prime cost.
189.93 = }% commission.
284.90 = #96 discount.
---
$38461.54 Ans. Total cost.
Operation to find the maximum purchasing rate of exchange.
37986.71 ) 200000 (5.2649, practically 5.26; maximum purchasing rate of exchange.
2. Suppose in the above problem, the price had been 5.25 and the rate of
New Orleans exchange on New York had been # per cent premium, instead of #
per cent discount, what would have been the full cost of the 200000 francs?
Ans. $38381.67.
4. FRENCH EXCHANGE. 763
SOLUTION.
Operation to find the cost of the 200000 francs at 5.25, the limited price, without considering the
# per cent premium or the # per cent commission,
Fr. 200000.00 - 5.25 = $38095.24 cost.
This cost includes the # per cent commission but is exclusive of the allowance
to be made for the # per cent premium. The following operation will allow for the
# per cent premium, and exclude or deduct the 4 per cent commission.
$100.00 prime cost assumed.
.50 = }% commission. 100
* 99.75 | 38095.24
$100.50 cost with #% commission included. --sºmº
5 = #% premium on $100 deducted. $38190.72 prime cost.
gºmº- 190.95 = }% commission added.
$99.75 cost of $100 including 4% commission --se *
and allowing for the #96 premium. $38381.67 Ans. Full cost.
286.43 = #% premium deducted.
$38095.24 net cost after paying #96 com-
mission and receiving #%
premium on exchange.
MISOELLANEOUS PRACTICAL PROBLEMS.
1330. 1. What cost fr. 2500.50, exchange 5.32% Ans. $470.02.
2. Find the value of 6400 francs at 5.214. Ans. $1227.23.
3. Find the value of 5940.75 francs at 5.18%. Ans. $1145.20.
4. How many francs can be bought for $10000 at 5.22; ? Ans. Fr. 52250.
5. Invest $4000 in francs at 5.254. Ans. Fr. 21010,
6. What are the proceeds of a draft on Paris for 32500 francs, exchange
5.30, and brokerage on exchange # per cent Ans. $6124.40.
OPERATION.
32500 - 5.30 = $6132.07. 96 on $6132.07 = $7.67. $6132.07 – $7.67 = $6124.40 Ans.
7. What is the cost of a draft on Paris for 32500 francs, exchange 5.30,
brokerage on exchange # per cent 3 Ans. $6139.74.
8. What will be the face of a bill on Havre that is bought for $5000,
exchange being 5.353% Ans. Fr. 26768.75.
9. Sold exchange for 12000 francs on Paris through a broker; exchange
5.23, brokerage 4 per cent. What were the proceeds of the bill? Ans. $2288.72.
10. A French invoice amounts to 5840.60 francs; the duty on same is 50 per
cent, the freight is $480. What is the importing cost? Ans, $2170.85.
OPERATION.
Fr. 5840.60 × $.193 = $1127.24 invoice -- 50% duty = $1690.85, -ī- $480 freight = $2170.85 import-
ing cost.
11. What cost a bill for the above fr. 5840.60, exchange at 5.23?
Ans. $1116.75.
764 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
12. Imported from France, merchandise amounting to fr. 8542.50. What is
the value in United States money? Ans. $1648.70,
13. Bought through a broker a bill for fr. 25800. Exchange 5.16, brokerage
+ per cent. What was the cost of the bill ? Ans. $5012.50.
NOTE.-25800.00 - 5.16 = $5000. 4% on $5000 = $12.50. $5000 + $12.50 = $5012.50.
14. Invested through a broker $7400 in French exchange at 5.24, paid per
cent brokerage. What was the face of the bill? Ans. Fr. 38727.58+.
NOTE.-$7400.00 – 100% = $7390.76 × 5.24 = fr. 38727.5824.
15. Received $1425 for a French bill that was sold through a broker at
5.25; the brokerage was # per cent. What was the face of the bill? *
Ans. Fr. 7500.
OPERATION INDICATED.
$1425 -- 99% = $1428.57+ x 5.25 = fr. 7499.99-H, practically fr. 7500.
1331. PARIS Exchange oN-
Fr. C.
25.20 per £1 Sterling.
London, short and 3 months,
3 & 4 99.50 ** 100 Francs.
e & 4 & 4
Belgium,
Italy, “ (: 3 é & tºs tº gº 89.50 “ 100 Lire (paper).
Netherlands, 3 months, º * * º 211.50 ** 100 Florins.
Germany, 3 * * tº tº º tº e tº 122.50 ** 100 Marks.
Austria, 3 * * * = g- gº º tº- 105 ** 100 Crowns.
Spain, 3 * * - - - - 505 ** 100 Pesos.
Portugal, 3 “ tº- †- tº gº 550 ** 100 Milreis.
Russia, 3 * * tºº. wº - - 312 ** 100 Roubles.
NotE.—These are average rates and may be more or less according to circumstances govern-
ing exchange.
EXCHANGE ON BELGIUM, SWITZERLAND, ITALY, SPAIN, GREECE,
AND FINLAND.
1332. The monetary unit of all these nations is the same, in value, as that of
France, and hence the operations of exchange would be the same as in French
exchange.
NOTE 1.-In Italy the monetary unit is called the lire, of 100 centesimi, in Spain the peseta,
of 100 centesimos; in Greece, the drachme of 100 lepta; and in Finland, the mark of 100 pfennigs
NOTE: 2.—The old monetary unit of Spain was the hard dollar or duro, valued at 99c. con-
taining 20 reales, each reale containing 34 maravedis,
GERMAN EXCELANGE.
NOTE.—See German Monetary Table, page 236.
1333. The present money of account throughout the German Empire is the
mark (Reichsmarks).
NOTE:-Following the German and French war, came the formation of the German Empire
and the unification of her Monetary Units, the Thaler, Florin and Marc Banco. By acts of the
German Parliament taking effect, one Dec. 4, 1871, and one July 9, 1873, the “MARK" was adopted
as the monetary unit of the German Empire.
The German Monetary Law of Dec. 4, 1871, ordains 1st, That “there shall be coined an
º: GERMAN EXCHANGE. 765:
imperial gold coin, 1394 pieces of which shall contain one pound of pure gold,” 2d, That the
“tenth of this gold coin shall be called MARK, and shall be divided into 100 pfennigs.” The same
law also provides that “besides the imperial gold coin of 10 marks, there shall be coined imperial
gold coins of 20 marks, of which 69% pieces shall contain one pound of pure gold.” By act taking
effect July 9th, 1873, 5 mark gold coin are also authorized, 279 of which must contain one pound
of pure gold. By this same act the following coins are also authorized: 1st. Silver coins, 5 mark,
2 mark, 1 mark, 50 penny and 20 penny pieces. 2d. Nickel coins, 10 penny and 5 penny pieces. 8d.
. Copper coins, 2 penny and 1 penny pieces. A pound of fine silver is coined into 20 five mark, 50
two mark, 100 one mark, 200 fifty penny and 500 twenty penny pieces.
The gold and silver coins are all ºf pure and To alloy.
The intrinsic and the United States Customhouse value of the mark in U. S.
gold is 23.8 cents; which is a smaller value than any of the previous German
monetary units; and because of its small value, exchange dealers and bankers
find it more convenient to base the rate of German exchange upon the equivalent
value of 4 gold marks expressed in United States money. Thus 23.82 x 4 = 95.2
cents, practically for exchange operations, 95% gold, which is quoted more or less
according as a premium is charged or a discount is allowed. To find the currency
or silver rate, the premium on gold is added to the gold rate.
TO CONVERT GERMAN MONEY INTO AN EXCELANGE EQUIVALENT
OF UNITED STATES DOLLAPS.
1334. 1. What is the gold cost of 15000 marks, exchange at 94%
Ans. $3525.
OPERATION.
C. Ea:planation.—The rate of exchange being the
4 # 000 price of 4 marks, we reason thus: Since 4 marks
Ö equal 94 cents, 1 mark is equal to the 4th part,
$3525.00 Ans. and 15000 marks, 15000 times as much.
2. What will a bill of exchange for 24832 marks cost; exchange at 95%, gold
11 per cent premium ? Ans. $6580.79.
OPERATION.
C. Ea.planation.—Here we reason thus: Since 4
* #sº marks cost 13+ cents, 1 mark will cost the #
*sº-º-º-º: part, and 24832 will cost 24832 times as much,
$5928.64 gold. te which is $5928.64 gold, to which we add the
652.15 = 11% premium on gold. premium on gold and produce the correct result
$6580.79 Ans. $6580.79.
3. Received instructions from a correspondent to purchase German exchange
to the amount of 4189. marks and 45 pfennigs. Exchange is 1073. What is the
cost? Ams. $1125.91.
OPERATION INDICATED.
C.
2 || 215
4 || 4189.45
766 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS *
ge
TO CONVERT UNITED STATES DOLLARS INTO AN EXCHANGE EQUIV-
ALENT OF MARKS.
1335. 1. What amount of German marks can be bought for $10000 gold,
exchange being 96? Ans. 41666.67 marks, or 41666. marks and 67 pfennigs.
OPERATION.
M. Ea:planation.—In this problem, we reason thus:
| 4 Since 96 cents buy 4 marks, 1 cent will buy the
96 |10000.00 96th part, and 1000000 cents will buy 1000000
41666.66% marks. times as many.
2. Invested $1260.55 silver in German exchange at 1054. What is the face
Of the bill ? Ans. 4790.69 ms.
NOTE.-The rate of exchange in this and the two following problems is based upon the
supposition that gold is at a premium over silver. *
FIRST OPERATION. SECOND OF ERATION.
M. M.
4 4
4.21 | 4 & 105.25 | 1260.65
1260.55 --
|ſº marks. Ans.
4790.69 marks. Ans.
3. A merchant invested for a correspondent $5000 silver, in German
exchange. The rate of exchange was 104; he charged 1 per cent commission on
the silver invested. What amount of marks did he buy ? Ans. 19040.37 ms. T
* FIRST OPERATION, SECOND OPERATION.
$ $ M,
| 100 100 | 4
º; 5000.00 101 || 5000 104 || 4950.495
1.04 || 4 mºmºmºsºm-º. ===- -
* = &ºm=º lsº invested. 19040.37 marks.
19040.37 marks. +
4. Invested for a correspondent $20000 silver, in German exchange at 1083.
How many marks did I buy, and how much does my correspondent owe me, allowing
1 per cent commission on the $20000 for investing 3 Ans. 73563.22 marks.
$20200. he owes.
1336. HAMBURG EXCHANGE ON.—
Portugal, 3 “
3 & 4 450 ** 100 Milreis.
Spain,
80.25 ‘‘ 100 Pesetas.
Ms. Ps.
London, short and 2 months, tº • * = 20.42 per £1 Sterling.
& 4 & 4 gº sº tº & 4
#, . . . . . . . ;as #:
Nº. é & º § * * * * * * sº 169,50 ** 100 Florins.
New Yor & 4 & º tº º * , 413 ** 100 Dollars.
Austria, 3 months, sº * > 4-> 175 “* 100 Florins.
Italy, & 4 sº ºp tºº 75 ‘‘ 100 Lire.
• NOTE.-These are average rates subject to an increase or decrease according to circumstances
governing exchange. -
& AUSTRIAN EXCHANGE. 767
MISCELLANEOUS PRACTICAL PROBLEMS.
1337. 1. Imported from E. Gretzner of Dresden, goods amounting to 85420.75
marks. If the duty is 60 per cent and the freight $340,30, what is the importing
cost 3 Ans. $32868.52.
OPERATION.
Marks 85420.75 × $.238 = $20330.14, Dresden cost of invoice; 60% duty = $12198.08 $20330.14 +
$12198.08 -- $340.30 = $32868.52.
NOTE.—The commercial student should make the proper entries for the above transactions.
2. Remitted to E. Gretzner, a bill of exchange for the above invoice of
85420.75 marks, exchange 964. What was the cost of the bill? Ans. $20607.76.
OPERATION INDICATED.
85420.75
4 | .965
NotE.—The commercial student should make the proper entries for the above transactions.
3. Imported from A. T. Kohn, Hamburg, Germany, merchandise amounting
to 4212.15 marks. What is the importing value in American money?
Ans. $1002.49.
4. Remitted to A. T. Kohn, a bill for 4212.15 marks bought at 94. What
did it cost 3 Ans. $989.86.
NOTE.-The commercial student should make the proper entries for the remittance of this
bill.
See pages 248, 249, 270, 271, 272 and 273, of Soulé's New Science and Practice of Accounts, for
the proper Invoice and Cash Book entries for such transactions as the above.
5. Invested for a correspondent $20000 in German exchange, at 95%; charged
A per cent brokerage. What was the face of the bill and what was my brokerage?
Ans. 83571.65 marks face of bill. $99.50 brokerage.
6. Bought for a correspondent 20000 marks at 954, and charged # per cent
brokerage. What was the cost of the marks, what was my brokerage, and what
was the total cost 3 Ans. $4762.50 cost of marks.
$11.91 brokerage.
$4774.41 total cost.
AUSTRIAN EXCHANGE.
1338. The kingdom of Austria-Hungary, by law of August 2, 1892, changed
its monetary standard from silver to gold, and by the same law it changed its
monetary unit from the florin to the krone (Crown) which is divided into one
hundred “hellers.”
Under this new law the coin of Austria-Hungary, is as follows:
(1). 20 and 10 gold crown pieces. (2). 1 crown silver pieces. (3). 20 and
10 nickle and heller pieces. (4). 2 and 1 bronze heller pieces.
768 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
The value of the gold crown is 20.3 cents American gold. At this valuation
the United States Customhouse estimates merchandise imported from Austria. See
Table of Foreign Coins for the value of the former monetary units.
Notwithstanding the new monetary unit and the new monetary standard
adopted by Austria in 1892, American and European bankers continue (August,
1895,) to base Austrian exchange on the value of the florin as it existed prior to
1892. This gold value of the florin is 48.2 cents; the silver value is 38 cents more
or less, according to the United States mint value of silver.
The ratio value between the new unit, the crown, and the old unit the florin,
is practically 100 crowns equals 42 florins gold; or, 100 fiorins equals 237.10 crowns.
The eacchange value of the florin varies according to the premium or discount
declared, and according to the value of silver in the United States.
NotE.—The Metric System of Weights and Measures is compulsory in Austria-Hungary.
PROBLEMIS.
TO CONVERT AUSTRIAN MONEY TO THE EXCELANGE EQUIVALENT
OF UNITED STATES DOLLARS.
1339. 1. What is the cost of 1480.45 florins, exchange at 414 cents?
Ans. $614.39.
OPERATION.
Fl. 1480.45
41%
$614,3867; Ans.
2. Bought 5862.70 florins, exchange 45, and paid 3 per cent commission.
What was the cost $ Ans. $2651.41.
TO CONVERT UNITED STATES DOLLARS TO THE EXCHANGE EQUIV-
ALENT OF AUSTRIAN MONEY.
1340. 1. How many florins can be purchased for $9703.45, exchange at 48%
cents 3 Ans. Fl. 20007.11 +.
OPERATION.
97.03.45
.485)#9703.450(Fl. 20007.11+ Ans. or thus: 97 || 2
2. Invested, through a broker, $20000 in florins, exchange 40 cents, broker-
age 4 per cent. What was the face of the bill? AnS. Fl. 49875.30.
B R A Z II, I A N E X C H A N G. E.
EXCHANGE ON RIO JANEIRO.
1341. The money of account of Brazil is the reis, 1000 of which make a
milreis.
BRAZILIAN EXCHANGE. 769
The INTRINSIC value of the milreis in United States money is 54.6 cents.
The exchange value is 54.6 cents for 1000 reis, more or less, according as
premium or discount is declared.
In the notation of accounts, the milreis are separated from the reis by a sign
thus (B, or thus $, called cifrao; and the milreis from the million by a colon thus:
Rs. 4: 500 $ 200, is read four thousand five hundred milreis, two hundred reis.
NOTE.-The French Metric System of Weights and Measures is the standard, and has been
compulsory in Brazil, since 1872. But the ancient weights and measures are still partly employed
by business men. The old Libra = 1.012 lbs. avoirdupois, and the arroba = 32.38 lbs. avoirdupois.
PROBLEMIS.
TO CONVERT BRAZILIAN MONEY TO THE EXCHANGE EQUIVALENT
OF UNITED STATES DOLLARS.
1342. 1. What is Rs. 2: 430 $ 360 worth, exchange at 55; 3
Ans. $1348.85.
OPERATION.
Rs. 2430.360 × 55} = $1348.8498.
2. What will Rs. 467 $ 000, Brazilian exchange cost, exchange 53; ?
Ans. $249.85.
TO CONVERT UNITED STATES DOLLARS TO THE EXCHANGE EQUIVA-
LENT OF BRAZILIAN MONEY.
1343. 1. How many milreis can be purchased for $2400, exchange at 51 ?
Ans. Rs. 4: 705 SS 882 +.
OPERATION.
Rs.
1000
.51 || 2400.00
Rs. 4: 705 S 882
2. Invested $20000 in exchange on Rio Janeiro, exchange at 52%. What is
the face of the bill ? Ans. Rs. 38 : 095 SS 238.
EXCELANGE ON BRAZIL IN ENGLISH MONEY.
NotE.—Most of the exchange on Brazil by United States Bankers and Merchants is drawn in
Pounds, SHILLINGs, and PENCE.
77o soul.E's PHILOSOPHIC PRACTICAL MATHEMATICS.
TO REDUCE BRAZILIAN MONEY TO ENGLISH.
PROBLEMS.
1344. 1. An invoice of coffee amounted to Rs. 77 : 426 $ 100, exchange at
What was the face of the bill in sterling?
21;d. per milreis, or per 1000 reis.
Ans. £6976 8S. 3d.
OPERATION.
Rs. 77 : 426 $ 100 x 21:#d. = 1674339,4124d.
12 | 1674339.4125d.
20
139528S. 3d.
36976. 8s.
2. What is the face of a bill for Rs. 62 : 418 $ 610, exchange at 24; d?
Ans. £6274 7s. 5d.
TO REDUCE UNITED STATES MONEY TO ENGLISH.
1345. 1. What is the face of a bill for $27646.98, exchange at $4.85?
Ans. £5700 8s. 2d.
OPERATION.
...sºssmoº
0
8.1640
12
1.968
2. An invoice of coffee amounts to $8490.16. What is the face of a sterling
bill for the amount, exchange at 4.86? Ans. £1746 18s. 11d.
NOTE.—For invoices of coffee as they are made in Rio de Janeiro, and for the exchange, and
the accounts current and interest accounts resulting from the importation of coffee from Brazil, see
Customhouse Business, pages 505 to 508.
EXCELANGE ON PORTUGAL.
1346. The monetary unit of Portugal is the MILREIS, the intrinsic and
Customhouse value of which is $1.08. (534d).
The exchange value is $1.08 more or less, according as premium or discount
is declared. The operations of exchange are the same in form as those in Brazilian
exchange.
The monetary table is as follows:
1 Milreis = 24 Crusados = 25 reales = 1000 Reis.
1 & & - 10 & 4 - 400 & 4
1 * * = 40 **
The almude of Lisbon = 3.7 gallons. The almude of Oporto = 5.6 gallons.
Yºk EXCHANGE ON VARIOUS COUNTRIES. 77 I
EXCHANGE ON HONDURAS, COSTA RICA, GUATEMALA, NICARAGUA,
SALVADOR AND COLOMBIA.
1347. The monetary unit of all these countries is the silver PEso, or dollar of
100 centavos, the value of which is, January 1, 1895, 45.5 cents United States gold.
This value of the silver PESO, is determined by the value of silver at the United
States Mints.
The rate of exchange is the CENTS charged for a peso, and varies according
to the general principles of exchange, the political and civil conditions of these
countries and the facilities of trade between them and the United States.
EXCHANGE ON PERU, ECUADOR AND CHILI.
1348. The monetary unit of Peru is the silver sol, valued at, January 1,
1895, 45.5 cents gold.
The monetary unit of Ecuador is the silver SUCRſ, the value of which is,
January 1, 1895, 45.5 cents gold.
The monetary unit of Chili is the gold and silver PESO of 100 centavos, valued
at 91.2 cents gold.
The rate of the exchange is the cents charged for a sol, or SUCRE, or PESO.
NotE.—See remarks above on Honduras Exchange.
EXCELANGE ON MEXICO AND THE ARGENTINE REPUBLIC.
1349. The monetary unit of Mexico is the silver PESO or DOLLAR of 100
centavos, the value of which is, January 1, 1895, 49.5 cents gold. The gold dollar
or peso of Mexico, is worth 98.3 cents.
The rate of exchange is the cents charged for a PESO.
The monetary unit of Argentine Republic is the gold and silver PESO of 100
centesimos, the intrinsic value of which is 96.5 cents gold.
The rate of exchange is the cents charged for a PESO.
CUBAN EXCELANGE.
1350. The monetary unit of Cuba is the gold and silver PESO of 100 centesi-
mos, the intrinsic value of which is 92.6 cents.
The rate of exchange is the cents charged for a PES.
NoTE.—See Customhouse Business, page 504, for problems in importing goods from Cuba.
R U S S I A N E X C H A N G. E.
EXCELANGE ON ST. PETERSBURG.
1351. The money of account of Russia is the ruble of 100 kopecks.
772 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. Yºr
The silver ruble is the legal unit and it contains 17.9961 grams, and is worth,
January 1, 1895, in American gold 36.4 cents. The gold ruble is worth January
1, 1895, in American gold 77.2 cents.
The exchange value is 36 or 77 cents, more or less, according to the premium
or discount declared, and according as the exchange is for silver or gold rubles.
Russia has an IMPERIAL gold coin worth $7.718, and a half IMPERIAL gold
coin worth $3.859.
Paper rubles are in general circulation in Russia, and their value is much
below that of the silver ruble.
Since the reign of Peter the Great, ten changes have been made in the
weights and fineness of Russian silver coin.
NoTE.—In Russia the Julian calendar, or old style, is still retained. This calendar is now,
and till the year 1900 will remain, 12 days later than the new style, and in leap years 13 days later.
Thus, January 15 in the United States, is January 3d in Russia. From the year 1900 to 2000, the
difference will be 13 days, and in leap years 14 days. This deviation commences March 1, each
year.
PROBLEMIS.
TO CONVERT RUSSIAN MONEY TO THE EXCELANGE EQUIVALENT
OF UNITED STATES DOLLARS.
1352. 1. What is the exchange equivalent or cost of 940 rubles, 28 kopecks,
exchange at 69.4% Ans. $651.14.
OPERATION.
940 rubles, 28 kopecks = 940.28 rubles. -
69+
$651.1439
2. What cost 31428.65 rubles, exchange at 73%, brokerage for buying # per
cent 3 º: Ans. $23157.81.
TO CONVERT UNITED STATES DOLLARS INTO AN EXCHANGE EQUIV-
ALENT OF RUSSIAN RUBLES.
1353. 1. What amount of exchange on St. Petersburg can I buy for $846.10,
exchange at 70 ? Ans. 1208.71 rubles.
OPERATION.
846.10 – 70 = 1208.71%.
2. Invested $5000 in exchange on Moscow at 68. What was the face of the
bill Ans. 7352.94 rubles.
OPERATION.
$5000 – 68 = 7352.94-H.
JAPANESE EXCHANGE.
1354. The monetary unit of Japan is the gold YEN or DOLLAR which contains
100 sens. It is worth intrinsically 99.7 cents gold. The value of the Silver YEN
X- EXCHANGE ON VARIOUS COUNTRIES. 773
is, January 1, 1895, 49.1 cents American gold. The gold coins consist of 20, 10, 5,
2, and 1 YEN pieces. The silver coins are 1 YEN and 50, 20, 10 and 5 SEN pieces.
The copper coins are 2 SENS, 1, § and #6 SEN pieces. Paper currency is in general
circulation, of which there are various denominations, corresponding to those in
Coins.
The rate of exchange is the CENTS charged for a YEN which varies according
to the general principles governing exchange, and according as the exchange is for
gold or silver YENS.
NOTE.—The metallic value of silver in foreign coins depends upon the mint market value of
bullion silver in the United States.
PROBLEMIS.
1. What cost a bill on Tokio, for Y. 2530.40, exchange at 984 g. *
Ans. $2492.44.
OPERATION.
2530.40 × 984 = 2491.4440.
2. How many YENs can be bought for $5000, exchange at $1.023%
Ans. 4878.04+ yens.
OPERATION.
$5000 — 1.02} = 4878.04 + yens.
3. An invoice of silk goods imported from Yokohama, amounts to 4210.60
YENS. If the Customhouse value is 99.7 cents per YEN and the rate of exchange is
$1, what is the importing cost of the goods, and what is the cost of the bill of
exchange, to remit in payment of the invoice? Ans. $4197.97 cost of goods.
4210.60 cost of bill.
OPERATION.
4210.60 × 99.7c. = $4197.96820
4210.60 × $1. = 4210.60
4. The United States duty on the above invoice is 50 per cent ad valorem.
Supposing the freight and insurance to be 10 per cent on the invoice, what would
be the total cost to the importer ? Ans. $6716.75.
NOTE.-Commercial students should make the Journal and Cash Book entries for all the
transactions in the above 3d and 4th problems. See Soule's New Science and Practice of Accounts,
pages 249 and 270 to 271, for the proper entries.
EXCELANGE ON SWEDEN, NORWAY AND DENMARK.
1355. The monetary unit of Sweden, Norway and Denmark, is the gold KRONE
or GRowN of 100 ore, the intrinsic value of which is 26.8 cents gold.
The rate of exchange is the cents charged for a crown, or, the value of 100
crowns in dollars.
774 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
CHINESE EXCHANGE.
1356. The money of account in China consists of the TAEL, MACE, CANDARINE
and CASH. The only official coinage of China is the copper CASH, of which about
1600 to 1700 equals 1 Haikwan or customs tael.
Large payments are made by weight of silver bullion, the standard or unit
of weight being the LIANG or TAEL. Hence the TAEL is a unit of Weight rather
than a monetary unit. By treaty, it is equal in weight to 13 ounces avoirdupois,
and an Haikwan tael is this weight of pure silver.
The following is the table of weight:
10 Sze = 1 Hu.
10 Hu = 1 Hao. f
10 Hao = 1 Li (nominal cash).
10 Li = 1 Fun (Candaren).
10 Fun = 1 Tsien (Mace).
10 Tsien = 1 Liang (Tael) 1% oz. avoirdupois by treaty.
16 Liang = 1 Kin (Catty) 14 lbs. { { { { “.
100 Kin = 1 Tan (Picul) 1334 lbs. { { & 4 & 4
NOTE.-The Chinese standards of weight and length vary throughout the Empire.
The intrinsic value of the silver tael of Shanghai is, January 1, 1895, 67.3%
American gold; and that of Haikwan or customs, is 74.9 cents gold.
NOTE.—By an imperial decree, issued in 1890, the silver dollar coined at the new Canton
Mint is made current all over the Empire. This dollar is equal in value to the United States and
Mexican silver dollar and to the silver Yen of Japan, taken at the United States Mint, market
value.
The rate of exchange is the variable dollar and cents charged for a tael.
PROBLEMS.
1. What is the cost of a bill for 2400 taels, exchange at 674 cents?
Ans. $1620.
OPERATION.
2400 × .67% = $1620.
2. What cost 1650 taels 840 cash, exchange at 75 cents? Ans. $1238.13.
OPERATION.
1650.840 at .75 = $1238.13.
3. Invested $3000 in taels, exchange at 72 cents. What was the amount
bought 3 Ans. 4166.6663 taels, or 4166 taels and 6663 cash, or 4166
taels, 6 mace, 6 eandarine, and 63 cash.
OPERATION.
$3000+ 72c. = 4166.666;.
ARBITRATION OF FOREIGN EXCELANGE.
1357. 1. A merchant in New Orleans wishes to remit 10000 marks to
Hamburg, direct exchange is 962. Exchange on London is 4.87; London
+% ARBITRATION OF FOREIGN EXCHANGE, 775
exchange on Hamburg is 20.50 m. for £1; New Orleans exchange on Paris is fr.
5.10; Paris exchange on London is fr. 25.10, and London exchange on Hamburg is
20.50 m. for £1. Which is the most advantageous way to remit?
Ans. Through London.
SOLUTION.
First operation to find the cost of direct exchange.
C.
96
4 || 10000
| $2400.00 cost of direct exchange.
Second operation to find the cost of exchange through London.
$
4.87 = 1f.
1:9 – m. 20.50 | 10000 marks.
| $2375.61 cost through London.
Third operation to find the cost of exchange through Paris and London.
Francs tº- 5.10 | 1.00 dollar.
Pound sterling 1 25.10 francs.
Marks - 20.50 | 1 pound sterling.
10000 marks.
$2400.77 cost through Paris and London.
2. I owe a correspondent in Paris fr. 25000, which I wish to settle by
remitting exchange. Direct exchange on Paris is 5.10. tº
But before making a purchase I find on inquiry that exchange on London is
4.87, and London exchange on Paris is £1 for fr. 25.20.
I also learn that exchange on New York is # per cent discount, and that New
York exchange on London is 4.86%, and London exchange on Paris as above, fr.
25.20. Allowing # per cent commission on the whole amount invested at each
intermediate place for re-investing, which is the most advantageous way to remit 3
º Ans. Through New York and London.
SOLUTION.
First operation to find the cost by direct exchange.
Francs 5.10 | 1
25000.00 francs.
*= assºsºme
$4901.96 cost, Ans.
Second operation to find the cost through London.
Proceeds of amount invested 99.50 100.00 assumed investment.
Francs given for £1 25.20 | 1 pound sterling.
Pound sterling 1 || 4.87 exchange value of 1:8.
25000 francs.
$4855.64 cost. Ans.
Or thus :
99.50 || 100 25.20 | ? = francs. ? = £.
25000.00 4.87
| ? = francs. ? = £. ? = $4855.63 Ans.
Third Operation to find the Cost through New York and London.
Assumed face of New York bill 100 | 99.25 cost of New York bill.
Amount invested in New York 99.50 | 100 cost of investment in New York.
Amount invested in London 99.50 | 100.00 cost of investment in London.
Francs given for £1 25.20 | 1 pound sterling.
Pound 1 || 4,865 exchange value of 1:6.
25000 francs.
$4838.46 cost. Ans.
776 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. jºr
3. Suppose in the operation through New York and London that commission
for reinvesting had been allowed only on the net amount invested, what would have
been the cost of exchange 3 Ans. $4838.21.
OPERATION.
Assumed face of New York bill 100 99.25 cost of New York bill.
Assumed investment in New York 100 || 100.50 cost of New York investment at , % commission.
Assumed investment in London 100 || 100.50 cost of investment in London at 3% commission.
Francs given for £1 25.20 | 1 pound sterling. -
Pound sterling 1 || 4.865 exchange value of 1.6.
25000 francs.
$4838.21 Ans.
THE UNIFICATION OF TEIE MONEY OF THE WORLD.
1358. The importance of a uniform monetary measure for the world is so
great that We cannot close our work on exchange without referring thereto.
y The great variety of monetary units of the different nations of the world, and
the consequent complications and intricacies of exchange, as well as the loss of time
tand increase of labor that are thereby entailed upon commercial men, proves most
clearly the need for the unification of the money of the world.
Since the downfall of the Roman empire, the various countries of Europe
have been afflicted with a heterogeneous coinage, and for ages past the loss of time
and money, resulting from this cause, has been seriously felt by travelers and
business men of all classes and climes.
Napoleon the First, looking down on the world from the rock of St. Helena,
declared that what Europe most needed was a common law, a common measure, and
a common money.
This solemn and wise utterance was a legacy not only to Europe but to the
whole family of nations. In 1821, the American Secretary of State, John Quincy
Adams, in his celebrated report to the Congress of the United States, pointed out
the incalculable advantages of a common measure and a common money, which
should, in his own comprehensive language, “overspread the globe from the equator
to the poles.”
With clear political sagacity, he saw, and said, that the object could only be
accomplished “by a general convention of nations to which the world shall be
parties,” and “in which the energies of opinion must precede those of legislation.”
In 1867, the delegates of nineteen nations, the United States being one, representing
320 millions of people, met at Paris to consider the important question of unifying
the money of the different nations, and in 1892 an international monetary conference
was held in Brussels, at which the question of a single monetary unit for the
leading nations of the earth was, with other financial matters, discussed; and
although much good has resulted from these conferences, the great object remains
unaccomplished. -
We, therefore, urge the commercial student and general reader to labor in
behalf of this cause until the dollar of the United States, the Pound of England,
the Franc of France, the Mark of Germany, the Crown of Austria, the Ruble of
Russia, the Milries of Brazil, the Tael of China, the Yen of Japan, and the other
monetary units of the world are all of the same metal, weight, and purity. The
names need not be the same, but the value and subdivisions should. This, the age
demands, commerce requires, and common sense approves; and but for the preju-
dice, the stupidity, bigotry, and the false national pride and fear of a few crowned
heads, it would have been accomplished years ago.
s TORAGE.
G). ---------------------R
A-4--a
vºy ºur
1359. Storage is the keeping of goods in a store, warehouse, or other place of
deposit, until required for use or shipment.
1360. Storage is also the price or charge made for the keeping of goods in a
. Storehouse.
1861. Cold Storage is storage in refrigerating chambers or other places
artificially cooled, to preserve articles liable to be damaged by heat.
1362. Bonded Warehouses are Government storehouses in which imported
goods on which duties have not been paid, are stored.
1363. A Grain Warehouse, in which grain is stored, is called an elevator.
1364. Cash Storage is storage that is computed and paid on each withdrawaſ
or delivery of goods.
1365. Storage Receipts or Warehouse Receipts are receipts showing the
amount of goods in the storehouse, and are frequently bought and sold, especially
those for grain and other produce, and are used as collaterals in discount and
exchange transactions.
1366. Credit or Time Storage is storage on one or several lots of goods
that may be received and delivered in different quantities at different times, and
the charges made periodically, or at the final withdrawal.
1367. The Rates of storage charges are generally fixed by the Boards of
Trade or Chambers of Commerce of the different cities, and are most usually made
by the month of 30 days, per box, barrel, bale, sack, etc. No deductions are
allowed when merchandise or property is withdrawn before the close of the month.
A fractional part of a month is counted as a full month.
In some cities and towns, if the property is taken out within fifteen days
after the expiration of the first month, a half month is charged, but if after fifteen
days a whole month.
The owners or agents of property pay all expenses incurred at the store or
warehouse for receiving, storing and delivering the property.
(777)
778 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. º
1368. Average Storage. Independent of the regular storage charges goods
are sometimes received, and, by special agreement, charges are made for the actual
time they were stored. In cases of this kind it is customary, in computing the
storage, to average the time, and charge a certain rate per month of thirty days.
PROBLEMS.
TO FIND THE AMOUNT OF STORAGE PER ARTICLE PER MONTH.
1369. 1. What is the cost of storage on 500 barrels molasses from November
3, to December 12, 1895, at 62 per barrel per month ? - Ans. $60.
OPERATION.
500 bbls. from Nov. 3, to Dec. 3, − 1 month - - - - - at 6c. = $30.00
500 bbls. from Dec. 3, to Dec. 9, - fraction of month - - - - at 6c. = 30.00
Total amount, - - - - - - - - - - - $60.00
Ea:planation.—The general custom, as above stated, is to charge for a whole month whenever
the goods are on storage a fraction of a month.
2. What would be the cost of storage on 300 bbls. flour received Oct. 10, and
delivered Oct. 28; 400 bbls. received Oct. 16, and delivered Nov. 15, and 600 bbls.
received Nov. 18, and delivered Nov. 30, at 5% per bbl. per month 3 Ans. $65.
OPERATION.
300 bbls. Oct. 10, to Oct. 28, fraction of a month º - º at 5c. = $15.00
400 bbls. Oct. 16, to Nov. 15, 1 month - - - - - - at 5c. = 20.00
600 bbls. Nov. 10, to Nov. 30, fraction of a month cº- tº- º at 5C. = 30.00
$65.00
TO FIND THE AMOUNT OF STORAGE FOR AVERAGE PERIOD OF
THIRTY DAYS PER, AFTICLE.
|
1370. 1. What is the cost of storage on 500 bbls, molasses received Nov. 3,
and delivered Dec. 12, 1895, at 6 cents per barrel; Ans. $39.
OPERATION. -
1895.
Nov. 3. 500 bbls. for 39 days = 19500 bbls. for 1 day. 6c. × 650 = $39.
19500 - 30 = 650 bbls. for 30 days, or 1 month.
TO FIND THE AMOUNT OF STORAGE PER AVERAGE PERIOD OF
THIRTY DAYS PER ARTICLE WEHEN DELIVERIES
HAVE BEEN MADE.
1371. 1. From the warehouse books we find the following receipts and deliv-
eries of flour, for the account of F. W. Ahsen & Co. By contract, storage is charged
for the actual time the flour was on storage, at the rate of 5% per barrel per month:
STORAGE DAILY BALANCE METHOD, 779
Receipts : Nov. 4, 1895, 200 bbls.; Nov. 5, 700 bbls.; Nov. 15, 1300 bbls.;
Nov. 29, 3800 bbls.; Nov. 30, 1200 bbls.
Deliveries: Nov. 7, 1895, 1000 bbls.; Nov. 11, 500 bbls.; Nov. 18, 600 bbls.;
Nov. 20, 100 bbls.; Nov. 27, 1000 bbls.; Dec. 5, 1400 bbls.; Dec. 7, 1200 bbls.;
Dec. 13, 800 bbls.; Dec. 16, 150 bbls.; Dec. 21, 300 bbls.; Dec. 24, 500 bbls.; Dec.
27, 300 bbls. What is the amount of storage due January 1, 1896, and how many
barrels remain in the warehouse? Ans. $234.15.
1150 barrels in warehouse.
FIRST OPERATION BY THE DAILY BALANCE METHOD.
F. W. AHSEN & CO.
* §3
1895. No. bbls. received. No. bbls. delivered. No. bbls. in warehouse. : g No. bºº for
2. aº
Nov. 4 2000 2000 1 2000
{ { 5 700 2700 2 5400
& 4 7 1000 1700 4 6800
** 11 500 1200 4 4800
** 15 1300 2500 3 7500
** 18 600 1900 2 3800
4 |20 100 1800 7 12600
tº 27 1000 800 2 1600
44 |29 3800 4600 1 4600
44 |30 1200 5800 5 29000
I)ec. 5 1400 4400 2 8800
& 4 7 1200 3200 6 19200
44 |13 800 2400 3 7200
{{ {16 150 2250 5 11250
4 & 21 300 1950 3 5850
4 & 24 500 1450 3 4350
« |27 300 1150 5 5750
9000 7850 30 ) 140500
4683;
9000 bbls. Teceived. Practically 4683 barrels.
7850 “ delivered. 5c.
1150 “ in warehouse. $234.15. Ans.
Ea:planation.—By the operation of the work, we see that there were 140500 barrels on storage
for 1 day, and allowing 30 days for the month, there were 4683% (practically 4683) barrels on
storage for one month, which, at 5c. per barrel per month, amounts to $234.15.
This method of equating the time and making up the account for storage is
applicable also to various other departments of business.
Butchers and live stock dealers frequently hire their stock pastured or fed,
with the privilege of entering and withdrawing stock as they may require, and pay
for the actual time that stock was pastured.
78o SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. ºr
SECOND OPERATION BY THE PRODUCT METHOD,
I)r. Or.
Date. Feceived. Days. Products. I)ate. Deliveries. Days. Products.
1895 ſ 1895
Nov. 4 2000) 58 116000||NOV. 7 1000 55 55000
& 4 5 700) 57 39900 ‘‘ 11 500 51 25500
** 15 1300|| 47 61100 ** 18 600| 44 26400
44 (29 3800 33 125400 “ 20 100. 42 4200
** 30 1200| 32 38400 “ 27 1000 35 35000
tº- Dec. 5 1400 27 37800
9000 & & 7 1200 25 30000
7850 - ** 13 800. 19 15200
* ‘‘ 16 150 16 2400
1150 bbls. on hand. “ (21 300 11 3300
** 24 500 8 4000
“ 27 300 5 1500 .
Total Dr. products, 380800 7850 Tot. Cr. prod’t 240300
‘‘ CT. { % 240300
Bal. of Dr. products, 140500
140500 + 30 = 4683% practically, 4683 barrels, 5c. x 4683 = $234.15 storage.
Explanation.—By this method, we assume that all items of receipt were in storage from the
date received till January 1, and find the equivalent storage on 1 barrel for such time for each item
of receipt, which gives 380800 days storage on 1 barrel. Then, since there were deliveries made,
we find the number of days that each item of delivery was made before January 1, and the
equivalent storage on 1 barrel for such time, for each item of delivery, which gives 240300 days
storage on 1 barrel. This is deducted from the total debit product and gives a net debit of 140500
days storage on 1 barrel, which we divide by 30, the storage period, and obtain the number of
barrels on which storage is due for one month at 5 cents.
2. The proprietor of a pasture received and delivered cattle for account of
John Gunn, as follows:
Received : Jan. 1, 1895, 900 head; Jan. 5, 350 head; Jan. 21, 1100 head.
Lelivered : Jan. 2, 1895, 120 head; Jan. 3, 120 head; Jan. 4, 150 head; Jan.
14, 360 head; Jan. 17, 200 head; Jan. 23, 250 head; Jan. 27, 400 head; Jan. 28,
120 head; Jan. 31, 100 head. What will be the cost for pasturing these cattle
from Jan. 1, to Feb. 1, 1895, counting the actual time that the stock was pastured,
at 75 cents per head per week, (7 days)? Ans. $2556.42%.
OPERATION BY THE DAILY BALANCE METHOD.
Account of Pasturage of Cattle, at seventy-five cents per head per week (seven days), for
JOHN GUNN, from January 1, to February 1, 1895. -
| Number head Number head Number head in No. days. Number head pastured
1895. received. delivered. pasturage. pastured. for one day.
Jan. | 1 900 900 1 900
2 * 120 780 1. 780
3 120 660 1. 660
4 150 510 1 510
5 350 860 9 7740
14 360 500 3 1500
17 200 300 4. 1200
21 1100 1400 2 2800
23 250 1150 4 4600
27 400 750 1 750
28 120 630 3 1890
31 100 530 1 530
| 2350 1820 23860


** = . Number received 2350 7 ) 23860 number pastured for one day.
Number delivered 1820
tº-ºm-º: 3408% average number pastured for one week.
Number in pasture 530 75e. × 3408} = $2556.42%.
A. STORAGE. 781
TO FIND THE STORAGE ON GOODS RECEIVED AND DELIVERED AT
DIFFERENT DATES, THE STORAGE TERM AND
RATE BEING GIVEN.
1872. 1. The following are the receipts and deliveries of coffee by a general
Warehouse: º
*
Received. Delivered.
1895. July 7, 800 bags. 1895. July 24, 700 bags.
“ 28, 250 “ Aug. 4, 100 “
Aug. 13, 300 “ & 4 8, 200 “
“ 26, 350 “
How much is due for storage August 26, at the rate of 3 cents per bag for
a period of 10 days, or any part of 10 days? Ans. $85.50.
OPERATION.
Received. Delivered. Storage Charges.
July 7, 800 July 24, 700 700 at 6c. - $42.00
“ 28, 250 Aug. 4, 100 100 at 90. - 9.00
Aug. 13, 300 “ 8, 200 200 at 6c. - 12.00
“ 26, 350 50 at 90. - 4.50
300 at 6c. - 18.00
1350 1350 $85.50 total storage.
NOTE.—All goods delivered are deducted from the oldest receipts in the warehouse.
Ea:planation.—By inspection, we see that the first delivery of 700 bags, July 24, was made 17
days after the first receipt of 800 bags, which is one term and a fraction of a term = 2 terms.
Hence a charge is made on the 700 bags for 2 terms at 6 cents per bag = $42. The remaining 100
bags received July 7, were delivered Aug. 4, which is 28 days, = 2*, terms = 3 terms. Hence a
charge is made on the 100 bags for 3 terms at 9 cents per bag = $9. Aug. 8, 200 bags were delivered
which are applied to the 250 received July 28, thus making 11 days storage = 1%; term = 2 terms
= 6 cents storage on the 200 bags = $12. Aug. 26, 350 bags were delivered. 50 of these bags were
applied on the receipt of July 28, and 300 on the receipt of Aug. 13. The 50 bags owe storage
from July 28, to Aug. 26, = 29 days = 2*, terms = 3 terms or 9 cents per bag = $4.50. The 300
#" owe storage from Aug. 13, to Aug. 26, = 13 days = 1%; terms = 2 terms at 6 cents per bag =
2. The following are the receipts and deliveries of corn by a grain elevator:
Received. Delivered.
1895. May 4, 2000 bushels. 1895. May 31, 1200 bushels.
“ 28, 1500 & 4 June 20, 2000 & 4
June 4, 1000 & 4 July 15, 4300 & 4
July 10, 3000 “
The storage on the above corn was 3 cents per bushel for the first 15 days
or fraction thereof, and 2 cents per bushel for each subsequent 15 days or fractional
part thereof. What is due for storage July 15, 1895? Ans. $379.
OPERATION.
Storage due on 1200 bu. May 4, to May 31, = 27 days = 2 terms at 5 cents = t- e- tº $60
Storage due on (2000-1200 =) 800 bu. May 4, to June 20, - 47 days = 4 terms at 9 cents = - 72
Storage due on 1200 bu. (amount taken from receipt of May 28, and added to balance of
May 4, to make delivery of June 20,) from May 28, to June 20, - 23 days = 2 terms
at 5 cents = º {- º - º - g- º º sº - tº sº g- º 60
Storage due on 300 bu. (balance of 1500-1200) from May 28, to July 15, - 48 days = 4 terms
at 9 cents = * = as * * * * * * * * * * cº º 27
Storage due on 1000 bu. from June 4, to July 15, – 41 days = 3 terms at 7 cents = - - 70
Storage due on 3000 bu. from July 10 to July 15, = 5 days = 1 term at 3 cents = -> * 90
782 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. Yºr
TO FIND THE STORAGE ON GRAIN SOLD IN THE EIDEVATOR OR
WAREHOUSE.
1373. 1. A grain warehouse received, stored and delivered wheat as follows:
Received. Delivered.
1895. May 3, 7000 bushels. 1895. May 13, 6000 bushels.
“ 20, 6000 { { June 9, 7000 4 &
“ 26, 9000 & 4 “ 13, 8000 ( &
June 8, 8000 & 4 “ 26, 9000 & &
30000 30000
Compute the total and the extra storage on this account based on the follow-
ing conditions: The seller to pay 2 cents per bushel for storage for the first 20
days, or part of 20 days; the buyer to pay 4 cent per bushel for each subsequent
10 days or part of 10 days; 5 days are to be allowed the purchaser without extra
storage during which time the wheat is to be delivered.
OPERATION.
Storage on
Date. Receipts || With ||Withdraw. Withdraw. | From To Days. 9. Rate. Amt. Total. Extra.
bushels. || drawals. ||als from als from
receipts. balances.
1895
May 3| 7000 1000 May 3|June 14 42 3# 3500 15|00
** 13 6000 6000 “ 3|May 18, 15 2 12000
** 20 6000
June 9 7000 6000 ** 20 June 14|| 25 2} | 15000 30|00
May 26 9000 1000 “ 26 July | 1| 36 3 30|00 10|00
June 13 8000 8000 ** 26 June 18 23 2} 20000 4000
** | 8 8000
** 26 9000 8000 June | 8|July | 1 23 2% 20000|$73500 4000
$135|00



Explanation.—First find storage on 6000 bu. delivered May 13, from May 3, to May 18, (May
13, +- 5 days) = 15 days or 1 term at 2 cents a bu. = $120.00.
Then since the balance of the 7000 – 6000 = 1000 bu. remained in storage from May 3, to the
next sale of 7000 bu. on June 9, we must find storage on same 1000 bu. from May 3, to June 14,
(June 9 + 5 days) = 42 days or 4 terms at 3% cents = $35. Balance of sale of June 9, must have
been taken out of the receipt of 6000 bu, on May 20, hence we find storage on 6000 bu. from May 20,
to June 14, (June 9 + 5 days) = 25 days or 2 terms at 24 cents = $150. Now find storage due on
the withdrawal of June 13, 8000 bu. from May 26, to June 18, (June 13 + 5 days) = 23 days or 2
terms at 24 cents = $200.
Since the withdrawal of 8000 bu. on the 13th of June, left a balance of 1000 bu. of the receipt
of May 26, we find storage on that 1000 bu. from May 26, to July 1, (5 days credit added to with-
drawal of June 26,) = 36 days or 3 terms at 3 cents = $30.
The final work is to find the storage on the last receipt of 8000 bu. on June 8, from June
8, to July 1, (June 26 + 5 days) = 23 days or 2 terms at 24 cents = $200. Thus producing $735
total storage and $135 extra storage, $735 — $135 = $600 storage due by the seller and $135 due by
the buyer. * *
f
STORAGE. 783
2. What is the storage on the following receipts and deliveries of flour at 1
cent per barrel for the first 10 days, or fraction thereof, and 3 cent for each subsequent
5 days or fraction thereof; Ans. $14.25.
Receipts. Deliveries,
1895. Feb. 3, 500 barrels. 1895. Feb. 11, 100 barrels.
21, 250 & 4 “ 18, 200 6 &
24, 300 & 4 “ 22, 225 & 4
28, 150 & 4 “ 24, 300 & 4
*mmº Mar. 5, 375 & 4
1200 *º-smº
1200
Storage on
Date. Receipts || Withdraw- || Withdrawals From TO Days. | Rate. Amt. Total.
barrels. als gross. from Balances.
receipts.
1895
Feb. | 3 500 400
** 11 100 100 Feb. 3|Feb. 11| 8 || 1 c. 1|00
** 18 200 200 200 “ 3| “.. [18] 15 1}c. 3|00
** 21 250 200 ** | 3| “.. [22] 19 2c. 4|00
** 22 225 25 ** 21| “ 22| 1 || 10, 25
225
44 (24 300 225 ** 21| ** |24|| 3 || 10. 2|25
** 124 300 75 ** [24] ** 24| 0
225 ** 24|Mar. || 5 || 9 || 1C. 2|25
** 28 150
Mar. 5 375 150 << |28 ‘‘ 5; 5 1c. 1|50
| | 1.425



3. Elevator “A” of the Mississippi Elevator Co., received and delivered for
B. Milmo & Co., wheat as follows:
1894. Nov. 18, received 545040 bushels, which were delivered Jan. 31, 1895.
Dec. 6, & 4 474020 & 4 & 4 & 4 & 4 & 4 31, & 4
{ { 11, & 4 326500 & 4 & 4 & 4 & & 4 31, & &
1895. Jan. 12, ** 692045. & & & 4 & & & 4 “ 31, “
& & 23, & 4 75000 6 & 6 & é & & 4 & 4 31, & &
{ { 26, & 4 52050 ( & “ & & é & 4 & & & (31, & 6
Total bushels, 2164725.
According to the Rules and Tariff of the Elevator Co., 4 cent for each 10
days or part of 10 days is charged for storage, and 1 cent per bushel is charged for
delivering. No charge is made on the fractions of a bushel.
due for storage and delivery 3
What is the amount
Ans. $483.30.
The student will make the bill for the above, as per the following form:
784 . SouLE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
FOFM OF BILL.
E L E W A T O R. “A.”
NEW ORLEANs, January 31, 1895.
Messrs B. MILMo & Co.,
To Mississippi Elevator Co. Dr.,
For storage charges on wheat.
Wheat. Corn. In Out Shipped on (a) || Amount.
1894 1895 ---
5450|30 Ex. El. A Bins. Nov. 18||Jan. 31||S. S. Pendarves, 3
4740/20 Dec. | 6 ( & ( & $11850
3265100 ** 11 ( & & 4 2}
1895
6920.45 Jan. [12 ( & & 4 103.80
750 QQ & 4 23 { { & & 1+
52015.0 ** 126 ( & & 4 650
21647|25 bushels. | $483|30
Received payment.
... NOTE.-In this bill the student will find the extensions of the first, third and fifth items of
delivery, and the rate of cents to charge for the second, fourth and sixth items.

º ſ
Nº ||
yº
º
Allllfacturing (Il Shārēs and Tūllage.
=N
—º-º-º-
1374. Manufacturing on shares means that the manufacturer receives a
Specified per cent or part of the articles or things manufactured in compensation
for services rendered in manufacturing the material supplied to him. This method
of paying for services is quite common in certain lines of business.
By tollage is meant the taking of a specified portion of goods or articles in
compensation for services bestowed upon the goods or articles, or upon the owner's
part thereof, after taking the toll out; as the grain taken by the miller in payment
for grinding.
PROBLEMS,
1. A planter sends to a miller to have ground 80 bushels of corn. The
miller charges } of the whole number of bushels for grinding. How much is his
toll ? Ans. 10 bushels.
OPERATION. Explanation.—Since the miller charges # of
the whole number of bushels for grinding, he is
8 ) 80 bushels. entitled to as many bushels as the 80 is times
sº-º-º: equal to 8, which is 10 times. Hence we see
10 bushels, Ans. that he charges 10 bushels for grinding 70 for
the planter or 1 bushel toll for every 7 bushels
ground for the planter.
2. Suppose, in the preceding example, that the miller had, by request, ground
the whole 80 bushels for the planter, and that the planter had agreed to send the
miller his toll the next day. How much corn should he send ?
Ans. 11% bushels, = 11 bush. 1 p.k. 5 qts. 13 pts.
FIRST OPERATION,
8 ) 80 Explanation.—By the exercise of our reason,
*== we see that, as the miller's toll, 10 bushels, is #
10. Toll for grinding 4. of the whole quantity, he actually charges the
toll for grinding # of the corn for the planter,
TOLL. and hence it is clear, as shown by the line
| 10 statement, that if the grinding of # of the corn
7 || 8 for the planter is worth 10 bushels toll, the
| *mºmº grinding of # is worth the seventh part, and #
11; bushels. Ans. or the whole, is worth 8 times as much, which is
11} bushels.



(785)
786
soul E's PHILOSOPHIC PRACTICAL MATHEMATICs. *
SECOND OPERATION.
8 ) 80 bushels corn.
10 bushels toll.
70 bushels ground corn.
Ezplanation.—In this solution we use numbers
representing the actual number of bushels of
corn brought to the mill and ground, instead of
the fractional numbers as in the first solution,
and in order to show clearly the relationship
TOLL. . numbers, and the manner of obtaining them,
10 we present in the first part of the operation the
70 80 figures showing the bushels of toll and ground
| *=- - corn or meal that the miller and planter would
11% bushels corn. Ans. have respectively received, had the usual custom
of taking toll from the amount of corn brought
to the mill been observed. By these figures we see that 70 bushels of ground corn require 10
bushels of toll, and as by the agreement the planter was to receive 80 bushels of ground corn, it is
clear, as shown by the line statement, that if 70 bushels ground corn require 10 bushels toll, 1
bushel will require the seventieth part, and 80 bushels 80 times as much, which is 11} bushels.
THIRD opFRATION.
100 bushels assumed as the quantity ground. B.
12} is # of same for grinding. 100
*-* • 874 80
87% is left after deducting toll.
913 bushels required to grind and deliver
80, and pay toll, and 80 deducted
from 91% gives 11}.
Or,
To find the toll on 91% bushels, we operate thus:
8 ) 91%
11}. Ans.
3. A planter agrees to gin and bale the cotton of his neighbor for 'g
under this agreement he ginned, baled and delivered 20 bales, weighing
How many pounds of cotton is due him for his labor?
pounds.
FIRST OPERATION.
1 pound = #3 — I'o - † = 9200 pounds the
owner's share of an unknown quantity to be
ginned.
Hence the following statement:
9200
9 10
102223 pounds = +3 the total quantity to gin
as both the owner's and the ginner's
shares.
Then 102223 – 9200 = 10223 pounds the ginner's
share, or the quantity yet to be ginned.
; and
9200
Ans. 10223 pounds.
Explanation.—By the conditions of the prob-
lem, the planter who gins and bales the cotton
is entitled to fºr of the whole quantity ginned,
and had he taken his pay out of the 9200 pounds
he would have received 920 pounds. But as the
whole 9200 pounds were delivered to the owner,
the planter must now gin such an additional
quantity as will pay him according to the con-
tract. By the terms of the contract he is to
receive ºf of all that is ginned, and hence as
shown by the operation, for each pound ginned,
Tº is for himself and Yº for the owner. There-
fore the 9200 pounds is ºr of the total quantity
that must be ginned, on which I'ſ may be com-
puted as the share or compensation of the
planter. Hence, since ºr = 9200 pounds +8 =
102223 pounds; "6 of which is 10223 pounds yet to beginned by and for the planter.
SECOND OPERATION.
1 pound ginned, – I'd pound ginner's share, - ſº, pound owner's share.
9200 pounds ginned for owner.
º's of 9200 =
920 pounds ginner's share, to be paid in cotton
to be ginned. 9200 pounds – 920 pounds = 8280 pounds.
Then the following proportional statement:
8280 : 9200 :: 920 : 10223 pounds yet to be ginned.
NoTE.—The explanation for this operation is similar to that of the second solution, problem 2.
STORAGE. * 787
THIRD OPERATION.
9200 – 10 = 920 pounds due for ginning 9200 pounds.
1 pound = #8 - To = 1°, pounds quantity ginned for the owner to earn ºr pound. Hence for
each pound ginned the planter receives To pound and the owner #. Since it is worth 920 pounds
to gin nº of the cotton for the owner, to gin fºr it is worth the 4th part, and to gin +3 it is worth
ten times as much, which is 10223 pounds
4. A planter contracts with his laborers to give them # of all the corn they
can raise on his plantation. The division to be made on the stalk. When the corn
becomes ripe, he contracts to pay them gº of his share for gathering the same.
The entire crop amounted to 14800 bushels. What was the planter's share #
- Ans. 7104 bushels.
OPERATION.
# ) 14800 total number of bushels.
º; ) 7400 bushels = planter's #.
296 bushels = ºr paid for gathering.
7104 bushels = planter’s share.
5. A rice planter contracts to have his rice harvested, thrashed and sacked
for # or 25 per cent of the yield.
The contractors harvested, thrashed, sacked and delivered to the planter
16500 sacks of 162 lbs. each of rough rice. It is then agreed that they shall
continue to harvest, thrash and sack enough more rice to pay them according to
the contract, and that the planter shall purchase the same at $1.80 per sack. What
sum will the planter owe the contractors? Ans. $9900.
FIRST OPERATION.
1 sack — 4 = + = 16500 sacks the owner's proportion or share of an unknown number to be
harvested, thrashed and sacked. Hence the following statement:
22000 sacks to be harvested, thrashed and sacked. # of 22000 = 5500 sacks, the share of the
contractors or harvestors.
5500 × $1.80 = $9900 Ans.
SECOND OPERATION.
16500 – 4 = 4125 sacks due contractors.
1 — # = } 4125
3 || 4
| 5500 sacks to be harvested, thrashed and sacked for the contractors.
5500 × $1.80 = $9900 Ans.
NoTE.—See second and third operations of problem 3.
THIRD OPERATION.
# = 25%; 100% — 25% = 75%; then 16500 sacks = 75% of a number; hence 1% will be the
75th part, and 100%, or the number will be 100 times as much. Thus, as 75: 100 : ; 16500 : 22000.
25% of 22000 is 5500 × $1.80 = $9900,
NoTE.—Rice that has been harvested is thrashed and sacked for 8 to 10 cents per sack or
barrel of 162 pounds.
788 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
6. The owner of timbered land engaged laborers to make and bunch
shingles; and for their services they were to receive # of what they manufactured.
They manufactured 240000 shingles. How many of the number are they entitled
to according to the terms of the agreement 3 Ans. 80000.
7. Suppose, in the preceding example, that the laborers had been required
to manufacture, in excess of the 240000, the shingles to which they were entitled.
How many would they have manufactured for themselves in order to make a just
Settlement 3 Ans. 120000.
8. A planter agrees to gin his neighbor's cotton, and receive in payment for
his labor and use of machinery ſº of the cotton ginned. He received 70000 pounds
of seed cotton, which made 20000 pounds of ginned cotton. How many pounds
of ginned cotton is he entitled to ? Ans. 2000 pounds.
9. The same planter referred to in the preceding problem agreed with
another neighbor to gin his cotton, and receive in payment for his services and use
of machinery seed cotton to the amount of Tº of the seed cotton ginned. He
received 70000 pounds of seed cotton. How many pounds must he deduct before
ginning to pay him according to contract, and how many pounds of ginned cotton
did he deliver to his neighbor, allowing 34 pounds of seed cotton to make one pound
of ginned? Ans. 6363 ºr pounds to be deducted.
* 18181;ºr pounds he delivered.
OPERATION.
To + 1 or +} = }} 3}) 63636 ºr pounds to gin.
11) 70000 total amount received. smass
6363 ºr pounds of seed cotton due the ginner. 18181*r pounds ginned cotton delivered to
the planter.
Explanation.—In this operation we see, by the exercise of our judgment, that, as the planter
gins the cotton for ºf of the amount ginned, we cannot compute the T'ſ on the 70000 pounds
received, for that would be charging "I'm on the amount ginned, and on the cotton charged for
ginning. By careful thought we see that if ºr of what is ginned is charged for ginning, then the
amount received to gin is equal to Thy plus 1, or 43, which is +}, and hence if the amount received is
+% of the amount ginned, ºr is the amount to be deducted for ginning.
SECOND OPERATION.
100 pounds assumed to gin. P.
10 pounds due for ginning. 100
Cºsmºs 110 || 70000
110 pounds of seed cotton required to gin 100. tºº-º-º-º-º-
63636 ºr pounds to gin.
70000
6363 ºr pounds of S. C. due the ginner.
34 ) 63636 ºr
1818.13+ pounds of ginned cotton.
Eacplanation.—The principles of this second solution have been so repeatedly explained in
Per Centum, Commission and Brokerage operations, that we deem it unnecessary to occupy space
with them.
10. A housekeeper wishes to supply his family with 30 bushels of flour. His
miller charges ſº toll for grinding. How many bushels of wheat must he send to
mill in order to obtain the exact 30 bushels of flour?
Ans. 32.Ér bush. = 32 bu. 2 p. 7 qts, ºr pts.

3 LLIGATION.
Øo).957-EEEEEEE EcºSEcº-EEEEEEEEEEEEEEEEEàNS
1375. Alligation, which signifies a tie or binding, is a process for solving
Questions relating to the mixing or compounding of different kinds of ingredients,
or of combining different qualities or values.
It is generally treated under two heads or divisions, viz: Alligation Medial
and Alligation. Alternate.
1376. Alligation Medial, or Medial Proportion, is the process of finding
the mean or average price or rate of a mixture of different kinds or qualities of
ingredients, when the quantity and value of each kind are given.
PROBLEMS.
1. A merchant mixes 400 lbs. of sugar worth 9 cents per lb., and 600 lbs.
worth 14 cents, with 1000 lbs. worth 15 cents; what is the mixture worth per pound?
- Ans. 13; cents.
OPERATION.
400 lbs. at 9e. = $36.00 Explanation.—In this example, we have only
‘‘ ‘‘ 14c. – 84.00 to multiply each given quantity by its price to
1000 ‘‘ ‘‘ 15C. - 150.00 , obtain its full value; then it is evident that the
cº- sum of these values, $270.00, divided by the
2000 lbs. = $270.00 whole number of pounds, 2000, will give the
mean value, or average price per pound of the
Hence 1 lb. = Ans. 13}c. mixture.
2. If I mix 50 lbs. of tea at 60 cents per lb., with 40 lbs. at 75 cents and 100
lbs. at 80 cents, what is the average value of the mixture per pound 3
Ans. 73}; cents.
3. What per cent of Alcohol in a mixture of 40 gallons, 85 per cent strong;
60 gallons, 90 per cent strong; 50 gallons, 98 per cent strong; and 30 gallons, 95
per cent strong? Ans. 9144%.
OPERATION INDICATED.
40 gals. (a) 85% = 3400
60 ( &
“ 90% = 5400
50 “ “ 98% = 4900
30 “ “ 95% = 2850
180 ) 16550 (
4. What is the purity of a mixture of 20 pwt. of gold, 14 carats fine; 30
pwt. 16 carats fine; 50 pvt. 12 carats fine; 40 pvt. 18 carats fine; and 10 pvt. pure
gold 3 Ans. 15+ carats fine.
ALLIGATION AITERNATE.
1377. Alligation Alternate is a kind of Indeterminate Reciprocal Proportion,
and is the process of finding the ratios of the quantities of the articles, whose
values or qualities are given, that must be taken to form a compound or mixture
of a given mean rate of value or quality.
(789)
790 . soule's PHILOSOPHIC PRACTICAL MATHEMATICs. *
It is the reverse of Alligation Medial, and may be proved by it.
The solution of questions in Alligation Alternate is based on the principle
of an equality of gains and losses; that is, in the composition of a mixture, the
gains on the articles whose values are less than the mean value, must be equal to
the losses on the articles whose values are greater than the mean value. Thus, if
we mix 1 pound of sugar worth 10 cents with 1 pound worth 12 cents, we will have
2 pounds worth 22 cents, or an average of 11 cents per pound. We here see that
the gain of 1 cent on the first pound is just equal to the loss on the second pound;
the amounts used being equal, the gain and loss are relatively equal. But suppose
we had 1 pound worth 10 cents that we wished to mix with sugar worth 13 cents,
so as to make a compound worth 12 cents per pound, we see that there will be a
gain of 2 cents on 1 pound of the first kind, and a loss of 1 cent on 1 pound of the
second kind, and hence to produce an equilibrium of gains and losses, we must
take enough of the kind at 13 cents to give a loss of 2 cents, that is 2 pounds.
Here we see that the quantity required of each kind is inversely proportional to
the gain or loss on 1 pound of that kind, as compared with the other, and if there
were several other kinds used, the same reasoning would apply, as shown by the
following illustrations:
The subject is treated under four cases or conditions.
CASE I.—Analytic Method.
TO FIND THE PROPORTIONAL PARTS TO BE USED WHEN THE
MEAN OR AVERAGE PRICE OF A MIXTURE AND TEIE PRICES
OF TEIE DIFFERENT ARTICLES ARE GIVEN.
1378. 1. What relative quantities of sugar, at 5 cents per lb. and 8 cents
per lb., must be used to make a compound worth 6 cents per pound?
Ans. 2 lbs. at 5 cents. 1 lb. at 8 cents.
OPERATION. Eacplanation.—If a compound of two grades
of sugar, one worth 5 cents and the other worth
1 || 2 8 cents, be made in certain proportions, and
- - -sms - - the mixture be priced or valued at 6 cents, there
5|| 1 |2 will be a gain on the 5 cent grade and a loss on
Ans the 8 cent grade. This gain and loss in the
8| 3 |1 whole mixture must be exactly equal, and hence
we must take such quantities of each as will
PROOF. give a gain and a loss of equal amount in the
unit of value, here 1 cent; the result will be
2 lbs. at 5c. = 10c. the correct quantities to be compounded.
1 lb. at 8c. = 8c. By selling 1 pound of sugar worth 5 cents for
tº- s=- 6 cents, there will be a gain of 6 — 5 = 1 cent
3 lbs. at 6c. = 18C. hence we place 1 pound opposite the 5.
By selling 1 pound of sugar worth 8 cents for 6 cents, there will be a loss of 8 — 6 = 2 cents,
and to lose 1 cent will require # of a pound, hence we place 3 pound opposite the 8. Therefore, 1
pound at 5 cents and # pound at 8 cents are the proportional quantities for the compound or mixture.
Now, since the gain and loss are equal in these proportional quantities, it is clear that they will be
equal in any number of times the same. Therefore, we may multiply the proportional quantities
by any number and produce other proportional numbers.
In this problem, we multiply by 2, the least common multiple of their denominators, and
thus obtain 2 pounds at 5 cents and 1 pound at 8 cents, for the mixture.
NOTE.—See page 159, for the Least Common Multiple.
* ALLIGATION. 791
2. What proportions of tea at 48 cents, 60 cents, 66 cents, and 72 cents a
pound, must be mixed together so that the mixture shall be worth 64 cents per
pound? Ans. 1 lb. at 48 cents, 2 lbs. at 66 cents,
1 lb. at 60 cents, 2 lbs. at 72 cents.
FIRST OPERATION. PROOF.
1 || 2 || 3 || 4 || 5 1 lb. at 48c. = 48c.
* | *-* | *-* mm-- *-* | * *== 1 ‘‘ ‘‘ 60C. — 60C.
ſ 48] T's 1 1 2 lbs. “ 66c. = $1.32
l ( & & 4 -
64: ; #| || Ans. 2 72c. = 1.44
(#| || 2 2 6 “ “ 64c. = $3.84
Ea:planation.—To maintain the equality of gains and losses, we must always compare the
prices of two articles or ingredients, offe that is greater and one that is less than the average or
mean price, and treat each pair or couplet thus compared as a separate problem. In the first
operation, we formed two couplets and compared 48 and 72, and 60 and 66. In the second operation,
we compared 48 and 66, and 60 and 72.
In the first operation, we find 1, By selling 1 pound of tea worth 48 cents, for 64 cents, there
will be a gain of 16 cents; and to gain 1 cent, would require I's of a pound, which we write in the
first column opposite the 48 cents. 2. By selling 1 pound of tea worth 72 cents, for 64 cents, there
is a loss of 8 cents; and to lose 1 cent, would require # of a pound, which is written in the first
column opposite the 72 cents. 3. By selling 1 pound of tea worth 60 cents, for 64 cents, there is a
gain of 4 cents; and to gain 1 cent, would require + of a pound, which we write in the second
column, opposite the 60 cents. 4. By selling 1 pound of tea worth 66 cents, for 64 cents, there is a
loss of 2 cents; and to lose 1 cent would require # of a pound, which we write in the second
column, opposite the 66 cents.
Now having finished with the couplets and produced the correct proportional quantities, we
reduce the fractional quantities to integers or whole numbers, which is done by multiplying the
quantities in the respective columns of gain and loss proportionals, by the least common multiple
of their respective denominators. Thus, Yºg and * were multiplied by 16, giving 1 and 2 pounds,
which we write in the third column. The + and # were multiplied by 4, giving 1 and 2 pounds
which we place in the fourth column. The proportional quantities are then arranged in column
five.
NotE 1.--When the prices of the ingredients and the mean rate are whole numbers, the
intermediate steps, as shown in columns 1 to 4, may be omitted, and the same results obtained by
taking the difference between each article used as a couplet and the mean rate, and placing it
opposite the other number of the couplet. See the linking method a few pages beyond.
NotE 2.—There will be as many columns of proportional quantities as there are couplets.
SECONID OPERATION,
1 2 3 || 4 5
- *- I -mº i t "-sºme NOTE.-This operation shows that by com-
§ Th; 1 2 : paring different prices, different proportional
643 à # 8 8 Ans. quantities are obtained.
72 # 1 1
NOTE. 1.--When there is an odd quantity, compare the same with one that has been already
compared and take the sum of the two numbers opposite the one used twice.
NotE 2.—A Common Factor may be used in any couplet or omitted from it without changing
the proportional parts. Hence it is obvious that there may be any number of answers in the same
proportion.
79.2 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
1379. Loss AND GAIN METHOD,
THIRD OPERATION AND PROOF.
Proportional
Ratio. quantities.
1 lb. at 48c. = + 16c. & *** — 1 lb. at 48c. = 48c.
64. 1 lb. at 69g. = + 49. }= 20c. gain = i. 1 lb, at 60c. = 60c.
1 lb. at 66c. = — 20. . 10c, loss = 1 2 lbs. at 66c. = 132c.
1 lb. at 72c. = — 8c. T º - 2 lbs. at 72c. = 144c.
6 lbs. at 64c. = 384c.
Explanation.—In this operation, we first write the prices of the given commodities in a
column in regular order, beginning with the smallest, and the mean price, 64 cents, on the left, as
a basis of comparison. We then see that when we use 1 pound of tea at 48 cents, in a mixture to
be worth 64 cents, we have a gain of 16 cents; we place this to the right of the price that produced
it, and mark it plus, to indicate a gain. ... We then compare the next price, 60 cents, with the mean
price, 64 cents, as in the first case, and find a gain of 4 cents, which we place as before to the right
of its producing term, 60, and mark it plus also. Then proceeding with the next term, 66, we find
by comparing it with the mean 64, that in using 1 pound there will be a loss of 2 cents; we place
this to the right of 66, and mark it minus (—) to indicate a loss. Comparing the fourth term, 72,
in the same manner, we find a loss of 8 cents, which we place and mark as in the last. Having
now found the gains and losses on all the terms taken singly, as compared with the mean term on a
basis of 1 lb. of each kind, we next add all the gains, marked plus, together, giving a total gain of
20 cents, and set to the right. We then add all the losses, marked minus, giving a total loss of 10
cents, and place to the right. Then, as these two amounts have a common measure, 10, we divide
them both by 10, and set the quotients still to the right. These results have the same ratio to each
other as the whole gain and loss, and hence we have the ratio of the gain to the loss as 2 is to 1, or
a relative gain of 2 cents to a loss of 1 cent in the admixture of 1 pound of each kind of tea. But
as the gains and losses in the whole mixture must be the same, we must equalize them by changing
the relative proportions of the ingredients used; in order to do this, we must multiply the quanti-
ties used, 1 pound of each, by the reciprocal ratio of its gain or loss. Now, the ratio of gain to
loss is as 2 to 1, and their reciprocal ratio is as 1 to 2. Therefore we multiply the given quantities,
1 pound of each ingredient, that gave a gain, or that stands opposite the sign plus, by the relative
Inumber or ratio of loss, 1, and place the product to the right in a column headed “Proportional
Quantities;” then we multiply the units, 1 pound, opposite the minus sign, by the relative number
or ratio of gain, 2, and set the results in the column of proportionals. These proportionals are the
quantities required to be taken of each kind of the ingredients at the prices opposite to which
they stand.
To prove the correctness of the work, multiply these proportional quantities by their
respective prices, and set the products to the right. The sum of the products divided by the sum
of the proportionals will give for a quotient the mean price, as in the example. By this method
the proportional quantities of all the commodities that give a gain are always equal, and all those
that give losses are equal. These proportionals may be multiplied by any number, and hence, an
indefinite number of answers having the same ratios will be found.
LINKING METHOD,
1380. We will now present another method by which the reciprocation of
ratios is more readily effected, and the results may be different from those of the
first two methods, though both are correct. The reason however is not so apparent,
though this method is often used, and has given the name to this branch of Arith-
metic Alligation, as the process is indicated by alligating, or linking, one rate that
is less than the mean rate with one that is greater, as shown below:
OPERATION AND PROOF.
48T 8 lbs. at 48c. = $3.84 Explanation.—We here write the simples as in
64 60 2 lbs. at 60c. = 1.20 the first method, and link one that is less with
66 4 lbs. at 66c. = 2.64 one that is greater than the mean rate; that is,
72_| 16 lbs. at 72c, - 11.52 we link 48 with 72, and 60 with 66. We then
-*. *- take the first couplet, 48 and 72, and comparing
30 lbs. at 64c. = $19.20 each term with the mean rate 64, we find there
will be a gain on 1 pound costing 48 cents, when
* ALLIGATION, 793
used in a compound valued at 64 cents, of 16 cents, and a loss of 8 cents on 1 pound costing 72
cents. Now, as the gain is to the loss as 16 is to 8, in this couplet where 1 pound of each is used,
and as the mixture must give an equal combination of gains and losses from each couplet, we know
from reason and previous explanations, that the quantities used must be taken in an inverse or
reciprocal ratio to the gain and loss, that is, as 8 is to 16. We therefore place the difference
between 64 and 48 = 16, not opposite the 48, to represent the gain as in the second method, but
opposite the 72, to represent the relative quantity to be taken at that price; and for the same
reason we place the difference between 64 and 72 = 8, opposite the 48. The terms of the other
couplet, 60 and 66 linked together, are treated in the same manner, by placing the difference
between the mean rate, 64, and each term, opposite to the one with which it is linked. These
differences thus arranged constitute the proportional quantities that are to be taken of the several
ingredients that form the compound, at the prices opposite to which they stand. The proof is the
same as in the first method. When there are more than three terms, in which two or more are
less, and two or more are greater than the mean, they can be linked in various ways, each variation
giving a new set of ratios, and as each set of ratios admit of indefinite multiples, the answers to
questions of this character, with those conditions, may be infinite. When any rate has several
others linked with it, each alligated term will produce separate differences, and their sum will be
the proportional required for that rate.
PROBLEMS WORKED BY THE ANALYTIC METEIOD,
1381. 1. A wine merchant has wine worth $1.10, $1.80, $2.50, per gallon,
which he wishes to mix with water so that the compound will be worth $1.50 per
gallon. What will be the proportional quantities of wine and water ?
Ans. 10 gallons water.
3 “ wine at $1.10.
4 { { { % {{ $1.80.
15 & 4 {4 {{ $2.50.
OPERATION.
| 1 || 2 || 3 || 4 || 5
0.0% ºf 10 10 \
$1.5% # i ; ; Ans.
2.5%l #6 15 15
2. A grocer has three grades of coffee worth respectively 10, 11, and 152 per
pound. In what proportions must they be mixed so that the compound would be
worth 122 per pound? Ans. In equal quantities.
OPERATION. PROOF.
| 1 || 2 || 3 || 4 || 5 3 lbs. at 10c. = $.30
3 lbs. at 11c. = .33
10| + 3 3 3 Ibs. at 15c. = .45
12 & 11 1. 3 || 3 × Ans. * ºsmºmº
15| | | + || 2––| 1 || 3 9 lbs. at 12c. = $1.08
Explanation.—Here we have two articles less than the mean rate, and one, greater. First
compare 10 and 15 as the first couplet with 12, thus: 10 from 12, 2, 33 12 from 15, 3, #. Second com-
pare 11 and 15 as the second couplet with 12, thus: 11 from 12, 1; 12 from 15, 3, #. Then, multiply
the proportional numbers in column 1 by 6, and those in column, 2 by 3. The 6, and 3 being
respectively the least common multiple of the denominators of the numbers in the respective
columns. Then arrange these results in the 5th column.
794 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS. º
3. A planter has oats worth 40 cents, corn worth 60 cents, and barley
worth 70 cents per bushel. He wishes to mix them so as to make a compound worth
b0 cents per bushel. What proportions must he use? Ans. 3 bushels of oats.
t 1 bushel of corn.
1 bushel of barley.
OPERATION.
| 1 | 2 | 3 | 4 | 5 Explanation.—Here we have two articles greater
than the mean rate and one less. First compare 40
40. To | I'ſ 2+| 1 || 3 and 70, as the first couplet, with 50, the mean price.
50 & 60 To 1 || 1 Ans. Second compare 40 and 60, as the Second couplet,
70 ºf 1 1 with 50. Then proceed as in the previous problems.
4. What relative qualities of flour worth $43, $5, and $7+ per barrel, must
be sold to realize an average price of $6+?
Ans. 4 bbls. at $44; 4 bbls. at $5; and 12 bbls. at $74.
OPERATION. Ea:planation.—First, use 43 and 74 as the first couplet,
| 1 || 2 || 8 || 4 || 5 which gives 1}=4=# and 1.
Second, use 5 and 7+ as the second, couplet, which
4}| # 4 4 gives 1+=#=# and 1. Then multiply the proportional
64 & 5 # 4 || 4 Ans. numbers in column 1 by 7 and those in column 2 by 5.
7#| 1 || 1 || 7-i-| 5 || 12 See problem 1, page 790, and problem 2, page 793.
5. If I mix wines worth respectively 60 cents, 80 cents, $1.00, $1.30, and
$1.50 per gallon, how many gallons of each kind will be required to form a mixture
worth $1.10 per gallon ?
Ans. 4 gals. at 60%, 2 gals. at 80g, 2 gals. at $1, 4 gals. at $1.30, 5 gals. at $1.50.
And other answers by different coupling.
| | | * | * | * | * | *
60 go 4 4
80– * 2 2
1103 1.00] Y's 2 || 2 X Ans.
1.30— #5 go 3-H 1 || 4
1.50 || zºo 5 5
6. Mixed oats at 40 cents, corn at 50 cents, rye at 75 cents, barley at 80
cents, and wheat at $1 per bushel. How many bushels will it take to make a
mixture worth 70 cents per bushel? *s
Oats. Corn. Rye. Barley. Wheat.
Ans. By the 1st, or reciprocal method, 1 bu. 2 bu. 4 bu. 2 bu. 1 bu.
Ans. By the loss and gain method, 9 bu. 9 bu. 10 bu. 10 bu. 10 bu.
Ans. By the linking method. 6 bu. 3 bu. 4 bu. 4 bu. 6 bu.
And other ratios by different methods of linking.
7. A goldsmith would mix gold 12 carats fine, with some of 13, 14, 16, 18,
and 20 carats, and alloy valued at nothing. How much of each kind will be
required to make a compound of 15 carats fineness?
Ans. By the 1st or reciprocal method, 5 parts of 12 carats; 3 parts
of 13 carats; 1 part each 14 and 16 carats; 2 parts of 18
carats; and 3 parts of 20 carats.
And other ratios by different method of work.



2} ALLIGATION. 795
CASE—II.
1382. When the quantity of one of the commodities composing a mixture of
a given mean value is given, to find the quantity of each of the others.
1. A merchant has molasses worth 35 cents, 52 cents, and 70 cents a gallon.
He wishes to form a mixture worth 60 cents per gallon, and to have the same
contain 120 gallons of the 35 cents per gallon molasses. How many gallons of each
grade of molasses must he take 3 Ans. See operation.
OPERATION.
1 || 2 || 3 4 || 5 Explanation.—We first find, as in case 1,
*-º- ºmº-E i = | *-m-tº-º sºm-ºmº || * article 1378, the proportional quantities to be 2
35 ºf 2 2 × 60 = 120 gallons of the 35 cents, 5 gallons of the 52 cents,
60 & 52 + 5 5 × 60 = 300), Ans. and 9 gallons of the 70 cents molasses. But
70 Y'ſ I'd 5-Hl 4 || 9 × 60 = 540 since 120 gallons of the 35 cents molasses,
instead of 2, are to be used, we must take 1%0.
= 60 times each of the other grades or priced molasses, in order that the gain and loss may be
equal. Hence we have 2 × 60 = 120 gallons at 35 cents, 5 × 60 = 300 gallons at 52 cents, and 9 ×
60 = 540 gallons at 70 cents.
2. A grocer has 100 pounds of coffee, worth 10 cents per pound, which he
wishes to mix with other grades of coffee worth respectively 12 cents, 14 cents, 15
cents, 16 cents, and 18 cents per pound, so that the mixture will be worth 15 cents
per pound. How much of each kind will be required? Ans. See operation.
OPERATION.
1 || 2 3 | 4 | 5 || 6 || 7
10- || || 3 3 × 33} = 100 pounds.
#- # 1. 1 × 33} = ; º
1 1 || 1 × 334 = 33% ‘
15 i; any quantity at 15c. Ans.
16:- 1 || 1 3-H 1 || 4 × 33} = 133 pounds.
18—|| # 5 5 × 33} = 1663. “
Explanation.—We first find the relative or proportional quantities, as in case 1, these
quantities are shown in column 7. Then, since 100 pounds of coffee worth 10 cents per pound are
to be used, we divide 100 by 3, the relative number for the 10 cents coffee, and in the quotient 33%
we have the ratio with which to multiply each of the proportional quantities, as shown in column
7, to obtain the quantity required of each. X.
NOTE:-The 15 cent grade of coffee being of the same value as the mean rate, it is not
considered in the operation. It may be omitted from the list of prices, if desired.
3. A merchant has 280 pounds of coffee, worth 16 cents per pound, which
he wishes to mix with other qualities worth 18 cents, 21 cents, 22 cents, and 24
cents respectively. How much of each additional kind will be required to compose
a mixture that will be worth 20 cents per pound?
Ans. By 1st method, 280 lbs. each at 16, 22, and 24 cents, and 560
lbs. each at 18, and 21 cents.
And other ratios by different methods of work and different linkings.
Operation by the 1st method.
1 || 2 || 3 || 4 || 5 || 6 |
16| # 1 1 × 280 = 280
18 # | } 1+| 1 || 2 × 280 = 560
20 & 21 1 2 || 2 × 280 = 560
22 # 1 1 × 280 = 280
24 + 1 1 × 280 = 280
280 + 1 = 280 ratio number to multiply each of the proportional quantities.
796 soul E's PHILOSOPHIC PRACTICAL MATHEMATICs. *
4. A stock broker has 140 shares of R. R. stock that cost him $60 per share.
How many more shares must he buy at $80 per share, so that he can sell the whole
at $75 per share and gain 20 per cent on his investment? Ans. 20 shares at $80.
OPERATIONS.
Statement to find average cost By the 1st method.
of 1 share sold at 20% gain. 1 2 3
100 * * : *= I tºms
120 75 62.É. # | 14 || 7
*º- 80 * 2 1 × 20 = 20 Ans.
$62+ cost.
140 - 7 = 20 ratio number by which
to multiply proportional quantity
of railroad stock.
By the linking method. By the loss and gain method.
60, 174 = ** = 7 60 24 gain = 1 = 7
62.É. 2} = } = 1 62}{3}}}#. = 7 = 1
1 × 20 = 20 Ans. 1 × 20 = 20 Ans.
1383. , When the quantity of two or more of the ingredients or articles is
given or limited, to find the quantity of each of the others.
º 1. How many pounds of Naples walnuts at 30 cents, and of almonds at 27
cents, must be mixed with 300 pounds Tenn. peanuts at 5 cents, 200 pounds of
Italian chestnuts at 8 cents, 150 pounds of Brazil nuts at 9 cents, and 250 pounds
of Louisiana pecans at 16 cents, so that the average price of the mixture may be
worth 15 cents
OPERATION,
300 lbs. Tenn. peanuts at 5c. = $15.00
200 lbs. Italian chestnuts at 8c. = 16.00
150 lbs. Brazil nuts at 90. = 13.50
250 lbs. Louisiana pecans at 16c. = 40.00
900 lbs. $84.50 ( 9, sc. = average or mean price of the given articles.
1 || 3 || 3 || 4 || 5
91s] ºr | Pº 270 |-|-216 || 486 900 pounds.
15 & 27 101 || 101 X x 13% = { 1872, “
30 101 101 1874, “
900 – 486 = 1% == ratio of increase for each proportional quantity.
Explanation.—In all problems of this kind, we * find the average or mean price of the
articles or ingredients given, according to the principles of Alligation Medial. See Article 1376.
Second, we find the proportional quantities of the average price of the articles given and of the
articles required, according to the principles of case 1. . See Article 1378. Third, we find the ratio
number with which to multiply the proportional quantities so that 900 pounds of the given articles
may be used.
As shown in the operation, the proportional quantities are 486 pounds of the given articles
and 101 pounds each of the ungiven quantities, the walnuts and the almonds. And since 900 pounds
of the given articles are to be used, we divide 900 by 486 and obtain 1% as the ratio number with
which to multiply the proportional numbers.
ALLIGATION. 7974
2. A goldsmith has 11 oz. of gold, 14 carats fine, 12 oz. of 15 carats, and 17
oz. of 18 carats. He wishes to mix with these, gold of 17 carats, 20 carats, and 22
Carats, so as to make the compound 19 carats fine. How much of each of the three.
last will be required?
11 oz
12 oz
17 Oz
40 oz.
640 – 40 = 1 oz., average 16 carats.
Ans. See operations.
OPERATIONS.
FIRST STEP.
. at 14 carats = 154 carats.
– 180 **
= 306 “ ſ
=640 6 &
STEP.
e= i <===s = ºmº || *====s* | *s-s-s-s Oz. Oz. Carats. Carats.
16| # l 1 40 at 16 = 640
19 17 # 1 Proportional 1 40 = 40 at 17 = 680
20 1 2 quantities. 27 * * * * 80 at 20 = 1600
22| || 1. 40 at 22 = 880
3860
40 – 1 = 40 = ratio multiplier.
Or thus, for the second step :
Oz. Carats. OE.
16–1 1 × 40 = 40 16- 3 × 13+ = 40
1944- | 3 × 40 = 120 Or 19| || | | < j} = }
20-l-l 3 × 40 = 120 thus: 20 2 x 13+ = 26}
22– 2 × 40 = 80 22—l 3 × 13% = 40
, 40 - 1 = 40 ratio number.
40 -i- 3 = 13% ratio number.
3. A broker has 60 shares of stock that cost $40 per share, and 40 shares
that cost $75 per share. How many more shares must he buy at $90 per share, so
that he can sell the whole at $50 per share and lose 33% per cent
Ans. 140 shares at $90.
OPERATIONS.
Statement to find average By the 1st method.
cost to lose 33}%. 40 || 3's 3
100 75 & 75
66; 50 90 || || || 7 × 20 = 140 Ans.
| $75 cost. 60 – 3 = 20 ratio multiplier.
By loss and gain method By the linking method.
£i 15 = 3
75
40 35 gain = 7 = 3
75
§ 15 loss = 3 = 7 × 20 = 140 Ans.
60 - 3 = 20 ratio multiplier. *
75 |
901 35 = 7 × 20 = 140 Ans.
60 – 3 = 20 ratio multiplier.
798
SouLE's PHILOSOPHIC PRACTICAL MATHEMATICs.
CASE—IV.
1884. When the quantity and rate of the mixture to be made is given, to find
the several quantities composing it, whose rates are given. *
1. A manufacturer has cotton worth 5 cents, jute worth 9 cents, and wool
worth 20 cents per pound. He desires to make a mixture of 533 pounds worth 12
cents per pound. How many pounds of each must he use ?
Ans. 164 each of cotton and jute, and 205 of wool.
OPERATION.
| | Explanation.—We first find the proportional
quantities according to Case 1, Article 1378.
5| + 8 8 × 20% = 164 Then we add the proportional quantities
12 & 9 + 8 8 × 20% = 164 together which gives 26. Then, since the re-
20 # 7–H 3 || 10 × 20+ = 205 quired mixture is to contain 533 pounds, we
*sº * *s divide 533 by 26 and obtain 20% as the ratio
6 533 multiplier for all the proportional quantities,
thus producing 164 pounds each of cotton and
jute, and 205 pounds of wool.
2. A wine merchant has an order for a cask of wine containing 126 gallons,
to cost just $1.20 per gallon. Not having any wine of this price, he makes a
mixture of other wines worth respectively 80 cents, $1.00, $1.30, $1.50, $1.80, and
Water worth nothing. How many gallons of each kind will be required?
* Ans. See operation.
FIRST OPERATION.
Proportional
numbers. gals.
ſ 0| T40 º 1. ...] 91% water at 0 = $000.
| 1: £o 3. 3 3 23# wine ºs 80c. = 23.26%
1. * 1 j} e-º-º: 9% “ “$1.00 = 9.69%
1203 i. . 5|3} < 91% = 195, “ “ 1.30 = 25.20
| 150 3% 4 4 38}} “ “ 1.50 = 58.15A,
U 180 go 2 3] 19é, “ “ 1.80 = 34.89%
13 126 gals. at $1.20 = $151.20
126 — 13 = 91% the ratio multiplier.
NOTE:- Different answers may be produced by different methods forming couplets, or by
linking, or by the method shown in Article 1379.
Operation and proof by the loss and gain method as elucidated in Article 1379.
Gal. Cts. Ratio. Gals. Cts. $
ſ 1, water at 0 = + 120 5 × 3 = 15 at 0 = 0.00
1, wine at 80 = + 40 X = 180 G. = 9 5 × 3 = 15 at 80 = 12.00
$ 1, “ at 100 = + 20 5 × 3 = 15 at 100 = 15.00 *
1.20) 1, “ at 130 = — 10 9 × 3 = 27 at 130 = 35.10
1, “ at 150 = — 30 X = 100 L. = 5 9 × 3 = 27 at 150 = 40.50
(i. * { at 180 = — 60 9 × 3 = 27 at 180 = 48.60
126 at 120 = 15.20
126 - 42 = 3 proportional multiplier.
Operation and proof by the linking method as elucidated in Article 1380.
0—I 60 6 × 4} = 27 gals. at 0c. = $ . .00
80–, 30 3 × 4} = 134 “ at 80c. = 10.80
*..., | 103 10 1 × 4} = 4; “ at 100c. = 4.50
1.20 | OT – 4 & – ©
130 20 2 × 44 = 9 at 130c. = 11.70
150–1 || 40 4 × 4} = 18 “ at 150c. = 27.00
18 12 × 4 + = 54 “ at 180c. = 97.20
28)126(4; 126 “ at 120c. = $151.20

+): * ALLIGATION. - 799
3. A liquor dealer has whiskey of 1000 proof, and another grade of 65° proof.
How many gallons of each must he take to maké a barrel of 46 gallons of 90°
proof? Ans. 32% gallons of 1000 proof. 13% gallons of 65° proof.
OPERATION,
Gal. Gal.
- * * * *m i wºme 46 46
90% 65] gº || 2 = % of 65° 7 2 7 5
100 ſo | 5 = } of 100°
-mºmºmº | —
| 13} 32%
4. A farmer carried 120 fowls to market, consisting of turkeys, geese, ducks,
and chickens, and sold the turkeys for $2.25 a piece, the geese for $1.50 a piece, the
ducks for 50 cents a piece, and the chickens for 25 cents a piece. He received $120
for the lot, averaging $1 a piece for all. How many of each kind were there?
Ans. See operation.
Operation by the analytic method.
1 || 2 || 3 || 4 || 5
-
25. As 5 5 60 chickens.
100) .5% 3% 1 || 1 × 12 = 12 ducks.
150 8% 1 || 1 T ) 12 geese.
225. His 3 3 36 turkeys.
10 120
120 – 10 = 12 = ratio multiplier.
Answer by loss and gain method. Answer by the linking method.
Chickens, - - se 35 Chickens, 50 20
Ducks iº- - º 35 e Ducks, 20 Or 50
Geese, tº- - - 25 Geese, 20 30
Turkeys, - º - 25 Turkeys, 30 20
5. A contractor hired 5 men to work. He paid per day to each as follows,
viz: to A. $5, to B. $4, to C. $3, to D. $2, to E. $1. Altogether they worked 120
days, for which he paid them $420. How many days did each work, and what
amount did each receive 3
Operation and answer, by the analytic method.
| 1. 2 | 3 4 || 5 || 6 || 7
E ſ 1 | } 6 6 ) 25% × 1 = $ 25%. E.
D 2 # 2 3 | 83 × 2 = 17+ D.
C 3} { 3 2 2 || 2 X x 4} = { 84 × 3 = 25% C.
B 4 2 2 6+| 3 || 8 | 34% x 4 = 13.7% B.
A 5 || 3 10 | 10 J 42% x 5 = 21.4% A.
2 Days, 120 $420
3%
$420
120 – 28 = 4% multiplier.
$420 - 120 = $34 per day.
Answer by the loss and gain method.
A worked 36 days at $5 per day = $180.
& 4 36 { { $4 ( & 144.
B >
C & 4 16 4 & $3 & 4 3– 48.
D & 4 16 ( & $2 { { ~ 32.
E & 4 16 “ $1 ** = 16.
120 $420.
8OO SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
6. A stock raiser sold 100 animals for $100; he sold pigs at $4 each, calves
at $1}, and sheep at $34. How many of each kind did he sell?
Ans. 42 pigs, 54 calves, 4 sheep.
OPERATION.
1 | 2 3 | 4 Ea:planation.—By the analytic method of Case
1, we find that the first couplet, as shown in
Pigs #| 3 | # ||15–H (3) = 1 column 3, requires 15 pigs to 4 sheep; and that
Calves | 1% 3 (6) = 2 second couplet, as shown in column 4, requires
$1 1 pig to 2 calves. Then by inspection, we see
that the sum of these proportional columns, 3
Sheep. U33| # 4 and 4, which is (19 -i- 3) 22, will not divide
===º sºmeºmºsºmsºmºms 100, the number of animals sold, without a
19 3 remainder, and as any remainder would give a
fraction of an animal, we cannot, therefore, in
100 — 19 = 81 a problem where animals are the articles or
ingredients use a multiplier containing a frac-
81 - 3 = 27 the ratio multiplier of the 4th tidn. Hence we must determine some number
column of proportionals. without a fraction with which to multiply the
proportional numbers in one of the proportional
columns to produce the 100 animals. This we do by trial as follows: First subtract 3, the sum of
the 4th column of proportionals, from 100 = 97 remainder. Now as this remainder cannot be
divided by 19, the sum of the 3d column of proportionals, without a remainder, our first trial fails
and we must try another way. We now subtract 19, the sum of the 3d column of proportionals,
from 100 = 81, which can be divided by 3, the sum of the 4th column of proportionals, giving a
quotient of 27. This 27 is the ratio multiplier of the 4th column of proportionals. Thus: 1 (Pig)
× 27 = 27 pigs; to which we add 15 pigs shown in the 3d column of proportionals, = 42 pigs; then,
2 (calves) × 27 = 54 calves. Then, as shown in the 3d column of proportionals, there are 4 sheep.
NoTE. 1.--Whenever the division of the remainder by the sum of the unused proportional
column, does not give an integer as a multiplier, we must, by multiplication of one of the other of
the proportional columns and a trial at division, try to find some integral multiplier.
NOTE: 2. —It should be remembered, as stated in note on page 791, that a common factor may
The used or omitted in any couplet or column of proportional quantities without changing the pro-
portional parts.
7. A man purchased 100 killed and dressed animals consisting of cows,
goats and pigs, for $100. The price of each was as follows: Cows $10 a piece, goats
$3 a piece, and pigs 50 cents a piece. How many of each kind did he purchase ?
Ans. 91% pigs, 4% goats, 4% cows.
Operation giving fractional numbers.
| | | * | * | * ||_
Average # * | * ||18-H 4 ||22) 913 pigs at $# = $45;
Cost 3 # 1 || 1 ſ
$1 1 : x 4} = { 44 goats at $3 = $123
10| | 1 | | 4; cows at $10 = $41;
2. 100 animals $100
100 - 24 = 4; ratio multiplier.
8. Suppose in the above problem that the animals were alive, how many of
each kind did he purchase ? Ans. 94 pigs, 1 goat, and 5 cows.
OPERATION. te
| 1 2 3 4 100 — 5 = 95
* 95 – 19 = 5 ratio multiplier.
#| 2 || 2 |18-H 4
$1 & 3 # 1 18 × 5 = 90 + 4 = 94 pigs at $4 = $47
10| || 1 1 × 5 = 5 cows at $10 = $50
*====º mºsºmºs 1 goat at $3 = $ 3
19 5 *s-sº
100 animals $100
* ALLIGATION. *g 8o I
9. Suppose in the above problem the price of the cows had been $10, the
price of the goats had been $1, and the pigs 124 cents. How many of each would
he have purchased ? Ans. 7 cows, 21 goats, and 72 pigs.
10. A person bought 100 head of cattle for $100 to-wit: Cows $10; yearlings
$5; calves $2; and goats at 50 cents each. How many were there of each kind?
Answers 1 and 2 as per operations below. Answers 3, 4, 5, 6, 7, 8,
9, and 10 may be produced according to the manner
of forming couplets and variations in finding ratio
multipliers.
Cows, 1 4 1. 1 1 1 1 1 1 4
Yearlings, 1 1 2 3 4 5 6 7 8 2
Calves, 24 5 21 18 15 12 9 6 3 2
Goats, 74. 90 76 78 80 82 84 86 88 92
100 100 100 100 100 100 100 100 100 100
OPERATION.
1 || 2 || 3 4 || 5 || 6
Cows ( 10|| $ 1. × 1 = 1 COW at $10 = $10
Yearlings || 5 + 1. × 1 = 1 yearling at 5 = 5
Calves $1 & 2 1. , 1 × 24 = 24 calves at 2 = 48
; × 1 = 18
Goats #|| } # # || 18 || 8 || 2 × 1 = 8) = 74 goats at # = 37
—|—|—X 24 = 48 *E=mº Eºmºmºmº
19 || 9 || 3 100 animals $100
100 — (19 + 9) = 72
72 – 3 == 24 ratio number for proportional column 6. The ratio number for columns 4 and 5 is 1.
Or, we may vary the work as follows:
5 times 3, the sum of column 6, = 15
To which add sum of column 5, 9
Gives, 24
Then, 100 — 24 = 76. 76 – 19 = 4 the integer ratio multiplier for column 4. 1 is the
multiplier for column 5, and 5 is the multiplier for column 6. See the following partial operation:
4 || 5 || 6
1. × 4 = 4 cows.
1 × 1 1 yearling.
1 × 5 = 5 calves.
× 4 = 72 R 90 goats.
18 8 2 × 1 = 8 *E*
× 5 = 10) 100 animals.
11. A drover bought 100 head of sheep, for which he paid $100, paying $3
a piece for the wethers; $1.50 a piece for the ewes; and 50 cents a piece for the
lambs; how many of each did he buy?
8O2
SouLE's PHILOSOPHIC PRACTICAL MATHEMATICS. jºr
OPERATION.
| 1 || 2
2 || 2
si'i 2
{ #
4
-º- 4 x 5 = 20.
2
2 100 — 20 = 80.
80 – 5 = 16.
4 |
Proportional columns.
3 || 4
4 2 × 16 = 64
* 2}: 5 = 10
X 5 =
1. × 16 =
Ea:planation.—Having found the proportional
numbers shown in columns 3 and 4, according
to Case 1, and Method, 1, then, in trial for
integer multipliers, multiply 4, the sum of
column 4, by 5 = 20, and subtract this product
from 100. Then divide the remainder 80 by 5,
the sum of column 3, - 16 which is the ratio
multiplier for column 3. The ratio multiplier
for column 4 is 5, the sum of column 3.
74 lambs at $ # = $37
10 ewes at 13 = 15
16 wethers at 3 = 48
100 animals $100
Several other answers may be produced by different methods and by variations in finding
ratio multipliers.

Partnership.
§ --→ =N
1385. Partnership is a contract whereby two or more persons agree to combine
their property, labor, knowledge, credit, good will, or skill, (one or more of these)
for the purpose of a common undertaking and the acquisition of a common profit.
The partnership or business is called Firm, House, Concern, or Company.
The money, merchandise, or other kind of property that is contributed or
brought into the partnership by the members thereof, is called Capital, or Capital
Stock, or Capital in Trade.
Sharing Gains and Losses.
The gains and losses of a partnership are shared most generally in certain
fractional or percentage parts agreed upon at the time the partnership is formed.
In determining these parts, the money or property investment, the business
experience, the skill, the commercial standing or credit of each partner, and the
service to be rendered by each, are all considered as elements in fixing the respective
shares of gains and losses.
Sometimes one partner furnishes the money capital and the other supplies
the business knowledge and credit, and each shares the gains and losses equally.
In other cases in which the business knowledge and experience are about equal and
the monetary capital invested is unequal, an Account Current and Interest Account
is kept with each partner, and the gains and losses are divided equally.
In a Commercial or a General Partnership in which no agreement has been
made regarding the division of gains and losses, they will be divided equally, by law.
NOTE.—In the practice of accounts, it is quite common to keep two Ledger Accounts with each
partner; one is called Capital Account or Stock. Account or Investment Account. To this account,
all investments are credited. The other account is called Private Account or Current Account. To
this account all withdrawals are charged and salary, if any, is credited.
When the books are closed, the loss or gain of each partner is carried to his Private or Current
Account, and then this account is closed into his Capital or Stock Account.
1386. Partnership is divided into different classes, as shown in the following
synopsis:
1387. Commercial or General Partners are such as conduct any regular and
lawful business for the purpose of gain.
PARTNERSHIPS IN LOUISIANA.
In Louisiana, the only Civil Law State in the Union, there are the following
classes of Partnerships: 1. Commercial Partners. 2. Ordinary Partners. 3. Part-
ners In Commendam. 4. Limited Partners.
1388. Commercial Partnership embraces the following lines of business: 1.
For the purchase of any personal property and the sale thereof, either in the same
state or changed by manufacture. 2. For buying or selling any personal property, as
factors or brokers. 3. For carrying personal property for hire in ships or other
vessels. Commercial Partnerships are divided into General and Special.
(803).
8O4 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. Yºk
1389. Every commercial partner is the accredited agent of his firm, and each
member of a firm is liable to the creditors thereof, without limit.
1390. Community of Profit is the critical test to prove the existence of a part-
nership.
1391. Ordinary Partners are all that are not commercial, as mechanics,
farmers, lawyers, hotel-keepers, etc. Ordinary Partners are divided into Univer.
Sal and Particular.
1892. Universal Partnership is one in which the parties make common stock
of all the property they respectively possess, real and personal, or personal only.
1393. Particular Partnership is one formed for a special business not of a
Commercial nature.
1894. Ordinary Partners are liable to the creditors of the firm pro rata, i. e.
in proportion to the number of partners, without reference to the proportion of the
Stock or profits to which each is entitled.
1395. Partners In Commendam. In Louisiana, a Partner In Commendam is
one who contracts with some house or firm, either Commercial or Ordinary, which is
already established, to invest a specified amount and to be exempt from any liability
exceeding the sum or amount invested or agreed to be invested. A Partner In
Commendam can take no part in the business; and the business must not be carried
on in the name of such partner, though it may be used after the firm name, with the
Words, Partner In Commendam. The violation of either of these conditions will
make the Partner In Commendam a Commercial Partner, liable without limit.
1396, Limited Partners in Louisiana. By an Act of the Legislature of
Louisiana, passed June 29, 1888, it is made lawful for any number of persons, not
less than three, who shall contribute not less than $5000, to organize a corporation
for the purpose of carrying on any lawful business or enterprise, not otherwise spe-
cially provided for, excepting the stock jobbing business of any kind.
All Limited Partners of this kind shall place the word “Limited,” as the last
word of the corporate name whereveritis written or used. Failure to so use the word
“Limited,” renders every member participant in such omission, liable for any
indebtedness, damage or liability arising therefrom.
PARTNERSEIIPS IN TELE COMMON LAW STATES.
1397. Partnerships, under the Common Law, are as follows: 1. Trading.
2. Non-Trading.
1398. Trading or Commercial Partnerships are such as are formed for buy-
ing, selling, or manufacturing, or otherwise procuring for sale and selling articles
for a profit. All such partners are responsible to creditors, without limit.
1399. Non-Trading Partners are those engaged in the prosecution of Some
occupation or calling, not of a trading or commercial character, as lawyers, mechan-
ics, physicians, etc. In Non-Trading Partnerships, a partner does not possess
general power to bind the firm, and, as a general rule, has no power to contract
partnership debts and issue commercial paper.
* PARTNERSHIP. 8O5
1400. Trading Partnerships are sub-divided into the following classes:
1. Universal. 2. General. 3. Particular. 4. Limited.
1401. A Universal Partnership is one in which all property of every nature
owned by the parties is contributed, and all profits, however made, are for joint
benefit.
1402. A General Partnership is one in which the parties carry on all their
trade and business for the joint benefit and profit of all, whether the investments
are equal or unequal.
1403. A Particular Partnership is one formed for a special branch of trade,
as distinguished from the general business or employ of the partners, or one of them.
1404. A Limited Partnership is one wherein one or more of the partners are,
by the terms of the contract, liable to creditors for such an amount only as he or
they shall have invested, or shall have agreed to invest in the partnership. The fol-
lowing points must be observed in all partnerships of this class: 1. There must be at
least one Commercial, General, or Unlimited Partner. 2. The contract of partner-
ship must be duly recorded with the Clerk of the County Court, or with some officer
specified by law.
NOTE.—In some States, Limited Partners are not permitted to take an active part in the busi-
ness, or to have their names appear in the firm name.
SPECIAL NAMES GIVEN TO PARTNERS.
1405, The following special appellations are given to partners in their differ-
ent relations:
1403. A Nominal or 0stensible Partner is one who holds himself out to the
world as such, but who has no interest in the business; he merely lends his name
and credit to the firm. All such partners are liable for the debts of the firm,
although they receive none of the profits.
1407. A Silent, a Secret, a Sleeping, or a Dormant Partner is one who
is not announced or known to the public as a partner, but who participates in the
profits. All such partners are liable, without limit, to the creditors of the firm.
1408. A Sub-Partner is one who agrees with one of the partners to partici-
pate in that partner's share of the gain, and to bear a part of his losses.
1409, Partners are responsible to one another for any violation of the articles
of agreement among themselves.
FORMATION OF PARTNERSHIPS.
1410. Partnerships may be formed by three different methods: 1. By Articles
of Agreement formally executed and delivered. 2. By a Verbal Agreement. 3.
IBy the Acts of the Parties.
When a contract of partnership is evidenced by written articles, which, under
all circumstances, should be done, great care should be taken to express clearly all
of the provisions, stipulations and terms that the parties desire to have enforced
against their co-partners.
8O6 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
NoTE:-The Law of Partnership is a subject of general interest. With the duties, obligations,
and responsibilities of this peculiar connection, every mercantile man should be acquainted.’ It is a
species of connection which pervades the mercantile world—which has existed and will exist in all
ages and nations. It is a word of most extensive usage; it comprehends equally the union of a day,
and those venerable guilds and ancient associations, which, in the country whence our laws havé
been derived, have existed for centuries; it applies at once to the ephemeral operations of advent-
urers whose projects commence and terminate in the compass of a few hours, and to the mighty
transactions of that company of merchants, which, commencing with a factory on the shores of India,
has overthrown kingdoms, and from their ruins cemented a splendid empire. The inducement to
its formation results from the combination of qualities and the concurrence of circumstances neces-
Sary to successful commercial adventure. It is a connection than which none more close and intimate
can exist among men. It places in the power of him with whom you form it, your property and
your reputation. His virtues and his skill, in unison with your own, may raise you to the pinnacle
of prosperity. His folly, or his crimes, may strip you of every flourishing branch and leaf, and
leave you a naked, withered, and dishonored trunk. It should be formed, therefore, with little
less care than the marriage tie, to which it has been, by a late chancellor, not inaptly compared.—
Law Journal, Vol. 1. *.
DISSOLUTION OF PARTNERSEIIPS.
1411. Partnerships may be dissolved for any one of the following reasons:
1. By the expiration of the time for which the partnership was formed. 2. By the
mutual consent of all the partners. 3. By death, or the physical or mental incapac
ity of a partner to perform his duties in the partnership. 4. By the personal bank-
ruptcy of a partner. 5. By a decree of the court for intemperance, immorality,
violation of the articles of agreement, fraudulent conduct, etc. 6. By the comple-
tion of the work or objects for which it was formed.
1412. The following is one of the most common forms for written instruments
of Partnership Agreement, or Articles of Partnership:
ARTICLES OF PARTNERSHIP, OR ARTICLES OF AGREEMENT.
Articles of Partnership made and entered into this first day of January, A. D. 1893, between
BATTLE BELL, N. L. BARFIELD and R. M. CLEMENT, all of the City of New Orleans and State of
‘Louisiana, Witnesseth : That the said parties hereto, having mutual confidence in each other, do
hereby form with each other a Partnership Agreement on the terms and conditions following, that
is to say:
1st. The partnership shall be for carrying on the Grocery, Western Produce and General
Commission Business, from the first day of January, A. D. 1893, for the term of three years, thence
ensuing with the condition that any party may be at liberty to dissolve the partnership at the
expiration of each and every six months, from the period of its commencement, by serving upon the
others, at least three months previous to such expiration, a written notice specifying that dissolu-
tion will take place at such time.
2d. The said partnership shall be conducted and carried on under the partnership name,
style and firm of BELL, BARFIELD & Co., and the place of business shall be in the City of New
Orleans, State of Louisiana.
3d. That the said BATTLE BELL shall contribute to the capital of said partnership twenty
thousand dollars legal currency. That N. L. BARFIELD shall contribute to the capital of the
partnership fifteen thousand dollars legal currency. That R. M. CLEMENT shall contribute to the
capital of said partnership ten thousand dollars legal currency.
These investments by the respective parties to the said partnership, shall constitute the capi-
tal of said partnership, for the prosecution of the said business. But each partner is entitled to
draw out of the profits or capital of said business, for his own separate use, a sum not exceeding one
hundred and fifty dollars per month, while the said partnership continues.
4th. It is agreed that the gains and losses shall be shared equally, i.e. one-third each by
each partner; and that an interest account shall be kept with each partner.
X- PARTNERSHIP. 807
5th. That each of the parties hereto shall diligently employ his time and capacity in the
business of said partnership, and be faithful to the others in all transactions relating to the same;
and that none will, without the written consent of the others, employ either the capital or credit of
the partnership in any other than partnership business.
6th. That in all matters respecting the transactions of the partnership and the management
of the business, the expressed opinion of the majority of the parties to this agreement shall govern
and be binding on all of said parties; and in cases of difficulty they shall have power to settle up
or sell the concern.
7th. That the books of account shall be kept according to the principles of Double Entry
by the said firm, and correct entries made therein of all moneys, goods, effects, debts, sales, pur-
chases, receipts, payments, and all other transactions of the said partnership.
8th. That none of the partners, during the continuance of this partnership, shall assume any
liability for another or others, by means of indorsing or of becoming guarantor or surety, without
first obtaining the written consent of the other parties thereto, nor shall any partner lend any of
the funds of the partnership without such consent of each of the other partners.
9th. That at the expiration of each and every year, from the commencement of this partner-
ship, an account of stock, effects, credits, debits, and all partnership transactions shall be taken,
the books balanced, and the net gain or loss carried to the respective partners' accounts, according
to article four of this agreement.
10th. That at the close of the partnership from whatever cause, the parties hereto agree to
make a true, just and final account of all things, truly adjust the same, and after all the partnership
business is adjusted, that they will equitably settle with each other, according to the conditions of
this agreement.
IN TESTIMONY WHEREOF, the parties to these presents have hereunto set their hands and seal,
the day and year first above written. º
- BATTLE BELL, {l. s. ;
WITNESS: a-ANº.
J. N. BRADFORD. N. L. BARFIELD, {L. s.3
JAMES HANCOCK. *-*N*
W. B. CUNYUS. R. M. CLEMENT. { L. s.;
1413. By article four of the above Articles of Partnership Agreement, we
See that although the investments by the different partners are not the Same,
the gains and losses are to be shared equally, and that an interest account is to be
kept. The keeping of an interest account between the partners, at the same rate
per cent, is in effect equalizing the capital of the different partners; i. e., it remu-
nerates the partners having the excess of capital out of the private funds, or the
individual gain of the partners who have the lesser capital.
1414. The money invested by each partner is in effect a loan to the firm, and
which the firm holds in trust for the protection of the creditor, and at the same
time uses to carry on the business.
Any other articles or clauses that the parties may desire, and which are not
inconsistent with established law, may be inserted in Articles of Partnership.
In all partnership contracts, each partner should have, and should carefully
preserve, a copy of the Written evidence of contract.
Partnership Settlements.
-
2f
v-v wº, ºr Tw
1415. What Quadratics is to Algebra, what Differential Calculus is to Mathe-
matics, what the Supreme Court is to our System of Jurisprudence, so is Partner-
ship Settlements to the Science of Accounts.
By Partnership Settlements we mean: 1. The finding and the adjustment of
the financial standing of partners at the close of the business, i.e., finding what is
the monetary interest of net capital, or the net insolvency of each partner in the
business at the time the statement is made and the adjustment thereof on the books
Or among the partners. 2. The finding and the adjustment of the unsettled finan-
cial affairs of business men who are not partners, but who have had business trans-
actions with each other, which involve monetary debits and credits. e
The adjustment of these relationships under all the multitudinous conditions
which arise among business men, requires on the part of the accountant a clear
mind, skill in accounts, thoroughness in practical mathematics, and an intimate
acquaintance with the principles of commercial law, business customs, and com-
mercial ethics. His attention should be given first, to the account current, or the
monetary debit and credit standing of each partner, or of each party in interest;
and secondly, to the law governing the business of the parties modified by the spe-
cial contracts entered into by them. Where no specific contracts were made, and
where law or business custom does not cover the case, equity must be applied, and,
strict justice between man and man must be rendered.
The imperative necessity for partnership settlements is of frequent occur-
rence. In the regular course of business, they should be made as often as once a year.
Whenever a new partner is admitted, or an old partner retires, and whenever a
change of interest or share in the gains and losses of the business takes place
among the partners, and at the final dissolution of the firm, partnership settlements
must in justice to all parties be made.
GAINS AND LOSSES.
1416. The most important question, and the first to be ascertained, in settle-
ments between partners, is the determination of the gain or loss of the business.
TERMS AND DEFINITIONS.
1417. The following terms and definitions are necessary to be known in Work-
ing Partnership Settlements:
1418. Resources, Assets or Effects are the property of all kinds possessing
value, and belonging to a firm, corporation, or individual.
NotE.—The amount of withdrawals of cash or property by the proprietor or partners, is to
the business or firm, a resource.
1419, Liabilities are the debts or obligations of all kinds owed by a firm,
corporation, or individual.
NoTE.—The amount of investment by the proprietor or partners, is to the business or firm, a
liability.
(808).
jºr PARTNERSHIP SETTLEMENTS, 809
1420. Investment is the amount of money and property contributed to the
capital of a firm, or appropriated for the transaction of business on private account.
1421. Net Investment is the excess of investments over the withdrawals; or
in case of no withdrawals, it is the total investment.
1422. Average Investment is a sum which, invested for the whole term of the
business, will produce an amount of interest equal to the interest on the different
actual investments for the time that they were respectively in the business.
1423. Capital is the total value of the resources of a firm or individual.
1424. Net Capital, or Present Worth is the excess of the resources over
the liabilities at the close of business, or at any time during business.
NOTE.-The net investment of the proprietor or partners, is not to be included in the liabilities
when the net capital is to be found.
1425. Insolvency is the indebtedness of a firm or individual in excess of the
I’éSOULI'CeS.
NOTE.—The withdrawals by the proprietor or partners are not to be included in the resources
when the insolvency is to be found.
1426. Net Gain is the excess of the resources over the liabilities, including the
withdrawals and investments by the proprietor or partners.
1427. Net LOSS is the excess of the liabilities over the resources including
the withdrawals and investments by the proprietor or partners.
1428. The Net Gain or Net LOSS is also the difference between the total gain
and total loss.
TWO METEHODS OF DETERMINING THE GAIN OR LOSS OF A BUSINESS.
1429. There are two methods of determining the gain or loss at the close of
a business, or at any time during the business. They are as follows:
1st. To collect and classify all the various items of gain and loss, and then
find the difference between their respective sums. This method can only be used
when the books have been correctly kept by Double Entry.
2d. To collect and classify all the various items of resources and liabilities
of the firm, and then find the difference between their réspective sums or aggregates.
This difference is a gain, when the resources exceed the liabilities, and a loss, when
the liabilities exceed the resources.
This fact is clear, for the reason that when the business was commenced, no
matter what was the net capital or net insolvency, the resources and liabilities were
in balance, and it is obvious to the accountant that nothing but a gain or a loss could
destroy that balance.
N. B. When classifying the items of the resources and liabilities, to find the
gain or loss of a business as above directed, include as resources the withdrawals of
the proprietor or partners, and include as liabilities, the investments of the proprie-
tor or partners.
NOTE.-Instead of including both the investments and withdrawals of the partners in the
list of resources and liabilities, the net investment or net withdrawal may be used, thus abridging
the work.
1430. The principles of Accounts upon which this is based, are as follows :
1st. That the firm owes each partner for what he invests, the same as it would
OWe any other person for money or property deposited with the firm
8 IO soul E's PHILOSOPHIC PRACTICAL MATHEMATICs. Yºr
2d. That each partner owes the firm of which he is a member, for all with-
drawals, the same as any other person would owe the firm who received money or
goods from it.
NOTE: 1.--To the critical accountant, there is a clear distinction between the firm itself and
the members composing it.
NOTE: 2.—In the case of a single proprietorship business, it is considered that the business
owes the proprietor for his investments, and that he owes the business for his withdrawals. Thus a
single proprietor of a business possesses two individualities, one as an individual, and one as a
business man.
The second method of determining the gain or loss, as above described, can
be used under all circumstances where it is possible to find the resources and lia-
bilities, and it is the only method that can be used when the books have been kept
by Single Entry, or have been Incorrectly kept by Double Entry, or when no books
have been kept—only memorandums have been made.
1431. ITEMS OF GAIN AND LOSS NOT TO BE INCLUDED IN THE STATE-
MENT OF RESOURCES AND LIABILITIES.
When making the statement of resources and liabilities, care must be taken
not to include any items of gain or loss.
All of the various items of gain and loss are represented in the list of resources
and liabilities. This is evident from the fact 1st, that a loss is the result of a
resource disposed of, without receiving property value therefor, and hence is absent
from the list of resources, and thus the amount of resources is decreased and
the loss is correspondingly increased. 2d, that a gain is the result of a resource
received, without giving property value therefor, and hence is present to increase the
list of resources, and thereby augments the gain.
1432. ORAI, JEXERCISES.
A. commenced business with a capital of $8000, and closed with an insolvency
of $1500. What was his loss? Ans. $9500.
B. commenced business with an insolvency of $3000, and closed with a capital
of $2000. What was his gain? Ans. $5000.
C. commenced business with a capital of $2000, and closed with a capital of
$400. What was his loss? Ans. $1600.
D. commenced business with a capital of $4000, and gained $1400 during the
business. What was his capital at closing? Ans. $5400.
E. commenced business with an insolvency of $3500, and gained $1500 during
the year. What was his indebtedness at the close of the year? Ans. $2000.
F. commenced business with a capital of $6200. He lost during business
$2100. What was his capital at closing? Ans. $4100.
G. closed business with an insolvency of $400. He gained $1800 during busi-
ness. What was his insolvency at commencing % Ans. $2200.
H. commenced business with an insolvency of $1700. Gained during the year
$2500. What was his capital at closing? Ans. $800.
1433. WEITTEN PROBLEMIS.
1. A. and B. are partners equal in gains and losses. At the close of business
A. has a net credit of $6500, and B. a net credit of $3500. There is no property on
hand, and no other liabilities. What settlement should they make?
Ans. B. must pay A. $1500 out of his private funds.
* PARTNERSHIP SETTLEMENTS. 8II
OPERATION.
Resources. I Liabilities. i IXEBIT. CREDIT.
$ 6500 * &
3500 A’s credit $6500–$5000 loss= - - - - - - $1500
#) $10000 loss. B's loss $5000–$3500 credit= - - $1500
$ 5000–A’s 3 loss.
5000–B's # loss. *
Explanation.—Considering that all investments of partners are liabilities of the firm, as shown
above, and that these investments amount to $10000, with no resources to offset them, it is there-
fore clear that there has been a loss of this amount, which must be shared by the members of the
firm as agreed in their articles of partnership. See Articles 45 and 46.
2. A. and B. are partners, equal in gains and losses. At the close of the busi-
ness, A. has a net debit of $6500, and B. a net debit of $3500. There are no other
resources and no liabilities. What settlement should they make?
Ans. A. must pay to B. $1500 out
of his private funds.
OPERATION. W
Resources. Liabilities. DEBIT. CREDIT,
$6500 A's debit $6500–$5000 gain = - $1500
3500
B's gain $5000–$3500 debit = - - - - - $1500
2)$10000
$5000 = A's net gain.
$5000 = B's net gain.
Explanation.—Since, as above shown, the withdrawals of the partners are to the firm a resource,
and since the resources of the firm are $10000, with no liabilities, it is therefore clear that there has
'been a gain to the firm of $10000, which is to be shared by the partners according to the agreement
in the articles of partnership.
3. A. and B. are partners, equal in gains and losses. A. has a net credit of
$6500 and B. a net debit of $3500. There are no other resources or liabilities.
What settlement should they make? Ans. B. must pay to A. $5000 out
of his private funds.
OPERATION.
Resources. | Liabilities. DEBIT. CREDIT.
$3500 $6500
a' 3500 * A’s credit $6500–$1500 loss = - - - - - $5000
$)$3000 loss. B's debit $3500 + $1500 loss = - $5000
$1500 = A’s loss.
$1500 = B’s loss.
Explanation.—In this problem, one partner owes his firm, and the firm owes one of its partners,
The excess of the liabilities over the resources is $3000, which is the loss of the firm, and which must
be shared according to agreement, equally. The operation shows clearly, that B. must pay to his
partner $5000.
4. C. A. Hines and J. L. Hall are partners. Hines ; and Hall 4 in gains and
losses. During the business year C. A. Hines invested $24000 and drew out $1700.
J. L. Hall invested $6000 and drew out $900. They have Mdse. $32000, Cash $4000,
Bills Receivable $5000, Personal Accounts $11000, Bank Stock $2000. They owe
on Bills Payable $3000, and on Personal Accounts $6000. They gained on Mdse.
$23308, on Interest $162, and on Bank Stock $320. Expense shows a loss of $5800,
Insurance a loss of $360, and Exchange a loss of $30. What is the net gain or
loss, and what is the net capital of each partner? Ans. $17600 net gain.
$35500 C. A. Hines' net capital. $9500 J. L. Hall's net capital.
8 I 2 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
&
Operation to find gain Operation to find gain by the Resource
by the Loss and and Liability Method.
Gain Method. *º. Liabilities.
se 2000 C. A. Hines $22300
Լols; 4000 || J. L. Hall 5100
360 | 162 #. §
30 320 2000
$6190 $23790 Resources $54000 $36400
Losses | 6190 Liabilities 36400
Net gain $17600 $17600 = net gain.
PARTNERS’ NET CAPITAL.
C. A. Hines' # net gain ($17600) is - - $13200 J. L. Hall's 3 net gain ($17600) is - - - $4400
do net investment - - - - 22300 do net investment - - - - 5100
do net capital at closing - - $35500 do net capital at closing - - $9500
IExplanation.—The operations of this problem and the general statements and directions above
given, regarding the two methods of finding the gain or loss of a business, make it clear without
repeating general principles.
5. A., B. and C. are partners, A. #, and B. and C. each #, in gains and losses.
At the close of business, their accounts stand as follows: A. is debited $3000 and
credited with $15000; B. is debited with $6000 and C. is credited with $10000.
They have cash in bank $28000. There are no other resources or liabilities. What
is the financial standing of each partner, and what settlement should be made?
Ans. B. will pay $3000 from his private funds.
A. will receive $18000 and C. will receive $13000.
OPERATION.
Resources. Liabilities.
B’s net withdrawal - gº - $ 6000 A’s net investment - gº se sº $12000
Cash - * - tº e Gº * - 28000 C’s net investment - wº- s tºº 10000
Total resources gº e tº – $34000 $22000
Total liabilities - gº º sº - 22000
Net gain me tº-º * * º - $12000
A’s 3 net gain is - - - $ 6000 || B's 4 net gain is - - - - $3000 C's # net gain is - - - $ 3000
A’s net investment is - - 12000 || B's net withdrawal is - - 6000 | C's net investment is - 10000
A's net capital is - - - $18000 | B's net insolvency is - - $3000 | C's net capital is - - $13000
Eacplanation.—From these figures, it is clear that B. must pay $3000 out of his private funds,
which, with the $28000 cash in bank, will make $31000; of which A. will receive $18000 and C. $13000.
6. Suppose, in the preceding problem, that B. has no private property and
proposes to give A. and C. each, his note. What should be the face of the note
given to each 3 Ans. A. $1741.94. C. $1258.06.
OPERATION.
First divide the $28000 cash between A. and C. in proportion to their respective net capitals,
thus:
A’s Share. C’s Share.
A’s capital - - - - - $18000 $28000 00 * $28000 00
C’s capital - - - - - 13000 $31000 | 18000 $31000 || 13000
A's and C's Capital - - $31000 $16258 06 $11741 94
A’s net capital - sº º sº tºº $18000 00 C's net capital º $º º tº $13000 00
A’s proportion of cash - - - 16258 06 || C's proportion of cash - - - 11741 94
A's face of note - alº dº jº tº 3 $ 1741 94 | C’s face of note | tº ſº tºº $ 1258 06
X- PARTNERSHIP SETTLEMENTS. 813
7. A. and B. were commercial partners, equal in gains and losses; no interest
to be allowed on investments. October, 1892, A. invested $8000, and B. $5000; May
1st, 1893, the firm was dissolved, and from an incorrectly kept set of books, the fol-
lowing facts and figures were taken by the partners and submitted for settlement:
A. has a net debit of $2100; B. a net credit of $3500; cash on hand, #349;
10 shares of Bank Stock, market value $1100; Expense Account is debited $5100;
Profit and Loss is debited $3000 and credited $500; the firm owe on a note $2000, and
interest on same $60. Approved, #.
Make, in practical accounting form, the statement showing the correct settle-
ment of the partnership affairs of this firm.
STATEMENT SETTLEMENT.
STATEMENT SHOWING THE RESOURCES AND LIABILITIES, THE NET GAIN, THE CON-
DITION OF THE PARTNERS’ ACCOUNTS AND THE FINAL SETTLEMENT,9. THE
COMMERCIAL FIRM OF A. & B., BASED UPON A STATEMENT OF FACTS APPROVED
BY THEM MAY 1, 1893.


ſ
RESOURCES :
A's net withdrawal tº- tº- - º - - - - sº #% º
Cash on hand - - - º - - - - -- - tº- 1100|00
Bank Stock, market value - - wº- - - -*. - * | -
Total amount of resources - wº- e º - º 6600|00
LIABILITIES :
500|00
B's net investment - * - *- - * - - gº 3500
Bills Payable - - - - - - - - - face $2000 } 206000
Int, due - * - 60
Total amount of liabilities - - - - - - - 5560|00
Net gain of the firm - - - - * - - * 1040|00
DIVISION OF GAIN : 0
A's # net gain - e- - *- - gº - - *- tº- - 5200
B’s net § - - - * - - - * * * º 520,00|| 1040 00
A A’S NET INSOLVENCY :
A's net withdrawal - - - s - * - - &º 2100.00
A's # net gain, deducted - - - - - eºs tº - º 52000
A's present net insolvency - - tº- - - - - - 1580|00
B’s NET CAPITAL :
B’s net investment - - - º - - - - - 350000
B's # net gain added - * - tº * * * * * 52000
B's present net capital - - - - - sº -> - - 402000
Net capital of firm am me me - ºne as º º _|_2440 00
FINAL SETTLEMENT : -
In final settlement, A. must pay to B., out of his private funds, or
give his individual note for - - - tº ºs º ºs I58000
B. must receive the cash - - - " - -. * tº- 340000
and the bank stock - - - - - - - - t- 110000
making a total of - - - * - tº m ºn as 6080|00
B. must then pay the note - - - - - - - - 200000
and interest on same * - * - sº - - - 6000|| 206000
B. will then have, as above, a net capital of * * * * 4020|00
NEw ÖRLEANs, May 1, 1893. GEO, SOULÉ, Accountant.
814 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
8. The following is a Trial Balance of a Double Entry Ledger, and the
inventory of stock at the close of the fiscal year, July 1, 1893:
TRIAL BALANCE.

E. P. Heald, (partner) - - - - - - - - - 2050,00|| 15500|00
A. Harris, (partner) gº - º - - - - tº- 2550|00|| 480000
Cash e tº- tº- - - as ºn tº tº- - fº º- 18567,00|| 16892|00
Mdse. tº - - - - sm as tº *E - - -> 6972|00|| 585250
J. M. Swoop - - - - - - - - - - - 40000|| 900|00
W. Voges * * * * * * * * * * = 967|50
Expense - - - * - ºn tº -> º - - - 52000
Profit and Loss º sº - * - º * , - º 4- 6743|00
City Railroad Stock 4- - ºs ºs º- tº º -> - 5500|00|| 12:500
Commission - - - - - - - - - - - 20000
| 44269||50|| 44269|50
Inventory of property unsold: Merchandise $680.40, Railroad Stock $5450.
NoTE.—The gains and losses are divided as follows: E. P. Heald two-thirds, and A. Harris
one-third.
9. Required, the financial standing of the partners by the two methods of
determining gains and losses. -
FIRST OPERATION BY THE RESOURCE AND LIABILITY METHOD.
STATEMENT OR BALANCE SHEET SHOWING THE RESOURCES AND LIABILITIES, THE
NET LOSS OF THE FIRM, THE NET CAPITAL OF E. P. HEALD, AND THE NET
INSOLVENCY OF A. HARRIS, JULY 1, 1893. W


RESOURCES :
Mdse. as per inventory - - - - - º - º sº 680|40
Railroad Stock, as per inventory - - - e - - sº 545000
Cash on hand - - * - * tº- - * - º ſº 1675|00
Personal Accounts, Dr. - - º- - - * - - º 967|50
Total amount of resources – - - º - º *- 87.7290
LIABILITIES :
E. P. Heald's net investment sº tº- - tº - sº &= 13450|00
Archie Harris’ net investmentſ tº- -, - * : *- tº gº 2250|00
Personal Accounts, Cr. - - - - - * * - ſº - 50000||.
Total amount of liabilities - - - g - - sº *- 16200|00
Net loss of business - * - - is sº - º - - 7427|10
PARTNERS’ PRESENT NET CAPITAL:
E. P. Heald's net Investment º - - tº- - º º 13450|00
E. P. Heald’s two-thirds net loss - - - s - - º 4951|40
E. P. Heald's net capital at closing - - - - - - 8498.60
Archie Harris’ net investment •ºme º- ſºy dºg º tºp * -º 225000
Archie Harris’ one-third loss - º * > - º º º ſº 2475||70
Archie Harris' net insolvency at closing * - - - - 22570
ºf PARTNERSHIP. SETTLEMENTS. - 815
SECOND OPERATION BY THE Loss AND GAIN METHOD.
MERCHANDISE. RAILROAD STOCK. PROFIT AND LOSS.
Cost $6972|Sales $5852.50|Cost $5500||Dividend $ 125 Net Debut $6743.00|$ 75 gain R. R. Stock
Invent. 680.40|P. & L. 75|Invent. 5450
$5575 $5575 Loss on Mdse. 439.10
P. & L. 439.10 i Loss by Exp. 520.00, 200 gain on Com.
$6972 $6972.00 Total loss - - $7702.10|$275 total gain.
275.00
$7427.10 net loss.
E. P. Heald’s two-thirds net loss is $ 4951 40 A. Harris' one-third net loss is - - $2475 70
do net investment is - – 13450 00 do net investment is - - - 2250 00
do net capital is - - - $8498 60 do net insolvency is - - - $ 225 70
10. Two physicians, Jones and Smith, formed a partnership for one year,
‘equal in gains and losses. They opened two offices, one in the upper and one in the
lower part of their city. Jones took charge of the upper office and had a practice
amounting to $4510; of this amount he collected $3600, leaving $910 outstanding.
The expenses of his office were $1100, which he paid out of the money collected.
He also paid $300 from the money collected on account of the expenses contracted
by his partner for the lower office. -
Smith took charge of the lower office and had a practice to the amount of
$3812; of this sum he collected and used personally, $3050. He also collected and
paid on account of expenses $400, leaving $362 outstanding, of which $200 is worth-
less. He contracted expenses amounting to $1641, and gave the firm's note on account
of expenses for $350 which is still unpaid.
At the close of the year Smith retires and Jones continues the practice.
They engaged an accountant to make a correct statement showing how they stood
financially with their business, and with each other. How should that statement be
made 3 Answer, as follows:
FIRST STATEMENT.
:STATEMENT OF SETTLEMENT OF THE PARTNERSHIP AFFAIRS OF JONES & SMITH.


*–
Amount earned by the practice of Jones - - - - - 4510|00
Amount earned by the practice of Smith - - - $3812 00
Less worthless accounts º - - - - 200 00 3612|00
Total amount earned - - - - - - - - 81.2200
Expenses of Jones' office - - - - - - - - 1100|00
* “ Smith's office - - - - - - - - 164100
Total expenses - - - - - - - - - 2741|00
Net gain or revenue - - • * * *- - - 5381|00
Jones' one-half net gain - -: º * - --> * - wº- 263050
Smith's one-half net gain * - - - - - - * 2690|50
5381|00|| 5381|00
JONES' ACCOUNT. T *
By one-half net gain * -* - - - *- *- g- sº 2690/50
To collections -> º - - * - - - - - 3600|00
Less amount paid for expenses $1100+300 = - - - tº- 140000|| 2200.00
Balance due Jones - - - - - - - - - - || _|_49050
SMITH's ACCOUNT. - *-
By one-half het gain * * * * * * * * * 2690|50
To collections used personally * * * * * tº º 3050|00
35950
l
i_Balance due by Smith to Jones - - - - - - -
(continued.) • ' -
SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS.
(Continued from the preceding page.)

RESOURCES :
Personal accounts due on Jones' practice
Personal accounts due on Smith's practice
Less worthless accounts
Smith's indebtedness as above
Total resources of the firm
LLABILITIES OF FIRM :
As shown above the total expenses were
Of this amount there was paid by Jones
By Smith $400 cash, $350 note =
Total expenses paid
Balance due for expenses
Face of note due
Total amount due by the firm
Net capital of firm (due to Jones)
FINAL SETTLEMENT.
In settlement, Smith must pay to Jones from his
indebtedness as above
Jones will collect the resources due, whi
thus placing in his hands
From this sum we will pay th
and the note outstanding
Total amount to be paid by Jones
Balance in his possession is
which is his met capital as above.
Personal accounts due Jones' office
Personal accounts due Smith $362 00
Less worthless accounts
Jones' withdrawals -
Total resources
Total liabilities
Net gain - -
$1100+300=
ch are
e balance due on expenses
910|00
162|00
107200
359|50
1431|50,
2741|00
2150
591
350
00
00
00
941
490
s
private funds, his
º 359150
1072100
1431/50,
gº - -
591
350
º sº
941|00,
490
SECOND OPERATION TO FIND THE GAIN.
RESOURCES :
910 00
$
200 00
162 00
2200 00
3050 00
$6322 00
941 00
$5381 00
é & - º
g- º
LIABILITIES :
$ 591 00.
350 00'
Due for expenses
Due for note -
Total liabilities $ 941 00:
Having the gain, the statement settlement would be made as shown above-
¥ - * PARTNERSHIP SETTLEMENTS. 8 I 7
11. The following facts and figures are submitted by the members of a com-
mercial firm for adjustment:
Jones and Smith are commercial partners. January 1, 1892, Jones Invested
$20000, and Smith $5000. The business was continued until January 1, 1893.
During the business, each partner devoted his services to the business, and neither
drew out, nor added to, his capital. There was no agreement regarding their
respective shares of gain or loss. January 1, 1893, a dissolution took place, and
after the goods were sold, the liabilities to third parties paid, and the resources col-
lected, there was cash in bank $35000.
According to law, custom, and equity, what is the amount due each partner?
STATEMENT OF SETTLEMENT.
STATEMENT OF THE RESOURCES AND LIABILITIES, NET GAIN, INTEREST ON INVEST-
MENTS, AND THE PARTNERS’ NET CAPITAL OF THE COMMERCIAL FIRM OF
JONES & SMITH, JANUARY 1, 1893:

RESOURCES :
Cash on hand - &=} * * tº Eº tº ſº * ge gº & 3500000
LIABILITIES :
Jones' net investment - * * gº * * ſº º tº º º º 2000000
{ Smith’s net investment - - - - - - - - - 5000|00
Total liabilities † : gº {-> is ſº tº º gºe tº 2500000
Net gain of the firm - - - - - - - - - 1000000
Jones’ one-half net gain - - - - - - - - - 500000
Smith's one-half net gain * * * * * me sº * 500000
10000|00|| 10000 00.
PARTNERS’ INTEREST ACCOUNT :
Jones' investment $20000. 1 yr. (a) 8% tº gº tºe tº sº 1600|00
Smith's investment $5000. 1 yr. (a) 8% tº sº tº dº * > 400|00
Balance of interest favor of Jones & tº º tº tº a gºs 1200.00,
Smith's one-half debit of interest is gº º tº tº † - * : 600|00
Jones' one-half credit of interest is tº e * > & ºt tºº ; : 600|00i
1200 00 1200.00
JONES' NET CAPITAL AT CLOSING:
Jones’ net investment as above tºº gº wº gº * * tº º 20000|00
“ one-half net gain as above - tº tº gº tº ſº tº 500000
‘‘ one-half credit of interest - gº tº sº e gº ſº 60000
Jones’ present net capital gº º ºs is sº º sº tº 2560000
* SMITH's NET CAPITAL AT CLOSING:
Smith’s net investment as above - s gº º * gº * 500000
“ one-half gain as above * º sº tº tº dº tº 500000
“ total investment and gain tº-3 tº sº tº ſº. * - 10000|00
& 4 one-half debit of interest deducted tº a * > sº * 600|00
“ present net capital - - - - - - - - 940000
Total net capital of firm at closing as ºn tº º ºs 3500000
NEW ORLEANs, January 1, 1893. GEO. SOULF, Accountant.
NOTE.—According to law and custom, commercial partners, in the absence of any agreement
pertaining to the division of the gains and losses, share the same equally. And in accordance with
equity, if no agreement is made to the contrary, interest is allowed on the investments and with-
drawals of each partner. * s
3.18 soul E's PHILosophic PRACTICAL MATHEMATICs. ¥
sECOND FORMULA of statement.



RESOURCES :
Cash on hand - sº - * sº- º - º - º - 3500000
LIABILITIES :
Jones’ net investment - A- g- tº --> º - $20000
Interest on same, one year at 8% - - - tº ºr 1600
Jones' total investment - - º ºn tº e de - ºn 21600|00
Smith's net investment - - - - - - - $5000
“ interest on same, one year at 8% * * * 400
“ total investment tº º cº- ºn º tº gº 540000|| 2700000
Net gain of the firm, allowing interest on investments - tº 8000|00
Jones’ one-half net gåin - * - - tº tº - º 4000|00
Smith's one-half net gain - sº ºn - sº tº - * 400000
8000|00|| 8000|00
PARTNERS’ NET CAPITAL AT CLOSING:
Jones’ net investment as above º gº - - - $21600
“ one-half net gain - º º - - - 4000
“ present net capital º e- º- - tº gº - E-3 25600|00
Smith's net investment as above - - - as sº $5400
“ one-half net gain * tº ºs - tº g- 4000
“ present net capital - * - sº- - - ºn 900 00
Total net capital of firm at closing - * =y - º $3500000
12. The following statement was received from a book-keeper:
“Adams and Lee were equal partners in business, the amount of capital
was $2500 each. By bad management the business was unsuccessful, and on closing
their books, they found that the money and goods on hand would just pay cer-
tain debts which they owed in New Orleans. This being done, the firm had no
debts due them nor did they owe any. But each partner had a special cash account
on the books, as each had had, from time to time, small amounts placed to his
credit, which amounts were used for the benefit of the firm, and on closing
the books, Adams had a cash balance due him of $800, and Lee a balance of $200.
They now desire to settle this between themselves by a note, and asked my advice,
and I proposed this settlement: that Adams is to give his note to Lee for $200, and
Lee to give his note to Adams for $800, or to make one note from Lee to Adams for
$600. I am not positive that I am right, and so refer the matter to higher authority
and request your decision whether or not I am correct.”
Is the solution of the book-keeper correct?
- Ans. No. Lee should give Adams his note for $300.
See problem 1,
13. The following statement was received from a book-keeper:
“Will you be kind enough to explain to me how to close my books, the
Trial Balance of which I hand you herewith ?
“The merchandise on hand as per Stock Book is $38116.23. Please also make
statement showing gain of business.”
(Continued.)
Y. PARTNERSHIP SETTLEMENTS. 819,
The following is the
TRIAL BALANCE:
Cash on hand - - - - - - $ 443 49 Stock - - - - - - - - - $24012 52"
Mdse. as per Ledger - - - - 14232 01 Bills Payable - - - - - - - 4728 24
Profit and Loss - - - - - 22187.87 Sundry accounts - - - - - - 18642 64
Sundry accounts - - - - - 10520 03
$47383 40 $47383 40
SOLUTION. a
Statement of the resources and liabilities and the net gain of the commercial
firm of , Galveston, Tex., August 1, 1893:
RESOURCES :
Cash on hand - - - - - - - - was tº gº 443/49
Mdse. as per account stock - - * & º g- s gº 38116123
Sundry personal accounts gº *E= †-8 * £ºs gº º *-. 1052003
Total resources of firm - gº ame tº ſº tº sº tº- 4907975
LIABILITIES :
Stock—net investment - gº * sº &_i ſº º -> wº 24012|52
Bills payable - # = *- * > tºº º tº i- * > -> sº 472824
Sundry personal accounts - - - - - tº gº tº 18642|64
Total liabilities of firm - - - - - * , sº gº 4.738340
Net gain of firm i º s gº tº º ºs is tº tº º 1696.35.

To close the books, first credit Mdse. account with the stock on hand, and then
close the account to profit and loss; then, there being, as per statement submitted,
no other loss and gain account, close profit and loss to stock account. Then close
stock account and all other open accounts to or by balance; then rule and add the
accounts, bring the balances down and take a trial balance of the balances.
14. The following statement and question were presented by a book-keeper:
“A., B., C. and D. are partners, equal in gains and losses. During the time
business was continued, A. drew $260; B. drew $155; C. drew $180; and D. drew
$10. There is due A. for rent $150; to B. for services $150; to C. for services $240,
and to D. for services $120. º,
“The net gain of the business independent of salaries due was $1860. What
is the amount due each partner?” Ans. A. $190, O. $360.
B. 295, D. 410.
OPERATION.
Net gain, per statement, is - tº- tº-8 gº sº tº gº sº $1860 00
Less amount due for rent and salaries º tº asp - sº 660 00
Amount to be equally divided - --> &= tºº tºº tº sº 1200 00
Each partner's share of gain tº e ºs º º tº ºs 300 00
GAINS. WITHDRAWALS. BALANCE, RENTS & SALARIES. AMT. D.U.E.
A. $300.00 — $260.00 == $ 40.00 + $150.00 F. $190.00 A.
B. 300.00 — 155.00 - 145.00 + 150.00 F: 295.00 B.
C. 300.00 — 180.00 F. 120.00 + 240.00 – 360.00 C.
D. 300.00 — 10.00 F. 290.00 + 120.00 = 410.00 D.
15. The following statement of facts occurred with two joint owners of a.
steamboat and was submitted for settlement as per conditions named below.
“J. Jones owns five-eighths of the steamboat Star Light, and S. Smith
owns three-eighths of said steamer. The boat is valued at $16000. The boat.
82O soul E's PHILOSOPHIC PRACTICAL MATHEMATICs. Yº
owes liabilities amounting to $6324.28; $3490.13 of the liabilities are owing to
J. Jones, one of the joint owners. The remainder of the liabilities are due to various
parties and amount to $2834.15.
“J. Jones holds a note of S. Smith for $1000, now due.
“It is agreed that one of the joint owners shall buy the other's interest, and
they therefore desire a statement showing how much Jones should pay Smith should
he purchase his interest, and how much Smith should pay Jones should he purchase
his interest in the boat.” What is the statement to be made 3
NOTE.—Joint owners of steamboats are responsible as to themselves for the liabilities of the
boat in proportion to their respective interests in the boat.
f
STATEMENT BASED UPON THE AIBOVE FACTS SHOWING AMOUNT DUE TO EITHER
OWNER OF THE STEAMBOAT STAR LIGHT BY THE OTHER IN CASE EITHER
SHOULD PURCHASE OF THE OTHER HIS INTEREST IN THE SAID BOAT AND
ASSUME ALL LIABILITIES TO THIRD PATRTIES.


1.
Steamboat Star Light, valued at - *- :- º gº * - * 16000|00
J. Jones' five-eighths interest is - gº tº -- «º sº * -º 10000|00
S. Smith's three-eighths interest is gº º •º - ºg wº 6000|00
& 1600000|| 1600000
- 2. -
Liabilities of the boat are - " - * - ſº tº a *g, * sº 6324|28
J. Jones' five-eighths of liabilities are sº &- tº- tº e- 3952|68
S. Smith's three-eighths of liabilities are - - - - - 2371|60
632428|| 632428
3.
Should J. Jones buy S. Smith's interest in the boat he would pay as
follows:
For S. Smith's three-eighths interest - - tº º qº -> 600000
For his, (J. Jones') five-eighths of liabilities *E* sº tº gº 3952|68
Total amount to be paid & tº º º sº tºº tº a 9952|68
Less the amount advanced by him and included in the liabilities 3490|13
Less S. Smith's note held by him as above - sº tº- <= * 100000
Total amount of deductions - tº º gº tº s s 4490|13
Net amount to be paid by J. Jones {-> tº tº ºs tº-ºº: gºs 5462 55
* Liabilities to third parties assumed by him - sº tº tº gº 2834|15
Net amount to be paid to S. Smith & sº - º sº g-r 2628|40
4.
Should S. Smith buy J. Jones’ interest in the boat he would pay as
follows: *
For J. Jones' five-eighths interest - - - - - - 10000|00
For his, (S. Smith's) three-eighths of liabilities * = s gº 2371|60
For his, (S. Smith's) note held by J. Jones * * tº sº 1000|00
Total amount to be paid by S. Smith - sº º sº gº 13371/60
Less Liabilities due third parties gº * = * - sº * * º 2834.15
Net amount to be paid to J. Jones tº º tº tº º * = º 10537 45
16. The following question was received from a book-keeper:
“Messrs. A. and B. entered into partnership, share and share alike in the
gains and losses. They agreed to adjust their current accounts at the end of each
year, so that each shall draw equally. Now, at the close of the first year, the debit of
x- PARTNERSHIP SETTLEMENTS. 82 I
A's current account is $627.24 in excess of B's. What entry should be made to
close their current accounts?” Answer as follows:
B. must draw from the firm $627.24, or A. must pay to B., out of his private funds, one-half of
the $627.24 which he has drawn in excess. In case B. draws the $627.24, the entry would be, B. to
Cash for the amount.
In case A. paid B. one-half of his excess of withdrawals, the entry would be made thus:
B. - - - - - - - - $313 62
To A. - - - - - - - - $313 62
In case it is desired that A. shall pay B., and in case A. has no private funds and is obliged to
draw the necessary amount from the firm to equalize the current accounts, the entries would be as
follows:
FIRST.
A. - - - - - - - - $627 24
To Cash -> -> º cº- sº º sº e- $627 24
SECOND.
- - - - - - - - $627 24
To A. - - - - - - - - - $627 24
In place of the two entries above, one resultant entry may be made thus:
- - - - - - - - $627 24
To Cash e * -º º * * sº - $627 24
With proper explanations.
17. Smith and Brown have 42000 logs to drive down a certain river. Of
this number, 15000 have been taken for other parties at the agreed rate of $6 per
100. Whatever the balance costs to drive, they are to pay in proportion to the
amount of logs each of them owns. Smith has 12000 logs and Brown 15000. Smith
pays in wages and supplies $900; Brown pays in wages and supplies $800; Smith
collected $600 for 10000 logs at $6 per 100; Brown collects $300 for 5000 logs at $6
per 100; Smith's books show the account of Smith and Brown charged with $900
and credited with $600. The question is: What further Journal Entry or entries
has Smith to make to close this account and show his cost or expense of driving his
12000 logs? What has he to pay Brown, or what to receive from Brown, as the
case may be 3
NOTE.-The above problem was submitted by a correspondent to the readers of “BUSINESS.” The
following is our solution.
OPERATION :
1. The expense of driving the 42000 logs was $1700, of which Smith paid
$900 and Brown $800. We therefore credit Smith and Brown respectively, with the
amount paid.
Dr. Smith. Cr. Dr. Brown. Cr.
$600 00 $900 00 $300 00 || $800 00
355 56 444 44
2. Of the $900 collected for driving the logs of third parties, Smith collected
$600 and Brown $300. We debit each for the amount he collected.
3. By deducting the $900 received for driving the logs belonging to third
parties from the $1700, the cost of shipping 42000 logs, there is a balance of $800,
which, by the terms of the problem, is to be borne by Smith and Brown, in proportion
to the logs owned by them respectively.
OPERATION TO FIND THE AMOUNT EACH OWIES OF THE EXPENSE.
Smith 12000 logs. Smith. Brown.
Brown 15000 “ $ 800 00 $ 800 00
27000 || 12000 00 27000 || 15000 00
27000 $ 355 56 Tº IIIAI
822 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
4. We now charge each in the above accounts with the amount he owes for
expense. The accounts then show that Smith owes Brown $55.56. When this is
paid, Brown will be debited and Smith credited.
Smith had debited in his own books, the account of Smith & Brown with
the $900 paid by him, and credited the same account with the $600 collected. The
entries for these transactions were as follows:
1. Smith & Brown *-*. tº- * $900 00
To Smith (representing the cash or property paid out) - $900 00
2. Smith (representing the cash received) - T - - $600
To Smith & Brown tº- ©º tº- 4- º sºs tº- tº- - $600 00
These entries produce the following accounts:
Dr. Smith & Brown. Cr. Dr. Smith. Cr.
$900 00 || $600 00 $600 00 $900 00
NOTE.—To close these accounts, Smith is to credit Smith & Brown with $355.56, which is the
i. it cost to ship his logs, and to debit Smith & Brown with $55.56, the amount he paid to.
I’OWIl.
The following journal entries should be made to effect the object:
1. Log Expense (or Smith) - - - - - - $355 56
To Smith & Brown *- sº gº sº *- tº-º sº - $355 56
2. Smith & Brown *s º * * * «-» º $ 55 56
To Smith (representing cash) 4.- º gº - - - $ 55 56
When these entries are posted, the accounts will stand as follows:
Dr. Smith & Brown. Cr. Dr. Smith. Cr. Dr. Log Expense. Cr.
$900 00 || $ 600 00 $600 00 $900 00 $355 56
55 56 355 56 55 56
$955 56|s 95556
By closing Log Expense to Smith, will produce a balance of both accounts.
NOTE 1.-It will be observed that the account of Brown in the first part of the statement.
does not belong to the books of Smith.
NOTE 2.-If it is preferred, the debit item of Log Expense may be charged directly to Smith.
18. Jones, a planter, agreed with his neighbor, Smith, to manufacture shin-
gles from timber furnished by Smith, and to receive for his services and the use of
machinery, one-fourth of the number manufactured. He manufactured and delivered
to Smith 300,000 shingles. How many must be manufactured for himself to pay
him according to contract 3 Ans. 100,000 shingles.
FIRST OPERATION.
1,–4 =#. Then 300,000 is three-fourths of the number to be manufactured; and since three-
fourths of the number is 300,000, one-fourth of the number would be the third part, which is 100,000.
SECOND OPERATION.
According to the terms of the contract 1 = the whole number manufactured.
and + = “ number paid for manufacturing.
and the difference, #- “ number manufactured for Smith.
Hence the 300,000 = # of the number to be manufactured and conversely # = 300,000; then
since # = 300,000 + = the third part which is 100,000 and 4 or the whole number = 4 times as much
§." 400,000, # of which is 100,000, the number to be manufactured to pay for manufacturing
5 e *
19. A merchant received from a planter 500,000 pounds of seed cotton to gin
and bale. He 1s to receive for the service is of the seed cotton ginned and baled.
Yºr PARTNERSHIP SETTLEMENTS. 823
How many of the 500,000 pounds of seed cotton received must the merchant gin for
the planter? Ans. 454545+ pounds seed cotton to gin for the planter.
OPERATION.
1 or 100 = assumed pounds to gin.
To or 10 = pounds due for ginning.
175 or 110 = pounds of seed cotton required to gin 1 or 100 pounds of seed cotton.
Then: 11%; : 1 :: 500,000 : x or, 110 : 100:: 500,000: × ; or thus: | 1 100
11 || 10 110 || 500,000
500,000 —
454545 ºr fås.
454545 ºr fös. Ans.
20. A father wishes to divide $14 between his son and daughter, so that his
daughter will receive one-third more than his son. How much will he give to each?
Ans. $6 to son. $8 to daughter.
OPERATION.
$1 = son's share. º Son Daughter.
1} = daughter's share. $
sº- 14 14
2} = } = the sum of the proportions 7 3 7 3
according to the condition. 3 || 4
$6 $8
21. A father willed his estate valued at $40000 to his three children, in pro-
portion, as follows: John one-third, Henry one-fourth, and Katie one-fifth. Before
the settlement was made, Henry died. What sum should John and Katie each
receive 3 Ans. John $25000. Katie $15000.
OPERATION.
# -- # = # = the sum of the proportional John. Katie.
interests of John and Katie. 40000 40000
8 || 15 8 || 15
3 5
$25000 $15000
NOTE.-Henry having died, his one-fourth proportional interest is therefore to be divided
between John and Katie in proportion to their proportional interests, as is also the remainder of the
estate.
EXAMPLES.
22. B. A. Root and J. H. Knox have been doing a partnership business
under the firm and style of Root & Knox. They now wish to settle their business
and dissolve co-partnership. The following is a list of their resources and liabilities,
obtained by taking an account of stock, and from their memorandum books: cash
on hand and in bank, $2600; merchandise in store, $7400; bank stock, market
value, $1400; bills receivable, $2100; personal accounts due the firm, $3000. They
owe on notes $800, and on open accounts $500. B. A. Root invested $5000 and
withdrew $700. J. EI. Knox invested $4500 and withdrew $500. What was the
gain or loss of the business? Ans. $6900 gain.
OPERATION.
RESOURCES : LIABILITIES :
Cash - * - - $2600 B. A. Root's net investment - $4300
Merchandise - s -> - - 7400 J. H. Knox’s “ & 4 • ems 4000
Bank stock - s - - tº- 1400 Bills payable mºs - * tº- 800
Bills receivable - - - - 2100 Personal accounts - sº 500
Personal accounts * - - 3000
Total resources - - - $16500
** liabilities - - - 9600 Total liabilities - - - $9600.
Net gain - - - - $6900 Ans.
Explanation.—In the operation of this question, we simply classify and add the resources and
liabilities of the firm, and then find the difference between their sums, which, because the resources
are the greater, is a gain.
824 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
23. Suppose, in the foregoing example, that the partners had shared the
gains equally, what would have been the net capital of each partner on closing?
Ans. $7750 Root's capital. $7450 Knox's capital.
OPERATION.
Net gain as shown above, $6900.
B. A. Root's one-half net gain - - - - - - - - - - - - $3450
& 4 investment - - - - - - tº - - - $5000
Less his withdrawals - - - - sº gº - º 700
B. A. Root's net investment - - s - - - - - - 4300
& & net capital - - - - - - - - - - - - , $7750
J. H. Knox's one-half net gain - - - - - - - - - - - $3450
4 & investment - - - dº - - 3-8 - - * $4500
Less his withdrawals - * - - - - - - 500
J. H. Knox's net investment - - * - - º * - *s {- - 4000
é & net capital - - - - - E - - - ºn - $7450
-ms-
24. M. W. Conoly, Chas. Block, and C. Riggs have been doing business under
the commercial firm and name of Conoly, Block & Riggs. The gains and losses are
to be shared as follows: Conoly one-half, Block one-fourth, and Riggs one-fourth.
During the business, Conoly invested $12000 and drew out $1200; Block invested
$5000 and drew out $1600; and Riggs invested $6000 and drew out $700.
The following is a list of the resources and liabilities of the firm at the time of
closing: cash on hand and in bank, $3600; merchandisein store, per inventory, $10850;
house and lot, No. 1428 St. Charles street, valued at $4800; bank stock of various
banks, market value, $2250; office and store fixtures, $1300; horse and wagon, $540;
bills receivable, $3200; personal accounts due the firm, $2500, on which it is agreed
to allow 10% for doubtful debts. The liabilities are: bills payable, $5000; mort-
gage on the real estate for $2000; personal accounts, $4500. What is each partner's
net capital and the present worth of the firm on the closing of the business?
Ans. Conoly’s net capital, $9695.00.
Block’s {{ 2847.50
Biggs' {{ 4747.50
Present worth of firm, $17290.00.
OPERATION TO FIND THE GAIN.
RESOURCES : LIABILITIES :
Cash tº º - -> sº - $ 3600 M. W. Conoly invested - $12000
Merchandise - - - tº *- 10850 & 4 withdrew - 1200
Real estate - - sº * * - 4800
Bank stock - - - - - 22:50 M. W. Conoly’s net invest. $10800
Office and store fixtures & - 1300 Chas. Block invested – $5000
Horse and wagon º- tº - 540 & 4 Withdrew - 1600
Bills receivable - * º - 3200
Personal accounts º - $2500 Chas. Block's net invest. 3400
Less 10% cº- ºr º 250 C. Riggs invested - - $6000
& 4 withdrew - tº- 700
2250 , -
C. Riggs' net investment 5300
Bills payable tº- tº 5000
Mortgage - m am 2000
Personal accounts - tº- 4500
Total resources º º - $28790 Total liabilities - $31000
“ resources tº- 28790
Net loss - - - $2210
PARTNERSHIP SETTLEMENTS. 825
EACH PARTNER'S SHARE OF LOSS. M. W. CONOLY'S NET CAPITAL.
M. W. Conoly's # of which is $1105 00 M. W. Conoly’s net investment $10800 00
Chas. Block's # of which is 552 50 4 & # net loss - - 1105 00
C. Riggs' + of which is 552 50 *-ºs
é & net cap’l at closing $9695 00
C. BLOCK's NET CAPITAL. C. RIGGS' NET CAPITAL.
Chas. Block's net investment $3400 00 C. Riggs’ net investment - - $5300 00
& 4 + net loss - - - 552 50 & 4 + net loss - - - - 552 50
é & net capital at closing $2847 50 “ net capital at closing $4747 50
The net capital of the firm at closing is $9695.00 + $2847.50 + $4747.50 = $17290.00.
NOTE.-In the following questions, we have the resources, liabilities (except the partners’
investments) and the net gain or loss given to find the net capital at commencing.
25. J. A. V. Barton has just closed his business with a gain of $5280. His
present resources and liabilities are as follows: cash, $9000; merchandise unsold,
$1500; bills receivable, face value, $6000; personal accounts, $4500. He owes on
notes $5000, and to sundry persons $3500. What was his net capital at com-
mencing 3 AnS. $7220.
| OPERATION.
RESOURCES : LIABILITIES .
Cash -> tº- º - - sº - $9000 Bills payable * - - * – $5000
Merchandise - -> - - sº º 1500 Personal accounts - - º -*. 3500
Bills receivable - - - sº - 6000
Personal accounts - - - sº - 4500 Total liabilities - - me - $8500
Total resources - - º - $21000
Resources - - me tº me - * tº º * * * * - * - $21000
Liabilities - º º - - - - " - - º tº 4- * - - º 8500
Present net capital - - - º sº - -- - --> * - tº ºs - $12500
From which we deduct the net gain tº - º - º º º ſº- º - 5280
And obtain - º - º - - &- º º sº t- gº º $7220
which is the amount of net capital at commencing.
Explanation.—In the solution of this example, we first find the total amount of resources and
liabilities; second, we deduct the liabilities from the resources and obtain $12500 as the present net
capital. Then, since the present net capital includes the gain, it is clear that by deducting the gain,
$5280, from the present net capital, we have in the remainder the net capital at commencing, which
is as above $7220.
26. A. and Z. lost, during business, $8400. Their resources at closing are:
cash, $10000; merchandise, $8000; bills receivable, $6000; personal accounts, $3500.
They owe on notes $12000, and personal accounts, $7500. A. invested three-
fourths and Z. one-fourth of the capital. What was the amount invested by each
at the commencement of the business? Ans. $12300 by A. $4100 by Z.
º OPERATION.
RESOURCES : LIABILITIES :
Cash - - - - - - - $10000 || Bills payable - - - - - $12000
Merchandise - - tº º - - 8000 Personal accounts *- -> º - 7500
Bills receivable - - gº --> - 6000
Personal accounts 3500 Total liabilities - - - - $19500
Total resources - º - - $27500
826 SOULE's PHILOSOPHIC. PRACTICAL MATHEMATICs. *
Resources wº †º gº gº s tºº tº gº ge gº tºº tº * . gº tº sº $27500
Liabilities tº º * * * * * * * * = * * * * * 19500.
Present net capital or worth - - - - - - - - - - - - $8000
To which we add the loss - sº tº dº &= * * wº tº tº º sº tºº 8400
And obtain * * * tº-º * wº # = * > sº tº $16400
which is the net capital of the firm at commencement.
Then, as A. invested three-fourths of the capital and Z. one-fourth, it is clear that three-fourths.
of $16400 is A's capital, and one-fourth of $16400 is Z's capital. &
# of $16400 = $12300, A's capital. # of $16400 = $4100, Z's capital.
27. A. and Z. were partners. They were both insolvent when they com-
menced business, and the firm assumed the liabilities of each. The proportion of
insolvency of the partners was, A. § and Z. §. They were equal in gains and losses.
During the business their profits were $7100, and at the close of the business their
resources were: cash, $2800; merchandise, $11400; and personal accounts, $1800.
Their liabilities at closing were: bills payable, $12800; and personal accounts, $5400.
What was the net insolvency of the firm and of each partner at commencing?
What was the capital or insolvency of each partner at closing?
Ans. $9300, insolvency of firm at commencing.
$6200, {{ Of A. “ 4%
$3100, {{ Of Z. “ {{
$2650, {{ of A. “ closing.
$450, net capital of Z. “ 4%
OPERATION.
RESOURCES : LIABILITIES :
Cash -, - - - - - - $ 2800 Bills payable - - - - - $12800
Merchandise Gºe sº ſº- * º 11400 Personal accounts *- tº tº Eº 5400
Personal accounts gº $º gº tº- 1800
Total liabilities - * - º-s, ſº $18200
Total resources gº tº gº - $16000 Af
Resources º tº * sº - - - dº i. e. sº gº * > s tº ſº tºº $16000
Liabilities º agº, tº * * - g- gº gº sº sº * sº tº gº tº sº 18200
Net insolvency at closing tº º º ºs º ºs º ºs º ºs º º $2200
Gain during business added - - E tº ſº sº gº - - - - - 7100
Net insolvency.of firm at commencing - 4-º - - $9300
# of $9300=$6200, A’s insolvency at commencing. # of $9300=$3100, Z's insolvency at commencing.
Total gain, $7100.
A's one-half of which is #º $3550 Z's one-half of which is sº $3550.
A's insolvency at commencing - - $6200 | Z's insolvency at commencing - - $3100
“ # net gain tº gº tº sº gº 3550 “ # net gain * * * * * * 3550.
“ present net insolvency - - $2650 “ present net capital sº s = $450
Explanation.—The difference between the resources and liabilities at closing, not including the
partners' accounts, gives us the insolvency of the firm at closing, $2200. To this amount, we add
the gain and obtain the insolvency at commencing, $9300. Had the difference between the resources
and liabilities at closing been a net capital, not exceeding the gain, which by the terms of the ques-
tion it could not have been, we would, for obvious reasons, have deducted the same from the gain in
order to have found the insolvency of the firm at commencing.
28. W., X. and Y. have dissolved partnership, and from their irregularly
kept books and memoranda, they submit the following statement of facts and figures
for a final settlement. They were equal in gains and losses. They have cash on
* PARTNERSHIP SETTLEMENTS. 827
hand, $14500; merchandise in store, per inventory, $24200; bank stock, market
value, $2100. They own the store that they occupy, which is valued at $31000. The
store and office furniture is valued at $3800. The personal accounts due them
amount to $9500, on which it is agreed to allow 20% for doubtful accounts.
They hold bills receivable amounting, face value, to $8000, on which there is
$540 interest due. They owe on notes, $5000, on which there is $150 interest due,
and on open accounts they owe $4500.
During the business W. invested $25000 and withdrew $3250.
4% 4% X. 4. 22500 {{ 2810.
4% 44 Y. & 4 21000 &é 4100.
They have kept a private book which shows the expense to have been $5768.
W. desires to withdraw from the business, and hence wishes to know his present
net capital. How much is it? * Ans. $29666,67.
OPERATION.
RESOURCES : LIABILITIES :
Cash - - - - - - - $14500 || W’s net investment sº tº gº - $21750
Merchandise º tº sº * - wº 24200 | X’s “ & 6 &=> cº- iº $º 19690
Bank stock - tº sº sº sº dº 2100 Y’s “ é & wº tº- {-} gº 16900
Real estate - - tº sº * > - 31000 || Bills payable * * *—- ºg tº: 5000
Store and office furnitur wº- * = cº 3800 | Interest on same - * = tºº gº sº 150
Personal accounts - - $9500 Personal accounts - - - - - 4500
Less 20% - - - - 1900
— 7600 Total liabilities - - - - $67990
Bills receivable - sº s sº <--> 8000
Interest on same - sº tº tº * 540
Total resources tº a ſº tº - $91740
Resources * = gº gº wº tºº * * : sº tº tº e º cº sº tºº (º tº º $91740 00
Liabilities - † = sº tº ſº * - sº dº º ſº * e ºs tº ºs 67990 00
Total net gain tº *º tº tº tº tº gº © tº 3 gº ºs º * - + )#23750 00
W’s one-third net gain is - - - - - - - - - - - - $ 7916 67
“ net investment wº tº tº º gº tº º tº tº * … º tºo tº-e tºº 21750 00
“ present net capital - - - - - - - - - - - - $29666 67
29. A., B. and C. enter into co-partnership and agree to share the gains
and losses equally. It is also agreed that A. shall keep the books and receive there-
for a salary of $1800 per year; that B. shall act as chief salesman and receive there-
for a salary of $1500; and that C. shall act as salesman and receive a salary therefor
of $1200. At the close of the year, there was a net gain, exclusive of the $4500 sala-
ries, of $3600. How much of the gain is due to each partner?
Ans. A. $1500. B. $1200. C. $900,
OPERATION.
CR. DR. CR.
A’s salary - - - - $1800 – $300 = $1500
B’s “ tº E - & º 1500 — 300 = 1200
C’s “ tº gº gº sº 1200 — 300 = 900
Total liabilities - - dº $4500
Less gain - - - Gº 3600
Net loss sº * * * * - + ) $900 A
•º
Each partner's one-third loss $300
828 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. y
30. W. S. Meek, doing business in Lewisville, Ark., had his store and goods
burned. His books were also destroyed, and from a trial balance, taken the day
that the fire occurred, and preserved by being at his residence, the face value of the
ledger is shown to be as follows:
DR. CR.
W. S. Meek me sº ſº tº e tºº $ 1200 $14400
Merchandise ſº e- tº *† 32600 24800
Cash - tº age gº º tº º tº tº 50524 47215
Bills receivable - sº * = tº tº º 1600 600
Bills payable tº tº tº sº gº 1100 3100
Interest º sº * > gº tº tº 120 50
M. & T. R. R. stock - * - tº -> * > 1000
Expense * > º º - - gº 780
Personal accounts due gº & # sº 4500
Personal accounts owing - - - 3259
$934.24 $934.24
His goods were fully insured, and in settling with the insurance company, it
was agreed, after thorough investigation and close figuring, to allow 35% gain on
the sales of merchandise. The R. R. stock is valued at cost, making an allowance
of 10% on the personal accounts due for worthless accounts. What was the net gain
of W. S. Meek during business, and what was his present net capital?
Ans. $5129.63 gain. $18329.63 capital.
OPERATION.
RESOURCES : LIABILITIES :
Insurance Co. for Mdse. burned, al- W. S. Meek's net investment - $13200 00
lowing 35% gain on sales - " - $14229 63 || Bills payable tº gº tº my 2000 00
Cash on hand $º tºº &s sº 3309 00 | Personal accounts ſº tº ſº 3259 00
Bills receivable º tº & gº 1000 00 sºmºmºmº-
M. & T. R. R. stock &º tºº gº 1000 00 Total liabilities $º tº tº $18459 00
Personal accounts - $4500
Less 10% - - - 450
4050 00
Total resources sº * gº $23588 63
Resources sº tº tº- tº tº dº sº tºº te tº tº tº > º gº – $23588 63
Liabilities iº sº gº dºe * : gº & ge º * ſº º sº tº 4. It * > 18459 OO
Net gain during business - - - - - - - - - - - - $ 5129 63
Net investment added gº gº tº ºs º iº wº gº iº ſº * - º tºº 13200 00
Net capital at closing - - - - - - - - - - - - - $18329 63
Ea:planation.—It will be observed that in the statement of resources and liabilities, we did not
include the expense or interest accounts. They were omitted for the reason that they do not here
show any available resource or real liability, and hence could not be embraced in a statement of this
character to obtain gain or loss. It is true that the gains and losses of the business were affected
by these accounts, but it was done through some property account at the time the debits and credits
were created. This is the case with all accounts of this character, such as: Premium, Discount,
Commission, Profit and Loss, etc., which appear on the face of a double entry Ledger, and which
must not therefore be classed as resources or liabilities. In problem No. 28, we included the
interest due on the matured bills receivable and bills payable because the amounts there represented
actual resources and liabilities.
To find the amount of merchandise on hand, we first find the cost of the amount sold, and
then by deducting the cost of the sales from the total cost, we obtain the amount lost by fire.
XF
PARTNERSHIP SETTLEMENTS. 829
31. The following is a trial balance of W. B. Taylor's Ledger:
DR. CR.
W. B. Taylor - - - - - $ 1600 $15000
Cash gº tº sº sº tº ſº tº 60000 55000
Real estate - - - iºg sº tº 4000 5000
Merchandise - * * * $º ſº tº 21000 27500
Expense tº º ºs tº tº tº 4500
Bank stock - - - - - - 800 1050
R. R. stock - - - - tº gº 1000
Interest * > tº º sº tºº. tº 350 100
Commission - - - - - - 1000
Personal accounts, Dr. tº tº tº 13900
Personal accounts, Cr. tº tº º º- 2500
$107150 $107150

The net gain of the business has been $9214.60. The only property on hand,
unsold, is merchandise and R. R. stock. What is the value of the merchandise?
Ans. $5214,60.
OPERATION.
RESOURCES : LIABILITIES :
Cash - gº gºs º s * > - $ 5000 W. B. Taylor’s net investment - $13400 00
R. R. stock - - - - - - 1000 Personal accounts tº º º 2500 00
Personal accounts - - - - 13900 *===ºse
*==mº Liabilities - - - - $15900 00
Resources, not including merchan- Net gain added - - - 9214 60
dise - - - - - $19900 *===
Total Hiabilities and gain - - - - $25,114 60
Less known resources e- tºº ſº tº 19900 00
Value of merchandise - - - - $ 52.14 60
Or,
Known resources, as above - - - - - - - - - - $19900 00
Liabilities, as above sº gº º -, - dº gº º tº - - 15900 00
Gain - - - - - - - - - - - - - - $4000 00
Which deducted from the whole gain tº º tº tº tº tº ſº 9214 60
Gives a gain balance of - - - - - - - - - $5214 60
which must be the value of the merchandise on hand in order to produce the whole gain.
32. A. and Z. are partners in business with equal investments and equal in
gains and losses. During the business their total net gain is $4410. A. proposes to
draw out 25% of his share of the gain, allowing the balance to remain as an addi.
tional investment. Z. proposes to draw out 20% of his gain, leaving the balance as
an additional investment. The total net gain is equal to 12.4% of the net invest-
ment. What is the net investment of each after the gains have been adjusted
according to the above basis # Ans. A's, $19293.75.
Z’s, $19404,00.
OPERATION.
| 100 = assumed net investment of A. and Z.
12# 4410
$35280 total original investment by A. and Z.
$17640 A's one-half of original investment.
17640 Z’s 6 & & 4 6 &
836 soule's PHILosophic PRACTICAL MATHEMATICS. *
A's one-half of net gain is - - $2205 00, Z's one-half of net gain is - - $2205 60
** 25% withdrawals - - - 551 25 “20% withdrawals - - - 441 00
“ additional investment tº-e - $ 1653 75 “ additional investment - - $ 1764 00
“ original net investment - - 17640 00 “ original net investment - - 17640 00
“ present net investment - - $19293 75 “ present net investment - - $19404 00
33. A dishonest book-keeper closed a merchant's books and found the net
gain to be $13412,45. After he had closed the books, he destroyed the Cash Book
and the cash account in the Ledger, and ran away with the cash on hand. The
resources of the house, not including the cash, amount to $85418.20, and the liabili-
ties amount to $91300.15. What amount of cash did the book-keeper steal?
Ans. $19294.40.
OPERATION.
Resources - tº º - º º º - tº- $85418 20
Liabilities * º - g- º sº - - 91300 15
Loss, by reason of not including cash - - - - $ 5881 95
Net gain - sº- - - . - º - - - 13412 45
Amount of cash stolen - - *- sº º - - $19294 40
34. A. J. Gibson, D. Kinsella, and J. W. Sandidge associate themselves
in business with equal investments and equal interests in the gains and losses. It
is agreed that each partner shall be charged $3 per day for absence from duty.
During the year A. J. Gibson was absent 41 days; D. Kinsella, 65 days; and J. W.
Sandidge, 34 days. How shall the partners adjust the matter between themselves
without making any entries in their books? Ans. D. Kinsella must pay to A. J.
Gibson $17, and to J. W. Sandidge $38.
OPERATION.
DR. CR. DR. CR.
A. J. Gibson owes for 41 days at $3 = $123 $140 – $17
D. Kinsella ** ** 65 “ “ 3 = 195 140 = $55
J. W. Sandidge “ “ 34 “ “ 3 = 102 140 = 38
Total amount due the partners - $420 $55 $55
One-third of which due to each is $140
35. X., Y. and Z. are partners. X. one-half, Y. one-fourth, and Z. one-fourth
interest in gains and losses. On closing business, after all the property has been
divided, it is found that X's account has $500 excess of debit, and that Y’s account
has $150 excess of debit. What settlement is necessary to properly adjust the
matter between the partners? Ans. X. must pay Y. $12.50, and Z. $162.50.
OPERATION. 2
RESOURCES : X? h tº $325 00
y g - - * º - - 500 s one-half of resources is - º 5
#. º: - - * ºr - - sº Y’s one-fourth “ & 4 - E- 162 50
* * Z's one-fourth “ & 4 * * 162 50
Total resources * * - - $650
DR. CR. DR. CR.
X’s - - $500 $325.00 = $175
Y’s tº- e. 150 162.50 = $ 12.5
Z's tº cº- 162.50 = 162.50
* PARTNERSHIP SETTLEMENTS. 831
36. Geo. J. Siegel and Chas. Herzog contracted with B. W. Brown & Co. to
erect a cotton factory for the sum of $18000. Not wishing to incur the expense of
a book-keeper, they agreed that each should keep a correct account of all receipts
and disbursements on account of the contract, and when the work was completed
they would have a general settlement. &
On the completion of the factory, their accounts and business affairs pertain-
ing thereto stand thus:
G. J. Siegel has paid for material and wages $7124, and received from B. W.
Brown & Co., at various times, on account of the contract, $5250,
Chas. Herzog has paid for material and wages $5418, and received from B.
W. Brown & Co., on account of the contract, $4870. They owe to various parties
for material furnished and labor performed, $1985. What has been the profit? How
much is due from B. W. Brown & Co., and how much of it is due to Siegel and how
much to Herzog, after paying the amount due for the material and labor?
Ans. $3473 profit; $3610.50 due to Siegel;
$2284.50 due to Herzog; $7880 due by B. W. Brown & Co.
OPERATION.
Contract price to build the factory * - º - - tº- - tºg - * $18000
Contractors’ costs to build the same are:
Amounts paid by G. J. Siegel - - - - - - - - - $7124
& 4 & 4 Chas. Herzog, - - s - - º * G- - 5418
& 4 unpaid - -- - - sº - º e- - - ‘- 1985
Total contractors’ cost - º - †-> t- º *- ſº- sºme tº- 14527
Net profit º - - me - º - - - - dº - $3473
G. J. Siegel’s one-half net gain is . $1736.50 Chas. Herzog's one-half net gain is - $1736.50
& 4 credit (am’t paid) is - 7124.00 credit (am’t paid) is - 5418.00
4 & total credit is - - - $8860.50 $ & total credit is - - - $7154.50
& 4 debit (am’t received) is 5250.00 & 4 debit (am’t received) is 4870.00
& & loalance of credit is - $3610.50 & 8 balance of credit is - $2284 50
B. W. Brown & Co. contracted to pay - - - nº- * - - * - $18000
& 4 & 4 paid to Siegel - - - - - tº - - - $5250
& 4 & 4 paid to Herzog tº- - - gº ºn s - * - 4870
10120
* { & 4 owe a balance of - - - - - - - - - $7880
-
-
RECAPITULATION OF BALANCES DUE,
Balance due to G. J. Siegel - - - - - - $3610.50
& 4 “ Chas. Herzog - - - - - - 2284.50
& 4 “ for material and labor - - - - 1985.00
Which added gives - - - - - $7880.00
which is the balanº due tº Bºw Brown & Co.
37. T. R. Winn, of New Orleans, and C. P. Johnson, of Texas, enter into a
contract to buy and sell cattle, and share the gains and losses equally. To com-
mence the business, T. R. Winn remitted to C. P. Johnson a sight check for $20000.
Johnson purchased at different times to the amount of $35421.50, and shipped to
832 soulE's PHILOSOPHIC PRACTICAL MATHEMATICs. #
Winn at different times, from which he has sold to the amount of $28740. Johnson
has sold to the amount of $7196.
It is now proposed and agreed to dissolve the partnership. In making up
their memorandum accounts for final settlement, it is found that Winn has paid for
expenses of the business $785.30, and Johnson $1014.65. That Winn has cattle on
hand, valued at New Orleans market price, amounting to $2116, and that Johnson
has on hand cattle that cost $1871. It is agreed in making the settlement that each
shall retain the cattle in his possession at the valuation above given and pay cash
for the balance due the other. What was the gain or loss? What is the cash
balance due by one partner to the other, and by which partner is it due #
Ans. $2701,55, gain.
$8719.93, bal. due by Winn to Johnson.
OPERATION.
T. R. WINN’S ACCOUNT: C. P. JOHNSON'S ACCOUNT:
DR. CR. DR. CR.
By cash, original investment $20000.00 | By cash investment - $15421.50
“ paid for expenses 785.30 “ paid for expenses 1014.65
To cash, rec’d from sale of cattle $28740 To cash rec’d from sale of cattle $7196
‘‘ cattle, value on hand º 2116 “ cattle, value on hand - 1871
$30856, $20785.30 $9067 $16436.15
RECAPITULATION.
RESOURCES : LIABILITIES :
T. R. Winn Withdrew - - º $30856.00 T. R. Winn’s investment - - º $20785.30
C. P. Johnson “ tº- - º 9067.00 || C. P. Johnson’s “ - - º 16436.15
Total resources - - - $39923.00 Total liabilities - - - $37221.45
37221.45
Net gain - - - - * $2701.55
T. R. Winn’s investment - †- $20785.30 C. P. Johnson's investment - - $16436.15
& 4 one-half net gain e- 1350.77 é & one-half net gain - 1350.78
& 4 total Interest - - $22136.07 & 4 total interest - $17786.93
& 4 withdrawals - - 30856.00 & 4 withdrawals - 9067.00
& 4 balance of indebtedness $8719.93 & 6 balance of capital $8719.93
which he owes to his partner. which is due to him by his partner.
38. “A co-partnership existed between C., D., E. & F., of New York; they
agreed to put into the business $25000 each. C. was to have one-eighth, D. one-sixth,
E. one-fourth, and F. one-third; and at the expiration of the year, the balance of
gain in excess of such proportions, was to be placed to profit and loss account for
the next year, as well as to cover any loss that might arise after balancing their books.
On adjusting their accounts, they found a gain of $95676.15, but that, instead of the
partners fulfilling their contract, C. and D. had put in on account of their capital
only $15000 each. In this case, what sum or proportion of the gain must be placed
to their respective credits?”
The foregoing problem we take from a recent treatise on Commercial Calcu-
lation, wherein the author made, by an approximation solution only, 1296 figures.
Yºr PARTNERSHIP SETTLEMENTs. 833
In our solution, working all our statements and counting all the figures, we make
but 269. The following is
OUR SOLUTION.
By the first condition' of the co-partnership, each partner, C., D., E. and F., was required to
invest $25000; and by the second condition, they were to share the gains, respectively $, $, 4, #,
(which added, equals g,) leaving 4 undivided, which was to be placed to the credit of their profit
and loss account.
Now, since the first condition of the co-partnership was not complied with by C. and D.,
they having invested less than they were required to invest by the first condition, their respective
shares of the gain, ; and #, must be reduced in the same proportion as the amounts invested by
them were reduced from the amounts required to be invested by the first condition of the co-partner-
ship. And as C. invested but $15000; instead of $25000 as required by the first condition, and since
$15000 is # of $25000, C. is therefore to receive # of 4 of the gain, according to the second condition,
Which is ºs for his proportional share.
D. having invested only $15000, which is but # of the amount required by the first condition,
he is therefore to receive but 3 of the # of the gain specified in the second condition, which is ºr for
his proportional share.
E. and F. having complied with the first condition of the co-partnership, their respective
shares, and also the # of the gain to be credited to profit and loss, according to the second condition,
remain unchanged.
From the foregoing deductions, we have the following proportional interests of the gain:
C. ſº, D. Tº E. #, F. , and P. and L. §.
Having ascertained the respective proportional interests of the gain at the close of the year,
we have but to divide the whole gain in accordance there with.
By adding C's #, D's Tºi, E's #, Fs #, and P. and L’s $, we find they sum up but +43. The
deficit of Tºfti of unity is occasioned by reason of the decrease of C's and D's shares of gain and will
result in augmenting the share of gain of each partner and of profit and loss, in proportion to their
respective shares.
These increased proportions are respectively:
C’s #5 of 120 - 9 - T#5
D’s To of 120 -: 12 - Tºg NOTE.--The 106 is the sum
E’s + of 120 - 30 - *}; of the respective shares as
F's # of 120 - 40 - Yº; shown above.
P. and L's # of 120 F. 15 - Yºs
The whole gain of the business being $95676.15, Tºs of the gain is :
106) 95676.15 (902.6052
C’s Share of the gain is 9 times = $ 8123.45
ID's & 4 & 4 & 4 12 “ ſ F. 10831.26
E’s & 4 4 & & 4 30 “ $902.6052 = 2.7078.15
F's “ . . . . 40 “ | •= 36104.21
P. and L’s “ & 4 & 4 15 “ = 13539.08
$95676.15
39. A. contracted with Z. to serve him in the capacity of agent in the pur-
chase and sale of goods. July 1, Z. delivered to A. merchandise amounting to
$820.70, and cash $600. A. bought at various times merchandise valued at $4039.20,
and sold at various times merchandise amounting to $3615.55. At the end of the
year, Z. Wishes to close the agency and Settle with A. A. is allowed $300 for services.
There is $1508.40 worth of merchandise on hand, which A. returns to Z. What
834 soulE's PHILosophic PRACTICAL MATHEMATICs. 4×
has been the gain or loss on the sales of merchandise, the gain or loss of the busi.
ness, and how much does Z. owe A. or A. owe Z. ?
Ans. $264,05, gain on sales.
$35.95, loss of the business.
Z. owes A. $123.65.
OPERATION OPERATION
To find how A. & Z. stand in a Dr. and Cr. sense. To find the Gain on the Sakes of Merchandise.
A. IN ACCOUNT WITH Z. MERCHANDISE ACCOUNT.
Dr. Cr.
Dr. Cr. To value of Mdse. invested in
To cash paid to him - - $ 600.00 the business - - $ 820.70
‘‘ rec'd from sales of Mdlse. 3615.55 “ value of Mdse. purchased
“Mdse on hand the day of set- during business - - 4039.20
tlement gº tº tº- 1508.40 By amount received from
By cash paid for Mdse. Mdse. sold gº $3615.55
purchased - e- $4039.20 * value of Mdse. On
“ Mdse. returned - 1508.40 hand * sº 1508.40
“ salary for services 300.00 || To Profit and Loss (for gain on
To Balance (amount due to A.) 123.65 sales) Eº gº gº dº 264.05
$5847.60 $5847.60 $5123.95 §5123.95
OPERATION
To find the Loss of the Business.
Expense incurred to conduct the business (A’s salary) {- * £º tº r Q = iº sº $300.00
Gain as above :- º º tº º sº * º º tº g-º tº- tº- sº 264.05
Net loss of the business - * s sº s $-º tº- sº &= sº sº º $ºn $35.95
SECOND OPERATION TO FIND IRALANCE DUE TO A.
A. in Account with Z.
e Dr. Cr.
To cash, paid to him - - - - $600.00 | By Mdse. returned - - - - $1508.40
“ Mdse. delivered to him - * sº 820.70 || “ salary - gºe º sº g- & ſº 300.00
** Gain on sales s sº * => sº 264.05
To Balance due to A. - sº *-* sº 123.65
$1808.40 $1808.40
40. J. Jones, a merchant in New Orleans, has a branch store in the country.
He places in charge of said store, H. Smith, at a salary of $600. At the time Smith
took charge, the books showed the following accounts: Goods on hand, $2100;
cash, $50; bills receivable, $130; personal accounts receivable, $340; bills payable,
$620. The rent of the store was $20 per month.
At the end of the year, Jones examines the business and finds goods in stock,
$1800; goods bought, $4500; goods sold, $6300; expenses paid, $72; rent paid, $50;
salary paid, $250; bills receivable in safe, $300; bills payable, outstanding, $750;
personal accounts receivable, $840; personal accounts payable, $120. Received for
interest, $36; for running a post office, $180; for other services, $24, Paid the
joroprietor, cash, $300.
* PARTNERSHIP SETTLEMENTS. 835
Required, 1. The net gain, allowing 10% of the bills receivable and personal
accounts receivable for bad debts. 2. The present worth of the country house.
3. The correct cash balance. Ans. $714 net gain.
$2414 present worth.
$998 cash on hand.
The above problem was presented to us with a request to supply the required
2.IlSW62TS.
The following are our solutions:
FIRST SOLUTION.
1. OPERATION TO FIND NET INVESTMENT OF J. JONES.
J. JONIES.
To bills payable - - - - - $ 620 | By Mdse. - - - - - $2100
** cash drawn tº- º - º tº- 300 | ** cash - tº- - * wº- - tº- 50
To balance, met investment º Gº tº 1700 ‘‘ bills receivable - - E - ſº- 130
“ personal accounts receivable - dº 340
$2620 $2620
2. OPERATION TO FIND CASH BALANCE.
J. Jones, Branch Store.
To Mdse. at opening - - - - $2100 | By net investment - - - - $1700
“ “ purchased me - = - 4500 || “ Mdse. Sales - - * - was 6300
“ cash, for expenses - - sº 4- 72 ‘‘ bills payable at closin g- - -> 750
* * * * * * rent - nº º g-º º 50 || “ personal accounts at closing - - 120
“ “ “ salary - - - - - 250 | ** interest received - se - - 36
“ personal accounts receivable at closing 840 “ post Öffice salary -> * - - 180.
“bills receivable at closing - - tº 300 “ service, cash - - - - - 24
- Total credits - - * > - - $9110
Total debits - - *- - º tº $8112 “ debits - - * dº ſº. 8112
Cash balance - - - - - - $ 998

3. OPERATION TO FIND TEIE NET GAIN AND THE NET WORTH OF THE BUSINESS.
Resources. Liabilities.
Cash on hand - - - - - - $998 || Bills payable - - - - - - $ 750
Mdse. ** ** - i- *- - º ſº 1800 | Personal accounts - - * º - 120
Bills receivable - º - $300 270 Due for rent - - - - *- - 190
Less 10% - tº me * 30 “ salary - m tº tº º 350
Personal accounts - º - 840 756 J. Jones' net investment tº tº- º- 1700
Less 10% º me sm - 84
Total resources * * tº- $3824 Total liabilities - gº - $3110
** liabilities sº- - - tº 31.10
Net gain - - - - * tº $ 714
J. Jones' net investment - - sº º tº- • * ſº- - * tº- - ſº- - $1700
J. Jones’ net gain * , ~ * * * tº gº tº tº * * me tº mº - 714.
J. Jones' net worth - - * gº - sº gº - - sº - - sº g- - $2414
Explanation.—In problems where the conditions of the business are given at the opening and
at the closing, the correct cash balance may be determined as follows: 1. Open an account with
the proprietor and find his net investment by crediting his account with the cash and all other pro-
836 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. X-
perty advanced on commencing or during the business; and then debiting the same with all liabili-
. at commencing, and for all withdrawals during the business, as shown in the first operation
a UOV6.
2. Open an account with the Branch Store and debit the same with the balance of all real
accounts except cash which is not known, and also with the total of the representative or loss and
gain accounts, such as merchandise, real estate, expense, rent, salary, etc. Then credit the Branch
Store account as follows: 1st.—With the proprietor's net investment as shown in the account first
opened with him. 2d.—Credit it with the credit balances of all real accounts, and with the totals
of all representative accounts, such as merchandise, interest, etc. When this has been done, take the
difference between the aggregate debits and the aggregate credits, and the same will be the correct
cash balance, as shown in the operation. Now, having the cash balance, to find the net gain and
the net capital make the usual statement of resources and liabilities as shown in the third opera-
tion.
NOTE.-In case of partnerships, first open accounts with each partner and find the net invest-
ment, as above.
SECOND SOLUTION.
J. J. ONES.
To bills payable - - - º - $ 620 | By Mdse. - - ºn - - - $2100
** cash drawn * - - g- - 300 | ** Cash - - º - - - 50
To balance met capital - - gº ºn £414 || “ bills receivable * * * * 130
“ personal accounts receivable - º 340
“ net gain from P. and L. tº º 714.
$3334 $3334
Merchandise.
To inventory at commencing sº - $2100 | By sales - - - º * - º $6300
“ purchases - - - - - - 4500 || “ inventory at closing ºn tº º 1800
To profit and loss - - - - - 1500
$8100 $8100
C. A. S. H. .
To amount on hand at commencing - $ 50 | By expenses - - tº- -: - fºr $ 72
‘‘ interest - - - * * * 36 || “ rent, paid on account - - - 50
“ P. O. receipts - - - - - 180 || “ salary “ “ & 4 º - º 250
“ sundries - - - - 24 || “. J. Jones ‘‘ ‘‘ & 4 sº - - 300
“ Mdse. sales * * - - - $6300 “Mdse. purchases - - º $4500
*** * * * *|†º º º
$4922
By balance on hand as sº - * 998
$5920 $5920
Bills Receivable.
Dr. at commencing * * $130
“ at closing - - - - - - 300
• * increase during business - - - $170
X} PARTNERSHIP SETTLEMENTS. 837
Personal Accounts EReceivable.
—A-i-
Dr. at commencing - - - - - $340
“ at closing - * = 4- 4-> sº 840
“ increase during business - - - $500
IBills Payable.
Cr. at commencing - - - - - $620
* “ at closing - - - - - 750
“ increase during business - - - $130
OPERATION TO FIND THE NET GAIN BY THE RESOURCE AND LIABILITY METHOD.
Besources and Liabilities.
Mdse. inventory at closing tºº – $1800 | Due for rent - - - - - - $ 190
Cash, at closing - - - - - 998 “ salary ess tº e º º ſº 350
Bills receivable at closing - - $300 270 | Bills payable sº º tº tº tº 750
Less 10% for bad debts - - 30 Personal accounts - - - - - 120
Personal accounts at closing - 840 756 J. Jones' net investment - - - - 1700
Less 10% for bad debts - $º 84 * -ºssº
tº-ºº- Total liabilities - - - - $31.10
Total resources - - * wº $3824
“ liabilities - - * wº 31.10
Net gain tº º ºs º ºs $ 714 *
J. Jones' net investment - - 1700
& 4 “ capital - sº -> $2414
OPERATION TO FIND THE NET GAIN BY THE LOSS AND GAIN METHOD,
LProfit and LOss.
To rent - - - - - - - $ 240 | By Mdse. - - - - - - - $1500
“ salary of clerk - wº- s gº sº- 600 | ** interest sº * sº s e- sº 36
“expenses - tº- *º * gº * - 72 “ P. O. salary s º gº tº-> * 180
“10% loss on personal accounts -g 84 “sundry services wº as ºs & 24
“10% loss on bills receivable - tº- 30
Total loss {º º tº sº, sº $1026 Total gain <--> - - - - $1740
“ loss &- gº sº sº º 1026
Net gain - - - - - - $ 714
NOTE.-This form of statement can be made only in cases where all the items of gain and loss
are given, or where they can be found from the facts presented.
Ea:planation.—In this solution we have opened accounts with the business man, and with the
resources and liabilities, and thus show the details of the business more fully than were shown by
the first solution. In the solution, we have shown the two methods of determining the gains of a
business, and two methods of showing the proprietor's net capital or networth.
THIRD METHOD OF SOLUTION.
The problem may be solved by first finding the net gain of the business as shown in the second
solution by the loss and gain method. Then, having the net gain, make the resource and liability
statement as shown. In the second solution, except the item of cash, which is yet unknown. This
statement will give resources $2826 and liabilities $31.10. Then since there was a gain of $714, it is
evident, from the principles given in article 1429 that the resources must exceed the liabili-
ties by that sum, $714, which would make the º of the resources $3110 –– $714 = $3824,
Then, since the resources with the item of cash omitted, amount only to $2826, therefore the differ-
ence between $3824 and $2826, which is $998, must be the correct cash balance. Having the cash
balance and the net gain, it remains only to add the net gain to the proprietor's net investment
and thus produce his net worth at closing.
838 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. x-
41. S. Hicks established a branch store in a country town and placed W.
Wood in charge, at a salary of $40 per month, and a commission of 10% of the
profits. Rent of store was $360 per year and salary of assistant clerk, $25 per
month. S. Hicks placed in the store merchandise amounting to $2210, and handed
W. Wood $100 for account of the business. At the expiration of sixteen months, Mr.
Wood having become dissipated and untrustworthy, a settlement was made. From
the irregularly kept books and the inventory of goods, the following facts were
learned: Merchandise on hand, $3440; merchandise bought during the business,
$15200; merchandise sold during the business, $17760; bills receivable on hand,
$3100; personal accounts due the house, $1050; three shares of leather factory
stock, $270; bills payable outstanding, $4820; personal accounts payable, $2300,
not including the clerks’ salaries. Cash on hand, $1280; paid on account of
rent, $400; paid assistant clerk on account of salary, $350; paid S. Hicks, $1200;
W. Wood drew $900. What has been the gain of the business? What is the
net capital of S. Hicks? What is the condition of W. Wood's account?
Ans. 1. Net gain, $1040. 2. S. Hicks' net capital, $2046.
3. W. Wood owes S. Hicks, $156.
SOLUTION.
OPERATION TO FIND S. HICKS' NET INVESTMENT.
S. H I C E S -
To cash, drawn out {- gº gº ſº- $1200 | By Mdse. invested sº gº gº * * $2210.
To balance met investment gº º =} 1110 || “ cash & 4 e- &= * tº sº 100
$2310 $2310.
OPERATION TO FIND THE NET GAIN OF THE BUSINESS.
Besources. Liabilities.
Mdse. on hand º gºe gº; e- t- S. Hicks’ net investment tº- gº gº $1110.
Bills receivable as º gº me sº 3100 | Bills payable - - * * * 4820
Personal aceounts receivable tº- * 1050 | Personal accounts - sº * tº s &= 2300
Leather factory stock - gº tº-º tº 270 | Due for rent tº e * s - $480 } 80.
Cash on hand tºº {-º * sº- tº 1280 | Less amount paid - cº- cº- - 400
W. Wood Eº i__ * gº – 900 260 Assistant's salary - sº sº - $400 } 50.
Less salary, 16 months - - - 640 Less amount paid - - - - 350
Total resources 4- gº gº tºº $9400
Total liabilities * = gº gº tº- 8360
Net gain - - - - - - $1040 Total liabilities - - - - $8360,
OPERATION TO FIND THE BALANCE OF OPERATION TO FIND S. HICKS' NET
W. WOOD'S ACCOUNT, CAPITAL.
W. Wood, salary 16 months @ $40 $640 S. Hicks’ net investment - - - $1110.
& 4 10% of $1040 profit - 104 & 4 90% of $1040 profit - gº •º 936.
Total amount due sº º sº $744
W. Wood has drawn - a- * -º sº 900
“ owes S. Hicks - - - $156 | S. Hicks' net capital - - - - $2046.
42. Jones of New Orleans, employs A. to conduct his branch house at New
Iberia, which had been in operation for a year, and the books showed at the last
t
* PARTNERSHIP SETTLEMENTS. 839
closing the following results, viz: Cash on hand, $500; Mdse., $2000; bills receiv-
able, $1600; personal accounts receivable, $900, and bills payable, $700; and per-
Sonal accounts payable, $1300.
The agent manages the business for another year, and presents the following
statement of the Ledger accounts at the closing of the current year, except the
cash account not given. Mdse. Dr., $15000; Cr., $14000; balance on hand, $2400;
real estate on hand, $1500; bills receivable, $1200; personal accounts receivable,
$1000; interest account Dr., $80; bills payable account Cr., $2000; personal accounts
payable, $1600; commission account Cr., $50. What was the cash balance, net
gain, and present worth ? Ans. cash, $1870. Net gain, $1370.
Proprietor's net worth, $4370.
OPERATION TO FIND THE NET INVESTMENT.
Jones’ Account.
Bills payable - - - - - - - $700 | Cash - - - - - - - $ 500
Personal accounts payable - sº º 1300 || Mölse. wº- tº- - e- - º sº 2000
Bills receivable - - - º sº 1600
Personal accounts receivable - º - 900
Total CT. - - s - º - $5000
‘‘ Dr. - * sº º &- 4- 2000
Total debit - - - - - - $2000 | Net investment - - - - - $3000
OPERATION TO FIND THE CASH BALANCE.
Branch. Account.
Mdse. Dr. - - - - - - $15000 || Mdse. Cr. - - as ºs º- - $14000
Real estate - - º - - º 1500 | Bills payable - - tº- - º 2000
Bills receivable - - - - * 1200 | Personal accounts payable - - º 1600
Personal accounts receivable * sº 1000 l Commissions - - •- - - {- 50
Interest and discount - - -> tº- 80 | Proprietor's net investment * ºn 3000
Total - - s - - - $18780
To cash balance - º -> - º 1870
$20650 $20650.
OPERATION TO FIND THE NET GAIN.
Resources. Liabilities.
Cash as above * * * * * $1870 | Proprietor's net investment - tº $3000
Real estate - - * - - º 1500 | Bills payable - - - sº- gº 2000
Hills receivable - sº - - g- 1200 | Personal accounts payable - - - 1600
Personal accounts receivable - - 1000
Mdse. on hand - e - - sº 2400 Total liabilities t- * - sº $6600
“ reSources º - - tº 7970
Total resources - - - - 7970 Net gain * * * ºn 4 º $1370.
OPERATION TO FIND THE NET WORTH.
Jones’ net investment - tº- - - º - - $3000
Jones’ net gain - e- - º tº- - sº - º 1370
Jones’ net worth * - sº º - tº ſº º $4370
Explanation.—This problem may be solved in the same manner as shown in the several solutions
of problem 40, and of problem 41. But to abridge and simplify the work, we have solved it by
84o soul E's PHILOSOPHIC PRACTICAL MATHEMATICs. ºf
another method based on the ultimate results of the accounts, which method is applicable to all
similar problems.
The following directions render this method clear:
1st. Open an account for the proprietor, and credit the same with the cash, merchandise, per-
sonal accounts receivable, bills receivable and all other property, furnished at commencing, and
with any other advances made during the business. Then debit the account with the bills payable,
personal accounts payable, and any other obligations due on commencing, and for any withdrawals
during the business. The balance of his account will be his net investment.
2d. Open an account with the branch house and debit the same, first, with the debit balances
of all the real accounts, except cash, which is not known; second, with the totals of all the property
and representative accounts. Then credit the branch house account first, with the proprietor's net
investment; second, with the credit balances of all real accounts; and third, for the totals of all
property and representative accounts.
The sum of all the debits, subtracted from the sum of all the credits, should be the balance of
cash on hand. 4.
3d. Make a statement of all the given net resources including cash, and, of all the given net
liabilities, including the proprietor's net investment. Their difference will be the net gain of the
business. -
NOTE.-If allowances are to be made for doubtful debts, unpaid expenses, rents, salaries, etc.,
for which no entries have been made, and for licenses, taxes, insurance, etc., that have been paid,
but whose time has not expired, make the necessary adjusting entries in the list of resources and
liabilities
43. A. M. Perdue, a merchant in Pine Bluff, Arkansas, engages M. J. Mul-
ler to take charge of his branch store at Linnwood. The following assets and liabil-
ities are shown by a statement made the day M. J. Muller assumed charge : Cash,
$275; merchandise in stock, $2990; bills receivable, $180; personal accounts Cr.,
$340: bills payable, $210. For conducting the business, M. J. Muller is to receive
$40 per month; rent of store is to be $20 per month. At the end of fifteen months,
Perdue wishes to sell out his business, and on investigating his affairs, finds the
following state of things: Insurance paid, $162.50; sundry expenses paid, $135;
one Jefferson Co. bond ($100), worth $90; merchandise bought, $3450; merchandise
sold, $5905; merchandise on hand, $3245; rent paid, $300; salary paid, $500; per-
sonal accounts Dr., $475; personal accounts Cr., $375; bills receivable, $325; bills
payable, $290; paid interest and discount, $12.50; received interest and discount,
$10. Perdue has drawn out $200. He cancels his policy of insurance and is
allowed $55, return premiums. Find gain of business, present net capital of A. M.
Perdue, and cash on hand 3 Ans. Cash on hand, $890. Net gain, $1565.
Net capital, $4260.
First Operation by the Method shown in Problem 42.
Ferdue’s Account.
Personal accounts payable - - - $ 340 | Cash - - * - - * - $ 275
Bills payable - sº - -- - 210 || Merchandise - - - - - - 2990
"Withdrawal - - 4- +- - tº- 200 || Bills receivable sº - -> - * 180
Total credit 4- - - & - $3445
‘‘ debit - - - tº- - 750
Net investment - - sº - $2695
! Net gain as below. - *m tº - 1565
Total debit sº º tº º - $750 Net capital at closing * - $4260
* PARTNERSHIP SETTLEMENTS. 841
Branch. Account.
Insurance paid - - - - $ 162.50 | Insurance returned - - - - $ 55.00
Sundry expenses paid - - - - 135.00 || Merchandise *- - - - - 5905.00
Jefferson Co. bond - º - - 90.00 | Personal accounts Cr. - - - e- 375.00
Mdse. bought - - - - - 3450.00 | Bills payable - - - - - 290.00
Rent paid º - - * * - 300.00 || Interest – * - - - *- ºn 10.00
Salary paid - * * tº º - 500.00 | Perdue's net investment - - - 2695.00
Mdse. at commencing - - - - 2990.00
Personal accounts receivable gº - 475.00
Bills receivable - - - º - 325.00
Interest and discount - º- ºr cº- 12.50
Total º - - - - $8440.00
To cash balance - - - wº 890.00
$9330.00 $9330.00
Resources. Liabilities.
Cash as above - - - - - $ 890 | Perdue's net investment º - - $2695
Mdse. on hand -> - - E- - 3245 | Bills payable º - - - - 290
Personal accounts receivable - -> 475 | Personal accounts payable - - - 375
Jefferson Co. bond - - cº- sº º 90 | Salary due gº ſº sº & sºme 100
IBills receivable º - - sºs - 325
Total resources - t- gºe tº- $5025
Less liabilities - - <-> - 3460
Net gain - - * cº- cº- * - $1565 Total liabilities tº º Cº- - g- $3460
Second Operation by the Detailed Account Method.
Perdue's Capital Account.
OPERATION TO FIND CASH RECEIVED AND
PAID ON ACCOUNT OF MERCHANDISE.
Withdrawal - - $ 200|Investment - - $2895 Merchandise.
Net capital - - 4260 |Gain - - - - - 1565
- Mdse. bought $3450|Mdse. sold - - $5905
$4460 $4460|Fă by bińs pay. $80 Rec’d bills rec. $145
TDue on personal ‘‘ bond - 90
accounts - 35 Personal acct's
— 115 due - - - 475
tº — 710 .
Merchandise. *-
Paid in cash - $3345|Received cash $5195
At commencing - $2990|Sold - - - - - $5905 º
Bought - - - - 3450|Inventory - - - 3245 Bills Payable.
Profit and loss - - 2710
On commencing $210
$9150 $9150 “ closing - " - 290
Increase * - s so
Int, and Disc't. Ičent.
Cash. -
Paid $12.50 | Rec’d $10 $300
On hand - - - $ 275|Expenses paid $135.00 d
Insurance returned 55|Rent paid - - 300.00
Interest received - 10|Perdue withdrew 200.00 Insurance. Salary.
Rec’d from Mdse. Insurance paid ; s
sales - - - - *|Rºy Pº ºPd. $162.50 | Retd $55|Paid $500
Mdse. paid - 3345.00 Due 100
Interest paid - 12.50
Balance on hand 890.00 Bond. Lapense.
$5545 $5545.00 —I- -
*-*-* so siast
842 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. 3.
Bills Receivable. | Personal Accounts Or. | Personal Accounts Dr.
On commenc'g $180 On commenc'g $340||On closing - $475
At closing - 325 “ closing - 375
Increase - $145 Increase - $ 45
Profit and Loss. Resources and Liabilities.
Rent sº - $300.00 || Mdse. gain - $2710.00 || Cash s - $ 890 | Perdue's net inv. $2695
Salary - - 600.00 | Interest º 10.00 || Mölse. sº gº 3245 | Bills payable 296).
Interest - - 12.50 * Bills receivable 325 | Personal accts. Cr. 375
$162.50 } Bond - - 90 | Salary due 100
Insurance 55.50 (107.50 Personal accounts 475 $3460
Expenses - 135.00 Net aain tºº 1565
Perdue's gain 1565.00 et gav
44. A Temperance town placed in the hands of a druggist, who is to act as
agent, liquor amounting to $415 and cash $50. The agency continued fifteen
months. On settlement, it was found that the agent had purchased goods amount-
ing to $4600; sold goods $5810, and had on hand, which he turned over to the town
authorities, goods amounting to $1005.50. The agent was to receive a salary of $20
per month and 10% of the gross profits. What does the agent owe the town 3
What was the net gain of the town 3 Ans. 1. Agent owes the town $779.05.
2. Net gain of the town is $1320.45.
First Operation to find the Balance Due the Town.
DR. Druggist. CR.
To merchandise - - - - - $ 415.00 | By merchandise returned - - - $1005.50
‘‘ cash º sº sº gº º ºs 50.00 || “ 15 months’ salary - - - - 300.00.
“ gain on sales - - - - - 1800.50 || “.10% of profits - - - - - 180.05
$1485.55.
By balance due town - * gº tº 779.95.
$2265.50 $2265.50,
Second Operation to find the Balance Due the Town.
Druggist.
To cash, at commencing - - - 50.00 | By cash, merchandise bought - - $4600.00
** ** merchandise sales º-º º 5810.00 “ 15 months’ salary - - sº º 300.00
“ 10% of profits * * * * 180.05
By balance due the town es e- * 779.95
$5860.00 $5860.00
OPERATION TO FIND THE GAIN ON MERCHANDISE.
Merchandise.
Merchandise on commencing - a 3. - $ 415 || Merchandise sales gº tº gº is * $5810.00
& 4 purchased - gº gº tº 4600 44 inventory sº * iſ a 1005.50
Sales and inventory - sº *-* - $6815.50
Total cost - * = gº *…* gº 5015.00
e **—
Total cost - - - - - $5015 Gain on merchandise - - - $1800.50,
* PARTNERSHIP SETTLEMENTS. 843
OPERATION TO SHOW TOWN’S NET GAIN.
Profit and Loss.
Salary - - - - - - - $300.00 || Gain on merchandise - - - - $1800.50
10% commission on gross gain - - 180.05
Town’8 met gain - - - - - 1330.45
$1800.50 $1800.50
Explanation.—The above problem, like problem 39, merits the critical attention of the
accountant. The operations are made so explicit that we deem an extended explanation unnecessary,
for those familiar with accounts. Parties who are not familiar with accounts are not expected to
comprehend the work.
45. A., B. and C. bought a lot of merchandise on speculation for $3000. A.
paid $1500, B. paid $900 and C. paid $600. It was agreed that each one's interest in
the goods should be in proportion to the amount of money invested. They then sold
to D. one-fourth interest in the goods for $1800, and A., B. and C. agreed to so adjust
the matter that each of them should be one-third owner in the remainder of the
goods. What was the financial adjustment between A., B. and C.?
Ans. C. paid $360. A. received $1800.
|B. received $360.
SOLUTION.
OPERATION TO FIND THE INTEREST SOLD TO D., | OPERATION TO FIND WHAT C. MUST PAY FOR Tg5
THE INTEREST THAT A. AND B. SOLD RE- AT THE RATE OF $1800 FOR + INTEREST IN
SPECTIVELY AND THE ADDITIONAL INTER- $3000, IN ORDER. To own 1% or # INT.
EST ROUGHT BY C. IN THE REMAINING GOODS.
Amount Interest Interest Interest Interest
paid. owned. retained. sold % bought
"A. & B. by C. 1800
A. $1500 +3 — ºr = 1°, $ 4.
B. 900 ºr — ºr = Tâu 120 || 6
C. 600 #; Jºy = q Tân g -º-º:
sºme - $360
$3000 #}
#3 + 4 = 1% = the interest sold to D. or thus:
#3 — ºr = ** = int. retained by A., B. & Cº. # : Hºo :: $1800 : $360.
fºr – 3 = % = intofA.,B. & C. respectively
OPERATION TO FIND THE AMOUNT OF MONEY DUE A. AND B. RESPECTIVELY FOR
THE INTERESTS SOLD TO D. AND C.
Amount received from D. -> - - $1800 A. sold as above - s - º - #
Amount received from C. sº - - 360 IB. sold as above - - sº - tº º T#7
Total interest sold - - - - Tº
Total amount received is º tº- $2160 A - B
which is to be divided between A. and B. in pro- $2160 | $2160
portion to the interest each sold. 36 | 120 36 | 120
120 || 30 120 | 6
$1800 $360
844
+k
soule's PHILOSOPHIC PRACTICAL MATHEMATICS.
46. A manufacturing corporation has an agency in a distant city, for the
sale of its goods. At the close of the year, December 31st, 1891, the agency showed
the following inventory or statement:
Merchandise in stock - 4- - e-
Accounts receivable - - - - -
Cash on hand - gº - tº- - º
Store fixtures - -> - tº- - tº-
$256,897.00
102,105.00
2,220.40
4,500.00
*-
$365,722.40
December 31st, 1892, the agency presented the following resource statement:
Merchandise in store - *- - -
Accounts receivable - tº- -
Cash - - == - - - -
Store fixtures - - - º - -
$238,305.00
110,300.00
3,515.15
4,200.00
$356,320.15
During the fiscal year, the agency received from the manufacturing company
for expenses $30000. and remitted to the company $285,427. The sales of goods for
the year were $452,512. Allowing the expense of the business and the loss on sales
for bad debts to be equal to 18%% of the sales, what has been the net gain of the
agency 3 Ans. $161,178.75.
OPERATION
Agency at —
To Mdse. Dec. 31, 1891 sº - $256,897.00 | By Mdse. Dec. 31, 1892 tº- - $238,305.00
“ accounts receivable, Dec. 31, 1891 102,105.00 || “ accounts receivable, Dec. 31, 1892 110,300.00
** cash wº- - - º - 2,220.40 “ cash a- -> - º - 3,515.15
“ store fixtures - - me - 4,500.00 || “ store fixtures - - sº - 4,200.00
$365,722.40 $356,320.15
“ cash received from Mfg. Co. - 30,000.00 “ cash remitted to Mfg. Co. - 285,427.00
$395,722.40 $641,747.15
395,722.40
Gain - - - * = - $246,024.75
Expenses and loss, 184% on $452,512 = - tº- tº - &= - 84,846.00
Net gain of Agency
$161,178.75
Manufacturing Company Account at the Agency.
To net inventory, Dec. 31, 1892 - $356,320.15
• ‘‘ cash remitted - - -: - 285,427.00
Total debit - - - - $641,747.15
“ credit - - mº - 395,722.40
Net debit, - - - - - $246,024.75
Less expense as above - º - 84,846.00
Net debit - - - - - $161,178.75
By net inventory, Dec. 31, 1891
‘‘ cash received - - º
Total credit - - s
$365,722.40
30,000.00
$395,722.40
PARTNERSHIP SETTLEMENTS. 845
47. A firm sold goods at 70% discount on list price, the net sales amounting
to $200,000. The net gain on the sales was $60000. They now wish to allow 75%
discount on list price and increase the sales so as to produce the same amount of
gain. What will be the amount of net sales, the amount of list price, and the cost?
Ans. $375,000 net sales. $1,500,000 list price. $315,000 cost.
OPERATION.
1. Find the list price of $200,000 net sales at 70% discount. 30 : 100 :: 200,000 : $666,666; =
the, list price.
2. $666,666; list price–70% = $200,000 net sales, giving a gain of $60000, and a cost of
$140,000.
3. $666,666; list price—75% = $166,666; net sales, – $140,000 cost = $26666; gain.
Having now produced the gain, $266663, that would have been realized on the net sales of
$166,666; at a discount of 75% on list price, we divide the gain that is required, whatever it may be,
in this case $60000, by the amount of gain realized, in this case $266663, and in the quotient we have
the ratio between the gain realized and the gain required. Thus, to further elucidate, $60000 --
$266664 = 24; which shows that the gain realized at 75% discount ($266663) is but 24 part of the
gain required ($60000). It also shows that the net sales which produced the $266663 gain, are but
24 part of the net sales required to give 24 times as much gain. HENCE we MULTIPLY THE $166,666;
NET SALES BY 24 AND PRODUCE $375,000, THE NET SALES REQUIRED.
Now, since the net sales are only 25% of the list price, therefore 4 times the net sales will be
the list price, which is $375,000 x 4 = $1,500,000 or, thus: 100 — 75% = 25, and then 25 : 100 : :
$375,000; $1,500,000.
PROOF.
$1,500,000 list price — 75% = $375,000 net sales — $60000 gain = $315,000 cost.
SECOND PROOF.
$140,000 cost gained $60000 = 42%.
$315,000 cost gained $60000 = 1991%.
42% -- 1941% = 2+ which is the ratio between the gain when 70% discount was allowed and
the gain when 75% discount was allowed, as shown above.
ADJUSTMENT OF INTEREST ON PARTNERS’ ACCOUNTS.
Many of the following questions involve in their solution the principles
of the science of Double Entry Book-keeping as well as the principles of the science
of numbers, and are therefore of special importance to young accountants and mer-
chants who aspire to eminence and success in the line of their professions. -
48. J. M. Bracey, T. A. Hammons, and T. G. Mackie are partners. They shar
gains and losses equally, and by the conditions of the partnership, an account cur-
rent and interest account is to be kept with each partner, at 8%; i.e., interest at 8%
is to be allowed on the investments, and charged on the withdrawals of each part-
ner, by which means the investments of each partner are equalized.
At the end of the year, when they wish to close their books, it is found that
there is interest on the debits and credits of the partners' accounts, as follows:
Dr. Cr. Net Cr.
J. M. Bracey - - $750.00 $2350.00 => $1600.00
T. A. Hammons - tº- 515.50 1740.00 = 1224.50
T. G. Mackie wº gº 160.75 641.20 =
480.45
Total credit of interest - tº * * * > sº $3304.95
846 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. X-
What is the correct Journal Entry to be made in the books of the firm, in
order to properly adjust the interest between the partners?
OPERATION.
Amount of credit interest - tº * tºº * $3304.95
Each partners' one-third of which is - - $1101.65
IDr. Cr. Dr. Cr.
J. M. Bracey e- * tº- $1101.65 $1600.00 = $498.35
T. A. Hammons - tº- &- 1101.65 1224.50 = 122.85
T. G. Mackie sº- sº- * 1101.65 480.45 = $621.20
$621.20 $621.20
From these figures, the following entry is made:
T. G. Mackie To Sundries $621.20
To J. M. Bracey - - - - - - - - $498.35
“T. A. Hammons gº *g tº-º tº me sº * tºº 122.85
The above entry is the only correct and proper one that can be made to
adjust the interest that accrues on the partners' investments. But the general Way
Of adjusting partnership interest of this kind is by passing it through profit and
loss account. This is done by the following entry:
Interest (or Profit and Loss) To Sundries - - $3304.95
To J. M. Bracey - - sº me tº tº sº sº ºn tº º $1600.00
“ T. A. Hammons &º tº sº sº sº sº sº * * - s 1224.50
“ T. G. Mackie - * * * -> tºº, sº * * tºº sº ſº * - tº- 480.45
This entry produces the same result, so far as the partners are concerned, as
the correct one, but by it a false loss is represented, and hence it is not a proper
entry to make.
It is clear to any accountant that no loss can be sustained by a firm unless
the capital of that firm is thereby decreased, and it is also clear that while partner-
ship interest effects changes in the investments of the different members of the
firm, it does not decrease the capital of the firm, and therefore no loss has occurred
by reason of partnership interest, and consequently the books should not represent
a loss.
It is a Sorrowful fact, not generally known, that many merchants of reputed
respectability, for the purpose of showing small profits in their business, charge their
interest accounts with 12, 15, or 20% interest on capital, and their expense account
With partners' salaries sufficiently large to absorb nearly all of the gain.
This practice has been indulged in to such an extent by the non-observers of
the principles of ethics, as to bring disgrace upon honorable merchants in all
Sections of the country. The practice should be discontinued.
49. The following problem was presented by a teacher for solution:
A., B. and C. form a co-partnership under the following conditions: A. is to
manage the business, and to receive therefor $2400 per annum, which amount is to
be credited on July 1st. He is to receive interest on his salary and to pay interest
on sums withdrawn at the rate of 6% per annum. B. and C. are to furnish the cap-
ital, and to receive interest therefor at the rate of 6% per annum; the net gain or
loss to be divided equally. B. invests, January 1st, $10000; April 1st, $5000. C.
Invests, January 1st, $10000; July 1st, $5000, and draws out, September 16th, $500.
A draws out, February 1st, $200; March 1st, $400; July 11th, $500; October 1st,
X-
847
PARTNERSHIP SETTLEMENTS.
$200; November 21st, $100. At the end of the year the gain—without taking into
account either the salary to be paid to A. or the interest on the partners' accounts—
is $8437.16. What will be the balance of each partner's account when all the items
have been properly entered ?
NOTE.—Interest to be computed in
months of 30 days.
Ans. A's credit balance, $2505.75. B's credit balance, $17307.58.
C’s credit balance, $16723.83.
OPERATION.
ACCOUNT CURRENT AND INTEREST ACCOUNT WITH PARTNERS.
DR. A. CR.
February 1, 11 months $11.00 interest $200 July 1, 6 months, - $72.00 int. $2400.00
March 1, 10 “ 20.00 & 4 400 By balance of interest 48.84 23.16
July 11, 5; “ 14.17 “ 500
October 1, 3, “ 3.00 é & 200 ‘‘ one-third net gain - - - 1482.59
Nov. 21, 1: “ 67 & 100. - *-m-mº
mºst tº-ººººº. $3905.75
$48.84 $1400 1400.00
$2505.75
DR. B. CR.
January 1, 12 months $600 interest $10000.00
April 1, 9 ** 225 « 5000.00
By interest - gº * «s sº 825.00
“ net gain - - - - - 1482.58
$17307.58
DR. C. CR.
September 16, 33 months $8.75 interest $500.00 | January 1, 12 months, $600 interest $10000.00
* July 1, 6 & 4 150 & 4 5000.00
$750
By balance of int. 8.75 741.25
“one-third net gain - - 1482.58
Y $17223.83
500.00
$16723.83
OPERATION TO FIND THE NET GAIN.
Total gain of the business is sº º sº gº tº tº * sº - - - - $8437.16
From this deduct as follows:
Balance of interest due A. $ 23.16
& 4 & 4 && B. – 825.00
& & & 4 4 C. - 741.25 $1589.41
Salary paid to A. - - - - - - - 2400.00 3989.41
Wet gain of business - - - - - - - - - - $4447.75
A’s one-third net gain is * s =º sº tº - - - - $1482.59
B’s one-third of net gain is - & sº sº *- * sº sº gºe 1482.58
C’s one-third of net gain is - - - - - - * * *
1482.58
-848 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. º
Second Operation to adjust the Interest and Cain.
Int. Dr. Int. Cr. Net, Int. Cr. Dr. Cr.
A. $48.84 $ 72.00 $ 23.16 529.80 $506.64
B. 825.00 825.00 529.80 $295.20
C. 8.75 750.00 741.25 529.81 211.44
Total interest Cr. # $1589.41 $506.64 $506.64
Each partner's one-third int. $529.80%
Total gain - - - - * - - tº- * $8437.16
From which deduct salary of A. - - - - - 2400.00
Net gain, exclusive of interest on partners' accounts # $6037.16
Each partner's one-third of net gain - - - - 2012.38%
By this method of adjusting the matter, the partners' accounts will stand as follows:
Dr. A. Cr. Dr. B. Cr. Dr. C. Cr.
$200 $2400 $10000 $500 $10000
400 2012.39 5000 5000
-*-* - 295.20 211.44
500 2012.38 2012.39
200 44.12.39 - *- I ememº-mºs
100 1906.64 $17307.58 Cr. $17223.83
506.64 500.00
$1906.64 $2505.75 Cr. $16723.83 Cr.
50. A. and B. form a partnership, each investing $15,000. A. is to manage
the business, and B., who is to be absent, agrees to pay a book-keeper to take his
place in the house at $1200 per annum. The gains and losses are to be divided
equally. The business was conducted for fifteen months, during which time the book-
keeper drew his salary and charged the same to expense. The books were closed
at the expiration of the fifteen months. What entry should be made to adjust the
salary of the book-keeper? Ans.
FIRST FORM OF ENTRY. SECOND FORM OF ENTRY.
B. - - - - - - $750 B. - - - - - - $1500
To A. - - - - - - $750 To Profit and Loss, (or Expense) - $1500
For one-half of $1500 salary of book-keeper, For salary of book-keeper, the same having
charged to expense instead of to B. been charged to expense when it should have
been charged to B. f
51. C. and D. are partners, equal in gains and losses. , C. is to keep the books
and manage the financial affairs, for which he is to receive $200 per month. D. is
to superintend the purchasing and selling of goods, for which he is to receive $150
per month. At the close of two years, there was a cash gain of $9000, exclusive
of the amount paid on account of salaries. How much is due to each partner, C.
having drawn $2000 salary, and D. having drawn $2500 salary?
Ans. Due to C. $5350. Due to D. $3650.
OPERATION.
Gain as above † - - * *- º - -> * - - - sº tº- - ‘º $9000
C’s salary 24 months @ $200 = * - - - sº - $4800
Less amount received on account - - gºs - 2000
C’s balance due on salary - tº- tº - - E-l - $2800
D’s salary 24 months @ $150 == - - - * - $3600
Less amount received on account - - º - 2590
D's balance due on salary - - - -
Total liability of the firm for salaries - - - - - - - - - $3900
Net gain to be equally divided between C. and D. - - - - - - - $5100
Yºr
PARTNERSHIP SETTLEMENTS. 849
C’s one-half net gain is - - - $2550 | D's one-half net gain is - - - $2550
“ balance of salary is - - - - 2800 || “ balance of salary is sº tº m 1100
‘‘ interest in the $9000 is - º - $5350 | “ interest in the $9000 is - - - $3650
C’s interest $5350 —H D's interest $3650 = $9000.
52. W., X, Y, and Z. are partners. An interest account is to be kept with
each partner, for his investments and withdrawals. W. is to share one-half of the
gains and losses, and X, Y, and Z. are to share equally the remaining one-half.
At the close of the fiscal year, the account current and interest accounts of
the partners show the following interest balances: W. $1000 credit; X. $500 debit;
Y. $1700 credit; and Z. $400 debit. What is the proper Journal entry on the books
of the firm to adjust the matter? Ans.
FIRST FORM OF ENTRY, SECOND FORM OF ENTRY.
Sundries to Sundries Interest to Sundries - sº $2700
X. gº - º sº - $800 TO W. - * - t- - $1000
Z. Eº - s tº- º 700 ** Z. - *- - sº tº- 1700
To W. se - - º $ 100
“ Y. º - &- º 1400 | Sundries to Interest - tº- tº- 900
X. - - tº- - º 500
Z. - Gºs e- º - 400
OPERATION.
RESOURCES : i.IABILITIES : PARTNERS’ INT, ACCOUNTS:
Due by X. - - - $500 | Due to W. - - $1000 |ws int. credit - - $1000
** Z. - - - 400 ** Y. – Gºs - 1700 “ “ liability = 900
Total interest resources $900 Total int. liabilities $2700 || “ net credit of int. $100
‘‘ ‘‘ resources 900 s º
— |X’s inf. debit. - * $500
Net “ liabilities $1800 “ “ liability - 300
W’s int. liabilities $900 “ total debit of int. $800
& 4 & 4
§: & 4 & 4 § ys int. credit º $1700
Z's # “ & 4 300 liability - 30
“ net credit of int. $1400
Z's int. debit - tº- $400
“ “ liability º 300
“ debit of int. º $700
NOTE.-We regard the first form of entry as the better method of adjusting the interest on
partners' accounts. See problem 48.
53. Jones and Smith are partners, Jones two-thirds interest and Smith one-
third interest in the gains and losses. An account current and interest account is
to be kept with each partner. At the close of the fiscal year, Jones has $800 net
debit of interest, and Smith has $100 net debit of interest. 1. What entry is
required on the books to adjust the matter between the partners? 2. In case the
firm had dissolved and settled all matters except the interest, what settlement
should they make between themselves?
Ans. 1. Jones to Smith $200. $200.
2. Jones should pay Smith, out of his private funds, $200,
OPERATION.
| Dr. Cr. * Dr. Cr.
Jones owes - - - - $800 — $600 = $200
Smith owes * - º - 100 - 300 - $200
Total resources º cº- $900
Jones' two-thirds interest is $600
Smith's one-third interest is 300 -
85O SOULE'S PHILOSOPHIC PRACTICAL MATHEMATICS
54. Four mechanics, W., X, Y. and Z., form a partnership, for carrying on
their trade. They are to share gains and losses as follows: W. 3, X. #, Y. 3, Z. 4.
They have agreed that for all lost time they shall be charged respectively as follows:
W. $4 per day, X. and Y. $3.50 each per day, and Z. $3 per day. At the end of the
first year, it is found that they have lost time respectively as follows: W. 42 days,
X. 31 days, Y. 53 days, Z. 27 days. What is the correct entry on the books of the
firm to equitably adjust the matter among the partners? Ans.
Sundries To Sundries,
Y. - - - - - - - - $49.75
Z. ſº sº egº, * $º sº sº e 13.12
To W. & º sº tº gº ºne sº $35.62
To X. tº as * * * * * 27.25
OPERATION.
W. owes for 42 days at $4 == $168.00 | W’s three-eighths of amount due is - $203.62
X. ‘‘ ‘‘ 31 “ 34 - 108.50 | X's two-eighths 4 & & 4 - 135.75
Y. 4 & & 4 53 & 4 3# F 185.50 Y’s { { & 4 & & * 135.75
Z. “ “ 27 << 3 == 81.00 Z's one-eighth & 4 & & tºº, 67.88
Total amount due the partners, $543,00 $543.00
Dr. Cr. Dr. CT.
W. 8-, tº- #º Eº $168.00 $203.62 - $35.62
X. tº- sº º - 108.50 135.75 F 27.25
Y. tº- sº E-º #º- 185.50 135.75 - $49.75
Z. me *-> gº sº- 81.00 67.88 == 13.12
From these figures, we make the foregoing correct adjusting entry as shown in
the answer.
55. A. and Z. are partners, equal in gains and losses. After business had
been conducted for four months, it was agreed to admit W., and not wishing to close
the books at that time it was agreed, after careful investigation and calculation, to
allow A. and Z. $3000, as the net gain of the business up to that time; and that the
gains and losses of the new partnership shall be shared equally (#) by each partner.
What entry is necessary to adjust the $3000 gain between A. and Z. ?
Ans.
Profit and Loss To Sundries tº- - sº $3000
To A. tº gº- 3- sº $1500
To Z. tº- -> - sº 1500
(With explanation.)
56. A., B. and C. are partners. At the close of the year, A. retires from the
business. Preparatory to closing the books and making a final settlement with A.,
it is agreed to allow 10% discount on all personal accounts due the firm, and 5%
discount on all the bills receivable, as doubtful or bad debts. The personal accounts
due amount to $21281. The bills receivable amount to $14500. What entry is
necessary to adjust this anticipated loss? Ans.
Profit and Loss, old account es; sº * tº tº tº sº. - $2853.10 *
To Profit and Loss, new account - - ſº gº $2853.10
For 10% allowed as worthless, on $21281 personal accounts $2128.10
“ 5% allowed as worthless on $14500 bills receivable 725.00
Total amount sº tº- * e tº - $2853.10
From these figures, we make the adjusting entry as above.
* PARTNERSHIP SETTLEMENTS. 85 I
57. X., Y. and Z. are partners in the wholesale grocery and commission
business. At the end of their fiscal year, when they desire to close their books and
ascertain their actual gain or loss, they find that “Com. Sales” account has a credit
of $48500, resulting from the partial sales of consignments which are unclosed and
for which no account sales have been rendered. They also find that “Charges
Account” has been debited with $4050 for various charges paid on open consign-
ments unsold and partially Sold. They also find that there is $1812 accrued interest
due on matured bills receivable, and $280 due on bills payable.
Allowing 24% commission on the $48500, what are the necessary entries to
properly adjust the commission, charges and interest ? Ans.
FNTRY FOR ADJUSTING THE COMMISSION.
Commission new account tº ſº- * tºº e- gº g- $1212.50
To Commission old account *- * & * * * - $1212.50
For 24% on $48500 sales of unclosed consignments, per Com. Sales Ledger.
ENTRY FOR ADJUSTING THE CHARGES.
Charges new account - - - - - - - - - $4050
To Charges old account - - * =: Eºs º ºs º- $4050
For amount paid on unclosed consignments, per Com. Sales Ledger.
ENTRY FOR ADJUSTING THE INTEREST.
Interest new account - tº- * - sº gº sº es sº - ºr $1532
To Interest old account - - . - es $º º tº $º $1532
For Interest on bills receivable $1812
Less “ “ “ payable 280
Interest in favor of firm $1532
NOTE.-In all entries of this kind, whether new account is debtor to old account, or old
account to new account, the new account part of the entry is not to be posted until the old account
side is posted and the old account closed and ruled up. See Higher Work of Closing Ledgers in
Soulé's New Science and Practice of Accounts.
58. The following question was sent to us by the cashier of a banking insti-
tution in Texas, where accounts are kept in both coin and currency:
“Our Bank receives deposits in currency and gold. In the regular course of business we have
sold $15000 or $20000 gold belonging to depositors, and credited currency exchange account with the
premium realized. We now wish to ascertain our net gain for the purpose of declaring a dividend,
and as we will have to buy gold to pay our gold depositors when they demand their funds, it is
clear that to close our books as the accounts now stand, we would show a false gain equal to the
premium paid for gold. Under these circumstances, what entry must we make to adjust the matter
and keep our books in correct balance? Gold is now ruling at 114% premium.”
Ans.
Bxchange old account
To Exchange new account.
The amount of debit and credit of this entry would be the premium on the
gold sold, belonging to depositors, at the current rate per cent. The credit side of
this entry would not be posted until the old account was closed and ruled up.
59. The following questions were sent to us by a Texas merchant:
“I am doing a general merchandising business and keep accounts in both coin and currency.
It often happens that a customer who owes a balance in coin wishes to pay it in currency, and that a
852 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. ºt
Customer who owes a balance in currency wishes to pay it in coin. Now, suppose the first customer
to be Smith and the second Jones, and the balance in each case to be $550, and gold 10% premium,
what would be the correct entries to reduce the coin balance to currency and the currency balance to
coin, and to settle both accounts?” º
Ans.
JourNAL ENTRY TO REDUCE THE Coin BALANCE TO Currency.
Smith, currency account To Sundries, - - - - - $605
To Smith, coin account - - - tº - $550
To Exchange (10% premium on gold) 55
CASH ENTRY TO SETTLE WITH SMITH.
To Smith, (enter in currency cash column,) - - * - * * 605
JOURNAL ENTRY TO REDUCE THE Currency BALANCE TO Coin.
Sundries To Jones, currency account - - - - 550
Jones, coin account - - - - - - - - - - - - $500
Exchange (10% premium on gold), - º- - - - - * *- 50
CASH ENTRY TO SETTLE WITH JONES.
To Jones, (enter in gold cash column), - - - - - - - 500
A DIFFICULT SETTLEMENT AND JOURNAL ENTRY.
60. Smith, Jones and Brown are commercial partners, Smith one-half, and
Jones and Brown one-fourth each in gains and losses. In the regular course of busi-
ness, Smith, without consulting Jones and Brown, sells $1800 worth of merchandise,
and receives a note for the same at four months, which the book-keeper entered in
the books in the regular manner. Jones and Brown, on learning of the transaction,
disapproved of it on the ground that the maker of the note is an irresponsible person,
and proposed to Smith to sell to him their respective interests at 25% discount, to
which proposition Smith agreed. Jones and Brown then indorsed the note over to
Smith, who leaves it in the business, and the book-keeper is instructed to make the
necessary entry. What is the correct resultant entry 3 Ans.
Sundries To Smith - - - - - - º - $225
Jones - - - - - - - - - - - - - $112.50
Brown - - - - - - - - - - - - - 112.50
Statement of the several entries which, when the debits and credits are can-
celled to the lowest amounts, give the resultant entry.
First entry on receiving the note:
Bills receivable gº +- - tº- - - e. - tº- a- $1800
To Merchandise - - - - sº - - &= $1800
Second entry, when Jones and Brown each take from the business $450 worth
of property to sell to Smith at 25% discount:
Sundries to Bills receivable - - - - - - tº º $900
Jones g- - º - - º - - tº - tº º $450
Brown ºp - •º - & º e- sº Acº --> º º 450
PARTNERSHIP SETTLEMENTS. 853
Third entry, when Jones and Brown sell to Smith at a discount of 25% the
property, bills receivable, taken out of the business:
{
- - - - - $675
- $337.50
º º sº - 337.50
Fourth entry, when Smith invests and guarantees the $900 worth of property
bought for $675.
Smith to Sundries - tº {- º -
“ Jones -> º - -> - º º - º-
** Brown - º - º - ae
Bills receivable - 4- - - - -> *- - º - $900
To Smith tº-> - - - - - - º - $900
NOTE.-If it is preferred, the whole note may be taken out of the firm b
º y the partners, before
making the sale to Smith, and the other entries made to correspond. º
Without making the above entries, the 4 proper debits and credits may be
determined by the following process of reasoning:
Jones and Brown, fearing they may lose $450 each at the maturity of the note,
now agree with their partner, Smith, to allow him $112.50 each, ($25% of $450) if
he will secure them against this anticipated loss of $450 each. In consideration of
the $112.50 from each, Smith agrees or guarantees to protect them against the antic,
ipated loss, and hence each owes him $112.50.
NOTE.-In case the note proves worthless, Smith will be debited and Bills Receivable credited.
61. W., X, Y. and Z. were partners, each one-fourth in gains and losses.
They dissolved partnership, paid all their outstanding liabilities, and divided all
their resources. They then examine their books and find the following balances to
the credit of each partner: W. $1100; X. $240; Y. $810; Z. $680. What settle-
ment must be made between the partners to adjust the matter, and what entry must
be made on the books when the final settlement is made 3 Ans.
Sundries To Sundries,
W. - - - - - - - - - - - - - - $392.50
Y. - - - - - - - - - - - - - 102.50
To X. - - * - s º º - -- $467.50
To Z. - - - - - - - - - 27.50
OPERATION.
Cr. Dr. Dr. Cr.
Amount due W. * - * - $1100.00 $707.50 -: $392.50
& 4 “ X. º - º 240.00 707.50 > $467.50
& 4 “ Y. º - - 810.00 707.50 – 102.50
<< * * Z. * - º 680.00 707.50 – 27.50
Total liabilities is - +) $2830.00 $495.00 $495.00
Each partner's 4 of liabilities $ 707.50
By this operation we see that there is due to W. $392.50, and to Y. $102.50,
of which X. owes $467.50 and Z. $27.50.
When x. and Z. pay to W. and Y. the respective amounts due by them, it
is clear that W. and Y. will then owe them, and hence the above adjusting entry.
854 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. X-
OPENING ENTRY TO CHANGE SINGLE ENTRY BOOKS TO DOUBLE
ENTRY, AND USE THE SINGLE ENTRY LEDGER.
62. A. S. Blaffer and P. Birba are commercial partners, equal in gains and
losses. They are keeping their books by the Single Entry System, but desire to
change them to the Double Entry System, and use the present Single Entry Ledger.
The following formula of work accomplishes the desired object :
NEW ORLEANs, May 1, 1893.
For the purpose of changing our books from single entry to double entry, we
make the following abstract of our commercial affairs this date:

IT E M S. RESOURCES, LIABILITIES.


Merchandise per stock book tº- * = tºº g- - 17000|00
Cash on hand and in bank - s - e - -º- gº * - 7800|00
Bills receivable, Nos. 1, 2, 3 and 4 sms º º me * 2500|00
Canal Bank stock - sº tº sº tº º m s * 100000
Bills payable, Nos. 1 and 2 – * 185000
Our resources and liabilities shown by our Single Entry Ledger,
are as follows:
A. S. Blaffer's account - -3 º 1150.00 850000
P. Birba's account - gº -: º 180000 1200000
tºº Dr. 910000
231 personal accounts - tº e ſº sº
67 personal accounts - ºn tº - - - Cr. 4400|00
Total amount of resources - s tº gº tºº & 403:50.00
Total amount of liabilities tº e º ºr tº º tº 267:50.00 26750|00
Net gain to date - sº tºº sº * * e tº e tº 13600|00
A. S. Blaffer's one-half net gain is gº ºs º gº me 680000
P. Birba's one-half net gain is * tº &= * * *- 680000
Approved :
A. S. BLAFFER,
P. BIRBA.
From the foregoing statement, the following Journal entry is made, which
converts the Single Entry Ledger into a Double Entry Ledger:
FIRST FORM OF ENTRY. SECOND FORM OF ENTRY.
Sundries to Sundries Sundries to Sundries
v/A. S. Blaffer, (amount posted) $ 1150 v.A. S. Blaffer, (amount posted) $ 1150
1/P. Birba, (amount posted) - 1800 VP. Birba, (amount posted) - 1800
1/231 personal accounts, (posted) 9100 1/231 personal accounts, (posted) 9100
Mdse. per inventory - - - 17000 Mdse. per inventory - - - 17000
Cash on hand tº dº º me tº 7800 Cash on hand - - - - - - 7800
Bills receivable - - - - - 2500 Bills receivable - - - - - 2500
Canal Bank stock - - - - 1000 Canal Bank stock - - - - 1000 *
1/ To A. S. Blaffer, (am’t posted) $ 8500 | V To A. S. Blaffer, (am’t posted) $8500
V “ P. Birba, (amount posted) 12000 4 & 6 & “ # net gain 6800
w/ “ 67 personal acc’ts, (posted) 4400 | V “ P. Birba, (am’t posted) - 12000
“ P. and L., net gain to date 13600 & & 4 & # net gain 6800
“ Bills payable - - - - - 1850 | V “ 67 personal acc’ts, (posted) 4400
“Bills payable - - - - - 1850
NoTE.—The debits and credits, in either of the forms of entry, that are checked and marked
“posted,” must not be reposted.
*:
SETTLEMENTS. 855
PARTNERSHIP
TO OPEN NEW BOOKS FROM AN OLD SET WHICH HAS BEEN
INCORRECTLY REPT.
63. A. and Z. are partners, doing a mercantile and manufacturing business,
and their books are, in their own language, “in an awfully mixed condition.”
They therefore desire a new set opened, and accordingly employ an accountant to
perform the duty. How should he proceed? Ans. He should first, on loose paper,
from the old books, memorandum records of all kinds, and from the partners,
ascertain the present resources and liabilities of the firm.
them as follows:
He should then arrange
List of the Resources and Liabilities of A. and Z., January 1st, 1893, as shown by
their old irregularly kept books, the inventory of stock on hand, and memoran-
dum accounts; the same being agreed to by the members of the firm, and
arranged as per subjoined statement to facilitate the opening of a new set
of books according to the principles of double entry.
RESOURCES :
Mdse. per account of stock - - *
Cash on hand - - - * - -
10 shares Germania Bank stock, valued
$14218
9210
at $130 º - - - - tºs 1300
5 shares N. O and T. R. R. stock, valued
at $90 * e- * sº - *- 450
Mill, machinery, tools, etc. - - – 11280
Bills receivable per B. B. º * º 4210
Personal accounts due the firm, as per the
following Journal entry, the names
being omitted to economize space - 7460
A’s withdrawals in excess of his invest-
ments - - - e- - †- 3100
Total amount of resources - - $51228
From the foregoing statement, the following Journal entry is made,
opens the new set of books:
Sundries To Sundries
Molse. - -> - - - $14218
Cash - - - - - - 9210
Germania Bank stock º *- 1300
N. O. & T. R. R. stock - tº- 450
Mill, machinery, tools, etc. - 11280
Bills receivable - * - 4210
81 personal accounts, 7460
(Insert the 81 names)
To A. * * * $ 7115
** Z. º -> º 37058
“ bills payable - 1280
“ personal accounts 2675
(Insert the names of the
33 persons owed.)
LIABILITIES :
Z's net investment - * - º - $26843
9 personal accounts due by the firm as
per old Ledger - - - tº- - 1750
24 personal accounts due by the firm as
per memorandums - - †- - 925
Bills payable as per B. B. - tº - 1280.
Total liabilities - - - - - $30798
Which deducted from the resources
as above - sº - º - 51228
Gives the net gain to date - - - $20430
which
A's one-half of net gain is - * - - $10215
LErom which deduct his net withdrawals 3100
Gives his present net capital - {-} - $ 7115
Z's one-half net gain is - * * - 10215
To which add his net investment - - 26843
Gives his present net capital - - - $37058
NOTE.—After this entry is made, the amount of cash should be entered in the Cash Book as
amount on hand.
64. A. and B. are partners, equal in gains and losses.
A. Owns one-fourth
and B. owns three-fourths of the net capital.
The resources and liabilities at commencing, not including the partners'
investments, were as follows:
Cash -> - $ 5000
Mdse. -> sº 15000
RESOURCES: { Bills receivable 2000
Bank stock - 3000
Personal accounts 8000
Total resources - E- $33000
ſ Bills payable – $15000
LIABILITIES: t Personal accounts 6000
Total liabilities $21000
Total resources 33000
Net capital of the firm - $12000
(Continued.)
856 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. jºr
The resources and liabilities at the close of the fiscal year, not including the
Partners' investments, were as follows: *
Cash - - - $ 3000
Mdse. - - * 12000
#. . º 9000 ſ Bills payable - $12000
e SOD1
RESOURCES: Bank it. *ounts º º LIABILITIES: | Personal accounts 7000
Real estate - G- 4000 Total liabilities $19000
Total resources $40000
What has been the net gain, and what is each partner's net capital *
Ans. Net gain $9000. A’s net capital $7500.
B’s net capital $13500.
OPERATION.
Resources at commencing as above - $33000 |A's one-fourth capital at commencing $3000
Liabilities at commencing as above - 21000 || B's three-fourths capital at commencing 9000
Net capital at commencing as above $12000 Firm’s capital - - - - $12000
Resources at closing as above - - $40000 || A's one-fourth net capital at commencing $3000
Liabilities at closing as above - - . 19000 || A's one-half met gain - - - - 4500
Net capital at closing - - - 21000 || A's net capital at closing - - - $7500
Less net capital at commencing ' - 12000
B’s three-fourths net capital at commencing $9000
Net gain during business - - $9000 || B's one-half net gain - - - - 4500
B’s net capital at closing - º- - - $13500
A's net capital $7500 + B's net capital $13500 = $21000 = the net capital of the firm at closing.
65. J. Jones, J. E. Soulé and B. B. Euston are partners, doing business
under the name and firm of Jones, Soulé & Euston. They commenced business
January 1, 1893. The gains and losses are to be shared equally, each one-third.
July 1, 1893, Jones retires from the concern, and Soulé and Euston continue the busi-
ness and liquidate the affairs of the old firm. In order to settle with J. Jones, July
1, the books are to be adjusted so that his account may exhibit the cash balance due
to him on that date. With a view to effect this, the following facts have been ascer-
tained and agreed to by all of the partners:
The merchandise on hand per account of stock is valued at $49528.
The merchandise on hand belonging to Mdse. in joint account with H. Davis,
each one-half, is valued at $4236. Charges incurred, but not posted to said account,
amount to $210. The commission on sales is 24%. The charges incurred, but not
posted, on “Commission Sales Account,” amount to $687.65, and the commission on
the sales effected, but not posted, amounts to $915.18. It is agreed to allow the new
firm 10% on all personal accounts receivable and bills receivable, to cover doubtful
debts. Of the cash on hand, $500 is gold, which is worth 10% premium.
The balance of S. Brown's account is payable in San Francisco, and exchange
is 2% discount. Interest is to be charged on all accounts due the firm, and allowed
jºk
857
PARTNERSHIP
SETTLEMENTS.
on all accounts that the firm owe, at 6%; and also on the investments and with-
drawals of the partners, at the same rate.
The following shows the face of the Ledger June 30:

Due by Due b
A. Dr. Cr. A.
J. Jones’ capital account - - - - - - $3000000|| Jan. 1
May 4|| $1850,00|| “... “ current account - - - - - -
J. E. Soulé's capital account - º - º- - 30000|00|| Jan. | 1
April 20 2408:00|| “. current account - tº- º tº- -
B. B. Euston's capital account - * - tº tº- 30000|00|| Jan., | 1
June | 3 1320|00 & 4 current account - 4- - - -
Bills payable - * *- vº - º - 10400;00|| Nov. 4
Aug. 20) 5084000|| Bills receivable - - - - * - tº sº
14250|60|| Cash sº tº- tº º tº tº cº-
7141085|| Merchandise - tº- - tº - tº - º - 31581|50
9810|00|| Commission sales, - ºt - º - - - 42138||00
(The balance $32328, due by average August 19.)
8621|00|| Merchandise in joint account with H. Davis - - 14140,00|| July 10
(The debit is our one-half investment.)
7412,40|| Expense º - º - º - º -
810|30|| Interest tº- - - - vº - º - 142070
2130,00| Exchange - - º - tº - º - 3816|40
641815|| Charges * *- sº º sº - º º 451.1|80
230|00|| Commission º º- tº º tº tº * - 7913|10
5450|00|| Profit and LOSS - - ſº a- - - - tº- 800|00
H. Davis sºs &- º - tº tº- - 1200000|| Aug. 17
May 28|| 1340000|| S. Brown º - se - - - * -
July 16|| 17815,00|| J. Lewis e- ſº- * - * - E - |
W. Wood ºs - sº - tº - * - 445000|| April | 1
Jan. [15 5000|00|| G. B. Bracket e - tº - tº - º -
Mar. 3 9405|00|| A. Williams sº- - º - * - ºs -
M. Moses º - º ºs * - sº - 5409|80|| Sept. 11
|$228581130 |$228581|30
What are the Journal entries to be made, in accordance with the foregoing
agreements and exhibit of facts, to effect the desired settlement with the retiring
partner, J. Jones?
Mdse. in Joint Account with H. Davis, new account,
To Mdse. in Joint Account with H. Davis, old account,
For one-half value of Mdse. unsold.
Mdse. in Joint Account with H. Davis
To Charges
“Commission at 24% on sales
“ H. Davis, his one-half net proceeds
“Profit and Loss, our one-half net gain
tentry to credit Charges with the amount due,
but not posted, on “Com. Sales”).
Charges, new account, $687.65
Ans.
(Entry to credit Merchandise in Joint Account with H. Davis).
$2118
$2118
(Entry to close the account of Merchandise in Joint Account with H. Davis),
To Charges, old account,
To Sundries,
Commission, new account,
To Commission, old account,
$7637
$210.00
353.50
6788.25
285.25
(Entry to credit Com. with the amount due, but
not posted, on “Com. Sales”).
- $915.18
$915.18
858
Yºr
SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS.
(Entry to allow 10% on all Personal Accounts
Receivable and Bills Receivable due the Firm).
(Entry to adjust the Premium on Gold and the
Discount on S. Brown's account).
Profit and Loss, old account, - $9646 Exchange, old account, i-e * $218
To Exchange, new account, $218
To Profit and Loss, mew account, $9646
For 2% disc. on $13400 = $268
Less 10% prem. on 500 gold = 50
(Entry to adjust the Interest on the Partners' | Or, as many accountants would make the entry.
Investments and Withdrawals). thus:
Sundries To B. B. Euston - su. to Sundries To Interest - - $52.00
J. Jones * gº º sº. sº .24 J. Jones, 57 ds. (a) 6% on $1850 $17,57
J. E. Soulé - - - - 11.16 J. E. Soulé, 71 “ “6% “ 2408 28.49
B. B. Euston, 27 “ “6% “ 1320 5.94
# Or, considering the interests on the investments for 181 days at 6%, and on
the withdrawals as in the preceding entry, thus:
Interest To Sundries - tº-3 4- tº º tº ºt sº * * $2663
** J. Jones tº- :- as tºº s wº †-º $887.43
** J. E. Soulé - sº iº 4- gº gº- sº 876.51
** B. B. Euston & ſº tºº gº aº gº tº 899,06
(Entry to adjust the Interest or Discount due the Firm on Personal Accounts matured and
unmatured, and on Bills Payable unmatured).
Interest, new account, - - - - - - - - - $1061.22
To Interest, old account, - &= - dº e- $1061.22
For accrued interest on the following credit balances:
S. Brown, $13400.00 for 33 ds. (a) 6% $ 73.70 | H. Davis, 6788.25 for 10 ds. at 6%, 11.31
G. B. Brackett, 5000.00 “ 166 “
“6% 138.33 || M. Moses,
A. Williams, 9405.00 ** 119 “
“6% 186.53 Com. Sales,
5409.80 “ 73 “
“ 6%. 65.82
32328.00 “ 50 “
“6%. 269.40
H. Davis, 12000.00 “ 48 “ “6% 96.00 | Bills Payable 10400.00 “ 127 “ “ 6%. 220.13
(Entry to adjust the Interest or Discount due by the Firm on Personal Accounts and Bills
Receivable).
Interest, old account, - - - - - - - - - - $546.40
To Interest, new account, - - - - - - $546.40
For accrued interest on the following debit balances:
J. Lews, $17815 for 16 days a 6% - - $ 47.51
Bills receivable, 50840 “ 51 “ “6% - - 432.14
W. Wood, 4450 “ 90 “ “ 6% - - 66.75
NOTE.-When the foregoing entries shall have been posted, except the new account debits and
credits, the gains and losses will be ascertained in the usual way, and the books closed. Profit and
Loss should be closed to the partners' current account, and their current account should be closed
to their capital account.
TEIE ADMISSION OF A NEW PARTNER INTO A COMMERCIAL FIRM.
66. There are many reasons for admitting new partners into commercial houses.
i. Some partners are admitted for their money. 2. Some for their reputation. 3.
Others for their knowledge or skill.
There are also many ways of admitting partners. 1. Some are admitted to
an agreed upon interest in the gains and losses, by making a specified cash or note
investment. 2. Some, by paying to the old members of the firm, in cash or notes,
jk PARTNERSHIP SETTLEMENTS. 859
a stated price for 4, 3, #, or § interest, of the gains and losses. 3. By paying to the
members of the firm, a specified sum for a part or share of their respective interests.
4. By paying to the firm a specified sum for 3, #, +, or § interest or ownership in the
entire stock or capital of the firm, and a certain share of the gains and losses of
the new firm.
There are also other ways of admitting new partners, but these are the most
general, and we shall briefly consider the entries to be made under the four cases
named.
In the first case, where a specified cash investment was made for a certain
share in the gains and losses, cash would be debited and the party credited for the
amount invested.
In the second case, where a specified bonus of cash or note payment was made
to members of the old firm for a certain interest in the gains and losses, no entry
would be made in the books for the cash or notes received by the old members,
unless they should make an investment of the same in the firm. They sold only an
interest of their respective interests in the gain and losses of the business, and the
only thing necessary to be done is to divide the amount received between the old
partners, or invest it to their respective credits in proportion to the interest sold,
and then to draw new articles of agreement Specifying each partner's interest.
To illustrate, let us suppose that Jones and Smith are partners, Jones 3 and
Smith # in gains and losses, and that Brown is admitted as a partner with # interest
in the gains and losses of the business on the payment of a bonus of $10000 cash,
and that Jones and Smith Sell to Brown the # interest in proportion to their
respective shares, § and #, and invest the same in the business. . What is the entry,
and what are the respective interests of the three partners in the new firm 3
The cash entry in Journal form, is as follows:
Cash to Sundries º - tº- a- tº- gº * - º e $10000
To Jones, his two-thirds of $10000 - - - *g $6666.67
“Smith, his one-third of $10000 - - * ge 3333.33
The partners' respective interests are now as follows:
Jones four-ninths, Smith two-ninths, and Brown three-ninths of the
gains and losses of the new business.
In the third case, the old partners who sold, should be debited each for the
proportion sold and the new partner credited for the amount bought.
To illustrate this, we will again suppose that Jones and Smith are partners,
Jones 3 and Smith # in gains and losses, and that Brown pays to Jones and Smith
$12000 for a partnership, and # interest in the gains and losses, and that they sell
the # interest in proportion to their present shares. What is the entry, and what
is the respective interest of each partner in the gains and losses 3
The entry is as follows:
Sundries to Brown º -> - º - - « » - - - - $12000
Jones, his two-thirds of $12000 - - - - - - - - - $8000
Smith, his one-third of $12000 - º - - sº - * tº- - 4000
The partners' interests are, Jones four-ninths, Smith two-ninths, and Brown
three-ninths. Should Jones and Smith invest this money in the firm, Cash would be
debited and they credited. •
86O SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. H
º
In the fourth case, where a specified amount was paid for an interest or owner-
ship of a certain part of the whole capital of a firm, the amount paid for the interest
should be divided between the old partners in proportion to the net capital of each;
and then the amount of the one-fourth interest sold should be credited to the pur.
chaser, and the old partners debited for the same in proportion to their respective
net capital.
To illustrate this, we will suppose Soulé and Weiss to compose a firm that
sells to Harris # of the net capital of their firm for $30000 cash, and that, while they
are now, Soulé # and Weiss + in gains and losses, in the new firm the interests will
be Soulé #, Weiss +, and Harris 4.
We will further suppose that the books have been closed and exhibit the
following statement of accounts:
SouL.E.' ' WEISS. CASH. MDSE.
| $100000 $10000 $9000 $90000
BILLS RECV. BILLS PAY. H. KENT.
$5000 $6000
$10000
This exhibit shows the net resources to be $110000, 4 of which is $27500.
For this $27500 and a partnership with one-fourth interest in gains and losses,
Harris pays $30000. This $30000 divided between Soulé and Weiss, in proportion
to their net capital, gives Soulé $27272.73, and Weiss $2727.27. * 3.
The $27500 being the one-fourth of the net capital purchased by Harris, must
be credited to him, and Soulé and Weiss debited for the same in proportion to their
respective net capital. The following is the adjusting entry: w
Sundries to Harris - gº tº- * * * ſº- $27500
Soulé, his proportion - - - $25000
Weiss, his proportion - - - 2500
Articles of Agreement should be drawn by the new firm specifying the invest-
ments, interests in gains and losses, whether interest is to be allowed on the part-
ners' investments and charged on their withdrawals, etc., etc. See Articles of Agree-
ment a few pages further On.
Should Soulé and Weiss invest the money received for the sale of their one-
fourth net capital, cash would be debited and they credited in the usual manner.
A party desiring to purchase an interest in any business should carefully
examine the accounts and the goods in stock. He should give close attention to the
loss and gain accounts and to the personal accounts due the firm. He should have
an account of stocktaken in the presence of some person who is thoroughly informed
on the value of the line of goods or material in which the house deals.
*IIIsi, and Parlieri Atmº.
=N
* A Ama A_t
v--~~~ *--w
1426. For a definition of Partnership, and a statement of the different classes
of Partnership, the Formation of Partnerships, the Dissolution of Partnerships,
Articles of Partnerships, Partnership Settlements, etc., see pages 1 to 5.
1427. Partnership Average is the process of finding what proportional
amount of the gain and loss is to be given to each member of the firm on final set-
tlement, or at any specified time.
Sharing Gains and Losses.
The amount of gain or loss to be apportioned to each partner, depends upon
the following conditions: 1. The gain or loss may be divided in proportion to the
net investment or capital of each partner, or some other agreed upon proportional
basis. 2. By allowing each partner a salary, and dividing the remainder of the
profits according to a fixed proportion. 3. By allowing each partner interest on his
net investment, and dividing the remainder of the profits according to a fixed pro-
portion. 4. By allowing each partner interest on his net investment, and a salary
for services rendered, for skill possessed, or for credit or good-will, and then divid-
ing the remainder of the profits according to some fixed proportion. See Partner-
ship, page 1.
CLASSIFICATION.
The subject is classified into simple and compound partnership average.
1428. Simple Partnership Average is the process of finding the proportional
amount of gain or loss to be apportioned to each member of the firm when the net
capital of each has been invested for the same length of time; or, in brief, it is the
proportional division of the gain or loss between the partners, according to the capi-
tal invested by each.
1429. Compound Partnership Average is the process of finding the propor.
tional amount of gain or loss to be apportioned to each member of the firm when the
net capital of each has not been invested for the same length of time; or, in brief,
it is the proportional division of the gain or loss between the partners, according to
their average met investments and the periods of time that they were in use by the
firm.
SIMPLE PARTNERSEIIP AVERAGE PROBLEMS.
67. Jones, Smith and Brown contracted to do business as partners for
one year. Jones invested $5000, Smith $4000, and Brown $1500. The gain was
$3150. What was the proportional share of each partner ?
Ans. Jones $1500. Smith $1200. Brown $450.
(Continued.)
(861).
862
soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. 4×
FIRST OPERATION.
Statement to find Statement to find Statement to find
Jones' share. Smith's share. Brown's share.
Jones’ capital - $5000
Smith’s “ - 4000 $ GAIN. $ GAIN. $ GAIN.
Brown’s “ - 1500 3150 3150 3150
- 10500 || 5000 10500 | 4000 10500 | 1500
Total capital $10500 -
$1500 $1200 $ 450
Explanation.—By adding the individual capital of each partner, we obtain a total capital of
$10500, which is the amount that gained and is to receive the $31.50; and as the gain is to be divided in
proportion to the respective capital of each partner, we place the gain on our statement line and
reason thus: If $10500 capital gains $3150, $1 will gain the 10500th part, and $5000 (Jones’ capital)
will gain 5000 times as much. The result of this statement gives us $1500 as Jones' share of the
gain. The reasoning for the statements to obtain Smith's and Brown's respective shares is the same
as that in the statement to obtain Jones' share, and hence is omitted.
SECOND OPERATION.
º Statement to find Statement to find | Statement to find
Jones' capital - $5% $10 Jones' share. Smith’s share. Brown’s share.
Smith’s “ +- 49% 8 $ *... $...
Brown’s “ - I 5% 3 21 #0 t 1 | #so 21 gº
Total capital - $21 $1500 $1200 $150
Explanation.—In this solution, we cancel the investments of the respective partners by divid-
ing each amount by 500, which lessens the figures in the operation without changing the ratios,
The same reasoning is applied as given in the first Solution, and hence is omitted.
THIRD OPERATION.
Statement to find the
rate per cent. gain. Jones' capital - - - - $5000 30% of which is $1500
$ GAIN. Smith’s “ - - - - 4000 30% “ “ 1200
3150 Brown's “ - - - - 1500 30% “ “ 450
10500 | 100
3) Total capital - - - $10500
30% gain.
68. A. and Z. formed a partnership for the purpose of speculating in
various kinds of business. A. invested $8400, and Z. $6000. At the close of the
business, the net loss amounted to $1728. What was each partners' proportional
share" Ans. A’s $1008. Z's $720.
OPERATION
Statement to find A’s loss. Statement to find Z's loss.
A’s investment - $84%
Z’s 4% º 6000 1728 1728
º 144 | 84 144 60
$144 $1008 $720
69. W., X, Y, and Z. entered into partnership for the purpose of conduct-
ing a general merchandising business. W. invested $3500, X. $3000, Y. $2500, and
Z. $1000. At the expiration of the partnership, the net gain was $15000. What
was each partner's proportional share ? Ans. W. $5250. X. $4500.
Y. $3750. Z. $1500.
70. Three young gentlemen, A., O. and P., contracted for the use of a carriage
for thirty days for $100, which they agreed to pay in proportion to the number of
*
# .* PARTNERSHIP SETTLEMENTS. 863
days that each had the use of it. A. used it 9 days; O. 14 days; and P. 6 days,
and 1 day it was not in use. What is the amount due by each 3
Ans. A. $31,03. O. $48.28. P. $20.69.
71. A father has $2000 which he wishes to divide among his three children,
in proportion to their ages, which are 6, 8 and 11 years. What amount will each
receive? Ans. 6 years, $480. 8 years, $640. 11 years, $880.
OPERATION.
6 + 8 + 11 = 25 years.
Statement to find the amount | Statement to find the amount | Statement to find the amount
for the child of 6 years. for the child of 8 years. for the child of 11 years.
$ $ $
2000 2000 2000
25 | 6 25 || 8 25 || 11
$480 $640 $880
COMPOUND PARTNERSEIIP AVERAGE PROBLEMIS.
72. Jones, Smith and Brown associated themselves together in business, and
agreed to share the gains and losses in proportion to their respective investments
and the time the same were employed by the firm.
Jones invested $4000 for 12 months; Smith invested $3500 for 9 months;
Brown invested $2000 for 8 months. The gain during the business was $9550.
What was each partner's share ? Ans. Jones $4800; Smith $3150.
|Brown $1600.
OPERATION.
Mos. Investment for i mo.
Jones invested - tº- tº- - $4000 × 12 = $48000
Smith. “ <- tº- gº tº 3500 × 9 = 31500
BrOWn. “ * > gº g- • * 2000 × 8 = 16000
$95500
Statement to find Jones' share. [Statement to find Smith's share. Statement to find Brown’s share.
$ $ $
9550 9550 9550
95500 || 48000 95500 || 31500 95500 | 16000
-$ 4800 $ 3150 | $ 1600
Ea:planation.—In this problem, the gain being shared in proportion to the money invested
and the time for which it was invested, and the amounts and periods of time of each partner being
different, we must, therefore, find the respective equivalent investment of each partner for 1 month.
This is done, as shown in the operation, by multiplying the amount invested by the time that it
was invested.
The reasoning for the work is as follows: $4000 invested for 12 months is equivalent to 12
times $4000 for 1 month, which is $48000; then $3500 invested for 9 months is equivalent to 9 times
as many dollars for 1 month, which is $31500; and then $2000 invested for 8 months is equivalent to
8 times as many dollars for 1 month, which is $16000. By this work, we obtain the respective
investments of each partner for 1 month. We then add the same, to obtain the total capitalinvested
for 1 month, and then proceed to find each partner's share of the gain by our line statements, in the
same manner and with the same reasoning that is presented in the first problem of Simple Part-
nership Average.
73. A. commenced business January 1, 1893, with a capital of $2500; on
the 1st of May, he entered into partnership with B. who invested $3000; and on the
*
3.
864 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. § {
1st of October, C. was admitted as a third partner and invested $4000. At the close -
of the year, there was a gain of $5500. What was each partner's apportionment of
the gain? Ans. A’s $2500. B's $2000. C's $1000.
FIRST OPERATION.
1893. Mos. Am’t for 1 mo.
January 1, A. invested - - - $2500 × 12 = $30000
May 1, B. & 4 +- g- s 3000 × 8 = 24000
October 1, C. & & - tº- ºr 4000 × 3 = 12000 ,
Total capital for 1 month, - - - - - $66000
Statement to find A’s share. Statement to find B's share. Statement to find C’s share
GAIN. $ GAIN. $ G.
5500 e 5500 5500
66000 || 30000 66000 || 24000 66000 || 12000
$2500 $2000 $1000
SECOND OPERATION.
To find the respective shares of each partner by first finding the rate per cent.
gain, we proceed thus:
#500 A’s capital for 1 month is $30000, 84% of which is - $2500
66000 || 100 B’s “ “ 1 “ “ 24000, 84% of which is * - 2000
C’s “ “ 1 “ “ 12000, 84% of which is tº- 1000
84% gain.
Explanation.—In the first part of the operation of this problem, we could have shortened the
work by cancellation, but in order to more fully elucidate this subject, we did not cancel.
74. W., X, Y, and Z. engaged the services of a teacher for 10 months for
$600, and agreed to pay the same in proportion to the number of children sent and
the time that they attend School. -
W. sent 3 for 200 days; X. sent 2, 1 for 200 days and 1 for 120 days; Y. sent
5, 3 for 180 days, 1 for 150 days, and 1 for 90 days; Z. sent 4, 1 for 200 days, and 3
for 160 days. What amount was due from each?
Ans. W. $151.26. X. $80,67. Y. $196.64. Z. $171.43.
OPERATION.
No. sent. aiº. No. º
W. e - - - 3 200 600
1 200 = 200
X. - - - - 1 #} = {}} 300
3 180 = 540
Y. e - - * > 1 150 - 150 780
1 90 = 90
1. 200 = 200
Z. - - - - 3 160 = 480 } 680
Total number for 1 day -> - - - 2380
Statement to find the Statement to find the Statement to find the Statement to find the
amount due by W. amount due by X. amount due by Y. amount due by Z.
$ $ $
600 600 600 600
2380 600 2380 || 320 2380 || 780 2380 | 680
$151.26 $80.67s $196.64 $171.43
Explanation.—In problems of this character, where there are several parties to receive the
#. or contribute to the loss or expense, it would be better first to find the rate per cent. as shown
preceding problems.
PARTNERSHIP SETTLEMENTS. 865
75. A. and B. formed a partnership in which A. invested $8000 for 10
months, B. $5000 for 12 months. They lost, during the partnership, $700. What
was the loss of each partner? Ans. A. $400. B. $300.
76. X., Y. and Z. executed a piece of work for which they received $500. X.
worked 15 days of 8 hours each; Y. worked 15 days of 10 hours each; and Z. worked
12 days of 6 hours each. How should the $500 be divided among them ?
Ans. X. $175.44. Y. $219.30. Z. $105.26.
77. Jones, Smith and Brown, on the 1st of January, 1893, associated them-
selves together as partners for the term of 1 year.
January 1, Jones invested $10000; May 1, he drew out $4000; October 1, he
drew out $2000; and December 1, he invested $1000. *
January 1, Smith invested $6000; July 1, he invested $2000; November 1,
he drew out $1000.
March 1, Brown invested $4000; April 1, he invested $1000; August 1, he
drew out $3000; November 1, he drew out $2500.
The gain at the close of the year amounted to $9700. What was each part-
ner's share of the gain, and what is Brown's indebtedness to the firm 3 --
Ans. Jones' gain, $4150. Smith's gain, $4100;
Brown's gain, $1450, and indebtedness, $500.
. FIRST OPERATION.
Am’t Invested
Mos. for 1 month.
January 1, Jones invested - - - $10000 4 $40000
May 1, “ withdrew - ſº º 4000
º $6000 5 30000
October 1, “ withdrew - a º 2000
$4000 2 8000
December 1, “ invested & ſº º sº 1000
$5000 1 5000
Jones’ capital for 1 month tº gº º sm º ºr ºl $83000
Am't Invested
s Mos. #. 1 .
January 1, Smith invested - - - $6000 6 $36000
July 1, 4 & é & gº tº ſº 2000
$8000 4 32000
November 1, “ withdrew - - tºº 1000
$7000 2 14000
Smith's capital for 1 month - - - - - - - 82000
Am't Invested
Mos. for 1 .
March 1, Brown invested - - tºº $4000 1 $ 4000
April 1, & 4 4 & * > tº 1000
$5000 4 20000
August 1, ** withdrew - - tºs 3000
$2000 3 6000
November 1, ** withdrew - sº g- 2500
Excess of withdrawals, or amount due
the firm by Brown {-º º sº * º $500 2
$30000
Excess of withdrawals for 2 months deducted - - 1000
Brown's capital for 1 month - - - - - 29000
Total capital for 1 month gº ºn tº sº º $194000
366 soulE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
Having the total capital for 1 month, to find each partner's share of the gain,
We have but to proceed as shown in preceding problems, either by the line state-
ment or per centum.
SECOND OPERATION.
In problems of this character, where there are several investments and with-
drawals by the different members of the firm, it is best first to average the account
of each partner. The following operation will elucidate this method:
Dr. JONIES. - CR;
Mos. Mos. Jollars
1893. OS *...] 1893. OS zºº.
May 1, $4000 8 = - - $32000 January 1, $10000 12 = - - $120000
October 1, 2000 3 = - 6000 | December 1, 1000 1 = tº 1000
$38000 $121000
38000
Balance of dollars for 1 month gº º * = tº $83000
DR. SIMITEI. CR.
Mos. I)oll & ID
1893. OS fº. 1893. Mos tºº.
November 1, $1000 2 - - - $2000 || January 1, $6000 12 = - - $72000
July 1, 2000 6 = - - 12000
$84000
2000
Balance of dollars for 1 month - - - - - - $82000
DR. BFOWN. CR.
Mos. I)olla Mos. g Dollars
1893. OS fº. 1893. OS f.
August 1, $3000 5 = - - $15000 |March 1, $4000 10 =s - - $40000
November 1, 2500 2 = - - 5000 |April 1, 1000 9 = - - 9000
$5500 $20000| $5000 * $49000
5000 20000
Due the firm, $500 Balance of dollars for 1 month - - - - - - $29000.
RECAPITULATION.
Jones’ capital for 1 month - - - - - - $83000
Smith’s “ for 1 month - - - - • - 82000
Brown’s “ for 1 month, - - - - - - 29000
Total capital for 1 month - - - $194000
Having thus obtained the total capital of the firm for one month, we would
proceed as in previous examples to find each partner's share.
78. A. and Z. bought a house for $10000. A. paid $6000 and Z. $4000. The
house rents for $150 per month. To how much of the monthly rent is each entitled?
Ans. A. $90. Z. $60.
* PARTNERSHIP SETTLEMENTs. 867
79. X, Y, and Z. hired a pasture for the season for $90. X. pastured 9 head
of mules for 150 days; Y. pastured 11 head of mules for 110 days, and Z. pastured
24 head of mules for 160 days. How much is each to pay?
Ans. X. $18.98. Y. $17.02. Z. $54.
80. Four persons, A., B., C. and D., residing contiguous to each other, agree
to build a schoolhouse, and contribute to the payment of the same in the reciprocal
ratio of their respective distances from the schoolhouse. The cost was $1800. The
schoolhouse was kocated # of a mile from A's residence; 1 mile from B's residence;
14 miles from C's residence, and 2 miles from D's residence. How much did each
contribute % Ans. A. $837.2043. B. $418.60%.
C. 334.8843. D. 209.30}}.
NoTE.—The reciprocal of a number is the quotient of 1 divided by the number.
OPERATION.
A. # mile distant. The reciprocal of # = 2
B. 1. & 4 & & 4 & & 4 4 & 1 - 1
& & & & & 4 & 4 4 & *-
$. # & & & 4 & & ( & & 4 # = :
Sum of reciprocals - ºp e. -> 41% = #3.
Statement to find A’s Statement to find B’s Statement to find C’s Statement to find D's.
contribution. contribution. contribution. contribution.
$ $ $ $
1800 1800 1800 1800
43 || 10 43 || 10 43 || 10 43 || 10
2 1 5 || 4 2
$837.2043 $418.60% $334.884; $209.30}}
AVERAGING SALES IN COMMISSION.
81. A commission merchant received from three different correspondents,
X, Y. and Z., consignments as follows: from X., of Jefferson, Texas, he received
200 bbls. beef; from Y., of Shreveport, La., he received 120 bbls. beef; and
from Z., of Galveston, Tex., he received 210 bbls. beef. A sale of the three lots was
effected at $11. But on inspection, it was found that Y’s beef was 25% better than
Z’s, and X's was 10% better than Y’s. What amount of money is to be credited to
the sales account of each correspondent?
Ans. X. $2524.80. Y. $1377.17. Z. $1928.03.
OPERATION.
IBbls. Bbls, of the quality of Z's.
X. 200 + 25% = 250 + 10% = 275
Y. 120 + 25% = 150
Z. 210 210
530 635 of truality of Z's.
11
$5830
Operation to find X's credit. Operation to find Y’s credit. Operation to find Z's credit.
$ $ $
635 | 5830 635 5830 - 5830
275 150 635 | 210
$2524.80 $1377.17 $1928.03
Explanation.—The difference in the quality of the beef being given in per cent, we, there-
fore, to find the equivalent number of barrels of X’s and Y’s beef according to the standard quality
868 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *.
of Z’s, add to Y's 25% of the number of barrels, and to X's, as his is worth 10% more than Y’s, we
first add 25% of the number of barrels, which gives 250 of the quality of Y’s, and then to this we
add 10% of the number of barrels, and obtain 275 barrels of the quality of Z's. The balance of the
operation is too simple to require explanation, and hence we omit it.
82. A commission merchant received on consignment from A., $300 barrels
X flour; from B., 240 barrels XX flour; from C., 50 barrels XXX flour; and from
D., 100 barrels choice flour. The current market price for Xis $6.00; for XX, $6.50;
for XXX, $7.00; and for choice grade $8, with a market tending upward. The four
lots were sold at the round figure, $7.00. What was the correct amount to credit to
the Sales account of each consignor ? Ans. A. $1927.72. B. $1670.69.
C. 374,83. D. 856.76.
OPERATION.
Bbls. Market Market
rate. value.
A. 300 (2) 6 == $1800
B. 240 (2) 6} = 1560
C. 50 (2) 7 F 350
D. 100 (2) 8 == 800
690 $4510 Total market value.
7
$4830 Amount of sales.
STATEMENT TO FIND A's, B's, C's AND D's RESPECTIVE CREDITs.
A’s. B’s. C’s. D’s.
$ $ $
4830 4830 4830 4830
4510 | 1800 4510 | 1560 4510 || 350 4510 || 800
$1927.72 $1670.69 $374,83 $856.76
Ea:planation.—In the operation of this problem, we must find the amount of sales, then the
amount of the market value Öf each grade, and then, having these two amounts, we divide the
sales between the different consignors in proportion to their respective amounts at the market rate.
. #. the price per barrel, we wonld divide the total amount due each by the number of barrels
sold for each.
COTTON AVERAGE PEROBLEMS.
83. A cotton factor has four bales of cotton, weighing as follows: 450, 460,
398 and 486 pounds. What is the average weight per bale?
Ans. 4484 pounds.
OPERATION.
450
460
398
486
4) 1794
448# pounds. Ans.
84. A cotton factor has four lots of cotton belonging to four different parties.
The first lot consists of 3 bales, weighing 1400 pounds, and is classed as Ordinary,
the quoted market price of which is 16 g. The second lot consists of 8 bales, weigh-
ing 3500 pounds, and is classed as Low Middling, the quoted market price of which
is 1842. The third lot consists of 4 bales, weighing 1800 pounds, and is classed as
Strict Middling, the quoted market price of which is 1942. The fourth lot consists
Yºr PARTNERSHIP SETTLEMENTS. 869
of 5 bales, weighing 2500 pounds, and is classed as Good Middling, the quoted
market price of which is 202. What is the average price per pound?
OPERATION Ea:planation.—This method, although sufficiently correct for
By the practical method. practical purposes, is not quite accurate, for the reason that the
Bales. Cts. weight of different bales of cotton is not the same. But in prac-
: % ;# = 1. tice the weight is not often known at the time the average price is
4 @ 19+ = 77% required, and hence the necessity in those cases of using the num-
5 (a) 20 = 100% ber of bales which really represent pounds. To have added the
20 ) $3.73 four prices, 16%, 1849, 1949, and 20% together, and divided by 4,
1843% Ans. would have been a woeful error.
OPERATION BY THE EXACT METEIOD.
Bales. Pounds.
3 - 1400 (2) 16 gº := $224.00
8 = 3500 (a) 18% - 647.50
4 = 1800 (a) 194% - = 346.50
5 = 2500 (a) 20 g := 500.00
9200 ) $1718.00 (1834%, Ans.
79800
tº 73600
6200
85. Suppose, in the foregoing problem, that a sale of the four lots of cotton
had been effected at 192, what would have been the correct increased price for
each lot and grade of cotton ?
OPERATION.
Pounds. Market Price. "Amount.
1400 (2) 16 g. = $224.00
3500 (2) 1849. = 647.50
1800 (2) 194%. - 346.50
2500 (2) 20 g. F 500.00
9200 $1718.00
199.
$1748.00
Statement to find price Statement to find price | Statement to find price | Statement to find price
of the Ordinary. of the Low Middling. of the Strict Middling. of the Good Middling.
g .*
16 2 || 37 4 || 77 20
1718 || 1748 1718 1748 1718 1748 1718 1748
16.279;#3%. Ans. 18,823#9. Ans. 19.586;#39. Ans. 20.349%$3%. Ans.
Ea:planation.—In the operation of this problem, we first find the value of each lot of cotton at
the quoted market price, and then by adding the same we obtain the total market value, $1718.
Then, by the use of our reason, we see that 16, 183, 194, and 20% prices gave $1718 value, and by trans-
position that $1718 value required 16, 184, 19+, and 20% prices, and as the whole cotton was sold for
$1748, we have this simple proportional question: if $1718 value requires 16, 184, 19+, and 20%
prices, what will $1748 value require? The line statements give the exact answers to this question,
and complete the solution. *
Had the cotton been sold at a price less than the average, the same method
of work would be used to find the decreased prices.
87o soulE's PHILOSOPHIC PRACTICAL MATHEMATICS. 3C
It has been said by many that questions like the preceding one could not be
solved, and it is claimed by others, in whose business similar questions are of daily
occurrence, that because the correct results often contain such inconvenient frac-
tions, the correct solution is of no advantage. The method in general use is the
guessing method; a method of solving arithmetical questions that we cannot indorse.
By the guessing method of finding the increased prices, injustice, unintentional of
course, is done to some of the owners of the different lots of cotton, for the reason
that the guessing cannot be made exactly in the correct proportion; and hence, even
though the guessing prices absorb the exact amount received, one owner will have
more and another less than his correct proportional increase.
By our correct solution of this character of questions, we admit that the
fractions in the results are often impracticable according to custom, but by these
correct results, extending the figures to two or three places of decimals, we can
readily see what the nearest practical fractions are, and hence do the nearest possi-
ble justice to all the owners.
By our correct method, we avoid the repeated guessing operations that are
often necessary to arrive at a satisfactory approximate result, and never have any
considerable excess or deficit of money by reason of the increase or decrease in
price.
There is another objection that is sometimes raised in opposition to the cor-
rect method of work. It is, that because there is more cotton of a high grade than
there is of a low grade, or vice versa, the increase or decrease of price should not.
be in exact proportion to the quoted market rates. If this circumstance is taken
into consideration in fixing the selling price of the different lots, then we would
increase or decrease the quoted market rate of such cotton as influenced the selling
price, and proceed by the exact method of work.
To show the practical prices of the above problem, we present the following:
1400 (a) 16+ = $227.50 Or, 1400 (a) 16+ = $227.50 Or, 1400 (a) 164 = $227.50
3500 (a) 18+3 = 658.43 3500 a 18; = 660.62 3500 (a) 18:# = 658.43.
1800 (a) 19% = 352.12 1800 (a) 19% = 351.00 1800 (a) 193 = 353.25
2500 (a) 20:# = 509.37 2500 (a) 20:# = 509.37 2500 (a) 20:# = 509.37.
$1747.42 $1748.49 $1748.55.
The following table of 16ths and 32nds will facilitate the changing of the
decimals of the cents in the price answers to practical equivalent common fractions,
in all problems of this character:
TABLE OF 16THS AND 32NDS AND THEIR EQUIVALENTS IN DECIMALS.
# = .03125 | # = .28125 | }} = .53125 # = .78125.
# = Tº = .0625 | }} = # = .3125 | }} = |* = .5625 | #3 = +3 = .8125
# = .09375 | }} = .34375 | }} = .59375 | #} = .84375.
# = # = .125 # = # = .375 #} = # = .625 # = g = .875
# = .15625 | }} = .40625 | }} = .65625 | # = .90625.
# = * = .1875 ## = ** = .4375 | ## = +} = .6875 #9 = H = .9375
** = .21875 | }# = .46875 | }} = .71875 # = .96875.
# = + = .25 # = } = .5 # = # = .75 # = 1 = 1.
PARTNERSHIP SETTLEMENTS. 871
DIVISION AND PROPORTIONAL DIVISION.
86. A benevolent man has $600, of which he wishes to give to A. $, to B. #,
to C. § and to D. #. What amount will each receive?
Ans. A. $200. B. $150. C. $120. D. $100.
OPERATION.
A's one-third of $600 = $200
B’s one-fourth of 600 = 150
C’s one-fifth of 600 = 120
D’s one-sixth of 600 = 100
$570 total amount donated.
30 remaining on hand.
$600
87. Another benevolent man has $600 which he wishes to give to W., X, Y,
and Z., in proportion to #, +, + and #. What amount would each receive 3
Ans. W. $210.53. X. $157.89. Y. $126.32. Z. $105.26.
OPERATION.
# = W’s proportion. Statement to find W’s propor- || Statement to find X’s propor-
+ = X’s & 4 tional share. tional share.
+ — Y’s & 4 &
# = Z's & 4 $
gºmers 600 600
## = the sum of the propor- 19 20 19 || 20
tions which are to re- 3 4
ceive the amount of tº mºs *-
$600. $210.53 Ans. $157.89
Statement to find Y’s | Statement to find Z's Explanation.—The operation of this problem
proportional share. proportional share. is so simple that special explanation is deemed
unnecessary. The preceding example, No. 86, is
often given, when the answer to this example is
600 600 required. To show clearly the difference between
19 20 19 || 20 giving certain parts of a sum of money to differ-
5 6 ent parties, and giving the whole sum of money
sº- in proportion to certain parts, we have presented
$126.32 $105.26 the two preceding questions.
88. A planter has seventeen horses which he wishes to give to his three sons,
Henry, William and Robert, in proportion to 3, § and #. How many horses will
each receive 3 . Ans. Henry, 9. William, 6, Robert, 2.
OPERATION.
# = Henry’s proportion. HENRY. WILLIAMs } ROBERT.
# = William’s { { -
# = Robert's & & Horses. Horses. Horses.
tº-ººp 17 17 17
+} = the sum of the pro- 17 | 18 17 | 18 17 | 18
portional s h are s 2 3 9
which are to re- *-º-º-º- * *E=
ceive the 17 horses. 9 Ans. 6. Ans. | 2. Ans.
89. What are the respective amounts resulting from the division of $77 pro.
portionally to #, #, 4, # 3 Ans. $30, $20, $15, $12.
87.2 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. *
SETTLEMENTS WITH SPECIAL CONDITIONS.
90. A. and Z., residents of New Orleans, hired a team to go to the lake, a
distance of five miles and return. The charge for the use of the horses and car.
riage was $10. On arriving at the lake, Z. proposed to return by some other convey-
ance, to which proposition A. agreed. They also agreed to settle the charges for the
use of the horses and carriage in proportion to the distance that each rode. What
must each pay ? Ans. A. $6.663. Z. $3.33}.
OPERATION.
* To find the amount A. | To find the amount Z.
A. rode to the lake and returned = 10 miles. must pay. must pay.
Z. ‘‘ ‘‘ lake = 5 miles. $ $
fººms 10 10
Total number of miles ridden 15 || 10 15 || 5
by both {- ſº-> sº - = 15 miles. $6 66# Ans $3 33} Ans.
91. Suppose in the above problem, that they had agreed to settle the charge
for the use of the horses and carriage justly, what would each have paid?
Ans. A. $7.50, Z. $2.50.
SOLUTION.
The distance to the lake and return being ten miles, and the charge for the use of the horses
and carriage being $10, it is clear that ten miles' ride costs $10, and if ten miles’ ride cost $10, one
mile ride will cost the 10th part, which is ($10 – 10) $1. Then A. and Z. having ridden together to
the lake, a distance of five miles, in justice, they are required to settle equally the expense of the
ride that they have equally enjoyed. And the expense for the five miles at $1 per mile is $5, which,
settled equally, is ($5 - 2) $2.50 that A., and $2.50 that Z. must pay. This settles the expense to the
lake, where Z. retires and A. becomes solely responsible. The expense for the five miles’ ride by A.
from the lake to the city, at $1 per mile, is $5, which justice requires A. to settle, as he enjoyed the
ride alone. We thus obtain $2.50 + $5 = $7.50 that A. would have paid, and $2.50 that Z. would
have paid, had they settled justly.
Or, thus:
A. and Z. having ridden together to the lake, which is ºr of the whole distance, it is clear
that they should, equally, pay ºr of the whole cost. The whole cost being $10, ºr of it is $5, and #
of $5 is $2.50, the amount each should pay for riding to the lake. Then, as A. rode the remaining
Tº of the whole distance alone, it is clear that he should alone pay the remaining ºr of the whole
expense; ºr of $10 is $5, which added to the $2.50 first obtained makes $7.50 that A. is to pay. This
completes the solution, and gives $7.50 as A's share of the expense, and $2.50 as Z's share of the
expense.
92. A. and Z. bought merchandise to the amount of $800, of which sum A.
paid $500 and Z. paid $300. Soon after they sold to G. § of the whole stock for
$400. How much of the $400 must A. and Z. receive respectively, in order to con-
stitute each, A., Z. and G. § owner of the goods? Ans. A. $350. Z. 50.
FIRST OPERATION.
Capital Am't of invoice Am't of in- Statement to find Statement to find
contrib- each retained. voice each amount due A amount due Z.
uted. sells. tº
A. $500 – $266; = $233} $ $
Z. 300 — 266# = 33# | 400 400
} of 800 $266; amount of invoice sold. sº % sº io
$266; $350. Ans. $50. Ans.
* - PARTNERSHIP SETTLEMENTS. 873
SECOND OPERATION.
Statement to find Statement to find
Amount paid. Part owned. Part retained. Part sold amount due A. amount due Z.
Z. - 300 = } – # = ºr 400 400
*- sº- 8 || 24 8 || 24
# of )}(# # 24 7 24
$350 | $5
93. Two newsboys, James and Henry, on taking their seats to eat some
cakes that they had purchased, were requested by another member of their frater-
nity named “Dick,” to allow him to dine with them. To this request, they cheer.
fully assented, and the three ate the cakes. When they had finished, Dick laid on
the table 402 for his share of the dinner. How much of this 40g must James
receive, who contributed 5 cakes to the dinner, and how much must Henry receive,
who contributed 3 cakes 3 Ans. James 352. Henry 52."
OPERATION.
JAMES, HENRY.
James 5 — 23 = 2* = } g ſº
Henry 3 — 23 = } = } 40 40
º - 8 || 3 8 || 3
# of 8 § 3 7 3
2} 35% 52
94. A cotton factor insured :
200 bales of cotton for 3 months at 14%;
150 - 4 {{ 44. 4, 2 4% at 13% ;
100 és {{ {{ << 1 { { at #% ;
What was the average rate per cent. of insurance, supposing that each bale
was of the same value # Ans. 1,29#%.
FIRST OPERATION.
Bales. Per cent. on 1 bale.
200 a 14% = 300.00
150 a 13% = 206.25
100 (a) #% = 75.00
450 ) 581.25 (1.29#96 Ans.
SECOND OPERATION.
Bales. Assumed value. Insurance. Operation to find the rate per cent.
200 a $100 = $20000 (a) 14% = $300.00 $
150 (a) 100 = 15000 (a) 13% = 206.25 581.25
100 (a) 100 = 10000 (a) #96 = 75.00 45000 || 100
$45000 $581.25 1.295% Ans.
95. A. contracted with Z. to deliver to him :
78 bales, 38531 pounds, Good Middling cotton, at 154%,
and 72 “ 36500 “ Middling “ at 142;
(Continued.)
874 soule's PHILOSOPHIC PRACTICAL MATHEMATICs. Yºr
but A. having disposed of the foregoing cotton to other parties, by subsequent
agreement A. delivered and Z. received:
2475 pounds of Inferior at 92;
6681 “ “ Ordinary at 102 ;
12219 “ “ Good Ordinary at 122;
7218 “ “ Low Middling at 1342;
Z. then demanded the balance due him in Middling and Good Middling cotton.
How many pounds of each grade, Middling and Good Middling, were due him 7
Ans. 25527# pounds Middling.
26947# “ Good Middling.
OPERATION.
Bales.
78 - 38531 pounds Good Middling at 15.4% - $5972.30
72 - 36500 “ Middling at 140 - 5110.00
Total value - * - -> • - $11082.30
RECEIPTS.
2475 pounds Inferior at 9% $222 75
6681 “ Ordinary at 10% 668.10
12219 4 & Good Ordinary at 12% 1466.28
7218 “ Low Middling at 133% 974 43
Total receipts - * - 3331.56
Balance due - * - $7750.74
which is to be paid in Good Middling and Middling at contract price,
in proportion to $5972.30 and $5110.
Statement to find the amount to be paid in Statement to find the amount to be paid in
Good Middling. Muddling.
$ $
7750.74 \ 7750.74
11082.30 || 5972.30 11082.30 || 5110
$4176.91 $3573.83
$4176.91 - 154% = 26947# lbs. of Good $3573.83 – 14% = 25527 ºr lbs. of
Middling cottom. Middling cotton.
SETTLING ESTATES.
96. A man had two sons and three daughters; to the youngest son, he gave
# of 14 times $3600, which was # of the elder son's share, and what the elder son
received was # of g of the whole estate; the balance was divided among his three
daughters in reciprocal proportion to their ages, 8, 10, 12 years. What was each
daughter's share 3 Ans. $7114.87 for the one 8 years old.
5691.89 % {{ 10 $4
4743.24 “ “ 12 {{
SOLUTION.
# of 14 = #, and # of $3600 = 2700 = the younger son's share; then if $2700 is # of the elder
son's share, # is the 4 of it, which is $1350, and # or the whole share is 5 times as much, which is
$6750; then if $6750 is # of g = } of the estate, 3 or the whole estate is 4 times as much, which is
Yºr PARTNERSHIP SETTLEMENTS. 875
$27000. Then the two sons' shares, $2700 + $6750 = $9450, deducted from the whole estate, $27000,
leaves $17550, belonging to the three daughters in reciprocal proportion to their ages, 8, 10 and 12
years.
The reciprocal of 8 is #
& & & 4 & 4 10 is To
& & ( &
# -- I'm -H T's = ºr
“ 12 is fºr
Having now the reciprocals, the following statements give the respective
shares of each daughter:
17550
17550 17550
37 || 120 37 120 37 120
8 10 12
$7114.87 $5691.89 $4743.24
To perform the first part of the solution without the reasoning, we would make
the following statements:
Statement to find the younger Statement to find the elder Statement to find the whole
son's share. son's share. estate.
$ $ $
3600 2700 6750
2 2 5 2 || 7
2 || 3 e-mº-º-º-º-º: 7 || 8
| $6750
$2700 $27000
97. A., B. and C. were to receive $6000 in proportion to #, 4 and #; but B.
having died, it is required to divide the money between A. and C. What sum
should each receive?
Ans. A. $4000. C. §2000.
OPERATION.
# -H # = }
$ Explanation.—B. having died, his # pro-
6000 6000 portional interest is therefore divided be-
3 tween A. and C. in proportion to their pro-
2 2 portional interests, # and #; we therefore
$4000 A's share.
$2000 C's share.
have, as shown in the operation, but to di-
vide the whole sum in proportion to # and #.
98. A father willed his estate valued at $40000 to his three children in pro-
portion as follows: John #, Henry # and Katie #.
Defore the settlement was made,
Henry died. What sum should John and Katie each receive 3
Ans. John $25000. Katie $15000.
.* OPERATION.
KATIE.
$
# -- + = fºr the sum of the proportional 40000 40000
interests of John and Katie. Then 8 15 : 15
3
$25000 $15000
NoTE.—Henry having died, his 4 proportional interest is therefore to be divided between John
and Katie in proportion to their proportional interests, as shown in the operation.
876 soule's PHILOSOPHIC PRACTICAL MATHEMATICs. Yºr
99. A man at his death left an estate amounting to $35000, to his wife and
two children, a son, and daughter. His children being absent in Europe, he directed
by will, that if his son returned, his wife should have # of the estate, and the son
the remainder; but if the daughter returned, his wife should have 3 and the daugh-
ter the remainder. Now, it so happened that they both returned. What sum
should each receive 3
Ans. $20000 son. $10000 wife. $5000 daughter.
SOLUTION.
By the first condition of the will, the son was to receive twice as much as the wife; and by the
second condition the wife was to receive twice as much as the daughter. Hence to comply with both
conditions, we assume 1 to represent the daughter's share; then twice 1, or 2 will be the wife's share,
and twice 2, or 4 will be the son's share; and 1 + 2 + 4 = 7 = the sum of all the shares. Then } of
$35000 = $5000, the daughter's share; # of $35000 = $10000, the wife's share; and # of $35000 =
$20000, the son's share.
100. A father leaves a number of children and a certain sum of money to be
divided among them as follows: The first is to receive $100 and ſº of the remain-
der; the second is to receive $200 and I's part of what then remains; again, the
third is to receive $300 and ºr of the residue, and so on; each succeeding child is
to receive $100 more than the one immediately preceding, and then I's part of what
still remains. At last it is found that all the children have received equal sums.
What was the amount left, and how many children were there?
Ans. The sum divided was $8100 and the number of children 9.
SOLUTION.
After giving the first $100, there remained the sum minus $100, and ſo of this is ºn of the sum
minus $10; hence the first received $100 + ºr of the sum minus $10 or $90+ tº of the sum, and
there remained of the whole # of the sum minus $90; giving the second $200, there remained nº of
the sum minus $290 and ſo of this is r&d of the sum minus $29; hence the second received $200 +
rán of the sum minus $29 or $171 + rào of the sum.
Now according to the conditions of the question, the first received the same as the second,
hence $90 + ºn of the sum = $171 + +}o of the sum or ſº of the sum = 130 of the sum + $81, or
again, rºw of the sum = $81; hence the sum = $8100. Now the first received $100 + To of what
remained or $100 + $800 = $900.
But since each child is to receive an equal sum, there must be as many children as $8100 is
equal to $900 which is 9.
PROPORTIONAL DIVISION.
101. Divide 18 oranges between Katie and Sallie, so that Katie will have +
more than Sallie. What number will each receive 3
Ans. Katie 10. Sallie 8.
FIRST OPERATION.
1 = the assumed number given to Sallie. Statements to find the proportional
1+ = & & & & é & Katie. share of each.
2+ = the sum of the ratio numbers by which SALLIE. KATIE.
the oranges are to be divided. 18 | 18
9 || 4 9 || 4
sº 4 || 5
8 *mºsºs
PARTNERSHIP SETTLEMENTS. 877
SECOND OPERATION.
4 = the assumed number given to Sallie. SALLIE, KATTE.
5 = é & 4 & & & Katie. 18 18
-sº 9 || 4 9 || 5
9 = the sum of the ratio numbers. - mº-
8 10
102. A young husband has one and a fourth dozen apples which he wishes to
divide between his wife and his mother-in-law, so that his mother-in-law will have
# less than his wife. How many will he give to each?
Ans. Wife 11+. Mother-in-law 33.
OPERATION.
1 = the assumed number given to his wife. Statements to find the share of each.
# = & 4 (£ & 4 mother-in-law.
- WIFE. MOTHER-IN-LAW.
1} = the sum of the ratio numbers, by which the ap- 15 15
ples are to be divided. 4 || 3 4 || 3
* 3 l —
11+ 3#
NOTE.-If preferred, 3 and 1 may be assumed as the ratio numbers, in place of 1 and #.
103. Divide $2000 between A., B. and C., so that A's part will be to B's as 2
to 3, and that C. will have as much as A. and B. together, less $200. What amount
will each receive? Ans. A. $440. B. $660. C. $900.
N SOLUTION.
As C. is to have as much as A. and B. together, less $200, A. and B, will therefore have # of
$2000 + 4 of $200, which is $1100. Then, as A. and B. are to receive this in the proportion of 2 to
3, A. will therefore receive # and B. # of $1100. # of $1100 = $440 A's share; # of $1100 = $660 B's
share; and lastly, as C is to receive as much as A. and B. together, less $200, he will therefore
receive $1100 — 200 = 900.
104. A. and B. made an investment in real estate, A. invested $26217.50 and
B. invested $1707.40.
From the proceeds of a sale of a portion of said real estate, A. drew out
$15289.25 and B. drew out $2908.95.
B. now desires to be an equal owner with A. in the remainder. How much
must B. pay A. to be an equal owner?
OPERATION,
A. invested, - * > - º - * = º - º - ſº º se - as $26217.50
A. drew out, - * * is ºn tº * * * * * - * * = 15289.25
A's net investment of which B. owes A. one-half, - - - - - - - - $10928.25
B, invested, - --> - * - º - * - º - - $1707.40
B. drew out, - * - sº - sº - gº - :- & -> 2908.95
B’s net withdrawals of which he owes one-half to A. * * * * gºs º- tº 1201.55
Total amount of which B. owes A. one-half, º º º -> º - º * # ) $12129.80
Amount B. must pay to A. sº º - *- gº tº ſº º º - º $6064.90
PROOF.
A’s total investment, - - - $26217.50 | B. invested, sº º ºs - - $1707.40
A’s withdrawals, - $15289.25 B. paid to A. - - - - - 6064.90
A. received from B. - 6064.90 *messammº
-º-º- B’s total investment, - - º $7772.30
21354.15 B. drew out, * - - - º 2908.95
—º-
A's # net investment, tº º ºs $4863.35 | B's net investment, - - - - $4863.35
878 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. 3.
THE LAND AND THE DITCH PROBLEM.
105. A. and B. paid $600 ($300 each) for 300 acres of land. In dividing the
land according to quality, it was agreed that A. should have the best quality, and
pay therefor 75 cents per acre more than B. How many acres did each receive, and
what price per acre did each pay ?
Ans. A. received 122.8 acres and paid $2.443 + per acre.
B. 64 177.2 4 {{ 1.693 + éé
OPERATION.
300 2 75 g = $225; and $600 – 225 = $375 reserved remainder. Then $225
× 4 × 300 = $270000 to which is added 375* = 140625; 270000 + 140625 =
v4.10625 = 640.8 +
Then 640.8 + 375 = 1015.8, which -- (300 × 2) = 600 = $1.693 + the price
per acre paid by B. Then $1.693 + 75 cents = $2.443 + the price per acre paid
by A.
Then $300 + 1.693 = 177.2 acres that B. receives; and $300 + $2.443 = 122.8
acres that A receives.
Explanation.—This and the following problem belong strictly to algebra, and although they
may be solved by arithmetical operations deduced from algebraic equations, they cannot be as
analytically explained as regular arithmetical problems. The solution of all problems of this
character may be performed by observing the following directions: First find the cost of the whole
number of acres at the difference between the prices per acre; then subtract this from the price paid
for the whole land and reserve the remainder. Then multiply the value produced by the difference
in price by 4 times the whole amount paid by him who paid the least per acre, and to the product,
add the square of the reserved remainder; then extract the square root of this sum, and to it add
the reserved remainder. Then divide this sum by twice the whole number of acres, and the quotient
will be the price per acre paid by him who paid the least Then to this price add the difference
between the prices paid and the sum will be the price paid by him who paid the greater price.
Then divide the amount paid by each by the respective price paid by each, and the quotients will
be the number of acres that each is to receive.
It will be observed that there is in the problem an interminable decimal which,
because it arises in extracting a root, cannot be expressed as a vulgar fraction. It
may, however, be approximated to any extent.
Problems of this character are often misstated by giving either the price that
each is to pay or the number of acres that each is to receive. To give either, ren-
ders the problem absurd and insolvable.
106. A. and B., together, agree to dig 100 rods of ditch for $100. The part
of the ditch on which A. was employed was more difficult of excavation than the
part on which B. was employed; and it was therefore agreed that A. should receive
for each rod 25 cents more than B. received for each rod that he dug. How many
rods must each dig, and at what price so that each may receive just $50?
Ans. B. receives $ .8903 + per rod, and digs 56.16 rods.
A. “ $1.1403 per rod, and digs 43.84 rods.
OPERATICN,
100 x 25g = $25; and $100 – $25 = $75, reserved remainder. Then $25 ×
4 × 50 = 5000 to which we add 75* = 5625; 5000 + 5625 = 10625, the square root
of which is 103.07 -H
Then 103.07 -- 75 = 178.07, which -- (100 × 2) = 200 = $.8903 + the price per
rod received by B. Then $.8903+ 25 g = $1.1403 the price per rod received by A.
Then $50 + $.8903 = 56.16 rods that B. digs; and $50 + $1.1403 = 43.84
rods that A. digs.
* PARTNERSHIP SETTLEMENTS. 879
107. A., B., C. and D. agree to do a piece of work for $5500. A., B. and C.
can do it in 20 days; B., C. and D. in 24 days; C., D. and A. in 30 days; and D.,
A. and B. in 36 days. In how many days can all do it, working together; in how
many days can each do it, working alone; and what part of the pay ought each to
receive # Ans. See the solution.
SOLUTION.
As A., B. and C. can do the work in 20 days, they can do gº of it in 1 day; as B., C., and D.
can do it in 24 days, they can do ºr of it in 1 day; as Q, D. and A. can do it ; 30 days, they Call
do gº of it in 1 day; and as D.; A. and B. can do it in 36 days, they can do #5 of it in 1 day; there-
fore, A., B., C. and D., by working 3 days each, will do gº + g + 3 + 35 = } of the work, and
in 1 day they will do # of #} = #6; and as they all do gº of the work in 1 day, it will require as
many days to do the whole work as 1 is times equal to #6, which is (1 -— gºs =) 19%r days.
Now, by deducting successively the work that three can do in 1 day, when they work
tººther from what the four can do in 1 day, we obtain the work that each one will do in 1 day;
U18
#5 – 35 = Tºro the work that D. can do in 1 day.
#5 — ºr = Tºg the work that A. can do in 1 day.
# — ºr = +&#5 the work that B. can do in 1 day.
#6 — ºf = gº the work that C. can do in 1 day.
Having now the work that each can do in 1 day, it is clear that it will require each as many
days to do the work alone as 1 is equal to the part of the work performed by each; thus
1 — Tºso gives 1080 days that it would require D.
1 - Tºg gives 108 days that it would require A.
1 - Tºšo gives 56+3 days that it would require B.
1 -— gº gives 43% days that it would require C.
In dividing the money, it is evident that each should receive such
amount paid as he did of the whole work. uch a part of the whole
This is shown by multiplying the amount of work performed b h dail
of days that all worked; thus p y each daily by the number
A. Tºg × 19 ºr = ſºr A's proportion.
B. Tºšo X 1911 = }; B's & &
C. ºf X 1911 = ºr C's & f
D. Tºso X 191+ = ºr D's & 4
According to these proportions the money is divided thus:
*r of $5500 is $1000, A’s share.
# of 5500 is 1900, B’s “
fºr of 5500 is 2500, C's “
ºr of 5500 is 100, D's “
Total amount, - - - - - $5500
111. A young hoodlum, a modern evolvement of the human race, stole a bas-
ket of peaches and divided them among three brother hoodlums and himself as
follows: To the first, he gave 4 of the whole number and + of a peach more; to the
second, he gave # of What remained and # of a peach more ; to the third, he gave #
of what remained and # of a peach more. The stealer retained what was then left,
for himself, which was # the number he gave to the first hoodlum. What Was the
number of peaches stolen, aud how many did each hoodlum receive?
Ans. 7 peaches were stolen. 1st hoodlum received 2.
2d 46 44 2.
3d {{ 4% 2.
The stealer had 1.
SOLUTION.
In all problems of this kind, the following principle governs: That when the fractional parts
of the successive portions are consecutively increasing, each of the parts or shares of the several
88O SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. jºr
persons are equal, except the last one, which, from the nature of things, must be one less than the
average. Hence in the given problem, the fact stated that the last one's share is half of the first
(and each of the others also) and from the above principle that it must be one less, it must neces-
sarily be ONE, and since each of the others is one greater, they must be two each, and since there
are three persons having equal portions, there will be 6 + the 1 for the last, making 7 in all.
Hence, to solve problems of this kind: add 1 to the last number, (ONE), and multiply by the num-
ber of persons, and then subtract 1 from the product.
112. Three persons, say A., B. and C., buy a piece of land for $4569, and the
parts for which they pay bear the following proportions to each other, viz: the sum
of the first and second, the sum of the first and third, and the sum of the second
and third, are to each other as #, #, and #3. How much did each pay, and what part
did each own 7 Ans. A. paid $1015; and owns #.
B. 46 1523 “ #.
C. “ 20303 “ 4.
OPERATION.
A. and B. = } = To = 5 shares.
A. and C. = } = 1% = 6 “
B. and C. = Pºſ = To = 7 “
**
18 shares; but as each party’s share has been
counted twice, we therefore divide the same by 2 and obtain 9 shares.
Then, to find the respective shares of A., B. and C., we deduct from the sum of the shares.
the number of shares owned by each two of the parties, thus:
A., B. and C. own 9 shares. A., B. and C. own 9 shares. A., B. and C. own 9 shares.
A. and B. own 5 “ A. and C. own 6 * * B. and C. own 7 : «
C. owns 4 * * B.’ owns 3 * * A. owns 2 * *
Having the respective shares of each party, we make the following state-
ments and produce the amount due to each:
A’s 3. B's 3. C's #.
A. 2 shares $ $ $
B. 3 * * 4569 4569 4569
C. 4 * * 9 || 2 9 || 3 9 || 4
9 * * $1015; $1523 $2030;
113. Three men take an interest in a mining company. A. put in $480 for 6
months; B. a sum not named for 12 months; and C. $320 for a time not named:
when the accounts were settled, A. received $600 for his stock and profits; B. $1200
for his; and C. $520 for his. What was B's stock, and C's time?
Ans. B's Stock $800. C's time 15 mos.
Investment. Time invested. Am’t rec'd in settlement. Gain.
A. $480. 6 mos. $ 600 $120 = }#} = + of stock.
B. : 12 mos. 1200 º
C. 320. º 520 200
OPERATION.
Since A's investment of $480 gained $120 in 6 months, in 12 months, the time B's was invested,
it would gain twice as much or $240, which is #8 = } of his investment.
Then, since the gain of each is in like proportion, the $1200 received in settlement by B. for
12 months is # of his investment of stock; and since $1200 is 3 of B's investment, 3 is the third part
or $400, and two halves or the whole investment is $800, and his gain is $1200 — $800 = $400.
Now having B's investment $800, and his gain for 12 months $400, and also C's investment
* PARTNERSHIP SETTLEMENTS. 88.1
$320, and his gain $200, to find C's time, we make the following classification and proportional state-
ment :
CLASSIFICATION. STATEMENT.
Months. º
12 Explanation.—In making the
Investment. Gain. Months. 320 800 statement, We .*. as º:
lows: Since an II]. We St-
B’s $800 $400 12 400 | 200 º º
*- nt requires 12 months time
C’s 320 200 ! ment req
15 mos. to produce $400 gain, $1 would
require 800 times as many
months, and $320 investment the 320th part. Then, since $400 gain requires 12 months time to pro-
duce it, $1 gain will require the 400th part, and $200 gain will require 200 times as many months.
114. Jones, Smith, Brown and Wood formed a partnership. It was agreed
that Jones should receive # of the gains, Smith 4, Brown #, and Wood #. The
remainder of the gains was to be credited to Reserve Fund, to be used for the Con-
tingencies of the business. It was agreed that each partner should invest $15000,
on condition of receiving the above proportional share of profits. Jones however
invested only $12000 and Smith invested only $10000. The net gain amounted to
$16500. Since Jones and Smith failed to invest the required amount, how should
the gain be divided ? Ans. Jones $5225; Smith 3265.62%;
Brown $3918.75; Wood $3265.624.
Beserve Fund $825.
SOLUTION.
1. By the terms of the partnership, each partner was to invest $15000 which sum was to
entitle each to the following shares of the profits: Jones + + Smith 4 + Brown 4 + Wood # = #3;
thus leaving go to be credited to Reserve Fund.
2. Since Jones and Smith failed to invest the full amount of $15000 each, the respective
shares of their gain must be reduced in the same proportion as their respective investments were
reduced from the $15000 required by the terms of the agreement. The following statements give
their reduced shares:
$15000 : 12000:: * : *r = Jones' share of gain for his $12000 investment.
$15000 : 10000:: + : # = Smith’s share of gain for his $10000 investment.
3. The respective shares of the partners in the #} of the distributable gain are now as fol-
lows: Jones 1%, Smith #, Brown ; and Wood # = # which is the sum of the interests or shares that
are to receive ## of the gain.
4. The total gain is $16500
*if deducted for Reserve Fund is 825
Amount to be divided is $15675
5. The following statements give the share of each partner:
JONES. SMITH. BROWN. WOOD.
$15675 $15675 $15675 $15675
4 5 5 4 5 4 5
15 || 4 6 5 6
$5225 | $3265.62% | $3918.75 $3265.62%
NOTE.-By the terms of the agreement, the partners were to receive in different proportions
## of the gain, and the remainder, ºo, was to be credited to Reserve Fund. The respective interest
of each partner was conditioned upon an investment of $15000; hence those partners who invested
less than $15000 are to have their original shares reduced, as shown above, and then the # of
the gain is to be divided among the partners in proportion to the revised shares, as shown in the
operation. Reserve Fund thus receives ºf of the gain, as per agreement. By this method, the part-
ners received the #3 of gain, in equitable proportion to their actual investments, based upon the
terms of agreement. This we believe to be the proper method of adjusting the matter.
In Problem 38, pg. 832, which involves the same conditions as this problem, the # interest,
which was to be credited to Profit and Loss, was allowed to participate in the gain resulting from
the decrease in the respective shares of C. and D. because they failed to make full investments as
they had agreed. By thus giving Profit and Loss a part of this gain, more than # of the gain of the
882 PARTNERSHIP SETTLEMENTS, *
business is credited to this account, and consequently less than g of the gain is distributed among
the partners. In the solution of Problem 38, which problem was selected from another work, we
adopted the method of the author of the problem, by which he credited PROFIT and Loss with more
than # of the gain.
We now present the above problem and give a different solution and submit the two methods
for the consideration of accountants. *
115. Suppose, in the above problem, that the partners had made the following
investments: Jones $12000, Smith $25000, Brown $15000 and Wood $15000. How
should the gain have been divided ?
Ans. Jones, $3980.95 +. Smith, $6220.24 –.
Brown, 2985.71 +. Wood, 2488.10 —.
Reserve Fund, $825.
OPERATION.
$15000 : 12000 : : * : *r = Jones' share of the gain for his $12000 investment.
15000 : 25000 : : + : # = Smith’s “ “ “ “ 25000 “
Then Jones’ ºr + Smith's ºr + Brown's # -- Wood's # = # which is the sum of the shares or
interests that are to receive #8 of the gain.
The total gain is - - $16500
25 deducted for reserve fund 825
Gain to be divided is - $15675
The following statements give the proportional share of each partner:
Jones’ ſº. Smith's #. Brown’s #. Wood's #.
$15675 $15675 $15675 $15675
63 || 60 63 60 63 60 63 || 60
15 4 12 || 5 5 6
$3980.95+ $6220.24— $2985.71+ $2488.10—
116 A., B., and C. are general partners in business, equal in gains and losses. They conclude to
dissolve their commercial or trading partnership and form a stock company. Accordingly they
close their books and find the net capital of each to be as follows: A. $32238, B. $28712, and C.
$28644. Total $89594.
The stock company is organized with a capital of 1000 shares at $100, amounting to $100000,
of which each of the partners of the former business is to subscribe for 333% shares.
The excess of the capital of the stock company over the total capital of the old firm is
($100000 — $89594) $10406.
It is agreed that to offset or balance this excess, they will value some patents which the old
firm possess at $10406, and place the same in the list of assets.
What amount of money must B. and C. pay to A. respectively, to adjust the matter equitably
among the three ? Ans. See operation.
OPERATION.—First Step.
The valuation of the patents creates a new asset, which belongs to the members of the old
firm in proportion as they shared the gains and losses, # each. Hence $10406 equals $3468.66# due
respectively to A., B., C.
Second Step.
A’s credit $32238 + $3468.67 = $35706.67 — $33333.34 = A's Cr. $2373.33
B’s credit 28712 -- 3468.67 = 32180.67 – 33333.33 = B's Dr. 1152.66
C’s credit 28644 + 3468,66 = 32112.66 – 33333.33 = C's Dr. 1220.67
$89594 $100000.00 $2373.33
10406 valuation placed on patents.
*=ºº
$100000 -- 3 = $33333.33} = each partner's interest in the stock company.
$ From these statements it is clear that B. and C. must pay to A. respectively, $1152.66 and
1220.67.

nvolution.
eSºº - - - - - - SEC EE C - C - EC Cº - Cº-ºººººº-ºº: ===N
1426. Involution is the process of multiplying a number by itself a certain
number of times. The several products are called the powers of the number.
The number used as a factor is called the root or first power.
The second power or square is the root used twice as a factor.
The third power or cube is the root taken three times as a factor.
The fourth power or biquadrate is the root taken four times as a factor; and
all higher powers indicate the number of times the given number or root enters as
a factor.
The number of times the root is used as a factor is called a degree, as the
1st, 2d, 3d, and 4th degrees; and the number denoting that degree is called the
index, or exponent of the power. Thus, the cube of 5 is represented by 5°. The
exponent should be a small figure, and written to the right of the upper part of the
root or number. A fraction should be inclosed in a parenthesis before writing the
exponent, to indicate that it belongs to the whole expression, otherwise it might be
considered as belonging to the numerator alone. Thus the square of # is denoted
(#)”.
The following arrangement will illustrate the formation of degrees and
powers for the number 5; thus:
59 – 5 – 5 = 1 = zero power of 5.
5* = 1 × 5 = 5 = 1st power of 5.
5* = 5 × 5 = 25 = 2d power or square of 5.
5* = 5 × 5 × 5 = 125 = 3d power or cube of 5.
5* = 5 × 5 × 5 × 5 = 625 = 4th power or biquadrate of 5.
We here see that there are three things to be noted in Involution, viz.: 1st.
The root or number; 2d. The exponent; and 3d. The power. Considering the 1st
power as a multiplicand, we see the number of multiplications will be one less than
expressed by the exponent. Any power of 1 is 1; any power of a number greater
than 1, is greater than the given number; and any power of a number less than
1, is less than the number itself.
When the required power is of a high degree, instead of multiplying by the
root continually, we can involve any of the subordinate powers, the sum of whose
degrees amounts to the degree of the required power; always observing that the
product of the powers corresponds to the sum of the exponents. Thus, to find the
7th power of the number 3, we may multiply its 4th power (81) by its 3d power
(27), or 37 = 34 × 33.
f When the exponent of the required power is a composite number, we may
raise the root to a power indicated by one of the factors of the given exponent, and
then raise this result to a power indicated by the other factor. Thus, to raise 5 to
the 12th power, or 5*, we first find the 4th power of 5 to be 625; we then raise the
625 to the 3d power, because 5** = (5*)*. * * ~ *
(883)
884 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
The difference between the last and the preceding proposition should be
clearly understood to prevent confusion, their principles being in nowise analogous.
To raise a vulgar fraction to any power, raise both numerator and denomina-
tor to that power. Mixed numbers should be changed to improper fractions. To
find any power of a decimal, or mixed decimal, multiply as in whole numbers, and
point off in the result as many places for decimals as will equal the number of
decimal places in the root, multiplied by the exponent of the power.
EXAMPLES.
1. What is the 2d power of 3465? Ans. 12006225.
2. What is the 5th power of 256% Ans. 1099511627776.
3. What is the 3d power of #3 Ans. ###.
4. What is the 6th power of 33? Ans. 2430%r.
5. What is the 4th power of .125? Ans. .000244140625.
6. What is the cube of 3.024% - Ans. 27.653197824.
7. What is the 7th power of .0043 Ans. .000000000000000016384.
8. What is the 10th power of 7 ? Ans. 282475249.
The 2d power of numbers ending in 5, when not too large, may be very
readily found by taking all the figures except the 5, and then multiply the number
thus taken by itself increased by 1, and to the product annex 25, the square of the
5. Thus, to find the square of 25, we omit the 5, and multiply the 2 by 3 = 6, then
annexing 25, we obtain 625. *
Find the square of 15, of 35, of 55, of 75, of 105, of 225, of 575.
To find any power of a number that is an aliquot part, or aliquot multiple of
a hundred or thousand, we form a fraction by taking the given number for a
numerator, and 100, or 1000, for a denominator, and reduce to its lowest terms; then
raise the numerator of this fraction to the required power, and annex as many
naughts as would equal the number of naughts in the 100 or 1000 used, multiplied
by the exponent of the power; then divide by the denominator of the reduced
fraction, as many times as there are units in the exponent. Thus, to find the cube
of 375, we have º
# = 4; then, 15° = 3375000000 + 4 cubed, = Ans. 52734375.
Or,
†† = }; then, 3% = 27000000000 -- 8 cubed, = Ans. 52734375.
EXAMPLES.
2. Find the square of 875. Ans. 765625.
3. Find the cube of 625. Ans. 244140625.
4. Find the 4th power of 375. AnS. 1977539.0625.
5. Find the 5th power of 225. Ans. 576650390625.
6. Find the 6th power of 125. AnS. 381469.7265625.
7. Find the 7th power of 75. Ans. 13348388671875.
8. Find the 8th power of 25. Ans. 1525878.90625.
9. Find the 6th power of 163. Ans. 214334.70%.
10. Find the 5th power of 33}. Ans. 41152263;ºr.
11. Find the 4th power of 6663. Ans. 197530864.1974+.
12. Find the product of the cube of 663 by the square of 33333.
Ans. 3292.181069958###.
volution.
A-a-/-/*********A*/-/-/-/=NO
=N
1427. Evolution, or the extraction of roots, is the reverse of Involution, and
is the process of finding the equal factors, or roots, of numbers or quantities.
A root of a number, as illustrated in involution, is a factor which, multiplied
by itself a certain number of times, will produce that number.
The name or degree of the root is correlative with that of the power. Thus,
the Square root is the correlative term of the square or second power, and the cube
root that of the third power or cube.
The root to be extracted is indicated by placing the radical sign (V) before
the number or power, with the index of the root over the sign. Thus, VI25 denotes
the cube root of 125. The index of the square root is generally omitted, the radical
sign itself always indicating the square root. The root is also sometimes indicated
by means of a fractional exponent; thus, 36% denotes the square root of 36; and
64* indicates the cube root of the square of 64, or the square of the cube root of 64.
A number that is a perfect power is called a rational number, and its root a
rational root ; thus 27, being a perfect cube, is rational. A number that is not a
perfect power is called an irrational number or surd, and its root an irrational or
surd root; thus 24, being an imperfect power, is a surd, and also its root. A num-
ber, however, may be rational to one degree and a surd to another. Thus 27 is a
rational cube, but an irrational square.
All the rational roots of whole numbers are whole numbers.
All the powers of fractional numbers are fractional numbers.
All prime numbers, except, 1, have no rational roots.
All composite numbers that have rational roots, must have all the exponents
of their prime factors divisible by the exponent of the required root. Thus, ‘VºI6
= V35 x 23 = VG = 6.
The number of rational roots is comparatively small.
The extraction of roots higher than the square and cube is not usually given
in arithmetics, though roots of any degree may be easily found by an extension of
the method given in this work; but all higher roots are more readily obtained by
the use of logarithms.
EXTRACTION OF THE SQUARE ROOT.
1428. The extraction of the square root of a number is the process of finding
one of its two equal factors.
The following numbers in the first line are the square roots of those in the
second.
Roots: 1
1
3, 4, 5, 6, 7, 8, 9, 10, 100.
Squares: 9
16, 25, 36, 49, 64, 8i, 100, 10,000.
7
2,
4
7 7 7
(885)
886 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. #.
We here see that the square root of a number less than 100, is less than 10;
the square root of a number less than 10000, is less than 100, etc. Therefore the
square of any single figure will always be found in the first two right hand places
of the power, and the square of a number of two figures, in the first four places of
the power, every two figures in the square giving one figure in the root. For this
reason, in the extraction of the square root, we separate the numbers into periods of
two figures each, from the right, though the left hand period will sometimes contain
but one figure, when its root is less than 4.
To investigate the principles for the extraction of the square root, let us take
a number of two places of figures as 54, and represent the digit in the tens' place
by a (5), and the digit in the units' place by u (4); then the number 54 or (50 + 4)
will be represented by 10a; + u. We assume this binomial for the basis of our
operations, instead of the number itself, because first, principles can be more clearly
demonstrated by generalizing their formula than by using ordinary figures, as these
special figures would serve to obscure the exhibition of general principles; and we
prefer the above form of the binomial (10a; + w) to the one generally used, viz., (a,
+ w), because the relative position of these factors, aſ and u, in the expansion, is
more easily perceived, and the reason for certain operations, and the location of
their results in the extraction of roots, can be more readily explained.
EXPANSION,
10a; -- u Explanation.—We here square 10a: + w, i. e.,
10a; -- w multiply it by itself. The process is so plain
*=== that no explanation is necessary. We find the
10xu w” result to be 1002.” -- 20xu + w”, or in common
100x2 + 10 cu. language would be read thus: One hundred
* times the square of the tens, plus twenty times
100.c” -- 20aw —- wº the tens multiplied by the units, plus the square
of the units.
Let us now arrange the formula in a vertical column on the left, then substitute the digits
of the number 54 for their representatives aſ and
100ac2 = 5 × 5 × 100 = 2500 or 25 w, and perform the operations indicated, and we
20acw = 5 × 4 × 20 = 400 or 40 have for the several terms these partial results,
w” = 4 × 4 = 16 or 16 viz.: For 100x2 = 2500; for 20xu = 400; for wº
= 16; and their sum = 2916 for the square of
100.cº. + 20xu + w” = 2916 = 2916 54. Also observe that the coefficient of ac” is
100, and contains two ciphers; the coefficient of
acu is 20, and contains one cipher less than the first, and w? has no cipher, or one less than the last.
By virtue of this constant arrangement, we can cancel or discard the ciphers altogether, and
arrange the results as in the last column, where we see that after setting down the value of 2* =
25, we place the next term 2a:w = 40, one place to the right, and wº = 16, one place further to the
right. This arrangement gives the ordinary binomial square of a -- w = z*-H 2xu + w”, and in
this form we will use it for convenience after the first illustrations.
Let us now find the square root of 2916, from the above formula, by resolving it into its
factors.
EVOLUTION, 887
THEORETICAL OPERATION.
100a:* -i- 20xu + w” = 29.16 ( 5 4
Q. -- ?!,
1003:2 25 00 +
20a = 100 ) 4 16
Q!, E 4
X -*-
w = 4 × 104
20aw -- w? = 4 16
0
PRACTICAL OPERATION.
29.16 ( 54
25
104 ) 4 16
4 16
*-*.
0
Explanation.—Since evolution is, the reverse
of involution, we will commence with the term
on the left, 100a:”; we first divide by the 100, by,
pointing off from the right, two places; now
we know that a, the square root of acº, must be
found wholly in the figures to the left of the
point, viz., 29, the greatest perfect square of
which is 25, whose square root is 5; we place
this to the right like a quotient in division, and
designate it by a ; we then subtract its square
25 from 29, leaving a remainder 4, to which we
annex the second period 16. This new dividend
416, we see is what remains after subtracting
the value of 100.cº from 2916, and hence the 416
is equal to the (20ru + w”); now, as we do not
yet know the value of u, we will eliminate wº
from both terms of 20xu + w”, which gives us
(202 -- wy w = 416, which we arrange vertically
as in the operation; and as we do know...the
value of 20x = 20 x 5 = 100, and that it is
many times greater than w, we can consider it
as a trial divisor, and say, 20a or 100 divided
into 416 gives approximately a quotient 4 = w,
which we place in the root. Having now found w = 4, we add it to 100, making 104 = 202 + u, for
the true divisor; we now multiply this by w = 4, and obtain 20a;u + w” = 416; subtracting this
from the dividend, and finding no remainder, completes the work, and gives the exact root, 54 = x:
–H u, or rather 10a; -- w.
In the practical operation we omit the two naughts of the 100, and find the first figure of the
root as before, subtract its square, and bring down the next period for a new dividend. We then
omit the naught of the 20, and use only the 2 to multiply into the tens’ figure of the root 5, thus, 5
× 2 = 10, for a trial divisor; to compensate for the elision of the divisor, we cut off the right
hand figure of the dividend before dividing to find the units’ figure of the root, or w, that is, we
say 10 into 41 = 4; we then annea, this 4 to the 10, making 104 for the true divisor, which being
multiplied by 4, and the product subtracted from the dividend, leaves no remainder; we then
place the 4 in the root, which completes the work.
The philosophy of the theoretical process should be thoroughly understood,
and then the reason for the contractions in the practical method will be apparent.
Since all numbers above units may be considered as composed of tens and
units, thus 456 may be called 45 tens and 6 units, therefore when there are several
figures in the required root we may, after having found the first two left hand
figures of the root, consider them as tens, and the next figure to be found as the
units' figure, and use the same formula and process for finding the third figure as
we did for the second; and the same principle and application may be extended
indefinitely to any number of places, so that, by the same kind of operation, the
square root of any number may be found, or approximated, as illustrated in the
following example:
2. Find the square root of 56475225.
OPERATION,
Root.
56'47.5225 ( 7515
49
7 × 2 = 145 ) 747
5 725
75 × 2 = 1501 ) 2252
1 1501
751 × 2 = 15025 ) 75.125
75.125
0
Ea:planation.—We first point off into periods
of two figures each from the right, them find the
greatest perfect square in the first left hand
period, 56, to be 49, the square root of which is
7, which we place to the right for the first
figure of the root, and its square, 49, we subtract
from 56; to the remainder 7 we bring down or
annex the second period, 47, giving us 747 for
the first dividend; we then multiply the figure
in the root 7, by 2 = 14 for the first trial divisor,
and ascertain how often it is contained in the
dividend exclusive of the right hand figure,
i. e., in 74, (for the reason explained in the
previous example, having multiplied the root
'by 2 instead of 20); we find the nearest quotient
888 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. 3.
to be 5, which we place in the root, and also at the right of the trial divisor 14, giving 145 for the
true divisor, which we multiply by the 5, and subtract the product 725 from the dividend 747,
leaving a remainder of 22, to which we annex the third period, 52, giving 2252 for the second
ºlividend; we then multiply the figures in the root, 75 by 2, or add the last figure of the root, 5, to
the last divisor, 145, giving 150 for the second trial divisor, which we find is contained in the new
dividend exclusive of the right hand figure (225), 1 time; we place this figure 1 in the root, and
also to the right of the trial divisor 150, giving 1501 for a true divisor, which we multiply by the
l, and subtract the product from the dividend; to the remainder, 751, we annex the fourth period,
25, giving 75.125 for a new dividend, then double the figures of the root, 751, or add the last root
figure, 1, to the last divisor, 1501, giving us 1502 for a new trial divisor, and operating as before
explained, we find the next root figure to be 5, which we place in the root and at the right of the
*: then after multiplying and subtracting, as there is no remainder, we have the exact root
o,
If there was a remainder, we could annexperiods of ciphers, and continue the operations in
the same manner until sufficiently exact, each period of ciphers annexed giving one decimal place
in the root. Whenever the significant figures of the given number do not give an exact root, the
number is a surd, and the root Indeterminate, and cannot be exactly expressed by any combination
of fractions, either vulgar or decimal, but admits of an infinitesimal approximation by decimals.
When the product of any divisor by its root figure is greater than its corresponding dividend,
We must diminish the last figure of the root and divisor.
}. When the dividend, exclusive of its right hand figure, is less than the trial divisor, a cipher
must be placed in the root, and also at the right of the divisor, then bring down the next period to
the right of the dividend, and proceed as before.
When the given number is a decimal, or a whole number and decimal, we point off the
decimal part into periods, towards the right from the decimal point, annexing a cipher if necessary
to make the number of places even.
When the given number is a fraction, reduce it to its lowest terms; if it is a mixed number,
change it to an improper fraction; then if both terms of the fraction are perfect powers, extract
the root of each; if one or both terms are surds, multiply them together and extract the root of
the product to a sufficient approximation, then divide the result by the denominator; or reduce
the fraction to a decimal and extract the root.
Thus, the square root of #4, or Vă = vº. = 3, Ans,
And the square root of 124 = Vºſ = } = 34, Ans.
And the square root of g = V5 × 8 = V40 = 6.3245 +, divided by 8 = .7905
+, Ans.
Or, Vă = v.6250 = .7905 +, Ans.
EXAMPLES.
1. What is the square root of 13778944? Ans. 3712.
2. What is the square root of 2106.81? Ans. 45.9.
3. What is the square root of .00007225? Ans. .0085.
4. What is the square root of 7 ºr ? Ans. 23.
5. What is the square root of 4257+?, ? Ans. 654.
6. What is the value of 59049? º Ans. 243.
7. What is the value of V813 - Ans. 729.
8. What is the square root of 781260581179521? Ans. 27951039.
APPLICATIONS OF SQUARE ROOT.
1429. 1. What is the length of the side of a square field, in yards, that
contains 40 acres? Ans. 440 yards.
2. What is the length and width of a rectangular field, in yards, that is 24
times as long as wide, and contains 25 acres? Ans. 550 yards long.
220 yards wide.
º: EVOLUTION. 889
SOLUTION.
25 acres = 121000 square yards; then 121000 – 24 = 48400 = square of shortest side; then
v.484.00 = 220 yards, shortest side.
It is demonstrated in geometry that the square of the hypotenuse of a right-angled triangle
is equal to the sum of the squares of the other
C two sides. In the annexed diagram we see that
the square of the hypotenuse ac, is equal to the
sum of the squares of the base ab and perpen-
dicular be, and hence when any two of the sides
are given, the third can be found; thus, when
the base and perpendicular are given, we find
the square root of the sum of their squares for
the hypotenuse; and when the hypotenuse and
|
Base. B one side are given, we find the square root of
the difference of their squares for the other
ac” = ab? -- bcº side.
i. e., 5* = 4” + 32
Or, 25 = 16 -– 9
We also observe that the areas of all similar triangles, with the same base, have the same
ratio to each other as their altitudes, and
All similar triangles have their areas proportional to the squares of their corresponding
sides.
3. A wall 40 feet high stands on the side of a moat 30 feet wide; find the
length of a ladder that will reach from the opposite side of the moat to the top of
the Wall ? Ans. 50 feet.
4. In the center of a garden 400 feet square, there is a pole 100 feet high.
What is the distance from each corner of the garden to the foot of the pole, and
also to the top of the pole?
Ans. Distance from each corner to foot of pole, 282.84+ feet.
Distance from each corner to top of pole, 300 feet.
SOLUTION.
# of 400 = 200; then, V2002 × 2 = V/80000 = 282.84+
And V10O2. T80000 = 300.
5. A square plot of ground contains 24 acres; what will be the area, the
length of sides, and the diagonal, in yards, of another square field whose sides are
ten times as great 3 Ans. Area, 250 acres.
Length of side, 1100 yards.
Length of diagonal, 1555.63+, yards.
SOLUTION.
2} acres × 10% = 250 acres.
Then, V250×4840 = 1100 yards on each side.
And V2 × 1100° = 1555.63+, yards diagonal.
6. A room is 18 feet wide, 24 feet long, and 124 feet high ; what is the length
of the diagonal drawn from one of the lower corners to the upper corner farthest
from it # , Ans. 32% feet.
SOLUTION.
V18” -- 24?–E (12})* = 324.

890 soul E's PHILOSOPHIC PRACTICAL MATHEMATICs. *
7. Two ships sail from the same port; one due east at the rate of 300 miles
per day, the other due south at the rate of 125 miles per day. After a certain time
they were 1625 miles apart; allowing the ocean to be a plane, how many days were
they out 3 Ans. 5 days.
OPERATION INDICATED,
300? – 90000
1252 – 15625
V105625 = 325 miles apart the 1st day.
Then, 1625 – 325 = 5 days.
8. A rectangular field is 23 times longer than it is wide, and the length of
its diagonal is 520 rods. How many acres does it contain } AnS. 600 acres.
SOLUTION.
V1°-F (2})* = 23, proportional diagonal.
And 520 – 23 = 200, ratio of diagonals.
Then, 1 rod x 200 = 200 rods wide.
And 2% rods X 200 = 480 rods long.
480 × 200
Then ** = 600 acres.
9. A ladder 15 feet long, with its foot a certain distance from a wall, just
reaches the top of the wall; another ladder 20 feet long, will reach the top of the
wall when its foot is 7 feet farther from the wall than the first. What is the height
of the wall, and the distance of the foot of each ladder from the wall?
Ans. Shortest base, 9 feet.
Longest base, 16 feet.
Height of wall, 12 feet.
solutios.
20* — (15* + 7°) = 126 -- (2 × 7) = 9 feet shortest base.
The other conditions are now easily found.
The above solution depends on the following principle, viz, : The difference
between the squares of the hypotenuses of two right-angled triangles, whose
altitudes are the same, diminished by the square of the difference of their bases, is
equal to-twice the rectangle (or product) of the difference of the bases by the shorter
side.
10. A pole 10 feet from a wall just reaches the top; when removed 10.8 feet
farther from the wall, the upper end descends 8.4 feet. What is the length of the
pole and height of the wall? Ans. Pole, 26 feet; Wall, 24 feet.
SOLUTION.
(10 + 10.8)* — (10)” -- (8.4)” -- (2 × 8.4) = 15.6
Then, 15.6 –H-8.4 = 24 feet, height of wall.
Or by principle of next example, thus:
(20.8 + 10) × 10.8 = 332.64 - 8.4 = 39.6
39.6 –- 8.4
Then, **** = 24 feet, height of wall.
ºr'.' . CUBE ROOT. 891.
11. The foot of a ladder placed 7 feet from the middle of a street, the upper
end will reach a point on a wall on the nearer side, 60 feet from the ground, and if
turned to the other side, will reach a point 52 feet from the ground. What is the
length of the ladder, the width of the street, and the distance of the foot of the
ladder from each side of the street 2
Ans. Length of ladder, 65 feet.
Width of street, 64 feet.
Distance of ladder from nearer wall, 25 feet.
Distance of ladder from farther wall, 39 feet.
SOLUTION.
60 + 52 = 112 × 8 = 896 -- (2 × 7) or 14 = 64 feet, width of street; then, 64 -- 2 = 32 feet, half
the width of the street; and 32 — 7 = 25 feet, distance to nearest wall; and 32 + 7 = 39
feet, distance to the farther waii. -
Then, W60?-E25% or V52* T393 = 65 feet, length of ladder.
The above solution depends on the following principle, viz.: In two right-
angled triangles with the same hypotenuse, the difference of the squares of two
similar sides, is equal to the difference of the squares of the other two similar sides,
hence 60° — 52* = 896 is also equal to the difference of the squares of the two
portions of the width of the street; but the difference of the squares of two num-
bers is equal to the sum of the two numbers multiplied by the difference, hence 60
+ 52 = 112 × 8 = 896; and since the difference of the two portions of the street
is equal to twice the distance of the ladder from the centre, or 2 × 7 = 14, and as
the sum of the two portions of the street multiplied by their difference is, by the
above principle, equal to 896, if we divide 896 by 14, the quotient, 64 feet, will be
the width of the street; the remaining part of the operation is self-evident.
CUBE ROOT.
1434. Since the cube of a number is the continued product of the number
taken three times as a factor, or raised to the third power, as explained in involu-
tion, the cube root of a number is the converse of the cube, or one of the three
equal factors of the given number.
The following comparison of the digits, etc., and their cubes, will illustrate
the relation between numbers in general and their third powers, or cubes.
The cubes of the nine digits should be committed to memory.
Numbers, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 99.
Cubes, 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 970299.
By comparing the above, we observe that the cube of the smallest number of
a single figure (1), contains one figure (1), and the cube of the largest single figure
(9) contains three figures (729), or two more than the smallest; also the cube of 10
= 1000, and the cube of 99 = 970299, or two places more than the 10. Therefore
we see that the cube of a single figure will contain from One to three places; the
cube of a number of two figures will contain from four to six places; and in
892 soul.E's PHILOSOPHIC PRACTICAL MATHEMATICs. * .
general, the number of places in the cube will be at most three times the given
number of places, and at least three times the given number, less two; and con-
versely, we may deduce the general principle that a number, or cube, consisting of
one to three figures, will give one figure in its root; when of four to six figures, its
root will contain two figures; from seven to nine figures, gives three figures in the
root, etc. This is the reason why, in extracting the cube root, we point off the
number into periods of three figures each, from the right, each period giving one
place in the root. The left hand period may contain one, two, or three figures, and
its root is always found by inspection, or from the tabular cubes of the nine digits
given above.
To demonstrate the formation of a cube, let us take a number of two places
of figures, as 54, and represent them, as in the elucidation of the square root, by
10a: + u, respectively. Then, raising this to the third power by simple multiplica-
tion, we obtain the following expansion, i. e., (10a: + u) * = 1000a:* + 300a:*u +
30aw” + w”. From this quantity we see that the third power of a is taken 1000
times, and hence it is necessary, before finding the value of a, to divide by 1000,
because the third power of 10a; has no representation in the first three right hand
figures of the entire cube of 10a: + u, but lies wholly to the left. From the above
formula we see that the cube of any number consisting of tens and units, is equal
to 1000 times the cube of the tens, plus 300 times the square of the tens multiplied
by the units, plus 30 times the tens multiplied by the square of the units, plus the
cube of the units.
We will now expand the number 54 to the third power, by the formula, and
for convenience will arrange the terms in columns, to correspond with the numerical
denominations,
Thus, (10x + w)* = (50 – 4)*
Or, 1000a:8 = 5 × 5 × 5 × 1000 = 125000
300acºw – 5 × 5 × 4 × 300 = 30000
30acu” = 5 × 4 × 4 × 30 = 2400
w° = 4 × 4 × 4 == 64
1000x3 + 300ac”w – 30aw” -- w8 = 157464
We here make a = 5 in ten’s place, and u = 4 in unit's place; then perform-
ing the operations indicated in the formula, we obtain the several results corres-
ponding to the equivalent terms of the formula. We see from these partial results
that the third power of the tens (5) has been multiplied by 1000, giving 125000;
therefore in extracting the cube root, we must first divide this number by 1000
before proceeding to resolve the 125 into its three equal factors; the first root figure
or factor must always be found by inspection or trial, and in this case we find it to be
5 tens, the 5 being the cube root of 125, and the tens the cube of the 1000, which
was rejected by division, or pointing off. If our number was just 125000, then 5
tens or 50 would be the required root, but as the whole number is 157464, we know
there must be a unit figure to produce the other partial results, and with the 125000
make up the total 157464, whose cube root is required.
* CUBE ROOT. - 893
To find the unit's figure, let us resume our analysis by taking the entire
number to operate on in regular order, instead of the partial results, thus:
157464 ( 54
1000a:3 – 53 125
300ac2 = 52 × 300 – 7:500 ) 32464
} 30a:w = 5 × 4 × 30 = 600
w” = 4 × 4 = 16
X *sº
w or 4 × 8116 = 32464
We first place the formula on the left, observing to eliminate u from the three
last terms, as Indicated by the brace, and place the w at the bottom, as a final
multiplier when found; we then point off the number into periods of three figures,
as before explained, and find by trial the largest cube in the left hand period to be
125, the root of which is 5 tens; we place this root figure to the right, as a quotient
in division, and its cube 125 under the first period, and after subtracting, annex the
second period. This constitutes the first dividend, and from the formula we see
that this amount 32464, contains 300a:*u + 30aw” + w8, but as we do not yet know
the unit figure (w), we will eliminate, or take out, (w) from each term, and obtain
(300a:* + 30aw -- wº) u = 32464. Now if we knew the value of the quantities con-
tained in the parenthesis, we could find u by dividing 32464 by that value. We do
know, however, the value of the first term 300a,”, which is 7500, and from a com-
parison of the several terms, we see that the first term 300a,” is many times greater
than all the other terms; we will therefore use this value, 7500, for a trial divisor,
and find that it is contained in the dividend, 4 times, with a remainder which we
ignore; we place the 4 in the root, and then perform the operations indicated by
the remaining terms in the parenthesis or brace (30aw -- u%), that is, we add to the
trial divisor, 7500, 30 times the ten's figure 5 multiplied by the unit's figure 4, or 5
x 4 x 30 = 600, and also the square of the unit's figure 4, or 4 x 4 = 16; the sum
of these, 8116, is the complete divisor, and this multiplied by the last root figure, 4,
and the product extended and subtracted from the dividend, leaves no remainder;
therefore 54 is the exact cube root.
2. Find the cube root of 2797.26264.
279,726'264 ( 654
216
1000.03 – 63 -
300ac2 = 62 × 300 = 1 ( 10800 ) 63726 1st dividend.
30a:w - 6 × 5 × 30 = 2 900
X *=s* * <!
w or 5 × 11725 = 58625
2d trial div. 65° X 300 = 1267500 ) 5101264. 2d dividend.
65 × 4 × 30 = 7800
42 - 16
2d complete divisor, 1275316 – 5101264
0 remainder,
After pointing off into periods, we find the greatest cube in the first period
to be 216, the cube root of which is 6; we place this to the right for the first figure
894 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. jºr
of the root, and subtract its cube 216, from the first period 279, and to the remainder
03 annex the second period, making 63726 for the first dividend; then, according to
the formula, we take 300 times the square of the root, 6, or 36 x 300, giving 10800
for the first trial divisor, and then find that this is contained in the dividend 5
times; we place the 5 for the second figure of the root, and then to complete the
divisor, we add to the trial divisor 30 times the product of the previous root figure
by the last, or 6 × 5 × 30 = 900, and also the square of the last root figure 5 =
25, the sum of all giving 11725 for the first complete divisor, which we multiply by
the last figure of the root, 5, and subtract the product 58625 from the first dividend,
and to the remainder bring down the next period for the second dividend.
We then take 300 times the square of all the figures in the root, or 65° x 300
= 1267500, for the second trial divisor, which we find is contained in the dividend
4 times; we place the 4 in the root, and then proceed to complete the divisor as
before, that is, we add to the trial divisor 30 times the product of the previous
figures of the root, 65, by the last, 4, or 65 × 4 × 30 = 7800, and the square of 4
= 16; the sum of these, 1275316, gives the 2d complete divisor, which we multiply
by 4 and subtract the product from the dividend, and as there is no remainder, 654
is the exact cube root. º
If there was a remainder, we would annex periods of three ciphers each, and
continue the operation in the same manner, until sufficiently exact, each period of
ciphers giving one decimal place in the root. *
NoTE.—The following new and original methods of finding the trial divisors, after the first,
embrace a beautiful principle, applicable to roots of every degree. Thus, multiply the numbers
connected by the brackets on the left, by their respective positions, counting downwards, as 1st,
2d, 3d, as indicated by the small figures to the left, and their sum with two ciphers annexed gives
the next trial divisor, which is written below. . This is most easily done by multiplying, first the
right hand figure of the 3d line by 3, and the significant figures if there be any, above it, by 2 and
1 respectively, and set their sum below the last complete divisor, then multiply, and add the
successive left hand columns in the same manner. Or we may find the trial divisor by adding the
three numbers as shown connected by the right hand bracket, observing to multiply by 2, or
double the figures of the middle line or square of units, in adding, and annex two ciphers to the
result, in either case, for the trial divisor.
3. What is the cube root of 58050510848?
OPERATION.
Power = 58 050'510848 (3872 root.
33 - 27
1st trial divisor, 3° X 3 – 27 ) 31050 = 1st dividend.
3 × 3.x 3 = 72 )
8 = 64!/2
1st complete divisor, aſsiſ – 27872
2d trial divisor, 4332 ) 3178510 = 2d dividend.
38 × 7 × 3 = 798
72 – 49 .2
2d complete divisor, 44.1229 || = 3088603
3d trial divisor, 449307 ) 89907848 = 3d dividend.
387 × 2 × 3 = 2322 )
22 - #.
3d complete divisor, 44953924 j = 89.907848
0 = remainder.
& - CUBE ROOT. * 895
In this illustration we dispense with all unnecessary figures, by setting the
numbers that form the complete divisors one place successively to the right, because
we omit the ciphers in the multipliers 300 and 30, and in dividing by the trial
divisors, observing to exclude the two right hand figures of the dividend to find the
next root figure.
To find the trial divisors, after the first, we add the last complete divisor and
the two numbers immediately above, as shown by the brace on the right, observing
to double the figures of the middle number in adding; or the first method mentioned
in the note to last example, is the most scientific.
If the number is a mixed or pure decimal, we point off as in square root, only
in periods of three figures each, to the left of the decimal point for the integers, and
to the right for the decimals, annexing ciphers if necessary to complete the last
period.
When we wish to limit the decimal places of approximation in the root to a
certain number, we may extract half, or one more than half, the required number of
root figures by the regular method, and then continue to divide the last remainder
by the last complete divisor, until the required number of root figures is obtained.
This last process may be further abbreviated by cutting off, or canceling a figure
from the right of the divisor at each consecutive division, no ciphers being annexed
to the remainders; the last decimal root figure, by this process, may be in error.
When the divisor is very large, we need use of it only one more than the number of
root figures yet to be found, and also cut off from the dividend one less than from
the divisor, observing in multiplying by the root figure to carry for the figure cut
Off.
In the following example, we illustrate the above processes, and also present
another modified method of operating for finding the trial and complete divisors.
Thus, after fiading the first trial divisor by taking 3 times the square of the first
figure of the root, and obtaining the second figure of the root, we then take 3 times
the first root figure, and to the product annex the second root figure, multiply the
result by the second root figure and set the product two places to the right of the
trial divisor, and by adding we obtain the first complete divisor.
Then to find the 2d trial divisor, we add together, as shown by the braces on
the right, the last complete divisor, the number just above, and the square of the
last root figure, which may be written below or added mentally; then, after finding
the 3d root figure, we take 3 times the preceding figures of the root and annex the
last root figure, and multiply the result by the last root figure, and extend the prod-
uct two places to the right of the trial divisor and add for the second complete
divisor, and proceed in the same manner as far as requisite. This is perhaps the
most concise method of operations, when the general theory is first well understood.
SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS.
4. Find the cube root of 166.9545 to eight places of decimals.
OPERATION.
'166.954 500 ( 5.50637828+
53 - 125
52 × 3 = 75 ) 41954
155 × 5 = 775
1st complete divisor, 8275 ſ = 41375
5? – 25 *-
2d trial divisor, 90.7500 57.9500.000
16506 × 6 = 9.9036
2d complete divisor, 90849036 | = 545094216
6* = §§)
3d trial divisor, 90948108 34.0578 4000
165183 × 3 = 495.549
* – 2728591 9047
711986|4953
636671
75315
72762
2553
1819
734
727
7 = remainder.
When any product is greater than the dividend above it, diminish the last
figure of the root.
When any trial divisor is greater than its corresponding dividend, place a
To find the cube root of common fractions, we may
cipher in the root, and two ciphers to the right of the trial divisor, then bring down
the next period to the right of the dividend and continue as before.
1st, Extract the cube root of each term, if both are rational; or,
2d, If surds, multiply the numerator by the square of the denominator, then
extract the cube root of this product to any approximation, and divide the result by
the given denominator; or,
3d, Reduce the fraction to a decimal and extract the cube root.
EXAMPLES AND APPLICATIONS.
What is the cube root of 618470208%
What is the cube root of 3484156096%
What is the cube root of 19.617462873?
Find the value of V72 to nine decimal places.
Find the value of V3744.
What is the cube root Of # t
:
Ans. 852.
Ans. 1516.
Ans. 2.697.
Ans, 4,160167646.
Ans. 196.
Ans. #.
º: CUBE ROOT. 897.
7. What is the cube root of ºf Ans. }.
8. What is the cube root of 3 to five decimal places? Ans. .87358.
9. What is the cube root of # to six decimal places? Ans. .629961--.
10. How many quarter-inch solid blocks are equal in volume to one solid
inch; also to one solid foot? Ans. 64, and 110592.
11. A box is 12 feet long, 103 feet wide, and 4 feet deep; what is the side of
a cubical box of equal capacity ? Ans. 8 feet.
12. A vat holds 6776 hogsheads; if it is twice as wide as deep and twice as
long as wide, what are its dimensions?
Ans. 19 ft. 3 in. deep; 38 ft. 6 in. wide; and 77 ft. long.
SOLUTION.
Extract the cube root of (6776 × 63 × 231) divided by (1 × 2 × 4); and multiply the result
by the proportional parts, 1, 2, and 4, for the depth, width and length. Thus,
231 in. deep.
462 in. wide.
9
*/576 A 68 x 281 = 231 : * =
× 4 = 924 in. long.
1 X 2 x 4
Geometry demonstrates that,
Similar solids have the same ratio to each other as the cubes of their like
dimensions, or as the cubes of the ratio of their like dimensions; and
The like dimensions of similar solids have the same ratio as the cube roots of
their solidities, or as the cube root of the ratio of their solidities.
13. A block of marble is 12 inches long, and contains 576 cubic inches; what
are the contents of a similar block 18 inches long? Ans. 1944 cubic inches.
14. The solidities of two similar blocks are 162 and 2058 cubic inches; the
length of the first is 9 inches, what is the length of the second Also, what are
the other dimensions of each, whose ratios are as 1, 2, 3?
Ans. 21 in., length of second block.
First block, 6 in. wide and 3 in. thick.
Second block, 14 in. wide and 7 in. thick.
SOLUTION TO FIRST QUESTION.
*# = 12}}, ratio of solidities; then.
3 -— 3 -—
VI2; = Vā; - , = 24, ratio of lengths; then, 9 × 2 = 21 in.
The solution of the second question is the same as in example 12 or by direct
proportion from the lengths.
15. A rectangular reservoir is 25 feet deep; the ratio of the other two sides
is as 3 to 4, and the longest transverse diagonal, or distance from one lower corner
to the opposite upper corner, is 65 feet; what are the other dimensions of the
reservoir; how many gallons does it contain, and what would be the side and
contents of a cubical reservoir of 5 times the capacity ?
Ans. Longest side, 48 feet.
Shortest side, 36 feet.
Contents, 323158% gallons.
Side of cubical reservoir, 60 feet.
Contents of cubical reservoir, 1615792}} gals.
398 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs.
SOLUTION.
V65°-25* = 60 feet, diagonal of sides, or bottom.
And, W3*-F4* = 5, diagonal ratio; then, 60 - 5 = 12, multiple of ratio of sides.
Then, 3 x 12 = 36, width; and 4 × 12 = 48, length. e
The other conditions are now easily found, thus:
v/48 x 36 S&T25 × 5 = 60 feet, side of cubical reservoir. .
And, (48 × 36 × 25 × 1728) — 231 = 323158% gals, in first reservoir,
16. A man has two rectangular plots of ground of equal extent; the first is
twice as long as the second, and the second is twice as wide as the first, and the
Squares of their diagonals are 59189 yards and 17141 yards respectively. He
exchanges the two plots for a square garden of equal area to the two plots. He
then sells the garden for as many mills per square yard as the garden is yards
square, and receives $1331, thereby gaining 10 per cent on original valuation. What
were the dimensions and contents of each piece of ground; what did the garden
sell for per square yard, and what was the original value of each piece?
Ans. First plot, 242 yards long and 25 yards wide, = 14 acres, cost $605.
Second plot, 121 yards long and 50 yards wide, = 14 acres, cost $605.
Garden, 110 yards square, = 2; acres, cost $1210, or 10 cents per
square yard, and sold for 11 cents per square yard.
SOLUTION.
3 -—
i/$1331,000 = 110 mills; then side of garden = 110 yards, and area of garden = 12100 square
yards, and 12100 - 2 = 6050 square yards, the area of each plot. 110 mills selling price at 10%
gain, gives 100 mills or 10 cents for cost per square yard.
To find the sides of the two plots, we first find the sum and the difference of
the square of the diagonals, and operate as follows:
OPERATION.
5918.9 square of 1st diagonal 59189
17141 square of 2d diagonal 17141
2* + 1* = 5) 76330 sum of squares 2" — 1* = 3) 42048 difference of squares.
-*m-
15266 = sum of squares of shorter 14016 = difference of squares of shorter length
length and width. [and width,
Then to 15266 And from 15266
add 14016 subtract 14016
2) 29282 2 ) 1250
Square of shorter length V11641 V625 square of shorter width.
Extract square root, - 121 shorter length; 25 shorter width.
2 2
242 greater length; 50 greater width.
cube Root. 899.
MISCELLANEOUS EXAMPLES IN SQUARE AND CUBE ROOT.
1435. 1. Two men formed a copartnership; the first invested $3600 for 864
days, and received # of the gains; the second invested as many dollars as he
continued days in business. What was his capital? Ans. $2160.
OPERATION INDICATED.
$3600 x 864 days = 3110400 = # gains and # gains = 4665600.
Then extract square root, for second capital and number of days = V/3600 × 864 × # = 2160.
2. If it take 8 planks to build one panel of fence, and there are just 25.
panels on One side of a square acre, how many miles square must the tract be, that.
Will contain as many acres as it will take planks to fence it? Also, if it cost $1
per acre for the work, how many men must be employed at $1% per day, to do the
Work in as many days as there are men?
Ans. Side of the square tract, 31.622776-H miles.
Number of men required, 600.
SOLUTION.
25 × 8 × 4 = 800 planks for one acre; then, 800° = 640000 acres.
And 640000 – 640 = 1000 square miles.
Then, V1000 = 31.622776+ miles square.
SOLUTION TO SECOND QUESTION.
$640000 - 1% = 360000 days of work.
And V360000 = 600, number of men.
3. A farmer has a rectangular field, containing 30 acres, whose sides are as
3 to 4; what will it cost to dig a ditch 6 feet wide and 4 fºr feet deep diagonally
through it, at 40 cents per cubic yard? Also, how many men will it take to do the
Work, each man working twice as many days as there are men, and receiving twice.
as many cents per day as each works days?
Ans. $640, cost. 20 men. 40 days each. 80 cents per day.
SOLUTION.
__ & 30 acres = 1306800 square feet.
4 2 . 3 × 4 = 12 ratio divisor.
— `
52- | 3 Gives quotient = 108900.

Extract square root = 330 feet, multiple of ratios.
Then, 3 × 330 = 990 feet width of field.
4 × 330 = 1320 feet length of field.
5 × 330 = 1650 feet length of diagonal of field.
Then, 1650 × 6 × 4 ºr = 43200 cubic feet of ditch.
Divided by 27 gives 1600 cubic yards of ditch.
1600 × 40 cents = $640.00 cost of work.
Divide by 1 × 2 × 4 = 8 = 8000.
Extract cube root = 20 multiple of 2d ratios.
Man 1 × 20 = 20 men.
Days 2 × 20 = 40 days.
Cents 4 × 20 = 80 cents per day.
90O soul E's PHILOSOPHIC PRACTICAL MATHEMATICs. º
4. A man bought a square piece of land, paying as many cents per square
rod as the piece was rods square; he then sold it at an advance of 17% per cent,
and gained $880. How many acres in the piece, and how many rods square ? What
was the total cost and selling price, and the cost and selling price per square rod?
Ans. Area, 40 acres,
Or, 80 rods square.
Total cost, $5120.
Total selling price, $6000.
Cost per square rod, 80 cents.
Selling price per square rod, 93% cents.
OPERATION INDICATED.
To find cost.
100 Since the land was a certain number of rods
275 | 16 square and the price paid was as many cents
880 g
per square rod as the price was rods square, the
$5120 cost. cost of the whole in cents was the cube of the
880 gain.
price per square rod.
$ºss-sº
$6000 Sales.
Hence väI3000 = 80 cents.
= 80 rods square.

*ºmºmºmº
I
jy
i
|
rithmetical Progression.
---------------------------------N
1436. Arithmetical Progression, sometimes called Progression by Difference,
or Equidifferent Series, is a series, or succession of numbers or quantities that
continually increases or decreases from term to term, by a common difference.
The numbers constituting a Progression are called the terms of the series, of
which the first and last are denominated the Eartremes, and all the intermediate
terms, the Means.
An Arithmetical Progression is an ascending series, when each successive
term is formed by adding the common difference to the preceding term ; and a
descending series, when the common difference is subtracted consecutively from the
first to the last term ; thus:
1, 3, 5, 7, 9, is an ascending series, common difference + 2
13, 10, 7, 4, 1, is a descending series, common difference — 3
In an arithmetical progression, there are five things or parts to be considered,
any three of which being given, the others can be determined.
These parts are named and represented as follows:
1st. The First Term, represented by 0.
2d. The Last Term, {{ { % l.
3d. The Common Difference, {{ {{ d.
4th. The Number of Terms, 4% {{ 70.
5th. The Sum of the Series, 4% {{ S.
Since any one of these five parts may be found from any three of the four
remaining parts, there will arise just 5 x 4 = 20 cases of problems, producing as
many formulas and resulting solutions.
Only 5 or 6 of these 20 cases are usually treated in arithmetics, the other
cases requiring more elaborate algebraic demonstrations and formulas, and are of
less practical application.
PROBLEM I.
1437. To find the last term, when the first term, common difference, and
number of terms are given.
1. The first term is 2, the common difference 3, and the number of terms
5; what is the last term 3
Illustration. Thus, 2, (2 + 3), (5 + 3), (8 + 3), (11 + 3).
Gives 2, 5, 8, 11, 14.
(901)
902 . SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
Here we see that the common difference, 3, is added as many times, less one,
as there are terms; that is, 5 — 1 = 4 times, which gives 12 to be added to the first
term 2, - 14 for the last term, or thus: 2 + (5 – 1) × 3 = 14.
Hence the general formula, l = a + (n − 1)d.
Or, expressed in common language, the last term is equal to the first term,
plus or minus the product of the number of terms, less one, by the common differ-
CIlC62.
The sign plus or minus, it, is used to show that the quantity to the right, (n
— 1)d, is to be added to the first term when the series is ascending, but to be sub-
tracted from the first term when the series is descending.
2. The first term of an increasing arithmetical progression is 5, the common
difference is 6, and the number of terms 16; what is the last term 2
Ans. 5 + (15 × 6) = 95, last term.
3. The first term of a decreasing arithmetical progression is 450, and the
common difference is 10; what is the 33d term 3
Ans. 450 — (32 × 10) = 130 for the 33d term.
4. If a man's salary is $20 the first month, $30 the second month, and so
on, increasing $10 each month, what will it be the 20th month ? Ans. $210.
5. What will be the amount of $1000 in 25 years, at 6 per cent per annum,
simple interest? Ans. $2500.
SOLUTION.
$106 + (24 × 6) = $250, amount of $100; then $1000 will give $2500.
6. The first term of an ascending series is #, and the common difference is
#; what is the 1000th term 3 Ans. 500+.
7. The first term of a descending series is 1000, the common difference fºr,
and the number of terms 9000; what is the last term 3 Ans. 100.1.
PROBLEM II.
1438. To find the common difference, the extremes and number of terms being
given.
Since the number of common difference is always one less than the number of
terms, and the sum of all the common differences is equal to the difference between
the first and last terms, if we divide the difference of the extremes by the number
of terms less one, the quotient will be the common difference; hence the
Formula, d = # for ascending, and d = #=} for descending.
1. The first term of an arithmetical series is 2, the last term 74, and the
number of terms 25; what is the common difference?
Ans. #F# = # = 3, common difference.
* ARITH METICAL PROGRESSION. 903
2. A man travels 15 days, going 10 miles the first day; after which he
increases the distance an equal number of miles each day; the last day's journey
Was 80 miles; what was the daily increase? Ans. 5 miles.
3. In a company of 101 persons, whose ages are in arithmetical progression?
the youngest is 1 year old, and the oldest is 51 years; what is the common difference
of their ages } Ans. 6 months.
4. The first term of an arithmetical series is 21, the last 3, and the number
of terms 83; what is the common difference? AmS. #.
To insert any number of arithmetical means between two given numbers, is
simply to find the common difference by the above formula, and then form the
series, observing that the whole number of terms is two more than the given number
of means, viz., the two extremes.
5. Insert 5 arithmetical means between 10 and 40.
SOLUTION.
d :- *=} = 5, common difference; then 10, 15, 20, 25, 30, 35, 40 = series.
6. The extremes are 24 and 52, and the whole number of terms is 100; what
is the second term, and the next to the last term 3
Ans. Second term, 3; next to the last term, 51#.
PROBLEM III.
1439. To find the number of terms, when the extremes and common difference
are given.
Since the difference of the extremes, divided by the common difference, is
equal to the number of common differences, then by adding one we obtain the num-
ber of terms; hence the
Formula, n = ** + 1
That is, the number of terms is equal to the difference of the extremes
divided by the common difference, with the quotient increased by 1.
1. The extremes of an arithmetical series are 7 and 127, and the common
difference is 5; what is the number of terms?
SOLUTION,
in- + 1 = 24 + 1 = 25, number of terms, a Ans.
2. How long will it take a man to make a journey, if he travels 12 miles the
first day and increases the distance 3 miles each succeeding day, the last day’s
journey being 72 miles? Ans. 21 days.
3. A person settles his account by paying 25 cents the first day, with 10
cents additional each succeeding day; the last day's payment was $36,65; how
many days was he in paying? AnS. 365 days.
904 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. Yºr
PROBLEM IV.
1440. To find the sum of all the terms, or series, when the extremes and num-
ber of terms are given.
If we take any series and write under it the same series in an inverse order,
the sum of each couplet of terms must be equal, and the summation of all the
couplets will be twice the sum of the given series.
Thus, 3 + 5 + 7 -- 9 + 11 = 35, sum of given series.
And 11 + 9 + 7 -- 5 + 3 = 35, sum of given series inverted.
Then, 14 + 14 + 14 + 14 + 14 = 70, = twice the sum of the given series.
Here we see that the sum of the extremes, 14, is equal to the sum of any two
terms that are equally distant from the extremes; therefore, if we multiply the sum
of the extremes, 14, by the number of terms, 5, the product, 70, will be twice the
sum of the series; then, this divided by 2 will give 35 for the sum of the given
series. But we may divide either factor, 14 or 5, by 2, with the same result; hence
the
Formula, s = ** x ºn, or, . × a + l
That is, the sum of the series is equal to half the sum of the extremes multi-
plied by the number of terms; or half the number of terms multiplied into the sum
of the extremes.
1. The extremes of an arithmetical progression are 5 and 165, and the num-
ber of terms 33; what is the sum of the series?
Ans, s = ** x 33 = 85 × 33 = 2805, sum.
2. The extremes are 0 and 560, and the common difference is 7; what is the
sum of the series 3 Ans. 22680.
We first find the number of terms by Problem III, and then the sum of the
Series.
3. If a person pay $50 the first month, and $50 additional each subsequent
month, what amount of debt will be discharged in one year? Ans. $3900.
OPERATION INDICATED.
! = 50 + (11 × 50) = 600.
a 50 + 600# * x 12 = 3900.
4. If 1000 oranges be laid in a line five feet from each other, the first being
five feet from a basket, how far will a person travel who gathers them up singly,
returning with them one by one to the basket? Ans. 947 miles, 1613.4 yards.
SOLUTION.
5 feet X 2 10 feet = 1st term.
5000 feet X 2 = 10000 feet = last term.
10010 feet = sum of extremes.
500 = } number of terms,
Divided by 5280) 1005000 feet = whole distance = 947 miles, 1613+ yards.
ARITHMETICAL PROGRESSION. 905
We append a few additional formulas, without explanations, for some of the
Other simpler cases.
PROBLEM W.
1441. To find the last term, when the first term, number of terms, and sum of
the series are given.
Formula, l = #– (l,
That is, the last term is equal to twice the sum of the series divided by the
number of terms, and the quotient diminished by the first term.
PROBLEM VI.
1442. To find the first term, when the last term, number of terms, and sum of
the series are given.
Formula, a = º — l
That is, the first term is equal to twice the sum of the series divided by the
number of terms, and the quotient diminished by the last term.
PROBLEM VII.
1443. To find the number of terms, when the extremes and sum of the series
are given.
28
a + i
That is, the number of terms is equal to twice the sum of the series divided
by the sum of the extremes.
Formula, n =
PROBLEMI VIII.
1444. To find the sum of the series, when the first term, common difference,
and number of terms are given.
Formula, 8 = : [2a + (n − 1) x d
That is, the sum of the series is equal to the number of terms less one,
multiplied by the common difference, this product added to twice the first term, and
the sum multiplied by half the number of terms.
If the last term were given in Problem VIII, instead of the first, the follow-
ing would be the
Formula, s = ; [2, — (n − 1) x d
Here we subtract the product of the number of terms less one by the common
difference, from twice the last term, and multiply the remainder by half the number
of terms, for the sum of the series.
906 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. º
MISOELLANEOUS EXAMPLES.
NoTE.-Questions in 8imple interest are solved by arithmetical progression.
1445. 1. What will $10000 amount to in 50 years, at 8 per cent simple
interest ? * Ans. $50000.
SOLUTION.
, The amount of $100 at 8 per cent for one year, $108, is the first term, and the amount for 50
years is the last term; the interest for 1 year, $8, is the common difference, and 50 years the num-
ber of terms; hence, to find the last term, we have $108 + (49 × $8) = $500, amount of $100 for
50 years; then by proportion $10000 will amount to $50000.
2. What sum must be loaned at 6 per cent simple interest for 20 years, to
amount to $4400% Ans. $2000.
SOLUTION.
$106 + (19 × $6) = $220, amount of $100; then by proportion $4400 require $2000.
3. At what rate per cent simple interest must $5000 be loaned to amount to
$20000 in 30 years? Ans. 10 per cent.
SOLUTION.
$5050 + (29 × $50) = 6500 — 5000 = 1500, amount of interest at 1 per cent; then $15000 –
1500 = 10 per cent,
4. How long a time must $7000 be loaned at 7 per cent simple interest, to
amount to $10430% - Ans. 7 years.
SOLUTION,
$7490 + (0 × $490) = $7490 – 7000 = $490 interest for 1 year; then $3430 - 490 = 7 years.
5. A heavy body falling unobstructed, moves through a distance of 16; feet
the first second, with an accelerating velocity of 324 feet for each additional second;
what distance will it fall in 31 seconds 3 Ans. 15456 ſº feet.
w SOLUTION.
This question may be solved by Problems I and IV, or directly by Problem VIII, thus:
8 = ** [2 × 16** + (31 — 1) × 32}] = 15456H,
6. A man owes $4510 to 41 different persons, in arithmetical progression; to
the first he owes $10, how much does he owe the last, and what is the common
difference of the debts? Ans. Amount due the last, $210.
- Common difference of debts, $5.
Solution of first question by Problem V, -
l = axlºw – 10 = 210, last term.
7. Two persons start at the same time from two cities, 207 miles apart, to
meet each other; the first travels 10 miles the first day, and 20 miles the last day;
the second travels 12 miles the first day, and 27 miles the last day. In how many
days will they meet; and how far will each travel; and what is the daily increase
of each 3 Ans. 6 days.
First travels 90 miles; with a daily increase of 2 miles.
Second travels 117 miles; with a daily increase of 3 miles.
ARITHMETICAL PROGRESSION. 907
Solution of first question by Problem VII,
First day's travel of both, 10 + 12 = 22, and last day’s, 20 + 27 = 47 n = # = 6 days.
8. A man earns 10 cents the first day, and 10 cents additional each succeed-
ing day, for 30 days; his expenses are $1 a day; when will he be most in debt;
when will the sum of his earnings and expenses be equal; and what will he make
clear at the end of the month ?
Ans. Most in debt, $4.50, on the 9th and 10th.
Sum of earnings and expenses equal on the 19th, $19.
Clear earnings for 30 days, $16.50.
9. Two persons, A. and B. start from the same place at the same time; A.
goes due north, traveling 3% miles the first day, with a regular daily increase
afterward; the last day's journey being 26% miles. B. goes due west, traveling 4;
miles the first day, and 35% miles the last day. In a certain time they are 500 miles
apart, on a straight line; how many days have they traveled; how many miles
each; and what was their respective daily increase?
Ans. 20 days.
A. traveled 300 miles; daily increase, 1% miles.
B. traveled 400 miles; daily increase, 13 miles.
SOLUTION.
V(3})* + (43)* = Vº = ** = 6 miles apart the 1st day.
Then by proportion,
6 : 500::3% : 300 miles, distance A. traveled.
6 : 500 : : 4; : 400 “ & 4 B. {{
Then find the number of days by Problem VII, thus:
2 × 300 or 2 × 400
3} + 26% Aff-F35; = 20 days.


geometrical Progression.
*===mºsºmsºmºmºmº-me
1446. A Geometrical Progression is a series of numbers or quantities,
Whose formation results from the law of a constant multiplier, either integral or
fractional, called the Common Ratio, which bears the same relation to the common
difference of an arithmetical progression that multiplication does to addition. The
Series of a geometrical progression is ascending, when each successive term increases
by a constant factor, that is when the ratio is greater than one ; when the ratio is
less than one, or a proper fraction, the terms will decrease from the first, and the
Series is descending.
Thus, 1, 3, 9, 27, 81, etc., is an ascending series, whose ratio is 3,
And 4, 2, 1, 3, 4, etc., is a descending series, whose ratio is #.
A descending series may be transformed into an ascending series by reversing
the order of its terms, substituting the last for the first, and inverting, the terms
of the fractional ratio. *. -
When a descending series is continued to infinity, that is when the number
of terms is infinite, the last term must be considered equal to zero (0).
In geometrical progression, as in arithmetical, there are five parts or quanti-
ties, any three of which being given, the others can be determined; they are:
1st. The First Term, represented by A.
2d. The Last Term, 46 4 L.
3d. The Common Ratio, - 44 “ R, or r when an index.
4th. The Number of Terms, {{ “ N, or n when an exponent.
5th. The Sum of all the Terms, {{ {{ S.
Since any one of these five parts may be required from any three of the four
remaining parts, there will result by combination, as in arithmetical progression, 5
× 4 = 20 cases of problems, requiring 20 distinct formulas for their solution.
Only a few of the 20 cases in geometrical progression admit of solution by
ordinary arithmetical operations, because the common ratio (R) enters into the
series by continued involution, and its value must be found by the extraction of the
higher roots; and the number of terms (N) is represented in the series as an
exponent, and its value, in most cases, can only be found by logarithmic computa-
tions.
When a geometrical progression terminates, or is limited between two quan-
tities, the first term and the last term are called the extremes, and the intermediate
terms, the geometrical means,
PROBLEM I.
1447. To find the last, or any proposed term, when the first term, common
ratio, and number of terms are given.
(908)
* GEOMETRICAL PROGRESSION. - 96.9
1. If the first term of a geometrical progression is 2, the ratio 3, and the
number of terms 5, what is the last term? *
Illustration. Thus, 2 (2 x 3) (2 x 3°) (2 x 3°) (2 x 4*), etc.
Gives 2 6 18 54 162 last term.
We here see that the second term is equal to the product of the first term, by
the first power of the ratio; the third term is equal to the product of the first term
by the second power of the ratio, and in general any term is equal to the product of
the first term by that power of the ratio whose index is one less than the number
of the required term; hence the
Formula, L = A x Rº-1
That is, the last term is equal to the first term multiplied by the ratio raised
to a power one less than the number of terms.
PROBLEM II.
1448. To find the first term, when the last term, ratio and number of terms
are given.
Since the last term, as shown in Problem I, is equal to the product of the
first term by the ratio raised to a known power, if we divide the last term by that
Known power of the ratio, the quotient will necessarily be the first term; hence the
Formula, A = L -- Rº-" or A = #.
71–1
That is, the first term is equal to the last term divided by that power of the
ratio whose exponent is one less than the number of terms.
2. The first term of an ascending geometrical series is 7, the ratio is 4, and
the number of terms is 6; what is the last term.
SOLUTION.
L = 7 x 4" = 7 × 1024 = 7168, last term.
3. The first term of a descending series is 4, the ratio # and the number of
terms 5; what is the last term 3
SOLUTION.
L = 4 × (#)* = 4 × ºr = ºr, last term.
4. The last term is 26244, the ratio 3, and the number of terms 8; what is
the first term 3
SOLUTION.
A = 26244 -i- 37 = 26244 - 2187 = 12, first term.
5. The last term is #g, the ratio 3, and the number of terms 11; what is the
first term 3
SOLUTION.
A = rºg -- (3)** = gº -- rººz = 12, first term.
9 IO soul E's PHILOSOPHIC PRACTICAL MATHEMATICs. º:
6. Bought 20 acres of land, giving 5 cents for the first acre, 10 cents for
the second, 20 cents for the third, and so on; what did the last acre cost?
Ans. $26214.40.
OPERATION INDICATED,
! = 5 × 219 = $26214.40.
PROBLEM III.
1449. To find thé sum of a series, when the extremes and ratio are given.
If we take a series, as 2, 10, 50, 250, 1250, whose ratio is 5, and multiply each
term by the ratio, we will obtain a new series 5 times greater than the first, as
indicated below.
5, 10, 50, 250, 1250, 6250 = 5 times given series.
2, 5, 10, 50, 250, 1250 = 1 time given series.
– 2 6250
2
5 — 1 = 4) 6248 = 4 times the sum of given series.
1562 = sum of the given series.
We here observe that all the terms of both series are identical, or common to
each, except the first term of the first series and the last term of the second; hence,
if we arrange them as above, so that the terms of the first series will fall under
similar terms of the second, and subtract, we see that the like terms will cancel each
other, and the difference between the first term of the first series and the last term
of the second; but since the second series is 5 times greater than the first, their
difference is 5 — 1, or 4 times greater than the sum of the first series; then, if we
divide this difference by 4 (the ratio less one) the quotient will be the sum of the
given series; hence we deduce the following th
S (L X R)-A
Formulas, for an ascending series.
R – 1
S = *=º for a descending series.
sº That is, in common language for either formula, the sum of the series is
equal to the difference between the product of the last term by the ratio, and the
first term, divided by the difference between the ratio and (1).
When the series is descending, instead of using the second formula, we may
transform it into an ascending series by reversing the order of its terms, and
taking the reciprocal of the ratio. *
1. The extremes of a geometrical progression are 5 and 32805, and the ratio
is 3; what is the sum of the series?
SOLUTION.
32805 × 3 = 98415 – 5 = 984.10 - 2 = 49205, sum of series.
2. The extremes of a descending series are 81 and gºs, and the ratio is #;
what is the sum of the series?
SOLUTION.
81 – (##3 × #) = 81 — rig = 80%}} – } = 121}}}, sum of series.
Or,
81 × 3 = 243 – g = 242}## - 2 = 121}#}, sum of series.
º: GEOMETRICAL PROGRESSION. 9 II
3. How large a debt can be discharged in one year by paying $5 the first
month, and twice as much each subsequent month as the preceding, and what will
be the last payment? Ans. $10240, last month’s payment.
$20475, total debt.
First find the last term by Problem I, and then the sum of the series.
4. If a man travel 5 miles the first hour, and increases his speed 10 per cent
each subsequent hour, how far will he go in 10 hours, and how far the last hour?
Ans. Last hour, 11.789738455 miles.
Whole distance, 79.687123005 miles.
Find the last hour's travel by Problem I, then the whole distance by Problem
III. sº
5. If a rifle ball moves 1000 feet the first second, and diminishes its velocity
10 per cent each subsequent second, how far will it go in 10 seconds, and how far the
last second? Ans. 387.420489 feet, last second.
*- 6513,215599 feet, whole distance.
OPERATION INDICATED,
* L = 1000 × (4%).9 = 387.420489 = last term.
s — 1000 - (387.420489 × #)
Tº = 6513.215599 = sum of series.
PROBLEM IV.
1450. To find the sum of the series, when the first term, number of terms and
ratio are given, without finding the last term.
LR — A
In Problem I we have L = AR”, and in Problem III we have S = #–#;
by substituting the value of L, in Problem I, (AR-1), for L in Problem III, we
(ARn–4 × R) — A
R – 1
obtain, S = which reduced gives the
A*-A or S = *(*~12
Formula, S = #–H R – 1
That is, the sum of the series is equal to the product of the first term by the
ratio raised to the power indicated by the number of terms, minus the first term,
and the result divided by the ratio minus 1. Or, by second formula, the sum is
equal to the known (nth) power of the ratio, less one, multiplied by the first term,
and the product divided by the ratio less one.
1. The first term is 10, the ratio 4, and the number of terms 5; what is the
Sum of the Series?
SOLUTION.
anxº-w OT tox{-p = 3410, sum of series.
2. A person bought 10 acres of land, agreeing to pay 5 cents for the first
9I 2 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. Yºr
acre, 25 cents for the second acre, and so on in a quintuple ratio; what did the
whole cost 3 Ans. $122070.31.
OPERATION INDICATED.
sºme ºmºmºmºmºsºmsºmºsºmºsº º' --mm smºs-smºsºms-tº-sº sº-sº
NoTE.—To find the 10th power of 5, divide the 10th power of 10, or 1 wiih 10 ciphers
annexed, by 4 five times. ge
3. If a man travel 81 miles the first day, and diminish his rate of travel so
that each succeeding day's journey shall be # of the previous day, how far will he
, travel in 7 days?
SOLUTION.
* × [l - (8)'] = 8 × ### = 228 miles.
1 — #
4. If a man earn 1 cent the first day, 2 cents the second, 4 cents the third,
and so on, how much will he earn in 30 days? AnS. $10737418.23.
OPERATION INDICATED.
S = 1x*=1 = $10737418.23.
To find the 30th power of 2, we may cube the 10th power of 2.
Find the second answers to examples 3 and 4, in Problem III, by Problem
IV.
PROBLEM V.
1451. To find the sum of a decreasing geometrical eeries, when the number of
terms is infinite, the first term and ratio being given.
When a descending geometrical progression is carried to infinity, the last
term must be considered as 0; hence if we take the second formula, Problem III,
S = **** and substitute for L, its value 0, there will result for an infinite
series, the
iſ:
A.
1 — R.
Formula, S =
That is, the sum of an infinite series is equal to the first term, divided by
one minus the ratio.
1. What is the sum of the infinite series, 9, 3, 1, #, #, etc.?
SOLUTION.
ºf © GEOMETRICAL PROGRESSION. 9 I 3
2. What is the sum of an infinite series, whose first term is one million, and
ratio ºn 3 Ans. 111111.1%.
If the ratio was #5% Ans. 10000000.
If the ratio was #3 Ans. 11428.57%.
If the ratio was Tºp 3 Ans. 1010101gº.
This process may be used for determining the value of circulating decimals,
because their repetends are in geometrical progression, the common ratios of which
are +º, ++o, Tºgo, etc., according to the number of figures in the repetend; thus, the
circulate .333, etc., gives the series #3 + # + #. etc.; and 232323, etc., gives #
*y
23 23 g - - * . . ;
+ i.; + i., etc., to infinity; hence the value of the circulate .3 is equal to 1 — I'd
= , -- ", = }; and the value of 23 is equal to #5 -- # = #; and as the
operation is uniform for the summation of all circulates to infinity, we have the
principle demonstrated that the value of any circulating decimal is equal to a frac-
tion with the given repetend for a numerator, and as many 9’s for a denominator, as
there are places in the repetend.
3. What is the value of the circulates .54, '003, .6984.12, .42857 it
Ans. #, #3, #4, #.
4. If a cannon ball move 2000 feet the first second of time, and each
succeeding second #5 as far as the preceding, until it comes to a state of rest, how.
far will it move? Ans. 20000 feet.
OPERATION INDICATED.
s = * = 20000.
Po
5. The hour and minute hands of a clock are together at 12 o'clock or noon,
when will they be together again 3 Ans. 1 h. 5 m, 27 ºr Sec. P. M.
This question gives the infinite series, 1 + +3 + H+, etc.
PROBLEMI VI.
1452. To find the ratio, when the extremes and number of terms are given.
In Problem I it is demonstrated that the last term of a geometrical progres-
sion is equal to the first term, multiplied by the ratio raised to a power one less
than the number of terms; therefore if we divide the last term by the first term,
the quotient will be equal to the ratio raised to the n—1th power, and extracting
the m—1th root will give the ratio; hence the
Formula, R = y*—
1. The first term is 6, the last term 750, and the number of terms 4; what is
the ratio #
SOLUTION.
- — 3 —
R = Ví. = V125 = 5, ratio.
The extremes are 4 and 2916, and the number of terms 7, what is the
Ans. 3.
4)
a de
ratio 3
9I4. SouLE's PHILOSOPHIC PRACTICAL MATHEMATICs.
To find the 6th root we may take the cube root of the square root.
- 3. The first term of a decreasing geometrical series is 16, the last term gº,
and the number of terms 10; what is the ratio? AnS. #.
To find the 9th root, extract the cube root of the cube root.
NotE.—When the number of terms is large, the extraction of the higher roots should be
made by logarithms.
The above Problem is used to insert any number of geometrical means, or
mean proportionals, between two given numbers.
EXAMPLES.
1. Insert three geometrical means between 6 and 3750.
Explanation.—Since there are three terms to be inserted between two extremes, there will be
five terms in the series, and we therefore have the extremes and number of terms given to find the
ratio by Problem VI; that is, we divide the last term, 3750, by the first term, 6, giving 625; then,
sir,se the number of terms is 5, we extract the 5–1 or 4th root of 625, which gives 5 for the ratio;
then to complete the series, we multiply the first term by the ratio for the second, and the second
by he ratio for the third, and so on, which gives for this example the series, -
6, 30, 150, 750, 3750.
To find a single mean proportional between two given numbers, we may
extract the square root of the product of the extremes for the mean term.
2. Find and insert a mean proportional betweeli 4 and 324.
Ans. 4, 36, 324, series.
3. Find and insert two mean proportionals between 5 and 320.
Ans. 5, 20, 80, 320, series.
4. Find and insert five geometrical means between 4 and 2916.
Ans. 4, 12, 36, 108, 324, 972, 2916, series.
5. Find eight geometrical means between 3 and 1536.
Ans. 6, 12, 24, 48, 96, 192, 384, 768, means.
To find the 9th root of 512, extract the cube root of the cube root.
6. Insert seven geometrical means between 9 and ###, descending series.
Ans. 9, 6, 4, 23, 1%, 12%, #4, ###, ###, series.
To find the 8th root of ºr, extract the square root three times consecutively.
PROBLEM VII.
1453. To find the ratio, when the extremes and sum of the series are given.
Let us take a series, as 4, 8, 16, 32, 64, whose ratio is 2, and write it twice,
omitting the first term the first time, and the last term the second time, thus:
8, 16, 32, 64 = 120 = sum of series, less the first term.
4, 8, 16, 32 = 60 = sum of series, less the last term.
We see by conparing the two series, that the terms and sum of the first series
Yºr GEOMETRICAL PROGRESSION. 9 I5
are 2 times as large as the corresponding terms and sum of the second series; hence
2 must be the common ratio. Or we may demonstrate it thus: Take the first
formula, Problem III, S = # = * , and clear it of fractions, giving RS — S = RL
- A ; then by transposition we obtain, R(S — L) = S – A, and to find R, we have
by division the
Formula, R = S – A
---
S — L
That is, the ratio is equal to the sum of the series minus the first term, divided
by the sum minus the last term. t
1. The extremes of an ascending series are 4 and 2500, and the sum of the
Series is 3124; what is the ratio? Ans. 5, ratio.
SOLUTION.
R = _3124 – 4 = % = 5, ratio.
T 3124 C 2500
2. The extremes of a descending series are 9 and #, and the sum of the
series is 134%; what is the ratio? Ans. }, ratio.
SOLUTION.
13; – ºr 13; # = }, ratio.
3. The first term of an infinite series is 16, and the sum of the series is 64;
what is the ratio? Ans. #, ratio.
SOLUTION.
__ 64 – 16 as e
R == GTO T # = #, ratio.
4. The distance between two places by rail is 260.868.155 miles; if a train
moves 50 miles the first hour, and 26.57 ºr miles the last hour, what will be the ratio
of the decreasing progression of the rate of travel? Ans. Hº, ratio.
OPERATION INDICATED.
_ 260.868.155 – 50. _ 210.868155
- - = # = ratio.
260,868.155 — 26.570205 234.297.950
PROBLEM VIII.
1454. To find the number of terms, when the extremes and ratio are given.
* By Problem I we see that L = A x Rº"; dividing both numbers by the
first term A, we obtain the
Formula, # = Rº-1
That is, the quotient of the last term by the first, is equal to a power of the
ratio one less than the number of terms, and by adding one to that power we have
the number of terms. - -
9 I 6 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICs.
Or we may indicate it thus:
Formula, Rn =# × R.
From this we see, that the number of terms (n) is that power of the ratio
which is equal to the quotient of the last term by the first, multiplied by the ratio.
1. The extremes are 3 and 375, and the ratio is 5; what is the number of
terms ?
SOLUTION.
** = 125 = 5°; then, 3 + 1 = 4, number of terms.
Or, * x 5 = 625 = 5*, gives 4, number of terms.
2. The first term is 125, the last Tºg, and the ratio #; what is the number of
terms? 2 * *
SOLUTION.
Tº — 125 = Tºgr = (#)"; then, 6 + 1 = 7, number of terms.
3. If in traveling 2623 miles, a locomotive runs 32 miles the first hour, and
78% miles the last hour, what was the ratio of the increasing rate of speed, and how
many hours were occupied ? Ans. 14, ratio. 5 hours.
First find the ratio by Problem VII, and then the number of terms.
4. If I pay $5 the first month, and $984.15 the last month, in geometrical
progression, the common ratio being 3, what amount will be disbursed, and how
many months will it require? Ans. Amount, $147620. Time, 10 months.
OPERATION INDICATED.
S = ºus;*)=s = $147620 = amount.
× 3 = 59049 = nth power of ratio 3.
3n - 98415
5
Find the amount by Problem III, and the time by Problem VIII.
PROBLEM IX.
1455. To find the last term, when the first term, common ratio, and sum of the
Series are given.
If we take the formula of Problem VII, R = := #. and clear it of fractions,
we obtain RS — RL = S — A ; then by transposing we have, RL = RS — S +
A, and reducing gives the
_ S(R – 1) + A
Formula, L R
That is, the last term is equal to the product of the sum by the ratio less
one, plus the first term, and the result divided by the ratio.
1. The first term is 5, the ratio 4, and the sum 6825; what is the last term?
SOLUTION.
6825 × 3 = 20475 + 5 = 20480 – 4 = 5120, last term.
4% GEOMETRICAL PROGRESSION. 917
2. If a merchant owes $98410, and pays $10 the first month, and triple the
amount each succeeding month, how long will it take to pay the debt, and what will
be the last payment? Ans. $65610, last payment. Time, 9 months.
Find the last payment by Problem IX, then the number of months by Prob.
lem VIII.
APPLICATIONS OF GEOMETRICAL PROGRESSION TO COMPOUND
INTEREST AND COMPOUND DISCOUNT.
1456. Any principal at compound interest, for a number of years, will produce
for the successive years, or other intervals, a series in geometrical progression.
Thus if we put $1 at compound interest at 6 per cent for 4 years, we will have
$1 for the first term, and the amount of $1 for one period ($1.06) for the ratio, and
4 years for the number of terms, which gives the series $1, (1.06), (1.06%), (1.06%),
(1.064).
The last term, 1.06%, gives the amount of $1 for 4 years at 6 per cent.
We here see the number of terms, 5, is one greater than the given number of
years, because the first term, $1, existed previous to any interval of time, and
hence if we raise the ratio (the amount of $1 for a specified interval) to a power
indicated by the number of years or periods, we will have the compound amount of
$1 for that time; and since all amounts, for the same time and rate, are proportional
to their principals, if we multiply the amount of $1 by any principal, we will obtain
the compound amount for that principal, and by deducting the principal from the
amount we have the compound interest. From the above, and principles previously
explained in Problems VI and VIII, Geometrical Progression, we may deduce the
following conclusions, viz.:
That in compound interest, the amount and interest vary directly, or in the
same ratio as the principal, when the time and rate are the same;
Whereas the rate varies as the nth root of the quotient of the amount by the
principal ;
And the time varies as the nth power of the ratio;
That is, increasing the rate or time, increases the interest in a still greater
ratio.
1. What is the amount of $2500 for 3 years at 8 per cent compound interest ?
Ans. $3149.28.
SOLUTION.
$2500 × (1.08)* = $2500 × 1.259712 = $3149.28
Here we have the principal, $2500, for the first term, the amount of $1 for
one year at 8 per cent ($1.08) for the ratio, and 3, the given number of years (one
less than the number of terms), for the power of the ratio; which corresponds to
Problem I, Geometrical Progression.
918 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. ºr
2. What is the amount of $10000 for 10 years at 10 per cent compound
interest ? Ans. $25.937.425.
3. What is the compound amount and interest on $5000 for 4 years at 6 per
cent, compounded semi-annually 3 Ans. Amount, $6333.8504.
Interest, $1333.8504.
SOLUTION.
$5000 × (1.03)*
When the intervals are less than a year, we take such a part of the rate as
the given interval is of 12 months, which gives (1.03) for the ratio in the last
example, to be raised to the 8th power, because there are 8 semi-annual intervals in
4 years.
4. What is the compound interest on $50000, at 8 per cent compounded
quarterly, for 4 years? Ans. Interest, $18639.285.
SOLUTION.
From Table, (1.02)16 = 1.3727857
PRESENT WORTH, AND COMPOUND DISCOUNT.
1457. To find the present worth at compound interest, of any sum due at a
future time. Since different amounts are proportional to their principals, it is plain
that if we divide any given amount by the amount of $1, for the given rate and
time, the quotient will be the required principal or present worth ; and the difference
between the given amount and present value, will be the compound discount.
The above principle of compound discount furnishes the only correct method
of finding the present value of government or corporation bonds, due at a future
time, or of estimating the value of investments for long periods of time, without
gain or loss, except the interest.
1. What principal will amount to $11576.25 in 3 years, at 5 per cent com-
pound interest ? Ans. $10000.
SOLUTION.
By Problem II, Geometrical Progression, $11576.25 + 1.05% = $10000.
Or by Table, page 611. Divide the given amount by the tabular amount of $1, for the same
time and rate, because the amounts in the table are the different powers of the rates at top, for
the number of years opposite the amounts. -
2. What principal will produce $724.50 compound interest at 7 per cent in 2
years? & Ans. $5000.
SOLUTION.
$724.50 + (1.07*) –1 = $5000. ---
e Or by Table. Divide the given compound interest by the compound interest of $1, for the
given time and rate.
4% GEOMETRICAL PROGRESSION. - 9 I9
3. In how many years will $12000 amount to $17569.20 at 10 per cent per
annum, compound interest? Ans. 4 years.
SOLUTION.
By Problem VIII, Geometrical Progression, 17569.20 - 12000 = 1.4641 = (1.10)*
Or by Table. In the column of the given rate per cent, find the quotient of the amount by
the principal, and then find the year opposite.
4. At what rate in 3 years will $50000 amount to $62985.60%
Ans. 8 per cent.
SOLUTION.
By Problem VI, $62985.60- 50000 = VI.2597E = 1.08 = 8%
Or by Table. Opposite the given year, find the quotient of the amount by the principal, the
column at top will show the rate per cent
5. What would be the present value of $25.907.10 in State bonds due in 10
years, and not bearing interest, allowing 8 per cent compound interest on money
invested ? Ans. $12000,
Solution by Problem II, Geometrical Progression.
Or by Table. Divide amount of bonds, $25.907.10 by the tabular amount of $1, at 8 per cent
for 10 years, viz., by 2.158925. Y
|
6. What would be the present value of $134550 in government bonds, due in
20 years, allowing 10 per cent compound interest on investment # Ans. $20000.
SOLUTION,
Divide $134550 by tabular amount, $6.7275.

nnuities.
-----------------N
A. Aſ a
-ºr * * *-s ºr e = * *
1458. An Annuity is properly a sum of money payable annually; but the
term is also applied to payments made at the end of regular intervals of time, as
monthly, quarterly, semi-annually, etc., and to continue for a given period, for life,
or forever,
Annuities are of three kinds, viz.: Certain, Contingent, and Perpetual.
T459. A Certain Annuity is one that commences at a specified time and
continues for a definite period.
1460. A Contingent Annuity is one whose commencement or termination
depends on some contingent event, as the life of a person or persons, hence its
continuance is also uncertain; when depending on the life of an individual, it is
called a Life Annuity.
1461. A Perpetual Annuity, or Perpetuity, is one which is to continue
forever. -
Each of these three kinds of Annuities is subject to the three following
conditions, viz.: It may be,
1st. An Annuity in Possession, or an Immediate Annuity, when it begins at
once, or has already commenced.
2d. A Deferred Annuity, when it is to commence at some certain future time.
3d. An Annuity in Reversion, when it is to commence at the death of one
person and revert to another, or when its commencement depends on the occurrence
of some other certain event.
1462. An Annuity Foreborne, or in Arrears, is one whose periodical pay-
ments have not been made when due, but allowed to accumulate with interest on
each payment in arrears, until final maturity.
1463. The Amount or Final Walue of an annuity, or the value of one in
arrears, is the sum of all the payments due, with interest on each, from the time.
they became due to the end of the given period, or until the termination of the
annuity.
1464. The Present Worth or Walue of an annuity is equal to the sum of
the present values of all its payments, on such a sum as would amount to its final
value, at the given rate and time. The present worth of a Perpetual Annuity is a
sum, whose periodical interest would be equal to the annuity.
When simple interest is allowed on annuties, the computations may be direct
on the several payments, or by the principles of Arithmetical Progression.
When compound interest is allowed, we may employ the principles of Geo.
metrical Progression, or use the Tables calculated for that purpose.
(920)
X. ANNUITIES. 92I
REMARKS ON ANNUITIES. It is not always easy to determine when an
annuity begins. It should be remembered that it begins, not at the time the first
payment is made, but one interval of time before. Hence in problems where the
annuity is paid in advance, the tabular number to be used should be for one year
more than the number of annuities.
NOTE.—The interval of time is the period between the payments, and may be a year, half-
year, quarter or month.
PROBLEM I.
1465. Annuities at Simple Interest.
To find the amount of an annuity foreborne or in arrears, at simple interest.
1. What would be the amount of an annuity of $1000, in arrears for 4 years,
payable annually, at 8 per cent, simple interest? Ans. $4480.
SOLUTION. tº
Sum of four annual payments, $1000 × 4 = $4000 .--
Interest on 1st payment of $1000 for 3 years
& 4 & 4 2d & 4 & 4 1000 { { 2 & 4
& & { { 3d ( & & & 1000 & 4 1. & 4
Equals interest on $1000 for 6 years at 896 480
Amount, $4480
Explanation.—Since the first payment does not become due until the end of the first year, it
will draw interest for only 3 years; the second payment for 2 years; the third payment for 1 year;
and the last payment not bearing interest; the sum of the interests on the several payments, for
their respective times, is equal to the interest on one payment for the sum of the several periods
producing interest, or for 3 + 2 + 1 = 6 years, which at 8 per cent gives $480 interest to be added
to the sum of the 4 annual payments of $1000, giving for the amount $4480,
SOLUTION BY ARITHMETICAL PROGRESSION,
Problem No. IV, Arithmetical Progression, for finding the sum of a series, may be used: The
first payment, $1000, being the first term; the amount of the first payment for the number of years
or periods less one, the last term; and 4 years the number of terms; hence we have
wife, × 4 = 1120 × 4 = $4480, the amount or sum.
2. A person deposited in a savings bank, for the benefit of his son, $100 on
each birthday; what is due him on his 21st birthday, and how much when he is 21
years of age, interest at 6 per cent 3
SOLUTION.
(100 + 220) × 4 = $3360, amount due on 21st birthday.
And (100 + 226) x * = $3586 — $100 = $3486, amount due when 21 years of age,
Or, $3360 + interest on $2100 for 1 year $126 = $3486, as above.
922 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. †
* PROBLEM II.
1466. Annuities at Compound Interest.
The amounts of the several payments of an annuity at compound interest,
form a descending geometrical series, and by reversing, or considering the series as
ascending, we have the annuity or periodical payment, for the first term; the
amount of $1 for one year, or $1 plus the rate of interest expressed decimally, for
the ratio; the time in years or other intervals, for the number of terms; and the
required amount for the sum of the series. Hence, to find the amount or final value
of an annuity in arrears, or for any given time, at compound interest, we may use
the formula of Problem IV, Geometrical Progression, as illustrated in the following:
1. What will a certain annuity of $800, in arrears for 4 years, amount to at
5 per cent compound interest? Ans. $3448.10.
SOLUTION.
1.054 — 1
T05 TT × $800 = 4.310.125 × $800 = $3448.10 General formula.
s — A X (R* - 1)
R – 1
Or, by a modification of the above formula, thus:
(800 × 1.054) — 800
.05
When the number of years is large, instead of involving the ratio to that power, take the
amount of $1 for the given number of years and rate, from the Compound Interest Tables, pages
611-12, which corresponds to the required power of the ratio. When Annuity Tables are accessible,
we simply multiply the amount of $1, for the given time and rate, by the given annuity; thus,
= $3448.10, as above.
From Table No. 1, page 925, $4.310125 x 800 = $3448.10
2. What is the final value of a life annuity certain, of $500, in 20 years,
payable annually at 6 per cent compound interest? Ans. $18392.7955.
3. A person, at the age of 23, deposits in a savings bank $250 each year,
until he is 60 years old; what will it amount to at 7 per cent compound interest?
Ans. $40084.35.
PROBLEM III.
1467. To find the Present Worth, or Value of a Certain Annuity.
The present worth or value of a certain annuity, is that principal which, at
the same rate of interest till the termination of the annuity, would be increased to
the amount of the annuity foreborne for the same time.
Therefore, if we find the final value, or amount of an annuity by the last
problem, and divide it by the amount of $1 for the same time and rate, (as found
from the Compound Interest Tables,) the quotient will be the present worth of the
annuity. Or, having Annuity Tables, we simply multiply the present worth of $1
for the given time and rate, as found in Table No. 2, by the given annuity.
1. What is the present value of an annuity of $400, payable annually for 10
years, at 6 per cent compound interest? Ans. $2944.0348.
2: ANNUITIES. 923
SOLUTION.
We first find the amount of the annuity by Problem I, to be $5272,318; then this amount
divided by the amount of $1 for the same rate and time, as per Compound Interest Table, or by
1.06%, gives the present worth ; thus,
1.790848 ) 5272,318 ( $2944.0348, present worth.
Or by Annuity Table No. 2, $7.360087 × 400 = $2944.0348
2. What is the present value of a pension of $100, payable annually for 25
years, at 4 per cent compound interest ? Ans. $1562,208.
3. What is the present value of an annuity of $500, to commence 10 years
hence, and continue 20 years thereafter, at 6 per cent compound interest?
Ans. $3202,372.
A
SOLUTION.
Amount of $500 annuity for 20 years = $18392.7955.
Amount of $1 for 30 years = $5.743491.
Then, $18392.7955 – 5.743491 = $3202.372
Or by Problem V for annuities in reversion.
PROBLEM IV.
1468. To find the Present Value of a Perpetual Annuity.
The present worth of a perpetuity, or perpetual annuity, is that sum of
money, whose interest is equal to the given annuity.
1. What is the present worth of a perpetuity of $400 per annum, allowing
compound interest at 8 per cent 3 Ans. $5000.
OPERATION.
Explanation.—At 8 per cent, it will require
100 $100 capital to produce $8 interest, and hence
8 || 400 $1 interest will require as much capital, and
40000 $400 interest will require 400 times as much, or
tºº-º-º: $5000 capital, which is equal to the present,
$5000 value of a perpetual annuity of $400.
2. What is the present worth of a perpetual annuity of $1500 a year, at 6
per cent compound interest? Ans. $25000.
PROBLEM V.
1469. To find the Present Worth of an Annuity Deferred, or in Reversion.
1. What is the present worth of an annuity of $500, to commence in 4
years, and to continue 6 years, allowing 6 per cent compound interest ?
es” Ans, $1947.4905.
924 soulE's PHILOSOPHIC PRACTICAL MATHEMATICS. º:
SOLUTION.
$7,360087 from Table 2 present worth of $1 for 10 years.
3.465106 & & 4 & 6 & & 4 ( & ‘‘ ‘ ‘ ‘‘ 4 * {
$3.894981 value of $1 for the 6 intervening years.
500 given annuity.
$1947.490500 present worth of the annuity.
Explanation.—We first find, from the Tables, the present worth of an annuity of $1 for the
whole period, or until the annuity terminates, viz., for 4 + 6 = 10 years, we then find from the
Table the present value of $1 to the commencement of the reversion, viz., for 4 years; we then
find the difference between these present values of an annuity of $1, and multiply it by the num-
ber of dollars in the given annuity, for the present worth required.
2. What is the present worth of a reversionary annuity of $1000, deferred
for 10 years, and to continue 15 years thereafter, at 5 per cent compound interest?
Ans. $6372.21.
PROBLEM VI.
1470. To find the Annuity, when the Present Worth, Time, and Rate per cent
are given.
1. What annuity continued for 11 years, at 6 per cent compound interest,
would give a present worth of $6309.50% Ans. $800.
SOLUTION.
$6309.50 – 7.886875 = $800
Since the present worth of $1 multiplied by the annuity would give the full present worth,
it follows, that if we divide the given present worth by the present worth of $1 (as found in Table
No. 2), the quotient will be the annuity.
PROBLEMI VII.
1471. To find the Annuity, when the Amount, Time, and Rate per cent are given.
1. What annuity will amount to $5172.15, in 4 years, at 5 per cent compound
interest ? Ans. $1200.
SOLUTION.
$5172.15 — 4,310.125 = $1200
We here divide the given amount by the amount of $1 (from Table No. 1), for the same time
and rate; the reason is the same as in the preceding problem. w
The subject of annuities is of a very comprehensive nature, and of great
practical importance, but a full discussion of all its principles and applications
would be too extended for this treatise. The number of problems here given is
greater than is usually presented in arithmetics, and with the aid of the accom-
panying Tables, which give the present and final values of $1 at the ordinary rates
of interest, from 1 to 50 years, its applications to reversions, leases, rents, etc., may
be readily comprehended and applied. Its employment in problems of life insur-
ance, life estates, dowers, pensions, etc., would require in addition the use of
Mortality Tables, and the special regulations determined by actuaries, to whose
works the student is referred for more elaborate dissertations on the subject.
ANNUITY TABLE NO, I, 925
1472. TABLE No. 1.
SHOWING THE AMOUNT OF AN ANNUITY OF ONE DOLLAR PER ANNUM, IMPROVED
AT COMPOUND INTEREST FOR ANY NUMBER OF YEARS NOT EXCEEDING FIFTY.

Yrs. || 3 per cent. 3% per cent. 4 per cent. 5 per cent. 6 per cent. 7 per cent.
1 1.000 000 1.000 000 1.000 000 1.000 000 1.000 000 1.000 000
2 2.030 000 2,035 000 2.040 000 2.050 000 2.060 000 2.070 000
3 3.090 900 3.106 225 3.121 600 3.152 500 3.183 600 3.214 900
4 4.183 627 4.214 943 4.246 464 4,310 125 4,374 616 4.439 943
5 5.309 136 5.362 466 5,416 323 5.525 631 5,637 093 5,750 739
6 6.468 410 6.550 152 6.632 975 6.801 913 6,975 319 7,153 291
7 7.662 462 7.779 408 7.898 294 8.142 008 8,393 -838 8,654 021
8 8.892 336 9.051 687 9.214 226 9.549 109 9,897 468 10,259 803
9 10.159 106 10.368 496 10.582 795 11.026 564 11,491 316 11.977 989
10 11.463 879 11.731 393 12.006 107 12.577 893 13.180 795 13.816 448
11 12.807 796 13,141 992 13,486 351 14.206 787 14,971 643 15,783 599
12 14.192 030 14.601 962. 15.025 805 15,917 127 16,869 941 17.888 451
13 15.617 790 16.113 030 16,626 838 17,712 983 18.882 138 20,140 643
14 17.086 324 17,676 986 18.291 911 19.598 632 21.015 066 22.550 488
15 18.598 914 19.295 681 20.023 588 21.578 564 23.275 970 25.129 O22
16 20.156 881 20,971 030 21.824. 531 23.657 492 25.670 528 27.888 054
17 21.761 588 22 705 O16 23.697 512 25,840 366 28.212 880 30.840 217
18 23.414 435 24,499 691 25.645 413 28.132 385 30.905 653 33.999 033
19 25.116 868 26.357 180 27.671 229 30.539 OO4 33.759 992 37.378 965
20 26,870 374 28,279 682 29.778 079 33.065 954 36.785 591 40.995 492
21 28.676 486 30.269 471 31.969 202 35.719 252 39.992 727 44,865 177
22 30.536 780 32.328 902 34.247 970 38.505 214 43.392 290 49,005 739
23 32.452 884 34.460 414 36,617 889 41.430 475 46.995 828 53.436 141
24 34.426 470 36,666 528 39,082 604 44, 501 999 50.815 577 58.176 671
25 36.459 264 38.949 857 41.645, 908 47.727 099 . 54.864 512 63.249 030
26 38.553 042 41.313 102 44.311 745 51.113 454 59.156 383 68.676 470
27 40,709 634 42.759 060 47,084 214 54,669 126 63.705 766 74.483 823
28 42.930 923 46.290 627 49,967 583 58.402 583 68.528 112 80.697 691
29 45.218 850 48.910 799 52.966 286 62.322 712 73.639 798 87.346 529
30 47.575 416 51,622 677 56.084 938 66.438 848 79.058 186 94.460 786
31 50.002 678 54,429 471 59.328 335 70.760 790 84.801 677 102.073 041
32 52.502 759 57.334 502 62.701 469 75,298 829 90.889 778 110.218 154
33 55.077 841 60.341 210 66.209 527 80,063 771 97.343 165 118.933 425
34 57.730 177 63.453 152 69.857 909 85.066 959 104.183 755 128.258 765
35 60.462 082 66,674 013 73.652 225 90.320 307 111.434 780 138.236 878
36 63.271 944 70,007 603 77.598 314 95.836 323 119.120 867 148.913 460
37 66.174 223 73.457 869 81.702 246 101.628 139 127.268 119 160,337 400
38 69.159 449 77.028 895 85.970 336 107.709 546 135.904 206 172.561 020
39 72.234 233 80.724. 906 90.409 150 114.095 023 145.058 458 185.640 292
40 75.401 260 84.550 278 95.025 516 120.799 774 154,761. 966 199.635 112
41 78.663 298 88.509 537 99.826 536 127,839 763 165.047 684 214,609 570
42 82,023 196 92.607 371 104.819 598 135,231 751 175.950 645 230,632 240
43 85.483 892 96.848 629 110.012 382 142.993 339 187.507 577 247.776 496
44 89.048 409 101.238 331 115.412 877 151.143 006 199,758 032 266.120 851
45 92.719 861 105.781 673 121,029 392 159.700 156 212.743 514 285.749 311
46 96,501 457 110.484 031 126.870 568 168.685. 164 226.508 125 306.751 763
47 100,396 301 115.350 973 132.945 390 178.119 422 241.098 612 329.224 386
48 || 104,408 396 120.388 297 139.263 206 188,025 393 256.564 529 353.270 093
49 || 108 540 648 125.601 846 145.833 734 198.426 663 272.958 401 378.999 000
50 || 112.796 867 130 999 910 152.667 ()84 209,347 976 290.335 905 406.528 929
926 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
1473. TAIBLE NO. 2.
SHOWING THE PRESENT WORTH OF AN ANNUITY OF ONE LOLLAR PER ANNUM,
TO CONTINUE FOR ANY NUMBER OF YEARS NOT EXCEEDING FIFTY.

Yrs. 3 per cent. || 3% per cent. 4 per cent. || 5 per cent. || 6 per cent. || 7 per cent. 8 per cent. Yrs.
1 || 0.970 874 0.966 184 0.961 538 0.952 381 || 0,943 396 || 0.934 579 0.9259 1
2 | 1.913 470 | 1.899 694 | 1.886 095 | 1.859 410 | 1.833 393 | 1.808 ()17 1.7833 2
3 2.828 611 || 2.801 637 || 2.775 091 2 723 248 2.673 012 2.624 314 2.5771 3
4 || 3.717 098 || 3.673 079 || 3.629 895 || 3.545 951 | 3,465 106 || 3.387 209 3.3121 4
5 || 4.579 707 || 4,515 052 || 4.451 822 || 4,329 477 || 4.212 364 || 4,100 195 3.9927 5
6 || 5.417 191 || 5.328 553 || 5.242 137 5.075 692 || 4.917 324 || 4,766 537 4,6229 6
7 || 6.230 283 || 6.114 544 || 6,602 O55 || 5.786 373 || 5.582 381 || 5.389 286 5.2064 7
8 || 7.019 692 || 6.873 956 | 6.732 745 6.463. 213 || 6.209 744 || 5,971 295 5.7466 8
9 || 7.786 109 || 7.607 687 || 7.435 332 || 7.107 822 6.801 692 || 6.515 228 6.2469 9
10 || 8,530 203 || 8.316 605 || 8.110 896 || 7.721 735 | 7.360 087 7.023 577 6.7101 10
11 || 9.252 624 9.001 551 || 8.760 477 || 8.306 414 || 7.886 875 || 7.498 669 7.1390 11
12 || 9.954 004 || 9,663 343 || 9.385 074 || 8.863 252 | 8.383 844 || 7.942 671 7.5361 12
13 || 10.634. 955 | 10.302 738 || 9,985 648 || 9,393 573 || 8.852 683 || 8.357 635 8.9038 13
14 | 11.296 073 || 10.920 520 | 10.563 123 || 9,898 641 9.294. 984 8.745 452 8.2442 14
15 11.937 935 | 11.517 411 || 11.118 387 || 10.379 658 9.712 249 || 9,107 898 8.5595 15
16 || 12.561 102 || 12.094. 117 | 11.652 296 || 10.837 7.70 || 10.105 895 || 9,446 632 8.8514 16
17 | 13.166 118 || 12.651 321 | 12.165 669 || 11,274 O66 || 10.477 260 || 9,763 206 9.1216 17
18 13.753 513 || 13.189 682 | 12.659 297 11.689 587 | 10.827 603 || 10.059 (070 9.3719 18
19 || 14.323 799 || 13.709 837 || 13.133 939 12.085 321 || 11.158 116 || 10.335 578 9.6036 19
20 14.877 475 14.212 403 || 13.590 326 12.462 210 | 11.469 421 | 10.593 997 9,8181. 20
21 | 15.415 024 || 14.697 974 || 14.029 160 12.821 153 || 11.764 077 10.835 527 | 10.0168 21
22 15.936 917 | 15.167 125 || 14.451 115 || 13.163 003 || 12.04.1 582 11.061 241 10.2007 22
23 | 16.443 608 || 15.620 410 || 14.856 842 | 13.488 574 || 12.303 379 || 11.272 187 | 10.3711 23
24 | 16.935 542 | 16.058 368 || 15.246 963 || 13.798 642 | 12.550 358 || 11.469 334 || 10.5288 24
25 | 17,413 148 || 16.481 515 | 15.622 080 || 14.093 945 12,783 356 || 11.653 583 || 10,6748 25
26 17.876 842 | 16.890 352 15,982 769 14.275 185 | 13.003 166 11.825 779 || 10.8100 26
27 | 18.327 031 || 17.285 365 | 16.329 586 || 14.643 034 13.210 534 11.986 709 || 10.9352 27
28 || 18.764 108 || 17.667 O19 || 16.663 063 || 14,898 127 | 13.406 164 | 12.137 111 || 11.0511 28
29 || 19.188 455 | 18.035 767 | 16.983 715 15.141 074 || 13.590 721 || 12.277 674 11.1584 28
30 | 19,600 441 | 18.392 045 || 17.292 033 15.372 451 | 13.764 831 | 12.409 041 || 11.2578 30
31 | 20.000 428 | 18.736 276 || 17.588 494 | 15.592 811 || 13.929 086 12.531 814 11.3498 31
32 20.338 766 | 19.068 865 17.873 552 | 15.802 667 || 14,084 043 | 12.646 555 | 11.4350 32
33 || 20.765 792 || 19.390 208 || 18 147 646 | 16.002 549 || 14.230 230 | 12,753 790 || 11.5139 33
34 21.131 837 | 19.700 684 || 18.411 198 || 16.192 204 || 14,368 141 | 12.854 009 || 11.5869 34
35 | 21.487 220 20 000 661 | 18.664 613 | 16.374 194 || 14.498 246 || 12.947 672 | 11.6546 35
36 21 832 252 20.290 494 | 18.908 282 | 16.546 852 14.620 987 13.035 208 || 11.7172 36
37 22.167 235 | 20.570 525 | 19. 142 579 | 16.711. 287 || 14.736 780 13,117 017 | 11.7752 37
38 22.492 462 20.841 087 | 19.367 864 | 16.867. 893 14,846 019 || 13.193 473 || 11.8289 38
39 22.808 215 21.102 500 | 19.584 485 || 17.017 ()41 || 14,949 075 || 13.264 928 11.8786 39
to 23.114 772 21.355 072 | 19,792 774 17.159 086 15.046 297 || 13.331 709 || 11.9246 40
41 || 23.412 400 || 21.599 104 || 19.993 052 17.294 368 || 15.138 016 || 13.394 120 | 11.9672 41
42 23. 701 359 || 21.834, 883 20.185 627 17.423 208 15.224 543 || 13.452 449 || 12.0067 42
43 23.981 902 22.062 689 || 20.370 795 17.545 912 15.306 173 13. 506 962 12.0432 43
44 24.254 274 22.282 791 | 20.548 841 || 17,662 773 | 15.383 182 | 13.557 908 12.0771 44
45 24,518 713 22.495 450 20,720 040 || 17.774 070 | 15.455 832 13.605 522 | 12.1084 45
46 24.775 449 22.700 918 20.884 654 17.880 067 15.524 370 13.650 020 | 12.1374 46
47 || 25.024 708 22.899 438 21.042 936 || 17.981 016 || 15.589 O28 || 13 691 608 ; 12.1643 47
48 25.266 707 || 23 091 244 || 21.195 131 || 18.077 158 || 15.650 027 | 13.730 474 || 12.1891 48
49 || 25.501 657 || 23.276 564 || 21.341 472 | 18.168 722 || 15.707 572 || 13.766 799 || 12.2122 49
50 || 25,729 764 23,455 618 || 21.482 185 | 18.255 925 | 15.761 861 | 13,800 746 12.2335 50
NoTE.—This table gives the accurate result when payments are made at the end of the stated
periods of time, but when the annuity or other fund is paid in advance, we must multiply the
amount by the annuity value of $1 for one period more than the number of periods, and subtract
from this product the amount of one payment; or, the answer may be found when payments are
made in advance, by multiplying the annuity value of $1 for the number of years, and adding to
this product interest on same for one of the given periods of time.
ANNUITY TABLE NO. 3.
927
1474.
TABLE NO.
TABLE OF PRESENT worTH, or VALUE OF $1, DUE AT THE END OF ANY
3.
NUMBER, OF YEARS NOT IEXCEEDING 50, ALLOWING COMPOUND INTEREST.
Jind of 1 per 2 per 3 per 4 per 5 per 6 per I per 8 per 9 per 10 per | End of
Year. cent. cent. | cent. cent. cent. cent. cent. cent. cent. Cent. Year
1 .9901 .9804 ,9709 | .961538 .952381 |.943396 || 934579 | .925926 .917431 .909091 1
2 .9803 .9612 .9426 | .924556 | .907029 . .889996 | .873439 || .857339 .841680 | .826446 2
3 .9706 .9423 .9151 .888996 | .863838 |.8396.19 .816298 ||.793832 .772183 | .751315 3
4 .9610 .9238 .8885 .854804 || 822702 | .792004 || ,762895 ||.735030 .708425 | .683013 4
5 .9515 .9057 .8626 |.821927 | .783526 .747258 |.712986 || ,680583 .649931 | .620921 5
6 .9420 .8880 ,8375 .790315 .746215 .704961 .666342 . .630170 ,596.267 | .564474 6
7 .9327 .8706 .8131 | .7599.18 .710681 .665057 .622750 | .583490 | .547034 .513158 7
8 .9235 .8535 .7894 | .730690 | .676839 .627412 | .582009 | .540269 .501866 ,466507 8
9 .9143 .8368 .7664 | .702587 | .644609 | .591898 |.543934 | .500.249 |.460428 .424098 9
10 .9053 .8203 .7441 |.675564 .613913 | .588395 | .508349 |,463193 ) .4224.11 | .385543 || 10
11 .8968 .8043 .7224 .649581 | .5846.79 | .526788 .475093 .428883 | .387533 .350494 || 11
12 .887.4 .7885 .7014 | .624597 | .556837 | .496969 | .444.012 | .397114 | .355.535 | .318631 || 12
13 .8787 .7730 .6810 | .600574 | .530321 | .468839 || .414964 .367698 || .326179 | .289664 || 13
14 .8700 .7579 .6611 .577475 | .505068| .442301 . .387817 | .340461 .299.246 | .263331 || 14
15 .8613 .7430 .6419 |.555265 | .481017 | .417265 | .362446 | .315242 | .274538 | .239392 || 15
16 ,8528 .7284 .6232 | .533908 .458112 | .393646 .338735 | .291890 | .251870 | .217629 || 16
17 .8444 .7142 .6050 | .513373 | .436297 .371364 | .316574 .270269 .231073 .197845 || 17
18 .8360 .7002 .5874 .493628 .415521 | .350344 | .295864 | .250249 | .211994 | .179859 18
19 .8277 .6864 .5703 | .474642 | .395734 . .330513 | .276508 || .231712 .194490 .163508 || 19
20 .8195 .6730 .5537 .456387 | .376889 | .311805 | .2584.19 | .214.548 .178431 .148644 20
21 .8114 .6598 .5375 .438834 .358942 | .294.155 .241513 .198656 .163698 || .135131 || 21
22 .8034 .6468 .5219 .421955 .341850 | .277505 | .225713 | .183941 | .150182 | . 122846 || 22
23 .7954 .6342 .5067 .405726 .325571 | .261797 | .210947 .176315 .137781 ...111678 || 23
24 .7876 .6217 .4919 | .390121 .310068 .246979 | .197147 | . 157699 | .1264.05 | .101526 24
25 .7798 . 6095 .4776 | .375117 | .295.303 | .232999 | .184249 .146018 .115968 .092296 || 25
26 .7720 .5976 .4637 | .360689 .281.241 .219810 | .172195 | .135202 | .106393 | .083905 26
27 .7644 .5859 .4502 .346817 | .267848 .207368 .160930 | .125.187 | .097608 .076278 27
28 .7568 .5744 .4371 | .333477 .255094 | .195630 | .150402 | .115914 | .089548 .069343 28
29 .7493 .5631 .4243 | .320651 | .242946 .184557 | .140563 | . 107328 .082155 .063039 || 29
30 .7419 .5521 .4120 | .308318 . .231377 | .174110 | .131367 | .099377 .075371 | .057309 || 30
31 .7346 .5412 .4000 | .296460 | .220359 .164255 | . 122773 | .09.2016 | .069148 .052099 31
32 .7273 .5306 .3883 | .285058 | .209866 | .154957 | .114741 || 085200 | .063438 .047362 32
33 .7201 .5202 .3770 .274094 | .1998.73 .146186 .107235 | .078889 .058200 | .043057 33
34 .7130 .5100 .3660 | .263552 | . 190355 | .137912 .100219 |.073045 .053395 | .039143 34
35 .7059 .5000 .3554 .253415 | .181290 | .130.105 | .093663 | .06.7635 | .048986 .035584 35
36 .6989 .4902 .3450 | .243669 .172657 .122741 .087535 .062625 | .044941 |.032349 || 36
37 .6920 ,4806 .3350 | .234297 | .164436 | .115793 | .081809 | .057986 | .041231 .0294.08 || 37
38 .6852 .4712 .3252 | .225285 | .156605 | . 109239 .076457 .053690 .037826 |.026735 | 38
39 .6784 .4619 .3158 .216621 | .1491.48 || 103056 | .071455 | .049713 | .034703 |.024.304 || 39
40 ,6717 .4529 .3066 | .208289 |.142046 | .097.222 | .066780 | ,046031 |.031838 |.02.2095 || 40
41 .6650 .4440 .2976 .200278 || 135282 | .091719 |.062412 .042621 |.029209 | .020086 || 41
42 .6584 .4353 .2890 | .1925.75 .128840 | .086527 | .0583.29 ,039.464 | .026797 .018260 || 42
43 .6519 .4268 .2805 | .185.168 |.122704 .08.1630 | .054513 | .036541 | .024584 .016600 43
44 .6454 .4184 .2724 . .178046 | .116861 .077009 | .050946 .033834 .022555 . .015091 || 44
45 .6391 .4102 .2644 .171198 | .111297 | .07.2650 | .047613 | .031328 .020692 | .013719 || 45
46 .6327 . 4022 .2567 .164614 | .105997 | .06.8538 |.044499 || .029007 | .018984 .012472 || 46
47 .6265 .3943 .2493 .158283 | .100949 |.064658 . .04.1587 | .026859 |.017416 | .011338 || 47
48 .6203 .3865 .2420 | .152195 | .096142 | .06.0998 || .038867 | .024869 | .015978 | .010307 || 48
49 .6141 .3790 .2350 | .146341 . .091564 | .057546 | .036324 .023027 | .014659 | .0093.70 || 49
50 .6080 .3715 .2281 | .140713 | .087204 | .054288 . .033948 .021321 | .013449 | .008519 || 50
get P. W., multiply by the number given.
NOTE: 1. –The table shows what per cent. of any principal the present worth is: Hence to
Thus the present worth of $10000 due in 15 years at
4% = $10000 × .555265 = $5552.65.
NOTE: 2. —The per cent. of present worth for any number of years may be found by º:
$1.00 by the amount of $1.00 at compound interest as given in the table, on page 611.
of $1.00 for 55 years at 3% = $1.00 - 5.08214859.
Thus P.
Slſ(S.
*_º_-_-_º_-_-_-_-_-_* - - - - e s • * * * * * * * * * * * * * * * * * * * * * * * * ~ * * * * * ess e ere e º 'º - e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. =N
1475. The Metric System of Weights and Measures derives its name from
the metre, which is the primary base or unit from which all of the other units of the
system are derived. This metre, when adopted by the French nation, was thought
to be, and is very nearly, the (rogºrgan) one ten-millionth part of the distance from
the equator to either pole, measured on the earth's surface. Its length, in American
measure, is 39.3708 inches, very nearly.
NOTE. —The term meter is from the Greek metron, a measure.
1476. The Standard Meter is a bar of platinum and is kept among the
national archives in Paris; but duplicates of it have been furnished to the United
States and other nations.
HISTORICAL.
1477. The Metric System is one of the products of the French revolution
of 1789, which resulted in the establishment of the French Republic, on the 22d
day of September, 1792. The progressive spirits of the revolutionists claimed that
every thing needed reforming. They demanded a change in the calendar, a new
classification of the seasons, a new order of months, with new names and a change
of days, a new arrangement for Sunday and festive days, and above all they
demanded a decimal system of weights, measures, and values.
March 17, 1791, a report was presented to the French National Assembly
proposing the adoption of the Metric System. By the laws of 1793 and 1795, a
temporary meter and kilogramme were adopted, and in 1795 a commission was
appointed, under the direction of the Academy of Sciences, for the purpose of
perfecting the system.
The first and most important duty was to determine an invariable standard
unit for all measures of length, area, solidity, capacity, and weight. For the con-
Summation of this object, a trigonometrical survey was made, by two eminent
mathematicians, Delambre and Mechain, of the arc of the meridian through Paris,
from Dunkirk, France, to Mont Jouy, near Barcelona, Spain, a distance of 93.
degrees, more than one-tenth of the quadrant of the meridian. From the survey of
this meridian, the length of the quadrant from the equator to the pole, measured on
the earth's surface, was computed.
The length of the Paris meridian quadrant, thus obtained, was divided into
10,000,000 equal parts, and one of these parts was called a meter and taken as the
primary unit of the French system of measures. This meter was adopted as the
(928)
ºx METRIC SYSTEM OF WEIGHTS AND MEASURES, 929
unit of length, and from it all the other units of measure are derived by the appli-
cation of the decimal scale. It is equal to 39.37079, practically 39.37 inches in
length, and at the time of its adoption, it was thought to be exactly the one ten-
millionth part of the distance from the equator to the pole, on the meridian of
Paris. But by more accurate measurement, based upon the fact that the earth's
equator is not a perfect circle, but slightly elliptical, a fact not considered by the
French mathematicians and astronomers, it is found to be ris part of an inch too
short. This very small error, almost imperceptible in a single meter, amounts to
5124 feet in the length of the quadrant measured. But this slight discrepancy does
no injury to the system. The meter has a fixed length and is decimally divided, and
that is all that is required.
In 1799, the French nation formally adopted the revised metric system; but
by decree of February 12, 1812, the old system was re-established and continued in
use until July 4, 1837, when the decree of February 12, 1812, was repealed and the
metric system was again brought in use. A royal decree, April 17, 1839, determined
the denominations, form, and dimensions of all instruments and measures, for trade
and use. The metric system is now compulsory throughout France. It has also
been adopted and is in general use by several other nations.
Previous to the adoption of the metric system by the French nation, their
units of weights and measures were, in many respects, quite similar to those of the
English, many of which came down the centuries through Egypt, Greece, and
Rome, and were originally largely derived from different parts of the human body;
such as the foot, the length of the foot of Hercules, or of a king; the ulna or
yard, the distance from the middle of the chest or lips, to the tip of the middle
finger; the palm or hand,-the width of the hand; the span,—the distance from
the tip of the thumb to tip of the little finger when extended; the digit or the
finger,--the length of the first finger; the cubit, the distance from the elbow to
the tip of the middle finger; the fathom, the distance from the tips of the middle
fingers of the two hands when extended in opposite directions, etc. Other units
were taken from familiar objects in nature, such as the barley corn, the grain of
wheat, shells, horns, etc.
These once variable ancient and English units of length have now certain
fixed values based upon the Imperial Standard Yard, the length of which is 36
inches. This standard yard is such that a pendulum equal in length to 39.13929 of
its inches, will vibrate seconds, in a vacuum, at the level of the sea, in the latitude
of London, the thermometer being 620 Fahrenheit. The Standard Yard of the
United States is a copy of the English Standard Yard, marked upon a brass scale,
and deposited in the Treasury Department at Washington. Copies of this yard
have been supplied to all the States.
By reason of the decimal divisions of the metric system, it possesses many
advantages over all other systems and is now used wholly, or in part, by nearly all
civilized countries.
93O soule's PHILOSOPHIC PRACTICAL MATHEMATICs. º
ADOPTED BY THE UNITED STATES.
1478. The Metric System was legalized and adopted, but not made compul-
sory in the United States, by an act of Congress passed in 1866. And it is the
only system ever authorized by the Government of the United States.
It is adopted by the United States Coast Survey, is used in the Mints and
the Post Offices, and largely by all scientists, colleges, and universities.
1479. The Meter is the unit of Length, and from it are derived the Are, the
Stere, or Cubic Meter, the Liter, and the Gram. From these five units, all others
are formed. *
1480. The Are (air), is the unit of surface, or square measure, and consists
of a square whose side is 10 meters; hence, it contains 100 square meters.
NOTE. —The Are is from the Latin, area, a surface.
1481. The Cubic Meter, or Stere (stair), is the unit of solidity, and consists
of a cube whose edge is one meter.
1482. The Liter (Leeter), is the unit of the capacity of vessels, etc., and is
a vessel whose volume is equal to a cube whose edge is one-tenth of a meter.
1483. The Gram is the unit of weight, and is the weight of a cube of
distilled water (weighed in a vacuum, 39.20 F., or 40 C.) whose edge is one-hundredth
of a meter.
Each of these five units has its multiples and sub-multiples, or its higher and
lower metric denominations.
1484. The Multiple Units, or Higher Denominations, are formed by prefixing
to the name of the base units, the Greek numerals, Deka, (10), Hecto, (100), Kilo,
(1000), and Myria, (10000).
They are used as follows:
Dekameter, 10 meters. Rilometer, 1000 meters.
Electometer, 100 meters. Myriameter, 10000 meters.
1485. The Sub-Multiple Units, or lower denominations, are formed by
prefixing to the name of the base units, the Latin numerals, Deci (#6), Centi (+}a),
and Milli (Toºrs). g
They are used as follows:
e Y ~~~ 4-3 e - sº e
Decimeter, ſº, meter. Centimeter, Tºg meter. Millimeter, Tºrg meter.
NOTE.-The student should memorize these Greek and Latin prefixes, and also the names of
the primary or base units, before proceeding further.
º: METRIC SYSTEM OF WEIGHTS AND MEASURES. 93 I
METRIO TABLES.
1486. TABLE OF LINEAR MEASURE, OF WHICH THE METER Is THE BASE UNIT.
tº Equivalents in
Metric Units. English Measures.
1 Millimeter (Tºgo of a M.) = gº sº * : *- gº gº {- .03937–1– inch.
10 mm. = 1 Centimeter (Tºp of a M.) F: ge tº $º & sº gº gºe .3937 -- inch.
10 cm. = 1 Decimeter (I'ſ of a M.) - tº sº gº * * - - 3.93707–H inches-
10 dr.m. = 1 Meter (1 meter) = 39.37079 in., practically 39.37 in., or 3.28089–1- feet.
10 M. = 1 Dekameter (10 meters) F º * , ſº sº * * = - 32.80899–1- feet.
10 Dm. = 1 Hectometer (100 meters) 2– º * * tº- * * *g gºs - 19.88423-1- rods.
10 Hm. = 1 Kilometer (1000 meters) -: &º * * 4- tº tº-e - tº- .62138-|- mile.
10 Km. = 1 Myriameter [Mm.] (10000
meters) = tº <- * tº 4- 8- tº 6.21382–H miles.
1487. The Meter, like our yard, is used in measuring short distances, cloths,
etc.
1488. The Kilometer is used in measuring long distances and is about 3 of
a COmmon mile.
1489. The Centimeter and Millimeter are used by artisans and scientists,
in measuring very small lengths.
1490. Methods of Reading. The number 25365.897 metres, is read, in
English,
Twenty-five thousand three hundred and sixty-five metres, and 897 thou-
Sandths of a metre. But in the language of the metric system, it may be read,
Two myriametres, 5 kilometres, 3 hectometres, 6 decametres, 5 metres, 8.
decimetres, 9 centimetres, and 7 millimetres. It may also be read, beginning with
the lowest denomination, 7 millimetres, 9 centimetres, etc., etc.
In reading, remember that the unit of any place is one-tenth of the unit in the
place next at the left, and ten times as great as the unit of the place next at the
right. Hence, the change from one unit to another, and the methods of reduction
and reading, are identical with those in the system of decimal currency.
1491. TABLE OF SURFACE OR squarE MEASURE, OF WHICH THE ARE IS THE.
BASE UNIT.
100 sq. Millimeters, (sq. mm.) = 1 sq. Centimeter, (sq. cm.) = .155–1– sq. in.
100 sq. Centimeters = 1 sq. Decimeter, (sq. dm.) ~ 15.5-H Sq. in.
or . 1076–H sq. ft.
1.19603–H sq. yds.
119.6034-H sq. yds.
or 3.95383-H sq. rods.
2.47114-H acres.
.3861–H sq. mile.
100 sq. Decimeters = 1 sq. Meter, = 1 Centare (Ca.) or (sq. M.)
100 sq. Meters == 1 sq. Dekameter, (sq. Dm.) or Are, (A.)
100 sq. Dekameters, or Ares, = 1 sq Hectometer, (sq. Hm, or Hectare,) (Ha.)
100 sq. Hectometers or Hectares, = 1 sq. Kilometer
=
NOTE.—This measure is used for measuring land, flooring, ceilings, etc. The lower denom--
inations are seldom used. -
932 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. Yºr
1492. TABLE OF CUBIC, OR SOLID MEASURE, OF WHICH THE CUBIC METER OR
STERE IS THE BASE UNIT.
1000 cu. Millimeters, (cu. mm.) = 1 cu. Centimeter, (cu. cm.) = .061027–H cu. in.
1000 cu. Centimeters = 1 cu. Decimeter, (cu. dm.) = 61.02705-- cu. in.
1000 cu. Decimeters = 1 Cubic Meter, (cu. M.) or STERE, (St.) = 35.31658+ cu. ft.,
or 1.30802–H cu. yds., or .2759-H cord.
NOTE.-The Cubic Meter is the unit used for measuring ordinary solids, as boxes, excava-
tions, etc. The Cubic Centimeter and Cubic Millimeter are used for measuring very minute bodies.
#
When the Cubic Meter is applied to the measurement of wood, it is called the
Stere, and has the following units:
TABLE. --.
1 Decistere, (dst.) = 3.53165-H cu. ft.
10 Decisteres, (dst.) = 1 STERE, = 35.31658+ cu. ft.
10 Steres, (St.) = 1 Dekastere, (Dst.) = 353.1658+ cu. ft.
or 13.0802–H cu. yds.
1493. TABLE FOR MEASURING THE CAPACITY OF VESSELs, ETC., of which
THE LITER IS THE BASE UNIT.
1 Milliliter, (rºom of a liter) .06102–H cu. in.
10 ml. = 1 Centiliter, (T′,′) of a liter) = .61027-- “
10 cl. = 1 Deciliter, (I'o of a liter) = 6.1027–1– & &
10 dl. = 1 Liter, (1 liter) == 61.02705-- “
10 L. = 1 Dekaliter, (10 liters) = 610.2705–- & 4
10 Dl. = 1 Hectoliter. 100 liters) = 6102.70502–- “
10 Hl. = 1 Kiloliter, (1000 liters) 61027.05024-H. “
10 Kl. = 1 Myrialiter Ml. (10000 lit.) = 610270.5024-H. & 4
NOTE.-The Liter, or the cube of a decimeter, is the unit of capacity for both Liquid and
Dry Measures. The Hectoliter is the unit for measuring liquids, grain, or fruits.
The following table shows the equivalents of the Liter units, in United
States measures:
1494. TABLE OF EQUIVALENTS.
Metric Denominations. Dry Measure. Liquid Measure.
1 Milliliter = .001816–H pts. = .0338 fl. Oz.
or .00845-H gi.
1 Centiliter F .01816–H pts. = .338 fl. oz.,
or .084539-H gi.
1 Deciliter = .181625-H pts. = .84539-H gi.
1 Liter F. .908128-H qts. = 1.056745-H qts.
1 Dekaliter = 9.08128-1- qts. = 2.64186+ gals.
1 Hectoliter = 2.8379-1- bus. = 26.4186+ “
1 Kiloliter, or Stere = 28.379-H bus. = 264.186+ “
1 Myrialiter = 283.79–H bus. = 2641.86+ & &
1495. TABLE FOR MEASURING WIEIGHT, OF WHICH THE GRAM Is THE BASE
UNIT.
10 Mg. or 100 Kg.
1 Quintal (100000 grams) 220.46212-- & 4
10 Q. or 1000 Kg.
1 Tonneau or Ton (T.) (1,000,000 grams) = 2204 62124+ & 4
or 1.10231-H+ Tons.
1 Milligram (Tºdo of a gram) F. .015432–H gr. Troy.
10 mg. = 1 Centigram (Thºr of a gram) = .15432–H & 4
10 cg. = 1 Decigram (ºr of a gram) = 1.54324-H. & 4
10 dg. = 1 Gram (1 gram) = 15.43248–H & 4
10 G. = 1 Dekagram (10 grams) F .35273-H oz. Avoir.
10 Dg. = 1 Hectogram (100 grams) = 3.52739–1– & 4
10 Hg. = 1 Kilogram or Kilo (1000 grams) = 2.20462–H lbs. Avoir.
10 Kg. = 1 Myriagram (10000 grams) = 22.04621+ “
METRIC SYSTEM OF WEIGHTS AND MEASURES.
933
1496. The Gram is the unit used in weighing gold, silver, jewels, and letters,
and in compounding Inedicines.
It is a little less than 15; grains Troy.
1497. The Kilogram, or Kilo, is the unit used in weighing common articles
in trade; as grain, Sugar, butter, etc.
dupois.
It is a very little more than 24 lbs. Avoir-
1498. The Tonneau, or Ton, is used for weighing very heavy articles, and
is a little in excess of 16 more than the United States ton.
NOTE.—The pound of Germany, Austria, and Denmark is # of a Kilogram.
TABLE OF EQUIVALENTS.
1499. By which Units of the American measure may be easily changed to
metric units, and vice versa.
1 inch = 2.54 cm. 1 drin. - .328 ft.
1 foot = 3.0487 dra. 1 M. = 1.0936 yards = 3.2808 feet =
1 yard = .9144 M. 39.3707 In.
1 rod = 5,029 Dm. 1 Dm. = 1.9884 rods.
1 mile = 1.6095 Km. 1 Km. - .6213 mi.
1 sq. in. = 6.4516 sq. cm. 1 Sq. cm. = .155 sq. in.
1 sq. ft. = 9.2936 sq. dm. 1 Sq. dm. = .1076 sq. ft.
1 sq. yd. = .8361 sq. M. or centare. 1 sq. M. = 1.196 sq. yds.
1 sq. rā. = .2531 sq. Dm. or Are. 1 Are = 39538 sq. rās.
1 acre = .4048 Hectare, or sq. Hm. = 119.6034 sq. yds.
1 sq. mi. = 2.59 sq. Km. 1 Hectare = 2.4711 acres.
1 cu. in. = 16.3934 cu. cm. 1 sq Km. = .3861 sq. mi.
1 cu. ft. = 28.3286 cu. dm. 1 cu. cm. = .061 cu. in.
1 cu. ya. = .7645 cu. M. 1 cu. dim. = .0353 cu. ft.
1 cord. = 3.6281 Steres. 1 cu. M. = 1.308 cu. yds.
1 fl. oz. = 2.9585 cl. 1 Stere - .2759 cord. *
1 1. qt. = .9463 L. 1 cl. = .338 fl. Oz.
1 gal. = .3785 DI. 1 L. = 1.0567 l. qts.
1 dry qt. = 1.1012 L. 1 D1. = 2,6418 gals.
1 peck = .8809 Dl. 1 L. - .9081 dry qts.
1 bushel = .3523 HI. 1 Dl. = 1.1351 pecks.
1 Troy gr. = 64.935 mg. 1 HI. = 2.8379 bushels.
1 oz. Troy = 31.1526 G. 1 mg. = .0154 gr. Troy.
1 Troy lb. = .3732 Kilo. 1 Gram = 15.4324 grs. Troy.
1 oz. A.V. = 28.409 G. - .0321 oz. Troy.
1 lb. Av. = .4536 Kilo. 1 Kllo = 2.6791 lbs. Troy.
1Ton (2000 lbs.) = .9072 Tonneau. 1 Gram - .0352 oz. Av.
1Ton (2240 lbs.) = 1.01605 Tonneau. 1 Kilo = 2 20:46 lbs. Av.
1 lb. Av. = .9070 German lb. 1 Tonneau = 1.1023 ton = 2204,6212 lbs Av.
1 cm. = .3937 in. 1 Ger. lb. = 1.1023 lbs. Av.
GENERAL PRINCIPLES AND DIRECTIONS.
1500. Since the metric system is based upon the decimal scale, wherein 10
units of a lower order or denomination make 1 of the next higher, therefore all
operations in addition, subtraction, multiplication, and division are performed in
the same manner and are governed by the same numerical law as the operations in
decimals or with dollars, cents, and mills.
934 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
Abbreviations of the base unit and of its higher denominations begin with
a capital ; the lower denominations begin with a small letter. All abbreviations
should be read in full. Thus 6 M., 7 dm., 5 cm., 3 mm., should be read 6 meters. 7
decimeters, 5 centimeters, 3 millimeters.
The units of length, capacity, and weight increase and decrease accord, 1g
to the decimal scale; hence, Teach order of units must occupy one place, when
written in full.
Thus, 127.4215 meters would be written, 1 Hm., 2 Dm., 7 M., 4 dm., 2 cm., Y.5
IIllſle
The scale of square or surface measure is (10 × 10) 100; hence, each order
of units must have two places of figures.
Thus, 43 Ha., 8 A., 6 ca. may be written 43.0806 Ha.; and read, 43 hectaxes
and 806 centares. Or, they may be written 4308.06 A., and read, 4308 ares and 6.
centares. Or thus, 430806 ca., and read, 430806 centares.
The units of cubic or solid measure increase by the scale of (10 × 10 × 10)
1000; hence, each order of units must have three places of figures.
Thus, 36 cu. M., 8 cu. dm., 25 cu. cm., may be written, 36.008025 cu. M.; or
thus, 36008.025 cu. dm.; or thus, 36008025 cu. cm.
TO WIRITE AND READ METRIC NUMBERS.
1501. 1. Write and read in the unit of meters and the lower denominations.
127.4215 meters. Ans. 127 M. 4 dm. 2 cm. 1.5 mm.
2. Write and read in higher units 210524683 millimeters.
Ans. 21 Mm. 0 Km. 5 Hm. 2 Dm. 4 M. 6 dm. 8 cm. 3 mm.
3. Write and read 46286 mm. as decimeters. Ans. 462.86 drin.
4. Write and read 46286 mm. as dekalmeters. Ans. 4.6286 Dm.
5. Write in lower units 5.46732 Kg. wº
Ans. 5 Kg. 4 Hg. 6 Dg. 7 G. 3 dg. 2 cg.
6. Write 5.46732 Kg. as grams. Ans. 5467.32 G.
7. Write 604 L. in higher units. Ans. 6 Hl. 0 Dl. 4 L.
8. Write 604 L. as centiliters. Ans. 604U0 cl.
9. Write 604 L. as myrialiters. Ans. .0604 Ml.
10. Write 6 Dm. 3 dm. 4 mm. as meters and decimals. Ans. 60.304 M.
NOTE.--When there are omissions in any of the denominations of the number given, their
places must be filled with naughts.
11. Write as Kilograms and decimals, 434278 dg. Ans. 43.4278 Kg.
TO REDUCE METRIO NUMBERS FROM EIIGHER UNITS OR DENOMIN-
ATIONS TO LOWER.
1502. 1, Reduce 8 meters to millimeters.
OPERATION.
8 M. Ea:planation.—Remembering the table for the
10 Metric Linear Measure, we reason as follows:
10 e Since 1 meter = 10 dm., there are 10 times as
10 many dm. as M.; then since 1 dm. = 10 cm.,
*sº there are 10 times as many cm. as dim. ; then
8000 mm. Ans. since 1 cm. = 10 mm., there are 10 times as many
mm. as crim., which is, as shown by the opera-
tion, 8000 mm. Or we might reason thus: Since 1 M. = 1000 mm., there are 1000 times as many
mm. as meters.
METRIC SYSTEM OF WEIGHTS AND MEASURES. 935
GENERAL DIRECTIONS.
1503. From the foregoing elucidation, we derive the following general direc-
tions, to reduce Metric Numbers from Higher to Lower denominations: *
For the Linear, Capacity, and Weight Measures, multiply by 10 (annea 1
maught) for each lower denomination to which the given number is to be reduced.
For Surface Measure, multiply by 100 instead of 10; and for Cubic Measure,
multiply by 1000 instead of 10.
NotE.—In reducing the denominations of the Stere, in wood measure, 10 is the multiplier.
12.
13.
14.
15.
PROBLEMS.
Reduce 42 Him. to meters. Ans. 4200 M.
Reduce 33 Dm. to decimeters. Ans. 3300 dim.
Reduce 8 L. to milliliters. Ans. 8000 ml.
Reduce 1 Kl. to liters. Ans. 1000 L.
Reduce 4 G. to milligrams. Ans. 4000 mg.
Reduce 21 Kg. to centigrams. Ans. 2100000 cg.
Reduce 9.82 M. to millimeters. Ans. 98.20 mm.
Reduce 16 A. to centares. Ans. 1600 ca.
Reduce 25 sq. M. to sq. millimeters. Ans. 25000000 sq. mm.
Beduce 14 cu. M. to cu. millimeters. * Ans. 14000000000 cu. Imm.
Peduce 3 cu. dm. to cu. Centimeters. Ans. 3000 cu. cm.
Reduce 22 Ha. to centares. Ans. 220000 ca.
Reduce 54 DS. to decisters. Ans. 5400 ds.
Reduce 1 millier or tonneau to milligrams. Ans. 1000000000 mg.
TO REDUCE METRIC NUMBERS FROM LOWER UNITS OR DENOMINA-
1504.
TIONS TO EIIGHER.
1. Reduce 44505 mm. to meters.
OPERATION.
44505 Explanation.—Here again remembering the
10 Metric Linear Table, we reason as follows:
10 Since 10 mm. = 1 cm. there are I'd as many cm.
10 as mm. ; then since 10 cm. = 1 dra. there are ºf
*=== as many dm. as crim. ; then since 10 dm. = 1 M.
44.505 M. Ans. there are nº as many M. as dn., which is, as
shown by the operation, 44.505 M. Or we may
reason thus: Since 1000 mm. = 1 M. there are Tºwn as many M. as mm.
1505.
GENERAL DIRECTIONS.
From the foregoing elucidation, we derive the following general direc-
tions to IReduce Metric Numbers from Lower to Higher denominations:
For the Linear, Capacity, and Weight Measures, divide by 10 (point off 1
place) for each higher denomination to which the given number is to be reduced.
936 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
For Surface Measure point off 2 places instead of 1.
And for Cubic Measure point off 3 places instead of 1, for each higher denom-
ination to which the given number is to be reduced.
NotE.-In reducing the denominations of the Stere, in wood measure, 10 is the divisor.
PROBLEMS.
2. Reduce 48752 cm. to meters. Ans. 487.52 M.
3. Reduce 12307 drin. to Km. Ans. 1.2307 K.m.
4. Reduce 120 M. to Hm. Ans. 1.2 Hm.
5. Reduce 333444 cl. to liters. Ans. 33.34.44 L.
6. Reduce 10234 di. to Kl. Ans. 1.0234 Kl.
7. Reduce 24 ml. to Ml. Ans. .0000024 Ml.
8. Reduce 484500 mg. to G. Ans. 484.5 G.
9. Reduce 2389 cg. to kilograms. Ans. .02389 Kg.
10. Reduce 10 G. to Mg. Ans. .001 Mg.
11. Reduce 1 mg. to tonneaus. Ans. .000000001 T.
12. Reduce 3414864 sq. mm. to sq. M. Ans. 3.414864 sq. M.
13. Reduce 8.27 sq. cm. to sq. dm. Ans. .0827 sq. dm.
14. Reduce 55544.433366 cu. mm. to cu. M. Ans. 55.544.433366 cu. M.
15. Reduce 897568 cu. dm. to steres. Ans. 897.568 S.
16. Reduce 2246 cu. M., or Steres, to dekasteres. Ans. 224.5 Ds.
17. Reduce 447 ds. to dekasteres. Ans. 4.47 Ds.
TO ADD METRIC NUMBERS.
1506. 1. What is the sum of 462 mm. 28 cm. 406 dm. and 16 Dm.
Ans. 201.342 M., or 201342 mm.
FIRST OPERATION. SECONIO OPERATION. -
.462 462 Ea:planation.—In all problems of this kind,
.28 280 the numbers should be written in the base unit
40.6 40600 of the table and added as in decimals. Accor-
160. j60000 dingly, as shown in the first operation, we wrote
&=mºmºmº sºmmemº mºmsº *E---, -º-º-º-º: the numbers in meters and decimals of meters
201.342 M. Ans. 201342 mm. and then added.
In the second operation, we wrote and added the numbers as mm.
GENERAL IDIRECTIONS.
1507. From the foregoing elucidation, we derive the following general direc-
tions for Adding Metric Numbers:
Write the numbers to be added in the BASE UNIT, and decimals of the same, of
the table to which they belong and then add as in decimals. Art. 445, page 223.
* PROBLEMS.
2. Add 68 M. 42 dm. 3204 mm. and 63 Hm. Ans. 6375.404 M.
3. Add 7 Mm. 2 Km. 5 Hm. 7 Dm. 8 M. 3 dm. 4 cm. and 8.07 mm. -
Ans. 72578.34807 M.
METRIC SYSTEM OF weights AND MEASURES. 937
4. Find the sum of 21 L. 16 dl. 4 cl. and 27 Hl. Ans. 2722.64 L.
5. A grocer has four boxes containing as follows: 1st, 8.5 L.; 2d, 7 Dl.; 3d,
21 cl. ; 4th, 1 M. and 8 ml. How many L. in all? . Ans. 79.718 L.
6. What is the sum of 12 G. 3 dg. 9 cg. and 6 mg 3 Ans. 12.396 G.
7. What is the weight in Kg. of 3 bales of cotton which weigh respectively:
204.6 Kg.; 205 Kg. 4 Hg.; and 208 Kg. 8 G. 3 Ans. 618.008 Kg.
8. Add in Kg., 14 T. 6 Q. 8 Mg. 7 Kg. 4 Hg. 5 Dg. 9 G.
--> Ans. 14687.459 Kg.
9. Add the above problem in the unit of Tonneaus. Ans. 14.687459 T.
10. What is the sum of 124 sq. M., 6 sq. dm., and 37 sq. cm. ?
Ans. 124.0637 sq. M.
OPERATION INDICATED.
124.
,06
.0037
- *-
11. Add 42.8 sq. M., 21.65 sq. M., 28 sq. dm., and 4 sq. cm.
Ans. 64,7304 sq. M.
12. Add in the unit of Ares, 39.5 A., 25 Ha., and 84 ca. Ans. 2540.34 A.
13. What is the sum of 14.5 c.u. M., 23 cu. dm., 123 cu. cm., and 24 cu. mm. 3
Ans. 14.523123024 cu. M.
OPERATION INDICATED.
14.5
.023
.000123
.000000024
14. Add in the unit of Steres, 22 S., 12 Ds., and 15 ds. Ans. 143.5 S.
TO SUBTRACT METRIC NUMBERS.
1508. 1. What is the difference between 75.6 M. and 85 mm. 3
Ans. 75.515 M.
OPERATION. \
75.6 — M. Explanation.—In subtraction as in addition,
.085 = M. we write the numbers to be subtracted in the
base unit of the measure table to which the
75,515 M. Ans. numbers belong, and subtract as in decimals.
GENERAL DIRECTIONS.
1509. From the foregoing elucidation, we derive the following general direc-
tions for Subtracting Metric Numbers:
Write the numbers to be subtracted in the base unit and decimals thereof, of the
table to which the numbers belong, and then subtract as in decimals. Art, 447, page
224.
PROBLEMIS.
2. From 4324.08 Km. take 123.5 M. in the unit of M. and also of Km.
Ans. 4323956.5 M. 4323.9565 Km.
938 -- SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
3. Find the difference between 274.25 L. and 44.5 cl. Ans. 273.805 L.
4. A barrel contained 151.44 L., and 6 L. and 7 dl. leaked out. How many
liters still remain # Ans. 144.74 L.
5. What is the difference in L. between 1 Ml. and 1 ml. 3
AnS. 9999.999 L.
6. From 264.5 G. take 28.4 cg. Ans. 264.216 G.
7. From 1428 Kg. take 16.5 Hg. Ans. 1426.35 Kg.
8. What is the difference in T. and Kg. between 2 tonneaus and 2 Kg. ?
* Ans. 1.998 T.; 1998 Kg.
9. What is the difference between 88.21 sq. M. and 38 sq. dm.”
Ans. 87.83 sq. M.
10. A plantation contains 2471.14 Ha. If 2471.14 A. are sold, how many
hectares will remain } Ans. 2446.4286 Ha.
11. From 1528 cu. M. take 1 cu. M. 1 cu. dm. 1 cu. cm. and 1 cu. mm.
* AnS. 1526.998998999 Cu. M.
12. A wood dealer bought 150 steres of wood; he sold 70.25 cu. M. How
much has he remaining 3 * Ans. 79.75 S., or cu. M.
TO MULTIPLY METRIC NUMBERS.
1510. 1. How many meters are there in 11 pieces of cloth, each containing
36.25 meters ? “ Ans. 398.75 M.
OPERATION.
36.25 Explanation.—In multiplication of Metric
11 - Numbers, we proceed in the same manner as in
simple and decimal numbers. Hence no extend-
398.75 M. Ans. ed educidation and general directions are deemed
necessary.
2. What will 204.5 meters cost at $2 per meter? Ans. $409.
3. If a man walks 20000 meters per day, how many kilometers will he walk
in 60 days? Ans. 1200 Km.
4. What will 484 dm. cost at $1.50 per meter? Ans. $72.60.
5. What will 484 dm. cost at $1.50 per dm * Ans. $726.
6. What cost 75.2 liters of milk at 52 per L. Ans. $3.76.
7. What cost 200.52 Hl. of corn at $1.60 per Hl.? Ans. $320.83.
8. A fruit dealer bought 6 Hl. of pecans, at $9 per Hl., and sold them at 152
per liter. How much did he gain? Ans. $36.
9. Sold 26.4 Kg. of grapes at 452 per kilo. How much was received for
them ż * Ans. $11.88.
10. What cost 27.25 tonneaus of hay at $20.50 per T. * Ans. $558.625.
11. What cost 416.3 Kg. of sugar at 20g Ans. $83.26.
12. Bought 362,3122 G. of silver at 34% per gram. What did it cost?
Ans. $12.68–H.
13. How many kilograms in 49000 pills, the weight of each being .5 dg. 8
Ans. 2.45 Kg.
20. METRIC SYSTEM OF WEIGHTS AND MEASURES. 939
14. How many sq. meters in a yard 40.5 M. long and 15.24 M. wide?
*. Ans. 617.22 sq. M.
15. A plantation is 2.42 Km. long and 1500 M. wide. How many hectares
does it contain 3 Ans. 363 Ha.
16. How many cubic meters in a box 2.8 M. long, 2.1 M. wide, and 8.5 dm.
deep 3 Ans. 4.998 cu. M.
17. What will be the freight on 4 boxes, each measuring 3 M. by 26 M. by
.9 M. at $5.25 per cu. M. ? Ans. $147.42.
18. How many cubic meters of earth in a levee 140 M. long, 2.3M. deep, and
20 M. wide at the base and 15.2 M. wide at the top 3 Ans. 5667.2 cu. M.
19. How many steres in a pile of wood 28.5 M. long, 3.2 M. high, and 4.3
M. Wide 3 Ans. 392.16 S., or cu. M.
20. How many cubic meters of earth will it take to fill a lot .5 meter deep,
the lot being 60.2 meters long, and 25 meters wide? Ans. 752.5 cu. M.
TO DIVIDE METRIC NUMBERS.
1511. 1. A man walked 1600 meters in 20 minutes. How many meters did
he walk in 1 minute 3 Ans. 80 M.
OPERATION.
1600 — 20 = 80 M. Ans.
Explanation.—In division of Metric Numbers, we proceed with the operations in the same
manner as in simple and decimal numbers. Hence no extended elucidation and general directions
are deemed necessary.
2. The steamer Katie ran 500 Km. in 22 hours. How many kilometers did
she run per hour? Ans. 22.727–H Km.
3. If 6.5 meters make a suit, how many suits can be made from 195 meters?
Ans. 30 suits.
4. In 425 liters, how many hectoliters? g Ans. 4.25 Hl.
5. Bought 45 liters of strawberries for $3.375. What was the price per L. "
Ans. 732.
6. If you divide 180 hectoliters of potatoes equally among 24 persons, what
will each receive 3 Ans. 7.5 EIl.
7. 21 kilograms of sugar cost $4.62. What was the cost of 1 Kg. ?
Ans. 222.
8. How many days will 42.5 T. of coal last a family, if they burn 150 Kg.
per day? Ans. 283.33+ days.
9. A druggist has 2.45 Kg. of medicine which he wishes to make into pills,
each to contain .5 dg. How many pills will there be? Ans. 49000 pills.
10. A garden contains 900 sq. M., and is 22.5 M. wide. How long is it 3
Ans. 40 M.
11. A side-walk is 80.4 M. long by 4.2 M. wide. How many tiles, each 2.4
dm. long and 1.2 dm. wide, will be required to pave the side-walk 3
Ans. 11725 tiles.
94O soul E's PHILOSOPHIC PRACTICAL MATHEMATICs. ºr
12. How many ares in a piece of land 60 meters long and 42.2 meters wide?
* Ans. 25.32 A.
13. A box is 2.5 dm. long, 2 dm. wide, and 1.5 dm. deep. How many of such
boxes may be put in a larger box which is 2.5 M. long, 2 M. wide, and 1.5 M. deep
Ans. 1000 boxes.
14. If you buy 5.6 dekasteres of wood and use 1.1 cu. meters per day, how
long will it last : Ans. 50+} days.
TO REDUCE METRIC TO AMERICAN WEIGHTS AND MEASURES.
1512. 1. Reduce 2.6 meters to feet. Ans. 8.53–H feet.
FIRST OPERATION.
1 M. = 39.37+ inches, practically. Explanation.—Referring to the table of equiv-
tºmº 2.6 alents, we find that 1 M. = (practically) 39.37
12 ) 102.362 inches. in. We then reason as follows: Since 1 M. =
8.530+ feet. Ans. 39.37 in., 2.6 M. are equal to 2.6 times as many,
or thus: which is 102.362 in. ; then since 12 in. = 1 ft.,
12 ;" & there are ſº as many feet as inches, which is
dºmºrºss-sºrs -- *- 8.53–H feet.
SECOND OPERATION.
3.2808–1– = feet, the equivalent of a meter.
2.6
8.53008-H = feet, Ans.
GENERAL IDIRECTIONS.
1513. From the foregoing elucidations, we derive the following general direc-
tions for Reducing Metric to American Weights and Measures:
Multiply the equivalent value of the metric unit given, by the metric number to be
ºreduced, and them, if necessary, reduce the product to the denomination required.
NOTE.-In working the following problems, four decimal places have been taken as a stand-
ard for all unit equivalents of more than that many decimals. The calculations are made directly
upon the equivalency of the denominations to be reduced.
2. Reduce 408.2 kilometers to yards. Ans. 446407.52 yds.
3. In 24.5 cm. how many inches : Ans. 9.6456-H in.
4. Reduce 70 M. to yards. Ans. 76.552-- yds.
5. Reduce 25.5 liters to dry quarts. Ans. 23.1565-H d. Qts.
6. In 40 kiloliters how many gallons? Ans. 10567.44+ gals.
7. In 200 hectoliters how many bushels? Ans. 567.58+ bus.
8. Reduce 25 grams to Troy grains. Ans. 385.81+ grs.
9. Reduce 75.2 kilograms to pounds Avoirdupois. Ans. 165.7859-H lbs.
10. In 444 tonneaus how many tons? Ans. 489.4256+ tons.
11. Reduce 5.5 mg. to Troy grains. Ans. .0847–H grs.
4. METRIC SYSTEM OF WEIGHTS AND MEASURES. 94. I
12. Reduce 24 ares to square rods. Ans. 94.8912–H Sq. rds.
13. Reduce 85.5 hectares to acres. Ans. 211.279-H acres.
14. In 500 hectares, how many sq. miles? Ans, 1.9305+ Sq. mi.
15. Reduce 150 cu. meters to cu. feet. Ans. 5297.475-H Cu. m.
16. Reduce 7.9 cu. mm. to cu. inches. Ans. .0004819-H cu. in.
TO REDUCE AMERICAN TO METRIC WEIGHTS AND MEASURES.
1514. 1. Reduce 40 feet to meters. Ans. 12.192–H M.
()PERATION.
§ ft. Ea:planation.—Remembering that a meter =
39.37) 480 in. (12.192–H. M. 39.37 inches, practically, we first reduce the
or thus: *-*m-mºsºs feet to inches, and then reason as follows: Since
40 in 39.37 in, there is 1 meter, in 480 in, there are
39.37 || 12 as many meters as 480 in, are times = to 39.37
*-* | *=º in., which is 12.192—H.
SECOND OPERATION.
3.048 decimeters = the equivalent of 1 foot.
40
121.920 decimeters, which divided by 10 = 12.192 meters, Ans.
2. Reduce 1 mile, 8 rods, 10 feet, and 6 inches to kilometers.
Ans. 1.652781–H PGm.
OPERATION.
1 mi., 8 rôls., 10 ft., 6 in. 39.37)65070 inches.
320 *** *-*-g
&=ºmº 1652.781–H. M.
328 rods. 1.652781–H Km.
16#
** == Or,
5422 feet. 65070 in. × 2.54 cm. = 165278.1+ cm. = 1.652781–H. Km.
12
65070 inches.
Explanation.—In this problem, we first reduce the given number to inches, then to meters, as
above explained; and then to kilometers by dividing by 1000, the number of meters in a kilometer.
GENERAL DIRECTIONS.
1515. From the foregoing elucidations, we derive the following general direc-
tions for reducing American to Metric Weights and Measures:
1O. Reduce the given number to the lowest unit named, or to a convenient
denomination, of which the equivalent value of the metric unit is known ; then divide
the same by the equivalent value of the metric unit, and when necessary reduce the
quotient to the required denomination.
Or 20. Multiply the given number, reduced to its lowest denomination, by the
equivalent value of such denomination in the metric unit.
94.2 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS.
3. In 70 yards, how many meters? Ans. 64.008-H M.
4. Reduce 35 gallons, 3 quarts, and 1 pint of molasses to liters.
#. Ans. 135.8001-H liters.
5. Reduce 5 bushels, 2 pecks, 5 quarts, and 1 pint to liters.
Ans. 199.8678-H liters.
6. In 20 gallons of milk, how many liters? Ans. 84,539-H L.
7. In 400 bushels, how many hectoliters? . Ans. 140.9493— Bil.
8. In 3 gills, how many centiliters? Ans. 41.4364-H cl.
9. Reduce 200 pounds Avoirdupois to kilograms. Ans. 90.7194+ Kg.
10. Reduce 15 ounces Avoirdupois to hectograms. Ans. 4.2525+ Hg.
11. Reduce 10 ounces Troy to grams. Ans. 311.0339-H G.
12. Reduce 50 tons to tonneaus. Ans. 45.3597–H T.
13. Reduce 20 grains Troy to centigrams. Ans. 129.6176+ cg.
14. Reduce 240 sq. yards to sq. meters. Ans. 200.6689— sq. M.
15. In 500 sq. feet, how many sq. decimeters? Ans. 4646.8401-H sq. dm.
16. In 200 acres, how many hectares? Ans. 80.9356 Hectares.
17. Reduce 1610 cu. feet to cu. meters. Ans. 45.5877–H cu. M.
18. Reduce 2500 cu. yards to cu. meters. Ans. 1911.3149–H cu. M.
19. In 190 cords, how many steres? Ans. 688.6553–H Steres.
MISOELLANEOUS PROBLEMS.
1516. 1. A yard is 21.4 M. long and 15.12 M. wide. How many bricks will
it require to pave the yard, each brick being 10.32 cm. long, and 9.16 cm. wide?
2. How many cubic meters in a piece of marble 3.5 M. long, 3.21 M. wide,
and 1.85 M. thick? *
3. How many liters in a cylindrical vessel 2.3 M. deep and 8.2 de.. in diame-
ter ?
4. What is the weight of 22 bags of wheat averaging 68.42 Kg. ?
5. If 4 men eat 9.5 Kg. a week, how much would 25 men eat at the same
rate 3
6. If 4 men can dig 245 M. of a ditch in 24 days, how long will it take 10
men to dig 3200 M. at the same rate 3
7. A room is 6.5 M. long and 4.25 M. wide. How many square meters of
carpet will be required to carpet the floor, the carpet being 75 cm. wide, no allow-
ance being made for matching patterns or in turning under ?
8. A room is 8.4 M. long, 6.25 M. wide, and 4.5 M. high. How many liters of
air does it contain }
9. If the breathing of one person pollutes the air of a room at the rate of
.1855 com. per minute, how long will it take, at the same rate, for 50 persons to
pollute the air of a room 30 M. long, 20 M. wide, and 8.25 M. high 3
10. If 42 Kg. of cotton are required to make 150 M. of cloth .75 M. wide,
how many kilometers of cotton will it take to make 5000 M. of cloth 1.2 M. wide?
11. How many cubic meters in a cylindrical piece of marble 2.1 M. in
diameter and 6.42 M. long? -
12. How many square meters, board measure, in 24 pieces of timber 9.32
M. long, 48 cm. wide, and 32 cm. thick?
º
1517. Combinations are the different groups which can be made of n things
taken m in a group, m being equal to or less than n.
Or Combinations are the various ways that different things, the number of
which is known, can be chosen or selected taking them one by one, two by two,
three by three, etc., without regard to their order.
PROPOSITION.
1518. Any number of things being given, to determine in how many ways
they may be combined two and two, three and three, etc., without regard to the
order.
PROBLEMS.
1. If there are four routes 1, 2, 3, 4 leading to New York, in how many
different ways can a person go and return ? Ans. 16 ways.
Explanation.—Since a person has the choice of going by any route and returning by the same
or another route, four choices are offered to him, and since he has the same number of choices with
each route (there being four of them) he must therefore have the choice of 16 ways, viz.:
Go. Return. Go. Return. Go. Return. Go. Return.
1 1. 2 1 3 1 4 1
1 2 2 2 3 2 4 2
1 3 2 3 3 3 4 3
1 4 2 4. 3 4 4 4
sm-m. g-ºº-ºº- *-mºs ºmºm-º:
TAT + TaT + TT + TT– 16
{
2. Suppose in the above problem, the tourist does not desire to return by the
same route, how many combinations or choices of route has he? Ans. 12.
Explanation.—If he goes by any route, then he has the choice of returning by three others
or three ways of making the round trip for each different route leading to New York. Since there
are four routes leading there, he has 3 × 4 = 12 choices in the matter, viz.:
Go. Return. Go. Return. Go. Return. Go. Return.
\ } § 1 1 .
1 & 3 2 & 3 s: 2 *} 2
& 4 & 4 3
3 + 3 + 3 + 3 = 12
From the above the following principle is evolved:
If one thing can be done in a different ways, and (when it has been done in
any one of these ways) another thing can be done in b different ways, then both can
be done in a × b different WayS.


(943)
944 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
3. In how many ways can the letters from the word marble be taken with
the letters from the word home 3 Ans. 24 Ways.
Explanation.—Each letter of the word marble (there are 6) can be combined with each letter
of the word home (there are 4). Therefore we have 6 x 4 = 24 ways they may be taken. Extend-
ing the preceding principle, we derive the following: If one thing can be done in a ways, and then
a. Second thing can be done in b ways, and a third in c ways, and a fourth in d ways, etc., the
number of ways of doing all the things will be a × b X c X d, etc.
4. In how many ways can 5 books be given to five persons? Ans. 120.
Explanation.—Since the first book may be given to any one of the five persons, there are five
ways of bestowing it.
The second book may be given to any one of four persons; the third book to any one of
three persons; the fourth book to any one of two persons, and the last book to the fifth person.
Hence the books may be bestowed in one of 5 × 4 × 3 × 2 × 1 = 120 ways.
5. In how many ways can a vowel and a consonant be chosen out of the
Word logarithms? Work according to first principle. Ans. 21 ways.
6. In how many ways can 4 dolls be given to 3 girls? Ans. 81 Ways.
Explanation.—Each doll can be given to any one of the three girls. Therefore the number of
ways of disposing of them is 3 × 3 × 3 × 3 = 81 ways.
PERMIUTATIONS.
—eſpe—
1519. In the definition of combinations, we found that the order of the
elements in combination was disregarded.
A Permutation of any number of elements or things means a group of that
number of elements or things put together with reference to their order or sequence.
PROBLEMIS.
1. In how many ways can the letters in the word hound be arranged, taken
all at a time Ans. 120 ways.
Ea:planation.—The reason given in problem 4 in combinations applies here, and hence is
omitted. º
Formula 1. The number of permutations of n diifferent elements taken all
at a time is *
n (n − 1) (n − 2) . . . . . . × 2 × 1.
The continued product of this formula is the whole number of permutations
of n elements taken all at a time.
For convenience this formula may be written | n and is read factorial n.
2. In how many ways can 10 different colored lights be arranged, taken 3 at
a time? Ans. 720 ways.
Formula 2. n (n – 1) (n — 2) ...... to r factors or n (n − 1) (n − 2) . . . . . .
(n – r + 1). Substituting the figures for the letters we have 10 (10 – 1) (10 – 8
+ 1) = 10 × 9 x 8 = 720, Ans.
ºt PROBABILITIES OR CHANCE. 945
In the formula, n = number of things given to be arranged and r = number
of things taken at a time.
Formula 3. The number of arrangements of n elements, of which p are
alike, q others are alike, and r others are alike etc., is
| p X | g × r
3. How many permutations can be made out of the letters of...the Word
tintinnabulation? Ans. 12,108,096,000.
OPERATION.
| 16
| 4 || 3 || 3
4. How many arrangements or permutations can be made out of 8 books, 3
being Arithmetics, 2 being Histories, 2 being Grammars, and 1 being a Dictionary?
|2
Ans. 1680.
OPERATION.
| 8
| 3 || 2 || 2
-
-smº
-
For other and higher cases of permutations, the learner is referred to
Wentworth's Complete Algebra.
PROBABILITIES OR CHANCE.
—eſpe—
1520. This subject has engaged the attention of writers on mathematics since
Galileo first wrote on it in 1642. Many mathematicians however, were prejudiced
against it on account of its ready applicability to games of chance with cards,
dice, etc.
When an event can happen in a certain number of ways, and it fails in a
certain other number of ways, the chance of its happening can be expressed by a
fraction whose numerator expresses the number of chances favorable to the event
happening and whose denominator expresses the whole number of cases.
Simpson defines the probability of an event happening to be the ratio of the
chances by which the event in question can happen to all the chances by which it
cannot happen or fail. If m + n be the whole number of cases that can happen, and
m represents the number of cases favorable to the happening of the event, the
probability of the event happening would be expressed by the fraction H and
946 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. Yºr
the chances against its happening would be 770, + n" Therefore we see that the sum
of the chances of an event happening or not happening is unity or certainty.
In tossing a coin once, what chance is there that it will turn up a head or a
tail? Since there are only two things possible to turn up, either a head or a tail, it
is evident that the probability of its being a head is H = } and the probability
that it will be a tail is 1 # i = }. The sum of the chances for and against is 3 + 4
= 1 = certainty. That is, it is certain that one or the other will turn up.
What chance is there in two throws of turning a head up twice?
According to the subject of combinations or choice, we find that with 2
throws there will be 4 combinations possible.
1st explanation for Problem 1, in Combinations, is applicable, viz:
1 head and head.
2 head and tail.
3 tail and head.
4 tail and tail.
Of the four ways in which it is possible for the coins to fall, there is only 1
way in which the event will happen. The probability of the event happening is
hence #. It might be shown that the chance of throwing 3 heads or 3 tails in
succession, will be $, or 4 × 4 × 4. The chance of two or more single events being
known, the chance of their all taking place together, may be found by multiplying
the probabilities of the events, considered singly.
To find the chance of throwing in one throw all heads or all tails with 2, 3, or
4 pieces, may be found as above. When two pieces are tossed up there are 4 com-
binations possible, hence the chance is 4. When three pieces are thrown up there
are 8 combinations, for each coin may fall in one of two ways, and there being 3
coins there are 2 × 2 × 2 = 8 combinations possible. The chance is therefore #.
Tossing up three pieces at once, is just the same as tossing up the same piece 3
times. e
1. What odds should be offered to make an even bet that in 2 throws a
head will fall at least once? Ans. 3 to 1.
As we have seen before, in two throws there is only 1 throw in which we find
no head, while we have a head in three of them.'
2. What chance is there in one throw of throwing an ace with one die?
Ans... }
3. What chance is there in one throw of throwing double aces with two
dice? Ans. #'s.
PROBABILITIES OR CHANCE, 947
1521. TABLE TO SHow How MANy ways A GIVEN SUM MAY BE THROWN
WITH ONE, TWO, THREE, OR MoRE DICE.
Number of dice.
1 2 3 4 5 6
| 1 l
2 || 1 || 1 Ea.planation of Table.—If it be required to find how
many ways 8 could be thrown with 4 dice, we first look
© 3 || 1 || 2 | 1 in the 1st column on the left for 8, and at the top of
tº
# 4 || 1 || 3 || 3 1 the table for 4, and in the square below opposite to the
3. ETTT. To Tal ITT 8 we find 35, the number of ways in which 8 may be
3 |—|— thrown with 4 dice. The probability of throwing 8
§ 6 || 1 5. 10. 10| 5 || 1 points with 4 dice is 1355. The probability of throwing
É 7 6 || 15 20| 15| 6 any number of points with a given number of dice is
2. 8 5 21 || 35| 35| 21 found by dividing the number of ways the points can
9 4 || 25 6| 70| 56 be thrown with the given number of dice by that
5 56 7 power of 6 indicated by the number of dice.
10 3 27 | 80|| 126|| 126
11 2 27 | 104| 205 252
12 1 || 25 | 125, 305| 456

IPROBLEMS
1. Two persons sit down to play for a certain sum of money; and agree
that he who first gets three games shall be the winner. One of them won 2 games
and the other, won 1; but being unwilling to continue their play, they resolve to
divide the stake; how much ought each person to receive?
If the play were to be continued, and the second should win, then each one
having only 1 game to win out, would be entitled to half of the money.
But if the first one should win the game, the whole money (#) deposited
would be his; and if he lost it, he would have a right to the half. But since one
case is as probable as the other, the first player has a right to one-half of the sum
of the two parts (3 + 3 = }) = }, and the second player only #.
2. Suppose the first player needs one game to win, and the second three, how
should the money be divided ?
If 'the first player should win one game, he would be entitled to all (#) the
money; and if he lost the game he would be entitled to #, according to the case
above. But as both are equally probable, the first player should have # of the two
sums (4 + # = }) = }, and the Second player #.
3. Suppose the first player needs two games to be out, and the second three,
how should the money be divided ? -
If the first player won the game, he would be entitled to , as shown in the
948 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. Yºr
“...
problem above. If the second player won, each would be entitled to #. But as
both are equally probable, the first player is entitled to # of the sum of the two
parts (; + 4 = ##) = +}, and the second to #.
4. If 4 cards are drawn from a pack of 52 cards, what is the chance that
four kings will be drawn f
FIRST OPERATION.
* * * * * * * = 270725 = number of ways four cards can be drawn from the pack.
1 × 2 × 3 × 4 *
SECOND OPERATION.
Four, cards can be selected so as to be kings in 4 × 3 × 2 × 1 = 24 ways. Therefore the
chance of drawing four kings from a pack is grážg or 1 chance in 11280.
The first operation depends upon the principle, that “Out of n different
elements, the number of selections of r elements is equal to the number of arrange-
ments of n elements divided by | rº' The formula for which is
n (n − 1) (n — 2) . . . . . . (n – r + 1)
| r 2
S =
The second operation depends upon the principle, that “The number of
arrangements or permutations of n different elements taken all at a time is n (n —
1) (n − 2). . . . . . × 2 × 1.” This formula is represented by Ln.
5. If 2 letters are selected at random, out of the alphabet, what is the
chance that both will be vowels? Ans. 1 in 32.
OPERATION,
Two letters can be selected from 26 in *::: = 325 ways; and 2 vowels can be selected
from 5 vowels in ; X : = 10 ways. Therefore the chance of drawing 2 vowels is ºr = ºr or 1
chance in 32.
6. In a box of chess are 16 pawns, 4 knights, 4 bishops and 2 queens. What
is the chance of drawing one of each in 4 draws? Ans. 1 in 700.
OPERATION,
26 × 25 X 24 × 23
1 × 2 × 3 × 4
1 pawn, ºan be selected from 16 pawns in 16 ways; 1 knight from 4 knights in 4 ways; 1 bishop
from 4 bishops in 4 ways, and 1 queen from 2 queens in 2 ways. Hence 1 pawn, 1 knight, 1 bishop,
and 1 queen can be drawn in 16 × 4 × 4 × 2 = 512 ways, Hence the chance is gºt or 1 in 700,
4 pieces of chess can be selected from 26 pieces in = 358800 ways. Also
Çontinued Fractions.
----------------------------Rs
—sº a dº-
1522. A continued fraction is a fraction with one for a numerator, and whose
denominator is an integer plus a fraction with one for a numerator, and whose
denominator is an integer plus a fraction, etc.
NoTE.—For a full elucidation of fractions, see pages 164 to 167, and 216 to 219.
Continued fractions were first used by Cataldi, in 1613, and by Lord Brounker,
in 1670. Continued fractions are used to express approximately, ratios that by
reason of the large numbers used, do not carry with them appreciated values.
They are also used to determine approximations to roots of surds, and are also
applied to the Solution of indeterminate problems.
Reduce # to a continued fraction.
1.
2.7
# 5
-as
-
3 + 1
6T1
1 + +
DIRECTIONS.—Divide both numerator and denominator by 27 and we have
#x ; divide the numerator and the denominator of the second fraction by 4, and
1
we have 3 + I_; divide the numerator and the denominator of the third fraction
6 - #
by 3, and we have 1
3-E I
6 —H 1
I-Ei Ans. This completes the work, as the numerator
of the last fraction is unity.
Reduce the continued fraction
1
3 + 1
6
1
I-E to a common fraction.
First Method.
1.
1 + # º
by multiplying the numerator and the denominator by 3, adding to the denominator
* (949)
Reduce the last two terms of the continued fraction to a simple fraction
95o SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. º:
f. 3.&
the numerator of the last fraction, and we obtain #. Take this result and the preceding
partial fraction, and we have *—. multiply as before, and we have the simple
6 + #7
fraction #1. Take this result and the preceding partial fraction, and we have ; Tº
multiply as before, and we have #, the simple fraction.
Second Method.
1st step = } = 1st approximation.
2d step = *H = +% = 2d approximation.
3d step = :g-H = # = 3d approximation.
4th step = #H = # = original fraction.
Explanation.—Write the first term (#) of the continued fraction as the first approximate
fraction. Write this approximate fraction with the next term of the continued fraction and
reduce to a simple fraction, and we have : = }; second approximate fraction. Write this
; multiply this fraction
l
by the denominator of the second term as above, and add to both terms the corresponding terms of
1 × 6 + 1
1 × 19 + 3
Write the third approximate fraction with the last term of the continued fraction, and we have
fraction with the next term of the continued fraction and we have :
the first approximate fraction ($), and we have = y; the third approximate fraction.
# # ; multiply both terms by 3, and add to the product the corresponding terms of the second
+
approximate fraction (#), and we have ####, = ##, the answer.
To illustrate the application of continued fractions, let us find the approxi-
mate ratios of a circle to a circumscribed square. The ratio is ###. Reducing
this to a continued fraction, we have
1
1 + 1
3
}
*—
Now obtain the approximate values according to the 1st or 2d methods, and
4 7 1 7.2 18:3, 3 5.5 - 89.3 3 927 7.8 & 4
we have +, #, Tây 32 ++, 2 197 ###, 4 52.7 TI37; 5 JO (, ) iſ Ó O Ú ().
NoTE.—Fractions should be reduced to their simplest form before beginning the operation.
1523. 1. The product of three numbers is 71757; one of the numbers is 51,
and another 21. What is the third number? Ans. 67.
2. What is that number which being divided by 25, the quotient increased
by 6* + 2, the sum diminished by the difference between 41 and 25, the remainder
multiplied by 9, and the product divided by 27, the quotient will be 10?
- Ans. 200.
3. Henry owes James $28.50 which he wishes to pay with an equal number
of picayunes, dimes, quarters, halves and dollars. How many of each will it
require ? Ans. 15.
4. What is the fraction, which being divided by gº, will produce #3
Ans. Zºo.
5. What number, multiplied by 3, will give a product of 22? Ans. 33.
6. What fraction, to which if you add +, the sum will be #3 Ans. #.
7. What quantity, to which if you add 3 of itself, the sum will be 75?
Ans. 45,
OPERATION.
# -- 1 = 1% = }, then if # = 75, # will = 15 and # = 45. Ans,
8. What quantity, from which if you subtract 3 of itself, the remainder will
be 15% Ans. 24.
OPERATION.
1 — # = }, then 15 — # = 24. Ans
9. Divide 18 oranges between A. and B. so that A, will have + more than
B. What number will each have? - Ans. A. 10. B. 8.
OPERATION. -
B. A.
18 18
9 || 4 9 || 4
1 assumed for B. &=ºmº | * 4 5
1+ ( & & 4 A. 8 - -*
e- 10
2# = # the sum of the proportions according to the conditions of the question
10. Divide 18 oranges between A. and B. so that A. will have + less than
B. What number will each have? Ans. A. 7# B. 10%.
OPERATION.
B. A.
18 ; 18
7 || 4 7 || 4
1 assumed for B. *-mº | -- 4 || 3
# & 4 ** A. º 10% - -->
7}
1} = 4 the sum of the proportions according to the conditions of the problem.
(951)

952 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
- 11. Divide $2000 among A., B. and C. so that A’s part will be to B's as
2 to 3, and that C. will have as much as A. and B. together, less $200. What
amount will each receive?
SOLUTION.
As C. is to have as much as A, and B. together, less $200, A. and B. will therefore have 3 of
$2000 + 3 of $200, which is $1100. Then, as A. and B. are to receive this in the proportion to 2 and
3.A. will therefore receive # and B. 3 of $1100. # of $1100 = $440 A's share; # of $1100 = $660
B’s share; and lastly, as C. is to receive as much as A. and B. together, less $200, he will therefore
receive $1100 — 200 = $900.
12. Divide 1254 acres among A., B. and C. giving C. 74 acres more than
y B. and B. 12# acres more than A.
OPERATION INDICATED.
B & 4 & & { { { %
&
7+ + 12# = ; = number acres C. had more than A.
1 # - & 4 & 4
32# = number acres over an equal division of the balance. .
1253 — 31# = 92% balance.
92.75 –– 3 = 30.91% acres, A’s share.
12.75 -- 30.913 = 43.66: “ B’s “
20 -i- 30.91} = 50.913 “ C's “
125.50 acres, total.
13. A. and B. have $9000. 5 times A's money is to 6 times B’s money as 5
to 9. How much has each.
OPERATION INDICATED.
5 × A's share = # × 6 times B's share. 1 × A's share = # of B's share.
1 = B's share of the whole. } = A's share of the whole.
1 + 3 = 13 = sum of A. and B's shares. Then,
; : 1: ; $9000: $5400 B's share. : : #: ; $9000: $3600 A's share.
14. A merchant sold 384 bbls. flour, part at $7.25 and part at $5.50 per bbl.
He received for the whole $42 more than if he had sold all at $63 a bbl. How
many bbls. were sold at each price?
OPERATION INDICATED.
$7.25 = value 1st quality = - $7.25
5.50 = ** 2d & & - * 5.50
$1.75 = difference in price of 1st 2 | 12.75
quality over 26. $6,37} = average price if half of each
quality were sold.
2 | 384
192 = half of amount sold.
1.75)42.00 24 blols. more of 1st quality sold than 192.
192 + 24 = 216 bbls. 1st quality.
192 – 24 == 168 “ 2d & 4
MISCELLANEOUS PROBLEMS.
953
15. A man engaged to work a year for $540, and a suit of clothes. He
Worked 9 months and received $396 and the suit of clothes.
What was the Suit
OPERATION INDICATED.
$540 + 1 suit.
45 -- Tº “
9 months = $396 -- 1 “
44 + “
$45 + "g, monthly average for twelve months.
44 + # 4 & & 4 {
Worth 3
12 months =
1 month =
1 month =
$45 – 44 = $1.
* nine & 4
§ – I's = #. Hence $1 = 3; value of the suit and # = $36 value of suit.
16. A man rides a certain distance at the rate of 6 miles an hour, and
Walks back the same distance at the rate of 34 miles an hour. If it takes him 4;
hours to go both ways, what is the distance between the two places?
OPERATION INDICATED.
1 hour's ride = 6 miles.
1 hour.
7 || 2
6
or 33 : 6 : : 1 : 1% hours.
1} = hours to walk back.
1 = hour to ride there.
2} = hours to ride and walk.
19 || 7
4
6 miles.
9 or 2% : 6: : 43 : 10}
1
10# = number miles places are apart.
17. Two trains, one 210 feet long and the other 230 feet long, were going in
the same direction on parallel tracks.
Having started together with engines
Opposite, one of them passed the other in 15 seconds.
When going in opposite
directions at the same rate of speed they passed each other in 33 seconds. How
fast were they going 3
OPERATION
When the shorter train is the faster.
210 feet = length of faster train.
230 feet = length of other train.
440 = number of feet heads of engines were
apart at end of 3% seconds.
- Seconds. Feet. Feet.
3# 440 | 440
15 QC 15 4
15
1760 = feet travelled by
trains in 15 Sec.
1760
210 = feet faster train went over the other.
2 | 1550
775 = feet travelled by slower train in 15
210 seconds.
985 = feet travelled by faster train in 15
seconds.
Mi. Mi.
775 985
15 || 60 15 60
5280 || 60 5280 || 60
443 mi, per hour
faster train runs.
35; mi. per hour
slower train runs.
INDICATED.
When the longer train is the faster.
1760
230 = feet faster train went over the other.
2 | 1530
765 = feet travelled by slower train in 15
230 seconds
995 = feet travelled by faster train in 15
seconds.
Mi.
765
15 || 60
5280 | 60
34}% miles per hour slower train runs.
Mi.
995
15 || 60
5280 || 60
45; miles per hour faster train runs.
954 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
*
18. A column of troops, 25 miles long is ordered to a point 25 miles distant.
A courier starts simultaneously with the rear of the column, and reaches the head
thereof. Returning, he meets the rear of the column at the point where the head
originally Was. . Both the troops and the courier are to travel at a uniform rate of
speed. How many miles does the courier travel?
It is evident from the conditions of the problem that the courier had to travel the whole
length of the column (25 miles) plus the distance which the head of the column moved while the
courier was overtaking it (a miles). It is also evident that when the courier reached the head of
the column, the distance that the rear of the column travelled must have been the extra distance
the courier had to travel to overtake the head of the column, viz.: a miles. Then, since the
courier; in returning arrives at the end of the column when it had reached the point where the
head first lay he must have travelled a miles in returning from the head of the column.
C
K_ 25 NAILEs . > X Mill-Es K
/ X MILEs 'N
N Z
&
D C
A%– × Mil-Es X 25OMILEs X B
rºz 25 N/IIL.E.S. X 25 N/IILES >

A = rear of column.
O = head of column.
B = point head of column must reach.
C = point courier reached when he overtook the head of the column.
C’ = point where head of column is when overtaken by the courier.
D = point rear of column reached when courier had reached the head of the column.
From the above explanations and diagram, we obtain the following proportion
which, when simplified, gives the desired result:
25-H ac : æ : : a 25 – 3:
ac2 = 625 — ac”
2 ac2 = 625
ac2 = 312.5
a = 17.6776–1– miles.
Since a = the extra distance the courier travelled to overtake the head, and as he had to
i. the same distance to meet the rear, he must have travelled (2 × 17.6776–H) + 25 = 60.3552
19. If a man counts $1 every second and counts 12 hours every day, without
stopping, how many days will it take him to count $1000000, and how many to
count a billion dollars?
Ans. 23# days to count a million.
23148# days = 63 years, 1374; days to count a billion.
NotE.—365+ days were considered a year, in reducing the days to years.
20. How many years, of 365+ days each, will it require to count a trillion, by
counting 12 hours every day, and counting $2 every second 3
Ans. 31688 years, 32% days.
2} MISCELLANEOUS PROBLEMS. * 955
º
21. § of A's money and # of B's money is $14, and A’s money is $4 more
than B's money. How much has each 3 Ans. A. $40; B. $36.
OPERATION.
$13 w! of $4 = 80c. part of A's $4 excess in the $14; $14.00 – 80c. = $13.20; then } + k = }} =
a $
13.20 - 13.20
11 || 30 11 || 30
5 1 = A's part. 6 || 1 = B's part.
$7.20 + .80 A's excess = $8 = } of A's money, $6 = # of B's money, which is,
which is, therefore, $40. therefore, $36.
22. What is the interest on £50 12s. 6d. for 93 days at 6 per cent, allowing
365 days to the year 3 º Ans. 15s. 6d.
NOTE.-See interest on English Money.
23. A benevolent man gave $5 to each of several beggars and had $15
remaining on hand. Had he given each $10, it would have taken all the money he
had. How many beggars were there? Ans. 3.
OPERATION,
$10 — 5 = $5; $15 - 5 = 3. Ans.
24. A., B. and C. are to receive $26 in proportion to 4, §, #. What will each
receive? Ans. A. $12; B. $8; C. $6.
OPERATION.
A. B. C.
$26 $26 $26
13 | 12 13 | 12 13 | 12
# -- # -- + = }} 2 3 4
$12 $8 | $6
25. A. and B. bought at auction an invoice of merchandise for $800, of
which sum A. paid $500 and B. $300. They then sold to C. § interest in the
merchandise for $400. How much of the $400 must A. and B. receive respectively
in order to constitute each, A., B., C. § owner of the goods?
Ans. A. $350; B. $50.
NOTE.-For the solution of this and similar questions, see pages 872 and 873.
26. A California miner has a spherical ball of gold 2 inches in diameter
which he wishes to exchange for spherical balls 1 inch in diameter. How many of
the smaller spheres should he receive? Ans. 8.
NoTE.—For the solution of all similar questions, see pages 377 and 394.
27. A boy bought a certain number of peaches at the rate of 4 for 5 cents
and paid for them with apples at the rate of 2 for 3 cents. How many peaches did
he buy, providing it required 300 apples to pay for the peaches? Ans. 360.
OPERATION.
P. 300
300 5 || 4 2 || 3
2 | 3 | $4.50 Or, 5 || 4
- | $450 value of apples. 360 Ans. 360
956 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. Yºr
28. Two little girls, Susie and Katie, were employed by a fruit dealer to
peddle oranges; the first day they received 30 oranges each, Susie's being of superior
Quality, which they sold as follows:
Susie sold 2 for 5 cents, and received 75 cents.
Katie { { 3 {{ 5 € $ { % 50 { %
*
Total receipts, $1.25
The next day Susie was sick, and Katie took the 60 oranges and sold them at
the rate of 5 for 10 cents. At night, when she made her returns to the fruit dealer,
she had but $1.20; and as 60 oranges brought $1.25 the day before, when sold 2 for
5 cents and 3 for 5 cents, or seemingly 5 for 10 cents, the fruit dealer charged little
Katie with stealing the 5 cents. Did little Katie steal the 5 cents, and if not, how
do you account for its loss?
Ans. Little Katie did not steal it. It was lost because the ratio of 5 for
10 cents ceased when 10 sales had been effected.
SOLUTION.
As Susie sold 2 oranges at a sale, it is evident that to sell 30 she made 15 sales, which, at 5
cents a sale, amounted to 75 cents; and as Katie, the first day, sold 3 oranges at a sale, it is evident
that to sell 30 she made 10 sales, which, at 5 cents a sale, amounted to 50 cents, making, as above,
$1.25 for 60. The second day Katie mixed the 60 oranges, and by selling 5 at a sale made 12 sales,
which, at 10 cents a sale, amounted to $1.20. By this we see that the 5 cents was lost by the
manner of effecting sales; and hence, little Katie is innocent of the crime with which she is
charged. The error in making the sales was in mixing and selling all of the oranges at the rate
of 5 for 10 cents; for, as shown above, the inferior oranges, 3 for 5 cents, were all sold in 10 sales,
and hence the ratio of 3 inferior and 2 superior making 5 for 10 cents, then ceased, and the remain-
ing 10 oranges should have been sold 2 for 5 cents. By selling the 60 oranges at the rate of 5 for
10 cents, there must have been 2 sales which consisted wholly of superior oranges.
29. If 5% times 54 yards of cloth, which is 14 times 13 yards wide, cost 23
times 2% dollars, what will 103 times 103 yards cost that are 14 times 14 yards
Wide # Ans. $25.
30. A planter hired a laborer for a term of 120 days, on condition that for
every day he worked he should receive $1.00 and his board, and that for every day
he did not work he should pay 50 cents for his board. At the expiration of the 120
days a settlement was made and he received $76.50. How many days did he work?
Ans. 91.
OPERATION.
Had he worked the 120 days he would have received (at $1 per day) $120.
But as he only received * = tº tº gº -*. tº- tº- &= se 76.50
He lost by reason of not working * as * * sº gº tº $43.50
And as he lost each day that he did not work ($1.00, amount of services, + 50 cents, amount
paid for board) = $1.50, it is clear that he lost as many days as $43.50 are times equal to $1.50,
which is 29; and if he lost 29 days, he therefore worked 120 – 29 = 91 days.
31. A. can do a piece of work in 6 days, and B. can do the same work in 8
days. How many days will it take for both to do the work? Ans. 3%.
OPERATION.
If A, can do the work in 6 days, in 1 day he can do of it; and if B. can do the work in 8
days, in 1 day he can do # of it; therefore # -H # = }ſ of the work is done in 1 day ; and if ºr of
the work is done in 1 day, by transposition #4 of the work require one day, and if ºr of the work
require 1 day, ºr will require # of 1 day, and ##, or the whole work, will require 24 times as many
days, which is 3}.
4× MISCELLANEOUS PROBLEMs. 957
32. A. and B. can do a piece of work in 10 days; A. alone can do it in 15
days. How many days will it take B. to do it? Ans. 30 days.
OPERATION.
y’s – ºr = sºr of the work done in 1 day by B.
1 -- a's = 30 days. Ans.
33. A. and B. can do a piece of work in 14 days. A. can do # as much as
B. How many days will it take each to do it, working alone? g
Ans. 24; days for B. 323 days for A.
OPERATION.
(B.) # -- (A.) # = 4 work done by A. and B. in one day.
Hence in 1 day B. does # of the work.
& 4 & 4
And “ 1 “ A. “ #
14 – # = 244 days for B.
14 — 3 = 323 ** A.
34. A cistern has 3 pipes by which it may be filled, and 2 by which it may
be emptied. The first of the three filling pipes can fill it in 12 hours, the second in
20 hours and the third in 2 days. The first of the emptying pipes can empty it in
24 hours and the second in 30 hours. If all of the pipes are left open, how many
hours will it require to fill the cistern ? Ans. 1243 hours.
OPERATION.
* + 38 + # = % h.
* + k, == +$o 19 ło
#6 — T30 = % * 12+3 hours. Ans.
35. If it takes 5 boys 20 days to do a piece of work, and 6 girls 30 days to
do the same work, how many days will it take a boy and a girl to do it?
Ans. 64% days.
OPERATION.
d.
5 × 20 = 100. Tºo + Tºo = +&G 1
7 || 450
6 × 30 = 180. wº- =&m-º-º-º-º:
64% days.
36. A barrel of flour will last a man, wife and servant 50 days; but when
the man is absent, it will last his wife and servant 90 days. How long would it last
the man 3 Ans. 1124 days.
OPERATION.
|;
* — ſo = x3; 2 225
112%
37. A merchant bought 5 bbls. flour for $40 and sold the same at a gain of
25%. He then bought 5 bbls. more for $40, and sold the same at a loss of 25%.
Did he gain or lose, and if so, how much 3 Ans. He neither gained nor lost.
38. A merchant sold 5 bbls. of inferior flour for $40 and gained 25%. He
then sold 5 bbls. of superior flour for $40 and lost 25%. Did he gain or lose in the
two transactions, and if so how much 3 Ans. He lost $54.
958 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. º:
A NOTE WITH COLLATERAL SECURITY AND A MARGIN.
39. A merchant discounted his note for $1200 with a banker and gave him
insurance stock as collateral Security.
The market value of the stock was $112. A margin of 20% was to be kept
good on the stock. How many shares of stock did the merchant deliver and what
instrument of writing did he execute and deliver to the banker besides the note?
Ans. 1. 14 shares of stock.
2. A special power of attorney to sell the stock.
FIRST OPERATION. SECONI) OPERATION.
$100 $112 market value of shares.
80 | 1200 22.40 = 20% margin.
$1500 = amount of collaterals. $ 89.60 net security per share.
1500 – 112 = 13% shares, practically 14 shares. $1200--89.60 = 13% shares, practically 14 shares.
NoTE.—It is the custom, when giving stock collaterals to give a special power of attorney
to sell the stock, so that in case the stock should decline and the margin should not be made good,
a sale of the stock can be made to protect the payment of the note.
40. A. can do # of a piece of work in 10 days; B. can do # of it in 6 days,
and C. can do # of it in 3 days. How many days will it take them all to do the
Work # Ans. 34°1's days.
OPERATION.
§ X To = T's the part of the work done by A. in 1 day. d.
# X # = & 4 & 4 & 4 B. : : 1 * * | 1
# X # = ºr & 4 { { “ . C. “ 1 “ 113 360
#3 the part of the work done by A., B. and C. in 1 day. 3 ºf Ans.
41. Three persons, A., B. and C., do a piece of work; A. and B. together do
3 of it, and B. and C. do ºr of it. What part of the work is done by B. ?
Ans. #.
SOLUTION.
As A. and B. do à of it, it is clear that C. does the remaining 3 ; and as B. and C. do ºr of it,
it is clear that A. does the remaining 1°F. Then, as A. does ºr and C. §, they, together, do ºr + 3
= $3; and if A. and C. do 35 of a piece of work done by A., B. and C., it is clear that B. does the
difference between $3 and ##, which is #.
42. A., B., C. and D. agree to do a piece of work for $5500. A., B. and
C. can do it in 20 days; B., C. and D. in 24 days; C., D. and A. in 30 days; and
D., A. and B. in 36 days. In how many days can all do it, working together; in
how many days can each do it, working alone; and what part of the pay ought each
to receive # Ans. For the solution see page 879.
43. A planter being asked how many cattle he had, replied that he had # of
them in one pasture, 4 in another, and 25 in a third pasture. How many had he?
Ans. 60.
SOLUTION.
* + 4 = "g the number in the first and second pastures. Hence 's from all of his cattle, #
= #, the number in the third pasture; 25 is therefore ſº of his cattle; and if & are 25, ſº is the #
past, which is 5, and 43, or the whole number, are 12 times 5, which is 60. -
jºr dr, MISCELLANEOUS PROBLEMS. 959
44. If A, 4 and # of a man's money is $780, what amount has he?
Ans. $800.
SOLUTION.
# -- + -i- # = }}. Then if #3 equal $780, #5 is the 39th part, which is $20, and #3, or the
whole amount, is equal to 40 times ºf, or $20, which is $800.
45. A grocer bought a barrel of molasses, and after # of it had leaked away
he drew out 6 gallons, when it was found to be # full. How many gallons did the
barrel hold 3 Ans. 40 gals.
SOLUTION,
3 — 4 = }; then # — # = #5, then ºr = 6 gallons, and 35 = # of 6 gallons, which is 2, and #3
= 20 times as much, which is 40.
46. A speculator after losing # of his money, and # of the remainder, ha&
$1750. How many dollars had he at first. Ans. $7000.
SOLUTION.
# = his first loss.
# X + = + = “ second loss.
+ = # “ total loss, which, deducted from his whole amount,
leaves +, hence $1750 is + of his money, and the whole amount, 4 times as much, which is $7000.
47. A patron of the “wheel of fortune” (misfortune) in the lottery, lost by
his first investment # of his money; by his second investment, # of the remainder,
and by his third investment, à of the second remainder; he then donated $10 to
aid in the destruction of such disreputable institutions, and had $75 left. What
Sum had he at first 3 Ans. $680.
SOI,UTION.
Having lost by the 1st investment 4 of his money, he had # remaining; then having lost by
the 2d investment, # of the remainder, he therefore lost # of # = 4; and 4 déducted from # = } as
the second remainder; then having lost by his 3d investment # of the 2d remainder, he therefore
lost # of # = #; and # deducted from # = # remaining on hand after his 3d loss. Hence, the $10
ºt tº on hand = $85, is # of his first sum, and consequently the whole amount was 8
timeS $85 = g -
48. A merchant deposited his money in 4 banks; in the first, he deposited
#; in the second, #; in the third, #; and in the fourth, $2400 more than g of the
whole. How much money had he? Ans. $72000.
OPERATION.
# -- # -- # = nºr = proportion deposited in three banks, hence the difference between ſº, and
+3 = ſo is the proportion deposited in the fourth bank. And as the amount deposited in the fourth
bank was $2400 more than 1% of the whole, the difference between Ho and ſº, is therefore the pro-
portion that $2400 is to the whole sum. ſo — "... = 5%, hence $2400 is 3% of the whole sum, and #3
or the whole sum is 30 times $2400, which is $72000.
49. If 3 men and 5 women can do a piece of work in 14 days, 6 women and
8 boys can do the same work in 104 days, and 4 men and 7 boys can do it in 12;
days. How long will it take 2 men, 10 boys and 3 women to do it?
Ans. 944 days.
NOTE.-For the solution of this problem, see page 331.
96o SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS: *
50. A farmer sold 60 fowls, a part were turkeys and a part chickens; for the
turkeys he received $1.10 a piece, and for the chickens, 50 cents a piece, and for
the whole, he received $51.60; how many were they of each 3 Ans. 24 chickens.
36 turkeys.
SOLUTION.
Had all the fowls been turkeys and sold at $1.10 each, he would have received ($1.10 × 60),
$66.00; but because some were chickens, and sold at 50 cents each, he only received $51.60, hence
the difference, ($66–$51,60) = $14.40 is the deficit by reason of having sold some chickens, and as the
turkeys were sold at $1.10, and the chickens at 50 cents, there is a deficit on each chicken of ($1.10
— 50c. = 60); therefore there were as many chickens as $14.40, are equal to 60c. which is ($14.40
- 60c.) = 24; and as there were 60 in all, it is clear that 60 — 24 = 36, is the number of turkeys.
51. A planter being asked how many head of mules he had, replied, that if
he had as many more, one-half as many more, and seven mules and a half, he would
then have 100. How many had he? Ans. 37.
SOLUTION.
According to the conditions of the question the 100 head is the result of as many more, one-
half as many more, and 8even and one-half added to the true number; and hence it is evident, that
by deducting 7% from 100 we shall have in the remainder the true number, plus as many more, and
one-half as many more. 100 — 7# = 92%; now as 92% is the true number plus the true number, and
one-half of the true number, it is evident that it is 2% or 3 times the true number, and hence 92+
— 23 = 37, the true number.
52. A father bought a certain number of oranges, which he divided among
the members of his family, as follows: To his wife he gave # of all that he had, and
# an orange more; to his daughter he gave # of the remainder, and # an orange
more; to his son he gave # of the second remainder, and # an orange more; and to
his servant, he gave 1, and had 1 left for himself; how many oranges did he buy,
and how many did each receive?
SOLUTION.
(Problems of this character are most easily solved by working in the reverse direction or
order from which the occurrences of the problem took place).
It is evident, on inspection, that he had 2 oranges after having made the third division (with
his son), and as he gave his son # an orange more than 3 of what he had before, making the third
division, it is evident that he must have had 2 times 2 (the remainder) plus 1, which is 5; then as
5 was what he had before making the third division, it is what he had after making the second
division; and as he gave in the second division 3, an orange more than ; of what he had before
making the second division, it is also evident that he had 2 times 5, plus 1, which is 11. Then as
11 was what he had left before making the second division, it is what he had after making the first
division; and as he gave in the first division 3 an orange more than % of what he had before
making the first division, it is likewise evident that he had 2 times 11, plus 1, which is 23.
Then 23 - 2 = 11} + 4 = 12, the wife's part.
23 — 12 = 11; 11 — 2 = 5% + 3 = 6, the daughter's part.
11 – 6 = 5; 5 - 2 = 2+ + 4 = 3, the son's part.
5 – 3 = 2; 2 — 1 = 1, the servant's part.
2 — 1 = 1 remaining, the father's part.
53. A. and B. had the same income; A. saved of his, but B. by spending
$50 more per annum than A. found himself at the end of ten years $300 in debt;
how much was the income of each. Ans. $160.
SOLUTION.
B. having spent $300 in 10 years, at $50 per year, it is plain that he spent ſº of it in ºne
ear; ºn of $360 is $30. Now as B. spent $50 more per year than A., the difference between what
}. spent per year, and $50 is what A. saved. This difference is 50 – 30 = $20. \ -
60 Therefore $20 is 4 of their yearly income, and # or the whole income is 8 times $20, which is
$160,
x. MISCELLANEOUS PROBLEMS. 961
54. There is a fish whose head is 9 inches long, his tail is as long as his head
and half of his body, whilst the body is as long as his head and tail both; what is
the length of the fish? Ans. 72 inches.
SOLUTION.
Since the tail is as long as the head and # the body, it is equal to 9 inches + 4 of the body,
but as the body = the head and tail, it (the body) = 9 inches + 9 inches + 4 of the body. That
is the body = 18 inches + # the body. Hence # the body is 18 inches, and # or the whole body 36
inches; and as the body is ; the length of the fish, 3 or the whole length is 2 times 36 inches,
which is 72 inches.
55. A hare is 40 leaps before a hound and takes 4 leaps while the hound
takes 3, but 2 of the hound's leaps are equal to 3 of the hare's; how many leaps
must the hound take to catch the hare ? Ans. 240.
SOLUTION.
Since 2 leaps of the hound = 3 leaps of the hare, 1 leap of the hound = 1% leaps of the
hare; then, 3 leaps of the hound = 4% of the hare; hence, in making 3 leaps the hound gains #
leap of the hare, and in making 1 leap, (; – 3) # ; then 6 leaps of the hound gain 1 leap of the
hare, and 6 × 40 = 240, number of leaps.
56. A general formed his men into a square, that is, an equal number in
rank and file, and found that he had 59 men over; and increasing the number in
both rank and file by 1 man, he wanted 84 men to complete the square; how many
men had he? Ans. 5100.
SOLUTION.
To increase the number in both rank and file by 1 man, it is evident, that it would require
twice the number of men in rank or file at first, plus 1 (for the man occupying the corner). And
as it require to effect this, 59 -- 84.5- 143 men; then (143 – 1) – 2 = 71, the number of men
in rank and file at first. Therefore, 71* + 59 = 5100 men in the army.
57. Three footmen, A., B. and C. start together and travel in the same direc-
tion around an island 72 miles in circumference; A. travels 6 miles a day, B. 8, and
C. 12. In what time will they all be together again? Ans. 36 days.
SOLUTION.
72 – 6 = 12 days required for A. to travel around the island.
72 - 8 - 9 { % é & B. & 4 { { & 4
72 —- 12 - 6 { { & 4 C. { { & 4 & 4
Then the least common multiple of 12, 9 and 6 is the number of days required.
3 ) 12 9 6
2 ) 4 3 2
2 3 1
3 × 2 × 2 × 3 = 36 days, Ans.
or, thus:
B. gains on A, 2 miles a day, and hence will pass A. every 73 days, = 36 days. C. gains on
A. 6 miles a day, and hence will pass A. every # days = 12 days...Therefore, the least common
multiple of 35 and 12 will give the number of days that B. and C. will first pass A, together.
12 ) 36 12
3 1
12 × 3 = 36 days, Ans.
962 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
58. A lake is 56 miles in circumference. A. traveling 30 miles per day, B.
24, C. 16, D. 12 and E. 10, all commence to travel at the same time and in the same
direction around the lake. In how many days will they all be together again?
Ans. 28 days.
SOLUTION.
A. gains on E. 20 miles per day. 2 ) {} }} hºº º
B. ** E. 14 “ é &
C. 44 E. 6 “ ( { 3 ) 21 30 70 210
D. & 4 E. 2 & 4 & 4
56 - 20 = 23 = 2+} = {}
56 - 14 = 4 = 4 = }}
56 - 6 = 9+ = 9 * = ***
56 - 2 = 28 = 28 = 4%
*
7 ) 7 10 70 70
10 ) 1 10 10 10
1 1 1 1
2 × 3 × 7 × 10 = ** = 28 days, Ans.
59. How many gallons in a tank of the following description and dimen-
sions: The upper portion is in the form of a vertical rectangular prism, 13 ft. 5 inches
deep, 11 feet 9 inches long, and 9 feet 3 inches wide. The bottom is a segment of a
horizontal cylinder, the chord being 9 feet 3 inches, and the height or versed sine
4 feet Ans. 13372.404 gallons.
NOTE.—For the solution of this problem, see page 399.
60. A live stock dealer sold two horses, each at the same price. On one
horse he gained 25 per cent, and on the other he lost 25 per cent. His net loss on
the two sales was $12. What was the cost of each horse?
Ans. $72 cost of 1st horse, $120 cost of 2d horse.
NOTE.-For the solution of this problem, see page 477.
61. A man, at his death, left an estate amounting to $35000, to his wife and
two children, a son and a daughter. His children being absentin Europe, he directed
by will, that if his son returned his wife should have of the estate, and the son
the remainder; but if the daughter returned, his wife should have 3 and the
daughter the remainder. Now it so happened that they both returned. What sum
should each receive? Ans. $20000 son, $10000 wife. $5000 daughter.
NOTE.-For the solution, see page 876.
62. A man had 2 sons and 3 daughters; to the youngest son he gave 4 of 14
times $3600, which was # of the elder son's share, and what the elder son received
was # of g of the whole estate; the balance was divided among his 3 daughters in
reciprocal proportion to their ages, 8, 10, 12 years; what was each daughter's
share 7 Ans. $7114.87 for the one 8 years old.
$5691.89 “ (: 10 44
$4743.24 “ “ 12 46
Nore.—For the solution, see page 874,
º
ºf MISCELLANEOUS PROBLEMS. 963
63. A., B. and C. were to receive $6000, in proportion to 3, 4 and #; but B.
having died, it is required to divide the money between A. and C. What sum
should each receive # Ans. A. $4000. C. $2000.
NOTE.-For the solution, see page 875.
64. A. and B. paid $600 ($300 each) for 300 acres of land. In dividing the
land according to quality, it was agreed that A. should have the best quality, and
pay therefor 75 cents per acre more than B. How many acres did each receive, and
what price per acre did each pay ?
Ans. A. received 122.8 acres and paid $2.443 per acre.
B. { % 177.2 4 { % 1,693 (4
NOTE. – For the solution, see page 878.
65. A merchant receives from France, an invoice valued at port of shipment
at 4550.25 francs.
In the invoice are 100 dozen pairs of kid gloves that cost 18.50 francs per
dozen. Freight and insurance on the whole invoice to New Orleans, amounted to
$87.82. Duty was paid on the gloves at 33% per cent. For what price per pair
must the merchant mark these gloves so that he may discount 10 per cent from his
asking price and yet gain 163 per cent on full importing cost 3 Ans. 55-H 2.
OPERATION INDICATED.
% of charges on invoice. $ import cost 1 dozen gloves.
87.82 s
4550.25 | 18.50
.193 || 100 *º-
$3.57
10% practically. 1.19 = duty at 33}.
.36 = proportion of charges at 10%.
$ selling price per pair. -
5.12 $5.12 full N. O. cost per dozen.
6 || 7
9 : 10
12
|ºre
66. August 30, 1895, No. 7, coffee was quoted in Rio at 138850, exchange
103d. What is the equivalent value per pound in New Orleans, the rate of English
Exchange being $4.88, no allowance being made for time, freight or insurance?
Ans. 9.24+ 2.
FIRST OPERATION. SECOND OPERATION.
13$850 13$850 × 10#d = 147.15625d.
32.38 10;
240 || 4.88 147.15625 - 32.38 = 4.545 – d per pound.
9.24-H c. Ans. (4.545 – d x $4.88) — 240 = 9.24-H c. Ans.
Explanation.—The rate, 138850, is 13 milries and 850 reis per arroba, which is equal to 32.38
pounds avoirdupois. The 10#d, exchange is the English gold value of a milreis of silver of Brazil
including exchange. The $4.88 is N. O. gold rate of exchange on England.
67. A. and B. together agree to dig 100 rods of ditch for $100. The part of
the ditch on which A. was employed was more difficult of excavation than the part
on which B. was employed; and it was therefore agreed that A. Should receive for
each rod 25 cents more than B. received for each rod that he dug. How many rods
must each dig and at what price so that each may receive just $50?
Ans. B. receives $ .8903 -H per rod, and digs 56.16 rotis,
A. receives $1.1403 per rod, and digs 43.84 rods.
NOTE.—For the solution, see page 878.
964 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. *
68. A person being asked the hour of the day, said, the time past noon is #
of the time from now to midnight. What is the hour?
Ans. 3 hrs. 25 m. 42% s.
OPERATION.
12 hrs. = 1 + # = % of time to midnight.
If 12 = }, + = } X 12 and # = } X 12 = 24 hrs.
*# = 3% hrs. = 3 hrs. 25 m. 42% s. Ans.
Ea:planation.—Since the time past noon is # of the time to midnight, we may consider the
time to midnight as a whole, or #, and the two periods of time taken together = } + # = }. Ther.
as the time from noon to midnight = 12 hrs., we have 12 hrs. = % of the time to midnight; and if
# = 12 hrs., + = } x 12, and # = } x 12 = ** = 3 hrs. 25 m. 42% s. Ans.
69. What time after 5 o'clock will the hour and minute hands of a watch be
together ? Ans. 27 m. 16++ S.
OPERATION.
Tºr x 60 = 5 ºr m. = number minute spaces passed by hour hand at first crossing.
5 × 5*r = 27 ºr m. = number minute spaces after 5 crossings, i. e., between 5 and 6 o'clock.
27 ºr m. = 27 m. 1614, s.
Ea:planation.—Since the hands are together at 12 M., the beginning of the hour circle, and
cross each other 11 times during the passage of the hour hand (with a uniform motion) once around
the circle of 60 minute spaces, we see that the first crossing of the hour hand will require h of the
60 spaces, or 5*, spaces, and at the fifth crossing, occurring between 5 and 6 o'clock, the hour hand
will have passed over 5 times 5 #1 m. = 27.4 m. = 27 m. 16#, s. Ans.
70. On the dial of a town clock the end of the minute hand requires 4
minutes and 42 seconds to pass over a space of 6 inches. What is its length ?
Ans. 12.19 + inches.
OPERATION.
42 s. = ## = ſo of a minute. Ea:planation.—We first find what part of the
41% dial circle is 4 m. and 42 s. The circle being
= fl O = #7 - divided into 60 spaces (i. e., minutes) we find it
41% m. F5 – 500 of a circle. to be gºſo of the whole. This fraction represents
6 inches of length of perimeter. If 6 ſo = 6 in.,
47 600 gºry = 4% of 6 and {}}}, or whole circle, - 600
sing 6 times A, of 6 or 90%; g. Having thus found the
circumference, we find the radius, i. e., the
| 12.19056 in. Ans. minute hand, by dividing by 3.1416 and taking
# the quotient.
71. A lady being asked how old she was at the time of her marriage,
replied, that the age of her oldest son was 13 years; that he was born 2 years after
her marriage; that when she married, the age of her husband was three times her
own, and that now her husband was twice as old as herself. How old was she and
her husband when married? Ans. 15 and 45.
SOLUTION.
Since the son, aged 13, was born 2 years after the marriage, she is 15 years older than when
married. Her husband's age then was to her’s as 3:1; with 15 years added to both it was as 2 : 1,
or 4: 2, i. e., she is now twice as old as when married. But as the addition of 15 produced this
<duplication, 15 years must have been her age and 3 × 15 = 45, the age of her husband at marriage.
40: MISCELLANEOUS PROBLEMs. 965
72. A father was 34 years old when his son was born; his son is now one-
half as old as his father, less 4 years. How old is he? Ans. 26 years.
SOLUTION,
We see that if the son were 4 years older, he would be # the father's age, therefore twice his
age plus twice 4 years would equal the father’s age; but as the father's age equals 34 + the son's
age, 2 × 4 = 8, must be the difference between 34 and the son's age, or 34 – 8 = 26. Ans.
73. A person being asked his age, replied, that if his age were increased by
§ of itself and # of itself, and 16 more, the sum would equal 3 times his age. What
was his age 3 27 yrs., 5 mos., 4} days, Ans.
SOLUTION.
Assuming unity or 1 as his age, and increasing this by # -- # we have 1 + 3 + # = 2%. Since
the addition of 16 to this sum equals thrice his age, or 3, we see that 16 must be the difference
between 2.É, and 3, i. e., 3 — 25, or 's = 16. If 16 is ſº of his age, ſº = } x 16, and the whole, or
# = 12 × 4 × 16 = ** = 27% years, which reduced gives 27 years, 5 months, 4} days. Ans.
74. A cistern drains a roof 64 feet by 42, how many gallons will be caught,
Supposing a rain-fall of , an inch 3 Ans. 837.81 gallons.
OPERATION,
2 || 64 Ea:planation.—We first find the area of roof in
12 sq. inches, then as the fall is # inch, there will
231 | 42 be # as many cub. inches of water as there are
12 sq. inches of surface. Dividing this quotient
* *s by 231 cub, inches = 1 gallon, we obtain the
837.81 + required gallons.
75. I can insure a piece of property worth $25000, by paying 1; per cent
premium annually, or effect a permanent insurance by paying 13 annual premiums
down; money being worth 7 per cent, which ought I to prefer, and how much will
I gain? Ans. $85,714 +.
OPERATION.
25000 × 1% 96 = 300 = annual premium.
300 – 7 96 = 4285.714 –- = principal.
(300 × 13) + 300 = 4200.
4285,714 – 4200 = 85.714 –H = difference in favor of permanent insurance.
Explanation.—In order to pay every year a premium of $300, a sum must be placed at interest
at 7 per cent, sufficient to produce it, i. e., 4285,714; the difference between this and 13 annual
premiums ( = 3900) plus the premium for the first year in advance, (i. e., 3900 + 300 = 4200) will
be the gain in favor of paying in one payment.
76. A., B., and C. bought a grindstone, each paying # of the cost. The
diameter of the stone is 32 inches, allowing 6 inches waste for aperture, how many
inches must each grind off, to get an equal share.
Ans. 1st, 2.821 + in. 2d, 3.621 + in. 3d, 6.556 + in.
966 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. *
SOLUTION.
Diameter of grindstone, 32 in., and 32* = 1024 proportional area.
& ( & & aperture, 6 & & & 4 62 -: 36 & 4 4 &
3 ) 988 proportional area for use.
Each person’s proportion, - ge 329% square inches.
And 1024 square of A's diameter, 32 A.’s diameter,
Less 329; A's portion,
Equal, 694; square of B's diameter, and V694; = 26.3565 B's diameter,
Difference bet. A’s and B's diameter divided by 2) 5.6434
gives thickness A. grinds off, - - *. 2.8217 -i- inches.
And 6943 square of B's diameter.
Less 329% B's portion, 26.3565 B's diameter,
Equals 365; square of C's diameter, and V365 = 19.1136 C's “
Difference bet. B's and C’s diameter, divided by 2) 7.2428
gives thickness B. grinds off, - gº - 3.6214 inches.
Less 19.1136 C's diameter.
And: 6. diameter of aperture.
2 ) 13.1136 difference between C's diameter, and aperture –- 2,
Gives 6.5568 inches, thickness C. grinds off.
Eacplanation.—Since the areas of circles are proportional to the squares described on their
diameters, we can most easily operate by using the squares of the diameters, instead of the exact
areas of the circles themselves. The remaining operations will be apparent from an inspection of
the solution as given above.
77. A cylindrical column is 40 feet high and 64 feet in circumference; what
length of rope will it require to wind round this column from the bottom to the top,
by leaving a space of 2 feet between each coil, making no allowance for the thick-
ness of the rope? Ans. 136.014 + feet.
SOLUTION.
2 ) 40 feet of cylinder.
20 number of coils of rope.
64 feet circumference of cylinder.
130 sum of bases of the 20 rectangles made by the rope.
Then 130° = 16900 square of base.
And 40° = 1600 square of altitude.
V18500 = 136,014 + feet, hypotenuse or length of rope.
Explanation.—A rope wound round a cylinder in a spiral line, will trace at each circuit, the
bypotentise of a right angled triangle, or diagonal of a rectangle, of which the base would be the
circumference of the cylinder, and the altitude, the distance between the coils of the rope. This
is easily demonstrated by taking a rectangular piece of paper and drawing a diagonal with a
pencil. Then by joining the two opposite sides of the paper, we have a spiral line around a
cylinder, and as there will be as many such diagonals as there are coils round the cylinder, or as
many as the number of times the height of the cylinder is greater than the height of one coil;
therefore, if we divide the altitude of the cylinder by the distance between the coils, we will have
the number of circuits made by the rope, then the circumference of the cylinder, multiplied by the
number of circuits, will give the sum of the bases of all the rectangles. Then if to the square of
this base, we add the square of the height of the cylinder, and extract the square root of the sum,
the result will be the sum of all the diagonals or length of the rope.
ºr MISCELLANEOUS PROBLEMS. 967
78. A stone weighs 121 pounds; what is the least number of pieces into
which it may be broken, and what is the weight of each, in order that we can weigh
with the aid of a pair of scales and number of pounds, from 1 to 121 ? *
Ans. 1, 3, 9, 27, 81.
SOLUTION,
In any geometrical series proceeding in a triple ratio each term is 1 more than twice the sum
of all the preceding, and hence, the above series might proceed to any extent
In using the weights, they must be put in one or both sides of the scales, as may be required;
thus to weigh 2 pounds, we would put 1 pound in one side and 3 pounds in the opposite side.
79. A nurseryman had 6000 trees which he wished to set out, in a rectangular
orchard, whose sides were as 3 to 5; if he placed them 5 yards apart, each way, how
many trees would there be in each row, any how many square yards of ground
would they stand on ?
OPERATION.
3 x 5 = 15 trees in proportional rectangle.
Then 15) 6000
400 = multiple of proportional number of trees or area;
And V400 = 20 multiple of ratio of sides. *
Then 20 × 3 = 60 number of trees on shorter side;
and 20 × 5 = 100 number of trees on longer side.
Proof 100 × 60 = 6000, whole number of trees.
Also (60 – 1) x 5 = 295 yards wide;
and (100 – 1) x 5 = 495 yards long.
Then 495 x 295 = 146025 square yards.
Baplanation.—Since the number of trees on each side are in the ratio of 3 to 5, if we take 3
trees one way, and 5 the other, we will have 3 x 5 = 15 trees in the proportional rectangle; but
there are 6000 trees to be set out, hence the required rectangle will be as many times greater than
the proportional rectangle, as 6000 is greater than 15, or 400 times greater; and since the sides of
similar rectangles are as the square roots of their areas, by extracting the square root of 400, we
have 20 for the multiple of the given ratios 3 and 5, giving 60 trees on one side and 100 on the
other. Then to find the length of each side, since the number of spaces is one less than the number
of trees in a row, we multiply the distance 5 yards by the number of trees less one.
80. A drover bought 100 head of sheep, for which he paid $100, paying $3
a piece for the wethers; $1.50 a piece for the ewes; and 50 cents a piece for the
lambs; how many of each did he buy?
NOTE. –For the solution, see page 801.
81. If 12 oxen consume 3% acres of grass in 4 weeks, and 21 oxen consume
10 acres in 9 weeks, how many oxen will consume 24 acres in 18 weeks, the grass
being at first equal on every acre, and growing uniformly and constantly 2
Ans. 36.
The analysis of this problem naturally divides itself into two distinct parts,
viz.: 1st, the number of oxen the grass already grown will support for a given time;
and 2d, the number of oxen the increase only will support. In the first case the
number of oxen supported varies directly as the number of acres grazed and inver-
968 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
sely as the time. In the second case the number of oxen supported varies directly
only as the number of acres grazed, time being irrelevant.
SOLUTION.
If 3} acres with 4 weeks increase support 12 oxen for 4 weeks, 10 acres would support for the
same time 10 × 12 = 36 oxen, and if 10 acres (with 4 weeks increase) support 36 oxen for 4 weeks
3#
they would support for 9 weeks 3 × 36 = 16 oxen. But by the problem, 10 acres with 9 weeks
increase would support 21 oxen, hence the 5 weeks difference of increase must support 21 — 16 = 5
oxen during the 9 weeks, and the whole 9 weeks increase of 10 acres would support # x 5 = 9 oxen
for same time. Then since the grass already grown plus the 9 weeks increase supports 21 oxen and
the increase alone supports 9 oxen, the Inumber supported by that already grown will be 21–9 =
12 oxen.
Now since the grass already grown of 10 acres will support 12 oxen for 9 weeks, the grass of
24 acres will support #3 × 12 = 283 oxen for 9 weeks, and for 18 weeks 3 × 28} = 14% oxen.
And again, since the increase of 10 acres supports 9 oxen, the increase of 24 acres will, by
the principal in the second case, support #3 × 9 = 213 oxen any length of time.
Summing up, we have the already grown grass of 24 acres supporting 14% oxen for 18 weeks,
and the increase of 24 acres supporting 21% oxen for same time; by addition 143 + 21% = 36 oxen,
Ans.
82. If two wagon wheels, one 54 and the other 6 feet in diameter be fixed
upon an axle, seven feet apart, and moved round on their resulting line of direction;
how large circles will they describef Ans. 168 feet diameter of large circle.
154 {{ ** Small é%
SOLUTION.
6 — 5} = } foot, difference of diameter.
Then } : 6 :: 7 : 84 radius × 2 = 168 feet diameter of large circle,
and less 7
Or, : 5} :: 7: 77 radius, x 2 = 154 feet diameter of small circle.
Explanation.—Since the wheels are fixed on the axle, they will make the same number of
revolutions, and since one is smaller than the other, they will describe concentric circles, and the
circumferences of these circles will be proportional to the circumferences of the wheels, but the
diameters of the wheels are proportional to their circumferences, bence we may operate directly
from the diameters. Therefore, if we draw a line from the top of the larger wheel to the top of
the smaller, and extend it, it will touch the ground at the centre of the circles, and form the
hypotenuse of two right angled triangles, whose bases are the radii of the circles, and altitudes
the diameter or height of the wheels; then if we draw a straight line from the top of the lower
wheel to the higher wheel, parallel to the ground or base, we will have a similar triangle, whose
base is the distance between the wheels, and altitude the difference between the heights of the
wheels. Then, since the sides of similar triangles are proportional, we have the proportion as
given above, viz: The difference in the height of the wheels, is to the height of either wheel, as
the distance between the wheels, is to the distance from that wheel to the centre of the circle.
From the above demonstration we see, that in questions of this kind, the diameter of the circles is
equal to twice the product of the height of the corresponding wheels by the distance between
them, divided by the difference in their diameters.
83. Three persons bought a sugar loaf in the form of a perfect cone 25 inches
high, and agreed to divide it equally by sections parallel to the base; what was the
slant height of each one's share, the base being 12 inches in diameter ?
- Ans. Upper section, 17.826 -- inch.
Middle 44 4.633 + “
Lower 44 3.250 + “
26. MISCELLANEO US PROBLEMS. 969
The analysis of this problem is based upon the Geometrical principles that
the volumes of similar cones vary as the cubes of the radii of their bases and also
as the cubes of their altitudes. *.
solution.
Since the altitude of the cone constituting the upper # of the loaf deducted from the whole
3 , — tºº
altitude gives the height of the lower section, we have for the altitude of the upper cone Va: v.
3 -—
= 25 : required altitude, or required altitude = V2 × 25, from which we have,
—H
iſ,
1,25992 × 25 - 1.44224 = 21.839 -- in. altitude of 3.
25 — 21.839 = 3.161 + in. altitude of lower section.
In same way for the cone forming the upper section we have,
— 3 2–
*/. ; VT = 25: required altitude, or,
3 2– *
Required altitude = V × 25, which gives 25 × 1.44224 = 17.334 + inches, altitude of
V.
upper section.
And 21.839 — 17.334 = 4.505 -- altitude of middle section.
Then by means of the ratios of the cubes of the radii we find in the same manner:
Slant height upper section = 17.826 inches.
& 4 ** middle ** = 4.633 {{
& & ‘‘ lower “ = 3.250 { { Ans.
84. A father leaves a number of children and a certain sum of money to be
divided among them as follows:
The first is to receive $100 and #5 of the remainder; the second is to receive
$200 and +3 part of what then remains; again, the third is to receive $300 and ºf
of the residue, and so on; each succeeding child is to receive $100 more than the
one immediately preceding, and then ſo part of what still remains. At last it is
found that all the children have received equal sums. What was the amount left,
and how many children were there?
NOTE.—For the solution, see page 876.
85. The diameter of a cylindrical cistern is 6 feet, 8 inches, what must be
its height to hold 3000 gallons? Ans. 137,8676 in. height.
OPERATION.
3000 gallons.
231 cu. in. 1 gallon.
693000 cu. in. in a cistern to hold 3000 gallons.
80 in. diameter of cistern.
80 é & & 4 & 4 & 4
6400 square of diameter.
.7854 ratio of the area of a square to the area of a circle.
5026.5600 area of the cistern,
693000 - 5026.56 = 137.8676 inches height.
80 3000
80 || 231
or thus: .7854
137.8676 in. = 11 feet, 5.8676 inches.
970 soule's PHILOSOPHIC PRACTICAL MATHEMATICS. &
86. The height of a cylindrical cistern is 7 feet, 6 inches, what must be the
diameter so that it will hold 2000 gallons? Ans. 80.84 inches diameter.
OPERATION.
90 | 2000
.7854 || 231
| 6535.9477
W6535,947 = 80.84 + inches. Ans.
4 NotE.—For more extended work of this character, see the Measurement of Solids, pages 370
to 418.
87. A merchant sold goods to the amount of $10, and in payment therefor
the purchaser handed him a $50 bill, but not having change, he stepped into the
adjoining store and had the bill changed. He then returned $40 to the purchaser
and placed $10 in his cash drawer. A few hours afterwards the party that had
changed the $50 bill returned it as a counterfeit, and to redeem it the merchant
took the $10 from the drawer and borrowed $40 from a third party. How much did
he lose by reason of the counterfeit bill?
Ans. $10 worth of goods and $40 in money.
SOLUTION.
By the question, we see that he first gave $40 in money and $10 in goods to the man who gave
him the counterfeit bill, then borrowed $40 from a third party and paid $50 to the man that
ehanged the bill.
When the merchant sold the goods and made change through a third party there was no loss
sustained, for he gave but an equivalent for a bill supposed to be good.
When he redeemed the counterfeit bill he lost $40, the amount borrowed, and the $10 received
for the goods, but as the $10 was received for the goods, it was not, therefor, a loss of $10 in money,
but of $10 worth of goods. Hence his loss was $40 in money and $10 in goods, or thus:
After redeeming the counterfeit bill the merchant owed $40 borrowed money and had
disposed of $10 worth of goods. And as he had no money on hand before selling the $10 worth of
goods, or after redeeming the counterfeit bill, it is hence clear, that he lost $40 in money and $10
worth of goods.
88. The price of corn is 80 cents per bushel. What is it worth, per cental?
- Ans. $1.42%.
NOTE.-For the solution of this and the following problem, see page 279.
89. The price of wheat is $2.50 per cental. What is it worth per bushel ?
Ans. $1.50.
90. A merchant in Mobile has a balance of $5000 due him in Louisville, and
he instructs the Louisville merchant to remit by a draft, payable 60 days after
sight, exchange being 3 per cent premium. The Louisville merchant, through error,
remitted a sight draft, which, when received in Mobile, is accepted and paid after
the expiration of three days of grace. Now if the Mobile merchant loans this
money at 8 per cent, and allowing that money is worth 9 per cent in Louisville, will
he gain or lose by the error of the Louisville merchant? Ans. Loses $12.87.
MISCELLANEOUS PROBLEMS. 971
OPERATION
To find the face of the sight draft sent.
$
| 100
100.50 || 5000.00
$4975.124
OPERATION
To find the value of the sight draft at the
expiration of 60 ds. at 8 per cent.
$
| 4975,124
360 || 8
60
$66.3349 interest.
4975.124
$5041.4589
OIPERATION
To find the face of a 60 day draft, sight exchange
being # per cent premium and money
worth 9 per cent.
100
63
$
360
| $1.57;
$100
1.573
$98.42%
.50 = } % premium.
$98.92%
100
197.85 2
5000.00
$5054.33
OPERATION
To find the loss.
Value of 60 day draft, {- $5054,33
“ “ sight draft, - - 5041.46
Loss, - tº G º $ 12.87
NotE.—Sight Drafts in Mobile are entitled to 3 days grace.
For more extended explanations on this character of problems, see pages 730 to 736.
91. Sold a watch for $17.16 and lost as much per cent as the watch cost.
What was the cost 3
1st Ans. $78. 2d Ans. $22.
NOTE: 1.--It is generally believed that an arithmetical problem can have but one correct
answer; but this belief is clearly erroneous. For if we apply the conditions of the sale to both of
the answers to this problem, we prove their correctness.
NotE 2.—All problems of this kind, provided the selling price does not exceed $25 or 25 of
any other monetary unit, may be solved and the two answers obtained by the following method,
which is derived from an Algebraic solution :
Subtract the selling price from 25, and to the square root of the remainder add 5; then
multiply this sum by 10 and the product will be the greater answer. Or, from 5 subtract the square
root above obtained and multiply the difference by 10; the result will be the lesser answer.
OPERATION.
25 — 17.16 = 7.84; then the square root of 7.84 is 2.8; then 2.8 + 5 = 7.8; 7.8 × 10 = $78
the greater cost. And 5 — 2.8 = 2.2; then 2.2 × 10 = $22 the lesser cost.
$78 cost — (78% = $60,84) = $17.16 selling price.
$22 cost — (22% = $4.84) = $17.16 selling price.
92. Sold goods for $12.75 and lost as much per cent as they cost. What did
they cost?
Ans. $85; or $15.
972 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. A.
93. Sold goods for $39 and gained as much per cent as they cost. How
much did they cost? Ans. $30.
NotE.—All problems of this kind may be solved by the following method:
To the selling price add $25; from the square root of this sum subtract 5; then multiply the
remainder by 10, and the product will be the cost.
* OPERATION.
$39 + $25 = $64; the square root of $64 is $8; $8 – $5 = $3; $3 × 10 = $30 cost, Ans.
94. Sold goods for $56,904, and gained as much per cent as they cost. What
did they cost # Ans. $404.
95. If an article had cost 20 per cent less, the gain would have been 30 per
cent more. What was the per cent gain } Ans. 20%.
NOTE.—This and the following problem are purely Algebraic and can only be worked arith-
metically, by processes deduced or abbreviated from Algebraic equations. Hence, in the operations
given, no arithmetical reasoning is presented.
OPERATION.
$100 cost assumed.
20% less.
$20.00 = 20% less.
100.00 = cost as above.
--- 4
$80.00 = decreased cost. 20 100
30% more. tº- -º
-** *- $20 gain = 20% gain. Ans.
$24.00 = 30% more.
20.00 = 20% less.
$ 4 = gain on $20 less.
96. If the cost had been 12 per cent more, the gain would have been 15 per
cent less. What was the per cent gain # Ans. 40%.
OPERATION,
100 cost assumed.
12% more.
$12.00 = 12% more.
100.00 = cost as above.
sºsºme as--ºs , 4.80
$112.00 = increased cost. 12.00 100
15% less. *m-
40% gain, Ans.
$16.80 = 15% less.
12.00 = 12% more.
$ 4.80 = gain on $12 more.
97. What is the value of 30 — (2.4 – 8.37 -- 21.625) + 3.46 × .12?
Ans. 14.7602.
5. OPERATION.
2.4 – 8.37 = — 5.97; — 5.97+ 21.625 = 15.655, the result of the parenthesis; 3.46 × .12 =
.4152. Then 30 — 15.655 = 14.345; 14.345 ––.4152 = 14.7602, Ans.
- Or,
2.4 + 21.625 = 24.025; 24.025 – 8.37 = 15.655, the result of the parenthesis. 30 — 15.655 =
14.345; 3.46 × .12 = .4152; 14.345 -- .4152 = 14.7602, Ans.
NOTE.-The above and the following problem are Algebraic equations, and the Algebraic
principles involved in their solution are as follows:
* MISCELLANEOUS PROBLEMS. 973
g In all Algebraic and arithmetical equations or operations indicated by the +, −, X, or --
signs, and in simplifying fractions, the application of the signs of operation should be well under-
stood, as was explained on page 134. But to understand thoroughly, the elucidation of the pre-
ceding problem, as well as that of the following and all others of a similar character, we deem a
repetition of the principles necessary and state them as follows:
f th The signs + and — affect only the number or 'expression which immediately follows either
Of the Iſl.
The signs X and — indicate an operation to be performed with the numbers or expressions
between which either may be, and this operation must be performed before that of any + or —
which may immediately precede or follow.
The parenthesis, ( ), or the vinculum, , indicate that the quantities which they include
must be simplified into one expression before they can be connected to any other quantity.
The operation of combining and simplifying always begins with the ( ) or if any, and
then with the operations indicated by the X and – signs. The operations of + and — are then
performed and are generally considered from left to right.
It will be observed that the result of the parenthesis in the above equation was subtracted
from the quantity before it, as indicated by the minus sign between them; this was done because
the result of the parenthesis was a plus quantity. Had it been a minus quantity, i. e., had the
result been a quantity preceded by a minus sign, it would then have been added to the quantity
before the parenthesis, for the reason that the minus sign before the parenthesis would cancel the
minus sign before the result of the parenthesis and thereby produce a plus quantity. Should a
plus sign precede the parenthesis, the operation to be performed with the result of the parenthesis
§: !. º accordance with the sign before the resulting quantity and independent of the + sign
efore the ( ).
98. 20 — 30 + 2 × 6 = what? Ans. 2.
99. Show that 3 is greater than + and less than ###
e ## 18 g 13 — 3 13 -- 3
100. Express decimally the value of ##### Ans. 2.4293-H.
# of (3 – Tºg)
º “º º 1 + #y . 5
101. Simplify (1++++) + (1+ Tii) Ans. #.
102. A cylindrical oil tank, 25 feet 11 inches long, and having an inside
diameter of 6 feet 4% inches, was full of oil. When the tank was measured it
was found that oil to the depth of 6 inches of the diameter had leaked out. How
many gallons of oil were lost?
OPERATION.
inches diameter 76.75) 6.0000000 (.07817-1-
7 I
7 inches full.
6.75
0.75
6 inches empty.
.078 = .02835 .02835
.079 = .02889 .00009
.00054 .02844
.17
.0000918 .02844
76.75
76.75
231 || 311
311
4 || 307 225.544 gallons leaked out.
4 || 307
231 | .7854
&==ºmº || mºmº sº
6228.680+ gallons if tank were full.
NOTE.- See Problem 1, page 398, for explanation of similar problems.
974. soule's PHILOSOPHIC PRACTICAL MATHEMATICS. *
103. The diameter of a safety valve is 43, inches, and it is placed 44 inches
from the fulcrum. The weight of the lever and safety valve is 140 pounds. What
weight must be attached to the end of a 48 inch lever, operating the safety valve, to
withstand a pressure of 160 pounds to the square inch Ans. 98.565+ pounds.
OPERATION INDICATED.
2 9 15,90435
2 | 9 160
.7854 2 || 9
º - ºmºmº-
11451.132 lbs. pressure on valve.
15.90435 square inches in surface of valve.
11451.132 - 48 = 238.565-H pounds weight to withstand pressure of 160 pounds to the square
inch. 238.565 — 140 = 98.565-H pounds to be added to the end of the lever.
NOTE.-See Mechanical Powers, page 419, and problem 2, page 422.
104. Allowing the journals of a car to be 8 inches long, and the projected
surface on the brass to be 7# inches long by 3% inches wide, what is the actual weight
per square inch on each of the 8 journals of a car which weighs 26000 pounds and
is laden with 32 tons of freight 3 Ans. 414.7-i- pounds per Sq. in.
FIRST OPERATION. second OPERATION.
7# × 3% = 27# sq. in. Surface bearing on each journal. | 90000
64000 + 26000 = 90000 pounds, car and freight. º :
90000 - (27# X 8) = 414.7+ pounds per sq. in. Ans. -
414.7–H lbs. Ans,
CURIOUS AND AMUSING QUESTIONS.
1524. 1. A sea captain on a voyage, had a crew of 30 men, half of whom
were blacks. Being becalmed on the passage for a long time, their provisions began
to fail, and the captain became satisfied, that unless the number of men was greatly
diminished, all would perish of hunger before they could reach any friendly port.
He therefore proposed to the sailors that they should stand in a row on deck, and
that every ninth man should be thrown overboard until one-half of the crew were
thus destroyed. To this they all agreed. How should they stand so as to save the
Whites ? Ans. WWww.bbbbbww.bwww.bwbbww.bbbwbbww.b.
NOTE.-This result might have been easily ascertained before hand by the captain by the
following method. Place 30 letters, as the letter w, it being the initial of white in a row; then
commence counting off every ninth one which may be cancelled, when 15 have in this way been
cancelled, their places may be supplied with the letter b, the initial of black.
2. A. and B. have an eight gallon cask full of wine, and wish to make an
equal division of it, but have only two empty vessels with which to do it, one of
which contains 5 gallons, and the other 3. By what means shall they accomplish
this division ?
SOLUTION.
Fill the 3 and pour it into the 5; then fill it again and from it fill up the 5, which will leave
one gallon in the 3 gallon keg, empty the 5 into the eight and pour the one from the 3 into the 5;
fill the 3 again and empty it into the 5. Then there will be 4 gallons in the 5 gallon keg, and the
same left in the 8.
T CURIOUS AND AMUSING QUESTIONS. 975.
3. A countryman having a fox, a goose and a peck of corn, came to a river,
where it so happened that he could carry but one over at a time. Now as no two
were to be left together that might destroy each other, he was at his wit’s end, for
says he: “Though the corn can’t eat the goose, nor the goose eat the fox; yet the
fox can eat the goose, and the goose eat the corn.” How shall he carry them over,
that they shall not destroy each other?
SOLUTION.
Let him first take over the goose, leaving the fox and corn, then let him take over the fox
and bring the goose back, then take over the corn, and lastly take over the goose again.
4. A man, his wife, and two sons desire to cross a river. They have a boat
that will carry but 100 pounds. The man weighs 100 pounds, the woman weighs
100 pounds and each boy weighs 50 pounds. How can they all cross the river in
the boat?
OPERATION.
Let the 2 boys go over, and 1 boy return with the boat.
Let the man go over, and the boy return with the boat.
Let the 2 boys go over, and 1 boy return with the boat.
Let the woman go over, and the boy return with the boat.
Let the 2 boys go over.
:
5. Three jealous husbands with their wives are to pass over a river in a boat
which can carry but two at a time. How can these six persons row themselves
over two at a time, so that none of the wives may be found in the company of one
or two men unless her husband be present?
SOLUTION.
This may be effected in two or three ways; the following may be as good as any : Let A. and
wife go over; let A. return; let B's and C’s wives go over; A's wife returns; B. and C. go over; B.
and wife return; A. and B. go over; C's wife returns, and A’s and B's wives go over; then C. comes
back for his wife.
Simple as this question may appear, it is found in the works of Alcium, who flourished a
thousand years ago, hundreds of years before the art of printing was invented.
6. How can you arrange the 9 significant figures so that when added the
Sum Will be 100% Ans. *
83.
9
100.
. Prove that a pound and a half a cheese weighs more than 2 pounds of
7
butter.
8. How much heavier is a pound of iron than a pound of gold 1
Ans. 1240 grains Avoirdupois. 1020;# grains Troy.
9. How can 5 be taken from 5 and leave a remainder of 5?
976 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. *
10. How can 45 be taken from 45 and leave a remainder of 45%
OPERATION.
###################
8 + 6 + 4 + 1 + 9 + 7 -- 5 + 3 + 2 = 45 Remainder.
11. What can you prefix to IX to make it 6. Ans. S.
12. A student of mathematical logic proves that a cat has 3 tails by the
following process of reasoning:
1. No cat has two tails;
2. One cat has 1 tail more than no cat;
3. Now since no cat has 2 tails and since 1 cat has 1 tail more
than no cat, therefore 1 cat has 3 tails.
Where is the error in the logic 3
13. A young hoodlum, a modern evolvement of the human race, stole a basket
of peaches and divided them among 3 brother hoodlums and himself as follows: To
the first, he gave # of the whole number and # of a peach more; to the second, he
gave # of what remained and # of a peach more; to the third, he gave 3 of what
remained and # of a peach more. The stealer retained what was then left for
himself, which was 4 the number he gave to the first hoodlum. What was the
number of peaches stolen, and how many did each hoodlum receive?
Ans. 7 peaches were stolen.
1st hoodlum received 2.
2d {{ {{ 2.
3d {{ { % 2.
The stealer had 1.
SOLUTION.
In all problems of this kind the following principle prevails: That when the fractional parts
of the successive portions are consecutively increasing, each of the parts or shares of the several
persons are equal, except the last one, which from the nature of things must be one less than the
average. Hence in the given problem, the fact stated that the last one’s share is half of the first,
(and each of the others also) and from the above principle that it must be one less, it must
necessarily be ONE, and since each of the others is one greater, they must be 2 each, and since there
are three persons having equal portions, there will be 6 + the 1 for the last, making 7 in all.
Hence, to solve problems of this kind:
Add 1 to the last number, (ONE), and multiply by the number of persons, and then subtract
1 from the product.
14. If 3 men eat 3 dozen oysters in 3 minutes, how many dozen oysters can
100 men eat in 100 minutes? Ans. 3333; doz. oysters.
15. If 14 cats catch 1; rats in 14 minutes, how many cats will it take to catch
200 rats in 50 minutes? Ans. 6 cats.
16. If 1; hens lay 14 eggs in 14 days, how many eggs will 6 hens lay in 7
days? Proportional Answer, 28 eggs. Logical Answer, 24 eggs.
NOTE.-For the solution of the three preceding problems, see page 320.
APPENDIX.
—eſpe—
Building and loan Associations.
&
---------------------------------------------R
^
—-º-º-Aº-
-ºr-wr-wº-
1525. Building and Loan Associations are co-operative corporations, institu-
ted for the purpose of raising a fund by periodical subscriptions or deposits,
interest on loans, premiums or discounts, etc., for the following purposes: 1. To
assist members in building or purchasing homes, or to loan to members on mortgages
for any purpose. 2. To serve as a depository for such members as desire simply
to invest their money in its shares and to participate in the profits.
NOTE.—In some of the States, Building Associations are permitted, under certain restrictions,
to loan money to persons who are not members. It is claimed by some of the wisest financial men
in these Associations that this privilege should be general, under certain equitable regulations in
behalf of members.
Associations of this character are known under various names, such as
Building Associations, Loan and Investment Companies, Mutual Homestead Asso-
ciations, Saving Fund Associations, Mutual Benefit Societies, etc., etc.
These institutions are practically Co-operative Banking Institutions, having
within their charter rights, special privileges and advantages as to the manner of
loaning their funds at high rates of discount or premium, and of taking mortgage
securities, etc.
The usury laws do not apply to Building and Loan Associations in many of
the States of the union.
These institutions had their origin in England in 1795, were established in
Bhiladelphia in 1831, and have become eminently popular throughout the United
States and Europe. They have done and are doing a great financial and economic
work for private citizens, and through them, as self-supporting, tax-paying and
happy people, the resultant good to States and Nations is clearly apparent.
Through these Associations, many persons of very moderate earnings are
encouraged and enabled to profitably invest their savings, and by the aid of the
Associations to buy or to build homes.
1526. The Capital, or the authorized capital of an Association, is the sum of
the shares as set forth in the Charter, the capital clause of which generally reads
as follows: The capital of this corporation shall be Five Hundred Thousand Dollars,
and shall be issued in shares of $200 each, making 2500 shares.
NOTE. –The amount of capital or authorized capital varies in different Associations from
$50000 to $50000000.
1527. The Members of an Association are the parties who subscribe for
shares, and are of two classes, BORROWING and DEPOSITING MEMBERS.
(977)
978 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
1528. Borrowing Members are those subscribers who borrow money from
the Association. *
1529. Depositing Members are those who subscribe for shares as an invest.
Iment.
1530. The Shares are generally issued in series of a specified sum for each
Series, and at periodic times according to the demands of the business and the
judgment of the Board of Directors. Each series issued is generally closed when-
ever the installments and profits have brought the shares of the series to par. In
Some cases the series is closed before the shares reach par. When a new series is
issued no more shares of a prior series are issued.
In some associations the full paid shares are held as deposits, and draw a
specified rate of interest in favor of the owner.
Members are limited to a specified number of shares, which each may own;
in Some associations they are subject to fine for certain delays in making payment
of subscriptions, and for failure to make payments of installments and interest for
a specified time, their shares may become forfeited to the Association.
NOTE.—Regarding fines and the forfeiture of shares, the Charters and By-Laws of the
various associations prescribe different regulations. -
1531. The Dues are payments for subscriptions for shares, and are generally
made payable weekly or monthly. Monthly payments are $1 per share, and weekly
payments are 25 cents or 50 cents, on each share of $200 of stock. In some of the
plans, members may, however, pay any multiple of the monthly or weekly payment,
or any part of their subscriptions in advance. Such payments will be credited
on the shares and will participate in the profits. i
1532. Premium or Discount is the percentage that the borrowing members-
bid, and under regulations, varying in different associations, pay as a bonus for the
privilege and preference of borrowing and using the funds of the Association.
Some associations fix a minimum premium below which rate the funds will
not be loaned. But loans are generally made by auction, and all members have an
equal chance to bid for the loan.
The following figures taken from a table of the literature of a prosperous and
well managed association, will illustrate how premiums or discounts are made on
loans:
1533. TABLE showing discount and net loan on ONE and on FIVE shares at 15,
18, 20, 224 and 25 per cent, also the amount of weekly dues and weekly interest.
ON ONE SHARE. ON FIVE SHARES.
|
p *::: Discount. |Net Loan. Duºeekly Int. Discount. |Net Loan. I... weekly Int
15 $30 $170 25 cents. | 23 cents. $150 $850 $1.25 $1.15
18 36 164 180 820
20 40 160 200 800
22} 45 155 225 775
25 50 150 250 750

- NoTE. 1.-The full table extends from 15 to 25 per cent by 3 per cent increase of rate.
NOTE: 2.--To find the net loan on any number of shares, multiply the net on ONE SHARE at
the given rate by the number of shares.
* BUILDING AND LOAN ASSOCIATIONS. 979.
The Charters of some Building Associations fix the minimum premium at 15
per cent. In some States, there is no minimum limit to the premiums; this seems
more business like and more ethical, for thereby the rate of premium will adjust
itself according to the law of supply and demand. -
1534. Dividends are the profits of the Association, declared monthly, quar-
terly, Semi-annually or annually, and distributed among all shares in force. Various
regulations are made by different associations regarding the manner of declaring
dividends, according to the kind of shares, “Current” or “Full Paid,” the dues
paid thereon, the cash value of the capital, profits, etc.
1535. Withdrawals. Members may withdraw, under stated regulations of
the Association, partially paid or full paid shares on the surrender of their certifi-
cates and receive in return the amount paid in as dues, and a fair share of the
profits, less any delfnquencies or pro rata of expenses that they may owe.
NOTE.-There is a great diversity of method or plan by Associations in different States, and
also in the same state, governing Withdrawals.
TERMINATING, SERIAL, PERMANENT AND DAYTON PLANS,
There are four plans for Building and Loan Associations which are known as
the Terminating, the Serial, the Permanent, and the Dayton Plans.
1536. The Terminating Plan, as originally adopted and carried out, required
all the stock to be issued at the beginning of the Association; or in case shares
were subscribed at any time during the progress of the Association, the shares
should date back to the time of first payment of dues, and the subscriber should
pay in a sum equal to all the dues on the shares subscribed, for the full time elapsed
since the date of organization, together with an additional sum equal to a pro rata
share of the accumulated earnings of the Association, up to the date of his subscrip-
tion. By this plan, all profits and earnings were retained until such time as they,
together with the payments made by the subscribers for dues, amounted to the
$200 par value of the stock. When this result was reached, the Association and
membership terminated, with a division of the capital.
1537. The Serial Plan is a modification or extension, or evolution of the
terminating plan. While it is conducted in the main on the same general principles,
it differs in the manner of issuing shares subsequent to the organization. This
plan requires the same monthly payments of dues and the same observance of
charter regulations, but the distinguishing feature is the issuing of stock at different
periods during the life of the Association. Each series issued bears the date of the
issue, and while the shares of each separate series are equal in value, they differ in
value from the shares of every other series. New members pay only from the date
of the series they enter. It will be remembered, that in the terminating plan, all the
shares have at all times an equal Value.
By the Serial Plan, each issue or series runs its course independently, and
without reference to others issued before or after it. Thus, through the Serial issues,
98o soulE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
the association may continue indefinitely, or as long as there are persons who desire
to subscribe for the shares.
The periods for the issue of the different series may be monthly, quarterly,
semi-annually, annually or biennially, according to the amount of business and the
judgment of the directors.
1538. The Permanent Plan is closely related to the serial plan. It is claimed
to be the highest development of the Co-operative Building Association idea yet
attained. By this plan, there are no stated periods for the issue of its stock; stock
is subscribed and payments made at any time. Each member's shares are treated
as a Separate series. Dividends are declared on the actual amount of dues paid in
by each member, and as each member's stock matures or reaches par, it is paid off.
One of the features of this plan is the net premium system, by which only earned
premiums become the visible profits. $
1539. The Dayton or Ohio Plan is the same in all essential particulars as
the Permanent Plan. The stock face value (usually $100) is issued at any time,
and generally no fines or forfeitures are prescribed for non-payment, the members
making payments at such times and in such amounts as may suit their own con-
venience. One of the special features of this plan is, that the borrowing and the
Inon-borrowing members pay the same, the weekly payments of the former being
first applied to the payment of the interest upon the loan, the balance being placed
as a credit on the stock; credits commence to draw dividends at once and are
compounded semi-annually. The stock is matured by the payment of the weekly
dues and the dividends. There is no premium or bonus in this plan, and it becomes
more nearly a Banking Institution than any of the others, and has less restrictions
upon its members. The popularity of this system seems to be growing in all parts
of the country, and in modified forms, has been adopted by many of the Building
Associations.
NOTE.-There is much discussion among building association men regarding the respective
merits of these different systems; and changes and modifications thereof are made by different
associations.
DIFFERENT PLANS OF CHARGING PEEMIUMS AND INTEREST.
1540. There are four general plans of charging premiums and interest on
loans, which constitute the chief items of profit.
In the equitable adjustment of the profits among the different serial accounts
constituting the capital, it is necessary to understand the nature and effects of
these different plans. They are as follows: The Gross Plan, the Net Plan, the
Installment Plan and the Interest in Advance Plan.
1541. The Gross Plan is that in which the interest is charged on the gross
amount of the loan and the payment of the interest is required to be paid in
advance. Thus, a member borrows on 1 share $200 at 20 per cent premium, he
receives the net sum of $160, and pays interest on the gross sum of $200. Hence
the name, Gross Plan.
XF Bl]ILDING AND LOAN ASSOCIATIONS. 981
1542. The Net Plan is that in which the interest is charged only on the net
sum received by the borrower, but as in the Gross Plan, it is paid or deducted in
advance. Hence the name, Net Plan.
1543. The Installment Plan is one of recent adoption, and is a modification
of the Gross and Net Plans.
By this plan, the borrower receives the full ultimate value of $200 per share,
and pays the premium in monthly or weekly installments, in the same manner that
dues and interest are paid. From this fact, it is named the Installment Plan. It is
more favorable to the borrower than the Gross or Net Plan, and it is claimed that
its merits entitle it to universal adoption. By it, many of the complications
resulting from the deduction of premiums paid in advance are avoided, and as the
premium is paid only as it falls due, there is no premium to return to a repaying
borrower for unexpired time, and no discount on premiums bid on loans advanced
upon stock of the older series. Thus the premium account by this plan shows the
actual profit from this source.
1544. The Interest in Advance Plan. This plan has been recently introduced
into many associations throughout the country. It does not, like the Gross Plan,
require the premium in advance, but it substitutes interest in advance. Regarding
this plan Mr. Wrigley remarks:
“The interest is made payable as many months in advance of the date of the loan, as the
borrower chooses to bid, and the one bidding the greatest number of months in advance will be
the successful bidder. This interest is the same on each share for each month ($1.00) as under the
Gross Plan, the difference consisting only in paying it in advance for a given time. The mode of
obtaining a loan on this plan is as follows:
“A borrower wishing to realize about $800 net, and bidding 100 months interest in
advance, would take eight shares, which at $200 per share would be - wº - &= º $1600
Deduct 100 months' interest in advance *E. gº º sº * tº-º; sº ſº- cº 800
Net sum realized gº gº tº-> s {- g wº tº- tº- tº * . s $800
“Having paid the interest in advance for 100 months, or eight years and four months, of
course he has no interest to pay in monthly installments, as by the Gross Plan; and there being no
premium to pay or deduct, there remains but the $8 per month dues to pay on the eight shares he
has borrowed upon. He therefore pays no more per month as a borrower than he would pay as a
non-borrower. If the shares should mature in eight years, the borrower having paid four months'
interest in advance of the time of winding up, the four months’ excess of interest would be
returned to him with his satisfied mortgage. On the other hand, if the shares should exceed eight
years and four months in running out, and reach as far as, say nine years, the borrower would, on
and after the 101st month, have to pay monthly interest in addition for eight months longer.
“If one borrows on an old series, he is allowed a deduction or rebate from his interest for
each month of the past age of the series.
“If he desires to repay a loan, he is allowed a return of interest, paid in advance, for each
unexpired month.
“The foregoing is an outline of the plan of working.
“The apparent gain of the association by this method results from the greater amount of
profit deducted at once on making a loan as compared with the Gross Plan.
“For instance: On a loan, under the new plan, on eight shares at 100 months' interest in
advance, $800 would be retained as a profit, the borrower receiving the net sum of $800; while
under the Gross Plan of deducting premiums on a loan of five shares at 20 per cent premium, but
982 soulE's PHILOSOPHIC PRACTICAL MATHEMATICs. ºr
$200 will be retained as profit, while the borrower receives the same net amount of money ($800)
as the borrower under the new plan receives.
“It will thus be seen that the association apparently gains the advantage of $800 immediate.
profit under the new plan, against $200 under the Gross Plan, while the borrower receives the same
amount of money in each case; and, under the Interest in Advance Plan, pays but $8 a month,
against $10 a month under the Gross Plan.”
Other advantages are also claimed for this plan.
STATE AND UNITED STATES LEAGUES.
1545. A Building and Loan Association League is a chartered organization
or confederation of different associations, for the promotion of joint interests, com-
mon purposes, and mutual Support. -
1546. State Leagues are those formed by associations domiciled within the
respective States.
1547. A United States League is one formed by associations domiciled in
the various States. These leagues are of great benefit to the cause in which they
serve. Through them, all matters relating to the cause are discussed by the ablest,
minds engaged in the service, and the interest of each is made the good of all.
NATIONAL BUILDING ASSOCIATIONS.
1548. The name “Nationals” is applied to a class of Building and Loan:
Associations that is patterned, organized, and conducted somewhat according to
the general methods of the original or parent Building and Loan Associations as:
above described. They are a growth (branches or children, as it were) of the parent,
institutions, and have individualized themselves by incorporating some new
features, the practical advantages of which are seriously questioned by the original
Associations. The new leading features of the Nationals are the establishment of
agencies in different cities and States and the loaning of money to whomsoever
may wish to borrow on their terms. By reason of these features, they are denomin-
ated “Nationals” in contradistinction to the parent associations which loan money
only in their respective localities, and as a rule only to their respective members,
and are hence denominated “Locals.” The “Nationals” are of quite recent birth,
and in pressing their claims for patronage have, in Some cases, introduced non-
ethical methods in their management, and used decoying statements through their
agents and in their circular and advertising literature.
In many cases, they masquerade and do business on $20,000,000 or $50,000,000.
authorized capital, when in truth they have not as many thousand in net assets. To
open offices throughout the country, they incur expenses for salary of agents, rent
of offices, commission of solicitors, and advertising that consume much of, and
sometimes, all of the profits and all the dues, and then dishonorable bankruptcy
follows. The leading special features of the Nationals make some of them of
greater service to their army of officials than to their subscribers. These officials.
* BUILDING AND LOAN ASSOCIATIONS. 983
of the “Nationals” are often protected by a special expense fund, which is raised
by a direct tax on the members. It is believed by those who have had the largest
experience with Co-operative Building Associations, that the business of loaning
money is conducted on too small a margin of profit to admit of the various expenses
necessary to carry on Building Associations on the expensive agency plan of the
Nationals.
Great prudence should be exercised by parties who wish to receive the
benefits of a well and ethically conducted Building Association. “All is not gold
that glitters.” And experience proves that some locals as well as many Nationals
have proved unfaithful.
PLANS OF LOANING. MONEY OR PREMIUM PLANS.
1549. The six thousand Building and Loan Associations in the United States
adopt different plans with various combinations, of loaning their money. The
Commissioner of Labor, Washington, D.C., in his Ninth Annual Report on Building
and Loan Associations, 1894, gives 68 different plans covering 38 pages. It is not
the function of this work to present these plans. Those who are interested in the
subject, we refer to the report of the Commissioner of Labor, to be found at almost
every Building and Loan Association.
PLANS OF WITH DRAWAL.
1550. Different Associations adopt different plans for the withdrawal of
members. This subject is also fully treated by the Commissioner of Labor, in his
Ninth Annual Report, 1894, wherein he presents 12 plans, with over 100 variations.
We refer those interested in the subject to said report.
PLANS @F DISTRIBUTION OF PROFIT.
1551. To adjust the profits and to find the value of each series and of each
share at the close of each gain declaring period, is one of the most important and
complex operations which arises in the accounting Work of Building and Loan
Associations. As in the case of premium plans and plans of withdrawing, there
are various plans for the distribution of profits. Of these numerous plans, the Com-
missioner of Labor, in his Ninth Annual Report on Building and Loan Associations,
1894, presents 25 plans with numerous variations, covering 35 pages.
We give space and credit for the 1st, 8th and 9th plans presented in said
report:
PLAN 1-(Partnership Ptan).
“This plan apportions the profits among series just as profits among partners are apportioned
in a firm where the partners enter at different dates, each series representing a partner,
984 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
1. Multiply the dues paid in on the shares in force in each series by the equated time of
investment.
2. Take the sum of these products and then find what fractional part each product is of the
SUllrl.
3. These fractions are the parts of the total net profits belonging to each series.
To illustrate the rule, let us suppose that an association whose monthly dues are $1 per share
had three series in force at the end of the third year, and that the number of shares in each series
and their value per share were as follows: º
First series, 500 shares, value per share $38.87; second series, 600 shares, value per share
$25.27; third series, 400 shares, value per share $12.32; that a fourth series of 500 shares is then
issued; the net profits for the fourth year are $3,000, and the total net profits for the four years are
$5,325. Required: The value of a share of each series at the end of the fourth year.
The first series above alluded to has run four years, or forty-eight months. Forty-eight $1
payments have therefore been made on each share of stock. The first dollar paid has been invested
forty-eight months; the second dollar paid, forty-seven months; the third dollar paid, forty-six
months, etc., the last dollar of the forty-eight having been invested one month. The times of
investment thus form a decreasing arithmetical series, with forty-eight for the first term, one for
#he last term, and forty-eight for the number of terms. The total investment is thus equal to $1
invested for 1,176 months (the sum of the series), equivalent to $48 invested for 24; months.
Treating the other series in the same way, we find that $36 paid per share in the second
series has been invested for 184 months; $24 paid per share in the third series, for 12# months; and
$12 paid per share in the fourth series, for 64 months; then—
$48 × 500 × 24% = $588,000, first series’ investment for one month.
$36 × 600 × 18} = $399,600, second series’ investment for one month.
$24 × 400 × 12# = $120,000, third series’ investment for one month.
$12 × 500 × 6+ = $39,000, fourth series’ investment for one month.
$1,146,600, total investment for one month.
Bence the total net profits are divided as follows:
******, or ºr of the total profits belong to the first series.
*% or ºr of the total profits belong to the second series.
******, or ºr of the total profits belong to the third series.
T####30 or TÉr of the total profits belong to the fourth series.
Total profits to be divided are $5,325.
*r of $5,325 = $2,730.77, first series' share of the profits,
Mººr of $5,325 = $1,855.81, second series' share of the profits.
*ºr of $5,325 = $557.30, third series' share of the profits.
T#r of $5,325 = $181.12, fourth series' share of the profits.
$2,730.77 – 500 = $5.46, profit of a share of the first series.
$1,855.81 – 600 = $3.09, profit of a share of the second series.
$557.30 – 400 = $1.39, profit of a share of the third series.
$181.12 – 500 = $0.36, profit of a share of the fourth series.
$48.00, dues paid, + $5.46, profit, = $53.46, value of a share of the first series.
$36.00, dues paid, + $3.09, profit, = $39.09, value of a share of the second series.
$24.00, dues paid, +- $1.39, profit, - $25.39, value of a share of the third series.
$12.00, dues paid, +- $0.36, profit, - $12.36, value of a share of the fourth series.
In the above example, all the net profits made during the four years have been apportioned
to the several series, but some associations apportion only each year's profits in this way. Other
associations using this plan simplify the process, but obtain the same results, by dividing the total
Investment for one month (or for one week, as the case may be), into the profits to be apportioned,
for the profit on $1 invested for one month, which is then multiplied successively by the sum of the
number of weeks, months, or other periods of time for which each dollar of dues in each series
has been invested. The products will be the amount of the profits belonging to a share in each
series.
* BUILDING AND LOAN ASSOCIATIONS. 985
M
A few associations have been found that arrive at the same results by using the following
method, which is known as Clark's plan: wº
1. Multiply the number of shares in force in each series by the quotient obtained by
dividing the sum of the number of weeks, months, or other periods of time for which each dollar
of dues in each series has been invested by the product obtained by multiplying the dues paid in
on one share during the first year by the average time of investment, for the equalized results for
each series.
2. Take the sum of these results and divide it into the total profits since the beginning of
the associations, for the rate per cent of profit.
3. Multiply the quotients already found by the rate per cent of profit, for the profit of a
share in each series.
We have before seen that $48 dues per share in the first series have been invested for 244
months, which is equal to $1 invested for 1,176 months. In like manner $36 dues per share in the
second series have been invested for 184 months, which is equal to $1 invested for 666 months; $24
dues per share in the third series, for 124 months, which is equal to $1 invested for 300 months; and
$12 dues per share in the fourth series for 64 months, which is equal to $1 invested for 78 months.
The average time of first year's payments is simply half the time of investment, which is 6
months. Twelve dollars invested for an average period of 6 months is equal to $1 invested for 72
months.
Then : &
1,176 – 72 = 16.333.
666 – 72 = 9.250.
*. 300 – 72 = 4.166.
78 – 72 = 1,083. *
500 x 16.333 = 8,166.50, equalized result for the first series.
600 × 9.250 = 5,550.00, equalized result for the second series.
400 × 4.166 = 1,666.40, equalized result for the third series.
500 × 1.083 = 541.50, equalized result for the fourth series.
*E**-ºº-ºº.
15,924.40, equalized result for all series.
$5,325, the total profits, – 15,924.40 = 33.4392, the rate per cent of profit.
$0.334392 × 16.333 = $5.46, profit of a share of the first series.
$0.334392 × 9.250 = $3.09, profit of a share of the second series.
$0.334392 × 4.166 = $1.39, profit of a share of the third series.
$0.334392 × 1.083 = $0.36, profit of a share of the fourth series.
For the value of each share, add the dues as above.
There is a modification of plan 1, which follows the same general method as that shown in
the first illustration, but differs in certain particulars and gives a different result. The modifica-
tion is as follows: Instead of finding the exact equated time of investment, many associations
arrive at an approximate equated time by taking one-half the number of months a series has run.
Using the same data as in the above illustrations, we get 24, 18, 12, and 6 as the average number
of months the series have run. It is this modification that is commonly, but erroneously, called
the partnership plan.
ILLUSTRATION.
$48 × 500 × 24 = $576,000, first series’ investment for one month.
$36 × 600 x 18 = $388,800, second series’ investment for one month.
$24 × 400 × 12 = $115,200, third series’ investment for one month.
$12 × 500 × 6 = $36,000, fourth series’ investment for one month.
$1,116,000, total investment for one month.
986 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. *
The total net profits are then divided in proportion to each series' investment for one month,
thus:
#7% or #9, of $5,325 = $2,748.39, first series' share of the profits.
#% or ºr of $5,325 = $1,855.16, second series' share of the profits.
tºº, or ºr of $5,325 = $549.68, third series' share of the profits.
Tºg}} or rāg of $5,325 = $171.77, fourth series' share of the profits.
$2,748.39 - 500 = $5.50, profit of a share of the first series.
$1,855.16 -i- 600 = $3.09, profit of a share of the second series.
$549.68 - 400 = $1.37, profit of a share of the third series.
$171.77 - 500 = $0.34, profit of a share of the fourth series.
This modification of plan 1 has been simplified, the principle consisting in casting out come
mon factors in the process of multiplication.
The first series has run 48 months; the second, 36 months; the third, 24 months; and the
fourth, 12 months. The average time of investment, as we have before seen, is 24, 18, 12, and 6.
respectively. Then we proceed thus:
48, age in months, x 24, average time, × 500 shares.
36, age in months, x 18, average time, × 600 shares.
24, age in months, x 12, average time, X 400 shares.
12, age in months, x 6, average time, × 500 shares.
It will be readily seen that 12 is a factor common to all the numbers of the first column, and
that 6 is a factor common to all the numbers in the second column. Casting out these factors, We
have— t
4 × 4 × 500 = 8,000. Hence #% of the total profits belong to the first series.
3 × 3 × 600 = 5,400. Yºr of the total profits belong to the second series.
2 × 2 × 400 = 1,600. Tºr of the total profits belong to the third series.
1 × 1 × 500 = 500. Tºg of the total profits belong to the fourth Series.
Total, 15,500.
Briefly put, then, the simplification is as follows: Multiply the number of shares in force in
each series by the square"of the time of investment expressed in terms or periods corresponding to
the intervals between the series, and then divide the profits in proportion to these products.
The foregoing simplification has been still further simplified by finding the profit of a share
in each series directly, instead of finding each series' share of the profit, as follows:
1. Multiply the number of shares in force in each series by the square of the time of invest-
ment expressed in terms or periods corresponding to the intervals between the series.
2. Divide the sum of these products into the results obtained by multiplying the total net,
profit by the square of the time of investment expressed as above.
The total of the products, as in the last illustration, is 15,500; then—
$5,325, total profits, × 4 × 4 + 15,500 = $5.50, profit of a share of the first series.
$5,325, total profits, x 3 × 3 – 15,500 = $3.09, profit of a share of the second series.
$5,325, total profits, x 2 × 2 - 15,500 = $1.37, profit of a share of the third series.
$5,325, total profits, × 1 × 1 — 15,500 = $0.34, profit of a share of the fourth series.
A share of the first series receives 16 times as much profit as a share of the fourth series; a
share of the second series, 9 times as much; and a share of the third series, 4 times as much. This
method, therefore, reveals the fact that, by multiplying the number of shares in force in each
series by the square of the time each series has been invested, expressed in years, half years, quarter
years, etc., corresponding to the intervals between the series, a correct basis of calculation is
reached. These simplifications, however, are practicable only where series are issued at regular
intervais, as fractions complicate the operation. This simplification is known as Rice's rule.
BUILDING AND LOAN ASSOCIATIONS. 987:
A few associations arrive at the same results by dividing the total investment for one month
into the profits, for a rate per cent of profit, and then applying the rate to each series' investment
for one month for each series' share of the profits. The process is also varied in the following
manner: Find what annual rate of interest the profits are equivalent to on the amount of dues
paid for one-half the time that all the dues have been invested, and apply this rate on the dues
paid per share for one-half the time of investment, for the profit of a share in any series.
Other variations are the following:
1. The profits are distributed on the amount of dues actually paid in on the shares in force
in each series (not what the regular payments should have amounted to), multiplied by one-half
the time of investment.
2. The profits are distributed on the total amount of dues standing to the credit of the
shareholders in the loan fund multiplied by one-half the time of investment.
3. The series are not allowed to participate in the profits for the term in which they were
issued.
4. The profits are distributed on the amount of dues actually paid in on all shares in force
that are three months old or over, multiplied by one-half the time of investment, shares less than
three months old not participating.
5. The profits are distributed to the free shares only, dues on shares borrowed on being
credited on loans.
6. Profits arising from withdrawals are divided equally among the shares of the respective
series from which the shares were withdrawn.
7. Profits arising from entrance fees are divided equally among the shares of the respective
Series in which the shares are taken.
8. A profit of $1 is given to all shares six months old or over. The remainder of the profits
is distributed on the dues paid in on the shares in force six months old or over multiplied by one-
half the time of investment.
9. Profits arising from premiums are divided equally among all the shares in force at the
end of the period during which the loans were made. Profits from all other sources are distributed
in accordance with the modified rule. -
10. A fixed rate of interest is given on the total amount of dues paid on the shares in force
at each apportionment. This interest is deducted from the profits for the term and the remainder
distributed according to the modified rule.
11. A fixed rate of interest is given on the value of the shares in force as declared by the
last report. This interest is deducted from the profits for the term and the remainder is distributed
according to the modified rule.
12. A portion of the total amount of premiums received by and due the association is
arbitrarily determined upon, and held in reserve to be applied in future dividends; the amount
thus determined upon is deducted from the total profits, and the remainder of the profits is dis-
tributed as follows: The interest and dividends allowed on free shares withdrawn are added to the
dues paid in on such shares, and the sum total of said interest and dividends is deducted from the
amount of distributable profits; the balance is distributed among all the shares in accordance with
the foregoing modified rule. *
There is still another modification of plan 1, as follows: Multiply each series' investment
(that is, the dues paid in on the shares in force) by one-half the number of months invested plus
one and apportion the profits in proportion to these products.
Using the same data as before we proceed as follows:
ILLUSTRATION.
$48 × 500 × 25 = $600,000, first series' investment for one month.
$36 × 600 × 19 = $410,400, second series' investment for one month.
$24 × 400 × 13 = $124,800, third series' investment for one month.
$12 × 500 × 7 = $42,000, fourth series’ investment for one month.
$1,177,200, total investment for one month.
988 SouLE's PHILOSOPHIC PRACTICAL MATHEMATICs. &
Then, proceeding as before, we find that—
#9% or §§ of $5,325 = $2,714.07, the first series' share of the profits.
Yº, or # of $5,325 = $1,856.42, the second series' share of the profits.
***ś or ### of $5,325 = $564.53, the third series' share of the profits.
T#####5 or ºr of $5,325 = $189,98, the fourth series' share of the profits.
$48, dues, + ($2,714.07 – 500) = $53.43, value of a share of the first series.
$36, dues, + ($1,856.42 – 600) = $39.09, value of a share of the second series.
$24, dues, + ( $564.53 - 400) = $25.41, value of a share of the third series.
$12, dues, + ( $189,98 - 500) = $12.38, value of a share of the fourth series.”
NoTE. 1.-1280 local associations and 81 Nationals throughout 45 States adopt this plan with
some of its variations.
NOTE: 2. —1 association in Louisiana adopts this method with variations.
&
PLAN 8.—(Brooks' Plan).
1. Find the legal rate of interest on the values of the shares as declared by the last report
and deduct it from the profits of the term for the net profit. *
2. Divide the net profit by the sum of the dues paid in during the term and the interest on
the previous series, for the rate per cent of profit.
3. Multiply the sum of the interest and dues for the term on one share of each series by the
rate per cent of profit, for the profit on one share.
4. Add the previous value of a share of each series, the legal interest on this value, the
dues paid in during the term, and the profit on the share to find the present value of a share of
each series.
Let the legal rate of interest be 6 per cent, and using the same data as in the preceding
illustrations, we proceed thus:
ILLUSTRATION.
$38.87 × .06 = $2.33, interest on a share of the first series.
$25.27 × .06 = $1.52, interest on a share of the second series.
$12.32 × .06 = $0.74, interest on a share of the third series.
$2.33 × 500 = $1,165, interest on first series.
$1.52 × 600 = $912, interest on second series.
$0.74 × 400 = $296, interest on third series.
$2,373, total interest on old series.
$3,000, the profits for the term, - $2,373, - $627, the net profits.
$12 × 2,000, the number of shares in force, = $24,000, the dues paid during the term.
$24,000 + $ 2,373 = $26,373, the active capital.
$627 –– $26,373 = 2.3774, the rate per cent of profit.
($12 + $2.33) × .023774 = $0.34, the profit of a share of the first series
($12 + $1.52) × .023774 = $0.32, the profit of a share of the second series.
($12 + $0.74) × .023774 = $0.30, the profit of a share of the third series.
$12 × .023774 = $0.29, the profit of a share of the fourth series.
$38.87 -- $2.33 + $12 + $0.34 = $53.54, value of a share of the first series.
$25.27 -i- $1.52 -- $12 + $0.32 = $39.11, value of a share of the second series.
$12.32 + $0.74 + $12 + $0.30 = $25.36, value of a share of the third series.
$12 + $0.29 = $12.29, value of a share of the fourth series.
Some associations using this rule vary it by allowing less than the legal rate of interest on
the old values, but no separate classification of such associations has been made.
Another variation from the rule consists in allowing the legal rate of interest, not only on
the values of the old series at the beginning of the term, but also on the equated amount of dues
paid in on all the shares during the term.
Yºr BUILDING AND LOAN ASSOCIATIONs. 989
Another variation from the rule is as follows: After the legal rate of interest for the term on
the values of the old series has been deducted from the profits for the term. the remainder of the
profit is divided by the sum of the value of all shares at the beginning of the term, the dues paid
in during the term, and the interest for the rate per cent of profit. This rate is then applied to the
sum of the previous value of each share, the dues paid in during the term, and the interest on the
share for the profit on one share.
NOTE.-25 associations adopt this plan.
PLAN 9.
The profits are divided equally among all the shares in force. This plan is used principally
by terminating associations.
The following variations of this rule have been found:
1. The profits for each term are divided equally among all the shares in force.
2. The profits for each term are divided equally among all shares three months old or over,
six months old or over, or nine months old or over, as the rules of the association may provide. In
such cases shares less than three months, six months, or nine months old do not participate in the
profits.
3. Shares less than a year old are given such portion of the profits as the board of directors
may allow. The remainder of the profits is divided equally among all the shares one year old or
OWOI’.
4. The profits for each term are divided equally among the free shares.
5. Shares one year old or over receive equal amounts of the profits for the term. Shares
nine months old receive three-fourths as much as those a year old; shares six months old one-half
as much, etc.
6. The profits made by each series are kept separate, and are divided equally among all the
shares of the series.
7. Profits arising from premiums and fines are divided equally among the shares of the
series in which such profits were made. All other profits are divided equally among all the shares.
NOTE: 1.—1062 locals and 2 Nationals adopt this plan with variations.
NOTE 2.—2 associations in Louisiana adopt this method with variations.
PROBLEM BY THE GROSS PLAN.
1552. To elucidate, in detail, the manner of adjusting profits and finding the
value of the different series and the separate shares by the Gross plan, we present
the following problem and work the same in full, according to the method of Edmund
Wrigley.
PROBLEM, FIRST TERM.
An association which issues yearly series and loans on the Gross plan, Coms
mencing business with an issue of 500 shares of $200 each, and collecting dues 25
cents per week, would, during the first fiscal year, receive dues amounting to $6500.
Bresuming that the dues had been loaned at 25 per cent premium, that the
gains by interest etc., were, at the close of the fiscal year, $400, and the expenses
990 soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. * .
$300, what would be the net gain for the term, the value of the series, and the
value per share ? Ans. See operation.
*.
< OPERATION.
500 shares at 25 cents for 52 weeks = $13 per share, = $6500 dues.
$6500 loaned at 25 per cent premium = * * * m = * - $1625 prem.
Gains by interest , etc. tº gº º ºs sº s ſº tº ſº * 400
Total gain, tº a gº sº tº º * , sº s & gº - $2025
Less expenses, say, - - º gº {- se tº-8 &= 300
Net gain for the term, sº gº sº tº º ºs - - - 1725
$8225
Total value of first series, * * sº gº *- tº- $º wº
$8225 – 500 shares = $16.45 value per share.
SECOND TERMI.
The association issued at the beginning of the second year or term, a second
series of, say, 800 shares.
The following statement shows the business of the second term or year:
Passive capital $8225. Dues on first series for second year $6500. Dues on
second series for first year, 52 weeks at 25 cents on 800 shares, = $10400. Total
dues $16900. Interest on $8225 passive capital or value of first series at 6 per cent
for second term; $493.50. Premium on $6500 dues of first series at 25 per cent
$1625. Premium on $10400 dues on second series at, say, 30 per cent, $3120. Total
premium $4745. ſº
Average interest on active capital, $22138.50, at, say 6 per cent for 3 term or
average time, $664.15. Total gain, $5902.65, expenses, say $402.65. Net gain for
second term, $5500. What is the value per share of each series and what is the
value of each series?
OPERATION.
Dues for Second Term. e **
Dues for second term on first series, 500 shares, 52 weeks at 25 cents = s tº § wº $6500
Dues for first term on second series, 800 shares, 52 weeks at 25 cents = - - - - 10400
Total dues for both series, second term, sº 3- tºº tº tº es sº º $16900
Gains for Second Term.
Interest on $8225, passive capital, at 6 per cent for second term, - tºº $493.50
Premium on $6500, at 25 per cent for second term, - - - - - 1625.00
Premium on $10400, at 30 per per cent for first term, tº E tº gº 3.120.00
Average interest on active capital, $22138.50 at 6 per cent for 3 term, y
or average time, sº tº gº tº ſº- * - sº e- s º 644.15
Total gain, - - - - - - - - - - - - - $5882.65
Less expenses, *- s sº s sº sº dº sº gº sº tº . sº- * 402.65
- - - - $5480.00.
Net gain for second term, - º sº - - sº- g
X- BUILDING AND LOAN ASSOCIATIONS. 99 I
PRESENT VALUE OF ONE SHARE OF FIRST SERIES AND TOTAL VALUE OF FIRST SERIES.
Value of one share, first series, at beginning of term, $16.45. Interest on same at 6 per cent
for the term is 98.7 cents.
.987 × 500 shares first series = $493.50, as shown above.
$5480 met gain — $493.50 = $4986.50 gain to be divided among the 1300 shares of both
series. $4986.50 – 1300 = $3.83576-H per share.
Dues on one share of first series for two years, $26
Gain on one share, first term, ($1725 – 500) - 3.45
Gain by interest, second term, - .987 4.82276-H
Gain by interest, first term, - 3.83576–H $ —
Total gain in two terms or years, - $34.27276-H value of each share.
$34.27276 × 500 shares = $17136.38 value of first series.
PRESENT VALUE OF ONE SHARE OF SECOND SERIES AND TOTAL VALUE OF SECOND SERIES.
Dues on one share, second series, for one term or year, * - $13.
Gain per share for second term, as above, - º - -> - 3.83576–H
Value of each share of second series, - - sº - wº- - $16,83576–H
$16.83576 × 800 shares = $13468.61 value of second series.
$17136.38 + $13468.61 = $30604.99 total value of both series, 1300 shares, at the close of second
term or year.
NOTE: 1.-257 associations adopt this plan with variations. The variations are shown on
page 437 of the report of the Commissioner of Labor, 1894, and pages 83 to 100 of Wrigley’s How to
Manage Building Associations.
NOTE 2.—8 associations in Louisiana adopt this method with variations.
METHODS AIDOPTED BY SOME OF TELE NEW ORLEANS BTUILDING
º AND LOAN ASSOCIATIONS.
1553. In response to inquiries made of several Building and Loan Associations
in New Orleans, regarding their method of determining and adjusting profits, we
received replies from four of the most popular, prosperous and ably managed
associations in the city. They are as follows:
OFFICE OF
MUTUAL BUILDING AND HOMESTEAD ASSOCIATION,
No. 337 (Old No. 87) ST. CHARLES STREET, MASONIC TEMPLE.
NEW ORLEANs, Aug. 2, 1895.
GEO. SOULF, Esq.,
City.
IXEAR SIR:
Replying to your favor of the 29th ult, I am pleased to furnish you with the information
requested by you, and hope it may prove serviceable.
Our profits are apportioned upon the partnership plan, and the dividend is a percentage based
992 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
upon the amount each shareholder has invested for the time it is invested. While we issue shares
in what we nominally term series, said shares do not mature at the one time, because the amounts
deposited thereon may exceed the minimum installment, and hence the accumulated capital
invested, and consequent profits vary and cause shares to mature at uneven dates.
The rule we follow may be formulated thus:
1. Find the net profits to be apportioned.
2. Find the cash value at the commencement of the period just ended, of all shares then ir
force, which are still in force.
3. Find the total of installments paid in on all shares within the period Just ended.
4. Add to item (2), the amount of item (3), reduced to its equivalent capital for an invest-
ment covering the whole period. e
5. Divide item (1), by item (4), and the result will be the per cent of profit.
6. At the rate found apportion each share (or batch of shares under one account) with its
individual profit, following in each case the method described above.
EXAMPLE.
On 10 shares, the dues for 52 weeks at $2.50 equals, - - - - - - - - $130.00
These being paid monthly, it is the equivalent of Sixty-five (65) Dollars invested for one
year, and if profit is six (6) per cent, the dividend equals, gº tº gº tº * = tº 3.90
Cash value of shares at second year, - {º tº sº. sº tº se *- & wº - $133.90
Add dues for year, - gº gºs tº tº sº tº $º º ſº gº e- & tº- º ſº 130.00
The capital then invested is, - - - tº gº º ºs - - - $133.90
Plus average of $130, tºº tº- sº s dºe * * tº- - * sº 65.00
Equivalent capital for year, º gº tº Eº tº tº - - - $198.90
On which a six (6) per cent dividend would be, - sº tº 4-3 sº tº * > tº º º 11.93
Cash value of shares at third year, tº gº *s gº * º sº * tº gº gº $275.83
Add dues for year, - tº- tº * gº $º wº e- tº- * Ǻs * = sº gº 4-º 130.00
Following same method, $275.83, plus $65, equals $340.83 at six (6) per cent, - tº-e tº 20.44
Cash value at fourth year, º *-i- dº * = fº tº gº * * &= } tº tº iºs $426.27
And so on.
Very respectfully,
ESPY W. H. WILLIAMS, Secretary.
P. S.—This company also issues shares on the Dayton plan. These shares are $100 face
value, payable in installments of not less than 25 cents per week, but as much more as desired by
holder. The shares are issued at any time, and begin to participate in profits at once. Full paid
shares are issued on payments of $100 each, and dividends are paid thereon in cash every six
months.
Loans are also made without any premium or bonus, and without bidding, simply upon
application filed.
E. W. H. W., Sec'y.
—cºe—
OFFICE OF
UNION HOMESTEAD ASSOCIATION,
NEW 336 CARON DELET STREET.
NEW ORLEANs, August 7, 1895.
Col. GEO. SOULE,
City. *
DEAR SIR:—In response to your circular letter of July 29th, I note with interest, that you
intend giving representation in your forthcoming work on “Philosophic Practical Mathematics,”
to the problems of Building and Loan, or Homestead Associations. * * * †. º $
As to our own method of distributing profits to members, I reply as follows:
The member stands to us in the relation of a series, of which you speak. Our system con-
* BUILDING AND LOAN ASSOCIATIONS. 993
templates the calculation of dividend on every pass-book, independently of every other. No two
need be exactly the same. We do not issue stock in series, except nominally; we call all the issue
of any one month the series of that month, and give it a number; but the books or shares in that
series are each independent of every other.
For example: In series 38, issued last January, books 521, 529, and 536, each call for 10
shares, on which the regular deposit would be $5.00 per month. But on June 30th, the standing of
the three books was, respectively, $75, $30 and $70.
We do not declare a dividend of so much per share, but a percentage dividend, this time 5
per cent for the six months. This enables us to give to each of these books its fair (right) portion
of earnings. Where the deposits are made regularly, month by month, we take half the amount
as the average investment; so on book 529, on which 6 payments of $5 were made, total $30, we
have an average investment of $15 at 5 per cent dividend, 75 cents. On book 521, there was a
deposit of $50 in February, on which we allow interest from the time it was put in, and give that
book $2.68 dividend.
This method of averaging is not mathematically exact, but is close enough for all practical
purposes, especially in view of the fact that the surplus or undivided profit, is held and belongs to
the stockholders, and may be subsequently divided among them, when ordered by the directors.
In case of a book in force prior to January, for example, No. 412, issued July 1894, it showed a
value, Jan. 1, of $76,85, (6 × 12.50 plus dividend) and during the six months six payments of $12.50
were made. We add to the capital, at the beginning of the term, half the deposits of the term,
$37.50, and (dropping fractional parts of $1) have an invested capital for the term of $114, at 5 per
cent, gives dividend $5.70, and cash value of the book, July 1, 1895, $157.55.
In case but 5 monthly payments had been made during the term, I would allow but 5/6ths
of the regular dividend.
This method is simple, and can be made as accurate as desired. Its beauty lies in its
flexibility, adapting itself to the needs of every book, or every share, or every series, if applied to
the series capital. It can be made as exact as any method in use, if one wishes to determine to a
minute fraction the rate of earnings or dividend. It is applicable to any so-called “permanent”
associations, the specific name adopted by most associations issuing stock continuously, and
treating each pass-book independently.
Trusting, that from the foregoing, you will understand clearly our method of division of
profits, and holding myself in readiness to answer promptly any questions you wish to ask in
further connection with this subject, I am,
Very truly,
JNO. E. HUFFMAN,
Secretary.
—sºe—
THE
SECURITY BUILDING AND LOAN ASSOCIATION,
No. 5 Connrnercial Place, Roorn 1.
NEW ORLEANs, August 12, 1895.
GEORGE SOULF, Esq.
City.
DEAR SIR:
Your favor, 29th, was duly received, and allow me to apologize for not answering sooner.
Our association, the Security Building and Loan Association does not issue stock in series.
The shares are $100 each, and are issued at any time to suit the subscriber. Our method of divid-
ing the profits is much more simple than that of serial associations, as we conduct our business on
the cash basis, or Dayton plan.
The following extract from Article 4, of our Charter will help to explain:
“On January 1, and July 1, of each year, the Board of Directors shall ascertain the earnings
994 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
of the association for the preceding six months, and out of such earnings they shall (1) deduct all the
expenses of the association for such time; (2) set aside to the fund for contingent losses such an
amount as they may deem best, until such contingent fund equals 5 per cent of the entire out-
standing mortgage loans, which fund shall be invested as directed by the board; (3) the balance
of said earnings shall be declared as a dividend to the shareholders, in proportion to the value of
stock of each member, at the time the preceding dividend was declared, and all payments made
between such dividend periods shall begin to earn dividends from the beginning of the next semi-
annual term.”
“The dividends declared on ‘running stock’ shall be credited on the pass books of the
members semi-annually, and dividends on “paid up stock’ shall be paid in cash on the fourth
Monday of January and July of each year.”
The Security has declared its third semi-annual dividend of 4 per cent.
I beg to enclose copy of our charter, also our last semi-annual statement.
The serial system is not being adopted by the new associations; it is complicated and not
easy for the average stockholder to understand.
Any further information that I can furnish will be cheerfully given.
Very truly,
W. E. DODSWORTH,
Secretary.
—eſpe—
OFFICE OF
THIRD DISTRICT BUILDING ASSOCIATION,
CORNER ROYAL STREET AND LAFAYETTE AVENUE.
NEw ORLEANs, August 13, 1895.
PROF. GEO. SouLé,
City.
DEAR SIR:
Enclosed please to find example of the method of distribution of profits as adopted by this
association. Trusting same will be of use to you, I remain,
Yours truly,
JOS. L. SPORL, Secretary.
EXAMPLE,
Entitled to full
Series No. Shares. Paid in on dividend for Pro rata on
A. 708 $9204.00 12 months or 52 weeks, one year, - * * * $9204.00
B 109 981.00 9 4 & ** 39 “ # of year, tº a º 735.50
C 327 2125.50 6 “ “ 26 “ 4 “ gº tºº 1062.75
D 841 2943.50 3 & 4 “ 13 “ + “ * 735.87
Total paid in $15254.00 $11738.12
The net profits for distribution are $2969.
Divide net profits $2969 × 100 by amounts prorated, $11738, gives 25.29-H 96, multiply $9204
amount paid in by series A by 25.29%, gives $2327.69 gain for the whole number of shares of series
A. $2327.69 - 708 shares of series A = $3.28 dividend per share of series A.
Then to find the dividend per share for each of the other series, we proceed thus:
series B $ 735.50 prorated value x 25.29% = $186.01 – 109 No. shares = $1.71 dividend per share.
48 C 1062.75 “ “ × 25.29% = 268.77 – 327 “ “ = .82+ “ & 6
48 D 735.87 “ “ × 25.29% = 186.10 - 841 “ “ = .22+ “ é &
NoTE.—The payments on stock in this company are 25 cents per week, or $13 per year, of 52
weeks. Hence the different series are credited as follows:
º: BUILDING AND LOAN ASSOCIATIONS. 995
series A is credited with the full $13.00 on each share as paid in.
* B “ 4 & “ only 9.00 * * * * ‘‘ ‘‘ “ instead of $9.75 because some are
tº C « {{ ** the full 6.50 “ (: & & ( & & 4 [delinquent.
“ D tº & 4 & 4 3.50 < * * * & ( & & “ instead of $3.25 because some are
[paid in advance.
According to the foregoing work, the value of the stock or series at the end of the year would
be as follows:
Series A, amount paid in $9204.00+ dividend $2327.69 = <--> - & - - º $11531.69
“ B, 4 & “ “ 981.00-- * { 186.01 = º - gº º - Gº 1167.01
“ C, { { “ “ 2125.50+ { { 268.77 – º - sº - - º 2394,27
“ D, { { “ “ 2943.50+ ( & 186.10 = * - = - - tºs 3129.60
Total value, tº º sº; - * * --> © - - - - - $18222.57
Balance undivided by reason of the non-exactness of the gain per cent, tº .43
$18223.00
MISCELLANEOUS PRACTICAL PROBLEMS.
To find the rate per cent of interest received by a non-borrowing member on maturing
shares.
1554. What rate of interest is gained on shares of building associations, the
maturity value of which is $200, the dues being $1 per month, supposing the series
to mature in 84 years? Ans. 22.39 — ‘ſ.
oPERATION,
84 years = 102 months at $1 = $102 dues paid.
$200 maturity value of share — $102 = $98 gain on the monthly investment of $1.
102, first term + 1, last term = 103.
103 × 51, half the number of terms = 5253 months or (5253 – 12) 437; years, equated time
of interest on $1; or conversely, interest on $437# for 1 year.
Then since $437% gain $98 in 1 year, the rate per cent is 437; ; 98 : : 100 : 22.39 – 96, Ans.
Or thus, to find the rate per cent:
$
98
437; 100
22.39 – 9%, Ans.
SECOND OPERATION TO FIND RATE PER CENT.
Interest on $437# for 1 year at 1 per cent is $4.37%.
Then as $4,374 : $98 : : 1 per cent : 22.39 – 96 Ans.
Explanation.—Since $1 dues were paid each month for 102 months, the first dollar paid bears
interest in favor of the investor, for 102 months; the second payment is at interest 101 months,
etc., to the last payment, which is at interest for 1 month. Hence, the interest on the different
payments is equivalent to the interest on $1 for a number of months equal to the sum of an arith-
metical series, whose first term is 102, last term 1, and number of terms 102. This number is, as
shown in the operation, 5253 months, or 437# years. Then since the interest on $1 for 437+ years is
the same as the interest on $437# for 1 year, we may use either statement in computing the interest
or in finding the rate per cent.
996 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
To find the rate of interest paid by a borrower, on the Net Plan.
1555. A member bought a loan in a building association conducted on the
net plan, and paid $60 premium on exchange of 8 shares, new issue. What will be
his rate per cent of interest if the series expires in 84 years, and 6 per cent interest
is charged on the loan 3 Ans. 6.25+ 7%.
OPERATION. -
8 shares at $200, = - * - $1600 par of shares.
8 shares at 60 premium = - - 480 premium.
Amount of loan, - * - $1120
Interest on $1120 for 1 month is enºw, = $5.60
Dues, 1 month on 8 shares at $1, = º - 8.00
Monthly payment is - - - - - $13.60
$13.60 × 100 months period of loan = $1360 total payment.
Then the interest on the $13.60 monthly payments for the different periods, is equivalent to
the interest on $1 for a number of months equal to the sum of an arithmetical series, whose first
term is 100, last term 1, and number of terms 100, which is, 100 + 1 = 101 × 50, half the number
of terms, 5050 months. Interest on $13.60 for 5050 months is $343.40 + $1360 = $1703.40, aggregate
payments.
$1703.40 — $1120 = $583.40 interest on the loan for 84 years.
$583.40 - 8+ = $70.008 interest for 1 year.
Then what per cent of $1120 is $70.008? It is ($70.008 × 100) -- $1120 = .0625-H or 6.25-H%
Ans.
NotE.—By the net plan the premium is deducted from the par value, and interest is charged
on the net amount of the loan.
To find the gain per cent realized by a Building and Loan Association, on its invest-
ments of dues, at simple interest.
1556. If the dues of a Building and Loan Association are $1 per month, and
the shares mature in 83 years, what would be the annual average gain per cent of
the association, provided it invested monthly the dues as received, at 6 per cent
simple interest? - Ans. 6.06 — ?).
* OPERATION.
84 years = 102 months at $1 = $102 dues received in 83 years.
The interest on the $1 monthly dues, or receipts for the 102 months, is equivalent to the
interest on $1 for the equated time, which is (102 + 1 = 103 × 51) 5253 months or 437# years.
Interest on $1 for 437+ years, at 6 per cent, is $26.26}.
This $26.264 interest is the gain on $1 monthly dues for 84 years.
Then $26.264 – 84 years = $3.09 gain per year on $102 dues, or average investment of ($102
- 2) $51. g
$3.09 – $51 = $.0606 — or 6.06 – 96 average gain per annum.
Or thus, to find the per cent as per page 599:
Interest on $51 for 1 year, at 1 per cent, is 51 cents.
X} BUILDING AND LOAN ASSOCIATIONS. 997
Then since 1 per cent for the time gives 51 cents interest, conversely 51 cents required 1 per
cent; and since 51 cents interest required 1 per cent, 1 cent will require the 51st part and 309 cents
will require 309 times as much, which is 6.06 – 96.
Or thus, 51 : 100 :: 3.09: 6.06 – 96.
PROOF.
Interest on $51 for 84 years at 6.06 – 96 = $26.27 —.
To find the per cent realized by a Building and Loan Association on its investments of
dues, at simple interest invested weekly.
1557. If the dues of an association are 50 cents a week, and the shares mature
in 43 years, what would be the annual average gain per cent to the association, if
it invested weekly the dues received at 6 per cent, simple interest?
Ans. 6.03 — 7%.
OPERATION.
44 years × 52 weeks = 234 weeks at 50 cents = $117 dues received in 44 years.
The interest on the 50 cents weekly dues for the 234 weeks is equivalent to the interest on
the 50 cents for the equated time, which is (234 + 1 = 235 × 117) 27495 weeks, or 52833 years of 52
weeks each.
The interest on 50 cents for 27495 weeks, at 6 per cent, is $15.864.
This $15.864 interest is the gain on 50 cents monthly dues for 43 years. Then 15.8625 – 4} =
$3.525 gain per year on $117 dues or average investment of (117 –– 2) $58.50.
Then $3.525 -- $58% = .0603 — or 6.03 – 96 average gain per annum of 52 weeks.
Or thus, 584 : 100 : ; 3.525 : 6.03 – 96.
NOTE.—In the above operation 52 weeks are used as a year.
To find the average annual gain per cent realized on weekly payments, simple interest
being allowed on the weekly payments for each year, and the annual amounts
to be put at compound interest to the close of the period.
1558. Using the figures of the preceding problem, what would be the com.
pound interest, and the average annual gain per cent, allowing 6 per cent simple
interest on the weekly payments for each year of 52 weeks, and the annual amounts
to be put at compound interest at 6 per cent, to the close of the period or the
maturity of the shares, 44 years of 52 weeks each?
-*. OPERATION.
The dues being 50 cents per week, the sum received and loaned is $26 for the first year, at 6
per cent.
The interest on the 50 cents weekly payment of dues for 52 weeks, of the first year, is equiva-
lent to the interest on 50 cents for the equated time, which is (52 + 1 = 53 × 26) 1378 weeks.
The interest on 50 cents for 1378 weeks, at 6 per cent, is 79.5 cents. Then 26 + 79.5 cents =
$26,795 yearly value of the 50 cents weekly payments.
This $26,795 is a yearly recurring amount to be placed at compound interest; thus forming a
series of annuities, each term of which is $26,795.
998 soul E's PHILOSOPHIC PRACTICAL MATHEMATICs. º:
Then to find the sum of these annuities, we have the following formula:
S = A X Rºl — A = S = $26,795 × 1.06% – $26,795 $26,795 × 1.2624770 – $26,795
R – 1 -p .06 - .06 *-
$117.21 = the sum of an annuity of $1 at compound interest for 4 years at 6 per
cent.
NOTE.—The 1.2624770 is the fourth power of 1.06, and is obtained from the Compound Interest
Table, page 612.
To this $117.21 we must add: (1). The interest thereon for 6 months, at 6 per cent, which is
$3.51. (2). The $13 dues for 26 weeks. (3). The interest on the $13 dues for the equated time,
which is 20 cents. Thus making a total sum of $133,93.
See the close of the first solution for the full details of finding gain and gain per cent.
SECOND SOLUTION BY COMPOUND ANNUITY TABLE.
$4.374616
26,795
$117.217835.720
3.51
13.00
.20
$133,927–1–
To find the total cost of a loan to a borrower at 6 per cent simple interest, Gross plan,
monthly payments, $1 per share.
1559. A member of a building association bought a loan on 10 shares in a
new series, at $15 and the stated premium, with 6 per cent interest. What will be
the full cost of his loan if the series matures in 110 months? Ans. $2810.50.
OPERATION.
NOTE.-By the Gross and Installment Plans, the dues and the interest at 6 per cent on each
share of $200, are each $1 per month.
$1 dues + $1 interest = $2 monthly payment per share.
$2 × 10 shares = $20 monthly installments to be made and which earn interest as follows:
The first installment will earn interest for 110 months, the second installment, 109 months,
and so on, until the payment of the last installment which earns no interest. Hence the interest
on the 110 installment payments for the different periods of time is equivalent to the interest on
$20, one installment for the equated time which is (110 – 1 = 111 × 55) 6105 months.
The interest on $20 for 6105 months, at 6 per cent, is $610.50. $20 each payment x 110 payments
= $2200, sum of payments.
$2200 + $610.50 = $2810.50 total cost of loan to the borrower.
NOTE.-This method is known as the 6 per cent method.
To find the actual premium charged on loans and refunded on their payment.
1560. A member bought a loan on 8 shares of stock 1 year and 4 months old,
and paid $80 premium. What is the actual premium if the value of a share was
$18 at the last annual report 3 Ans. $71.20.
º: BUILDING AND LOAN ASSOCIATIONS. - 999
OPERATION.
$18, value of 1 share at last report + $4 dues for four months = $22 present accumulated value
of each share.
$200 unaccumulated value — $22 accumulated value = $178 unaccumulated value. Hence
### of the par is unaccumulated; therefore #$$ of the $80 premium bid is the actual premium paid
by the borrower. §§ of $80 = $71.20, actual premium.
NOTE 1.--To find the met loan, deduct the actual premium.
Nors 2.—To find the total cost of a loan, multiply monthly payments by the number of
months.
To find the average annual gain per cent realized on monthly payments, simple interest
being allowed on the monthly payments for each year, and the annual amounts
to be put at compound interest to the close of the period.
1561. 1. If the dues of a Building and Loan Association are $1 per month,
and the shares mature in 84 years, what would be the compound interest and the
average annual gain per cent allowing 6 per cent simple interest on the monthly
payments for each year, and the annual amounts to be put at compound interest at
6 per cent to the close of the period or the maturity of the shares?
Ans. Compound interest $30,42; 7.016+ %.
REMARKS.—The solution of this problem forms the basis for solving many
problems of interest to Building'and Loan Associations, and to all parties concerned
in loaning or borrowing money. We therefore give three solutions of the problem
in order to fully elucidate the principles involved, and to make clear the methods of
Solution.
FIRST SOLUTION,
1st year. 12 months at $1 = $12 dues received 1st year. One dollar the first receipt was at interest
for 12 months, the second receipt of $1 was at interest for 11 months, etc. Hence the
interest on the receipts for dues for the 12 different periods, is equivalent to the interest on $1
for a number of months represented by the sum of an arithmetical series, whose first term is
12, last term 1, and number of terms 12. Thus 12 + 1 = 13 × 6, half the number of terms,
= 78 months equated time of interest on $1 at 6 per cent gives, * , - - $ .39
Dues received and loaned the 1st year, * * & gº * - wº * * * fº ę 12.00
Amount to loan the 2d year, - - - " - - - - - - - $12.39
£d year. Interest on $12.39 for 12 months, at 6 per cent, * * * * * * = .74
Dues 2d year and interest for 78 months, equated time, on $1 as in the first year, - 12.39
Amount to loan the 3d year, - gº gº * > tº i- * º º tº tº - $25.52
3d year. Interest on $25.52 for 12 months, at 6 per cent = - tº sº º tº º ºs 1.53
Dues the 3d year and interest, 78 months, on $1 as in the first year, - e- ge tºº 12.39
*Amount to loan the 4th year, º $º tº- gº * : sº tº- tº † = tº º - $39,44
4th year. Interest on $39.44, 12 months, at 6 per cent = * tºº º * * = tº 2.37
Dues and interest, 4th year, as in the first year, - tº º tº tº gº sº sº E- 12.39
Amount to loan, 5th year, sº tº - - - - - - - - - - $54.20
Bth year. Interest on $54.20, 12 months, at 6 per cent = tº ſº * * * * * 3.25
Dues and interest, 5th year, as in first year, * > tº a ſº tº e tº * , : º tº 12.39
Amount to loan, 6th year, gº tºº & Cº º º • * * - - - $69.84
IOOO SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS.
3.
6th year. Interest on $69.84, 12 months, at 6 per cent, e - se º - tº - $ 4.19
Dues and interest, 6th year, as in first year, - * - - tº - tº - 12.39
Amount to loan, 7th year, º - - tº - -e - - º - º ºg $86.42
7th year. Interest on $86.42, 12 months, at 6 per cent, sº - º - - * - 5.19
Dues and interest, 7th year, as in first year, - eº - tº- tº * sº {- 12.39
Amount to loan, 8th year, º - º - - º - º - - &= - $104.00
8th year. Interest on $104.00, 6 months, at 6 per cent, - º - º tº - º - 6.24
Dues and interest, 8th year, as in first year, - sº -* - º - tº - 12.39
Amount to loan, one-half of 9th year, sº - º - - º - eme - $122.63
84th year. Interest on $122.63, 6 months, at 6 per cent, º - tº - 3.68
Dues and interest, for 6 months are equivalent to the interest on $1 for the equated
time, which is (6 -- 1 = 7 × 3) 21 months. Interest on $1 for 21 months, at 6 per
cent, is $.11 + $6 dues = - tº- - * -* t- º tº- 6.11
Compound amount of $102 for 84 years, at 6 per cent, - - - -> - " - $132,42
I)ues deducted, º - º - - - - --> - º º - º - 102.00
Interest received in 84 years, - - sºme - * 4- º tº- º & - $30,42
$30.42 + 84 = $3.5782–H gain per year on an average investment of (102 − 2) $51.
51 : 100 : : 3.5782–H : 7.016+ 9% average gain.
SECOND SOLUTION.
By the Compound Interest Table, page 611.
The payments being $1 per month, the sum received and loaned is $12 for the first year. The
interest on the $1 monthly dues or receipts for the 12 months of the first year is equivalent to the
interest on $1 for the equated time, which is, as fully shown in the first solution, (12 + 1) = 13 ×
6 = 78 months. The interest on $1 for 78 months at 6 per cent, is 39 cents. $12 + 39c. = $12,39
the yearly value of the monthly payments.
This $12.39 is a yearly recurring amount to be placed at compound interest; thus forming a
series of annuities, each term of which is $12.39.
Then to find the sum of these annuities, we have the following formula:
S = A × Rn — A Or S = $12.39 × 1.06% – $12.39 $12.39 × 1.5938481 – $12.39
- R – 1 *s .06 - .06 *-
$122.62962852 the sum of an annuity of $1 at compound interest for 8 years, at 6
per cent. A
NotE.—The 1.5938481 is the 8th power of 1.06, and is obtained from the Compound Interest
Table, page 612.
To this $122.63 we must add: (1). The interest thereon for 6 months at 6 per cent, which is
$3.68. (2). The $6 dues for 6 months. (3). The interest on the $6 dues for the equated time, which
is, as shown in the first solution, 11 cents. Thus making a total sum of $132.42.
See the close of the first solution for the full details of finding gain and gain per cent.
NoTE.—See the remarks on Annuities, on page 921.
THIRD SOLUTION.
By the Annuity Table, page 925.
We first produce the annuity of $12.39 as shown in the first and second solutions, then to
find the sum of the annuities at the end of the given time, we have but to multiply the amount of
the annuity of $1 per annum, at compound interest for 8 years, by the annual annuity $12.39.
jºr BUILDING AND LOAN ASSOCIATIONS. IOOI
OPERATION.
$9.897468 = sum of an annuity of $1 per annum at compound interest for 8 years at 6 per
12.39 [cent. See Table, page 925.
$122.62962852 = the sum of an annual annuity of $12.39 for 8 years, at 6 per cent compound interest.
To this sum $122.63, we add interest and dues for 6 months, as shown at the close of the
first and second solutions.
NOTE 1.-See the close of the first solution, for the full details of finding gain and gain per
cent.
NOTE 2.-See the remarks on Annuities, page 921.
FOURTH SOLUTION
By Logarithms.
All problems of this kind can be worked by the Algebraic formula for
finding the amount of Sinking Funds, or the amount of Annuities.
The formula is A = sº-0. Applying this formula to the above problem,
and using figures instead of letters, we have amount = $12.39 gº — 1)
To raise $1.06 to the 8th power by the use of Logarithm Tables proceed as follows: (1). Find
the log of the number 1.06 = .025306. (2). Multiply this log by 8 = .202448. (3). Look for this
log in the Table of Logarithms. If you cannot find it exactly, find the log just less than it (.202216)
and the one just greater than it (.2024.88). The difference between the two extremes is 272, and
the difference between the given log (.202448) and the lesser log (.202216) is 232. Write these
differences into a fraction as, #}}. (4). Find the number corresponding to log .202216 = 1.59300;
find the number corresponding to log .2024.88 = 1.59400. (5). Take the ### part of 100 = 85+ and
add it to 1.59300, writing it in the place of the naughts = 1.59385. Multiply difference between
1.59385 and 1 = .59385 by 12.39 = 7.3578015. (6). Divide this product by .06 = $122.63 = amount
for 8 years.
Now having the compound amount for 8 years, we proceed for the remaining 6 months as
shown in the first and second solutions.
If the monthly dues in the above problem were to be compounded monthly,
the following would be the formula for finding the amount at the end of the 102nd
month :
_ S(Rn--1–1) $1 (1.005”—1) _S (Rn–1) $1 (1.005%–1)
A = ** – 4–S = .005 — $1, or A = * º ,005
= $132.63, to this amount add interest for one month = 66% = $133.29.
OPERATION FOR FIRST FORMUL.A. USING LOGARITHMS TO FACILITATE MULTIPLICATIONS.
Log 1.005 = .002166 × 103 = log .223098. The number corresponding to this log is $1.67147.
From this subtract $1 = $.67.147. Divide this amount by $.005 = $134.29. Less $1 = 133.29, final
value.
OPERATION BY SECOND FORMUL.A. USING LOGARITHMS TO FACILITATE MULTIPLICATION.
Log 1.005 = .002166 × 102 = log .220932. The number corresponding to this log is $1.66315.
Subtract $1 = $.66315. Divide this by $.005 = $132.63. Add + per cent on $132.63 = 66c. =$133.29.
final value.
IOO2 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
2. A Land and Investment Company sells to B. a piece of property for
$4000, at 8 per cent interest, on condition that B. shall pay for the property at the
rate of $90 per month, which monthly payments are to include the interest on the
successive monthly balances. How long will it take B. to pay for the property,
the first payment being made one month after the sale? How many payments did
B. make 3 What was the total interest? What was the average rate per cent, per
annum on the $4000? What did the company gain and B. lose in excess of the 8
per cent interest charged on the successive monthly balances?
Answers. (1). It requires 53 months to pay for the property. (2). B. made
53 payments, 52 of $90 each and 1 of $79.66. (3). B. paid $4759.66. (4). Total
interest was $759.66. (5). The average rate per annum on the $4000 was 4.3 per
cent. (6). The company gained and B. lost in excess of the 8 per cent on the
successive monthly balances which bore interest, the interest (simple, annual or
compound, according to manner of making investments), on each $90 monthly
payment, except the last payment of $79.66, from the time each payment was
made until the expiration of 53 months.
PARTIAL OPERATION
By the United States Partial Payment Method.
OPERATION OPERATION
For the first two payments. For the last two payments.
Amt. at 896 int. for the 1st month, - $4000.00 | Bal. at 896 int. for the 52d month, - $168.01
Int. on same for the 1st month, - º 26.67 || Int. on the same for the 52d month, - 1.12
Amount due at close of 1st month, - $4026.67 || Amt. due at the close of 52d month, - $169.13
1st payment for the 1st month, - gº 90.00 52d payment, e * * º tº gº 90.00
Bal. at 896 int. for the 2d month, - $3936.67 | Bal. at 89% int. for the 53d month, - $79.13
Int. on same for the 2d month, - e 26.24 Int, on same, for the 53d month, - s .53
Amount due, at close of 2d month, - $3962.91 Amt, due at the close of 53d month, - $79.66
2d payment, - sº tº º sº º º 90.00 53d payment, - * * tº e iº tº º 79.66
Bal. at 896 int. for the 3d month, - $3872.91 *
We have omitted in this partial operation the several items of interest,
amount, payments and balances, from the third item of interest to the 51st payment,
both inclusive. The beginning and closing work of the problem which we have
presented is sufficient to clearly elucidate the same.
A. made 52 payments of $90 each = $4680, and 1 payment, the last, of $79.66,
making a total of $4759.66, and 53 payments in 53 months or 4 years and 5 months.
The interest paid was $759.66. $759.66 interest on $4000, for 53 months = 4 3
per cent per annum, as per Article 1166, page 599.
Though the average annual interest is but 4.3 per cent on the $4000, the
company made and B. lost 8 per cent on the successive monthly amounts that bore
interest. And as the $90 monthly payments were received by the company and
paid out by B. the use or interest on said payments to the company would be equal
to about 4.3 per cent, thus making nearly 8.6 per cent.
* BUILDING AND LOAN ASSOCIATIONS. IOO3
It will be remembered that by the United States system of partial payments,
the shorter the intervals of time between payments, the greater the gain is to the
party receiving the payments.
3. A Loan and Land Company effects loans or sells lands on the following
conditions: 1. To charge 8 per cent interest on the monthly principal loaned, or on
the amount due for land. 2. To divide the sum loaned or the amount due for land
by the number of months the loan is to run, and to take notes, secured by mortgage
payable in successive months, each for an amount equal to the quotient obtained
by this division, plus the monthly interest on the successive principals loaned.
According to this method, what would be the face of each note for a loan of
$3600 for 6 years at 8 per cent? Ans. See partial operation.
PARTIAL OPERATION.
$3600 loaned -- 72 months = $50 partial face of each of the 72 notes.
Amount at interest the first month, - - - - - - - - - - - $3600.00
Interest on same, 8 per cent, 1 month, * = º º is sº º sº * * 24.00
Amount due at the close of the first month, sº nº me tº º ºs - - - $3624.00
Face of first note, $50 + $24 = $74, - - - - - - - - - - - 74.00
Balance at interest the second month, tº ºs º ºs º º sº - - - $3550.00
Interest on same, 8 per cent, 1 month, * * * * * * * * * * 23.67
Amount due at the close of the second month, - tº tº tº * * tº gº tº - $3573.67
Face of second note, $50 + $23.67 = $73.67, sm tº º ºs º º º sº. gº 73.67
Balance at interest the third month, - ſº gº {º tº gº { } { } - - - $3500.00
Interest on same, 8 per cent, 1 month, ſº * , dº ſº. tº gº gº ºn tº tº ºr 23.33
Amount at the close of the third month, tº º tº e tº s tº tº º - - - $3523.33
Face of third note, $50 + $23.33 = $73.33, - - - - - - - - - - 73.33
Balance at interest the fourth month, - - - - - - - - - - $3450.00
And in like manner we continue the operation to find the face of the 72
notes.
It will be observed that the operation involves the United States method of
partial payments as elucidated in the preceding problem.
To find the interest and the rate per cent of interest realized by a Loan and Investment
Company on certain conditions.
1562. 1. A Loan and Investment Company loans $3000 for 5 years, at 8 per
cent per annum. The interest ($1200) is added in advance, and the Whole amount,
$4200, is made payable in 60 notes of $70 each, payable monthly, without interest.
Allowing that the company may re-invest the monthly collections of the notes at 8
per cent simple interest, what would be the total interest received by the company,
and what would be the rate per cent per annum ?
Ans. $2026 total interest; 13% rate per cent per annum.
NotE.—For the solution of this and the two following problems, see page 675.
IOO4. soul E's PHILOSOPHIC PRACTICAL MATHEMATICS. Yºr
2. Suppose in problem 1, above, that the monthly note and interest collec-
tions were re-invested at simple interest as therein stated, what would have been
the respective interest gain for each year? tº
3. Suppose in problem 1, above, that the monthly collection of interest on
the notes and the interest on such interest were used, at the close of each year, as
a part of the dividend fund, what would be the amount of interest earned yearly 3
4. April 8, 1894, a party purchased from a real estate holder a house and lot
for $7500, on the following terms and conditions: $300 was paid in cash. 36 notes
of $200 each bearing 7 per cent interest were given; three of the notes were made
payable in 3 months, July 8, 1894. The other 33 notes were made payable monthly
after July 8, 1894. After the three notes payable July 8, and 19 notes payable
monthly after July 8 had been paid, the holder sold the 14 remaining notes at 7 per
cent per annum discount. Counting 30 days to the month without grace or discount
day, what were the proceeds of the 14 notes, Bankers' Discount Method
Ans. $3136.74.
NotE.—For the solution of this problem, see page 590.
5. A party holds $2700 of stock, which pays $80 per month revenue, (35%
per cent interest per annum). The stock has 27 years to run. What is the present
value of the stock, the current rate of interest being 6 per cent, allowing 6 per cent
interest on all the monthly payments of the revenue or income 3 Ans. $12614.02.
NoTE.—For the solution of this and the following problem, see page 591.
6. Suppose in the above problem, interest is allowed only on the annual pay-
ments of the monthly revenue instead of the monthly payments, what would be the
present value of the stock # Ans. $12575.34.
7. A Homestead Association sold to Levin Cooper Soulé, a house and lot
for $5000, on the following conditions of payment: Cash $1000; the remainder in
80 notes of $50 each, bearing 7 per cent interest and payable monthly, as the
months expire, in 80 consecutive months.
In this transaction, what is the rate per cent interest that the homestead
makes and what per cent interest does Levin Cooper Soulé pay on the $4000?
Ans. The homestead makes and Levin Cooper Soulé loses,
3.543% + 3.45% = 7% + .5512+ ſº use of inter-
est on interest payments, = 7.5512+ 7%.
NOTE. 1.-For the solution of this problem, see pages 592 to 596.
NoTE_2.—The following are valuable works on Building and Loan Associations: How to
Manage Building Associations, by Edmund Wrigley. Treatise on Building Associations, by N. C.
Thompson. H. S. Rosenthal’s Manual for Building and Loan Associations. G. A. Endlish on the
Laws of Building Associations. Ninth Annual Report of the Commissioner of Labor.

==".
6.
wº
3% , CHECK FIGURE SYSTEM OF PROVING POSTINGS. IOO5
*-
Multiples of 18. TEIE THIRTEEN CHECK FIGURE SYSTEM OF PROW-
To be memorized ING POSTINGS OR TRANSFERS.
or used on a
card.
* gº GENERAL DIRECTIONS FOR. USING THIS SYSTEMI.
13 351 689
26 364 702 *E--->
; § # 1. Separate the number when consisting of more than three figures,
65 403 741 into periods of threes.
º § # 2. When there is more than one period of odd or of even periods in the
104 442 780 number, add the like periods together.
117 455 793
130 468 806 3. When the number contains less than, or only three figures, then
#. # § subtract therefrom some multiple of 13 which is less than the number.
; § #: 4. When the odd period or the sum of the odd periods, is more than 13 less
195 533 871 than the even period or the sum of the even periods, then increase the odd period or
; : ; periods by adding thereto some multiple of 18, 80 as to produce a sum not exceeding
234 572 910 13 greater than the sum of the even period or periods. Then subtract the even
% ; ; period from the odd and the difference will be the Check Figure.
273 611 949 g = i e g
286 624 962 5. When the even period is more than 13 less than the odd period, then add
299 637 975 some mutiple of 13 thereto 80 as to produce a sum less than the odd period and
; § 988 which, when subtracted from the odd period, will leave a remainder less than 13,
338 676 which will be the Check Figure.
A SHORT PRACTICAL METHOD OF FINDING TEIE 13 CHECK FIGURE.
Multiples of 13 In practice it is necessary to memorize only the first ten multiples of 13,
To be memo- and to acquire quickness in seeing the products thereof when multiplied by
rized. the numbers from 2 to 10. By this knowledge, the various multiples of 13
13 may be seen instantly by the mind and applied in a manner as shown in the
; following illustrative examples:
52 1st. When there is but one period, mentally see what multiple of 13
; must be subtracted from it to leave a remainder less than 13; this remain
. der is the Check Figure.
104 Thus: To find the Check Figure of 57, we subtract 52 from 57 and obtain
# a remainder of 5, which is the Check Figure.
The Learner should also 2d. When there are two periods, odd and even, mentally see
memorize the products of the what multiple of 13, if any, must be added to the lesser period so
most of these multiples by that the difference between them will be less than 13.
2, 3, 4, 5, 6, 7, 8, 9 and 10. Thus: 1st. To find the Check Figure of 27453, we mentally
Thus: 65 × 3 195 : add 416 to 27, the even period, making 443, which we subtract from
78 × 4 313: 453, the odd period, and produce 10, as the Check Figure.
gig 5 iss; Or 2d. To find the Check Figure of 352865, we add 520 to the
52 × 10 520 'ete even period, making 872; from which we subtract the odd period
* y *
and produce 7 remainder; then as 872 was larger than the odd
period, we subtract 7 from 13 and produce 6 as the Check Figure. If we had added 507 to the
even period, the operation would have been shorter. But to produce 507 from the table of the first
10 multiples of 13, more mental work is required than when we use 520.
* * Find the Check Figure of 242,723,492,168. (891)= sum of odd periods, 734 = sum of
even periods, plus 156 (2 times 78) = 890. 1 Check Figure.
IOO6 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICs. *
Find the Check Figure of the following numbers by mental operations. No
figures to be written except the Check Figure.
NOS. OPERATIONS.
(1) 13 13 from 13 - * , - & Check Fig.
(2) | 15 13 from 15 - *- - - {{ {&
(3) 29 26 from 29 -> e- 3- - {{ {{
(4) | 84 78 from 84 •º {{ {{
(5) 142 130 from 142 - º º 1 {{ {{
(6) 568 520 (10 times 52)+39=559 from 568 = {{ {{
(7) || 794 780 (10 times 78)+13=793 from 794 F % - 4%
(8) 907 780 (10 times 78)+117=897 “ 907 F 1 {{ {{
or thus: 910 less 807=3, and 3 from 13 = l {{ {{
(9) 2,640 652 (10 times 65+2 in. 2d period), less
640=12, and 12 from 13, - - 44
or thus: 520-H2+ 117 =639 from 640
{ % {{
(10) || 6,028 13+6=19 from 28 - º F. {{ {&
(11) || 2,972 910–H 65+2=977 less 972=5 & 5 from 13 = {{ {4
or thus: 912–H52=964 from 972 F {{ {{
(12) 53,415 351 (3 times 117)+53 in 20 period=404
{{ {{
{{ 66
{{ & 4
and 404 from 415 ge -> ->
(13) | 178,213 178+26–204 from 213 - ſº ſº
(14) 249,007 234 (2 times 117)+7-H13=254 less 245
or thus: 234-H 7=241 from 249–8 and
8 from 13 gºs †- - - º
or thus: 260—13-H17 =254 less 249
(15) || 7,210,586 |390 (10 times 39)+210, in 2d period,
=600 less 593 (sum of odd periods)
1.
:;i
{{ 44
{{ {{
=7 from 13 - º º * = • 6 {{ {{
or thus: 351 (3 times 117)+210+26
=587 and 587 from 593 - º - s 6 {{ {{
(16) 324,808,410|734 (sum of odd periods) + 78 = 812
less even period, º tº tº- * = s 4 {{ {{
(17) | 1,540,017 | 18 (sum of odd periods) + 520 (10 times *
52)=538 from 540=2; and 2 from 13 = - 11 {{ {{
Or thus: 18+533=551 less 540 - = • 11 {{ {{
(18) 96,244,503 |599 (sum of odd periods) less (244+351)
595 from 599 - - - º - - 4 {{ {{
(19) 25,000 26 (2 times 13) less 25 (even period) - - 1. {{ {{
(20) | 1,302 312 (4 times 78)+1=313 less 302=11
and 11 from 13 - - tº- º := • 5 {{ {{
(21) 100 91 from 100 sº - - - º - * 9 {{ {{
(22) $ 3.60 |351 s & - tº- tº = 9 cf.
(23) | $ 360.00 39 * * - * - es = 3 cf.
(24) $ 425.00 |455+42=497 - - tº = 3 cf.
(25) $ 204.50 |416+13+20=449 from 450 - = 1 cf.
(26) $6028.15 (208-1-602) from 815 º = 5 cf.
(27) $ 400.00 52 less 40 - - - - = 12 cf.
(28) $ .43 39 from 43 - tº tº = 4 Cf.
By memorizing perfectly the first 10 multiples of 13, and becoming quick in producing mentally
the products thereof, by the numbers 2 to 10, any person, who can add and subtract with facility,
may see mentally and instantly the successive steps which produce the Check Figure.
An hour's practice per day, for a few days, on the foregoing, and on similar problems which
may be assumed or taken from any book of account, will enable any bookkeeper or clerk to write
the Check Figure almost at sight.
ffi This knowledge of the 13 Check Figure system is of great value to all bookkeepers and to
Oil Cô Iſle Il.
sk CHECK FIGURE SYSTEM OF PROVING POSTINGS, ETC. roo?
APPLICATION OF THE 13 CHECK FIGURE METHOD OF LOCATING
ERRORS AND OF PROVING POSTINGS AND TRANSFERS.
Post or transfer the following items from the Sales Book to the Ledger:
Ck.
SALES BOOK ITEMS. Fig. LEDGER ACCOUNTS.
J. P. Hayter, tº- - $ 120 00| 1 J. P. HAYTER.
D. M. Smith, - - - 2400 00 7 CK
F. B. Denson, 17284 50| 9 Fig,
J. C. Rimes, - - - 79| 1 $120 00 || 1
|
Total, - - $19805 29, 18 D. M. SMITH.
Check Fig. of the total is tº-e 5 F.
66 ** ** 18, the Sum
of the Check Figures is tº 5
which proves the cor-
*
$2400 00 7
rectness of the addition.
F. B. BENSON.
By inspection, we observe that the #.
respective sums of the Check Figures in $17284 50 | *
the Sales Book and on the debit of the
Ledger are the same, 18, and that 5 is the
Check Figure of each. J. C. RIMES.
This proves that the items were posted §:
to the correct side of the Ledger Ac- .79 || "i"
COuntS. 18 sum of C. fs.
In like manner the posting of the Invoice Book, Cash Book, Journal or any
other book that is used as a posting medium, would be made and the work proved
by the 13 Check Figure.
NotE.--When posting, set the Ledger check figures on a slip containing debit and credit
check figure columns, and make this test for each book posted.
TO PROVE ADDITION BY TEIE 13 CHECK FIGURE METEIOD.
Pirst. Find the check figure of each of the numbers added. 2d. Find the
check figure of these check figures. 3d. Find the check figure of the sum of the
numbers. If the check figure of the sum of the numbers is the same as the check
figure of the check figures of the different numbers, the work is correct as far as
this method of proof can determine.
PROBLEMI. 345 = 7 c. f.
87 = ** f \ 21 — 13 = 8 c.f
Prove the addition of the follow- 4357 = 2 c. f. — 13 = 8 c. f. V
796 = 3 C. f.
ing numbers by the 13 Check Figure
method. 5585 = 8 c. f. V
TO PROVE SUBTRACTION BY THE 13 OHECK FIGURE METEIOD.
First. Find the check figures of the minuend, subtrahend and remainder.
2d. Subtract the check figure of the subtrahend from the check figure of the
minuend and if the difference is the same as the check figure of the remainder, the
work is correct as far as this method of proof can determine. º
NOTE: 1.-Increase the check figure of the minuend by 13, when it is less than the check
figure of the subtrahend.
NOTE: 2. —By adding the check figures of the subtrahend and the remainder, the sum should
equal the check figure of the minuend, and thus prove the work. When this sum is in excess of
13, then subtract 13 theref, on and the difference will equal the check figure of the minnend.
PROBLEMIS.
Prove the subtraction Minuend, - 857 = 12 c. f. $230 50 c. f. =1
of the following numbers Subtrahend, 465= 10 c. f. 87 25 C. f. = 2
by the 13 Check Figure tºmºmºsºms - gºmºmº-
method. Remainder, 3.92= 2 c. f. $143 25 = 12 c. f.
}12 C. f.
IOO8 SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. ºr
TO PROVE MULTIPLICATION BY THE 13 CHECK FIGURE METHOD,
First. Find the check figure of the multiplicand, the multiplier, and the
product. 2d. Find the check figure of the product of the check figures of the
multiplicand and the multiplier. If this check figure is the same as the check
figure of the product, the operation is correct, as far as this proof can determine.
PROBLEMS,
Prove the multipli-|Multiplicand, 6428 = 6 451 = 9 — as -
cation of the follow- Multiplier, . 4357 = 2 } =12 C. f. 64 = 12 } =108–4 c. f.
ing numbers by the 13
Check Figure method Product, 28006796 = 12 C. f. 28864 = 4 c. f.
TO PROVE DIVISION BY THE 13 CEIECK FIGURE METHOD.
First. Find the check figures of the divisor, dividend, quotient, and remain-
der, if any. 2d. Multiply together the check figures of the divisor and quotient,
and find the check figure of the product. 3d. To this check figure add the check
figure of the remainder and find the check figure of this sum; if this check figure
is the same as the check figure of the dividend, the work is correct, as far as this
method of proof can determine.
PROBLEMI.
Prove the division of DIVISOR. I)IVII) END. QUOTIENT.
the following number by (9 c. f.) 3207 )467582 = (11 c. f.) (145 = 2 c. f.
the 1 3 Check Figure
method. Remainder, 2567 = 6 c. f.
9× 2 = 18 = 5 c. f. -- 6 = 11, same as c. f. of the dividend.
NOTE.-Tbere are certain errors that cannot be detected by the 13, the 11 and the 9 Check
Tigure methods of proof.
THE CENT ET. UNE, OR 101 METHOD FOR FINDING THE CHECK FIGURE IN POSTING.
Separate the given number into periods of two figures or places each, from the right.
From the sum of the odd place periods, subtract the sum of the even place periods, adding
101 to the sum of the odd place periods, when smaller than the sum of the even place periods.
If the remainder contains three or more figures, proceed as before, until there remains not
more than two figures. Then subtract the tens figure from the units figure, adding 10 to the
units figure when the smaller, as in ordinary subtraction. The remainder, or final differential,
will be the check figure required. The two places or figures for cents must always be considered
the first odd period.
THE MILLE ET UNE, OR 1001 METHOD FOR FINDING THE CHECK FIGURE IN POSTING.
Separate the given number into periods of three figures each, from the right. Then from the
sum of the odd place periods, subtract the sum of the even place periods, the difference will be the
remainder, if the whole number had been divided by 1001. Then reduce the remainder by the 9,
11, 13, or 101 method for the final differential or check figure.
THE 99 METHOD of FINDING THE CHECK FIGURE.
Separate the number into periods of two figures from the right. Then add all the periods
and reduce the sum by the same method to two figures; then reduce to one figure by adding,
as for 9’s. The final result will be the check figure.
THE 999 METHOD OF FINDING THE CHECK FIGURE.
Separate the number into periods of three figures from the right. Then add all the periods,
and reduce the sum to two figures by the 99 method, and reduce this result by the 9 method, for
the final check figure.
See pages 590 to 596 of Soulé's New Science and Practice of Accounts for full explanation
and application of the 9 and 11 check figure systems.
º: EQUATION OF PAYMENTS WITH SPECIAL FEATURES.
IOO9
PROBLEM 1.
Jones bought of Brown the following bills of goods on 60 days credit and
2% discount in 10 days, with the privilege of 60 days additional credit, in case the
discount privilege is not availed of: Jan'y 4, '03, $300; Jan'y 26, $150; Feb'y 10,
$800; Mar. 20, $1200; Apr. 1, $90; May 25, $500.
If payment of this account is made May 30, '03, what is the amount due to
Brown, the rate of interest being 6%
Ans. $3019.73,
NOTE.—See pp. 627 to 653, for full elucidation of Equation of Accounts.
FIRST OPERATION.—FIRST STEP
To equate the sales and find when the time
expires for the 2% discount privilege.
1903.
Jan'y 4. $ 300 × 0 = 0
4; 26. 150 × 22 = 3300
Feb'y 10. 800 × 37 = 29600
Mar. 20, 1200 x 75 = 90000
Apr. 1. 90 × 87 = 7830
May 25, 500 x 141 = 70500
$3040 ) 201230 (66 ds
SECOND OPERATION.—FIRST STEP
To find the Equated Date of the account, allow-
ing 120 days credit on all sales.
1903.
Jan'y 4. $ 300x120= 36000
46 26. 150 × 142= 21300
Feb'y 10. 800 × 157 = 125600
Mar. 20. 1200 x 195– 234000
Apr. 1. 90 × 207 = 18630
May 25. 500 × 261 = 130500
$3040 ) 566030 (186 ds.
566030 ds. -- $3040 = 186 days after
Jan'y 4, '03, gives July 9, '03, as the
Bquated Date for the payment of the
account without discount.
NotE.-By adding 120 days to the 66 days,
equated date of the sales, as shown in the sec-
ond step of the first operation, we could have
found the 186 days credit on the whole account,
without making the second operation.
SECOND STEP OF FIRST OPERATION.
201230 ds. -- $3040 = 66 days.
66 ds. equated date of sales after Jan.
4, 1903.
credit on each sale.
discount privilege.
60 ds.
10 ds.
136 ds. after Jan'y 4, '03, is May 20, '03,
the Equated Date for the pay.
ment of all the purchases at
2% discount. But as pay-
ment was not made until May
30, '03, the 2% discount is not
allowed, but 60 ds. additional
credit is allowed in place
thereof.
Ea:planation.—From the above we see that
the discount privilege ceased May 20, '03; that
the equated date of the account, allowing 120
days credit on each sale, is July 9, '03; and
that the account was settled May 30, '03.
Hence Jones is entitled to interest on the
amount purchased from May 30 to July 9, 03,
which is 40 days.
SECOND STEP OF SECOND OPERATION
To find the interest and balance due
to Brown.
$
3040 $3040 Amount.
60 | 40 20.27 Int.
| $20,263=int. $3019.73 due Brown.
IOIO SOULE's PHILOSOPHIC PRACTICAL MATHEMATICS. *
PROBLEM 2.
A. sold to B. the following invoices of goods: May 10, $500; May 20, $700;
June 1, $400, July 3, $1000. The terms are 60 days credit and 2% discount in 10.
days, or 60 days additional credit, in case the 2% discount is not availed of.
B. desires to settle the account on July 20, 1904. What amount does he owe
A. on this date 3 AnS. $2536.96.
1st Operation to equate the sales and find the 2d Operation to find amount due July 20, '04,
date that the 2% discount privilege ceases. when the settlement was made.
1904. By the above work we see that the
May 19. 3500 x 9 = 0 - discount privilege extended to Aug. 15,
me". * . § = §. '04, and as the settlement was made
July 3. 1000 × 54 = 54000 July 20, '04, B. is entitled to the 2% dis-
count, and to the interest on the balance.
$2600 ) 69800 (27 days | from July 20, to Aug. 15, '04. Thus:
To this 27 days credit after May 10, '04,
we add 60 days credit on each sale,
and also 10 days credit for dis’nt option
and thus — & 2 TAS
ſº 2.548 Balance.
produce 97 days credit after May 10, '04, § 2.) g
which gives Aug. 15, '04, as the date OIl 11.04 hº º
which the 2% discount privilege expires. o, "U4, ays at 6%.
$2600 Amount of purchases.
52=2% discount.
*
NOTE. –By adding 120 days credit to the 95
above 27 days, we produce 147 days credit after $2536.96 Net, balance due A. July
May 10, which gives Oct. 4, '04, as the equated 20, '04.
date of the whole account, which, however, is
not required in this problem, as the settlement
was made within the 2% discount limit.
PROBLEM 3.
Suppose in the preceding problem that B. had settled the account August
28, '04, what amount would have been due to A.3 Ans. $2583.97.
Explanation.—From the work of the preceding problem, we see that the privilege of 2% dis-
count expired Aug. 15, '04. Then, by the terms of the contract, B. was entitled to 120 days credit.
which gave Oct. 4, '04, as the equated date of the account, without considering the element of
2% discount. Hence, if he settles the account Aug. 28, '04, he is entitled to the interest thereon
from Aug. 28, to Oct. 4,-37 ds, a 6% = $16.03 from $2600 = $2583.97 balance.
PROBLEM 4.
The firm of E. P. Singeltary, sold to A. Barbier & Co., the following
invoices of merchandise :
1904. The terms for each sale are as follows: 60 days.
Feb. 19, $372.30 credit and 2% discount in 10 days, or 60 days additional
“ 23, 53.41 credit in case the 2% discount privilege is not accepted
Mar. 3, 181.39 by the purchaser.
“ 16, 365.79 A settlement of this account was made Sept. 1, 1904.
“ 25, 240.25 (1). What is the date on which the 2% discount privi-
“ 29, 109.12 lege expires 3 (2). What is the average date of the
Apl. 7, 57.09 account ? (3). What is the amount due to E. P. Sin-
“ 20, 130.00 geltary on settlement, allowing 6% interest ?
e-sºmsºmº sºmºsºmº Ans. 1, May 22/04. Ans. 2, July 11/04.
$1509.35 Ans. 3, $1522.43.
:}: EQUATION OF PAYMENTS WITH SPECIAL FEATURES. IOIII
PROBLEM 5.
Smith sold to Jones on 60 days credit, with the privilege of 3% discount in
10 days, the following invoices of goods: May 10, $500; May 20, $700; June 1,
$400; July 3, $1000.
Jones has paid on account, as follows: June 9, $800; Aug. 25, $900. A final
settlement was made Sept. 15, '04. Allowing interest at 6%, and 360 ds. as an
interest year, what does Jones owe Smith on final settlement 3 Ans. $877.57.
FIRST SOLUTION,
OPERATION Or thus : 20 OPERATION.
To find the Equated Date of Sales and the time
tº e 1904
limit of the di t privilege. o
loo." O e discount privilege May 10. $ 500 × 70 – 35000
May 10. $ 500 × 0= 0 4 20. 700 x 80 = 56000
{{ 20. 700 × 10= 7000 June 1. 400 x 92 = 36800
June 1. 400 × 22= 8800 July 3. 1000 x 124 = 124000
July 3. 1000× 54= 54000 -
$2600 ) 251800 (97 ds.
$2600 ) 69800 (27 ds. af.
ter May 10. after May 10, '04, = August 15, '04, as
the Equated Date of the sales, allowing
60 days credit and 10 days discount,
privilege.
27 ds. after May 10, '04.
60 ds. Cr. On each sale.
10 ds. Dis. privilege.
97 ds. after May 10, = Aug. 15, '04, =
limit of dis. privilege.
OPERATION TO EQUATE THE ACCOUNT, BOTH DEBIT AND CREDIT ITEMS.
1904. 1904.
May 10. $ 500 x 60= 30000 June 9. $800 x 30 = 24000
4 20. 700 × 70= 49000 Aug. 25. 900x107 = , 96300
June 1. 400x 82= 32800
July 3. 1000×114=114000
$2600 225800 $1700 120300
1700 120300
$ 900 ) 105500 (1173 ds. after May 10, '04, = Sept. 4, '04, as the
Equated date or time for the payment of the balance.
By inspection, we see that the $800 paid June 9, was paid before the expi-
ration of the discount privilege, August 15, and hence the debtor is entitled to
discount thereon, which, at 3%, is $24. This deducted from the $900 balance of
account, leaves a net balance of $876 due Sept. 4%, ’04.
As settlement was not made until Sept. 15, '04, the creditor is therefore
entitled to interest on the net balance, from Sept. 4%, ’04, to Sept. 15, '04, 10% days
at 6% = $1.57 interest. $876 + $1.57 = $877.57, amount due Sept. 15, '04.
NOTE: 1. –In practice, it is not the custom to use the exact fraction of a day, but to state
the day to the nearest unit. In this problem we have used the fraction of a day in order to obtain
the exact interest, so as to compare the same with the interest produced by the solution of the
problem, by the Account Current and Interest Account Method, which is given on the following
3,976.
pag NOTE: 2. —Since the purchaser is allowed 3% discount on all payments paid within the dis-
count time privilege, his discount credit on the $300 payment would be in strict accuracy and
justice, $24.74, i. e. in the ratio $100 credit for every $97 paid. But this degree of accuracy is not
observed in calculations of this character,
(Continued. )
IOI 2 SOULE'S. PHILOSOPHIC PRACTICAL MATHEMATICS. jºr
NOTE 3.—Problems containing conditions of credit and of discount, like the five preceding
problems, are of quite recent origin, and have given rise to considerable discussion among account-
ants. These discussions result from the different interpretations given to the conditions, and
these interpretations are often reached because a solution in accordance there with will give a
financial result in favor of the party holding such an interpretation. As an example of this point
note the following operation of the above problem.
VARIATION OF PROBLEM.
EQUATION OF THE ACCOUNT ALLOWING 70 DAYS CitFDIT ON EACH ITEM OF PURCHASE.
1904. 1904.
May 10, $ 500 x 70– 35000 June 9. $800 × 30 = 24000
64 20. 700 × 80= 56000 Aug. 25. 900 × 107 = 96300
June 1. 400 × 92= 36800
July 3. 1000 × 124=124000
$2600 251800 $1700 120300
1700 120300
$ 900 ) 131500 (146 ds. after May 10, '04, = Oct. 3, '04, as the
Equated Date for the payment of the balance.
This equation is made on the interpretation that the debtor has really 70
days credit on each item of purchase, because he had 10 days discount privilege.
This, we think, is an erroneous construction of the conditions of the contract.
The 10 days discount privilege were conditioned on the payment of cash as the
respective sales came due. But as the cash payments were not made, the 10 days
privilege cannot, in justice, be claimed.
A difference of opinion exists among accountants and also among business
men, regarding the proper manner of adjusting accounts like, or similar, to the
above. In some cases, special agreements between the parties, and in Some cases
the custom of the trade will govern the method of adjustment or settlement.
SECOND SOLUTION OF PROBLEM 5.
BY THE PARTIAL PAYMENT METHOD, OR BY MAKING PERIODIC PARTIAL SETTLEMENTS
CORRESPONDING WITH THE DATES OF PAYMENTS,
1904. 1904.
May 10. $500×60 = 30000 June 9. $800 × 30=24000
“ 20. 700 × 70= 49000
June 1. ‘ 400 x 82= 32800
$1600 111800
800 24000
$ 800 ) 87800 (1093 =110 ds. after May 10 = Aug. 28, '04.
3% dis’t, 24
Balance, $776
1904. 1904.
June 28, $ 776 × 80=62080 Aug. 25. $900 x 77=69300
July 3. 1000x84=84000
$1776 146080
900 69300
$ 876 ) 76780 (87 ds. after June 9, '04, = Sept. 4, '04. .
NotE.—The $776 balance from the first equation is not due until August 28, '04. Hence, as
the Focal date 1s June 9, there is 80 days credit on this balance.
By this method of solution, we obtain the same equated date as in the first solution, Sept. 4,
1904. The remaining part of the operation, to find the balance due to the creditor on the day of
settlement, is performed in the same manner as in the first solution.
*
* Account cURRENT AND INT. Account WITH SPECIAL CONDITIONS IOI3
THIRD SOLUTION BY THE ACCOUNT CURRENT AND INTEREST ACC’T METHOD,
joNEs in Account Current and Interest Account at 6% to SEPT. 15, '04, with SMITH. 60 ds, credit
is allowed on all sales, and 3% discount is allowed on all payments made within 10
days after the expiration of the 60 days credit. *
I)”, Or.
When When
Ds || Prodc't| Amt. Ds.|Prod'ct Amt.
due, due.
1904. * 1904. ººm-->
May |10|Mdse. 60 ds. &|July 9, 68|| 34000|50000 June 9|Cash, - - June 9| 98|| 78400 80000
3% dis. in 10 ds. g
“ 20Mdse. “ “ || “ |19||58||40600|70000|Aug. 25. “ tº º Aug.25 21|| 18900 90000
June 1. “ { % { % ** 31|| 46|| 18400|| 40000
July 3. “ & 4 ‘‘ ||Sept | 1 14|| 14000||100000
Dr. Int. on $1.
To Bal. of ds on 107000 *==
$1, which di- 97.300 Ds. Int, on $1. 97300
vided by the *-*E*
6% int. divisor 9700
6000, gives
int. $1.617
Less int. on 2400
$24 for 10; Sept|4}|By3%dis.on $800 4.
ds. = .043 Sept. 7; By ićiance due 87.757
*º-º- 157 the creditor, tº-º
Int. due the *E*m. I ºm-
creditor $1.574 2601|57 2601|57
Explanation—By this method of working the problem, interest is computed on all debit and
credit items from the time due to date of settlement; hence interest has been computed on the
$900 balance of account, from the EQUATED DATE to the date of payment, which is 105 days
shown in the solution above by the equation method. This $900 balance of account is entitled to
a credit of $24, the same being 3% discount on $800 paid within the discount period, thus leaving
a net debit balance of $876, which bears interest from the equated date to the date of settlement.
Now, as interest has been computed on $900. instead of $876 from the equated date to the date of
settlement, it is evident, therefore, that the interest produced by this method of solution is in
excess to a sum equal to the interest on the amount of DISCOUNT CREDIT, (in this case $24) from
the equated date to the date of settlement. This interest, in this problem, amounts to $,043, and
is deducted from the full amount of interest, $1.617, as is shown in the operation.
NotE 1.--To find the limit of time of the discount privilege, equate the sales as shown in
the preceding solution.
NOTE 2.-For a full elucidation of Equation of payments, see pages 627 to 653. For a full
elucidation of Accounts Current and Interest Accounts, see pages 653 to 670,

le renew the tharge, book must be broughl io ſhe desk.
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