REESE LIBRARY OF THK UNIVERSITY OF CALIFORNIA, Received t^^&Zx^ /^ ^ Accessions No..^f^f^if.S? Shelf No.... THE NATIONAL 'ARITHMETIC, ON THE INDUCTIVE SYSTEM; COMBINING THE ANALYTIC AND SYNTHETIC METHODS, TOGETHER WITH THE CANCELLING SYSTEM; FORMING A. COMPLETE MERCANTILE ARITHMETIC. BY BENJAMIN GREENLEAF, A. M. PRINCIPAL OP BRADFORD TEACHERS' SEMINARY. tteto REVISED, ENLARGED, AND MUCH IMPROVED. BOSTON: ROBERT S. DAVIS, AND GOULD, KENDALL, & LINCOLN. NEW YORK: PRATT, WOODFORD, & Co., AND CADY & BURGESS. PHILADELPHIA : THOMAS, COWPERTHWAIT, & Co. BALTIMORE : GUSHING & BROTHER. And sold by the trade generally. 1848. Entered according to Act of Coiigres, in th year 1847. BY BENJAMIN GKEENLEAF, In the Clerk'* Office of the District Court of the District of Massachusetts. STANDARD BOSTON SCHOOL BOOKS. GREENLEAP'S SERIES OP ARITHMETICS. 1. MENTAL ARITHMETIC, upon the Inductive 1'lan, designed for beginners. By Benjamin Greenleaf, A. M., Principal of Bradford (Mass.) Teachers' .Seminary. 2. INTRODrCTl6N TO THE NATIONAL ARITHMETIC, designed lor Common Schools. Twelfth Improved stereotype edition. 3. THE NATIONAL ARITHMETIC, for advanced Scholars in Common Schools and Academies. Eighteenth improved stereotype edition. COMPLETE KEYS TO THE INTRODUCTION AND NATIONAL ARITHMETICS, containing Solutions and Explanations, for Teachers only. (In separate volumes. ) %* The attentlbn of Teachers and Superintendents of Schools generally is respect fully Invited to this popular system of Arithmetic, which is well adapted to a^classes of students. The 'National Arithmetic' has been extensively introduced in various sections of the United States, and is highlv recommended by many distinguished teachers who have used it, for its adaptedncss to give students a thorough practical knowledge of the science. It is the text-book in the Normal Schools in Massachusetts, and New York city, and the best schools in Boston, New York, Philadelphia, Rich mond, Charleston, .Savannah, Mobile, New Orleans, and other cities. PARKER'S PROGRESSIVE EXERCISES In English Composition Forty -fifth improved stereotype edition. Price 34 cents. THE CLASSICAL READER, A Selection of Lessons in Prose and Verse, from the most esteemed English and American writers. Intended for the use of the higher classes in public and private seminaries. By Rev. F. W. P. Greenwood, D. D. and G. B. Emerson, A. M. of Bos- ton. Tenth edition, stereotyped. With an engraved Frontispiece. SMITH'S CLASS BOOS OP ANATOMY, Explanatory oi the nrst principles of Human Organization as the basis of Physical Education ; with numerous Illustrations, a full Glossary, or explanation of Technical Terms, and practical Questions at the bottom of the page. Designed for Schools and Families. Ninth stereotype edition, revised and enlarged. A GRAMMAR OP THE GREEK LANGUAGE. By Benjamin Franklin Kisk. Twenty-sixth improved stereotype edition. Fist't Greek Grammar is used in Harvard University, and in many oth^r distinguished Col- legiate and Academic Institutions, in various parts of the United Slates. FISK'S GREEK EXERCISES (New Edition.) Greek Exercises; containing the substance of the Greek Syntax, illustrated by passages from the best Greek authors, to be written out from the words given in their simplest form, by Benjamin Franklin Fisk. ' Consuetudo et exercitutio facilitatem niaxime parit.' Quintil. Adapted to the Author's 'Greek Grammar.' Sixteenth stereotype edition. Fist's ' Greek Exercises ' are v>/{ adapted to illustrate the rules of the Grammar, and constitute a very useful-accompaniment thereto. LEVERETT'S CJESAR'S COMMENTARIES. Call Julii Csesaris Commentaril de Bello Gallico ad Codices Parisinos recensiti, a N. L. Achaintre et N. E. Lfinaire. Accesserimt N' tula: Anglicse, atque Index Uisto- ricus et Geographicus. Curavit F P. Leverett, A. M. FOLSOM'S CICERO'S ORATIONS. M. T. Ciceronis Orationes Qiuudain Stlectae, Notis Illustrate. [By Charles Fol- som, A. M.] In Usuin Academiaj Exoniensis. Editio stereotypa, Tabulis Analytic!* instructa. I have examined with some attention Csesar's Commentaries, edited by Leverett, and Cicero's Orations, edited by Folsom, and am happy to recommend them to clas- sical teachers, as being, In my estimation, far superior to any other editions of those works to which students in this country have general access. 'JOHN J. OWKN, Editor of 'Xenophon't Anabasis,' and Principal of Cornelius Institute, A. Y. City. ALGER'S MURRAY'S GRAMMAR, AND EXERCISES. ALGERB MURRAY'S READER AND INTRODUCTION. Published by ROBERT S. DA VIS, School- Book Publisher, BOSTON, and fold by all the principal Booksellers throughout the United State*. G^" Also constantly for sale,, (hi addition to his mun publications,) a complete assortment of School Books and Stationery, which are offered to Booksellers, Scliool Committees, and Teachers, on very liberal terms. PREFACE. THE National Arithmetic has now been before the public for nearly twelve years, and has met with an acceptance far be- yond the original expectation of the author. The demand for it has constantly increased, and such has been the encourage- ment which both the author and the publisher have received from teachers of the highest character, and from the public gen- erally, that the work has been thoroughly revised and very con- siderably enlarged, particularly in the department of demon- stration, and is now presented in a form which, it is believed, will greatly increase its value. In the work of revision and enlargement, the author has availed himself of important sug- gestions from many practical teachers, and has had the direct assistance of gentlemen intimately acquainted, not only with the business of teaching Arithmetic, but also with the higher branches of Mathematics. His own labors in this work have been hardly less than in the original preparation of the book, and he is confident that the improvements introduced into the present edition will be seen and appreciated by all who may compare it with preceding editions. It has been the author's privilege, for more than thirty years, to be engaged in the business of instruction. He has been acquainted with the methods of communicating knowledge which were formerly practised, and has endeavoured to make himself familiar with all the improvements in this respect which distinguish the present age from the past. The present work is offered to the public, as one constructed on a plan which appears to the author better adapted to meet the wants of the times than any other now in use. The end to be sought in the study of Arithmetic he regards as twofold, a prac- tical knowledge of numbers and the art of calculation, and the discipline of the mental powers ; and the present work, it is believed, will be found fitted to these two objects. It is in- tended to be comprehensive in its principles, and sufficiently extensive in its details ; and while the road to a knowledge of 4 PREFACE. the science is not designed to be unnecessarily steep and rug- ged, the author does not desire to relieve the learner of all occasion for effort, nor make him feel that the " Hill of Sci- ence " is no hill at all, but only a fiction of former ages. The author's idea is, that, in order to become a thorough and accom- plished arithmetician, one must study, and the National Arith- metic proposes no substitute for mental exertion. Still, it is not designed to be difficult beyond the necessities of the case, and no pupil, who is faithful to himself, will, it is thought, find reason to complain that enough is not done by way of suitable illustration to facilitate his progress. It is the opinion of some teachers, that no rules should be furnished the pupil to aid him in performing arithmetical questions, but that every pupil should form his own rules by the process of induction. But the author's experience has led him to a different conclusion, nor does he think that the inser- tion of proper rules, in a work like the present, interferes in the least with the necessity of study, or a thorough knowledge of the different numerical processes. The National Arithmetic is intended to be complete hi itself; but the smaller works of the author will prepare the pupil for an easy entrance upon the study of it. The learner can omit the more difficult parts of the present work until he reviews it, if thought advisable by the teacher. A few rules, which are omitted in some works on Arithmetic at the present day, the author has thought best to retain, such as Practice, Progression, Position, Permutation, &c. For, though these rules may not in themselves be of great practical utility, yet, as they are well adapted to improve the reasoning powers, and give interest to the higher departments of arith- metical science, it is deemed desirable to place them within reach of the student. In closing these prefatory remarks, the author would earnest- ly recommend that the pupil be required to give a minute and thorough analysis of every question he performs, at least until he has proved himself familiar with all the principles involved in the rule under consideration, and also the manner of their application. He would further recommend a frequent and thorough review of the parts of the work which the pupil has gone over, the exercise having respect mainly to the principles involved in the preceding rules and examples, Bradford Seminary, September 1, 1847. CONTENTS. PAGES SECT. INTRODUCTION 7-11 1. Numeration . . . . . . . . 13-17 2. Addition 18-21 3. Subtraction 21-24 4. Multiplicatior 24-29 5. Division 29-37 6. Contractions in Multiplication 37-39 7. Contractions in Division 39-41 8. Miscellaneous Examples 41-43 9. Tables of Money, Weights, and Measures . . 44 - 50 10. Compound Addition 50-54 11. Compound Subtraction . . . . . . 55-58 Exercises in Compound Addition and Subtraction . .58-60 12. Reduction 60-67 13. United States Money. Addition, Subtraction, Multipli- cation, and Division of ; Bills in . . . . 67-77 14. Compound Multiplication 78-81 Bills in English Money 81-83 15. Compound Division . . . . . . .84-88 Questions to be performed by Analysis . . . 88-89 16. Vulgar Fractions 89-109 17. Addition of Vulgar Fractions 110-114 18. Subtraction of Vulgar Fractions . . . . 114-121 19. Multiplication of Vulgar Fractions .... 121-125 20. Division of Vulgar Fractions .... 125-129 21. Questions to be performed by Analysis . . . 127-135 22. Decimal Fractions. Numeration of Decimal Fractions 135 - 137 23. Addition of Decimals 137-138 24. Subtraction of Decimals 138-139 25. Multiplication of Decimals . . . . 139-141 26. Division of Decimals 141-142 27. Reduction of Decimals 142-145 28. Miscellaneous Examples 145 - 147 29. Exchange of Currencies 149-152 30. Infinite or Circulating Decimals . , , . 152-153 Reduction of Circulating Decimals . . . 153-156 31. Addition of Circulating Decimals .... 156-157 32. Subtraction of Circulating Decimals . . . 157-158 33. Multiplication of Circulating Decimals . . . 158-159 34. Division of Circulating Decimals .... 159 35. Mental Operations in Fractions, &c. . . . 159-161 36. Questions to be performed by Analysis . . 162 - 164 37. Simple Interest 164-172 1* 6 CONTENTS. 38. Partial Payments 173-181 39. Miscellaneous Problems in Inteiest . . . 181 - 182 40. Compound Interest 183-186 41. Discount 187-188 42. Percentage 188-189 43. Commission and Brokerage 189-191 44. Stocks 191-192 45. Insurance and Policies 192-193 46. Banking 193-194 47. Barter 195 48. Practice 196-198 49. Equation of Payments 199-201 50. Custom-House Business 201 - 205 51. Ratio 205-207 52. Proportion . . . . . . .207-217 53. Compound Proportion, or Double Rule of Three . 217-221 54. Chain Rule 221-223 55. Partnership, or Company Business . . . 223-225 56. Partnership on Time 225-227 57. General Average 227-229 58. Profit and Loss 229-234 59. Duodecimals 234-238 60. Involution ; Evolution, or the Extraction of Roots ; Table of Powers 238-240 61. Extraction of the Square Root . . . .241-248 62. Extraction of the Cube Root . . . . 248 - 256 63. Arithmetical Progression 257-261 64. Geometrical Series, or Series by Quotient . . 261 -267 65. Infinite Series 267-268 66. Discount by Compound Interest .... 67. Annuities at Compound Interest .... 269-272 68. Assessment of Taxes 272-275 69. Alligation 275-279 70. Permutations and Combinations .... 279-282 71. Life Insurance 282-285 72. Position 286-290 73. Exchange 290-305 74. Value of Gold Coins 305-309 75. Geometry (Definitions) 309-313 Geometrical Problems 313-316 Mensuration of Solids and Superficies . . .316-327 76. Gauging 327-328 77. Tonnage of Vessels 328 - 329 78. Mensuration of Lumber 330-331 79. Philosophical Problems 331-335 80. Mechanical Powers 335-340 81. Specific Gravity 340-341 82. Strength of Materials . . . . . 341-344 83. Astronomical Problems 345-347 84. Miscellaneous Questions 347-354 APPENDIX. Weights and Measures . . . 355 - 360 INTRODUCTION. HISTORY OF ARITHMETIC. THE question, who was the inventor of Arithmetic, or in what age or among what people did it originate, has received different answers. In ordinary history we find the origin of the science attributed by some to the Greeks, by some to the Chaldeans, by some to the Phoe- nicians, by Josephus to Abraham, and by many to the Egyptians. The opinion, however, rendered most probable, if not absolutely cer- tain, by modern investigations is, that Arithmetic, properly so called, is of Indian origin, that is, that the science received its first defi- nite form and became the regular germ of modern Arithmetic in the regions of the East. It is evident, from the nature of the case, that some knowledge of numbers and of the art of calculation was necessary to men in the ear- liest periods of society, since without this they could not have per- formed the simplest business transactions, even such as are incidental to an almost savage state. The question, therefore, as to the invention of Arithmetic deserves to be considered only as it respects the origin of the science as we now have it, and which, as all scholars admit, has reached a surprising degree of perfection. And in this sense the honor of the invention must be awarded to the Hindoos. The history of the various methods of Notation, or the different means by which numbers have been expressed by signs or characters, is one of much interest to the advanced and curious scholar, but the brevity of this sketch allows us barely to touch upon it here. Among the ancient nations which possessed the art of writing, it was a natu- ral and common device to employ letters to denote what we express by our numeral figures. Accordingly we find, that, with the Hebrews and Greeks, the first letter of their respective alphabets was used for 1, the second for 2, and so on to the number 10, the latter, however, inserting one new character to denote the number 6, and evidently in order that their notation might coincide with that of the Hebrew's, the sixth letter of the Hebrew alphabet having no corresponding one in the Greek. The Romans, as is well known, employed the letters of their al- phabet as numerals. Thus, I denotes 1 ; V, 5 ; X, 10 ; L, 50 ; C, 100 ; D, 500 ; and M, 1000. The intermediate numbers were ex- pressed by a repetition of these letters in various combinations ; as II for 2 ; VI for 6 ; XV for 15 ; IV for 4 ; IX for 9, &c. They fre- 8 INTRODUCTION. quently expressed any number of thousands by the letter or letters de- noting so many units, with a line drawn above ; thus, V, 5,000 ; VI, 6,000 ; X, 10,000 ; T, 50,000 ; Cf, 100,000 ; M, 1,000,000. In the classification of numbers, as well as in the manner of ex- pressing them, there has been a great diversity of practice. While we adopt the decimal scale and reckon by tens, other nations have adopted the vicenary, reckoning by twenties ; others the quinary, reckoning by fives ; and others the binary, reckoning by twos. The adoption of one or another of these scales has been so general, that they have been regarded as natural, and accounted for by referring them to a common and natural cause. The reason for assuming the binary scale probably lay in the use of the two hands, which were em- ployed as counters in computing ; that for employing the quinary, in a similar use of the five fingers on either hand ; while the decimal and vicenary scales had respect, the former to the ten fingers on the two hands, and the latter to the ten fingers combined with the ten toes on the naked feet, which were as familiar to the sight of a rude, uncivil- ized people as their fingers. It is an interesting circumstance that in the common name of our numeral figures, digits (digiti) or fingers, we preserve a memento of the reason why ten characters and our present decimal scale of numeration were originally adopted to express all numbers, even of the highest order. It is now almost universally admitted that our present numeral char- acters, and the method of estimating their value in a tenfold ratio from right to left, have decided advantages over all other systems, both of notation and numeration, that have ever been adopted. There are those who think that a duodecimal scale, and the use of twelve numeral figures instead often, would afford increased facility for rapid and ex- tensive calculations, but most mathematicians are satisfied with the present number of numerals and the scale of numeration which has attained an adoption all but universal. It was long supposed, that for our modern Arithmetic the world was indebted to the Arabians. But this, as we have seen, was not the case. The Hindoos at least communicated a knowledge of it to the Arabians, and, as we are not able to trace it beyond the former people, they must have the honor of its invention. They do not, however, claim this honor, but refer it to the Divinity, declaring that the inven- tion of nine figures, with device of place, is to be ascribed to the benefi- cent Creator of the universe. But though the invention of modern Arithmetic is to be ascribed to the Hindoos, the honor of introducing it into Europe belongs unques- tionably to the Arabians. It was they who took the torch from the East and passed it along to the West. The precise period, however, at which this was done, it is not easy to determine. It is evident that our numeral characters and our method of computing by them were in common use among the Arabians about the middle of the tenth cen- tury, and it is probable that a knowledge of them was soon afterwards communicated to the inhabitants of Spain and gradually to those of the other European countries. Their general adoption in Europe would not seem to have been earlier than the twelfth or thirteenth century. HISTORY OF ARITHMETIC. 9 The science of Arithmetic, like all other sciences, was very limited and imperfect at the beginning, and the successive steps by which it has reached its present extension and perfection have been taken at long- intervals and among different nations. It has been developed by the necessities of business, by the strong love of certain minds for mathematical science and numerical calculation, and by the call for its higher offices by other sciences, especially that of Astronomy. In its progress we find that the Arabians discovered the method of proof by casting out the 9's, and that the Italians early adopted the practice of separating numbers into periods of six, for the purpose of enumera- tion. To facilitate the process of multiplication, this latter people also introduced, probably from the writings of Boethius, the long neg- lected Table of Pythagoras. The invention of the Decimal Fraction was a great step in the ad- vancement of arithmetical science, and the honor of it has generally been given to Regiornontanus, about the year 1464. It appears, how- ever, more properly to belong to Stevinus, who in 1582 wrote an ex- press treatise on the subject. The credit of first using the decimal point, by which the invention became permanently available, is given by Dr. Peacock to Napier, the inventor of Logarithms ; but De Mor- gan says that it was used by Richard Witt as early as 1613, while it is not shown that Napier used it before 1617. Circulating Decimals received but little attention till the time of Dr. Wallis, the author of the Arithmetic of Infinites. Dr. Wallis died at Oxford, in 1703. The greatest improvement which the art of computation ever re- ceived was the invention of Logarithms, the honor of which is un- questionably due to Baron Napier, of Scotland, about the encT of the sixteenth or the commencement of the seventeenth century. The oldest treatises on Arithmetic now known are the 7th, 8th, 9th, and 10th books of Euclid's Elements, in which he treats of proportion and of prime and composite numbers. These books are not contained in the common editions of the great geometer, but are found in the edition by Dr. Barrow, the predecessor of Sir Isaac Newton in the mathematical chair at Cambridge. Euclid flourished about 300 B. C. The next writer on Arithmetic mentioned in history is Nicomachus, the Pythagorean, who wrote a treatise relating chiefly to the distinc- tions and divisions of numbers into classes, as plain, solid, triangular, &c. He is supposed to have lived near the Christian era. The next writer of note is Boethius, the Roman, who, however, copied most of his work from Nicomachus. He lived at the begin- ning of the sixth century, and is the author of the well-known work on the Consolation of Philosophy. The next writer of eminence on the subject is Jordanus, of Namur, who wrote a treatise about the year 1200, which was published by Joannes Faber Stapulensis in the fifteenth century, soon after the invention of Printing. The author of the first printed treatise on Arithmetic was Pacioli, or, as he is more frequently called, Lucas de Burgo, an Italian monk, who in 1484 published his great work, entitled Summa de Arithmetica, &c., in which our present numerals appear under very nearly their modern form. 10 INTRODUCTION. In 1522, Bishop Tonstall published a work on the Art of Computa- tion, in the Dedication of which he says that he was induced to study Arithmetic to protect himself from the frauds of money-changers and stewards, who took advantage of the ignorance of their employers. In his preparation for this work, he professes to have read all the books which had been published on this subject, adding, also, that there was hardly any nation which did not possess such books. About the year 1540, Robert Record. Doctor in Physic, printed the first edition of his famous Arithmetic, which was afterward augmented by John Dee, and subsequently by John Mellis, and which did much to advance the science and practice of Arithmetic in England in its early stages. This work, which is now quite a curiosity, effectually de- stroys the claim to originality of some things of which authors much more modern have obtained the credit. In it we find the celebrated case of a will, which we have in the Miscellaneous Questions of Webber and Kinne, and which, altered in language and the time of making the testament, is the llth Miscellaneous Question in the pres- ent work. This question is, by his own confession, older than Record, and is said to have been famous since the days of Lucas de Burgo. In Record it occurs under the " Rule of Fellowship." Record was the author of the first treatise on Algebra in the English language. In 1556, a complete work on Practical Arithmetic was published by Nicolas Tartaglia, an Italian, and one of the most eminent mathema- ticians of his time. From the time of Record and Tartaglia, works on Arithmetic have been too numerous to mention in an ordinary history of the science. De Morgan, in his recent work (Arithmetical Books), has given the names of a large number, with brief observations upon them, and to this the inquisitive student is referred for further information in regard both to writers and books on this subject since the invention of Print- ing. It is remarkable that De Morgan knew next to nothing of any American works on Arithmetic. He mentions the " American Ac- countant" by William Milns, New York, 1797, and gives the name of Pike (probably Nicholas Pike) among the names of which he had heard in connection with the subject. He had also seen the Memoir of Zerah Colburn. Of the compilation of Webber and the original works of Walsh and Warren Colburn, he seems to have been entirely ignorant. The various signs or symbols, which are now so generally used to abridge arithmetical as well as algebraical operations, were introduced gradually, as necessity or convenience taught their importance. The earliest writer on Algebra after the invention of Printing was Lucas de Burgo, above mentioned, and he uses p for plus and m for minus, and indicates the powers by the first two letters, in which he was fol- lowed by several of his successors. After this, Steifel, a German, who in 1544 published a work entitled Arithmetica Integra, added considerably to the use of signs, and, according to Dr. Hutton, was the first who employed -f- and to denote addition and subtraction. To denote the root of a quantity he also used our present sign ^/, origi- nally r, the initial of the word radix, root, The sign = to denote HISTORY OF ARITHMETIC. 11 equality was introduced by Record, the above-named English mathe- matician, and for this reason, as he says, that " noe 2 thynges can be moar equalle," namely, than two parallel lines. It is a curious cir- cumstance that this same symbol was first used to denote subtraction. It was also employed in this sense by Albert Girarde, who lived a little later than Record. Girarde dispensed with the vinculum em- ployed by Steifel, as in 3-J-4, and substituted the parenthesis (S-f-4), now so generally adopted. The first use of the St. Andrew's cross, X , to signify multiplication is attributed to William Oughtred, an Englishman, who in 1631 published a work entitled Clavis Mathe- matiaR, or Key of Mathematics. It was intended to notice several other works, ancient and modern, but the length to which this sketch has already extended forbids it. We must not, however, "omit to mention two American works, which have done much for the cause of practical Arithmetic in this country. These are the large work of Nicholas Pike, first published about 1787, and the little unpretending " First Lessons " in Arithmetic, by War- ren Colburn. From the former of these many later authors have bor- rowed much that is useful, and the latter has exerted an influence on the method of studying Arithmetic greater, perhaps, than any other modern production. No better elementary work than that of Col- burn has ever, it is believed, appeared in any language. We had thought of alluding to the ancient philosophic Arithmetic, and the elevated ideas which many of the early philosophers had of the science and properties of numbers. But a word must here suffice. Arithmetic, according to the followers of Plato, was not to be studied " with gross and vulgar views, but in such a manner as might enable men to attain to the contemplation of numbers ; not for the purpose of dealing with merchants and tavern-keepers, but for the improve- ment of the mind, considering it as the path which leads to the knowl- edge of truth and reality." These transcendentalists considered per- fect numbers, compared with those which are deficient or superabun- dant, as the images of the virtues, which, they allege, are equally remote from excess and defect, constituting a mean between them ; as in the case of true courage, which, they say, lies midway between audacity and cowardice, and of liberality, which is a mean between profusion and avarice. In other respects, also, they regard this anal- ogy as remarkable: perfect numbers, like the virtues, are "few in number and generated in a constant order ; while superabundant and deficient numbers are, like vices, infinite in number, disposable in no regular series, and generated according to no certain and invariable law." We conclude this brief sketch in the earnest hope that the noble science of numbers may ere long find some devoted friend who shall collect, arrange, and bring within the reach of ordinary students, much more fully than we have done, the scattered details of its long- neglected history. ARITHMETICAL SIGNS. Sign of equality ; as 12 inches = 1 foot signifies that 12 inches are equal to one foot. -|- Sign of addition ; as 8 -J- 6 = 14 signifies that 8 added to 6 is equal to 14. Sign of subtraction ; as 8 6 = 2, that is, 8 less 6 is equal to 2. X Sign of multiplication ; as 7 X 6 42, that is, 7 multiplied by 6 is equal to 42. -j- Sign of division ; as 42 -i- 6 = 7, that is, 42 divided by 6 is equal to 7. -3^ 2 - Numbers placed in this manner imply that the upper num- ber is to be divided by the lower one. : : : : Signs of proportion ; thus, 2 : 4 : : 6 : 12, that is, 2 has the same ratio to 4 that 6 has to 12 ; and such numbers are called proportionals. 12 3-|-4=z 13. Numbers placed in this manner show that 3 is to be taken from 12, and 4 added to the remainder. The line at the top is called a vinculum, and connects all the numbers over which it is drawn. 9 2 implies that 9 is to be raised to the second power ; that is, multiplied by itself. 8 3 implies that 8 is to be multiplied into its square, or to be raised to the third power. \f This sign prefixed to any number shows that the square root is to be extracted. \/ This sign prefixed to a number shows that the cube root is to be extracted. Sometimes roots are designated by fractional indices, thus ; 9* denotes the square root of 9 ; 27* denotes the cube root of 27. ( ) [ ] Parentheses and brackets are often used instead of a vinculum. Thus, (7 3) x 5 = 60 -=- 3. 03= An edition of this work, without ansicers, is published for the ac- commodation of those teachers who prefer that the pupil should not have access to them. A KEY, containing solutions and explanations, is also published for the convenience of teachers. ARITMTETIC, SECTION I. ARITHMETIC is the science of numbers, and the art of com- puting by them. The operations of Arithmetic are performed principally by Addition, Subtraction, Multiplication, and Division. NUMERATION. NUMERATION teaches to express the value of numbers, either by words or characters. Numbers in Arithmetic are expressed by the ten following characters, which are called numeral figures ; viz. 1 (owe), 2 (two], 3 (three), 4 (four), 5 (five), 6 (six), 7 (seven), 8 (eight), 9 (nine), Q (cipher, or nothing). The first nine of these figures are called significant, as dis- tinguished from the cipher, which is of itself insignificant. Besides this value of the rfumerical figures, they have another value, dependent on the place which they occupy, when con- nected together. This is illustrated by the following table and its explanation. 2 14 NUMERATION. [SECT. i. I J.I i H 4 " o S3 c a i? 2 o i i U3 | CO fl d CO C 1 *-^ 3 QJ gH ^3 Q '3 3 ffi H H ffi H 7 6 5 4 3 2 1 8 7 6 5 4 3 2 9 8 7 6 5 4 3 9 8 7 6 5 4 9 8 7 6 5 9 8 7 6 9 8 7 9 8 9 Here any figure occupying the first place, reckoning from right to left, denotes only its simple value or number of units. But the figure standing in the second place denotes ten times its simple value ; that occupying the third place, a hundred times its simple value, and so on to any required number of places ; the value of any figure being always increased tenfold by its removal one place to the left. Thus, in the number 1834, the 4 in the first place denotes only four units, or simply 4 ; the 3 in the second place signi- fies three tens, or thirty ; the 8 in the third place signifies eighty tens, or eight hundred ; and the 1 in the fourth place, one thou- sand ; so that the whole number is read thus, one thousand eight hundred thirty-four. Although the cipher has no value of itself, when standing alone, yet, being joined to the right hand of significant figures, it increases their value in a tenfold proportion ; thus, 5 signifies simply five, while 50 denotes five tens, or fifty ; 500, five hun- dred, and so on. NOTE. The idea of number is the latest and most difficult to form. Before the mind can arrive at such an abstract conception, it must be familiar with that process of classification, by which we successively re- mount from individuals to species, from species to genera, from genera to orders. The savage is lost in his attempts at numeration, and significant- ly expresses his inability to proceed by holding up his expanded fingers or pointing to the hair of his head. See Lacroix. SECT. I.] NUMERATION. 15 NUMERATION TABLE. The following is the French method of enumeration, and is in general use in the United States and on the continent of Europe. In order to enumerate any number of figures by this method, they should be separated by commas into divisions of three figures each, as in the annexed table. Each division will be known by a different name. The first three figures, reckon- ing from right to left, will be so many units, tens, and hun- dreds, and the next three so many thousands, and the next three so many millions, &c. g Vigintillions. Jj Novemdecillions. jg Octodecillions. J Septendecillions. i Sexdecillions. jl Quindecillions. S Quatuordecillions. Tridecillions. i Duodecillions. J3 Undecillions. 8 Decillions. i Nonillions. i Octillions. I Septillions. % Sextillions. jl Quintillions. j8 Quadrillions. I Trillions. E Billions. I Millions, jl Thousands. JS Units. The value of the numbers in the an- nexed table, expressed in words, is One hundred twenty-three vigintillions, four hundred fifty-six novemdecillions, seven hundred eighty-nine octodecil- lions, one hundred twenty-three septen- decillions, four hundred fifty-six sexde- cillions, seven hundred eighty-nine quin- decillions, one hundred twenty-three quatuordecillions, four hundred fifty-six tridecil lions, seven hundred eighty-nine duodecillions, one hundred twenty-three undecillions, four hundred fifty-.six de- cillions, seven hundred eighty-nine no- nillions, one hundred twenty-three oc- tillions, four hundred fifty-six septil lions, seven hundred eighty-nine sextillions, one hundred twenty-three quintillions, four hundred fifty-six quadrillions, seven hundred eighty-nine trillions, one hun- dred twenty-three billions, four hundred fifty-six millions, seven hundred eighty- nine thousands, one hundred twenty- three units. 16 NUMERATION. [SECT. NUMERATION TABLE. The following is the old English method of enumeration, but it has be- come almost obsolete in this country. In order to enumerate any number of figures by this method, they should be separated by semicolons into divisions of six figures each, and each division separated in the middle by a comma, as in the annexed table. Each division will be known by a different name. The first three figures, in each division, reckoning from right to left, will be so so many units, tens, and hundreds of the name belonging to the division, and the three on the left will be so many thousands of the same name. The value of the numbers in the annexed table, expressed in words, is Three hundred and seventeen thousand, eight hundred and ninety-seven tridecillions ; four hundred and thirty-one thousand, thirty-two duodecillions ; six hundred thirty-nine thousand, eight hundred six- ty-four undecillions ; three hundred six- ty-one thousand, three hundred sixteen decillions ; four hundred sixty-one thou- sand, three hundred fifteen nonillions ; one hundred twenty-three thousand, six hundred seventy-five octillions; eight hundred sixteen thousand, one hundred thirty-one sepfillions ; one hundred twen- ty-three thousand, four hundred fifty-six sextillions; one hundred twenty-three thousand, six hundred fourteen quintil- lions ; three hundred fifteen thousand, one hundred thirty-one quadrillions; three hundred ninety- eight thousand, eight hundred thirty-two trillions; five hun- dred sixty -three thousand, eight hundred seventy-one billions; three hundred fifty-one thousand, six hundred fifteen millions ; one hundred twenty-three thousand five hundred sixty-one. NOTE. The student must be familiar with the names, from units to tridecillions, and from tridecillions to units, so that he may repeat them with facility either way. S- Thousands. | Tridecillions. S Thousands. Duodecillions. 1 Thousands. J Undecillions. ^ Thousands. 5 Decillions. J2 Thousands. S Nonillions. 8 Thousands. Jl Octillions. js Thousands. js Septillions. jg Thousands. jg Sextillions. jS Thousands. Quintillions. j Thousands. js Quadrillions. 1 Thousands. J Trillions. 1 Thousands. " Billions. J2 Thousands, jj; Millions. $ Thousands. 2 Units. SECT, i.] NUMERATION. 17 Let the following numbers be written in words : 706 313,461 604,021 3,607,005 607,081,107 470,803,020 7,801,410,909 322,172,517,101 607,100,001,070 407,000,010,703,801 200,070,007,801,000 670,812,000,170,063,891 478,127,815,016,666,060,707 800,800,800,800,800,800,800,800 127,081,061,071,081,010,009,007,007 407,144,140,070,060,700,007,101,800,808 Let the following numbers be written in figures : * 1. Twenty-nine. 2. Four hundred and seven. 3. Twenty-three thousand and seven. 4. Five millions and twenty-seven. 5. Seven millions, two hundred five thousand and five. 6. Two billions, two hundred seven millions, six hundred four thousand and nine. 7. One hundred five billions, nine hundred nine millions, three hundred eight thousand two hundred and one. 8. Nine quintillions, eight billions and forty-six. 9. Fifteen quintillions, thirty-one millions and seventeen. 10. Five hundred seven septillions, two hundred three tril- lions, fifty-seven millions and eighteen. 11. Nine nonillions, forty-seven trillions, seven billions, two millions, three hundred ninety-two. 12. Fifteen duodecillions, ten trillions, one hundred twenty- seven billions, twenty-six millions, three hundred twenty thou- sand four hundred twenty-six. * To express numbers by figures, begin at the left hand with the highest order mentioned, and, proceeding to units, write in each succes- sive order the figure which denotes the given number in that order. If any of the intervening orders are not mentioned in the given number, supply their places with ciphers. 2* 18 ADDITION. [SECT. H. SECTION II. ADDITION. ADDITION is the collecting of numbers to find their sum. 1. A man has three farms ; the first contains 378 acres, the second 586 acres, and the third 168 acres. How many acres are there in the three farms ? In this question, the units are first added, and OPERATION, their sum is found to be 22 ; in 22 units there are Acres. two tens and two units. The two units are writ- ten under the column of units, and the 2 (tens) 586 are carried to be added with the tens, which are found to amount to 23 tens 2 hundreds and 1132 3 tens. The 3 is written under the column of tens, and the 2 (hundreds) is carried to the column of the hundreds, which amount to 11=1 thousand 1 hundred. The whole of which is set down. Hence the pro- priety of the following RULE. Write units under units, tens under tens, 4444 2341 31. Divide 9594004321 by 78000 4321 32. Divide 162068563389 by 40506 6789 33. Divide 5427563776896 by 808070 7896 34. Divide 475065610503 by 481007 8967 35. Divide 8794170278 by 470 278 36. Divide 70006876 by 7000 6876 37. Divide 204060808062747 by 2020202 2345 38. Divide 700003456 by 10000 3456 39. Divide 7207276639 by 9009 4567 40. Divide 63126068678 by 70070 5678 41. Divide 3394240208391 by 706007 6785 42. Divide 169233137936 by 10080 7856 43. Divide 915527086796874 by 3011101 8567 44. Divide 454115186870257 by 500123 8765 45. Divide 12032109124169380 by 6007023 1357 CASE II. To divide by any number with ciphers annexed. Cut off the ciphers from the divisor, and the same number of figures from the right hand of the dividend. Then divide the remaining fig- ures of the dividend by the remaining figures of the divisor, and the re- sult will be the quotient. To complete the work, annex to the last re- mainder found by the operation the figures cut off from the dividend , and the whole will form the true remainder. EXAMPLE. 1. Divide 36378967 by 31000. 31,000)36378,967(1173 31 ~53 31 227 217 108 93 15967 Remainder. 36 DIVISION. [SKCT. y. Quotients. Rem. 2 Divide 32100 by 6000 5 2100 3. Divide 3167810 by 160000 19 127810 4. Divide 12345678 by 1400000 8 1145678 5. Divide 1637851 by 500000 3 137851 6. Divide 3678953 by 326100 11 91853 7. Divide 41111111 by 1100000 37 411111 CASE III. To divide by a unit with ciphers annexed. Cut off as many figures from the right hand of the dividend as there are ciphers in the divisor, and the figures on the left hand of the separa- trix will be the quotient, and those on the right hand the remainder. Quotients. Rem. 1. Divide 123456789 by 10 12345678 9 2. Divide 987654321 by 100 9876543 21 3. Divide 1221 12347800 by 1000 122112347 800 4. Divide 89765432156 by 1000000 89765 432156 CASE IV. To divide by a composite number, that is, a number pro- duced by the multiplication of two or more numbers. Divide the dividend by any one of the factors, and the quotient thus found by anotJier, and thus proceed till every factor has been made a di- visor, and the last quotient will be the true quotient required. NOTE. To find the true remainder, we multiply the last remainder by the last divisor but one, and to the product add the next preceding remainder ; we multiply this sum by the next preceding divisor, and to the product add the next preceding remainder; and so on, till we have gone through all the divisors and remainders to the first. This rule will be better understood by the pupil, after he has become acquainted with fractions. EXAMPLES. 1. Divide 47932 by 72. As 72 is equal to 9 times 8, we first 9)47932 divide the dividend by 9, and the quotient 8^5325 7 thence arising by 8 ; and to find the true ' remainder, we multiply the last remainder, 665 5 r= 52 5^ by the first divisor, 9, and to the prod- uct add the first remainder, 7 ; and find the amount to be 52, the true remainder. SECT, vi.] CONTRACTIONS IN MULTIPLICATION. 37 2. Divide 5371 by 192. We find 192 equal to the product of 4 4)5371 times 6 times 8, = 4x6x8= 192. We 6^1342 3 therefore divide by these factors, as in the last example. To find the true remainder, 8)223 4 we mu ltiply the last remainder, 7, by the 27 7 = 187 last divisor but one, 6 ; and to the product add the last remainder but one, 4 ; this sum we multiply by the first divisor, 4 ; and to the product add the first remainder, 3 ; and find the amount to be 187. Quotients. Rem. 3. Divide 7691 by 24= 4 X 6 320 11 4. Divide 8317 by 27 = 3 X 9 308 1 5. Divide 3116 by 81 = 9 X 9 38 38 6. Divide 61387 by 121 = 11 x 11 507 40 7. Divide 19917 by 144 = 12 X 12 138 45 8. Divide 9 1746 by 336= 6x 7X8 273 18 9. Divide 376785 by 315 = 5 X 7x9 1196 45 SECTION VI. CONTRACTIONS IN MULTIPLICATION. I. To multiply by 25. RULE. Annex two ciphers to the multiplicand, and divide it by 4, and the quotient is the product required. Rationale. By annexing two ciphers, we increase the mul- tiplicand one hundred times, and by dividing this number by 4, the result will be an increase of the multiplicand only twenty- five times, because 25 is one fourth of 100. 1. Multiply 785643 by 25. OPERATION. 4)78564300 19641075 Product. 2. Multiply 9876543 by 25. Ans. 3. Multiply 47110721 by 25. Ans. II. To multiply by 33 J. RULE. Annex two ciphers to the multiplicand, and divide it by 3, and the quotient is the product required. 4 38 CONTRACTIONS IN MULTIPLICATION. [SECT. vi. Rationale. As in the last case, by annexing two ciphers, we increase the multiplicand one hundred times ; and by dividing the number by 3, we only increase the multiplicand thirty-three and one third times, because 33 is one third of 100. 4. Multiply 87138942 by 33. OPERATION. 3)8713894200 2904631400 Product. 5. Multiply 66666993 by 33. Ans. 6. Multiply 12336723 by 33. Ans. EL To multiply by 125. RULE. Annex three ciphers to the multiplicand, and divide by 8, and the quotient is the product. NOTE. By annexing three ciphers, the number is increased one thou- sand times; and, by dividing by 8, the quotient will be only one eighth . of 1000, that is, 125 times. 7. Multiply 12345678 by 125. OPERATION. 8)12345678000 1543209750 Product. IV. To multiply by any number of 9's. RULE. Annex as many ciphers to the multiplicand as there are 9's in the multiplier, and from this number subtract the number to be multi- plied, and the remainder is the product required. 8. Multiply 87654 by 999. OPERATION. By annexing three ciphers, we make the 87654000 number one thousand times larger. If from 87654 ^ s nu ^ber, with the ciphers annexed, we p^j , subtract the multiplicand, we make the prod- Product. uct one thousandth part less; that the product will be only 999 times the multiplicand. Q. E. D. 9. Multiply 7777777 by 9999. Ans. 77769992223. 10. Multiply 5555 by 999999. Ans. 555499-1445. I* ,!*""" 1 ? m " lti P 1 y b 7 an y number of 3's, proceed as above and di- vide the product by 3; but if it be required to multiply by 6's, proceed as above and then multiply the product by 2, and divide the" result by 3, and the quotient is the product. SECT, vii.] CONTRACTIONS IN DIVISION. 11. Multiply 987654 by 333333. OPERATION. 987654000000 987654 3)987653012346 329217670782 Product, Ans. 12. Multiply 32567895 by 3333. Ans. 13. Multiply 876543 by 66666. Ans. OPERATION. 87654300000 876543 87653423457 2 3)175306846914 58435615638 Product, Ans. 14. Multiply 345678 by 6666666. Ans. V. When the multiplier can be separated into periods, which are multiples of one another, the operation may be contracted in the following manner. 15. Multiply 112345678 by 288144486. OPERATION. 112345678 288144486 674074068 = the product by 6. 5392592544 = the foregoing product X by 8 for 48. 16177777632 = the lest product x by 3 for 144. 32355555264 = the last product X by 2 for 288. 32371787641631508 Product. SECTION VII. CONTRACTIONS IN DIVISION. I. To divide by 5. RULE. Multiply the dividend by 2, and tJie product, except the last figure at the right, is the quotient. NOTE. The remainder will be tenths. 40 CONTRACTIONS IN DIVISION. [SECT. VH. 1. Divide 67895 by 5. Ans. 13579. OPERATION. 67895 2 13579,0 Quotient. II. To divide by 25. RULE. Multiply the dividend by 4, and the product, except the last two figures at the right, is the quotient. NOTE. The two figures at the right are hundredths. 2. Divide 8765887 by 25. Ans. 350635^. OPERATION. 8765887 4 350635,48 Quotient. HI. To divide by 33 J. RULE. Multiply the dividend by 3, and the product, except the last two figures at the right, is the quotient, and the last two are hundredths. 3. Divide 876735 by 33. Ans. 26302^. OPERATION. 876735 3 26302,05 Quotient. IV. To divide by 125. RULE. Multiply the dividend by 8, and the product, except the last three figures, is the quotient, and these last three figures will be thou- sandths. 4. Divide 1234567 by 125. Ans. OPERATION. 1234567 8 9876,536 Quotient. 5. Divide 8786789 by 125. Ans 6. Divide 1234567 by 125 Ans. V. A short method of performing Long Division. 7. Divide 16294996 by 24. Ans. 6789M. SECT, vin.] MISCELLANEOUS EXAMPLES. 41 OPERATION. This method differs from the 24)16294896(678954, Ans. common way byplacing the right- hand figure of every product im- 181229 mediately under the dividend. 169129 121 121 8. Divide 3545304 by 47. Ans. 75432. 9. Divide 45005091 by 57. Ans. 789563. VI. To divide by any number of 9's, when their number is not less than half the number of places that will be in the quo- tient, and when there is no remainder. RULE. Annex as many ciphers to the dividend, as there are 9's in the divisor. Then write the proper dividend under the number thus found, and subtract it from the number to which ciphers have been annexed; and, as many places of the remainder at the right hand as there were ciphers annexed, are so many figures for the right hand of the quotient ; and, for t/ie remaining numbers of the quotient, a competent number must be taken from the left hand of the above remainder. 10. Divide 123332544 by 999. OPERATION. By- examining the dividend and 123332544000 divisor, we know there will be 6 123332544 places in the quotient. We there- 123 209211 456 ta ^ e t ^ lree ^ mese figures from I90d*fi Our t An the ri ht hand of the remainder for > Quotient, Ans. ^ ^ right . hand figureg of the quotient, and the other three we take from the left hand of the remainder. 11. Divide 12332655 by 999. Ans. 12345. 12. Divide 987551235 by 9999. Ans. 98765. 13. Divide 9123456779876543211 by 999999999. Ans. 9123456789. SECTION VIII. MISCELLANEOUS EXAMPLES. 1. What number multiplied by 1728 will produce 1705536 ? Ans. 987. 4* 42 MISCELLANEOUS EXAMPLES. [SECT. vm. 2. If a garrison of 987 men are supplied with 175686 pounds of beef, how much will there be for each man ? Ans. 178 Ibs. 3. In one dollar there are 100 cents ; how many dollars in 697800 cents ? Ans. $ 6978. 4. In one pound there are 16 ounces ; how many pounds are in 111680 ounces ? Ans. 6980 Ibs. 5. A dollar contains 6 shillings ; how many dollars are in 5868 shillings ? Ans. $ 978. 6. The President of the United States receives a salary of $ 25,000 ; what does he receive per month ? Ans. $ 2083. 7. A man receiving $ 96 for 8 months' labor, what does he receive for 1 month ? Ans. $ 12. 8. The distance from Haverhill to Boston is 30 miles ; and, if a man travel 6 miles an hour, how long will he be in going this distance ? Ans. 5 hours. 9. The annual revenue of a gentleman being $8395, how much per day is that equivalent to, there being 365 days in a year? Ans. $23. 10. The car on the Liverpool railroad goes at the rate of 65 miles an hour ; how long would it take to pass round the globe, the distance being about 25,000 miles ? Ans. 384^ hours. 11. How much sugar at $ 15 per cwt. may be bought for $ 405 ? Ans. 27 cwt. 12. In 6789560 shillings how many pounds, there being 20 shillings in a pound ? Ans. 339478 pounds. 13. The Bible contains 31,173 verses ; how many must be read each day, that the book may be read through in a year ? Ans. 85-^|f verses. 14. In 123456720 minutes how many hours ? Ans. 2057612 hours. 15. A gentleman possessing an estate of $ 66,144, bequeathed one fourth to his wife, and the remainder was to be divided be- tween his 4 children ; what was the share of each ? Ans. $ 12,402. 16. A man disposed of a farm containing 175 acres at $ 87 per acre ; of the avails he distributed $ 1234 for charitable pur- poses ; $ 197 was expended for the purchase of a horse and chaise ; the remainder was divided between 6 gentlemen and 8 ladies, and each lady was to receive twice as much as a gentle- man ; what was the share of each ? Ans. $ 627 for a gentleman, and $ 1254 for a lady. 17. If there are 160 square rods in an acre, how many acres are in 1086240 square rods ? Ans. 6789 acres. SECT, vin.] MISCELLANEOUS EXAMPLES. 43 18. If 144 square inches make one square foot, how many square feet in 14222160 square inches ? Ans. 98765 feet. 19. What number is that, which being multiplied by 24, the product divided by 10, the quotient multiplied by 2, 32 sub- tracted from the product, the remainder divided by 4, and 8 subtracted from the quotient, the remainder shall be 2 ? Ans. 15. 20. What is the difference between half a dozen dozen, and six dozen dozen ? Ans. 792. 21. Bought of F. Johnson 8 barrels of flour at $ 7 per barrel, and 3 hundred weight of sugar at $ 8 per hundred. What was the amount of his bill ? Ans. $ 80. 22. Sold S. Jenkins my best horse for $ 75, my second-best chaise for $ 87, a good harness for $ 31. He has paid me in cash $ 38, and has given me an order on Peter Parker for $ 12. How many dollars remain my due ? Ans. $ 143. 23. T. Webster has sold his wagon to J. Emerson for $ 85. He is to receive his pay in wood at $ 5 per cord. How many cords will it require to balance the value of the wagon ? Ans. 17 cords. 24. Purchased a farm of 500 acres for $ 17,876. I sold 127 acres of it ut $ 47 an acre, 212 acres at $ 96 an acre, and the remainder at $ 37 an acre. What did I gain by my bargain ? Ans. $14,402. 25. A tailor has 938 yards of broadcloth ; how many cloaks can be made of the cloth, if it require 7 yards to make one cloak ? Ans. 134 cloaks. 26. Bought 97 barrels of molasses at $ 5 a barrel. Gave 17 barrels to support the poor, and the remainder was sold at $ 8 a barrel. Did I gain or lose, and how much ? Ans. $ 155 gain. 27. There are 12 pence in one shilling ; required the num- ber of pence in 671 shillings. Ans. 8052 pence. 28. Twelve inches make one foot in length ; required the number of inches in 5280 feet, it being the length of a mile. Ans. 63360 inches. 29. In one pound avoirdupois there are 16 ounces ; required the ounces in 1728 pounds. Ans. 27648 ounces. 30. Required the number of shillings in 8136 pence. Ans. 678 shillings. 31. It requires 1728 cubic inches to make one cubic foot ; required the number of cubic inches in 3787 cubic feet. Ans. 6543936 inches. 44 MONEY AND WEIGHTS. [SECT. ix. SECTION IX. TABLES OF MONEY, WEIGHTS, AND MEASURES. T7NITED STATES MONEY. 1 Cent, marked c. 1 Dime, " d. 1 Dollar, " $. 1 Eagle, " E. 10 10 10 10 Mills. 10 100 1000 10000 Mills Cents Dimes Dollars n Cents. 1 10 100 1000 Dimes. 1 Dollars. 10 = 1 Eagle. 100 = 10 = 1 4 Farthings 12 Pence " 20 Shillings " 21 Shillings sterling " 28 Shillings N. E. " ENGLISH MONEY. make 1 Penny, 1 Shilling, 1 Pound, 1 Guinea, 1 Guinea, NOTE. One pound sterling is equal to $ 4.44|- 4 48 960 d. 1 12 240 marked a. 1 20 d. s. . G. G. 100 Centimes FRENCH MONEY. make 1 Franc : .186 dollar. TROY WEIGHT. 24 Grains make 1 Pennyweight,^ marked dwt. 20 Pennyweights " 1 Ounce, " oz. 12 Ounces " 1 Pound, Ib. gfr 24 480 5760 By this weight are weighed gold, silver, and jewels. NOTE. " The original of all weights used in England was a grain or corn of wheat, gathered out of the middle ofthe ear ; and, being well dried, 32 of them were to make one pennyweight, 20 pennyweights one ounce, and 12 ounces one pound. But in later times, it was thought sufficient to divide the same pennyweight into 24 equal parts, still called grains, being the least weight now in common use j and from hence the rest are com- puted." make dwt. 1 20 240 1 Pennyweight,^ 1 Ounce, 1 Pound, oz. = 1 = 12 SECT, ix.] WEIGHTS AND MEASURES. APOTHECARIES' WEIGHT. 20 Grains S\ 3 Scruples 8 Drams S 12 Ounces 60 480 5760 make 1 3 24 288 1 Scruple, 1 Dram, 1 Ounce, 1 Pound, dr. 1 8 = v 96 marked sc. or 9 " dr. or 5 " oz. or " Ib. or oz. 1 12 Apothecaries mix their medicines by this weight ; but buy and sell by Avoirdupois. The pound and ounce of this weight are the same as in Troy Weight. AVOIRDUPOIS WEIGHT. 16 Drams make 1 Ounce, marked oz. 16 Ounces " 1 Pound, Ib. 28 Pounds , 1 Quarter " qr. 4 Quarters | 1 Hundrec Weight, " cwt. 20 Hundred Weight H i 1 Ton, " ton. dr. oz. ' 16* = 1 Ib. 256 = 16 = 1 qr. 7168 = 448 = | 28 = cwt. 28672 = 1792 = 1 112 = 4 = 1 ton. 573440 = 35840 = 2240 = 80 = 20 = 1 I3y this weight are weighed almost every kind of goods, and all metals except gold and silver. By a late law of Massachusetts, the cwt. contains 100 Ibs. instead of 112 Ibs. A ton is reckoned at the custom-houses of the United States at 2240 Ibs. LONG MEASURE. 3 Barleycorns, or 12 Lines make 1 Inch, 12 Inches " 3 Feet " 6 Feet " 54 Yards, or 16$ Feet " 40 Rods " 8 Furlongs " 3 Miles " 694 Miles nearly " 360 Degrees " in. ft. 12 =, 1 y d . 36 = 3 = 1 198 = 164 = 54 7920 = 63360 = 660 5280 1760 1 Inch, marked in. Foot, " ft. Yard, Fathom, " yd. fth. Rod, or Pole, " rd. Furlong, " fur. Mile, m. League lea. Degree, " Deg. Circle of the Earth. or rd. = 1 fur. = 40 = 1 m. = 320 = 8 == 1 46 MEASURES. [SECT. ix. CLOTH MEASURE. 24 Inches make Nail marked na. 4 Nails u Quarter of a yard, " qr. 4 Quarters Yard, " yd. 3 Quarters ii Ell Flemish, " E. F. 5 Quarters u Ell English, 4 Quarters 1| inch Ell Scotch, " E. S. SQUARE ME iSURE. 144 Square inches make Square foot, marked ft. 9 Square feet H Square yard, " yd. 304 Square yards " 2724 Square feet " 40 Square rods or poles " Square rod or pole, ' p. Square rod or pole, " p. Rood, " R. 4 Roods 11 Acre, " A. 640 Acres Square mile, " S. M. in. ft. 144=. 1 yd. 1596 = 9 = 1 P. 39204 = 2724 = 304 = 1 R. 1568160 = 10890 = 1210 = 40 = 1 A. 6272640 = 43560 = 4840 = 160 = 4=1 s.M. 4014489600 = 27878400 = 3097600 = 102400 = 2560 = 640 = 1 DRY MEASURE. 2 Pints make Quart, marked qt. 4 Quarts " Gallon, " gal. 2 Gallons " Peck, . " pk. 4 Pecks Bushel, " bu. 36 Bushels " Chaldron, " ch. pts. gal 8 = 1 pk. 16 = 2 = 1 bu. 64 = 8 = 4=1 ch. 2304 = 288 = 144 = 36 =1 NOTE. This measure is applied to all goods that are not liquid and are sold by measure, as corn, fruit, salt, coals, &c. A Winchester Bushel is 18 inches in diameter, and 8 inches deep. The standard Gallon Dry Measure contains 268| cubic inches. 2 Pints 4 Quarts 32 Gallons 54 Gallons 2 Hogsheads 2 Butts ALE AND BEER MEASURE. make Quart, Gallon, Barrel, Hogshead, Butt, Tun, marked qt. " gal. bar. " hhd. " butt. " tun. SECT. IX.] MEASURES AND TIME. 47 pts. 8 256 432 864 qt. 4 128 216 432 32 54 108 bar. = 1 - 1 = 3 hhd. 1 butt. = 1 NOTE. By a law of Massachusetts, the Barrel for cider and beer shall contain 32 gallons, but in some other States it is of different capa- city. The Ale Gallon contains 282 cubic or solid inches. Milk is sold by the Beer Gallon. WINE MEASURE. 4 Gills 2 Pints 4 Quarts 42 Gallons 63 Gallons, or 1 Tierces 2 Tierces 2 Hogsheads 2 Pipes, or 4 Hhds. ptS. 2 8 336 504 672 1008 2016 qt. 4 168 252 336 504 1008 1 42 63 84 126 252 make Pint, Quart, Gallon, Tierce, Hogshead, Puncheon, Pipe or Butt, Tun, marked pt. " qt. " gal. " tier. " hhd. " pun. " pi. " tun. tier. = 2 hhd. 1 1J = 3 = 2 = 6 = 4 = pun. 1 tun. 1 NOTE. The Wine Gallon contains 231 cubic inches. Water, wine, and spirits are measured and sold by this measure. A cubic foot of distilled water weighs 158 ounces Avoirdupois. The English Imperial Gallon contains 277 cubic inches, and weighs 10 Ib. Avoirdupois, or 12 Ib. 1 oz 16 dwt. 16gr. Troy. There is no legal measure in the United States for tierce, hogshead, puncheon, pipe, or butt. 60 Seconds, or 60" 60 Minutes 24 Hours 7 Days 4 Weeks 13 Months, 1 day, 6 hours, or 365 days, 6 hours, 12 Calendar months OF TIME. make Minute, (4 Hour, ( Day, (4 Week, Month, ' or I 1 Julian Year, " 1 Year, marked m. h. d. w. mo. y. y- 48 MOTION AND DISTANCES. [SECT. ix. sec. m. 60 = 1 h. 3600 = 60 = 1 d. 86400 = 1440 = 24 = 1 w . 604800 = 10080 = 168 = 7=1 mo. 2419200 = 40320 = 672 = 28 = 4 = 1 y . 31557600 = 525960 = 8766 = 365| = 1 NOTE. The true solar year is the time measured from the sun's leav- ing either equinox or solstice, to its return to the same again. A period- ical year is the time in which the earth revolves round the sun, and is 365 d. 6 h. 9 m. 14 sec., and is often called the Sidereal year. The civil year is that which is in common use among the different nations of the world, and contains 365 days for three years in succession, but every fourth year contains 366 days. When any year can be divided by four, without any remainder, it is leap year, and has 366 days. w. d. h. mo. d. h. Or, 52 1 6 =5 13 1 6 = 1 Julian Year. d. h. m. sec. But, 365 5 48 57 = 1 Solar Year. And, 365 6' 9 144=1 Sidereal Year. The days in each month are as follows : January, March, May, July, August, October, and December have 31 days each ; April, June, September, and November have 30 days each ; February has 28 days, excepting leap year, when it has 29. CIRCULAR MOTION. 60 Seconds, or 60 " make 1 Prime minute, marked / 60 Minutes ' 1 Degree, " 30 Degrees 1 Sign, s. 12 Signs, or 360 Degrees, the whole great circle of the zodiac. 6'6 = 1 3600 = 60 = 1 8 108000 = 1800 = 30 = 1 1296000 = 21600 = 360 = 12 = zodiac. MEASURING DISTANCES. 7$ Inches make 1 Link. 25 Links " 1 Pole. 100 Links 1 Chain. 10 Chains " 1 Furlong. 8 Furlongs " 1 Mile. SECT. IX.] MISCELLANEOUS TABLE. 49 Inches. Link. *ll = 1 Pole. 192 = 25 __ 1 Chain. 792 100 _. 4 = 1 Furlong. 7920 1000 40 = 10 = 1 Mile. 63360 = 8000 = 320 = 80 == 8 = 1 SOLID MEASURE. 1728 Inches 27 Feet 40 Feet of round timber 128 Feet, i. e. 8 in length, 4 in breadth, > and 4 in height, f make 1 Foot. 1 Yard. 1 Ton. 1 Cord of wood. NOTE. One ton of round timber, as usually surveyed, contains solid feet. MISCELLANEOUS TABLE. A gallon of train oil A stone of butcher's meat A gallon of molasses A stone of iron A tod A firkin of butter A firkin of soap A quintal of fish A weigh A sack A puncheon of foreign prunes A last A fother of lead A barrel of anchovies " raisins flour pork or beef soap " shad or salmon in Connect- ) icut or New York fish in Massachusetts cider and beer herrings in England " salmon or eels do. 8 bushels of salt, measured on board the vessel, 7i do. measured on shore, 5 weighs M s 7A pounds. 8 " 11 " 14 " 28 56 94 100 182 364 1120 4368 " 19 cwt. 30 pounds. 112 " 196 " 200 " 256 " 200 " 30 gallons. 32 " 32 " 42 " 1 hogshead. 50 COMPOUND ADDITION. [SECT. x. 3 hoops make 40 casts 10 hundred 12 units, or things, 12 dozen 144 dozen " cast, hundred, thousand, dozen. great gross. SECTION X. COMPOUND ADDITION. WHEN numbers are applied to things, the measure or value of which is expressed by different denominations, they lose their abstract character, and become subject to restrictions, imposed upon them by the denomination to which they are applied. Thus, when we say six cents, ten days, or three inches, we have not only the idea of number, but also the idea of a certain value or measure, which subjects the number in connection with it to certain limitations. And, when used in such connections, we call numbers denominate. Thus in . 4s. 7d. the num- bers 6, 4, and 7 are denominate numbers, so called, because they are applied to express each a particular denomination. When now we have several numbers of different denomina- tions, which we wish to add together, we call the process by which this is done Compound Addition ; which we define by saying, ^ That it consists in adding together two or more numbers of different denominations to find the sum total. RULE. Write all the given numbers of the same denomination under each other ; as dollars under dollars, cents under cents, <5fC. Then add to- gether the numbers of the lowest denomination and divide the sum by the number which it takes of that denomination to make one of the denom- ination next above it, and set the remainder directly under the column that has been added. Carry the quotient to the column of the next de- nomination, and add as before, dividing by the number which it takes of this denomination to make one of the denomination next above it, setting down the remainder and carrying the quotient as before, and thus pro- ceed till the column of the highest denomination is added, under which place its whole sum, and the numbers expressing the several denomina- tions will be the sum total required. SECT, x.] COMPOUND ADDITION. 51 EXAMPLES. i. UNITED STATES MONEY. 2. 3. ft. cts. m. ft. cts. m. E. ft. cts. m. 325 67 3 28 15 6 71 3 41 5 186 35 8 16 16 3 61 6 82 6 161 89 9 63 81 5 16 1 96 2 987 15 8 14 61 6 41 7 82 1 891 61 6 38 74 5 54 8 36 3 176 81 3 16 16 8 41 9 48 5 2729 51 7 ENGLISH MONEY. 4. 5. 6. . 8. d. . s. d. . s. d. qr. 471 16 9 28 6 9^ 31 17 9 2 147 17 8 15 16 ni 16 16 6 1 613 13 11 31 13 HJfc 16 11 11 1 115 11 7 14 16 9 19 19 9 3 41 19 6 17 17 7| 61 17 1 3 48 12 2 32 18 8* 14 14 4 2 1439 11 7 TROY WEIGHT. 7. 8. Ib. oz. dwt. & Ib. oz. dwt. gr. 16 11 19 23 123 9 7 13 31 10 18 16 98 11 17 14 63 9 12 15 49 7 13 21 17 8 13 12 13 10 10 20 61 7 12 16 47 9 19 23 17 6 17 22 51 5 15 15 209 7 15 8 APOTHECARIES' WEIGHT. 9. 10. Ib 3 B gr. ft 3 9 gr. 27 11 7 2 19 37 9 6 1 18 16 10 6 1 13 14 4 4 2 11 41 9 3 2 16 61 6 3 2 6 38 10 5 2 14 41 4 7 2 16 41 4 4 1 11 39 8 4 1 12 16 6 6 2 6 51 11 7 2 19 183 6 3 1 19 52 COMPOUND ADDITION. [SECT. x. AVOIRDUPOIS WEIGHT. 11. Ton. cwt. qr. Ib. oz. dr. 61 19 3 27 15 15 63 13 3 16 11 11 51 12 3 17 7 6 61 16 1 11 12 12 13 13 3 12 13 15 71 18 2 13 14 14 324 15 2 16 12 9 12. cwt. qr. Ib. oz. dr. 61 2 11 11 14 16 3 15 15 11 41 3 13 9 9 38 2 11 10 10 42 1 9 8 13 31 3 27 11 12 LONG MEASURE. 13. 14. Deg. m. fur. rd. ft. in. br. m. fur. rd. yd. ft. In. br. 17 69 7 39 16 11 2 69 7 31 5 2 11 1 61 62 3 17 12 9 1 16 6 16 4 1 6 2 16 16 6 16 13 10 2 61 7 32 3 2 10 1 48 19 3 15 15 6 1 73 3 16 4 2 9 2 17 58 6 33 14 7 1 19 4 14 1 1 8 2 33 35 5 19 9 9 2 75 5 25 5 2 7 1 195 54j h 1 24 j r 7 T .=4 i h=6 195 54 5 24 1 NOTE. As half a mile is equal to 4 furlongs, we add them to the 1 furlong, which make 5 furlongs. And as half a foot is equal to 6 inches, we add them to the 7 inches, which make 13 inches ; and these are equal to 1 foot 1 inch. By the same method, we obtain the result in the 14th and 17th questions. CLOTH MEASURE. 15. yd. qr. na. in. 37 3 3 2 61 3 1 1 13 2 2 2 32 1 1 1 61 2 2 2 22 1 3 229 3 3 1 16. E.E. qr. na. in. 671 1 1 1 161 3 3 2 617 3 1 2 178 321 717 2 1' 2 166 3 2 1 SECT, x.] COMPOUND ADDITION. 53 LAND OR SQUARE MEASURE. 17. 18. A. R. p. ft. in. A. R. p. ft. 761 3 37 260 125 38 1 39 272 131 2 16 135 112 61 3 38 167 613 1 14 116 131 35 3 19 198 161 3 13 116 123 47 3 16 271 321 2 31 97 96 86 2 13 198 47 3 19 91 48 46 1 14 269 2038 1 13 2 95 SOLID MEASURE. 19. 20. Ton. ft. in. Cord. ft. in. 29 36 1279 61 127 1161 69 19 1345 37 89 1711 67 18 1099 61 98 1336 71 14 1727 43 56 1678 43 35 916 91 119 1357 53 17 1719 81 115 1129 335 23 1173 WINE MEASURE. 21. 22. Tun. 61 hhd. 3 gal. qt. pt. 62 3 1 hhd. 67 fi qt. pt. O J. 39 2 16 1 1 16 16 3 68 3 57 2 1 39 16 3 87 3 45 3 1 47 62 1 1 47 2 59 3 1 43 57 3 47 3 39 2 1 71 61 3 1 354 30 1 - ALE AND BEER MEASURE. 23. 24. Tun. hhd. gal. qt. hhd. gal. qt. pt. 46 3 50 3 161 c o Do 3 1 91 2 48 3 371 52 3 1 17 3 18 98 19 1 81 3 38 2 47 43 1 41 1 47 1 61 43 1 1 37 2 29 3 42 27 3 1 317 2 17 5* 54 COMPOUND ADDITION. [SECT. x. DRY MEASURE. bu. 37 25. f qt. pt. 5 1 ch. 16 26. bu. 31 t I 1 ' 61 3 7 1 39 31 3 1 32 2 2 14 16 3 1 71 1 6 1 55 15 3 61 1 3 1 71 17 3 1 32 3 3 1 42 14 3 1 298 4 1 TIME. 27. 2 3. y. mo. w. d, h. m. 8. y. mo. d. h. 57 11 3 6 23 29 55 13 5 29 17 31 11 1 3 19 19 39 61 11 17 21 46 9 2 2 17 28 56 15 9 19 16 43 10 1 1 18 17 48 61 10 25 23 32 9 1 3 16 23 28 41 4 16 17 14 1 1 5 22 28 16 18 5 9 6 227 2 3 21 28 2 CIRCULAR MOTION. 29. 30. S. O i it s. o . 4 29 59 59 11 11 16 51 6 17 17 29 6 6 6 16 11 16 56 58 9 14 56 56 9 13 46 51 3 29 29 49 5 27 16 42 9 17 18 58 % 25 17 17 6 13 13 52 5 10 35 16 NOTE. We divide the sura of the signs, in these questions, by 12, and write down the remainder, because it is Circular Motion. MEASURING DISTANCES. 31. 32. m. fur. ch. p. 1. m. fur. ch. p. 1. 17 7 9 3 24 27 4 3 1 21 16 3 4 1 15 29 3 1 3 23 27 4 6 2 17 67 3 3 1 19 18 6 3 3 21 21 7 1 3 16 61 7 7 2 16 16 7 9 3 13 17 1 8 2 19 31 4 8 1 20 160 1 1 12 SECT, xi.] COMPOUND SUBTRACTION. 55 SECTION XI. COMPOUND SUBTRACTION. COMPOUND SUBTRACTION teaches to find the difference be- tween two numbers of different denominations. RULE. Write the smaller compound number under the greater, in the order of the different denominations, as pounds under pounds, shillings under shillings, <5fc. Begin with the lowest denomination and subtract each lower number from the one above it, and write the difference underneath. If, in any denomination, the lower number be greater than the one above it, add to the upper number the number required of this denomination to make one of the next higfier ; and, from the number thus obtained, subtract the lower number and set down the remainder underneath. Carry one to the next denomination in the subtrahend, and proceed in like manner with the subtraction, till the operation has been performed in all the columns, setting down the entire difference, between the upper and lower numbers of the highest denomination, and the result will be the difference required. NOTE. The reason for increasing the number of the minuend by the number required of a lower denomination to makeone of the next higher, is precisely the same as that for which we add ten to a figure of the minuend, in Simple Subtraction. In both cases we add the number de- noting the ratio between the denomination in question, and the next higher number of the minuend. In Simple Subtraction, the ratio of in- crease from right to left being uniformly tenfold, we add ten, while in the case of farthings, pence, and shillings, we add 4, 12, and 20. EXAMPLES. UNITED STATES MONEY. 1. 2. $ cts. m. $ cts. m. 169 81 3 681 16 7 85 93 8 189 43 8 83 87 5 ENGLISH MONEY. . s. d. . B. 87 16 3J 617 11 19 17 9 181 15 67 18 6j 56 COMPOUND SUBTRACTION. [SECT. xi. 5. TROY WEIGHT. 6. lb oz. dwt. CT. lb. oz. dwt. gr 71 3 12 15 58 5 12 10 16 10 17 20 19 9 17 21 54 4 14 19 APOTHECARIES' WEIGHT. 7. 8. fe 71 S 5 9 gr. ft S 3 1 14 15 2 2 B 15 18 6 7 2 19 99 1 1 18 52 6 3 1 14 E AVOIRDUPOIS WEIGHT. 9. 10. T. 71 cwt. 18 T lb. oz. dr. cwt. 13 1 13 73 T lb. 15 oz. 13 19 19 2 16 8 5 19 i 19 15 51 18 2 24 9 8 CLOTH MEASURE. 11, 12. yd. qr. 67 1 na. 1 in. 1 E.E. 51 na. 3 18 2 2 2 19 3 1 48 2 2 i LONG MEASURE. 13. 14. m. 16 fur. 7 rd. 18 ft. 3 in. bar. deg. m. fur. 21 38 41 3 rd. 29 yd. 2 ft. in. bar. 172 9 7 19 16 82 29 36 5 31 3 1 9 1 6 7 38 2 5 2 = 6 6 7 38 2 11 2 NOTE As half a foot is equal to 6 inches, we add them to the 5 inches, which make 11 inches. The same principle is adopted in the 14th, 15th, and 16th examples. LAND OR SQUARE MEASURE. 15. 16. A. R. p. ft. In. A. R. p. yd. ft. in. 56 1 19 119 110 13 1 15 19 1 17 17 3 13 127 113 9 3 16 30 5 19 38 2 5 264 33 SECT. XI.] COMPOUND SUBTRACTION. 57 SOLID MEASURE. 17. 18. Tons. ft. in. Corda. ft. in. 49 13 1611 361 47 1178 18 15 1719 197 121 1617 30 37 1620 WINE MEASURE. 19. 20. Tun. hhd. gal. qt. 79 3 19 1 pt. hhd. gal. qt. pt. 1 16 1 1 11 1 28 2 1 9221 68 1 53 3 ALE AND BEER MEASURE. 21. 22. Tun. hhd. gal. qt. pt. hhd. gal. qt. 63 1 15 1 769 18 1 19 3 16 3 1 191 19 3 43 1 52 1 1 DRY MEASURE. 23. 24. ch. bu. pk. qt. 56 2 1 1 ch. bu. pk. qt. 39 12 2 1 38 3 1 2 12 25 3 5 17 34 3 7 TIME. 25. 26. mo. d. h. m. s. y. m. w. d. h. m. s. 6 16 13 27 19 48 2 5 19 27 31 1 22 16 41 37 19 10 3 7 21 38 56 4 23 20 45 42 CIRCULAR MOTION. 27. 28. S. o in S. o i ,, 6 11 12 48 4 19 41 22 9 8 15 56 1 22 19 28 9 2 56 52 58 COMPOUiND ADDITION. [SECT. w. MEASURING DISTANCES. 29. 30. m. fur. ch. p. 1. m. fur. ch. p. 1. 21 1 3 2 19 28 6 1 2 18 19 2 1 3 21 15 7 3 1 19 1 7 1 2 23 EXERCISES IN COMPOUND ADDITION AND SUBTRACTION. 1. What is the sum of 16^. 5s. 8d. 2qr. 3l. 16s. lid. 3qr. 2L. 11s. Iqr. 19c. Os. lOd. 3qr. 13. 13s. 7d. 3qr. and28<. 17s. 5d. Iqr. ? Ans. 13 1. 5s. 8d. 2. Bought of a London tailor a vest for !. 13s. 4d., a coat for 7<. 12s. 9d., pantaloons for 2. 3s. 9d., and sur- tout for 9<. 8s. Od. ; what was the whole amount ? Ans. 20JE. 17s. lOd. 3. Bought a silver tankard, weighing lib. 8oz. 17dwt. 14gr., a silver can, weighing lib. 2oz. 12dwt, a porringer, weighing lloz. 19dwt. 20gr., and three dozen of spoons, weighing lib. 9oz, 15dwt. lOgr. ; what was the whole weight ? Ans. 51b. 9oz. 4dwt 20gr. 4. What is the weight of a mixture of 31b 45 23 29 14gr. of' aloe, 2ft> 7 63 IB 13gr. of picra, and lib 105 13 29 17gr. of saffron ? Ans. 7tb 105 33 19 4gr. 5. Add 321b 95 13 29 14gr. ; 13 fo 75 63 19 13gr. ; and 16ft 115 73 19 12gr. together. Ans. 63fc 45 73 29 19gr. 6. Sold 4 loads of hay ; the first weighed 27cwt. 3qr. 181b. ; second, 31cwt. Iqr. 151b. ; third, 19cwt. Iqr. 151b. ; and fourth, 38cwt. 2qr. 271b. ; what is the weight of the whole ? Ans. 117cwt. Iqr. 191b. 7. Bought 5 pieces of broadcloth ; the first contained 17yd. 3qr. 2na. ; second, 13yd. 2qr. Ina. ; the third, 87yd. Iqr. 3na. ; the fourth, 27yd. Iqr. 2na. ; and the fifth, 29yd. Iqr. 2na. ; what was the whole quantity purchased ? Ans. 175yd. 2qr. 2na. 8. A pedestrian travelled, the first week, 371m. 3fur. 37rd. 5yd. 2ft. lOin. ; the second week, 289m. 2fur. 18rd. 3yd. 1ft. 9in.; and the third week he travelled 399m. 7fur. 3ft. llin. ; how many miles will he have travelled ? Ans. 1060m. 5fur. 16rd. 5yd. 1ft. 9. A man has 3 farms ; the first contains 186A. 3R. 14p. ; SECT. XL] COMPOUND SUBTRACTION. 59 the second, 286A. 17p. ; and the third, 115A. 2R. ; how much do they all contain ? Aris. 588 A. IE,. 31p. 10. The Moon is 5s. 18 14' 17" east of the Sun ; Jupiter is 7s. 10 29' 28" east of the Moon ; Mars is 11s. 12 11' 56" east of Jupiter ; and Herschel is 7s. 18o 38' 15" east of Mars ; how far is Herschel east from the Sun ? Ans. 7s. 29o 33' 56". 11. I have 4 piles of wood; the first contains 7 cords, 76ft. 1671m. ; the second, 16c. 28ft. 56in.; the third, 29c. 127ft. 1000 in. ; and the fourth, 29c. 10ft. 1216in. ; how much is there in all? Ans. 82c. 115ft. 487in. 12. A vintner sold at one time 73hhd. 43gal. 3qt. Ipt. of wine ; at another, 27hhd. 3gal. ; at another, 15hhd. 3qt. Ipt. ; and at another, 161hhd. and 2qt. ; how much did he sell in all ? l Ans. 276hhd. 48gal. Iqt. 13. A man has 3 sons ; the first is 14y. 3mo. 2w. 5d. old ; the second is 9y. lOmo. 3w. 4d. 23h. 12m. 15sec. ; and the third is 2y. Imo. 3w. 2d. 7m. ; what is the sum of their ages ? and how much older is the first than the second ? Ans. 26y. 3mo. Iw. 4d. 23h. 19m. 15sec. " 4y. 5mo. 3w. Od. Oh. 47m. 45sec. NOTE. When 4 weeks are reckoned as a month, it requires 13 months to make one year. 14. I have 73A. of land ; if I should sell 5A. 3R. Ip. 7ft., how much should I have left ? Ans. 67A. OR. 38p. 265ft. 15. A. owes B. 100<, ; what will remain due after he has paid him 3s. 6^d. ? Ans. 99 . 16s. 5d. 16. It is about 25,000 miles round the globe ; if a man shall have travelled 43m. 17rd. 9in. how much will remain to be travelled ? Ans. 24,956m. 7fur. 22rd. 15ft. 9in. 17. Bought 7 cords of wood ; and 2 cords 78ft. having been stolen, how much remained ? Ans. 4c. 50ft. 18. I have 15 yards of cloth ; having sold 3yd. 2qr. Ina., what remains ? Ans. 1 lyd. Iqr. 3na. 19. If a wagon loaded with hay weighs 43cwt. 2qr. 181b., and the wagon is afterwards found to weigh 9cwt. 3qr. 231b., what is the weight of the hay ? Ans. 33cwt. 2qr. 231b. 20. Bought a hogshead of wine, and by an accident 8gal. 3qt. Ipt. leaked out ; what remains ? Ans. 54gal. Ipt. 21. I had 10A. 3R. lOp. of land ; and I have sold two house- lots, one containing 1A. 2R. 13p., the other, 2A. 2R. 5p. ; how much have I remaining ? Ans. 6A. 2R. 32p. 22. The Moon moves 13 10' 35'' in a solar day, and the 60 REDUCTION. [SECT. xn. Sun 59' 8" 20"' ; now supposing them both to start from the same point in the heavens, how far will the Moon have gained on the Sun in 24 hours ? Ans. 12 11* 26" 40'". 23. A farmer raised 136bu. of wheat ; if he sells 49bu. 2pk. 7qt. Ipt., how much has he remaining ? Ans. 86bu. Ipk. Oqt. Ipt. 24. If from a stick of round timber, containing 2T. 18ft. 1410in., there be taken 38ft. 1720in., how much will be left ? Ans. IT. 19ft. 1418in. 25. If from lib. of ipecacuanha there be taken at one time 4S 23 13gr., and at another, 35 13 29 14gr., how much will be left? Ans. 4 33 29 13gr. 26. A brewer has in one cellar 18bbl. 3gal. 2qt. of beer, and in another, 13bbl. Ip. ; what is the whole quantity, and how much more is in one cellar than the other ? Ans. 31bbl. 3gal. 2qt. Ipt. " 5bbl. 3gal. Iqt. Ipt. 27. If from $ 100.00 there be paid at one time $ 17.28,5, at another time $ 10.00,5, and at another $ 37.15, how much will remain? Ans. $35.56. SECTION XII. REDUCTION. THE object of Reduction is to change the denomination of numbers without altering their value. It consists of two parts, Descending and Ascending. The former is performed by Mul- tiplication, the latter by Division. "'Reduction Descending teaches to bring numbers of .a higher denomination to a lower ; as, to bring pounds into shillings, or tons into hundred weights. Reduction Ascending teaches to bring numbers of a lower denomination into a higher ; as, to bring farthings into pence, or shillings into pounds. REDUCTION DESCENDING. EXAMPLE. 1. In 48. 12s. 7d. 2qr., how many farthings ? SECT, xii.] REDUCTION. 61 In this example, we multiply the 48,. by 20, because it takes 20 shillings to make a pound ; and to this product we add the 12s. in the question. Then we multiply by 12, because it takes 12 pence to make one shil- ling ; and to the product we add the 7 pence in the question. We then multiply by 4, the number of farthings in a penny, and to the product we add the 2 farthings, and the work is done. From the above example and illustration, we deduce the following RULE. Multiply the highest denomination given by the number required of the next lower denomination to make one of the denomination next above it, and add to the product thus obtained the corresponding denomina- tion of tfie multiplicand. Proceed in this ivay, till the reduction is brought to the denomination required by the question. NOTE 1. To multiply by i, we divide the multiplicand by 2, and to multiply by , we divide by 4. NOTE. 2. The answers to Reduction Descending will be found in the questions of Reduction Ascending. 2. In 127<. 15s. 8d. how many farthings ? 3. In 28. 19s. lid. 3qr. how many farthings? 4. In 378. how many pence ? 5. How many grains in 281b. lloz. 12dwt. 15gr. troy ? 6. In 171b. 12dwt. troy, how many pennyweights ? 7. If a silver tankard weigh 31b. lloz., how many grains will it be ? 8. How many scruples in 231b. apothecaries' weight ? 9. If a load of hay weigh 3T. 16cwt. 2qr. 181b., how many ounces will it be ? 10. Required the number of drachms in a hogshead of sugar, weighing 2T. 17cwt. 3qr. 161b. 15oz. 13dr. 11. In 57yd. how many nails ? 12. In 83947E.E. 4qr. how many nails ? 13. In 2263E.F. 2qr. how many quarters ? 14. How many feet in 79 miles ? 15. How many inches in 396 furlongs ? 16. How many inches from Haverhill to Boston, the distance being 30 miles ? 17. How many barleycorns will it take to reach round the world ? 6 62 REDUCTION. [SECT. xn. 18. In 403m. 7fur. 35rd. 2yd. Oft. Oin. Ibar. how many barleycorns ? 19. In 413Le. 2m. 2fur. 38rd. lyd. Oft. 7in. how many inches ? 20. In 144m. Ifur. 8rd. lyd. 1ft. how many feet ? 21. How many inches in 1051yd. 2ft. 5in. ? 22. In 3576fur. 12rd. 3yd. how many yards ? 23. How many square feet in 25 acres ? 24. How many square rods in 365 square miles ? 25. The surface of the earth contains 196563942 square miles. What would it be in square inches ? 26. Required the number of feet in 10A. 3R. 38p. 6yd. 5ft. 72in. 27. In 2R. Op. 24yd. 3ft. how many inches ? 28. In 1A. 3R. 34p. 27yd. 4ft. 54in. how many inches ? 29. In 17 cords of wood, how many inches ? 30. In 19 tons of round timber, how many inches ? 31. How many cubic feet of wood in 128 cords ? 32. In 4899hhd. 4gal. 3qt. how many quarts ? 33. In 1224 tuns Ip. Ihhd. 19gal. Iqt. Opt. Igi. how many gills? 34. How many pints in 790p. Ohhd. 58gal. Oqt. Ipt. ? 35. In 460 butts Ihhd. 31 gal. of beer, how many gallons ? 36. In 36hhd. 26gal. 3qt. Ipt. how many pints? 37. In 16 tons of round timber, how many inches ? 38. How many seconds from the deluge, it being 2348 years B. C., to the year 1836 ? 39. How many days did the last war continue, it having com- menced June 18, 1812, and ended Feb. 17, 1815? 40. How many pecks in 676 chaldrons ? 41. In 657 cents, how many mills? 42. In 3165 dimes, how many mills ? 43. In 63 dollars, how many cents ? 44. In 27 eagles, how many mills ? REDUCTION ASCENDING. EXAMPLE. 1. In 76789 farthings, how many pounds ? OPERATION. Ans. 79<. 19s. 9d. 4)76789 We first divide by 4, because 4 far- 12)19197 r Tqr things make a penny. We then di- OHM^QQ QJ ' vide b y 12 ' because 12 P ence make a 20)1599 9d. shilling. Lastly we divide by 20, the 79. 19s. 9d. number of shillings in a pound. SECT, xii.] REDUCTION. 63 From the preceding illustration and example, we deduce the following RULE. Divide the lowest denomination given by the number which it takes of that denomination to make one of the next higher ; so proceed, until it is brought to the denomination required. Any remainders occurring in the successive divisions will be of the same denominations with the divi- dends to which they respectively belong. NOTE 1. To divide by 5^, we multiply the multiplicand by 2, and divide the product by 11 ; to divide by 16^, we multiply by 2 and divide by 33 ; and to divide by 272^, we multiply by 4 and divide by 1089. NOTE 2. The answers to Reduction Ascending are the questions in Reduction Descending. NOTE 3. When we divide by 11 or 33, and there is a remainder, we divide that remainder by 2 to get the true remainder; and, when we di- vide by 1081), we must divide the remainder by 4. 2. In 122672 farthings, how many pounds ? 3. In 27839 farthings, how many pounds ? 4. In 90720 pence, how many pounds ? 5. In 166S63 grains, how many pounds troy ? 6. How many pounds in 4092 pennyweights ? 7. How many pounds troy in 22560 grains ? 8. In 6624 scruples, how many pounds ? 9. In 137376 ounces, how many tons ? 10. In 1660157 drachms, how many tons ? 11. How many yards in 912 nails ? 12. Required the ells English in 1678956 nails. 13. Required the ells Flemish in 6791 quarters. 14. Required the miles in 417120 feet. 15. Required the furlongs in 3136320 inches. 16. Required the miles in 1900800 inches. 17. How many degrees in 4755801600 barleycorns ? 18. How many miles in 76789567 barleycorns ? 19. How many leagues in 78653167 inches ? 20. Required the miles in 761116 feet. 2 1 . Required the yards in 37865 inches. 22. Required the furlongs in 786789 yards. 23. How many acres in 1089000 square feet ? 24. How many square miles in 37376000 square rods ? 25. How many square miles in 789,103,900,894,003,200 square inches ? 26. In 478675 square feet, how many acres ? 27. In 3167856 square inches, how many roods ? 28. How many acres in 12345678 square inches ? 29. How many cords in 3760128 cubic inches ? 64 REDUCTION. [SECT. in. 30. How many tons of round timber in 1313280 cubic inches ? 31. How many cords in 16384 cubic feet? 32. How many hogsheads of wine in 1234567 quarts ? 33. How many tuns of wine in 9877001 gills ? 34. In 796785 pints of wine, how many pipes ? 35. How many butts of beer in 49765 gallons ? 36. In 15767 pints, how many hogsheads of beer ? 37. In 1105920 inches, how many tons of round timber ? 38. In 132005440800 seconds, how many years ? 39. In 974 days, how many years and calendar months ? 40. How many chaldrons in 97344 pecks ? 41. How many cents in 6570 mills ? 42. How many dimes in 316500 mills ? 43. How many dollars in 6300 cents ? 44. How many eagles in 270000 mills ? COMPOUND REDUCTION. 1. In 57. 15s. how many dollars ? Ans. $ 192.50cts. 2. In 67 . 14s. 9d. how many crowns at 6s. 7d. each ? Ans. 205cr. 5s. 2d. 3. How many pounds and shillings in 678 dollars ? Ans. 203. 8s. 4. How many ells English in 761 yards ? Ans. 608E. E. 4qr. 5. How many yards in 61 ells Flemish ? Ans. 45yd. 3qr. 6. How many bottles, that contain 3 pints each, will it take to hold a hogshead of wine ? Ans. 168. 7. How many steps, 2ft. Sin. each, will a man take in walk- ing from Bradford to Newburyport, the distance being 15 miles ? Ans. 29700. 8. How many spoons, each weighing 2oz. 12dwt., can be made from 51b. 2oz. Sdwt. of silver ? Ans. 24. 9. How many times will the wheel of a coach revolve, whose circumference is 14ft. 9in. in passing from Boston to Washing- ton, the distance being 436 miles ? Ans. 156073fV 9 T . 10. I have a field of corn, consisting of 123 rows, and each row contains 78 hills, and each hill has 4 ears of corn ; now if it take 8 ears of corn to make a quart, how many bushels does the field contain ? Ans. 149bu. 3pk. 5qt. Opt. 11. If it take 5yd. 2qr. 3na. to make a suit of clothes, how many suits can be made from 182 yards ? Ans. 32. 12. A goldsmith wishes to make a number of rings, each SECT, xii.] REDUCTION. 65 weighing 5dwt. lOgr., from 31b. loz. 2dwt. 2gr. of gold ; how many will there be ? Ans. 137. 13. How many shingles will it take to cover the roof of a building, which is 60 feet long and 56 feet wide, allowing each shingle to be 4 inches wide and 18 inches long, and to lay one third to the weather? Ans. 20160. 14. There is a house 56 feet long, and each of the two sides of the roof is 25 feet wide ; how many shingles will it take to cover it, if it require 6 shingles to cover a square foot ? Ans. 16800. 15. If a man can travel 22m. 3fur. 17rd. a day, how long would it take him to walk round the globe, the distance being about 25000 miles ? Ans. 1114ffff days. 16. If a family consume 71b. lOoz. of sugar in a week, how long would lOcwt. 3qr. 161b. last them ? Ans. 160 weeks. 17. Sold 3 tons 17cwt. 3qr. 181b. of lead at 7d. a pound ; what did the lead amount to ? Ans. 254c. 10s. 2d. 18. What will 5cwt. Iqr. lOlb. of tobacco cost, at 4d. a pound? Ans. ll. 4s. 3d. 19. What will 7 hogsheads of wine cost, at 9 cents a quart ? Ans. $ 158.76. 20. What will 15 hogsheads of beer cost, at 3 cents a pint ? Ans. $ 194.40. 21. What will 73 bushels of meal cost, at 2 cents a quart ? Ans. $ 46.72. 22. A merchant has 29 bales of cotton cloth ; each bale con- tains 57 yards ; what is the value of the whole at 15 cents a yard ? Ans. $ 247.95. 23. A merchant bought 4 bales of cotton ; the first contained 6cwt. 2qr. lllb. ; the second, 5cwt. 3qr. 161b. ; the third, 7cwt. Oqr. 71b. ; the fourth, 3cwt. Iqr. 171b. He sold the whole at 15 cents a pound ; what did it amount to ? Ans. $ 385.65. 24. A merchant having purchased 12cwt. of sugar, sold at one time 3cwt. 2qr. lllb. and at another time he sold 4cwt. Iqr. 151b. ; what is the remainder worth, at 15 cents per pound ? Ans. $ 67.50. 25. Bought 4 chests of hyson tea ; the weight of the first was 2cwt. Iqr. 71b. ; the second, 3cwt. 2qr. 151b. ; ihe third, 2cwt. Oqr. 201b. ; the fourth, 5cwt. 3qr. 171b. ; what is the value of the whole, at 37 cents a pound ? Ans. $ 589. 12^. 26. Purchased a cargo of molasses, consisting of 87 hogs- heads ; what is the value of it, at 33 cents a gallon ? Ans. $ 1808.73. 6* 66 REDUCTION. [SECT. XH. 27. From a hogshead of wine, lOgal. Iqt. Ipt. 3 gills leaked out. The remainder was sold at 6 cents a gill ; to what did it amount ? Ans. $ 100.86. 28. A man has 3 farms ; the first containing 100A. 3R. 15rd. ; the second, 161 A. 2R. 28rd. ; the third, 360 A. 3R. 5rd. He gave his oldest son a farm of 1 12A. 3R. 30rd. ; his second son a farm of 316A. 1R. 18rd. ; his youngest son a farm of 168 A. 3R. 13rd. ; and sold the remainder of his land at 1 dollar and 35 cents a rod ; to what did it amount ? Ans. $ 5436.45. 29. A grocer bought a hogshead of molasses, containing 87gal. Iqt., from which 13 gallons leaked out ; what is the remainder worth, at 1 cent a gill ? Ans. $ 23.76. 30. A man bought 4 loads of hay ; the first weighing 25cwt. Oqr. 171bs. ; the second, 37cwt. 2qr. 171b. ; the third, 18cwt. 3qr. 141b. ; and the fourth, 37cwt. Iqr. 171b. ; what is the value of the whole, at 2 cents a pound ? Ans. $266.74. REDUCTION OF THE OLD NEW ENGLAND CUR- RENCY TO UNITED STATES MONEY. The original currency of N. E. was pounds, shillings, pence, and farthings ; but, on the adoption of the Constitution of the United States, it was changed to dollars, cents, and mills. It is frequently necessary to reduce the former to the present cur- rency of the United States ; for which we have the following RULE. If pounds only are given, annex three ciphers and divide by 3, and the quotient will be the sum required in cents. If pounds and an even number of shillings are given , annex to the pounds half the number of shillings and two ciphers, and divide as before. If the number of shillings be odd, take half of the largest even number of shillings and annex it to the pounds with the figure 5 and one cipher , instead of two as above, and proceed as in the former instances. If pounds, shillings, pence, and farthings are given, annex to the pounds and shillings, as before, and find the number of farthings con- tained in the given pence and farthings, taking care to increase their number by I, if they exceed 12, and by 2, if they exceed 36. Annex the number thus obtained to the pounds in such a way that the units of the farthings shall occupy the third place from the pounds, and divide by 3, as before, and the quotient will be the result in cents. NOTE. A demonstration of this rule will be found in Sec. XXIX. SECT. XIII.] UNITED STATES MONE 1. Reduce 162<. 2. Change 319=. 3. Change 176<, 4. Reduce 315 <. EXAMPLES. to United States Money 3)162000 $ 540.00 Ans. $ 540. 17s. to United States Money. 3)319850 $ 1066:i6 Ans. $ 1066.16f . 17s. 8Jd. to United States Money. 176850 36 3)176886 f589^62 Ans. $589.62. to U. S. Money. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. SECTION XIII. UNITED STATES MONEY. THE denominations of Federal or United States Money being in the ratio of 10, 100, and 1000 to each other, operations in- 619 =. to cc 166 . to cc 318. to cc 101 . to cc 144 <. to cc 161 <. 18s. to cc 361^. 17s. to cc 99 . 11s. to cc 100 <. 9s. to cc 661=. 7s. to cc 47 rf. Us. to (C 109 <. Is. to cc 16 . 17s. 6Jd. to cc 69 . Is. 3Jd. to cc 87,. 16s. lid. to cc 14 . 7s. 7^d. to cc 73 . 3s. 4d. to cc 47^. 12s. lOd. to cc 187 . 5s. Ofd. to cc 10<. Os. 3|d. to cc Ans. $ 1050.00 cc 2063.331 cc 553.33 cc 1060.00 cc 336. 66f cc 480.00 cc 539.66| cc 1206. 16 cc 331.83 cc 334.83 cc 2204.50 cc 158.50 cc 363.50 cc 56.25 cc 230.2 1 cc 292.82 cc 47.93 cc 243.89 1 158.80f cc 624.17| cc 33.38 68 UNITED STATES MONEY. [SECT. xm. volving dollars, cents, and mills are performed, when the num- bers have been properly set down, as under the rules for sim- ple numbers. ADDITION. RULE. Write dollars under dollars, cents under cents, and mills under mills, and then proceed as in Simple Addition, and tfie result will be obtained in the several denominations added. NOTE. In all operations of United States Money, it must "be borne in mind that a cent is one hundredth of a dollar, and hence, in arranging a column of cents or annexing any number of cents to dollars, 1 cent, 2 cents, &c., must be written .01, .02, &c., which denote one hundredth, two hundredth^, &c. EXAMPLES. I. 2. 3. 4. ft els. m. ft cts. m. ft cts. m. ft cts. m. 375.87,5 78.19,3 171.01,3 861.07,3 671.12.7 18.01,4 382.09,4 516.71,6 387.14,3 91.03,8 999.90,0 344.67,3 184.18,9 16.81,7 155.06,8 617.81,4 147.75.8 81.47,6 48.15,3 169.97,3 63.07,2 43.18,4 49.61,9 810.42,6 1829.16,4 5. Add the following sums, $18.16,5, $701.63, $ 151.16,1, $ 375.08,9, and $ 471.01,7. Ans. $ 1717.06,2. 6. Bought a horse for eighty-seven dollars nine cents, a pair of oxen for sixty-five dollars twenty cents, and six gallons of molasses for two dollars six cents five mills ; what was the amount of my bill ? Ans. $ 154.35,5. 7. Sold a calf for three dollars eight cents, a bushel of corn for ninety-seven cents five mills, and three bushels of rye for three dollars five cents ; what was the amount received ? Ans. $7.10,5. SUBTRACTION. RULE. Write the several denominations of the subtrahend under the corre- sponding ones of the minuend, and then proceed as in Simple Subtraction, and the result will be the difference in the several denominations sub- tracted. SECT, xiii.] MULTIPLICATION. 69 EXAMPLES. 1. 2. 3. 4. $ cts. m. $ eta. m. 9 eta. m. 8 cts. m 871.16,1 478.47,7 167.16,3 163.16,7 89.91,8 199.99,1 98.09,7 9.09,8 781.24,3 5. Bought a farm for $ 1728.90, and sold it for $ 3786.98, what did f gain by my bargain? Ans. $2058.08. 6. Gave $ 79.25 for a horse, and $ 106.87,5 for a chaise, and sold them both for $ 200 ; what did I gain ? Ans. $ 13.87,5. 7. Bought a farm for $ 8967, and sold it for nine thousand eight hundred seventy-six dollars seventy-five cents ; what did I gain? Ans. $909.75. 8. Bought a barrel of flour for $ 7.50, three bushels of rye for $ 2.75, three cords of wood at $ 5.25 a cord ; I sold the flour for $ 6.18, the rye for $ 3.00, and the wood for $ 6.75 a cord ; what was gained by the bargain ? Ans. $ 3.43. 9. A young lady went a " shopping." Her father gave her a twenty-dollar bill. She purchased a dress for $ 8.16, a muff for $ 3.19, a pair of gloves for $ 1.12, a pair of shoes for $ 1.90, a fan for $0.19, and a bonnet for $ 3.08 ; how much money did she return to her father ? Ans. $ 2.36. MULTIPLICATION. RULE. With the dollars, cents, (Sf-c.for the multiplicand, proceed as in Simpk Multiplication, and the result will be the product in the terms of. the lowest denomination contained in the multiplicand. If the multipli- cand consists of dollars only, the product will be dollars ; if there are cents, either with or without dollars, the product will be cents, and the two right hand figures must be separated by the appropriate point. If there are mills, the product will be mills, and tfie three right hand figures must be pointed off. The figures on the left of t/ie point will denote dollars, the next two following it will denote cents t and the third mills. EXAMPLES. 1. What will 365 barrels of Genesee flour cost, at $ 5.75 a barrel ? $5.75 365 2875 3450 1725 $2098.75 Ans. 70 UNITED STATES MONEY. [SECT. xin. 2. What will 128 pounds of sugar cost, at 13 cents ~7 mills a pound ? $.13,7 128 1096 274 137 $ 17.53,6 Ans. 3. What will 126 pounds of butter cost, at 13 cents a pound ? Ans. $16.38. 4. What will 63 pounds of tea cost, at 93 cents a pound ? Ans. jj 58.59. 5. What will 43 tons of hay cost, at 13 dollars 75 cents a ton? Ans. $591.25. 6. If 1 pound of pork is worth 7 cents 3 mills, what are 46 pounds worth ? Ans. $ 3.35,8. 7. If Icwt. of beef cost 3 dollars 28 cents, what are 76cwt. worth ? Ans. $ 249.28. 8. What will 96,000 feet of boards cost, at 11 dollars 67 cents a thousand ? Ans. $ 1120.32. 9. If a barrel of cider be sold for 2 dollars 12 cents, what will be the value of 169 barrels ? Ans. $358.28. 10. What will be the value of a hogshead of wine, contain- ing 63gals., at 1 dollar 63 cents a gallon ? Ans. $ 102.69. 11. Sold a sack of hops, weighing 396 pounds, at 11 cents 3 mills a pound ; to what did it amount ? Ans. $ 44.74,8. 12. Sold 19 cords of wood, at 5 dollars 75 cents a cord ; to what did it amount ? Ans. $ 109.25. 13. Sold 169 tons of timber, at 4 dollars 68 cents a ton ; what did I receive ? Ans. $ 790.92. 14. Sold a hogshead of sugar, weighing 465 pounds ; to what did it amount, at 14 cents a pound ? Ans. $ 65.10. 15. What will be the amount of 789 pounds of leather, at 18 cents a pound ? Ans. $ 142.02. 16. What will be the expense of 846 pounds of sheet lead, at 5 cents 7 mills a pound ? Ans. $ 48.22,2. 17. When potash is sold for 132 dollars 55 cents a ton, what will be the price of 369 tons ? Ans. $ 48910.95. 18. What will 365 pounds of beeswax cost, at 18 cents 4 mills a pound ? Ans. $ 67.16. 19. If 1 pound of tallow cost 7 cents 3 mills, what are 968 pounds worth? Ans. $70.66,4. SECT, mi.] MULTIPLICATION. 71 20. What will a chest of souchong tea be worth, containing 69 pounds, at 29 cents 9 mills a pound ? Ans. 820.63,1. 21. If 1 drum of figs cost 2 dollars 75 cents, what will be the price of 79 drums ? Ans. $ 217.25. 22. If 1 box of oranges cost 6 dollars 71 cents, what will be the price of 169 boxes ? Ans. 8 1133.99. 23. Purchased 796 pounds of cocoa, at 11 cents 4 mills a pound ; what did I have to pay ? Ans. 8 90.74,4. 24. A farmer sold 691 bushels of wheat, at 1 dollar 25 cents a bushel ; what did he receive for it ? Ans. 8 863.75. 25. What are 97 pounds of madder worth, at 17 cents 6 mills a pound ? Ans. $ 17.07,2. 26. A merchant sold 73 hogsheads of molasses, each con- taining 63 gallons, for 44 cents a gallon ; how much money did he receive ? Ans. 8 2023.56. 27. A drover has 169 sheep, which he values at 2 dollars 69 cents a head ; what is the value of the whole drove ? Ans. 8454.61. 28. A farm containing 144 acres is valued at 69 dollars 74 cents 8 mills an acre ; what is the amount of the whole ? Ans. $ 10043.71,2. 29. An auctioneer sold 48 bags of cotton, each containing 397 pounds, at 13 cents 7 mills a pound ; what is the value of the whole ? Ans. $2610.67,2. 30. If 1 yard of broadcloth cost 5 dollars 67 cents, what will be the value of 48 yards ? Ans. 8 272.16. 31. A wool-grower has 179 sheep, each producing 4 pounds of wool ; what will be its value, at 59 cents 3 mills a pound ? Ans. 8424.58,8. 32. A house having 17 rooms requires 6 rolls of paper for each room ; now, if each roll cost 1 dollar 17 cents, what will be the expense for all the rooms ? Ans. $ 119.34. 33. What will 89 yards of brown sheeting cost, at 17 cents a yard ? Ans. 815.13. 34. What will 47,000 of shingles cost, at 3 dollars 75 cents a thousand ? Ans. 8 176.25. 35. Bought 47 hogsheads of salt, each containing 7 bushels, for 1 dollar 12 cents a bushel ; what did it cost ? Ans. 8368.48. 36. What will a ton of hay cost, at 1 dollar 17 cents a hun- dred weight ? Ans. 8 23.40. 37. If 1 foot of wood cost 63 cents, what will 39 cords cost ? Ans. $ 196.56. 72 UNITED STATES MONEY. [SECT. mi. 38. What will 163 buckets cost, at 1 dollar 21 cents a bucket ? Ans. $ 197.23. 39. What will 78 barrels of shad cost, at 3 dollars 89 cents a barrel? Ans. $303.42. 40. If 1 pound of salmon cost 17 cents 5 mills, what will 789 pounds cost ? Ans. $ 138.07,5. 41. Bought 163 grindstones, at 6 dollars 79 cents each ; what did the whole cost ? Ans. 1106.77. 42. Sold 49 green hides, at 1 dollar 95 cents each ; what did they all amount to ? Ans. 895.55. 43. If a man's wages be 1 dollar 19 cents a day, what are they for a year ? Ans. $ 434.35. 44. Hired a horse and chaise to go a journey of 146 miles, at 16 cents a mile ; what did it cost ? Ans. $ 23.36. 45. If it be worth 3 dollars 68 cents to plough one acre of land, what would it be worth to plough 79 acres ? Ans. $290.72. 46. What will 148 tons of plaster of Paris cost, at 2 dollars 28 cents a ton ? Ans. 8 337.44. 47. Bought 79 tons of logwood, at 49 dollars 75 cents a ton ; what did it cost ? Ans. 8 3930.25. 48. Bought 5 gross bottles of castor oil, at 37 cents a bottle ; what did it cost ? Ans. $ 266.40. 49. Sold 19 dozen pair of men's gloves, at 47 cents a pair ; to what did they amount ? Ans. 8 107.16. 50. Bought a hogshead of wine for 97 cents a gallon, and sold it for 1 dollar 75 cents a gallon ; what did I gain ? Ans. 849.14. 51. Bought 75 barrels of flour, at 5 dollars 75 cents a barrel, and sold it at 6 dollars 37 cents ; what did I gain ? Ans. $46.50. 52. Bought 17 score of penknives, at 17 cents each ; what did they all cost ? Ans. $ 57.80. 53. What will 17 tons of coal cost, at $ 9.62 per ton ? Ans. 8 168.35. 54. What will 19 barrels of cider cost, at 8 1.37 per barrel ? Ans. 826.12,5. DIVISION. RULE. With the sum given for the dividend, proceed as in Simple Division, and the result will be the quotient in the lowest denomination contained in the dividend. SECT, xin.] DIVISION. 73 The rule for pointing off cents and mills is the same as in Multiplication. If the dividend consist of dollars only, and be either smaller than tJie divisor, or not divisible by it without -a remainder, annex two or three ciphers, as the case may require, and tJie quotient will be cents or mills accordingly. 1. If 97 bushels of wheat cost $ 147.82,8, what is the value of one bushel ? Ans. $ 1.52,4. OPERATIpN. 97) 147.82,8($ 1.52,4 97_ 508 485 232 194 388 388 2. Bought 1789 acres of land for $ 1699.55 ; what cost one acre? Ans. $0.95. 3. A trader sold 425 pounds of sugar for $ 51.00 ; what was the cost of one pound ? Ans. $ 0. 12. 4. When rye is sold at the rate of 628 bushels for $ 471.00, what is that a bushel ? Ans. $ 0.75. 5. A merchant bought 329 yards of broadcloth for $ 904.75 ; what cost one yard ? Ans. $ 2.75. 6. When a chest of tea containing 42 pounds can be bought for $ 31.50, what cost one pound ? Ans. $ 0.75. 7. If it cost $1460 to support a family 365 days, what would be the expense per day ? Ans. $ 4.00. 8. A shoe-dealer sold 125 cases of shoes for $ 2500 ; what was the cost per case ? Ans. $ 20.00. 9. A flour-merchant sold 475 barrels of flour for $2018.75; what cost one barrel ? Ans. $ 4.25. 10. Bought 42 barrels of pears for $ 73.50 ; what cost one barrel ? Ans. $ 1.75. 11. If 1624 pounds of pork cost $97.44, what cost one pound ? Ans. $ 0.06. 12. If 47000 shingles cost $ 176.25, what is the cost per thousand ? Ans. $ 3.75. 13. Bought 148 tons of plaster of Paris for $ 337.44 ; what was it per ton ? Ans. $ 2.28. 14. If 78 barrels of fish cost $ 303.42, what will one barrel cost? Ans. $3.89. 7 74 UNITED STATES MONEY. [SECT. xin. 15. A farmer sold 691 bushels of wheat for $ 863.75 ; what was it per bushel ? Ans. $ 1.25. 16. If a man earn $ 434.35 in a year, what is that per day ? Ans. $1.19. 17. Sold 169 tons of timber for $ 790.92 ; what cost one ton ? Ans. $ 4.68. 18. What cost one pound of leather, if 789 pounds cost $142.02? Ans. $0.18. 19. If 369 tons of potash cost $48910.95, what will be the price of one ton ? Ans. $ 132.55. 20. Bought 47 hogsheads of salt, each hogshead containing 7 bushels, for $ 368.48 ; what cost one bushel ? Ans. $ 1.12. 21. If 19 cords of wood cost $ 106.97, what cost one cord ? Ans. $5.63. 22. When 19 bushels of salt can be bought for $ 30.87,5, what cost one bushel? Ans. $ 1.62,5. 23. If 17 chests of souchong tea, each weighing 59 pounds, cost $ 672.01, what cost one pound? Ans. $ 0.67. 24. Sold 73 tons of timber for $ 414.64 ; what did I receive per ton ? Ans. $ 5.68. 25. Bought oil at the rate of 144 gallons for $ 234.00 ; what did I give per gallon ? Ans. $1.62,5. 26. A landholder sold 47 acres of land for $ 1774.25 ; what did he receive per acre ? Ans. $ 37.75 27. What is the price of one yard of broadcloth, if 163 yards cost $1106.77? Ans. $6.79. 28. If a farm, containing 144 acres, is valued at $ 10043.71,2, what is one acre worth ? Ans. $ 69.74,8. BILLS. 1. Boston, July 4, 1835. Mr. James Dow, Bought of Dennis Sharp, 17 yds. Flannel, at .45 cts. 19 " Shalloon, " .37 " 16 " Blue Camlet, " .46 " 13 " Silk Vesting, " .87 " 9 " Cambric Muslin, " .63 " 25 " Bombazine, " .56 " 17 " Ticking, " .31 " 19 " Striped Jean, .16 " ~~$61.33 Received payment, Dennis Sharp. SECT. XIII.] 2. Mr. Samuel Smith, 13 Ibs. Tea, 16 " Coffee, Sugar, BILLS. 75 36 47 12 7 13 " Cheese, " Pepper, " Ginger, Haverhill, May 5, 1835. Bought of David Johnson, .98 cts. .15 " .13 " .09 " .19 " " Chocolate, Received payment, at u it u tl u II .17 " .61 " $35.45. David Johnson. 3. Mr. John Dow, 17 yds. Broadcloth, Salem, February 29, 1835. Bought of Richard Fuller, 29 60 49 18 27 75 36 49 Cassimere, Bleached Shirting, Ticking, Blue Cloth, . Habit do. Flannel, Plaid Prints, Brown Sheeting, Received payment, $5.25 1.62 .17 .27 3.19 2.75 .61 .75 .18 "8372.90. Richard Fuller. 4. Mr. John Rilley, 10 pair Boots, 19 " Shoes, 83 Hose, 47 Ibs. Ginger, 91 " Chocolate, 47 " Pepper, 68 " Flour, 27 pair Gloves, Received payment, Baltimore, January 20, 1835. Bought of James Somes, at $2.75 " 1.25 " 1.29 " .17 " .39 " .23 .13 " 1.39 ~$ 258.98. James Somes. UNITED STATES MONEY. [SECT. xin. 5. Philadelphia, June 11, 1835. Mr. Moses Thomas, Bought of Luke Dow, 27 National Spelling-Books, at $ 0. 19 25 Parker's Composition, .27 17 National Arithmetics, .75 9 Greek Lexicons, 3.75 8 Ainsworth's Dictionaries, " 4.50 27 Greek Readers, " 2.25 18 Folio Bibles, " 9.87 75 Leverett's Caesar, " .31 67 Fisk's Greek Grammar, .75 15 Folsom's Cicero's Orations, " 1.12 $423.09. Received payment, Luke Dow, by Timothy True. 6. Boston, June 26, 1835. Dr. Enoch Cross, Bought of Maynard & Noyes, 14 oz. Ipecacuanha, at $ 0.67 23 " Laudanum, " .89 17 " Emetic Tartar, " 1.25 25 " Cantharides, " 2.17 27 Gum Mastic, " .61 56 " Gum Camphor, 8 136.94. Received payment, Maynard & Noyes, by Timothy Jones. 7. Newburyport, June 5, 1835. Mr. John Somes, Bought of Samuel Gridley, 7 yds. Broadcloth, at $4.50 16flbs. Coffee, - " .16 18J " Candles, " .25 30 " Soap, " .17 3 " Pepper, " .19 Ginger, " .18 " Received payment, Samuel Gridley. SECT. XIII.] BILLS. 77 8. Mr. Benjamin Treat, 37 Chests Green Tea, 41 " Black do. 40 " Chests of Imperial Tea, 13 Crates Liverpool Ware, Received payment, Boston, May 1, 1835. Bought of John True, at $ 25.50 16.17 97.75 169.37 $7718.28. John True. 9. Mr. John Cummings, 97 bbl. Genesee Flour, 167 " Philadelphia do. 87 " Baltimore do. 196 " Richmond do. 275 Howard St do. 69 bu. Rye, 136 " Virginia Corn, 68 " North River do. 169 Wheat, 76TonLehighCoal, 89 " Iron, 49 Grindstones, 39 Pitchforks, 197 Rakes, 86 Hoes, 78 Shovels, 187 Spades, 91 Ploughs, 83 Harrows, 47 Handsaws, 35 Millsaws, 47 cwt. Steel, 57 " Lead, Received payment, New York, July 11, 1835. Bought of Lord & Secomb. at $6.25 5.95 6.07 5.75 7.25 1.16 .67 .76 1.37 9.67 69.70 3.47 1.61 .17 .69 1.17 .85 11.61 17.15 3.16 18.15 9.47 6.83 17315.32. Lord & Secomb. 78 COMPOUND MULTIPLICATION. [SECT. xiv. SECTION XIV. COMPOUND MULTIPLICATION. CASE I. COMPOUND MULTIPLICATION consists in multiplying numbers of different denominations by simple numbers. I. What will 6 bales of cloth cost, at 7. 12s. 7d. per bale ? 8 d In this question, we multiply 7d. by 6, and find 7 12 7 ^e product to be 42d. This we divide by 12, the g number of pence in a shilling, and find it contains JJT TjT 3s. and 6d. We write the 6d. under the pence, and carry 3 to the product of 6 times 12, and find the amount to be 75s., which we reduce to pounds by dividing them by 20, and find them to be 3. 15s. We write down the shillings under the shillings, and carry 3 to the product of 6 times 7<. ; and we thus find the answer to be 45. 15s. 6d. From the above illustration we deduce the following RULE. When the multiplier is less than 12, multiply by the multiplier and carry as in Compound Addition. NOTE. For the answers in Multiplication, see Section XV., in Di- vision. 2. What cost 9yds. of cloth, at 1 . 3s. 8d. per yard ? Ans. 10. 13s. Od. 3. What cost 7bbls. of flour, at !<. 8s. 7d. per barrel ? 4. What cost 81bs. of Cayenne pepper, at 7s. 9d. per Ib. ? 5. Multiply 10yd. 3qr. 3na. by 5. 6. Multiply 3cwt. Iqr. 81b. by 9. 7. Multiply 7T. llcwt. Iqr. 201b. by 5. 8. Multiply 7 days 15h. 35m. 18sec. by 10. 9. Multiply 18<. 16s. 74-d. by 4. Ans. 75<. 6s. 6d. 10. Multiply 15=. 11s. 8|d. by 8. Ans. 124c. 13s. lOd. II. Multiply 27. 19s. lld. by 9. Ans. 251,. 19s. 7d. 12. Multiply 19<. 5s. 7d. by 11. Ans. 212. Is. 7|d. 13. Multiply 81. 14s. 9d. by 8. Ans. 653^. 18s. Od. 14. Multiply 15. 18s. 5d. by 7. Ans. 111. 8s. lid. 15. Multiply 13=. 5s. 4|d. by 12. Ans. 159^. 4s. 9d. 16. Multiply 171b. 7oz. 13dwt. 13gr. by 9. 17. Multiply 151b. lloz. 19dwt. 15gr. by 7. SECT, xiv.] COMPOUND MULTIPLICATION. 79 18. Multiply 16T. 12cwt. 3qr. 131b. 12oz. by 11. 19. Multiply 13T. 3cwt. Iqr. 141b. 13oz. by 8. 20. Multiply 21b 5i 53 19 16gr. by 8. 21. Multiply 47yd. 3qr. 2na. 2in. by 7. 22. Multiply 17m. 7fur. 36rd. 13ft. 7in. by 12. 23. Multiply 16deg. 39m. 3fur. 39rd. 5yd. 2ft. by 9. 24. Multiply 16deg. 20m. 7fur. 12rd. 8ft. llin. IJ-bar. by 6. 25. Multiply 16A. 2R. 4p. 19yd. 7ft. 79in. by 11. 26. Multiply 7 cords 1 16ft. 1629m. by 4. 27. Multiply 29hhd. 61gal. 3qt. Ipt. 3gi. by 7. 28. Multiply 3 tuns 3hhd. 56gal. 2qt. by 9. 29. Multiply 7hhd. 5gal. 2qt. Ipt. by 8. 30. Multiply 19bu. 2pk. 7qt. Ipt. by 6. 31. Multiply 36ch. 18bu. 3pk. 7qt. by 7. 32. Multiply 13y. 316d. 15h. 27m. 39sec. by 8. 33. If a man gives each of his 9 sons 23A. 3R. 19p., what do they all receive ? 34. If 12 men perform a piece of labor in 7h. 24m. 30sec., how long would it take 1 man to perform the same task ? 35. If 1 bag contain 3bu. 2pk. 4qt., what quantity do 8 bags contain ? CASE II. When the multiplier is more than 12, and is a composite num- ber, that is, a number which is the product of two or more numbers, the question is performed as in the following EXAMPLE. 36. What will 42 yards of cloth cost, at 6s. 9d. a yard ? . e. d. In this example, we find that 6 069 multiplied by 7 will produce the 6 quantity 42 yards. We therefore 206 = price of 6 yds. multiply 6s 9d. first by the 6, and y then its product by 7 ; and the last uct, 14. 3s. 6d. is the answer A o 14 3 6 = The pupil will now see the propriety of the following RULE. Multiply by one of the factors of the composite number, and tlte prod- uct thus obtained by the other. 37. What will 16 yards of velvet cost, at 3s. 8d. per yard ? 80 COMPOUND MULTIPLICATION. [SECT. sir. 38. What will 72 yards of broadcloth cost, at 19s. lid. per yard ? 39. What will 84 yards of cotton cost, at Is. lid. per yard ? 40. Bought 90 hogsheads of sugar, each weighing 12cwt. 2qr. 1 lib. ; what was the weight of the whole ? 41. What cost 18 sheep, at 5s. 9d. a piece ? 42. What cost 21 yards of cloth, at 9s. lid. per yard ? 43. What cost 22 hats, at 11s. 6d. each ? 44. If 1 share in a certain stock be valued at 13. 8s. 9Jd., what is the value of 96 shares ? 45. If 1 spoon weigh 3oz. 5dwt 15gr., what is the weight of 120 spoons ? 46. If a man travel 24m. 7fur. 4rd. in 1 day, how far will he go in 1 month ? 47. If the earth revolve 15' per minute, how far per hour ? 48. Multiply 39A. 3R. 17p. 30yd. 8ft. lOOin. by 32. 49. If a man be 2d. 5h. 17m. 19sec. in walking 1 degree, how long would it take him to walk round the earth, allowing days to a year ? CASE III. When the multiplier is such a number as cannot be pro- duced by the product of two or more numbers, we should pro- ceed as in the following EXAMPLE. 50. What is the value of 53 tons of iron, at I&. 17s. lid. a ton? . a. d. . s. d. 18 17 11 18 17 11 _ 5 __ 3 94 9 7 = price of 5 tons. 56 13 9 = price of 3 tons. _ 10 944 15 10= price of 50 tons. Because 53 is a prime 56 13 9 = price of 3 tons, number, that is, it cannot be 1001 9 7 = price of 53 tons. P rod f uced b \ the P rodu ^ of any two numbers ; we there- fore find a convenient composite number less than the given number, viz. 50, which may be produced by multiplying 5 by 10. Having found the price of 50 tons by the last Case, we then find the price of the 3 remaining tons by Case I., and add it to the former, making the value of the whole quantity 9s. 7d. SECT, xiv.] BILLS. 81 The pupil will hence perceive the propriety of the following RULE. Take for successive multipliers two or more numbers, whose continued product will be nearest the proper multiplier, and then find the value of the remainder by Case I., and the sum of the last two products ivitt be the answer. 51. What will 57 gallons of wine cost, at 8s. 3d. per gallon ? 52. Bought 29 lots of wild land, each containing 117A. 3R. 27p. ; what were the contents of the whole ? 53. Bought 89 pieces of cloth, each containing 37yd. 3qr. 2na. 2in. ; what was the whole quantity ? 54. Bought 59 casks of wine, each containing 47gal. 3qt. Ipt. ; what was the whole quantity ? 55. If a man travel 17m. 3fur. 13rd. 14ft. in one day, how far will he travel in a year ? 56. If a man drink 3gal. Iqt. Ipt. of beer in a week, how much will he drink in 52 weeks ? 57. There are 17 sticks of timber, each containing 37ft. 978in. ; what is the whole quantity ? 58. There are 17 piles of wood, each containing 7 cords 98 cubic feet ; what is the whole quantity ? 59. Multiply 2hhd. 19gal. Oqt. Ipt. by 39. 60. Multiply 3bu. Ipk. 4qt. Ipt. Igi. by 53. 61. Multiply 16ch. 7bu. 2pk. Oqt. Opt. by 17. BILLS. 1. London, July 4, 1835. Dow, Vance, & Co., of Boston, U. S., Bought of Samuel Snow, 45 yds. Broadcloth, at 8s. 4d. 50 " " " 10s. 6d. 56 " " " 3s. 7d. 63 " " " 12s. llfd. 72 19s. lid. 81 " " " 9s. 3d. 35 " " " 19s. 7d. 99 " " " 16s. 0d. 66 " " " 8s. lid. 33 " " 16s. ~376<. 7s. Of d. Received payment, Samuel Snow. 82 COMPOUND MULTIPLICATION. [SECT. xiv. 2. Quebec, Jan. 8, 1835. Mr. John Vose, Bought of Vans & Conant, 46 Ivory Combs, at 3s. 5d. 47 Ibs. Colored Thread, " . 6s. 9jd. 51 yds. Durant, " Is. 8d. 52 Silk Vests, " 6s. 7d. 53 Leghorns, " 11s. 9^d. 57 ps. Nankin, " 8s. 58 Ibs. White Thread, " 9s. ~~128. 16s. Received payment, Vans & Conant. 3. Montreal, July 4, 1835. Mr. James Savage, Bought of Joseph Dowe, 83 gals. Lisbon Wine, at 6s. 7d. 85 " Port do. " 3s. 9d. 86 " Madeira do. " 4s. lld. 87 " Temperance do. " 3s. 6d. 89 " Oil, " 5s. 3d. 91 Leghorns, " 19s. 10d. 92 Ibs. Green Tea, " 3s. ld. $3 pair Thread Hose, " 4s. 4d. 94 " Silk Gloves, " 3s. 3d. 95 " Silk Hose, " 6s. 6d. 97 yds. Linen, " 5s. 5Jd. 98 gals. Winter Strained Oil, 7s. 7d. ^33a. 19s. Received payment, Joseph Dowe. 4. Montreal, June 17, 1835. Mr. Samuel Simpson, Bought of Lackington, Grey, & Co. 19yds. Cloth, at Is. 6d. 23 " Worsted, " 7s. 8d. 26 " Baize, 3s. ll^d. 29 Camlet, 6s. 10|d. 31 " Bombazine, " Is. 5d. 34 " Linen, " 3s. 7d. 37 " Cotton, " 11s. 9d. 38 " Flannel, " 6s. lid. 39 " Calico, " 3s. lO^d. 41 " Broadcloth, " 6s. 9^d. 43 " Nankin, " 7s. 5fd. ~106c. Is. Received payment, Lackington, Grey, & Co. SECT, xiv.] BILLS. 83 5. Liverpool, June 2, 1835. John Jones, of Philadelphia, U. S., Bought of Thomas Hasseltine, 297 yds. Black Broadcloth, at 17s. 3d. 473 " Blue do. " 9s. lld. 512 " Red do. " 15s. lOd. 624 " Green do. " 12s. 8d. 765 " White do. " 19s. 9|d. 169 " Black Velvet, tc 13s. 5d. 698 " Green do. " 15s. 6fd. 315 " Red do. " 14s. 3d. 713 " White do. " 11s. 7d. 519 " Carpet, " 13s. 6d. 147 " Black Kerseymere, " 16s. 7fd. 386 " Blue do. " 14s. 3d. 137 " Green do. " 19s. 9d. 999 " Black Silk, " 15s. 8d. ~~5012<. Os. lljd. Received payment, Thomas Hasseltine. 6. London, May 11, 1846. Messrs. Kimball, Jewett, & Co., of Boston, U. S., Bought of Benjamin Fowler, . s. d. 2345 yds. Red Broadcloth, at l. 17s. 9d. = 4428 12 7 7186 " Green do. " 3. 15s. 8d. = 27202 1 8011 " Black do. " 2. 18s. 10d. = 23574 8| 6789 " Blue do. " l. 6s. 9d. = 9080 5 9 3178 " White do. " 2. Is. 7fd. = 6617 10 5^ 2365 " Pongee Silk, l. 2s. S^d. = 2685 5 2^ 5107 " Black do. 4< l. 7s. 5|d. = 7006 3 3| 4444 bales Cotton Cloth, 3. 16s. 8^d. = 17039 19 3 7777 there will be T V as many shil- lings as pence ; therefore T V of f of a penny is 3 = -fa of a shilling. Again, as 20 shillings make a pound, there will be ^V as many pounds as shillings ; there- fore ^ of <% of a shilling is ^^ of a pound. Q. E. D. This question may be abridged. Thus, f X A X A = Tinnr = *iv Ans - RULE. Multiply the denominator of the given fraction by all tJte denominations between it and tlie one to which it is to be reduced, and over the product write the given numerator. 2. Reduce f of a farthing to the fraction of a pound. 3. Reduce f of a grain troy to the fraction of a pound. 106 VULGAR FRACTIONS. [SECT. XTI. 4. Reduce of a scruple to the fraction of a pound. 5. Reduce T 6 T of an ounce to the fraction of a hundred weight. 6. Reduce of a pound to the fraction of a ton. 7. Reduce of an inch to the fraction of an ell English. 8. Reduce f of an inch to the fraction of a mile. 9. Reduce of a barleycorn to the fraction of a league. 10. Reduce of an inch to the fraction of an acre. 11. Reduce of a quart to the fraction of a tun, wine meas- ure. 12. Reduce f of a pint to the fraction of a bushel. 13. Reduce of a minute to the fraction of a year (365J days). CASE X. To reduce fractions of a higher denomination to a lower. 1. What part of a penny is $%-$ of a pound ? OPERATION. We explain this question in cn X = &TF = i&s. the followin g manner. As i v j_2 - J.2 - 3r\ A no shillings are twentieths of a w X - |d. Ans. * . as many parts of a shilling in ^^ of a pound, as there are parts of a pound ; therefore, ^^ of a pound is equal to ^^ of ^ = -^j = 1^7 of a shilling. And as pence are twelfths of a shilling, there will be twelve times as many parts of a penny in -2 of a shilling, as there are parts of a shilling ; therefore, ^2- of a shilling is equal to -$ of -^ = f = of a penny, Ans. The operation of this question may be facilitated in the fol- lowing manner. OPERATION. X - 2 T - X V = f f g = f d. Ans. RULE. Let the given numerator be multiplied by all' the denomi- nations between it and tJie one to which it is to be reduced ; then place the product over the given denominator and reduce the fraction to its lowest terms. 2. Reduce 12 T 00 of a pound to the fraction of a farthing. 3. Reduce 96 * 00 of a pound troy to the fraction of a grain. 4. Reduce 7^7 of a pound, apothecaries' weight, to the fraction of a scruple. 5. Reduce Tv \ v of a cwt. to the fraction of an ounce. 6. Reduce p^jpj f a ton to tne fraction of a pound. SECT, xvi.] VULGAR FRACTIONS. 107 7. Reduce 3^-5- of an ell English to the fraction of an inch. 8. Reduce Tr^w of a mile to the fraction of an inch. 9. Reduce TTT^FU f a league to the fraction of a barley- corn. 10. Reduce srtr/irsvv f an acre to tne fraction of an inch. 11. Reduce yrVs f a tun f wine measure to the fraction of a quart. 12. Reduce ^f -$ of a bushel to the fraction of a pint. 13. Reduce i^rfirsTr f a y ear to tne fraction of a minute. CASE XI. To find the value of a fraction in the known parts of the integer. 1. What is the value of T 3 T of a <. ? By Case IX. &. = ^.X^-= f ?s. = 5 T 5 T s. ; and T \s. =. r 5 T X V 2 = d. = 5&d. ; and T * T d. - T 5 T X * - ftqr. = This question may be analyzed thus: If l. is 20s., T 3 T of a . is T 3 T of 20s. = 5 T 5 T s. ; and if Is. is 12d., T 5 T of a shil- ling is T 5 T of 12d. = 5 T 5 T d. ; and if Id. is 4qr., T 5 T of a penny is T 5 T of 4qr. = 1-^j-qr., Answer, as before. Or, it may be performed in the following manner. OPERATION. In this operation, it will be per- ceived that we multiply the numer- 11)60 (5s. ator f tne fraction by the successive 55 lower denominations, beginning with g the highest, and divide each product j 2 by the denominator. ll)"60(5d. j>5 5 _4 ll)20(l T 9 T qr. From these illustrations, we derive the following 9 RULE. Multiply the numerator by the next lower denomination of the integer, and divide tJte product by the denominator ; if any thing remain, multiply it by the next less denomination, and divide as before, and so continue as far as may be required; and the several quotients will be the answer. 108 VULGAR FRACTIONS. [SECT. xvi. 2. What is the value of ^ of a shilling ? Ans. 3d. 3. What is the value of $ of a guinea, at 28 shillings ? Ans. 21s. 9d. lqr. 4. What is the value of / T of a cwt. ? Ans. 2qr. 151b. 4oz. 5 T 9 T dr. 5. What is the value of f of a Ib. avoirdupois ? Ans. 7oz. l|dr. 6. What is the value of f of a Ib. troy ? Ans. lOoz. 13dwt. 8gr. 7. What is the value of ^ of a Ib. apothecaries' weight. Ans. 3 53 19 8. What is the value of ^ of a yard ? Ans. 2qr. Ona. 1 9. What is the value of of an ell English ? Ans. 2qr. 3na. 10. What is the value of | of a mile ? Ans. 6fur. 30rd. 12ft. 11. What is the value of f of a furlong ? Ans. 35rd. 9ft. 2in. 12. What is the value of fa of an acre ? Ans. 2R. 6rd. 4yd. 5ft. 127^in. 13. What is the value of T 9 T of a rod ? Ans. 144ft. 19 T yin. 14. What is the value of -^ of a cord ? Ans. 9ft. 1462fin. 15. What is the value of & of a hhd. of wine ? Ans. 6gal. 2qt. Ipt. 0-^gi. 16. What is the value of of a hhd. of beer ? Ans. 42gal. 17. What is the value of of a year (365 days) ? Ans. 174d. 16h. 26m. 5/ 3 sec. 18. What is the value of 7^ o f a dollar ? Ans.f7.74AV CASE XII. To reduce any mixed quantity of weights, measures, &c., to the fraction of the integer. 1. What part of a shilling is Id. ? is 2d. ? is 3d. ? is 4d. ? 2. What part of a pound is 2s. ? 3s. ? 4s. ? 5s. ? 6s. ? 9s. ? 3. What part of a furlong is 3rd. ? 4rd. ? 5rd. ? 8rd. ? 9rd. ? 4. What part of a hogshead is 5gal. ? 8gal. ? lOgal. ? 5. What part of a foot is 2 inches ? 3in. ? 4in. ? 5in. ? 8in. ? SECT, xvi.] VULGAR FRACTIONS. 109 6. What part of a . is 5s. 5d. l T 9 T qr. ? s. d. qr. 20 5 5 1-^f In this question, the shillings, 12 12 pence, farthings, &c. are reduced 240 65 to elevenths of farthings for the nu- 4 4 merator of a fraction. A pound is 2g} also reduced to the same denomi- nation, for a denominator. The fraction is then reduced to its lowest 10560 2880 termg- RULE. Reduce the given number to the lowest denomination it con- tains for a numerator ; and then reduce the integer to the same denom- ination, for the denominator of the fraction required. 7. Reduce 3d. to the fraction of a shilling. Ans. / . 8. Reduce 21s. 9d. 1^-qr. to the fraction of a guinea. Ans. J. 9. Reduce 2qr. 151b. 4oz. 5 T 9 T dr. to the fraction of a cwt. Ans. -j7 T . 10. What part of a pound are 7oz. ldr. ? Ans. . 11. What part of a pound troy are lOoz. 13d wt. 8gr. ? Ans. f . 12. What part of a pound apothecaries' weight are 3 55 r. ? 19 12 T ^gr. ? Ans. T V 13. What part of a yard are 2qr. Ona. 1-^in. ? Ans. T 7 jj-. 14. What part of an ell English are 2qr. 3na. 0|in. ? Ans. f . 15. What part of a mile are 6fur. 30rd. 12ft. Sin. 0|br. ? Ans. H- 16. Reduce 35rd. 9ft. 2in. to the fraction of a furlong. Ans. f . 17. What part of an acre are 2R. 6rd. 4yd. 5ft. 127 T Vn. ? Ans. A. 18. What part of a square rod are 144ft. 19^^. ? Ans. ^-. 19. What part of a cord are 9ft. 1462f^in. ? Ans. &. 20. What part of a hogshead of wine are 6gal. 2qt. Ipt. T \gi. ? Ans. T \. 21. What part of a hhd. of beer are 42gal. ? Ans. f. 22. What part of a year (365 da.) are 174d. 16h.26m. sec. ? Ans. % 10 O VULGAR FRACTIONS. [SECT. XTII. SECTION XVII. ADDITION OF VULGAR FRACTIONS. CASE I. To add fractions that have a common denominator. 1. What part of an apple is and ? and ? f and ? 2. What part of a dollar is | and f ? T 2 ^ and & ? ^ and T ^ ? 3. What part of a shilling is T 3 j and -r ? T V> T V, and & ? 4. What part of an orange is | and f ? and ? f and f ? 5. Add T 3 z, ^, A. and ri together. OPERATION. 7 ll = = = 2 Ans. In this question, we add the numerators, and divide their sum by the denominator. RULE. Write the sum of the numerators over the common denom- inator, and reduce the fraction if necessary. 6. Add TV, TV, TV, i If, and ff together. Ans. 3ff . 7. Add 7 V &> H it. and if together. Ans. 2JJ. 8. Add ^, |f, |f, |f, and Jf together. Ans. 2/ T . 9. Add T%, T V 9 T, 1%, and l together. Ans. If. 10. Add fH, |Sf , ^Jf , and ^ T together. Ans. 2||f . 11. Add fff , f |f , fff , and f |f together. Ans. 3?f f. 12. Add |ff|, ff, and |f together. Acs. 13. Add !f|, Jf, and f f J together. Ans. 14. Add T^ftfr, ^%, and f^ together. Ans. CASE II. To add fractions that have not a common denominator. 1. What is the sum of , T ^-, f i and $ ? First Method. OPERATION. 4)8 12 16 20 4x2x3x2x5 = 240 common denominator. 2)2 3 4 5 8 1325 12 16 20 30X 7 = 210 20X 5=100 = 165 12X13 = 156 Having found a common de- 631 nominator by Case VIII. , we 240 ~ ^A proceed as in the last Case. Ans. SECT. XVII.] VULGAR FRACTIONS. Ill Second Method. OPERATION. 7 X 12 X 16 X 20 = 26880 5x 8X 16X20=12800 11 X 8X 12x20 21120 13 X 8 X 12 X 16 = 19968 80768 8 X 12 X 16 X 20 = 30720 Let the pupil exam- ine the second method of reducing fractions to a common denomi- nator in Case VIII., Sec. XVI. RULE. Reduce mixed numbers to improper fractions, and compound fractions to simple fractions ; then reduce all the fractions to a common denominator, and the sum of their numerators written over the common denominator will be the answer required. 2. Add , , fa and | together. Ans. 2*-f . 3. Add T 7 T , A, I, and together. Ans. l|f. 4. Add T 3 , T * F , fa and J together. Ans. 2Jg$. 5. Add T 5 T , -jj^, y 1 ^-, and ^ together. Ans. 1. 6. Add T f , ft , J$, and T 9 ; together. Ans. 3$. 7. Add , , , , , and | together. - Ans. 1 T 8 3 . 8. Add f, f , and 5f together. Ans. 6f & . 9. Add ^-, /2-, and 9 T 3 T together. Ans. 9f . 10. Add f , f ^ and 4^ together. Ans. 6^. 11. Add f, 7^, and 8J together. Ans. 12. Add J, 3|, and 5f together. Ans. 13. Add 6f , 7f , and 4f together. Ans. 18f . NOTE. If the quantity be a mixed number, -the better way is to add their fractional parts separately, as in the following example. 14. What is the sum of llf , 15|, 12yV, and 17f ? 4)4 8 12 6 3)1 2 3 6 2)1 2 1 2 1111 OPERATION. 4x3x2 = 2 llf Ans. = 57^ 4 6 X 3 = 18 3X7 = 21 2 X 5 = 10 4 X 5 = 20 69 24 15. What is the sum of llf, 19^, and 23f ? Ans. 16. What is the sum of 18|, 27-^, and 49 J ? Ans." 96. 17. What is the sum of 21f, 18f, and 26 ? Ans. 66 T |. 1-4 VULGAR FRACTIONS. [SECT. xvn. 18. What is the sum of 17f, 14, and 13f ? Ans. 45 T B . 19. What is the sum of 16f , 8J, 9f , 3, and If ? Ans. 40 T V 20. What is the sum of 371 ^, 614^, and 81| ? Ans. 1068f. 21. Add f of 18 T 3 r , and j of f of 6 T 3 T together. Ans. 12ff-}. 22. Add | of 18, and ^ of J4- of 7^ together. Ans. 13%. 23. Add 4 of 15, and f of 107f together. Ans. 93f 24. Add of f of 28 to a. Ans. 6 T 9 ^. 25. Add*,2| f --, and together. Ans. 8jH|H- CASE III. To add any two fractions, whose numerators are a unit. RULE. Place the sum of the denominators over their product. EXAMPLE. 1. Additof 4 + 5= 9 NOTK. The truth of this rule is evident from the fact, that this pro- cess reduces the fractions to a common denominator, and then adds the numerators. If the numerators of the given fractions be alike, and more than a unit, multiply the sum of the denominators by one of the numerators for a new numerator, then multiply the denomina- tors together for a new denominator. SECT, xvn.] VULGAR FRACTION 10. Add f to f . 4 + 5 = 9 X3 4 X 5 11. Add fctoM to M to A,f to f, f to f, f to A, ^ to f 12. Add f to T 3 r , f to f , I to f , f to |, f to f , | to A- 13. Add f tO f , f tO T 6 r , f tO -ft, T 6 T tO Ai A tO A T 6 T tO A- 14. Add f tO A, T 8 T tO T 8 T. TT tO A, A tO T 8 5 . A to T 8 T , A to A- 15. Add A tO T 9 T , A ^ T 9 *, T 9 * tO A, T 9 T7 to T 9 T , A * A- 16. Add I- to I , ^ to A> I to T 7 T , I- to A I- to A> f to A- 17. Add f to Ai f to A. f to A> t to A, f to A t to A- 18. Add to ^, if to , to it, if to ^, it to if 19. Add i to ii, i to ii, . to ii, to ii, to i|. NOTE. The preceding rule may be found very useful, because all similar questions may be readily performed mentally. CASE IV. To add compound numbers. 1. Add A of a . to T 9 T of a . Value of A of a . == 10s. 9d. Oiqr. This question is per- Value of T 9 T of a .= 16s. 4d. l T 5 T qr. formed by finding the 1. 7s. Id. 2 T 5 Aqr. values of A of a - and A of a . by Case XL, Sec. XVI. The fractions if and T 5 T are added by Case II. of Addition of Fractions. The following questions are per- formed in the same manner. The above question may be performed by first adding the fractions of the pounds together, and then finding their value by Case XL ; thus : OPERATION. A*- + A& = if^- = !< 7s - ld - *&s A- 3. Take from , , , , |, , | ; T V from T V, T V, iV 4. Take | from |, ,,; T V from , , 1, f , |. 5. Take | from , , ; } from J, , J, |. 6. Take T T from T V, |, -, i, |, i, f, i, ^. 7. Take ^ from ; from , -J ; ^ from . 8. Take T V from , ^, ^, ^, ^, |, 1, i, ^, 1 V. 9. Take T V from & $, ^ |, ^ }, , ^ ^, ^ 10. Take T V from ^, ^, , -i, J, |, 4, . NOTE. If the numerators of the given fractions be alike, and more than a unit, multiply the difference of the denominators by one of the nu- merators for a new numerator, then multiply the denominators together for a new denominator. OPERATION. II. Take f from f. 7 3 = 4;4X2_ 12. Take f from f ; f from f- ; -& from f ; -& from f . 13. Take f from f ; T 4 T from f ; T 4 T from f ; T * T from T V 14. Take % from ; T 5 T from ; T 5 T from f ; ^\- from -fa- 15. Take T 6 T from f ; T 6 T from f ; T 6 7 from f- ; & from T 6 T . 16. Take T \ from ^ ; -^ from f ; T ^ from f ; T 4 ^ from T 4 T . 17. Take T 8 T from | ; T 8 T from f ; -^ from f ; T 8 T from f . 18. Take f T from f ; T 2 T from f ; ^- from f ; T 2 T from f . 19. Take |$ from ^ ; |f from ^ ; ^ from if ; | from tf. 20. Take -^ from J ; V 2 - ^om J T 2 - ; J^ from -'/ ; J- from 4^-. NOTE. The above questions, and those of a similar kind, may readily be performed mentally. 120 VULGAR FRACTIONS. [SECT. xvin. CASE V. To subtract compound numbers. 1. From /y of a . take f of a . OPERATION. 33 common denominator 24 To perform this question, Value of A . = 12s. Value of f . = 4s. 5 d. Ans. 8s. TO we find by Case XI., Sect. i| XVI.,the value of ^.= 12s. 33 8y 8 T d. ; and also of f . = 4s. 5^d. ; we then find a com- mon denominator of the fractional part, by multiplying together their denominators, 1 1 X 3 33. We then proceed as in Case II., Sect XVIII. This question can be performed by first subtracting the frac- tion f of a <. from -/y of a ., and then reducing the remainder by Case XL, Sect. XVI. ; thus : /y . f . = (^. = 0<. 8s. 3d. lqr. Ans. 2. From 7^- of a ton take of a cwt. OPERATION. cwt. qr. Ib. oz. dr. & of a ton = 1 1 4 8 4 of a cwt. = 3599 Ans. 1 26 14 This question may also be performed by first reducing of a cwt. to the fraction of a ton by Case IX., Sect. XVI., and sub- tracting it from TJ\ of a ton, and then reducing the remainder to its proper terms by Case XI., Sect. XVI. Thus : ^ 2^ = ^ of a ton = Iqr. 261b. 14oz. lO^dr. Ans. RULE. Find the value of the fractions in integers ; then subtract as in the foregoing rules. 3. From of an ell English take f of a yard. Ans. 3qr. Ona. 2^in. 4. From f of a mile take -/y of a furlong. Ans. Ifur. 5rd. 10ft. lOin. 5. From f of a degree take f of a mile. Ans. 49m. Ofur. 13rd. lift. 9in. l^bar. SECT, xiii.] DIVISION. 73 The rule for pointing off cents and mills is the same as in Multiplication. If the dividend consist of dollars only, and be either smaller than the divisor, or not divisible by it without a remainder, annex two or three ciphers, as the case may require, and the quotient will be cents or mills accordingly. 1. If 97 bushels of wheat cost $ 147.82,8, what is the value of one bushel ? Ans. $ 1.52,4. OPERATION. 97) 147.82,8($ 1.52,4 97_ 508 485 232 194 388 388 2. Bought 1789 acres of land for $ 1699.55 ; what cost one acre? Ans. $0.95. 3. A trader sold 425 pounds of sugar for $ 51.00 ; what was the cost of one pound ? Ans. $ 0.12. 4. When rye is sold at the rate of 628 bushels for $ 471.00, what is that a bushel ? Ans. $ 0.75. 5. A merchant bought 329 yards of broadcloth for $ 904.75 ; what cost one yard ? Ans. $ 2.75. 6. When a chest of tea containing 42 pounds can be bought for $ 31.50, what cost one pound ? Ans. $ 0.75. 7. If it cost $1460 to support a family 365 days, what would be the expense per day ? Ans. $ 4.00. 8. A shoe-dealer sold 125 cases of shoes for $ 2500 ; what was the cost per case ? Ans. $ 20.00. 9. A flour-merchant sold 475 barrels of flour for $2018.75; what cost one barrel ? Ans. $ 4.25. 10. Bought 42 barrels of pears for $ 73.50 ; what cost one barrel ? Ans. $ 1.75. 11. If 1624 pounds of pork cost $97.44, what cost one pound ? Ans. $ 0.06. 12. If 47000 shingles cost $ 176.25, what is the cost per thousand ? Ans. $ 3.75. 13. Bought 148 tons of plaster of Paris for $ 337.44 ; what was it per ton ? Ans. $ 2.28. 14. If 78 barrels of fish cost $ 303.42, what will one barrel cost ? Ans. $ 3.89. 7 74 UNITED STATES MONEY. [SECT. xm. 15. A farmer sold 691 bushels of wheat for $ 863.75 ; what was it per bushel ? Ans. $ 1.25. 16. If a man earn $ 434.35 in a year, what is that per day? Ans. $ 1.19. 17. Sold 169 tons of timber for $ 790.92 ; what cost one ton ? Ans. $ 4.68. 18. What cost one pound of leather, if 789 pounds cost 8142.02? Ans. $0.18. 19. If 369 tons of potash cost $48910.95, what will be the price of one ton ? Ans. $ 132.55. 20. Bought 47 hogsheads of salt, each hogshead containing 7 bushels, for $ 368.48 ; what cost one bushel ? Ans. $ 1.12. 21. If 19 cords of wood cost $ 106.97, what cost one cord ? Ans. $5.63. 22. When 19 bushels of salt can be bought for $ 30.87,5, what cost one bushel ? Ans. $ 1.62,5. 23. If 17 chests of souchong tea, each weighing 59 pounds, cost $ 672.01, what cost one pound? Ans. $ 0.67. 24. Sold 73 tons of timber for $ 414.64 ; what did I receive per ton ? Ans. $ 5.68. 25. Bought oil at the rate of 144 gallons for $ 234.00 ; what did I give per gallon ? Ans. $1.62,5. 26. A landholder sold 47 acres of land for $ 1774.25 ; what did he receive per acre ? Ans. $ 37.75 27. What is the price of one yard of broadcloth, if 163 yards cost $1106.77? Ans. $6.79. 28. If a farm, containing 144 acres, is valued at $ 10043.71,2, what is one acre worth ? Ans. $ 69.74,8. BILLS. 1. Boston, July 4, 1835. Mr. James Dow, Bought of Dennis Sharp, 17 yds. Flannel, at .45 cts. 19 " Shalloon, " .37 " 16 " Blue Camlet, " .46 " 13 " Silk Vesting, " .87 " 9 " Cambric Muslin, " .63 " 25 " Bombazine, " .56 " 17 " Ticking, " .31 " 19 " Striped Jean, " .16 " $61.33 Received payment, Dennis Sharp. SECT. XIII.] 2. Mr. Samuel Smith, BILLS. 13 Ibs. Tea, 16 " Coffee, 36 " Sugar, 47 " Cheese, 12 " Pepper, 7 " Ginger, 13 " Chocolate, 75 Haverhill, May 5, 1835. Bought of David Johnson, at Received payment, Dayid .98 cts. .15 " .13 " .09 .19 " .17 " .61 " $35.45. 3. Mr. John Dow, 17 yds. Broadcloth, Salem, February 29, 1835. Bought of Richard Fuller, 29 60 49 18 27 75 36 49 Cassimere, Bleached Shirting, Ticking, Blue Cloth, Habit do. Flannel, Plaid Prints, Brown Sheeting, Received payment, $5.25 1.62 .17 .27 3.19 2.75 .61 .75 .18 8372.90. Richard Fuller. 4. Mr. John Rilley, 10 pair Boots, 19 " Shoes, 83 " Hose, 47 Ibs. Ginger, 91 " Chocolate, 47 " Pepper, 68 " Flour, 27 pair Gloves, at H Received payment, Baltimore, January 20, 1835. Bought of James Somes, $2.75 1.25 1.29 .17 .39 .23 .13 1.39 ft 258.98. James Somes. 76 UNITED STATES MONEY. [SECT. xm. 5. Philadelphia, June 11, 1835. Mr. Moses Thomas, Bought of Luke Dow, 27 National Spelling-Books, ' at $0.19 25 Parker's Composition, .27 17 National Arithmetics, " .75 9 Greek Lexicons, 3.75 8 Ainsworth's Dictionaries, " 4.50 27 Greek Readers, " 2.25 18 Folio Bibles, " 9.87 75 Leverett's Caesar, " .31 67 Fisk's Greek Grammar, .75 15 Folsom's Cicero's Orations, " 1.12 "423.09. Received payment, Luke Dow, by Timothy True. 6. Boston, June 26, 1835. Dr. Enoch Cross, Bought of Maynard & Noyes, 14 oz. Ipecacuanha, at $ 0.67 23 " Laudanum, " .89 17 " Emetic Tartar, " 1.25 25 " Cantharides, " 2.17 27 " Gum Mastic, " .61 56 " Gum Camphor, " ~~$ 136.94. Received payment, Maynard & Noyes, by Timothy Jones. 7. Newburyport, June 5, 1835. Mr. John Somes, Bought of Samuel Gridley, 7 yds. Broadcloth, at $ 4.50 16flbs. Coffee, " .16 18J " Candles, " .25 30 " Soap, " .17 3 Pepper, " .19 7J " Ginger, " .18 $ 48.01 J. Received payment, Samuel Gridley. SECT. XIII.] BILLS. 77 8. Mr. Benjamin Treat, 37 Chests Green Tea, 41 " Black do. 40 " Chests of Imperial Tea, 13 Crates Liverpool Ware, Received payment, Boston, May 1, 1835. Bought of John True, at $25.50 16.17 97.75 169.37 $7718.28. John True. 9. Mr. John Cummings, 97 bbl. Genesee Flour, 167 " Philadelphia do. 87 " Baltimore do. 196 " Richmond do. 275 " Howard St. do. 69 bu. Rye, 136 " Virginia Corn, 68 " North River do. 169 " Wheat, 76TonLehighCoal, 89 " Iron, 49 Grindstones, 39 Pitchforks, 197 Rakes, 86 Hoes, 78 Shovels, 187 Spades, 91 Ploughs, 83 Harrows, 47 Handsaws, 35 Millsaws, 47 cwt. Steel, 57 " Lead, Received payment, 7* New York, July 11, 1835. Bought of Lord & Secomb. at $6.25 5.95 6.07 5.75 7.25 1.16 .67 .76 1.37 9.67 69.70 3.47 1.61 .17 .69 1.17 .85 11.61 17.15 3.16 18.15 9.47 6.83 $ 17315.32. Lord & Secomb. 78 COMPOUND MULTIPLICATION. [SECT. xiv. SECTION XIV. COMPOUND MULTIPLICATION. CASE I. COMPOUND MULTIPLICATION consists in multiplying numbers of different denominations by simple numbers. 1. What will 6 bales of cloth cost, at 7. 12s. 7d. per bale ? 8 . a. In this question, we multiply 7d. by 6, and find 7 12 7 me product to be 42d. This we divide by 12, the Q number of pence in a shilling, and find it contains jc TpT ~ 3s. and 6d. We write the 6d. under the pence, and cany 3 to the product of 6 times 12, and find the amount to be 75s., which we reduce to pounds by dividing them by 20, and find them to be 3. 15s. We write down the shillings under the shillings, and carry 3 to the product of 6 times 7. ; and we thus find the answer to be 45<. 15s. 6d. From the above illustration we deduce the following RULE. When the multiplier is less than 12, multiply by the multiplier and carry as in Compound Addition. 2. What cost 9yds. of cloth, at l. 3s. 8d. per yard ? Ans. 10. 13s. Od. 3. What cost 7bbls. of flour, at !<. 8s. 7d. per barrel ? 4. What cost 81bs. of Cayenne pepper, at 7s. 9d. per Ib. ? 5. Multiply 10yd. 3qr. 3na. by 5. 6. Multiply 3cwt. Iqr. 81b. by 9. 7. Multiply 7T. llcwt. Iqr. 201b. by 5. 8. Multiply 7 days 15h. 35m. 18sec. by 10. 9. Multiply IS. 16s. 7d. by 4. Ans. 75^. 6s. 6d. 10. Multiply 15. 11s. 8|d. by 8. Ans. 124.. 13s. lOd. 11. Multiply 27 . 19s. lld. by 9. Ans. 251,. 19s. 12. Multiply 19<. 5s. 7d. by 11. Ans. 212<. Is. 7|d. 13. Multiply Sl. 14s. 9d. by 8. Ans. 653<. 18s. Od. 14. Multiply 15. 18s. 5d. by 7. Ans. 111. 8s. lid. 15. Multiply 13. 5s. 4f d. by 12. Ans. 159^. 4s. 9d. 16. Multiply 171b. 7oz. 13dwt. 13gr. by 9. 17. Multiply 151b. lloz. 19dwt. 15gr. by 7. BECT. xiv.] COMPOUND MULTIPLICATION. 79 IS. Multiply 16T. 12cwt. 3qr. 131b. 12oz. by 11. 19. Multiply 13T. 3cwt. Iqr. 141b. 13oz. by S. 20. Multiply 21b 51 53 19 16gr. by 8. 21. Multiply 47yd. 3qr. 2na. 2in. by 7. 22. Multiply 17m. 7fur. 36rd. 13ft. 7in. by 12. 23. Multiply 16deg. 39m. 3fur. 39rd. 5yd. 2ft. by 9. 24. Multiply 16deg. 20m. 7fur. 12rd. 8ft. llin. lbar. by 6. 25. Multiply 16A. 2R. 4p. 19yd. 7ft. 79in. by 11. 26. Multiply 7 cords 1 16ft. 1629m. by 4. 27. Multiply 29hhd. 61 gal. 3qt. Ipt. 3gi. by 7. 28. Multiply 3 tuns 3hhd. 56gal. 2qt. by 9. 29. Multiply 7hhd. 5gal. 2qt. Ipt. by 8. 30. Multiply 19bu. 2pk. 7qt. Ipt. by 6. 31. Multiply 36ch. 18bu. 3pk. 7qt. by 7. 32. Multiply 13y. 316d. 15h. 27m. 39sec. by 8. 33. If a man gives each of his 9 sons 23A. 3R. 19p., what do they all receive ? 34. If 12 men perform a piece of labor in 7h. 24m. 30sec., how long would it take 1 man to perform the same task ? 35. If 1 bag contain 3bu. 2pk. 4qt., what quantity do 8 bags contain ? CASE II. When the multiplier is more than 12, and is a composite num- ber, that is, a number which is the product of two or more numbers, the question is performed as in the following EXAMPLE. 36. What will 42 yards of cloth cost, at 6s. 9d. a yard ? . a. d. In this example, we find that 6 069 multiplied by 7 will produce the 6 quantity 42 yards. We therefore 206 = price of 6 yds. multiply 6s 9d. first by the 6, and y then its product by 7 ; and the last - ^ . f . product, 14=. 3s. 6d. is the answer = price of 42 yds. or price of the 42 yards> The pupil will now see the propriety of the following RULE. Multiply by one of the factors of the composite number, and tlie prod- uct thus obtained by the other. 37. What will 16 yards of velvet cost, at 3s. 8d. per yard ? 80 COMPOUND MULTIPLICATION. [SECT. XIT. 38. What will 72 yards of broadcloth cost, at 19s. lid. per yard ? 39. What will 84 yards of cotton cost, at Is. lid. per yard ? 40. Bought 90 hogsheads of sugar, each weighing 12cwt. 2qr. 1 lib. ; what was the weight of the whole ? 41. What cost 18 sheep, at 5s. 9^-d. a piece ? 42. What cost 21 yards of cloth, at 9s. lid. per yard ? 43. What cost 22 hats, at 11s. 6d. each ? 44. If 1 share in a certain stock be valued at 13. 8s. 9d., what is the value of 96 shares ? 45. If 1 spoon weigh 3oz. 5dwt 15gr., what is the weight of 120 spoons ? 46. If a man travel 24m. 7fur. 4rd. in 1 day, how far will he go in 1 month ? 47. If the earth revolve 15' per minute, how far per hour ? 48. Multiply 39A. 3R. 17p. 30yd. 8ft. lOOin. by 32. 49. If a man be 2d. 5h. 17m. 19sec. in walking 1 degree, how long would it take him to walk round the earth, allowing 365 days to a year ? CASE III. When the multiplier is such a number as cannot be pro- duced by the product of two or more numbers, we should pro- ceed as in the following EXAMPLE. 50. What is the value of 53 tons of iron, at 18^. 17s. lid. a ton? . 8. d. . a. d. 18 17 11 18 17 11 5 3 94 97 = price of 5 tons. 56 13 9 = price of 3 tons. 10 944 15 10 = price of 50 tons. Because 53 is a prime 56 13 9 = price of 3 tons, number, that is, it cannot be TTwvi n c- KO A produced by the product of 9 7 = price of 53 tons. ny , wo nu * bers ^ ^ fore find a convenient composite number less than the given number, viz. 50, which may be produced by multiplying 5 by 10. Having found the price of 50 tons by the last Case, we then find the price of the 3 remaining tons by Case I., and add it to the former, making the value of the whole quantity 100L. 9s. 7d. SECT, xiv.] BILLS. 81 The pupil will hence perceive the propriety of the following RULE. Take for successive multipliers two or more numbers, whose continued product will be nearest the proper multiplier, and then find the value of the remainder by Case I. , and the sum of the last two products will be the answer. . 51. What will 57 gallons of wine cost, at 8s. 3|d. per gallon ? 52. Bought 29 lots of wild land, each containing 117A. 3R. 27p. ; what were the contents of the whole ? 53. Bought 89 pieces of cloth, each containing 37yd. 3qr. 2na. 2in. ; what was the whole quantity ? 54. Bought 59 casks of wine, each containing 47gal. 3qt. Ipt. ; what was the whole quantity ? 55. If a man travel 17m. 3fur. 13rd. 14ft. in one day, how far will he travel in a year ? 56. If a man drink 3gal. Iqt. Ipt. of beer in a week, how much will he drink in 52 weeks ? 57. There are 17 sticks of timber, each containing 37ft. 978in. ; what is the whole quantity ? 58. There are 17 piles of wood, each containing 7 cords 98 cubic feet ; what is the whole quantity ? 59. Multiply 2hhd. 19gal. Oqt. Ipt. by 39. 60. Multiply 3bu. Ipk. 4qt. Ipt. Igi. by 53. 61. Multiply 16ch. 7bu. 2pk. Oqt. Opt. by 17. BILLS. 1. London, July 4, 1835. Dow, Vance, & Co., of Boston, U. S., Bought of Samuel Snow, 45 yds. Broadcloth, at 8s. 4d. 50 " " " 10s. 6d. 56 " " " 3s. 7d. 63 " " " 12s. llfd. 72 " " " 19s. lid. 81 " " " 9s. 3d. 35 " " " 19s. 7d. 99 " " " 16s. 0d. 66 " " " 8s. lid. 33 " " 16s. lld. ~376. 7s. Ofd. Received payment, Samuel Snow. > COMPOUND MULTIPLICATION. [SECT. xiv. 2. Quebec, Jan. 8, 1835. Mr. John Vose, Bought of Vans & Conant, 46 Ivory Combs, at 3s. 5d. 47 Ibs. Colored Thread, " 6s. 9d. 51 yds. Durant, " Is. 8d. 52 Silk Vests, " 6s. 7d. 53 Leghorns, " 11s. 9d. 57 ps. Nankin, " 8s. 3d. 58 Ibs. White Thread, " 9s. lljd. ~128. 16s. Received payment, Vans & Conant. 3. Montreal, July 4, 1835. Mr. James Savage, Bought of Joseph Dowe, 83 gals. Lisbon Wine, at 6s. 7d. 85 " Port do. " 3s. 9d. 86 " Madeira do. " 4s. lld. 87 " Temperance do. " 3s. 6jd. 89 " Oil, " 5s. 3d. 91 Leghorns, " 19s. 10d. 92 Ibs. Green Tea, " 3s. ld. 93 pair Thread Hose, " 4s. 4d. 94 Silk Gloves, " 3s. 3d. 95 " Silk Hose, " 6s. 6d. 97 yds. Linen, " 5s. 5]d. 98 gals. Winter Strained Oil, 7s. 7^d. ~338<. 19s. Received payment, Joseph Dowe. 4. Montreal, June 17, 1835. Mr. Samuel Simpson, Bought of Lackington, Grey, & Co. 19 yds. Cloth, at 23 " Worsted, " 26 " Baize, " 29 u Camlet, " 31 " Bombazine, " 34 " Linen, " 37 " Cotton, " 38 Flannel, " 39 " Calico, " 41 " Broadcloth, " 43 Nankin, " Received payment, Lackington, Grey, & Co. SECT, xiv.] BILLS. * 5. Liverpool, June 2, 1835. John Jones, of Philadelphia, U. S., Bought of Thomas Hasseltine, 297 yds. Black Broadcloth, at 17s. 3d. 473 " Blue do. " 9s. 512 " Red do. " 15s. H 624 " Green do. " 12s. 8d. 765 " White do. " 19s. 9d. 169 " Black Velvet, . is done by adding 40 to the denominator and 1 to the nume- rator, and in the fraction gf by adding 2 to the numerator and 40 to the denominator. Q. . D. CASE IV. To find the value of a decimal in integral or whole numbers. 1. What is the value of .790625. ? OPERATION. .790625 Now it is evident, that .790625. expressed 20 in terms of a shilling must be the product of 15 812500 .790625 . multiplied by 20, and that to con- J2 tinue the reduction to the lowest terms we must h lyRnnnr multiply by the same number as in common re- duction. 4 aoooooo RULE. Multiply tlie given decimal by the number which will bring it to the next lower denomination, and cut off for a remainder as many places on the right as there are places in the given decimal. SECT, xxvni.] DECIMAL FRACTIONS. 145 Multiply this remainder by the number which will bring it to the next lower denomination, cutting off for a remainder as before, and thus pro- ceed till the reduction is carried to the denomination required. The sev- eral integral numbers, standing at the left hand, will be the answer sought in the different lower denominations. 2. What is the value of .625 of a shilling ? Ans. 7d. 3. What is the value of .6725 of a cwt. ? Ans. 2qr. 191b. 5^z. 4. What is the value of .9375 of a yard ? Ans. 3qr. 3na. 5. What is the value of .7895 of a mile ? Ans. 6fur. 12rd. 10ft. 6f in. 6. What is the value of .9378 of an acre ? Ans. 3R. 30p. 13ft. 9 T 9 ^m. 7. Reduce .5615 of a hogshead of wine to its value in gal- lons, &c. Ans. 35gal. Iqt. Opt. 3|ff gi. 8. Reduce .367 of a year to its value in days, &c. Ans. 134da. Ih. 7m. 19sec. 9. What is the value of .6923828125 of a cwt. ? Ans. 2qr. 2 lib. 8oz. 12dr. 10. What is the value of .015625 of a bushel ? Ans. 1 pint. 11. What is the value of .55 of an ell English ? Ans. 2qr. 3na. 12. What is the value of .6 of an acre ? Ans. 2R. 16p. SECTION XXVIII. MISCELLANEOUS EXAMPLES. 1. What is the value of 7cwt. 2qr. 181b. of sugar, at $ 11.75 per cwt. ? Ans. $ 90.01, 3. 2. What cost 19cwt. 3qr. 141b. of iron, at $ 9.25 per cwt. ? Ans. $ 183.84,3}. 3. What cost 39A. 2R. 15p. of land, at $ 87.37,5 per acre ? Ans. $ 3459.50,3f f . 4. What would be the expense of making a turnpike 87m. 3fur. 15rd., at $ 578.75 per mile ? Ans. $ 50595.41 V- 5. What is the cost of a board 18ft. 9in. long, and 2ft. 3^-in. wide, at $ .05,3 per foot ? Ans. $ 2.27,7^. 6. Goliath of Gath was 6^ cubits high ; what was his height in feet, the cubit being 1ft. 7.168in. ? - Ans. 10ft. 4.592in. 13 146 DECIMAL FRACTIONS. [SECT, 7. If a man travel 4.316 miles in an hour, how long would he be in travelling from Bradford to Boston, the distance being 29 miles ? Ans. 6h. 50m. 6sec.+ 8. What is the cost of 5yd. Iqr. 2na. of broadcloth, at $ 5.62 per yard ? Ans. $ 30.23,4f . 9. Bought 17 bags of hops, each weighing 4cwt. 3qr. 71b., at $ 5.87 per cwt. ; what was the cost ? ~ Ans. $ 480.64,8-^. 10. Purchased a farm, containing 176A. 3R. 25rd., at $ 75.37 per acre ; what did it cost ? Ans. $ 13334.3Q,8]f . 11. What cost 17625 feet of boards, at $12.75 per thou- sand? Ans. $ 224.7 J,8. 12. How many square feet in a floor 19ft. 3in. long, and 15ft. 9in. wide ? Ans. 303ft. 27in. 13. How many square yards of paper will it take to cover a room 14ft. 6in. long, 12ft. 6in. wide, and 8ft. 9in. high ? Ans. 52yd. 14. How many solid feet in a pile of wood 10ft. 7in. long, 4ft. wide, and 5ft. lOin. high ? Ans. 246 T ft. 15. How many garments, each containing 4yd. 2qr. 3na., can be made from 1 12yd. 2qr. of cloth ? Ans. 24. 16. Bought Igal. 2qt. Ipt. of wine for $ 1.82 ; what would be the price of a hogshead ? Ans. $70.56. 17. Bought 125yd. of lace for $ 15.06 ; what was the price of 1 yard? Ans. $0.12. 18. What cost 17cwt. 3qr. of wool, at $ 35.75 per cwt. ? Ans. $ 634.56,2. 19. What cost 7hhd. 47gal. of wine, at $ 87.25 per hogs- head ? Ans. $ 675.84^. 20. How many solid feet in a stick of timber 34ft. 9in. long, 1ft. Sin. wide, and 1ft. 6in. deep ? Ans. 65.15625ft. 21. How many cwt. of coffee in 17f- bags, each bag contain- ing 2cwt. Iqr. 71b. ? Ans. 41cwt. Oqr. 5lb. 22. If 18yd. Iqr. of cloth cost $36.50, what is the price of 1 yard ? Ans. $2.00. 23. If $ 477.72 be equally divided among 9 men, what will be each man's share ? Ans. $ 53.08. 24. A man bought a barrel of flour for $ 5.375, 7gal. of mo- lasses for $1 .78, 9gal. of vinegar for $1.1875, Igal. of wine for $1.125, 141b. of sugar for $1.275, and 51b. of tea, for $2.625; what did the whole amount to ? Ans. $ 13.36,7. 25. A man purchased 3 loads of hay ; the first contained 2f tons, the second 3| tons, and the third ly 1 ^ tons ; what was the value of the whole, at $ 17.625 a ton ? Ans. $ 128.88,2 T . SECT, xxix.] EXCHANGE OF CURRENCIES. 147 26. At $ 13.625 per cwt., what cost 3cwt. 2qr. Tib. of sugar ? Ans. $ 48.53,9 T V 27. At $125.75 per acre, what cost 37 A. 3R. 35rd. ? Ans. $ 4774.57,0 T V 28. At $11.25 per cwt., what cost 17cwt. 2qr. 211b. of rice? Ans. $198.98,4$. 29. What cost 7J- bales of cotton, each weighing 3.37cwt., at $ 9.37 per cwt. ? Ans. $ 244.85, 1 T V 30. What cost 7hhd. 49gal. of wine, at $ 97.625 per hogs- head ? Ans. $ 759.30,5f % . 31. What cost 7yd. 3qr. 3na. of cloth, at $4.75 per yard ? Ans. $ 37.70,3. 32. What cost 27T. 15cwt. Iqr. 3lb. of hemp, at $183.62 per ton ? Ans. $ 5098.03,7^. 33. What is the cost of constructing a railroad 17m. 3fur. 15rd., at $1725.87,5 per mile ? Ans. $ 30067.97 ,8. 34. When $624.53125 are paid for 17A. 3R. 15p. of land, what is the cost of one acre ? Ans. $ 35. 35. Paid $494.53125 for 19T. 15cwt. 2qr. 141b. of hay; what was the cost per ton ? Ans. $ 25. 36. How much land, at $40 per acre, can be obtained for $1004.75 ? Ans. 25A. OR. 19p. 37. Bought of Queen Victoria 9 acres of land, for which I paid 157.753 125<. Required the price per acre. Ans. 17<. 10s. 6d. 38. If $198.984375 are paid for 17cwt. 2qr. 211b. of rice, what is the value of Icwt. ? Ans. $11.25. SECTION XXIX. EXCHANGE OF CURRENCIES. IT is well known, that, in different States of the Union, the American dollar has a different value as expressed in shillings and pence. The origin of this difference is thus explained. Previous to the formation of the Constitution, all accounts in this country were kept in the currency of Great Britain, and the dollar was reckoned at 4s. 6d. sterling. Owing, however, to the want of money, several States under the colonial gov- ernment issued Bills of Credit, which were not received by the 148 EXCHANGE OF CURRENCIES. [SECT. ixix. British merchants in payment for goods at their par value. Holders of those bills were therefore obliged to pay a larger nominal amount than though they had paid in sterling. Thus eight shillings in the bills of New York passed for one dollar, or 4s. 6d. sterling. In the bills of the New England Colonies, where the depreciation was less, six shillings made a dollar, and in South Carolina and Georgia, four shillings and eight pence. In the ordinary reckonings of the people, shillings and pence are still considerably used, and their estimated value in different States is as follows. In New England, Indiana, Illinois, Missouri, Virginia, Ken- tucky, Tennessee, Mississippi, Texas, Alabama, and Florida, the dollar is valued at 6 shillings, $ 1 = ^j. = -^. In New York, Ohio, and Michigan, the dollar is valued at 8 shillings, $ 1 &. = f <. In New Jersey, Pennsylvania, Delaware, and Maryland, the dollar is considered 7 shillings and 6 pence, $ 1 = In North Carolina the dollar is reckoned at 10 shillings, In South Carolina and Georgia 4 shillings 8 pence is the value of a dollar, $ 1 = */&. = &. In Canada and Nova Scotia the dollar is valued at 5 shillings, The following table exhibits the legal rates of interest in the United States, and the penalty of usury. STATES. RATE OF INTEREST. PENALTY OF USURY. Maine, 6 pr. ct. Forfeit of the debt or claim. N.Hampshire, 6 " Forfeit of threefold the usury. Vermont, 6 " Recovery in an action, with costs. Massachusetts, 6 Forfeit of threefold the usury. Rhode Island, 6 Forfeit of the usury and interest on the debt. Connecticut, 6 Forfeit of the whole debt New York, 7 Usurious contracts void. New Jersey, 6 Forfeit of the whole debt. Pennsylvania, 6 Forfeit of the whole debt. Delaware, 6 Forfeit of the whole debt. Maryland, 6 On tobacco contracts, 8 per ct. Usurious contracts void. Virginia, 6 Forfeit double the usury. North Carolina, 6 Contracts for usury void. Forfeit double the usury. South Carolina, 7 " Forfeit of interest and usury, with costs. SECT, xxix.] EXCHANGE OF CURRENCIES. 149 STATES. Georgia, Alabama, Louisiana, Tennessee, Kentucky, Ohio, Indiana, Illinois, Missouri, Michigan, Arkansas, Florida, Wisconsin, Iowa, Texas RATE OF INTEREST. PENALTY OF USURY. 8 per ct. Forfeit of three times the usury, and con- tracts void. 8 " Forfeit of interest and usury. 8 " By contract as high as 10. Recovery in action of debt. 5 " Bank 6 ; by agreement as high as 10; con- tracts void. Contracts void. Recovery with costs. Contracts void. A fine of double the usury. By agreement as high as 12. Forfeit of threefold whole amount of interest. By agreement as high as 10. Forfeit of the interest and usury. Forfeit of the usury and one fourth the debt. By agreement as high as 10. Usury re- coverable ; contracts void. Forfeit of interest and usury. By agreement as high as 12. the excess. By agreement as high as 12. the excess. Forfeit treble Forfeit treble 8 Dist. Columbia, 6 " Contracts void. NOTE. On debts or judgments in favor of the United States, interest is computed at the rate of 6 per cent, per annum. In order, therefore, to change any of the above currencies to United States money, the shillings, pence, and farthings, if there be any, must first be reduced to decimals of a pound, and annexed to the pounds. RULE. Divide the pounds by the value of a dollar in the given cur- rency, EXPRESSED BY A FRACTION OF A POUND ; that is, to change the old Neuo England currency to United States money, divide by T 3 ; be- cause 6 shillings is T 3 of a pound. To change the old currency of New York, 1f has been given in all our arithme- tics as the value of the pound sterling in United States Money. It is time the error was corrected. The nominal par of exchange with London, as expressed in reports of exchange, is 109.496-|-> or very nearly 1095, being 9| above the comput- ed par of $ 4.444|, represented by 100. The Bank of England was established in 1694, by a company who ad- vanced a loan of > 1,200,000 sterling to government. Specie payment was suspended in 1797, and resumed, by act of Parliament, May 1, 1823. The term Sterling is derived from the Eastcrli.ngs, who were expert re- finers from the eastern part of Germany, who came into England and first established the standard proportion of silver, lloz. 2dwt. fine silver, and 18dwt. alloy. The first sterling was coined in 1216. In the reign of Charles the Second (1666) a new gold coinage was minted, called Guin- eas, from the country from which the gold was originally brought. In 1816 (150 years after) the guinea was superseded by a new coin, called the sovereign, which represents the pound sterling. The guinea (old coin- age) weighs 129gfgr., standard. The sovereign (new coinage) weighs 123|pgr., standard. The standard legal fineness of gold in England is 22 carats, or f|$j. The standard of silver is lloz. 2dwt. = fg = $jfo The sovereign contains precisely HSggggr. of pure gold. The coinage of the United States is regulated by Congress. By the last act of Congress, January 18, 1837, the standard for both gold and silver was fixed at fggg, that is, suppose any of our gold or silver coin to be di- vided into 1000 equal parts, 900 of those parts are pure gold or silver, and 100 parts are alloy. By this act the eagle weighs 258gr. Troy, standard. Containing, . . 232.2gr. pure gold, And . . * 25.8gr. alloy. Our gold coinage, then, is 2l| carats fine ; our silver coinage is lOoz. 16dwt. fine, 6dwt. short of sterling fineness, which.is lloz. 2dwt. The American dollar weighs 412gr., standard, containing 371 |gr. pure silver, and 414gr- alloy. The alloy in our gold coins is mostly silver, and in our silver coin it is copper. The ratio of gold to silver in our coinage is 15|g|| to 1, that is, what- ever an ounce of silver may be worth, an ounce of gold is worth 15l|| times as much. The pound sterling under the above act, as represented by the sovereign of legal weight and fineness, is ^SIP exactly, = $4.866+, which is the real gold par with London. TO REDUCE STERLING MONEY TO UNITED STATES MONEY. RULE. Express the shillings, pence, and farthings decimally; then multiply sterling by 3 ?jS, and the product will be dollars, <3fC. NOTE. This rule supposes the sovereign, which represents the pound sterling, to be f^p fine, and to contain USslggr. of pure gold. But the sovereign falls a little short of its legal weight and fineness. So that its 152 CIRCULATING DECIMALS. [SECT. xxx. real value in our currency does not vary essentially from $4 84. This is the value assigned to it by act of Congress, in calculating ad valorem duties in our custom-houses on goods imported from England, which are invoiced in sterling money. Therefore, multiply sterling by $4.84 and we shall have the custom-house and market par value of sovereigns or pounds sterling. N. B. $4.444f never represented the true value of the pound sterling in the United States currency. Under the act of Congress of the 2d of April, 1792, establishing the mint and regulating the coins of the United States, the value of the pound ster- ling was $4 56-f- By the act of Congress of the 28th of June, 1834, called the Gold, Bill, the value of the pound or sovereign was $4 STj^Vs- By the act of Con- gress of the 18th of January, 1837, supplementary to the act of 1834, the value of the pound sterling becomes ft TOKOS 4- = $4. 86723303 Sov- ereigns are usually valued at $4.85 at the banks. SECTION XXX. INFINITE OR CIRCULATING DECIMALS.* DEFINITIONS. 1. DECIMALS produced from Vulgar Fractions, whose denom- inators do not measure their numeratprs, and distinguished by the continual repetition of the sapie f gure or figures, are called infinite or circulating decimals. 2. The circulating figures, that is, those that are continually repeated, are called repetends. If only the same figure is re- peated, it is called a single repetend, as .11111 or .5555, and is expressed by writing the figure repeated with a point over it. Thus . 1 1 1 1 1 is denoted by . 1 , and .5555 by .5. 3. If the same figures circulate alternately, it is called a com,' pound repetend, as .475475475, and is distinguished by putting a point over the first and last repeating figures ; thus, .475475475 is written .475. 4. When other figures arise before those which circulate, it is called a mixed repetend ; as .1246, or .17835. 5. Similar repetends begin at the same place ; as .3 and .6 ; or 5. 123 and 3.478. * Infinite or circulating decimals being less important for use than many other rules, and somewhat difficult in their operation, the student can omit them until he reviews the Arithmetic. SECT, xxx.] CIRCULATING DECIMALS. 153 6. Dissimilar repetends begin at different places ; as .986 and .4625. 7. Conterminous repetends end at the same place ; as .631 and .465. 8. Similar and conterminous repetends begin and end at the same place ; as .1728 and .4987. REDUCTION OF CIRCULATING DECIMALS. CASE I. To reduce a simple repetend to its equivalent vulgar fraction. If a unit with ciphers annexed to it be divided by 9 ad infini- tum, the quotient will be one continually ; that is, if ^ be re- duced to a decimal, it will produce the circulate .1 ; and since .1 is the decimal equivalent to , .2 will be equivalent to f , .3 to f , and so on, till .9 is equal to f or 1. Therefore every single repetend is equal to a vulgar fraction, whose numerator is the repeating figure, and denominator 9. Again, ^, or ^^, being reduced to decimals, makes .01010101, and .001001001 ad in- Jinitum, .01 and .001 ; that is, F V = .61, and ff fa = .001 ; consequently 2 F 3= .02, and ^f F = .002 ; and, as the same will hold universally, we deduce the following RULE. Make the given decimal the numerator, and let the denomi- nator be a number consisting of as many nines as there are recurring places in the repetend. If there be integral figures in the circulate, as many ciphers must be annexed to the numerator as the highest place of the repetend is distant from the decimal point. EXAMPLES. 1. Required the least vulgar fraction equal to .6 and .123. .6 = f = f Ans. .123 = iff = #& Ans. 2. Reduce .3 to its equivalent vulgar fraction. Ans. . 3. Reduce 1.62 to its equivalent vulgar fraction. Ans. lf. 4. Reduce .769230 to its equivalent vulgar fraction. Ans. if. CASE II. To reduce a mixed repetend to its equivalent vulgar fraction. 154 CIRCULATING DECIMALS. [SECT. XM. 1. What vulgar fraction is equivalent to .138 ? OPERATION. .138 = TVS, + yfo = iU + irfcr = * = A Ans. As this is a mixed circulate, we divide it into its finite and circulating parts ; thus .138 = .13, the finite part, and .008 the repetend or circulating part ; but .13 = T \y^ ; and .008 would be equal to f, if the circulate began immediately after the place of units ; but, as it begins after the place of hundreds, it is f of T*IT = ** Therefore .138 = Jfo + ^ = $tf + Q. E.D. RULE. To as many nines as there are figures in the repetend, an- nex as many ciphers as there are finite places for a denominator ; mul- tiply the nines in the denominator by the finite part, and add the repeat- ing decimal to the product for the numerator. If the repetend begins in some integral place, the finite value of the circulating must be added to the finite part. 2. What is the least vulgar fraction equivalent to .53 ? Ans. T^-. 3. What is the least vulgar fraction equivalent to .5925 ? Ans. f . 4. What is the least vulgar fraction equivalent to .008497133 ? Ans. 5. What is the finite number equivalent to 3.62 ? Ans. 31Jf CASE III. To make any number of dissimilar repetends similar and conterminous. 1. Dissimilar made similar and conterminous. OPERATION. Any given repetend whatever, 9.167 = 9.61767676 whether single, compound, pure, or 14 g __ 14 60000000 m i xe( ^ mav De transformed into an- other repetend, that shall consist of : 3.16555555 a n equal or greater number of figures 12.432 = 12.43243243 at pleasure . thus 4 may ^ chan?0(] 8.181 8.18181818 into ,44 or .444 ; and .29 into - 1.307 = 1.30730730 or ^^ And as some of the circu . lates in this question consist of one, some of two, and others of three places ; and as the least common multiple of 1, 2, and 3 SECT, xxx.] CIRCULATING DECIMALS. 155 is 6, we know that the new repetend will consist of 6 places, and will hegin just so far from unity as is the farthest among the dissimilar repetends, which, in the present example, is the third place. RULE. Change the given repetends into other repetends , which shall consist of as many figures as the least common multiple of the several number of places found in all the repetends contains units. 2. Make 3.671, 1. 007 i, 8.52, and 7.616325 similar and con- termio us. 3. Make 1.52, 8.7156, 3.567, arid 1.378 similar and conter- minous. 4. Make .0007,. 14 14 14, and 887.1 similar and conterminous. CASE IV. To find whether the decimal fraction equal to a given vulgar fraction be finite or infinite, and of how many places the repe- tend will consist. RULE. Reduce the given fraction to its least terms, and divide the denominator by 2, 5, or 10, as often as possible. If the whole denomi- nator vanish in dividing by 2, 5, or 10, the decimal will be finite, and will consist of so many places as you perform divisions. If it do not vanish, divide 9999, cf-c., by the result till nothing remain, and the num- ber of 9 "s used will show the number of places in the repetend; which will begin after so many places of figures as there are 10 '5, 2's, or 5's used in dividing. NOTE. In dividing 1.0000, &c., by any prime number whatever, except 2 or 5, the quotient will begin to repeat as soon as the remainder is 1. And since 9999, &c., is less than 10000, &c., by 1, therefore 9999, &c., divid- ed by any number whatever, will leave a for a remainder, when the re- peating figures are at their period. Now whatever number of repeating figures we have when the dividend is 1, there will be exactly the same number when the dividend is any other number whatever. For the prod- uct of any circulating number by any other given number will consist of the same number of repeating figures as before. Thus, let 378137813781, &c., be a circulate, whose repeating part is 3781. Now every repetend (3781), being equally multiplied, must produce the same product. For these products will consist of more places, yet the overplus in each, being alike, will be carried to the next, by which means each product will be equally increased, and consequently every four places will continue alike. And the same will hold for any other number whatever. Hence it ap- pears, that the dividend may be altered at pleasure, and the number of places in the repetend will be still the same ; thus, n = .09, and f* T = .27, where the number of places in each are alike ; and the same will be true in all cases. 156 CIRCULATING DECIMALS. [SECT. MXI. EXAMPLES. 1. Required to find whether the decimal equal to -f^f be finite or infinite ; and if infinite, of how many places the repe- tend will consist. (2) (2) (2) fffi = 2)T % = 8 = 4 = 2=1; therefore, because the de- nominator vanishes in dividing, the decimal is finite, and con- sists of four places ; thus, 16)3 ;${$$. 2. Required to find whether the decimal equal to //<& be finite or infinite ; and, if infinite, of how many places that rep- etend will consist. JM, = -& 2)112 = 56 = 28 = 14 = 7. Thus, ''ffff Jf ; therefore, because the denominator, 112, did not vanish in di- viding by 2, the decimal is infinite ; and as six 9's were used, the circulate consists of six places, beginning at the fifth place, because four 2's were used in dividing. 3. Let T V be the fraction proposed. 4. Let ^4j- be the fraction proposed. SECTION XXXI. ADDITION OF CIRCULATING DECIMALS. EXAMPLE. 1. Let 3.5 + 7.651 + 1.765 + 6.173 + 51.7 + 3.7+27.631 and 1.003 be added together. OPERATION, Dissimilar. Similar and Conterminoui. 3.5 = 3.5555555 rj -gg'j 7 6516516 Having made all the numbers similar ". . . and conterminous by Sect. XXX., Case 1.765 = 1.7657657 ni., we add the first six columns, as in 6.173r= 6.1737373 Simple Addition, and find the sum to 51.7 = 51.7777777 be 3591224 = W?Ws = 3.591227. 3.7 = 3.7000000 The repeating decimals .591227 we 27.631 = 27.6316316 write in their P r P er place, and carry 3 - Q~' _ nAonrvnA to me next column, and then proceed as in whole numbers. 103.2591227 ECT. xxxn.] CIRCULATING DECIMALS. 157 RULE. Make the repetends similar and conterminous, and find their sum, as in common Addition. Divide this sum by as many 9's as there are places in the repetend, and the remainder is the repetend of the sum, which must be set under the figures added, with ciphers on the left when it has not so many places as the repetends. Carry the quotient of this division to the next column, and proceed with the rest as with finite decimals. 2. Add 27.56 + 5.632 + 6.7 + 16.356 + .71 and 6.1234 together. Ans. 63.1690670868888. 3. Add 2.765 + 7.16674 + 3.671 + .7 and .1728 together. Ans. 14.55436. 4. Add 5.16345 + 8.6381 + 3.75 together. Ans. 17.55919120847374090302. 5. Reduce the following numbers to decimals, and find their sum : , f , and . Ans. .587301. SECTION XXXII. SUBTRACTION OF CIRCULATING DECIMALS. EXAMPLE. 1. From 87.1645 take 19.479167. OPERATION. Having made the numbers similar 87.164*5 87.164545 and conterminous, we subtract as in IQ A^Q-ia^f 10. AiQifvi whole numbers, and find the remain- Xc'. i /l7 J.O/ lU.'db/cHO / , - , . ', , cor-ro c : der of the circulate to be 5378, from 67.685377 which we subtract 1, and write the remainder in its place, and proceed with the other part of the question as in whole numbers. The reason why 1 should be added to the repetend may be shown as OPERATION ftt ws - The minuend may be considered 16f|ff, an( ^ ^ e subtrahend 7ff|- ; we then proceed with 79 if t these numbers as in Case II. of Subtraction of Vulgar [fit* Fractions ; and the numerator 5377 will be the re- 8J-fi& peating decimal. Q. E. D. RULE. Make the repetends similar and conterminous, and subtract as usual ; observing, that if the repetend of the subtrahend be greater than the repetend of the minuend, then the remainder on the right must be less by unity than it would be if tJie expressions were finite. 14 156 CIRCULATING DECIMALS. [SECT. xxxm. 2. From 7.1 take 5.t)2. Ans. 2.08. 3. From 315.87 take 78.0378. Ans. 237.838072095497. 4. Subtract | from f . Ans. .079365. 5. From 16.1347 take 11.0884. Ans. 5.0462. 6. From 18.1678 take 3.27. Ans. 14.8951. 7. From 3.123 take 0.71. Ans. 2.405951*. 8. From take f Ans. .246753. 9. From f take f. Ans. .158730. 10. From T 9 T take T V Ans. .1764705882352941. 11. From 5.12345 take 2.3523456. Ans. 2.77 1 105582 1666927777988888599994. SECTION XXXIII. MULTIPLICATION OF CIRCULATING DECIMALS. 1. Multiply .36 by .25. First Method. In the first method, OPERATION. we reduce the num- ofi_36_ 4. 9^2 _i 5 23 bers to vulgar frac- 36 - if A, -25 - TIT + w - w tiongj and tl f en muki . A X f = F 9 ^ = .0929 Answer. ply and reduce them 2. Multiply 582.347 by .08. Second Method. In the second method, OPERATION. we multiply as in whole 582.1347 X .08 = 46.58778 Answer, numbers, but we add two units to the product ; for 8 x 347 = 2776 = % = 2 Ht- Thus we see the repeating number is 778. RULE. Turn both the terms into their equivalent vulgar fractions, and find the product of those fractions as usual. Then change the vul- gar fraction expressing the product into an equivalent decimal, and it will be the product required. But, if the multiplicand ONLY has a rep- etend, multiply as in whole numbers, and add to tfte right-hand place of the product as many units as there are tens in the product of the left- hand place of the repetend. Tlie product will then contain a repetend whose places are equal to those in the multiplicand. BBCT. xxxv.] MISCELLANEOUS. 159 3. Multiply 87.32586 by 4.37. Ans. 381.6140338. 4. Multiply 3. 145 by 4.297. Ans. 13.5169533. 5. What is the value of .285714 of a guinea ? Ans. 8s. 6. What is the value of .461607142857 of a ton ? Ans. 9cwt. Oqr. 261b. 7. What is the value of .284931506 of a year ? Ans. 104da. SECTION XXXIV. DIVISION OF CIRCULATING DECIMALS. 1. Divide .54 by .15. OPERATION. .54 |* 6 Having reduced the num- 1 k _ _j I B _i4_ 7 t> ers to vulgar fractions, we e T JL^Jr >7v =Tw divide ne by the ther ' and TT ' ? l r . T ~T TT * change the quotient to a de- *?f = 3ff = 3.506493 Ans. cima j; RULE. Change both the divisor and the dividend into their equiva- lent vulgar fractions, and find their quotient as usual. Change the vul- gar fraction expressing the quotient into its equivalent decimal, and it will be the quotient required. 2. Divide 345.8 by .6. Ans. 518.83. 3. Divide 234.6 by .7. Ans. 301.714285. 4. Divide .36 by .25. Ans. 1.4229249011857707509881. SECTION XXXV. MENTAL OPERATIONS IN FRACTIONS, &c. IF any number be divided into two equal parts, and into two unequal parts, the product of the two unequal parts togeth- er with the square of half the difference of the two unequal parts is equal to the square of one of the equal parts. Also, The product of any two numbers is equal to the square of 160 MISCELLANEOUS. [SECT. xxxr. half their sum, less the square of half their difference. See Euclid's Elements, Book Second, Proposition Fifth. NOTE. A number is said to be squared when it is multiplied by itself; thus, the square of 5 is 5 X 5 = 25. From the above proposition we deduce the following rules. To multiply any number containing a half by itself. RULE 1. Multiply the whole number given in the question by the next larger whole number, and to the product add the square of the half 1. Multiply 5 by 5J. OPERATION. :=30 Ans. NOTE. The whole number given is 5, and the next larger whole num- ber is 6. 2. Multiply 7 by 7. Ans. 56 J. 3. Multiply 3| by 3. Ans. Iftf, 4. Multiply 9 by 9. Ans. 90, . 5. Multiply 11 by 11. Ans. 132. 6. Multiply 2(4 by 20. Ans. 420 T . 7. Multiply 30| by 30. Ans. 930, . le will hold good if we multiply any num- NOTE The same principl ber by itself whose unit is a 5. RULE 2. Take the next least number that ends in a cipher, and mul- tiply it by the next larger number ending in a cipher, and add to the product the square of 5 = 25, and the result will be the product. 8. Multiply 25 by 25. Ans. 625. The next less number ending in a cipher is 20, and the next larger is 30; 30x20 = 600; 5x5 = 25; 600 + 25 = 625 Ans. 9. Multiply 35 by 35. Ans. 1225. 10. Multiply 85 by 85. Ans. 7225. 11. Multiply 95 by 95. Ans. 9025. To find the product of two mixed numbers, whose fractional part is a half, and whose difference is a unit. RULE 3. Multiply the larger number idthout the fraction by itself, and from the product subtract the fractional part multiplied by itself, and the result will be tfte product. 12. Multiply 6 by 7f Ans. 48J. SECT, xxxv.] MISCELLANEOUS. 161 OPERATION. 7 X 7 = 49 ; x ; 49 = 48} Ans. 13. Multiply 8 by 9. Ans. 80}. 14. Multiply 11 by 12J. Ans. 143f. 15. Multiply 19^- by 20. Ans. 399. 16. Multiply 89i by 90 J. Ans. 8099}. NOTE. If the fractional parts of the numbers approach within a J, , or |, &c., of the larger number, the principle is the same. 17. What is the product of 4f multiplied by 5. Ans. 24f . OPERATION. 5X5 = 25; ix = i; 25 - J = 24| Ans. 18. Multiply 7} by 8. Ans. 63|f . 19. Multiply 9$ by lOf Ans. 99ff . 20. Multiply 8| by 9f . Ans. 80^. 21. Multiply 19^ by 20 T V Ans. 399^. To find the product of two numbers, one of which is as much less than either 20, 30, 40, &c., as the other is more than either of these numbers. RULE 4. Multiply the 20, 30, or 40, as the case may be, by itself, and subtract from the product the square of half the difference of the two numbers to be multiplied, and the result will be the product. 22. Multiply 28 by 32. Ans. 896. We find that 28 is as much less than 30 as 32 is more than 30 ; we therefore multiply 30 by 30 = 900, and from this prod- uct we subtract the square of 2 = 4 ; 900 4 = 896 Ans. 23. What is the product of 75 by 85 ? Ans. 6375. 24. What is the product of 83 by 77 ? Ans. 6391. 25. What is the product of 97 by 103 ? Ans. 9991. 26. What is the product of 17 by 23 ? Ans. 391. 27. What cost 18cwt. of steel, at $22 per cwt. ? Ans. $ 396. 28. What cost 27 tons of hay, at $ 33 per ton ? Ans. $891. 29. What cost 64 gallons of oil, at $ 0.56 per gallon ? Ans. $35.84. 30. What cost 28 tons of hay, at $ 32 per ton ? Ans. $ 886. 31. What cost 49 tons of iron, at $ 51 per ton ? Ans. $2499. 14* 162 MISCELLANEOUS. [SECT, MXVI. SECTION XXXVI. QUESTIONS TO BE PERFORMED BY ANALYSIS. - I. If a man travel 48 miles in 12 hours, how far will he travel in 17 hours ? Ans. 68 miles. [The following is the most obvious solution of this question. If he travel 48 miles in 12 hours, in ] hour he will travel 1*5 of 48 miles, which is 4 miles. Then if be travel 4 miles in 1 hour, he will in 17 hours travel 17 times as far, which is 68 miles, the answer.] 2. If 72 pounds of beef cost $ 6.48, what will 1 pound cost ? what will 675 pounds cost ? Ans. $ 60.75. 3. If of a dollar buy 1 pound of sugar, how much may be bought for $ 1 ? how much for $ 29 1 ? Ans. 2391b. 4. If a hogshead of wine cost $ 73.50, what is the value of 1 gallon ? what cost 17hhd. 45 gallons ? Ans. $ 1302.00. 5. Bought 11 bushels of rye for $ 9.00 ; what cost 25 bush- els ? Ans. $ 20.45^ T . 6. If a crew of 15 hands consume in 3 months 1620 pounds of beef, how much would be sufficient for 27 hands for the same time ? Ans. 29161b. 7. Bought 9 yards of flannel for $ 7.00 ; what would be the value of 37| yards ? Ans. $ 29. 3 If 8. If a certain field will pasture 8 horses nine weeks, how long will it pasture 23 horses ? Ans. 3^ weeks. 9. If 7 T 3 2- dozen of hats cost $ 318.50, what will 19| doz- en cost ? Ans. $ 874.95f f 10. If 1 ton of hay cost $ 25.00, what will 17 tons 13cwt. 19 pounds cost ? Ans. $ 441.46-J-. II. What part of 9f is 25 ? Ans. 2ff. 12. What part of 25 is 9 ? Ans. T ^ . 13. If $ 25 will pay for 7-jj- yards of broadcloth, what would be the price of 97 yards ? Ans. $ 328.8 If . 14. If $ 47.25 will pay for the keeping of 7 horses 2 months, what would it cost to keep 43 horses for the same time ? Ans. $290.25. 15. If 7 yards of cloth a yard wide is sufficient to make a cloak, how many yards would it take if the cloth was If yards wide ? Ans. 4f yards. 16. If a barrel of beer will last 10 men a week, how long would it last 1 man ? how long 37 men ? Ans. 1^ days. 17. If 5 calves are worth 9 sheep, how many calves will purchase 108 sheep ? Ans. 60 calves. SECT, xxxvi.] MISCELLANEOUS. 163 18. If 11 yards of cotton 3 quarters wide are sufficient to line a garment, how many yards would it require that were 5 quarters wide ? . Ans. 6f yards. 19. If 7 pairs of shoes will purchase 2 pairs of boots, how many pairs of shoes would it take to buy 18 pairs of boots ? Ans. 63 pairs. 20. If four gallons of vinegar be worth 7 gallons of cider, what quantity of vinegar would it take to buy 47 gallons of cider ? Ans. 26f gallons. 21. If a man travel 377 miles in 15 days, how far would he travel in 1 day ? how far in 100 days ? Ans. 2513 miles. 22. If 12 men can dig a ditch 50 feet long, 4 feet wide,and 3 feet deep, in30 days, how long would it take 1 man ? how long 47 men ? Ans. 7f \ days. 23. If 5 barrels of flour cost $25.75, what will 39 barrels cost ? Ans. $ 200.85. 24. Bought 17 acres of land for $791.01 ; what would 98 acres 3 roods and 14 perches cost ? Ans. $4598.90,8. 25. Bought 97f yards of cloth for $ 275.20 ; what is the price of 7 yards ? Ans. $ 19.78-f . 26. If a pole 7 feet long cast a shadow of 5 feet, how high is that steeple whose shadow is 97 feet? Ans. 135f feet. 27. Gave 3cwt. of sugar, at $ 9, for -f^f of an acre of land ; how much sugar would it have required to purchase an acre ? Ans. 5f f cwt. 28. If a vessel sail 47f miles in 3f hours, how far would it sail in 1 week ? Ans. 2425 2 1 1J m. 29. James can mow a field in 7 days, by laboring 10 hours a day ; how many days would it take him to perform the work by laboring 12 hours a day ? Ans. 5| days. 30. If -jZj- of a lot of land is worth $42.12, what is the value of f of it? Ans. $29.4 If 31. If a man can earn $ 10.27 in f of a week, how much would he earn in a month ? Ans. $71.89. 32. If 18 men can reap 72 acres in 5 days, how long would it take 6 men to perform the labor ? Ans. 15 days. 33. If 19 gallons of wine can be bought for $ 25, how many gallons will $ 71.25 buy ? Ans. 54/g- gallons. 34. If a penny loaf weighs 8 ounces when wheat is $ 1 a bushel, what should it weigh when wheat is sold for $ 1.25 a bushel ? Ans. 6f ounces. 35. If a basket which contains 1 bushels must be filled with apples 7 times to make one barrel of cider, how many barrels may be made by its being filled 126 times ? Ans. 18bbl. 164 SIMPLE INTEREST. [SECT. XXXTII. 36. If a cloak can be made of 4| yards of cloth that is If yards wide, how many yards would it take of cloth that is of a yard wide ? Ans. 6yd. Iqr. 3f na. 37. If a box 4 feet long, 2 feet wide, l feet high, contains 300 pounds of sugar, how much will a box that is 8 feet long, 4 feet wide, and 3 feet high, contain ? Ans. 24001b. 38. How much in length, that is 12f rods in breadth, will make an acre ? Ans. 12y 8 T rd. 39. Sound, uninterrupted, moves 1 142 feet in a second ; how long, after a cannon's being discharged at Boston, is the time before it is heard at Bradford, the nearest distance being 25^ miles ? Ans. 2 minutes. 40. Bought ^ T of a ton of potash, and sold -f of it for $ 46.70 ; what was the value of a ton ? Ans. $ 93.40. 41. If -j 4 T of a lot of land be worth $97, what would the whole lot be worth ? Ans. $ 266.75. 42. If 19 pounds of salmon be worth 50 pounds of beef, how much salmon would buy 77 pounds of beef? Ans. 43. What part of 17 T \ is 4 ? Ans. 44. What part of lls. 3d. is 15s. ? Ans. |. 45. What part of 7 yards is 3| ells English ? Ans. SECTION XXXVII. SIMPLE INTEREST. INTEREST is the compensation which the borrower of money makes to the lender. PRINCIPAL is the sum lent. AMOUNT is the interest added to the principal. PER CENT., a contraction of per centum , is the rate establish- ed by law, or that which is agreed on by the parties, and is so much for a hundred dollars for one year. CASE I. GENERAL RULE. Let the per cent, be considered as a decimal of a hundred dollars, and multiply the principal by it, and the product is the interest for one year. But, if it be required to find the interest for more than one year, multiply the product by the number of years. NOTE. The decimal for 6 per cent, is 06 ; for 7 per cent., .07 ; for 8 per cent., .08 ; for 9$ per cent., .0925 ; for 2 per cent., .025, &c. The decimal must be pointed off as in Multiplication of Decimal Fractions. SECT, xxxvii.] SIMPLE INTEREST. 165 This rule is obvious from the fact that the rate per cent, is such a part of every hundred dollars. Thus 6 per cent, is T F of the principal. NOTE. When no particular per cent, is named, 6 per cent, is to be understood, as it is the legal interest in the New England States. See page 148. 1. What is the interest of 8 144 for 1 year ? Ans. $ 8.64. OPERATION. Qg There being two places of decimals in the . multiplier, we point off two in the product. 8 8.64 Ans. 2. What is the interest of $ 78.78 for 3 years ? OPERATION. $78.78 .06 There being two places of decimals in the 4 72 68 multiplicand, and two in the multiplier, we point 3 off four places in the product. $ 14. 18,04 Ans. NOTE. It is a custom with merchants to reject mills in their compu- tations, but when the decimal of a cent exceeds 5 they add 1 to the num- ber of cents. Thus, they would reckon $81.93,8 to be in value $ 81.94. 3. What is the interest of 8 675 for 1 year ? Ans. 8 40.50. 4. What is the interest of 8 1728 for 1 year ? Ans. 8 103.68. 5. What is the interest of $ 19.64 for 2 years ? Ans. $ 2.35,6. 6. What is the interest of 8 896.28 for 3 years ? Ans. 8 161.33. 7. What is the interest of $ 349.25 for 10 years ? Ans. 8 209.55. 8. What is the interest of 8 3967.87 for 2 years ? Ans. $476.14,4. 9. What is the interest of 8 123.45 for 6 years ? Ans. $ 44.44,2. 10. What is the interest of $ 89.25 for 50 years ? Ans. $267.75. 11. What is the interest of $ 17.25 for 7 years ? Ans. $7.24,5. 12. What is the interest of $ 29.19 for 9 years ? Ans. 8 15.76,2. 13. What is the interest of 8 617.56 for 25 years ? Ans. 8 926.34. 166 SIMPLE INTEREST. [SECT, xxxvn. 14. What is the amount of $ 31.75 for 100 years ? Ans. $ 222.25. 15. What is the amount of $ 76.47 for 7 years ? Ans. $ 108.58,7 16. What is the amount of $ 716.57 for 4 years ? Ans. $ 888.54,6. 17. What is the amount of $ 178.56,5 for 30 years ? Ans. $499.98,2. 18. What is the interest of $ 97.06 for 9 years ? Ans. $52.41,2. 19. What is the interest of $ 0.75 for 75 years ? Ans. $3.37,5. 20. What is the interest of $ 750 for 12 years ? Ans. $ 540. CASE II. To find the interest for months at 6 per cent. RULE. Multiply the principal by half the number of months, ex- pressed decimally as a per cent. ; that is, for 12 months multiply by .06 ; for 8 months multiply by .04 ; for 7 months, .035 ; for 1 month, .005 ; and point for decimals as in the last rule. NOTE 1. It is obvious, that, if the rate per cent, were 12, it would be 1 per cent a month. If, therefore, it be 6 per cent, it will be a half per cent, a month, that is, half the months will be the per cent. NOTE 2. If any other per cent, is wanted, proceed as above, and then multiply by the given rate per cent, and divide by 6, and the quotient is the interest. 1. What is the interest of $ 368 for 8 months ? $368 .04 = half the months. $ 14.72 = Answer. 2. What is the interest of $ 637 for 10 months ? Ans. $ 31.85. 3. What is the interest of $1671.32 for 14 months ? Ans. $116.99. 4. What is the interest of $ 891.24 for 9 months ? Ans. $40.10. 5. What is the interest of $ 819.75 for 11 months ? Ans. $ 45.08,6. 6. What is the interest of $ 3671.25 for 13 months ? Ans. $238.63. 7. What is the interest of $ 61.18 for 15 months ? Ans. $ 4.58. 8. What is the interest of $ 3181.29 for 18 months ? Ans. $ 286.31. SECT, xxxvn.] SIMPLE INTEREST. 167 9* What is the interest of $ 11.39 for 19 months ? Ans. $ 1.08. 10. What is the interest of $ 9.98 for 23 months ? Ans. $ 1.14. 11. What is the interest of $87.19 for 27 months ? Ans. $11.77. 12. What is the interest of $ 32.18 for 36 months ? Ans. 8 5.79. 13. What is the interest of $ 167.18 for 50 months ? Ans. $41.79. 14. What is the interest of $ 386.19 for 100 months ? Ans. $ 193.09. CASE III. To find the interest of any sum for months and days, at 6 per cent. RULE.* Find the interest for the number of months, as under Case II. Then to find it for the number of days,multiply the principal by one sixth the number of days, and, if the principal be dollars, cut off three figures from the right hand, and those at die left ivill be the interest in dollars, and those at the right will be cents and mills. But if the prin- cipal be dollars and cents, five figures must be cut off from the right hand, and those at the left will be the interest, <^c., as before. * The reason for this rule, so far as it relates to the interest for any num- ber of months, was explained above. But the reason for the operation in the case of days is not so obvious. It will be seen, however, that it is nothing but an abridgment of the general rule for calculating interest, as given on page 164, which will appear from what follows. To find the interest for any number of years, we multiply the principal by the annual per cent., and the product thus obtained by the number of years, and cut off two figures, &c. In like manner, it is evident that to find the interest for any number of days, we have only to multiply by the daily per cent., and the product thus obtained by the number of days, and cut off as in the case of years. But 6 per cent, per annum is 55 per cent per diem, allowing 360 days to a year. Now, to multiply the princi- pal by c of the days, and cut off one figure from the product at the right, is the same as to multiply by 55 (the daily per cent.), and the product thus obtained by the whole number of days. Where the principal is dollars only, the rule directs to cut off three figures. Two of them are cut off ac- cording to the general rule, on page 164, and the other, because in the op- eration we have multiplied by | the number of days, instead of $$ of them, which would have been the proper multiplier. The above may be illustrated conciselv by the following operation. Let it be required to find the interest of 75 dollars for 180 days. By the gen- eral rule we have 75 x go -*- 100 X 180 = $ 2.25. By the rule under Case III. we have 75 X s of 180 = $2.25. 168 SIMPLE INTEREST. [SECT, xixrn NOTE 1. If one half the number of months be expressed by a single figure, we have only to annex to this figure | the number of days, and multiply the principal by the number thus found, cutting off as above, and we obtain, by a single operation, the interest for the months and days. NOTE 2. If any other per cent, than 6 is given, we may proceed as above, and then multiply by the given rate and divide by 6, and the result will be the interest sought. 1. What is the interest of $ 68.25 for 8 months and 24 days? Ans. $3.00,3. OPERATION. $68.25 .044 The first 4 in the multiplier is half of the 8 27300 months ; the second 4 is one sixth of the 24 27300 2. What is the interest of $ 637.28 for 17 months and 19 days, at 8 per cent. ? Ans. $74.91,5. 3. What is the interest of $396.15 for 13 months and 9 days ? Ans. $ 26.34,3. 4. What is the interest of $ 16.75 for 7 months and 17 days, at 7 per cent. ? Ans. $ 0.73,9. 5. What is the interest of $976.18 for 29 months and 23 days, at 9 per cent. ? Ans. $ 217.93,2. 6. What is the interest of $ 36.18 for 3 months and 7 days ? Ans. $0.58,4. 7. What is the interest of $ 51.17 for 9 months and 29 days, at 4 per cent. ? Ans. $ 1.69,9. 8. What is the interest of $365.19 for 33 months and 4 days, at 2 per cent. ? Ans. $ 20.16,6. 9. What is the interest of $ 125.75 for 5 months and 4 days ? Ans. $ 3.22,7. 10. What is the interest of $ 35.49 for 1 month and 2 days, at 7 per cent. ? Ans. $ 0.23,6. 11. What is the interest of $ 112.50 for 3 months and 1 day, at 9 per cent. ? Ans. $ 2.70,1. 12. What is the interest of $97.15 for 35 months and 27 days ? Ans. $ 17.43,8. 13. What is the interest of $47.15 for 1 month and 19 days, at 13 per cent. ? Ans. $ 0.86,6. 14." What is the interest of $ 678.75 for 87 -months and 20 days? Ans. $ 297.51,8. 15. What is the interest of $ 86 for 99 months and 29 days, at 25 per cent. ? Ans. $ 179.10,6. SBCT. xxxrii.] SIMPLE INTEREST. 169 16. What is the interest of $33.35,8 for 15 months and 17 days ? Ans. $ 2.59,6. 17. What is the interest of $ 144 for 5 days ? Ans. $0.12,0. CASE IV. When the interest is required on any sum, from a certain day of the month in a year, to a particular day of a month in the same, or in another year. RULE. Find the time, by placing the latest date in an upper line, and the earliest date under it. Let the year be placed first ; the number of months that have elapsed since the year commenced at its right hand, and the day of the month next ; tJien subtract the earlier from the latest date, and the remainder is the time for which the interest is required. Then proceed as in the last rule. Or, the months may be reckoned by their ordinal number, as in Operation Second. NOTE. Many practical men prefer reckoning interest by the second method. EXAMPLES. 1. What is the interest of $84.97, from Sept. 25, 1833, to March 8, 1835 ? Ans. $ 7.40,6. Operation First. Operation Second. First Method. Yrs. m. d. Yre. m. d. & 04. 07 1835 2 8 1835 3 8 7is7i 1833 8 25 1833 9 25 __'^il \ iTT^ i K~T^ 59479 1 5 13 67976 1416 $7.40,655 Second Method. It is evident that 4 months' interest $ 84.97 is -^ of a year's interest ; and for the 06 same reason, 1 month's interest is of 1 year, = 5.0982 4 months' interest ; and as 10 days is * 4 months, ^ ==. 1.6994 of a month, the interest for that time is 1 month, .4248 "of a month's interest ; and if the in- 10 days, \ = .1416 terest of 10 days be divided by 5, the 2 days, = 283 quotient will be 2 days' interest, and 1 day, = 141 the half of this will be 1 day's interest. $7^40,64, interest as before. 2. What is the interest of $786.75, from Dec. 9, 1831, to May 11, 1833 ? Ans. $ 67.13,6. 15 170 SIMPLE INTEREST. [SKCT. xxxvn. 3. What is the interest of $ 98.25, from July 4, 1826, to Oct. 19, 1829 ? Ans. $ 19.40,4. 4. What is the interest of $76.89,5, from Jan. 11, 1822, to July 27, 1833 ? Ans. $ 53.26,2. 5. What is the interest of $22.76,3, from Feb. 19, 1806, to July 18, 1830 ? Ans. $33.34,4. 6. What is the interest of $ 76.35, from August 17, 1830, to May 5, 1832 ? Ans. $7.86,4. 7. What is the interest of $97.86, from May 17, 1821, to Dec. 19, 1828 ? Ans. $ 44.55,8. 8. What is the interest of $ 1728.75, from Nov. 19,1823, to June 18, 1826 ? Ans. $267.66,8. 9. What is the interest of $ 99.99,9, from Jan. 1, 1800, to Feb. 29, 1832 ? Ans. $ 192.96,4. 10. What is the interest of $ 16.76, from Dec. 17, 1811, to June 17, 1822 ? Ans. $10.55,8. 11. What is the interest of $35.61, from Nov. 11, 1831, to Dec. 15, 1833 ? Ans. $4.47,4. 12. What is the interest of $ 786.97, from Oct. 19, 1827, to August 17, 1831, at 7 per cent. ? Ans. $ 225.92,5. 13. What is the interest of $96.84, from Nov. 27, 1829, to July 3, 1832, at 7 per cent ? Ans. $ 18.88,3. 14. What is the interest of $ 11.10,5, from April 17, 1832, to Dec. 7, 1832, at 7 per cent. ? Ans. $ 0.49,6. 15. What is the interest of $ 117.21, from June 19, 1806, to June 17, 1819, at 8 per cent. ? Ans. $ 129.46,1. 16. What is the interest of $ 17869.75, from Feb. 7, 1830, to Jan. 11, 1832, at 5 per cent. ? Ans. $ 1722.44,5. 17. What is the interest of $71.09,1, from July 29, 1823, to June 19, 1827, at 12 per cent. ? Ans. $ 33.17,5. 18. What is the interest of $ 83.47, from Nov. 8, 1830, to Ju- ly 11, 1833, at 8f per cent. ? Ans. $ 19.53,7. 19. What is the interest of $79.25, from Dec. 8, 1831, to July 17, 1833? Ans. $7.64,7. 20. What is the interest of $ 175.07, from Jan. 7, 1825, to Oct. 12, 1829 ? Ans. $ 50.04. 21. What is the interest of $ 12.75, from June 16, 1831, to August 20, 1833 ? Ans. $ 1.66,6. 22. What is the interest of $ 197.28,5, from Dec. 6, 1932, to Jan. 11, 1834 ? Ans. $ 12.98,7. 23. What is the interest of $ 12.69, from Jan. 2, 1833, to August 30, 1834, at 7 per cent. ? Ans. $ 1.47,5. 24. What is the interest of $79.15, from Feb. 11, 1831, to June 10, 1833, at 7 per cent. ? Ans. $ 13.37,3. SECT, xxxvii.] SIMPLE INTEREST. 171 25. What is the amount of $83.33, from March 11, 1831, to Jan. 1, 1833, at 7 per cent. ? Ans. $94.61,4. 26. What is the amount of $100.25, from March 2, 1831, to June 1, 1831, at 4 per cent. ? Ans. 8101.24,1. 27. What is the amount of 8 369.29, from April 30, 1830, to July 31, 1832, at 9 per cent. ? Ans. 8 444.16,3. 28. What is the interest of 8769.87, from Jan. 1, 1830, to June 17, 1835, at 9 per cent. ? Ans. 8 399.41,2. 29. What is the interest of 869.75, from Jan. 11, 1833, to June 29, 1833, at 17 per cent. ? Ans. 8 5.69,6. 30. What is the interest of 8 368.18, from April 2, 1816, to June 19, 1835, at 2 per cent. ? Ans. 8141.48,3. 31. What is the interest of 816.16, from March 3, 1831, to Dec. 6, 1833, at 1 per cent. ? Ans. 8 0.44,5. 32. What is the interest of 81728.19, from May 7, 1824, to July 17, 1830, at per cent. ? Ans. $ 26.76,2. 33. What is the interest of 8397.16, from Dec. 29, 1831, to June 30, 1833, at 5 per cent. ? Ans. 8 32.82,6. 34. What is the amount of $1760.07, from Feb. 17, 1831, to Dec. 19, 1832, at $ per cent. ? Ans. 81776.25,2. 35. G. K. M., of Bradford, has sent shoes at several times to Spofford & Tileston, New York, as follows : January 16, 1834, were sent shoes to the value of 8 865.00 Feb. 17, 1834, " " " 386.27 March 29, 1834, " " " 769.25 May 25, 1834, " " " 183.75 June 19, 1834, " " " 396.81 The above were sold on six months' credit. G. K. M. has re- ceived of Spofford & Tileston as follows : Sept. 1, 1834, 81000; Oct. 19, 1834, 8375.25; Nov. 15, 1834, $681.29; Dec. 8, 1834, 8100 ; March 12, 1835, 8275.28. Required the balance at the time of settlement, Sept. 9, 1835. Ans. Spofford & Tileston owe to G. K. M. $195.51+. CASE V. To find the interest for any sum for days. RULE.* If the rate per cent. be. 06, multiply the principal by the number of days, and divide by 6083J, and the quotient is the interest ; but if the rate per cent, be .05, divide by 7300. * This rule is but an abridgment of the obvious method for obtaining the interest for any number of days, which is, first to find it for one year, then divide by 365, which gives it fbr one day, and then multiply the 172 SIMPLE INTEREST. [SECT, XMTII. EXAMPLES. 1. What is the interest of $1835 for 35 days ? Ans. $10.55,7. 2. What is the interest of $165.37 for 165 days? Ans. $ 4.48,5. 3. What is the interest of $16.87 for 79 days, at 5. per cent.? Ans. $0.18,2. 4. What is the interest of $167 for 87 days, at 5 per cent.? Ans. $1.99. 5. What is the interest of $ 761.81 for 165 days ? Ans. $ 20.66,2. 6. What is the interest of $76.18,5 for 315 days, at 5 per cent. ? Ans. $ 3.28,7. 7. What is the interest of $178.69,7 for 271 days ? Ans. $ 7.96. 8. What is the interest of $1728.79 for 318 days, at 5 per cent ? Ans. $ 75.30,8. 9. What is the interest of $73 for 73 days, at 5 per cent. ? Ans, $ 0.73. 10. What is the interest of $ 96.10 for 54 days ? Ans. $ 0.85,3. 11. What is the interest of $144.50 for 144 days, at 5 per cent. ? Ans. 6 2.85. 12. What is the amount of $1728 for 200 days ? Ans. $1784.81. result by the number of days. The full operation of finding the interest of any sum, say $1835 for 3o days, will be as follows: 1835 3( ^- 06 X 35 = 10.55,7+. Now it is evident, that, instead of multiplying 1835 (the numerator) by .06, we may divide 365 (the denominator) by it, without affecting the re- sult. But 365 -7- .06 = 6083-+-, so that we have only to multiply the principal by the number of days, and divide the product by 6083J, when the rate is o per cent., and we have the interest required. For the same reason, we divide by 7300 when the rate is 5 per cent. In the same way a divisor may be found, answering to any given rate. All the foregoing rules are on the principle, that there are only 360 days in a year ; yet they are adopted by all mercantile men, and also by banks. But if a note be written for 144 days, the holder of it can obtain interest for only J^f of a year by the law of Massachusetts. If a note be written for months, calendar months are understood, whether the months have 28 or 31 days. If a note be dated April 21st, for one month, it will be due May 21st; but if it be dated Jan. 28th, 29th, 30th, or 31st, it being for one month, it will be due Feb. 28th, it being the last day of the month, unless it be leap year. SECT, xxxviii.] SIMPLE INTEREST. SECTION XXXVIII. PARTIAL PAYMENTS. WHEN notes are paid within one year from the time they be- come due, it has been the usual custom to find the amount of the principal from the time it became due until the time of payment ; and then to find the amount of each indorsement from the time it was paid until settlement, and to subtract their sum from the amount of the principal. EXAMPLES. (1.) $1728.00. Baltimore, January 1, 1833. For value received, I promise Riggs, Peabody, & Co. to pay them, or order, on demand, one thousand seven hundred and twenty-eight dollars, with interest. John Paywell, Jr. On this note are the following indorsements. March 1, 1833, received three hundred dollars. May 16, 1833, received one hundred and fifty dollars. Sept. 1, 1833, received two hundred and seventy dollars. De- cember 11, 1833, received one hundred and thirty-five dollars. What was due at the time of payment, which was Decem- ber 16, 1833 ? OPERATION. Principal, ... Interest for 11 months and 15 days, First payment, - - $300.00 Interest for 9 months and 15 days, 14.25 Second payment, ... 150.00 Interest for 7 months, - - 5.25 Third payment, - - 270.00 Interest for 3 months and 15 days, 4.72 Fourth payment, ... 135.00 Interest for 5 days, - - .11 Ans. $ 948.03. $1728.00 99.36 $1827.36 $ 879.33 Balance remaining due, Dec. 16, 1833, $ 948.03 (2.) $700.00. Concord, Feb. 4, 1834. For value received, we jointly and severally promise James Thomas to pay him, or order, on demand, seven hundred dol- lars, with interest. Sampson Phillips. Attest, Henry Dix. Richard Fletcher. 15* 174 SIMPLE INTEREST. [SECT, xxxvni. On this note are the following payments. March 18, 1834, received one hundred and sixty dollars. June 24, 1834, received two hundred dollars. September 11, 1834, received one hundred and twenty dollars. October 5, Iti34, received sixty dollars. What was due on this note Nov. 28, 1834 ? Ans. $ 180.43. (3.) $ 600.00. Portland, May 16, 1834. For value received, we, Luke A. Homer, as principal, and Daniel D. Snow and Ichabod Frost, as sureties, promise John Webster to pay him, or order, in one year, six hundred dollars, with interest after three months. Luke A. Homer. Daniel D. Snow. Attest, M. Peters. Ichabod Frost. On this note are the following indorsements. Sept. 18, 1834, received one hundred and 'thirty-six dollars. December 5, 1834, received one hundred and ninety-seven dollars. February 11, 1835, received two hundred dollars. April 19, 1835, received forty dollars. What was due August 1, 1835 ? Ans. $ 40.31,2. In the United States courts, and in most of the courts of the several States, the following rule is adopted for estimating in- terest on notes and bonds, when partial payments have been made. RULE. Compute the interest on the principal sum, from the time when the interest commenced to the first time when a payment was made, which exceeds, either alone or in conjunction with the preceding payments, if any, the interest at that time due ; add that interest to the principal, and from the sum subtract the payment made at that time, together with the preceding payments, if any, and the remainder forms a new principal ; on whicJi compute and subtract the interest, as upon the first principal, and proceed in the same manner to the time of judgment. The following example will illustrate the above rule. (4.) $165.18. Boston, June 17, 1827. For value received, I promise James E. Snow to pay him, or order, on demand, one hundred and sixty-five dollars and eighteen cents, with interest. James Y. Frye. Attest, John True. On this note are the following indorsements. December 7, 1827, received eighteen dollars and thirteen cents of the within note. October 19, 1828, received twenty-eight dollars and sixteen cents. September 25, 1829, received thirty-six dollars and twelve cents. July 10, 1830, re- ceived three dollars and eighteen cents. June 6, 1831, received thirty- SECT. XXXVIII.] SIMPLE INTEREST. 175 six dollars and twenty-eight cents. December 28, 1832, received thirty- one dollars and seventeen cents. May 5, 1833, received three dollars and eighteen cents. September 1, 1833, received twenty-five dollars and eighteen cents. October 18, 1834, received ten dollars. How much remains due September 27, 1835 ? Ans. $15.41,7. METHOD OF OPERATION. Principal, carrying interest from June 17, 1827, Interest from June 17, 1827, to Dec. 7, 1827, 5mo. 20 days, Amount, First payment, Dec. 7, 1827, Balance for new principal, - Interest from Dec. 7, 1827, to Oct. 19, 1828, lOmo. 12da., Amount, Second payment, Oct. 19, 1828, - Balance for new principal, - Interest from Oct. 19, 1828, to Sept. 25, 1829, llmo. 6da., Amount, Third payment, Sept. 25, 1829, - Balance for new principal, - Interest from Sept. 25, 1829, to June 6, 1831, 20mo. llda., Amount, Fourth pay't, July 10, 1830, a sum less than int'st, 3.18 Fifth pay't, June 6, 1831, a sum greater than irit'st, 36.28 Balance for new principal, - Interest from June 6, 1831, to Dec. 28, 1832, I8mo. 22da., Amount, Sixth payment, Dec. 28, 1832, ... Balance for new principal, Interest from Dec. 28, 1832, to May 5, 1833, 4mo. 7da., Amount, Seventh payment, May 5, 1833, - Balance for new principal, - Interest from May 5, 1833, to Sept. 1, 1833, 3mo. 26da., Amount, Eighth payment, Sept. 1, 1833, - Balance for new principal, - - Interest from Sept. 1, 1833, to Oct. 18, 1834, 13mo. 17da., Amount, $165.18,0 4.68,0 169.86,0 18.13,0 151.73,0 7.88,9 159.61,9 28.16,0 131.45,9 7.36,1 138.82,0 36.12,0 102.70,0 10.45,8 113.15,8 39.46,0 73.69,8 6.90,3 80.60,1 31.17,0 49.43,1 1.04,6 50.47,7 3.18,0 47.29,7 91,4 48.21,1 25.18,0 23.03,1 1.56,2 24.59,3 176 SIMPLE INTEREST. [SECT, xxxvm. Amount brought forward, $ 24.59,3 Ninth payment, - Balance for new principal, - 14.59,3 Interest from Oct. 18, 1834, to Sept. 27, 1835, llmo. 9da., .82,4 Balance due at the time of payment, - - $ 15.41,7 (5.) $ 769.87. Salem, June 17, 1829. For value received, I promise L. Swan to pay him, or order, on demand, seven hundred and sixty-nine dollars and eighty- seven cents, with interest. Samuel Q. Peters. , Attest, Moses Haynes. On this note are the following payments. March 1, 1830, received seventy-five dollars and fifty cents. June 11, 1831, received one hundred and sixty-five dollars. September 15, 1831, received one hundred and sixty-one dollars. Jan. 21, 1832, received forty-seven dollars and twenty- five cents. March 5, 1833, received twelve dollars and seventeen cents. December 6, 1833, received ninety-eight dollars. July 7, 1834, received one hundred and sixty -nine dollars. What remains due Sept. 25, 1835 ? Ans. $ 226.29,7. (6.) $ 300.00. Haverhill, April 30, 1831. For value received, I promise Kimball & Hammond to pay them, or order, on demand, three hundred dollars, with in- terest. Simpson W. Leavet. Attest, James Quintire. The following partial payments were made on this note. June 27, 1832, received one hundred and fifty dollars. December 9, 1832, re- ceived one hundred and fifty dollars. What was due, Oct. 9, 1833 ? Ans. $ 26.73,5. (7.) $ 54.18. New York, Feb. 11, 1832. For value received, I promise John Trow to pay him, or order, on demand, fifty-four dollars and eighteen cents, with interest. Luke M. Sampson. On this note are the following payments. July 11, 1833, received twelve dollars and twenty-five cents. August 15, 1834, received two dollars and ten cents. July 9, 1835, received three dollars and twelve cents. August 21, 1835, received thirty-seven dollars and eighteen cents. What was due Dec. 17, 1835. Ans. $10.22,2. (8.) $1728.00. Boston, Jan. 7, 1831. For value received, we jointly and severally promise Jones, Oliver, & Co. to pay them, or order, on demand, one thousand seven hundred and twenty-eight dollars, with interest. John Bountiful. Attest, Timothy True. James Trusty. . SECT, xxxviii.] SIMPLE INTEREST. 177 On this note are the following payments. February 9, 1832, received seven hundred and sixty dollars and twenty-eight cents and five mills. March 5, 1833, received sixty-eight dollars and fifty cents. December 28, 1833, received eight hundred and seventy-six dollars and twenty-eight cents. July 17, 1834, received sixty dollars. What was due at the time of payment, which was Oct. 1, 1834 ? Ans. $ 209.22,9. (9.) $ 500.00. Philadelphia, May 7, 1829. For value received, I promise John Jordan to pay him, or order, on demand, five hundred dollars, with interest. Thomas C. True. The following partial payments were indorsed on this note. June 29, 1830, received one hundred dollars. December 5, 1831, received one hundred dollars. March 12, 1832, received five dollars. July 4, 1833, received ninety-five dollars. December 1, 1834, received two hundred dollars. What was due Jan. 1, 1836 ? Ans. $141.50,4. (10.) $ 89.75. Newburyport, March 19, 1831. For value received, we jointly and severally promise John Frost to pay him, or order, on demand, eighty-nine dollars and seventy- five cents, with interest. Henry Augustus. Attest, James Snow. Marcus T. Cicero. On this note are the following indorsements. December 6, 1831, re- ceived twelve dollars and twelve cents. February 17, 1832, received twelve dollars and twelve cents. March 19, 1833, received three dollars and sixteen cents. December 28, 1834, received two dollars and eighteen cents. January 1, 1835, received twenty-five dollars and twenty-five cents. March 11, 1835, received thirty-one dollars arid eighteen cents. July 17, 1835, received five dollars and eighteen cents. September 1, 1835, received six dollars and twenty-nine cents. What was due Dec. 29, 1835 ? Ans. $10.57. (11.) $ 1000.00. New York, January 1, 1840. For value received, I promise to pay James Johnson, or or- der, on demand, one thousand dollars, with interest at seven per cent. Samuel T. Fortune. Attest, Job Harris. On this note are the following indorsements. Sept. 28, 1840, received one hundred and forty-four dollars. March 1, 1841, received twenty dol- lars. July 17, 1841, received three hundred and sixty dollars. Aug. 9, 1841, received one hundred and ninety dollars. Sept 25, 1842, received one hundred and seventy dollars. Dec. 11, 1843, received two hundred dollars. July 4, 1845, received seventy-five dollars. What was due June 1, 1847 ? Ans. $ 7.61. 178 SIMPLE INTEREST. [SECT, xxxvm. The following is the rule established by the Supreme Court of the State of Connecticut RULE. Compute the interest 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 there 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 man- ner from one payment to another, till all the payments are absori>ed ; provided the time between one payment and another be one year or more. But if any payments be made before one year's interest hath accrued, then compute the interest on the principal sum due on the obligation for one 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 together.* If any payments be made of a less sum than the interest arisen at the time of such payment, no interest is to be computed, but only on the principal sum for any period. (12.) $ 900.00. New Haven, June 1, 1828. For value received, I promise J. D. to pay him, or order, nine hundred dollars, on demand, with interest. James L. Emerson. On this note are the following indorsements. June 16, 1829, received two hundred dollars of the within note. Aug. 1, 1830, received one hun- dred and sixty dollars. Nov. 16, 1830, received seventy-five dollars. Feb. 1, 1832, received two hundred and twenty dollars. What was due August 1, 1832 ? Ans. $ 417.82,2. OPERATION. Principal, ...... $900.00 Interest from June 1, 1828, to June 16, 1829, 12 months, 56.25 956.25 First payment, ..... 200.00 756.25 Interest from June 16, 1829, to Aug. 1, 1830, 13 months, 51.04,6 807.29,6 Second payment, ... . 160.00,0 647.29,6 Interest for one year, 38.83,7 686.13,3 * If the year extends beyond the time of payment, find the amount of the remaining principal to the time of payment; find also the amount of indorsement, or indorsements, and subtract their sum from the amount of the principal. SECT, XXXYIII.] SIMPLE INTEREST. 179 Amount brought forward, $ 686.13,3 Am't of 3d pay't, from Nov. 16, .1830, to Aug. 1, 1831, 8mo., 78.18,7 607.94,6 Interest from Aug. 1, 1831, to Aug. 1, 1832, 12 months, 36.47,6 644.42,2 Am't of 4th pay't, from Feb. 1, 1832, to Aug. 1, 1832, 6mo., 226.60,0 Balance due August 1, 1832, - $417.82,2 Method of computing annual interest in the State of Ver- mont. When the contract is for the payment of interest annually, and no payments have been made, we adopt the following RULE. Find the interest of the principal for each year, separately, up to the time of payment ; we must then find the simple interest of these interests, severally , from the time they become due up to the time of pay- ment, and the sum of all the interests added to the principal will be the amount. Should payments have been made, we find the amount of the principal and from this we subtract the amount of the indorsement or indorse- ments to the end of the first year. We proceed in the same manner to the time of payment , But when the contract is for a sum payable at a specified time, with, annual interest, and payments are made before the debt becomes due, we find the interest of the principal up to the time of the first payment, and this interest we reserve ; we then subtract the payment from the princi- pal, and find the interest of the remainder up to the time of the next payment, which interest we add to the other interest, and so continue up to- the time the debt becomes due, and the sum of the interests, added to the last principal, ivill be the amount due at that time. After the debt becomes due, the interest is to be extinguished annually, if the payments are sufficient for that purpose. 13. J. Jones has J. Smith's note, dated January 1, 1840, for $ 500, with interest, to be paid annually, at 6 per cent. What was due January 1, 1844 ? Ans. $ 630.80. OPERATION. 1st year, $ 500 x .06 $ 30. $ 30 X . 18 = $ 5.40 2d ' " $ 500 x. 06 = $30. $30 x .12 =$3.60 3d " $ 500 x. 06 = $30. $30 x .06 = $1.80 4th $ 500 X. 06 = $30. $1080 interest of $ 120 interest of principal. [interest. Principal $ 500 00. Th - g . g ^ method when Interest of principal, 120.00. ^ no indorsements> Interest of interest, Amount, $ 630,80 Ans. 180 SIMPLE INTEREST. [SECT. ZXXVIH. (14.) $ 300.00. Thetford, Vt., January 1, 1843. For value received, I promise to pay Hiram Orcutt, Esq., or order, on demand, three hundred dollars, with interest an- nually. M. T. Cicero. Attest, P. M. Virgil. On this note are the following indorsements. Sept. 1, 1843, received eighty dollars. July 1, 1844, received one hundred dollars. .April 1, 1845, received fifty dollars. What was due January 1, 1847 ? OPERATION. Principal, Interest for one year, First payment, ... Interest on first payment, 4 months, - Amount of payment, New principal, ... Interest on new principal, one year, Second payment, Interest on second payment, 6 months, Amount of payment, - New principal, ... Interest on new principal, one year, Third payment, Interest on third payment, 9 months, $300.00 18.00 Amount, 318.00 80.00 1.60 - sTeo 236.40 14.18 Amount, 250.58 100.00 3.00 103.00 - 147.58 Amount,, 50.00 2.25 52.25 104.18 6.25 - $ 1 10.43 Amount of payment, .... New principal, ..... Interest on new principal, one year, Remains due January 1, 1847, ... The above is the method of computing annual interest, when there are indorsements on the note, and it is not payable at a specified time. (15.) 8 600.00. Montpelier, Vt., June 1, 1844. For value received, I promise to pay John Smith, or order, six hundred dollars, in three years, with interest annually. Attest, S. Morse. John Y. Jones. SECT, xxxix.] SIMPLE INTEREST. 181 On this note are the following indorsements. Sept. 1, 1845, received two hundred and fifty dollars. Nov. 1, 1846, received two hundred dollars. What was due June 1, 1847 ? Ans. $ 242.75. OPERATION. $600 x .105 = 63.00, interest from June 1, 1844, to Sept. 1, 250 [1845. 350 X. 07 =24.50, interest from Sept. 1, 1845, to Nov. 1, 200 [1846. 150 x .035= 5.25, interest from Nov. 1, 1846, to June 1, 92- 75 $ 2.75, amount of interest. [1847. $242.75 remains due June 1, 1847. The above is the method of computing annual interest, when the contract is for a specified time. NOTE. The above methods of computing annual interest are not only practised in the courts of Vermont, but a similar method has been adopted in some of the courts of New Hampshire j but it is not the method in Massachusetts. SECTION XXXIX. MISCELLANEOUS PROBLEMS IN INTEREST. PRINCIPAL, interest, and time given, to find the rate per cent. 1. At what rate per cent, must $500 be put on interest to gain $ 120 in 4 years ? OPERATION. BY ANALYSIS. $500 The interest of $1 for the .0 1 given time at one per cent, is 4 5 00 cents. $ 500 will be 500 times as 4 much, = $ 500 X. 04 = $20.00. 2000) 120.00(6 per cent. Ans. ^ If. *. 20 1 f7 e l P er cen *' 120.00 2 Wl11 & lve W* 6 P er cent - RULE. Divide the given interest by the interest of the given sum at 1 per cent, for the given time, and the quotient will be the rate per cent, required. 2. At what rate per cent, must 8120 be on interest to amount to $133.20 in 16 months ? Ans. 8J- per cent. 16 182 SIMPLE INTEREST. [SECT, xxxix. 3. At what rate per cent, must $ 280 be on interest to amount to $ 411.95 in 6 years ? Ans. 7 per cent. Principal, interest, and rate per cent, given, to find the time. 4. How long must $500 be on interest at 6 per cent, to gain $120 ? OPERATION. BY ANALYSIS. $ 50 We find the interest of 8 1 .00 at the given rate for one year is 6 30.00) 120.00(4 years, Ans. cents. $ 500 will, therefore, be 500 120.00 times as much, = $ 500 X -06 = $ 30.00. Now, if it take 1 year to gain $ 30, it will require -^ to gain $ 120, = 4 years, Ans. RULE. Divide the given interest by the interest of the principal for 1 year, and the quotient is the time. 5. How long must $120 be on interest at 8 percent, to amount to $133.20? Ans. 16 months. 6. How long must $ 280 be on interest at 7 per cent, to amount to $ 411.95 ? Ans. 6 years. Interest, time, and rate per cent, given, to find the principal. 7. What principal at 6 per cent, is sufficient in 4 years to gain $120? OPERATION. BY ANALYSIS. $1 .06 The interest of $1 for the given ^Qg rate and time is 24 cents. If 24 4 cents, then, give $1 for principal, ^4)120.00(1 500, Ans. $12 ' 00 *g 1 W* times as 120 00 much, $ 500, Ans. RULE. Divide the given interest or amount by the interest or amount of $ 1 for the given rate and time, and the quotient is the principal. 8. What principal at 8 per cent, is sufficient in 16 months to gain $ 13.20 ? Ans. $ 120.00. 9. What principal at 7 per cent, is sufficient in 6J- years to amount to $ 41 1.95 ? Ans. $ 280.00. I SFCT. XL.] COMPOUND INTEREST. 183 SECTION XL. COMPOUND INTEREST. THE law specifies that the borrower of money shall pay a certain number of dollars, called per cent., for the use of one hundred dollars for a year. Now, if this borrower does not pay to the lender this per cent, at the end of the year, it is no more than just that he should pay interest for the use of it, so long as he shall keep it in his possession ; and this is called Compound Interest. 1. What is the compound interest of $ 300 for 3 years ? Ans. 9 57.30,4. First Method. Second Method. $ 300, principal. 5 = 1 S )300 1.06 l=s) 15 TOO", interest for 1 year. ? 300 5 = ^)318 318.00, amount for 1 year. 1== 3 ) 15.90 1.06 3 " 18 19.08, int. for second year. !* 200, $100 of which I was to pay down, and the other $100 in two years, the equated time for the payment of both sums would be one year. It is evident, that, for deferring the payment of the first $ 100 for 1 year, I ought to pay the amount of $ 100 for that time, which is $ 1 06 ; but for the other $ 100, which I pay a year before it is due, I ought to pay tine present worth of $ 100, which is $94.33fg, where- as, by Equation of Payments, I only pay $200. Strict justice would there- fore demand that interest should be required on all sums from the time they become due until the time of payment, and the present worth of all sums paid before they are due. The better rule would be, to find the pres- ent worth on each of the sums due, and then find in what time the sum of these present worths would amount to the payments. 200 EQUATION OF PAYMENTS. [SECT. XLIX. 3. There is owing to a merchant $ 1000 ; $ 200 of it is to be paid in 3 months, $ 300 in 5 months, and the remainder in 10 months. What is the equated time for the payment of the whole sum ? Ans. 7 months 3 days. 4. A owes B $150, $ 50 to be paid in 4 months, and in 8 months. B owes A $ 250 to be paid in 10 months. It is agreed between them that A shall make present payment of his whole debt, and that B shall pay his so much sooner as to bal- ance the favor. I demand the time at which B must pay the $ 250. Ans. 6 months. 5. A merchant has $144 due him, to be paid in 7 months, but the debtor agrees to pay one half ready money, and one third in 4 months. What time should be allowed him to pay the remainder ? Ans. 2 years 10 months. 6. There is due to a merchant $ 800, one sixth of which is to be paid in 2 months, one third in 3 months, and the remain- der in six months ; but the debtor agrees to pay one half down. How long may the debtor retain the other half so that neither party may sustain loss ? Ans. 8 months. 7. I have purchased goods of A. B. at sundry times and on various terms of credit, as by the statement annexed. When is the medium time of payment ? Jan. 1, a bill amounting to $ 375.50 on 4 months' credit. " 20, " " 168.75 on 5 months' credit. Feb. 4, u " 386.25 on 4 months' credit. March 11, " " 144.60 on 5 months' credit. April 7, " 386.90 on 3 months' credit. FORM OF STATEMENT. Due May 1, $375.50 June 20, 168.75 x 50= 843750 June 4, 386.25 x 34 = 1313250 Aug. 11, 144.60 x 102 = 1474920 July 7, 386.90 x 67 = 2592230 $1462.00 )6224150(42fda. Ans. 584800 376150 292400 83750 The medium time of payment will therefore be 42J-ff days, that is, 43 days from May 1, which will be June 12. ECT. i.] CUSTOM-HOUSE BUSINESS. 201 S. I have sold to C. D. several parcels of goods, at sundry times, and on various terms of credit, as by the statement an- nexed. Jan. 1, a bill amounting to $ 600 on 4 months' credit Feb. 7, " 370 on 5 months' credit. March 15, " " 560 on 4 months' credit. April 20, " " 420 on 6 months' credit. When is the equated time for the payment of all the bills ? Ans. July 11. 9. Purchased goods of John Brown, at sundry times, and on various terms of credit, as by the statement annexed. March 1, 1845, a bill amounting to $ 675.25 on 3 months. July 4, " " ' 376. 18 on 4 months. Sept. 25, " " " 821.75 on 2 months. Oct. 1, " " 961.25 on 8 months. Jan. 1, 1846, " 144.50 on 3 months. Feb. 10, " " " 81 1.30 on 6 months. March 12, " 567.70 on 5 months. April 15, " 369.80 on 4 months. What is the equated time for the payment of the above bills ? Ans. March 16, 1846. SECTION L. CUSTOM-HOUSE BUSINESS. IN every port of the United States where merchandise is either exported or imported, there is an establishment called a Custom-house. Connected with this are certain officers, ap- pointed by government, called custom-house officers, whose business is to collect the duties on various kinds of merchandise, &c., imported into the United States. The following article on Allowances, &c., was very politely furnished the author by the officers of the Boston custom-house, and may therefore be relied on as perfectly correct. Allowances. Draft, is an allowance made by the officers of the United States government in the collection of duties on merchandise liable to a specific duty, and ascertained by weight, and is also 202 CUSTOM-HOUSE BUSINESS. [SECT. L. given by the usage of merchants in buying and selling. It is a deduction from the actual gross weight of the article paying du- ty by the pound or sold by weight. For example, a box of sugar actually weighs 500 pounds. The draft upon this weight is 4 pounds. 500 gross. 4 draft. 496 difference. Upon this difference is made a further allow- ance of fifteen per cent, as tare, or as the actual weight of the box before the sugar was put into it This tare is allowed by the government in the collection of the duty, and by the mer- chant in buying and selling. Take, then, the box of sugar, say 5001b. gross. 4 draft. 496" difference. 74 tare. 422 net weight, upon which a duty is paid to the government, or price is paid to the merchant in his sale. This tare of 74 pounds, or 15 per cent., is usually more than the actual tare, but is assumed as the probable or actual tare, by reason of the im- possibility of " starting " every box to ascertain the actual weight of the sugar, and the actual weight of the box which contains it. This tare is sufficiently correct for the collector of the duty, and the merchant who deals in the article. It is intended to be a liberal allowance, and varies but little from the actual tare. Drafts allowed at the custom-house in the collection of du- ties, and by the merchants in their purchases and sales, are as follow : Allowance for Draft. Draft, is another name for Tret^ which is an allowance in weight for waste. lb. lb. On 112 1 Above 112 and not exceeding 224 2 " 224 " " 336 3 " 336 " " 1120 4 " 1120 " " 2016 7 2016 9 EXPLANATION. ^ Many articles of merchandise are weighed separately ; for example, boxes and casks of sugar, chests of SECT. L.] CUSTOM-HOUSE BUSINESS. 203 tea and indigo. Upon each box or cask, or chest, an allowance should be made for draft, according to its weight, as by the above rule. Bags of sugar and coffee, or bars of iron and bun- dles of steel, might be weighed together ; say, 10 bags of coffee at one draft might weigh 1121 pounds ; from this gross weight must be deducted 7 pounds as draft ; 35 bars of Russia iron might be weighed at one draft, weight 2250 pounds, upon which would be an allowance of 9 pounds draft, and by law and usage there can be no greater allowance than 9 pounds for draft. A greater or less number of bags of coffee, or bars of iron, or any other article of merchandise, is weighed, and the deduction is according to the weight of each draft. An old rule, and probably a better one, among merchants was the allowance of per cent, on the gross weight of all merchandise weighed, as draft. Allowance for Leakage. Two per cent, is allowed on the gauge of ale, beer, porter, brandy, gin, molasses, oil, wine, and rum, and other liquors in casks, besides the real wants of the cask ; for example, a cask of molasses may gauge 140 gallons, gross gauge ; from this first deduct 5 gallons, the actual wants, or the quantity ne- cessary to fill the cask, we have 140 gross. 5 out. 135 difference. 3 two per cent, for leakage. 132 gallons net. Tare is an allowance made for the actual or supposed weight of the cask, box, case, or bag, which contains the article of merchandise. The usage of merchants is in conformity with the law and usage of the officers of the customs in their allowance for tare, directed by law, or found to be correct by their examination and experience. The tariff of the United States being in its details so unset- tled, it is deemed advisable not to insert any table. EXAMPLES. 1. Find the net weight of a hogshead of sugar, weighing gross 12281b., tare 12 per cent. Ans. 10751b. 204 CUSTOM-HOUSE BUSINESS. [BKCT. i. OPERATION. 12281b. gross weight. JTlb. draft. 1221 12 per cent, of 122 lib. 1461b. tare. 10751b. net weight. NOTE. For draft let the pupil examine page 202. 2. Required the net weight of 6 boxes of sugar, weighing gross as follows, the tare being 15 per cent. : - No. 1, 450 Ibs. No. 2, 470 do. OPERATION. No. 3, 510 do. 29141b. gross. No. 4, 496 do. 6x4 = 241b. draft No. 5, 468 do. 2890 No. 6,520 do. Tare 15 per cent. 433 Gross, 2914 24571b. net weight. 3. What is the net weight of 4 chests of tea, which weigh as follows, tare 22 per cent. ? Ans. 3841b. OPERATION. No. 1, 120 Ibs. 4801b. gross. No. 2, 116 do. 81b. draft. No. 3, 126 do. 472 No. 4, US do. 22 x 4 =_881b. tare. 480 do. 3841b. net. OPERATION. 4. What is the net weight of 5 bags 53 }' of pepper, weighing as follows : 1081b., 1 laib., lOOlb., 120lb., and 981b., tare 2 535 per cent. ? Ans. 524. 2 P er cent^lllb. tare. 5241b. net. OPERATION. 5. What is the net weight of 4 , ... t> ,, 51b. arait. kegs of mace, weighing as follows : 1121b., 1201b., 1181b., and llOlb., tare 33 per cent. ? Ans. 3051b. 33 P er cent - 15Qlb - **Te. 3051b. net. NOTE. In making allowances, if there be a fraction of more than half a pound, 1 pound is added to the tare. IECT. LI.] RATIO. 205 AMERICAN DUTIES. The duties on merchandise imported into the United States are either specific or ad valorem duties. Specific duty is a certain sum paid on a ton, hundred weight, pound, square yard, gallon, &c. ; but when the duty is a certain per cent, on the actual cost of the goods in the country from which they are imported, it is called an ad valorem duty, that is, a duty according to the value of the article. 6. What is the duty on 6 hogsheads of sugar, weighing gross as follows: No. 1, 1276lb., No. 2, 12801b., No. 3, 11781b., No. 4, 13781b., No. 5, 15701b., No. 6, 1338lb. ; duty 2J- cents per lb., tare 12 per cent. ? Ans. $175.52,5. 7. What is the duty on an invoice of woollen goods, which cost in London 986 sterling, at 44 per cent, ad valorem, the pound sterling being $ 4.84 ? Ans. $ 2099 78-|-. 8. Required the duty on 5 pipes of Port wine, gross gauge as follows : No. 1, 176 gallons, No. 2, 145 gallons, No. 3, 128 gallons, No 4, 148 gallons, No. 5, 150 gallons ; wants of each pipe, 4 gallons ; duty 15 cents per gallon. Ans. $106.80. 9. Required the duty on a cargo of iron, weighing 270 tons, at $ 30 per ton ? Ans. $ 8100. 10. Compute the duty on 7890 pounds of tarred cordage, at 4 cents per pound ; duty If per cent. Ans. $310.08. 11. What duty should be paid on 10 casks of nails, weighing each 4501b. gross, at 4 cents per lb. ? Ans. $164.12. SECTION LI. RATIO. RATIO is the relation which one quantity bears to another of the same kind with respect to magnitude ; and the comparison is made by considering how often the one is contained in the other, or how often the one contains the other. Thus, the ratio of 12 to 3 is expressed by dividing 12 by 3 -\ = 4, ratio ; or it may be expressed by dividing 3 by 12 = T 3 2- = ^, ratio. The former is the method by which the English mathematicians express ratio, and the latter is the French method. The former of these quantities is called the antecedent, and the latter the consequent. 18 206 RATIO. [SECT. LI. When the antecedent is equal to the consequent, it is called a ratio of equality ; thus the ratio of 6 to 6 = f = 1. But if the antecedent be larger than the consequent, it is a ratio of greater inequality ; and if the antecedent be less than the con- sequent, it is a ratio of less inequality. The antecedent and consequent are called the terms of the ratio ; and the quotient of the two terms is the index or expo- nent of the ratio. Compound ratio is made up of two or more ratios, by multi- plying their terms and exponents together. The ratio of 8 to 6 and of 4 to 2 may be compounded ; thus, 8 to 6 = |; 4 to 2 ; 8 x 4 to 6 x 2 = g| ; 32 to 12 = f } , If a ratio be compounded of two equal ratios, it is called a duplicate ratio ; of three ratios, it is called a triplicate ratio, dec. Thus, if the ratio of 4 to 2 be 2, and the ratio of 6 to 3 be 2, the ratio of 4 X 6 to 2 X 6 will be 2 X 2, that is, the ratio of 24 to 12 will be 2 2 , &c. If the terms of a ratio be prime to each other, no quantities can be found in the same ratio but what would be multiples thereof. Numbers that are prime to each other are the least of all numbers in the same ratio. If, therefore, we wish to ascertain whether the ratio of 3 to 7 is greater or less than the ratio of 4 to 9, since these ratios are represented by the fractions y and $-, we reduce them to a common denominator, f and f f ; and, since the latter of these is greater than the former, it is evident that the ratio of 3 to 7 is less than the ratio of 4 to 9. If we have the terms of a ratio given in large numbers, that are prime to each other, and we wish to find a ratio nearly equivalent, whose terms are expressed by smaller numbers, we adopt the following RULE. Divide the greater term by the Jess, and that divisor by the remainder, as in Sect. XVI. , Case I., of Vulgar Fractions. Then, if the antecedent be greater than the consequent, the first quotient divided by 1 gives the first ratio ; if less, a unit divided by the first quotient will express the first ratio. Multiply the terms of the first ratio by the second quotient, and add a unit to the numerator or denominator, according as the antecedent of the original terms is greater or less than its consequent, and ice have the second ratio. Then, as a general principle, we multiply fhe terms of the ratio last SECT. LII.J PROPORTION. 207 found by tlie next succeeding quotient, and to the product we add the cor- responding terms of the preceding ratio, and we have the next succeed- ing ratio ; and thus we proceed until there is no remainder, or until we have arrived at a sufficient approximation. 1. Let it be required to find a series of ratios in less num- bers, constantly approaching to the ratio of 314159 to 100000, which is nearly the ratio of the circumference of- a circle to its diameter. OPERATION. 100000)314159(3 300000 ^ 14159)100000(7 99113 887~) 14159(15 13305 "854)887(1 854 33, &c. 3 = f , the first ratio. f 3 * 7)+1 = 2 T 2 -, the second ratio, being the approximation of Ar- [chimedes. third ratio. fourth ratio, the approximation of Metius. -f- }, &c., in a continued frac- tion. SECTION LII. PROPORTION. PROPORTION is the likeness or equalities of ratios. Thus, be- cause 5 has the same relation or ratio to 10 that 8 has to 16, we say such numbers are in proportion to each other, and are therefore called proportionals. If any four numbers whatever be taken, the first is said to have the same ratio or relation to the second, thakthe third has to the fourth, when the first number or term contains the second 208 PROPORTION. [SECT. LII. as many times as the third contains the fourth, or when the second contains the first as many times as the fourth does the third. Thus, 8 has the same ratio to 4 that 12 has to 6, be- cause 8 contains 4 as many times as 12 does 6. And 3 has the same relation to 9 that 4 has to 12, because 9 contains 3 as many times as 12 does 4. Ratios are represented by colons, and the equalities of ratios by double colons. 3 : 9 : : 8 : 24 is read thus : 3 has the same ratio or relation to 9 as 8 to 24. The first and third numbers of a proportion are called antecedents, and the second and fourth are called consequents ; also, the first and fourth are called extremes, and the second and third are called means. Whatever four numbers are proportionals, if their antece- dents or consequents be multiplied or divided by the same num- bers, they are still proportionals ; and if the terms of one pro- portion be multiplied or divided by the corresponding term of another proportion, their products and quotients are still propor- tionals. This will appear evident from the various changes that the following example admits. 4 : 8 : : 3 : 6 Directly. 8 : 4 : : 6 : 3 By inversion. 4 : 3 : : 8 : 6 By permutation. 4-}-8:8::3-f-6:6By composition. 4:4-f8::3:3-|-6By composition. 4:8 4::3:6 3 By division. 8 4:8:: 6 3:6 By division. 4X4:8x8::3x3:6x6By compound ratios. : f : : : f By division. That the product of the extremes is equal to that of the means is evident from the following consideration. Let the fol- lowing proportionals be taken. 12 : 3 : : 8 : 2. From the def- inition of proportion, the first term contains the second as many times as the third does the fourth ; therefore, -^ f ; but -^ = *gt, and J = ^t ; and if 24, the numerator of the first fraction, which is a substitute for the first term, be multiplied by 6, the denominator of the second fraction, and a substitute for the fourth term, the product will be the same as if 6, the denomi- nator of the first fraction, and a substitute for the second term, be multiplied by 24, the numerator of the second fraction, and a substitute fot the third term. Thus 24 x 6 = 6 X 24. There- fore the product of the extremes is, in all cases, equal to that of the means. SECT, ni.] PROPORTION. 209 If, then, one of the extremes be wanting, divide the product of the means by the extreme given ; or, if one of the means be wanting, divide the product of the extremes by the means given, and the result will be the term sought. To apply this, we will take the following question. If 5 yards of cloth cost $15, what will 7 yards cost ? It is evident that twice the quantity of cloth would cost twice the sum, and that three times the quantity, three times the sum, &c. ; that is, the price will be in proportion to the quantity purchased. We then have 'three terms of a proportion given, one of the extremes and the two means, to find the other extreme. Thus, 5:7:: 15. Therefore, to find the other extreme by the rule above stated, we multiply the two means, 7 and 15, and divide their product by the extreme given, and the quotient is the extreme required. 7 X 15 = 105. 105 -H 5 = 21 dollars, the answer required. To perform this question by analysis, we reason thus. If 5 yards cost 15 dollars, 1 yard will cost one fifth as much, which is 3 dollars ; and if 1 yard cost 3 dollars, 7 yards will cost 7 times as much, which is 21 dollars. RULE.* State the question by making that number which is of the same name or quality of the answer required the third term; then, if the answer required is to be greater than the third term, make the second term greater tftan the first ; but if the answer is to be less than the third term, make the second less than the first. Reduce the first and second terms to the lowest denomination men- tioned in either, and the third term to the lowest denomination mentioned in it. Multiply the second and third terms together, and divide their product * This rule was formerly divided into the Rule of Three Direct, and the Rule of Three Inverse. The Rule of Three Direct included those ques- tions where more required more and less required less ; thus, If 51b. of coffee cost 60 cents, what would be the value of lOlb .? would be a ques- tion in the Rule of Three Direct, because the more coffee there was the more money it would take to purchase it. But if the question were thus : If 4 men can mow a certain field in 12 days, how long would it take 8 men ? it would be in Inverse, because the more men the less would be the time to perform the labor, that is, more would require less. The method for stating questions was this : To make that number which is the demand of the question the third term, that which is of the same name the first, and that which is of the same name as the answer re- quired, the second term . If the question was direct, the second and third terms must be multiplied together, and their product divided by the first ; but if it was inverse, the first and second terms must be multiplied together, and their product di- vided by the third. 18* 210 PROPORTION. [SECT. LII. by tlw first, and the quotient is the answer, in the same denomination to which the third is reduced. If any thing remains after division, reduce it to the next lower de- nomination, and divide as before. If either of tfte terms consists of fractions, state the question as in whole numbers, and reduce the mixed numbers to improper fractions, compound fractions to simple ones, and invert the first term, and then multiply me three terms continually together, and the product is the . answer to the question. Or the fractions may be reduced to a common denominator- and their numerators may be used as whole numbers. For tv hen fractions are reduced to a common denominator, their relative value is as their numerators. NOTE 1. In the Rule of Three, the second term is the quantity whose price is wanted ; the third term is the value of the first term ; when, therefore, the second term is multiplied by the third, the answer is as much more than it should be, as the first term is greater than unity ; therefore, by dividing by the first term, we have the value of the quan- tity required". Or, multiplying the third by the number of times which the second contains the first will produce the answer. NOTE 2. The pupil should perform every question by analysis, pre- vious to his performing it by Proportion. EXAMPLES. 1. If a man travel 243 miles in 9 days, how far will he travel in 24 days ? Ans. 648 miles. e a cube. To effect this, we must find the superficial contents of three sides of the cube, and with these we must divide the remain- ing number of cubic feet or blocks, and the quotient will show the thickness of the additions. As the length of a side is 30 feet, the superficial contents will be 30 X 30 = 900 square feet, and this multiplied by 3, the number of sides, will be 900 X 3 = 2700 feet With this as a divisor, we inquire how many times it is contained in 19,656, and find it to be 6 times (one or two units, and sometimes three, must be allowed on account of the other deficiencies in enlarging the cube). This 6 is the thickness of the additions to be made to the three sides of the cube, and by multiplying their superficial contents by it, we have the solid contents of the additions to be made 2700 X 6 = 16200 ; that is, we multiply the triple square by the last quotient figure, and this may be represented by the three superficies A B C D, E F G H, and I K L M. (See figure 2.) Having applied these additions to our cube, we find there are three other deficiencies, a b c <2, efg h, ij k /, the length of which is equal to that of the additions, 30 feet, and the height and breadth of each are equal to the thickness of the ad- ditions, 6 feet. To find the contents of these, we multiply the product of their length, breadth, and thickness by their num- ber ; thus, 6x6x30x3 = 3240 ; or, which is the same thing, we multiply the triple quotient by the square of the last quo- tient figure ; thus, 90 X 6 X 6 = 3240. See rule. Fig. 2; Having made these additions to the cube, we still find one other deficiency, N. (See figure 2 ) The length, breadth, and thickness of which are equal to the thickness of the former additions, viz. 6 feet. The contents of this are found by multiplying its length, breadth, and thickness together ; that is, cubing the last quotient figure ; thus, 6x6 X 6 =: 2 16. By making this last addition, we find that our cubical monument is finished, and that the first figure together with the several additions is equal to the cu- bical blocks, 46,656. 'a b/ / J*/ ) C e J S BA SECT. LXH.] CUBE ROOT. 251 Proof. 27000 = contents of fig. 1. 16200= " " first additions. 3240 = " " second additions. 216 = " " third addition. 46656 = contents of the whole monument. I. Required the cube root of 77308776. OPERATION. 77308776(426 root. 4 X 4 X 300 4800 64 4x 30 120 4920) 13308 = 1st dividend. 1st divisor = 4920 9600 4800x2 = 9600 480 120 x 2 x 2 = 480 8 2x2x2= 8 10088 = 1st subtrahend. 1st subtrahend = 10088 530460)1*220776 = 2d dividend. 42 x 42 x 300 = 529200 3175200 42 x 30 = 1260 45360 2d divisor = 530460 216 529200 x 6 = 3175200 3220776 = 2d subtrahend. 1260 x 6 X 6 = 45360 6x6x6= 216 2d subtrahend = 3220776 2. What is the cube root of 34965783 ? Ans. 327. 3. What is the cube root of 436036824287 ? Ans. 7583. 4. What is the cube root of 84.604519 ? Ans. 4.39. 5. Required the cube root of 54439939. Ans. 379. 6. Extract the cube root of 60236288. Ans. 392. 7. Extract the cube root of 109215352. Ans. 478. 8. What is the cube root of 116.930169 ? Ans. 4.89. 9. What is the cube root of .726572699 ? Ans. .899. 10. Required the cube root of 2. Ans. 1. 25994-. II. Find the cube root of 11. Ans. 2.2239+. 12. What is the cube root of 122615327232 ? Ans. 4968. 13. What is the cube root of f $ ? Ans. f. 14. What is the cube root of ff ? Ans. tf. 15. What is the cube root of f ? Ans. f . 16. What is the cube root of ||f f ? Ans - H- To find the cube root of any number mentally, less than 1 ,000,000, when the number has an exact root. 252 CUBE ROOT. [SECT LXII. RULE. As there will be two figures in the root, the first may easily be found mentally, or by the table of powers ; and if t)ie unit figure of the power be 1, the unit figure in the root will be 1 ; and if it be 8, the root will be 2 ; and if 7 it will be 3 ; and if the unit of the power be 6, the unit of the root will be 6 ; and if 5, it will be 5 ; if 3, it will be 7 ; if 2, it will be 8 ; and if the unit of the power be 9, the unit of the root will be 9. This will appear evident by inspecting the table of powers. 17. What is the cube root of 97336 ? Ans. 46. Explanation. By examining the left-hand period, we find the root of 97 is 4, and the cube of 4 is 64. The root cannot be 5, because the cube of 5 is 125. The unit of the power is 6 ; therefore, by the above rule, the unit figure in the root is 6. The answer, therefore, is 46. 18. What is the cube root of 132651 ? Ans. 51. 19. What is the cube root of 148877 ? Ans. 20. What is the cube root of 175616 ? Ans. 21. What is the cube root of 185193 ? Ans. 22. What is the cube root of 238328 ? Ans. 23. What is the cube root of 262144 ? Ans. 24. What is the cube root of 389017 ? Ans. 25. What is the cube root of 405224 ? Ans. 26. What is the cube root of 531441 ? Ans. 27. What is the cube root of 24389 ? Ans. 28. What is the cube root of 42875 ? Ans. SECOND METHOD OF EXTRACTING THE CUBE ROOT. The following rule for the extraction of the root of the third power, though it is essentially the same with the former, may yet serve to make the reasons for the several steps of the op- eration more intelligible to the learner. RULE. Separate tJie number whose root is to be found into periods, as under the former rule, and find by trial the greatest root in the left- hand period, and put it in the place of the quotient. Subtract the third power of this root from the period to which it be- longs, and to the remainder bring down the next period for a dividend. Then, to find a divisor, annex a cipher to the root already found, and multiply twice the number thus formed by the number itself, and to the product add the second power of this number.* * This is the same as multiplying the square of the radical figure by 300, as in the former rule. SECT. MIL] CUBE ROOT. 253 Ascertain how many limes this divisor is contained in the dividend, and write the result in the quotient.* T/ien, to find the subtrahend, multiply this divisor by its quotient, and write the product under the dividend. To this add three times the pre- ceding radical figure ivith a cipher annexed,^ multiplied by the second power of the figure last obtained, and also the third power of this last figure. Subtract the sum of their several products from the dividend above them, and to the remainder bring down the next period for a new dividend. With the parts of the root already found proceed to find a divisor and subtract as above, and so on, till the successive figures of the root are all obtained. The rationale of the above rule may be made to appear by the solution of the following question. Let it be required to find the cube root of 17576. OPERATION. 20 X 2 = 40 20 800 17^76 (20 + 6 = 26 Ans. 20 X 20 = 400 8000 Divisor 1200 ) 9576 dividend. 7200 2160 216 ~9576 subtrahend. We now raise the quantity 20 + 6 to the third power. 20 + 6 20 + 6 400 + 120 120 + 36 400 + 240 + 36 20+6 8000 + 4800 + 720 2400+1440 + 216 8000 + 7200 + 2160 + 216 = 17576. * This quotient figure must sometimes be less than the one indicated by the divisor, and in extreme cases the divisor may give a quotient too large by several units. The quotient required can, of course, never ex- ceed 9. t This accounts for -multiplying by 30, in the foregoing rule, which is a factor in the triple quotient in finding the subtrahend. 22 254 CUBE ROOT. [SECT. LXII. Now, by observing this operation, and remarking what would be lost in the course of it by omitting the second figure of the root, 6, taking 20 instead of 26, we see that when we have found the 20 the next inquiry is, what number must be added to 20, so that, if we multiply it into itself once and into 20 twice, and the sum of these products, together with the second power of twenty, by twenty plus this number, the result will be 17576, or 8000 + 9576. Then, in order to obtain this num- ber, or an approximation to it, we take twice the part of the root found, 20, and multiply the result, 40, by 20, as we should do in raising it to the third power, and make this, which is 800, a part of the divisor, and the product of which by 6 was lost in the operation for want of the 6 added to 20. But this is not all the loss. There is also the second power of 20 by 6, and therefore 400 to be added to the 800 for a divisor. There still remains the further loss of the third power of 6 (216), and also of 6 times 240 and 20 times 36 ; but these we neglect in the formation of the divisor. The divisor is contained in the dividend 7 times ; but, making the allowance of a unit for the neglect of the numbers above named, we take 6 for the quo- tient figure, and proceed to find the subtrahend, which, accord- ing to the rule and the foregoing operation of raising 20 -|- 6 to the third power, must be 1200x6+1440 + 720 + 216=: 17576. APPLICATION OF THE CUBE ROOT. PRINCIPLES ASSUMED. Spheres are to each other as the cubes of their diameters. Cubes, and all similar solid bodies, are to each other as the cubes of their diameters, or homologous sides. 29. If a ball, 3 inches in diameter, weigh 4 pounds, what will be the weight of a ball that is 6 inches in diameter ? Ans. 321bs. 30. If a globe of gold, one inch in diameter, be worth $120, what is the value of a globe 3A inches in diameter ? Ans. $ 5145. 31. If the weight of a well-proportioned man, 5 feet 10 inches in height, be 180 pounds, what must have been the weight of Goliath of Gath, who was 10 feet 4f inches in height ? Ans. 1015.1+lbs. 32. If a bell, 4 inches in height, 3 inches in width, and of SECT. LXII.] GENERAL RULE FOR ROOTS. 255 an inch in thickness-, weigh 2 pounds, what should be the dimen- sions of a bell that would weigh 2000 pounds ? Ans. 3ft. 4in. high, 2ft. Gin. wide, and 2in. thick. 33. Having a small stack of hay, 5 feet in height, weighing Icwt., I wish to know the weight of a similar stack that is 20 feet in height. Ans. 64cwt. 34. If a man dig a small square cellar, which will measure 6 feet each way, in one day, how long would it take turn to dig a similar one that measured 10 feet each way ? Ans. 4.629+ days. 35. If an ox, whose girth is 6 feet, weighs GOOlbs., what is the weight of an ox whose girth is 8 feet ? Ans. 1422.2+lbs. 36. Four women own a ball of butter, 5 inches in diameter. It is agreed that each shall take her share separately from the surface of the ball. How many inches of its diameter shall each take ? Ans. First, .45+^ inches ; second, .62-}- inches ; third, .78-}- inches ; fourth, 3.15-)- inches. 37. John Jones has a stack of hay in the form of a pyramid. It is 16 feet in height, and 12 feet wide at its base. It contains 5 tons of hay, worth $ 17.50 per ton. Mr. Jones has sold this hay to Messrs. Pierce, Rowe, Wells, and Northend. As the upper part of the stack has been injured, it is agreed that Mr. Pierce, who takes the upper part, shall have 10 per cent, more of the hay than Mr. Rowe ; and Mr. Rowe, who takes his share next, shall have 8 per cent, more than Mr. Wells ; and Mr. Northend, who has the bottom of the stack, that has been much injured, shall have 10 per cent, more than Mr. Wells. Re- quired the quantity of hay, and how many feet of the height of the stack, beginning at the top, each receives. Ans. Pierce receives 27 5 - 5 T 4 7 r cwt. and 10.365-j- feet in height ; Rowe, 24ffcwt. and 2.293+ feet; Wells, 22ff$cwt. and 1.646+ feet ; Northend, 25^cwt. and 1.494+ feet. A GENERAL RULE FOR EXTRACTING THE ROOTS OF ALL POWERS. RULE. 1. Prepare the given number for extraction, by pointing off from the unit's place, as the required root directs. 2. Find the first figure of the root by trial, or by inspection, in the table of powers, and subtract its power from the left-hand period. 256 GENERAL RULE FOR ROOTS. [SECT. LXII. 3. To the remainder, bring down the first figure in the next period, and call it the dividend. 4. Involve the root to tJie next inferior power to that which is given, and multiply it by the number denoting the given power for a divisor. 5. Find how many times the divisor is contained in the dividend, and the quotient will be another figure of the root. An allcnoance of two or three units is generally made, and for the higher powers a still greater allowance is necessary. 6. Involve the whole root to the given power, and subtract it from the given number, as before. 7. Bring down the first figure of the next period to the remainder for a new dividend, to which find a new divisor, as before ; and in like manner proceed till the whole is finished. 1. What is the cube root of 20346417 ? OPERATION. 20346417(273 2 3 = _8_ = l*t subtrahend. 2 2 X 3 = 12) 123 = 1st dividend. 273 19683 = 2d subtrahend. 2T 2 X 3 = 2187) 6634 = 2d dividend 273 3 = 20346417 = 3d subtrahend. 2. What is the fourth root of 34828517376 ? OPERATION". 34828517376(432. Ans. 4 4 = 256 = 1st subtrahend. 4 3 X 4 = 256) 922 = 1st dividend. 43 4 = 3418801 = 2d subtrahend. 43 3 X 4 = 318028) 640507 = 2d dividend. 432< = 34828517376 = 3d subtrahend. 3. What is the 5th root of 281950621875 ? Ans. 195. 4. Required the sixth root of 1178420166015625. Ans. 325. 5. Required the seventh root of 1283918464548864. Ans. 144. 6. Required the eighth root of 218340105584896. Ans. 62. SECT. LXIII.] ARITHMETICAL PROGRESSION. 257 SECTION LXIII. ARITHMETICAL PROGRESSION. WHEN a series of quantities or numbers increases or de- creases by a constant difference, it is called arithmetical pro- gression, or progression by difference. The constant difference is called the common difference or the ratio of the progression. Thus, let there be the two following series : 1, 5, 9, 13, 17, 21, 25, 29, 33, 25, 22, 19, 16, 13, 10, 7, 4, 1. The first is called an ascending series or progression. The second is called a descending series or progression. The numbers which form the series are called the terms of the progression. The first and last terms of the progression are called the extremes, and the other terms, the means. Any three of the five following things being given, the other two may be found : 1st, the first term, 2d, the last term, 3d, the number of terms, 4th, the common difference, 5th, the sum of the terms. PROBLEM I. The first term, last term, and number of terms being given, to find the common difference. To illustrate this problem, we will examine the following series, 3, 5, 7, 9, 11, 13, 15, 17, 19. It will be perceived that in this series 3 and 19 are the ex- tremes, 2 the common difference, 9 the number of terms, and 99 the sum of the series. It is evident, that the number of common differences in any number of terms will be one less than the number of terms. Hence, if there be 9 terms, the number of common differences will be 8, and the sum of these common differences will be equal to the difference of the extremes ; therefore if the differ- ence of the extremes, 19 3 16, be divided by the number of common differences, the quotient will be the common differ- ence. Thus 16 -r- 8 = 2 is the common difference. 22* 258 ARITHMETICAL PROGRESSION. [SECT. LXTII. RULE. Divide the difference of the extremes by the number of terms less one, and the quotient is the common difference. 1. The extremes are 3 and 45, and the number of terms is 22. What is the common difference ? OPERATION. 45 3 ^-^ = 2 Answer. 2. A man is to travel from Albany to a certain place in 11 days, and to go but 5 miles the first day, increasing the dis- tance equally each day, so that the last day's journey may be 45 miles. Required the daily increase. Ans. 4 miles. 3. A man had 10 sons, whose several ages differed alike ; the youngest was 3 years, and the oldest 48. What was the common difference of their ages ? Ans. 5 years. 4. A certain school consists of 19 scholars ; the youngest is 3 years old, and the oldest 39. What is the common differ- ence of their ages ? Ans. 2 years. PROBLEM II. The first term, last term, and number of terms being given, to find the sum of all the terms. Illustration. Let 3, 5, 7, 9, 11, 13, 15, 17, 19, be the series, and 19, 17, 15, 13, 11, 9, 7, 5, 3, the same series inverted. 22, 22, 22, 22, 22, 22, 22, 22, 22, sum of both series. From the arrangement of the above series, we see that, by adding the two as they stand, we have the same number for the sum of the successive terms, and that the sum of both series is double the sum of either series. It is evident that if, in the above series, 22 be multiplied by 9, the number of terms, the product will be the sum of both series, 22 X 9 = 198, and therefore the sum of either series will be 198 -r- 2 = 99. But 22 is also the sum of the ex- tremes in either series, 3 -f- 19 = 22. Therefore, if the sum of the extremes be multiplied by the number of terms, the prod- uct will be double the sum of the series. RULE. Multiply the sum of the extremes by the number of terms, and half the product will be the sum of tlie series. 5. The extremes of an arithmetical series are 3 and 45, and the number of terms 22. Required the sum of the series. SECT. LXHI.] ARITHMETICAL PROGRESSION. 259 OPERATION. 45 + 3 X 22 = 528 Answer. 6. A man going a journey travelled the first day 7 miles, the last day 51 miles, and he continued his journey 12 days. How far did he travel ? Ans. 348 miles. 7. In a certain school there are 19 scholars ; the youngest is 3 years old, and the oldest 39. What is the sum of their ages ? Ans. 399 years. 8. Suppose a number of stones were laid a rod distant from each other, for thirty miles, and the first stone a rod from a basket. What length of ground will that man travel over who gathers them up singly, returning with them one by one to the basket ? Ans. 288090 miles 2 rods. PROBLEM III. The extremes and the common difference being given, to find the number of terms. Illustration. Let the extremes be 3 and 19, and the common difference 2. The difference of the extremes will be 19 3 == 16 ; and it is evident, that, if the difference of the extremes be divided by the common difference, the quotient is the num- ber of common differences ; thus 16 -=- 2 = 8. We have de- monstrated in Problem I. that the number of terms is one more than the number of differences ; therefore 8 -f- 1 = 9, the number of terms. RULE. Divide the difference of the extremes by the common differ- ence, and the quotient increased by one toill be the number of terms re- quired. 9. If the extremes are 3 and 45, and the common difference 2, what is the number of terms ? OPERATION. 45 3 -[- 1 22 Answer. 10. In a certain school the ages of all the scholars differ alike ; the oldest is 39 years, the youngest is 3 years, and the difference between the ages of each is 2 years. Required the number of scholars. Ans. 19. 11. A man going a journey travelled the first day 7 miles, the last day 51 miles, and each day increased his journey by 4 miles. How many days did he travel ? Ans. 12. 260 ARITHMETICAL PROGRESSION. [SECT. LXIII PROBLEM IV. The extremes and common difference being given, to find the sum of the series. Illustration. Let the extremes be 3 and 19, and the com- mon difference 2. The difference of the extremes will be 19 3 16 ; and it has been shown in the last problem, that, if the difference of the extremes be divided by the common dif- ference, the quotient will be the number of terms less one ; therefore the number of terms less one will be 16 -5- 2 = 8, and the number of terms 8 -f- 1 = 9. It was demonstrated in Problem II. that if the number of terms was multiplied by the sum of the extremes, and the product divided by 2, the quo- tient would be the sum of the series. RULE. Divide the difference of the extremes by the common differ- ence, and add I to the quotient; multiply this quotient by the sum of the extremes, and half the product is the sum of the series. 12. If the extremes are 3 and 45, and the common differ- ence 2, what is the sum of the series ? Ans. 528. 13. A owes B a certain sum, to be discharged in a year, by paying 6 cents the first week, 18 cents the second week, and thus to increase every week by 12 cents, till the last payment should be $ 6. 18. What is the debt ? Ans. $ 162.24. PROBLEM V. The extremes and sum of the series being given, to find the common difference. Illustration. Let the extremes be 3 and 19, and the sum of the series 99, to find the common difference. We have before shown, that, if the extremes be multiplied by the number of terms, the product would be twice the sum of the series ; there- fore, if twice the sum of the series be divided by the extremes, the quotient will be the number of terms. Thus, 99 X 2 198 ; 3 -f- 19 = 22 ; 198 -~ 22 = 9 is the number of terms. And we have before shown, that, if the difference of the ex- tremes be divided by the number of terms less one, the quo- tient will be the common difference ; therefore, 19 3 = 16 ; 9 1 zz: 8 ; 16-T-8 2is the common difference. RULE. Divide twice the sum of tJte series by the sum of the ex- tremes, and from the quotient subtract 1 ; and with this remainder divide the difference of the extremes, and the quotient is the common difference. 14. The extremes are 3 and 45, and the sum of the series 528. What is the common difference ? Ans. 2. BECT. Liiv.j GEOMETRICAL SERIES. 261 PROBLEM VI. The first term, number of terms, and the sum of the series being given, to find the last term. Illustration. Let 3 be the first term, 9 the number of terms, and 99 the sum of the series. By Problem II. it was shown, that, if the sum of the extremes were multiplied by the number of terms, the product was twice the sum of the series ; therefore, if twice the sum of the series be divided by the num- ber of terms, the quotient is the sum of the extremes. If from this we subtract the first term, the remainder is the last term ; thus 99 X 2 = 198 ; 198 -7- 9 = 22 ; 22 3 = 19, last term. RULE. Divide twice the sum of the series by the number of terms ; from the quotient take the first term, and the remainder will be the last term. 15. A merchant being indebted to 22 creditors $ 528, ordered his clerk to pay the first $ 3, and the rest increasing in arith- metical progression. What is the difference of the payments, and the last payment ? Ans. Difference 2 ; last payment, $ 45. SECTION LXIV. GEOMETRICAL SERIES, OR SERIES BY QUOTIENT. IF there be three or more numbers, and if there be the same quotient when the second is divided by the first, and the third divided by the second, and the fourth divided by the third, &c., those numbers are in geometrical progression. If the series increase, the quotient is more than unity ; if it decrease, it is less than unity. The following series are examples of this kind : 2, 6, 18, 54, 162, 486. 64, 32, 16, 8, 4, 2. The former is called an ascending series, and the latter a descending series. In the first, the quotient is 3, and is called the ratio ; in the second, it is J. 262 GEOMETRICAL SERIES. [SECT. LIIT. The first and last terms of a series are called extremes, and the other terms means. PROBLEM I. One of the extremes, the ratio, and the number of terms being given, to find the other extreme. Let the first term be 3, the ratio 2, and the number of terms 8, to find the last term. Illustration. It is evident, that, if we multiply the first term by the ratio, the product will be the second term ; and, if we multiply the second term by the ratio, the product will be the third term ; -and, in this manner, we may carry the series to any desirable extent. By examining the following series, we find that 3, carried to the 8th term, is 384 ; thus, (1.) (2.) (3.) (4.) (5.) (6.) (7.) (S.) 3, 6, 12, 24, 48, 96, 192, 384, ascending series. But the factors of 384 are 3, 2, 2, 2, 2, 2, 2, and 2 ; there- fore the continued product of these numbers will produce 384. But, by multiplying 2 by itself six times, and that product by 3, is the same as raising the ratio, 2, to the seventh power, and then multiplying that power by the first term. Hence the fol- lowing RULE. Raise the ratio to a power whose index is equal to the num- ber of terms less one ; then multiply this power by the first term, and the product is the last term, or other extreme. Illustration. In the above question the number of terms is 8 ; we therefore raise 2, the ratio, to the seventh power, it being one less than 8 ; thus, 2x2x2x2x2x2 X 2 = 128. We then multiply this number by 3, the first term, and the 'product is the last term ; thus, 128 x 3 = 384, last term. The above rule will apply in a descending series. Let the following numbers be a geometrical descending series : 384, 192, 96, 48, 24, 12, 6, 3, descending series. Let the first term be 384, the number of terms 8, and the ratio , to find the other extreme. By the above rule, we raise the ratio, , to the seventh power, it being one less than the number of terms, 8. Thus, X XXXXX= T!F- We then multiply this power by the first term ; thus, TS- X -f- 4 - = iff = 3, the extreme required. But as the last term, or any term near the last, is very tedious to be found by continual multiplication, it will often be necessary, in order to ascertain it, to have a series of numbers SECT. LXIV.] GEOMETRICAL SERIES. 263 in arithmetical proportion, called indices or exponents, begin- ning either with a cipher or a unit, whose common difference is one. When the first term of the series and ratio are equal, the indices must begin with a unit, and in this case the prod- uct of any two terms is equal to that term signified by the sum of their indices. rpi ( 1, 2, 3, 4, 5, 6, &c., indices or arithmetical series. u ' J 2, 4, 8, 16, 32, 64, &c., geometrical series. Now 6 + 6 = 12 = the index of the twelfth term, and 64 X 64 = 4096 = the twelfth term. But when the first term of the series and the ratio are differ- ent, the indices must begin with a cipher, and the sum of the indices made choice of must be one less than the number of terms given in the question ; because 1 in the indices stands over the second term, and 2 in the indices over the third term, &c. And, in this case, the product of any two terms, divided by theirs/, is equal to that term beyond the first signified by the sum of their indices. T , ( 0, 1, 2, 3, 4, 5, 6, &c., indices. ' { 1, 3, 9, 27, 81, 243, 729, &c., geometrical series. Here, 6 -f 5 = 11, the index of the 12th term. 729 X 243= 177147, the 12th term, because the first term of the series and ratio are different, by which means a cipher stands over the first term. Thus, by the help of these indices, and a few of the first terms in any geometrical series, any term whose distance from the first term is assigned, though it were ever so remote, may be obtained without producing all the terms. 1. If the first term be 4, the ratio 4, and the number of terms 9, what is the last term ? OPERATION. 1. 2. 3. 4 + 4 = 8 4. 16. 64. 256 X 256 = 65536 = power of the ratio, whose exponent is less by 1 than the number of terms. 65536 X 4, the first term = 262144 == last term. Or, 4 X 4 = 262144 = last term, as before. ae 262144, the ratio J, E last term? x au^ ^i# =4, the last term. 2. If the first term be 262144, the ratio J, and the number of terms 9, what is the last term ? Ans. 4. 364 GEOMETRICAL SERIES. [SECT. LXIV. 3. If the first term be 72, the ratio , and the number of terms 6, what is the last term ? Ans. /y. 4. If I were to buy 30 oxen, giving 2 cents for the first ox, 4 cents for the second, 8 cents for the third, &c., what would be the price of the last ox ? Ans. $10737418.24. 5. If the first term be 5, and the ratio 3, what is the seventh term ? Ans. 3645. 6. If the first term be 50, the ratio 1.06, and the number of terms 5, what is the last term ? Ans. 63.123848. 7. What is the amount of $160.00 at compound interest for 6 years ? Ans. $ 226.96,305796096. 8. What is the amount of $ 300.00 at compound interest at 5 per cent, for 8 years ? Ans. $ 443.23,6+. 9. What is the amount of 8100.00 at 6 per cent, for 30 years ? Ans. $ 574.34,9117291325011626410633231080264584635- 7252196069357387776. PROBLEM II. The first term, the ratio, and the number of terms being given, to find the sum of all the terms. In order that the pupil may understand the following rule, we will examine a question analytically. Let the following be a geometrical series, and we wish to obtain its sum : 1, 3, 9, 27, 81. Illustration. By examining this series, we find the first term to be 1, the last term 81, the ratio 3. If we multiply each term of the following series, 1, 3, 9, 27, 81, by 3, the ratio, their product will be 3, 9, 27, 81, 243, and the sum of this last series will be three times as much as the first series. The dif- ference, therefore, between these series will be twice as much as ihejirst series. 3, 9, 27, 81, 243 = second series. 1, 3, 9, 27, 81, = first series. ~Q, 0, 0, 0, 243 1 = 242, difference of the series. As this difference must be twice the sum of the first series, therefore the sum of the first series must be 242 -4- 2 = 121. By examining the above series, we find the terms in both the same, with the exception of the first term in the first series, and the last term in the second series. We have only, then, to SECT. LXIV.] GEOMETRICAL SERIES. 265 subtract the first term in the first series from the last term in the second series, and the remainder is twice the sum of the first series ; and half of this being taken gives the sum of the series required. RULE. Find the other extreme, as before, multiply it by the ratio, and from the product subtract the given extreme. Divide the remainder by the ratio less 1 (unless the ratio be less than a unit, in which case the ratio must be subtracted from 1 ) , and the quotient multiplied by the other extreme will give the sum of the series. See operation, ques- tion 10. Or, raise the ratio to a power whose index is equal to the num- ber of terms ; from which subtract 1 , divide the remainder by the ratio less 1, and tJue quotient, multiplied by the given extreme, will give the sum of the series. See operation, question 11. But if the ratio be a fraction less than a unit, raise the ratio to a power whose index shall be equal to the number of terms ; subtract this power from 1 , divide the remainder by the difference between 1 and the ratio, and the quotient, multiplied by the given extreme, will give the sum of the series required. See operation, question 12. 10. If the first term be 10, the ratio 3, and the number of terms 7, what is the sum of the series ? Ans. 10930. OPERATION. 3X3X3X3X3X3X10 = 7290, last term. _ 7290X3 = 21870; 21870 10 = 21860; 21860-*- 31 = 10930 Ans. 11. If the first term be 4, the ratio 3, and the number of terms 5, what is the sum of the series ? Ans. 484. OPERATION. 3X 3X3X3X3= 243; 243 1 = 242; 242 -f- 3 1 = 121 ; 121 X 4 = 484 Ans. 12. If the first term be 6, the ratio f , and the number of terms 4, what is the sum of the series ? Ans. OPERATION. f X I X f X f = i& 6 5 ; 1 - f f I ; f f I - *V 6 5 - X * = ^ = 9^ Ans. 13. How large a debt may be discharged in a year, by pay- ing $ 1 the first month, $ 10 the second, and so on, in a tenfold proportion, each month ? Ans. $111111111111. 14. A gentleman offered a house for sale, on the following terms ; that for the first door he should charge 10 cents, for the second 20 cents, for the third 40 cents, and so on in a geo- 23 266 GEOMETRICAL SERIES. [SECT. LXIY metrical ratio, there being 40 doors. What was the price of the house ? Ans. $ 109951 162777.50. 15. If the first term be 50, the ratio 1.06, and the number of terms 4, what is the sum of the series ? Ans. 218.7308. 16. A gentleman deposited annually $10 in a bank, from the time his son was born until he was 20 years of age. Required the amount of the deposits at 6 per cent., compound interest, when his son was 21 years old. Ans. $ 423.92,2+. 17. If the first term be 7, the ratio J, and the number of terms 5, what is the sum of the series ? Ans. 9 2 8 / F . 18. If one mill had been put at interest at the commence- ment of the Christian era, what would it amount to at com- pound interest, supposing the principal to have doubled itself every 12 years, January 1, 1837 ? Ans. $ 114179815416476790484662877555959610910619- 72.99,2. If this sum was all in dollars, it would take the present inhab- itants of the globe more than 1,000,000 years to count it. If it was reduced to its value in pure gold, and was formed into a globe, it would be many million times larger than all the bodies that compose the solar system. PROBLEM III. To find the sum of the second powers of any number of terms, whose roots differ by unity. RULE. Add one to the number of terms, and multiply this sum by the number of terms ; then add one to twice live number of terms, and multiply this sum by the former product, and the last product, divided by 6, will give the sum of all the terms. 19. What is the sum of 10 terms of the series I 2 , 2 2 , 3 2 , 4 2 , 5 2 , 6 2 , 7 2 , 8 2 , 9 2 , 10 2 ? OPERATION. IPX - L g - = 385 Ans. 20. What is the sum of 100 terms of the series I 2 , 2 2 , S 2 , 4 2 , 5 2 , 6 2 , 7 2 , 8 2 , 9 2 , 10 2 , &c., to 100 2 ? Ans. 338350. 21. Purchased 50 lots of land ; the first was one rod square, the second was two rods square, the third was three rods square, and so on, the last being 50 rods square. How many square rods were there in the 50 lots ? Ans. 42925. 22. Let it be required to find the number of cannon shot in a square pile, whose side is 80. Ans. 173880. . SECT. LXV.] INFINITE SERIES. 267 NOTE. A square pile is formed by continued horizontal courses of shot laid one above another, and these courses are squares, whose sides decrease by unity from the bottom of the pile to the top row, which is composed of only one shot. PROBLEM IV. To find the sum of the third power of any number of terms, whose roots differ by unity. RULE. Add one to the number of terms, and multiply this sum by half the number of terms; the square of this product is the sum of all the series. 23. Required the sum of the following series : I 3 , 2 3 , 3 3 , 4 3 , 5 3 , 6 3 , 7 3 , 8 3 , 9 3 , 10 3 , II 3 , 12 3 . OPERATION. 12 + 1=13; 12-f-2 = 6; 13X6 = 78; 78 X 78 = 6084 Ans. 24. I have 10 blocks of marble, each of which is an exact cube. A side of the first cube measures one foot, a side of the second 2 feet, a side of the third 3 feet, and so on to the 10th, whose side measures 10 feet. Required the number of cubical feet in the blocks ? Ans. 3025 cubic feet. 25. What is the sum of 50 terms of the series I 3 , 2 3 , 3 3 , 4 3 , 5 3 , 6 3 , 7 3 , &c., up to 50 3 ? Ans. 1625625. SECTION LXV. INFINITE SERIES. AN INFINITE SERIES is such as, being continued, would run on ad infinilum ; but the nature of its progression is such, that, by having a few of its terms given, the others to any extent may be known. Such are the following series : 1, 2, 4, 8, 16, 32, 64, 128, &c., ad infnitum. 125, 25, 5, 1, , 2-V T , ^, &c., ad infinitum. To find the sum of a decreasing series. RULE. Multiply the first term by the ratio, and divide the product by the ratio less 1, and the quotient is the sum of an infinite decreasing series. 1. What is the sum of the series 4, 1, , T ^, ^V> &c., con- tinued to an infinite number of terms ? OPERATION. 4 v 4 = 5 Answer. 268 DISCOUNT BY COMPOUND INTEREST. [SECT. LXVI. 2. What is the sum of the series 5, 1, , ^, dec., continued to infinity ? Ans. 6. 3. If the following series, 8, f , T %, -yf 7 , &c., were carried to infinity, what would be its sum ? Ans. 9. 4. What is the sum of the following series, carried to in- finity : 1, , , 2V, -g\, &c. ? Ans. 1. 5. What is the sum of the following series, carried to infin- ity : ll,-YSH> & c.? Ans. 12 f. 6. If the series f , , J, -j^-, 2^, &c., were carried to infinity, what would be its sum ? Ans. 1. ^ SECTION LXVI. DISCOUNT BY COMPOUND INTEREST. 1. What is the present worth of $ 600.00, due 3 years hence, at 6 per cent, compound interest ? OPERATION. 1.06) 3 = 1.191016)600.00($ 503.77+ Ans. By analysis. We find the amount of $ 1 at compound interest for 3 years to be $ 1.191016; therefore $ 1 is the present worth of $ 1.191016 due 3 years hence. And if $ 1 is the present worth of $ 1.191016, the present worth of 600 = $503.77,1+ RULE. Divide the debt by the amount of one dollar for the given time, and the quotient is the present worthy which, if subtracted from the debt, will leave the discount. 2. What is the present worth of $ 500.00, due 4 years hence, at 6 per cent, compound interest ? Ans. $ 396.04,6+. 3. What is the present worth of $1000.00, due 10 years hence, at 5 per cent, compound interest ? Ans. $ 613.91,3-)-. 4. What is the discount on $800.00, due 2 years hence, at 6 per cent, compound interest ? Ans. $ 88.00,3-)-. 5. What is the present worth of $ 1728, due 5 years hence, at 6 per cent, compound interest ? Ans. $ 1353.93. 6. What is the discount on $ 3700, due 10 years hence, at 5 per cent, discount, compound interest? Ans. $2271.47. 7. What is the present worth of $ 7000, due 2 years hence', at 5 per cent, compound interest ? Ans. $ 63492.10. SECT. LXVII.] ANNUITIES AT COMPOUND INTEREST. 269 SECTION LXVII. ANNUITIES AT COMPOUND INTEREST. AN annuity is a certain sum of money to be paid at regular periods, either for a limited time or for ever. The present worth or value of an annuity is that sum which, being improved at compound interest, will be sufficient to pay the annuity. The amount of an annuity is the compound interest of all the payments added to their sum. To find the amount of an annuity at compound interest. RULE. Make $ 1.00 the first term of a geometrical series, and the amount of $1.00 at the given rate per cent, the ratio. Carry the series to so many terms as the number of years, and find its sum. Multiply the sum thus found by the given annuity, and the product will be the amount. EXAMPLES. 1. What will an annuity of $60 per annum, payable yearly, amount to in 4 years, at 6 per cent. ? 1 + 1.06 + 1.06+ 1.06 = 4.374616. 4.374616 X 60 = $ 262.47,6+ Answer. f ' L< ^~ \ X 60 = $262.47,6+ Answer. 1.06 1 2. What will an annuity of $ 500.00 amount to in 5 years, at 6 per cent. ? Ans. $ 2818.54,6+. 3. What will an annuity of $1000.00, payable yearly, amount to in 10 years ? Ans. $13180.79,4+. 4. What will an annuity of $ 30.00, payable yearly, amount to in 3 years ? Ans. $ 95.50,8 +. To find the present worth of an annuity. As the first payment is made at the end of the year, its pres- ent worth or value is a sum that will amount in one year to that payment ; and as the second payment is made at the end of the second year, its value is a sum that will, at compound interest, amount in two years to that payment ; and the same principle is adopted for the third year, fourth year, &c. This may be illustrated in the following question. 23* 270 ANNUITIES AT COMPOUND INTEREST. [SECT. LXVJI. 5. What is the present worth of an annuity of $1.00, to con- tinue 5 years, at compound interest ? The present worth of $ 1.00 for 1 year = $ 0.943396 The present worth of $ 1.00 for 2 years = $ 0.889996 The present worth of $ 1.00 for 3 years $ 0.839619 The present worth of $ 1.00 for 4 years = $ 0.792094 The present worth of $ 1.00 for 5 years = $ 0.747258 $1.212363 By the above illustration, we perceive that the present worth of an annuity of $ 1, to continue 5 years, is $ 4.2 1 >-{-. Hence, having found the present worth of an annuity of $ 1 for any given time by Section LXVL, the present worth of any other sum may be found by multiplying it by the present worth of $1 for that time. RULE. Multiply the present worth of the annuity of one dollar for the given time by the given annuity, and the product is the present worth required. Or, find the amount of the annuity by tlie last rule, and then find its present worth. 6. What is the present worth of an annuity of $ 60, to bo continued 4 years, at compound interest ? First Method. The present worth of $ 1.00 for 1 year = $ 0.943396 The present worth of $ 1.00 for 2 years = $ 0.889996 The present worth of $ 1.00 for 3 years = $ 0.839619 The present worth of $ 1.00 for 4 years = $ 0.792093 8 3.465104 $ 3.465104 X 60 = $ 207.90,6+ Answer. 4 Second Method. -1= $ 4.374616 x .792093 = $ 3.465102 x 60 = $ 207.90,6-f Ans. 7. A gentleman wishes to purchase an annuity, which shall afford him, at 6 per cent, compound interest, $ 500 a year for ten years. What sum must he deposit in the annuity office to produce it ? Ans. $ 3680.04-}-. 8. What is the present worth of an annuity of $ 1000, to continue 10 years? Ans. $7360.08. 9. What is the present worth of an annuity of $ 1728, to continue 3 years? Ans. $4618.96. SECT. LXVII.] ANNUITIES AT COMPOUND INTEREST. 271 By the assistance of the following tables, questions in annui- ties may be easily performed. TABLE I. Showing the amount of $ 1 annuity from 1 year to 40. Years. 5 per cent. 6 per cent. 1 Years. 5 per cent. 6 per cent. 1 1.0000UO 1.000000 21 35.719252 39.992727 2 2.050000 2.060000 22 38.505214 43.392290 3 3.152500 3 183600 23 41.430475 46.995828 4 4.310125 4.374616 24 44.501999 50.815577 5 5.525631 5.637093 25 47.727099 54.864512 6 6 801913 6.975319 26 51 113454 59.156:583 7 8.142008 8.393838 27 54.669126 63.705766 8 9.549109 9.897468 28 58.402583 68.528112 9 11.026564 11.491316 29 62.322712 73 639798 10 12.577893 13 180795 30 66.438847 79 058186 11 14.206787 . 14.971643 31 70.760790 84.801677 12 15.917127 16.869941 32 75.298829 90.889778 13 17.712983 18.882138 33 80.063771 97.343165 14 19.598632 21.015066 34 85.066959 104.183755 15 21578564 23.275970 35 90.220307 111.434780 16 23.657492 25.672528 36 95.836323 119.120867 17 25.840366 28.212880 37 101.628139 127.268119 18 28132385 30 905653 38 107.709546 135.904206 19 30 539004 33.759992 39 114.095023 145.058458 20 33.065954 36.785591 40 120.799774 154.761966 TABLE II. Showing the present value of an annuity of $ 1 from 1 year to 40. Years. 5 per cent. 6 per cent. Years. 5 per cent. 6 per cent. 1 952381 0.943396 21 12.821153 11764077 2 1.859410 1.833393 22 13.163003 12.041582 3 2.723248 2.673012 23 13.488574 12.303379 4 3 545950 3.465106 24 13.798642 12.550358 5 4.329477 4.212364 25 14.093945 12.783356 6 5075692 4.917324 26 14 375185 13.003166 7 5 786373 5.582381 27 14.643034 13 210534 8 6.463213 6.209794 28 14.898127 13.406164 9 7.107822 6.801692 29 15.141074 13.590721 10 7.721735 7.360087 30 15.372451 13.764831 11 8 306414 7.886875 31 15.592810 13.929086 12 8.863252 8.388844 32 15.802677 14.084043 13 9.393573 8.852683 33 16.002549 14.230230 14 9.898641 9.294984 34 16.192904 14.368141 15 10379658 9.712249 35 1 16.374194 14.498246 16 10.837770 10.105895 36 16.546852 14.620987 17 11.274066 10 477260 37 16.711287 14 736780 18 11.639587 10.827603 38 16.867893 14.846019 19 12.085321 11.158116 39 17.017041 14.949075 20 12.462216 11.469921 40 17.159086 15.046297 272 ASSESSMENT OF TAXES. [SECT. LXVIII. 10. What is the present worth of an annuity of $ 200, at 5 per cent, compound interest, for 7 years ? Ans. $1157.27-}-. 1 1. What is the present worth of an annuity of $ 300, to con- tinue 8 years, at 6 per cent, compound interest ? Ans. $ 1862.93,8-f . 12. What is the present value of* an annuity of $ 100, at 6 per cent, for 9 years ? Ans. $ 680.16,9+. Questions to be performed by the preceding tables. 13. What will an annuity of $ 30 amount to in 1 1 years, at 6 per cent. ? By Table I., the amount of $ 1 for 11 years is $14.971643 ; therefore, $ 14.971643 X 30 = $ 449.14,9+ Answer. 14. What is the present worth of an annuity of $ 80 for 30 years, at 5 per cent. ? By Table II., the present worth of $ 1 for 30 years is $15.372451, therefore $15.372451 X 80 = $ 1229.79,6+ Ans. 15. What will an annuity of $800 amount to in 25 years, at 5 per cent. ? Ans. $ 38181.67,9+. 16. What will an annuity of $40 amount to in 30 years, at 6 per cent. ? Ans. $ 3162.32,7+. 17. Required the present worth of an annuity of $ 500, to continue 40 years, at 6 per cent. Ans. 8 7523.14,8+. 18. A certain parish in the town of B., having neglected for 6 years to pay their minister's salary of $ 700, what in justice, provided he has preached the truth, should he receive ? Ans. $ 4882.72,3. SECTION LXVIII. ASSESSMENT OF TAXES. A TAX is a duty laid by government, for public purposes, on the property of the inhabitants of a town, county, or State, and also on the polls * of the male citizens liable by law to as- sessment. A tax may be either general or particular ; that is, it may affect all classes indiscriminately, or only one or more classes. * Poll is said to be a Saxon word, meaning head. In the constitution it means a person ; that is, a person who is liable to taxation. SECT. LXTIII.] ASSESSMENT OF TAXES. 273 Taxes may be either direct or indirect ; that is, they may either be imposed on the incomes or property of individuals, or on the articles on which these incomes or property are ex- pended. The method of assessing town taxes is not precisely the same in all the States, yet the principle is virtually the same. In some of the States the poll tax is more than in others. The following is the law regulating taxation in Massachusetts (see Revised Statutes, page 79) : " The assessors shall assess upon the polls, as nearly as the same can be conveniently done, one sixth part of the whole sum to be raised ; provided the whole poll tax assessed in any one year upon any individual for town and county purposes, except highway taxes, shall not exceed one dollar and fifty cents ; and the residue of said whole sum to be raised shall be apportioned upon property " ; that is, on the real and personal estate of individuals which is taxable. RULE FOR ASSESSING TAXES. First take an inventory of all the taxable property, real and personal, in the town or county, and then the number of polls liable to taxation. Multiply the sum assessed on each poll by the number of taxable polls in the town. Subtract this amount from the sum to be raised by the town. TJien as the whole valuation of the town is to the sum to be raised, after having deducted the amount to be paid by the polls, so is the amount of each man's real and personal estate to his tax. Or, if the sum to be raised on property be divided by the valuation of t/ie town, the quotient will be the sum to be paid on each dollar of an individual" 1 s real or personal estate. Multiply each man's property by this sum, and the product will be the amount of his taxes. The town of B. is to be taxed $4109. The real estate of the town is valued at $ 493,000, and the personal property at $ 177,000. There are 506 polls, each of which is taxed $1.50. What is John Smith's tax, whose real estate is valued at $ 3700, and his personal at $ 2300, he paying for 6 polls ? And what will be the tax on $ 1.00 ? OPERATION. $ 1.50 X 506 $759, amount assessed on the polls. $ 493,000 -f- $177 ,000m $ 670,000, amount of taxable property. $ 4109 $ 759 = $ 3350, amount to be assessed on property. $ 670,000 : $ 3350 ::$!:$ .005, to be assessed on each dollar. $3700 X .005= $18.50, tax on Smith's real estate. $ 2300 X .005 = $ 11.50, tax on Smith's personal estate. $ 1.50 X 6 = $ 9.00, tax on 6 polls. $ 18.50 + $ 11.50 + $ 9.00 = $ 39.00, amount of Smith's tax. 274 ASSESSMENT OF TAXES. [SECT. MVIII. What will be the amount of taxation on each of the following individuals of the above town, their taxable property being as annexed to their names ? Persons. James Dow, John Brown, Samuel Foster, James Emerson, A. C. Hasseltine, Real Estate. $4780 7500 1135 8960 7140 Personal Estate. No. of Polls. $1720 2120 175 5000 3720 9 Form of a tax-list committed to the collector, containing the answers to the above questions. Names. No. of Polls. Poll Tax. ITscT 1.50 10.50 3.00 0.00 9.00 Tax on Real Estate. Tax on Personal Estate. $8.60 10.60 .87 25.00 18.60 11.50 Total. $37.00 49.60 17.05 72.80 54.30 39.00 Time when paid. James Dow, John Brown, Samuel Foster, James Emerson, A. C. Hasseltine, John Smith, 3 1 7 2 6 $23.90 37.50 5.67 44.80 35.70 18.50 Having found the amount to be raised on the dollar, the op- eration of assessing taxes will be much facilitated by the use of the following TABLE. $ $ $ $ $ $ 1 gives 0.005 40 gives 0.20 700 gives 3.50 2 " 0.010 50 " 0.25 800 " 4.00 3 " 0.015 60 " 0.30 900 " 4.50 4 " 0.020 70 " 0.35 1000 " 5.00 5 " 0.025 80 " 0.40 2000 10.00 6 " 0.030 90 " 0.45 3000 15.00 7 " 0.035 100 " 0.50 4000 " 20.00 8 " 0.040 200 " 1.00 5000 " 25.00 9 " 0.045 300 " 1.50 6000 " 30.00 10 " 0.050 400 " 2.00 7000 " 35.00 20 " 0.100 500 " 2.50 8000 " 40.00 30 " 0.150 600 " 3.00 9000 " 45.00 By the aid of the above table the amount of any person's tax may be found. Required the amount of James Dow's tax, his real estate being $ 4780, his personal $ 1720, and he paying for 3 polls. SICT. LXIX.] ALLIGATION. 275 1. To find the amount on his real estate. OPERATION. Dow's tax on $ 4000 is $ 20.00 " " 700 " 3.50 " " 80 " .40 r , [estate. $ 4780 $ 23.90 Amount of Dow's tax on real 2. To find the amount on his personal estate. OPERATION. Dow's tax on $ 1000 is $ 5.00 " " 700 " 3.50 " " 20 " .10 r [sonal estate. $ 1720 $ 8.60 Amount of Dow's tax on per- Tax on real estate, $ 23.90 Tax on personal estate, 8.60 Tax on 3 polls, 4.50 Dow's whole tax, $ 37.00 NOTE. It will be necessary to construct a different table, although on the same principle, when a different per cent, is paid on the dollar. SECTION LXIX. ALLIGATION. ALLIGATION teaches how to compound or mix together sev- eral simples of different qualities, so that the composition may be of some intermediate quality or rate. It is of two kinds, Al- ligation Medial and Alligation Alternate. ALLIGATION MEDIAL. Alligation Medial teaches how to find the mean price of sev- eral articles mixed, the quantity and value of each being given. RULE. As the sum of the quantities to be mixed is to their value, so is any part of the composition to its mean price. EXAMPLES. 1. A grocer mixed 2cwt. of sugar at $ 9.00 per cwt., and Icwt. at $7.00 per cwt., and 2cwt. at $ 10.00 per cwt. ; what is the value of Icwt. of this mixture ? 276 ALLIGATION. [SECT. LXIX. 2cwt. at 8 9.00 = $ 18.00 1 7.00= 7.00 _2 " 10.00 = 20.00 5 " : 845.00:: Icwt. : 89.00 Answer. 2. If 19 bushels of wheat at $ 1.00 per bushel should be mixed with 40 bushels of rye at $0.66 per bushel, and 11 bushels of barley at $ 0.50 per bushel, what would a bushel of the mixture be worth ? Ans. $ 0.72,7f . 3. If 3 pounds of gold of 22 carats fine be mixed with 3 pounds of 20 carats fine, what is the fineness of the mixture ? Ans. 21 carats. 4. If I mix 20 pounds of tea at 70 cents per pound with 15 pounds at 60 cents per pound, and 80 pounds at 40 cents per pound, what is the value of 1 pound of this mixture ? Ans. NOTE. If an ounce, or any other Quantity, of pure gold be divided into 24 equal parts, these parts are called carats. But gold is often mixed with some baser metal, which is called the alloy ; and the mixture is said to be so many carats fine, according to the proportion of pure gold con- tained in it ; thus, if 22 carats of pure gold and 2 of alloy be mixed to- gether, it is said to be 22 carats fine. ALLIGATION ALTERNATE. This rule teaches us how, from the prices of several articles given, to find how much of each must be mixed to bear a cer- tain price. CASE I. RULE. Place the prices under each other, in the order of their value; connect the price of each ingredient, tvhich is less in value than the in- tended compound, with one which is of greater value than the compound. Place tJie difference between the price and that of each simple, opposite to the price with which they are connected. EXAMPLES. 5. A merchant has spices, some at 18 cents a pound, some at 24 cents, some at 48 cents, and some at 60 cents. How much of each sort must he mix that he may sell the mixture at 40 cents a pound ? Ihs. cts. 20 at 18 ^ Mean rate 40 < ^ ' ' 16 " 48 ( Answers - 22 " 60 ) SECT. LXIX.] Mean rate 40 ALLIGATION. cts. Ibs. c 18 i 8 at ] 24 i | 20 " 24 48 l I 22 " 48 n 1 60 16 Mean rate 40 cts. Ibs. Ibs. Iba. 28 28 38 cts. at 18 " 24 " 48 15 zu - 20- 22 J - o - 8 - 16 24, 48 i Answers. Explanation. By connecting the less rate with the greater, and placing the differences between them and the mean rate al- ternately, the quantities resulting are such, that there is precise- ly as much gained by one quantity as is lost by the other, and therefore the gain and loss upon the whole must be equal, and the compound will have the value of the proposed rate ; the same will be true of any other two simples managed according to the rule. In like manner, let the number of simples be what they may, and with how many soever every one is linked, since it is always a less with a greater than the mean price, there will be an equal balance of loss and gain between every two, and consequently an equal balance on the whole. It is obvious, from the rule, that questions of this sort admit of a great variety of answers ; for having found one answer, we may find as many others as we please by only multiplying or dividing each of the quantities found by 2, 3, or 4, &c., the reason of which is evident ; for if two quantities of two sim- ples make a balance of loss and gain with respect to the mean price, so must also the double or treble, the half or third part, or any other equimultiples or parts of these quantities. 6. How much barley at 50 cents a bushel, and rye at 75 cents, and wheat at $ 1.00, must be mixed, that the composi- tion may be worth 80 cents a bushel ? Ans. 20 bushels of rye, 20 of barley, and 35 of wheat. 7. A goldsmith would mix gold of 19 carats fine with some of 15, 23, and 24 carats fine, that the compound may be 20 carats fine. What quantity of each must he take ? Ans. 4oz. of 15 carats, 3oz. of 19, loz. of 23, and 5oz. of 24. 8. It is required to mix several sorts of wine at 60 cents, 80 cents, and $ 1.20, with water, that the mixture may be worth 75 cents per gallon ; how much of each sort must be taken ? Ans. 45gals. of water, ogals. of 60 cents, 15gals. of 80 cents, and75gals. of $1.20. 24 278 ALLIGATION. [SECT. LXII. CASE II. When one of the ingredients is limited to a certain quantity. RULE. Take the difference between each price and the mean rate, as before; then say, as the difference of that simple whose quantity is given is to the rest of the differences severally, so is the quantity given to the several quantities required. EXAMPLES. 9. How much wine at 5s., at 5s. 6d., and at 6s. a gallon, must be mixed with 3 gallons at 4s. per gallon, so that the mixture may be worth 5s. 4d. per gallon ? 8 + 2=10 Then 10 : 10 : : 3 : 3 8 + 2 = 10 10 : 20 : : 3 : 6 16 -f 4 = 20 10 : 20 : : 3 : 6 16 -|- 4 = 20 Ans. 3gals. at 5s., 6 at 5s. 6d., and 6 at 6s. 10. A grocer would mix teas at 12s., 10s., and 6s. per pound, with 20 pounds at 4s. per pound ; how much of each sort must he take to make the composition worth 8s. per pound ? Ans. 201bs. at 4s., lOlbs. at 6s., lOlbs.at 10s.,and201bs. at 12s. 11. How much port wine at $ 1.75 per gallon, and temper- ance wine at $ 1.25 per gallon, must be mixed with 20 gal- lons of water, that the whole may be sold at $ 1 .00 per gallon ? Ans. 20gals. port wine, and 20gals. temperance wine. 12. How much gold of 15, 17, and 22 carats fine must be mixed with 5 ounces of 18 carats fine, so that the composition may be 20 carats fine ? Ans. 5oz. of 15 carats, 5oz. of 17, and 25oz. of 22. CASE III." When the sum and quality of the ingredients are given. RULE. Find an ansiver as before, by linking ; then say, as the sum of the quantities or differences, thus determined, is to the given quantity, so is each ingredient found by linking to the required quantity of each. * To this case belongs the curious fact of King Hiero's crown. Hiero, King of Syracuse, gave orders for a crown to be made of pure gold ; but suspecting that the workmen had debased it, by mixing it with silver or copper, he recommended the discovery of the fraud to the famous Archimedes, and desired to know the exact quantity of alloy in the crown. Archimedes, in order to detect the imposition, procured two other mass- es, the one of pure gold, the other of copper, and each of the same weight of the former ; and by putting each separately into a vessel full of water, the quantity of water expelled by them determined their specific gravi- SECT. LSI.] PERMUTATIONS AND COMBINATIONS. 279 EXAMPLES. 13. How many gallons of water must be mixed with wine at $ 1.50 per gallon, so as to fill a vessel containing 100 gallons, that it may be sold at $ 1.20 per gallon ? gals. gals. gals. gala. 19ft / 1 30 150:100:: 30 : 20 water ) A U { 150 I 120 150 : 100 : : 120 : 80 wine J * 150 14. A merchant has sugar at 8 cents, 10 cents, 12 cents, and 20 cents per pound ; with these he would fill a hogshead that would contain 200 pounds. How much of each kind must he take, that he may sell the mixture at 15 cents per pound ? Ans. 33lbs. of 8, 10, and 12 cts., and lOOlbs. of 20 cts. SECTION LXX. PERMUTATIONS AND COMBINATIONS. THE permutation of quantities is the showing how many dif- ferent ways the order or position of any given number of things may be changed. The combination of quantities is the showing how often a less number of things can be taken out of a greater, and com- bined together, without considering their places or the order they stand in. CASE I. To find the number of permutations or changes that can be made of any given number of things, all different from each other. RULE. Multiply all the terms of the natural series of numbers, from ties ; from which, and their given weights, the exact quantities of gold and alloy in the crown may be determined. Suppose the weight of each crown to be 10 pounds, and that the water expelled by the copper was 92 pounds, by the gold 52 pounds, and by the compound crown 64 pounds : what will be the quantities of gold and al- loy in the crown ? The rates of the simples are 92 and 52, and of the compound 64 ; there- fore, 92 j 12 of copper, 40 : 10 : : 12 : 31bs. of copper \ An , wpr 62 _J 28 of ^ 40 : 10 : : 28 : Tibs, of gold 5 Ans 280 PERMUTATIONS AND COMBINATIONS. [SECT. LXX. 1 up to the given number, continually together, and the last product will be the answer required. This rule may be illustrated by inquiring how many different numbers may be formed from the figures of the following number, 789, making use of the three figures in each number. OPERATION. 789, 798, 879, 897, 978, 987. It will be perceived, that six are all the permutations the above number will admit of. By adopting the rule, we find the same answer. 1X2X3 = 6 Ans. 1. How many changes may be rung on 6 bells ? OPERATION. 1X2X3X4X5X6 = 720 changes. Ans. 2. For how many days can 10 persons be placed in a differ- ent position at dinner ? Ans. 3628800. 3. How many changes may be rung on 12 bells, and how long would they be in ringing, supposing 10 changes to be rung in one minute, and the year to consist of 365 days, 5 hours, and 49 minutes ? Ans. 479001600, and 91y. 26d. 22h. 41m. 4. How many changes or variations will the letters of the alphabet admit of? Ans. 403291461126605635584000000. CASE II. Any number of different things being given, to find how many changes can be made out of- them, by taking any given number of quantities at a time. RULE. Take a series of numbers, beginning at the number of things given, and decreasing by I, to the number of quantities to be taken at a time ; the product of all the terms will be the answer required. Illustration. The above rule may be illustrated by inquiring how many different numbers can be made by a selection of any three figures from the following number, 1234. OPERATION. 123, 124, 132, 134, 142, 143, 213, 214, 231, 234, 241, 243, 312, 314, 321, 324, 341, 342, 412, 413, 421, 423, 431, 432. By examining the above, it will be perceived that there are 24 different numbers or permutations ; and this number may be obtained by multiplying the number of things given, 4, by the next lower number, and that product by the next lower, and SECT. LXX.] PERMUTATIONS AND COMBINATIONS. 281 so continuing the multiplications as many times as there are things to be taken at a time ; and the things to be taken at a time are 3 ; we therefore find the continued product of the first three numbers, beginning at 4 ; thus, 4 X 3 X 2 = 24 Answer. NOTE. The questions under this rule refer to permutations and not combinations. 5. How many changes can be rung with 4 bells out of 8 ? OPERATION. 8X7X6X5= 1680 changes, Answer. 6. How many words can be made with 6 letters out of 26 of the alphabet, admitting a word might be made from consonants ? Ans. 165765600. CASE III. To find the number of combinations of any given number of things, all different one from another, taking any given number at a time, without reference to their arrangement. RULE. Take the series 1, 2, 3, 4, <3fC., up to the number to be taken at a time, and find the product of all tJie terms. Take a series of as many terms, decreasing by 1 from the given num- ber out of which the election is to be made, and find the product of all the terms. Divide the last product by the former, and the quotient will be tJie number required. 7. How many combinations can there be of any three let- ters of the alphabet out of any four letters, without reference to their arrangement or permutations ? Illustration. By examining four letters, a, b, c, d, we find they will admit of only four combinations. Thus, abc, abd, acd, bed. And this number can be ascertained in the following manner : 1X2X3=6; 4X3X2 = 24 ; 24 -s- 6 = 4, the number of combinations. 8. How many combinations can be made of 7 letters out of 10, the letters all being different ? Ans. 120. OPERATION. 1X2X3X4X5X6X7=: 5040 ; 7 being the number taken at a time. 10X9X8X7X6X5X4 = 604800 = same number from 10. 604800 5040 = 120 Ans. Number of combinations. 24* 282 LIFE INSURANCE. [SECT. LXXI. CASE IV. To find the compositions of any number of things, in an equal number of sets, the things being all different. RULE. Multiply the number of things in every set continually to- gether t and tJie product will be the answer required. 9. Suppose there are 4 companies, in each of which there are 9 me.n ; it is required to find how many ways 4 men may be chosen, one out of each company. 9x9x9x9 = 6561 Ans. 10. There are 4 companies, in one of which there are 6 men, in another 8, and in each of the other two 9 men. What are the choices, by a composition of 4 men, one out of each com- pany ? Ans. 3888. 11. How many changes are there in throwing 5 dice ? Ans. 7776. SECTION LXXI. LIFE INSURANCE. INSURANCE on life is a contract, which stipulates for the pay- ment of a certain sum of money on the death of an individual, in consideration of the immediate payment of a specified sum, or more frequently of an annuity, or annual premium, to be continued during the existence of the life insured. . Contracts of this kind are of immense importance to society. Every man whose income depends on his own life or exertions, and on whom others are dependent for support, must be sensible of the advantages of arrangements by means of which, at a small sacrifice of immediate comfort, he is enabled effectually to provide against the casualties of life. Though nothing can be more uncertain than the continuance of an individual life, yet nothing is more invariable than the duration of life in the mass ; consequently, the exact value of life insurances can be calcu- lated without any uncertainty whatever, and a man by effecting an insurance secures to his family, against risk of accident, the advantages they would have from enjoying his exact proportion of the average duration of life. Such transactions provide against destitution, and tend directly to the accumulation of capital. No wise man will therefore neglect to provide against contingencies. SECT. LXXI.] LIFE INSURANCE. 283 The following table from Milne's Treatise on the Valuation of Annuities and Assurances (Vol. II. p. 565) shows the " ex- pectation of life " at every age, according to the law of mor- tality at Carlisle.' Expectation of Life. Age. Expectation. Age. Expectation. Age. Expectation. Age. Expectation. 36.72 26 37.14 52 19.68 78 6.12 1 44.68 27 36.41 53 18.97 79 5.80 2 47.55 28 35.69 54 18.28 80 5.51 3 49.82 29 35.00 55 17.58 81 5.21 4 50.76 30 34.34 56 16.89 82 4.93 5 51.25 31 33.68 57 16.21 83 4.65 6 51.17 32 33.03 58 15.55 84 4.39 7 50.80 33 32.36 59 14.92 85 4.12 8 50.24 34 31.68 60 14.34 86 3.90 9 45.57 35 31.00 61 13.82 87 3.71 10 48.82 36 30.32 62 13.31 88 3.59 11 48.04 37 29.64 63 12.81 89 3.47 12 47.27 38 28.96 64 12.30 90 3.28 13 46.51 39 28.28 65 11.79 91 3.26 14 45.75 40 27.61 66 11.27 92 3.37 15 45.00 41 26.97 67 10.75 93 3.48 16 44.27 42 26.34 68 10.23 94 3.53 17 43.57 43 25.71 69 9.70 95 3.52 18 42.87 44 25.09 70 9.19 96 3.46 19 42.17 45 24.46 71 . 8.65 97 3.28 20 41.46 46 23.82 72 8.16 98 3.07 21 40.75 47 23.17 73 7.72 99 2.77 22 40.04 48 22.50 74 7.33 100 2.28 23 39.31 49 21.81 75 7.01 101 1.79 24 38.59 50 21.11 76 6.69 102 1.30 25 37.86 51 20.39 77 6.40 103 0.83 By the above table it will be perceived, that tj*e average age to which 100 persons will live from birth will be 38 T 7 ^ years, and that 100 persons having attained the age of 35 years will live on an average 31 years longer ; and the average time which 100 persons will continue to live, after the age of 100 years, is 2-^ years. Taking the above table as the basis of their calculations, various life-insurance companies have been formed, whose tables or rates of insurance vary according as they charge a greater or less per cent, on their capital. We have selected two tables from those used by the numerous cor- porations which have been formed in the United States. The first is the one adopted by the Massachusetts Hospital Life In- surance Company, and the other by the New York Life Insur- ance Company. 284 LIFE INSURANCE. [SECT. LXXI. Age. Massachusetts. New York. 1 Year 7 Years. 10 Years. 1 Year. 7 Years. For Life. 14 .82 .84 .85 .72 .86 .53 15 .83 .85 .86 .77 .88 .56 16 .84 .86 .87 .84 .90 .62 17 .85 .87 .88 .86 .91 .65 18 . .86 .88 .89 .89 .92 .69 19 .87 .89 .90 .90 .94 .73 20 .88 .90 .92 .91 .95 .77 21 .89 .91 .93 .92 .97 .82 22 .90 .92 .94 .94 .99 .88 23 .91 .94 .96 .97 .03 .93 24 .92 .95 .98 .99 .07 .98 25 .93 .97 .99 1.00 .12 2.04 26 .95 .98 .01 1.07 .17 2.11 27 .96 1.00 .03 1.12 .23 2.17 28 .98 '1.02 .05 1.20 .28 2.24 29 .00 1.04 .07 1.28 .35 2.31 30 .01 .06 .09 1.31 .36 2.36 31 .03 .08 .11 1.32 .42 2.43 32 .05 .10 .14 1.33 .46 2.50 33 .07 .13 .16 1.34 .48 257 34 .09 .15 .19 1.35 .50 2.64 35 .12 .18 .22 1.36 .53 2.75 36 .14 .20 .25 1.39 .57 2.81 37 .16 .23 .29 1.43 .63 2.90 38 .19 .27 .34 1.48 .70 3.05 39 .22 .31 .39 1.57 .76 3.11 40 .25 .35 .44 1.69 .83 3.20 41 .28 .40 .50 1.78 .88 3.31 42 .31 .46 .58 1.85 .89 . 3.40 43 .35 .53 1.66 1.89 .92 3.51 44 .40 .61 1.75 1.90 .94 3.63 45 .47 .70 1.85 1.91 .96 3.73 46 .54 .80 1.95 1.92 .98 3.87 47 .62 .90 2.07 1.93 .99 4.01 48 .71 2.01 2.20 1.94 2.02 4.17 49 .81 2.14 2.34 1.95 2.04 4.49 50 1.91 2.27 2.49 1.96 2.09 4.60 51 2.03 2.42 2.65 1.97 2.20 4.75 52 2.16 2.58 2.83 2.02 2.37 4.90 53 2.29 2.75 3.03 2.10 2.59 5.24 54 2.44 2.94 3.24 2.18 2.89 5.49 55 2.60 3.14 3.47 2.32 3.21 5.78 56 2.78 3.36 3.72 2.47 3.56 6.05 57 2.96 3.61 400 2.70 4.20 6.27 58 3.17 3.87 4.29 3.14 4.31 6.50 59 3.39 4.17 4.62 3.67 4.63 6.75 60 3.64 4.48 4.97 4.35 4.91 7.00 SECT. LXXI.J LIFE INSURANCE. 285 The preceding table shows at what rate a hundred dollars may be insured on a person's life for one year, for seven years, ten years, and for life, at any age from 14 to 60 years. Thus, a man who is 42 years old, and who wishes to obtain a life insurance in Massachusetts for one year, for $ 100, must pay $1.31, and in the same proportion for a larger sum. And, if he would obtain a life insurance in New York, he must pay annu- ally $ 3.40 on every hundred dollars for which he obtains insur- ance. EXAMPLES. 1. What premium will the Massachusetts Hospital Life Insur- ance Company require for the insurance of a life for 1 year for $ 1728, the person being 30 years of age ? Ans. $ 17.45. OPERATION. $ 1728 X .0101 = $ 17.45 Ans. NOTE. As the premium is $ 1.01 on $ 100, the sum insured must be multiplied by ffi .0101. 2. What amount of premium must James Kimball pay at the above office to effect an insurance on his life for 7 years for $ 8000, his age being 53 years ? Ans. $ 220. OPERATION. $ 8000 x .0275 = $ 220 annually, Ans. 3. John Smith, 60 years of age, wishes to engage in a very profitable speculation, and, being destitute of the necessary funds, he effects an insurance on his life for 10 years, for $78,000, at the office of the above Company. Required the amount of the annual premium. Ans. $3736.20. 4. What will be the premium per annum for insuring a per- son's life, who is 15 years old, for 82000 for 7 years, at the New York Life Insurance Company ? Ans. $ 17.60. 5. A gentleman 45 years of age, being bound on a long and dangerous voyage, and wishing to secure a competence for his family, obtains an insurance for life at the above office in New York, for $ 12,000. By an act of Providence, he dies before the end of the third year. What is the net gain to his family ? Ans. 8 10,657.20. 6. John Swan, 20 years old, effects an insurance for life at New York, for $10,000, for which he pays an annual premium .of ^J per cent. If the Insurance Company loan the premium at 6 per cent, compound interest, and Swan should die at the age of 60 years, who will gain by the insurance ? Ans. Company will gain $ 17,382.86+. NOTE. All premiums are paid annually and in advance. 286 POSITION. [SECT. LXXII. SECTION LXXH. POSITION. POSITION is a method of performing such questions as cannot be resolved by the common direct rules, and is of two kinds, called single and double. SINGLE POSITION. SINGLE POSITION teaches us to resolve those questions whose results are proportional to their suppositions. RULE. Take any number and perform the same operations with it as are desirable to be performed in the question. Then say, as the result of the operation is to the position, so is the result in the question to the number required. EXAMPLES. 1. A schoolmaster being asked how many scholars he had, replied, that if he had as many more as he now had, and half as many more, he should have 200; of how many did his school consist ? Suppose he had 60 As 150 : 200 : : 60 Then, as many more 60 60 Half as many more _30 15Q j^J ( 8Q ^^ Ang> 150 12000 By analysis. By having as many more, and half as many more, he must have 2 times the original number ; therefore, by dividing 200 by 2, we obtain the answer, 80, as before. NOTE. Having performed all the following questions by position, the student should then perform them by analysis. 2. A person after spending and J of his money had $ 60 left ; what had he at first? Ans. $ 144. 3. What number is that, which, being increased by , , and i of itself, the sum shall be 125 ? Ans. 60. 4. A's age is double that of B, and B's is triple that of C, and the sum of all their ages is 140. What is each person's age ? Ans. A's 84, B's 42, C's 14 years. 5. A person lent a sum of money at 6 per cent., and at the end of 10 years received the amount $ 560. What was the sum lent ? Ans. $ 350. SECT. LXXII.] POSITION. 287 6. Seven eighths of a certain number exceed ^ by 81 ; what is the number ? Ans. 120. 7. What number is that whose f exceed |- by 2-J~ ? Ans. 87. DOUBLE POSITION.* DOUBLE POSITION teaches to resolve questions, by making two suppositions of false numbers. Those questions in which the results are not proportional to their positions belong to this rule. RULE. Take any two convenient numbers, and proceed with each ac- cording to the conditions of the question. Find how much the results are different from the result in the question. Multiply each of the errors by the contra supposition, and find the sum and difference of the prod- ucts. If the errors are alike, divide the difference of the products by the difference of the errors, and the quotient will be the answer. If the er- rors are unlike, divide the sum of the products by the sum of the errors, and the quotient will be the answer. . NOTE. The errors are said to be alike when they are both too great, or both too small ; and unlike when one is too great and the other too little. EXAMPLES. 1. A lady purchased a piece of silk for a gown at 80 cents per yard, and lining for it at 30 cents per yard ; the gown and lining contained 15 yards, and the price of the whole was $ 7.00. How many yards were there of each ? Suppose 6 yards of silk, value $ 4.80 She must then have 9 yards of lining, value 2.70 Sum of their values, $ 7.50 Which should have been 7.00 So the first error is 50 too much, -j- .50 Again ; suppose she had 4 yards of silk, value $ 3.20 Then she must have 1 1 yards of lining, value 3.30 Sum of their values, $ 6.50 Which should have been 7.00 So that the second error is 50 too little, .50 * This rule is founded on the supposition, that the first error is to the second as the difference between the true and first supposed number is to the difference between the true and second supposed number. When this is not the case, the exact answer to the questions cannot be found by this rule. 288 POSITION. [SECT. LXIII. First supposition multiplied by last error, 6 X 50 := 3.00 Last supposition multiplied by first error, 4 X 50 = 2.00 Add the products, because unlike^ $ 5.00 500 -f- 50 +50 = 5 yards of silk, ) A 5 X 80 = $ 4.00 15 5 = 10 yards of lining, f# 10 X 30 = 3.00 Proof $17.00 By Analysis. As the silk and lining contain 15 yards, and cost $7.00, the average price per yard is 46 ; and this taken from 80 leaves 33 ; and 30 taken from 46f leaves 16f ; and as the quantity of lining will be to that of the silk as 33 to 16f , it is therefore evident that the quantity of lining is twice the quantity of silk. Wherefore, if 15, the number of yards, be divided into three parts, two of those parts (10) will be the number of yards for the lining, and the other part (5) will be the yards for the silk, as before. NOTE. The student should perform each question by analysis. 2. A and B invested equal sums in trade ; A gained a sum equal to of his stock, and B lost $ 225 ; then A's money was double that of B's. What did each invest ? Ans. $ 600. 3. A person being asked the age of each of his sons, replied, that his eldest son was 4 years older than the second, his second 4 years older than the third, his third 4 years older than the fourth, or youngest, and his youngest half the age of the oldest. What was the age of each of his sons ? Ans. 12, 16, 20, and 24 years. 4. A gentleman has two horses and a saddle worth $ 50. Now, if the saddle be put on the first horse, it will make his value double that of the second horse ; but if it be put on the second, it will make his value triple that of the first. What was the value of each horse ? Ans. The first $ 30, second $ 40. 5. A gentleman was asked the time of day, and replied, that f of the time past from noon was equal to ^ of the time to midnight. What was the time ? Ans. 12 minutes past 3. 6. A and B have the same income. A saves T V of his, but B, by spending $100 per annum more than A, at the end of 10 years finds himself $ 600 in debt. What was their income ? Ans. $ 480. 7. A gentleman hired a laborer for 90 days on these condi- tions : that for every day he wrought he should receive 60 cents, and for every day he was absent he should forfeit 80 SECT. LXXII.] POSITION. 289 cents. At the expiration of the term he received $ 33. How- many days did he work, and how many days was he idle ? Ans. He labored 75 days, and was idle 15 days. The following question, with some variation in the language, is taken from Fenn's Algebra, page 62. It is believed, howev- er, that Sir Isaac Newton was the author of it. 8. If 12 oxen eat 3 acfes of grass in 4 weeks, and 21 oxen eat 10 acres in 9 weeks, how many oxen would it require to eat 24 acres in 18 weeks, the grass to be growing uniformly ? Ans. 36 oxen. OPERATION BY ANALYSIS. Each ox eats a certain quantity in each week, which we may suppose to be 100 pounds ; and of the whole quantity eaten in each case, a part must have already grown during the time of eating. Then, by the first conditions of the question, 12 X 4 X 100 = 48001bs. = whole quantity on 3 acres for 4 weeks. ' 4800 -T- 3^ = 14401bs. = whole quantity on 1 acre for 4 weeks. By the second conditions of the question, 21 X 9 X 1QQ 189001bs. = whole quantity on 10 acres for 9 weeks. 18900 -;- 10 = 18901bs. = whole quantity on 1 acre for 9 weeks. 1890 1440 4501bs. = the quantity grown on an acre for 9 4 = 5 weeks. 450 -r- 9 4 = 901bs. = the quantity which grows on each acre for 1 week. 90 X 3^ X 4 12001bs. = quantity grown on 3 acres for 4 weeks. 4800 1200 = 36001bs. = original quantity of grass on 3 acres. 3600 -r- 3 = lOSOlbs. original quantity on 1 acre. Then, by the last condition of the question, 24 x 1080 = 259201bs. = original quantity on 24 acres. 24 x 90 x 18 388801bs. = quantity which grows on 24 acres in 18 weeks. 25920 -f- 38880 = 648001bs. = whole quantity on 24 acres for 18 weeks. 64800 - 18 = 36001bs. = quantity to be eat from 24 acres each week. 3600 -H 100 = 36 = number of oxen required to eat the whole, and the answer to the question. 25 290 EXCHANGE. [SECT LXXIII. 9. There is a fish whose head weighs* 15 pounds, his tail weighs as much as his head and as much as his body, and his body weighs as much as his head and tail. What was the weight of the fish ? Ans. 721bs. 10. Suppose a clock to have an hour-hand, a minute-hand, and a second-hand, all turning on the same centre. At 12 o'clock all the hands are together ancf point at 12. (1.) How long will it be before the second-hand will be be- tween the other two hands, and at equal distances from each ? Ans. 60-j^V seconds. (2.) Also before the minute-hand will be equally distant be- tween the other two hands? Ans. 61f^ seconds. (3.) Also before the hour-hand will be equally distant be- tween the other two hands ? Ans. 59 seconds. SECTION LXXIII. EXCHANGE. EXCHANGE is the act of paying or receiving the money of one country for its equivalent in the money of apother country, by means of Bills of Exchange. This operation, therefore, com- prehends both the reduction of moneys and the negotiation of bills. It determines the comparative value of the currencies of all nations, and shows how foreign debts are discharged, loans and subsidies paid, and other remittances made from one coun- try to another, without the risk, trouble, or expense of transport- ing specie or bullion. BILLS OF EXCHANGE. A Bill of Exchange is a written order for the payment of a certain sum of money, at an appointed time. It is a mercantile contract, in which four persons are mostly concerned ; viz. 1. The drawer, who receives the value, and is also called the maker and seller of the bill. 2. The person upon whom the bill is drawn is called the drawee. He is also called the acceptor, when he accepts the bill, which is an engagement to pay it when due. 3. The person who gives value for the bill, who is called the buyer, taker, and remitter. 4. The person to whom it is ordered to be paid, who is called SECT. LXX.III.] EXCHANGE. 291 the payee, and who may, by indorsement, pass it to any other person. Most mercantile payments are made in Bills of Exchange, which generally pass from hand to hand, until due, like any other circulating medium ; and the person who at any time has a bill in his possession is called the holder. When the holder of a bill disposes of it, he writes his name on the back, which is called indorsing ; and the payee should be the first indorse r. If the bill be indorsed in favor of any particular person, it is called a special indorsement ; and the person to whom it is thus made payable is called the indorsee, who must also indorse the bill if he negotiates it. Any person may indorse a bill, and every indorser (as well as the acceptor, or payee) is a security for the bill, and may therefore be sued for payment. The term of a bill varies according to the agreement between the parties, or the custom of countries. Some bills are drawn at sight ; others, at a certain number of days, or months, after sight or after date ; and some, at usance, which is the customa- ry or usual term between different places. te Days of grace are a certain number of days granted to the acceptor, after the term of a bill is expired. Three days are usually allowed. In reckoning when a bill, payable after date, becomes due, the day on which it is dated is not included ; and if it be a bill payable after sight, the day of presentment is not included. When the term is expressed in months, calendar months are understood ; and when a month is longer than the preceding, it is a rule not to go in the computation into a third month. Thus, if a bill be dated the 28th, 29th, 30th, or 31st of Janu- ary, and payable one month after date, the term equally expires on the last day of February, to which the days of grace must, of course, be added ; and therefore the bill becomes due on the 3d V7, 1 6.55,7 164 .59,8 .63,5 BRUNSWICK. Pistole, double in proportion, - 4 214 4.27,1 4.54,8 Ducat, 2 5| 2.09,2 2.23,0 COLOGNE. Ducat, ------ 2 5| 2.12,5 2.26,7 COLOMBIA. Doubloons, - 17 9 14.56,0 15.53,5 DENMARK. Ducat, current, - - - - 2 1.70,5 1.81,2 Ducat, specie, - 2 5| 2.12,5 2.26,7 Christian d'Or, - ... 4 7 3.77,0 4.02,1 EAST INDIES. Rupee, Bombay, 1818, 7 11 6.65,4 7.09,6 Rupee, Madras, 1818, - - - 7 12 6.66,7 7.11,0 Pagoda, Star, - - - - 2 4| 1.68,9 1.79,8 ENGLAND. Guinea, half in proportion, 5 8* 4.79,9 5.07,5 Sovereign, - 5 24 4.57,0 4.84,6 Seven Shilling Piece, - - - 1 19 1.60,0 1.69,8 FRANCE. Double Louis, coined before 1786, 10 11 9.08,7 "9.69,7 Louis, coined before 1786, 5 54 4.54,1 4.84,6 Double Louis, coined since 1786, 9 20 8.59,0 9.15,3 Louis, coined since 1786, 4 22 4.29,5 4.57,6 Double Napoleon, or 40 francs, - Napoleon, or 20 francs, 8 7 4 34 7.23,2 3.61,6 7.70,2 3.85,1 FRANKFORT ON THE MAINE. Ducat, 2 5| 2.13,7 2.27,9 GENEVA. Pistole, old, ... - 4 74 3.73,7 3.98,5 Pistole, new, - - - - 3 155 3.23,2 3.44,4 HAMBURG. Ducat, double in proportion, 2 5| 2.13,7 2.27,9 GENOA. K3 2 15 8 Q Qn Q HANOVER. J 5 -> . L *s arch at D ; with the same extent cross the arch again from D to E ; L then with one foot of the dividers " in D, and, with any extent above the half of D E, describe an arch a ; take the dividers from D, and, keeping them at the same extent, with one foot in E, intersect the former arch a in a ; from thence draw a line to the point B, which will be a perpendicular to A B. PROBLEM IV. From a given point, a, to let fall a perpendicular to a given line AB. Set one foot of the dividers in the point a, extend the other so as to reach beyond the line A B, and describe an arch to cut the line \ A B in C and D ; put one foot of T r?~~ the dividers in C, and, with any .. -~r\ extent above half C D, describe an arch b ; keeping the dividers at the X. same extent, put one foot in D, and intersect the arch b in b ; through which intersection, and the point fl, draw aE, the perpendicu- lar required. PROBLEM V. To draw a line parallel to a given line A B. Set one foot of the dividers in E_ F any part of the line, as at c; ex- /'"' b~~^ /-'"a """ * N tend the dividers at pleasure, un- less a distance be assigned, and de- A B scribe an arch ; with the same SECT. LXXV.] GEOMETRY. 315 extent in some other part of the line A B, as at e, describe the arch a ; lay a rule to the extremities of the arches, and draw the line E F, which will be parallel to the line A B. PROBLEM VI. To make a triangle whose sides shall be equal to three given lines, any two of which are longer than the third. Let ABC be the three given lines; ^ draw a line, A B, at pleasure ; take the r> line C in the dividers, set one foot in A, ^ and with the other make a mark at B ; then take the given line B in the dividers, and, setting one foot in A, draw the arch C ; then take the line A in the dividers, and, setting one foot in B, intersect the arch C in C ; lastly, draw the lines A C and B C, and the triangle will be completed. PROBLEM VII. To make a square whose sides shall be equal to a given line. Let A be the given line ; draw a line, A B, equal to the given line ; from B raise a perpendicular to C, equal to A B ; with the same extent, set one foot in C, and describe the arch D ; also, with the same extent, set one foot in A, and intersect the arch D ; lastly, draw the line A D and C D, and the square will be completed. In like manner may a parallelogram be constructed, only at- tending to the difference between the length and breadth. PROBLEM VIII. To describe a circle, which shall pass through any three given points, not in a straight line. Let the three given points be A B C, through which the circle is to pass. Join the points A B and B C with right lines, and bisect these lines ; the point D, where the bisecting lines cross each other, will be the centre of the circle required. Therefore, place one foot of the dividers in D, extending the other to either of the given points, and / f-> ~ *-, / A A 316 GEOMETRY. [SECT. LXXT. the circle described by that radius will pass through all the points. Hence it will be easy to find the centre of any given circle ; for, if any three points are taken in the circumference of the given circle, the centre will be found as above. The same may also be observed when only a part of the circumference is given. MENSURATION OF SOLIDS. DEFINITIONS. 1. Solids are figures having length, breadth, and thickness. 2. A prism is a solid, whose ends are any plane figures which are equal and similar, and its sides are parallelograms. NOTE. A prism is called a triangular prism, when its ends are triangles; a square prism, when its ends are squares ; * pentagonal prism, when its ends are pentagons ; and so on. 3. A cube is a square prism, having six sides, which are all squares. 4. A parallelepiped is a solid, having six rectangular sides, every opposite pair of which is equal and parallel. 5. A cylinder is a round prism, having circles for its ends. 6. A pyramid is a solid, standing on a triangular, square, or polygonal basis, and its sides are triangles, whose vertices meet in a point at the top, called the vertex of the pyramid. SECT. LXXV.] GEOMETRY. 317 7. A cone is a solid figure, having a circle for its base, and its top terminated in a point or vertex. 8. A sphere is a solid, bounded by one continued convex surface, every point of which is equally distant from a point within, called the centre. The sphere may be conceived to be formed by the revolution of a semicircle about its diam- eter, which remains fixed. A hemisphere is half a sphere. 9. The segment of a pyramid, sphere, or any other solid, is a part cut off the top by a plane parallel to the base of that figure. 10. A frustum is the part that remains at the bottom after the segment is cut off. 11. The sector of a sphere is composed of a segment less than a hemisphere, and of a cone having the same base with the segment, and its vertex in the centre of the sphere. 12. The axis of a solid is a tine drawn from the middle of one end to the middle of the opposite end ; as between the op- posite ends of a prism. The axis of a sphere is the same as a diameter, or a line passing through the centre, and terminating at the surface on both sides. 13. The height or altitude of a solid is a line drawn from its vertex, or top, perpendicular to its base. MENSURATION OF SUPERFICIES AND SOLIDS. PROBLEM I. To find the area of a square or parallelogram. RULE. Multiply the length by the breadth, and the product is the superficial contents. 1. What are the contents of a board 15 feet long and 2 feet wide ? Ans. 30 feet. 2. The State of Massachusetts is about 128 miles long and 48 miles wide. How many square miles does it contain ? Ans. 6144 miles. 27* 318 GEOMETRY. [SECT. LXXY. 3. The largest of the Egyptian pyramids is square at its base, and measures 693 feet on a side. How much ground does it cover ? Ans. 1 1 acres 4 poles. 4. What is the difference between a floor 40 feet square, and 2 others, each 20 feet square ? Ans. 800 feet. 5. There is a square of 3600 yards area ; what is the side of a square, and the breadth of a walk along each side of the square, and each end, which may take up just one half of the square ? A $ 42.42+ yards, side of the square. s - I 8.78+ yards, breadth of the walk. PROBLEM II. To find the area of a rhombus or rhomboid. RULE. Multiply the length of the base by the perpendicular lieight. 6. The base of a rhombus being 12 feet, and its height 8 feet, required the area. Ans. 96 feet. PROBLEM III. To find the area of a triangle. RULE. Multiply the base by half the perpendicular height ; or, add the three sides together; then take half of that sum, and out of it sub- tract each side severally ; multiply the half of the sum and these remain- ders together, and the square root of this product will be the area of the triangle. 7. What are the contents of a triangle, whose perpendicular height is 12 feet, and the base 18 feet ? Ans. 108 feet. 8. There is a triangle, the longest side of which is 15.6 feet, the shortest side 9.2 feet, and the other side 10.4 feet. What are the contents ? Ans. 46. 139-|- feet. PROBLEM IV. Having the diameter of a circle given, to find the circumference. RULE. Multiply the diameter by 3.141592. NOTE. The exact ratio of the diameter of a circle to its circumference has never yet been ascertained. Nor can a square, or any other right- lined figure, be found, that shall be exactly equal to a given circle. This is the famous problem, called the squaring the circle, which has exercised the abilities of the greatest mathematicians for ages, and has been the oc- casion of so many endless disputes. Van Ceulen, a Dutchman, was the first who ascertained this ratio to any great degree of exactness, which he extended to thirty-six places of decimals ; * and it was effected by * This is said to have been thought so curious a performance, that the numbers were cut on his tombstone, in St. Peter's church-yard, at Leyden. SECT. LXXV.] GEOMETRY. 319 means of the continual bisection of an arc of a circle. This process was exceedingly troublesome and laborious ; but since the invention of Flux- ions and the Summation of Infinite Scries, there have been several meth- ods found for doing the same thing with less labor and trouble, and far more expedition. If, therefore, the diameter of a circle be 1 inch, the circumference will be 3.1415926535897932384626433832795028841971693- 9937510582097494459230781640628620899862803482534211706798214808- 65132823066470938446460955051822317253594081284802 inches nearly. 9. If the diameter of a circle is 144 feet, what is the circum- ference ? Ans. 452.389248 feet. 10. If the diameter of the earth be 7964 miles, what is its circumference ? Ans. 25019.638688+ miles. PROBLEM V. Having the diameter of a circle given, to find the area. RULE. Multiply half the diameter by half the circumference, and the product is the area ; or, which is the same thing, multiply the square of the diameter by .785398, and the product is the area. Demonstration. If we suppose a circle to be divided into an infinite number of triangles, by lines drawn from the centre of the circle to the circumference, we may find the contents of each triangle by multiplying its perpendicular height by half its base, but its perpendicular height is half the diameter of the circle, and half its base is half a certain portion of the circum- ference ; and all the bases of all the triangles united form the whole circumference. Again, if multiplying half the circumference by half the di- ameter give the area of a circle, it is evident that the area will be obtained by multiplying one fourth the circumference by the whole diameter ; and, as the circumference of a circle, whose diameter is 1, is 3.141592, therefore by multiplying 1 by one fourth of 3.141592 we shall obtain the area of a circle whose diameter is 1. Thus, 3.141592 X i .785398. And as cir- cles are to each other as the squares of their diameters (see page 246), therefore, if we wish to obtain the area of a circle whose diameter is 20 feet, we make the following statement. As I 2 foot : 20 2 feet : : .785398 : 314.1592 feet, Ans. And this process is equivalent to multiplying the square of the diameter of the given circle by .785398. Q. E. D. 11. If the diameter of a circle be 761 feet, what is the area ? Ans. 454840.475158 feet. 12. There is a circular island, three miles in diameter ; how many acres does it contain ? Ans. 4523.89-f- acres. 320 GEOMETRY. [SECT. LXXV. PROBLEM VI. Having the diameter of a circle given, to find the side of an equal square. RULE. Multiply the diameter by .886227, and the product is the side of an equal square. Demonstration. We have seen in Problem V. that the area of a circle, whose diameter is 1, is .785398163397 ; if, there- fore,, we extract the square root of this number, we shall obtain the side of a square of a circle whose diameter is 1. Thus, v/.785398163397:= .886227. And since, as we have before stated, the diameters of circles are to each other as the sides of their similar inscribed figures, therefore, as 1, the diame- ter of the given circle, is to the diameter of the required circle, so is .886227, the side of a square equal to the given circle, to the side of a square equal to the required circle. If, therefore, the diameter of a circle were 20 feet, and it was required to find the side of a square that would contain that quantity, we should make the following statement : As 1 foot : 20 feet : : .886227 : 17.72454 feet, Ans. We see, from this process, that multiplying the diameter of the required circle by .886227 gives the side of an equal square. Q. E. D. 13. I have a round field, 50 rods in diameter ; what is the side of a square field, that shall contain the same area ? Ans. 44.31135+ rods. PROBLEM VII. Having the diameter of a circle given, to find the side of a square inscribed. RULE. Multiply the diameter by .707106, and the product is the side of a square inscribed. Demonstration. Let A B C D represent a square inscribed in a circle whose diam- eter is 1. D B is the diameter of the circle, and it is also the diagonal of the square. As DAB is a right-angled triangle, the squares of D A and A B are equal to the square of B D ; but AD is equal to A B, therefore the square of D A is equal to half the square of B D. The square of B D is 1, therefore the square of DA is .5 ; and the square root of .5 is v.5 = .707106 A D, the side of the square, whose diameter is 1. Q. E. D. There- SECT. LXXV.] GEOMETRY. 321 fore, to find the side of a square inscribed in any circle, we say, as 1 is to the diameter of any required circle, so is .707106 to the side of a square inscribed in the required circle. 14. I have a piece of timber 30 inches in diameter ; how large a square stick can be hewn from it ? Ans. 21.21-f- inches square. 15. Required the side of a square, that may be inscribed in a circle 80 feet in diameter. Ans. 56.56848-f- feet. PROBLEM VIII. In a given circle to describe a hexagon and an equilateral triangle, and to find the length of one of the sides of the in- scribed triangle. RULE. Multiply the diameter by .8660254 and the product is the side of an inscribed equilateral triangle. Demonstration. Let A B C D E F be the given circle, and G the centre. From the point B in the circumference apply the radius B G six times to the circumference, and join B C, CD, D E, EF, FA, and AB, and the figure A B C D E F, thus formed, is an equi- lateral inscribed hexagon. Join the alternate angles AE, EC, and C A, and the figure A E C, thus formed, is an equilateral triangle inscribed. It is equilateral because the three sides subtend the equal arches of the circum- ference. A B C G is a rhombus, and the diagonal B G is equal to either side of the rhombus. If, therefore, the diameter of the circle A D is 1, the semidiameter A G or B G will be .5 ; and B H, which is half of BG, will be .25. AHB is a right-angled triangle, and therefore A H is equal to the square root of the difference of the squares of A B and B H. Thus A H = -V/AB^^B H 2 = A/&^& = V.25 .0625 V.1875 .4330127. Now if AH be .4330127, AC, which is twice AH, will be .8660254. But AC is the side of the equilateral triangle inscribed ; and as we have before shown, page 246, that all similar figures are in proportion to their homologous sides, it will therefore follow, that as the diameter of the given circle, which is 1, is to the side of its inscribed equilateral triangle 322 GEOMETRY. [SECT. LXXV. .8660254, so is the diameter of any other circle to the side of its inscribed equilateral triangle. 16. Required the side of an equilateral triangle, that may be inscribed in a circle 80 feet in diameter. Ans. 69.28-|- feet. 17. Required the side of an equilateral triangle, that may be inscribed in a circle 50 inches in diameter. Ans. 43.3-f-in. 18. There is a certain piece of round timber 30 inches in di- ameter ; required the side of an equilateral triangular beam, that may be hewn from it. Ans. 25.98-|- inches. PROBLEM IX. Having the circumference of a circle given, to find the di- ameter. RULE. Multiply the circumference by .3183098, and the product is the diameter. Rationale. As we have seen in Problem IV., the ratio of the circumference of a circle to its diameter is as 3.141592 to 1 ; and as the ratio of all circumferences of circles to their di- ameters is the same, therefore, if the circumference of a circle be 1, its diameter may be found by the following proportion : As 3.141592 :!::!:: .3183098. Wherefore, if we multiply the circumference of any circle by .3183098, the product is the diameter. Q. E. D. 19. If the circumference of a circle be 25,000 miles, what is its diameter ? Ans. 7957.74-f- miles. 20. If the circumference of a round stick of timber be 50 inches, what is its diameter ? Ans. 15.91549+ inches. PROBLEM X. Having the circumference of a circle given, to find the side of an equal square. RULE. Multiply the circumference by .282094, and the product is the side of an equal square. We have demonstrated in a previous problem, that, when the diameter of a circle is 1, the side of an equal square is .886227 ; but when the circumference is 1, the side of an equal square must be as much less than this number, as 1 is less than 3.141592; that is, it will be equal to the number of times .886227 will contain 3.141592 ; .886227 -f- 3. 141592^.282094 ; therefore, if we multiply the circumference of any circle by .282094, the product is the side of an equal square. Q. E. D. SECT. LXXV.] GEOMETRY. 323 21. I have a circular field 360 rods in circumference; what must be the side of a square field, that shall contain the same quantity? Ans. 101.55-|- rods. 22. John Smith had a farm, which was 10,000 rods in cir- cumference, and which he sold at $ 71.75 per acre, and he purchased another farm containing the same quantity of land in the form of a square ; required the length of one of its sides. Ans. 2820.94-f rods. PROBLEM XI. Having the circumference of a circle given, to find the side of an equilateral triangle inscribed. RULE. Multiply the circumference by .2756646, and the product is the side of an equilateral triangle inscribed. Rationale. We have seen in Problem VIII. that the ratio of the diameter of a circle to its inscribed triangle is as 1 to .8660254 ; but the ratio of the circumference of a circle to its diameter is as 3.141592 to 1 ; therefore, the ratio of the cir- cumference of a circle to its inscribed equilateral triangle is as 3.141592 to .8660254 ; that is, it will be equal to the number of times .8660254 will contain 3.141592. Thus .8660254 -i- 3.141592 = .2756646. Therefore, by multiplying the circum- ference of any circle by .2756646, we obtain the side of an equilateral triangle inscribed. Q. E. D. 23. How large an equilateral triangle may be inscribed in a circle, whose circumference is 5000 feet ? Ans. 1378.323ft. 24. Required the side of an equilateral triangular beam, that may be hewn from a round piece of timber 80 inches in cir- cumference. Ans. 22.05-J- inches. PROBLEM XII. Having the circumference of a circle given, to find the side of an inscribed square. RULE. Multiply the circumference by .225079, and the product is the side of a square inscribed. Rationale. We have seen in Problem VII. that the ratio of the diameter of a circle to its inscribed square is as 1 to .707106 ; we have also seen in Problem IV. that the ratio of the circum- ference of a circle to its diameter is as 3.141592 to 1; there- fore, the ratio of the circumference of a circle to its inscribed square is as 3.141592 to .707106. That is, it will be equal to the number of times .707106 will contain 3.141592. Thus g2 4 GEOMETRY. [SECT. LXXV. .707106 -T- 3.141592 = .225079 ; therefore, if the circumfer- ence of any circle be multiplied by .225079, the product is the side of a square inscribed. Q. E. D. 25. I have a circular field, whose circumference is 5000 rods ; what is the side of a square field that may be made in it ? Ans. 1125.395+ rods. 26. How large a square stick may be hewn from a piece of round timber 100 inches in circumference ? Ans. 22.5+ inches square. NOTE. If we wish to find the circumference of a tree, which will hew any given number of inches square, we divide the given side of the square by .225079, and the quotient is the circumference required. 27. What must be the circumference of a tree that will make a beam 10 inches square ? Ans. 44.42+ inches. 28. What must be the circumference of a tree, that, when hewn, it may be 18 inches square ? Ans. 79.97+ inches. 29. I have a garden which is 20 rods square ; required the circumference of a circle, in feet, that will inclose this garden. Ans. 1466.15+ feet PROBLEM XIII. To find the contents of a cube or parallelopipedon. RULE. Multiply the length, height, and breadth continually to- gether, and the product is the contents. 30. How many cubic feet are there in a cube, whose side is 18 inches ? Ans. 3| feet. 31. What are the contents of a parallelopipedon, whose length is 6 feet, breadth 2 feet, and altitude If feet ? Ans. 26 feet. 32. How many cubic feet in a block of marble, whose length is 3 feet 2 inches, breadth 2 feet 8 inches, and depth 2 feet 6 inches ? Ans. 21 feet. PROBLEM XIV. To find the solidity of a prism. RULE. Multiply the area of the base, or end, by the height. 33. What are the contents of a triangular prism, whose length is 12 feet, and each side of its base 24- feet ? Ans. 32.47+ feet. 34. Required the solidity of a triangular prism, whose length is 10 feet, and the three sides of its triangular end, or base, are 5, 4, and 3 feet. Ans. 60 feet. SECT. LXXV.] GEOMETRY. 325 PROBLEM XV. To find the solidity of a cone or pyramid. RULE. Multiply the area of the base by J of its height. 35. What is the solidity of a cone, whose height is 12 feet, and the diameter of the base 2 feet ? Ans. 20.45-}- feet. 36. What are the contents of a triangular pyramid, whose height is 14 feet 6 inches, and the sides of its base 5, 6, and 7 feet ? Ans. 71.035+ feet. QUESTIONS TO EXERCISE THE ABOVE PROBLEMS. 37. I have a round garden, containing 75 square rods ; how large a square garden can be made in it ? Ans. 47.7464-)- square rods. 38. I have a circular garden containing 75 square rods; what must be the side of a square field that would contain it ? Ans. 9.772+ rods. 39. There is a small circular field, 25 rods in diameter; what is the difference of the areas of the inscribed and circum- scribed squares, and how much do they differ from the areas of the field ? Ans. 312.5 rods, the difference of the squares; 134.12625+ rods, the circumscribed square, more than the area ; 178.373 rods, inscribed square less than the area. PROBLEM XVI. To find the surface of a cone. RULE. Multiply the circumference of the base by half its slant height. 40. What is the convex surface of a cone, whose side is 20 feet, and the circumference of its base 9 feet ? Ans. 90 feet. PROBLEM XVII. To find the solidity of the frustum of a cone. RULE. Multiply the diameters of the two bases together, and to thf product add $ of the square of the difference of the diameters ; then mul- tiply this sum by .785398, and the product will be the mean area between the two bases ; lastly, multiply the mean area by the length of the frus- tum, and the product will be the solid contents. Or, find tlie area of the two ends of the frustum, multiply those two areas together, and extract tfte square root of their product. To this root add the two areas, and 326 GEOMETRY. [SKCT. LXXV. multiply their sum by one third of the altitude or length of the frustum, and the product will be the solidity of the frustum of a cone, or pyramid. NOTE. These are the exact rules for measuring round timber, and should be adopted. See page 330. 41. What are the contents of a stick of timber, whose length is 40 feet, the diameter of the larger end 24 inches, and the smaller end 12 inches ? Ans. 73 feet, nearly. PROBLEM XVIII. To find the solidity of a sphere or globe. RULE. Multiply the cube of the diameter by .5236. 42. What is the solidity of a sphere, whose axis or diameter is 12 inches ? Ans. 904.78-j- inches. 43. Required the contents of the earth, supposing its circum- ference to be 25,000 miles. Ans. 263858149120.06886875 miles. PROBLEM XIX. To find the convex surface of a sphere, or globe. RULE. Multiply its diameter by its circumference. 44. Required the convex surface of a globe, whose diameter or axis is 24 inches. Ans. 1809.55-f- inches. 45. Required the surface of the earth, its diameter being 7957 miles, and its circumference 25,000 miles. Ans. 198943750 square miles. PROBLEM XX. To find how large a cube may be cut from any given sphere, or be inscribed in it. RULE. Square the diameter of the sphere, divide that product by 3, and extract the square root of the quotient for the ansioer. Demonstration. It is evident, that, if a cube be inscribed in a sphere, its corners or angles will be in contact with the sur- face of the sphere, and that a line passing from the lower cor- ner of the cube to its opposite upper corner will be the diame- ter of the sphere ; and that the square of this oblique line is equal to the sum of the squares of three sides of the inscribed cube is evident, from the fact that the square of any two sides of the cube (suppose two sides at the base) is equal to the square of the diagonal across the base ; and that the square of SECT. LXXVI.] GAUGING. 327 this diagonal (which we have just proved to be equal to the square of two sides at the base) and the square of the height of the cube are equal to the square of the diagonal line, which passes from the lower corner of the square to the opposite up- per corner, which line is the diameter of the sphere. There- fore, the square of the diameter of any sphere is equal to the sum of the squares of any three sides of an inscribed cube ; or J of the square of the diameter of any sphere is equal to the square of one of the sides of an inscribed cube. Q. E. D. 46. How large a cube may be inscribed in a globe 12 inches in diameter ? Ans. 6.928-j-in. in the side of the cube. 12 v 12 - 48 ; v48 = 6.928+ Answer. o 47. How large a cube may be inscribed in a sphere 40 inch- es in diameter ? Ans. 23.09-f- inches. 48. How many cubic inches are contained in a cube, that may be inscribed in a sphere 20 inches in diameter ? Ans. 1539.6+ inches. SECTION LXXVI. GAUGING. GAUGING is the art of finding the contents of any regular ves- sel, in gallons, bushels, &c. PROBLEM I. To find the number of gallons, &c., in a square vessel. RULE. Take the dimensions in inches; then multiply the length, breadth, and height together ; divide the product by 282 for ale gallons, 231 for wine gallons, and 2 150.42 for bushels. 1. How many wine gallons will a cubical box contain, that is 10 feet long, 5 feet wide, and 4 feet high ? Ans. 1496 T 8 T gal. 2. How many ale gallons will a trough contain, that is 12 feet long, 6 feet wide, and 2 feet high ? Ans. 882|f gal. 3. How many bushels of grain will a box contain, that is 15 feet long, 5 feet wide, and 7 feet high ? Ans. 421.8bu. 328 TONNAGE OF VESSELS. [SECT. LXXVII. PROBLEM II. To find the contents of a cask. RULE. Take the dimensions of live cask in inches ; namely, the dl ameter of the bung and head, and the length of the cask. Note the dif- ference between the bung diameter and t/te head diameter. If the staves of the cask be much curved between the bung and the head, multiply the difference by .7 ; if not quite so much curved , by .65 ; if they bulge yet less, by .6 ; and if they are almost straight, by .55 ; add the product to the head diameter ; the sum will be a mean diameter by which the cask is reduced to a cylinder. Square the mean diameter thus found, then multiply it by the length ; divide the product by 359 for ale or beer gallons, and by 294 for wine gallons. 4. Required the contents in wine gallons of a cask, whose bung diameter is 35 inches, head diameter 27 inches, and length 45 inches. 35 __ 27 x .7 = 5.6 32.6 X 32.6 x 45 = 47824.20 27 + 5.6 =32.6 478242Q 294 162.66 wine gallons. 5. What are the contents of a cask in ale gallons, whose bung diameter is 40 inches, head diameter 30 inches, and length 50 inches ? Ans. 185.55+ ale gallons. PROBLEM III. To find the contents of a round vessel, wider at one end than the other. RULE. Multiply the greater diameter by the less; to this product add of the square of their difference, then multiply by the height, and divide as in the last rule. 6. What are the contents in wine measure of a tub, 40 inches in diameter at the top, 30 inches at the bottom, and whose height is 50 inches ? Ans. 209.75 wine gallons. SECTION LXXVII. TONNAGE OF VESSELS. CARPENTER'S RULE. For single-decked vessels, multiply the length, breadth at the main beam, and depth in the hold together, and divide the product by 95, and the quotient is the tons SECT. LXXVII.] TONNAGE OF VESSELS. 329 But for a double-decked vessel, take half of the breadth of the main beam for the depth of the hold, and proceed as before. 1. What is the tonnage of a single-decked vessel, whose length is 65ft., breadth 20ft., and depth 10ft. ? Ans. 136f| tons. 2. What is the tonnage of a double-decked vessel, whose length is 70 feet, and breadth 24 feet ? Ans. 212 T 4 g- tons. GOVERNMENT RULE. If the vessel be double-decked, take the length thereof from the fore part of the main stem to the after part of the stern-post above the upper deck ; the breadth thereof at the broad- est part above the main wales, half of which breadth shall be accounted the depth of such vessel, and then deduct from the length % of the breadth ; multiply the remainder by the breadth , and the product by the depth, and divide this last product by 95, the quotient whereof shall be deemed the true contents or tonnage of such ship or vessel ; and if such ship or vessel be single-decked, take the length and breadth, as above directed, deduct from the said length f of the breadth, and take the depth from the under side of the deck-plank to the ceiling in the hold, and then multiply and divide as aforesaid, and the quotient shall be deemed the tonnage. 3. What is the government tonnage of a single-decked ves- sel, whose length is 70 feet, breadth 30 feet, and depth in the hold 9 feet ? Ans. 147|f- tons. 4. What is the government tonnage of a single-decked ves- sel, whose length is 75 feet, breadth 22 feet, and depth in the hold 12 feet? Ans. 171ff tons. 5. What is the government tonnage of a double-decked ves- sel, which has the following dimensions : length 98 feet, breadth 35 feet ? Ans. 496 tons. 6. Required the government tonnage of a double-decked ves- sel, whose length is 180 feet, and breadth 40 feet. Aris. 1313|f tons. 7. Required the government tonnage of a single -decked ves- sel, whose length is 78 feet, width 21 feet, and depth 9 feet. Ans. 130 5 T \ tons. 8. What is the government tonnage of a double-decked vessel, whose length is 159ft., and width 30ft. ? Ans. 667 J- tons. 9. What is the government tonnage of Noah's ark, admitting its length to have been 479 feet, its breadth 80 feet, and its depth 48 feet. Ans. 1742 1 T 9 ^ tons. 10. What is the government tonnage of a vessel, whose length is 200 feet, and breadth 35 feet ? Ans. 1153/ F tons. 11. The new ship Montezuma is 280 feet in length, and 40 feet in breadth. Required the government tonnage. Ans. 2155-U tons. 28* 330 MENSURATION OF LUMBER. [SECT. LXXVIII. SECTION LXXVIII. MENSURATION OF LUMBER. PROBLEM I. To find the contents of a board. RULE. Multiply the length of the board, taken in feet, by its breadth taken in inches, divide this product by 12, and the quotient is the con- tents in square feet. 1. What are the contents of a board 24 feet long, and 8 inches wide ? Ans. 16 feet. 2. What are the contents of a board 30 feet long, and 16 inches wide ? Ans. 40 feet. PROBLEM II. To find the contents of joists. RULE. Multiply the depth and width, taken in inches, by the length in feet; divide this product by 12, and the quotient is the contents inject. 3. How many feet are there in 3 joisis, which are 15 feet long, 5 inches wide, and 3 inches thick ? Ans. 56 feet. 4. How many feet in 20 joists, 10 feet long, 6 inches wide, and 2 inches thick ? Ans. 200 feet. PROBLEM III. To measure round timber. We have inserted below the rule usually adopted by survey- ors of lumber ; but it is a very unjust rule, if it is intended to give only 40 cubic feet of timber for a ton. For, if a stick of round timber be 40 feet long, and its circumference be 48 inches, it is considered by surveyors to contain one ton, or 40 feet ; whereas, it in reality contains, according to the following correct process, 50^ cubic feet. OPERATION. 48 x .31831=15.27888 ; 15.27888 -^ 2 = 7.63944 ; 7.63944 X 24= 183.34656 ; 183.34656 X 40 = 7333.8624 ; 7333.8624 -f- 144 = 50 T 9 ^-, that is, it contains as many cubic feet as a stick SOy 9 ^ feet long and 1 foot square. RULE. Multiply the length of the stick, taken in feet, by the square SECT. LXXIX.] PHILOSOPHICAL PROBLEMS. 331 of the girth, taken in inches ; divide this product by 144, and the quo- tient is the contents in cubic feet. NOTE. The girth is usually taken about of the distance from the larger to the smaller end. 5. How many cubic feet in a stick of timber, which is 30 feet long, and whose girth is 40 inches ? Ans. 20 feet. 6. If a stick of timber is 50 feet long, and its girth is 56 inches, what number of cubic feet does it contain ? Ans. GSyV feet. 7. What are the contents of a log 90 feet long, and whose circumference is 120 inches ? Ans. 562 feet. SECTION LXXIX. PHILOSOPHICAL PROBLEMS.* PROBLEM I. To find the time in which pendulums of different lengths would vibrate ; that which vibrates seconds being 39.2 inches. The time of the vibrations of pendulums are to each other as the square roots of their lengths ; or their lengths are as the squares of their times of vibrations. RULE. As the square of one second is to the square of the time in seconds in which a pendulum would vibrate, so is 39.2 inches to the length of the. required pendulum. EXAMPLES. 1. Required the length of a pendulum that vibrates once in 8 seconds. I 2 : 8 2 : : 39.2in. : 2508.8in. = 209 T ^ feet, Ans. 2. Required the length of a pendulum that shall vibrate 4 times a second. Ans. 2^y inches. 3. Required the length of a pendulum that shall vibrate once a minute. Ans. 3920 yards. 4. How often will a pendulum vibrate whose length is 100 feet ? Ans. once in 5.53+ seconds. PROBLEM II. To find the weight of any body, at any assignable distance above the earth's surface. * For demonstrations of the following problems, the student is referred to Enfield's Philosophy, or to the Cambridge Mathematics. 332 PHILOSOPHICAL PROBLEMS. [SECT. LXXIX. The gravity of any body above the earth's surface decreases, as the squares of its distance in semidiameters of the earth from its centre increases. RULE. As the square of the distance from the earth's centre is to the square of the earth's semidiameter , so is the weight of the body on the earth's surface to its weight at any assignable distance above the surface of the earth, and vice versa. 5. If a body weigh 900 pounds at the earth's surface, what would it weigh 2000 miles above its surface ? Ans. 4001bs. 6. Admitting the semidiameter of the earth to be 4000 miles, what would be the weight of a body 20,000 miles above its sur- face, that on its surface weighed 144 pounds? Ans. 41bs. 7. How far must a body be raised to lose half its weight ? Ans. 1656.85-+- miles. 8. If a man at the earth's surface could carry 150 pounds, how much would that burden weigh at die earth, which he could sustain at the distance of the moon, whose centre is 240,000 miles from the earth's centre ? Ans. 540,0001bs. 9. If a body at the surface of the earth weigh 900 pounds, but being carried to a certain height weighs only 400 pounds, what is that height ? Ans. 2000 miles. PROBLEM III. By having the height of a tide on the earth given, to find the height of one at the moon. RULE. As the cube, of the moon's diameter, multiplied by its density, is to the cube of the earth's diameter, multiplied by its density, so is the height of a tide on the earth to the Jieight of one at the moon. 10. The moon's diameter is 2180 miles, and its density 494 ; the earth's diameter is 7964 miles, and its density 400. If, then, by the attraction of the moon, a tide of 6 feet is raised at the earth, what will be the height of a tide raised by the attrac- tion of the earth at the moon ? Ans. 236.8-f- feet. NOTE. The above question is on the supposition that the moon has seas and oceans similar to those on the earth, but astronomers at the pres- ent time doubt their existence at this secondary planet. PROBLEM IV. To find the weight of a body at the sun and planets, having its weight given at the earth. If the diameters of two globes be equal, and their densities different, the weight of a body on their surfaces will be as their densities. SECT. LXXIX.] PHILOSOPHICAL PROBLEMS. 333 If their densities be equal, and their diameters different, the weight of a body will be as ^ of their circumference. If their diameters and densities be both different, the weight of a body will be as f of their semidiameters, multiplied by their densities. Therefore, having the weight of a body on the surface of the earth given, to find its weight at the surface of the sun and the several planets, we adopt the following RULE. As of the earth's semidiameter, multiplied by its density, is to $ of the sun's or planet's semidiameter, multiplied by its density, so is the weight of a body at the surface of tlie earth to the weight of a body at the surface of the sun or planet. 11. If the weight of a man at the surface of the earth be 170 pounds, what will be his weight at the surface of the sun, and the several planets, whose densities, &c., are as in the following table ? Sun, Jupiter, Saturn, Earth, Moon, Density. Diameter. Semidiameter. Semidiameter. 100 94.5 67 400 494 883246 89170 79042 7964 2180 441623 44585 39521 3982 1090 294415 29723 26347 2654 726 Ans. Weight at the sun 4714.6-f-lbs., at Jupiter 449.7-j-lbs., at Saturn 282.6-j-lbs., at the moon 57.4+lbs. PROBLEM V. To find how far a heavy body will fall in a given time, near the surface of the earth. Heavy bodies near the surface of the earth fall 16 feet in one second of time ; and the velocities they acquire in falling are as the squares of the times ; therefore, to find the distance any body will fall in a given time, we adopt the following RULE. As the square of I second is to the square of the time in seconds that the body is falling, so is 16 feet to the distance in feet that the body will fall in the given time. 12. How far will a leaden bullet fall in 8 seconds ? I 2 : 8 2 : : 16ft. : 1024ft. = Answer. 13. How far would a body fall in 1 minute ? Ans. 10 miles 1600 yards. 334 PHILOSOPHICAL PROBLEMS. [SECT. LXXIX. 14. How far would a body fall in 1 hour ? Ans. 39,272 miles 1280 yards. 15. How far would a body fall in 9 days ? Ans. 1,832,308,363 miles 1120 yards. PROBLEM VI. The velocity given, to find the space fallen through to ac- quire that velocity. RULE. Divide the velocity by 8, and the square of the quotient will be the distance fallen through to acquire that velocity. 16. The velocity of a cannon-ball is 660 feet per second. From what height must it fall to acquire that velocity ? Ans. 6806 feet. 17. At what distance must a body have fallen to acquire the velocity of 1000 feet per second ? Ans. 2 miles 5065 feet. PROBLEM VII. The velocity given per second, to find the time. RULE. Divide tJte velocity by 8, and a fourth part of tlte quotient will be the time in seconds. 18. How long must a body be falling to acquire a velocity of 200 feet per second ? Ans. 6 seconds. 19. How long must a body be falling to acquire a velocity of 320 feet per second ? AJIS. 10 seconds. PROBLEM VIII. The space through which a body has fallen given, to find the time it has been falling. RULE. Divide the square root oftJie space in feet fallen through by 4, and the quotient will be the time in seconds in which it was falling. 20. How long would a body be falling through the space of 40,000 feet ? Ans. 50 seconds. 21. How long would a ball be falling from the top of a tow- er, that was 400 feet high, to the earth ? Ans. 5 seconds. PROBLEM IX. The weight of a body and the space fallen through given, to find the force with which it will strike. RULE. Multiply the space fallen through by 64, then multiply the square root of this product by the weight, and the product is the momen- tum^ or force with which it will strike. SECT. LXXX.] MECHANICAL POWE 22. If the rammer for driving the pi IPS of Warren Bridge weighed 1000 pounds, and fell through a spa^^^J^qt, tyfthi what force did it strike the pile ? : ^^ : =^ :. Vl6 X 64 = 32 32 X 1000 = 32,0001bs. Answer. 23. Bunker Hill Monument is 220 feet in height; what would be the momentum of a stone, weighing 4 tons, falling from the top to the ground ? Ans. 255,7761bs. SECTION LXXX. MECHANICAL POWERS. THAT body which communicates motion to another is called the power. The body which receives motion from another is called the weight. The mechanical powers are six, the Lever, the Wheel and Axle, the Pulley, the Inclined Plane, the Screw, and the Wedge. THE LEVER. The lever is a bar, movable about a fixed point, called its fulcrum or prop. It is in theory considered as an inflexible line, without weight. It is of three kinds ; the first, when the ^>rop is between the weight and the power ; the second, when the weight is between the prop and the power ; the third, when the power is between the prop and the weight. A power and weight acting upon the arms of a lever will balance each other, when the distance of the point at which the power is applied to the lever from the prop is to the distance of the point at which the weight is applied as the weight is to the power. Therefore, to find what weight may be raised by a given power, we adopt the following RULE. As the distance between the body to be raised, or balanced, and the fulcrum or prop, is to the distance between the prop and the point where the power is applied, so is the power to the weight which it will balance. 336 MECHANICAL POWERS. [SECT. LXXX. 1. If a man weighing 170 pounds be resting upon a lever 10 feet long, what weight will he balance on the other end, the prop being one foot from the weight ? Ans. 15301bs. 2. If a weight of 1530 pounds were to be raised by a lever 10 feet long, and the prop fixed one foot from the weight, what power applied to the other end of the lever would balance it ? Ans. 1701bs. 3. If a weight of 1530 pounds be placed one foot from the prop, at what distance from the prop must a power of 170 pounds be applied to balance it ? Ans. 9 feet. 4. At what distance from a weight of 1530 pounds must a prop be placed, so that a power of 170 pounds, applied 9 feet from the prop, may balance it ? Ans. 1 foot. 5. Supposing the earth to contain 4,000,000,000,000,000,000,- 000 cubic feet, each foot weighing 100 pounds, and that the earth was suspended at one end of a lever, its centre being 6000 miles from the fulcrum or prop, and that a man at the other end of the lever was able to pull, or press with a force of 200 pounds ; what must be the distance between the man and the fulcrum, that he might be able to move the earth 1 Ans. 12,000,000,000,000,000,000,000,000 miles. 6. Supposing the man in the last question to be able to move his end of the lever 100 feet per second, how long would it take him to raise the earth one inch ? Ans. 52,813,479,690y. 17d. 14h. 57m. 46sec. THE WHEEL AND AXLE. The wheel and axle is a wheel turning round together with its axle ; the power is applied to the circumference of the wheel, and the weight to that of the axle by means of cords. An equilibrium is produced in the wheel and axle, when the weight is to the power as the di- ameter of the wheel to the diam- eter of its axle. To find, therefore, how large a power must be applied to the wheel to raise a given weight on the axle, we adopt the following SECT. LXXX.] MECHANICAL POWERS. 337 RULE. As the diameter of the wheel is to the diameter of the axle, so is the weight to be raised by the axle to the power that must be applied to the wheel. 7. If the diameter of the axle be 6 inches, and the diameter of the wheel 4 feet, what power must be applied to the wheel to raise 960 pounds at the axle ? Ans. 1201bs. 8. If the diameter of the axle be 6 inches, and the diameter of the wheel 4 feet, what power must be applied to the axle to raise 120 pounds at the wheel ? Ans. 9601bs. 9. If the diameter of the axle be 6 inches, and 120 pounds applied to the wheel raise 960 pounds at the axle, what is the diameter of the wheel ? Ans. 4 feet. 10. If the diameter of the wheel be 4 feet, and 120 pounds applied to the wheel raise 960 pounds at the axle, what is the diameter of the axle ? Ans. 6 inches. THE PULLEY. The pulley is a small wheel, movable about its axis by means of a cord, which passes over it. When the axis of a pulley is fixed, the pulley only changes the direction of the power ; if movable pulleys are used, an equilibrium is produced when the power is to the weight as one to the number of ropes applied to them. If each movable pulley has its own rope, each pulley will be double the power. To find the weight that may be raised by a given power. RULE. Multiply the power by the number of cords that support the weight, and the product is the weight. 11. What power must be applied to a rope, that passes over one movable pulley, to balance a weight of 400 pounds ? Ans. 2001bs. 12. What weight will be balanced by a power of 10 pounds, attached to a cord that passes over 3 movable pulleys ? Ans. 601bs. 29 338 MECHANICAL POWERS. [SECT. LXXX. 13. What weight will be balanced by a power of 144 pounds, attached to a cord that passes over 2 movable pulleys ? Ans. 576lbs. 14. If a cord, that passes over two movable pulleys, be at- tached to an axle 6 inches in diameter, whose wheel is 60 inches in diameter, what weight may be raised by the pulley, by ap- plying 144 pounds to the wheel ? Ans. 57601bs. THE INCLINED PLANE. An inclined plane is a plane which makes an acute angle with the horizon. The motion t)f a body descending an inclined plane is uni- formly accelerated. The force with which a body descends an inclined plane, by the force of attraction, is to that with which it would descend freely, as the elevation of the plane to its length ; or, as the size of its angle of inclination to radius. To find the power that will draw a weight up an inclined plane. RULE Multiply the weight by the perpendicular Iteight of the plane, and divide this product by t/te length. 15. An inclined plane is 50 feet in length, and 10 feet in perpendicular height; what power is sufficient to draw up a weight of 1000 pounds ? Ans. 2001bs. 16. What weight, applied to a cord passing over a single pul- ley at the elevated part of an inclined plane, will be able to sustain a weight of 1728 pounds, provided the plane was 600 feet long, and its perpendicular height 5 feet ? Ans. 14f Ibs. 17. A certain railroad, one mile in length, has a perpendic- ular elevation of 50ft. ; what power is sufficient to draw up this elevation a train of cars weighing 20,0001bs. ? Ans. 189^-. 18. An inclined plane is 300 feet in length, and 30 feet in perpendicular height ; what power is sufficient to draw up a weight of 2000 pounds ? Ans. 200lbs. 19. An inclined plane is 1000 feet in length, and 100 feet in perpendicular height ; what power is sufficient to draw up this elevation a weight of 5000 pounds ? Ans. SOOlbs. 20. What weight applied to and passing over a single pulley, at the elevated part of an inclined plane, will be able to sustain a weight of 70001bs., provided the plane is 300 feet long and its perpendicular height 30 feet ? Ans. 7001bs. SECT. LXIX.] MECHANICAL POWERS. 339 THE SCREW. The screw is a cylinder, which has either a prominent part or a hollow line passmg round it in a spiral form, so inserted in one of the opposite kind that it may be raised or de- pressed at pleasure, with the weight upon its upper, or suspended beneath its lower surface. In the screw the equilibrium will be produced, when the power is to the weight as the distance between the two contiguous threads, in a di- rection parallel to the axis of the screw, to the circumference of the circle described by the power in one revolution. To find the power that should be applied to raise a given weight. RULE. As the distance between the threads of the screiv is to the cir- cumference of the circle described by the power, so is the power to the weight to be raised. NOTE. One third of the power is lost in overcoming friction. 21. If the threads of a screw be 1 inch apart, and a power of 100 pounds be applied to the end of a lever 10 feet long, what force will be exerted at the end of the screw ? Ans. 75,398.20+lbs. 22. If the threads of a screw be an inch apart, what pow- er must be applied to the end of a lever 100 inches in length to raise 100,000 pounds ? Ans. 79.5774-^-lbs. 23. If the threads of a screw be an inch apart, and a pow- er of 79.5774-j- pounds be applied to the end of a lever 100 inches in length, what weight will be raised ? Ans. 100,0001bs. 24. If a power of 79.5774-)- pounds be applied to the end of a lever 100 inches long, and by this force a weight of 100,000 pounds be raised, what is the distance between the threads of the screw ? Ans. an inch. 25. If a power of 79.5774+ pounds be applied to the end of a lever, raising by this force a weight of 100,000 pounds, what must be the length of the lever, if the threads of the screw be an inch apart ? Ans. 100 inches. 340 SPECIFIC GRAVITY. [SECT. LXXXI. WEDGE. The wedge is composed of two inclined planes, whose bases are joined. When the resisting forces and the power which acts on the wedge are in equilibrio, the weight will be to the power as the height of the wedge to a line drawn from the middle of the base to one side, and parallel to the direction in which the re- sisting force acts on that side. To find the force of the wedge. RULE. As half tlie breadth or thickness of the head of the wedge is to &ne of its slanting sides, so is the power which acts against its head to the force produced at its side. 26. Suppose 100 pounds to be applied to the head of a wedge that is 2 inches broad, and whose slant is 20 inches long, what force would be affected on each side ? Ans. 20001bs. 27. If the slant side of a wedge be 12 inches long, and its head 1 inches broad, and a screw whose threads are of an inch asunder be applied to the head of this wedge, with a pow- er of 200 pounds at the end of tlie lever, 16 feet long, what would be the force exerted on the sides of the wedge ? Ans. 5147184.2-f-lbs. SECTION LXXXI. SPECIFIC GRAVITY.* To find the specific gravity of a body. RULE. Weigh the body both in water and out of water, and note the difference, which will be the weight lost in water; then, as the weight lost in water is to the whole weight, so is the specific gravity of water to the specific gravity of the body. But if the body whose specific gravity is required is lighter than water, affix to it another body ht-nrii r than water, so that the mass compounded of the two may sink together. Weigh the dense body and the compound mass separately, both in water and out of it ; then find how much each loses in water by subtracting its weight in water from its weight in air ; and subtract tlie less of these remainders from the greater ; then say, as the last remainder is to the * The specific gravity of a body is its weight compared with water ; the water being considered 1000. SECT. LXXXII.] STRENGTH OF MATERIALS. 341 weight of the body in air, so is the specific gravity of water to the spe- cific gravity of the body. NOTE. A cubic foot of water weighs 1000 ounces. 1. A stone weighed 10 pounds, but in water only 6 pounds. Required the specific gravity. Ans. 2608.6-J-. 2. Suppose a piece of elm weigh 15 pounds in air, and that a piece of copper, which weighs 18 pounds in air and 16 pounds in water, is affixed to it, and that the compound weighs 6 pounds in water. Required the specific gravity of the elm. Ans. 600. SECTION LXXXII. STRENGTH OF MATERIALS. THE force with which a solid body resists an effort to sepa- rate its particles or destroy their aggregation can only become known by experiment. There are four different ways in which the strength of a solid body may be exerted ; first, by resisting a longitudinal tension ; secondly by its resisting a force tending to break the body by a transverse strain ; thirdly, in resisting compression, or a force tending to crush the body ; and, fourthly, in resisting a force tending to wrench it asunder by torsion. We shall, however, only consider the strength of materials as affected by a trans- verse strain. When a body suffers a transverse strain, the mechanical ac- tion which takes place among the particles is of a complicated nature. The resistance of a beam to a transverse strain is in a compound ratio of the strength of the individual fibres, the area of the cross section, the distance of the centre of gravity of the cross section from the points round which the beam turns in breaking. The following are the facts and principles on which mechan- ics make their calculations. 1. A stick of oak one inch square and twelve inches long, when both ends are supported in a horizontal position, will sus- tain a weight of 600 pounds ; and a bar of iron of the same dimensions will sustain 2190 pounds. 2. The strength of similar beams varies inversely as their 29* 342 STRENGTH OF MATERIALS. [SECT. LXXXII. lengths ; that is, if a beam 10 feet long will support 1000 pounds, a similar beam 20 feet long would support only 500 pounds. 3. The strength of beams of the same length and depth is directly as their width ; that is, if there be two beams, each 20 feet long and 6 inches deep, and one of them is 6 inches wide and the other but 3 inches, the former will support twice the weight of the latter. 4. The strength of beams of the same length and width is as the squares of their depths ; that is, if there be two beams, each of which is 20 feet long and 4 inches wide, but one is 6 inches deep and the other is 3 inches deep, their strength is as the squares of -these numbers. Thus, 6x6 = 36; 3x3 = 9; that is, the strength of the former is to the latter as 36 to 9. It will, therefore, sustain four times the weight of the latter. Thus, 36 -r- 9 = 4. 5. To compare the strength of two beams of the same length, but of different breadth and depth, we multiply their widths by the squares of their depths, and their products show their com- parative strength. Thus, if we wish to ascertain how much stronger is a joist that is 2 inches wide and 8 inches deep, than one of the same length that is 4 inches square, we mul- tiply 2 by the square of 8, and 4 by the square of 4 ; thus, 2 X 8 X 8 = 128; 4 X 4 X 4 = 64 ; 128 H- 64 = 2. Thus, we see that although the quantity of material in one joist is the same as in the other, yet the former will sustain twice the weight of the latter. Hence " deep joists " are much stronger than square ones, which have the same area of a transverse section. 6. To compare the strength of two beams of different lengths, widths, and depths, we multiply their widths by the squares of their depths, and divide their products by their lengths, and their quotients will show their comparative strength. Therefore, if we wish to ascertain how much stronger is a beam that is 20 feet long, 8 inches wide, and 10 inches deep, than one 10 feet long, 6 inches wide, and 5 inches deep, we adopt the following formulas : 8X10X10 6X5X5__ ~20~ ~IO~~ The strength of the former, therefore, is to the latter as 40 to 15 ; that is, if the first beam would sustain a weight of 40cwt., the latter would sustain only 15cwt. 7. Having all the dimensions of one beam given, to find SECT. LXXXII.] STRENGTH OF MATERIALS. 343 another, part of whose dimensions are known, that will sustain the same weight. We multiply the width of the given beam by the square of its depth, and divide this product by the length, and the result we call the reserved quotient ; then, if we have the length and breadth of the required beam given to find the depth, we multiply the reserved quotient by the length of the required beam, and divide the product by its width, and the quotient is the square of the depth of the required beam. If the length and depth of the required beam were given to find the width, we multiply the reserved quotient by the length of the required beam, and divide this product by the square of the depth of the required beam, and the quotient is the breadth. But if the width and depth of the required beam were given to find the length, we multiply the width of the required beam by the square of the depth, and divide this product by the reserved quotient, and the result is the length of the required beam. 8. A triangular beam will sustain twice the weight with its edge up that it will with its edge down. Hence split-rails hav e twice the strength with the narrow part upward, which they have with the narrow part downward. 9. In making the above calculations, we have not noticed the weight of the beam itself, and in short distances it is of but little consequence ; but where a long beam is required, its weight is of importance in the calculation. 10. A beam supported at one end will sustain only one fourth part the weight which it would if supported at both ends. 11. The tendency to produce fracture in a beam by the ap- plication of a weight is greatest in the centre, and decreases towards the points of support ; and this ratio varies as .the square of half the length of the beam to the product of any two parts where the weight may be applied. Hence the tendency of a weight to break a bar 8 feet long, when applied to the centre, to that of the same weight, when applied 3 feet from one end, is as 4 X 4 = 16 to 3 X 5 = 15. QUESTIONS TO BE PERFORMED BY THE PRECEDING RULES. 1. If a stick of oak 1 inch square and 12 inches long, when both ends are supported in a horizontal position, will sustain a weight of 600 pounds, how many pounds would a similar stick sustain, that was 36 inches long ? Ans. 200 pounds. 2. If a beam 4 inches square and 12 feet long would support a weight of 1000 pounds, how many pounds would a similar beam support, that was 3 feet long ? Ans. 4000 pounds. 344 STRENGTH OF MATERIALS. [SECT. LXXXII. 3. If a beam 20ft. long, 4in. wide, and 6in. deep, will sustain a weight of 2000lbs., how many pounds would a similar beam sustain, that was Sin. wide ? Ans. 15001bs. 4. If a beam 10 feet long, 4 inches wide, and 3 inches deep, will sustain a weight of 1000 pounds, how many pounds would a similar beam support, that was 6 inches deep ? Ans. 4000 pounds. 5. If a beam 10 feet long, 2 inches wide, and 4 inches deep, will sustain a weight of 1000 pounds, how many pounds would a beam, that is 10 feet long, 4 inches wide, and 6 inches deep, sustain ? Ans. 4500 pounds. 6. If a beam 2 feet long, 2 inches wide, and 3 inches deep, will sustain a weight of 4000 pounds, what will a beam, that is 4 feet long, 3 inches wide^ and 6 inches deep, sustain ? Ans. 12000 pounds. 7. If a beam that is 10 feet long, 4 inches wide, and 6 inches deep, will sustain a weight of 4 tons, what weight will a beam sustain, that is 20 feet long, 8 inches wide, and 10 inches deep ? Ans. 1 1 tons. 8. If a beam 6 inches square and 8 feet long will support a weight of 2000 pounds, what weight will a beam 10 feet long and 10 inches square sustain ? Ans. 7407| pounds. 9. If a beam 15 feet long and 5 inches square will sustain a weight of 1200 pounds, required the length of a beam, that is 8 inches square, that will sustain a weight of 2000 pounds. Ans. 36 ff f feet. 10. If a beam 8 feet long and 7 inches square will sustain a weight of 3000 pounds, how many inches square must be the beam, that is 6 feet long, that will sustain 2000 pounds ? Ans. 5.5-f- inches. 11. If a bar of iron 10 feet long, 2 inches wide, and 3 inches deep, will sustain 10 tons, what must be the depth of a bar, that is 12 feet long and 3 inches wide, that will sustain 30 tons ? Ans. 4.64-f- inches. 12. If it require 1000 pounds to break a certain beam, that is 24 feet long, when placed in its centre, required the weight necessary to break it, when placed within 4 feet of one end of the beam? Ans. 13. If a beam 6 inches square and 10 feet long will sustain a weight of 2000 pounds from its centre, what weight would a beam of the same material sustain that is 10 inches square and 12 feet long, if the weight were suspended 2 feet from the centre ? Ans. SECT. Lxxxiii.j ASTRONOMICAL PROBLEMS. 345 SECTION LXXXIII. ASTRONOMICAL PROBLEMS. PROBLEM I. To find the dominical letter for any year in the present cen- tury, and also to find on what day of the week January will begin. RULE. To the given year add its fourth part, rejecting the frac- tions ; divide this sum by 7 ; if nothing remains, the dominical letter is A ; but if there be a remainder, subtract it from 8, and the residue will show the dominical letter, reckoning = A, 2 = B, 3= C, 4 = 1), 5 =E, 6 = F,7=G. These letters will also show on what day of the week January begins. For when A is the dominical letter, January begins on the Sabbath; when B is the dominical letter, January begins on Saturday ; C begins it on Friday ; D, on Thursday; E, on Wed- nesday; F, on Tuesday ; G, on Monday. NOTE. If it be required to find the dominical letter for the last centu- ry, proceed as above ; only, if there be a remainder after division, sub- tract it from 7, and the remainder shows the dominical letter, reckoning 1=A, 2 = B,3=C, 4 = D, 5=E, 6=F, = G. 1. Required the dominical letter for 1835. OPERATION. 4) 1835 8 4 := 4 z= D =i dominical letter. - As D is the dominical letter, January began 7)229* on Thursday, and the fourth day was the 327-4 Sabbath. 2. Required the dominical letter for 1836. OPERATION. 4) 1836 8 6 = 2 = B and C dominical letters. 459 7)2295 I* 1 l ea P years there are two dominical letters. 327 -6 ^ e ^ ast l etter > ^ * s f r J anuar 7 an d February, and B for the remainder of the year. As C is the dominical letter, January began on Friday, and the third day was the Sabbath. 3. Required the dominical letter for 1841 ? Ans. C. 4. Required the dominical letter for 1899. Ans. A. 5. Required the dominical letters for 1896. Ans. D and E. 6. What is the dominical letter for 1786 ? Ans. A. 7. What is the dominical letter for 1837 ? Ans. A. 346 ASTRONOMICAL PROBLEMS. [SECT. LXXXIII. PROBLEM II. To find on what day of the week any given day of the month will happen. RULE. Find by the last problem the dominical letter for the given year, and on what clay in January will be the first Sabbath ; and the corresponding day in the succeeding months will be as follows : Wed- nesday for February ; Wednesday for March ; Saturday for April; Monday for May ; Thursday for June ; Saturday for July ; Tuesday for August ; Friday for September ; Sabbath for October; Wednesday for November ; Friday for December. Having found the day of the week for any day in the month, any other day may be easily obtained, as may be seen in the following example. 8. Let it be required to ascertain on what day of the week was the 25th day of September, 1838. The dominical letter for 1838 is G ; therefore, the 7th of January was the Sabbath, and, by the above rule, the 7th of February was Wednesday, the 7th of March was Wednesday, the 7th of April was Saturday, the 7th of May was Monday, the 7th of June was Thursday, the 7th of July was Saturday, the 7th of August was Tuesday, the 7th of September was Friday. If the 7th was Friday, the 14th, the 21st, and the 28th were Fridays. And if the 21st was Friday, the 22d was Saturday, the 23d was the Sabbath, the 24th was Monday, and the 25th, the day required, was Tuesday. The following distich will assist the memory in finding the corresponding days of the month : At Dover Dwells George Brown Esquire, Good Christian Friend, And David Friar. It will be recollected, that the initial A is for the Sabbath, B for Monday, C for Tuesday, D for Wednesday, E for Thurs- day, F for Friday, and G for Saturday. NOTE. When it is leap year, the days of the week, after February, will be one day later than on other years. 9. Required the day of the week for the 4th of July, 1836. We find the dominical letters to be B and C, the 3d day of January therefore was on the Sabbath, and by the above rule the 3d day of July would have been on Saturday ; but as it was leap year, the 3d of July was one day later ; that is, it was on the Sabbath ; and, if the 3d was the Sabbath, the 4th of July was on Monday. 10. On what day of the week will be Dec. 8, 1849 ? Ans. Saturday. SECT. LXXXIV.J MISCELLANEOUS QUESTIONS. 347 11. On what day will happen July 4, 1857 ? Ans. Saturday. 12. On what day will March begin in the year 1890 ? Ans. Saturday. 13. On what day of the week was our Independence de- clared ? Ans. Thursday. 14. There will be a " Transit of Venus," December 8, 1874 ; on what day of the week will it happen ? Ans. Tuesday. 15. On what day of the week were you born ? SECTION LXXXIV. MISCELLANEOUS QUESTIONS. 1. What number must 7f be multiplied by, that the product may be 6f ? Ans. |f . 2. What number is that, which, being multiplied by half it- self, the product shall be 4 ? Ans. 3. 3. What fraction is that, which, being divided by llf , the quotient shall be 5 ? Ans. 4. What part of 7f is 9f- ? Ans. f f- f- or 7 5. Reduce to a simple fraction. Ans. T 5 T . 19f 6. Add f- of a ton to & of a cwt. Ans. 12cwt. Iqr. 8f Ib. 7. If the earth make one complete revolution in 23 hours 56 minutes 3 seconds, in what time does it move one degree ? Ans. 3m. 59" 20'". 8. Multiply f- of 9^ by | of |- of 8}. Ans. 449f . "B" T 9. Divide 12"^ of -f- of 100 by A of 7f TD " * Ans. 363j|ff . 10. What number is that to which if f of f- be added the sum will be 1 ? Ans. %%. 11. A certain gentleman, at the time of his marriage, agreed to give his wife of his estate, if, at the time of his death, he left only a daughter, and, if he left only a son, she should have of his property ; but, as it happened, he left a son and a daughter, by which the widow received in equity $ 2400 less than if there had been only a daughter. What would have been his wife's dowry if he had left only a son ? Ans. $ 2100. 12. A gentleman being asked what o'clock it was, said that 348 MISCELLANEOUS QUESTIONS. [SECT. LXXXIV. it was between 5 and 6 ; but, to be more particular, he said that the minute hand had passed as far beyond the 6 as the hour hand wanted of being to the 6 ; that is, that the hour and minute hands made equal acute angles with a line passing from the 12 through the 6. Required the time of day. Ans. 32m. 18 T 6 7 sec. past 5. 13. A, B, and C are to share $ 100,000 in the proportion of , , and -^, respectively ; but C's part being lost by his death, it is required to divide the whole sum, properly, between the other two. Ans. A's part is $ 57,142f , and B's $ 42,857f 14. A father devised T 7 of his estate to one of his sons, and T 7 F of the residue to the other, and the remainder to his wife. The difference of his sons 1 legacies was found to be 257<. 3s. 4d. What money did he leave for his widow ? Ans. 635<. Os. 10f$d. 15. In the walls of Balbec, in Turkey, the ancient Heliop- olis, there are three stones laid end to end, now in sight, that measure 61 yards in length; one of which is 63 feet long, 12 feet thick, and 12 feet broad ; what is its weight, supposing its specific gravity to be 3 times that of water ? Ans. 759 tons. 16. A burden of 2001bs., suspended on a pole 4ft. in length, the point of suspension being 6in. from the middle, is carried by two men, the ends of the pole resting on their shoulders ; how much of this load is borne by each man ? Ans. 1251bs. and 751bs. 17. The new court-house in Boston has 8 pillars of granite, each 25ft. 4in. in length, 4ft. 5in. in diameter at one end, and 3ft. 5in. in diameter at the other end. How many cubic feet do they contain, and what is their weight, allowing a cubic foot to weigh 3000 ounces ? Ans. 2455.03 cubic feet, 205.4 tons. 18. A father, dying, left his son a legacy, of which he spent in 8 months ; $- of the remainder lasted him 12 months longer; after which he had only $410 left. What did his father bequeathe him ? Ans. $ 956.66. 19. A butcher, wishing to buy some sheep, asked the owner how much he must give him for 20 ; on hearing his price, he said it was too much ; the owner replied, that he should have 10, provided he would give him a cent for each different choice of 10 in 20, to which he agreed. How much did he pay for the 10 sheep, according to the bargain ? Ans. $ 1847.56. 20. A merchant sold goods to a certain amount, on a com- mission of 4 per cent., and, having remitted the net proceeds to the owner, he received per cent, for prompt payment, which amounted to $ 15.60. What was the amount of his com- mission ? Ans. $ 260.00. SECT. LXXXIV.] MISCELLANEOUS QUESTIONS. 349 21. A, of Boston, remits to B, of New York, a bill of ex- change on London, the avails of which he wishes to be invested in goods on his account. B having disposed of the bill at 7^ per cent, advance, he received $ 9675.00, and having reserved for himself per cent, on the sale of the bill, and 2 per cent, for commission, what will remain for investment, and for how much was the bill drawn ? Ans. For investment, $9461.58^; the bill was ,2025. 22. Bunker Hill Monument is 30ft. square at its base, and 15ft. square at its top ; its height is 220ft. From the bottom to the top, through its centre, is a conical avenue 15ft. in diameter at the bottom and about lift, at the top. How many cubic feet are there in the monument ? Ans. 86,068.518-f-ft. 23. A hired a house for one year for $ 300 ; at the end of 4 months he takes in M as a partner, and at the end of 8 months he takes in P. At the end of the year what rent must each pay ? Ans. A pays $ 183 ; M pays $ 83 ; P pays $ 33^. 24. A merchant receives on commission three kinds of flour ; from A he receives 20 barrels, from P 25 barrels, and from C 40 barrels. He finds that A's flour is 10 per cent, better than B's, and that B's is 20 per cent, better than C's. He sells the whole at $ 6 per barrel. What in justice should each man receive ? Ans. A receives $ 139f|; B, $ 158f ; C, 8211Jf. 25. Bought 100 barrels of flour, at $ 5 per barrel, and im- mediately sold it on a credit of 6 months. The note which I received for pay, I got discounted at the Suffolk Bank, and, on examining my money, I found that I had gained 20 per cent, on my purchase. What did I receive per barrel for the flour ? Ans. $6.18!f|f. 26. Purchased for a cloak 5^ yards of broadcloth, that was 1|- yards wide ; to line this, I purchased flannel that was yard wide, but on being wet it shrunk 1 nail in width, and 1 yard in every 20 yards in length. How many yards of flannel was it necessary to buy ? Ans. 12|f f- yards. 27. How many bricks would it require to build the walls of a house 40 feet long, 30 feet wide, and 20 feet high, admitting the walls to be a foot thick, and that each brick was 8 inches long, 4 inches wide, and 2 inches thick ? Ans. 73,440 bricks. 28. How many feet of boards would it require to cover a house, that was 40 feet square, and whose height to the top of the plate was 20 feet, the roof projecting 1 foot over the plate, and coming to a point over the centre of the house, 15 feet above the garret floor ? Ans. 5367.7+ feet. 30 350 MISCELLANEOUS QUESTIONS. [SECT. LXXXIV. 29. Lent a friend $ 700, which he kept 20 months. Some years after I borrowed of him 300 ; how long should I keep it to balance the favor ? Ans. 46 months. 30. John Lee gave half of his estate to his wife, of the remainder to his oldest son, and ^ of the residue to his oldest daughter, and of what then remained to be distributed to his other children, who received $ 200 each ; required the number of his children, and the value of his estate. Ans. 12 children ; estate, $ 10,000. 31. A and B set out to travel round a certain island, which is 80 miles in circumference. A travels 5 miles a day, and B 7 miles a day. How far must B travel to overtake A ? Ans. 448 miles. 32. If 24.4 cubic inches of lead weigh 16 pounds, required the number of feet of lead pipe that can be made 'from 80 pounds of lead, the diameter of the pipe to be 1 inch, and the thickness of it of an inch. Ans. 123.24-f- feet. 33. How long a tube can be made from a cylinder of lead 8 inches long and 2 inches in diameter, and through the centre of which is a hole of an inch in diameter ; the tube or pipe to be of an inch in diameter, and f of an inch in thickness ? Ans. 39. 11+ feet. 34. Four men, A, B, C, and D, bought a stack of hay, con- taining 8 tons, for $ 100. A is to have 12 per cent, more of the hay than B, and B is to have 10 per cent, more than C, and C is to have 8 per cent, more than D. Each man is to pay in proportion to the quantity he receives. The stack is 20 feet high, and 12 feet wide at its base, it being an exact pyramid ; and it is agreed that A shall take his share first from the top of the stack, B is to take his share the next, and then C and D. How many feet of the perpendicular height of the stack shall each take, and what sum shall each pay ? Ans. A takes 13.22+ft., and pays $ 28.93ff f|- ; B takes 3.14-fft., and pays $25.83lff; C takes 1.94+ft. and pays $23.48fm ; D takes 1.68-fft., and pays S21.74|f. 35. Suppose the rails of a railroad to be 5 feet 4 inches apart at the place of the wheels bearing, and on a curve line of 1200 feet radius for the outer rail. Suppose the wheels of the car running on the rails to be firmly fixed to the axle, and that it is 5 feet from the outside of the flange of one wheel to the outside of the flange of the opposite wheel ; that from the outer side of one wheel to the outer side of its opposite wheel is 5 feet 8 inches ; that the diameter of each wheel at the outside SECT. LXXXIV.] MISCELLANEOUS QUESTIONS. 351 of the flange is 3 feet ; the face of the wheels to be so bevelled that at the outside of each wheel the diameter of the wheel is 2 feet 11.5 inches, and that the axle will always be in a posi- tion square across the two rails. In what part, between the two wheels, must the centre of gravity of the load be placed, so that the weight of the load shall bear equally on each rail ? OPERATION. 5ft. 4in. 5ft. rr 4in. ; and 4in. -J- 2 = 2in. = the place, be- yond the flange, on the face of the wheel, where the wheel bears on the rail. The perpendicular width of the face of the wheel is 4 inches, and in that distance the wheel diminishes in diam- eter 3ft. 2ft. 11. Sin. = .5in. Then, as 4in. : .5in. : : 2in. : .25in. ; and so the diameter of the wheel at the bearing is 3ft. .25in. = 2ft. 1 1.75in. As the radius of the curve of the outer rail is to that of the inner rail, so is the diameter of the outer wheel at the place of bearing to that of the inner wheel ; viz. as 1200ft. : 1200ft. 5ft. 4in. = 1194ft. 8in. : : 2ft. 11.75in. : 2ft. 11.59iin. Then the difference of the diameters of the two wheels at the places of bearing is 35.75in. 35.591 = .158in. Then, to see how much on the perpendicular face of the wheel this difference in diameter will vary the bearing ; as 3ft. 2ft. 11.5in. = .5in. : 4in. : : .158in. : 1.27iin. But this difference must be applied half to each wheel, viz. 1.27 lin. -5- 2=.635in., which brings the place of bearing .635in. nearer to the flange on the outer wheel, and .635in. further from the flange on the inner wheel. And the middle point between the two bearings will be .635in. from the centre of the axle towards the inner curve ; and in that place must the centre of gravity of the whole load be placed, to make the weight of the load bear equally on each rail ; viz. .635in. from the middle of the axle towards the inner rail, Answer. 36. The dimensions of a bushel measure are 18 inches wide, and 8 inches deep ; what should be the dimensions of a similar measure that would contain 8 bushels ? Ans. 37in. wide, 16in. deep. 37. What is the weight of a hollow spherical iron shell 5in. in diameter, the thickness of the metal being lin., and a cubic inch of iron weighing ^ of a pound ? Ans. 13.23871bs. 38. At a certain time between 2 and 3 o'clock, the minute hand was between 3 and 4. Within an hour after, the hour 352 MISCELLANEOUS QUESTIONS. [SECT. LXXXIT. hand and minute hand had exactly changed places with each other. What was the precise time when the hands were in the first position ? Ans. 2h. 15m. 56-^sec. 39. Required the contents of a cube, that will contain a globe 20 inches in diameter ; also of a cube, that may be inscribed in said globe. Ans. Larger cube 8000 cub. in. ; smaller, 1539 cub. in. 40. If in a pair of scales a body weigh 90 pounds in one scale, and only 40 pounds in the other, required the true weight, and the proportion of the lengths of the two arms of the balance- beam on each side of the point of suspension. Ans. the weight 601bs., and the proportions 3 to 2. 41. In turning a one-horse chaise within a ring of a certain diameter, it was observed that the outer wheel made two turns, while the inner wheel made but one ; the wheels were each 4 feet high ; and supposing them fixed at the distance of 5 feet asunder on the axletree, what was the circumference of the track described by the outer wheel ? Ans. 62.83-f- feet. 42. The ball on the top of St. Paul's church is 6 feet in di- ameter. What did the gilding of it cost, at 3d. per square inch? Ans. 237. 10s. Id. 43. There is a conical glass, 6 inches high, 5 inches wide at the top, and which is part filled with water. What must be the diameter of a ball, let fall into the water, that shall be im- mersed by it ? Ans. 2.445-|-in. 44. A certain lady, the mother of three daughters, had a farm of 500 acres, in a circular form, with her dwelling-house in the centre. Being desirous of having her daughters settled near her, she gave to them three equal parcels, as large as could be made in three equal circles, included within the periphery of her farm, one to each, with a dwelling-house in the centre of each ; that is, there were to be three equal circles, as large as could be made within a circle that contained 500 acres. How many acres did the farm of each daughter contain, how many acres did the mother retain, how far apart were the dwelling-houses of the daughters, and how far was the dwell- ing-house of each daughter from that of the mother ? Ans. Each daughter's farm contained 107 acres 2 roods 31.22+ rods. The mother retained 176 acres 3 roods 26.34-f- rods. The distance from one daughter's house to the other was 148.119817-f- rods. The mother's dwelling-house was distant from her daughters' 85.51-|- rods. SECT. LXXXIV.] MISCELLANEOUS QUESTIONS. 353 34 1 45. Required the cube of - . Ans. ? 8 T . 5*2f 78 46. Required the cube root of ^-^ 3 - Ans. . fifi 2 ftQ 5 47. Multiply the cube root of * T by the square of ~. C5/2 Ans. Jfr. 49. Three carpenters, J. Smith, J. Carleton, and John Jones, agree with T. Jenkins to build him a house and find the mate- rials for $ 1000, of which $ 600 were paid in advance, and the remainder when the work was finished. Carleton and Jones took $ 50 each of the first payment. When the work was completed, it appeared by Smith's account, who received the money and paid the bills, for which he was allowed a com- pensation of $ 10, that he had paid $ 648.95, exclusive of the payments to Carleton and Jones, and that he had labored 63 days. Carleton worked 51 days, and he was allowed $ 20 for the use of his shop, &c. Jones worked 60 days, and his bill for boarding the men they hired was $ 68.75. Smith, on set- tling with Jenkins and allowing him $23.15 charged to Carle- ton, and $ 17.48 charged to Jones, receives the balance in cash, and on exhibiting his statement of the business to Carleton and Jones, he pays to each the balance due. How much did they make per day, and how was the last payment disposed of? Ans. They received $ 1.45 per day, and Carteton received $ 20.80, Jones received $ 88.27, and Smith received 8 250.30. 50. A and B engaged to reap a field for 90 shillings ; and as A could reap it in 9 days, they promised to complete it in 5 days. They found, however, that they were obliged to call in C, an inferior workman, to assist them for the last two days, in consequence of which B received 3s. 9d. less than he other- wise would have done. In what time could B and C each reap the field ? Ans. B could reap it in 15 days, and C in 18 days. 51. Samuel Jenkins and Jarnes Betton, who have each an apprentice, engage to build a small house for 8 630. By agreement between them, Jenkins's apprentice is to be allowed $'0.62 per day, and Betton's $ 1.00. When the work was finished, it appeared that Jenkins had worked 120 days, and his apprentice 100. Betton worked 96 days, and his apprentice 30* 354 MISCELLANEOUS QUESTIONS. [SECT. LXXXIV. 135 days. While doing the work, they received each $ 210. What is each man's share of the remaining payment ? Ans. Due to Jenkins $ 92.50 ; to Betton $ 117.50. 52. A merchant tailor bought 40 yards of broadcloth, 2 yards wide ; but on sponging it, it shrunk in length upon every 4 yards half a quarter, and in width, one nail and a half upon every 1 yards. To line this cloth, he bought flannel 5 quar- ters wide, which, being wet, shrunk the whole width on every 20 yards in length, and in width it shrunk half a nail. Required the number of yards of flannel used in lining the cloth. Ans. 71 T 7 7j yards. 53. What is the square of .1 ? 54. If a stick of timber, 6 inches square and 10 feet long, will support from its centre 1000 pounds, how many pounds would a similar stick, that is 10 inches square and 20 feet long, support, if the weight were suspended 4 feet from the centre of the stick ? 55. A certain gentleman has an elegant lamp, to which are appended 72 brilliants, each of which had occupied a particular place. His daughter, one day, in preparing the lamp, dis- placed the brilliants, and in replacing them, found that she had put some of them in the wrong situation, and was at a loss to know how to correct the mistake. At length she told her father that, after dinner, she would begin and place the bril- liants in all the situations they would admit of, and then she would be sure of finding the correct way of adjusting them, and that she would not take her tea until she had effected it. Now, supposing she were to place them in all the various ways they would admit, how long would she be obliged to wait for her tea, provided she could make one change each minute ? Ans. 612344583768860868615240703852746727407780917 8469732898382301496397838498722168927420416000000000 0000000 minutes. In order that the pupil may have some general idea of the time denoted by the answer in figures to the above question, let him suppose a globe composed of fine sand, whose circumfer- ence shall be equal to the orbit of Herschel,each cubic inch of sand containing one million of particles ; and then suppose the lady at the end of every million of years' labor to remove one of these particles, and she will have removed the whole globe, particle by particle, millions of years before she can take her tea. APPENDIX.] WEIGHTS AND MEASURES. 355 APPENDIX. WEIGHTS AND MEASURES, WITH AN HISTORICAL ACCOUNT OF STANDARDS. IN commerce and in science, all bodies are estimated by number, weight, or measure. When reckoned by number, they are referred to a unit as a standard of comparison, and when estimated by weight or measure, there is always a reference to some certain fixed quantity, as a pound, a gallon, a mile, to which the quantity measured or weighed bears a specified and definite proportion. Weights are used to ascertain the gravity of bodies, and measures to determine their magnitude, or the space which they occupy. Standards of weights or measures are certain quantities of gravity or extension, which are fixed upon as those with which all other quantities of objects reckoned by weight or measure are to be compared ; and such standards have always been found necessary, and have existed in every age and nation. It has, however, been only in a highly civilized state of society that they have been such as to secure an accurate and equitable result in the transaction of business ; and few things are more indicative of a cultivated age and people, than the exactness with which science provides and adjusts the standards of com- parison, required by the sale and interchange of commodities. In the early stages of society, the ordinary standards of weight and measure were some simple objects or ideas, with which all in the community were supposed to be familiar. Thus, all measures of length were sometimes reckoned by comparing them with the human foot ; or, for the sake of greater definite- ness, as all human feet were not of the same length, they were referred to the king's foot as a standard. In some cases the length of the arm was used, and in other instances the length of a grain or corn of wheat or barley. In land measure, an acre was what could be ploughed by a yoke of oxen in a day, traces of which notion we have in the Hebrew and Latin words used to denote an acre, which properly signify a yoke or pair, that is, a yoke or pair of oxen. In early ages, also, weights as well as distances were meas- 356 WEIGHTS AND MEASURES. [APPENDIX. ured by grains of corn,* arid hence, in England and some other European countries, the lowest denomination of weight is still a grain. Originally, 32 of these grains were reckoned to a penny- weight, but in later times the number was fixed at 24. A scruple meant originally a small stone, which was regard- ed as equal to 20 grains, and a dram (Greek drachma) was literally what one could hold in his hand, the word being de- rived from a verb signifying to grasp with the hand. With standards thus variable and uncertain, it is evident that no people could advance far in commerce, and as facilities for commercial intercourse opened, they must adopt some more satisfactory methods of ascertaining the quantity and value of what they bought and sold. Some fixed and permanent stand- ards of comparison were needed, to which all might refer, and in which all should have confidence. Accordingly, kings and legislators early gave attention to this subject, and endeavoured, not only to provide such standards, but to preserve them from alteration. In Rome, the model weights and measures were kept for safety in the temple of Jupiter ; and among the Jews, they were preserved first in the Tabernacle, and afterwards in the Temple, and their custody committed to the sacerdotal fam- ily. In England, the standard yard, to which, as we shall see, all the legal measures and even weights of the kingdom are ul- timately referred, has for ages been kept with the greatest care by the government. This yard,t as history informs us, was in- troduced or revived as a standard of measure by Henry I., who lived at the beginning of the 12th century, and who ordered that the ulna or ancient ell, answering to the modern yard, should be of the exact length of his own arm, and all other measures founded upon it. This is the historical origin of the present English standard of measure, and it is said to have been preserved to the present time without any sensible variation! In 1742, the Royal Society caused a yard to be made from a very careful comparison of the standard ells or yards of the reigns of Henry VII. and Elizabeth, which had been kept at the Exchequer. In 1758, a committee of the House of Com- mons recommended that a rod which had been made by their order from that of the Royal Society, and marked " Standard Yard, 1758," should be declared the legal standard of all meas- ures of length. This rod consisted of a solid brass bar a little * Corn in the English sense, meaning wheat, barley, &c., and not our Indian corn or maize. The word yard meant originally a rod or shoot, and cannot be sup- posed to have had any very definite length. APPENDIX.] WEIGHTS AND MEASURES. 357 more than an inch square, and 39.06 inches long. At about an inch and a half from each end, there was inserted a gold pin or stud, and in these pins, at the distance of 36 inches from each other, are two points, which are intended to designate the exact length of the yard. The subject was further considered by another committee, and a second rod made by the same man, and as an exact copy of the above f which is known as the " Standard Yard of 1760," and which was declared by Parlia- ment in the " Act of Uniformity," in 1824, which took effect January 1, 1826, to be the legal standard of measure for Great Britain. As the distance between the two fixed points will vary with the temperature of the rod, the Act provides that it shall be at the temperature of 62 Fahrenheit. This standard yard, which was to be denominated the " Im- perial Yard," was intended to afford a fixed standard of meas- ure, which could only be lost with the destruction or mutilation of the standard rod. But this, it was foreseen, might easily occur, as in fact it did happen in 1834, at the burning of the two Houses of Parliament, when the imperial model shared the fate of the other valuable relics which were consumed in that ancient pile ; and to provide against such an accident, it was declared that the imperial or standard yard, as compared with a pendulum vibrating seconds in the latitude of London, should bear the proportion of 36 to 39.1393 inches. To procure fur- ther accuracy, this pendulum was to move in a vacuum, at the level of the sea, and at the temperature of 62 Fahrenheit. And thus was fixed an absolute and invariable standard of linear measure, by which, according to the Act above named, all measures of extension are determined, whether the same be lineal, superficial, or solid. A third part of this standard yard is a legal foot, a twelfth part of the foot a legal inch, 5 such yards a pole, 220 a furlong, and 1760 a mile. A legal rood of land contains 1210 square yards, and an acre 4840 square yards, or 160 square poles. By the same Act, all measures of capacity are determined by the imperial gallon, which contains 277.274 cubic inches. Two such gallons make a peck, one fourth of a gallon a quart, &c. As this standard gallon is determined by cubic inches, we see that it is ultimately referred to the standard yard as the basis on which it is founded. The Act provides, however, that it may also be determined by weight, in which case the measure to contain a gallon must be of a capacity to hold 10 pounds, avoirdupois weight, of distilled water, weighed in air, 358 WEIGHTS AND MEASURES. [APPENDIX. at the temperature of 62 Fahrenheit, the barometer being at 30 inches. For a standard of weights, the law of England makes the pound Troy* to contain 5760 grains, one cubic inch of dis- tilled water weighed as above, weighing 252.458 such grains, and the pound Avoirdupois to contain 7000 such grains. The Act of Parliament which brought the standards of Eng- lish weights and measures to their present state of comparative perfection was passed in 1835. This Act prohibits the use of the well-known Winchester bushel, and also of heaped measure. The subject of weights and measures is still under consideration by scientific men in England, and further improvements have 1 been proposed, and will probably be introduced. It has even been suggested by commissioners appointed by government, that the system of coinage be changed, and one having decimal pro- portions adopted. The system of weights and measures established by law in the United States is essentially the same as the English ; but there is not, even at this day, that uniformity of practice in this country, which the interests of extensive trade require. At the organization of the Federal Government, authority was given to Congress to regulate this important matter, but no laws have as yet been enacted by that body to secure a uniform system through all the States. By an order of Congress, in June, 1836, a set of standard weights and measures was prepared for the use of each custom-house, and for each State, and these consti- tute the legal standards of the nation, so far as any national standard can be said to exist. The standards thus provided were similar to those used in England prior to the Act of 1824, and they lack that accuracy and scientific character which are exceedingly desirable in a country so extensive and so commer- cial as the United States. Many of the States of the Union have attempted to reduce their weights and measures to a uni- form system, and not a little legislation has been directed to this end ; but generally with very little effect. And until Congress shall take up the matter in good earnest, and with the aid of sci- entific men, but little can be done to secure a result so important. h h tT n f ? T^t Tr 7 and Avoirdupois has been variously given but the most probable explanation is this : that the former is de- nved from avoirs (avend), the ancient name for goods or chattels, and poids (weight.) The word Troy has probably no reference to a toin in th? m ' I" T S former ^ u PPfd, hut, as applied to weight, originated in the monkish name of Troy Novant, which was given to London, and founded on an ancient legend ; so that Troy weight is properly London UNIVERSITY OF CALIFORNIA LIBRARY