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 University of California Berkeley 
 
 THE THEODORE P. HILL COLLECTION 
 
 of 
 EARLY AMERICAN MATHEMATICS BOOKS 
 
DAY AND THOMSON'S SERIES. ' 
 
 II I G n E E 
 
 ARITHMETIC; 
 
 OB THB 
 
 SCIENCE AND APPLICATION OF NUMBERS; 
 
 COMBINING THB 
 
 ANALYTIC AXD SYNTHETIC MODES OF IXSTPOJCTION. 
 
 DESIGNED FOB 
 
 ADVANCED CLASSES IN SCHOOLS AND ACADEMIES. 
 
 BY JAMES B. THOMSON, LL.D., 
 
 OB or MENTAL ARITHMETIC; EXERCISES m ARITHMETICAL ANALTBIS, 
 PRACTICAL ABITIIMKTIO; EDITOR OF DAY'fl SCHOOL ALGEJSRA; 
 LBGCNDRB'B OEOMETBT, ETC. 
 
 ONE HUNDRED AND TWENTIETH EDITION. 
 
 NEW YORK: 
 
 IYISON, PHINNEY & CO, 48 & 50 WALKER STREET. 
 
 CHICAGO: S. C. GRIGGS & CO., 39 & 41 LAKE ST. 
 BOSTON: BROWN, TAQGARD A CHASE. PHILADELPHIA: SOWEK, BARNES * oo n 
 
 AND J. B. LIPPINCOTT & CO. CINCINNATI; MOORR, WIL8TACII, KEYS A CO. 
 
 SAVANNAH: j. M. COOPER & co. BT. Loms: KEITH & WOODS. NEW 
 ORLEANS: BLOOAIFIEXJ), BTEEL & co. DETROIT :>. RAYMOND A co. 
 
 1862. 
 
DAY AND THOMSON'S MATHEMATICAL SERIES. 
 
 FOR SCHOOLS AND ACADEMIES. 
 
 I. MENTAL ARITHMETIC; or, First Lessons in Numbers? for Begin. 
 ers. This work commences with the simplest, combinations of numbers, and 
 gradually advances to more difficult combinations, as the mind of the learnci 
 expands and is prepared to comprehend them. 
 
 II. PRACTICAL ARITHMETIC; Uniting the Inductive with the 
 Synthetic mode of Instruction; also illustrating the principles ,of CANCEI.A- 
 Tiox. The design of this work is to make the pupil thoroughly acquainted 
 with the reason of every operation which he is required to perform. It 
 abounds in examples, and is eminently practical- 
 
 III. KEY TO PRACTICAL ARITHMETIC ; Containing the answers, 
 with numerous suggestions, &c. 
 
 IV. HIGHER ARITHMETIC; or, the Science and Application of Num- 
 bers; For advanced Classes. This work is complete in itself, commencing 
 with the fundamental rules, and extending to the highest department of the 
 science. 
 
 V. KEY TO HIGHER ARITHMETIC; Containing all the answers, 
 with many suggestions, and the solution of the more difficult questions. 
 
 VI. THOMSON'S DAY'S ALGEBRA; This work is designed to be a 
 Mcid and easy transition from the study of Arithmetic to the higher branches 
 of Mathematics. The number of examples is much increased ; and the work 
 is every way adapted to the improved methods of instruction in Schools and 
 Academies. 
 
 VII. KEY TO THOMSON'S DAY'S ALGEBRA; Containing the 
 answers, the solution of the more difficult problems, &c. 
 
 VIII. THOMSON'S LEGENDRE'S GEOMETRY ; With practical 
 notes and illustrations. This work has received the approbation of many 
 of the most eminent Teachers and Practical Educators. 
 
 IX. PLANE TRIGONOMETRY, AND THE MENSURATION OF 
 HEIGHTS AND DISTANCES; with a summary view of the Mature, and 
 Use of Logarithms ; Adapted to the method of instruction in Schools and 
 Academies 
 
 X. ELEMENTS OF SURVEYING ; Adapted both to the wants of th 
 (earner and the practical Surveyor. 
 
 Entered according to Act of Congress, in the year 1847, 
 
 BY JAMES B THOMSON, 
 In the Clerk's Office of the Northern District of New York. 
 
PREFACE. 
 
 THE Higher Arithmetic which is now presented to the public, is tho 
 third and last of a series of Arithmetics adapted to the wants of different 
 classes of pupils in Schools and Academies. The title of each explains 
 the character of the work. The series is constructed upon the principle, 
 fhat " there is a place for everything, and everything should be in its 
 proper place." Each work forms an entire treatise in itself ; the examples 
 in each are all different from those in the others, so that pupils who study 
 the series, will not be obliged to purchase the same matter twice, nor to 
 solve the same problems over again. 
 
 The Mental Arithmetic, is designed for children from six to eight years 
 of age. It is divided into progressive lessons of convenient length, begin- 
 ning with the simplest combinations of numbers, and advancing by grad- 
 ual steps, to more difficult operations, as the mind of the learner expands 
 and is prepared to comprehend them. 
 
 The Practical Arithmetic embraces all the subjects requisite for a 
 thorough business education. The principles and rules are carefully 
 analyzed and demonstrated ; the examples for practice are numerous, and 
 the observations and notes contain much information pertaining to busi- 
 ness matters, not found in other works of the kind. This is the FIRST 
 SCHOOL BOOK in which the Standard Units of Weights and Measures 
 adopted by the Government in 1834, were published. 
 
 The Higher Arithmetic is designed to give a full development of the 
 philosophy of Arithmetic, and its various applications to commercial pur- 
 poses. Its plan is the following : 
 
 1. The work is complete in itself. It commences with notation, and 
 illustrating the different properties of numbers, the principles of Cancela- 
 tion, and various other methods of contraction, extends to the higher 
 operations in mercantile affairs, and the more abstruse departments of 
 the science.- 
 
 %. Great pains have been taken to render the definitions and rules clear, 
 concise, exact, comprehensive. 
 
 3. It has been a cardinal point never to anticipate a principle ; and never 
 to use one principle in the explanation of another, until it has itself been 
 explained or demonstrated. 
 
 4. Nothing is taken for granted which requires proof. Every principle 
 therefore has been retnvestigated, and carefully analyzed. 
 
VI PREFACE. 
 
 5. The principles are at ranged consecutively and the dependence of 
 each on those that precede it, is pointed out by references. Treated in 
 this manner, the science of Arithmetic presents a series of principles and 
 propositions alike harmonious and logical ; and the study of it cannot fail 
 to exert the happiest influence in developing and strengthening the reason- 
 ing powers of the learner. 
 
 6. The rules are demonstrated with care, and the reasons of every oper 
 &tion fully illustrated. 
 
 7. The examples are copious and diversified; calling every principle 
 into exercise, and making its application thoroughly understood. 
 
 8. In the arrangement of subjects, the natural order of the science ha* 
 been carefully followed. Common Fractions have therefore been placed 
 immediately after Division, for two reasons. First, they arise from divi- 
 sion, and a connexion so intimate should not be severed without cause. 
 Second, in Reduction and the Compound Rules, it is often necessary to 
 multiply and divide by fractions, to add and subtract them, also to carry 
 for them, unless perchance the examples are constructed for the occasion 
 and with special reference to avoiding these difficulties. 
 
 For the same reason Federal Money, which is based upon the decimal 
 notation, is placed after Decimal Fractions ; Interest, Commission, &c., 
 after Percentage. To require a pupil to understand a rule before he is 
 acquainted with the principles upon which it is based, is compelling him 
 to raise a superstructure, before he is permitted to lay a foundation. 
 
 9. In preparing the Tables of Weights and Measures, no effort has 
 been spared to ascertain those in present use in our country ; and reject- 
 ing such as are obsolete, we havo introduced the Standard Units adopted 
 by the Government, together with the methods of determining and apply- 
 ing those standards. 
 
 10. Great labor has also been expended in preparing full and accurate 
 Tables of Foreign Weights and Measures, and Moneys of Account, and 
 in comparing them with those of the United States. 
 
 Such is a brief outline of the present work. In a word, it is designed 
 to be an auxiliary to the teacher, a lucid and comprehensive text-book for 
 the pupil, and an acceptable acquisition to the counting-room. It contains 
 many illustrations and principles not found in other works before the 
 public, and much is believed to be gained in the method of reasoning 
 and analysis. No labor has been spared to render it worthy of the 
 marked favor with which the former productions of the author have 
 been received. 
 
 J. B. THOMSON- 
 
 NewY<rrk, August, 1847. 
 
SUGGESTIONS 
 
 ON THE 
 
 MODE OF TEACHING ARITHMETIC. 
 
 I. QUALIFICATIONS/ The chief qualifications requisite in teaching Arith- 
 metic, as well as other branches, are the following : 
 
 1. A thorough knowledge of the subject. 
 
 2. A love for the employment. 
 
 3. An aptitude to teach. These are indispensable to success. 
 
 II. CLASSIFICATION. Arithmetic, like reading, grammar, &c., should be 
 taught in classes. 
 
 1. This method saves much time, and thus enables the teacher to devote 
 more attention to oral illustrations. 
 
 2. The action of mind upon mind, is a powerful stimulant to exertion, and 
 iannot fail to create a zest for the study. 
 
 3. The mode of analyzing and reasoning of one scholar, will often suggest 
 new ideas to others in the class. 
 
 4. In the classification, those should be put together who possess as nearly 
 equal capacities and attainments as possible. If any of the class learn quicker 
 than others, they should be allowed to take up an extra study, or be furnished 
 with additional examples to solve, so that the whole class may advance together. 
 
 5. The number in a class, if practicable, should not be less than six, nor 
 over twelve or fifteen. If the number is less, the recitation is apt to be defi- 
 cient in animation ; if greater, the turn to recite does not come round suffi- 
 ciently often to keep up the interest. 
 
 III. APPARATUS. The Black-board and Numerical Frame areas indispen- 
 sable to the teacher, as tables and cutlery are to the house-keeper. Not a reci- 
 tation passes without use for the black-board. If a principle is to be demon- 
 strated or an operation explained, it should be done upon the black-board, so 
 that all may see and understand it at once. 
 
 To illustrate the increase of numbers, the process of adding, subtracting, 
 multiplying, dividing, &c., to young scholars, the Numerical Frame furnishes 
 one of the most simple and convenient methods ever invented. 
 
 Every one who ciphers will of course have a slate. Indeed, it is desirable 
 that every scholar in school, even to the very youngest, should be furnished 
 with a slate, so that when their lessons are learned each one may busy himself 
 In writing and drawing various familiar objects. Idleness in school is the parent 
 tfriischirf, and employment is the best antidote against disobedience. 
 
 Geometrical diagrams and solids are also highly useful in illustrating many 
 points in arithmetic, and no school should be without them. 
 
 IV. RECITATIONS. The first object in a recitation, is to secure the attention 
 of the. class. This is done chiefly by throwing life <ind variety into Ihe eier* 
 citse. Children loaf-he dullness, while animation am variety are iheii delight,. 
 
Vlll SUGGESTIONS. 
 
 2. Every example should be analyzed ; the " why and the wherefore" of 
 every step in the solution should be required, till each member of the class be- 
 comes perfectly familiar with the process of reasoning and analysis. 
 
 3. To ascertain whether each pupil has the right answer, it is an excellent 
 method to name a question, then call upon some one to give the answer, and 
 before deciding whether it is right or wrong, ask how many in the class agree 
 with it. The answer they give by raising their hand, will show at once how 
 many are right. The explanation of the process may now be made. 
 
 V. OBJECTS OF THE STUDY. When properly studied, two important ends are 
 attained. 1st. Discipline of mind, and the development of the reasoning powers. 
 2d. Facility and accuracy in the application of numbers to business calculations. 
 
 VI. THOROUGHNESS. The motto of every teacher should be tJwroughness. 
 Without it, the great ends of the study of Arithmetic are defeated. 
 
 1. In securing this object, much advantage is derived from frequent reviews. 
 
 2. Every operation should be proved. The intellectual discipline and habits 
 of accuracy thus secured, will richly reward the student for his time and toil. 
 
 3. Not a recitation should pass without practical exercises upon the black- 
 board or slates, besides the lesson assigned. 
 
 4. After the class have solved the examples under a rule, each one should 
 be required to give an accurate account of its principles with the reason for each 
 step, either in his own language or that of the author. 
 
 5. Mental Exercises in arithmetic are exceeding Ly useful in making ready 
 and accurate arithmeticians ; hence, the practice of connecting mental with. 
 written exercises, throughout the whole course, is strongly recommended. 
 
 VII. SELF-RELIANCE. The habit of self-reliance in study, is confessedly in- 
 valuable. Its power is proverbial; I had almost said, omnipotent. " Where 
 there is a will, there is a way." 
 
 1. To acquire this habit, the pupil, like a child learning to walk, must be 
 taught to depend upon himself. Hence, 
 
 2. When assistance is required, it should be given indirectly ; not by taking 
 the slate and solving the example for him, but by explaining the 'meaning of 
 the question, or illustrating the principle on which the operation depends, by 
 supposing a more familiar case. Thus the pupil will be able to solve the 
 question himself, and his eye will sparkle with the consciousness of victory. 
 
 3. The pupil should be encouraged to study out different solutions, and to 
 adopt the most concise and elegant. 
 
 4. Finally, he should learn to perform examples independent of the answer. 
 Without, this attainment the pupil receives but little or no discipline from the 
 study, and acquires no confidence in his own abilities. W r hat though he cornea 
 to the recitation with an occasional wrong answer; it were better to solve me 
 question unde/'slaiidingly and alone, than to copy a score of answers from the 
 book. What would the study of mental arithmetic be worth, if the pupil had 
 the answers before him 1 What is a young man good for in the counting-room, 
 who cannot perform arithmetical operations without looking to the 
 
 Every on 3 pronounces him unjil to be tiusted with bus (lies? calculations. 
 
CONTENTS. 
 
 FAOt 
 
 INTRID JCTON, a brief survey of the science of Mathematics, 13 
 
 SECTION I. 
 NOTJ noN, ........20 
 
 Numeration, - - - 24 
 
 Formation of different systems of Notation, ... i*J 
 
 SECTION II. 
 
 GENERAL RULE for Addition, . . . . 3;} 
 
 Diflenent methods of proving Addition, - - - 3-1 
 
 Counting-room Exercises, -... 37 
 
 Adding nuiul>crs expressed by the Roman Nutation, - - 40 
 
 SECTION 111. 
 
 GENERAL RCLE for Subtraction, - - - - - 41 
 
 Different methods of proving Subtraction, - - . - 4i 
 
 Subtracting numbers expressed liy the Roman Notation, 47 
 
 SECTION IV. 
 
 GENERAL RULE for Multiplication, - - - - -53 
 
 Different methods of proving Multiplication, ... 54 
 
 Contractions in Multiplication, six methods, - - - 5(3 
 
 SECTION V. 
 
 GENERAL RULE for Division, - - . - - 71 
 
 Different methods of Proving Division, - - - - 74 
 
 Contractions in Division, nine methods, - - - 74 
 
 General principles in Division, - - - - HI 
 
 CA.NCEI.ATIHN, .... 8:5 
 
 Applications of the fundamental rules, - - b5 
 
 SECTION VI. 
 
 PROPERTIES OF NUMBERS, ... 80 
 
 Method of finding the prime numbers in any series, - - J>3 
 
 Table of prime numbers from I to 3-4 1 3. - - { 1 
 
 Numbers changed from the decimal to other scales of Notation, - !<.") 
 
 Analysis of Composite Numbers, ..... J)7 
 Greatest Common Divisor, two methods, .... ](}(\ 
 
 Least Common Multiple, three methods, - - - 102 
 
 SECTION VII. 
 
 FRACTIONS, - ... 10* 
 
 General principles pertaining to Fractions, - - < K>3 
 
 Reduction of Fractions, - - - - - .11 
 
 Addition of Fractions, - - - " " ' '** 
 Subtraction of Fractions, ... |-J-J 
 
 Multiplication of Fractions, common method, ... lx.l 
 
 Multiplication of Fractions by Cancelation, - - 130 
 
 Uonlractions in Multiplication of Fractions, four methods, - 131 132 
 
 1* 
 
X CONTENTS. 
 
 PAOB 
 
 Division af Fractions, common method, - - - - l.'J3 
 
 Division of Fractions by Cancelation, - - - 136 
 
 Contractions in Division of Fractions, three methods, - - l^H 
 
 Complex Fractions reduced to Simple ones, ... 139 
 
 SECTION VIII. 
 
 COMPOUND NUMBERS, ..... 144 
 
 Tables of Compound Numbers, ..... 144 
 
 The Standard for Gold and Silver Coin of the United States, - 145 
 
 The Standard UNIT of Weight of the United States, - 147 
 
 Origin of Weights and Measures, - - 147 
 
 The Avoirdupois Pound of the United States and Great Britain, - 148 
 
 The Standard UNIT of Length of the United States, - - 150 
 
 " " " of New York and Great Britain, - 151 
 
 The Standard UNIT of Liquid Measure of the United States, - - 154 
 
 " " of Dry Measure of the United States, 155 
 
 Method of determining whether a given year is Leap Year, - 157 
 
 French Money. Weights and Measures, - - l(l 
 
 Foreign Weights and Measures, compared with United States, - KJ3 
 
 GKNKRAL RULE for Reduction, ..... 166 
 
 Applications of Reduction, - ... 169 
 
 Cubic Measure reduced to Dry, or Liquid Measure, &c., - 172 
 
 Longitude reduced to Time, &c., - - - - - 175 
 
 Compound Numbers reduced to Fractions, ... 176 
 
 Fractional Compound Numbers reduced to whole numbers, - - 179 
 
 Compound Addition, ...... 181 
 
 Compound Subtraction, ...... 183 
 
 Comi.ouml Multiplication, ------ 186 
 
 Ccir.pound Division, - - - - - - -189 
 
 SECTION IX. 
 
 DECIMAL FRACTIONS, their origin, &c., - ... 191 
 
 Method of reading Decimals, - - - 193 
 
 Addition of Decimals, .____. 195 
 
 Subtraction of Decimals, ___._. 197 
 
 Multiplication of Decimals, ..... 199 
 
 Contractions in Multiplication of Decimals, two cases, - 201 
 
 Division of Decimals, - - 205 
 
 Contractions in Division of Decimals, .... 208 
 
 Decimals reduced to Common Fractions, - - - 210 
 
 Common Fractions reduced to Decimals, - - 2il 
 
 Compound Numbers reduced to Decimals, ... 215 
 
 SECTION X. 
 
 ADDITION of Circulating Decimals, .... 222 
 
 Subtraction of Circulating Decimals, ..... 223 
 
 MultJulication of Circulating Decimals, - - 224 
 
 Divis/m of Cijculating Decimals, - ... 225 
 
CONTENTS. XI 
 
 PA8B 
 
 SECTION XI. 
 
 FEDERAI MONEY, - ..... 226 
 
 Addition of Federal Money, - ... 229 
 
 Subtraction of Federal Money, 231 
 
 Multiplication of Federal Money, ..... 232 
 
 Division of Federal Money, ... 234 
 
 Counting-room Exercises, Contractions, &c. .... 237 
 
 To find the cost of articles bought and sold by the 100 or 1000, - 238 
 
 Bills, Accounts, &c., ... 239 
 
 SECTION XII. 
 
 PERCENTAGE, Percentage Table, &c., .... 241 
 
 Applications of Percentage, --.... 244 
 
 Commission, Brokerage, and Stocks, .... 244 
 
 Commission deducted in advance, and the balance invested, - - 2-17 
 
 INTEREST, - .... 2-19 
 
 General method for computing Interest, .... 252 
 
 Second method, multiplying the Prin. by the Int. of $1 for the time, 255 
 
 To compute Interest by half the number of months, ... 257 
 
 To compute Interest by the number of days, ... 258 
 
 Applications of Interest, remarks on Promissory Notes, &c., - - 258 
 
 Forms of Negotiable Notes, &c., ..... 2GO 
 
 Partial Payments j United States Rule, - - - - 2G1 
 
 Connecticut Rule, -----.. 263 
 
 Vermont Rule, ....... 2G4 
 
 Problems in Interest, ...... 2G5 
 
 Compound Interest, ....... 270 
 
 Discount, - - - - - - -. 272 
 
 Bank Discount, - ..... 274 
 
 To find what sum must be discounted to produce a given amount, 277 
 
 Insurance, four Cases, ... _ 278 
 
 The sum to be insured to recover the premium and the property, - 282 
 
 Life Insurance, ....... 282 
 
 Profit and Loss, four Cases, ...... 283 
 
 Duties, Specific and Advalorem, ..... 289 
 
 Assessment of Taxes, -.-.... 292 
 
 To find what sum must be assessed to raise a given net amount, - 295 
 
 Formation of Tax Bills, ..... 295 
 
 Rate Bills for Schools, - - 297 
 
 SECTION XIII. 
 
 ANALYSIS, - ... 298 
 
 Analyse so.utions of questions in Simple Proportion, - 299 
 
 " ' " Barter, - - - - 301 
 
 " " " Partnership, ... jju'J 
 
 " " General Average, - - - 303 
 
 Alligation, ... 304 
 
XI CONTENTS. 
 
 Analytic solutions >f questions in Compound Proportion, - 307 
 
 " " " Position, ... 308 
 
 " " " Practice, - - 309 
 
 SECTION XIV. 
 
 RATIO, general principles pertaining to it, - - - 313 
 
 Siirple Proportion, - - - - - 321 
 
 Siirple Proportion and its Proof by Cancelation, ... 325 
 
 Compound Proportion, - - . - - - 328 
 
 Compound Proportion and its Proof by Cancelation, - - 330 
 
 Conjoined Proportion, - ..... 332 
 
 SECTION XV. 
 
 DUODECIMALS, its principles, &c., ..... 334 
 
 Multiplication of Duodecimals, ..... 334 
 
 SECTION XVI. 
 
 EQUATION OP PAYMENTS, - .... ^38 ^ 
 
 Partnership, ........ 340 
 
 General Average, ----..- 343 
 
 Exchange of Currencies, ...... 345 
 
 Foreign Coins and Moneys of Account, .... 348 
 
 Exchange, Form of Bills of Exchange, &c., - - - - 351 
 
 Arbitration of Exchange, ...... 355 
 
 Alligation, Medial, and Alternate, .... 356 
 
 SECTION XVII. 
 
 INVOLUTION, and Evolution, - - ... . 3GO 
 
 Properties of Squares and Cubes, ..... 305 
 
 Extraction of the Square Root, ..... 3;7 
 
 Demonstration of the Square Root, ..... 308 
 
 Applications of the Square Root, ..... 370 
 
 Extraction of the Cube Root, Homer's Method, ... 374 
 
 Demonstration of the Cube Root, ..... 376 
 
 Extraction of Higher Roots, ---... 378 
 
 SECTION XVIII. 
 
 Arithmetical Progression, ...... 381 
 
 Geometrical Progression, ...... 384 
 
 Annuities, ----.... 3^6 
 
 Permutations and Combinations, ..... 3#g 
 
 SECTION XIX. 
 
 Application of Arithmetic to Geometry, .... 389 
 
 Mensuration of Surfaces and Solids, ..... 3S9 
 
 Measurement of timber, --.... 392 
 
 Gauging of Casks, - - . . . . . 393 
 
 Tonnage of Vessels, -.. 393 
 
 Mechanical Powers, ....... 394 
 
 Miscellaneous Examples, - . . . . . 395 
 
INTRODUCTION. 
 
 ART, 1 Anything which can be multiplied, divided , or measutvd, 
 is called QUANTITY. Thus, lines, weight, time, number, &c., are 
 quantities. 
 
 OBS. 1. A line is a quantity, because it can be measured in feet and inches; 
 weight can be measured in pounds and ounces; time, in hours and minutes; 
 numbers can be multiplied, divided, &c. 
 
 2. Color, and the operations of the mind, as love, hatred, desire, choice, &c., 
 cannot be multiplied, divided, or measured, and therefore cannot properly be 
 called quantities. 
 
 2. MATHEMATICS is the science of Quantity. 
 
 3* The fundamental branches of Mathematics are, Arithmetic t 
 Algebra, and Geometry. 
 
 4. Arithmetic is the science of Numbers. 
 
 5 Alyebra is a general method of solving problems, and of 
 investigating the relations of quantities by means cf letters and 
 signs. 
 
 OBS. Fluxions, or the Differential and Integral Calculus, may be considered 
 as belonging to the higher branches of Algebra. 
 
 6. Geometry is that branch of Mathematics which treats of 
 Magnitude. 
 
 7 The term magnitude signifies that which is extended, or 
 which has one or more of the three dimensions, length, breadth, and 
 tiiickness, Thus, lines, surfaces, and solids are magnitudes. 
 
 QUEST. 1. What is Quantity 7 Give some examples of quantity. O/ 1. Why is a lina 
 
 *k quantity 1 Weight 1 Time ? Numbers ? Are color and the operations of the mind 
 
 quantities 1 Why not? 2. What is Mathematics 1 3. What are the fundamental 
 
 branches of mathematics 1 4. What is Arithmetic 1 5. Algebra ? 6. Geometry 1 1. What 
 
 ! meant by magnitude 1 
 
14 INTRODUCTION. 
 
 OBS. 1. A line is a magnitude, because it can le extended in length; a 
 surface, because it has length and breadth ; a solid, because it has length, 
 breadth, and thickness. 
 
 I. Motion, though a quantity, is not, strictly speaking, a magnitude ; for it 
 has neither length, breadth, nor thickness. 
 
 3. The term magnitude is sometimes, though inaccurately, used as synony- 
 mous with quantity. 
 
 8 Trigonometry and Conic Sections are branches of Mathemat- 
 cs, in which the principles of Geometry are applied to triangles, 
 and the sections of a cone. 
 
 9, Mathematics are either pure or mixed. 
 
 In pure mathematics, quantities are considered, independently 
 of any substances actually existing. 
 
 In mixed mathematics, the relations of quantities are investi- 
 gated in connection with some of the properties of matter, or 
 with reference to the common transactions of business. Thus, in 
 Surveying, mathematical principles are applied to the measuring 
 of land ; hi Optics, to the properties of light ; and in Astronomy, 
 to the heavenly bodies. 
 
 OBS. The science of pure mathematics has long been distinguished for the 
 clearness and distinctness of its principles, and the irresistible conviction which 
 they carry to the mind of every one who is once made acquainted with them. 
 This is to be ascribed partly to the nature of the subjects, and partly to the 
 exactness of the definitions, the axioms, and the demonstrations. 
 
 1 0. A definition is an explanation of what is meant by a word, 
 or phrase. 
 
 OBS. It is essential to a complete definition, that it perfectly distinguish* the 
 thing defined, from everything else. 
 
 1 1 A proposition is something proposed to be proved, or 
 required to be done, and is either a Theorem., or a Problem. 
 
 1 2. A theorem is something to be proved. 
 
 1 3 A problem is something to be done, as a question to be 
 solved. 
 
 (It EST. Obs. Why is a line n magnitude ? A surface ? A solid 7 Is motion a magni- 
 tude ? Why not? &. Of how many kinds are mathematics ? In pure mathematics how 
 are quantities considered? How in mixed mathematics? Obs. For what is the science 
 of pure mathematics distinguished? 10. What is a definition ? Obs. What is essential 
 to a complete definition ? 11. What is a proposition ? 12. V theorem ? 13. A problem? < 
 
INTRODUCTION. 15 
 
 OBS. 1. In tie statement of every proposition, whether theorem or problem, 
 certain things must be given, or assumed to be true. These things are called 
 the data of the proposition. 
 
 2. The operation by which the answer of a problem is found, is called a 
 solution. 
 
 3. When tb.3 given problem is so easy, as to be obvious to every one without 
 explanation, i: is called a postulate. 
 
 14. One proposition is contrary, or contradictory to another, 
 when what is affirmed in the one, is denied in the other. 
 
 OBS. A proposition and its contrary, can never both be true. It cannct be 
 true, that two given lines are equal, and that they are not equal, at the same 
 time. 
 
 1 5 One proposition is the converse of another, when the order 
 is inverted ; so that, what is given or supposed in the first, be- 
 comes the conclusion in the last ; and what is given in the last, is 
 the conclusion, in the first. Thus, it can be proved, first, that if 
 the sides of a triangle are equal, the angles are equal ; and sec- 
 ondly, that if the angles are equal, the sides are equal. Here, in 
 the first proposition, the equality of the sides is given, and the 
 equality of the angles inferred ; in the second, the equality of the 
 angles is given, and the equality of the sides inferred. 
 
 OBS. In many instances, a proposition and its converse are both true, as in 
 the preceding example. But this is not always the case. A circle is a figure 
 bounded by a curve; but a figure bounded by a curve is not necessarily a 
 circle. 
 
 16. The process of reasoning by which a proposition is shown 
 to be true, is called a demonstration. 
 
 OBS. A demonstration is either direct or indirect. 
 
 A direct, demonstration commences with certain principles or data which are 
 admitted, or have been proved to be true ; and from these, a series of other 
 truths are deduced, each depending on the preceding, till we arrive at the truth 
 which was required to be established. 
 
 An indirect demonstration is the mode of establishing the truth of a propo- 
 sition by proving that the supposition of its contrary, involves an absurdity. 
 
 QPEST. Obs. What is meant by the data of a proposition? By the solation cf 
 problem ? What is a postulate 1 14. When is one proposition contrary to another 
 Obs. Can a proposition and its contrary both be true? 15. When is one proposition the 
 converse of another ? Obs. Can a proposition and its converse both be true ? 16. What 
 Is a demonstration ? Obs. Of how many kinds are demonstrations? What is a direct 
 demonstration ? An indirect demonstration ? 
 
16 INTRODUCTION. 
 
 This is commonly called redudio ad absurdum. The former is the more com- 
 mon method of conducting a demonstrative argument, and is the meet satisfac 
 tory to the mind. 
 
 1 7 A Lemma is a subsidiary truth, or proposition, demon- 
 strated for the purpose of using it in the demonstration of a 
 theorem, or the solution of a problem. 
 
 1 8. A Corollary is an inference or principle deduced from a 
 preceding proposition. 
 
 1 A ScJiolium is a remark made upon a preceding prop- 
 osition, pointing out its connection, use, restriction, or extension. 
 
 20. An Hypothesis is a supposition, made either in the state- 
 ment of a proposition, or in the course of a demonstration. 
 
 AXIOMS. 
 
 21. An Axiom is a self-evident proposition ; that is, a prop- 
 osition whose truth is so evident at sight, that no process of 
 reasoning can make it plainer. The following axioms are among 
 the most common : 
 
 1. Quantities wliich are equal to the same quantity, are equal 
 to each other. 
 
 2. If the same or equal quantities are added to equals, the 
 sums will be equal. 
 
 3. If the same or equal quantities are subtracted from equals, 
 the remainders will be equal. 
 
 4. If the same or equal quantities are added to unequals, the 
 sums will be unequal. 
 
 5. If the same or equal quantities are subtracted from unequals, 
 the remainders will be unequal. 
 
 6. If equal quantities are multiplied by the same or equal 
 quantities, the products will be equal. 
 
 7. If equal quantities are divided by the same or equal quan- 
 tities, the quotients will be equal. 
 
 8. If the same quantity is both added to and subtracted from 
 another, the value of the latter will not be altered. 
 
 QCTKST 17. What is a lemma? 18. What is a corollary? i9. Wha- is a scholium 1 
 SO. What Is an hypothesis? 21. What is an axiom 1 Name some of the most common 
 not* 
 
INTRODUCTION. 17 
 
 9. If a quantity is both multiplied and divided by the same or 
 an equal quantity, its value will not be altered. 
 
 10. The whole of a quantity is greater than a part. 
 
 11. The whole o f a quantity is equal to the sum of all its parts. 
 
 SIGNS. 
 
 22. Addition is represented by the sign (-{-), which is called 
 plus. It consists of two lines, one horizontal, the other perpen- 
 licular, forming a cross, and shows that the numbers between 
 which it is placed, are to be added together. Thus, the expression 
 6+8, signifies that 6 is to be added to 8. It is read, " G plus 8,"- 
 or " added to 8." 
 
 OBS. The term plus is a Latin word, originally signifying " more." hence 
 "added to." 
 
 23. Subtraction is represented by a short horizontal line ( ), 
 which is called minus. When placed between two numbers, it 
 shows that the number after it is to be subtracted from the one 
 before it. Thus, the expression 9 4, signifies that 4 is to bo 
 subtracted from 9 ; and is read, " 9 minus 4," or " 9 less 4." 
 
 OBS. The term minus is a Latin word, signifying kst. 
 
 24. Multiplication is usually denoted by two oblique lines 
 crossing each other (x), called the sign of multiplication. It 
 shows that the numbers between which it is placed, are to be 
 multiplied together. Thus, the expression (9x0), signifies that 
 9 and C are to be multiplied together, and is read, " 9 multiplied 
 by 6," or simply, "9 into 0." Sometimes multiplication is de- 
 noted by a point (.) placed between the two numbers or quanti- 
 ties. Thus, 9.6 denotes the same as 9X6. 
 
 OBS. It is Letter to denote the multiplication of figures by a cross than by a 
 point ; for the latter is liable to be confounded with the decimal point. 
 
 24. a. When two or more numbers are to be subjected to the 
 same operation, they must be connected by a line ( ) placed 
 
 show 1 Obs. What is the meaning of the term phis 7 23. How is subtraction represented 1 
 What is the sign of subtraction called 1 What does it show? Obs. What does the term 
 minus signify 7 24. How is multiplication usually denoted ? What does the sign of mul- 
 tiplication show 7 In what other way is multiplication sometimes denoted 1 
 
18 INTRODUCTION. 
 
 over them, called a vinculum, or by a parenthesis ( ). Thus the 
 expression (12-{-3)x2, shows that the sum of 12 and 3, is to be 
 multiplied by 2, and is equal to 30. But 12-J-3X2, signifies 
 that 3 only is to be multiplied by 2, and that the product is to 
 be added to 12, which will make 18. 
 
 25. Division is expressed in two ways : 
 
 First, by a horizontal line between two dots (-5-), called the 
 sign of division, which shows that the number before it, is to be 
 divided by the number after it. Thus, the expression 246 
 signifies that 24 is to be divided by 6. 
 
 Second, division is often expressed by placing the divisor under 
 the dividend, in the form of a fraction. Thus, the expression 
 3j, shows that 35 is to be divided by 7, and is equivalent to 
 35V7. 
 
 26. The equality between two numbers or quantities, is rep- 
 resented by two parallel lines (=), called the sign of equality. 
 Thus, the expression 5-|-3=8, denotes that 5 added to 3 are 
 equal to 8. It is read, " 5 plus 3 equal 8," or " the sum of 5 
 plus 3 is equal to 8." So 7+5=16 4=12. 
 
 QTTEST. 24. a. When two or more numbers are to be subjected to the same operation, 
 what must be done ? 25. In how many ways is division expressed 1 What is the first ? 
 What does this sign show ? What is the second ? 26. How is the equality between two 
 numbers or quantities represented? 
 
ARITHMETIC, 
 
 SECTION I. 
 NOTATION AND NUMERATION. 
 
 ART. 27, Any single thing, as a peach, a rose, a book, is 
 called a unit, or one ; if another single thing is put with it, the 
 collection is called two ; if another still, it is called three ; if an- 
 other, four ; if another, Jive, &c. 
 
 The terms, one, two, thtw, tec., by which we express how rwny 
 single things or units are under consideration, are the names of 
 numbers. Hence, 
 
 28, NUMBER signifies a unit, or a collection of units. 
 
 OES. 1. Numbers are divided into two classes, abstract and concrete. 
 
 When they are applied to particular objects, as two pears, Jive pounds, ten 
 dollars, &c., they are called concrete numbers. 
 
 When they do not refer to any particular object, as when we say four and 
 five are nine, they are called abstract numbers. 
 
 2. Whole numbers are often called integers. 
 
 3. Numbers have various properties and relations, and are applied to variou 
 computations in the practical concerns of life. These properties and applica- 
 tions are formed into a system, called Arithmetic. 
 
 29. ARITHMETIC is the science of numbers. 
 
 OBS. 1. The term Arithmetic is derived from the Greek word arithmgiiMt, 
 which signifies the art of reckoning by numbers. 
 
 2. The aid of Arithmetic is required to make and apply calculations not 
 only in faisiness transactions, but in tJmost every department of mathematics. 
 
 . What is a single thing called? If another is put with it, what is the col- 
 lection called 1 If another, what? What are the terms one, two, three, &c. ? 28. What 
 does number signify? Obs. Into how many classes are numbers divided? When are 
 they called concrete? When abstract? To what are numbers applied 1 29. What to 
 Arithmetic 1 Obs. In what is the aid of arithmetic require i ? 
 T.H. 2 
 
20 NOTATION. [SECT. I 
 
 Numbers are expressed by words, by letters, and by figures. 
 
 NOTATION. 
 
 30. The art of expressing numbers by letters or figures, ii 
 called NOTATION. There are two methods of notation in use, the 
 Roman and the Arabic. 
 
 3 1 . The Roman method employs seven capital letters, viz : I, 
 V, X, L, C, D, M. When standing alone, the letter I, denotes 
 one ; V, Jive ; X, ten ; L, fifty ; C, one hundred ; D, five hun- 
 dred ; M, one thousand. To express the intervening numbers from 
 one to a thousand, or any number larger than a thousand, we re- 
 sort to repetitions and various combinations of these letters. The 
 method of doing this will be easily learned from the following 
 
 TABLE. 
 
 I denotes 
 
 one. 
 
 XXX denote 
 
 thirty. 
 
 
 II 
 
 ft 
 
 two. 
 
 XL 
 
 forty. 
 
 
 III 
 
 tt 
 
 three. 
 
 L 
 
 fifty. 
 
 
 IV 
 
 it 
 
 four. 
 
 LX 
 
 sixty. 
 
 
 V 
 
 tt 
 
 five. 
 
 LXX 
 
 seventy. 
 
 
 VI 
 
 tt 
 
 six. 
 
 LXXX 
 
 eighty. 
 
 
 VII 
 
 tt 
 
 seven. 
 
 XC 
 
 ninety. 
 
 
 VIII 
 
 ft 
 
 eight. 
 
 C 
 
 one hundred. 
 
 
 IX 
 
 tt 
 
 nine. 
 
 CI 
 
 one hundred and 
 
 me. 
 
 X 
 
 
 
 ten. 
 
 CX 
 
 one hundred and 
 
 t.en. 
 
 XI 
 
 
 
 eleven. 
 
 cc 
 
 two hundred. 
 
 
 XII 
 
 tt 
 
 twelve. 
 
 ccc 
 
 three hundred 
 
 
 XIII 
 
 ft 
 
 thirteen. 
 
 COCO " 
 
 four hundred. 
 
 
 XIV 
 
 tt 
 
 fourteen. 
 
 D 
 
 five hundred. 
 
 
 XV 
 
 tt 
 
 fifteen. 
 
 DC 
 
 six hundred. 
 
 
 XVI 
 
 ft 
 
 sixteen. 
 
 DCC " 
 
 seven hundred. 
 
 
 XVII 
 
 tt 
 
 seventeen. 
 
 DCCC " 
 
 eight hundred. 
 
 
 XVIII 
 
 ft 
 
 eighteen. 
 
 DCCCC " 
 
 nine hundred. 
 
 
 XIX 
 
 tt 
 
 nineteen. 
 
 M 
 
 one thousand. 
 
 
 XX 
 
 ft 
 
 twenty. 
 
 MM 
 
 two thousand. 
 
 
 XXI 
 
 tt 
 
 twenty-one. 
 
 MDCCCLV, 
 
 one thousand e 
 
 lgh 
 
 XXII 
 
 tt 
 
 twenty-two, &c. 
 
 hundred 
 
 and. fifty-five. . 
 
 
 . How are numbers usually expressed 1 30. What is notation? 
 methods are there in use 1 31. What is employed by the Roman method ? 
 
 How nwm> 
 
ARTS. 30-33.] NOTATION. 21 
 
 OBS. 1. This method of expiessing numbers was invented by the Romans, 
 and is therefore called the Roman Notation. It is now seldom used, except to 
 denote chapters, sections, and other divisions of books and discourses. 
 
 2. The letters C and M, are the initials of the Latin words centum^ and 
 mille, the former of which signifies a hundred, and the latter a thousand: for 
 this reason it is supposed they were adopted to represent these numbers. 
 
 3 1 a. It will be perceived from the Table above, that every 
 time a letter is repeated, its value is repeated. Thus I, standing 
 alone, denotes one ; II, two ones, or two, &c. So X denotes ten ; 
 KX, twenty, &c. 
 
 When a letter of a less value is placed before a letter of a 
 greater value, the less takes away its own value from the greater ; 
 but when placed after, it adds its own value to the greater. 
 
 32. A line or bar ( ) placed over a letter, increases its 
 value a thousand times. Thus, V denotes five, V denotes five 
 thousand ; X, ten ; X, ten thousand, &c. 
 
 OBS. 1. In the early periods of this notation, four was written IIII, instead 
 of IV ; nine was written VIIII, instead of IX ; forty was written XXXX, 
 instead of XL, &c. 
 
 The former method is more convenient in performing arithmetical operations 
 In addition and subtraction ; while the latter is shorter and better adapted to 
 ordinary purposes. 
 
 2. A thousand was originally written CIO, which, in later times, was 
 changed into M ; Jive hundred was written 10 instead of D. Annexing O to 
 10 increased its value ten times. Thus, IOO denoted Jive thousand; IOOO, 
 fifty thousand, &c. 
 
 3. Prefixing C and annexing O to the expression CIO, makes its value ten 
 times greater: thus, CCIOO denotes ten thousand-, CCCIOOO, a hundred 
 thousand. According to Pliny, the Romans carried this mode of notation no 
 further. When they had occasion to express a larger number, they did it by 
 repetition. Thus, CCCIOOO, CCCIOOO, expressed two hundred thousand, &c. 
 
 33. The common method of expressing numbers is by the 
 Arabic Notation. The Arabic method employs the following ten 
 cJtaracters or figures, viz : 
 
 12 34567 '8 90 
 
 one, two, three, four, five, six, seven, eight, nine, zero. 
 
 QrKST.~O6s. Why is this method called Roman ? 31. a. What is the effect of repeating 
 a letter? If a letter of less value is placed before another of greater value, what is th 
 effect? If placed after, what? 32. When a line or bar is placed over a letter, how does 
 it affect its value ? 33. What is the common way of expressing numbers 1 How many 
 characters does this method employ 1 
 
22 NOTATION. [SECT. 1 
 
 The first nine are called significant figures, because each one 
 always has a value, or denotes some number. They are also 
 called digits, from the Latin word digitus, which signifies a 
 finger. 
 
 The last one is called a cipher, or naught, because when stand- 
 ing alone it has no value, or signifies nothing. 
 
 OBS. 1. It must not be inferred, however, that the cipher is useless; for when 
 placed on the right of any of the significant figures, it increases their value. 
 It may therefore be regarded as an auxiliary digit, whose office, it will be seen 
 hereafter, is as important as that of any other figure in the system. 
 
 2. Formerly all the Arabic characters were indiscriminately called ciphers ; 
 hence the process of calculating by them was called ciphering ; on the same 
 principle that calculating by figures is called figuring. 
 
 34. It will be seen that nine is the greatest number that can 
 be expressed by any single figure in the Arabic system of Nota- 
 tion. 
 
 All numbers larger than nine are expressed by combining to- 
 gether two or more of these ten figures, and assigning different 
 values to them, according as they occupy different places. For 
 example, ten is expressed by combining the 1 and 0, thus 10 ; 
 eleven by two Is, thus 11 ; twelve by 1 and 2, thus 12 ; twenty, 
 thus 20 ; thirty, thus 30 ; &c. A hundred is expressed by com- 
 bining the 1 and two Os, thus 100; two hundred, thus 200; a 
 thousand by combining the 1 and three Os, thus 1000; two thou- 
 sand, thus 2000 ; ten thousand, thus 10,000 ; a hundred thousand, 
 thus 100,000; a million, thus 1,000,000; tea millions, thus 
 10,000,000; &c. Hence, 
 
 35. The digits 1, 2, 3, <fcc., standing alone, or in the right 
 hand place, respectively denote units or ones, and are called units 
 of the Jirst order. * 
 
 When they stand in the second place, they express tens, or ten 
 ones ; that is, their value is ten times as much as when standing 
 
 QUEST. What are the first nine called 1 Why ? What else are they called ? What 
 fe the last one called? Why? Obs. Is the ciphe. useless? What may it be regarded ? 
 What is the origin of the term ciphering? 34. What is the greatest number that can be 
 expressed by one figure 1 How are larger numbers expressed? 35. What do the digits, 
 i, 2, 3, &c., denote, when standlig alone, or in the right hand place? What are they 
 then called? What do they den j to when standing in the second plaoe) 
 
ARTS. 34-37.J NOTATION 2.3 
 
 in the first or rigl.it hand place, and they are called units of the 
 second order. 
 
 When occupying the third place, they express hundreds ; that 
 is, their value is ten times as much as when standing in the sec- 
 ond place, and they are called units of the third order. 
 
 When occupying the fourth place, they express thousands ; that 
 is, then* value is ten times as much as when standing in the third 
 place, and they are called units of the fourth order, &c. Thus, it 
 will be seen that, 
 
 Ten units make one ten, ten tens make one hundred, and ten hun 
 dreds make one thousand ; that is, ten in an inferior order are equal 
 to one in the next superior order. Hence, universally, 
 
 36. Numbers increase from right to left in a tenfold ratio; 
 consequently each removal of a figure one place towards the left, in- 
 creases its value ten times. 
 
 . 1. The number which forms the basis, or which expresses the ratio 
 of increase in a system of Notation, is called the RADIX of that system. Thus, 
 the radix of the Arabic notation is ten. 
 
 2. The reason that numbers increase from right to left, instead of left to 
 right, is probably owing to the ancient practice of writing from the right hand 
 to the left. 
 
 37. The different values which the same figures have, are called 
 simple and local values. 
 
 The simple value of a figure is the value which it expresses 
 when it stands alone, or in the right hand place. Hence the sim- 
 ple value of a figure is the number which its name denotes. 
 
 The local value of a figure is the increased value which it ex- 
 presses by having other figures placed on its right. Hence the 
 local value of a figure depends on its locality, or the place which 
 
 QUEST. What Is their value then ? What are they called 1 What Is a figure called 
 when it occupies the third place 7 What is its value then ? What is it called when in 
 the fourth place ? What is its value 1 How many units are required to make one ten ? 
 How many tens make a hundred ? How many hundreds make a thousand 1 How many 
 of an inferior order are required to make one of the next superior order ? 36. What is the 
 general law by which numbers increase ? What is the effect upon the value of a figure 
 to remove it one place towards the left ? Note. What is the number called which forms 
 the basis jr the ratio of increase in a system of notation 1 What is the radix of the Arable 
 notation? Why do numbers increase from right to left? 37. What are the different 
 values of the same figure called 1 What Is the simple value of a figure ? What the local 1 
 
24 NUMERATION. [SECT. I. 
 
 it occupies in relation, to other numbers with which it is connected. 
 (Art. 35.) 
 
 OBS. 1. This system of notation is called Arabic, because it is supposed to 
 tave been invented by the Arabs. 
 
 2. It is also called the decimal system, because numbers increase in a ten- 
 fold ratio. The term decimal is derived from the Latin word decem, which sig- 
 nifies ten. 
 
 3. The early history of the Arabic notation is veiled in obscurity. It is th* 
 opinion of some whose judgment is entitled to respect, that it was invented by 
 the philosophers of India. It was introduced into Europe from Araoia abou 
 Jhs eighth century, and about the eleventh century it came into general use, 
 xrfii in England and on the continent. The application of the term digit to 
 the significant figures, affords strong presumptive evidence that the system had 
 its origin in the ancient mode of counting and reckoning by means of the 
 fingers ; and that the idea of employing ten characters, instead of twelve or 
 any other number, was suggested by the number of fingers and thumbs on both 
 hands. (Art. 33.) 
 
 NUMERATION. 
 
 38. The art of reading numbers when expressed by figures, tt 
 called NUMERATION. 
 
 The pupil will easily learn to read the largest numbers from the 
 following scheme, called the 
 
 NUMERATION TABLE. 
 
 I 
 
 
 685, 876, 389, 764, 391, 827, 218, 649, 853, 123, 234, 579, 793, 465, 623. 
 "xvt xrv! xin. xii. xi. x. ix. vin. vn. vi. v. iv. in. n. i. 
 
 39. The different orders of numbers are divided into periods 
 of taree figures each, beginning at the right hand. The first 
 period, which is occupied by units, tens, hundreds, is called uniti 
 
 Q.UI8T. Upon what does the local value of a figure depend 1 Obs. Why is this system 
 of notation called Arabic ? What else is it sometimes called 1 Why ? What do you 
 ay of its early history? When was it introduced into Europe ? What is the probable 
 origin of the system ? Why were ten characters, rather than any other number, adopted 7 
 *& What is Numeration 1 39. How are the orders of numbers divided 1 Wha is the 
 1*1 period called 1 By what is it occupied 1 
 
ARTS. 38, 39.] NUMERATION. 25 
 
 period ; the second is occupied by thousands, tens of thousands, 
 hundreds of thousands, and is called thousands' period ; the third 
 is occupied by millions, tens of millions, hundreds of millions, and 
 is called millions' period ; the fourth is occupied by billions, tens 
 of billions, hundreds of billions, and is called billions' period ; and 
 so on, the orders of each successive period being units-, tens, and 
 hundreds. 
 
 The figures in the table are read thus: 685 tredecillions, 876 
 duodecillions, 389 undecillions, 764 decillions, 391 nonillions, 827 
 octillions, 218 septillions, 649 sextillions, 853 quintillians, 123 
 quadrillions, 234 trillions, -579 billions, 793 millions, 465 thou- 
 sand, 6 hundred and twenty-three. 
 
 Note. 1. The terms thirteen, fourteen, fifteen, &c., are obviously derived 
 from three and ten, four and ten, five and ten, which by contraction become 
 thirteen, fourteen, fifteen, and are therefore significant of the numbers which 
 they denote. The terms eleven and twelve, are generally regarded as primitive 
 words ; at all events, there is no perceptible analogy between them and tho 
 numbers which they represent. Had the terms oneteen and twoteen been 
 adopted in their stead, the names would then have been significant of the 
 numbers one and ten, two and ten; and their etymology would have been 
 similar to that of the succeeding terms. 
 
 The terms twenty, thirty, forty, &c., were formed from two tens, three tens, 
 four tens, which were contracted into twenty, thirty, forty, &c. 
 
 The terms twenty-one, twenty-two, twenty-three, &c., are compounded of 
 twenty and one, twenty and two, &c. All the other numbers as far as ninety- 
 nine, are formed in a similar manner. 
 
 2. The terms hundred, thousand and million are primitive words, and bear 
 uo analogy to the numbers which they denote. The numbers between a hun- 
 dred and a thousand are expressed by a repetition of the numbers below a 
 hundred. Thus we say one hundred and one, one hundred and two, one 
 hundred and three, &c. 
 
 3. The terms billion, trillion, quadrillion, &c., are formed from million and 
 the Latin numerals bis, tres, quatuor, &c. Thus, prefixing bis to million, by a 
 slight contraction for the sake of euphony, it becomes billion ; prefixing tres to 
 million, it is easily contracted into trillion, &c. The Latin word bis signifies 
 two ; tres, three ; quatiwr, four ; quinque, five ; sex, six ; scptcm, seven ; octo t 
 eight ; novem, nine ; decem, ten ; undecim, eleven ; duodecim, twelve ; tredecvm, 
 thirteen. 
 
 Q.UEST. What is the second period called ? By what occujiod? What is the third 
 tailed ? By what occupied 1 What is the fourfl called 1 By w hat occupied ? Wha t H 
 he fifth called 1 By what occupied? Repeat the Numeration Table, beginning at the 
 ht hand. 
 
 3 
 
26 
 
 NUMERATION. 
 
 (.SECT. I. 
 
 Higher periods than those in the Table, may be easily formed by following 
 the above analogy. 
 
 4. The foregoing law, which assigns superior values to these ten characters, 
 according to the order or place which they occupy and the use of so many 
 derivative and compound words in forming the names of numbers, saves an 
 inconceivable amount of time and labor in learning Notation and Numeration, 
 as well as in their application. 
 
 4.-O. To read numbers which are expressed by figures. 
 
 Point them off into periods of three figures each ; tlien, beginning 
 t the left Jiand, read the figures of each period in tlie same manner 
 as those of tlie right hand period are read, and at the end of each 
 period, pronounce its name. 
 
 OBS. 1. The learner must be careful, in pointing off figures, always to begin 
 at the rigid hand ; and in reading them, to begin at the left hand. 
 
 2. Since the figures in the first or right hand period always denote units, 
 its name is not pronounced. Hence, in reading figures, when no period ia 
 mentioned, it is always understood to be the right hand, or units' period.j 
 
 EXERCISES IN NUMERATION. 
 
 Note. In numerating large numbers, it is advisable for the pupil first to 
 apply to each figure the name of the order which it occupies. Thus, beginning 
 at the right hand, he should say, " Units, tens, hundreds," &c., and point at 
 the same time to the figures standing in the order which he mentions. 
 
 Read the following numbers : 
 
 Ex. 1. 
 
 3506 
 
 11. 
 
 706305 
 
 21. 
 
 967058713 
 
 2. 
 
 6034 
 
 12. 
 
 1640030 
 
 22. 
 
 32100040 
 
 3. 
 
 5060 
 
 13. 
 
 830006 
 
 23. 
 
 106320000 
 
 4. 
 
 90621 
 
 14. 
 
 70900038 
 
 24. 
 
 780507031 
 
 5. 
 
 73040 
 
 15. 
 
 3067300 
 
 25. 
 
 4063107 
 
 6. 
 
 450302 
 
 16. 
 
 12604321 
 
 26. 
 
 29038450 
 
 7. 
 
 603260 
 
 17. 
 
 70003000 
 
 27. 
 
 1046347025 
 
 8. 
 
 130070 
 
 18. 
 
 161010602 
 
 28. 
 
 20380720000 
 
 9. 
 
 2021305 
 
 19. 
 
 80367830 
 
 29. 
 
 8503467039 
 
 10. 
 
 4506580 
 
 20. 
 
 400031256 
 
 30. 
 
 450670412463 
 
 QUEST. 40. How do you read numbers expressed by figures 7 Cbs. Where begin to 
 point them off"? Where to read theml Do you pronounce the natae of the right hand 
 period ? When no period is named, what is understood ? 
 
ARTS. 40, 41.] NUMERATION. 2*7 
 
 31. 130812000641 
 
 32. 5200240301000 
 
 33. 98760000216 
 
 34. 82600381000000 
 
 35. 403070003462000 
 
 36. 120340078910356 
 
 37. 43601000345000 
 
 38. 506302870045380 
 
 39. 42008120537062035 
 
 40. 653107843604893048 
 
 41. 210 256 031 402 385 290 845 381 467. 
 
 42. 361 438 201 219 763 281 572 829 318 278. 
 
 4 1 The method of dividing numbers into periods of three fig- 
 res, was invented by the French, and is therefore called the 
 French Numeration. 
 
 The English divide numbers into periods of six figures, in the 
 following manner : 
 
 II II 
 
 s a 
 
 % fg s . .g 
 
 S * a * g <s g 
 
 s ^ - 2 ^ I I 
 
 ^ s s g - s s 5 S < 
 
 g S S g g I S 
 
 i o I ; o i ^ . . 
 
 M ^H^^S ^H^ 
 
 sJS'si 'sJS'sl * J 
 
 rfi C_( ^ 75 i^A rf> t> r ^ W ^ ) 
 
 S *S 1 ? "S | S 'S g S "5 g S^2 
 
 " g 1 1 S -S " s I 'S a I i g I a j .-j 
 
 hP H H K H *"3 ^EHEH^nE' 1 ^ K E" 1 H ffi E" 1 *^ 
 
 423561, 234826, 479365 
 
 Period III. Period II. Period I. 
 
 According to this method, the preceding figures are read thus : 
 423561 billions, 234826 millions, and 479365. 
 
 OBS. 1. It will be perceived that the two methods agree as far as hundreds 
 of millions; the former then begins a new period, while the latter continues on 
 through thousands of millions, &c. 
 
 2. The French method is generally used throughout the continent of Europe, 
 as well as in America, and has been recently adopted by some English authors. 
 It b very genei illy admitted to be more simple and convenient than the Eng- 
 lish method. 
 
 . 41. What is the French method of numeration 1 \V~fat the English method f 
 Ois. Which is the more simple and convenient 1 
 
 2* 
 
28 NOTATION. [SECT. 1. 
 
 EXERCISES IN NOTATION. 
 
 4 2 To express numbers by figures. 
 
 Begin at the left hand, and write in each order the figure which 
 denotes the given number in that order. 
 
 If any intervening orders are omitted in the proposed number, 
 write ciphers in their places. (Art. 38.) 
 
 Write the following numbers in figures : 
 
 1. Two thousand, one hundred and nine. 
 
 2. Twenty thousand and fifty-seven. 
 
 3. Fifty-five thousand and three. 
 
 4. One hundred and five thousand, and ten. 
 
 5. Seven hundred and ten thousand, three hundred and one. 
 
 6. Two millions, sixty-three thousand, and eight. 
 
 7. Fourteen millions, and fifty-six. 
 
 8. Four hundred and forty millions, and seventy-two. 
 
 9. Six billions, six millions, six thousand, and six. 
 
 10. Forty-five billions, three hundred and forty thousand, and 
 
 11. Five hundred and fifty-six millions, three thousand, two 
 hundred and sixty-four. 
 
 12. Eight hundred and ten billions, ten millions, and seventy- 
 five thousand. 
 
 13. Ninety-six trillions, seven hundred billions, and fifty-four. 
 
 14. Three hundred and forty-nine quadrillions, five trillions, 
 seven billions, four millions, and twenty. 
 
 15. Nineteen quintillions. 
 
 16. Six hundred and thirty sextillions. 
 
 17. Two hundred and ninety-eight septillions. 
 18.Seventy-four octillions. 
 
 19. Four hundred and ten decillions. 
 
 20. Eight hundred and sixty-three duodecillions. 
 
 21. Nine hundred and thirty-five tredecillions. 
 
 22. Six hundred and seventy-three quintillions, seventeen quad 
 i vlions, and forty-five. 
 
 23. Twenty trillions, six hundred and forty-eight billions, and 
 trenty-five thousand. 
 
ARTS. 42-44. j NOTATION. 29 
 
 OBS. The great facility with which large numbers may be expressed both 
 in language and by figures, is calculated to give an imperfect idea of their real 
 magnitude. It may assist the learner in forming a just conception of a million, 
 a billion, a trillion, &c., to reflect, that to count a million, at the rate of a hun- 
 dred a minute, would require nearly seventeen days often hours each; to count 
 a billion, at the same rate, would require more than forty-Jive years; and /</ 
 count a trillion, more than 45,662 years. 
 
 43. From the preceding illustrations, the learner will per- 
 ceive that a variety of other systems of notation may be formed 
 upon the same principle, having different numbers for their 
 radices. Thus, if we wished to form a quinary system ; that is, 
 a system in which the numbers should increase in & five-fold ratio, 
 or has five for its radix, it would require four significant figures 
 and a cipher. Let the figures 1, 2, 3, 4, and 0, be the characters 
 employed ; then five would be expressed by 1 and 0, and would 
 be Avritten thus 10 ; six by 1 and 1, thus 11 ; seven by 1 and 2, 
 thus 12 ; eight by 1 and 3, thus 13 ; nine by 1 and 4, thus 14; 
 ten by 2 and 0, thus 20 ; eleven by 2 and 1, thus 21, &c. 
 
 4:4. In the binary or diadic system of notation developed by 
 Leibnitz, there are two characters employed, 1 and 0. The cipher 
 when placed at the right hand of a number, in this system, mul- 
 tiplies it by two. Thus the number one is expressed by 1 ; two 
 by 10; three by 11 ; four by 100; five by 101; six by 110; 
 seven by 111; eight by 1000; nine by 1001; ten by 1010; 
 eleven by 1011, &c. 
 
 OBS. 1. In like manner other systems of notation may be formed, having 
 three, four, six, eight, twelve, or any given number for their radix. 
 
 When the radix is two, the system is called bi.nary or diadic ; when three 
 it is called ternary ; when four, quaternary ; when five, quinary ; when six, 
 senary; when seven, septenary; when eight, odary; when nine, nonary, &c 
 
 2. It should be observed that every system of notation, formed upon the 
 foregoing principles, will require as many distinct characters, as there are urJts 
 In the radix, and that one of them must be a cipher, and another a unit. 
 
 For the method of changing numbers from the decimal to other scales of 
 notation, and the converse, see Arts. 162, 163. 
 
 QUEST. 43. Is the decimal notation the only system that can be formed on the same 
 Manciples! How would you form a quinary system of notation? Wiite six in the qui- 
 ary scale on the black-board. Write seven, nine, ten, eleven, twelve Ob$. How many 
 haracters will any system formed upon this principle require 1 
 
 3* 
 
30 NOTATION. [SECT. [. 
 
 45. About the commencement of the second century, Ptolemy 
 introduced the sexagesimal notation, which has sixty for its radix. 
 
 OBS. 1. It is said that the Chinese and some other eastern nations now em- 
 ploy this system in measuring time, using periods of sixties, instead of centuries. 
 Relics of the sexagesimal notation may also be seen in our division of the circle,, 
 and of time, where the degree and hour are each divided into 60 minutes, the 
 minute into (50 seconds, &c. 
 
 2. The Roman notation seems to have been commenced with V or five for 
 its radix, which was afterwards changed to X or ten. It may therefore be 
 regarded as a kind of combination of the quinary and decimal systems. 
 
 46. Since the number eight may be divided and sub-divided 
 so many times without a remainder, some contend that a system 
 of notation having eight for its radix, would be preferable to the 
 decimal system. 
 
 Others claim that the duodecimal notation ; that is, a system 
 with twelve for its radix, would be more convenient than either." 
 However this may be, the decimal system is so firmly rooted, it 
 were hopeless to attempt a change. 
 
 OBS. It may be doubted whether any other ratio of increase would, on the 
 whole, be more convenient, than that of the present system. If the ratio were 
 less, it would require more places of figures to express large numbers ; if the ratio 
 were larger, it would not indeed require so many figures, but the operations, 
 would manifestly be more difficult than at present, on account of the numbers 
 in each order being larger. Besides, the decimal system is sufficiently compre- 
 hensive to express with all desirable facility, every conceivable number, the 
 largest as well as the smallest ; and yet it is so simple, that a child may under- 
 stand and apply it. In a word, it is every way adapted to the practical ope- 
 rations of business, as well as the most abstruse mathematical investigations 
 In whatever light, therefore, it is viewed, the decimal notation must be re- 
 garded as one of the most striking monuments of human ingenuity, and iti 
 beneficial influence on the progress of science and the arts, on commerce and 
 civilization, must win for its unknown author the everlasting admiration and 
 gratitude of mankind. 
 
 * Barlow's Theory of Numbers, Leslie's Plalosophy of Arithmetic, Edi r CTji Eaej 
 elope dia. 
 
ARTS. 45-50.] ADDITION. 31 
 
 SECTION II. 
 
 ADDITION. 
 
 ART. 49. Ex. 1. A man bought three lots of land ; the first 
 contained 23 acres, the second 9 acres, and the third 15 acres : 
 how many acres did he buy ? 
 
 Solution. 23 acres and 9 acres are 32 acres, and 15 are 47 
 aeres. Ans. 47 acres. 
 
 OBS. It will be seen, that the solution of this example consists in finding a 
 tingle number, which will exactly express the value of the several given num- 
 bers untied together. 
 
 5O. The process of uniting two or more numbers together -, so 
 as to form a single number, is called ADDITION. 
 
 The answer, or the number thus found, is called the Sum or 
 Amount. 
 
 OBS. When the numbers to be added are all of the same denomination, as 
 all dpllars, all pounds, &c., the operation is called Simple Addition. 
 
 Ex. 2. A miller bought 7864 bushels of wheat of one man, 
 4952 bushels of another, and 3273 bushels of another : how 
 many bushels did he buy of all ? 
 
 Write the numbers under each other, so that Operation. 
 
 units may stand under units, tens under tens, 7864 
 
 &c., and draw a line beneath them. Then be- 4952 
 
 ginning at the right hand or units, add each 3273 
 
 column separately. Thus, 3 units and 2 units Ans. 16089 bu. 
 are 5 units, and 4 are 9 units. Write the 9 in units' place under 
 the column added. Next 7 and 5 are 12, and 6 are 18 tens. But 
 18 requires two figures to express it ; (Art. 34 ;) consequently it 
 cannot all be written under its own column. We therefore write 
 the 8 or right hand figure in tens' place under the column added, 
 and reserving the 1 or left hand figure, add it with the hundreds. 
 Thus, 1 which was reserved, and 2 are 3, and 9 are 12, and 8 are 
 20 hundreds. Set the or right hand figure under the column 
 
 QUEST. 50. What is Addition ? What is the answer called 1 Obs. When the nuiu 
 bers to be added are all of the same denomination, what is the opera don called 1 51. \Vha4 
 order,! of figures 4oyou add together ? 
 
32 ADDITION. [SECT. II. 
 
 add<i, and reserving the 2 or left hand figure, add it to the next 
 column as before. Thus, 2 which were reserved and 3 are 5, 
 and 4 are 9, and 7 are 16 thousands. Set the 6 under the col- 
 umn added ; and since there is no other column to be added, 
 write the 1 in the next place on the left. 
 
 5 1 It will be perceived in this example, that units are added 
 to units, tens to tens, &c. ; that is, figures of the same order are 
 added to each other. All numbers must be added in the same 
 manner. For, figures standing in different orders or columns ex- 
 press different values ; (Art. 35 ;) consequently, they cannot be 
 united together directly in a single sum. Thus, 3 units and 5 
 tens will neither make eight units, nor eight tens, any more than 3 
 oranges and 5 apples will make 8 apples, or 8 oranges. In like 
 manner it is plain that 7 tens and 2 hundreds will neither make 
 9 tens, nor 9 hundreds. 
 
 OBS. The object of writing units under units, tens under tens, &c., is to 
 prevent mistakes which might occur from adding different orders to each other. 
 
 52. When the sum of a column does not exceed 9, it will be 
 noticed, we set it under the column added ; but if it exceeds 9, 
 we set the units or right Jiand figure under the column added, 
 and reserving the tens or left hand figure, add it to the next 
 column. In adding the last column on the left, we set down the 
 whole sum. 
 
 OBS. The process of reserving tJie tens, or left hand figure, and adding it to 
 the next column, is called carrying tens. 
 
 53. The principle of carrying may be illustrated in the follow- 
 ing manner. 
 
 Take, for instance, the last example, 7864 
 
 and adding as before, write the sum of 4952 
 
 each column in a separate line. Thus, 3273 
 
 the sum of the units' column is 9 units ; 9 sum of units, 
 
 the sum of the tens' column is 18 tens, or 18* " " tens. 
 
 1 hundred and 8 tens ; the sum of the 19** " " hund. 
 
 hundreds' column is 19 hundred, or 1 14*** " " thou. 
 
 thousand 9 hundred; the sum of the 16089 Amount. 
 
 Q.uicsT.-Why not add figures of different orders together 1 
 
ARTS. 51 55. J ADDITION. 33 
 
 thousands' column is 14 thousand. Now, adding these results 
 together as they stand, units to units, tens to tens, <fcc., the amount 
 is 16089 bushels, which is the same as in the solution above. 
 
 Thus, it is evident, when the sum of a column exceeds 9, the 
 right hand figure denotes units of the same order as the column 
 added, and the tens or left hand figure denotes units of the next 
 higher order. Hence, 
 
 The reason we carry the tens or left hand figure to the next 
 column, is because it is of the same order as the next column, and 
 figures of the same order must always be added togetlier. (Art. 51.) 
 
 OBS. 1. The reason for setting down the whale sum of the last or left hand 
 column, is because there are no figures in the next order to which the left 
 hand figure can be added. It is, in fact, carrying it to the next column. 
 
 2. From the preceding illustration it will also be seen, that the object of 
 beginning to add at the right hand is, that we may carry t/ie tens, as we pro- 
 ceed in the operation. 
 
 54. From the preceding illustrations and principles we de- 
 rive the following 
 
 GENERAL RULE FOR ADDITION. 
 
 I. Write the numbers to be added, under each other / 50 that 
 units may stand under units, tens under tens, &c. (Art. 51. Obs.) 
 
 II. Begin at the right hand, and add each column separately. 
 When the sum of a column does not exceed 9, write it under the 
 column ; but if the sum of a column exceeds 9, write the units' 
 figure under the column added, and carry the tens to the next 
 column. (Arts. 52, 53.) 
 
 III. Proceed in this manner through all the orders, and set down 
 the whole sum of the last or left hand column. (Art. 53. Obs.) 
 
 55. PROOF. Beginning at the top, add each column down- 
 wards, and if tJie second result is the same as the first, the work is 
 supposed to be right. 
 
 GUEST. 54. How do you write numbers to be added 7 Why place units under units, &c. 1 
 Where do you begin to add ? When the sum of a column does no exceed 9, what do you 
 do with it ? When it exceeds 9, how proceed 1 What is mea:t by carrying the tens 1 
 Why carry the tens to the next column? Why begin to add a the right hand ? What 
 do you do with the sum of the last column ? 55. How is additio t proved 1 
 
34 ADDITION. [SECT. II. 
 
 Note. The object of beginning at the top and adding downwards, is that 
 the figures may be taken in a different order from that in which they were 
 added before. The order being reversed, the presumption is, that any mistake 
 which may have been made will thus be detected ; for it can hardly be sup- 
 posed that two mistakes exactly equal will occur. 
 
 56, Second Method Cut off the bottom line, and find the 
 sum of the rest of the numbers ; then add this sum and the bot- 
 tom line together, and if the second result is the same as the first, 
 the work is supposed to be right. 
 
 Note. 1. This method of proof depends on the axiom, that the whole of a 
 quantity is equal to the sum of all its parts. (Ax. 11.) 
 
 2. The method of cutting off the top line, and afterwards adding it to the 
 sum of the others, is objectionable on account of adding the numbers in the 
 same order as they were added hi the solution. (Art. 55. Note.) 
 
 5*7. Third Method. From the amount, subtract all the given numbers but 
 one, and if the remainder is equal to the number not subtracted, the work may 
 be supposed to be right. 
 
 Note. This method supposes the pupil to be acquainted with subtraction 
 before he commences this work. It is placed here on account of the con- 
 venience of having all the methods of proving the rule together. 
 
 58. Fourth Method* Cast the 9s out of each of the given numbers sepa- 
 rately, and place each excess at the right of the number. Then cast the 9a 
 out of the sum of these excesses ; also cast the 9s out of the amount ; and if 
 these two excesses are equal, the work may be supposed to be right. 
 
 Note. 1. This mode of proof is based on a peculiar property of the number 9. 
 For its illustration and demonstration, see Art. 161. Prop. 14. 
 
 2. To cast the 9s out of a number, begin at the left hand, add the digits, 
 together, and, as soon as the sum is 9 or over, drop the 9, and add the remain- 
 der to the next digit, and so on. For example, to cast the 9s out of 4626357, 
 we proceed thus : 4 and 6 are 10 ; drop the 9 and add the 1 to the next figure. 
 1 and 2 are 3, and 6 are 9 ; drop the 9 as above. 3 and 5 are 8, and 7 are 
 15 ; dropping the 9, we have 6 remainder. 
 
 EXAMPLES FOR PRACTICE. 
 
 59. Ex. 1. A man paid 2468 dollars for his farm, 1645 dollars 
 for a house, 865 dollars for stock, and 467 dollars for tools; how- 
 much did he pay for the whole ? 
 
 2. A produce merchant bought 5 cargoes of corn ; the first con- 
 
 QTJEST. Note. Why add the columns downwards, instead of upwards ? Can addition 
 be proved by any other methods ? 
 
 * Wallis' Arithmetic, Oxford 165T. 
 
ARTS. 56-59.] ADDITION. 35 
 
 tained 6725 bushels, the second 7208, the third 5047, the fourth 
 12386, and the fifth 10391 bushels: how many bushels did he 
 buy? 
 
 3. A tavern-keeper bought six loads of hay which weighed as 
 follows: 1725 pounds, 2163 pounds, 1581 pounds, 1908 pounds, 
 2340 pounds, and 1879 pounds: what was the weight of the 
 whole ? 
 
 4. A man gave 5460 dollars to his oldest son, to the next 4065, 
 o the next 6750, to the next 8000, and to the youngest 7276 
 
 dollars : how much did he give to all ? 
 
 5. A merchant, on settling up his business, found he owed one 
 creditor 176 dollars, another 841 dollars, another 1356 dollars, 
 another 2370 dollars, another 840 dollars : what was the amount 
 of his debts ? 
 
 6. The state of Maine contains 32400 square miles ; New Hamp- 
 shire, 9500; Vermont, 9700; Massachusetts, 7800; Rhode Island, 
 1251 ; and Connecticut, 4789 : how many square miles are there 
 in the New England States ? 
 
 7. The state of New York contains 46220 square miles ; New- 
 Jersey, 7948 ; Pennsylvania, 46215 ; and Delaware, 2068 : how 
 many square miles are there in the Middle States ? 
 
 8. The state of Maryland contains 10755 square miles; Virginia, 
 65700; North Carolina, 51632 ; South Carolina, 31565 ; Georgia, 
 61683; Florida, 56336; Alabama, 54G$4 ; Mississippi, 49356; 
 Louisiana, 47413 ; and Texas, 100000 : how many square miles 
 are there in the Southern States ? 
 
 9. The state of Tennessee contains 41752; Kentucky, 40023; 
 Ohio, 40500 ; Michigan, 60537 ; Indiana, 35626 ; Illinois, 56506 ; 
 Missouri, 70050: Arkansas, 54617; Iowa, 173786; and Wiscon- 
 sin, 92930 ; how many square miles are there in the Western 
 States? 
 
 10. What is the whole number of square miles in the United 
 States ? 
 
 11. What is the sum of 75234+41015 + 19075+1764-88350 
 + 10040? 
 
 12. What is the sum of 250120+30402+7850+465000 + 
 10046+65045? 
 
36 ADDITION. [SECT. II. 
 
 13. What is the sum of 85046 + 90045+412260+125781 + 
 4060 + 273048? 
 
 14. What is the sum of 1500267+45085+4652 + 4780400 + 
 90276 + 89760841? 
 
 15. What is the sum of 45702125 + 67070420+670856 + 
 4230825 + 750642 + 8790845 ? 
 
 16. What is the sum of 825760842 + 35620476 + 7800490 f 
 467243 + 98371 + 6425 + 740 ? 
 
 17. What is the sum of 2503 + 37621+475290 + 1223729 + 
 10671840 + 275600312? 
 
 18. What is the sum of 463270+2500 + 7200342 + 10271 + 
 426345 + 6200705? 
 
 19. What is the sum of 80429 + 7562345 + 700100 + 85261798 
 +4000101 + 3007002? 
 
 20. What is the sum of 756 + 849+934+680 + 720+843 + 
 657689 + 989876498 + 8045685+807266780? 
 
 21. What is the sum of 6457 + 29301 + 82406 + 7589 + 63489 
 +101364+46745? 
 
 22. Add together 786, 840, 910, 403, 783, 650, 809, 670, 408, 
 310, and 652. 
 
 23. Add together 16075, 250763, 7561, 830654, 293106, 
 2537104, and 316725. 
 
 24. Add together 25^ 40, 751, 302, 75, 831, 26, 43, 621, 340, 
 and 510. 
 
 25. Add together 493742, 56710607, 23461, 400072, 6811004, 
 8999003, and 26501. 
 
 26. Add together 629405, 7629, 31000401, 263012, 1300512, 
 390217, and 13268. 
 
 27. Add together 286013, 4016702, 1971342, 6894680, 28945, 
 and 2624302. 
 
 28. Add together 460167, 296345, 84634123,64205, 9673108, 
 and 1931456. 
 
 29. Add together 432678902, 310046734, 2167005, 327861 
 asid 293000428. 
 
ART. 60.J 
 
 ADDITION. 
 
 eOUNTING-ROOM EXERCISES. 
 
 GO, To the accountant as well as the mathematician, accuracy 
 and expertness in adding, are indispensable. These attainments can 
 be acquired only by frequent exercises in footing up long columns 
 of figures. 
 
 Note. 1. Instead of saying 4 and 8 are 12, and 2 arc 14, 
 and 7 are 21, and 4 are 25, &c., a skilful accountant, per- 
 forming the addition at a glance, simply pronounces the 
 results. Thus, four, twelve, twenty-one, thirty-one, (4-J-6 
 = 10,) thirty-seven, forty-seven, (7-f-3=;lO,) fifty-two. 
 
 2. When two or three figures taken together make 10, as 
 6 and 4, or 2, 3, and 5, &c., it accelerates the process to 
 add their sum at once. A little practice will enable the 
 student to run up a long column of figures with as much 
 facility almost as he can count. 
 
 3. When the columns are long, accountants sometimes 
 set the figure to be carried below the other figure under the 
 column added. Thus, the sum of the first column in the 
 example above being 52, set the 5 (the figure carried) be- 
 
 .ow the 2. The sum of the second column being 48, set the 4 below the 8, &c. 
 This method saves much time in reviewing an operation, and also enables us, 
 when interrupted, to resume the process where we left off. 
 
 Required the amount of each of the following e samples : 
 
 31. 
 
 Dollars. 
 2425 
 
 30. 
 
 
 86015 
 
 
 251 03 
 
 
 85057 
 
 
 12236 
 
 
 43026 
 
 
 67084 
 
 
 21167 
 
 
 54042 
 
 
 42158 
 
 
 240.34 
 
 
 459982 
 
 Aiis. 
 
 3045 
 
 Car. 
 
 3282 
 2793 
 2354 
 4262 
 9158 
 2653 
 3424 
 12G6 
 8742 
 2126 
 5387 
 
 Ans. 47872 
 Car. 465 
 
 32. 
 
 Dollars. 
 46,519 
 32,271 
 17,436 
 81,587 
 28,333 
 52,745 
 23,052 
 20,158 
 71,232 
 39,464 
 18,643 
 42,027 
 73,235 
 24,103 
 
 33. 
 
 Yards. 
 607,253 
 232,012 
 211,849 
 380,436 
 578,551 
 231,349 
 145,763 
 605,037 
 760,155 
 357,676 
 544,844 
 276,232 
 803,383 
 725,918 
 
 34. 
 
 Pounds. 
 421,536 
 310,101 
 797,019 
 233,680 
 124,402 
 255,353 
 852,057 
 618,041 
 100,266 
 971,134 
 536,920 
 703,352 
 420,503 
 312675 
 
3S ADDITION. [SECT. II. 
 
 85. 
 
 36. 
 
 37. 
 
 38. 
 
 318,037 
 
 460,375 
 
 963,172 
 
 849,652 
 
 272,465 
 
 841,681 
 
 300,725 
 
 361,728 
 
 530,634 
 
 239,724 
 
 463,248 
 
 412,381 
 
 109,871 4 
 
 763,256 
 
 721,003 
 
 635,403 
 
 693,036 
 
 437,891 
 
 387,356 
 
 872,545 
 
 764,543 
 
 825,432 
 
 241,653 
 
 406,223 
 
 323,638 
 
 285,678 
 
 603,280 
 
 294,867 
 
 428,432 
 
 310,720 
 
 532,176 
 
 811,236 
 
 389,763 
 
 403,521 
 
 278,321 
 
 576,037 
 
 210,045 
 
 687,489 
 
 829,248 
 
 213,744 
 
 760,806 
 
 324,061 
 
 171,320 
 
 764,368 
 
 636,215 
 
 530,724 
 
 206,782 
 
 305,216 
 
 253,734 
 
 623,452 
 
 461,027 
 
 436,720 
 
 251,600 
 
 487,638 
 
 589,203 
 
 823,284 
 
 575,453 
 
 290,731 
 
 248,639 
 
 217,436 
 
 807,720 
 
 803,256 
 
 730,461 
 
 592,301 
 
 930,046 
 
 731,463 
 
 672,398 
 
 243,762 
 
 174,173 
 
 379,574 
 
 246,175 
 
 731,445 
 
 626,245 
 
 823,156 
 
 928,340 
 
 429,374 
 
 342,734 
 
 928,348 
 
 731,629 
 
 684,569 
 
 61. Accountants often acquire the habit of adding two col- 
 umns of figures at a time. The power of rapid addition is easily 
 acquired, and is well worthy the attention of the student. The 
 following examples will illustrate the principle. 
 
 39. What is the sum of 312817+527236 + 141625+462415 
 + 251818 + 234112? 
 
 Operation. 
 
 Taking the two right hand columns, we 312817 
 
 say, 12 and 18 are 30, and 15 are 4C and 527236 
 
 25 are 70, and 36 are 106, and 17 are 123. 141625 
 
 Set down the 23 under the columns added, 462415 
 
 and carry the 1 or left hand figure to the 251818 
 
 column of hundreds. Proceed in the same 234112 
 
 manner with the other columns. Ans> 1930023 
 
ART. 61.] 
 
 ADDITION. 
 
 (41.) (42.) (43.) 
 
 21 
 30 
 11 
 13 
 20 
 15 
 34 
 18 
 12 
 17 
 23 
 
 22 
 13 
 40 
 25 
 14 
 11 
 33 
 45 
 12 
 20 
 18 
 
 44 
 20 
 25 
 17 
 50 
 14 
 16 
 28 
 11 
 14 
 37 
 
 (44.) 
 1325 
 1510 
 1314 
 3141 
 1016 
 2233 
 1224 
 2415 
 1830 
 1814 
 1621 
 
 (45.) 
 2610 
 1511 
 1021 
 1115 
 1513 
 4020 
 1316 
 1233 
 2515 
 1718 
 2142 
 
 (46.) 
 344235 
 402321 
 141511 
 201250 
 154036 
 132212 
 181714 
 213025 
 111817 
 161518 
 432733 
 
 What was the amount of exports and imports of the United 
 States in 1840, and of shipping in 1842 ? 
 
 
 (47.) 
 
 (48.) 
 
 (49.) 
 
 States. 
 
 Exports. 
 
 Imports. 
 
 Shipping. 
 
 Maine, . Dolls 
 
 . 1,018,269 
 
 Dolls. 628,762 
 
 T. 281,930 
 
 N. Hampshire, 
 
 20,979 
 
 114,647 
 
 23,921 
 
 Vermont, 
 
 305,150 
 
 404,617 
 
 4,343 
 
 Massachusetts, 
 
 10,186,261 
 
 16,513,858 
 
 494,895 
 
 Rhode Island, 
 
 206,989 
 
 274,534 
 
 47,243 
 
 Connecticut, 
 
 518,210 
 
 277,072 
 
 67,749 
 
 New York, 
 
 34,264,080 
 
 60,440,750 
 
 518,133 
 
 "New Jersey, 
 
 16,076 
 
 19,209 
 
 60,742 
 
 Pennsylvania, 
 
 6,820,145 
 
 8,464,882 
 
 113,569 
 
 Delaware, 
 
 37,001 
 
 802 
 
 10,396 
 
 Maryland, 
 
 5,768,768 
 
 4,910,746 
 
 106,856 
 
 Dist. of Columbia, 
 
 753,923 
 
 119,852 
 
 17,711 
 
 Virginia, 
 
 4,778,220 
 
 545,085 
 
 47,536 
 
 North Carolina, 
 
 387,484 
 
 252,532 
 
 31,682 
 
 South Carolina, 
 
 10,036,769 
 
 2,058,870 
 
 23,469 
 
 Georgia, 
 
 6,862,959 
 
 491,428 
 
 16,536 
 
 Alabama, 
 
 12,854,694 
 
 574,651 
 
 14,577 
 
 Louisiana, 
 
 34,236,936 
 
 10,673,190 
 
 144,128 
 
 Ohio, 
 
 991,954 
 
 4,915 
 
 24,830 
 
 Michigan, 
 
 162,229 
 
 148,610 
 
 12,323 
 
 Florida, 
 
 1,858,850 
 
 190,728 
 
 7,288 
 
40 ADDITION*. [SECT. II. 
 
 50. The appropriations of the Government of the United States, 
 for 1847, were as follows : for the Civil and Diplomatic expenses 
 4,442,790 dolls.; for the Army and Volunteers 32,178,401 
 dolls.; for the Navy "9,307,958 dolls.; for the Post Office De- 
 partment 4,145,400 dolls.; for the Indian Department 1,364,204 
 dolls.; for the Military Academy 124,906 dolls.; for building 
 Steam Ships 1,000,000 dolls. ; for Revolutionary and other Pen- 
 sions 1,358,700 dolls.; for concluding Peace with Mexico 3,000, 
 
 >00 dolls.; for Light Houses 518,830 dolls.; Miscellaneous 
 540.243 dolls. What was the amount of all the appropriations? 
 
 62. It may sometimes be convenient for the learner, as well 
 as gratifying to his curiosity, to be able to add numbers expressed 
 by the Roman Nutation. 
 
 51. A man paid MDCCCLXXXIII dollars for a farm, 
 DCCXXIIII dollars for stock, and CCCLXVIIII dollars for 
 tools : how much did he pay for all ? 
 
 Beginning at the right hand, we proceed thus : Operation. 
 
 four Is and four Is are eight, and three Is make MDCCCLXXXIII dolls, 
 eleven, which is equal to two Vs and I. We set DCCXXIIII dolls, 
 
 down the I, and adding the two Vs to one V CCCLXVIIII dolls, 
 
 makes fifteen, which is equal to X and V. Set- MMDCCCCLXXvT dolls, 
 ting down the V, we count in the X with the 
 
 other Xs, and find they make seven Xs or seventy, which is expressed by L 
 and XX. We set down the two Xs, and adding the L to the other Ls, il 
 makes three Ls, or one hundred and fifty, which is expressed by C and L. 
 Setting down the L, and counting the C with the other Cs, we have nine Cs 
 or nine hundred, which is expressed by D and CCCC. We set down the four 
 Cs, and counting the D with the other Ds, it makes three Ds or fifteen hun- 
 dred, which is expressed by M and D. We set down the D, and adding the 
 M to the other M, we have two Ms, which we set down on the left of the other 
 letters. Hence, 
 
 63. To add numbers expressed by the Roman Notation. 
 
 Beginning at the right hand, count all the letters of each kind to- 
 gether ; set down the result, and carry on the principle that five Is 
 make one V ; two Vs, one X ; five Xs, one L, <fec. . 
 
 OBS. The teacher can extend the exercises in the Roman Notation as fa. 
 as he may deem it expedient. A single example is sufficient to illustrate the 
 principle, and to show that the Roman is greatly inferior to the Arabic method 
 in its adaptation to business calculations. 
 
ARTS. 62-66.] SUBTRACTION. 41 
 
 SECTION III. 
 SUBTRACTION. 
 
 ART. 65. Ex. 1. A merchant bought 3*7 barrels of flour, and 
 afterwards sold 12 of them : how many barrels had he left? 
 Solution. 12 barrels from 37 barrels leave 25 barrels. 
 
 Am. 25 barrels. 
 
 OBS. It will be perceived, that the object in this example, is to find the dif- 
 ference between two numbers. 
 
 66. The -process of finding the difference between two numbers 
 is called SUBTRACTION. 
 
 The difference, or the answer to the question, is called the 
 
 Remainder. 
 
 OBS. 1. The number to be subtracted is sometimes called the subtrahend^ 
 and the number from which it is subtracted, the minuend. 
 
 2. Subtraction, it will be perceived, is the reverse of addition. Addition 
 unites two or more numbers into one single number ; subtraction, on the other 
 hand, separates a number into two parts. 
 
 3. When the given numbers are of the same denomination , the operation is 
 called Simple Subtraction. (Art. 50. Obs.) 
 
 Ex. 2. What is the difference between 5364 and 9387? 
 
 Write the less number under the greater, Operation. 
 units under units, tens under tens, &c. Then, 9387 
 
 beginning at the right hand, proceed thus : 5364 
 
 4 units frtom 7 units leave 3 units. Write 4023 Rem. 
 
 the 3 in the units' place, under the figure subtracted. 6 tens 
 from 8 tens leave 2 tens ; set the 2 in tens' place. 3 hundred 
 from 3 hundred leave hundred ; we therefore write a cipher in 
 hundreds' place. 5 thousand from 9 thousand leave 4 thousand; 
 set the 4 in the thousands' place. The answer is 4023. 
 
 QUEST. 66. What is subtraction? What is the difference or answer called 1 Ota 
 What is the number to be subtracted sometimes called ? The number from which it is 
 subtracted ? Of what is subtraction the reverse ? When the given numbers are of th 
 same denomination, what is the operation called 7 
 
42 SUBTRACTION. [SECT. III. 
 
 67. It will be observed, that we subtract units from units, 
 tens from tens, &c. ; that is, we subtract figures of the same order 
 from each other. This is done for the same reason that we add 
 figures of the same order to each other. (Art. 51.) 
 
 OBS. The less number is written under the greater, simply for convenience 
 in subtracting ; and units are placed under units, tens under tens, &c., to avoid 
 mistakes which might occur from taking different orders from each other. 
 
 68. It often happens that a figure in the lower number i 
 Icurger than that above it, and consequently cannot be taken 
 from it. 
 
 Ex. 3. What is the difference between 94 and 56 ? 
 
 Analytic solution. It is manifest that we cannot take 6 
 94=80 + 14 units from 4 units, for 6 is larger than 4. 
 56 = 50+6 To obviate this difficulty, we may take 
 
 Rem. 38=30+8 Iten from the 9 tens, and uniting it 
 
 with the 4 units, the upper number will become 8 tens and 14 
 units, or 80 + 14. Separating the lower number into the parts of 
 which it is composed, it becomes 5 tens and 6 units, or 50+6. 
 Now, subtracting as in the last example, 6 from 14 leaves 8, 50 
 from 80 leaves 30. The answer is 30 + 8, or 38. Or, we may 
 simply take 1 ten from the 9 tens, and adding it, mentally, to the 
 4 units, say 6 from 14 leaves 8 ; set the 8 under the figure sub- 
 tracted. Then, having taken 1 from the 9 tens, we have but 8 
 left, and 5 from 8 leaves 3. The answer is 38. 
 
 PROOF. 38 + 56=94; that is, the sum of the remainder and 
 smaller number being equal to the larger, the answer is right. 
 Hence, 
 
 69. When a figure in the lower number is larger than that 
 above it ; take 1 from the next higher order in the upper number, 
 and add it to the upper figure ; from the sum subtract the lower 
 figure, and diminishing the next upper figure by 1, proceed as 
 before. 
 
 OBS. 1. The process of taking one from the next higher order and adding it 
 to the figure from which the subtraction is to be made, is called borrmoing ten. 
 It is the reverse of carrying. 
 
 QUEST. 67. What orders of figures do you subtract from each other? Why no ftA- 
 tract different orders from each other ? 
 
ARTS. 67-71.] SUBTR ACTION. 43 
 
 2. This method of borrowu.g, it v^ill be seen, does not ajfea the difference 
 between the two given numbers ; for, it is simply transposing a part of one 
 order to another order in the same number, which, it is obvious, will neither 
 increase nor diminish its value. 
 
 3. It may be asked, how can we take one from the figure in the next higher 
 order, when that figure is a ciplier ? How can nothing lend anything, and how 
 can nothing be diminished by one ? The explanation of this apparent contra- 
 diction is this : when the next figure is a cipher, we go to the next higher 
 column still, and take one, which, added to the figure in the next lower or. ler, 
 makes ten ; we then take one from the ten and add it to the upper figure, and 
 proceed as before. 
 
 7 O. There is another method of borrowing, or rather of pay- 
 ing, which, though perhaps less philosophical than the preceding, 
 is more convenient in practice, especially when the figures in the 
 next higher orders are ciphers. Thus, in the last example, adding 
 10 to the upper figure, it becomes 14, and 6 from 14 leaves 8. 
 Set down the 8 as before. JSFow, instead of diminishing the next 
 upper figure by 1, if we add 1 to the next figure in the lower 
 number it becomes 6 tens ; and 6 from 9 leaves 3, which is the 
 same as 5 from 8. The answer is 38, the same as before. Hence/ 
 71. When a figure in the lower number is larger than that 
 above it, add 10 to the upper figure, and to compensate this, add 
 1 to the .next left hand figure in the lower number. 
 
 OBS. 1. This method of borrowing depends on the self-evident principle, 
 that if any two numbers are equally increased, their difference will not be 
 altered. Thj^the two given numbers are equally increased by this process, 
 is evident i|Hkthe fact that the 1 added to the lower number is of the next 
 superior order^o the 10 added to the upper number, and is therefore equal 
 to it. (Art. 35.) 
 
 2. The reason that we borrow 10, instead of 8, or 12, or any other number, 
 is because the radix, or ratio of increase, in the Arabic notation, is 10. (Art. 
 30.) If the radix of the system were 8, it would be necessary to borrow 8; 
 if 12. it would be necessary to borrow 12, &c. 
 
 3. On account of borrowing, the learner will perceive it is always necessary 
 to begin to subtract at the right hand. 
 
 Ex. 4. A man bought a house for 23006 dollars, and sold it for 
 21128 dollars : how much did he lose by his bargain ? 
 
 Operation. Proof. 
 
 Cost 23006 dolls. 21128 Less number. 
 
 Rec'd. 21128 dolls. 1878 Remainder. 
 
 Ans. 1878 dolls. 23006 Larger number. 
 
 T.H. Q 
 
44 SUBTRACTION. [SECT. Ill, 
 
 7 2. From the preceding illustrations and principles we derive 
 the following 
 
 GENERAL RULE FOR SUBTRACTION, 
 
 I. Write the less number under the greater, so tliat units may 
 stand under units, tens under tens, <&c. (Art. 67. Obs.) 
 
 ] I. Beginning at the right hand, subtract each figure in the lower 
 number from the figure above it, and set tJie remainder directly 
 wider the figure subtracted. (Art. 71. Obs. 3.) 
 
 III. When a figure in the lower number is larger than that above 
 it, add 10 tt, the upper figure ; then subtract as before, and add 1 
 to tlie next figure in the lower number, or consider the next upper 
 figure 1 less than it is. (Arts. 69, 70. Obs. 1, 2.) 
 
 73. PROOF. Add the remainder to the smaller number ; and 
 if the sum is equal to the larger number, the work is right. 
 
 OBS. This method of proof depends upon the principle, that the difference 
 between two numbers being added to the less, the sum must be equal to the 
 greater. For, the difference and the less number are the two parts into which 
 the greater is separated, and the whole of a quantity is equal to the sum of all 
 its parts. (Ax. 11.) 
 
 7 4. Second Method. Subtract the remainder from the greater 
 of the two given numbers ; and if the difference is equal to the 
 less number, the work is right. 
 
 75. Third Method. Cast the 9s out of the larger number, and place the 
 excess at the right. Next, cast the 9s out of the smaller number, and also 
 out of the remainder; then cast the 9s out of the sum of these two excesses, 
 and if this last excess is the same as the excess of the larger number, the work 
 may be supposed to be right. Thus, 
 
 Ex. 5. From 7843 Excess of 9s in the greater number is 4 
 Take 5675 " less " is 5 > 
 
 Rcm. 21G8 " " " remainder is 8 \ Now, 8-}-5=l3, 
 and the excess of 9s in 13 is 4, the same as that of the greater number. 
 
 QUEST. 72. How do you write numbers for subtraction ? Why write the less number 
 under the greater ? Why place units under units, &c. 1 Where do you begin to subtract ? 
 When a figure in the lower line is larger than that above it, how do you proceed 1 What 
 is meant by borrowing ten ? How many methods of borrowing are mentioned ? Illustrate 
 the first method upon the black-board. How does it appear that this method of borrowing 
 does not attest the difference between the two given numbers? Explain the second me- 
 thod. Upon what principle does this method depend 1 Why do yoi borrow 10, instead 
 of H, or 12, or any other number? Why do you nei;in to subtract at the right hand 1 
 73. How is subtraction proved ? O6.<r. Upon what principle does thi? method of proof de- 
 pend 1 Can subtraction be proved by any other methods ? 
 
ARTS. 7276.] SUBTRACTION. 45 
 
 Note. This method of proof depends on the same property of the number 
 9, as that in addition. (Art. 58. Note.) For, since the sura of the smaller 
 number and remainder is equal to the larger number, it follows that the 
 excess of 9s in the larger number must be equal to the excess of 9s in the 
 remainder and smaller number together. 
 
 EXAMPLES FOB PRACTICE. 
 
 76. Ex. 1. A merchant bought a ship for 35270 dollars, and 
 r >ld it for 42365 dollars : how much did he make by his bargain ? 
 
 2. A miller bought 46235 bushels of wheat, and ground 17251 
 bushels of it : how many bushels had he left ? 
 
 3. A speculator laid out 50000 dollars in wild land, and after- 
 wards sold it at a loss of 19046 dollars : how much did he get for 
 his land ? 
 
 4. A man owning a block of buildings worth 155265 dollars, 
 keeps it insured for 109240 dollars : how much would he lose in 
 case the buildings should be destroyed by fire ? 
 
 5. The distance from the Earth to the Sun is 95000000 of 
 miles ; the distance of Mercury is only 37000000 : how far is 
 Mercury from the Earth ? 
 
 6. The imports of Massachusetts in 1840, were 16,513,858 
 dollars, the exports were 10,186,261 dollars: what was the ex- 
 cess of her imports over her exports ? 
 
 7. The imports of New York in 1840, were 60,440,750 dol- 
 lars, the exports were 34,264,080 dollars : what was the excess 
 of her imports over her exports ? 
 
 8. The imports of Pennsylvania in 1840, were 8,464,882 dol- 
 lars, the exports were 6,820,145 dollars : what was the excess of 
 her imports over her exports ? 
 
 9. The imports of South Carolina in 1840, were 2,058,870 
 dollars, the exports were 10,036,769 dollars : what was tLe ex- 
 cess of her exports over her imports ? 
 
 10. The imports of Alabama in 1840, were 574,651 dollars, 
 the exports were 12,854,694 dollars : what was the excess of her 
 exports over her imports ? 
 
 11. The imports of Louisiana in 1840, were 10,673,190 dol- 
 lars, the exports were 34,236,936 dollars : what was the excess' 
 of her exports over her imports ? 
 
46 SUBTRACTION. [SECT. III. 
 
 12. The tonnage of the United States in 1842, was 2069857, 
 w 1846 it was 2500000 : what was the increase in 4 years? 
 
 13. 14. 15. 
 
 From 253760 3856031 54903670 
 
 Take 104523 462702 504089 
 
 16. 98761021050671. 28. 10000000999999. 
 
 17. 40067235001. 29. 99999999100000. 
 
 18. 36019001000000. 30. 83567000438567, 
 
 19. 5317004 3565. 31. 40600056 7632. 
 
 20. 1000000 456321. 32. 56409250 1057245. 
 
 21. 203502427040. 33. 20030000 72534. 
 
 22. 45563075460001. 34. 831756215256360. 
 
 23. 67030001300452. 35. 70301604250041. 
 
 24. 73256300436020. 36. 600503766849005. 
 
 25. 56037431 735671. 37. 34200591 8888888. 
 
 26. 80200430250. 38. 87035762 753017. 
 
 27. 96531768873625. 39. 95246300 9438675. 
 
 40. From 6764+3764 take 6500+2430. 
 
 41. From 2890+8407 take 4251+3042. 
 
 42. From 7395+4036 take 8297+1750. 
 
 43. From 8404+7296 take 3201 1562. 
 
 44. From 6008+9270 take 51362352. 
 
 45. From 9234 + 6850 take 93204783. 
 
 46. From 85642573 take 44311735. 
 
 47. From 72845362 take 60455729. 
 
 48. From 95614680 take 73526178. 
 
 49. From 8630 1763 take 2460 + 1743. 
 
 50. From 75612846 take 1734 + 2056. 
 
 51. From 96873401 take 3021 + 1754. 
 
 52. A man having 55000 dollars, paid 7520 dollars for a house,, 
 8260 dollars for furniture, 2375 dollars for a library, and in- 
 rested the balance in bank stock : how much stock did he buy ? 
 
 53. A gentleman worth 163250 dollars, bequeathed 15200 dol- 
 lars apiece to his two sons, 16500 dollars to his daughter, and to 
 his wife as much as to his three children, and the remainder to a 
 hospital ; how much did his wife receive, and how much the hos 
 pital ? 
 
ARTS. 76, 77.] SUBTRACTION. 47 
 
 54. A man bought three farms : for the first he paid 5260 dol- 
 lars, for the second 3585, and for the third us much as for the 
 first two. He afterwards sold them all for 15280 dollars: did he 
 make or lose by the operation ; and how much ? 
 
 55. What number is that, to which 3425 being added, the sum 
 will be 175250? 
 
 56. A man being asked how much he was worth, replied, if 
 you will give me 325263 dollars, I shall have two millions of dol- 
 lars : how much was he worth ? 
 
 57. A jockey gave 150 dollars for a horse, and meeting an ac- 
 quaintance swapped with him, giving 37 dollars to boot ; meeting 
 another, he swapped and received 28 dollars to boot ; he finally 
 swapped again and gave 78 dollars to boot, and then sold his last 
 horse for 140 dollars : how much did he lose by all his bargains ? 
 
 58. A speculator gained 3560 dollars, and afterwards lost 2500 
 dollars; at another time he gained 6283 dollars, and then lost 
 3450 dollars : how much more did he gain than lose ? 
 
 59. A man bought a house for MDCCCCXXXVII dollars, and 
 sold it for DCXVIIII dollars less than he gave: how much did 
 he sell it for ? 
 
 We perceive that the IIII in the lower number Operation. 
 
 cannot be taken from II in the upper number; MDCCCCXXXVII dolls, 
 we therefore borrow a V, which added to the II, DCXVIIII dolls, 
 
 makes IIIIIII ; then IIII from IIIIIII, leaves Ans. MCCCXVIII dolls. 
 Ill, which we set down. Now since we have 
 
 borrowed the V in the upper number, there are no Vs left from which we can 
 take the V in the lower number. We must therefore borrow an X ; but X ia 
 equal to VV; and V from VV leaves V, which we set down. Having bor- 
 rowed an X from the upper number, there are but XX left, and X from XX 
 leaves X. C from CCCC leaves CCC. D from D leaves nothing. And 
 nothing from M leaves M. Hence, 
 
 77. To subtract numbers expressed by the Roman Notation. 
 
 Write the less number under Hie greater ; then, beginning at the right hand, 
 take the number in the lower line from that expressed by the same tetters in the 
 upper line, and set the remainder below. If the number in the lower line it 
 larger than that expressed by the same letters in the upper line, bwrmt) a letter 
 next hig/tcr and add it to the number in the upper line ; then subtract as bcfcre, 
 observing to pay when you borrow as in subtraction of figures. (Art. 72.) 
 
 OBS. Other examples expressed by the Roman Notation, can be added by 
 the teacher, if deemed expedient. 
 
48 MULTIPLICATION | SECT. IV 
 
 SECTION IV. 
 
 MULTIPLICATION. 
 
 ARTT. 79* Ex. 1. What will 3 melons cost, at 15 cents apiece 
 
 Analysis. If 1 melon costs 15 cents, 3 melons will cost 3 times 
 15 cents; and 3 times 15 cents are 45 cents. Ans. 45 cents. 
 
 2. What will 4 sleighs cost, at 21 dollars apiece ? 
 
 Analysis. Reasoning as before, if 1 sleigh costs 21 dollars, 4 
 sleighs will cost 4 times as much; and 4 times 21 dollars are 
 84 dollars. Ans. 84 dollars. 
 
 OBS. It is obvious that 3 times 15 cents is the same as 15 cents-f-15 cents 
 -{-15 cents, or 15 cents added to itself 3 times ; and 4 times 21 dollars is the 
 same as 21 dolls.-f-21 dolls.-j-21 dolls.-f 21 dolls., or 21 dollars added to itself 
 4 times. 
 
 80. Thin repeated addition of a number or quantity to itself, is 
 called MULTIPLICATION. 
 
 The number to be repeated, or multiplied, is called the Multi- 
 plicand. 
 
 The number by which we multiply, is called the multiplier; 
 and shows Iww many times the multiplicand is to be repeated. 
 
 The number produced, or the answer to the question, is called 
 the product. Thus, when we say, 8 times 12 are 96, 8 is the 
 multiplier, 12 the multiplicand, and 96 the product. 
 
 81. The multiplier and multiplicand together are often called 
 factors, because they make or produce the product. 
 
 OBS. 1. The term factor is derived from a Latin word which signifies a 
 tgcnt, a doer, or prodiiccr. 
 
 2. When the multiplicand denotes things of one denomination only, the ope- 
 ration is called Simple Multiplication. 
 
 GUEST. 80. What is multiplication 7 What is the number to be repeated called? 
 What the number by which we multiply? What does the multiplier show! What la 
 the number produced called? 81. What are the multiplicand and multiplier togetbei 
 vailed 1 Why 1 Obs. What does the term factor signify 1 
 
. 79~82.| 
 
 MULTIPLICATION. 
 
 49 
 
 MULTIPLICATION TABLE. 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6| 7 
 
 S 
 
 9i 10 
 
 11 12 
 
 13 
 
 14 
 
 15 
 
 lo 17 
 
 18 
 36 
 
 19 
 
 _20 
 40 
 
 _|_4 
 
 6 
 
 8 
 
 10 
 
 12 
 
 14 
 
 16 
 
 18 20 
 
 22 
 
 24 
 
 26 
 
 28 
 
 30 
 
 32 
 
 31 
 
 38 
 
 9 
 
 12 
 
 15 
 
 18 
 
 21 
 
 24 
 
 27 
 
 3J 
 
 33 
 
 3:1 
 
 39 
 
 42 
 
 45 
 
 4* 
 
 51 
 
 54 
 
 5; 
 
 60 
 
 4 
 
 
 12 
 
 16 
 
 21 
 
 24 
 
 28| 32 
 
 36 
 
 40 
 
 44 
 
 48 
 
 52 
 
 56 
 
 60 
 
 64 
 
 68 
 
 72 
 
 76 
 
 80 
 
 loo 
 
 5 
 
 U 
 
 15 2~J 
 
 25 
 
 30 
 
 35 
 
 1 . 
 
 45 
 
 50 
 
 55 
 
 6U 
 
 i i."; 
 
 70 
 
 75 
 
 so 
 
 85 
 
 9d 
 
 95 
 
 r> 
 
 12 
 
 18 1 24 
 
 30 
 
 3J 
 
 42 1 48 
 
 54 
 
 60 
 
 66 
 
 72 
 
 78 
 
 HI 
 
 90 
 
 96 
 
 102 
 
 108 
 
 II4J120 
 
 7 
 8 
 
 11 
 16 
 
 211 28 
 
 35 
 
 42 
 
 49 
 
 56 
 
 63 
 
 70 
 
 77 
 
 HI 
 
 91 
 
 98 
 
 105 
 
 Ii2 119 
 
 126 
 
 133 
 
 140 
 
 24 
 
 32 
 
 40 
 
 48 
 
 56 
 
 64 
 
 72 
 
 H ! 
 
 88 
 
 96 
 
 104 
 
 112 
 
 120 128 136 
 
 144 
 
 1 52 
 
 160 
 
 9| 18 
 
 27 
 
 :! ; 
 
 45 
 
 54 
 
 63 
 
 ' - 
 
 81 
 
 g 
 
 9d 
 
 108 
 
 117 
 
 12o 135 I44JI53 
 
 162;17l|l8- 
 
 10 
 
 20 
 
 30 
 
 1 1 
 
 5!) 
 
 60 
 
 70 
 
 BO 
 
 90 
 
 1 .0 
 
 110 
 
 120il3;) 
 
 140 
 
 150 1160 '170 
 
 180 UK) 20L 
 
 11 
 
 72 
 
 *>-> 
 
 ** 1 
 
 33 
 
 44 
 
 55 
 
 I'M) 
 
 77 88 
 
 99 
 
 111! 
 
 121 132 143 
 
 154 MS 17(i J87|l!H 
 
 20! I 
 
 220 
 
 36 
 
 48 
 
 60 
 
 72 
 
 84 
 
 96 
 
 108! 12) 
 
 1321144 156 
 
 16S 
 
 180 1112 204 
 
 2ifii2-J8 
 
 240 
 
 13 
 
 26 
 
 39 
 
 52 
 
 6.1 
 
 78 
 
 yi 
 
 I04|117|13.j 
 
 143 15 ij 109 182 
 
 l:'5 : 2o8'221 
 
 234^247 
 
 Si 
 
 14 
 
 28| 42 
 
 56 
 
 7,1 
 
 84 
 
 U8|ll2!l26|!4!) 
 
 1 54 1 1 1 18 : 1 82 i I'.i: i 1 2 1 ' 224 \ 2:<S 
 
 252 26(i 
 
 15 
 16 
 
 31)1 45 
 
 60' 75 
 
 5)0| 105 
 
 120; 135' 150 1-.5 180 195 210 225 2401255 
 
 2 70 ! 285 
 
 300 
 
 32 48 
 
 64 1 80 
 
 9> 
 
 112 
 
 I2SJ144 I6o|l7<>jl!)2 2(Hi224i24 25!>!2;2 
 
 288; 304 
 
 320 
 
 17 
 
 34 
 
 51 
 
 66 85 
 
 102 
 
 119 
 
 136153 170 
 
 1 87 1 -204 22 1 238 ; 255 272 ' 2-<! 
 
 :M)t),:i23 
 
 340 
 3~iiO 
 
 -181 36 
 
 51 
 
 72 
 
 90 
 
 108 
 
 l~2ii 
 
 144! 1(52 IriO!l'.)8 21(5 234 252 2iO 288,306 
 
 324 342 
 
 19 38 
 
 57 
 
 7;;i 95 
 
 114 
 
 L33 
 
 152 I'll 190 '209 228 247; 2' 16 285 1 304 
 
 323 342 3til 
 
 380 
 
 2~jl 40i 60 1 80ilOJ 120 
 
 140 
 
 160il80 200 1 2 20, 240 260 280 1 300 320 
 
 340,3GO;380J4UO 
 
 Note. This Table was invented by Pythagar-as, and is therefore sometimes 
 called the Pythagorean Table. 
 
 The pupil will find assistance in learning the Multiplication Table ty ob 
 serving the following particulars. 
 
 1. The several results of multiplying by 10 are formed by simply adding a 
 cipher to the figure that is to be multiplied. Thus, 10 times 2 are 20, 10 times 
 
 3 are 30, &c. 
 
 2. The results of multiplying by 5 terminate in 5 and 0, alternately. Thus, 
 5 times 1 are 5, 5 times 2 are 10, 5 times 3 are 15, &c. 
 
 3. The first nine results of multiplying by 1 1 are formed by repeating the 
 figure to be multiplied. Thus, 11 times 2 are 22 ; 11 times 3 are 33, &c. 
 
 4. In the successive results of multiplying by 9, the right hand figure regu- 
 larly decreases by 1, and the left hand figure regularly increases by 1. Thus, 
 9 times 2 are 18; 9 times 3 are 27; 9 times 4 are 36, &c. 
 
 82. Multiplying by 1, is taking the multiplicand once: thus, 
 
 4 multiplied by 1 = 4. 
 
 Multiplying by 2, is taking the multiplicand twice : thus, 2 times 
 4, or 4+4 = 8. 
 
 Multiplying by 3, is taking the multiplicand three times : thus, 
 3 times 4, or 4+4 -f- 4 =12, &c. Hence, 
 
 QUEST. 82, What is it to multiply by 1 ? By 2 1 
 5 
 
 By 3 1 
 
50 MULTIPLICATION. [SfiCT. IV 
 
 Multiplying ly any whole number, is talcing the multiplicand as 
 many times, as there are units in the multiplier. 
 
 The application of this principle to fractional multipliers will 
 be illustrated under fractions. 
 
 OBS. 1. From the definition of multiplication, it is manifest that the product 
 is of the same kind or denomination as the multiplicand : for, repeating c num- 
 ber or quantity does not alter its nature. Thus, if we repeat dollars, they ara 
 still dollars ; if we repeat yards, they are still yards, &c. Consequently, if the 
 multiplicand is an abstract number, the product will be an abstract number; if 
 money, the product will be money ; if barrels, barrels, &c. 
 
 & Every multiplier is to be considered an abstract number. In familiar 
 *nguage it is sometimes said, that the price multiplied by the weight will give 
 the value of an article ; and it is often asked how much 25 cents multiplied by 
 i35 cents, &c., will produce. But these are abbreviated expressions,- and are 
 liable to convey an erroneous idea, or rather no idea at all. If taken literally, 
 they are absurd ; for multiplication is repeating a number or quantity a certain 
 number of times. Now to say that the price is repeated as many times as the 
 given quantity is lieavy, or that 25 cents are repeated 25 cents times, is non- 
 sense. But we can multiply the price of 1 pound by a number equal to the 
 number of pounds in the weight of the given article, and the product will be 
 the value of the article. We can also multiply 25 cents by the number 25 ; 
 that is, repeat 25 cents 25 times, and the product is 625 cents. Construed in 
 tljis manner, the multiplier becomes an abstract number, and the expressions 
 have a consistent meaning. 
 
 Ex. 3. What will 6 houses cost, at 2341 dollars apiece ? 
 
 Write the numbers on the slate as Operation. 
 
 in the margin, and beginning at the 2341 Multiplicand, 
 
 right hand, proceed thus : 6 times 1 6 Multiplier, 
 
 unit are 6 units ; write the 6 under the Ans. 14046 Dollars, 
 figure multiplied. 6 times 4 tens are 24 tens ; set the 4 or right 
 hand figure under the figure multiplied, and cany the 2 or left 
 hand figure "o the next product figure, as in addition. (Art. 52.) 
 6 times 3 hundreds, are 18 hundreds, and 2 to carry make 20 hun- 
 dreds ; set the under the figure multiplied, and carry the 2 to 
 the next product as before. 6 times 2 thousands are 12 thou- 
 sands, and 2 to carry make 14 thousands. Since there are no 
 
 QITKST. What is it to multiply by any whole number 7 Obs. Of what denomination is 
 the prodact? How does this appear 7 What must every multiplier be considered 7 Caj 
 you multiply by a given weight, a measure, or a sum of money ? 
 
ART. 83.] MULTIPLICATION. 51 
 
 more figures to be multiplied, set down the 14 in full as in addi- 
 tion. (Art. 53. Obs. 1.) The product is 14046 dollars. 
 
 83* The product of any two numbers will be the same, which- 
 ever factor is taken for the multiplier. Thus, 
 
 If an orchard contains 5 rows of trees, and ******* 
 each row has 7 trees, as represented by the ******* 
 stars in the margin, it is evident the whole ******* 
 umber of trees is equal either to the number ******* 
 of stars in a horizontal row repeated Jive times, ******* 
 or to the number of stars in a perpendicular row repeated seven 
 times, viz. 35. For, 7X5 = 35, also 5X7 = 35. 
 
 OBS. 1. It is more convenient and therefore customary to place the larger num 
 her for the multiplicand, and the smaller for the multiplier. Thus, it is easier 
 to multiply 8468946 by 3, than it is to multiply 3 by 8468946, but the product 
 would be the same. * 
 
 i* 
 
 Ex. 4. What will 237 coaches cost, at 675 dollars apiece? 
 
 Since it is not convenient to multi- Operation. 
 
 ply by 237 at once, we multiply first 675 Multiplicand, 
 
 by the 7 units, next by the 3 tens, 237 Multiplier, 
 
 then by the 2 hundreds, and place 4725 cost 7 coaches, 
 
 each result in a separate line, with 2025* cost 30 " 
 
 the first figure of each line directly 1350** cost 200 " 
 
 under that by which we multiply. 159975 cost 237 " 
 Finally, adding these results togeth- 
 er, units to units, &c., we have 159975 dollars, which is the whole 
 product required. (Ax. 1 1 .) 
 
 Note. When the multiplier contains more than one figure, the several pro- 
 ducts of the multiplicand into tne Keparate figures of the multiplier, are called 
 partial products. 
 
 OBS. 2. The reason for placing the first figure of the several partial products 
 undsr the figure by which we multiply, is to bring the same orders under each 
 other, and thus prevent mistakes in adding them together. (Art. 51.) 
 
 3. The several partial products are added together for the obvious purpose 
 of finding the whole product or answer required. (Ax. 11.) 
 
 QUEST. 83. Does it make any difference with the result, which of the given 
 i taKen for the multiplier 1 Obs, Which is usually taken ? Why 1 
 3* 
 
52 MULTIPLICATION. . [SECT. IV. 
 
 84* The principle of carrying the tens in multiplication is the 
 same as in addition, and may be illustrated in a similar manner. 
 (Art. 53.) Thus, 
 
 Ex. 5. 9382 Mult'd. Or, separating the multiplicand into 
 
 7 Mult'r. the orders of which it is composed, 
 14= units, 9382 = 9000 + 300 + 80 + 2, 
 
 56*=tens, and 9000X7 = 63000 
 
 21**=hunds. 300X7= 2100 
 
 63***=thou. 80X7= 560 
 
 65674 Product. 2x7 = 14 
 
 Adding these results together, we have 65674 Ans. 
 OBS. The reason for always beginning to multiply at the right hand of the 
 multiplicand, is that we may carry the tens as we proceed in the operation. 
 
 85. From this illustration it will be observed that units mul- 
 tiplied into units produce units ; tens into unjts, or units into tens, 
 produce" tens ; (Art. 83 f) hundreds into units, or units into hun- 
 dreds, produce hundreds, &c. Hence, 
 
 86 When units are multiplied into any order whatever, tlie 
 product will always be of the same order as the other figure. 
 
 And universally, the product of any two integers is of the order 
 next less than that denoted by the sum of the orders of tlie two given 
 figures. Thus, hundreds into tens produce thousands, or the 4th 
 ^rder, which is one less than the sum of the two given orders. 
 
 OBS. When the multiplier contains more than one figure, it is customary to 
 begin to multiply with its units' figure. The result however will be the same, 
 if we begin with its hundreds or any other order of the multiplier, and place 
 the first figure of the partial products, so that the same orders shall stand 
 under each other. 
 
 First Operation. 
 1357 
 
 Second Operation. 
 1357 
 
 3574 
 
 3574 
 
 407T~ 
 
 4071 
 
 6785 
 
 6785 
 
 9499 
 
 9499 
 
 5428 
 
 5428 
 
 4849918. Prod. 
 
 4849918. Prod. 
 
 QUEST. 85. What do units into units produce ? Units into tens, or tens into units 1 
 
ARTS. 84-89.] MULTIPLICATION. 53 
 
 Ex. 6. What is the product of 5690 into 3008 ? 
 
 After multiplying by the 8 units, we next Operation. 
 multiply by the 8 thousands, since there arc no 5G90 
 
 tens nor hundreds in the multiplier, and place 3008 
 
 the first figure of this partial product under the 45520 
 
 figure 3 by which we are multiplying. 17070 
 
 17115520 Ans. 
 
 87. From the preceding illustrations and principles we de- 
 rive the following 
 
 GENERAL RULE FOR MULTIPLICATION. 
 
 I. When the multiplier contains but one figure. 
 
 Write the multiplier under the multiplicand, units under units, 
 tens under tens, d'c. (Art. 83. Obs. 1.) 
 
 Beyin at the right hand and multiply each figure of the mul- 
 tiplicand by the multiplier, setting doivn- the result and carrying as 
 in addition. (Art. 84. Obs.) 
 
 II. When the multiplier contains more than one figure. 
 Multiply each figure of the multiplicand l)y each fiyure of tJie 
 
 multiplier separately, beyinniny with the units, and write the par- 
 tial products in separate lines, placing the first figure of each line di~ 
 rectly under the fiyure by which you, multiply. (Art. 8G. Obs. 2.) 
 Finally, add the several partial products toyetJter, and tJie sum 
 will be the whole product. (Art. 83. Obs. 3.) 
 
 88. PROOF. Multiply the multiplier by the multiplicand, 
 and if the product thus obtained is tlie same as the other product, 
 the work is supposed to be right. 
 
 OBS. This method of proof depends upon the principle, that the product of 
 any two numbers is the same, whichever is taken for the multiplier. (Art. 83.) 
 
 89. Second Method. Add the multiplicand to itself as many 
 
 QUEST. 86. When units are multiplied into any order, what order is the product? 
 When any two integers are multiplied together, of what order is the product ? 87. How 
 do you write the numbers for multiplication ? When the multiplier contains liut one fig 
 Ore, how proceed ? Why begin at the right hand of the multiplicand 1 When the multi 
 pUer contains more than one figure, how proceed? What is meant l>y partial products 1 
 Why place the first figure of each partial product under the figure by which yon multi- 
 ply? What is to be done with the partial products? Why add the several partial pro 
 dacts together? Why should this give the whole product? 88. How is multipUcaitoa 
 proved ? Obs. On what principle does this proof depend ? 
 
 5* 
 
54 MULTIPLICATION. [SECT. IV. 
 
 times as there are units in the multiplier, and if the product ob- 
 tained is equal to the amount, the work is right. 
 
 jy ole ^ when t he multiplier is small, this is a very convenient mode of proof. 
 
 90. Third Method. Cast the 9s out of the multiplicand and 
 multiplier ; multiply their remainders together, and casting the 
 9s out of their product, set down the excess ; then cast the 
 9s out of the answer obtained, and if this excess be the same as 
 that obtained from the multiplier and multiplicand, the work may 
 be considered right. 
 
 Ex 7. Multiply 565 by 356. 
 
 Operation. Proof. 
 
 565 The excess of 9s in the multiplicand is 7. 
 356 " 9s " multiplier is 5. 
 
 3390 7X5 = 35; and the excess of 9s is 8. 
 2825 
 1695 
 Prod. 201140. The excess of 9s in the Ans. is also 8. 
 
 91. Fourth Method. Divide the product by one of the fac- 
 tors, and if the quotient thus arising is equal to the other factor, 
 the work is right. 
 
 Note. This method of proof supposes the learner to be acquainted with 
 division before he commences this work. (Art. 57. Note.) It is simply re- 
 versing the operation, and must obviously lead us back to the number with 
 which we started : for, if a number is both multiplied and divided by the same 
 number, its value will not be altered. (Ax. 9.) 
 
 92. Fifth Method.* First, cast the 11s out of the multiplicand and multi- 
 plier; multiply their remainders together, cast the 11s out of the product, and 
 set down the excess ; then cast the 11s out of the answer obtained, and if the 
 excess is the same as that obtained from the multiplier and multiplicand, the 
 work is right. 
 
 Note. 1. This method depends on a peculiar property of the number 11. 
 For its further development and illustration, see Art. 161. Prop. 18. 
 
 2. To cast the 11s out of a number, begin at the right hand, mark the alter- 
 nate figures; then from the sum of the figures marked, increased by 11 if 
 necessary, take the sum of those not marked, and the remainder will be the 
 excess required. Thus to cast the 11s out of 39475025, mark the alter- 
 figuresj beginning at the right hand, 39475025, then the sum of 
 
 QUEST. Can multiplication be proved by any other methods? 
 * Leslie s Philosophy of Arithmetic. 
 
AKTS. 90-93.] MULTIPLICATION. 55 
 
 5_l_0-{-7-|-9=21. Again, the sum of the others, viz: 2+5-|-4-}-3= 14. Now, 
 114=7, the excess of 11s. 
 
 Or, as soon as the sum is 11 or over, we may drop the 11, and add the re- 
 mainder to the next digit. Thus, 5 and 7 are 12; dropping the 11, 1 and 9 
 are 10. Again, 2 and 5 are 7, and 4>.are 11 ; drop the 11, and there are 3 left. 
 Now, 10 3=7, the same excess as before. 
 Ex. 8. Multiply 237956 by 3728. 
 
 Operation. Proof. 
 
 237956 Excess of lls is 4. > Now, 4X10=40; the excess of 1U 
 
 3728 " " 10. $ in 40 is 7. 
 
 Ans. 887099908 Excess of lls in the answer is also 7. 
 
 EXAMPLES FOR PRACTICE. 
 
 93* Ex. 1. What will 435 acres of land cost, at 57 dollars 
 per acre ? 
 
 2. What cost 573 oxen, at 63 dollars per head ? 
 
 3. What cost 1260 tons of iron, at 45 dollars per ton ? 
 
 4. If a man can travel 248 miles in a day, how far can he 
 travel in 365 days ? 
 
 5. If an army consume 645 pounds of meat in a day, how 
 much will they consume in 1 1 5 days ? 
 
 6. If 1250 men can build a fort in 298 days, how long would 
 it take 1 man to do it ? 
 
 7. How many rods is it across the Atlantic Ocean, 'allowing 
 320 rods to a mile, and the distance to be 3000 miles ? 
 
 8. What is the product of 463X45? 
 
 9. What is the product of 348X62? 
 
 10. What is the product of 793X86 ? 
 
 11. What is the product of 75X42X56 ? 
 
 12. What is the product of 7198X216 ? 
 
 13. 31416X175. 22. 8320900X1328. 
 
 14. 8862X189. 23, 17500X732. 
 
 15. 7071X556. 24. 15607X3094. 
 
 16. 93186X4455. 25. 742215^X468. 
 
 17. 40930X779. 26. 9264397x9584. 
 
 18. 12345X686. 27. 46873J9X1987. 
 
 19. 46481X936. 28. 95073*0X7071. 
 
 20. 16734X708. 29. 39948123X6007. 
 
 21. 7^5X7575. 30. 73885246X6079. 
 
56 MULTIPLICATION. [SECT. IV. 
 
 31. 51902468X5008. 37. 58763718X6754. 
 
 32. 57902468X5080. 38. 73084163X7584. 
 
 33. 57902468X5800. 39. 144X144X144. 
 
 34. 12481632X1509. 40. 3851X3851X3851. 
 
 35. 79068025X1386. 41. 79094451X764094. 
 
 36. 92948789X7043. 42. 89548050X972800. 
 
 CONTRACTIONS IN MULTIPLICATION. 
 
 94. The general rule is adequate to the solution of all exam 
 pies that occur in multiplication. In many instances, however, 
 by the exercise of judgment in applying the preceding principles, 
 the operation may be very much abridged. 
 
 95. Any number which may be produced by multiplying two 
 or more numbers together, is called a Composite Number. 
 
 Thus, 4, 15, 21, are composite numbers; for 4=2X2; 15 = 
 5X3; 21 = 7X3. 
 
 OBS. 1. The factors which, being multiplied together, produce a composite 
 number, are sometimes called the component parts of the number. 
 
 y. The process of finding the factors of which a given number is composed, 
 is called resolving tJie number into factors. 
 
 Ex. 1. Resolve 9, 10, 14, 22, into their factors. 
 
 2. What are the factors of 35, 54, 56, 63 ? 
 
 3. What are the factors of 45, 72, 64, 81, 96? 
 
 96. Some numbers may be resolved into more than two fac- 
 tors; and also into different sets of factors. Thus, 12 = 2X2X3; 
 also 12 = 4X3 = 6X2. 
 
 4. What are the different factors and sets of factors of 8, 16, 
 18, 20, 24? 
 
 5. What are the different factors and sets of factors of 27, 32, 
 86, 40, 48? 
 
 9G. a. We have seen that the product of any two numbers is 
 the same, whichever factor is taken for the multiplier. (Art. 83.) 
 In like manner, it may be shown that the product of any three 01 
 
 QUKST. 95. What is a composite number ? Ob?. What are the factors which produce it 
 sometimes called ? What is meant by resolving a number into factors? l)fi. Are numbers 
 ever composed of more thnn two factors? 96. a. When three or more factors are to be 
 multiplied together, does * make any difference in what order they ate taken ? 
 
ARTS. 94-97.J MULTIPLICATION. 57 
 
 more factors will be the same, in whatever order they are multi- 
 plied. For, the product of two factors may be considered as one 
 number, and this may be taken either for the multiplicand, or the 
 multiplier. Again, the product of three factors may be consid- 
 ered as one number, and be taken for the multiplicand, or the mul- 
 tiplier, &c. Thus, 24 = 3X2X2X2 = 6X2X2 = 12X2 = 6X4 = 
 4X2X3 = 8X3. 
 
 CASE 1. When the multiplier is a composite number. 
 
 6. What will 27 bureaus cost, at 31 dollars apiece? 
 Analysis. Since 27 is three times as much as 9 ; that is, 27 = 9 
 
 X3, it is manifest that 27 bureaus will cost three times as much 
 as 9 bureaus. 
 
 Operation. 
 
 Dolls. 31 cost of 1 B. Having resolved 27 into the factors 
 
 9 9 and 3, we find the cost of 9 bureaus, 
 
 Dolls. 279 cost of 9 B. then multiplying that by 3, we have 
 
 3 the cost of 27 bureaus. 
 Dolls. "837 cost of 27 B. 
 
 7. What will 36 oxen cost, at 43 dollars per head ? 
 Solution. 36 = 9X4; and 43X9X4 = 1548 dolls. Ans. 
 
 Or, 36 = 3X3X4; and 43 X 3X3X4= 1548 dolls. Ans. Hence, 
 
 97. To multiply by a composite number. 
 
 Resolve the multiplier into two or more factors ; multiply the 
 multiplicand by one of these factors, and this product by another 
 factor, and so on till you have multiplied by all the factors. The 
 last product ivill be the answer required. 
 
 OBS: The factors into which a number may be resolved, must not be con- 
 founded with the parts into which it may be separated. (Art. 53.) The former 
 have reference to multiplication, the latter to addition ; that is, factors must be 
 multiplied together, but parts must be added together to produce the given 
 number. Thus, 56 may be resolved into two factors, 8 and 7 ; it may be sep- 
 arated into two parts, 5 tens or 50, and 6. Now, 8X!7=56, and 50-j-6=56. 
 
 8. What will 24 horses cost, at 74 dollars a head ? 
 
 QUEST. 97. When the multiplier is a composite number, how do you proceed ? Obs 
 What is the difference between the factors into which a number may be resolved and the 
 parts into which it may be separated 7 
 
58 MULTIPLICATION. | SfiCT IV, 
 
 9. What cost 45 hogsheads of tobacco, at 128 dollars a hogs 
 head? 
 
 10. What cost 54 acres of land, at 150 dollars per acre? 
 
 11. At 118 shillings per week, how much will it cost a family 
 to board 49 weeks ? 
 
 12. If a man travels at the rate of 372 miles a day, how far 
 will he travel in 64 days ? 
 
 13. At 163 dollars per ton, how much will 72 tons of lead cost ? 
 
 14. What cost 81 pieces of broadcloth, at 245 shillings apiece? 
 
 15. What cost 84 carriages, at 384 dollars apiece? 
 
 CASE II. When t/ie multiplier is 1 with ciphers annexed to it. 
 
 98* It is a fundamental principle of notation, that each re 
 moval of a figure one place towards the left, increases its value 
 ten times; (Art. 36;) consequently, annexing & cipher to a number 
 will increase its value ten times, or multiply it by 10; annexing 
 two ciphers will increase its value a hundred times, or multiply it 
 by 100 ; annexing three ciphers will increase it a thousand times, 
 or multiply it by 1000, &c. Thus, 15 with a cipher annexed, be- 
 comes 150, and is the same as 15X10; 15 with two ciphers an- 
 nexed, becomes 1500, and is the same as 15 X 100 ; 15 with thre? 
 ciphers annexed, becomes 15000, and is the same as 15X1000, 
 &c. Hence, 
 
 99. To multiply by 10, 100, 1000, &c. 
 
 Annex as many ciphers to the multiplicand as there are ciplien 
 in the multiplier, and the number thus^ formed will bi the product 
 required. 
 
 Note. To annex means to place after, or at the right hand. 
 
 16. What will ten boxes of lemons cost, at 63 shillings per 
 box? Ans. 630 shillings. 
 
 17. How many bushels of corn will 465 acres of land produce, 
 at 100 bushels per acre? 
 
 QUEST. 98. What is the effect of annexing a cipher to a number? Two ciphers 1 
 Three ? Four? 99. How do you proceed when the multiplier is 10, 100, 1000, &c. 1 JVot* 
 What Is the meaning of the term annex ? 
 
ARTS. 98-100 ] MULTIPLICATION. 59 
 
 18 Allowing 365 days for a year, how many days are there in 
 1000 yeais? 
 
 19. Multiply 153486 by 10000. 
 
 20. Multiply 3120467 by 100000. 
 
 21. Multiply 52690078 by 1000000. 
 
 22. Multiply 689063457 by 10000000. 
 
 23. Multiply 4946030506 by 100000000. 
 
 24. Multiply 87831206507 by 1000000000. 
 
 25. Multiply 67856005109 by 10000000000. 
 
 CASE Ill. WJien the multiplier has ciphers on the right hana. 
 
 26. What will 30 wagons cost, at 45 dollars apiece ? 
 
 Note. Any number with ciphers on its right hand, is obviously a composite 
 number; the significant figure or figures being one factor, and 1, with thl 
 given ciphers annexed to it, the other factor. Thus, 30 may be resolved into 
 the factors 3 and 10. We may therefore first multiply by 3 and then by 10, 
 by annexing a cipher as above. 
 
 Solution. 45X3 = 135, and 135X10=1350 dolls. Ans. 
 
 27. How many acres of land are there hi 3000 farms, if each 
 farm contains 475 acres ? 
 
 Analysis. 3000=3 X 1000. Now 475 X Operation. 
 
 3 = 1425 ; and adding three ciphers to this 475 
 
 product, multiplies it by 1000. (Art. 99.) 3 
 
 Hence, Ans. 1425000 acres. 
 
 1 OO. When there are ciphers on the right of the multiplier. 
 
 Multiply the multiplicand by the significant figures of the multi- 
 plier, and to this product annex as many ciphers, as are found on the 
 right of the multiplier. 
 
 OBS. It will be perceived that this case combines the principles of the two 
 preceding cases ; for, the multiplier is a composite number, and one of its fac- 
 tors is * with cipliers annexed to it. 
 
 28 How much will 50 hogs weigh, at 375 pounds apiece? 
 
 29 If 1 barrel of flour weighs 192 pounds, how much will 
 600 barrels weigh ? 
 
 30. Multiply 14376 by 25000. 
 
 QUEST. 100. When there are ciphers on the right of the multiplier, how do you pro 
 oeed 1 Obs. What principles does this case combine 1 
 
60 MULTIPLICATION. [SECT. IV 
 
 31. Multiply 350634 by 410000. 
 
 32. Multiply 4630425 by 6200000. 
 
 CASE IV. When the multiplicand has ciphers on the right hand, 
 
 33. What .will 37 ships cost, at 29000 dollars apiece? 
 Analysis. 29000 = 29X1000. But the Operation. 
 
 product of two or more factors is the same 29000 
 
 in whatever order they are multiplied. 37 
 
 (Art 96. a.) We therefore multiply 29 203 
 
 by 37, and this product by 1000 by adding 87 
 
 three ciphers to it. Ans. 1073000 dolls. 
 
 PROOF. 29000X37=1073000, the same as before. Hence, 
 1 1 When there are ciphers on the right of the multiplicand. 
 Multiply the significant figures of the multiplicand by the mul- 
 tiplier, and to the product annex as many ciphers, as are found on 
 the right of the multiplicand. 
 
 OBS. When both the multiplier and multiplicand have ciphers on the right, 
 multiply the significant figures together as if there were no ciphers, and to their 
 product annex as many ciphers, as are found on the right of both factors 
 
 34. Multiply 2370000 by 52. 
 
 35. Multiply 48120000 by 48. 
 
 36. Multiply 356300000 by 74. 
 
 37. Multiply 1623000000 by 89. 
 
 38. Multiply 540000 by 700. 
 
 Analysis. 540000=54X10000, and Operation. 
 
 700=7 X 1 00 ; we therefore multiply the 540000 
 
 significant figures, or the factors 54 and 7 00 
 
 7 together, (Art. 96. a,) and to this pro- Ans. 378000000 
 
 duct annex six ciphers. (Art. 99.) 
 
 39. Multiply 1563800 by 20000. 
 
 40. Multiply 31230000 by 120000. 
 
 41. Multiply 5310200 by 3400000. 
 
 42. Multiply 82065000 by 8100000. 
 
 43. Multiply 210909000 by 5100000. 
 
 QUEST. 101. When there are ciphers on the right of the multiplicand, how proceed 1 
 Oft*. How, when there are ciphers on the right both of the multiplier and multiplicand? 
 
ARTS. 101-104."] MULTIPLICATION. 61 
 
 102. There are oilier methods of contracting the operations in 
 multiplication, which, in certain cases, may be resorted to with 
 advantage. Some of the most useful are the following. 
 
 44. How many gallons of water will a hydrant discharge in 13 
 hou^s, if it discharges 2325 gallons per hour ? 
 
 Operation. Multiplying by the 3 units, we set the 
 
 2325 X 13 first figure of the product one place to ths 
 
 6975 right of the multiplicand. Now, since 
 
 Ant. 30225 gallons, multiplying by 1 is taking the multipli- 
 cand onct, (Art. 82,) we add together the multiplicand and tho 
 partial product already obtained, and the result is the answer. 
 
 PROOF. 2325X13=30225 gallons, the same as above. Hence, 
 
 103. To multiply by 13, 14, 15, &c., or 1, with either of the 
 other digits annexed to it. 
 
 Multiply by the units' figure of the multiplier, and write each 
 figure of the partial product one place to the right of that from 
 which it arises ; finally, add the partial product io the multipli- 
 cand, and the result will be the answer required. 
 
 Note. This method is the same, in effect, as if we actually multiplied by the 
 1 ten, and placed the first figure of the partial product under the figure by 
 which we multiply. (Art. 87. II.) 
 
 45. Multiply 3251 by 14. 46, Multiply 4028 by 17. 
 47. Multiply 25039 by 16. 48. Multiply 50389 by 18. 
 49. If 21 men can do a job of work in 365 days, how long 
 
 will it take 1 man to do it ? 
 
 Operation. We first multiply by the 2 tens, and set 
 
 365X21 the first product figure in tens' place, then 
 
 730 adding this partial product to the multipli- 
 
 Ans. 7665 days, cand, we have 7665, for the answer. 
 
 PROOF. 365X21 = 7665 days, the same as above. H(nce, 
 
 1.O4. To multiply by 21, 31, 41, &c., or 1 with either <&f the 
 other significant figures prefixed to it. 
 
 Multiply by the tens' figure of the multiplier, and write the first 
 
 
 
62 MULTIPLICATION. [SfiCT. IV. 
 
 figure of the partial product in tens' place / finally, add this par- 
 tial product to the multiplicand, and the result will be the answer 
 required. 
 
 Note. The reason of this method of contraction is substantially the same 
 as that of the preceding. 
 
 50 Multiply 4275 by 31. 51. Multiply 7504 by 41. 
 
 52. Multiply 38256 by 61. 53. Multiply 70267 by 81, 
 
 54. How much will 99 carriages cost, at 235 dollars apiece ? 
 
 Analysis. Since 1 carriage costs 235 Operation. 
 
 dollars, 100 carriages will cost 100 times 23500 price of 100 C 
 as much, which is 23500 dollars. (Art. 235 " of 1 C. 
 99.) But we wished to find the cost 23265 " of 99 C. 
 of 99 carnages only. Now 99 is 1 less 
 
 than 100 ; therefore, if we subtract the price of 1 carriage from 
 the price of 100, it will give the price of 99 carriages. Hence, 
 
 1O5. To multiply by 9, 99, 999, or any number of 9s. 
 
 Annex as many ciphers to the multiplicand as there are 9s in the 
 multiplier ; from, the result subtract tJie given multiplicand, and 
 tJie remainder will be the answer required. 
 
 Note. The reason of this method is obvious from the fact that annexing as 
 many ciphers to the multiplicand as there are 9s in the multiplier, multiplies it 
 by 100, or repeats it once more than is required; (Art. 99;) consequently, sub- 
 tracting the multiplicand from the number thus produced, must give the true 
 answer, 
 
 55. Multiply 4791 by 99. 56. Multiply 6034 by 999. 
 57. Multiply 7301 by 999. 58. Multiply 463 by 9999. 
 59. What is the product of 867 multiplied by 84 ? 
 
 Analysis. We first multiply by 4 in the usual Operation. 
 way. Now, since 8=4 X 2, it is plain, if the par- 867 
 
 tial product of 4 is multiplied by 2, it will give 84 
 
 the partial product of 8, But as 8 denotes tens, 3468X2 
 
 the first figure of its product will also be tens. 6936 
 (Art. 86.) The sum of the two partial products 72328 Ans. 
 will be the answer required. 
 
 Note. For the sake of convenience in multiplying, the factor 2 is placed at 
 the right of the partial product of 4, with the sign X> between them. 
 
ARTS. 105, 106.] MULTIPLICATION. (33 
 
 60. What is the product of 987 by 486 ? 
 
 Operation. 
 
 987 Since 48 = 6 X 8, we multiply the partial prod- 
 
 486 uct of 6 by 8, and set the first product figure 
 
 5922X8 in tens' place as before. (Art. 86.) 
 
 47376 
 
 479682 Ans. 
 
 PROOF. 987x486 = 479682, the same as above. Hence, 
 
 1 O6 When part of the multiplier is a composite number of 
 which the other figure is a factor. 
 
 First multiply by the fiyure that is a factor ; then multiply this 
 partial product by the other factor, or factors, taking care to write 
 the first figure of each frirtial product in its proper order, and 
 their sum will be tht ansit^r required. (Art. 86.) 
 
 OBS. When the figure m thousands, ten thousands, or any other column, is 
 a factor of the other par*, or ) arts of the multiplier, care must be taken to 
 place the first figure of its product under the factor itself, and the first figure 
 of each of the other partial products in its own order. (Art. 86.) 
 
 (61.) (62.) 
 
 2378 256841 
 
 936 85632 
 
 21402 X4 2054728 7x4 
 
 __85608 14383096 
 
 2225808 Ans. 8218912 
 
 21993808512 Ans. 
 
 63. Multiply 665 by 82. 64. Multiply 783 by 93. 
 
 65. Multiply 876 by 396. 66. Multiply 69412 by 95436, 
 67. 324325X54426. 68. 256721X85632. 
 
 69. What is the product of 63 multiplied by 45 ? 
 No1e. By multiplying the figures which produce the same order, and add- 
 ing the results mentally, we may obtain th* '-nswer without setting down the 
 partial products. 
 
 First, multiplying the units into units, we se 
 
 Operation. down the result and carry as usual. JSTow, since 
 
 63 the 6 tens into 5 mits, and 3 units into 4 tens will 
 
 45 both produce th same order, viz : .tens, (Art. 86,) 
 
 2835 Ans. we multiply them and add their products men- 
 
64 
 
 MULTIPLICATION. 
 
 [SECT. IV. 
 
 tally. Thus, 6X5 = 30, and 3 X 4 = 1 2 ; now, 30+ 1 2 = 42, and 1 
 (to carry) makes 43. Finally, 6X4 = 24, and 4 (to carry) make 28. 
 
 PROOF. 63X45 = 2835, the same as before. Hence, 
 
 1O7 To multiply any two numbers together without setting 
 down the partial products. 
 
 First multiply the units together / then multiply the figures 
 '(?"' ih produce tens, and adding the products mentally, set down the 
 tsult and carry as usual. Next multiply the figures which produce 
 Hundreds, and add the products, &c., as before. In like manner, 
 perform the multiplications which produce thousands, ten thou- 
 sands, (fcc., adding tJie products of each order as you proceed, and 
 thus continue the (peration till all the figures are multiplied. 
 
 70. What is the product of 12346789 into 54321 ? 
 
 Analytic Operation. 
 
 23456789 
 54321 
 
 2X5 
 
 2X4 
 3X5 
 
 2X3 
 3X4 
 4X5 
 
 2X2 
 3X3 
 4X4 
 5X5 
 
 2X1 
 3X2 
 4X3 
 5X4 
 6X5 
 
 3X1 
 4X2 
 5X3 
 6X4 
 7X5 
 
 4X1 
 5X2 
 6X3 
 7X4 
 8X5 
 
 5X1 
 
 6X2 
 7X3 
 8X4 
 9X5 
 
 6X1 
 7X2 
 8X3 
 9X4 
 
 7X1 
 
 8X2 
 9X3 
 
 8X1 
 
 9X2 
 
 9X1 
 
 6 
 
 6 9 
 
 Explanation. Having multiplied by the first two figures of the 
 multiplier, as in the last example, we perceive that there are three 
 .multiplications which will produce hundreds, viz : 7 X 1> 8X2, and 
 9X3 ; (Art. 86 ;) we therefore perform these multiplications, add 
 their products mentally, and proceed to the next order. Again, 
 there are four multiplications which will produce thousands, viz: 
 6 X 1, 7 X 2, 8 X 3, and 9 X 4. (Art. 86.) We perform these mul- 
 tiplications as before, and proceed in a similar manner through all 
 the remaining orders. Ans. 706296235269. 
 
 Note. 1. In the solution above, the multiplications of the different, figures aro 
 arranged in separate columns, that the various combinations which produce 
 the same order, may be seen at a glance. In practice it is unnecessary to de- 
 note these multiplications. The principle being understood, the process of 
 
ARTS. 107, 108.) MULTIPLICATION. 
 
 65 
 
 multiplying and adding may easily be carried on in the mind, while the final 
 product only is set down. 
 
 2. When the factors contain but two or three figures each, this method 
 is very simple and expeditious. A little practice will enable the student to 
 apply it with facility when the factors contain six or eight figures each, and 
 its application will afford an excellent discipline to the mind. It has sometimes 
 been used when the factors contain twenty- four figures each; but it is doubt-^. 
 ful whether the attempt to extend it so far, 'is profitable. 
 
 71, Multiply 25X25. 
 
 73 Multiply 81X64. 
 
 75. Multiply 194X144. 
 
 77. Multiply 4825X2352. 
 
 72. Multiply 54X54. 
 74. Multiply 45X92. 
 76. Multiply 1234X125. 
 78. Multiply 6521X5312. 
 
 1O8. By suitable attention, the critical student will discovei 
 various other methods of abbreviating the processes of multipli- 
 cation. 
 
 Solve the following examples, contracting the operations when 
 practicable. 
 
 79. 42634X63. 
 
 80. 50035X56. 
 
 81. 72156X1000. 
 
 82. 42000X40000. 
 
 83. 80000X25000. 
 
 84. 2567345X17. 
 
 85. 4300450X19. 
 
 86. 9803404X41, 
 
 87. 6710045X71 
 
 88. 3456710X18 
 
 89. 7000541X91. 
 
 90. 4102034X99. 
 
 91. 42304X999. 
 
 92. 50421X9999. 
 
 93. 67243X99999. 
 
 94. 78563X93. 
 
 95. 34054X639. 
 
 96. 52156X756. 
 
 97. 41907X54486 
 
 98. 26397X24648. 
 
 99. 12900X14000. 
 
 100. 64172X42432. 
 
 101. 26815678X81. 
 
 102. 85X85. 
 
 103. 256X256. 
 
 104. 322X325. 
 
 105. 5234X2435. 
 
 106. 48743000X637. 
 
 107. 31890420X85672. 
 
 108. 80460000X2763. 
 
 109. 2364793X8485672. 
 
 110. 1256702X999999. 
 6840005X91X61. 
 45067034X17X51, 
 788031245X81X16. 
 61800000X23000. 
 12563000X4800000. 
 91300203X1000000. 
 
 117. 680040000X1000000. 
 
 118. 4000000000X1000000, 
 
 111. 
 112. 
 113. 
 114. 
 115. 
 116. 
 
66 DIVISION. [SECT. V 
 
 SECTION V. 
 
 DIVISION. 
 
 ART. 11O. Ex. 1. How many barrels of flour, at 8 dollar 
 per barrel, can you buy for 56 dollars ? 
 
 Analysis. Since flour is 8 dollars a barrel, it is obvious you 
 can buy 1 barrel as often as 8 dollars are contained in 56 dollars; 
 and 8 dolls, are contained in 56 dolls. 7 times. Ans. 7 barrels. 
 
 Ex. 2. A man wished to divide 72 dollars equally among 9 beg- 
 gars : how many dollars would each receive ? 
 
 Solution. Reasoning as before, each beggar would receive as 
 many dollars as 9 is contained times in 72 ; and 9 is contained in 
 72, 8 times. Ans. 8 dollars. 
 
 OBS. The learner will at once perceive that the object in the first example, 
 is to find kow many times one number 'is contained in another; and that the 
 object of the second, is to divide a given number into equal parts, but its solu- 
 tion consists in finding how many times one number is contained in another, 
 and is the same in principle as that of the first. 
 
 Ill* The Process of finding how many times one number is 
 contained in another, is called DIVISION. 
 
 The number to be divided, is called the dividend. 
 
 The number by which we divide, is called the divisor. 
 
 The number obtained by division, or the answer to the question, 
 is called the quotient. It shows how many times the divisor is 
 contained in the dividend. Hence, it may be said, 
 
 112* Division is finding a quotient, which multiplied into the 
 divisor, will produce the dividend. 
 
 Note. The term quotient is derived from the Latin word guoties, which sig- 
 nifies how often, or how many times. 
 
 QUEST. 111. What is division ? What is the number to be divided called 1 The num- 
 ber by which we divide ? What is the number obtained called ? What does the quotient 
 show ? 112. What then may division be said to be ? 
 
ARTS. 110-114.] DIVISION. 67 
 
 113. The number which is sometimes left after division, is 
 called the remainder. Thus, when we say 5 is contained in 38, 
 7 times, and 3 over, 5 is the divisor, 38 the dividend, 7 the quo- 
 tient, and 3 the remainder. 
 
 OBS. 1. The remainder is of the same denomination as the dividend; for, It 
 IN a part of it. 
 
 2. The remainder is always less than the divisor ; for, if it were equzl to, 
 or greater than the divisor, the divisor could be contained once more !n the 
 dividend. 
 
 114. It will be perceived that division is similar in principle 
 to subtraction, and may be performed by it. For instance, to tind 
 how many times 7 is contained in 21, subtract 7 (the divisor) con- 
 tinually from 2 1 (the dividend), until the latter is exhausted ; then 
 counting these repeated subtractions, we shall have the true quo- 
 tient. Thus, 7 from 21 leaves 14 ; 7 from 14 leaves 7 ; and 7 from 
 7 leaves 0. Now by counting, we find that 7 has been taken from 
 21, 3 times ; consequently, 7 is contained in 21, 3 times. Hence, 
 
 Division is sometimes defined to be a short way of performing 
 repeated subtractions of the same number. 
 
 OBS. 1. It will be observed that division is the reverse of multiplication. 
 Multiplication is the repeated addition of the same number; division is the 
 repeated subtraction of the same number. The product of the one answers to 
 the dividend of the other ; but the latter is always given, while the former is 
 required. 
 
 2. When the dividend denotes things of one denomination only, the opera 
 tion is called Simple Division. 
 
 SHORT DIVISION. 
 
 Ex. 3. How many hats, at 2 dollars apiece, can be bought for 
 4862 dollars? 
 
 Operation. We write the divisor on the left of the divi- 
 
 Divisor. Divid. dend with a curve line between them; then, 
 
 beginning at the right hand, proceed thus : 2 is 
 Quot. 2431 contained in 4, 2 times. Now, since the 4 de 
 
 QUEST. 113. What is the number called which is sometimes left&fter division 1 Cbs. Of 
 what denomination is the remainder? Why? Is the remainder greater or less than the 
 divisor? Why? 114. Tfc what rule is division similar in principle? Obs. Of what is 
 division the reverse ? When the dividend denotes things of one denomination only, what 
 ts the operation called 1 
 
 T.H. 4 
 
68 DIVISION. [SECT. V 
 
 notes thousands, the 2 must be thousands ; we therefore wnte it 
 in thousands' place, under the figure divided. 2 is contained in 
 8, 4 times ; and as the 8 is hundreds, the 4 must also he hun- 
 dreds; hence we write it in hundreds' place, under the figure 
 divided. 2 in 6, 3 times ; the 6 being tens, the 3 must also be 
 tens, and should be set in tens' place. 2 in 2, once ; and since 
 the 2 is units, the 1 is a unit, and must therefore be written in 
 inits' place. The answer is 2431 hats. 
 
 1 1 5 When the process of dividing is carried on in the mind, 
 and the quotient only is set down, as in the last example, the opera- 
 tion is called SHORT DIVISION. 
 
 116. The reason that each quotient figure is of the same order 
 AS the figure divided, may be shown in the following manner : 
 
 Having separated the dividend 
 
 Analytic Solution. of the last example into the orders 
 
 4862 = 4000 + 800 + 60 + 2 of which it is composed, we per- 
 
 2)4000 + 800 + 60 + 2 ceive that 2 is contained in 4000, 
 
 2000+400 + 30 + 1 2000 times; for 2X2000 = 4000, 
 
 Again, 2 is contained in 800, 400 
 
 times; for 2X400 = 800, &c. Am. 2431. 
 
 Ex. 4. A man left an estate of 209635 dollars, to be divided 
 equally among 4 children : how much did each receive ? 
 
 Since the divisor 4, is not contained in 
 
 Operation. 2, the first figure of the dividend, we find 
 
 4)209635 how many times it is contained in the first 
 
 Ans. 52408-f dolls. two figures. Thus, 4 is contained in 20, 
 
 5 times ; write the 5 under the 0. Again, 
 
 4 is contained in 9, 2 times and 1 over ; set the 2 under the 9. 
 Now, as we have 1 thousand over, we prefix it mentally to the 
 6 hundreds, making 16 hundreds; and 4 in 16, 4 times. Write 
 the 4 under the 6. But 4 is not contained in 3, the next figure, 
 we therefore put a cipher in the quotient, and prefix the 3 to the 
 next figftre of the dividend, as if it were a. remainder. Then 4 ia 
 35, 8 times and 3 over ; place the 8 under the 5, and setting the re- 
 mainder over the divisor thus f, place it on the right of the quotient, 
 Note. To prefix means to place before, or at the left hand. 
 
ARTS. 115-118.] DIVISION. 69 
 
 117. When the divisor is not contained in any figure of the 
 dividend, a cipher must always be placed in the quotient. 
 
 OBS. The reason for placing a cipher in the quotient, is to preserve the true 
 local value of each figure of the quotient. (Art. 116.) 
 
 118. In order to render the division complete, it is obvious 
 that the whole of the dividend must be divided. But when there 
 is a remainder after dividing the last figure of the dividend, it 
 must of necessity be smaller than the divisor, and cannot be di- 
 vided by it. (Art. 113. Obs. 2.) We therefore represent the divi- 
 sion by placing the remainder over the divisor, and annex it to 
 the quotient. (Art. 28.) 
 
 OBS. 1. The learner will observe that in dividing we begin at the left hand, 
 instead of the right, as in Addition, Subtraction, and Multiplication. The rea- 
 son is, because there is frequently a remainder in dividing a higher order, 
 which must necessarily be united with the next lower order, before the division 
 can be performed. 
 
 2. The divisor is placed on the left of the dividend, and the quotient under 
 it, merely for the sake of convenience. When division is represented by the 
 sign -f-, the divisor is placed on the right of the dividend : and when repre- 
 sented in the form of a fraction, the divisor is placed under the dividend, 
 
 LONG DIVISION. 
 
 Ex. 5. At 15 dollars apiece, how many cows can be bought 
 for 3525 dollars? 
 
 Operation. 
 Having written the divisor on the left of Divisor. Divid. Quot. 
 
 the dividend as before, we find that 15 is 15) 3525 (235 
 
 contained in 35, 2 times, and place the 2 on 30 
 
 the right of the dividend, with a curve line 52 
 
 between them. We next multiply the di- 45 
 
 visor by this quotient figure, place the prod- 75 
 
 uct under the figures divided, and subtract 75 
 
 it therefrom. We ruw bring down the next 
 
 figure of the dividend, and placing it on the right of the remainder 
 5, we perceive that 15 is contained in 52, 3 times. Set the 3 on 
 the right of the last quotient figure, multiply the divisor by it, and 
 subtract the product from the figures divided as before. We then 
 
70 DIVISION." [SECT. V 
 
 briii** 1 down the next, which is the last figure of the dividend, to 
 the right oi this remainder, and finding 15 is contained in 75, 
 5 times, we place the 5 in the quotient, multiply and subtract as 
 before. The answer is 235 cows. 
 
 119. When the result of each step in the operation is written 
 down, as in the last example, the process is called LONG DIVISION. 
 Long Division is the same in principle as Short Division. The 
 only difference between them is, that in the former, the result of 
 each step in the operation is written down, while in the latter, 
 we carry on the process in the mind, and simply write the quotient. 
 
 OBS. 1. When the divisor contains but one figure, the operation by Slwrt 
 Division is the most expeditious, and therefore should always be practiced ; 
 but when the divisor contains two or more figures, it will generally be the most 
 convenient to use Long Division. 
 
 2. To prevent mistakes, it is advisable to put a dot under each figure of the 
 dividend, when it is brought down. 
 
 3. The French place the divisor on the right of the dividend, and the quo- 
 tient below the divisor,* as seen in the following example. 
 
 Ex. 6. How many times is 72 contained in 5904 ? 
 
 Operation. 
 
 5904 (72 divisor. The divisor is contained in 590, the 
 
 576 82 quotient. first three figures of the dividend, 8 
 
 144 times. Set the 8 under the divisor, 
 
 144 multiply, &c., as before. 
 
 Ex. 7 How many times is 435 contained in 262534 ? 
 
 Operation. Since the divisor is not contained 
 
 A35)262534(603fff Am. in the first three figures of the divi- 
 
 2610 '"' dend, we find how many times it is 
 
 1534 contained in the first four, the few- 
 
 1305 est that will contain it, and write 
 
 229 rem. the 6 in the quotient ; then multi- 
 
 VjrasT. -115. What is short division ? 119. What is long division ? What is the dif- 
 itttou\.e Uotween them 1 
 
 * Elements D'Ari thine tique, par M. Bourdon. Also, Lacroix's Arithmetic, translated by 
 Pri/e*$oi Farrar. 
 
ARTS. 119, 120.] DIVISION. 71 
 
 plying and subtracting as before, the remainder is 15. Bringing 
 down the next figure, we have 153 to be divided by 435. Bui 
 435 is not contained in 153 ; we therefore place a cipher in tin 
 quotient, and bring down the next figure. Then 435 in 1534, 9 
 times. Place the 3 in the quotient, and proceed as before. 
 
 Note. After the first quotient figure is obtained, for each figure of the ami 
 dend which is brought down, either a significant figure, or a cipher, must be pu. 
 in the quotient. (Art. 116.) 
 
 1 2O From the preceding illustrations and principles we de- 
 rive the following 
 
 GENERAL RULE FOR DIVISION. 
 
 I. When the divisor contains but one figure. 
 
 Write tlie divisor on tlie left of the dividend, with a curve lint 
 between them. Begin at the left hand, divide successively each 
 figure of the dividend by the divisor, and place each quotient figure 
 directly under the figure divided. (Arts. 116, 118. Obs. 1, 2.) 
 
 If there is a remainder after dividing any figure, prefix it to 
 the next figure of the dividend and divide this number as before ; 
 and if the divisor is not contained in any figure of the dividend, 
 place a ciplier in the quotient and prefix this figure to the next 
 one of the dividend, as if it were a remainder. (Arts. 117, 118.) 
 
 II. When the divisor contains more than one figure. 
 Beginning on the left of tJie dividend, find how many times the 
 
 divisor is contained in the fewest figures that will contain it, and 
 place the quotient figure on the right of the dividend with a curve 
 line between them. Tlicn multiply tlie divisor by this figure and 
 subtract tlie product from the figures divided ; to the right of tlte 
 remainder bring down the next figure of the dividend and divide 
 this number as before. Proceed in this manner till all tJie figure! 
 of the dividend are divided. 
 
 QUEST. 12f . How do you write the numbers for division "? When the divisor contelni 
 but one figure, how proceed ? Why place the divisor on the left of the dividend and th 
 quotient unde the figure divided? When there is a remainder after dividing a fig ire, 
 what is to be done with it ? When the divisor is not contained in any figure of the divl 
 dend how proceed ? Why ? Why begin to divide at the left hand ? When the divtaoi 
 contains more than one figure, how proceed 1 
 
 
72 DIVISION [SECT. V^. 
 
 Wlienever there is a remainder after dividing the last figure, 
 write it over the divisor and annex it to the quotient. (Art. 118.) 
 
 Demonstration. The principle on which the operations in Division depend, 
 Is that a part of the quotient is found, and the product of this part into the 
 divisor is taken from the dividend, showing how much of the latter remains to 
 be divided; then another part of the quotient is found, and its product into the 
 divisor is taken from what remained before. Thus the operation proceeds till 
 the whole of the dividend is divided, or till the remainder is less than the divisot 
 (Art. 113. Obs.2.) 
 
 . OBS When the divisor is large, the pupil will find assistance in determining 
 the quotient figure, by finding how many times the first figure of the divisor iat 
 contained in the first figure, or if necessary, the first two figures of the divi- 
 dend. This will give pretty nearly the right figure. Some allowance must, 
 however, be made for carrying from the product of the other figures of the di- 
 visor, to the product of the first into the quotient figure. 
 
 121* PROOF. Multiply the divisor by the quotient, to the 
 product add the remainder, and if the sum is equal to the dividend, 
 the work is right. 
 
 OBS. Since the quotient shows how many times the divisor is contained in 
 the dividend, (Art. Ill,) it follows, that if the divisor is repeated as many timeg 
 as there are units in the quotient, it must produce the dividend. 
 
 Ex. 8. Divide 256329 by 723. 
 
 Operation. Proof. 
 
 723)256329(354fff Ans. 723 divisor. 
 
 2169 354 quotient. 
 
 3942 2892 
 
 3615 3615 
 
 3279 2169 
 
 2892 387 rem. 
 
 387 rem. 256329 dividend. 
 
 122* Second Method. Subtract the remainder, if any, from 
 the dividend, divide the dividend thus diminished, by the quotient 
 and if the result is equal to the given divisor, the work is right. 
 
 When there is a remainder after dividing the last figure of the dividend, what 
 must be done with it? 121. How is division proved ? Obs. How does it appear that the 
 product of the divisor and quotient will be equal to the dividend, if the work Js right 1 
 Can division be proved by any other methods ? 
 
ARTS. 121- 127. J DIVISION. 73 
 
 123. Third Method. First cast the 9s out of the divisor 
 and quotient, and multiply the remainders together ; to the prod- 
 uct add the remainder, if any, after division ; cast the 9s out of 
 this sum, and set down the excess ; finally cast the 9s out of the 
 dividend, and if the excess is the same as that obtained from the 
 divisor and quotient, the work may be considered right. 
 
 Note. Since the divisor and quotient answer to the multiplier and muitipli- 
 umd, and the dividend to the product, it is evident that the principle of casting 
 out the 9s will apply to the proof of division, as well as that of multiplication. 
 (Art. 90.) 
 
 124. Fourth Method. Add the remainder and the respective products of 
 the divisor into each quotient figure together, and if the sum is equal to the 
 dividend, the work is right. 
 
 Note. This mode of proof depends upon the principle that the wlwle of a 
 fuantity is equal to the sum of all its parts. (Ax. 11.) 
 
 125. Fifth Method. First cast the 11s out of the divisor and quotient, and 
 .multiply the remainders together; to the product add the remainder, if any, 
 after division, and casting the lls out of this sum, set down the excess; 
 dually, cast the lls out of the dividend, and if the excess is the same as that 
 obtained from the divisor and quotient, the work is right. (Art. 92. Note 2.) 
 
 EXAMPLES FOR PRACTICE. 
 
 127, Ex. 1. A farmer raised 2975 bushels of wheat on 45 
 acres of land : how many bushels did he raise per acre ? 
 
 2. A garrison consumed 8925 barrels of flour in 105 days : 
 how much was that per day ? 
 
 3. The President of the United States receives a salary of 25000 
 dollars a year : how much is that per day ? 
 
 4. A drover paid 2685 dollars for 895 head of cattle : how 
 much did he pay per head ? 
 
 5. If a man's expenses are 3560 dollars a year, how much are 
 they per week ? 
 
 6. If the annual expenses of the government are 27 millions 
 of dollars, how much will they be per day ? 
 
 7. How long will it take a ship to sail from New York U 
 Liverpool, allowing the distance to be 3000 miles, and the ship 
 to sail 144 miles per day ? 
 
 8. Sailing at the same rate, how long would it take the same 
 ship to sail round the globe, a distance of 25000 miles ? 
 
 7 
 

 74 
 
 DIVISION. 
 
 [SECT. V 
 
 10. 47839 42. 
 
 11. 75043-52. 
 
 12. 93840 63. 
 
 13, 421645- 
 
 -74. 
 
 14 325000- 
 
 -85. 
 
 15, 400000- 
 
 -96. 
 
 16. 999999- 
 
 -47. 
 
 17. 352417- 
 
 -29. 
 
 18. 47081-7- 
 
 251. 
 
 19. 423405- 
 
 -485. 
 
 20. 16512-7- 
 
 344. 
 
 21. 304916- 
 
 -6274. 
 
 22. 12689-7- 
 
 145. 
 
 23. 145260- 
 
 -1345. 
 
 24. 147735- 
 
 -3283. 
 
 25. 1203033 327. 
 
 26. 1912500425. 
 
 27. 5184673 102. 
 
 28. 301140-7-478. 
 
 29. 889381037846. 
 
 30. 930268814356. 
 
 31. 9749320365. 
 
 32. 3228242 5734. 
 
 33. 75843639426-7-8593. 
 
 34. 65358547823^-2789. 
 
 35. 102030405060-7-123456. 
 
 36. 908070605040-7-654321. 
 
 37. 1000000000000000111. 
 
 38. 10000000000000001111. 
 
 39. 100000000000000011111 
 
 CONTRACTIONS IN DIVISION. 
 
 128. The operations in division, as well as those in multipli- 
 cation, may often be shortened by a careful attention to the appli- 
 cation of the preceding principles. 
 
 CASE 1. When the divisor is a composite number. 
 
 Ex. 1. A man divided 837 dollars equally among 27 persons, 
 who belonged to 3 familiet, each family containing 9 persons : 
 tow many dollars did each person receive ? 
 
 Analysis. Since 27 persons received 837 dollars, each one 
 must have received as many dollars, as 27 is contained times in 
 837. But as 27 (the number of persons), is a composite number 
 whose factors are 3 (the number of families), and 9 (the number 
 of persons in each family), it is obvious we may first find how 
 many dollars each family received, and then how many each per- 
 son received 
 
 Operation. 
 
 3)837 whole sum divided. 
 
 9)279 portion of each Fam. 
 Ans. 31 " " " person. 
 
 If 3 families received 337 
 dollars, 1 family must have 
 received as many dollars, as 
 3 is contained times ir 837 
 
 and 3 in 837, 279 times. That is, each family received 279 dollar! 
 
ARTS. 128, 129.] DIVISION. "75 
 
 Again, if 9 persons, (the number in each family,) received 279 dol- 
 lars, 1 person must have received as many dollars, as 9 is con- 
 tained times in 279 ; and 9 in 279, 31 times. Ans. 31 dollars. 
 PROOF. -31X27=837, the same as the dividend. Hence, 
 
 To divide by a composite number. 
 
 1. Divide the dividend by one of the factors of the divisor, then 
 iivids the quotient thus obtained by another factor ; and so on till 
 ill the factors are employed. The last quotient will be the answer. 
 
 II. To find the true remainder. 
 
 If the divisor is resolved into but two factors, multiply the last 
 remainder by the first divisor, to the product add the first remain- 
 der, if any, and the result will be the true remainder. 
 
 When more than two factors are employed, multiply each re- 
 mainder by all the preceding divisors, to the sum of their prod- 
 ucts, add the Jirst remainder, and the result will be the true re- 
 mainder. 
 
 OBS. 1. The true remainder may also be found by multiplying the quotient 
 by the divisor, and subtracting the product from the dividend. 
 
 2. This contraction is exactly the reverse of that in multiplication. (Art 97.) 
 The result will evidently be the same, in whatever order the factors are taken. 
 
 2. A man bought a quantity of clover seed amounting to 507 
 pints, which he washed to divide into parcels containing 64 pints 
 each : how rnary parcels can lie make ? 
 
 Note. Since 64=2x8X4, we divide by the factors respectively 
 Operation. 
 2)507 
 
 8)253~ 1 rem. ... = 1 pt. 
 4)31 5 rem. Now 5X2 =10 pts. 
 73 rem. and 3X8X2 = 48 pts. 
 Ans. 7 parcels, and 59 pts. over. 59 pts. True Rem. 
 
 Demonstration, 1. Dividing 507 the number of pints, by 2, gives 253 fo .to 
 
 quotient, or distributes the seed into 253 equal parcels, leaving 1 pint t AT 
 
 Now the units of this quotient are evidently of a different valiie from thos jf 
 
 the given dividend ; for since there are but half as many parcels as at fir it 
 
 QUBST. 129. How proceed when the, divisor is a composite number] How flu he 
 true remainder 1 
 
 4* 
 
76 DIVISION. [SECT. V. 
 
 is plain that each parcel must contain 2 pints, or 1 quart; that is, every unit 
 of the fi .-st quotient contains 2 of the units of the given dividend ; consequently, 
 every unit of it that remains will contain the same ; (Art. 113. Obs. 2 ;) there- 
 fore this remainder must be multiplied by 2, in order to find the units of the 
 given dividend which it contains. 
 
 2. Dividing the quotient 253 parcels, by 8, will distribute them into 31 other 
 equal parcels, each of which will evidently contain 8 times the quantity of the 
 preceding, viz : 8 times 1 quart =8 quarts, or 1 peck; that is, every unit of th 
 second quotient contains 8 of the units in the first quotient, or 8 times 2 of th 
 units in the given dividend ; therefore what remains of it, must be multiplied 
 by 8x2, or 16, to find the units of the given dividend which it contains. 
 
 3. In like manner, it may be shown, that dividing by each successive factor 
 reduces each quotient to a class of units of a higher value than the preced- 
 ing ; that every unit which remains of any quotient, is of the same value as 
 that quotient, and must therefore be multiplied by all the preceding divisors, in 
 order to find the units of the given dividend which it contains. 
 
 4. Finally, the several remainders being reduced to the same units as those 
 of the given dividend according to~ the rule, their sum must evidently be the 
 true remainder. (Ax. 11.) 
 
 3. How many acres of land, at 35 dollars an acre, can you buy 
 for 4650 dollars ? 
 
 4. Divide 16128 by 24. 5. Divide 25760 by 56. 
 
 6. Divide 17220 by 84. 7. Divide 91080 by 72. 
 
 * 
 
 CASE II. When the divisor is 1 with cipliers annexed to it. 
 
 13O It has been shown that annexing a cipher to a number 
 increases its value ten times, or multiplies it by 10. (Art. 98.) 
 Reversing this process ; that is, removing a cipher from the right 
 hand of a number, will evidently diminish its value ten times, or 
 divide it by 10 ; for, each figure in the number is thus restDred 
 to its original place, and consequently to its original value. 1 hus, 
 annexing a cipher to 15, it becomes 150, which is the same as 
 15X10. On the oth?r hand, removing the cipher from 150, it 
 becomes 15, which is the same as 150-f-10. 
 
 In the same manner it may be shown, that removing two ciphers 
 from the right of a number, divides it by 100; removing three, di- 
 vides it by 1000 ; removing four, divides it by 10000, (fee. Hence, 
 
 130. What is the effect of annexing a cipher to a number ? What is the effect 
 of removing a apher from the right of a number 1 How does this appear 1 
 
ARTS. 130-132.] DIVISION. 77 
 
 131. To divide by 10, 100, 1000, &c. 
 
 Cut off as *many figures from the right hand of the dividend as 
 tliere are ciphers in the divisor. The remaining figures of tlie div- 
 idend will lie the quotient, and those cut off the remainder. 
 
 d. In one dime there are 10 cents : how many dimes are there 
 in 200 cents ? In 340 cents ? In 560 cents ? 
 
 9. In one dollar there are 100 cents : how many dollars are 
 there in C5000 cents ? In 765000 cents ? In 4320000 cents 9 
 
 10. Divide 26750000 by 100000. 
 
 11. Divide 144360791 by 1000000. 
 
 12. Divide 582367180309 by 100000000. 
 
 CASE III. When the divisor has cipliers on tlie right Jiand. 
 
 13. How many hogsheads of molasses, at 30 dollars apiece, 
 can you buy for 9643 dollars ? 
 
 OBS. The divisor 30, is a composite numl>er, the factors of which are 3 and 
 10. (Arts. 05, 9(5.) We may, therefore, divide first by one factor and the 
 quotient thence arising by the other. (Art. 121). ) Now cutting off the right 
 hand figure of the dividend, divides it by ten ; (Art. 131 ;) consequently divid- 
 ing the -remaining figures of the dividend by 3, the other factor of the divisor, 
 Will give the quotient. 
 
 Operation. We first cut off the cipher on the right 
 
 3|0)964|3 of the divisor, and also cut off the right 
 
 321 -J-jj- Ans. hand figure of the dividend ; then divid- 
 ing 964 by 3, we have 1 remainder. 
 
 low as the 3 cut off, is part of the remainder, we therefore 
 nex it to the 1. Ans. '321-J-ft hogsheads. Hence, 
 
 1 32. When there are ciphers on the right hand of the divisor. 
 Cut off tlie ciphers, also cut off as many figures from the right 
 
 of the dividend. Then divide the other figures of the dividend by 
 ihe remaining figures of the divisor, and annex tlie figures cut off 
 from the dividend to tlie remainder. 
 
 14. How many buggies, at 70 dollars apiece, can you buy foi 
 7350 dollars ? 
 
 QUEST. 131. How proceed when the divisor is 10, 100, 1000, &c. ? ir2. When there art 
 tiphrs on the right hand of the divisor, how proceed 1 What is to bo done with figure* 
 cut off from the dividend ? 
 
78 DIVISION. 
 
 [SECT. V. 
 
 15. How many barrels will it take tc pack 36800 pounds of 
 pork, allowing 200 pounds to a barrel ? 
 
 16. Divide 3360000 by 17000. 
 
 133. Operations in Long Division may be shortened by sub- 
 tracting the product of the respective figures in the divisor into 
 each quotient figure as we proceed in the operation, setting down 
 the remainders only. This is called the Italian Method. 
 
 17. How many times is 21 contained in 4998 ? 
 
 Operation. 
 
 21)4998(238 This method, it will be seen, requires a much 
 
 79 smaller number of figures than the ordinary 
 
 168 process. 
 
 18. Divide 1188 by 33. 19. Divide 2516 by 37. 
 20. Divide 3128 by 86. 21. Divide 7125 by 95. 
 
 22. A merchant laid out 873 dollars in flour, at 5 dollars a 
 barrel : how many barrels did he get ? 
 
 Operation. We first double the dividend, and then di- 
 
 873 vide the product by 10, which is done by 
 
 2 cutting off the right hand figure. (Art. 131.) 
 
 110)174|6 But since we multiplied the dividend by 2, it 
 
 174 f Ans. is plain that the 6 cut off, is 2 times too large 
 
 for the remainder ; we therefore divide it by 
 
 2, and we have 8 for the true remainder. Hence, 
 
 134* When the divisor is 5. 
 
 Multiply the dividend by 2, and divide the product ~hy 10. 
 (Art. 131.)" 
 
 tfote. 1. When the figure cut off is a significant figure, it must be divided 
 by 2 for the true remainder. 
 
 2. This contraction depends upon the princvpto that any given divisor ig 
 contained in any given dividend, just as man^. times as twice that divisor is 
 contained in twice that dividend, three times that divisor in three times that divi- 
 dend, &c. For a further illustration of this principle see General Pr pciplcs 
 m Division 
 
 23. Divide 6035 by 5. 24. Divide 8450 by 5. 
 25. Divide 32561 by 5. 26. Divide 43270 ly 5. 
 
ARTS. 133-139.] DIVISION. 79 
 
 135* When the divisor is 15, 35, 45, or 55. 
 
 Double the dividend, and divide t/ie product by 30, 70, 90, or 
 110, as tlie case ma be. (Art. 132.) 
 
 Note. This method is simply doubling both the divisor and dividend. V\ o 
 must there tore divide the remainder, if any, by 2, for the true remainder. 
 
 ' 27. Divide 3256 by 15. 28. Divide 2673 by 35. 
 
 29. Divide 3507 by 45. 30. Divide 7853 by 55 
 
 136. When the divisor is 25. 
 
 Multiply tJie dividend by 4, and divide the product by 100 
 (Art. 131.) 
 
 Note. This is obviously the same as multiplying both the dividend and divi- 
 sor by 4. (Art. 134. Note 2.) Hence, we must divide the remainder, if any 
 "thus found, by 4, for the true remainder. 
 
 31. Divide 2350 by 25. 32. Divide 4860 by 25. 
 
 33. Divide 42340 by 25. 34. Divide 94880 by 25. 
 
 137. To divide by 125. 
 
 Multiply the dividend by 8, and divide the product by 1000. 
 (Art. 131.) 
 
 Note. This contraction is multiplying both the dividend and divisor by 8. 
 For the true remainder, therefore, we must divide the remainder, if any, by 8. 
 
 35. Divide 8375 by 125. 36. Divide 25426 by 125. 
 
 138. To divide by 75, 175, 225, or 275. 
 
 Multiply the dividend by 4, and divide the product by 300, 700 
 900, or 1100, as the case may be. (Art. 132.) 
 Note. For the t^ue remainder, divide the remainder, if any thus found, by 4 
 37. Divide 1125 by 75. 38. Divide 2876 by 175. 
 
 39. Divide 3825 by 225. 40. Divide 8250 by 275. 
 
 139. The preceding are among the most frequent and useful 
 modes of contracting operations in division. Various other 
 methods might be added, but they will naturally suggest them- 
 selves to the inventive student, as opportunities occur for their 
 application. 
 
 41. How long would it take a vessel sailing 100 miles per .lav 
 to cirtumnavigate the earth, whose circumference is 25000 miles? 
 
80 DIVISION. [SECT. V. 
 
 42. The distance of the Earth from the Sun is 95,000,000 of 
 miles : how long would it take a balloon going at the rate of 
 100,000 miles a year, to reach the sun ? 
 
 43. The debts of the several States of the Union, in 1840, 
 amounted to 171,000,000 of "dollars, and the number of inhabi- 
 tants was 17,000,000 : How much must each individual have beeUr 
 taxed to pay the debt? 
 
 44. The national debt of Holland is 800,000,000 of dollars, 
 And the number of inhabitants 2,800,000 : what is the amount 
 of indebtedness of each individual ? 
 
 45. The national debt of Spain is 467,000,000 of dollars, and 
 the number of inhabitants 11,900,000: what is the amount of 
 indebtedness of each individual ? 
 
 46. The national debt of Russia is 150,000,000 of dollars, and 
 the number of inhabitants 51,100,000 : what is the amount of 
 indebtedness of each individual ? 
 
 47. The national debt of Austria is 380,000,000 of dollars, 
 and the number of inhabitants 34,100,000 : what is the amount 
 of indebtedness of each individual ? 
 
 48. The national debt of France is 1,800,000,000 of dollars, 
 and the number of inhabitants 33,300,000 : what is the amount 
 of indebtedness of each individual ? 
 
 49. The national debt of Great Britain is 5,556,000,000 of 
 dollars, and the number of inhabitants 25,300,000 : what is the 
 amount of indebtedness of each individual ? 
 
 50. Divide 467000000000 by 25000000000. 
 
 51. 56824042. 62. 462156 75. 
 
 52. 785372 63. x 63. 3562189^-225. 
 '53. 896736 72. 64. 685726 32000 
 
 54. 67234568H-5. 65. 723564175. 
 
 55. 34256726-M5. 66. 892565 225. 
 
 56. 42367581^-45, 67. 456212 275. 
 
 57. 16753672 35. 68. 925673 125. 
 
 58. 3256385-7-55. 69. 763421 175. 
 
 59. 45672400^25. 70. 876240 275. 
 dO, 6245634-^45. 71. 7825600-^-80000. 
 
 '* 8245623125 72. 9200-1578-:- 100000. . 
 
Airis. 139-143.] DIVISION. 81 
 
 GENERAL PRINCIPLES 'iN DIVISION. 
 
 '. 4O. From the nature of division, it is evident, that the 
 value of the quotient depends both on the divisor and the divi- 
 dend. 
 
 141. If a given divisor ia contained in a given dividend a 
 certain number of times, the same divisor will obviously be con- 
 tained, 
 
 In double that dividend, twice as many times. 
 
 In three times that dividend, thrice as many times, <fec. Hence, 
 
 If the divisor remains the same, multiplying the dividend by any 
 number, is in effect multiplying the quotient by that number. 
 
 Thus, 6 is contained in 12, 2 times ; in 2 times 12 or 24, 6 is 
 contained 4 times ; (i. e. twice 2 times ;) in 3 times 12 or 36, 6 
 is contained 6 times ; (i. e. thrice 2 times ;) &c. 
 
 142. Again, if a given divisor is contained in a given divi- 
 dend a certain number of times, the same divisor is contained, 
 
 In half that dividend, half as many times ; 
 
 In a third of that dividend, a third as many times, &c. Hence, 
 
 If the divisor remains the same, dividing the dividend by any 
 number, is in effect dividing the quotient by that number. 
 
 Thus, 8 is contained in 48, 6 times ; in 48-7-2 or 24, (half of 
 48,) 8 is contained 3 times ; (i. e. half of 6 times ;) in 48-7-3 or 
 10, (a third of 48,) 8 is contained 2 times ; (i. e. a third of 6 
 times ;) &c. 
 
 143. If a given divisor is contained in a given dividend a 
 certain number of times, then, in the same dividend, 
 
 Twice that divisor is contained only half as many tunes ; 
 Three times that divisor, a third as many times, &c. Hence, 
 If tlie dividend remains the same, multiplying tlie divisor by any 
 number, is in effect dividing the quotient by that number. 
 
 Thus, 4 is contained in 24, 6 times ; 2 times 4 or 8 is ccn 
 
 QUEST. 140. Upon what does the value of the quotient depend 1 141. If the divisor rr 
 mains the same, what is the effect on the quotient to multiply the dividend? 142. Wh* 
 is the effect of dividing the dividend by any given number 7 143. If the divide id rema:: 
 the same, what is the effect of multiplying the divisor by any given number 1 
 
82 DIVISION. [SE*-T. V. 
 
 tained in 24, 3 times ; (i. e. half of 6 times ;) 3 times 4 01 12 is 
 contained in 24, 2 times ; (i. e. a third of 6 times ;) &o. 
 
 144* If a given divisor is contained in a given divicVad a 
 certain number of times, then, in the same dividend, 
 
 Half that divisor is contained twice as many times ; 
 
 A third of that divisor, three times as many times, &c. H^nce, 
 
 [f the dividend remains the same, dividing the divisor by any 
 tdmber, is in effect multiplying the quotient by tliat number. 
 
 Thus, 6 is contained in 36, 6 times ; 6-7- 2 or 3, (half of 6,) is 
 contained in 36, 12 times; (i. e. twice 6 times;) .6-r-3 or 2, (a 
 third of 6,) is contained in 36, 18 times ; (i. e. thrice 6 times ;) &c. 
 
 145* From the preceding articles, it is evident that any given 
 divisor is contained in any given dividend, just as many times as 
 twice that divisor is contained in twice that dividend ; three times 
 that divisor in three times that dividend, &c. 
 
 Conversely, any given divisor is contained in any given dividend 
 just as many times, as half that divisor is contained in half that 
 dividend ; a third of that divisor, in a third of that dividend, <3cc. 
 Hence, 
 
 1 46* If the divisor and dividend are both multiplied, or both 
 divided by the same number, the quotient will not be altered. 
 
 Thus, 6 is contained in 12, 2 times ; 
 
 2 times 6 is contained in 2 times 1 2, 2 times ; 
 
 3 times 6 is contained in 3 times 12, 2 times, &c. 
 Again, 12 is contained in 48, 4 times ; 
 
 12-r2 is contained in 48-1-2, 4 times ; 
 12-f-3 is contained in 48-r-3, 4 times ; &c. 
 
 1 47 . If the sum of two or more numbers is divided by Any 
 number, the quotient will be equal to the sum of tlie quotients 
 which will arise from dividing the given numbers separately. 
 
 Thus, the sum of 12+18 = 30 ; and 30-1-6 = 5. 
 Now, 12-r6 = 2; and 18-7-6 = 3; but the sum of 2 + 3= 5. 
 
 Again, the sum of 32 + 24+40=96 ; and 96-7-8 = ) 2. 
 Now, 32-7-8=4; 24-7-8 = 3; and 40-7-8=5; but 4+ 3+5= 12 
 
 144. What of dividing the divisor? 146. What is the effect m . i the quotient 
 If the divisor and dividend are both multiplied, or both divided by the sai.Mi number? 
 
\RTS. 144-150. J CANCELATION. 83 
 
 CANCELATION.* 
 
 148. We have seen that division is finding a quotient, ^hich, 
 multiplied into the divisor, will produce the dividend. (Art. .112.) 
 If, therefore, the dividend is resolved into two such factors that 
 one of them is the divisor, the other factor will, of course, be the 
 quotient. Suppose, for example, 42 is to be divided by 6. Now 
 the factors of 42 are 6 and 7, the first of which being the div : or, 
 he other must be the quotient. Therefore, 
 
 Canceling a factor of any number, divides the number by that 
 factor. Hence, 
 
 1 49. When the dividend is the product of two factors., one 
 of which is the same as the divisor. 
 
 Cancel the factor common to the dividend and divisor ; the 
 other factor of the dividend will be tJie answer. (Ax. 9.) 
 
 Note. The term cancel, signifies to erase or reject. 
 
 1. Divide the product of 34 into 28 by 34. 
 Common Method. By Cancelation. 
 
 34 
 
 28 28 Ans. 
 
 272 
 
 68 Canceling the factor 34, which is com- 
 
 34)952(28 Ans. mon both to the divisor and dividend, we 
 68 have 28 for the quotient, the same as be- 
 
 272 fore. 
 
 272 
 
 15O. The method of contracting arithmetical operations, by 
 rejecting equal factors, is called CANCELATION. 
 
 OBS. It applies with great advantage to that class of examples and problems, 
 which involve both multiplication and division ; that is, which require the pro- 
 duct of two or more numbers to be divided by another number, or by the product 
 of two or more numbers. 
 
 2 Divide 76X45 by 76. 3. Divide 03X81 by 81. 
 
 4. Divide 65X82 by 82. 5. Divide 95X73 by 05. 
 
 6. Divide the product of 45 times 84 by 9. 
 
 * Birk's Arithmetical Collections : London, 1764. 
 
84 APPLICATIONS OF [SECT. V. 
 
 Analysis. The factor 45 = 5X9> hence the dividend is com- 
 posed of the factors 84X5X9- We may therefore cancel 9, 
 which is common both to the divisor and dividend, and 84 X 5, 
 the )ther factors of the dividend, will be the answer required. 
 
 Operation. Proof. 
 
 0)81X5X0 84X5X9 = 3780 
 
 420 Ans. And 3780-7-9 = 420. 
 
 7. Divide the product of 45 X 6 X 3 by 18 X 5. 
 
 Operation. Proof. 
 
 *$X5)45X0X# 45X6X3 = 810; and 18X5 90 
 
 9 Ans. Now, 810-f-90 = 9 
 
 Note. We cancel the factors 6 and 3 in the dividend and 18 in the divi- 
 scr; for GX3=18. Canceling the same or equal factors in the divisor and 
 dividend, is dividing them both by the same number, and therefore does not 
 aQ'ect the quotient. (Arts. 14G, 148.) Hence, 
 
 151. When the divisor and dividend have common factors. 
 
 Cancel the factors common to both ; then divide the product of 
 those remaining in the dividend by the product of those remaining 
 in the divisor. 
 
 8. Divide 15X?X12 by 5X3X?X2. 
 
 9. Divide 27x3X4x7 by 9X12X6. 
 
 10. Divide 75X15X24 by 25X3X6X4X5. 
 
 Note. The further development and application of the principles of Cancela- 
 tion, may be seen in reduction of compound fractions to simple ones; in multi- 
 plication and division of fractions ; in simple and compound proportion, &c. 
 
 1 5 1 a. The four preceding rules, viz : Addition, Subtraction, 
 
 Multiplication, and Division, are usually called the FUNDAMENTAI 
 RULES of Arithmetic, because they are the foundation or basis or 
 all arithmetical calculations. 
 
 OBS. Every change that can be made upon the value of a number, must 
 necessarily either increase or diminish it. Hence, the fundamental operation! 
 in arithmetic are, strictly speaking, but two, addition and subtraction ; that is 
 v&cr-ease and decrease. Multiplication, we have seen, is an abbreviated form 
 of addition; division of subtraction. (Arts. 80, 114.) 
 
 151. a. Name the fundamental rules of Arithmetic. Why are these rules called 
 fundamental? 
 
ARTS. 151-154.] FUNDAMENTAL RULES. 85 
 
 APPLICATIONS OF THE FUNDAMENTAL RULES. 
 
 1 52. When the sum of two numbers and one of the numbers 
 are given, to find the other number. 
 
 From the given sum, subtract the, given number, and the remainder 
 will be the other number. 
 
 Ex. 1. The sum of two numbers is 87, one of which is 25: 
 what is the other number ? 
 
 Solution. 87" 25 = 62, the other number. (Art. 72.) 
 PN.OOF. 62+25 = 87, the given sum. (Ax. 11.) 
 
 2. A and B together own 350 acres of land, 95 of which be- 
 long to A : how many does B own ? 
 
 3. Two merchants bought 1785 bushels of barley together, one 
 of them took 860 bushels : how many bushels did the other have? 
 
 153* When the difference and the greater of two numbers are 
 given, to find the less. 
 
 Subtract the difference from the greater, and the remainder will 
 be the less number. 
 
 4. The greater of two numbers is 72, and the difference be- 
 tween them is 28 : what is the less number ? 
 
 Solution. 72 28 = 44, the less number. (Art. 72.) 
 PROOF. 44 + 28 = 72, the greater number. (Art. 73. Obs.) 
 
 5. A man bought a horse and chaise ; for the chaise he gave 
 265 dollars, which was 75 dollars more than he paid for the 
 horse : how much did he give for the horse ? 
 
 6. A traveler met two droves of sheep ; the first contained 
 1250, which was 125 more than the second had: l>ow many 
 Bheep were there in the second drove ? 
 
 1 5 t. When the difference and the less of two numbers ar 
 given, to find the greater. 
 
 QUEST. 152. When the sum of two numbers and one of them are given, how Is the other 
 found 1 153. When the difference and the greater of two numbers are {river, how is the 
 less found ? 154. When the difference and the less of two numbers are given, how is the 
 jreater found I 
 
 8 
 
86 APPLICATIONS OP [SECT. V 
 
 Add the difference and the less number together, and the sum will 
 be tlif. greater number. (Art. 73. Obs.) 
 
 7. The difference between two numbers is 12, and the less 
 number is 45 : what is the greater number ? 
 
 Solution, 45+12=57, the greater number. 
 PROOF. 57 45=12, the given difference. (Art. 72.) 
 
 8. A is worth 1890 dollars, and B is worth 350 dollars mcc 
 than A : how much is B worth ? 
 
 9. A man's expenses are 2561 dollars a year, and his in-line 
 exceeds his expenses 875 dollars : how much is his income \ 
 
 155* When the sum and difference of two numbers are given, 
 to find the two numbers. 
 
 From the sum subtract the difference, divide the remainder by 2, 
 and the quotient will be the smaller number. 
 
 To the smaller number thus found, add the given difference, and 
 the sum will be tlie larger number. 
 
 10. The sum of two numbers is 48, and their difference is 18 : 
 what are the numbers ? 
 
 Solution. 48 18=30, and 30^-2=15, the smaller number. 
 And 15 + 18=33, the greater number. 
 
 PROOF. 33 + 15=48, the given sum. (Ax. 11.) 
 
 11. The sfim of the ages of two men is 173 years, and the 
 difference between them is 15 years : what are their ages ? 
 
 12. A man bought a span of horses and a carnage for 856 
 dollars ; the carriage was worth 165 dollars more than the horses : 
 what was the price of each ? 
 
 156. When the product of two numbers and one of the 
 
 numbers are given, to find the other number. 
 
 Divide the given product by the given number, and the p>tient 
 will be Hue, number required. (Art. 91.) 
 
 QUEST. 155. When the sum and difference of two numbers are given, how are th 
 numbers found 1 156. When the product of two numbers and one of them arc given, now 
 la the other found ? 
 

 ARTS. 155-158.] FUNDAMENTAL RULES 87 
 
 13. The product of two numbers is 144, and one of the nun 
 bers is 8 : what is the other number ? 
 
 Solution. 144-r8=18, the required number. (Art. 120.) 
 PROOF. 18X8 = 144, the given product, (Art. 88.) 
 
 14. The product of A and B's ages is 3250 years, and B's age 
 ie 50 years : what is the age of A ? 
 
 15. The product of the length of a field multiplied by its 
 fcieadth is 15925 rods, and its breadth is 91 rods : what is its 
 length ? 
 
 157. When the divisor and quotient are given, to find the 
 dividend. 
 
 Multiply the given divisor and quotient together, and the product 
 will be the dividend. (Art. 121.) 
 
 16. If a certain divisor is 12, and the quotient is 30, what is 
 the dividend ? 
 
 Solution. 30X12=360, the dividend required. 
 PROOF. 360-^- 12=30, the given quotient. (Art. 120.) 
 
 17. If the quotient is 275 and the divisor 683, what must be 
 the dividend ? 
 
 18. If the divisor is 1031 and the quotient 1002, what must 
 be the dividend ? 
 
 158 When the dividend and quotient are given, to find the 
 divisor. 
 
 Divide the given dividend by the given quotient, and the quotient 
 thus obtained will be the number required. (Art. 122.) 
 
 19. A certain dividend is 864, and the quotient is 12 : what is 
 the divisor ? 
 
 Solution. 864-^12=72, the divisor required. (Art. 120.) 
 PROOF. 72X12=864, the given dividend. (Art. 121.) 
 
 20. A gentleman handed a purse containing 1152 shillings, to 
 
 QUK-ST. 157. When the divisor and quotient are given, how is the dividend fccnd 1 
 148. When the dividend and quotient are given, how is the divisor found ? 
 
88 PROPERTIES [SECT. V* 
 
 a company of beggars, which was sufficient to give them 24 shil 
 lings apiece : how many beggars were there ? 
 
 21. A farmer having 2500 sheep, divided them into flocks of 
 125 each : how many flocks did they make ? 
 
 159. When the product of three numbers and two of the 
 
 numbers are given, to find the other number. 
 
 Divide the given product by the product of the two oiven w*jn 
 ers, and the quotient will be the other number. 
 
 22. There are three numbers whose product is 288 ; one of 
 them is 8 and another 9 : it is required to find the other number. 
 
 Solution. 9X8=72 ; and 288-f-72=4, the number required. 
 PROOF. 9X8X4=288, the given product. 
 
 23. The product of three persons' ages is 14880 years ; the 
 rtge of the oldest is 31 years, and that of the second is 24 years : 
 what is the age of the youngest ? 
 
 24. If a garrison of 75 men have 18750 pounds of meat, 
 how long will it last them, allowing 25 pounds to each man per 
 month ? 
 
 25. The sum of two numbers is 3471, and the less is 1629: 
 what is the greater ? 
 
 26. The sum of two numbers is 4136, and the greater is 3074 : 
 what is the less ? 
 
 27. The difference between two numbers is 128, and the greater 
 is 760 : what is the less ? 
 
 28. The difference between two numbers is 340, and the less 
 is 634 : what is the greater? 
 
 29. The sum of two numbers is 12640, and their difference is 
 1608 : what are the numbers ? 
 
 30. The sum of two numbers is 25264, and their dkTerence 
 is 736 : what are the numbers ? 
 
 31. The sum of two numbers is 42126, and then* difference is 
 176 : what are the numbers ? 
 
 32. The product of two numbers is 246018, and one cr ? them 
 is 313 : what is the other number ? 
 
ARTS. 159, 160."] PROFER-TIES OF NUMBERS. 89 
 
 SECTION VI. 
 PROPERTIES OF NUMBERS.* 
 
 ART. 16O. The progress as well as the pleasure of the student 
 i& Arithmetic, depends very much upon the accuracy of his knowl- 
 edge of the terms, which are employed in mathematical reasoning. 
 Particular pains should therefore be taken to understand their 
 true import. 
 
 DBF. 1. An integer signifies a whole number. (Art. 28. Obs. 2.) 
 
 2. Whole numbers or integers are divided into prime and com- 
 posite numbers. 
 
 3. A composite number, we have seen, is one which may be 
 produced by multiplying two or more numbers together ; as, 4, 
 10, 15. (Art. 95.) 
 
 4. A prime number is one which cannot be produced by multi- 
 plying any two or more numbers together ; or which cannot be 
 exactly divided by any whole number, except a unit and itself. 
 Thus, 1, 2, 3, 5, 7, 11, 13, &c., are prime numbers. 
 
 OBS. 1. One number is said to be prime to another, when a unit is the only 
 number by which both can be divided without a remainder. 
 
 2. The learnei must be careful not to confound numbers which are prime 
 to ?ach other with prime numbers ; for numbers that are prime to each other, 
 may thenselves be composite numbers. Thus 4 and 9 are prime to each 
 other, while they are composite numbers. . 
 
 3. The number of prime numbers is unlimited. For those under 3, see 
 Table, page 94. 
 
 5. An even number is one which can be divided by 2 without 
 a remainder ; as, 4, 6, 8, 10. 
 
 QUEST. 160. Upon what does the progress and pleasure of the student in Arithmetic 
 ery much depend ? What is an integer 1 What is a composite number ? What is a 
 prime number? Are prime numbers divisible by other numbers? Obs. When is one 
 number said to be prime to another ? How many prime numbers are there ? What is an 
 Ten number? An odd number? Obs. Are even numbers prime or composite ? What 
 Is tru of odd numbers in this respect? 
 
 * Ba low on the Theory of Numbers ; also, Bonny< istle's Arithmetic. 
 

 . 
 
 90 PROPERTIES OF NUMBERS. [SECT. VI. 
 
 6. An odd number is one which cannot be divided by 2 with- 
 out a remainder ; as, 1, 3, 5, 7, 0, 15. 
 
 OBS. All even numbers except 2, are composite numbers ; an odd number ia 
 sometimes a composite, and sometimes a prime number. 
 
 7. One number is a measure of another, when the former is 
 contained in the latter, any number of times without a remainder. 
 T&us, 3 is a measure of 15 ; 7 is a measure of 28, &c. 
 
 8. One number is a multiple of another, when the former can 
 *b? divided by the latter without a remainder. Thus, 6 is a mul- 
 tiple of 3 ; 20 is a multiple of 5, &c. 
 
 OBS. A multiple is therefore a composite number, and the number thus con- 
 tained in it, is always one of its factors. 
 
 9. The aliquot parts of a number, are the parts by which it 
 can be measured or divided without a remainder. Thus, 5 and 7 
 are the aliquot parts of 35. 
 
 10. The reciprocal of a number is the quotient arising from 
 di viding a unit by that number. Thus, the reciprocal of 2 is % ; 
 the reciprocal of 3 is % ', &c. 
 
 11. The difference between a given number and 10, 100, 1000, 
 &c., that is, between the given number and the next higher order, 
 is called the ARITHMETICAL COMPLEMENT of that number. Thus, 
 3 is the complement of 7 ; 15 is the complement of 85. 
 
 OBS. The arithmetical complement of a number consisting of one integral 
 figure, either with or without decimals, is found by subtracting the number 
 from 10. If there are two integral figures, they are subtracted from 100; if 
 three, from 1000, &c. 
 
 12. A perfect number is one which is equal to the sum of all 
 its aliquot parts. Thus, 6 = 1 + 2 + 3, the sum of its aliquot parts, 
 and is a perfect number. 
 
 OBS. 1. All the numbers known, to which this property really belongs, are 
 the following: 6; 28; 496; 8128; 33,550,336; 8,589,869,056; 137,438,691,328 
 and 2,305,843,008,139,952,128.* 
 
 2. All perfect numbers terminate with 6, or 28. 
 
 QUEST. When is one number a measure of another ? What is a multiple Wha laiw 
 aliquot parts ? What is the reciprocal of a number 1 
 
 * Hutton's Mathematical Recreations. 
 
ARTS. 1G), 1GL] PROPERTIES OF NUMBERS. 91 
 
 161. By the term properties of numbers, is meant those 
 qualities or elements which are inherent and inseparable- frjm 
 them. Some of the more prominent are the following : 
 
 1. The sum of any two or more even numbers, is an even number. 
 
 2. The difference of any two even numbers, is an even number. 
 
 3. The sum or difference of two odd numbers, is even ; but the 
 sum of three odd numbers, is odd. 
 
 4. The sum of any even number of odd numbers, is even ; but 
 the sum of any odd number of odd numbers, is odd. 
 
 5. The sum, or difference, of an even and an odd number, is an 
 odd number. 
 
 6. The product of an even and an odd number, or of two even 
 numbers, is even. 
 
 7. If an even number be divisible by an odd number, the 
 quotient is an even number. 
 
 8. The product of any number of factors, is even, if any one of 
 them be even. 
 
 9. An odd number cannot be divided by an even number with- 
 out a remainder. 
 
 10. The product of any two or more odd numbers, is an odd 
 number. 
 
 31. If an odd number divides an even number, it will also 
 divide the half of it. 
 
 12. If an even number be divisible by an odd number, it will 
 also be divisible by double that number. 
 
 13. Any number that measures two others, must likewise 
 measure their sum, their difference, and their product. 
 
 14. A number that measures another, must also measure its 
 multiple, or its product by any whole number. 
 
 15. Any number expressed by the decimal notation, divided 
 by 9, will leave the same remainder, as the sum of its figures or 
 digits divided by 9. 
 
 Demonstration. Take any number, as 6357 ; now separating it into its seve- 
 ral parts, it becomes 6000+300+50+7. But 6000=6 XI 000 =6X(999+1) 
 =6X999+6. In like manner 300=3x99+3, and 50=5X9+5. Hence 
 6357=6X999-(-3x99+5X9+6+3+5+7; and 6357- 9 =(6x91 9+ 3X99-+ 
 
 Q.UEST. 161. What is meant by properties <f numbers ? 
 T.H. 5 
 
92 PROPERTIES OF NUMBERS. [SECT. VL 
 
 5X9+G+3+5+7)-i-9. But 6X999+3X99+5X9 is evidently divisible by 9 , 
 therefore if (5357 be divided by 9, it will leave the same remainder as G-j-3+5-J- 
 7-7- !). The same will be found true of any other .number whatever. 
 
 OBS. 1. This property of the number 9 affords an ingenious method of proving 
 each of the fundamental rules. (Arts. 90, 123.) The same property belong? to 
 the number 3; for, 3 is a measure of 9, and will therefore be contained an ex- 
 o 3t number of times in any number of 9s. But it belongs to no other digit. 
 
 2. The preceding is not a necessary but an incidental property of the num- 
 
 Ler 9. It arises from the law of increase in the decimal notation. If the radix 
 
 1 the system were 8, it would belong to 7 ; if the radix were 12, it would be- 
 
 cwg to 1 1 ; and universally, it belongs to the number that is one less than the 
 
 radix of the system of notation. 
 
 16. If the number 9 is multiplied by any single figure or digit 
 the sum of the figures composing the product, will make f. 
 Thus, 9X4 = 36, and 3 + 6 = 9. 
 
 17. If we take any two numbers whatever ; then one of them, 
 or their sum, or their difference, is divisible by 3. Thus, take 1 1 
 and 1 7 ; though neither of the numbers themselves, nor their sum 
 is divisible by 3, yet their difference is, for it is 6. 
 
 18. Any number divided by 11, will leave the same remainder, 
 as the sum of its alternate digits in the even places reckoning 
 from the right, taken from the sum of its alternate digits in the 
 odd places, increased by 11 if necessary. 
 
 Take any number, as 38405603, and mark the alternate fig- 
 ures-. Now the sum of those marked, viz: 8+0+6 + 3 = 17. 
 The sum of the others, viz : 3+4 + 5 + 0=12. And 17 12=5, 
 the remainder sought. That is, 38405603 divided by 11, will 
 leave 5 remainder. 
 
 Again, take 5847362, the sum of the marked figures is 14; 
 the sum of those not marked is 21. Now 21 taken from 25, 
 (14 + 11,) leaves 4, the remainder sought. 
 
 19. Every composite number may be resolved into prime factors. 
 For, since a composite number is produced by multiplying two or 
 more factors together, (Art. 160. Def. 3,) it may evidently be re- 
 solved into those factors ; and if these factors themselves are com- 
 posite, they also may be resolved into other factor?, and thus the 
 analysis may be continued, until all the factors are prime numbers. 
 
 20. The least divisor of every number is <i prime number. 
 For, every whole number is either prime, or composite ; (Art. 160. 
 
ART. 161. a.] PROPERTIES OF NUMBERS. 93 
 
 Def. 2 ;) but a composite number, we have just seen, can be re- 
 solved into prime factors ; consequently, the least divisor of every 
 number must be a prime number. 
 
 21. Every prime number except 2, if increased or diminished 
 by 1, is divisible by 4. See table of prime numbers, next page. 
 
 22. Every prime number except 2 and 3, if increased or 
 diminished by 1, is divisible by 6. 
 
 23. Every prime number, except 2 and 5, is contained without 
 remainder, in the number expressed in the common notation by 
 
 as many 9s as there are units, less one, in the prime number itself.* 
 Thus, 3 is a measure of 99 ; 7 of 999,999 ; and 13 of 999,999, 
 999,999. 
 
 24. Every prime number, except 2, 3, and 5, is a measure of 
 the number expressed in common notation, by as many Is as there 
 are units, less one, in the prime number. Thus, 7 is a measure 
 of 111,111 ; and 13 of 111,111,111,111. 
 
 25. All prime numbers except 2, are odd ; and consequently 
 terminate with an odd digit. (Art. 160. Def. 5.) 
 
 Note. I. It must not be inferred from this that all odd numbers are prime. 
 (Art. 1GO. Def. 6 Obs.) 
 
 2. It is plain that any number terminating with 5, can be divided by 5 with- 
 out a remainder. Hence, 
 
 26. All prime numbers, except 2 and 5, must terminate with 
 1, 3, 7, or 9 ; all other numbers are composite. 
 
 1 6 1 a. To find the prime numbers in any series of numbers. 
 
 Write in their proper order all the odd numbers contained in the 
 series. Then reckoning from 3, place a point over every third num- 
 ber in the series j reckoning from 5, place a point over every fifth 
 number reckoning from 7, place a point over every seventh num- 
 ber, and, so on. The numbers remaining without points, together 
 with the number 2, are the primes required. 
 
 Take the series of numbers up to 40, thus, 1, 3, 5, 7, 9, 11, 13, 
 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39 ; then adding tho 
 number 2, the primes are 1, 2, 3, 5, 7, 11, 13, &c. 
 
 AW^. This method of excluding the numbers which are not prime from a 
 series, was invented by Eratosthenes, and is therefore called Eratosthenes' Sieve, 
 
 * Thoie des Nombres, par M. Legendre 
 
94 PROPERTIES OF NUMBERS. [SECT. 
 
 TABLE OF PRIME NUMBERS FROM 1 TO 3413. 
 
 11173 
 
 409 
 
 659 941 
 
 12231511 
 
 1811 
 
 2129 
 
 2423 2741 
 
 3079 
 
 2 
 
 179 
 
 419: 
 
 661 
 
 947 
 
 1229 
 
 1523 
 
 1823 
 
 2131 
 
 2437 
 
 2749 
 
 3083 
 
 3 
 
 181 
 
 421 
 
 673 953 
 
 1231 
 
 1531 
 
 1831 
 
 2137 
 
 2441 
 
 2753 
 
 3089 
 
 5 
 
 191 
 
 431 
 
 677 
 
 967 
 
 1237 
 
 1543 
 
 1847 
 
 2141 
 
 2447 
 
 2767 
 
 3109 
 
 7 i 
 
 193 
 
 433 
 
 683 
 
 971 
 
 1249 
 
 1549 
 
 1861 
 
 2143 
 
 2459 
 
 2777 
 
 3119 
 
 11 
 
 197 
 
 439 
 
 691 
 
 977 
 
 1259 
 
 1553 
 
 1867 
 
 2153 
 
 2467 
 
 2789 
 
 3121 
 
 13! 
 
 199 
 
 443 
 
 701 
 
 983 
 
 1277 
 
 1559 
 
 1871 
 
 2161 
 
 2473 
 
 2791 
 
 3137 
 
 17 
 
 211 
 
 449 
 
 709 
 
 991 
 
 1279 
 
 1567 
 
 1873 
 
 2179;2477 
 
 2797 
 
 3163 
 
 19 
 
 223 
 
 457 
 
 719 
 
 997 
 
 1283 
 
 1571 
 
 1877 
 
 2203 
 
 25032801 
 
 3167 
 
 23 
 
 227461 
 
 727 
 
 1009 
 
 1289 
 
 15791879 
 
 2207 
 
 2521 
 
 2803 
 
 3169 
 
 29 
 
 229 
 
 463 
 
 733 
 
 1013 
 
 1291 
 
 1583)1889 
 
 2213 
 
 2531 
 
 28193181 
 
 31 
 
 233467J739 
 
 1019 
 
 1297 
 
 1597 
 
 1901 
 
 2221 
 
 2539 2833 
 
 3187 
 
 37 
 
 239 
 
 479 
 
 743 
 
 1021 
 
 1301 
 
 1601 
 
 1907 
 
 2237 
 
 2543 
 
 2837 
 
 3191 
 
 41 
 
 241 
 
 487 
 
 751 
 
 1031 
 
 1303 
 
 1607 
 
 1913 
 
 2239 
 
 2549 
 
 2843 3203 
 
 43 
 
 251 
 
 491 
 
 757 
 
 1033 
 
 1307 
 
 16091931 
 
 2243 
 
 2551 
 
 2851 
 
 3209 
 
 47257 
 
 499 
 
 761 
 
 1039 
 
 1319 
 
 1613 
 
 1933 
 
 2251 
 
 2557 
 
 2857 
 
 3217 
 
 53 
 
 263 
 
 503 
 
 769 
 
 1049 
 
 1321 
 
 1619 
 
 1949 
 
 2267 
 
 2579 
 
 2861 
 
 3221 
 
 59 
 
 269 
 
 509 
 
 773 
 
 1051 1327 
 
 1621 
 
 1951 
 
 226912591 
 
 2879'3229 ' 
 
 61 271 
 
 521 
 
 787 
 
 1061 
 
 1361 
 
 1627 
 
 1973 
 
 2273 
 
 2593 
 
 2887 
 
 3251 
 
 67|277523 
 
 797 
 
 1063 
 
 1367 
 
 1637 
 
 1979 
 
 2281 
 
 2609 
 
 2897 
 
 3253 
 
 71 
 
 281 
 
 541 
 
 809 
 
 1069 
 
 1373 
 
 1657 
 
 1987 
 
 22872617 
 
 2903 
 
 3257 
 
 73 
 
 283 
 
 547 
 
 811 
 
 1087 
 
 1381 
 
 1663 1993 
 
 2293 
 
 2621 
 
 2909 
 
 3259 
 
 79 
 
 293 
 
 557 
 
 821 
 
 1091 
 
 1399 
 
 1667 1997 
 
 2297 
 
 2633 
 
 2917 
 
 3271 
 
 83 
 
 307 
 
 563 
 
 823 
 
 1093 
 
 1409 
 
 1669 1999 
 
 2309 2647 
 
 2927 
 
 3299 
 
 89 
 
 311 
 
 569 
 
 827 
 
 1097 
 
 ! 1423 
 
 169312003 
 
 2311 2657 
 
 2939 
 
 3301 
 
 97 
 
 313 
 
 571 829 
 
 1103 
 
 J1427 
 
 1697 
 
 2011 
 
 2333 
 
 2659 
 
 2953 
 
 3307 
 
 101 
 
 317 
 
 577 839 
 
 1109 
 
 1429 
 
 1699 
 
 2017 
 
 2339 
 
 2663 
 
 2957 
 
 3313 
 
 103 
 
 331J587 
 
 853 
 
 1117 
 
 1433 
 
 1709 
 
 2027 
 
 2341 
 
 2671 
 
 2963 
 
 3319 ( 
 
 107 
 
 3371593 
 
 857 
 
 1123 
 
 1439 
 
 1721 
 
 2029 
 
 2347 
 
 2677 
 
 2969 3323 
 
 109 
 
 347 
 
 !599 
 
 859 
 
 1129 1447 
 
 1723 
 
 2039 
 
 2351 
 
 2683 
 
 2971 
 
 3329 
 
 113 
 
 349601 
 
 863 
 
 1151 
 
 ;1451 
 
 1733 
 
 2053 
 
 2357 
 
 2687 
 
 29993331 
 
 127 
 
 353 
 
 :607 
 
 877 
 
 1153 
 
 1453 
 
 1741 
 
 2063 
 
 2371 
 
 2689 
 
 3001 
 
 13343 
 
 131 
 
 359613 
 
 ;881 
 
 1163 1459 
 
 17472069 
 
 2377|2693 
 
 3011 
 
 '3347 
 
 137 
 
 367 
 
 ;ei7 
 
 883 
 
 1171 
 
 1471 
 
 1753 
 
 2081 
 
 2381 
 
 2699 3019 
 
 13359 
 
 139 
 
 373 
 
 ;ei9 
 
 ,887 
 
 1181 
 
 1481 
 
 1759 
 
 2083 
 
 2383 
 
 2707 
 
 3023 
 
 3361 
 
 149 
 
 379 
 
 !631 
 
 907 
 
 11871483 
 
 1777 
 
 2087 
 
 2389 
 
 ,2711 3037 
 
 3371 
 
 151 
 
 383 
 
 ! 641 
 
 1911 
 
 11931487 
 
 1783 
 
 2089 
 
 2393 
 
 2713 
 
 3041 
 
 3373 
 
 157 
 
 389643 
 
 919 
 
 1201 
 
 1489 
 
 1787 
 
 2099 
 
 2399 2719 
 
 30493389 
 
 163 
 
 397 
 
 |647 
 
 '929 
 
 1213 1493 
 
 1789 
 
 2111 
 
 241112729 
 
 3061 
 
 ; 3391 
 
 167 
 
 401 
 
 '6531937 
 
 12l7l499!l801 
 
 2113 
 
 2417 2731 
 
 3067|3407 
 
ART. 162.J PROPERTIES OF NUMBERS. 95 
 
 DIFFERENT SCALES OF NOTATION. 
 
 162. A number expressed in the decimal notation, may be 
 changed to any required scale of notation in the following manner. 
 Divide the given number by the radix of the required scale con- 
 tinually, till the quotient is less than the radix ; then annex to the 
 fast quotient the several remainders in a retrograde order; placing 
 iphers where there is no remainder, and the result will be t/te num~ 
 in the scale required. (Arts. 43, 44.) 
 
 Ex. 1. Express 429 in the quinary scale of notation. 
 
 Explanation. By Dividing the given number 5)429 
 by 5, it is evidently distributed into 85 parts, 5) 85 4 
 each of which is equal to 5, with 4 remainder. 5) 17 
 Dividing again by 5, these parts are distributed 3 2 
 
 into 17 other parts, each of which is equal to 5 Ans. 3204 
 
 times 5, and the remainder is nothing. Dividing by 5 the third 
 time, the parts last found are again distributed into 3 other parts, 
 each of which is equal to 5 times 5 into 5, with 2 remainder. 
 Thus, the given number is resolved into 3X5X5X5+2X5X5+ 
 0X5+4, or 3204, which is the answer required. 
 
 2. Change 7854 from the decimal to the binary scale. 
 
 Ans. 1111010101110. 
 
 3. Change 7854 from the decimal to the ternary scale. 
 
 Ans. 101202220. 
 
 4. Change 7854 from the decimal to the quaternary scale. 
 
 Ans. 1322232. 
 
 5. Change 7854 from the decimal to the quinary scale. 
 
 Ans. 222404. 
 
 6. Change 7854 from the decimal to the senary scale. 
 
 Ans. 100210. 
 
 7. Change 7854 from the decimal to the octary scale. 
 
 Ans. 17256. 
 
 8. Change 7854 from the decimal to the nonary scale. 
 
 Ans. 11680. 
 
 9. Change 7854 from the decimal to the duodecimal scale. 
 
 Ans. 4666. 
 
96 PROPERTIES OF NUMBERS. [SECT. VI. 
 
 10 Change 35261 from the decimal to the quaternary scale. 
 
 11. Change 643175 from the decimal to the octary scale. 
 
 12. Change 175683 from the decimal to the septenary scale. 
 
 13. Change 534610 from the decimal to the octary scale. 
 
 14. Change 841568 from the decimal to the nonary scale. 
 
 15. Change 592835 from the decimal to the duodecimal scale. 
 
 Note. Since every scale requires as many characters as there are units in 
 :1 e radix, we will denote 10 by t, and 11 by e. Ans. 2470 t e. 
 
 163* To change a number expressed in any given scale of 
 notation, to the decimal scale. 
 
 Multiply the left hand figure by the given radix, and to the 
 product add tlie next figure ; then multiply this sum by tlie radix 
 again, and to this product add tlie next figure ; thus continue the 
 operation till all tlie figures in t/ie given number Imve been employed, 
 and the last product will be the number in the decimal scale. 
 
 16. Change 3204 from the quinary to the decimal scale. 
 
 Operation. 
 
 Explanation. Multiplying the left hand figure 3204 
 
 by 5, the given radix, evidently reduces it to the 5 
 
 next lower order ; for in the quinary scale, 5 in 17 
 
 an inferior order make one in the next superior 5 
 
 order. For the same reason, multiplying this 85 
 
 sum by 5 again, reduces it to the next lower 5 
 
 order, <kc. 429 Ans. 
 
 OBS. This and the preceding operations are the same in principle, as reducing 
 compound numbers from one denomination to another. 
 
 17. Change 1522232 from the quaternary to the decimal scale. 
 
 Ans. 7854. 
 
 18. Change 2546571 from the octary to the decimal scale. 
 
 19. Change 34120521 from the senary to the decimal scale. 
 
 20. Change 145620314 from the septenary to the decimal scale. 
 
 21. Change 834107621 from the nonary to the decimal scale. 
 
 22. Change 403130021 from the quinary to the decimal scale. 
 
 23. Change 704400316 from the octary to the decimal scale. 
 
 24. Change 903124106 from the duod ecimal to the decimal scale. 
 
ARTS. 163-165.] PROPERTIES OF NUMBERS. 97 
 
 ANALYSIS OF COMPOSITE NUMBERS. 
 
 164. Every composite number, it has been shown, may be 
 resolved into prime factors. (Art. 1C 1. Prop. 19.) 
 
 Ex. 1. Resolve 210 into its prime factors. 
 
 Operation. We first divide the given number by 2, which 
 
 2)210 is the least number that will divide it with" 
 
 3)105 out a remainder, and which is also a prime 
 
 6)35" number. (Prop. 20.) We next divide by 3, 
 
 7 then by 5. The several divisors and the last 
 
 Am. 2, 3, 5, and 7. quotient are the prime factors required. 
 
 PROOF. 2X3X5X7 = 210. Hence. 
 
 1 G5 To resolve a composite number into its prime factors. 
 
 Divide the given number by the smallest number which will di- 
 vide it without a remainder ; then divide the quotient in the same 
 way, and thus continue the operation till a quotient is obtained 
 which can be divided by no number greater than 1. The several 
 divisors with the last quotient, will be thv prime factors required. 
 (Art. 161. Prop. 19.) 
 
 Demonstration. Every division of a number, it is plain, resolves it into two 
 factors, viz: the divisor and dividend. (Art. 112.) But according to the rule, 
 the divisors, in every case, are the smallest numbers that will divide the given 
 number and the successive quotients without a remainder; consequently they 
 are all prime numbers. (Art. 101. Prop. 20.) And since the division is con- 
 tinued till a quotient is obtained, which cannot be divided by any number 
 greater than 1, it follows that the last- quotient must also be a prime number; 
 for, a j rime number is one which cannot be exactly divided by any whole 
 number except a unit and itself. (Art. 100 Def. 4.) 
 
 OBS. 1. Since the least divisor of every number is a prime number, it is evi- 
 dent that a composite number may be resolved into its prime factors, by divid- 
 ing it continually by any prime number that will divide the given number and 
 rhe quotients without a remainder. Hence, 
 
 2. A composite number can be divided by any of its pr i me factors without u 
 remainder, and by the product of any two or more of them, but by 110 ofhcf 
 number. Thus, the prime factors of 42 are 2, 3, and 7. Now 42 can be di- 
 
 IBS- How do you resolve a composite number into its prime factors 1 Obi. Will 
 the aaiae result be obtained, if we divide by any of its prime factors ? 
 
98 GREATEST COMMON [SECT VI. 
 
 vided by 2, 3, and 7; also by 2x3, 2X7, 3x7, and 2x3X^5 but it (an be 
 divided by no other number. 
 
 2. Resolve 4 and 6 into their prime factouj. 
 Solution. 4 = 2X2 ; and 6 = 2X3. 
 
 3. Resolve 8 into its prime factors. Ans. 8=2X2X2. 
 
 Resolve the following composite numbers into their prime 
 actors : 
 
 4. 9. 22. 34. 40. 57. 58. 81. 
 
 5. 10. 23. 35. 41. 58. 59. 82. 
 
 6. 12. 24. 36. 42. 60. 60. 84. 
 
 7. 14. 25. 38. 43. 62. 61. 85. 
 
 8. 15. 26. 39. 44. 63. 62. 86. 
 
 9. 16. 27. 40. 45. 64. 63. 87. 
 
 10. 18. 28. 42. 46. 65. 64. 88. 
 
 11. 20. 29. 44. 47. 66. 65. 90. 
 
 12. 21. 30. 45. 48. 68. 66. 91. 
 
 13. 22. 31. 46. 49. 69. 67. 92. 
 
 14. 24. 32. 48. 50. 70. 68. 93. 
 
 15. 25. 33. 49. 51. 72. 69. 94. 
 
 16. 26. 34. 50. 52. 74. 70. 95. 
 
 17. 27. 35. 51. 53. 75. 71. 96. 
 
 18. 28. 36. 52. 54. 76. 72. 98. 
 
 19. 30. 37. 54. 55. 77. 73. 99. 
 
 20. 32. 38. 55. 56. 78. 74. 100. 
 
 21. 33. 39. 56. 57. 80. 75. 108. 
 
 76. Resolve 120 and 144 into their prime factors. 
 
 77. Resolve 180 and 420 into their prime factors. 
 
 78. Resolve 714 and 836 into their prime factors. 
 
 79. Resolve 574 and 2898 into their prime factors. 
 BO. Resolve 11492 and 180 into their prime factors. 
 
 81. What are the prime factors of 650 and 1728 ? 
 
 82. What are the prime factors of 1492 and 8032 ? 
 
 83. What are the prime factors of 4604 and 16806 ? 
 
 84. What are the prime factors of 71640 and 20324 ? 
 
 85. What are the prime factors of 84705 and 65948 ? 
 
 86. What are the prime factors of 92353 and 8 T 3784 ? 
 
ARTS. 166-168.] DIVISOR. 99 
 
 GREATEST COMMON DIVISOR. 
 
 166 A common divisor of two or more numbers, is a num- 
 ber which' will divide each of them without a remainder. Thus 
 
 2 is a common divisor of 6, 8, 12, 16, 18, &c. 
 
 167. The greatest common divisor of two or more numbers, 
 is the greatest number which will divide th em without a remainder. 
 Thus 6 is the greatest common divisor of 12, 18, 24, and 30. 
 
 OBS. A common divisor is sometimes called a common measure. It will be 
 eefcn that a common divisor of two or more numbers, is simply a factor which 
 is common to those numbers, and the greatest common divisor is the greatest 
 factor common to them. Hence, 
 
 1G8 To find a common divisor of two or more numbers. 
 
 Resolve each number into two or more factors, one of which sJiall 
 be common to all the given numbers. 
 
 Or, resolve the given numbers into their prime factors, then if 
 the same factor is found in each, it will be a common divisor. (Art. 
 1C5. Obs. 2.) 
 
 OBS. If the given numbers have not a common factor , they cannot have a 
 common divisor greater than a unit ; consequently they are either prime num- 
 bers, or are prime to each other. (Art. 1150. Def. 3. Obs. 2.) 
 
 Note. The following facts may assist the learner in finding common di- 
 visors : 
 
 1. Any numbe* ending in 0, or an even number, as 2, 4, 6, &c., may be 
 divided by 2. 
 
 2. Any number ending in 5 or 0, may be divided by 5. 
 
 3. Any number ending in 0, may be divided by 10. 
 
 4. When the two right hand figures are divisible by 4, the whole number 
 may be divided by 4. 
 
 5. If the three right hand figures of any number are divisible by 8, the 
 whole is divisible by 8. 
 
 Ex. 1. Find a common divisor of 6, 15, and 21. 
 
 Solution. 6 = 3X2; 15 = 3X5; and 21 = 3x7. The factor 
 
 3 is common to each of the given numbers, and is therefore a 
 common divisor of them. 
 
 <iuKST. 166. What is a common divisor of two or more numbers ? 167, What is the 
 greatest common divisor of two or more numbers ? Obs. What is a common divisor some- 
 times called! 168. How do you find a common divisor of two or more numbers] Obi. I" 
 two given numbers have not a common factor, what is true as to a common divisor ? 
 
 5* 
 
tOO GREATEST COMMON [SECT. VI. 
 
 2. Find a common divisor of 15, 18, 24, and 36. 
 
 3. Find a common divisor of 14, 28, 42, and 35. 
 
 4. Find a common divisor of 10, 35, 50, 75, and 30. 
 6. Find a common divisor of 82, 118, and 146. 
 
 6. Find a common divisor of 42 and 66. Ans. 2, 3, or 6. 
 
 169. It will be seen from the last example that two numbers 
 may have more than one common divisor. In many cases it is 
 highly important to find the greatest divisor that will divide two 
 or more given numbers without a remainder. 
 
 7. What is the greatest common divisor of 35 and 50 ? 
 
 Operation. Dividing 50 by 35, the remainder is 15, 
 
 35)50(1 then dividing 35 (the preceding divisor) by 
 
 35 15 (the last remainder) the remainder is 5 ; 
 
 15)35(2 finally, dividing 15 (the preceding divisor) by 
 
 30 5 (the last remainder) nothing remains ; con- 
 
 5)15(3 sequently 5, the last divisor, is the greatest 
 
 15 common divisor. Hence, 
 
 1 7 O. To find the greatest common divisor of two numbers. 
 
 Divide the greater number by the less ; then divide the preceding 
 divisor by the last remainder, and so on, till nothing remains. 
 The last divisor will be the greatest common divisor. 
 
 When there are more than two numbers given. 
 
 first find the greatest common divisor of any two of tliem ; 
 then, that of the common divisor tJms obtained and of another 
 given number, and so on through all the given numbers. The last 
 common divisor found, will be the one required. 
 
 Demonstration. Since 5 is a measure of the last dividend 15, in the preced- 
 ing solution, it must therefore be a measure of the preceding dividend 35; be- 
 cause 35 2xl5-(-5; and 35 is oae of the given numbers. Now, since 5 
 measures 15 and 35, it must also measure their sum, viz : 35-J-15, or 50, which 
 18 the other given number. (Art. 161. Prop. 13.) In a similar manner it may 
 be shown that the last divisor will, in all cases, be the greatest common divisor. 
 
 Note. Numbers which have no common measure greater than 1, are said to. 
 be incommensuro3)le. Thus 17 and 29 are incommensurable. 
 
 . 170. How find the greatest common divisor of two numbers 1 Of more than twol 
 
ARTS. 169-171.] DIVISOR. . *. 101 
 
 8. What is the greatest common divisor of 285 and 465 ? 
 
 9. What is the greatest common divisor of 532 and 1274 ? 
 
 10. What is the greatest common divisor of 888 and 2775 ? 
 
 11. What is the greatest common divisor of 2145 and 3471 ? 
 
 12. What is the greatest common divisor of 1879 and 2425 ? 
 
 13. What is the greatest common divisor of 75, 125, and 100 ? 
 
 Suggestion. Find the greatest common divisor of 75 and 125, 
 which is 25. Then that of 25 and 160. Ans. 5. 
 
 14. What is the greatest common divisor of 183, 3996, 108 ? 
 
 15. What is the greatest common divisor of 672, 1440, and 3472 ? 
 
 16. What is the greatest common divisor of 30, 42, and 66 ? 
 
 Analysis. By resolving the given num- Operation* 
 
 bers into their prime factors, (Art. 165,) 30 = 2X3X5 
 we find that the factors 2 and 3 are both 42 = 2X3x7 
 common divisors of them. But we have 66 = 2X3X11 
 seen that a composite number can be Now 2X3 = 6 Ans. 
 divided by the product of any two or 
 
 more of its prime factors ; (Art. 165. Obs. 2 ;) consequently 30, 
 42, and 66 can all be divided by 2X-3 ; for 2X3 is the product 
 of two prime factors common to each. And since they are the 
 onl\ r factors common to the given numbers, their product must 
 be the greatest common divisor of them. Hence, we deduce a 
 
 171* Second Method of finding the greatest common divisor 
 of two or more numbers. 
 
 Resolve the given numbers into their prime factors, and the con- 
 tinued product of those factors which are common to each, will be 
 the greatest common divisor. 
 
 OBS. If the given numbers have but one common factor, that factor itself ia 
 the greatest common divisor. 
 
 17. What is the greatest common divisor of 105 and 165 ? 
 
 18. What is the greatest common divisor of 36, 60, and 108 
 
 19. What is the greatest common divisor of 108, 126, and 162 ? 
 
 20. What is the greatest common divisor of 140, 210, and 315 ? 
 
 21. What is the greatest common divisor of 24, i2, 54, and 60 ? 
 
 22. What is the greatest common divisor of 5 C, 84, 1 40, and 168 ? 
 
102 LEAST COMMON [SECT. VI. 
 
 LEAST COMMON MULTIPLE. 
 
 172. One number is said to be a multiple of another, when 
 the former can be divided by the latter without a remainder 
 (Art. 160. Def. 8.) Hence, 
 
 173. A common multiple of two or more numbers, is a num- 
 ber which can be divided by eacl of them withe at a remainder. 
 Thus, 12 is a common multiple of 2, 3, and 4; 15 is a common 
 multiple of 3 and 5, (fee. 
 
 OBS. A comvwn multiple is always a composite number, of which each of 
 the given numbers must be t "actor ; otherwise it could not be divided by 
 them. (Art. 1G5. Obs. 2.) 
 
 174. The continued product of two or more given numbers 
 will always form a common multiple of those numbers. The same 
 numbers may have an unlimited number of common multiples ; 
 for, multiplying their continued product by any number, will form 
 a new common multiple, (Art. 161. Prop. 14.) 
 
 175. The least common multiple of two or more numbers, is 
 the least number which can be divided by each of them without a 
 remainder. Thus, 12 is the least common multiple of 4 and 6, for 
 it is the least number which can be exactly divided by them. 
 
 OBS. The least common multiple of two OT more numbers, is evidently 
 composed of all the prime factors of each of the given numbers repeated once, 
 and only once. For, if it did not contain all the prime factors of any one of 
 the given numbers, it could not be divided by tbat number. (Art. 165. Obs. 2.) 
 On the other hand, if any prime factor is employed more limes than it is re- 
 peated as a factor in some one of the given numbers, then it woild not be th 
 least common multiple. 
 
 Ex. 1. What is the least common multiple of 10 and 15 ? 
 
 Analysis. 10 = 2X5, and 15 = 3X5. The prime factors of 
 the given numbers are 2, 5, 3, and 5. Now since the .'actor 5 
 >ccurs once in each number, we may therefore cancel L in one 
 
 . 172. When is one number said to be a multiple of anothejf? 173 What *s a 
 common multiple? 174. How may a common multiple of two t>r more r mbers t 
 formed ? How many common multiples may there be of at y giver lumbers ? 176. What 
 if the least common multiple of two or more nuir oers 1 
 
ARTS. 1T2--176.J MULTIPLE. 103 
 
 instance, and the continued product of the remaining factors 2X3 
 X5, or 30, will be the least common multiple. 
 
 Operation. We first divide both the numbers by 5 
 
 5)10 " 15 in order to resolve them into prime fac- 
 
 2 " 3 tors. (Art. 175. Obs.) Thus, all the dif- 
 
 5X2X3=30^5. ferent factors of which the given num- 
 bers are composed, are found in the divisor and quotients onct, 
 and only once. Therefore the product of the divisor and quotients 
 5X2X3, is the least common multiple required. Hence, 
 
 176. To find the least common multiple of two or more 
 numbers. 
 
 Write the given numbers in a line with two points between them. 
 Divide by the smallest number which will divide any two or more 
 of them without a remainder, and set the quotients and the undivided 
 numbers in a line beloiv. Divide this line and set down the re- 
 sults as before j thus continue the operation till there are no two 
 numbers which can be divided by any number greater than 1. The 
 continued product of the divisors into the numbers in the last line, 
 will be the least common multiple required. 
 
 OBS. 1. We have seen that the least dimsor of every number isaprimt num- 
 ber ; hence, dividing by the smallest number which will divide two or more of 
 the given numbers, is dividing them by a prime number. (Art. 161. Prop. '20.) 
 
 The result will evidently be the same, if, instead of dividing by the smallest 
 number, we divide the given numbers by any prime number, that will divide 
 two or more of them, without a remainder. 
 
 2. The preceding operation, it will be seen, resolves the given numbers into 
 their prime factors, (Art. 1G5,) then multiplies all the different factors together, 
 taking each factor as many times in the product, as are equal to the greatest 
 number of times it is found in either of the given numbers. 
 
 3. If the given numbers are prime numbers, or are prime to each other, the 
 continued product of the numbers themselves will be their least common mul- 
 tiple. (Art. 168. Obs.) Thus, the least common multiple of 5 and 7 is 35; of 
 8 and 9 is 72. 
 
 UUKST. 176. How is the least common multiple of two or more numbers found t 
 Obs. If the Rivennumbers are prime, or are prime to each other, what is the least com 
 mon multiple of them 1 176. a. Upon what principle does this rule depend 1 Obs. Why 
 do you divide by the smallest number that will divide two or more of the given numben 
 without a remainder 1 
 
104 LEAST COMMON [SECT. VI 
 
 Ex. 2, What is the least common multiple of 6, 8, and 12 ? 
 
 Analysis. By resolving the given numbers Operation. 
 
 into their prime factors, it will be seen that 2 6 = 2X ; 3 
 is found once as a factor in 6 ; twice in 1 2 ; and 8 2x2X2 
 three times in 8. It must therefore be taken 12 = 2X2X3 
 three times'in. the product. Again, 3 is a lac- 2X2X2X3=24 
 tor of 6, and 12, consequently it must be taken only once in the 
 product. (Art. 176. Obs. 2.) Thus, 2X2X2X3 = 24 Ans. 
 
 Ex. 3. What is the least common multiple of 12, 18. and 36 ? 
 
 First Operation. Second Operation. Third Operation. 
 
 2)12 " 18 " 36 9)12 " 18 *' 36 12)12 " 18 " 36 
 
 2) 6 " 9 ""18 2)12 " 2 "~4 3) 1 " 18 " 3 
 3)~3~ 9~" 9 2) 6 " 1 " 2 1 " 6~" 1 
 
 3) 1 " 3~" 3 3 " r~" 1 And 12X3X6=216. 
 1 ' F~" T Now 9X2X2X3=108. 
 
 2X2X3X3 = 36 Ans. 
 
 Explanation. In the first operation, we divide by the smallest 
 numbers which will divide any two or more of the given numbers 
 without a remainder, and the product of the divisors, &c., is 36, 
 which is the answer required. 
 
 In the second and third operations, we divide by numbers that 
 will divide two or more of the given numbers without a remainder, 
 and in both cases, obtain erroneous answers. 
 
 Note. It will be seen from the second and third operations above, that 
 tl dividing by any number, which will divide two or more of the given num- 
 bers without a remainder," according to the rule given by some authors, Joes 
 not always give the least common multiple of the numbers. 
 
 176. a. The reason of the preceding rule depends upon the 
 principle that the least common multiple of any two or more num- 
 bers, is composed of all the prime factors of the given numbeis, 
 each taken as many times, as are equal to the greatest numle~ of 
 times it is found in either of the given numbers. (Art. 175. Obs.) 
 
 Note. 1. The reason for dividing by the smallest number, is because tha 
 divisor may otherwise be a composite number, (Art. 1G1. Prop. 20.) and have 
 a factor common to some one of the quotients, or undivided numbers in the 
 last line; consequently the continued product of them wotJd be too large for 
 
ARTS. 176. #. 177.] MULTIPLE. 105 
 
 the kast, common multiple. (Art. 175. Obs.) Thus, in t e second op:/ration the 
 divisor 9, is a composite number, containing the factor 3 common to the 3 in 
 the quotient; consequently the product is three times too lartfc. In the third 
 operation the divisor 12, is a composite number, and contains the factor 6 com- 
 mon to the 6 in the quotient; therefore the product is six times too large. 
 
 2. The object of arranging the given numbers in a line, is that all of them 
 may be resolved into their prime factors at the same time ; and also to present 
 at a giance the factors which compose the least common multiple required. 
 
 4. Find the least common multiple of 6, 9, and 15. 
 
 5. Find the least common multiple of 8, 16, 18, and 24. 
 G. Find the least common multiple of 9, 15, 12, 6, and 5. 
 
 7. Find the least common multiple of 5, 10, 8, 18, and 15, 
 
 8. Find the least common multiple of 24, 16, 18, and 20. 
 
 9. Find the least common multiple of 36, 25, 60, 72, and 36. 
 
 10. Find the least common multiple of 42, 12, 84, and 72. 
 
 11. Find the least common multiple of 27, 54, 81, 14, and 63 
 
 12. Find the least common multiple of 7, 11, 13, 3, and 5. 
 
 177. The process of finding the least common multiple 
 may often be shortened, by canceling every number which will 
 divide any other given number, without a remainder, and also 
 those which will divide any other number in the same line. The 
 least common multiple of the numbers that remain, will be the an- 
 swer required. 
 
 OBS. By attention and practice, the student will be able to discover, by in- 
 spection, the least common multiple of numbers, when they are not large. 
 
 13. Find the least common multiple of 4, 6, 10, 8, 12, and 15. 
 
 Operation. Since 4 and 6, will exactly di- 
 
 2)^ " " 10 " 8 " 12 " 15 vide 8, and 12, we cancel them. 
 
 2) " 4 " 6 " 15 Again, since 5 in the second line 
 
 3) 2 " 3 " 15 will exactly divide 1 5 in the same 
 
 2 " 1 " 5 line, we therefore cancel it, and 
 
 Now, 2X2X3 X2 X 5 = 120 Ans. proceed with the remaining num- 
 
 bers as before. 
 
 14. Find the least common multiple of 9, 12, 72, 36, and 144. 
 
 15. Find the least common multiple of 8, 12 20, 24, and 25. 
 
 16. Find the least common multiple of 1, 2, 3, 4, 5, 6, 7, 8, 9, 
 
106 COMMON [SECT. VI. 
 
 17. Find the least common multiple of 33, 12, 84, and 7. 
 
 18. Find the least common multiple of 54, 81, 63, and 14. 
 
 19. Find the least common multiple of 72, 120, 180, 24, and 36. 
 
 177. a. The least common multiple of two or more numbers 
 may also be found in the following manner. 
 
 First find the greatest common divisor of two of the given num- 
 bers ; by this divide one of these two numbers, and multiply the 
 quotient by the other. Then perform a similar operation on the 
 product and another of the given numbers ; thus continue the pro- 
 cess until all of the given numbers have been employed, and tJie 
 final result will be the least common multiple required. 
 
 20. What is the least common multiple of 24, 16, and 12 ? 
 
 Solution. By inspection, we find the greatest common divisor 
 of 24 and 16, is 8. Now 24-7-8 = 3; and 3X16 = 48. Again,' 
 the greatest common divisor of 48 and 12, is 12. Now 48-f-12 
 =4; and 4X12=48. Ans. 
 
 PROOF. Resolving the given numbers into their prime factors, 
 24 = 2X2X2X3; 16 = 2X2X2X2; and 12 = 2X2X3; (Art. 
 165 ;) consequently, 2X2X2x2X3=48, the least common mul- 
 tiple. (Art. 175. Obs.) 
 
 OBS. The reason of this rule depends upon the principle, that if the product 
 of any two numbers be divided by any factor which is common to both, the 
 quotient will be a common multiple of the two numbers. Thus, if 48, the 
 product of G and 8, be divided by 2, a factor of both, the quotient 24, will be 
 a multiple of each, since it may be regarded either as 8 multiplied by the quo- 
 tient of 6 by the factor 2, or as 6 multiplied by the quotient of 8 by the same 
 factor. Hence, it is obvious, that the greater the common measure is, the less 
 will be the multiple ; and, consequently, the greatest common measure will 
 
 oduce the least common multiple. 
 
 When the common multiple of the first two numbers is found, it is evident, 
 
 at any number which is a common multiple of it and the: third number, will 
 ?* a multiple of the first, second, and third numbers. 
 
 21. What is the least common multiple of 75, 120, and 300 ? 
 
 22. What is the least common multiple of 96, 144, and 720 ? 
 
 23. What is the least common multiple of 256, 512, and 1728 ? 
 
 24. What is the least common multiple of S'/5, 85 J, and 3400 ? 
 
ARTS. 77. a- 181. ] FRACTIONS. 107 
 
 SECTION VII. 
 FRACTIONS. 
 
 ART. 178. When a number or thing is divided into two equal 
 
 parts, one of those parts is called one half. If the number or 
 
 hing is divided into three equal parts, one of the parts is called 
 
 one third ; if it is divided into four equal parts, one of the parts 
 
 is called one fourth, or one quarter ; and, universally, 
 
 When a number or thing is divided into equal parts, the parts 
 take tJwir name from the number of parts into which the thing or 
 number is divided. 
 
 1 7 9 The value of one of these equal parts manifestly depends 
 upon the number of parts into which the given number or thing 
 is divided. Thus, if an orange is successively divided into 2, 3, 
 4, " , 6, &c., equal parts, the thirds will be less than the halves ; 
 the fourths, than the thirds ; the fifths, than the fourths, &c. 
 
 OBS. A half of any number is equal to as many units, as 2 is contained 
 times in that number ; a third of a number is equal to as many, as 3 is con- 
 tained times in the given number ; a fourth is equal to as many, as 4 is con- 
 tained in the number, &c. 
 
 1 8O When a number or thing is divided into tqual parts, 
 these parts are called FRACTIONS. 
 
 OBS. Fractions are used to express parts of a collection of things, as well as 
 of a single thing ; or parts of any number of units, as well as of one unit. 
 Thus, we speak of % of six oranges ; & o f 75 ; & Ct i n this case the collection, 
 or number to be divided into equal parts, is regarded as a whole. 
 
 181. Fractions are divided into two classes, Common and 
 Decimal. For the illustration of Decimal Fractions, see Sec- 
 tion IX. 
 
 QUEST. 178. What is meant by one half 1 ? What is meant by one third? What is 
 meant by a fourth ? What is meant by fifths ? By sixths ? How many sevenths make 
 a whole one ? How many tenths ? What is meant by twentieths ? By hundreds ? When 
 a number or thing is divided into equal parts, from what do the parts take their name 1 
 179. Upon what does the value of one of these equal parts depend 7 180. What are frao 
 tkras ? 181. Into how many classes are fractions divided 1 
 
108 FRACTIONS. [SECT. VII. 
 
 !2. Common Fractions are expressed by two numbers, one 
 plat ^d over the other, with a line between them. One half is 
 written thus ; one third, ^ ; one fourth, } ; nine tenths, -ft- ; 
 thirteen forty-fifths, $, &c. 
 
 The number below the line is called the denominator, and shows 
 into Jtow tnany parts the number or thing is divided. 
 
 The number above the line is called the numerator, and shows 
 Itow many parts are expressed by the fraction. Thus, in the frac 
 tion -f, the denominator 3, shows that the number is divided into 
 three equal parts ; the numerator 2, shows that two of those parts 
 are expressed by the fraction. 
 
 The denominator and numerator together are called the terms 
 of the fraction. 
 
 OBS. 1. The term fraction, is of Latin origin, and signifies broken, or sepa 
 rated into parts. Hence, fractions are sometimes called broken numltcrs, 
 
 2. Common fractions are often called vulgar fractions. This term, however, 
 is very properly falling into disuse. 
 
 3. The number below the line is called the denominator, because it gives the 
 name or denomination to the fraction ; as, halves, thirds, fifths, &c. 
 
 The number above the line is called the numerator, because it numbers the 
 parts, or shows how many parts are expressed by the fraction. 
 
 183* A proper fraction is a fraction whose numerator is less 
 than its denominator ; as, , -f , -f-. 
 
 An improper fraction is one whose numerator is equal to, or 
 greater than its denominator ; as, -f, f. 
 
 A mixed number is a whole number and a fraction expressed 
 together ; as, 4-f, 25-Ji. 
 
 A simple fraction is a fraction which has but one numerator and 
 one denominator, and may be proper, or improper ; as, f , f . 
 
 A compound fraction is a fraction of a fraction ; as, -f of -f- of f> 
 
 QUEST. 182. How are common fractions expressed ? What is the number below the 
 line called? What does it show? What is the number above the line called? What 
 does- it show ? What are the denominator and numerator, taken together, called ? 
 Obs. What is the meaning of the term fraction ? What are common fractions sometimes 
 called? Why is the lower number called the denominator 1 Why is the upper one 
 called the numerator ? 183. What is a proper fraction 7 An improper fraction 1 A Mixed 
 number 1 A simple fraction ? A compound fraction 1 
 
ARTS. 182-188.] FRACTIONS. 109 
 
 A complex fraction is one which has a fraction in its numerator 
 
 2^ 4 2- 1 
 
 or denominator, or in both ; as, , 
 
 5 5-3 84- 
 
 1 8 4-. Fractions, it will be seen both from the definition and 
 the mode of expressing them, arise from division, and may be 
 treated as exp:essions of unexecuted division. The numerator an- 
 swers to the dividend, and the denominator to the divisor. (Arts, 
 25, 182.) Hence, 
 
 1 85 The value of a fraction is the quotient of the numerator 
 divided by the denominator. Tims, the value of f is two ; of \ is 
 one ; of \ is one third, &c. Hence, 
 
 186 If the denominator remains the same, multiplying the 
 numerator by any number, multiplies the value of- the fraction bi 
 that number. For, since the numerator and denominator answe 
 to the dividend and divisor, multiplying the numerator is the sama 
 as multiplying the dividend. But multiplying the dividend, we 
 have seen, multiplies the quotient, (Art. 141,) which is the same 
 as the value of the fraction. (Art. 185.) Thus, the value of f=2 ; 
 now, multiplying the numerator by 3, the fraction becomes -^ 
 whose value is 6, and is the same as 2X3. 
 
 187. Dividing the numerator by any number, divides the value 
 of the fraction by that number. For, dividing the dividend, divides 
 the quotient. (Art. 142.) Thus, -J=2 ; now dividing the numera- 
 tor by 2, the fraction becomes -f, whose value is 1, and is the same 
 as 2-^2. Hence, 
 
 OBS. With a given denominator, the greater the numerator, the greater will 
 / oe the value of the fraction. 
 
 188. If the numerator remains the same, multiplying the de- 
 nominator by any number, divides the value of the fraction by that 
 number. For, multiplying the divisor, we have seen, divides th 
 
 QTTKST. What is a complex fraction ? 184. From what tin fractions arise ? 185. "That 
 is the value of a fraction ? 186. What is the effect of multiplying the nnmeritnr. <vhile 
 the denominator remains the same ? Explain the reason. 187. Wh. / is the effect of di 
 viding the numerator? Obs. With a given denominator, what is the effect ol increasing 
 *he numerator 7 188. What is the effect of multiplying the denominator 1 
 
110 FRACTIONS. [SECT. VII. 
 
 quotient. (A. 1.143.) Thus, -^=4; now multiplying the denom- 
 inator by 2, the fraction becomes -f, whose value is 2, and is the 
 same as 4-^-2. 
 
 189, Dividing the denominator by any number, multiplies the 
 value of the fraction by that number. For, dividing the divisor 
 multiplies the quotient. (Art. 144.) Thus, *-=4 ; now dividing 
 the denominator by 2, the fraction becomes -^S whose value is 3, 
 
 nd is the same as 4X2. Hence, 
 
 OBS. With a given numerator, the greater the denominator, the less wiii oe 
 the value of the fraction. 
 
 1 9O. It is evident from the preceding articles, that multiply- 
 ing the numerator by any number, has the same effect on the value 
 of the fraction, as dividing the denominator by that number. 
 (Arts. 186, 189.) And, 
 
 Dividing the numerator has the same effect, as multiplying the 
 denominator. (Arts. 187, 188.) 
 
 OBS. It will be observed, that multiplying or dividing the numerator of a 
 fiaction, has the same effect upon its value, as the same operation has upon 
 a whole number ; but, the effect of multiplying or dividing the denominator is 
 exactly contrary to that of the same operation upon a whole number. 
 
 191, If the numerator and denominator are both multiplied 
 or both divided by the same number, the value of the fraction will 
 not be altered. (Art. 146.) Thus, -^=3; now if the numerator 
 and denominator are both multiplied by 2, the fraction becomes 
 ^gS whose value is 3. If both terms are divided by 2, the frac- 
 tion becomes f, whose value is 3 ; that is, - L 4 - a =- 2 ^-=f =3. 
 
 192. Since the value of a fraction is the quotient of the 
 numerator divided by the denominator, it follows, 
 
 If the numerator and denominator are equal, the value is a unit 
 or one. Thus, =!, -f=l, &c. 
 
 QUEST. 189. W.iat is the effect of dividing the denomina OT ? Why? Obs, With a 
 given numerator, what is the effect of increasing the denominator 1 190. What may be 
 done to the denominator to produce the same effect on the value of the fraction, as mal 
 tiplying the numerator by any given number? What, to product the same effect as divid- 
 ing the numerator by any given number ? 191. What is the effect if the numerator and 
 denominator are both multiplied, or both divided by the same number? 192 When tho 
 numerator and denominator are equal, what is the value of the fraction? 
 
ARTS. 189-194.] nt ACTIONS. Ill 
 
 If the numerator is greater than the denominator, the value is 
 greater than one. Thus, =2, =l-f. 
 
 If the numerator is leas than the denominator, the value is less 
 than one. Thus, =1 third of 1, =4 fifths of 1. 
 
 193* FJ actions may be added, subtracted, multiplied, and 
 divided, as well as whole numbers. But, in order to perform 
 the se operations, it is often necessary to make certain changes in 
 llie terms of the fractions. 
 
 OBS It is evident that any changes may be made in the terms of a fraction, 
 which do not alter the quotient of the numerator divided by the denominator; 
 for, it' the quotient is not altered, the value remains the same. Thus. the. terms 
 of the fraction may be changed into -^, -f-, - L /-, &c., without altering its value j 
 for in each case the quotient of the numerator divided by the denominator is 2. 
 Hence, for any given fraction, we may substitute any other fraction, which 
 wih give the same quotient. 
 
 REDUCTION OF FRACTIONS. 
 
 194. The process of changing the terms of a fraction into 
 others, without altering its value, is called REDUCTION OF FRAC- 
 TIONS. 
 
 CASE I. 
 
 Bx. 1. Reduce # to its lowest terms. 
 
 First Operation. Dividing both terms of the 
 
 2K-u=-i 6 o : again, 5)tV \ Ans. fraction by 2, it becomes -{% : 
 
 again, dividing both by 5, we 
 
 obtain , whose terms are the lowest to which the given fraction 
 can be reduced. 
 
 Second Operation. If we divide both terms by 10, then 
 
 10)i- i Ans. greatest common divisor, (Art. 170,) the 
 
 given fraction will be reduced to its lowts 4 
 terms by a single division. Hence, 
 
 QUEST. When the numerator is larger than the denominator, what ? When smalles, 
 what 1 Obs. What changes may be made in the terms of a fraction 1 194. What la 
 meant by reduction of fractions 1 195. How is a fraction reduced to its lowest terms 1 
 
112 REDUCTION OP [SECT. VII. 
 
 195 To reduce a fraction to its lowest terms. 
 
 Divide the numerator and denominator by any number which 
 will divide them both witJwut a remainder ; and thus continue the 
 operation, till there is no number greater tfian 1 that will divide 
 them exactly. 
 
 Or, divide b^th the i.umerator and denominator by tlieir greatest 
 common divisor ; the two quotients thence arising will be the lowest 
 erms to which the given fraction can be reduced. (Art. 170.) 
 
 OBS. I. Since halves are larger than twentieths, it may be asked, how the 
 fraction , can be said to be in lower terms than --g-, It should be observed, 
 the expression lowest term, has reference to the number of parts into which the 
 unit or thing is divided, and not to the xalm or size of the parts. Thus, in J, 
 there are fewer parts than in ^--J-; in ^-. there are fewr parts than in -j^-, &c. 
 Hence, a fraction is said to be reduced to its l-owest terms, when its numerator 
 and denominator are expressed in the smallest numbers possible. 
 
 2. The value of a fraction is not altered by reducing it to its lowest terms ; 
 for, the numerator and denominator are both divided by the same number. 
 
 3. When the terms of the fraction are small, the former method will gen- 
 erally be found to be the shorter and more convenient ; but when the terms 
 are large, it is often difficult to determine whether the fraction is in its simplest 
 form, without finding the greatest common divisor of its terms. 
 
 2. Reduce -p- 8 to its lowest terms. Ans. -. 
 
 3. Reduce A. 11. Reduce -?ff. 
 
 4. Reduce A- 12. Reduce &*&. 
 
 5. Reduce if. 13. Reduce ff. 
 
 6. Reduce ff. 14. Reduce -fff. 
 
 7. Reduce 4ff. 15. Reduce $&. 
 
 8. Reduce ff. 16. Reduce if-ff. 
 
 9. Reduce -fifr. 17. Reduce itf-f. 
 10. Reduce iVr- IS. Reduce ff If. 
 
 CASE II. 
 
 19 Reduce ^ 7 a to a whole or mixed number. 
 
 Analysis. The object in this example, is to Operation. 
 *h.d a whole, or mixed number, whose value is 7)23 
 
 equal to the given fraction. Now, since 7 3f Ans 
 
 .Gbs. What is meant by the expression, lowest terms ? When is a fraction 
 said to be reduced to its lowest terms ? Is the value of a fraction altered by reducing, It 
 to its lowest terms ? Why not. 
 
.ARTS. 195-197. | FRACTIONS. 113 
 
 sevenths make 1 whole one, 23 sevenths will make as many 
 whole ones as 7 is contained times in 23. And 23-r-7 = 3f. 
 But the value of a fraction is the quotient of the numerator 
 divided by the denominator. (Art. 185.) Hence, 
 
 196* To reduce an improper fraction to a whole, or mixed 
 
 number. 
 
 Divide the numerator by the denominator, and the quotient w*H 
 be the whole, or mixed number required. 
 
 20. Reduce * to a whole or mixed number. Ans. 6*-. 
 
 Reduce the following fractions to whole or mixed numbers: 
 
 21. Reduce Y- 26. Reduce -W- 
 
 22. Reduce ^. 27. Reduce **. 
 
 23. Reduce if. 28. Reduce 
 
 24. Reduce V"- 29. Reduce 
 
 25. Reduce *-$. 30. Reduce 
 
 CASE III. 
 31. Reduce- the mixed number 27 to an improper fraction. 
 
 Operation. 
 
 Analysis. In 1 there are 5 fifths, and in 27 27- 
 
 there are 27 times as many. Now 5X27 = 135, 5 
 
 and 2 fifths make 137 fifths. Hence, 
 
 197* To reduce a mixed number to an improper fraction. 
 
 Multiply the whole number by the denominator of the fraction t 
 and to the product add the given numerator. The sum placed over 
 the given denominator, will form the improper fraction required. 
 
 OBS. 1. Any whole number may be expressed in the form of a fraction with- 
 out altering its value, by making 1 the denominator. 
 
 2. A whole number may also be reduced to a fraction of any denominator, 
 by multiplying the given number by the proposed denominates ; the product 
 will be the numerator of the fraction required. 
 
 QUEST. 196. How is an improper fraction reduced to a whole or mixed number? 
 197. How reduce a mixed number to an improper fraction? Obs. How express a whole 
 number in the form of a fraction ? How reduce it to a fraction of a given denominator 1 
 
114 REDUCTION OF [SfiCT. Vll 
 
 Thus, 25 may be expressed by *?-, -HP-, or W-, &c., for 2= 
 .^-L^A^ & c> So 12=^=-^=^=^ for the quotient of 
 each of these numerators divided by its denominator, is 12. 
 
 32. Reduce 14 to an impropei fraction. Ans. -^ a . 
 lleduce the following numbers to improper fractions : 
 
 33. Reduce I7f. 38. Reduce 856fg-. 
 
 34. Reduce 25-f. 39. Reduce 1304. 
 
 35. Reduce 48f 40. Reduce 4725|. 
 
 36. Reduce 70-^. 41. Reduce 445 to tenths. 
 
 37. Reduce 115-&-. 42. Reduce 672 to eighths. 
 
 43. Reduce 3830 to one hundred and fifteenths. 
 
 44. Reduce 5743 to six hundred and twenty-fifths. 
 
 CASE IV. 
 
 45. Reduce of to a simple fraction. 
 
 Analysis. f of % is 2 times as much as 1 third of -f-. Now 
 of -f- is g 3, or -fa ; for, multiplying the denominator divides the 
 
 value of the fraction. (Art. 188.) And 2 thirds is 2 times 2*4, or 
 
 7x2 
 
 -^-, which is equal to if, or -&. (Art. 195.) The answer is -fo. 
 
 OBS. This operation consists in simply multiplying the two numerators to- 
 gether and the two denominators. Hence, 
 
 198. To reduce compound fractions to simple ones. 
 Multiply all the numerators together for a new numerator, and 
 all the denominators together for anew denominator. 
 
 OBS. 1. That a compound fraction may be expressed by a simple one,. is evi- 
 dent from. the fact that a part of a part, must be equal to some part of the 
 whole. 
 
 2. The reason of the rule may be seen from the analysis of the preceding 
 example. 
 
 46. Reduce -f of -f- of f of -f to a simple fraction. 
 
 Ans. T^ph-, or ^. 
 
 47. Reduce -f of i of -f^ of ^ to a simple fraction. 
 
 48. Reduce \ of f of f of t of -fa to a simple fraction. 
 
 QUEST. 198. How are compound fractions reduced to simple ones? 
 
ARTS. 198, 199.] FRACTIONS. 115 
 
 49. Reduce f of f of -fa of -^ to a simple fraction. 
 
 50. Reduce i of f of -f of f of to a simple fraction. 
 
 Analysis. Since the product of Operation. 
 
 the numerators is to be divided 1 # $ 5 &_ 6 
 by the product of the denomina- jj $ 4 7 a~ &>* 
 tors, we may cancel the factors 2, 
 
 3, and 4, which are common to both ; for, this is dividing the 
 terms of the new fraction by the same number, (Art. 148,) and 
 therefore does not alter its value. (Art. 191.) Multiplying the 
 remaining factors together, we have -\, which is the answer re- 
 quired. Hence, 
 
 199. To reduce compound fractions to simple ones by 
 CANCELATION. 
 
 Cancel all the factors which arc common to the numerators and 
 denominators ; then multiply the rerr^initiy terms together as be- 
 fore. (Art. 198.) 
 
 OBS. 1. The reason of this rule depends upon the fact that the numerator 
 and denominator of the new fraction are, in effect, divided by the same num- 
 bers ; for, canceling a factor of a number divides the number by that factor. 
 (Art 148.) Consequently the value of the fraction is not altered. (Art. 191.) 
 
 2. This method not only shortens the operation of multiplying, but at the 
 same time reduces the answer to its lowest terms. A little practice will give 
 the student great facility in its application. 
 
 51. Reduce f of --f of f to a simple fraction. 
 
 Operation. 
 
 3 First we cancel the 3 and 8 in the 
 
 $ 1$ f $__3 . numerator, then the 24 in the denomina- 
 
 $ 7~~7 tor, which is equal to :he factors 3 into 8. 
 
 Finally, we cancel the 5 in the denomina- 
 tor and the factor 5 in the numerator 15, placing the other factor 
 3 above. We have 3 left in the numerator, and 7 in the denom 
 inator. Ans. f. 
 
 52. Reduce -f of -f of -f of if to a simple fraction. 
 
 53. Reduce | of ^ of | of -fa of f to a simple fraction. 
 
 QUEST. 199. How by cancelation ? How does it appear that this method will gtaj th 
 true answer ? Obs. What advantages does this method possess 1 
 T.H. 
 
116 REDUCTION OF fbfiCT. VIL 
 
 54. Roduce of f of -f of -f- of -ft- to a simple fr fiction. 
 
 55. Reduce f of -fc of -f of to a simple fraction. 
 
 56. Reduce of ^-f of f of 3*5- to a simple fraction. 
 
 57. Reduce f of -tf of -ff of -J-g- to a simple fraction. 
 
 58. Reduce -f- of -ft- of -ft- of f of -f to a simple fraction. 
 59 Reduce i of f of 4f of f of -f- to a simple fraction. 
 60. Reduce f of 3^ of f of i^ of - to a simple fraction. 
 Note. For reduction of complex fractions to simple ones, see Art. 239 
 
 CASE V. 
 
 Ex. 61. Reduce ^ ana -J- to a common denominator. 
 
 Aofe. Two or more fractions are said to have a common denominator, when 
 they have the same denominator. 
 
 Solution. If both terms of the first fraction -fc, are multi- 
 plied by the denominator of the second, it becomes -fa ; and if 
 both terms of the second fraction -J-, are multiplied by the de- 
 nominator of the first, it becomes -ft-. Thus the fractions -fa and 
 ft- have a common denominator, and are respectively equal to the 
 given fractions, viz : -fa=i, and T^=-}-. (Art. 191.) Hence, 
 
 2OO. To reduce fractions to a common denominator. 
 
 Multiply each numerator into all the denominators except its 
 own for a new numerator, and all tJie denominators together for a 
 common denominator. 
 
 62. Reduce \, f , and f to a common denominator. 
 
 Operation. 
 1X4X6=24 \ 
 
 3 X 3 X 6=54 > the tliree numerators. 
 5X3X4=60 ) 
 3X4X6=72 the common denominator. 
 
 Ans. ff, ft, and f . 
 
 OBS. The reason that the process of reducing fractions to a common denom 
 inator does not alter their value, is because the numerator and denominator ol 
 each of the given fractions, are multiplied by the same numbers ; and multiplying 
 
 QUEST. Note. What is meant by a common denominator 7 200. How are fractions re- 
 dnced to a common denominator ? Obs. Does the [focess of reducing fractions to a com 
 moa denominator alter their value 1 Why not ? 
 
ARTS. 200, 201.] FRACTIONS. 117 
 
 both tho numerator and denominator of a fraction by the same number, doe 
 not alter its value. (Art. 191.) 
 
 63. Reduce f, f, -J-, and -f- to a common denominator. 
 
 64. Reduce f , i, f, and f to a common denominator. 
 Reduce the following fractions to a common denominator: 
 
 65. Reduce f, i, , and -f. 69. Reduce if, ft, and ffr. 
 
 66. Reduce f , -f, -f, and f . 70. Reduce if, -j 2 ^, and ff. 
 
 67. Reduce f, -f, -ft, and T^/ 71. Reduce ff, -*, and i. 
 
 68. Reduce ft, f, if, and f. 72. Reduce , f, and iff- 
 
 CASE VI. 
 73. Reduce , -f-, and -| to the least common denominator. 
 
 Analysis. We first find the least Operation. 
 common multiple of all the given de- 2)3 " 4 " 8 
 nominators, which is 24. (Art. 176.) 2)3 " 2 " 4 
 The next step is to reduce the given 3 " 1 " 2 
 fractions to twenty-fourths without Now 2X2X3X2 = 24, the 
 altering their value. This may evi- least common denominator, 
 dently be done by multiplying both 
 
 terms of each fraction by such a number as will make its denom- 
 inator 24. (Art. 191.) Thus 3, the denominator of the first frac- 
 tion, is contained in 24, 8 times ; now, multiplying both terms of 
 the fraction i by 8, it becomes /f. The denominator 4, is con- 
 tained in 24, 6 times ; hence, multiplying the second fraction f by 
 6, it becomes if . The denominator 8, is contained in 24, 3 times : 
 and multiplying the third fraction f by 3, it becomes if. Ther- 
 fore 2 \, if, and if are the fractions required. Hence, 
 
 20 1. To reduce fractions to their least common denominator. 
 
 I. Find the least common multiple of all the denominators of 
 the given fractions, and it will be the least common denominator, 
 (Art. 176.) 
 
 II. Divide the least common denominator by the denominate 
 cf each given fra.ctlm, and multiply the quotient by the numerator * 
 the products will be tlie numerators of the fractions required. 
 
 GniflT 201. How are fractions reduced to the least common denominator I 
 
118 REDUCTION OF [SECT. VII, 
 
 OBS. 1. This process, in effect, multiplies both the numerator f.nd denomina- 
 tor of the given fractions by the same number, and consequently does not alte 
 their value (Art. 191.) 
 
 2. The rule supposes each of the given fractions to be reduced to its lowest 
 terms ; otherwise, the least common multiple of their denominators may not be 
 the least common denominator to which the given fractions are capable of being 
 reduced. Thus, the fractions ^, f-, and -^, when reduced to the least com- 
 mon denominator as they stand, become -fa, -j 6 ^, and -j 9 ^. But it is obvious 
 that these fractions are not reduced to* their least, common denominator; for, 
 fchey can be reduced to 1, |-, and |k Now, if the given fractions are reduced 
 to the lowest term*, they become $, , and |, and the leasl wmmon multiple of 
 their denominators, is also 4. (Art. 176.) 
 
 3. By a moment's reflection the student will often discover the least common 
 den -aiinator of the given fractions, without going through the ordinary pro- 
 ce of finding the least common multiple of their denominators. Take the 
 fra Hons i, f , and -fa ; the least common denominator, it will be seen at a 
 gl? >ce, is 4. Now if we multiply both terms of i by 2, it becomes -| ; and if 
 we livide both terms of -fa by 3, or reduce it to its lowest terms, it becomes {. 
 TJ is the given fractions are equal to |, |, and i, and are reduced to the least 
 co man deiwmiuator. 
 
 74. Reduce -f, f-, and to the least common denominator. 
 
 Operation. Now 2X2X3X2 = 24, the least com. denom. 
 2)4 " 6 " 8 Then 24-4 = 6, and 6X3 = 18, the 1st num. 
 2)2 " 3 " 4 24 6 = 4, and 4X5 = 20, the 2d " 
 
 1 " 3~ /r 2 24 8 = 3, and 3X7 = 21, the 3d " 
 
 Ans. if, fi, and ift. 
 
 T5. Reduce -f- and -fa to the least common denominator. 
 Reduce the following fractions to the least common denominator: 
 
 76. A, i, i, and -fa, 84. if, H, if, and ffj. 
 
 77. |, f, and i. 85. ^ H> it, and / 2t 
 
 78. i, |, 1, and H. 86. A, ii, if and -ff. 
 
 79. f, f, f, and -fV- 87. if, ffr, ^ and f. 
 
 80. f , i, f, and if. 88. ff , if, ff, and if. 
 
 81. i, 1*0, iHK and ffr. 89. fi, i, if, and . 
 
 82. -&, -H-, T^, and ff. 90. fft-, if, fi and -ffi-. 
 
 83. ^, I, if, and ii. 91. fi, -W, ti, and ^. 
 
 *iuEST. 9As. Does this process alter the value of the Riven fractions? Why ootf 
 r hat does this rule suppose respecting the given fractions 7 
 
ARTS. 201, 202. J FRACTIONS. 119 
 
 ADDITION OF FRACTIONS. 
 
 Ex. 1. A beggar meeting four persons, obtained f of a dollar 
 from the first, from the second, |- from the third, and from 
 the fourth : how much did he receive from all ? 
 
 Solution. Since the several donations are all in the same pans 
 of a dollar, viz : sixths, it is plain they may be added together in 
 the same manner as whole dollars, whole yards, &c. Thus, 1 
 sixth and 3 sixths are 4 sixths, and 4 are 8 sixths, and 5 are 13 
 sixths. Ans. J g a , or 2 dollars. 
 
 Ex. 2. What is the sum of -f and f- ? 
 
 OBS. A difficulty here presents itself to the learner ; for, it is evident, that 
 2 thirds and 3 fourths neither make 5 thirds, nor 5 fourths. (Art. 51.) This 
 difficulty may be removed by reducing the given fractions to a common de- 
 nominator. (Art. 200.) Thus, 
 
 Operation. 
 
 2X4 = 8 
 
 _ t the new numerators. 
 y 
 
 3X4=12, the common denominator. 
 
 The fractions, when reduced, are i\ and T\; now 8 twelfths X 
 9 twelfths=l7 twelfths. Ans. -j-f, or IT\. 
 
 2 O2. From these illustrations we deduce the following general 
 
 RULE FOR ADDITION OF FRACTIONS. 
 
 Reduce the fractions to a common denominator ; add their nu- 
 merators, and place t/ie sum over the common denominator. 
 
 OBS. 1. Cmnpmind fractions must, of course, be reduced to simple ones, be- 
 fore attempting to reduce them to a common denominator. (Art. 198.) 
 
 2. Mixed numbers may be reduced to improper fractions, and then be added 
 according to the rule; or, we may add the whole numbers and fractional parte 
 separately, and then unite their sums. 
 
 3. In many instances the operation may be shortened by reducing the give 
 fractions to the least common denominator. (Art. 120 ) 
 
 QUEST. 202. How are fractions added? Obs. What must be done with ccmpouni 
 fractions ? How are mixed n 'inbers added ? How may the operation frequently be short- 
 ened i 
 
120 ADDITION OF [SECT. VII 
 
 EXAMPLES. 
 
 3. What is the sum of i, , and f- ? Ans. V=2. 
 
 4. What is the sum of i, -f , -f , and f ? 
 
 5. What is the sum of f , f , i, and f ? 
 
 6. What is the sum of f, f , If, and ? 
 
 7. What is the sum of -f, i%, f , and -& ? 
 
 8. What is the sum of f, f , ft, and A ? 
 
 9. What is the sum of f, -fV, f, and f ? 
 
 10. What is the sum of f, -fo f, and V? 
 
 11. What is the sum of i, -J-, i, -f, and f ? 
 
 12. What is the sum of f of i, f of f , and f ? 
 
 13. What is the sum of i of f, -f of , and -ft ? 
 
 14. What is the sum of f- of | of $ of i, and | ? 
 
 15. What is the sum of f , -f of 3, f of i, and i ? 
 
 16. What is the sum of 4|, 8i, 2, 6f, and |? 
 
 17. What is the sum of i of 6, f of 2, 3, and 5? ? 
 
 18. What is the sum of f, if, ffr, ii, and -j-$ ? 
 
 19. What is the sum of 2l, 35i, -fi, and f of 1? 
 
 20. What is the sum of i of f , ^, 6|, If, and -| ? 
 
 21. What is the sum of i and 1*4 ? 
 
 Afofe. It is obvious, if two fractions, each of whose numerators is 1, are re* 
 duced to a common denominator, the new numerators will be the same as the 
 given denominators. (Art. 200.) Thus, if -| and -fa are reduced to a common 
 denominator, the new numerators will be 1 2 and 8, the same as the given de- 
 nominators. Now. the sum of the new numerators, placed over the product 
 
 1*2 UH 20 
 of the denominators, will be the answer; (Art. 202;) that is TS-T^=QT: or 
 
 l^Xo 'o 
 2?- , the answer required. Hence, 
 
 2O3 To find the sum of any two fractions whose numerators 
 are one. 
 
 Add tJie denominators together, place this sum over their prod- 
 uct, and tlue result will be the answer required. 
 
 OBS. 1. The reason of this rule may be seen from the fact that the opera 
 lion is the same as reducing the given fractions to a common denominator 
 then adding their numerators. 
 
 2. When the numerators of two factors are the same, their sum may be round 
 
 QUEST. 303. How is the sum of any two fractions found whose numerators are 1 1 
 Obs. How find the sum of two fractions whose numerators are the same 1 
 
ARTS. 203, 204.] FRACTIONS. 121 
 
 by multiplying the sum of the two denominators hy the common numerator, 
 and placing the result over the product of the given denominators Thus, the 
 
 , (4-J-5) X3 9X3 27 , - 
 sum of -3- and f is equal to ^^-^^ or 1-jfr. 
 
 22. What is the sum of - 2 -V and -gV? Of jV and 6 -* ? 
 
 23. What is the sum of - 5 -V and ^ ? Of & and -V ? 
 
 24. What is the sum of -g and gV ? Of -J^T and -g-ir ? 
 
 25. What is the sum of -| and -fa ? Of -& and -& ? 
 20 What is the sum of ^ and -fl ? Of * and if ? 
 
 27. What is the sum of if and ffr ? Of ffr and -^ft- ? 
 
 28. What is the sum of 5 and -f ? 
 
 Note. The design in tliis and the following examples, is to incorporate the 
 integers with the fractions, and express the answer fractionally. 
 
 Solution. 5=-^-. (Art. 197. Obs. 2.) Now -^-+f =Y Ans. 
 2O 4. Hence, to add a whole number and a fraction together. 
 
 O 
 
 Reduce the whole member to a fraction of the same denominator 
 as that of the given fraction ; then add their numerators together. 
 (Arts. 202, 197. Obs.' 1, 2.) 
 
 Nvtc. The process of incorporating a whole number with a fraction, is the 
 same as that of reducing a mixed number to an improper fraction. (Art. 197.^ 
 
 29. What is the sum of 45 and f ? 
 
 30. What is the sum of 320 and -fc ? 
 
 31. What is the sum of 452 and VW? 
 
 32. What is the sum of 6 3 5H +4 2 7^+16251? 
 
 33. What is the sum of 195H+600lHH-5630!-H-160? 
 
 34. What is the sum of 67ltl+483it+8421^-f 4325i? 
 
 35. What is the sum of 590ii+100fi+ 4005|f +3020^? 
 
 36. What is the sum of 239|f +644|i+1650|f +4500tV? 
 37 What is the sum of 6563i+1000i+1830f +83001? 
 
 38. What is the sum of 356H+46f+105i+000|+321-| ? 
 
 39. What is the sum of 41^-f 105|-h300f-f 241f +472-J-? 
 
 40. What is tin sum of 8G72!-f-163G45i-f-lSOOi-}-GG251fV? 
 
 41. What is the sum of 2G003|+19352-f + 92S31-f-68G93f ? 
 
 42. What is the sum of 1 9256-^+456001+1 of f of f ? 
 
 43. What is the sum of f of 28 + 6-^+45^ + of 300 ? 
 
 QUEST. 204. How add a whole number ami a fraction ? 
 
122 SUBTRACTION OF [bECT. 
 
 SUBTRACTION OF FRACTIONS. 
 
 2O 5 Ex. 1. A man bought -^ of an acre of land, and after- 
 "wards sold -^ of it : bow much land had he left ? 
 
 Solution. 7 tenths from 9 tenths leave 2 tenths. 
 
 Am. -fs of an acre. 
 
 2 A laborer having received - of a dollar for a day's work., 
 pent of a dollar for liquor : how much money had he left ? 
 
 Note. The learner meets with tHfe same difficulty here as in the second ei' 
 ample of adding fractions; that is, he can no more subtract fifflis from eighths^ 
 than he can add fifths to eighths; for, of a dollar taken from -J of a dollar will 
 neither leave 4 fifths, nor 4 eighths. The fractions must therefore be reduced 
 to a common denominator before the subtraction can be performed. 
 
 Operation, 
 
 3X8 = 24 ( ^ e numerators - (A rt - 200.) 
 8X5 = 40, the common denominator. 
 The fractions become -f-ft and -. Now {j H=ii 
 
 2OG From these illustrations we deduce the following general 
 
 RULE FOR SUBTRACTION OF FRACTIONS. 
 
 Reduce the given fractions to a common denominator ; subtract 
 the less numerator from the greater, and place the remainder over 
 the common denominator. 
 
 OBS. Compound fractions must be reduced to simple ones, as in addition of 
 fractions. (Art. 198.) 
 
 EXAMPLES. 
 
 3. From -f take ^. Ans. /$. 
 
 4. From -ff take A- 9- 'From ff take ff . 
 
 5. From -ff ^ke if. 10. From -f of f take | of f. 
 
 6. From if take f. 11. From.f of i take i of ^. 
 
 7. From H take |f. 12. From | of 40 take f of 20. 
 
 8. From ff take if. 13. From f of f of i take f of 
 
 QUEST. 206. How is one fraction subtracted from another 7 Ofs. Wtat is to be orone 
 with compound fraction* ? 
 
ARTS. 205-208.] FRACTIONS. 123 
 
 2O7. Mixed numbers may be reduced to improper fractions, 
 then to a common denominator, and be subtracted ; or, the frac- 
 tional part of the less number may be taken from the fractional 
 part of the greater, and the less whole number from the greater. 
 
 14. From 9-J- take 7^. 
 
 First Operation. Second Operation. 
 
 Ans. i=li, or l. Ans. If, or 1. 
 
 Note. Since we cannot take 3 fourths from 1 fourth, we borrow a unit jn 
 the second operation and reduce it to fourths, which added to the 1 fourth, 
 make 5 fourths. Now 3 fourths from 5 fourths leave 2 fourths : 1 to carry to 
 7 makes 8, and 8 from 9 leaves 1. 
 
 15. From 25f take 13-f. 17. From 178^ take 56-f. 
 
 16. From 230- 2 2 S - take 160-fV. 18. From 761fi take 482-^ 
 
 19. From 5 take f. 
 
 Suggestion. Since 3 thirds make a whole one, in 5 whole ones 
 there are 15 thirds; now 2 thirds from 15 thirds leave 13 thirds. 
 Ans. -f, or 4^. Hence, 
 
 2O 8. To subtract a fraction from a whole number. 
 
 Change the whole number to a fraction having the same denom- 
 inator as the fraction to be subtracted, and proceed as before. 
 (Art. 197. Obs. 2.) 
 
 OBS. If the fraction to be subtracted is a proper fraction, we may simply 
 borrow a unit and take the fraction from this, remembering to diminish the 
 whole number by 1. (Art. G9. Obs. 1.) 
 
 20. From 20 take f. ^^ns. 19f. 
 
 21. From 135 take 9|. 26.' From 720 take 125^. 
 
 22. From 263 take 24+-. 27. From 1000 take 25 ft. 
 
 23. From 168 take 30-|. 28. From 563 take 562f|. 
 
 24. From 567 take lOOffr. 29. From 9263 take 999-J-. 
 
 25. From 634 take 342f. 30. From 857 take 785 ii. 
 
 Q.UKST. 207. How are mixed numbers subtracted? 208. How is a fraction subtracter 
 from a whole number ? 
 6* 
 
124 MULTIPLICATION OP [Sp.CT. VII. 
 
 MULTIPLICATION OF FRACTIONS. 
 
 2O 9. We have seen that multiplying by a whole number, is 
 taking the multiplicand as many times as there are units in the 
 multiplier. (Art. 82.) On the other hand, 
 
 If the multiplier is only a part of a unit, it is plain we must 
 take only a part of the multiplicand. That is, 
 
 Multiplying by , is taking 1 half of the multiplicand once. 
 Thus, ]2Xi=6. 
 
 Multiplying by , is taking 1 third of the multiplicand once. 
 Thus, 12xi=4. 
 
 Multiplying by f, is taking 1 third of the multiplicand twice. 
 Thus, 12 X -3=8. Hence, 
 
 2 1 O. Multiplying by a fraction is taking a certain PORTION 
 of the. multiplicand as many times, as there are like portions of a 
 unit in the multiplier. 
 
 OBS. If the multiplier is a unit or 1, the product is equal to the multiplicand ; 
 if the multiplier is greater than a unit, the product is greater than the multi- 
 plicand ; (Art. 82 ;) and if the multiplier is less than a unit, the product is 
 less than the multiplicand. 
 
 CASE I. 
 
 211. To multiply a fraction and a whole number together. 
 Ex. 1 . If 1 man drinks f of a barrel of cider in a month, how 
 much will 5 men drink in the same time ? 
 
 Analysis. Since 1 man drinks ^ of a barrel, 5 men will drink 
 6 times as much; and 5 times 2 thirds are 10 thirds; that is, 
 fX5=Y> or 3|. (Art. 196.) Ans. 3 barrels. 
 
 Ex. 2. If a pound of tea costs | of a dollar, how much will 
 i pounds cost ? 
 
 Solution. 1x4=-^-; Imd ^^2-f, or 2i dons. Ans. 
 Or, since dividing the denominator of a fraction by any num 
 bei multiplies the value of the fraction by that number, (Art. 189,) 
 
 QUEST. 209. What is meant by multiplying by a whole number ? 210. What is meant 
 y multiplying by a fraction ? Obs. If the multiplier is a unit or 1, what is the product 
 equal to ? When the multiplier is greater than 1, how is the product, compared with th 
 multiplicand ? When less, how ? 
 
ARTS. 209-21 3. J FRACTIONS. 125 
 
 if we divide the denominator 8 by 4, the fraction will become f , 
 which is equal to 2-, the same as before. Hence, 
 
 To multiply a fraction by a whole number. 
 Multiply the numerator of the fraction by tlie whole number, 
 and write the product over the denominator. 
 
 Or, divide the denominator by the whole number, when this can 
 be done without a remainder. (Art. 180.) 
 
 OBS. 1. A fraction is multiplied into a number equal to its denominator by 
 canceling the denominator. (Ax. 9.) Thus ^X"7=4. 
 
 2. On the same principle, a fraction is multiplied into any factor m its de- 
 <n0miualf>r, by canceling that factor. (Art. 189.) Thus, -j^xS f . 
 
 3. Since multiplication is the repeated addition of a number or quantity to 
 ilsc/f, (Art. HO,) the student sometimes finds it difficult'to account for the fact 
 that the product of a number or quantity by- a proper fraction, is always kss 
 than the number multiplied. This difficulty will at once be removed by re- 
 flecting that mv.lt if tlyin^ by a fraction is taJring or repeating a certain portion 
 of the multiplicand as many times, as there are like portions of a unit in the 
 multiplier. (Art. '210.) 
 
 EXAMPLES. 
 
 3. Multiply -f-Jr by 15. Ans. **&*-, or 10. 
 
 4. Multiply i by 8. 9. Multiply fj- by 165. 
 
 5. Multiply by 30. 10. Multiply fff by 100. 
 
 6. Multiply |f by 27. 11. Multiply / 3 \ by 530. 
 
 7. Multiply -HHr by 45.- 12. Multiply If by 1000. 
 
 8. Multiply fi by 100. 13. Multiply Hi by 831. 
 14. Multiply 12| by 8. 
 
 Operation. 
 
 12 f 8 times -f are *-, which are equal to 5 and -J-. 
 
 8 Set down the i. 8 times 12 are 96, and 5 (which 
 
 Ans. 101^. arose from the frac^Hf make 101. Hence, 
 
 213. To multiply a mixed number by a whole one. 
 Multiply the fractional part and the whole number separately 
 and unite the products. 
 
 QUEST. 212. How multiply a fraction by a whole number ? Ols How is a fraction 
 multiplied by a number equal to its denominator ? How by any factol la its denominator * 
 213. How is a mixed number multiplied by a whole one ? 
 
126 MULTIPLICATION OF [SECT. VII 
 
 15. Multiply 45i by 10. Ans. 4511. 
 
 16. Multiply 811 by 9. 19. Multiply 127| by 35. 
 
 17. Multiply 31 f by 20. 20. Multiply 48^ by 47. 
 
 18. Multiply 148-J1 by 25 21. Multiply 250^ by 50. 
 
 2 1 4. Multiplying by a fraction, we have seen, is taking a 
 certain portion of the multiplicand as many times, as there are 
 like portions of a unit in the multiplier. Hence, 
 
 To multiply by : Divide the multiplicand by 2. 
 
 To multiply by -^ : Divide the multiplicand by 3. 
 
 To multiply by - : Divide the multiplicand by 4, &c. 
 
 To multiply by -f : Divide by 3, and multiply the quotient by 2. 
 
 To multiply by 1 : Divide by 4, and multiply the quotient by 3. 
 
 215. Hence, to multiply a whole number by a fraction. 
 
 Divide the multiplicand by the denominator, and multif>ly the 
 quotient by the numerator. 
 
 Or, multiply the given number by the numerator, and divide the. 
 product by the denominator. 
 
 Obs. 1. When the given number cannot be divided by the denominator 
 without a remainder, the latter method is generally preferred. 
 
 2. Since the product of any two numbers is the same, whichever is taken 
 for the multiplier, (Art 83,) the fraction may be taken for the multiplicand, 
 and the whole number for the multiplier, when it is more convenient. 
 
 22. If 1 ton of hay costs 21 dollars, how much -will f of a toE 
 cost? 
 
 Analysis. Since 1 ton costs 21 dollars, ^ of 4)21 
 
 a ton will cost -f- as much. Now, 1 fourth of 21 5 
 
 is -Vs an d i of 21 is Senes as much; but 3 
 
 s Scen 
 f cHlar 
 
 X3= j =-^-, or 15f clars. Ans. 15f dolK 
 
 23. Multiply 136 by i. Ans. 45 . 
 
 24. Multiply 432 by \. 26. Multiply 360 by 
 
 25. Multiply 635 by . 27. Multiply 580 by 
 
 UDEST. 215. How is a whole number multiplied ly a fraction I 216. How find a ftae 
 tional part of a number 1 
 
ARTS. 21J-217.J FRACTIONS. 127 
 
 28. Multiply 672 by f. 31. Multiply 660 by ^-. 
 
 29. Multiply 710 by f. 32. Multiply 840 by $-$. 
 
 30. Multiply 765 by H- 33. Multiply 975 by iff. 
 
 216. Since multiplying by a fraction is taking a certain por- 
 tion of the multiplicand as many times, as there are like portions 
 of a unit in the multiplier, it is plain, that the process of finding 
 a fractional part of a number, is simply multiplying the number 
 by the given fraction, and is therefore performed by the same rule. 
 
 Thus, -f of 12 dollars is the same as the product of 12 dollars, 
 multiplied by | ; and 12Xi=8 dollars. 
 
 OBS. The process of finding a fractional part of a number, is often a source 
 of confusion and perplexity to the learner. The difficulty arises from the 
 erroneous impression that finding a fractional part, implies that the given num- 
 ber must be divided by the fraction, instead of being multiplied by it. 
 
 34. What is -& of 457 ? Ans. 26G-&. 
 
 35. What is if of 16245 ? 38. What is iff of 5268 ? 
 
 36. What is f- of 25000 ? 39. What is -fff- of 45260 ? 
 
 37. What is & of 4261 ? 40. What is -jVoV of 452120 * 
 
 41. Multiply 64 by 5. 
 
 Operation. 
 
 2)64 We first multiply 64 by 5, then by , and the 
 
 5 sum of the products is 352. But multiplying by 
 
 320 is taking 'owe half of the multiplicand once. 
 
 32 (Arts. 82, 214.) Hence, 
 Ans. 3~52? 
 
 217* To multiply a whole by a mixed number. 
 Multiply first by the integer, then by the fraction, and add the 
 products together. (Art. 214.) 
 
 42. Multiply 83 by 7. Ans. 597f. 
 
 43. Multiply 45 by 8. . 47. Multiply 225 by 30$. 
 
 44. M iltiply 72 by 10^. 48. Multiply 342 by 20f. 
 
 45. Multiply 93 by 12f. 49. Multiply 432 by 35$. 
 
 46. Multiply 184 by 18|. 50. Multiply 685 by 42*. 
 
 QIT*ST. 217. How is a whole number multiplied by a mixed number 7 
 
128 MULTIPLICATION OF [SECT. VII 
 
 51 Multiply 125 by 10- 2 \. 56. Multiply 457 by 12f-. 
 
 52. Multiply 26 by 10 57. Multiply 107 by 47|i. 
 
 53. Multiply 256 by 17-ft-. 58. Multiply 510 by 85fjf-. 
 
 54. Multiply 196 by 41i. 59. Multiply 834 by 89-fV 
 65. Multiply 341 by 30^. 60. Multiply 963 by 95fi. 
 
 CASE II. 
 
 218. To multiply a fraction ly a fraction. 
 
 Ex 1. A man bought f of a bushel -of wheat, at 1 of a dolluf 
 per bushel how much did he pay for it ? 
 
 Analysis. Since 1 bushel costs of a dollar, i of a bushel 
 must cost i of -, which is 4^ of a dollar; for, multiplying the 
 denominator, divides the value of the fraction. (Art. 188.) Now, 
 if -- of a bushel costs -fa of a dollar, f of a bushel will cost 4 times 
 as much ; and 4 times -fa are f-g-, or -ft dolls. (Art. 195.) 
 
 Ans. y 7 ^ of a dollar. 
 
 Or, we may reason thus : since 1 bushel costs of a dollar, 
 f of a bushel must cost of $ of a dollar. Now -- of i is a com- 
 pound fraction, whose value is found by multiplying the numera- 
 tors together for a new numerator, and the denominators for a 
 new denominator. (Art. 198.) 
 
 Solution. f-Xf =-H, or -fo dollars, Ans. Hence, 
 
 219. To multiply a fraction by a fraction. 
 
 Multiply the numerators together for a new numerator, and the 
 denominators together for a new denominator. 
 
 OBS. 1. It will be seen that the process of multiplying one fraction by an- 
 other, is precisely the same as that of reducing compound fractions to simple 
 ones. (Art. 198.) 
 
 2. The reason of this rule may be thus explained. Multiplying by a fractica 
 is taking a certain part of the multiplicand as many times, as there are like 
 p(*,r(s of a unit in the multiplier. (Art. 210.) Now multiplying the denomina- 
 tor of the multiplicand by the denominator of the multiplier, gives the vjJue 
 if only one of the parts denoted by the given multiplier; (Art. 188;) we there- 
 fore multiply this new product by the numerator of the multiplier, to find the 
 number of parts denoiel by the given multiplier. (Art. 18G.) 
 
 QUEST. 219. How is a fraction multiplied by a fraction 1 Obs. To what is the proceM 
 of multiplying one fraction b> another similar? 
 
ARTS. 218-220.] FRACTIONS. 129 
 
 2. Multiply i by f . Am. -fa=-\. 
 
 3. Multinjy -f by -fr. 6. Multiply -f* by 
 
 4. Mul^Pii by if. 7. Multiply f by 
 
 5. Multiply if by ft. 8. Multiply ff by 
 
 9. What is the product of into -f into if into $ into $ ? 
 
 10. What cost 6| yards of cloth, at 4$ dollars per yard ? 
 
 Analysis. 4$ dollars =f, and 6f yards =- s y-. (Art. 197.) Now 
 ^- 1 !^, or 30. (Art. 196.) Ans. 30 dollars. Hence, 
 
 2 2O. When the multiplier and multiplicand are both mixed 
 numbers, they should be reduced to improper fractions, and then 
 be multiplied according to the rule above. 
 
 OBS. Mixed numbers may al^ be multiplied together, without reducing them 
 to improper fractions. 
 
 Take, for instance, the last example. We first multiply by 4, Operation. 
 
 the whole number. Thus, 4 times -f are -f-. equal to 2 and -| ; ** 6| 
 
 set down the |, and carry the 2. Next, 4 times 6 are 24, and 4J 
 
 2 to carry are 26. We then multiply by J, the fractional part. "2Gf 
 
 Thus, | of 6 is 3 ; and J of 2 thirds is \. The sum of the two 3j 
 
 partial products is 30 dollars, the same as before. 30 doll*. 
 
 11. Multiply 6f by 2-H-. 23. Multiply 246-^ by fi. 
 
 12. Multiply 8A by 6f. 24. Multiply 9 Iff- by -f off. 
 
 13. Multiply 13| by 17$. 25. Multiply 1475 by i of 21. 
 34. Multiply 15* by 20-f. 26. Multiply 34i by i of 68. 
 
 15. Multiply 30f by 44- 2 \. 27. Multiply 800 by f of 1000. 
 
 16. Multiply 63fi by 50|. 28. Multiply i of 75 by | of 28. 
 37. Multiply 17-ft by 25-H-. 29. Multiply 2-J- by -f of f of 85. 
 
 18. Multiply 47M by l7if. 30. Multiply f of 2i by f of 61. 
 
 19. Multiply ei^ by 32ff. 31. Multiply |- of 10$ by -f of 8i. 
 
 20. Multiply 7lff by 45- 6 \. 32. Multiply fof 16ibyf cf 9i. 
 
 21. Multiply 83-^ by 61^ i 33. Multiply fof^o of 20 by 25|-. 
 
 22. Multiply 96|f by 72-||. 34. Multiply ft of 65i ' y 40$. 
 
 35. What cost 125i bbls. of flour, at 7$ dollars per barrel? 
 
 36. What cost 2 5 Of acres of land, at 25i dollars per acre ? 
 
 37. If a man travels 40f miles per day, how far will he travel 
 in 135$ days? 
 
 Qv EST. 220. Wheu the multiplier and multiplicand are mixed numbers, how proceed 
 
130 MULTIPLICATION OF [SECT. VIL 
 
 CONTRACTIONS IN MULTIPLICATION OF FRACTIONS. 
 
 Ex. 1. Multiply f by and -ft- and i and f. Up 
 
 Operation. Since the factors 3, 5 and 8 'are 
 
 $ 17 7 common to the numerators and denom- 
 
 _. vx _ NX vx "v _ 
 
 001 1 $2 22 inators, we may cancel them; (Art. 
 191 ;) and then multiply the remain- 
 ing factors together, as hi reduction of compound fractions to sim- 
 ple ones. (Art. 199.) Hence, 
 
 221* To multiply fractious by CANCELATION. 
 
 Cancel all the factors common both to the numerators and de- 
 nominators ; then multiply together fa factors remaining in tJie 
 numerators for a new numerator, ancr those remaining in t/te de~ 
 nominators for a new denominator, as in reduction of compound 
 fractions. (Art. 199.) 
 
 OBS. 1. The reason of this process may be seen from the fact that the product 
 of the numerators is divided by the same numbers as that of the denominators, 
 and therefore the value of the answer is not altered. ^Art. 191.) 
 
 2. Care must be taken that the factors canceled in the numerators are ev~ 
 actly equal to those canceled in the denominators. 
 
 2. Multiply | by f and f. Ans. f . 
 
 3. Multiply f by -I into . 7. Multiply f of f by if. 
 
 4. Multiply f by tV into . 8. Multiply ft by if- of -fr- 
 
 5. Multiply f by i into . 9. Multiply ri of 4 by -ft. 
 
 6. Multiply | by f of f . 10. Multiply 3f by if of 8 
 
 11. Multiply | by $ and f andf and f. 
 
 12. Multiply i by -f- and -& and -fa and fg-. 
 
 13. Multiply if by -f and -f and -ft- and -ft . 
 
 L4. Multiply -ft into f into f into into 4^ into f . 
 
 15. Multiply -ft- into -ff into || into ff into f into i. 
 
 16. Multiply into -ft into % into -fe into -f^ into f. 
 
 17. What must a man pay for 3-^ barrels of flour,, when flour ; 
 worth 6 dollars a barrel ? 
 
 QUSST. 221. How are fractions multiplied by cancelation? Obs. How does it appe&. 
 that this process will give the true answer? What is necessary to be observet with re- 
 gard to canceling factors 1 
 
ARTS. 221-223.] FRACTIONS. 131 
 
 Analysis. 3-J- bbls. is % of 10 Operation. 
 
 bbls. ; now since 1 bbl. costs 6 dolls. 6 price of 1 bbl. 
 
 dollars, 10 Ws. will cost 10 times JLO^ 
 
 as much, or 60 dollars. But we 3)60 " of 10 bbls. 
 
 wished to find the cost of only 3i dolls. ~20 " of 3-J- bbls. 
 barrels, which is -J- of 10 bbls. 
 
 Therefore if we take of the cost of 10 bbls., it will of cours* 
 je the price of 3^ bbls. 
 
 PROOF. 6 dolls. X 3-^=20 dolls., the same as before. 
 
 Note. In like manner, when the multiplier is 33 J, 333 J, &c., if we multiply 
 by 100, 1000, &c., l of the product will be the answer. Hence, 
 
 222. To multiply a whole number by 3|, 33|, 333, (fee. 
 Annex as many ciphers w the multiplicand as there are 3s in th6 
 
 integral part of the multiplier ; then take of the number thus 
 produced, and the result will be the answer required. 
 
 OBS. 1. The reason of this contraction is evident from the principle that an- 
 nexing a cipher to a number multiplies it by 10, annexing two ciphers multi- 
 plies it by 100, &c. (Art. 98.) But 3$ is of 10 ; 33 is i of 100, &c. ; there- 
 fore annexing as many ciphers to th* multiplicand, as there are 3s in the in- 
 tegral part of the multiplier, gives a product 3 times too large ; consequently 
 J of this product must be the true ansvier. 
 
 2. When the multiplicand is a mixed number, and the multiplier is 3, 33 \ 
 &c., it is evident we may multiply by 10, 100, &c., as the case may be, and | 
 of the number thus produced will be the answer required. 
 
 18. Multiply 158 by 33^. Ans. 5266-f. 
 
 19. Multiply 148 by 3i. 22. Multiply 297 by 333|. 
 
 20. Multiply 256 'by 33i. 23. Multiply 56l by 3. 
 
 21. Multiply 1728 by 33^. 24. Multiply 426f by 33i. 
 
 223. To multiply a whole number by 6|, 66f, 666|, &o. 
 Annex as many ciphers to the multiplicand as there are 6s in the 
 
 integral part of the multiplier ; tJien take -f of the number thus 
 produced, and the result will be the answer required. 
 
 OBS. The reason of this contraction is manifest from the fact that 6| is > of 
 10 , 66| of 100, &c. 
 
 25. What will 6-f tons of iron cost, at 75 dollars per ton ? 
 
 QUEST. 222. How may a whole number be multiplied by 3$, 33$, &c.7 223 How 
 may a whol j number be multiplied by 6}, 66|, &c 
 
132 MULTIPLICATION OF [SECT. VII. 
 
 Analysis. -6-| tons is f of 10 Opeiation. 
 
 tons. Now if 1 ton costs 75 dol- dolls. 75, price of 1 ton 
 
 lars, 10 tons will cost 10 times as 10 
 
 much, or 750 dollars; and -f of 3)750 " of 10 ton& 
 
 750 dollars, (6-f=f of 10,) are 250 
 
 500 dollars, whnh is the answer 2 
 
 required. dolls. 500, " of 6f tons 
 
 PROOF. 75 dolls. X6f =500 dolls., the same as above. 
 
 26. Multiply 320 by 6f. 28. Multiply 837 by Gf. 
 
 27. Multiply 277 by 66f. 29. Multiply 645 by 666 J. 
 
 30. What will 12 acres of land cost, at 46 dollars per acre? 
 
 Analysis. 12 acres is -J- of lOOfc Operation. 
 
 *cres ; now since 1 acre costs 46 dol- dolls. 46, price of 1 A* 
 
 lars, 100 acres will cost 100 times as 100 
 
 much, or 4600 dollars. But we wished 8)4600 " 100 A. 
 
 to find the cost of only 1 2 acres, which dolls. 575 " 12 A 
 is i of 100 acres. Therefore i of the 
 cost of 100 acres, will obviously be the cost of 12 acres. 
 
 PROOF. 46 dolls. Xl2i=575 dolls., the same as before. 
 
 Note. In like manner, if the multiplier is 37J, 62 , or 87|, we may nv Jtip^y 
 by 100, t.nd f, f, or J of the product will be the answer. Hence, 
 
 224 To multiply a whole number by 12, 37, 62, or 87. 
 
 Annex two ciphers to the multiplicand, then take \, -f, -f, or -, 
 of the number thus produced, as tlie case may be, and tlie result 
 will be the answer required. 
 
 OBS. The reason of this contraction may be seen from the fact that 12 is J, 
 37J is I, 62 is f , and 87$ is 5 of 100. 
 
 31. Multiply 275 by 37. Ans. 10312-^. 
 
 32. Multiply 381 by 12. 34. Multiply 643 by 62. 
 
 33. Multiply 425 by 37. 35. Multiply 748 by 87. 
 
 225. To multiply a whole number by 1-f, 16|, 166|, &c. 
 
 Annex as many ciphers to the multiplicand as there are integral 
 figures in the multiplier., then ^- of the number thus produced will 
 be the product required. 
 
A.RTS. 224-226. J FRACTIONS. 133 
 
 OBS. The reason of this contraction is evident from the fact that 1| is -J 
 of 10; 1GI is | of 100; IGGf is of 1000, &c. 
 
 36. What will 16 f bales of Swiss muslin cost, at 735 dollars 
 per bale ? 
 
 Solution. Annexing two ciphers to 735 dolls., it becomes 
 72500 dolls. ; and 73500-7-6=12250 dolls. Ans. 
 
 37. Multiply 767 by If. 39 Multiply 489 by I6f. 
 
 38. Multiply 245 by 16t. 40. Multiply 563 by 166|. 
 
 Note. Specific rules might be added for multiplying by 1^, 11-^, lll, 8-^, 
 83j, 833J, 6$, &c., but they will naturally be suggested to the inquisitive mind 
 from the contractions already given. 
 
 DIVISION OF FRACTIONS. 
 
 CASE I. 
 226. Dividing a fraction by a whole number. 
 
 Ex. 1. If 4 yards of calico cost f of a dollar, what will 1 yard 
 cost? 
 
 Analysis. 1 is 1 fourth of 4 ; therefore 1 yard will cost 1 
 fourth part as much as 4 yards. And 1 fourth of 8 ninths of a 
 dollar, is 2 ninths. Ans. f of a dollar. 
 
 Operation. We divide the numerator of the fraction by 
 
 -7-4=3 Ans. 4, and the quotient 2, placed ovei the denomi- 
 nator, forms the answer required. 
 
 2. If 5 bushels of apples cost \% of a dollar, what will 1 bushel 
 cost? 
 
 Operation. Since we cannot dh' fle the numer 
 
 H_^.5 _ ii_ 11 A ator ^7 t ^ ie divisor 5, without a re- 
 
 12 12x5' mainder, we multiply the denomina* 
 
 tor by it, which, in effect, divides the fraction. (Art. 188.) 
 
 PROOF. H dolls. X5=ii dolls., the same as above. Henoe, 
 
134 DIVISION OP [SECT. Vll. 
 
 227, To divide a fraction by a whole number. 
 
 Divide the numerator by the whole number, when it can be don* 
 without a remainder ; but when this cannot be done, multiply the 
 denominator by the whole number. 
 
 3. What is the quotient of ^g- divided by 5 ? 
 
 First Method. Second Method. 
 
 15 3 15 15 15 3 
 
 ^ 5 = Ans ' + 5 
 
 4. Divide if by 9. 7. Divide -fi by 12. 
 
 5. Divide ff by 7. 8. Divide W by 25. 
 
 6. Divide ffr. by 16. 9. Divide -fJHJ- by 29. 
 
 CASE II. 
 228* Dividing a fraction by a fraction. 
 
 10. At -J- of a dollar a basket, how many baskets of peaches 
 can you buy for of a dollar ? 
 
 Analysis. Since f of a dollar will buy 1 basket, of a dollar 
 will buy as many baskets as is contained times in ; and % is 
 contained in , 4 times. Ans. 4 baskets. 
 
 11. At f of a dollar per yard, how many yards of cloth can be 
 bought for f of a dollar ? 
 
 OBS. 1. Reasoning as before, 5 of a dollar will buy as many yards, as | of 
 a dollar is contained in J. But since the fractions have different denomina- 
 tors, it is plain we cannot divide one numerator by the other, as we did in the 
 last example. This difficulty may be remedied by reducing the fractions to a 
 common denominator. (Art. 200.) 
 
 First Operation. 
 
 f and f reduced to a common denominator, become ^ and f . 
 (Art. 200.) Now fi-j-if =- H I and -?i=l-ft-. ^. 1-fr yards. 
 
 OBS. 2. It will be perceived that no use is made of the common denominator^ 
 after it is obtained. If, therefore, we invert the divisor, and then multiply the 
 two fractions together, we shall have the same result as before. 
 
 Second Operation. 
 (divisor in verted) =f, or IT\ yards, the same as 
 
 QUEST 227. How is a fraction divided by a whole number *? 
 
ARTS. 227-230 J FRACTIONS. J35 
 
 229. Hence, to divide a fraction by a fraction. 
 
 I. If the given fractions have a common denominator., divide tht 
 numerator of the dividend by the numerator of the divisor. 
 
 II. When the fractions have not a common denominator, invert 
 the divisor, and proceed as in multiplication of fractions. (Art. 219.) 
 
 OBS. 1. When two fractions have a common denominator, it is plain one 
 nuwrator can be divided by the other, as well as one whole number by an- 
 sike" ; for, the parts of the two fractions are of the same denomination. 
 
 2 When the fractions do not have a common denominator, the reason that 
 inverting the divisor and proceeding as in multiplication, will produce the triti 
 Answer, is because this process, in effect, reduces the two fractions to a com- 
 mon denominator, and then the numerator of the dividend is divided by the 
 numerator of the divisor. Thus, reducing the two fractions to a common de- 
 nominator, we multiply the numerator of the dividend by the denominator of 
 the divisor, and the numerator of the divisor by the denominator of the divi- 
 dend ; (Art. '200 ;) and, then dividing the former product by the latter, we have 
 ihe same combination of the same numbers as in the rule above, which will con- 
 sequently produce the same result. 
 
 We do not multiply the two denominators together for a common denomina- 
 tor ; for, in dividing, no use is made of a common denominator when found, 
 therefore it is unnecessary to obtain it. (Art. 228. Obs. 2.) 
 
 The object of inverting the divisor is simply for convenience in multiplying. 
 
 3. Compound fractions occurring in the divisor or dividend, must be re- 
 duced to simple ones, and mixed numbers to improper fractions. 
 
 230. The principle of dividing a fraction by a fraction may 
 also be illustrated in the following manner. Thus, in the last 
 example, 
 
 Dividing- the dividend by 2, the quo- Operation: 
 
 tient is -ft. (Art. 188.) But it is required --^2 = -ft 
 
 to divide it by 1 third of two ; consequently &X%=H 
 the -ft- is 3 times too small for the true And -^=1-^- Ans* 
 quotient ; therefore multiplying -ft by 3, 
 will give the quotient required ; and -ft-X3=-fi, or 1-^-. 
 
 Note. By examination the learner will perceive that this process is precisely 
 
 QTEST. 229. How is one fraction divided by another when they have a common de 
 nominator 1 How, when they have not common denominators ? Oba. When the fractions 
 have a common denominator, how does it appear that dividing any numerator by the other 
 will give the true answer 1 When the fractions have not a common denominator, how 
 does it appear that inverting the divisor and proceeding as in multiplication will give the 
 true answer 1 What is the object of inverting the divisor 7 How proceed when the divisor 
 or dividend are compound fractions or mixed numbers 7 
 
136 DIVISION OF [SECT. VII 
 
 the same in effect as the preceding; for, in both cases the denominator of thfc 
 dividend is multiplied by the numerator of the divisor, and the numerator of 
 the dividend, by the denominator of the divisor. 
 
 12. Divide f of J by 2. Ans. if, or ft. 
 
 13. Divide 8-f by 3. Ans. -|f, or 2. 
 
 14. Divide ? by ff-. 16. Divide 55 by 16f. 
 
 15. Divide ff by if. 17. Divide 461 by 68f. 
 
 231* The process of dividing fractions may often be con- 
 tf acted b* canceling equal factors in the divisor and dividend ; 
 (Art. 146 ;) or, after the divisor is inverted, by canceling factors 
 which are common to the numerators and denominators. (Art. 191.) 
 
 18. Divide \ of * of -ft by of -f of f. 
 
 Operation. For convenience we arrange the numera- 
 
 1 tors, (which answer to dividends,) on the 
 
 right of a perpendicular line, and the de- 
 ll $ nominators, (which answer to divisors,) on 
 the left ; then canceling the factors, 2, 3, 4, 
 and 7, which are common to both sides, 
 (Art. 151,) we multiply the remaining fac- 
 lYJ5=A- Ans ^ ors * n ^h e n um e ra tors together, and those 
 remaining in the denominators, as in the 
 rule above. Hence, 
 
 232. To divide fractions by CANCELATION. 
 
 Having inverted the divisor, cancel all the factors common both 
 to the numerators and denominators, and the product of those re- 
 maining on the right of the line placed over tlieprod.net of those 
 remaining on the left, will be the answer required. 
 
 OBS. 1. Before arranging the terms of the divisor for cancelation, it is always 
 necessary to invert them, or suppose them to be inverted. 
 
 2. The reason of this contraction is evident from the principle, that if the 
 numerator and denominator of a fraction are both divided by the same w ^Jf^r t 
 the value of the fraction is not altered. (Arts. 148, 191.) 
 
 19. Divide 18f by 6-f. Answer 3. 
 
 QI:ESI. 232. How divide fractions by cancelation? How arrange the terms of fh 
 given fractions "t Obs. What must be done to the divisor before arranging its terms ? Ho* 
 does it appear that this contraction will give the true answer 1 
 
ARTS. 231-234.] FRACTIONS. 137 
 
 20. Divide of -f by f of -ft- 23. Divide f of 7f- by f of f . 
 
 21. Divide f of ft by 6f. 24. Divide f of f of -f by f 
 
 22. Divide 15| by -ft- of -f. 25. Divide f-f of 7 by fi of 42. 
 26. Divide H of li of A of |i by -& of ff of f of 5. 
 
 CASE III. 
 
 233* Dividing a whole number' by a fraction. 
 
 27. How many pounds of tea, at -^ of a dollar a pound, sari be 
 oouglit for 15 dollars ? 
 
 Analysis. Since f of a dollar will buy 1 pound, 15 dollars will 
 buy as many pounds as f- is contained times in 15. Reducing the 
 dividend 15, to the form of a fraction, it becomes -4 s -; (Art. 197. 
 Obs. 1 ;) then inverting the divisor and proceeding as before, we 
 have -"^Xi^V-, or 20. Ans. 20 pounds. 
 
 Or, we may reason thus : -| is contained in 15, as many times 
 as there are fourths in 15, viz: 60 times. But 3 fourths will be 
 contained in 15, only a third as many times as 1 fourth, and 
 60-r-3 = 20, the same result as before. Hence, 
 
 234* To divide a whole number by a fraction. 
 
 Reduce the whole number to the form of a fraction, (Art. 197. 
 Obs. 1,) and then proceed according to the rule for dividing a 
 fraction by a fraction. (Art. 229.) 
 
 Or, multiply the whole number by the denominator, and divide 
 the product by the numerator. 
 
 OBS. 1. When the divisor is a mixed number, it must he reduced to an im- 
 proper fraction ; then proceed as above. 
 
 Or, reducing the dividend to a fraction having the same denominate, (Art. 
 197. Obs. 2,) we may divide one numerator by the other. (Art. 229. I.) 
 
 2. If the divisor is a unit or 1, the quotient is equal to the dividend ; if the 
 divisor is greater than a unit, the quotient is less than the dividend ; and if the 
 divisor is less than a unit, the quotient is greater than the dividend. 
 
 28. How much cloth, at 3 dollars per yard, can you buy for 
 28 dollars ? 
 
 QUEST. 234. How is a whole number divided by a fraction ? Obs How by a mixed 
 aumber 1 
 
'38 DIVISION OF L^ ECT ' 
 
 Operation. Since the divisor is a mixed number, 
 
 3^)28 we reduce it to halves ; we also reduce 
 
 the dividend to the same denominator ; 
 
 7) 56 halves. (Art. 197. Obs. 2 ;) then divide one nu> 
 
 Ans. 8 yards. merator by the other. (Art. 229. I.) 
 
 29. Divide 75 by f. 32. Divide 145 by 12|. 
 
 30 Divide 96 by f. 33. Divide 237 by 25f. 
 
 31 Divide 120 by 10^. 34. Divide 425 by 31$. 
 
 CONTRACTIONS IN DIVISION OF FRACTIONS. 
 
 235. When the divisor is 3|, 33i, 333^, &c. 
 
 Multiply the dividend by 3, divide the product by 10, 100, or 
 1000, as tlie case may be, and the result will be the true quotient. 
 (Art. 131.) 
 
 OBS. The reason of this contraction will be understood from the principle, 
 that if the divisor and dividend are both multiplied by the same number, the 
 quotient will not be altered. (Art. 14G.) Thus 3^X3=10; 33JX3=100; 
 333*X3=1000, &c. 
 
 35. At 3-^ dollars per yard, how many yards of cloth can be 
 bought for 561 dollars 9 
 
 Operation. We first multiply the dividend by 3, 
 
 dolls. 561 then divide the product by 10; for, mul- 
 
 3 tiplymg the divisor 3-^ by 3, it becomes 10. 
 
 1|0)1C8|3 (Art. 146.) 
 Ans. 168/0 yds. 
 
 36. Divide 687 by 33. Ans. 20-fljV. 
 
 37. Divide 453 by 33^, 38. Divide 2783 by 333. 
 
 236. When the divisor is If, 16|, 166|, &c. 
 
 Multiply the dividend by 6, and divide the product by 10, 100, 
 or 1000, as tJie case may be. 
 
 OBS. This contraction also depends upon the principle, that if the divisor 
 and dividend are both multiplied by the same number, the quotient will riat b 
 altered. (Art. 146.) Thus, lfX6=lO; 16|X6=100; 166X6=1000, &c. 
 
ARTS. 235-239.] FRACTIONS. J39 
 
 39. What is the quotient of 725 divided by 16| ? 
 Solution. 725X6 4350; and 4350-rlOO = 43i Ans. 
 
 40. Divide 367 by If. 42. Divide 849 by 16-f. 
 
 41. Divide 507 by 16-f. 43. Divide 1124 by 166-f. 
 
 237. V T hen the divisor is H, 1H, 11H, &c. 
 
 Multiply the dividend by 9, and divide the product by 10, 100, 
 *r 1000, as the case may be. 
 
 OBS. This contraction depends upon the same principle as the preceding 
 Thus, 1|X9=10; 11|X9=100; HlfX9=lOOO, &c. 
 
 44. Divide 587 by Hi. 
 
 Solution. 587X9 = 5283, and 5283-7-100 = 52^ Ans. 
 
 45. Divide 861 by H. Ans. 774-rV. 
 
 46. Divide 4263 by Hi. 47. Divide 6037 by llli- 
 
 Tvotc. Other methods of contraction might be added, but they will rjaturally 
 suggest themselves to the student, as he becomes familiar with the principles 
 of fractions. 
 
 238* From the definition of complex fractions, and the man- 
 ner of expressing them, it will be seen that they arise from di- 
 
 AL 
 
 vision of fractions. (Art. 183.) Thus, the complex fraction y-, is 
 the same as -f-7-f- ; for, the numerator, 4=f , and the denomina- 
 tor 1-J-^-f- ; but the numerator of a fraction is a dividend, and the 
 denominator a divisor. (Art. 184.) Now, -|-7-J=li-. which is a 
 simple fraction. Hence, 
 
 239* To reduce a complex fraction to a simple, one. 
 Consider the denominator as a divisor, and proceed as in dims- 
 ion of fractions. (Arts. 229, 232.) 
 
 OBS. The reason of this rule is evident from the fact that the denominator 
 of a fraction denotes a divisor, and the numerator, a dividend; (Art. 184; 
 nence the process required, is simply performing the division which is ex- 
 pressed by the given fraction. 
 
 QUEST. 238 P om what do complex fractions arise ? 239. How reduce them to sii- 
 plc fractions 7 
 
 T.H. t 
 
140 COMPLEX {SECT. VII 
 
 48. Reduce -| to a simple fraction. 
 Solutim. 4|=-KS and 7}=^. (Art. 197.) 
 
 NOW Y-r-^YX -gV, or -ff ^W5. 
 
 Reduce the following complex fractions to simple o nes : 
 
 8 5- 
 
 49. Reduce. 53. Reduce-^. 
 
 34 T6 
 
 5 1 7 
 
 50. Reduce y. 54. Reduce-^. 
 
 2-2 -2-5- 
 
 51. Reduce -|. 55. Reduce. 
 
 7 90 
 
 6i 44 
 
 52. Reduce . 56. Reduce . 
 
 24O. To multiply compbx fractions together. 
 
 First reduce the complex fractions to simple ones ; (Art. 39;) 
 then arrange the terms, and cancel the common factors, as in mul- 
 tiplication of simple fractions. (Art. 219.) 
 
 OBS. The terms of the complex fractions may be arranged for reducing them 
 to simple o/ies, and for multiplication at the same time. 
 
 57. Multiply g by g. 
 
 Operation. The numerator 3^=-|. (Art. 197.) Place 
 
 the 7 on the right hand and 2 on the left of 
 the perpendicular line. The denominator 2-f 
 = J s a , which must be inverted; (Art. 239 ;) 
 i. e. place the 12 on the right and the 5 on 
 the left of the line, 1-f =-V, and 4-=f, both 
 of wVich must be arranged in the same man- 
 ner as the terms of the multiplicand. Now, canceling the com- 
 mon factors, we divide the product of those remaining on the right 
 of the line by the product of those on the left, and the answer 
 is |. (Art. 219.) 
 
 QUKST. 240. How are complex fractions multiplied together ? 241. How is one com- 
 plex fraction divided by another ? 
 
ARTS. 240-242. J FRACTIONS. 141 
 
 58. Multiply ? by ~ 60. Multiply | by | into ~. 
 
 ^4 ^3 4 T f 
 
 59. Multiply i| by ^ . 61. Multiply ^ by ?| into j|. 
 
 241* To divide one complex fraction by another. 
 
 Reduce the complex fractions to simple ones, then proceed as in 
 '{vision, of simple fractions. (Arts. 229, 239.) 
 
 62. Divide byj. 
 
 4i 9 4 36 ,+ 144 
 *-5f= 5 X r - ig , and I 2=3X ? = 2 - I . (Art. 239.) 
 
 36 4 30 21 756 
 
 9X4 1X4 
 
 Or, since the given dividend = - and the divisor = - r 
 
 9X4 3 ^ Y 
 then X YTH ~ the ans wer - ( Art - 2 3 1 ) 
 
 9X4 3XV 0X^X3XV 21 
 But, (Art. 232.) x = => or 10- An* 
 
 63. Divide | by |. 64. Divide ^| by ?|. 
 
 APPLICATION OF FRACTIONS. 
 
 Ex. 1. A merchant bought 15- yards of domestic flan- 
 nel of one customer, 19^ of another, 12- of another, and 41-$^- of 
 another : how many yards did he buy of all ? 
 
 2. A grocer sold 16-J- Ibs. of sugar to one customer, 112 to 
 another, and 33^- to another : how many pounds did he sell ? 
 
 3. A clerk spent 26f dollars for a coat, 9f dollars for pants, 
 6f dollars for a vest, .5^- dollars for a hat, and 6-J- dollars for a 
 pair of boots : how much did his suit cost him ? 
 
 4. A man having bought a bill of goods amounting to 85-^g- dol- 
 lars, handed the clerk a bank note of 100 dollars: how much 
 change ought he *a receive back? 
 
142 DIVISION OF [SECT. VII. 
 
 5. A lady went a shopping with 135}- dollars in her purse* 
 she paid 17^ dollars for silk, 3f dollars for trimmings, 37 dol- 
 lars for a shawl, and 14f dollars foi a muff: how much money 
 had she left ? 
 
 6. A man having 1563^- dollars, spent 365f dollars, and lost 
 fif>2 dollars : how much had he left ? 
 
 7. What will 563 sLeep cost, at 2f dollars per head ? 
 
 8. What cost 748 barrels of flour, at 7-f dollars per barrel ? 
 
 9. What cost 378f yards of cloth, at 4 dollars per yard? 
 
 10. What cost 1121-^- Ibs. of tea, at 5 shillings per pound? 
 
 11. What cost 430 gallons of oil, at H dollar per gallon? 
 
 12. What cost ff of an acre of land, at 150 dollars per acre ? 
 
 13. A man worth 25000 dollars, lost H of it by fire : what 
 was the amount of his loss ? 
 
 14. A garrison had 856485 pounds of flour ; after being block- 
 aded 60 days, it was found that $ of it were consumed : how 
 many pounds of flour were left ? 
 
 15. At I7i dollars per ton, what cost 103-J- tons of hay? 
 
 16. How many bushels of corn will 115f acres produce, at 31-} 
 bushels per acre ? 
 
 17. What cost 675-|- tons of iron, at 45f dollars per ton? 
 
 18. If a ship sails 140-fV miles per day, how far will she sail 
 in 49 days? 
 
 19. If a Railroad car should run 4l miles per hour, how far 
 would it go in 12 days, running 10 hours per day? 
 
 20. A young man having a patrimony of 12234 dollars, spent 
 ~ of it in dissipation : how much had he left ? 
 
 21. At -f- of a dollar per yard, how many yards of satinet can 
 be bought for 124 dollars? 
 
 22. How many pounds of tea, at f of a dollar a pound can you 
 buy for 331 dollars? 
 
 23. How many gallons of molasses, at | of a dollar per gallon 
 can you buy for 235 dollars ?^ 
 
 24. At 8 pence a pound, how many pounds of sugar can you 
 buy for 163-J- pence? 
 
 25. At 5-J- pen; e a yard, how many yards of lace can be bought 
 for 279 pence ? 
 
ART. 242.] FRACTION?. 143 
 
 26. A dairy-man has 229 pounds of butter wlich he wishes 
 to pack in boxes containing 8% pounds each : how many boxes 
 will it require ? 
 
 27. A farmer wishes to put 384 bushels of apples into barrels, 
 each containing 2 bushels : how many barrels will it require ? 
 
 28. If 4 1 yards of cloth make a suit of clothes, how many suits 
 will 141 yards make ? 
 
 29. One rod contains 5- yards : how many rods are there in 
 210 yards? 
 
 30. A merchant paid 204^ dollars for 57 yards of cloth : how 
 much was that per yard ? 
 
 31. A grocer sold 50 barrels of flour for 31 1 dollars: what 
 did he get per barrel ? 
 
 32. A merchant wishes to lay out 657 dollars for wheat, which 
 is worth ! of a dollar a bushel : how much can he buy ? 
 
 33. At 18-f cents a dozen, how many dozen of eggs can you 
 buy for 87 cents ? 
 
 34. A grocer sold 15 pounds of coffee for 93f cents: how 
 much was that a pound ? 
 
 35. A shopkeeper sold 16 yards of satin for 163-fr shillings: 
 how much was that per yard ? 
 
 36. Bought 19 sacks of wool for 250f dollars: what was that 
 per sack ? 
 
 37. Paid 575f dollars for 96$ yards of cloth: what was the 
 cost per yard ? 
 
 38. Paid 1565^ dollars for iron, valued at 3 7-J- dollars per ton: 
 how many tons were bought ? 
 
 39. Paid 1315^ dollars for the transportation of 1286 barrels 
 of pork : what was that per barrel ? 
 
 40. Bought 37oi pounds of indigo for 65 2| dollars : what was 
 the cost per pound ? 
 
 41. Paid 1679i dollars for 475 kegs of lard: how much was 
 that per keg ? 
 
 4.2. If an army consumes 563^ pounds of meat per day, how 
 long will 150000 pounds supply it? 
 
 43. The cost of making 25i miles of Railroad was 856?35| dol 
 lars : what was the cost per mile ? 
 
144 COMPOUND NUMBERS. [SECT. VIII. 
 
 SECTION VIII. 
 COMPOUND NUMBERS. 
 
 ART. 243* Numbers which express things of tl e same kind 
 or denomination, are called Simple Numbers. Thus, 3 oranges, 
 7 books, 12 chairs, &c., are simple numbers. 
 
 Numbers which express things of different kinds or denomina- 
 tions, as the divisions of money, weight, and measure, are called 
 COMPOUND NUMBERS. Thus, 15 shillings 6 pence; 10 bushels 
 3 pecks, &c., are compound numbers. 
 
 OBS. The origin of Compound Numbers is ascribed to the wants and neces- 
 sities of the earlier ages of the world. Their divisions and subdivisions are 
 generally irregular, and seem to have been suggested by the caprice, or the lim- 
 ited business transactions of the rude ages of antiquity. It is much to be re- 
 gretted, both on account of simplicity and their adaptation to scientific pur- 
 poses, that their different denominations were not graduated according to the 
 law of increase in the decimal notation. 
 
 Note. Compound Numbers, by some authors, are called Denominate 
 Numbers. 
 
 FEDERAL MONEY. 
 
 214. federal Money is the currency of the United States. 
 The denominations are, Eagles, Dollars, Dimes, Cents, and Mills. 
 
 10 mills (m.) make 1 cent, marked ct. 
 
 10 cents " 1 dime, " d. 
 
 10 dimes " 1 dollar, " doll, or $. 
 
 10 dollars 1 eagle, E. 
 
 OBS. 1. Federal money was established by Congress, Aug. 8, 1786. It rt 
 based upon the principles of the decimal notation. The law of increase OL 
 radix, is the same as that of simple numbers, and it is confessedly one of th 
 most simple and comprehensive systems of currency in the civilized world. Pre- 
 vious to its adoption, English or sterling money was the principal currency of 
 the country. 
 
 Q.UEST. 343. What are simple numbers ? What are compound numbers ? 244. Wha 
 fc Federal money ? Recite the table. Obs. When and by whom was it established I 
 
ARTS, 243-246. J COMPOUND NUMBERS. 145 
 
 2. The names of the coins or denominations less than a dollar, are signifi- 
 cant of their value. The term dime, is derived from the French disme, which 
 signifies ten; the terms cent and mill, are from the Latin centum and mille. 
 the former of which signifies a hundred, and the latter a tkoitsand. Thus, 
 10 dimes*, 100 cents, or 1000 mills, make 1 dollar. 
 
 3. The sign ($), which is prefixed to Federal money, is called the Dollar 
 mark. It is said to be a ccntraction of " U. S.," the initials of United Slates^ 
 which were originally prefixed to sums of money expressed in the Federal 
 currency. At length the two letters were moulded or merged into a single char- 
 acter by dropping the curve of the U, and writing the S over it. Thus, th 
 um of seventy-five dollars, which was originally written " U. S. 75 dollars, 
 is now written $75. 
 
 245* The national coins of the United States are of three 
 kinds, viz : gold, silver, and copper. 
 
 1. The golds coin are the eagle, the double eagle* half eagle, 
 quarter eagle, and gold dollar* 
 
 The eagle contains 258 grains of standard gold ; the half eagle 
 and quarter eagle like proportions.! 
 
 2. The silver coins are the dollar, half dollar, quarter dollar, 
 the dime, half dime, and three-cent-piece. 
 
 The dollar contains 412 grains of standard silver; the others 
 like proportions.]- 
 
 3. The copper coins are the cent, and half cent. 
 
 The cent contains 1C8 grains of pure copper; the half cent, a 
 like proportion.^ Mills are not coined. 
 
 OBS. The fineness of gold used tor coin, jewelry, and other purposes, also 
 the gold of commerce, is estimated by the number of parts of gold which it 
 contains Pure gold is commonly supposed to be divided into 24 equal parts, 
 "alied carats. Hence, if it contains 10 parts of alloy, or some baser metal, it is 
 aid to be 14 carats fine ; if 5 parts of alloy, 19 carats fine ; and when abso- 
 lutely pure, it is 24 carats fine. 
 
 246. The present standard for both gold and silver coin o. 
 the United States, by Act of Congress, 1837, is 900 parts cf pur* 
 
 Of how many kinds are the coins of the United States ? What are they 
 What are the gold coins 1 The silver coins ? The copper? Obs. How is the fineness of 
 g>ld estimated ? Into how many carats is pure gold supposed to be divided 1 When i 
 contains 10 parts of alloy, how fine is it said to be ? 5 parts of alloy? 246. What is the 
 present standard for the gold am* silver coin of tho United States ? What is the alloy of 
 fold com t What of silver coin ? 
 
 * Added by Act of Congress, 1849. f According to Act of Congress, 1837. 
 
146 COMPOUND NUMBERS. [SECT. VIII 
 
 metal by weight to 100 parts of alloy. The alloy of gold coin is 
 composed of silver and copper, the silver not to exceed the cop- 
 per in weight. The alloy of silver coin is pure copper. 
 
 Note. The original standard for the gold coin of the United States by Act 
 of Congress, 1792, was 22 parts of pure gold to 2 parts of alloy ; the alloy 
 consisting of 1 part silver and 1 part copper. 
 
 The original standard for the silver coin was 1489 parts of pure silver to 1 79 
 parts of alloy; the alloy being of pure copper. 
 
 The tagle by the same act contained 270 grains of standard gold. The djl~ 
 ir contained 416 grains of standard silver. The cent contained 11 penny- 
 weights, or 264 grains of pure copper. 
 
 STERLING MONEY. 
 
 247* English or Sterling Money is the national currency of 
 Great Britain. 
 
 4 farthings (qr. or/ar.) make 1 penny, marked d. 
 
 12 pence " 1 shilling, " s. 
 
 20 shillings " 1 pound, or sovereign, . 
 
 21 shillings " 1 guinea. 
 
 OBS. 1. It is customary, at the present day, to express farthings in fractions 
 of a penny. Thus, 1 qr. is written i d. ; 2 qrs. J d. ; 3 qrs. f d. 
 
 2. The Pound Sterling is represented by a gold coin, called a Sovereign. 
 According to Act of Congress, 1842, its value is 4 dollars and 84 cents. Hence, 
 the value of a shilling is 24- cents ; that of a penny 2 cents, very nearly. 
 
 3. The letters . s. d. and q. are the initials of the Latin words, libra, soli- 
 dus, denarius, and quadrans, which respectively signifiy a pound, shilling, 
 penny, a.ndfartki7ig or quarter. The mark /, which is often placed between 
 shillings and pence, is a corruption of the longy. 
 
 Note. 1. Sterling money is supposed by some to have received its name 
 from the Easterlings, who it is said first coined it ; others think it is so called 
 to distinguish it from stocks, &c., whose value is nominal. 
 
 2. The pound is so called, because in ancient times the silver for it weighed 
 a pound Troy. A pound Troy of silver is now worth 66 shillings, or j3, 6s. 
 
 The Guinea is so called, because the gold of which it was originally made, 
 was brought from Guinea, on the coast of Africa. 
 
 248. The following denominations are frequently met with, 
 viz: the Groat = 4d. ; the Crown = 5s. ; the Nobler 65. Sd. ; 
 
 QUEST. 247. What is Sterling Money! Repeat the Tal le ? Obs. Hew are farthings 
 usually expressed 1 How is a pound sterling represented ? What is its value in dWait 
 and cents 1 
 
ARTS. 247-250.] COMPOUND NUMBERS. 147 
 
 the Angel=10s.; the Mark=13s. 4d. ; the Pistole=lGs. IQd. 
 
 the Moidore 27s. 
 
 OBS. The present standard gold coin of Great Britain, consists of 22 part* 
 pure gold, and 2 parts of copper.* 
 
 The weight of a Sovereign or , is 5 pwts., 3^-^ grains. 
 
 The standard silver coin consists of 37 parts of pure silver, and 3 parts of 
 copper. The weight of a shilling is 3 pwts. 15^ f grs. 
 
 In copper coin 24 pence weigh 1 pound avoirdupois. 
 
 TROY WEIGHT. 
 
 249. Troy Weight is used in weighing gold, silver, jewels, 
 liquors, &c., and is generally adopted in philosophical experiments. 
 
 24 grains (gr.) make 1 pennyweight, marked pwt. 
 20 pennyweights " 1 ounce " oz. 
 
 12 ounces " 1 pound, Ib. 
 
 OBS. 1. The abbreviation oz., is derived from the Spanish onza, which sig- 
 nifies an ounce. 
 
 2. The standard of Weights and Measures is different in different countries, 
 and in different States of the Union. In 1834, the Government of the United 
 States adopted a uniform standard, for the use of the several Custom-houses 
 and other purposes. 
 
 250. The standard Unit of Weight adopted by the Govern- 
 ment, is the Troy Pound of the United States Mint. It is 
 equal to 22.7^4422 cubic inches of distilled water, at its max- 
 imum density,f the barometer standing at 30 inches, and is 
 identical with the Imperial Troy pound of Great Britain, estab- 
 lished by Act of Parliament, in 1826.J 
 
 OBS. The weights and measures in present use, were derived from very im~ 
 'perfect and variable standards. A grain of wheat, taken from the middle of 
 the ear or head, and being thoroughly dried, was the original element of aL 
 weights used in England, and was thence called a grain. At first, a weight 
 
 QUEST. 249. In what is Troy Weight used 1 Repeat the Table 1 Ola. Do all the 
 States have the same standard of weights and measures 1 250. What is the standard 
 init of weight adopted by the Government of the United States ? JVbfe. When was Troy 
 Weight introduced into Europe ? From what was its name derived 1 
 
 * Hind's Arithmetic ; also, Hutton's Mathematics. 
 
 t The maximum density of water, according to Mr. Hassler, is at the temperature ot 
 39-83 deg. Fahrenheit. 
 
 t The Troy pound of the U. S. Mint, is an exact copy, by Captain Kater, of the Briti^fc 
 Imperial Troy pound.' Report of the Secretary of the Treasury, March 3, 1831, 
 
 7* 
 
148 COMPOUND NUMBERS. [SECT. VIII 
 
 equal to 32 grains, was called a pennyweight, from its being the weight of the 
 silver penny then in circulation. At a later period the pennyweight was di- 
 vided into 24 equal parts instead of 3*2, which are still called grains, being the 
 smallest weight now in common use. 
 
 Note. Troy Weight was formerly used in weighing articles of every kind, 
 It was introduced into Europe from Cairo in Egypt, about the time of the 
 Crusades, in the 12th century. Some suppose its name was derived from 
 Troyes, a city in France, which first adopted it ; others think it was derived 
 from Troy-novant) the former name of London.* 
 
 AVOIRDUPOIS WEIGHT. 
 
 251* Avoirdupois Weight is used in weighing groceries ana 
 all coarse articles ; as, sugar, tea, coffee, butter, cheese, flour, hay, 
 &c., and all metals except gold and silver. 
 
 16 drams {dr.) make 1 ounce, marked oz. 
 
 16 ounces " 1 pound, " Ib. 
 
 25 pounds " 1 quarter, " qr. 
 
 4 quarters, or 100 Ibs. " 1 hundred weight, " cwt. 
 
 20 hundred weight " 1 ton, " T. 
 
 Note. In weighing wool in England, 7 pounds make 1 clove ; 2 cloves, 1 
 stone; 2 stone, 1 tod; 6fc tods, Iwey; 2weys, 1 sack; 12 sacks, 1 last; 240 
 pounds, 1 pack. 
 
 OBS. 1. Formerly it was the custom to allow 112 pounds for a hundred 
 weight, and 28 pounds for a quarter ; but this practice has become nearly or 
 juite obsolete. In buying and selling all articles of commerce estimated by 
 weight, the laws of most of the States as well as general usage, call 100 
 pounds a hundred weight, and 25 pounds a quarter. 
 
 2. Gross weight is the weight of goods with the boxes, casks, or bags which 
 contain them. 
 
 Net weight is the weight of the goods only. 
 
 252* The Avoirdupois Pound of the United States, is equal 
 to 27.701554 cubic inches of distilled water, at the maximum 
 density, and at 30 inches barometer.f It is determined from the 
 Troy Pound, by the legal proportions of 5760 grains, which con- 
 
 QUEST. 251. In what is Avoirdupois Weight used ? Repeat the Table 1 CV;s. How majiy 
 pounds were formerly allowed for a hundred weight ? For a quarter ? What is gross weight! 
 Net weight 7 252. How is the Avomhipois pound of the United States determined ? 
 
 * Hind's Arithmetic, Art. 224. Also, North American Roview, Vol XLV. 
 t Reports of Secretary of Treasury, March 3, 1832: June 30, 1832. Also, Congressional 
 iKtcaments ot 1833. 
 
ARTS. 251-253. J COMPOLND NUMBERS. 149 
 
 jstitute the Troy pound, to 7000 grains Troy, which constitute the 
 Avoirdupois pound. That is, 
 
 5700 grains Troy make 1 pound Troy. 
 
 7000 grains " " 1 pound Avoirdupois. 
 
 437^ grains " " 1 ounce " 
 
 27i grains " " 1 dram " 
 
 OBS. 1, The British Imperial Pound Avoirdupois is equal to 27-7274 cubic 
 inches of distilled water, at the temperature of 62 Fahrenheit, when the 
 barometer stands at 30. It is determined from the Imperial Troy pound, 
 which contains 57(>0 grains, while the former contains 7000 grains. 
 
 2. Since the Troy pound of the United States is identical with the Troy 
 pound of England, the Avoirdupois pound of the former must be equal to that 
 of the latter ; for both bear the same ratio to the Troy pound. But the Eng- 
 lish avoirdupois pound is said to contain 27.7274 cu. in. of distilled water, 
 while that of the United States, according to Mr. Hassler, contains 27.701554 
 cu. in. This slight difference may be accounted for by the fact that the for- 
 mer was measured at the temperature of 02, while the latter was measured 
 at its maximum density, which is 39.83 degrees. 
 
 3. The standard of weight adopted by the State of New York, in 1827, is the 
 avoirdupois pound, whose magnitude is such that a cubic foot of distilled 
 water, at the maximum density, in a vacuum, will weigh 62 pounds, or 1000 
 ounces. 
 
 Note. The term avoirdupois, is thought by some to be derived from the 
 French avoir du poids, a phrase signifying to have weight. Others think it 
 is from avoirs, the ancient name of goods or c/tattels, and poids signifying 
 weight in the Norman dialect.* 
 
 APOTHECARIES' WEIGHT. 
 
 253. Apothecaries' Weight is used by apothecaries and phy- 
 sicians in mixing medicines. 
 
 20 grains (^r.) make 1 scruple, marked sc., or 3. 
 
 3 scruples " 1 dram, " dr., or 3. 
 
 8 drams " 1 ounce, " oz., or 5. 
 
 12 ounces " 1 pound, " tt>. 
 
 OBS. 1. The pound and ounce in this weight are the same, as the Troy 
 pound an<\ ounce; the other denominations are different. 
 
 2. Drugs and medicines are bought and sold by avoirdupois weight. 
 
 QricsT. 2.13. In what is Apothecaries' Weight used? Recitt the Table? Ohs. To 
 What are the aix>thecaries' ounce and pound equal ? How are drugs and medicines bought 
 and soldi 
 
 * President John Qaincy Adams on Weights and Measures ; also, Hind's Arithmetic. 
 
150 COMPOUND NUMBER8. [SECT. VIII 
 
 LONG MEASURE. 
 
 254* Long Measure is used in measuring distances where 
 length only is considered, without regard to breadth or depth. 
 It is frequently called linear or lineal measure. 
 
 12 inches (in.) make 1 foot, marked ft. 
 
 3 feet " 1 yard, " yd. 
 
 5j yards, or 16 feet " 1 rod, perch, or pole, " r. or p. 
 
 40 rods " 1 furlong, " fur. 
 
 8 furlongs, or 320 rods " 1 mile, " m. 
 
 3 miles " 1 league, " I. 
 
 60 geographical miles, or | j d ^ . 
 
 G9 statute miles S 
 
 360 degrees make a great circle, or the circumference of the earth. 
 
 Note. 4 inches make 1 hand ; 9 inches, 1 span ; 18 inches, 1 cubit ; 6 feet 
 1 fathom. 
 
 In measuring roads and land, surveyors use a chain which is 4 rods long, and 
 which is divided into 100 links. Hence, 25 links make 1 rod, and 7-f 2 ^ inches 
 make 1 link. This chain is commonly called G-unter's Chain, from the name 
 of its inventor. 
 
 OBS. 1. The inch is commonly divided either into eighths or tenths; some- 
 times, however, it is divided into twelfths, which are called lines. Formerly 
 the inch was divided into 3 barleycorns ; but the barleycorn is not employed as 
 a measure at the present day. The term barleycorn, is derived from a grain of 
 barley, which was the original element of Linear Measure. 
 
 2. The terms rod, pole, and -n^ch, from the French perche signifying a rod 
 are each expressive of the instrument, which was originally used as a measure 
 of this length. 
 
 255. The standard Unit of Length adopted by the Unite 
 States, is the Yard of 3 feet, or 36 inches, and is identical with 
 the British Imperial Yard. It is made of brass, at the temper- 
 ature of 62 Fahrenheit, from the scale of iighty-two inches pre- 
 \. pared by Trough ton, a celebrated English artist, for the survey 
 of the Coast of the United States. 
 
 OBS. 1. The Imperial standard yard of Great Britain is determined fromtha 
 end-idum which vibrates seconds in a vacuum, at the level of the sea. IB 
 
 QTTEST. 254. In what is Long Measure used? What is Long Measure sometimes 
 called? Recite the Table? Obs. How are inches usually divided? What is the 
 origin of the measure called barleycorn ? Is this measure now used ? 255. \V hat is tb 
 standard unit of Length adopted by the United States 1 
 
ARTS. 254-257.] COMPOUND NUMBERS. 151 
 
 Greenwich or London. This pendulum is divided into 391393 equal parts 
 aid 300000 of these parts are declared, by act of Parliament, to be the stand- 
 ard yard, at the temperature of 62 ; consequently, since the yard is divided 
 into 36 inches, it follows that the length of a pendulum vibrating seconds, un- 
 der these circumstances, is 39.1393 inches. 
 
 The English, yard is stud to have been originally determined by the length 
 of the arm of Henry I. King of England. 
 
 2. The standard of linear measure adopted by the State of New York, is 
 the pendulum which vibrates seconds, in a vacuum, at Columbia College, in the 
 city of New York, which is in the latitude of 40 42', 43". The yard is de- 
 clared to be -KHjiHHH" of this pendulum ; hence, the length of the .pendulum 
 is 39.101688 inches, at the temperature of 32. Should the standard yard 
 ever be lost, it could be recovered by resorting to the preceding experiment 
 
 CLOTH MEASURE. 
 
 2 56* Cloth Measure is used in measuring cloth, lace, and all 
 kinds of goods, which are bought and sold by the yard. 
 
 2$ inches (in.} make 1 nail, marked na. 
 
 4 nails, or 9 in. " 1 quarter of a yard, " qr. 
 
 4 quarters " 1 yard, " yd. 
 
 3 quarters, or f of a yard " 1 Flemish ell, " PI. e. 
 
 5 quarters, or 1 j yard " 1 English ell, " E. e. 
 
 6 quarters, or l yard " 1 French ell, " F. e. 
 
 OBS. Cloth measure is a species of long measure. Cloth, laces, &c., are 
 bought and sold by the linear yard, without regard to their width. 
 
 SQUARE MEASURE. 
 
 257* Square Measure is used in measuring surfaces, or 
 things whose length and breadth are considered without regard to 
 height or depth ; as, land, flooring, plastering, &c. 
 
 144 square inches (sq. in.} make 1 square foot, marked sq.ft. 
 
 9 square feet " 1 square yard, " sq. yd. 
 30| square yards, or ) 
 
 272i square feet \ l **' rod > P erch > or P ole > s ?' r ' 
 
 40 square rods 1 rood, " R. 
 
 4 roods, or 160 square rods " 1 acre, " A. 
 
 640 acres " 1 square mile, " M. 
 
 QTTEST. 256. In what is cloth measure used ? Repeat the Table Obs. Of what \M 
 cloth measure a, species 1 What is the kind of yard by which cloths, laces, &c , are bough} 
 and sold ? 257. In what is Square Measure used ? Recite the Table. 
 
152 
 
 COMPOUND NUMBERS. 
 
 [SECT. VIIL 
 
 Note. 16 square rods make 1 square chain ; 10 square chains, or 100,000 
 square links, make an acre. Flooring, roofing, plastering, &c., are frequently 
 estimated by the " square," which contains JOO square feet. 
 
 A hide of land, which is spoken of by ancient writers, is 100 acres. 
 
 OBS. 1. A. square is a figure which h&sfmir equal sides, and all its angles 
 right angks, as seen in the diagram. Hence, 
 
 A Sqibare Inch is a square, whose sides are each a 9 sq.ft.=\ sq. yd. 
 
 linear inch in length. 
 
 A Square Foot is a square, whose sides are each a 
 <#aeai foot in length. 
 
 A Square Yard is a square, whose sides are each a 
 linear yard, or three linear feet in length, and con- 
 tains 9 square feet, as represented in the adjacent 
 figure. 
 
 2. Square measure is so called, because its measuring unit is a square. Tho 
 standard of square measure is derived from the standard linear measure. Hence, 
 
 A unit of square measure is a square whose sides are respectively equal, in 
 length, to the linear unit of the same name. 
 
 CUBIC MEASURE. 
 
 258. Cubic Measure is used in measuring solid bodies, 01 
 things which have length, breadth, and thickness ; such as timber, 
 stone, boxes of goods, the capacity of rooms, ships, &c. 
 
 1728 cubic inches (cu. tw.) make 1 cubic foot, 
 
 27 cubic feet 
 40 feet of round, or 
 50 ft. of hewn timber 
 42 cubic feet 
 
 16 cubic feet 
 
 8 cord feet or 
 128 cubic feet 
 
 1 cubic yard, 
 1 ton, or load, 
 
 1 ton of shipping, 
 1 foot of wood, or ) 
 a cord foot, ) 
 
 1 cord, 
 
 marked cu.ft. 
 " cu. yd. 
 
 c. 
 
 A pile of wood 8 feet long, 4 feet wide, and 4 feet high, contains 1 
 For, 8X4X4;= 128. 
 
 ru, 
 
 . Obs. What is a square 1 What \r a square inch ? A square foot 1 A square 
 258. In what is Cubic Measure used Recite the Table. 
 
ARTS. 258, 259.] COMPOUND NUMBERS. 
 
 153 
 
 / 
 
 _- I 1 -?; 
 v- ; - - "~"~r-^ 
 
 Obs. 1. A Cube is a solid body bounded by six 
 equal sides. It is often called a hexaedron. Hence ? 
 
 A Cubic Inch is a cube, each of whose sides 
 is a square inch, as represented by the adjoin- 
 ing figure. 
 
 A Cubic Foot is a cube, each of whose sides 
 is a square foot. 
 
 2. Cubic Measure is so called, because its meas- 
 uring unit is a cube. It is often called solid meas- 
 ure. The standard of cubic measure is derived from the standard linear 
 measure. 
 
 A unit of cubic measure, therefore, is a cube whose sides are respectively 
 equal in length to the linear unit of the same name. 
 
 3. The cubic ton, sometimes called a load, is chiefly used for estimating the 
 cartage and transportation of timber. By a ton of round timber is meant, 
 such a quantity of timber in its rough or natural state, as when hewn, will 
 make 40 cubic feet, and is supposed to be equal in weight to 50 feet of hewn 
 timber. 
 
 The cubic ton or load, is by no means an accurate or uniform stand- 
 ard of estimating weight ; for, different kinds of timber, are of very different 
 degrees of density. But it is perhaps sufficiently accurate for the purposes to 
 which it is applied. 
 
 Note. For an easy method of forming models of the Cube and other regular 
 Solids, see Thomson's Legendre's Geometry, p. 222. 
 
 WINE MEASURE. 
 
 259* Wine Measure is used in measuring wine, alcohol, mo- 
 lasses, oil, and all other liquids except beer, ale, and milk. 
 
 4 gills (gi.) 
 
 2 pints 
 
 4 quarts 
 31 & gallons 
 42 gallons 
 63 gallons, or 2 barrels 
 
 2 hogsheads 
 
 2 pipes 
 
 marked pt. 
 
 make 1 pint, 
 " 1 quart, 
 " gallon, 
 " barrel, 
 " 1 tierce 
 
 hogshead, 
 
 pipe or butt, 
 
 tun, 
 
 OBS. 1. In England, 10 gallons make 1 anker; 18 gallons, 1 runlet; 9 
 tierces or 84 gallons, 1 puncheon. 
 
 2. Liquids are generally bought and sold by the gallon or its subdimnons, a* 
 
 QUKST.--O&S. What is a cube ? What is a cubic inch ? A cubic foot ? What is 
 meant by a ton of round timber ? 259. In what is Wine Measure used ? Recite the 
 Table. 
 
 bar. or bbl. 
 
 tier. 
 
 Mid. 
 
 pi. 
 
 tun. 
 
154 COMPOUND NUMBERS. [StiCT. VI11 , 
 
 the quart, pint, &c. Cider and a few cheap articles are bought and sold by 
 the barrel. The capacities of cisterns, vats, &c are sometimes estimated hi 
 hogsheads, and the quotations or prices-current of oils in foreign markets, are 
 usually made in tuns. But the tierce, and the pipe or bull are never used, as 
 such, in business transactions; their contents are given in gallons, quarts, &c. 
 
 26O. The standard Unit of Liquid Measure adopted by the 
 
 1 United States, is the Wine Gallon of 231 cubic inches, which is 
 
 equal to 58372.175^ grains of distilled water, at the maximum 
 
 density, weighed in air at 30 inches barometer, or 8.339 Ibs. 
 
 avoirdupois, very nearly.* 
 
 OBS. The British imperial standard measure of capacity, both for fluids and 
 dry goods, is the imperial gallon, which is equal to 10 pounds avoi -upois of 
 distilled water, at 02 thermometer and 30 inches barometer, ai contains 
 277.274 cubic inches. It is equal to 1.2 gal. wine measure U. S. 
 
 BEER MEASURE. 
 26 1 Beer Measure is used in measuring beer, ale, and milk 
 
 2 pints (pis.} make 1 quart', marked qt. 
 
 4 quarts " 1 gallon, " gal. 
 
 36 gallons " I barrel, " bar. or bbl 
 
 1 i barrels, or 54 gallons " 1 hogshead, " lihd. 
 
 OBS. 1. In England, 9 gallons make 1 frrkin; 2 firkins, 1 kilderkin; 2 kil 
 derkins, 1 barrel. 
 
 2. The beer gallon contains 282 cubic inches, and is equal to 10.1799321 
 pounds avoirdupois of distilled water, at the maximum density. In many place* 
 milk is measured by wine measure. 
 
 DRY MEASURE. 
 
 262* Dry Measure is used in measuring grain, fruit, cfec. 
 
 2 pints (pf.) make 1 quart, marked qt. 
 
 8 quarts " 1 peck, " pk. 
 
 4 pecks, or 32 qts, 1 bushel, " bu. 
 
 8 bushels " 1 quarter, " qr. 
 
 32 bushels, or 4 qrs. " 1 chaldron, " cli. 
 
 QUEST. 260. What is the standard unit of Liquid Measure of the Baited Sta/es? 
 How many cubic inches in a wine gallon ? 261. In what is Beer Measure isel 1 Recite 
 the Table. 262. In what is Dry Measure used 1 Repeat the Table. 
 
 * Reports of the Secretary of the Treasury, March, 1831, and June, ' 832. A so, Hassle* 
 on Weights and Measures. 
 
ARTS. 260-263.] COMPOUND NUMBERS. 155 
 
 OBS. In England flour is often sold by weight. A sack is equal to 280 
 lb?., and contains about five imperial bushels. 
 
 The following denominations, are sometimes used, viz : 2 quarts make 1 
 pottle; 2 bushels, 1 strike; 2 strikes or 4 bu., 1 coom; 2 cooms or 8 bu., 1 
 quarter ; 5 quarters, 1 wey or load ; 2 loads, 1 last. 
 
 In London 36 bushels of coal make a chaldron, but in New Castle 79J 
 bushels are said to be allowed for a chaldron. But coal in England aud in 
 this country, is now usually bought and sold by weight. 
 
 Note. Wine, Beer, and Dry Measures are often called capacity measure* 
 and are evidently a species of cubic measure. 
 
 263. The standard Unit of Dry Measure adopted by th 
 United States, is the Winchester Hushel, which is equal t 
 77.027413 pounds avoirdupois of distilled water, at the max 
 imum density, weighed in air at 30 inches barometer, and contain* 
 2150.4 cubic inches, nearly. 
 
 The Winchester bushel is so called, because the standard meas 
 ure was formerly kept at WincJiester, England. By statute, it '4 
 an upright cylinder, 18^ inches in diameter, and 8 inches deep. 
 
 OBS. 1 . The imperial bushel of Great Britain is equal to 80 Ibs. avoirdupois 
 of distilled water, at 62 Fahrenheit, and 30 inches barometer, and contains 
 2218. 192 cubic inches; consequently, it is equal to 1.032 bushel U. S., nearly. 
 It is an upright cylinder, whose internal diameter is 18.789 inches, and its 
 depth 8 inches. 
 
 The use of heaped measure was abolished by Act of Parliament, in 1835. 
 
 2. The standard bushel of the State of New York, is equal to 80 pounds 
 avoirdupois of distilled water, at the maximum density, at the mean pressure 
 of the atmosphere, and contains 2218.192 cubic inches.* 
 
 It is customary, at the present day, to determine capacity measures by the 
 weight of distilled water which they contain. This is evidently nv re accu- 
 rate than the former method of measurement by cubic inches. 
 
 3. In buying and selling grain, when no special agreement as to measure- 
 ment or weight, is made by the parties, a bushel, in the State of New York, by 
 Act of 1836, consists of 60 Ibs. of wheat, 56 Ibs. rye or Indian corn, 48 Ibs. cf 
 barley, and 32 Ibs. of oats. 
 
 There are similar statutes in most of the other States of the Union. This is 
 the most impartial method by which the value of grain can be estimated. 
 
 QUEST. 063. What is the standard unit of Dry Measure adopted by the Government 1 
 
 * By the same Act it was declared, that the standard liquid gallon should be 8 Ibs., and 
 the standard dry gallon 10 Ibs. avoirdupois of distilled water, at its maximum density 
 Butttis part of the statute was subsequently repealed, and the previous standard gallon 
 In the office of the Secretary of State, was continued m use. 
 
156 COMPOUND NUMBERS. [SECT. VIII 
 
 TIME. 
 
 204. Time is naturally divided into days and years ; the for- 
 mer are caused by the revolution of the Earth on its axis, the 
 latter by its revolution round the sun. 
 
 60 seconds 'see.) make 1 minute, marked min. 
 
 60 minutes " 1 hour, " hr. 
 
 24 noun " 1 day, " d. 
 
 7 days " 1 week wk. 
 
 4 weeks " 1 lunar month, " mo. 
 
 12 calendar months, or ) 
 
 365 days and 6 hrs, (nearly,) \ 1 civil y ear > * 
 
 The following are the names of the 12 calendar months into which the civj 
 or legal year is divided, with the number of days in each. 
 
 January, written (Jan.) the first month, has 31 days. 
 
 February, " (Feb.) " second " " 28 " 
 
 March, " (Mar.) " third " " 31 
 
 April, " (Apr.) " fourth " " 30 
 
 May, (May) fifth 31 
 
 June, " (June) " sixth " " 30 
 
 July, " (July) " seventh " 31 " 
 
 August, " (Aug.) eighth " 31 " 
 
 September, " (Sept.) " ninth " " 30 " 
 
 October, " (Oct.) " tenth " " 31 " 
 
 November, "" (Nov.) " eleventh " " 30 " 
 
 December, " (Dec.) " twelfth " " 31 
 
 The number of days m each month may be easily remembered from the fol- 
 lowing lines: 
 
 " Thirty days hath September, 
 April, June, and November ; 
 February twenty-eight alone, 
 All the rest have thirty-one ; 
 Except in Leap year, then is the time, 
 When February has twenty-nine." 
 
 OBS. 1. A Solar year is the exact time in which the earth revolves round 
 the sun, and contains 365'days, 5 hours, 48 minutes, and 48 seconds. 
 
 2. Since the civil year contains 365 days and 6 hours, (nearly,) it is plain 
 iha, in four years a whole da/ will be gained, and therefore every fourth year 
 must have 366 days. This day was originally added to the year, by repeating the 
 sixth of the Calends of March i\ the Roman calendar, which , corresponds with 
 
 QUEST. 264. 'low is time natural 7 divided ? Recite the Table. Obs What is a solai 
 year ? How is leap year occasioned*. To which month is the odd day added] 
 
ARTS. 264-266.] COMPOUND NUMBERS, 157 
 
 the 24th of February in ours. It was called the interculary day, from the 
 Latin intercalo, to insert. 
 
 The year in whch this day is added, is called Bissextile, from the Latin bis, 
 twice, and sextilis, the sixth. It is also called " Leap Year" because it leapa 
 over a day more than a common year. 
 
 3. The civil or kgal year is often called the Julian year, from Julius Caesar 
 emperor of Rome, who adapted the calendar or register of the civil year to 
 the supposed lengtl of the solar year, oy adding 1 day to every fourth year. 
 
 265* In process of time, as mathematical and astronomica 
 science advanced, it was found that the length of a solar yeai 
 was only 365 d. 5 hrs. 48min. 48 sec., or 11 min. 12 sec. less 
 than 365-}- days, which in 400 years amounted to about 3 days ; 
 consequently, the Julian calendar was behind the solar time. 
 This error at the time of Pope Gregory XIII., amounted to 10 
 days, which he corrected in 1582 by suppressing 10 days in the 
 month of October, the day after the 4th being called the 15th. 
 Hence this calendar is sometimes called the Gregorian calendar. 
 
 OBS. 1. This correction was not adopted in England till 1752, when the 
 'error amounted to 11 days. By Act of Parliament, 11 days, after the 2d of 
 September, were therefore omitted ; and the civil year by the same Act, was 
 made to commence on the 1st of January, instead of the 25th of March, as it 
 had done previously. 
 
 2. Dates reckoned by the old method or Julian calendar, are called Old 
 Style ; and those reckoned by the new method, are called New Style. 
 
 To change any date from Old to New Style, we must add 1 1 days to it ; and 
 if the given date in Old Style, is between the 1st of January and the 25th of 
 March, we must add 1 to the year in New Style. 
 
 Russia still reckons dates according to Old Style. The difference now 
 amounts to 12 days. 
 
 !(>(>. To ascertain whether a year is LEAP YEAK. 
 
 Divide the given year by 4, ana if there is no remainder, it is 
 Leap year. The remainder, if any, shows how many years have 
 elapsed since a Leap year occurred. Thus, dividing the year 1847 
 by 4, the remainder is 3 ; hence it is 3 years since the last leap 
 year, and the ensuing year will be leap year. 
 
 OBS. 1. To this rule there is an exception. For, we have seen, that a solzi 
 year is 11 min. and 12 sec. less than a Julian year, which is 365$ days, Tnia 
 rror, in 400 years, amounts to about 3 days ; consequently, if 1 day is added 
 
 QUEST. 266. How do you ascert <in whether a year is leap year? 
 
158 
 
 COMPOUND NUMBERS. 
 
 [SECT. VIII 
 
 every fourth year , that ia, if we have 100 leap years in 400 years, according to 
 the Julian calendar, the reckoning would fall 3 days behind the solar time. 
 Thus reckoning from the commencement of the Christian era, when it was 
 January 1st, 401 by the Julian time, it was January 4th by the solar time. 
 
 2. To remedy this error only 1 centennial year in four is regarded a kap 
 year ; or, which is the same in effect, whenever the centennial year, or the 
 number expressing the century, is not divisible by 4, that year is not a leap 
 year, while the other centennial years are. Thus, 17, 18, 19, denoting 1700, 
 1800, and 1900, are not divisible by 4, consequently they are not, leap yaars, 
 though according to the rule above they would be ; on the other hand 1 6 and 
 20, denoting 1600 and 2000, are divisible by 4, and are therefore leap years. 
 There is still a slight error, but it is so small that in 5000 years it scarcely 
 amounts to a day. 
 
 CIRCULAR MEASURE, OR MOTION. 
 
 267. Circular Measure is applied to the divisions of the cir- 
 cle, and is used in reckoning latitude and longitude, and the 
 motion of the heavenly bodies. 
 
 60 seconds (") make 1 minute, marked 
 
 60 minutes " 1 degree, " o 
 
 30 degrees " 1 sign, 5. 
 
 12 signs, or 360 " 1 circle, " c. 
 
 This measure is often called Angular Measure, and is chiefly usec by 
 astronomers, navigators, and surveyors. 
 
 OBS. 1. The circumference of every cir- 
 cle is divided or supposed to be divided, 
 into 360 equal parts, called degrees, as in 
 the subjoined figure. 
 
 2. Since a degree is -^^ part of the 
 circumference of a circle, it is obvious that 
 its length must depend on the size of the 
 circle. 
 
 270 
 
 Note. The division of the circumference of the circle into 36C equal part*, 
 took its origin from the length of the year, which, (in round numbers) wa* 
 supposed to contain 360 days, or 12 months of 30 days each. The 12 signs 
 
 QUXST 267. In what is Circular Measure used ? Repeat the Table. Obs. How is the 
 circumference of every circle divided ? On what does the length of a degree depend 1 
 
ARTS. 267-269.] COMPOUND NUMBERS. 
 
 159 
 
 corresponl to the 12 months. Thfc \errn minutes, is from the Latin 
 
 which signifies a small part. The term seconds, is an abbreviated expression 
 
 for second minutes, or minutes of the second order. 
 
 268. Since the earth turns on its axis from west to east once 
 in 24 hours, it evidently revolves 15 per hour; or 1 in 4 min- 
 utes, and 1' in 4 seconds of time. Hence, 
 
 When the difference of longitude between two places is V, the 
 difference in the time, or the hour of the day at these two places, is 4 
 teconds ; if the difference of longitude is 1 , the difference of time u 
 4 minutes ; if 2, the difference of time is 8 minutes, &c. 
 
 Thus, when it is noon at London, in Philadelphia, which is about 
 75 west from London, it is only 7 o'clock, A. M. For, if the 
 earth revolves 1 in 4 minutes, to revolve 75, it will require 75 
 times as long, and 4x75 = 300 min., or 5 hours. 
 
 OBS. 1. Since the earth revolves from west to east, it is manifest, that the 
 time is earlier as we go eastward, and later as we go westward. 
 
 2. This principle affords navigators and others a convenient and useful 
 method of ascertaining the difference of time between two places, when the 
 difference of their longitude is known ; also, for ascertaining the difference of 
 longitude between two places, when the difference in their time is known. 
 
 MISCELLANEOUS TABLE. 
 
 269. The following denominations no- included in the pre- 
 ceding Tables, are frequently used. 
 
 12 units ' 
 
 12 dozen, or 144 
 
 12 gross, or 1728 
 
 20 units 
 
 50 pounds 
 100 pounds 
 
 30 gallons 
 
 200 Ibs. of shad or salmon 
 19(! pounds 
 200 pounds 
 
 14 pounds of iron or lead 
 21 i stone 
 
 
 make 
 
 dozen, (doz?) 
 
 great gross. 
 
 score. 
 
 firkin of butter. 
 
 quintal of fish. 
 
 bar. of fish in Mass. 
 
 bar. in N. Y. and Conn. 
 
 bar. of flour. 
 
 bar. of pork. 
 
 stone. 
 
 1 fother. 
 
 Note. Formerly it was customary to allow 112 Ibs. for a quintal. 
 
 QPKST. 268. When the difference of longitude between two r laces is 1', what Is UM 
 iifference of time 7 Wher 1, what is the difference of time ? 
 
160 
 
 COMPOUND NUMBERS 
 
 [SECT. VIII. 
 
 PAPER AND BOOKS. 
 
 27CK The terms folio, quarto, octavo, &3., applied to books, 
 denote the number of leaves into which a sheet of paper is 
 bided. 
 
 24 sheets of paper 
 20 quires 
 
 2 reams 
 
 5 bundles 
 
 A sheet folded in two leaves 
 
 make 
 
 A sheet 
 A sheet 
 A sheet 
 A sheet 
 A sheet 
 
 " four leaves 
 " eight leaves 
 " twelve leaves 
 " eighteen leaves 
 " thirty-six leaves 
 
 1 quire. 
 1 ream. 
 1 bundle. 
 1 bale. 
 
 forms a 
 
 a quarto, or 4to. 
 an octavo, or 8vo. 
 a duodecimo, or 12mo. 
 an 18mo. 
 a 36mo. 
 
 DIMENSIONS OF DIFFERENT KINDS OF ENGLISH PAPER. 
 
 Names. 
 
 Writing. 
 
 Drawing. 
 
 Printing. 
 
 Pott, 
 
 15* by 12* in. 
 
 15* by 12* in. 
 
 
 Small Post, 
 
 16* by 13* in. 
 
 
 
 Fool's Cap, 
 
 16| by 13* in. 
 
 16f by 13* in. 
 
 
 Crown, 
 
 
 20 by 15 in. 
 
 20 by 15 in. 
 
 Demy, 
 
 20 by 15* in. 
 
 22 by 17 in. 
 
 
 Medium, 
 
 22* by 17* in. 
 
 
 23 by 18 in. 
 
 Royal, 
 
 24 by 19* in. 
 
 24 by 19i in. 
 
 26 by 20 in. 
 
 Super Royal, 
 
 27* by 19i in. 
 
 27* by 19i in. 
 
 
 
 Elephant, 
 
 
 28 by 23 in. 
 
 
 Double Crown, 
 
 
 
 30 by 20 in. 
 
 Imperial, 
 
 30} by 22 in. 
 
 30j by 22 in. 
 
 
 Atlas, 
 
 
 34 by 26 in. 
 
 
 Columbier, 
 
 
 34* by 23* in. 
 
 
 Double Demy, 
 
 
 
 38* by 26 in. 
 
 Double Elephant, 
 
 
 40 by 26} in. 
 
 
 Antiquarian, 
 
 
 52 by 31 in. 
 
 
 Double Atlas, 
 
 
 55 by 31* in. 
 
 
 Emperor, 
 
 * 
 
 68 by 48 in. 
 
 
 Note. American paper is usually rather larger than English paper at tie 
 game name. 
 
 QUEST. 270. What do the terms, folio, quarto, &c., denote, when applied to books I 
 What is a folio 1 A quarto ? An octavo 1 A duodecimo ? An 18ro. 7 A 36mo. 5 
 
ARTS. 270-273.] COMPOUND NUMBERS. 1 1 
 
 FRENCH MONEY, WEIGHTS, AND MEASURES. 
 
 271. The new system of Money, Weights, and Measures of 
 France, adopted in 1795, was formed according to the decimal 
 Notation. 
 
 FRENCH MONET. 
 
 272. The Franc is the unit money of the new system of 
 French currency. It is a silver coin, consisting of -fa pure sil- 
 ver, and iV of alloy. 
 
 10 centimes make 1 declme. 
 
 4 10 declines " 1 franc. 
 
 2V0fe. The value of a franc by Act of Congress in 1843, is $.186. The 
 value of the hvre lournois, the former unit of money, is $.185. 
 
 FRENCH LINEAR MEASURE. 
 
 273. The standard unit of the French Linear Measure, is 
 the Metre. Its length, according to the mean of the several com- 
 parisons of Troughton, Nicollet and Hassler, is equal to 39. 3809171 
 English, or United States inches. 
 
 10 metres make 1 decametre = 32.817431 U. S. feet. 
 
 10 decametres " 1 hectometre = 328.17431 " 
 
 10 hectometres " 1 kilometre = 3281.7431 " " 
 
 10 kilometres " 1 myriametre = 32817.431 " " 
 
 Note. 1. The standard by which the new French measures of length are 
 determined, is the quadrant of a meridian of the earth, or the terrestrial arc 
 from the equator to the pole, in the meridian of Paris. The ten-millionth\ part 
 of this arc is called a metre, which is equal to 39.378 U. S. in., nearly. 
 
 2. The metre is divided into 10 decimetres ; the decimetre into 10 centimetres ; 
 the centimetre into 10 millimetres. 
 
 3. The denominations of the old system of linear measure were the toise, 
 foot, inch, line, and point. 12 points=l line; 12 lines=:l inch; 12 in.=l 
 foot; 6 ft.=l toise. The old French foot was equal to 1.066 U. S. feet. 
 
 *. By a decree of 1812, the Toise, Aune, Foot, &c., are allowed to be used, 
 having the following ratios to the metre, viz: the toise =2 metres ; the foot= 
 J- metre ; the inch=- l - L 2 - metre ; the aune or ell=:l metre ; the bushel hec- 
 tolitre. 
 
162 COMPOUND CUMBERS. [SECT. VI ll. 
 
 FRENCH SQUARE MEASURE. 
 
 274. The unit of French Superficial Measure, is the Are, 
 Miose sides are each a decametre in length ; consequently, it con- 
 *-iins 100 square metres, or 119.6648496 U. S. sq. yds. 
 
 10 ares make 1 decare = 1196.64849G U. S. sq. yds. 
 
 tOdecares " 1 hectare =11966.48496 " " 
 
 10 hectares " 1 kilare = 119664.8496 " " 
 
 lOkilares " 1 myriare =1196648.496 " " 
 
 Note. The are is divided into 10 declares; the declare into 10 centiares; 
 the centiare into 10 milliares. 
 
 
 FRENCH CUBIC MEASURE. 
 
 275. The unit of French Cubic Measure, is the Stere, which 
 is a cubic metre, and is equal to 61074.1564445 cu. in. U. S. 
 
 10 decisteres make 1 stere = 35.34384 cu. ft. U. S. 
 10 steres " 1 decastere = 353.4384 " " 
 
 FRENCH LIQUID AND DRY MEASURE. 
 
 276. The unit of French Liquid and Dry Measures, is 
 called the Litre, which is a cubic decimetre, and is equal to 
 61.0741564445 cu. in. U. S., or 1.05756 qts. wine measure. 
 
 10 litres make 1 decalitre = 2.6439 gals, wine meas. 
 10 decalitres " 1 hectolitre = 26.439 " " " 
 10 hectolitres " 1 kilolitre = 264.39 " " " 
 
 Note. The litre is divided into 10 decilitres ; the decilitre into 10 centilitres; 
 the centilitre into 10 millilitres. 
 
 FRENCH WEIGHTS. 
 
 277. The unit of French Weights, is the weight of a cubic 
 centimetre of distilled water, at the maximum density, and is 
 called the Gramme. It is equal to 15.433159 grains Troy. 
 
 10 grammes make 1 decagramme = 154.33159 grs. Troy. 
 10 decagrammes " 1 hectogramme = 1543.3159 " " 
 10 hectogrammes " 1 kilogramme = 15433.159 " " 
 10 kilogrammes, " 1 myriagramme = 154331.59 " " 
 
Avrs. 275-279.] FOREIGN WEIGHTS, ETC. 163 
 
 . I. The gramme is divided into 10 decigrammes, the dctigramrfie 
 into 10 centigrammes ; the centigramme into 10 milligrammes. 
 
 ""2. The denomination chiefly used in making out invoices of goods sold by 
 weight, and in business transactions, is the Idhgramme, which is equal to 1000 
 grammes, or 2.21 Ibs. avordupois, very nearly. 
 
 3. In the old system of French weight, the livre-poids 2 marcs; the marc 
 =8 onces ; the once 8 gros ; the gros=72 grains. The livre is equal to one- 
 half the kilogramme. 
 
 FRENCH CIRCULAR MEASURE. 
 
 278. The c/rc/e is divided into 400 equal parts, called grades, 
 and the quadrant into 100 grades. The grade is again divided 
 into 100 equal parts, and each of these parts is subdivided into 
 100 other equal parts, according to the centesimal scale. Hence, 
 
 The seconde = .00009 English deg. 
 The minute = .009 " 
 
 The grade = .9 " " 
 
 Note. The names of the denominations larger than the unit in the French 
 Compound Numbers, are formed by prefixing to the name of the unit, the 
 Greek words, dcca, heclo, kilo, and myria; those less than the unit, are formed 
 by prefixing to the name of the unit, the Latin words, aeci, centi, and milLL 
 
 279. Foreign Weights and Measures compared with those of 
 the United States* 
 
 Amsterdam. 100 Ibs. (1 centner)=108.923 Ibs.f; 1 last=85.25 bu.; 1 ahm= 
 41 gals.; 1 foot Amsterdam 11-f in.; 1 foot Antwerp =11-}- in. ; 1 ell Am- 
 sterdam=2.26 ft.; 1 ell Brabant=2.3ft. ; 1 ell Hague=2.28 ft. 
 
 Daiai'ia. 1 picul=l36 Ibs.; 1 kann=.39 gal.; 1 ell=2.25 ft. 
 
 Bengal. I haut=l.5 ft.; 1 guz=3 ft.; 1 coss or mile =1.24 miles; 1 bazar 
 maud=82.14 Ibs. ; 1 factory maud=74.G(5 Ibs. 
 
 Bencoolen. 1 bahar=560 Ibs.; 1 bamboo =1 gal.; 1 coyangr=8 gals. 
 
 Bombay. 1 maud=28 Ibs. ; 1 covid=1.5 ft. ; 1 candy=25 bu. 
 
 Bremen. 1 pound =1.1 lb.; 1 centner^ 116 Ibs. ; 1 last=80.7 bu. ; 1 ft.=ll|in 
 
 Canton. 1 tael = l} oz. ; 1 catty =U Ibs. ; 1 picul=l33$ Ibs. ; 1 covid 14| in. 
 
 Denmark. 100 Ibs. (1 centner) = 110.25 Ibs.; 1 bbl. (toende)=3.95 bu.; 
 1 \iertel=2.04 gals. ; 1 foot Copenhagen, or Rhineland=12} in. 
 
 Florence and Leghorn. 100 Ibs. (1 cantaro)=74.8(j Ibs. ; 1 moggio= 16.59 bu.; 
 1 barile=12.04 gals.; 1 palmo=9f in. 
 
 * M'Culloch's Commercial Dictionary; also Kelly's Universal Cambist. 
 t The poutvls in this and the following comparisons are avoirdupois. 
 T.H. 8 
 
164 FOREIGN WEIGHTS, ETC. [SECT. VI11. 
 
 Genoa. 100 \')s. (1 peso grosso)=76f Ibs. ; t peso sottile=69.89 Ibs. ; 1 oina 
 
 =3.43 bu. ; 1 mezzarola=39.22 gals. ; 1 palmo=9-- in. 
 Hamburg*! foot=l 1.3 hi. ; 1 ell =22.6 in. nearly; 1 ell Brabant=27.6 in. ; 
 
 1 mile=4.68 miles ; lfass=Hbu. ; 1 last=89.64 bu. ; 1 ahm=38i gals. 
 Japan. I catti=1.3 Ibs.; 1 picul=l30 Ibs. ; 1 ichan=3 ft.; 1 inc or tetamy 
 
 =6^ ft.; 1 balec=16i gals. 
 Madras. 1 covid=l ft.; 1 catty =1J Ibs.; 1 picul=l33| Ibs.; 1 maud= 
 
 25 Ibs.; 1 candy =500 Ibs.; 1 garee=140bu. 
 
 Malta. 1 foot=10-- in.; 100 Ibs. (1 cantaro)= 174.5 Ibs. ; 1 salma 8.22 bu, 
 Manilla. 1 arroba=26 Ibs. ; 1 picul=143 Ibs. ; 1 palmo= 10.38 in. 
 Naples. 1 cantaro grosso=196.5 Ibs.; 1 cantaro piccolo=l06 Ibs.; 1 palmo=a 
 
 10$ in ; 1 tomolo=1.45 bu. ; 1 carro=52.24bu. ; 1 carro of wine =264 gals. 
 Netherlands. 1 ell =3.28 ft.; 1 mudde=2.84 bu.; 1 kan litre =2. 11 pints; 
 
 1 vat hectolitre=26.42 gals. ; 1 pond kilogramme 2.21 Ibs. 
 Portugal. 100 lbs. = 101.19 Ibs.; 1 arroba=22.:26 Ibs. ; 1 quintal =89.05 Ibs. ; 
 
 1 almude=4.37gals.; 1 alquiere=4f bu. ; 1 moyo=23.03 bu.; 1 last=70 bu. ; 
 
 1 peor foot=l2-k in.; 1 mile=l? mile. 
 Prussia. 100 lbs=103.11 Ibs. ; 1 quintal (110 lbs.)=H3.42 Ibs. ; 1 eimar= 
 
 18.14 gal.; 1 scheffel=1.56 bu. ; 1 foot=1.03 ft. ; I ell=2.l9ft.; 1 mile=: 
 
 4.68 miles. 
 Rome.lQQ libras=74.77 Ibs. ; 1 rubbio=9.3G bu. ; 1 barile=15.3l gals. ; 1 foot 
 
 =llfin.; 1 canna=(> ft- ; I mile=7| fur. 
 Russia 100 lbs.=90.26 Ibs.; 1 berquit=361.04 Ibs.; 40 Ibs. (1 pood)= 
 
 36 Ibs. ; 1 vedro=3i gals. ; 1 chetwert=5.95 bu. ; 1 foot Petersburg 1.18 ft.t; 
 
 1 foot Moscow=l.l ft; 1 arsheen=2 ft.; 1 mile (verst)=5.3 fur. 
 Sicily. 100 Ibs. (libras)=70 Ibs.; 1 cantaro grosso=192.5 Ibs.; 1 cantarf 
 
 sottile=l75 Ibs.; 1 salma generale=7.85 bu. ; 1 salma grossa=9.77 bu. ; 
 
 1 salma of wine=23.06 gals. ; 1 palmo=9 in. ; 1 canna=6 ft. 
 Spain. 1 arroba=25.36 Ibs.; 1 quintal=lOl.44 Ibs.; 1 arroba of wine = 
 
 4| gals. ; 1 moyo=G8 gals. ; 1 fanega =1.6bu. ; 1 foot=ll.l28 in.; 1 vara= 
 
 2.78 ft. ; 1 league (leagua)=4.3 m., nearly. 
 Sweden. 100 Ibs, ( -ictualie) =73.76 Ibs. ; "l foot=ll.G9 in. ; 1 ell=1.95 ft.; 
 
 1 mile=6.64 m. ; 1 kann=7.42 bu. ; 1 last=75 bu. ; 1 kann of wine= 
 
 69.09 gals. 
 Smyrna 100 Ibs. (1 quintal) =129.48 Ibs.; 1 oke=2.83 Ibs.; quillot^ 
 
 1.46 bu. ; 1 quillot of wine=l3.5 gals. ; 1 pic=2j- ft. 
 Trieste. 100 Ibs.=123.6 Ibs ; 1 stajo=2i bu.; 1 orna, or eimer=1494 gals.; 
 
 1 ell (for silk)=:2.1 ft. ; 1 ell (for woollen)=2.2 ft. ; 1 foot Austrian= 1.037 ft. ; 
 
 1 mile Austrian=4.6 m. 
 1 enice. 100 Ibs. (Ipesso grosso)=105.l8 Ibs. : 1 peso sottile=64.42 Ibs, ; 1 staja 
 
 =2.27 bu.; 1 moggio=9.08bu.; 1 anifora= 137 gals.; 1 foot=1.14ft.; 1 brae- 
 
 cio (for silk) =24.8 in. ; 1 braccio (for woollen) =26.6 in. 
 
 * New system of weights and measures adopted in 1843. 
 
 t In measuring timber English feet and inches are chiefly used throughout Russia. 
 
ART. 280.] REDUCTION. 165 
 
 REDUCTION. 
 
 28O The process of changing compound numbers from one 
 denomination into another, without altering their value, is called 
 REDUCTION. 
 
 Ex. 1. Reduce 5, 2s. Yd. and 3 far. to farthings. 
 
 Analysis. Since in l there are 20s., in 5 there are 5 times 
 as many, which is 100s., and 2, (the given shillings,) make 102s. 
 Again, since there are 12d. in Is., in 102s. there are 102 times 
 as many, which is equal to 1224d., and 7 (the given pence) make 
 123 Id. Finally, since in Id. there are 4 far., in 123 Id. there are 
 1231 times as many, or 4924 far., and 3, (the given far.,) make 
 4927 far. Ans. 4927 farthings. 
 
 Operation. 
 
 s. d. far. We first reduce the given pounds to shil- 
 
 5273. lings, by multiplying them by 20, because 
 
 20s. in 1. 20s. make l. (Art. 247.) We next re- 
 
 102 shillings. duce the shillings to pence, by multiply- 
 
 12d. in Is. ing them by 12, because 12d. make Is. Fi- 
 
 1231 pence. nally> we reduce t C pome to farthings by 
 
 4 far in Id. multiplying them by 4, because 4 far. 
 
 4927 far. Ans. make Id. 
 
 Note. 1. In this example it is required to reduce higher denominations to 
 lower ; as pounds to shillings, shillings to pence, &c. This is done by suc- 
 cessive multiplications. 
 
 2. In 4927 farthings, how many pounds, shillings, and pence? 
 
 Analysis. Since 4 far. make Id., in 4927 farthings, there are 
 as many pence as 4 is contained times in 4927, which is 123 Id., 
 and far. over. Again, since 12d. make Is., in 1231d. there are 
 as many shillings as 12 is contained times in 1231, which is 
 102s., and 7d. over. Finally, since 20s. make l, in 102s. there 
 
 QUEST. 280. What is Reduction 7 How are pounds reduced to shillings ? Why mul 
 tiply by 20 ? How are shillings reduced to pence ? Why 1 How pence to farthings ? Whv 
 
166 REDUCTION. [SlJCT. VIIL 
 
 are as many pounds as 20 is contained times in 102, which .'is 5, 
 and 2s. over. Am. 5, 2s. Yd. 3 far. 
 
 Operation. We first reduce the given farthings 
 
 4)4927 far. to pence, the next higher denomina- 
 
 12) 123 Id. 3 far. over. tion, by dividing them by 4, becaus 
 
 20) 102s. 7d. over. 4 far. make Id. (Art. 247.) Next we 
 
 5, 2s. over. reduce the pence to shillings by di- 
 
 Ans. 5, 2s. 7d. 3. far. viding them by 12, because 12d. 
 
 flLake 1 s. Finally, we reduce the shillings to pounds by dividing 
 
 them by 20, because 20s. make l. The last quotient with the 
 
 several remainders, constitute the answer. 
 
 Note. 2. The last example is exactly the reverse of the first ; that is, lower 
 denominations are reduced to higher, which is done by successive divisions. 
 
 281* From the preceding illustrations we derive the fol- 
 lowing 
 
 GENERAL RULE FOR REDUCTION. 
 
 I. To reduce compound numbers to lower denominations. 
 Multiply the highest denomination given, by that number which 
 
 it takes of the next lower denomination to make ONE of this higher ; 
 to the product, add the number expressed in this lower denomina- 
 tion in the given example. Proceed in this manner with each 
 successive denomination, till you come to the one required. 
 
 II. To reduce compound numbers to higher denominations. 
 Divide the given denomination by that n imber which it takes of 
 
 this denomination to make ONE of tlie next higher. Proceed in this 
 manner with each successive denomination, till you come to the one 
 tequired. The last quotient, with tJie (everal remainders, will be 
 the answer sought. 
 
 282 PROOF. Reverse the operation / that is, reduce back the 
 answer to the original denominations, and if the result corresponds 
 with the numbers given, the work is right. 
 
 QUEST. How are farthings reduced to pence ? Why divide by 4 ? How reduce pence 
 to shillings? Why? How reduce shillings to pounds ? Why? 281. How are compound 
 numbers reduced to lower denominations? How to higher denominations? 282 How 
 is Reduction proved ? 
 
ARTS. 281, 282. 1 REDUCTION. 167 
 
 OBS. 1. Each remainder is of the same denomination as the dividend from 
 which it arose. (Art. 113. Obs. 1.) 
 
 2. Reducing compound numbers to lower denominations may, with propri- 
 ety, be called Reduction by Multiplication ; reducing them to higher denom- 
 inations, Reduction by Division. The former is often called Reduction De* 
 vending ; the latter, Reduction Ascending. They mutually prove each other. 
 
 EXAMPLES FOR PRACTICE. 
 
 I. In 136 rods and 2 yards, how many feet? 
 
 Operation. Proof, 
 
 rods. yds. 3)2250 ft. 
 
 2)136 2 5i)750 yds. 
 
 __H yds. 1 r. _^ 
 
 682 11)1500 
 
 68 . T36f r. 4 rem.=2 yards. 
 
 750 yds. Now 136 r. 2 yds. is the 
 
 3 ft. 1 yd. given number. 
 2250 ft. Ans. 
 
 2. In 71, 13s. 6d., how many farthings? 
 
 3. In 90, Vs. 8d., how many farthings ? 
 
 4. In 295, 18s. 3|d., how many farthings? 
 
 5. In 95 guineas, 17s. 9fd., how many farthings? 
 
 6. How many pounds, shillings, &c., in 24651 farthings? 
 
 7. How many pounds, shillings, &c., in 415739 farthings? 
 
 8. How many guineas, &c., in 67256 pence? 
 
 9. In 36, 4s., how many six-pences ? 
 
 10. In 75, 12s. 6d,, how many three-pences ? 
 
 II. Reduce 29 Ibs. 7 oz. 3 pwts. to grains. 
 
 12. Reduce 37 Ibs. 6 oz. to pennyweights. 
 
 13. Reduce 175 Ibs. 4 oz. 5 pwts. 7 grs. to grains. 
 
 14. Reduce 12256 grs. to pounds, &c. 
 
 15. Reduce 42672 pwts. to pounds, <fec. 
 
 16. In 15 cwt. 3 qrs. 21 Ibs., how many pounds? 
 
 17. In 17 tons 12 cwt. 2 qrs., how many ounces? 
 
 QUEST. Obs. Of what denomination is each remainder 7 What may reducing compound 
 numbers to lower denominations be called? To higher denominational Which of ih 
 fundamental rules la employed by the formef ? Which by the latter 1 
 
168 REDUCTION. [SECT. 
 
 18. In 52 tons 3 cwt., how many pounds ? 
 
 19. In 140 tons, how many drams ? 
 
 20. In 16256 ounces, how many hundred weight, <fec. ? 
 
 21. In 267235 pounds, how many tons, <fec.? 
 
 22. In 5637 2 8 drams, how many tons, pounds, <fec. ? 
 
 23. Reduce 95 pounds (apothecaries' weight) to drams. 
 
 24. Reduce 130 pounds to scruples. 
 
 25. Reduce 6237 drams (apothecaries' weight) to pounds, <feu 
 
 26. Reduce 25463 scruples to ounces, &c. 
 
 27. How many feet in 27 miles ? 
 
 28. How many inches in 45 leagues ? 
 
 29. How many yards in 3000 miles ? 
 30 In 290375 feet, how many miles? 
 
 31. In 1875343 inches, how many leagues? 
 
 32. In 15 m. 5 fur. 31 r., how many rods ? 
 
 33. In 1081080 inches, how many miles, &c. ? 
 
 34. How many feet in the circumference of the earth ? 
 
 35. How many nails in 160 yards? 
 
 36. How many quarters in 1000 English ells? 
 
 37. In 102345 nails, how many yards, &c. ? 
 
 38. In 223267 nails, how many French ells? 
 
 39. In 634 yards, 3 qrs., how many nails ? 
 
 40. In 28 hhds. 15 gals, wine measure, how many quarts? 
 
 41. In 5 pipes, 1 hhd., how many gallons? 
 
 42. In 3 tuns, 1 hhd. 10 gals., how many gills? 
 
 43. In 12256 pints, how many barrels, wine measure? 
 
 44. In 475262 gills, how many pipes, &c. ? 
 
 45. In 50 hhds. 1 bbl. 10 gals., how many gills, wine measure . 
 
 46. In 45 bbls., how many pints, beer measure ? 
 
 47. How many barrels of beer in 25264 pints ? 
 
 48. How many hogsheads of beer in 136256 quarts? 
 
 49. How many pints in 45 hhds. 10 gals, of beer ? 
 60. In 15 bushels, 1 peck, how many quarts? 
 
 51. In 763 bushels, 3 pecks, how many quarts? 
 
 52. In 56 quarts, 5 bushels, how many pints ? 
 
 53. In 45672 quarts, how many bushels, &e. ? 
 64. In 260200 pints, how many quarts ? 
 
ARTS. 282, 283. J REDUCTION. 169 
 
 55. Reduce 25 days, 6 hours to minutes. 
 
 56. Reduce 365 days, 6 hours to seconds. 
 
 57. Reduce 847125 minutes to weeks. 
 58 Reduce 5623480 seconds to days. 
 59. How many seconds in a solar year? 
 
 GO. How many seconds in 30 years, allowing 365 days 6 houri 
 to a year? 
 
 61. How many years of Sabbaths are there in 70 years? 
 
 62. In 110 degrees, 20 minutes, how many seconds? . 
 
 63. In 11 signs, 45 degrees, how many seconds? 
 
 64. In 7654314 seconds, how many degrees? 
 
 65. In 1000000000 minutes, how many signs? 
 
 66. Reduce 1728 sq. rods, 23 yds. 5 feet to feet. 
 
 67. Reduce 100 acres, 37 rods to square feet. 
 
 68. Reduce 832590 sq. rods to sq. inches. 
 
 69. Reduce 25363896 sq. feet to acres, &c. 
 
 70. In 150 cubic feet, how many inches? 
 
 71. In 97 yds. 15 ft., how many cubic inches? 
 
 72. In 49 cords, 23 feet, how many cubic inches? 
 
 73. In 84673 cubic inches, how many feet ? 
 
 74. In 39216 cubic feet, how many cords? 
 
 75. In 65 tons of round timber, how many cubic inches? 
 
 76. In 4562100 cubic inches, how many tons of hewn timber? 
 
 APPLICATIONS OP REDUCTION. 
 
 283. To reduce Troy to Avoirdupois weight. 
 
 First reduce the yiven pounds, ounces, d'c., to grains ; then divide 
 oy the number of grains in a dram, and t/te quotient will be t/ie an- 
 swer in drams. (Art. 252.) 
 
 OBS. If the answer is required to be in pounds and a fraction of a j iu^4 
 divide the grains !)y 7000. 
 
 Ex. 1. In 175 pounds Troy, how many pounds avoirdupois? 
 Solution. 175X12X20X24 = 1008000 grs., and 1008000 
 grs.-~27ii=36864 drams, or 144 Ibs. avoirdupois. Ans. 
 
 " - > 
 
 How is Troy weight reduced to avoirdupois 1 
 
170 REDUCTION. [SECT. VIIL 
 
 2. In 700 Ibs. Troy of silver, how many pounds avoirdupois ? 
 
 3. In 840 Ibs. 6 oz. 10 pwts., how many pounds, &c., avoirdu- 
 pois ? 
 
 4. An apothecary bought 1000 Ibs. of opium by Troy weight, 
 and sold it by avoirdupois : how many pounds did he lose ? 
 
 6. A merchant bought 1500 pounds of lead Troy weight, and 
 sold it by avoirdupois : how many pounds did he lose ? 
 
 284* To reduce Avoirdupois to Troy weight. 
 
 First reduce the given pounds, ounces, &c. y to drams, then multiply 
 ty the number of grains in a dram, and the product will be the an- 
 swer in grains. (Art. 252.) 
 
 OBS. 1. When the given example contains pounds only, we may multiply 
 them by 7000, and the product will be grains. 
 
 2. If the answer is required to be in pounds and a fraction of a pound, di- 
 vide the grains by 57GO. 
 
 6. In 32 Ibs. avoirdupois, how many pounds Troy ? 
 Solution. 32X16Xl6X27ii=224000 g 1 " 8 -* and 224000 grs. 
 
 =38 Ibs. 10 oz. 13 pwts. 8 grs. Ans. 
 
 7. In 48 Ibs. avoirdupois, how many pounds Troy ? 
 
 8. A merchant bought 100 Ibs. 10 oz. of tea avoir., and sold it 
 by Troy weight : how many pounds did he gain ? 
 
 9. A druggist bought 1260 Ibs. of alum avoirdupois, and re- 
 tailed it by Troy weight : how many more pounds did he sell 
 than he bought ? 
 
 285. The area of a floor, a piece of land, or any surface which 
 has four sides and four right-angles, is found by multiplying its 
 length and breadth together 
 
 Note 1. The area of a figure is the superficial contents or space contained 
 within the line or lines, by which the figure is bounded. It is reckoned in 
 square inches, feet, yards, rods, &c. 
 
 2. A figure which has four sides and four right-angles, like the following 
 diagram, is called a Rectangle or Parallelogram. 
 
 . 284. How is avoirdupois weight reduced to Troy? 285. How do you find 
 the area or superficial contents of a surface having four side* and four right-angle* 1 
 JVbte. What is meant by the term area 7 How is it reckoned 1 What is a figure wnich 
 lias four sides and four right-angles called ? 
 

 ARTS. 284-286.] REDUCTION. 171 
 
 10. How many square yards of carpeting will it take to cover 
 a room, 4 yards long and 3 yards wide ? 
 
 Suggestion. Let the given room be 
 represented by the subjoined figure, the 
 length of which is divided into 4 equal 
 pans, and the breadth into 3 equal 
 parts, which we will call linear yards. 
 'Now it is plain that the room will con- 
 tain as many square yards as there are 
 squares in the given figure. But the 
 numbe of squares in the figure is equal to the number of equal 
 parts (linear yards) which its length contains, repeated as many 
 times as there are equal parts (linear yards) in its breadth ; that 
 is, it is equal to 4X3, or 12. Ans. 12 yds. 
 
 11 How many sq. feet in a floor, 20 feet long, 18 feet wide ? 
 
 12. How many acres in a field, 50 rods long, 45 rods wide ? 
 
 13. How many square yards in a ceiling, 35 feet long and 28 
 feet wide ? 
 
 14. How many acres in a farm, 420 rods long and 170 rods 
 wide ? 
 
 15. What is the area of a square field, whose sides are 80 rods 
 in length ? 
 
 16. How many yards of carpeting, a yard wide, will it take to 
 cover a floor 18 feet square. 
 
 17. How many yards of plastering are required to cover four 
 sides of a room, 18 ft. long, 15 feet wide, and 9 ft. highj? 
 
 18. How many square yards of shingling will cover both sides 
 of a roof, whose rafters are 20 feet, and whose ridge pole is 25 
 feet long ? 
 
 286. The epical contents, or solidity of boxes of goods, 
 piles of wood, &c., are found by multiplying the length, breadth, 
 and thickness together. 
 
 19. How many cubic feet in a box 5 feet long, ^ feet wide and 
 3 feet deep ? 
 
 Solution. 5X4=20, and 20X3=60. Ans. 60 cu. ft. 
 
 Q0S8T. 286. How are the cubical contents of a box of goods, a pile of wood %e. 
 found 1 
 
 8* 
 
172 REDUCTION [SECT. VIII 
 
 20. How many cubic feet in a Mock of granite, 65 in. long, 42 
 in. wide, and 36 in. thick? 
 
 21. How many cubic feet in a load of wood, 8 ft. long, 4^ ft. 
 high, and 3 ft. wide ? 
 
 22. How many cords of wood in a pile, 46 ft. long, 16 ft. high, 
 and 14if| feet wide? 
 
 23. How many cubic feet in a vat, 12 ft. long, 8 ft. wide, and 
 ?i ft. deep ? 
 
 24. How many cubic feet in a bin, 12 ft. long, 9 ft. deep, and 7 
 ft. wide ? 
 
 25. How many cubic yards in a cellar, 18 ft. long, 12 ft. wide, 
 and 9 ft. deep ? 
 
 26. How many cubic feet in a stick of timber, 2 ft. square, and 
 40 ft. long ? 
 
 27. How many cubic feet in a cistern 15 ft. long, 12 ft. wide, 
 and 10 ft. deep ? 
 
 287. To reduce Cubic to Dry, or Liquid Measure. 
 
 First reduce the given yards, feet, <kc., to cubic inches ; then 
 divide by tJie number of cubic indies in a gallon, or bushel, as 
 the case may be, and the quotient will be the answer required 
 (Arts. 260, 263.) 
 
 28. In 10752 cubic feet, how many bushels? 
 
 Solution. 10752 X 1728 = 18579456 cubic inches; and 
 185794564-2150-^=8640 bushels. 
 
 29. In 21504 cubic feet, how many bushels ? 
 
 30. In 462 cubic feet, how many wine gallons ? 
 
 31. In 1155 cubic feet and 33 inches, how many ^.vine gallons ? 
 
 32. In 846 cubic feet, how many beer gallons ? 
 
 33. In 1128 cubic feet and 141 in., how many beer gallons ? 
 
 34. How many bushels will a bin contain, which is 5 ft. long, 
 6 ft. wide, and 4 ft. deep ? 
 
 35. How many bushels will a bin contain, which is 8 ft. long, 
 4f ft. wide, and 3i ft. deep ? 
 
 QUEST 287. How reduce cubic to dry, or liquid measure ? 
 
ARTS. 287-289. J REDUCTION. 173 
 
 36. How many bushels will a bin contain, which is 14 ft. long, 
 10 ft. 8 in. wide, and 6 ft. 8 in. deep ? 
 
 37. How many wine gallons in a cistern, which is 6 ft. long, 
 5 ft. wide, and 4 feet deep ? 
 
 38. How many barrels of water (wine meas.) will a cistern 
 hold, which is 20 ft. long, 15 ft. wide, and 10 ft. deep ? 
 
 39. The distributing reservoir of the Croton Water Works in 
 the City of New York, is 436 ft. square and 40 feet high: how 
 many hogsheads of water will it hold ? 
 
 288. To reduce Dry, or Liquid, to Cubic Measure. 
 
 First find the number of bushels, if dry measure, or gallons, if 
 liquid measure, in the given example j then multiply by the num- 
 ber of cubic inches in a gallon, or bushel, as the case may be, and 
 the product will be the answer required. (Art. 263.) 
 
 40. How many cubic feet in a bin, which contains 100 bushels ? 
 
 Solution. 100x2150- 1 4 ir = 215040, and 215040 -J- 1728= 
 , or 124 cubic feet. Ans. 
 
 41. How many cubic feet in a lime kiln, which holds 100 
 bushels ? 
 
 42. How many cubic feet in the hold of a ship, which contains 
 1000 bushels of grain ? 
 
 43. How many cubic feet in 1 hogshead, wine measure? 
 
 44. How many cubic feet in a cistern, which holds 50 barrels 
 of water? 
 
 45. How many cubic feet in a vat, which contains 100 hogs- 
 heads wine measure ? 
 
 289. To reduce Liquid to Dry Measure, or Dry to Liquid 
 Measure. 
 
 First find the cubic inches in tlw given example ; then divid 
 them by the number of cubic inches in a gallon, or bus\el t as t/te case 
 may be, and the quotient will be the answer required. 
 
 QUIBT. 288. II >w reduce dry, or liquid measure to cubic? 289 How reduce liquid 
 to dry measure ? How dry to liquid measure 1 
 
174 REDUCTION. [SECT. YII1 
 
 46. In 40 gallons wine measure, how many bushels ? 
 Solution. 40X 231=9240 cu. in., and 9240 cu. in. H-2150-r*ff= 
 
 4-JH bushels. Ans. 
 
 47. In 6 hogsheads, 16 gallons, how many bushels ? 
 
 48. In 5 bushels, how many gallons wine measure ? 
 
 49. In 3200 quarts dry measure, how many hogsheads wine 
 measure ? 
 
 29O. To reduce Wine ti Beer Measure, or Beer to Wine 
 Measure. 
 
 First find the number of cubic inches in the the given example ; 
 then divide them by tlie number of cubic inches which it takes to 
 make a gallon in tJie required measure. 
 
 50. In 94 wine gallons, how many beer gallons ? 
 
 Solution. 94X231 = 21714 cu. in., and 21714 cu. in. 282 = 
 77 gallons. Ans. 
 
 51. In 1 hhd. wine measure, how many beer gallons? 
 
 52. A tavern-keeper bought 4 hhds. of cider wine measure, and 
 retailed it by beer measure : how many gallons did he lose ? 
 
 53. In 20 beer gallons, how many wine gallons? 
 
 54. A grocer bought 7238 gallons of milk beer measure, and 
 retailed it by wine measure : how many gallons did he gain ? 
 
 55. A druggist bought 10000 gallons of alcohol beer measure, 
 and sold it by wine measure : how many gallons did he gain ? 
 
 56. A grocer bought 65 hhds. 29 gals, and 2 quarts of milk by 
 Deer measure, and sold it to his customers by wine measure : how 
 many quarts more did he sell than he bought ? 
 
 57. A liquor dealer bought 120 pipes of wine which his clerk 
 retailed by beer measure : how many gallons more did he buy 
 than he sold ? 
 
 291 Since the earth revolves on its axis 1 in four minutes, 
 or I' in 4 seconds of time, (Art. 268,) it is evident that longitude 
 nay be reduced to time. That is, multiplying degrees of longi- 
 tude by 4 reduces them to minutes of time, multiplying minutes 
 of longitude by 4 reduces them to seconds of time, &c. 
 
 QDKST. 290. How reduce wine to beer measure 7 How oeer to wine measure 7 
 
ARTS. 290-293. J REDUCTION. 115 
 
 By reversing this process it is evident that time may be reduced 
 co longitude. Thus, dividing seconds of time by 4, will reduce 
 them to minutes of longitude ; dividing minutes of time by 4, will 
 reduce them to degrees, &c. Hence, 
 
 . To find the difference of time between two places from 
 the difference of their longitude. 
 
 Reduce the difference of longitude to minutes ; multiply tJiem by 4, 
 and the product will be the difference of time in seconds, which 
 may be reduced to hours and minutes. 
 
 OBS. When the difference oflongitude consists of degrees only, we may mul- 
 tiply them by 4, and the product will be the answer in minutes. 
 
 58. The difference of longitude between New York and Cin- 
 cinnati is 10 26' : what is the difference in their time? 
 
 Solution. 10 and 26' = G26'; (Art. 281;) now 626'X4= 
 ^504 seconds of time ; and 2504 sec. -r-60 = 41 min. 44 sec. Ans. 
 
 59. The difference of longitude between Albany and Boston is 
 2 9' : what is the difference in their time ? 
 
 60. The difference of longitude between Albany and Detroit is 
 9 45' : what is the difference in their time ? 
 
 61. The difference of longitude between New Haven and New 
 Orleans is 17*10' : what is the difference in their time? 
 
 62. The difference of longitude between Charleston, S. C. and 
 Mobile is 8 27' : what is the difference in their time? 
 
 63. The difference of longitude between New York and Canton 
 is 187 3' : what is the difference in their time ? 
 
 293* To find the difference of longitude between two places 
 from the difference in their time. 
 
 Reduce the given difference of time to seconds ; divide them by 4, 
 and the quotient will be the difference of longitude in minutes, which 
 may be reduced to degrees. (Art. 281.) 
 
 OBS. When there are no seconds in the difference of time, we may divide 
 the minutes by 4, and the quotient will be the answer in degrees. 
 
 Q.CIST, 292. How find the difference of time between two places from their differ 
 ence of longitude ? 293. How find the difference of longitude from the difference of time 1 
 
176 REDUCTION. [SECT. VIII. 
 
 64. A whip sailed from Boston to Liverpool ; on the fourth day 
 the master took an observation of the sun at noon, and found by 
 his chronometer that it was 1 hr. 5 min. and 40 sec. earlier than 
 the Boston time : how many degrees east of Boston was the ship ? 
 
 Solution. 1 hr. 5m. 40 sec.=3940 sec., (Art. 281,) and 3940 
 sec.- 4 = 985'. The ship had therefore sailed 985' east, which 
 i* equal to 16 25'. Ans. 
 
 65. The difference of time between Albany and Buffalo is 19 
 minutes : what is the difference of their longitude ? 
 
 66. The difference of time between Richmond and New Orleans 
 is 51 min. 4 sec. : what is the difference of their longitude ? 
 
 67. The difference of time between Boston and Cincinnati is 
 53 min. 32 sec. : what is the difference of their longitude ? 
 
 COMPOUND NUMBERS REDUCED TO FRACTIONS. 
 
 294. That one concrete number may properly be said to be a 
 part of another, the two numbers must necessarily express objects 
 of the same kind, or objects which can be reduced to the same 
 kind or denomination. Thus, 1 penny is Tj-Jr of a pound, but 1 
 penny cannot properly be said to be a part of a foot, or of a year ; 
 for, feet and years cannot be reduced to pence. So, 1 orange is 
 of 5 oranges ; but 1 orange cannot be said to be -^ of 5 apples, or 
 5 pumpkins; for apples and pumpkins cannot be reduced to 
 oranges. 
 
 Ex. 1. Reduce 2s. Yd. to the fraction of a pound. 
 
 Analysis. The object in this example is to find what part of 
 1 pound, 2s. Vd. is equal to. To ascertain this, we must reduce 
 both the given numbers to the same denomination, viz : pence. 
 Now 2s. 7d.=31d., and l=240d. (Art. 281. 1.) The question, 
 therefore, resolves itself into this : what part of 240 is 31 ? The 
 answer is ^Voj consequently 2s. 7d. (3 Id.) is -iftV of a pound. 
 Hance, 
 
 QUKBT -294 When can one concrete number be said to be a part of another? 
 
ARTS. 294, 296.] REDUCTION. 177 
 
 295. To red a compound number to a common fraction, 
 of a higher denomination. 
 
 First reduce the given compound number to the lowest denomina- 
 tion mentioned for tJie numerator / then reduce a UNIT of tlie de 
 nomination of the required fraction to the same denomination as the 
 numerator, and the result will be the denominator. (Art. 281.) 
 
 OBS. 1 The given number, and that of which it is said to be o.part t must, 
 in all cases, be reduced to the same denomination. (Art. 294.) 
 
 2. When the given number contains but one denomination, it of coarse re- 
 quires no reduction. 
 
 If the given number contains a fraction, the denominator of the fraction is 
 the lowest denomination mentioned. Thus, in 6|s., the lowest denomination 
 is fourths of a shilling; in -f far., the lowest denomination is fifths of a farthing 
 
 2. Reduce f of a penny to the fraction of a pound. 
 
 Solution. Since sevenths of a penny is the lowest and only 
 denomination given, we simply reduce l to sevenths of a penny 
 for the denominator. Now l = 240d., and 240d.x7=1680. 
 Ans. X-rcVo* or <-T8T' Hence, 
 
 296. To reduce a fraction of a lower denomination to an 
 equivalent fraction of a higher denomination. 
 
 Reduce a unit of the denomination of the required fraction to 
 the same denomination as the given fraction, and the result will be 
 the denominator. 
 
 Or, divide the given fraction by the same numbers as in reducing 
 whole compound numbers to higher denominations. (Art. 281. II.) 
 Thus in the last example, fd.-rl2=-&s., (Art. 227,) and A s --r- 
 20= rG -<W>==ldro. Ans. 
 
 OBS. When factors common to the numerator and denominator occur th* 
 operation may be shortened by canceling those factors. (Art. 221.) 
 
 3. Reduce -$ of a penny to the fraction of a pound. 
 
 Solution. By the last article, - = the answer. 
 
 7X12X20 
 
 By Cancelation -- = - - Ant 
 7X12X20 7X12X20,5 420 
 
 QTTKST. 295. How is a compound number reduced to a common fraction ? 296. How 
 Is a fraction of a lower denomination reduce 1 to the fraction of higher * 
 
178 REDUCTION. [Si:CT. VIII. 
 
 4. Reduce 4|s. tc the fraction of a pound. Ans. , or 
 
 5. Reduce 4s. Yd. to the fraction of a pound. 
 
 6. Reduce 9d. 2 far. to the fraction of a pound. 
 7 What part of l is 1 of 1 penny ? 
 
 8. What part of 1 Ib. Troy is 7 ounces ? 
 
 9, What part of 1 Ib. Troy is 16 pwts. 3 grs ? 
 
 10. What part of 1 Ib. avoirdupois is 8 oz. and 12 drams? 
 
 11. What part of 1 ton is 14 cwt. and 15 Ibs? 
 
 12. What part of 1 yd. is 2 ft. and 4 inches? 
 
 13. What part of 1 mile is 82i rods? 
 
 14. What part of 1 acre is 45 rods? 
 
 15. What part of 1 square rod is 63 square feet? 
 
 16. Reduce of 1 qt. to the fraction of a gallon. 
 
 17. Reduce 7 gallons to the fraction of a hogshead. 
 
 18. Reduce of 1 hour to the fraction of a day. 
 
 19. Reduce -f of 1 minute to the fraction of an hour. 
 
 20. Reduce -f of 1 second to the fraction of a week. 
 
 21. What part of 3, 5s. 6d. Ifar. is 2, Is. 3d. ? 
 
 Solution. Reducing both numbers to farthings, 3, 5s 6d. Ifar. 
 3145 far., and 2, Is. 3d. = 1980 far. (Art. 295. Obs.' .) No* 
 1980 is iff! of 3145, which is equal to fff. Ans. 
 
 22. What part of 2 is 7s. 6d. ? 
 
 23. What part of 7, 3s. is 3 ? 
 
 24. What part of 2 bushels is 3 pecks ? 
 
 25. What part of 10 bushels is XO quarts? 
 
 26. What part of 16 rods is 40 feet? 
 
 27. What part of 3 weeks is 2 days and 7 hours ? 
 
 28. What part of 2 hhds. 10 gals, is 45 gals.? 
 
 29. What part of 2 tons, 3 cwt. is 15 cwt. 65 Ibs. ? 
 
 30. What part of 1 ton is * Ibs. 10 ounces? 
 
 31. What part of 90 is lo 15' 30' ? 
 
 32. What part of 360 is 45 15' 10 ; - ? 
 
 33. What part of 3 Ibs. Troy is 1 Ib. 3 oz. ? 
 
 34. What part of 25 Ibs. Troy is 10 Ibs. 7 oi< 10 pwts t ? 
 
 35. What part of 1 acre is 40 rods ? 
 36 What nart of 5 acres is 14- acres ? 
 
ART 297, REDUCTION. 179 
 
 FRACTIONAL COMPOUND NUMBERS 
 
 REDUCED TO WHOLE NUMBERS OF LOWER DENOMINATIONS. 
 
 Ex. 1. Reduce -f of 1 to shillings and pence. 
 
 Analysis. f of ls.=i- of 5s. or -fs., consequently - of 20s. (l) 
 is 20 times as much, and -fs. X 20 =-4-^8. or 12s. and f of a shilling. 
 
 reasoning as before, f of ld.=i of 4d., or d., and % of 12d. 
 (Ir,.) is 12 times as much ; but ^d X 12=^d., or 6d. Therefore 
 | 12s. 6d. Ans. Hence, 
 
 * To reduce fractional compound numbers to whole num- 
 bers of lower denominations. 
 
 First reduce the given numerator to the next lower denomination ; 
 then divide the product by the denominator, and the quotient mil 
 be an integer of the next lower denomination. (Art. 281. I.) 
 
 Proceed in like manner with the remainder, and the several quo* 
 tients will be the whole numbers required. 
 
 OBS. This operation is the same in principle as reducing htgner denomina 
 tions of whole numbers to lower. (Art. 281. I.) Whenever the fraction be 
 comes improper, it is reduced to a whole or mixed number. (Art. 196.) 
 
 2. Reduce of l to shillings. Ans. 16s. 
 
 3. Reduce of l to shillings and pence. 
 
 4. Reduce f of Is. to pence and farthings. 
 
 5. Reduce f of 1 Ib. Troy to ounces, &c. 
 
 6. Reduce f of 1 ounce Troy to pennyweights. 
 
 7. Reduce f of 1 Ib. avoirdupois to ounces, &c. 
 
 8. Reduce f of 1 cwt. to pounds, <fec. 
 
 9. Reduce -| of 1 ton to pounds, <fcc. 
 
 10. Reduce i of 1 yard to feet and inches. 
 
 11. Red ace f of 1 rod to feet and inches. 
 
 12. Reduce $ of 1 mile to rods, feet, &c. 
 
 13. Reduce -f of 1 gallon wine measure to quarts, &c. 
 
 14. Reduce 1 of 1 hogshead wine measure to gallons, &c. 
 
 15. Reduce f of 1 peck to quarts, &c. Ans. 6 qts. ! pts. 
 
 16. Reduce % of 1 bushel to quarts, <fec. 
 
 17. Reduce of 1 hour to minutes and seconds. 
 
 dug ST.- -297. How are fractional compound numbers reduced to whole 
 
180 COMPOC,<D [SECT. VIII 
 
 18 R,duce -A of 1 day to hours, &c. 
 
 19. Reduce of 1 minute to seconds. 
 
 20. Reduce f of 1 degree to minutes, &c. 
 
 21. Reduce j-Jr$ to the fraction of a penny. 
 
 Solution. We reduce the numerator to pence, the denomina- 
 tion required, and divide it by the denominator, as in the last 
 article. Thus, 2X20X12=480; and 480-f-720=tf. There- 
 fore jiT>=$n<l.=^=t or id. Ans. Hence, 
 
 298. To reduce a fraction of a higher denomination to an 
 equivalent fraction of a lower denomination. 
 
 Reduce the given numerator to the denomination of the required 
 fraction, and place the result over the given denominator. 
 
 OBS. 1. This process is the same in principle as to reduce a.w7iole compound 
 r imber to a lower denomination. (Art. 281. I.) 
 
 1 & When factors common to the numerator and denominator occur, the op- 
 t \tloti may be shortened by canceling those factors. (Art. 221 <) 
 
 2 ^ 20 ^ 1** 
 Thus, in the last example, - -=the answer, 
 
 2X20X12 2X20 X*2 ., 
 By Cancel, __ = _ ? _ == id. Ans. 
 
 22. Reduce 7-}^ of 1 to the fraction of a penny. 
 
 2'3. Reduce -^? of 1 Ib. avoirdupois to the fraction of an ounce. 
 
 24. Reduce g7 g 64 of 1 mile to the fraction of a rod. 
 
 25. Reduce -f^- of a day to the fraction of an hour. 
 
 26. Reduce -j of 1 week to the fraction of 1 minute. 
 
 27. Reduce f of 1 yard to the fraction of a nail. 
 
 28. Reduce -^ of 1 bushel to the fraction of a quart. 
 
 29. Reduce i 9 ^ of 1 hhd. wine measure to the fraction of a quart. 
 
 30. Reduce -j 2 ^- of 1 Ib. Troy to the fraction of an ounce. 
 
 31. Reduce iVn^ of 1 pound Troy to the fraction of a pwfc. 
 
 32. Reduce ^V of an acre to the fraction of a rod. 
 
 33. Reduce -rfr of a square yard to the fraction of a foot. 
 84. Reduce -g-f^ of a degree to the fraction of a second. 
 
 QUEST. 298. How is a fraction of a higher denomination reduced to the fraction of a 
 fewer denomination 1 * 
 
ARTS. 298-300."] ADDITION. 181 
 
 ADDITION OF COMPOUND NUMBERS. 
 
 299. The process of adding numbers of different dtnomi' 
 nations, is called COMPOUND ADDITION. 
 
 1. What is the sum of 6, 11s. 5d. 1 far. ; 4, 9s. 6d..2 far. ; 
 ^3, 12s 8d. 3 far. ; and 8, Cs. 9d. 1 far. ? 
 
 Operation. Having placed the farthings under fai- 
 
 s. d. far. things, the pence under pence, &c., tt-d 
 
 6 '11 ' 5 " 1 add the column of farthings together, as 
 
 4 9 ' 6 ' 2 in simple addition, and find the sum is 7, 
 
 3 " 12 ' 8 3 which is equal to Id. and 3 far. over. 
 
 8 " 6 " 9 " 1 Set the 3 far. under the column of far- 
 
 23~" " 5 " 3 Ans. things, and carry the Id. to the column 
 
 of pence. The sum of the pence is 29, which is equal to 2s. and 
 
 5d. over. Place the 5d. under the column of pence, and carry 
 
 the 2s. to the column of shillings. The sum of the shillings is 
 
 40, which is equal to 2, and nothing over. Write a cipher under 
 
 the column of shillings, and carry the 2 to the column of pounds. 
 
 The sum of the pounds is 23. Ans. 23, Os. 5d. 3 far. 
 
 3 GO* Hence, we derive the following general 
 
 RULE FOR ADDING COMPOUND NUMBERS. 
 
 I. Write the numbers so that the same denominations shall staitd 
 under each other. 
 
 II. Beginning with the lowest denomination, find the sum yfeach 
 column separately, and divide it by that number which it requires 
 of the column added, to make ONE of the next higher denomination. 
 Set the remainder under the column added, and carry the quotum t 
 to the next column. 
 
 III. Proceed in this manner with all the other denominations 
 except the highest, whose entire sum is set down. 
 
 PROOF. The proof is the same as in Simple Addition. (Art. 55. 
 
 OBS. 1. Fractional compound numbers should be reduced to whole num 
 bers of lower denominations, then added as above. (Art. 1GG.) 
 
 QUEST. 299. What is Compound Addition 7 300. How do you write compound numbers 
 for addition ? Which denomination do you add first"? When the sum of any coluiun it 
 found, what is to be dene with it 1 What is done with the last column ? 
 
182 COMPOUND [SECT. VIII. 
 
 2. Compound Addi ion is the same in principle as Simple Addition. 
 In the latter, it is true, we uniformly carry the tens, and in the former we carry 
 for different numbers ; yet in each we always carry for that number which it 
 takes of the order or denomination we are adding to make one in the ne 
 higher order or denomination. 
 
 2. 3. 4. 
 
 s. d. s. d. s. d. far. 
 
 16 8 9 25 17 11 68 17 10 3 
 
 856 30 12 10 10 9 6 
 
 25 6 8 13 15 7 43 ID 11 2 
 
 50 11 Ans. 35 16 9 65 14 81 
 
 5. A farmer sold to one customer 3 tons, 5 cwt. 17 Ibs. 13 oz 
 of hay; to another, 4 tons, 7 cwt. 35 Ibs. 12 oz. ; to another, 
 
 1 ton, 15 cwt. 63 Ibs. 7 oz. : how much hay did he sell to all ? 
 
 6. What is the sum of 15 tons, 6 cwt. 45 Ibs. 5 oz. ; 3 tons, 
 17 cwt. 80 Ibs. 6 oz. ; 26 tons, 31 Ibs. 7 oz. V 
 
 7. What is the sum of 21 Ibs. 7 oz. 12 pwts. 10 grs. ; 28 Ibs. 
 5 oz. 8 pwts. 7 grs. ; 7 Ibs. 6 pwts. 15 grs. ; 41 Ibs. 6 oz. 20 grs. ; 
 9 Ibs. 7 grs. ? 
 
 8. What is the sum of 16 Ibs. 3 oz. 6 pwts. 19 grs. ; 100 Ibs. 
 8 oz. 16 pwts. ; 97 Ibs. 5 oz. 10 grs. ; 115 Ibs. 9 oz. ? 
 
 9. Add together 19 rods, 12 ft. 8 in.; 64 rods, 13 ft. 3 in.; 
 28 rods, 10 ft. 5 in. ; 60 rods, 9 ft. 11 in. 
 
 10. Add together 5 leagues, 2 m. 4 fur. 7 rods, 4 yds. ; 18 
 leagues, 2 m. 3 fur. 21 rods, 3 yds. ; 85 leagues, 6 fur. 10 rods, 
 
 2 yds. 1 ft. 
 
 11. Add together 19 yds. 3 qrs. 3 na. ; 21 yds. 2 qrs. 1 na. , 
 42 yds. 1 qr. 2 na. ; 30 yds. 3 qrs. 2 na. 
 
 12. Add together 65 yds. 3 qrs. 1 na. ; 81 yds. 2 qrs. 2 na; 
 100 yds. 3 qrs. 1 na. ; 95 yds. 1 qr. 1 na. ; 15 yds. 3 na. ; 28 
 yds. 2 qrs. 
 
 13. Add together 17 A. 25 r. 29 sq. ft. ; 49 A. 15 r 4 sq. ft; 
 62 A. 29 r. 31 sq. ft. ; 10 A. 45 r. 16 sq. ft. 
 
 14. Add together 100 A. 3 R. 12 r. ; 115 A. 2 R. 20 r. ; 160 
 A. 1 R. 15 r. ; 91 A. 2 R. 26 r. 
 
 QUEST. Oi . Does Compound Addition differ from simple Addition 1 
 
ART. 301. J SUBTRACTION. 183 
 
 15. One room in a Louse contains 15 sq. yds. 5 ft. 7 in. of plas- 
 tering; another 10 yds. 7 ft. 30 in. ; another 9 yds. 6 ft. 25 in.; 
 another 7 yds. 5 ft. 63 in. : how much plastering is there in all 
 of them ? 
 
 16. A merchant bought one cask of oil containing 73 gals. 3 
 qts. ; another 60 gals. 2 qts. ; another 40 gals. 1 qt. ; another 65 
 gals. 2 qts. : how much oil did he buy ? 
 
 17. What is the sum of 20 hhds. 41 gals. 3 qts, 2 pts. 3 gi. ; 
 81 hhds 20 gals. 1 qt. 1 pt. 3 gi. ; 48 hhds. 19 gals. 2 qts. 1 pt. 
 
 2 gi. ; 81 hhds. 40 gals. 1 gi. ? 
 
 18. What is the sum of 10 wks. 5 d, 12 hrs. 40 min. ; 21 wks. 
 
 3 d. 9 hrs. 15 min. ; 40 wks. 4 d. 17 hrs. 30 min. ; 42 wks. Id.? 
 
 19. What is the sum of 40 bu. 3$ pks. 4 qts. ; 63 bu. 2-J- pks. 
 
 5 qts. ; 80 bu. 7| pks. 1 qt. ; 45 bu. 2 pks. 3 qts. ; 90 bu. 1 pk. ? 
 
 20. What is the sum of 7 qrs. 6 bu. 1 pk. 3 qts. ; 27 qrs. 
 
 6 bu. 6 qts. ; 34 qrs. 1 bu. 6 qts. ; 65 qrs. 6 bu. 3 qts. ? 
 
 SUBTRACTION OF COMPOUND NUMBERS. 
 
 3O1* The process of finding the difference between numbers of 
 different denominations, is called COMPOUND SUBTRACTION. 
 
 1. From 35, 17s. 6d. 3 far., subtract 16, 9s. 8d. 2 far. 
 
 Operation. Having placed the less number under 
 
 s. d. far. the greater, with farthings under far- 
 
 35 " 17 " 6 " 3 things, pence under pence, <fec., we sub- 
 
 16 " 9 ' 8 " 2 tract 2 far. from 3 far., and set the 
 
 19 " 7 '' 10 " 1 Am. remainder 1 far. under the column of 
 
 farthings. But 8d. cannot be taken from 
 
 6d. ; we therefore borrow 1 from the next higher denomination, 
 which is shillings; and Is. or 12d. added to the 6d. make 18d. 
 Now 8d. from 18d. leaves lOd. Since we borrowed, we must 
 carry 1 to the next denomination in the lower number, as in sim- 
 ple subtraction. (Art. 72.) 1 added to 9 makes 10; and 10 from 
 17, leaves 7. Finally, 16 from 35, leaves 19. 
 
 Ans. 19, 7s. lOd. 1 fai. 
 
 QTTKBT. 301. What is Compound Subtraction? 
 
18 i COMPOUND [SECT. VIII 
 
 302 Hence, we derive the following general 
 
 RULE FOR SUBTRACTING COMPOUND NUMBERS. 
 
 1. Write the less number under the greater, so thai the same de- 
 nominations may stand under each other. 
 
 II. Beginning with the lowest denomination, subtract the num- 
 ber in each denomination of the lower Line from tlie number ibove 
 
 't, and set the remainder below. 
 
 III. When a number in any denomination of the lower line is 
 larger than the number above it, borrow one of tJie next higher de- 
 nomination and add it to t/te number in the upper line. Subtract 
 as before, and carry 1 to the next denomination in tlie lower line, as 
 in subtraction of simple numbers. (Art. 72.) 
 
 PROOF. The proof is the same as in Simple Subtraction. 
 
 OBS. 1. Fractional compound numbers should be reduced to whole numbers 
 of lower denominations, then subtracted as above. (Art. IGG.) 
 
 2. Compound Subtraction is the same in principle as Simple Subtrac- 
 tion In both cases, when the number in the lower line is larger than that 
 above it, we borrow as many units as it takes of the order or denomination we 
 are subtracting to make one of the next higher order or denomination, and in 
 both, we carry 1 to the next figure in the lower number. 
 
 2. From 48, 17s. 6d. 2 far., take 39, 14s. 9d. 3 far. 
 
 3. From l60i, 6|s. 3|d., take 100|, 8s. 
 
 4. From 1000, take 500, 6s. 7d. 2 far. 
 
 5. From 16 cwt. 3 qrs. 15 Ibs., take 8 cwt. 2 qrs. 8 Ibs. 6 oz. 
 
 6. From 85 tons 16 cwt. 39 Ibs., take 61 tons 14 cwt. 68 Ibs. 
 
 7. Subtract 69 m. 41 r. 12 ft. from 89 m. 10 r. 14 ft. 
 
 8. Subtract 17 1. 2 m. 3 fur. 4 r. 4 ft. from 191. 1 m. 2 fur. 15 r. 
 
 9. Subtract 49 bu. 3 pks. 6 qts. from 85 bu. 2 pks. 4 qts. 
 
 10. Subtract 95 qrs. 4 bu. 3 pks. from 115 qrs. 3 bu. 1 pk. 
 
 11. Subtract 29 yds. 2 qrs. 3 na. from 85 yds. 1 qr. 2 na. 
 
 12. Subtract 55 yds. 2 qrs. 1 na. from 100 yds. 
 
 13. Subtract 75 gals. 3 qts. 1 pt. from 82 gals. 2 qts. 
 
 QUEST. 302. How do you write compound numbers for subtraction ? Where oegm to 
 subtract? When the number in the lower line is larger than that above it, what is to be 
 doue ? Obs. Does Compound Subtraction differ from Simple Subtraction ? 
 
ARTS. 302, 303.] SUBTRACTION. 185 
 
 15. A man having 140 A. 17 r. of land, sold 5 4 A. 58 r. : how 
 much had he left ? 
 
 16. Two men having bought 465 A. 48 r. of land, one of them 
 wished to take '230 A. : how much would the other have? 
 
 17. A farmer having 144 cords. 55 ft. of wood, sold 87 c. 93 ft. : 
 how much had he left ? 
 
 18. In a certain village there are two public cisterns ; one con- 
 .,ains 446 cu. ft. 69 in., the other 785 cu. ft. 95 in. : what is the 
 lifference in their capacity ? 
 
 19. The latitude of the Cape of Good Hope is 30 55' 15' 
 and that of Cape Horn, 55 58' 30" : what is their difference ? 
 
 20. The latitude of the Straits of Gibraltar is 36 6' 30", and 
 *hat of the North Cape, 71 10' : required their difference. 
 
 21. The longitude of New York is 74 I', and that of Cincin 
 jiati 84 27' : required their difference. 
 
 22. From 160 yrs. 11 mo. 2 wks. 5 ds. 16 hrs. 30 min. 40 sec., 
 take 106 yrs. 8 mo. 3 wks. 6 ds. 13 hrs. 45 min. 34 sec. 
 
 23. What is the time from Feb. 22d, 1845, to May 21st, 1847 ? 
 
 Operation. May is the 5th month, and Feb. the 2d. 
 
 yr. mo. d. Since 22 days cannot be taken from 21 d., we 
 
 /847 " 5 " 21 borrow 1 mo. or 30 d. ; then say 22 from 51 
 
 1845 " 2 " 22 leaves 29. 1 to carry to 2 makes 3, and 3 
 
 (Lns. 2 " 2 " 29 from 5 leaves 2. 5 from 7 leaves 2. Hence, 
 
 3O3. To find the time between two dates. 
 
 Write the earlier date under the later, placing the years on the 
 left, the number of the month next, and tJie day of the month on the 
 right, and subtract as before. (Art. 302.) 
 
 OBS. 1. The number of the month is easily determined )y reckoning from 
 January, the 1st month, February the 2d, &c. (Art. 264.) 
 
 2. In finding the time between two dates, and in casting interest, 30 day 
 are considered a month, and 12 months a year. 
 
 3. Instead of setting down the ordinal number of the month, as in the 
 solution above, some prefer to write the number of whole months that have 
 
 QUEST. 303. How do you find the time between two dates ? Obs In finding tima be 
 tween two dates, and in casting interest, how many days are considered a month " ** 
 many months a year ? 
 
186 COMPOUND [SECT. VIIJ 
 
 in the given year. E. g., if the date is Feb. 22d, 1845, they would 
 write 1 in the place of months; because, it is said, 2 whole months have not 
 elapsed in the year 1845. But it may be doubted whether this method would 
 not ictid to frequent mistakes. 
 
 Besides, it may be urged with equal reason, that 1 ought to be deducted 
 from the day of the month, and 1 from the year ; for neither 22 whole days, nor 
 1845 whole years had elapsed at the time of the date, but the 22d day and the 
 1845th year were then passing. In this way. the subject, which in itself is 
 iniple, becomes intricate and perplexing. 
 
 24. General Washington was born Feb. 22d, 1732, and died 
 Dec. 14th, 1799 : how old was he ? 
 
 25. The Independence of the United States was declared, July 
 4th, 1776 : how long is it since? 
 
 26. A note was given Aug. 25th, 1840, and paid Feb. 6th, 
 1842 : how long did it run? 
 
 27. The United States Exploring Expedition sailed from Norfolk 
 on the 18th of Aug., 1838, and returned to New York on the 10th 
 of June, 1842 : how long was the voyage ? 
 
 COMPOUND MULTIPLICATION. 
 
 3O4 The process of multiplying numbers of different denom- 
 inations, is called COMPOUND MULTIPLICATION. 
 
 Ex. 1. What will 6 cows cost, at 5, 2s. 7|d. apiece? 
 
 Analysis. Since 1 cow costs 5, Operation. 
 
 2s. 7f d., 6 cows will cost 6 times as s. d. far. 
 much. Beginning with the lowest 5 " 2 " 7 " 3 
 denomination, 6 times 3 far. are 1 8 far., 6 
 
 equal to 4d. and 2 far. over. Set the 30 " 15 " 10 " 2 Ans. 
 2 far. under the denomination multi- 
 
 plied and carry the 4d. to the next product. 6 times 7d. are 42d. 
 and 4d. make 46d., equal to 3s. and lOd. Set the lOd. under the 
 pence, and carry the 3s. to the next product. 6 times 2s. are 12s 
 and 3s. make 15s. As the product 15s. does not make one in the 
 next denomination, we set it under the column multiplied. Fi- 
 nally, 6 times 5 are 30. The answer is 30, 15s. lO^d. 
 
 QUEST. 304. What is Compound Multiplication 1 
 
ARTS. 304, 305.] MULTIPLI ATIO*. IS/ 
 
 3O5* Hence, we deduce the following general 
 
 RULE FOR MULTIPLYING COMPOUND NUMBERS. 
 
 Multiply each denomination separately, beginning with tlie loivest, 
 and divide each product by that number which it takes of the denom- 
 ination multiplied; to make ONE of the next higher ; set down the 
 Wiainder, and carry the quotient to the next product, as in addition 
 ^ compound numbers. (Art 300.) 
 
 OBS. 1. When the multiplier is a composite number, ft is advisable to multi- 
 ply first by one factor and that product by the other. (Art. 97.) 
 
 2. Compound Multiplication is the same in principle as Simple Multiplica- 
 tion. In each we carry for that number which it takes of the order or denom- 
 ination we are multiplying, to make one of the next higher order or denorui 
 nation. 
 
 2. What will 28 horses cost, at 21, 3s, 7|d. apiece? 
 
 Operation. 
 s. d. far. 
 21 " 3 " 7 " 1 We multiply by the factors of 28, 
 
 7 which are 7 and 4, and set down each 
 
 148 " 5 " 2 " 3 result as above. 
 
 4 
 
 593 " " 11 " Ans. 
 
 3. What cost 7 acres of land, at 35, 6s. 7d. per acre? 
 
 4. What cost 18 barrels of flour, at l, 6s. 8d. per barrel? 
 
 5. A man bought 15 loads of hay, each weighing 1 T. 270|lbe, : 
 what was the weight of the whole ? 
 
 6. Multiply 16 tons, 3 cwt. 10f Ibs. by 25. 
 
 7. Multiply 12 Ibs. 3 oz. 16 pwts. by 56. 
 
 8. If 1 dollar weighs 17 pwts. 4 grs., how much will 96 dol- 
 lars weigh ? 
 
 9. Multiply 48 hhds. 15 gals. 2 qts. 1 pt. by 63. 
 10. Multiply 56 pipes, 1 hhd. 23 gals, by 100. 
 
 QUEST. 305. Where do you begin to multiply a compound number? What is dovo 
 with each product? Obs. When the multiplier is a compos'jte number, how 
 Does it differ from Simple Multiplication ? 
 T.H. 
 
188 COMPOUND [SECT. VIII. 
 
 11. Bought 72 pieces of cl:th, each containing 32f yards: 
 how much did they all contain ? 
 
 12. If 1 cloak requires 10 yds. 3 qrs., how much will 500 
 cloaks require ? 
 
 13. Multiply 175 miles, 7 fur. 18 rods by 84. 
 
 14. Multiply 40 leagues, 2 m. 5 fur. 15 r. by 50. 
 
 15. Multiply 149 bu. 12 qts. by 60. 
 
 16. Multiply 26 qrs. 7 bu. 3 pks. 5 qis. by 110. 
 
 17. Multiply 150 acres, 65 rods by 52. 
 
 18. Multiply 310 acres, 3 roods, 3 rods by 81. 
 
 19. Multiply 265 cu. ft. 10 in. by 93. 
 
 20. Multiply 148 cords, 29-r 3 - ft. by 650. 
 
 21. Multiply 365 d. 5 hrs. 48 min. 48 sec. by 35. 
 -22. Multiply 70 yrs. 6 mo. 3 wks. 5 d. by 17. 
 
 23. Multiply 75 40' 21" by 210. 
 
 24. If a ship sails 3 24' 10" per day, how far will she sail hi 
 60 days ? 
 
 25. If 1 acre produce 45 bu. 26 qts., how much will 100 acr< s 
 produce ? 
 
 26. If 1 barrel of flour requires 4 bu. 3 pks. 5 qts. of wheat, 
 how much will 500 barrels require ? 
 
 27. What cost a chest of tea containing 17 Ibs., at 6s. 10d. per 
 pound ? 
 
 28. What is the duty on 1000 gals, of brandy, at 13s. 7d. per 
 gallon ? 
 
 29. What is the duty on 10560 Ibs. of sugar, at 6d. 3 far. per 
 pound ? 
 
 30. What is the duty on 1500 yards of broadcloth, at 6s. 9|d. 
 per yd. ? 
 
 31. If 1 load of wood measures 117 ft. 110 in., how much will 
 40 loads of the same size measure ? - 
 
 32. If 1 quarter of beef weighs 216 Ibs. 7 oz., how much will 
 4 quarters weigh ? 
 
 33. If 1 bushel of salt weighs 72 Ibs. 10 oz., how much will 
 850 bushels weigh ? 
 
 34. If 1 cask of oil contains 86 gals. 2 qts. 1 pt., ho* much wiL 
 100 casks of the same size contain ? 
 
ARTS* 306, 307.] DIVISION. 189 
 
 COMPOUND DIVISION. 
 
 3O6 The process of dividing numbers of different deramina- 
 tions, is called COMPOUND DIVISION. 
 
 Ex. 1. Divide 25, 3s. 4d. 2 far. by 6. 
 
 Operation. Beginning with the pounds, we find 
 
 ? d far. 6 is contained in 25, 4 times and 1 
 
 6)25 " 3 " 4 " 2 over. Set the 4 under the pounds, 
 
 ~T~ 3 " 10 " 3 Ans. and reduce the remainder l to shil- 
 lings, which added to the 3s. make 
 
 23s. 6 in 23s. 3 times and 5s. over. Set the 3 under the shil- 
 lings, and reduce the remainder 5s. to pence, which added to the 
 4d. make 64d. 6 in 64d., 10 times and 4d. over. Set the 10 
 under the pence, reduce the 4d. to farthings, and divide as before. 
 Ans. 4, 3s. lOd. 3 far. 
 
 3O7. Hence, we deduce the following general 
 
 RULE FOR DIVIDING COMPOUND NUMBERS. 
 
 Begin with the highest denomination, and divide each separately. 
 Reduce the remainder, if any, to the next lower denomination, to 
 which odd the number of that denomination contained in the given 
 example, and divide the sum as before. Proceed in this manner 
 through all the denominations. 
 
 OBS. 1. Each partial quotient will be of the same denomination, as that part 
 of the dividend from which it arose. 
 
 2. When the divisor exceeds 12, and is a composite number, it is advisable 
 to divide first by one factor and that quotient by the other. (Art. 129.) If the 
 divisor exceeds 12. but is not a composite number, long division may be em- 
 ployed. (Art. 120. II.) 
 
 3. Compound Division is the same in principle, as Simple Division. Pre- 
 fixing the remainder to the next figure of the dividend in simple division, in 
 the same as reducing it to the next lower order or denomination, and adding 
 tne next figure to it. 
 
 QI-EST 306. What is Compound Division 1 307. Where do you begin to divide a com- 
 pound number ? What is done with the remainder? Obs. Of what denomination is each 
 partial quotient? When the divisor is a composite number, how proceed ? Does it dlflet 
 from Simple Division ? 
 
190 DECIMAL [SECT* VIII, 
 
 2. A man wished to divide 75 cwt. 2 qrs. 10 Ibs. of beef equally 
 among 35 families : how much could he give to each ? 
 
 Operation. 
 
 cwt. qrs. Ibs. We divide by the factors of 35, which 
 
 7)75 " 2 " 10 are 7 and 5, and set down each result as 
 
 5)1 " 3 " 5~ above. 
 
 2 " " 16 Ans. 
 
 8. Divide 312 Ibs. 9 oz. 18 pwts. by 43. 
 
 Ans. 7 Ibs. 3 oz. 6 pwta. 
 
 4. Divide 410 Ibs. 4 oz. 5 pwts. 6 grs. by 8. 
 
 5. Divide 786 bu. 18 qts. by 25. 
 
 6. A farmer raised 1000 hji. 3 pks. 16 qts. of wheat on 40 
 acres : how much was that per acre ? 
 
 7. A man bought 10 horses for 200, 15s. : how much did he 
 give apiece ? 
 
 8. Divide 87, 10s. 7 id. by 18. 
 
 9. A merchant tailor put 216 yds. 3 qrs. of cloth into 20 
 cloaks : how much cloth did each cloak contain ? 
 
 10. Divide 500 yds. 3 qrs. 2 na. by 54. 
 
 11. A man traveled 1000 miles in 12 days: at what rate did 
 he travel per day ? 
 
 12. Divide 1500 m. 2 fur. 30 r. 12 ft. by 7. 
 
 13. Divide 120 gals. 3 qts. 1 pt. by 72. 
 
 14. Divide 400 hhds. 10 gals. 2 qts. 1 pt. by 9. 
 
 15. Divide 365 d. 10 hr. 40 min. by 15. 
 
 16. Divide 111 yrs. 20 d 13 hrs. 25 min. 10 sec. by 11. 
 
 17. Divide 45 17' 10" by 25. 
 
 18. Divide 65 signs 12 47' by 41. 
 
 19. Divide 164 cords, 30 ft. by 17. 
 
 20. Divide 410 cords, 10 ft. 21 in. by 61. 
 
 21. If a chest of tea weighing 96 pounds cost 33, what will 
 I pound cost ? 
 
 22. If the duty on a pipe of wine is 50, Os. 6<1., what is the 
 duty per gallon ? 
 
 23. If a person spends 200 a year, what are his expenses per 
 day? 
 
ARTS. 308-3 10. J FRACTIONS. 191 
 
 SECTION IX.- 
 DECIMAL FRACTIONS. 
 
 308. Fractions which decrease in a tenfold ratio, or which ex- 
 press simply tenths, hundredt/is, thousandths, &c., are called DECI- 
 MAL FRACTIONS. 
 
 They arise from dividing a unit into ten equal parts, then di- 
 viding each of these parts into ten otlier equal parts, and so on. 
 Thus, if a unit is divided into 10 equal parts, 1 of those parts is 
 called a tenth. (Art. 1*78.) If a tenth is divided into 10 equal 
 parts, 1 of those parts will be a hundredth; for, T I o-7-10=ToT. 
 If a hundredth is divided into 10 equal parts, 1 of the parts will 
 be a thousandth ; for, 1^-5-7- 10=-niVo &c. (Art. 227.) 
 
 OBS. Fractions of this class are called decimals, because they regularly de- 
 crease in a tenfold ratio. (Art. 37. Obs. 2.) 
 
 Decimal fractions are said to have been invented by Lord Napier, in 1602. 
 
 309. Each order of whole numbers, we have seen, increases 
 in value from units towards the left in a tenfold ratio ; and, con- 
 versely, each order must decrease from left to right in the same 
 ratio, till we come to units' place again. (Art. 37.) 
 
 3 1 O. By extending this scale of notation below units towards 
 the right hand, it is manifest that the first place on the right of 
 units, will be ten times less in value than units' place ; that the 
 second will be ten times less than the first y the third ten times 
 less than the second, &c. 
 
 Thus we have a series of orders below units, which decrease in 
 a tenfold ratio, and exactly correspond in value with tenths, hun- 
 dred ths, thousandths, &c. (Art. 308.) 
 
 QUEST. 308. What are Decimal Fractions 1 From what do they arise ? Oft*. Why 
 called decimals? 309. In what manner do whole numbers Increase and decrease 1 
 310. By extending this scale below units, what would be the value of the first place on 
 the right of units 1 The second 1 The third ? With what do these orders correspond in 
 value 1 
 
192 DECIMAL [SECT. IX 
 
 311* Decimal Fractions are commonly expressed by writing 
 the numerator with a point ( . ) before it. 
 
 The point placed before decimals is called the Decimal Point, 
 or Separatrix. Its 'object is to distinguish the fractional parts 
 from whole numbers. 
 
 If the numerator does not contain so many figures as there are 
 ciphers in the denominator, the deficiency must be supplied by 
 prefixing ciphers to it. For example, -rV is written thus .1 ; & 
 thus .2 ; T^ thus .3 ; &c. T^ is written thus .01, putting the 
 1 in hundredths' place ; -rfo thus .05 ; &c. That is, tenths are 
 written in the first place on the right of units ; hundredths in the 
 second place ; thousandths in the third place, &c. 
 
 312* The denominator of a decimal fraction is always 1 with 
 as many cipliers annexed to it, as there are figures in the given nu- 
 merator. (Art. 308.) 
 
 313* The names of the different orders of decimals, or places 
 below units, may be easily learned from the following 
 
 DECIMAL TABLE. 
 
 . s i 
 
 L ! ! I ! 1 1 ! 1 1 
 
 .S-c*3 3 -5 ^3 .3 E "3 S-J= 
 
 3.267145986274 
 
 314* It will be seen from this table that the value of each 
 figure in decimals, as well as in whole numbers, depends upon the 
 place it occupies, reckoning from units. Thus, if a figure stands 
 in the first place on the right of units, it expresses tentJis ; if in 
 
 Q.ITE9T. 311. How are decimal fractions expressed ? What is the point placed before 
 decimals called? 312. What is the denominator of a decimal traction? 313. Repeat 
 the Decimal Table, beginning units, tenths, &c. 314. Upon what does the value of a de- 
 cimal depend 1 
 
ARTS. 311-316.] FRACTIONS. 193 
 
 the second, hundredths, &c. ; each successive place or order to- 
 wards the right, decreasing in value in a tenfold ratio. Hence, 
 
 315* Each removal of a decimal figure one place from units 
 towards the right, diminishes its value ten times. 
 
 Prefixing a cipher, therefore, to a decimal diminishes its value 
 ten times ; for, it removes the decimal one place farther from units* 
 place. Thus, .4=^; but .04=-r-o ; and .004=!-^, &c. ; for 
 ttie denominator to a decimal fraction is 1 with as many ciphers 
 annexe! to it, as there are figures in the numerator. (Art. 312.) 
 Annexing ciphers to decimals does not alter their value ; for, 
 each significant figure continues to occupy the same place from 
 units as before. Thus, .5=^; so .50 =-&-, or -flr, by dividing the 
 numerator and denominator by 10; (Art. 191,) and ,500=i\-,/V, 
 or -ft, &c. 
 
 OBS. 1. It should be remembered that the units' place is always the right 
 hand place of a whole number. The effect of annexing and prefixing ciphers 
 to decimals, it will be perceived, is the reverse of annexing and prefixing them 
 U> whole numbers. (Art. 98.) 
 
 2. A whole number and a decimal, written together, is called a mixed num- 
 ber. (Art. 183.) 
 
 316. To read decimal fractions. 
 
 Beginning at the left hand, read the figures as if they were whole 
 numbers, and to the last one add tlie name of its order. Thus, 
 
 .7 is read 7 tenths. 
 
 .36 " " 36 hundredths. 
 
 .475 " " 475 thousandths. 
 
 .6342 " " 6342 ten thousandths. 
 
 .57834 " " 57834 hundred thousandths. 
 
 .284648 " " 284648 millionths. 
 
 .8913629 " " 8913629 ten millionths. 
 
 OBS. In reading decimals as well as whole numbers, the units' place should 
 always be made the starting point. It is advisable for the learner to apply to 
 
 drKST. 315. What is the effect of removing a decimal one place towards the right 1 
 What then is the effect of prefixing ciphers to decimals 1 What, of annexing them 1 
 Obs. Which is the units' place ? What is a whole number and a decimal written to 
 gethcr, called ? 316. How are decimals read 1 Obs. In reading decimals, wha shook M 
 made the starting point 1 
 
194 DECIMAL [SECT. IX. 
 
 every figure the name of its order, or the place which it occupies, before at- 
 tempting to read them. Beginning at the units' place, he should proceed tow- 
 ards the right, thus units, tenths, hundredths, thousandUis, &c., pointing tc 
 each figure as he pronounces the name of its Drder. In this way he will be 
 able to read decimals with as much ease as he can whole numbers. 
 
 Read the following numbers : 
 
 (1.) (2.) (3.) (4.) 
 
 .32 .46274 42.008 2.403126 
 
 .03687 17.401 6.004534 
 
 .3624 .00368 23.07 1.100492 
 
 .82344 .00046 81.4389 9.000028 
 
 .13236 .00009 90.0104 8.001249 
 
 (5,) (6.) (7.) (8.) 
 
 12.683 6.00754 4.306702 9.2000076 
 
 20.064 3.0468 0.007006 8.0403842 
 
 35.0072 2.306843 1.13004 0.0000008 
 
 67.4008 1.710386 9.203167 4.3008004 
 
 Note. Sometimes we pronounce the word decimal when we come to tho 
 separatrix, and then read the figures as if they were whole numbers ; or, 
 simply repeat them one after another. Thus, 125.427 is read, one hundred 
 twenty-five, decimal four hundred twenty-seven ; or, one hundred twenty-fiv*,, 
 decimal four, two, seven. 
 
 Write the fractional part of the following numbers in decimals 
 (9.) (10.) (11.) (12.) 
 
 25^ 4r3hr 
 
 30 Mr 6Tffo 
 
 72-n/o 7-i 08 o 41 i H J o 9 
 
 13. Write 9 tenths ; 25 hundredths ; 45 thousandths. 
 
 14. Write 6 hundredths ; 7 thousandths ; 132 ten thousandth*, 
 
 15. Write 462 thousandths; 2891 ten thousandths. 
 
 16. Write 25 hundred thousandths ; 25 millionths. 
 
 17. Write 1637246 ten millionths; 65 hundred millionths. 
 
 18. Write 71 thousandths; 7 millionths. 
 
 19. Write 23 hundredths; 19 ten thousandths. 
 
 20. Write 261 hundred thousandths; 65 hundredths ; 121 
 lionths; 751 trillionths. 
 
 QUEST. Note. What other method of reading decimals Is ment oned * 
 
ARTS. 317-319.] FRACTIONS. 195 
 
 317. Decimal Fractions, it will be perceived, differ from 
 Common Fractions both in their origin and in the manner of ex- 
 pressing them. 
 
 Common Fractions arise from dividing a unit into any number 
 of equal parts ; consequently, the denominator may be any number 
 whatever. (Art. 182.) Decimals arise from dividing a unit into 
 ten .qual parts, then subdividing each of those parts into ten other 
 >qujl parts, and so on; .consequently, the denominator is always 
 10, 100, 1000, &c. (Arts. 308, 312.) 
 
 Again, Common Fractions are expressed by writing the numer- 
 ator over the denominator / Decimals are expressed by writing 
 the numerator only, with a point before it, while the denominator 
 is understood. (Arts. 182, 311.) 
 
 Decimals are added, subtracted, multiplied, and divided, 
 in the same manner as whole numbers. 
 
 OBS. The only thing with which the learner is likely to find any difficulty, 
 is pointing off the answer. To this part of the operation he should give par- 
 ticular attention. 
 
 ADDITION OP DECIMAL FRACTIONS. 
 
 319. Ex. 1. What is the sum 28.35; 345.329; 568.5; and 
 6.485? 
 
 Operation. Write the units under units, tenths under 
 
 28.35 tenths, hundredth* under hundredths, &c. ; 
 
 345.329 then, beginning at the right hand or lowest 
 
 568.5 order, proceed thus: 5 thousandths and 9 
 
 6.485 thousandths are 14 thousandths. Write the 
 
 948.664 Ans. 4 under the column added, and carrying the 1 
 to the next column, proceed through all the 
 orders in the same manner as in simple addition. (Art. 54.) Fi- 
 nally, place the decimal point in the amount directly under that 
 in the numbers added. 
 
 QUEST. 317. How do decimals differ from common fractions ? From what do common 
 fractions arise ? From what do decimals arise 1 How are common fractions expressed 1 
 Row are decimals ? 
 
 9* 
 
196 ADDITION OF [SECT. IX 
 
 3 2O. Hence, we deduce the following general 
 
 RULE FOR ADDITION OF DECIMALS. 
 
 Write he numbers so that the same orders may stand under each 
 other, placing units under units, tenths under tenths, hundred ths 
 under hundredths, t&c. Begin at the right Jiand or lowest order, 
 and proceed in all respects as in adding whole numbers. (Art. 54.) 
 
 Fr<M the right Imnd of tJie amount, point off as many figures' 
 for decimals as are equal to the greatest number of decimal places 
 in either of the given numbers. 
 
 PROOF. Addition of Decimals is proved in the same manner as 
 Simple Addition. (Art. 55.) 
 
 Note. The decimal point in the answer will always fall directly under the 
 decimal points in the given numbers. 
 
 EXAMPLES. 
 
 2. What is the sum of 25.7; 8.389; 23.056? Ans. 5*1.145. 
 
 3. What is the sum of 36.258 ; 2.0675 ; 382.45 ; and 7.3984 ? 
 
 4. What is the sum of 32.764; 5.78 ; 16.0037 ; and 49.3046 ? 
 
 5. What is the sum of 1.03041; 6.578034; 2.4178; and 
 4.72103 ? 
 
 6. Add together 4.25 ; 6.293; 4.612; 38.07; 2.056; 3.248; 
 and 1.62. 
 
 7. Add together 35.7603; 47.0076; 129.03; 100.007; and 
 JiO.32. 
 
 8. Add together 467.3004; 28.78249; 1.29468; and 3.78241. 
 
 9. Add together 21.6434 ; 800.7 ; 29.461 ; 1.7506 ; and 3.45. 
 
 10. Add together 45.001; 163.4234; 20.3045; 634.2104; 
 and 234.90213. 
 
 11. Add together 293.0072 ; 89.00301; 29.84567; 924.00369; 
 and 72.39602. 
 
 12. Add together 1.721341; 8.620047; 51.720345; 2.684; 
 and 62.304607. 
 
 13. Add together 1.293062; 3.00042 ; 9.7003146; 3.600426; 
 7.0040031 ; and 8.7200489. 
 
 Q.UKST. 330. How are decimals added ? How point off the answer 1 How is addition 
 of decimals proved 1 
 
ARTS. 320-321.] DECIMAL FRACTIONS. 197 
 
 14. Add together 394.61; 81.928; 3624.8103; 640.203; 
 6291.302; 721.004; and 3920.304. 
 
 15. Add together 25 hundredths, 8 tenths, 65 thousandths, 
 16 hundredths, 142 thousandths, and 39 hundredths. 
 
 16. Add together 9 tenths, 92 hundredths, 162 thousandths, 
 489 thousandths, and 92 millionths. 
 
 17. Add together 45 thousandths, 1752 millionths, 624 ten 
 millionths, and 24368 millionths. 
 
 18. Add together 29 hundredths, 7 millionths, 62 thousandths, 
 and 12567 ten millionths. 
 
 19. Add together 95 thousandths, 61 millionths, 6 tenths, 11 
 hundredths, and 265 hundred thousandths. 
 
 20. Add together 1 tenth, 2 hundredths, 16 thousandths, 7 
 millionths, 26 thousandths, 95 ten millionths, and 7 ten thou- 
 sandths. 
 
 21. Add together 96 hundred thousandths, 92 millionths, 25 
 hundredths, 45 thousandths, and 7 tenths. 
 
 22. Add together 85 thousandths, 17 hundredths, 36 ten thou- 
 sandths, 58 millionths, 363 hundred thousandths, 185 millionths, 
 and 673 ten thousandths. 
 
 SUBTRACTION OF DECIMAL FRACTIONS. 
 
 321. Ex. 1. From 425.684 subtract 216.96. 
 
 Operation. Having written the less number under the 
 
 425.684 greater, so that units may stand under units, 
 
 216.96 tenths under tenths, &c., we proceed exactly 
 
 208.724. Ans. as in subtraction of whole numbers. (Art. 72.) 
 
 Thus thousandths from 4 thousandths leaves 
 
 4 thousandths. Write the 4 in the thousandths' place. As the 
 
 next figure in the lower line is larger than the one above it, 
 
 we borrow 10. Now 9 from 16 leaves 7; sot the 7 under th 
 
 column and carry 1 to the next figure. (Art. 72.) Proceed in the 
 
 same manner with the other figures in the lower number. Finally, 
 
 place the decimal point in the remainder directly luider that ia 
 
 the given number. 
 
198 SUBTRACTION OF [Sfif T IX. 
 
 322* Hence, we deduce the following general 
 
 RULE FOR SUBTRACTION OP DECIMALS. 
 
 Write the less number under the greater, with units under 
 
 under tenths, hundredths under hundred ths, d'c. 
 as in whole numbers, and point off the answer as in addition <J/" 
 decimals. (Art. 320.) 
 
 PROOF. Subtraction of Decimals is proved in the same manurr 
 as Simple Subtraction. (Art. 73.) 
 
 A T r?te. When there are blank places on the right hand of the upper num- 
 ber, they may be supplied by ciphers without altering the value of the decimal, 
 (Art. 315.) 
 
 EXAMPLES. 
 
 2. From 456.0546 take 364.3123. Ans. 91.7423. 
 
 3. From 1460.39 take 32.756218. 
 
 4. From 21.67 take .682349. 
 
 5. From 81.6823401 take 9.163. 
 
 6. From 100.536 take 19.36723. 
 
 7. From .076345 take .009623478. 
 
 8. From 1 take .99. 
 
 0. From 10 take .000001. 
 
 10. From 65.00001 take .9682347. 
 
 11. From 24681 take .87623. 
 
 12. What is the difference between 25 and .25 ? 
 
 13. What is the difference between 3.29 and .999 ? 
 
 14. What is the difference between 10 and .0000001 ? 
 
 15. What is the difference between 9 and 9.99999 ? 
 
 16. What is the difference between 4636 and .4654 ? 
 
 17. What is the difference between 25.6050 and 567.392 ? 
 
 18. What is the difference between 76.2784 and 29.84234? 
 
 19. What is the difference between .0000001 and .0001 ? 
 
 20. What is the difference between .0000004 and .00004 ? 
 
 21. What is the difference between 32 and .00032 ? 
 
 Q.UKBT 322. How are decimals subtracted ? How point off the answer 1 Flow is t ib 
 traction of decimals proved ? 
 
ARTS 322, 323.] DECIMAL FRACTIONS. J99 
 
 22. \Vhat is the difference between .00045 and 45 ? 
 
 23. What is the difference between .00000099 and 99 ? 
 
 24. From 1 thousandth take 1 millionth. 
 
 25. From 7 hundred take 7 hundredths. 
 
 26. From 29 thousand take 92 thousandths. 
 
 27. From 256 millions take 256 thousandths. 
 
 28. From 46 hundredths take 46 thousandths. 
 
 29. From 95 thousandths take 999 ten thousandths. 
 
 30. From 1 billionth take 1 trillion th. 
 
 31. From 2874 millionths take 211 billionths. 
 
 32. From 6231 hundred thousandths take 154 millionths. 
 
 33. From 7213 ten thousandths take 431 hundred thousandths 
 
 34. From 8436 hundred millionths take 426 ten billionths. 
 
 MULTIPLICATION OF DECIMALS. 
 
 323* Ex. 1. If a man can reap .96 of an acre in a day, how 
 much can he reap in .5 of a day ? 
 
 Analysis. Since he can reap 96 hundredths of an acre in a 
 whole day, in 5 tenths of a day he can reap 5 tenths as much. 
 But multiplying by a fraction we have seen, is taking a part of the 
 multiplicand as many times as there are like parts of a unit in the 
 multiplier. (Art. 210.) Hence, multiplying by .5, which is equal 
 to T^ or , is taking half of the multiplicand once. Now .96, or 
 ^4-2=-^-. (Art. 227.) ButTW=.48. (Art. 311.) 
 
 Operation. We multiply as in whole numbers, and pointing 
 
 .96 off as many decimals in the product as there are 
 
 .5 decimal figures in both factors, we have 480, But 
 
 .480 Ans. ciphers placed on the right of decimals do not 
 
 affect their value ; the may therefore be omitted, 
 
 and w.e have .48 for the answer. 
 
 (2.) (3.) (4.) 
 
 ^Multiply 25.38 360.085 6843.02 
 
 *By .42 .0043 6.5 
 
 10.6596" Ans. 1.5483655 Ans. 44479.630 Ant 
 
200 MULTIPLICATION OF [SECT. IX 
 
 324. From the preceding illustrations we deduce the follow- 
 ing general 
 
 RULE FOR MULTIPLICATION OF DECIMALS. 
 
 Multiply as in whole numbers, and point off as many figures 
 from tJie right of the product for decimals, as there are decimal 
 places both in the multiplier and multiplicand. 
 
 If the product does not contain so many figures as there ar 
 dt'timals in loth factors, supply the deficiency by prefixing ciphers. 
 
 PROOF. Multiplication of Decimals is proved in the same man* 
 rer as Simple Multiplication. 
 
 OBS. The reason for pointing off as many decimal places in the product as 
 there are decimals in both factors, may be illustrated thus : 
 
 Suppose it is required to multiply .25 by .5. Supplying the denominators 
 .25=-i*\, and .5=-^. (Art. 312.) Now -^-xA^TWo- (Art. 215.) But 
 tVuV 125 5 ( Art - 311 ;) that is, the product of .25XA contains just as many, 
 decimals as the factors themselves. In like manner it may be shown that the 
 product of any two or more decimal numbers, must contain as many decimal 
 figures as there are places of decimals in the given factors. 
 
 EXAMPLES. 
 
 Ex. 1. In 1 rod there are 16.5 feet: how many feet are there 
 in 41.3 rods? 
 
 2. In 1 degree there are 69.5 statute miles : how many miles 
 are there in 360 degrees ? 
 
 3. In 1 barrel there are 31.5 gallons: how many gallons in 
 65.25 barrels ? 
 
 4. In 1 inch there are 2.25 nails : how many nails are there in 
 60.5 inches ? 
 
 5. In 1 square rod there are 30.25 square yards : how many 
 square yards are there in 26.05 rods ? 
 
 6. In one square rod there are 272.25 square feet : how many 
 square feet are there in 160 rods ? 
 
 QUEST. 324. How are decimals multiplied together 1 How do you point off the prod- 
 uct? When the product does not contain so many figures as there are decimals in both 
 (actors, what is to be done ? How is multiplication of decimals proved ? 
 
ARTS. 324, 325.] DECIMALS. 201 
 
 7. How many square rods are there in a field 60.5 rods long 
 wid 40.75 rods wide? 
 
 Multiply the following decimals : 
 
 8. 1.0013X.25. 21. 40.4369X1.2904. 
 
 9. 44.046 X. 43. 22. 100.0008 X-000306. 
 
 10. 3.6051X4.1. 23. 75.35060X62.3906. 
 
 11. 0.1003X6.12. 24. 31.50301 X 17.0352. 
 
 12. 8.0004X.004. 25. 0.000713X2.30561. 
 
 13. 35.601X1.032. 26. 42.10062X3.821013. 
 
 14. 213.02X4.318. 27. 1.0142034X0620034. 
 
 15. 0.0006 X. 00012. 28. 25067823 X. 0000001 
 
 16. 0.3005X-0035. 29. 64.301257X1.000402. 
 
 17. 10.2106X38.26. 30. 394.20023 X-00000003. 
 
 18. 164.023X1.678. 31. 2564.21035X4.300506. 
 
 19. 9.40061X15.812. 32. 840003.1709X112.10371. 
 
 20. 7.31042X10.021. 33. 0.834567834X-00000008, 
 
 CONTRACTIONS IN MULTIPLICATION OF DECIMALS. 
 CASE I. 
 
 325. When the multiplier is 10, 100, 1000, &c., the multi- 
 plication may be performed by simply removing flip decimal point 
 as many places towards the right, as there are ciphers in the mul- 
 tiplier. (Arts. 99, 3'24.) 
 
 1. Multiply 85.4321 by 100. Ans. 8543.21. 
 
 2. Multiply 42930.213401 by 10. 
 
 3. Multiply 1067.2350123 by 100. 
 
 4. Multiply 608.34017 by 1000. 
 
 5. Multiply 30.467214067 by 10000. 
 
 6. Multiply 446.3214032 by 100000. 
 
 7. Multiply 21.3456782106 by 100000. 
 
 8. Multiply 5 tenths by 1000. 
 
 9. Multiply 75 hundredths by 100000. 
 
 10. Multiply 65 ten thousandths by 1000 
 
 11. Multiply 48 hundred thousandths by 100000. 
 
 QUEST. 325. How proceed when the multiplier is 10, 100, fee. 1 
 
202 
 
 MULTIPLICATION OF 
 
 [SECT. IX. 
 
 12. Multiply 248 thousandths by 10000. 
 
 13. Multiply 381 ten thousandths by 10000. 
 
 14. Multiply 6504 ten millionths by 100000. 
 
 15. Multiply 834 thousandths by 1000000. 
 
 16. Multiply 1 millionth by 10000000. 
 
 CASE II. 
 
 326. When the number of decimal places in the multiplier 
 and multiplicand is large, the number of decimals in the product 
 must also be large. But decimals below the fifth or sixth ord^r, 
 express so small parts of a unit, that when obtained, they r 
 commonly rejected. It is therefore desirable to avoid the imno- 
 cessary labor of obtaining those which are not to be used. 
 
 17. It is required to multiply 1.3569 by .36742, and retain 
 five places of decimals. 
 
 First Operation. 
 1.3569 
 .36742 
 
 .40707 
 
 8141 
 
 949 
 
 54 
 
 83 
 
 276 
 
 7138 
 
 .49855 2198 Ans. 
 
 It is evident from the nature of decimal 
 notation, that if the partial product of 
 each figure in the multiplier is advanced* 
 one place to the right instead of the left, 
 the operation will correspond with the de- 
 scending scale, and at the same time will 
 give the true product. (Art. 86. Obs. ?.) 
 But since only five decimals are required, 
 those on the right of the perpendicular 
 are useless. Our present object i? tc 
 show how the answer can be obtained without them. 
 
 Beginning at the right hand, we will firsf 
 multiply the multiplicand by the tenths' figure 
 of the multiplier, and place the first figure of 
 the partial product under the figure multiplied. 
 In obtaining the second partial product, (i. e. 
 multiplying by 6,) it is plain we may omit the 
 right hand figure of the multiplican 1, for if 
 multiplied, its product will fall to the right of 
 the perpendicular line, and therefore will not 
 
ARTS. 326, 327.] DECIMALS. 203 
 
 Le used. But if we multiply 9 into 6, the product will be 54 ; 
 consequen ly there would be 5 to carry to the next product ; we 
 therefore 3arry 5 to 36, which makes 41. Again, in (hi third 
 partial product, (i. e. in multiplying by 7,) we may omit the two 
 right hand figures of the multiplicand ; for, their product will fall 
 to the right of the perpendicular line. But by recurring to the 
 rejected figures, it will be seen that the product of 7 into 6 is 42, 
 and 6 to carry make 48 ; we therefore add 5 to the product of 
 ? 'nto 5, because 48 is nearer 50 than 40 ; consequently it is 
 nearer the truth to carry 5 than to carry 4. In the fourth partial 
 product we may omit the three right hand figures, and in the fifth 
 or lant, the four light hand figures. 
 
 18 Multiply .235*6 by .3765, and retain 4 decimals in the 
 product. 
 
 Operation. Multiplying as before, the first figure of the 
 
 .2356 partial product must be set in the fifth order, 
 
 .3765 or one place to the right of the figure multi- 
 
 .0707 plied ; for, there are 4 decimals in the multipli- 
 
 165 cand and the one by which we multiply makes 
 
 14 5. (Art. 324.) But since we wish to retain 
 
 1 only 4 decimals in the product, we may omit 
 
 .0887 Ans. this figure, carrying 2 to the next product. 
 
 Proceed in the same manner with the other 
 
 figures in the multiplier. Finally, the sum of the partial products 
 
 which are retained, is the answer required. Hence, 
 
 327. To multiply decimals and retain only a given number 
 of decimal figures in the product. 
 
 Count off in the multiplicand as many decimal places less out,, 
 as are required in the product. Then beginning at the right hand, 
 figure counted off, multiply the multiplicand by tJie tenths or first 
 decimal figure of the multiplier, and set the first figure of the 
 partial product one place to the right of the figure multiplied, in- 
 creasing it by the nearest number of tens tJiat would arise from the 
 
 UUBST. -327. How multiiAj decimals, and retain a given number of figures in the product? 
 
201 MULTIPLICATION OF [SECT. IX 
 
 rejected j\ (jure if multiplied. Next multiply by the second decimal 
 figure., omitting the next right hand figure of the multiplicand and 
 carrying as before. Proceed in the same manner with all t/M figures 
 of the multiplier whose product will come under the decimal places 
 counted off, omitting an additional figure on the right of tJie mul- 
 tiplicand, as you multiply by each successive figure, and set the 
 first figure of each partial product under that of the preceding. 
 Finally, from the sum of the partial products, cut of the required 
 nvmber of decimals, and the result will be the answer. 
 
 OBS. 1. In order to determine where to place the decimal point in the prod- 
 uct, we have only to observe that the product of the right hand figure of the 
 multiplicand into the tenths of the multiplier is of the order denoted by the stun 
 of the orders of the two figures multiplied ; (Art. 324 ;) and when the multi- 
 plier is tenths it is of the order next lower than the figure multiplied. For this 
 reason the first partial product is set one place to the right of the figure multi- 
 plied. But since we count off one decimal less than is required in the prod- 
 uct, the right hand figure in the sum of the partial products must consequently 
 be the right hand decimal place in the answer. 
 
 2. If the multiplier contains units, lens, hundreds, &c., in multiplying by the 
 units, we must begin one figure to the right of those counted off, and set the 
 first figure of the partial product under the figure multiplied. In multiplying 
 by the tens, we must begin two figures to the right of those counted off, and 
 set the first figure of the partial product under that of the units; in multiply- 
 ing by the hundreds, we must begin three figures to the right, and set the first 
 figure of the partial product under that of the preceding, &c. This will bring 
 the same orders under each other. 
 
 19. Multiply .72543414 by .24826421 retaining 5 decimal 
 places in the product. 
 
 Operation. 
 
 .7254'3414 Having counted off 4 decimals in the mul- 
 
 .2482 6421 tiplicand, increase the product of 2 into 4 
 
 1450 9 by 1, because the product of the 3 rejected 
 
 290 2 into 2 is nearer 10 than 0. Set the 9 one 
 
 58 place to the right of the figure multiplied. 
 
 1 4 The 4 in the last partial product, is the 
 
 4 number which would be carried to this order, 
 
 .1800 9 Ans. if the V were multiplied by 6. 
 
ARTS. 327, 328.] DECIMALS. 205 
 
 20. Multiply 67.1498601 by 92.4023553 retaining fear deci- 
 mals in the product. 
 
 Operation. 
 
 67.149'8601 In this operation we multiply first by the 
 
 92.402 3553 tens figure of the multiplier, beginning two 
 
 6043.487 4 places to the right of those counted off in the 
 
 134 299 7 multiplicand. It is immaterial as to the re- 
 
 26 859 9 suit whether we multiply by the tenths first, 
 
 ] 34 3 or by the units, tens, or hundreds, provided 
 
 201 we set the first figure of the partial product in 
 
 3 4 its proper place. (Art. 327. Obs. 2.) 
 
 3_ 
 
 6204.805 1 Ans. 
 
 21. Multiply .863541 by .10983 retaining 5 decimal places. 
 
 22. Multiply 1.123674 by 1.123674 retaining 6 decimal places. 
 
 23. Mid ti ply .26736 by. 28758 retaining 4 decimal places. 
 
 24. Multiply .1347866 by .288793 retaining 7 decimal places. 
 
 25. Multiply .681472 by .01286 retaining 5 decimal places. 
 
 26. Multiply .053407 by .047126 retaining 6 decimal places. 
 
 27. Multiply .3857461 by .0046401 retaining 6 decimal places. 
 
 DIVISION OP DECIMAL FRACTIONS. 
 
 328* Ex. 1. How many bushels of oats, at .2 of a dollar a 
 6ushel, can you buy for .84 of a dollar ? 
 
 Analysis. Since 2 tenths of a dollar will buy 1 bushel, 84 
 hundredths of a dollar will buy as many bushels, as 2 tenths is 
 contained times in 84 hundredths. Now .84=T 8 ^ r ; and .2=-^, 
 ortW- (Art. 191.) And VW-r-iV^A, or 4-ft. But, (Art. 311,) 
 .4 1 2 -=4.2, which is the answer required. 
 
 Operation. 
 
 .2). 84 We divide as in whole numbers, and point off 
 
 4.2 Ans. one decimal figure in the quotient. 
 
 OBS. The reason for pointing off one decimal figure in the quotient may be 
 thus explained. 
 
 We have seen in the multiplication of decimals, that the product has aa 
 <nany decimal figures, as th< multiplier and multiplicand. \A.r " V M Now 
 
206 DIVISION OF [SECT. IX. 
 
 since the dividend is equal to the product of the divisor and quotient, (Art. 112,) 
 it follows that the dividend must have as many decimals as the divisor and 
 qtwtient together ; consequently, as the dividend has Iwo decimals, and the 
 divisor but one, we must point off one in the quotient. In like manner it may 
 be shown universally, that 
 
 329* The quotient must have as many decimal figures, as the 
 decimal places in the dividend exceed those in the divisor ; tJiat is, 
 *he decimal places in the divisor and quotient together, :nu$t be 
 qual in number to those in the dividend. 
 
 2. What is the quotient of 3.775 divided by 2.5 ? Ans. 1.51. 
 
 3. What is the quotient of .0072 divided by 2.4. 
 
 Operation. Since the dividend has three decimals 
 
 2.4).0072(.003 Ans. more than the divisor, the quotient must 
 
 72 have three decimals. But as it has but> 
 
 one figure, we prefix two ciphers to it to 
 
 make up the deficiency. 
 
 OBS. It will be noticed that 3, the first figure of the quotient, denotes 
 sandths; also the product of 2, the units figure of the divisor, into the first quo- 
 tient figure, is written under the thousandths in the dividend. Hence, 
 
 The first figure of tJie quotient is of the same order, as that 
 figure of the dividend under which is placed the product of the 
 units of the divisor into the first quotient figure. 
 
 33O. From the preceding illustrations we deduce the follow- 
 ing general 
 
 RULE FOR DIVISION OF DECIMALS. 
 
 Divide as in whole numbers, and point off as many figures for 
 decimals in the quotient, as the decimal places in the dividend exceed 
 thfise in tlie divisor. If the quotient does not contain figures enouoh t 
 supply the deficiency by prefixing ciphers. 
 
 PROOF. Division of Decimals is proved in the same manner of 
 Simple Division. (Art. 121.) 
 
 OBS. 1. When the number of decimals in the divisor is the same as thtt IB 
 the dividend, the quotient will be a whole number. 
 
 Q.VCST. 330. How are decimals divided How point off the quotient 1 How is division 
 of decimals proved ? 
 
ARTS. 329, 330.] 
 
 DECIMALS. 
 
 207 
 
 2. When there are more decimals in the divisor than in the dividend, annex 
 as many ciphers to the dividend as are necessary to make its decimal places 
 equal to th )se in the divisor. The quotient thence arising will be a whole 
 number. (Obs. I.) 
 
 3 After all the figures of the dividend are divided, if there is a remainder, 
 ciphers may be annexed to it and the invasion continued at pleasure. The 
 cipbsrs annexed must be regarded as decimal places belonging to the dividend. 
 
 Note. 1 For o binary purposes, it will be sufficiently exact to carry the 
 ^aotrnt *u three it four places of decimals ; but when great accuracy is re- 
 quire , ; must be carried farther. 
 
 2. W ^en there is a remainder, the sign -}-> should be annexed to the quo- 
 tient to show that it is not complete. 
 
 EXAMPLES. 
 
 4. How many boxes will it require to pack 71.5 Ibs. of butter, 
 if you put 5.5 Ibs. in a box ? 
 
 5. How many suits of clothes will 29.6 yds. of cloth make, al- 
 lowing 3.7 yds. to a suit? 
 
 6. If a man can walk 30.25 miles per day, how long will it take 
 him to walk 150.75 miles? 
 
 7. How many loads will 134642.156 Ibs. of hay make, allowing 
 1622.2 Ibs. for a load? 
 
 8. If a team can plough 2.3 acres in a day, how long will it 
 take to plough 63.75 acres? 
 
 9. How many bales of cotton in 56343.75 Ibs., allowing 375 Ibs. 
 to a bale ? 
 
 Divide the following decimals : 
 
 10. 46.84^-7.9. 
 
 11. 1.658H-.25. 
 
 12. 67234-T-.85. 
 
 13. 4.003346.31. 
 
 14. 73.8243 .061. 
 
 15 0.00033 .011. 
 
 16 236.041 1.75. 
 
 17 60.0001-1.0], 
 IF 300.40212.1. 
 
 4.32067 .001 
 
 20. 0.00006-.003. 
 
 21. 167342-^.002. 
 
 22. 684234.6-2682. 
 
 23. 0.000045 9. 
 
 24. 7.231068 .12. 
 
 25. 26.3845-K125. 
 
 26. 4-T-.00001. 
 
 27. 6-^.0000001. 
 
 28. 0.8 r.OOOOJ02. 
 
 29. 6541.234567-^21. 
 
 Obs. When the number of decimai places in the divisor is equal to that in the 
 dividend -hat is the quotient ? When there are more decimals in the divisor than in the 
 mvlaem' *ow proceed 1 When there is a remainder, what may be done ? 
 
208 
 
 DIVISION OF 
 
 [SECT. IX 
 
 CONTRACTIONS IN DIVISION OF DECIMALS. 
 CASE I. 
 
 331. When the divisor is 10, 100, 1000, &c., the dirision 
 may be performed by simply removing tJie decimal point in t~e 
 dividend as many places towards the left, as there are ciphers in 
 the divisor, and it will be the quotient required. (Arts. 131, 330.) 
 
 1. Divide 4672.3 by 100. Ans. 46.723. 
 
 2. Divide 0.8 by 10000. Ans. 0.00008. 
 
 3. Divide 672345.67 by 10. 
 
 4. Divide 10342.306 by 100. 
 
 5 Divide 42643.621 by 100000. 
 
 6. Divide 6723000.45 by 100UOOO. 
 
 *;. Divide 1.2300456 by 100000. 
 
 8. Divide 2.0076346 by 1000000. 
 
 CASE II. 
 
 332. When the divisor contains a large number of decimal 
 figures, the process of dividing may be very much abridged. 
 
 9. It is required to divide 3.2682 by 2.4736, and carry the 
 quotient to four places of decimals. 
 
 Common Method. 
 
 2.4736)3.2682(1.3212 
 
 2 4736 
 
 7946 
 
 7420 
 
 525 
 
 494 
 
 ~30 
 
 24 
 
 8 
 2~0 
 
 72 
 
 480 
 
 736 
 
 7440 
 9472 
 7968 
 
 Explanation. We perceive the first figure of the ijuotient Trill 
 be a whole number ; for the number of decimals in the divisor is 
 
 Qi STi 331. When the divisor is 10, 100, 1000, &c., how may the division be performed 1 
 
ARTS. 331-333. J DECIMALS. 209 
 
 equal to that of the dividend. (Art. 330. Obs. 1.) Now to obtain 
 the decimals required, instead of annexing a cipher to the several 
 remainders, which multiplies them respectively by 10, (Art. 98,) 
 we may cut off a figure on the right of the divisor at each division, 
 which is the same as dividing it successively by 10. (Art. 130.) 
 When we multiply the divisor by 3, the second quotient figure, 
 we carry 2 to the product of 3 into 3, because the product of 3 
 into 6, the figure omitted in the divisor, is nearer 20 than 10. 
 (Art 327.) We carry on the same principle to the first figure of 
 each product of the divisor into the respective quotient figures. 
 Hence, 
 
 333. To divide decimals, carrying the quotient to any re- 
 quired number of decimal places. 
 
 For the first quotient figure divide as usual ; then instead of 
 bringing down the next figure, or annexing a cipher to the remain- 
 der, cut off a figure on the right of the divisor at each successive 
 division, and divide by the other figures. In multiplying the divisor 
 by the quotient figure, carry for the nearest number of tens that 
 would arise from the product of the figure last cut off into the fig- 
 ure last placed in the quotient. (Art. 327.) 
 
 OBS. 1. The reason for this contraction may be seen from the principle, that 
 a tenth of the given divisor is contained in a.lenth of the dividend, just as many 
 times as the whole divisor is contained in the whole dividend; (Art. 145;) for, 
 cutting off a figure on the right of the divisor, and omitting to annex a cipher 
 to the dividend or remainder, is dividing each by ten. (Art. 130.) 
 
 2. When the divisor has more figures than the quotient is required to have, 
 including the whole number and decimals, we may take as many on the left 
 of the divisor as are required in the quotient, and divide by them as above. 
 
 3. If the divisor does not contain so many figures as are required in the quo- 
 tient, we must divide in the usual way, until we obtain enough figures to make 
 up this deficiency, and then begin the contraction. 
 
 10. Divide .4134 by .3243, and carry the quotient to four 
 places of decimals. 
 
 11 Divide .079085 by .83497, and carry the quotient to five 
 places of decimals. 
 
 12. Divide 2.3748 by 1.4736, and carry the quotient to three 
 places of decimals. 
 
210 REDUCTION OF [SECT. IX. 
 
 13. Divide .3412 by 8.4736, and carry the quotient to five 
 places of decimals. 
 
 14. Divide 1 by 10.473654, and carry the quotient to seven 
 places of decimals. 
 
 15. Divide .4312672143 by .2134123406, and carry the quo- 
 tient to four places of decimals. 
 
 16. Divide .879454 by .897, and carry the quotient to six 
 places of decimals. 
 
 REDUCTION OP DECIMALS. 
 
 CASE I. 
 
 334:* Decimals reduced to Common Fractions. 
 Ex. 1. Change the decimal .75 to a common fraction. 
 
 Suggestion. Supplying the denominator, ,75=- 1 -V L -. (Art. 311.) 
 Now -j 2 ^- is expressed in the form of a common*fraction, and, as 
 such, may be reduced to lower terms, and be treated in the same 
 manner as any other common fraction. Thus, T'J/V =-&, or -f-. 
 
 335. Hence, To reduce a Decimal to a Common Fraction. 
 
 Erase the decimal point ; then write the decimal denominator 
 under the numerator, and it will form a common fraction, which 
 may be treated in the same manner as other common fractions. 
 
 2. Change .225 to a common fraction, and reduce it to the 
 lowest terms. Ans. -f$. 
 
 3. Reduce .125 to a common fraction, &c. 
 
 4. Reduce .95 to a common fraction, &c. 
 
 5. Reduce .435 to a common fraction, <fec. 
 
 6. Reduce .575 to a common fraction, <fcc. 
 
 7. Reduce .656 to a common fraction, <fec. 
 
 8. Reduce .204 to a common fraction, &c. 
 
 9. Reduce .075 to a common fraction, &c, 
 10. Reduce .012 to a common fraction, &c. 
 
 "11. Change .0025 to a common fraction, &c. 
 12. Change .1001 to a common fraction, &c. 
 
 QUEST. 335. How are Decimals reduced to Common Fractions 1 
 
ARTS. 334-337.] DECIMALS. 21 1 
 
 13. Change .1844 to a common fraction, &c. 
 
 14. Change ,0556 to a common fraction, &c. 
 15 Change .1216 to a common fraction, &c. 
 
 16. Change .2005 to a common fraction, &c. 
 
 17. Change .0015 to a common fraction, &c. 
 
 CASE II. 
 
 336. Common Fractions reduced to Decimals. 
 Ex. 1 . Change f to a decimal. 
 
 Suggestion. Multiplying both terms by 10, tlie fraction be- 
 comes -f^-. Again, dividing both terms by 5, it becomes -fa. 
 (Art. 191 .) But -A=.8, which is the decimal required. (Art. 311.) 
 
 Note. Since we make no use of the denominator 10 after it is obtained, we 
 may omit the process of getting it ; for, if we annex a cipher to the numerator 
 and divide it by 5, we shall obtain the same result. 
 
 Operation. 
 
 5)4.0 A decimal point is prefixed to the quotient to 
 
 .8 Ans. distinguish it from a whole number. 
 2. Reduce -f to a decimal. Ans. 0.62*5, 
 
 337* Hence, to. reduce a Common Fraction to a Decimal. 
 
 Annex ciphers to the numerator and divide it by the denominator. 
 Point off an many decimal figures in the quotient, as you Imve an- 
 nexed ciphers to the numerator. 
 
 OBS. 1. If there are not as many figures in the quotient as you have an- 
 nexed ciphers to the numerator, supply the deficiency by prefixing ciphers to 
 the quotient. 
 
 2. The reason of this rule may be illustrated thus. Annexing a cipher to 
 the numerator multiplies the fraction by 10. (Arts. 98, 18*} ) If, therefore, the 
 numerator with a cipher annexed to it, is divided ' the \ nominator, the quo- 
 tient will obviously be ten times too large. (Art. 141.) Hence, in order to ob- 
 tain the true quotient, or a decimal equal to the given fraction, the quotient 
 thus obtained must be divided by 10, which is done by pointing off' one figure. 
 (Art. 131.) Annexing 2 ciphers to the numerator multiplies the fraction by 
 100; annexing 3 ciphers by 1000. (fee., consequently, when 2 ciphers are an- 
 nexed, the quotient will bv 1 , 100 times too large, and must therefore be divided 
 bi 100; when three ciphers are annexed, the quotient will be 1000 times too 
 I'ge, and must be divided by 1000, &c. (Art. 131.) 
 
 T.II. 
 
 Q.UKST. 337. How are Common Fractions reduced to Decimals ? 
 
 10 
 
4. . 
 
 8. i. 
 
 12. |. 
 
 5. i. 
 
 9. f . 
 
 13. i. 
 
 6. i. 
 
 10. f. 
 
 14. 1. 
 
 1. f- 
 
 11. f. 
 
 15. i 
 
 20. Reduce - t 
 
 } to a decimal. 
 
 Ans. 
 
 21. Reduce - 
 
 sVr to a decimal. 
 
 Ans. 
 
 212 REDUCTION OF [SECT. IX, 
 
 3. Reduce -HP to a mixed num^r. Ans. 1.0625. 
 
 Reduce the following fractions to decimals : 
 
 16. -\. 
 
 17. |. 
 
 18. f. 
 
 19. i. 
 
 . 0.166666, &c. 
 Ans. OJ 23123123, &c. 
 
 338. It will be seen that the last two examples cannot be 
 exactly reduced to decimals ; for there will continue to be a re- 
 mainder after each division, as long as we continue the operation. 
 
 In the 20th, the remainder is always 4 ; in the 21st, after ob- 
 taining three figures in the quotient, the remainder is the same as 
 the given numerator, and the next three figures in the quotient 
 are the same as the first three, when the same remainder will re- 
 cur again. The same remainders, and consequently the same fig- 
 ures in the quotient, will thus continue to recur, as long as tha : 
 operation is continued. 
 
 339* Decimals which consist of the same figure or set of fig- 
 ures continually repeated, as in the last two examples, are called 
 Periodical or Circulating Decimals ; also, Repeating Decimals, or 
 Repetends. 
 
 OBS. When only a single figure is repeated, it is more accurate to call them 
 repeating decimals ; but when two or more figures recur at regular intervals, 
 they are very properly called periodical, or circulating decimals. 
 
 34O When a common fraction can be exactly expressed by a 
 decimal, the decimal is said to be terminate, or finite ; but when 
 it cannot be exactly expressed by a decimal, it is said to be inter' 
 l linate, or infinite. 
 
 OBS. It seems to be incongruous to call a fraction infinite. (Art. 180.) The 
 term infinite, however, does not refer to the vafaie of the fraction, but 'o the 
 number Df decimal figures required to express its value. 
 
 QricsT. Obs. When there are not so many figures in the quotient as you have annexed 
 ciphers, what is to he done 1 339. What are periodic* i or circulating dec jiials 1 310. Wh 
 is a decimal terminate 1 When mterminate ? 
 
AIITS. 338-343.] DECIMALS. 213 
 
 3 4 1 If the denominator of a common fraction when reduced 
 to its lowest term, contains no prime factors but 2 and 5, its 
 equivalent decimal will terminate ; on the other hand, if it con- 
 tains any other prime factor besides 2 and 5, it will not terminate. 
 
 Thus e\ reduced to its lowest terms, becomes -fa, and the prime 
 factors of 20 are 2, 2, and 5; that is, 2(3=2X2X5. (Art. 165.) 
 We also find that ^=.05 ; it is therefore terminate. Again, 
 fs~--^s 5 an d the prime factors of 15 are 3 and 5 ; that is, 15 = 
 8 K 5 ; and -^-=.0666666, &c. ; it is therefore interminate. Hence, 
 
 342. To ascertain whether a common fraction can be exactly 
 expressed by decimals. 
 
 Reduce the given fraction to its lowest terms, and then resolve its 
 denominator into its prime factors. (Art. 341.) 
 
 OBS. The truth of this principle is evident from the consideration, that an- 
 nexing ciphers to the numerator, multiplies it successively by 10 ; but 2 and 5 
 are the prime factors of 10, and are the only numbers that can divide it without 
 a remainder. (Art. 165. Obs. 2.) But any number that measures another, 
 must also measure its product into any whole number; (Art. 161. Prop. 14;) 
 consequently, when the denominator contains no prime factors but 2 and 5, 
 the division will terminate ; but when it contains other factors, the division can 
 n<jt terminate. 
 
 22. Will 5^ produce a terminate or interminate decimal? 
 
 23. Will -f^- produce a terminate or interminate decimal ? 
 
 24. Will -- produce a terminate or interminate decimal ? 
 
 25. Will -V produce a terminate or interminate decimal? 
 20. Will 2-fir produce a terminate or interminate decimal? 
 
 27. Will -3^6- produce a terminate or interminate decimal? 
 
 28. Will -VV produce a terminate, or interminate decimal? 
 
 343* When the decimal is terminate, the number of figures it 
 contains, must be equal to the greatest number of times that either 
 of the prime factors 2 or 5, is repeated in its denominator, whea 
 the given fraction is reduced to its lowest terms. 
 
 Osa. The truth of this principle may be illustrated thus : * = .5; that is ie 
 decimal terminates with one place ; for, the denominator 2. is taken only ^nce 
 as a factor in 10, and therefore only one cipher is required to be annexed t^the 
 numerator to reduce it. Again, ^ = .25, which contains two decimal pl&ces. 
 Now the denominator 4 2X.2 ; and since 2 is contained only once as a factor 
 
214 REDUCTION OF [SECT. IX. 
 
 in 10. it is evident that 10 must be repeated as many times as a factor in the 
 numerator, as 2 is taken times as a factor in the denominator, in order to re- 
 duce the fraction. 
 
 For the same reason J will terminate with three places, and is equal to .125; 
 for, 8.=2X2X2. So ^-.2 ; that is, the decimal terminates with one place ; for, 
 since its denominator 5, is taken only mice as a factor in 10, it js necessary to 
 add only one cipher to its numerator in order to reduce it. In like manner it 
 may be shown that the number of figures contained in any terminate decimal, 
 w equal to the greatest number of times that either of the prime factors 2 or 5, 
 i* repeated in the denominator of the given fraction. 
 
 The same reasoning will evidently hold true when the numerator is 2, 3, 4, 
 &, &c., or any number greater than 1. In this case the decimal will be as 
 many times greater, as the numerator is greater than 1. 
 
 344. The number of figures in the period must always be owe 
 less than there are units in the denominator ; for, the number of 
 remainders different from each other which can arise from any 
 operation in division, must necessarily be one less than the units 
 in the divisor. For example, in dividing by 7, it is evident, the 
 only possible remainders are 1, 2, 3, 4, 5, and 6 ; and since in re- 
 ducing a common fraction to a decimal, a cipher is annexed to 
 each remainder, there cannot be more than six different dividends ; 
 consequently, there cannot be more than six different figures in 
 the quotient. Thus, -$-=.142857,142857, &c - 
 
 When the decimal is periodical or circulating, it is custom- 
 ary to write the period but once, and put a dot, or accent over 
 the first and last figure of the period to denote its continuance. 
 Thus, .46135135135, &c., is written .46135, and .633333 &c., .63. 
 
 Reduce the following fractions to circulating decimals : 
 
 31. i. 36. f. 41. f. 46, . 
 
 32. |. 37. |. 42, f. 47. f. 
 
 33. i. 38. f. 43. i. 48. . 
 
 34. f. 39. f. 44. f. 49. . 
 
 35. 'f. 40. -f. 45. f. 50. -f. 
 
 51. How many decimal figures are required to express -^7 
 
 52 How many decimal figures are required to express -fa ? 
 
 53 How many decimal figures are required to express ^$- 5 ? 
 54. How many decimal figures are necessary to express yj-j 
 
ARTS. 344-346.] DECIMALS. 215 
 
 55 How many decimal figures are necessary to express -J-J- ? 
 66. How many decimal figures are necessary to express ToW 
 
 57. Reduce -& to a decimal. 59. Reduce ^ir to a decimal. 
 
 58. Reduce -fe to a decimal. 60. Reduce f to a decimal. 
 
 Note. For the method of finding the value of periodical decimals, or of re- 
 ducing them to Common Fractions, also of adding, subtracting, multiplying t 
 and dividing them, see the next Section. 
 
 CASE III. 
 
 345* Compound Numbers reduced to Decimals. 
 Ex. 1. Reduce 13s. 6d. to the decimal of a pound. 
 
 Analysis. 13s. 6d.=162d., and l = 240d. (Art. 281.) Now 
 162d. is i^ of 240d. The question therefore resolves itself into 
 this : reduce the fraction -J^ to decimals. Ans. .6*75. Hence, 
 
 340. To reduce a compound number to the decimal of a 
 higher denomination. 
 
 First reduce the given compound number to a common fraction / 
 (Art. 295 ;) then reduce the common fraction to a decimal. (Art. 337.) 
 
 2. Reduce 4s. 9d. to the decimal of 1. Ans. .237+. 
 
 3. Reduce 10s. 9d. to the decimal of l. 
 
 4. Reduce 16s. 6d. to the decimal of l. 
 
 5. Reduce 17s. 7d. to the decimal of l. 
 
 6. Reduce 5d. to the decimal of a shilling. 
 
 7. Reduce 6d. to the decimal of a shilling. 
 
 8. Reduce 37 rods to the decimal of a mile. 
 
 9. Reduce ? fur. 2 "ods to the decimal of a mile. 
 
 10. Reduce j.o ittm. oO >ec. to tne decimal of an hour. 
 1 1 . Reduce 3 hrs. 3 min. to the decimal of a day. 
 
 12. Reduce 5 Ibs. 4 oz. to the decimal of a cwt. 
 
 13. Reduce 7 oz. 8 drams to the decimal of a pound. 
 
 14. Reduce 3 pks. 4 qts. to the decimal of a bushel. 
 
 15. Reduce 4 qts. 1 pt. to the decimal of a peck. 
 
 16. Reduce 4 qts. 1 pt to the decimal of a gallon. 
 
 QUEST. 346 How is a compound number reduced to the decimal of a b\gher dcnora 
 inatiim 1 
 
216 REDUCTION OF DECIMALS. [SECT. IX. 
 
 CASE IV. 
 347* Decimal Compound Numbers reduced to whole ones. 
 
 I. Reduce .387 to shillings, pence and farthings. 
 
 Operation. Multiply the given decimal by 20, because 
 .387 20s. make l, and point off as many figures 
 20 for decimals, as there are decimal places in 
 shil. 7.740 the multiplier and multiplicand. (Art. 330.) 
 12 The product is in shillings and a decimal 
 pence 8.880 of a shilling. Then multiply the decimal 
 4 of a shilling by 12, and point off as be- 
 far. 3.520 fore, &c. The numbers on the left of the 
 Ans. 7s. 8d. 3 far. decimal points, viz : 7s. 8d. 3 far., form the 
 answer. Hence, 
 
 348* To reduce a decimal compound number to whole num- 
 bers of lower denominations. 
 
 Multiply the given decimal by that number which it takes of the 
 next lower denomination to make ONE of this higher, as in reduction, 
 and point off the product, as in multiplication of decimal fractions. 
 (Art. 330.) Proceed in this manner with the decimal figures of 
 each succeeding product, and the numbers on the left of the decimal 
 point of the several products, will be the whole number required. 
 
 2. Reduce .725 to shillings, pence and farthings. 
 
 $, Reduce .1325 to shillings, &c. 
 
 4 Reduce .125s. to pence and farthings. 
 
 5. Reduce .825s. to pence and farthings. 
 
 6. Reduce .125 cwt. to pounds, &c, 
 
 7. Reduce .435 Ibs. to ounces and drains. 
 
 8. Reduce .275 miles to rods, <fec. 
 
 9. Reduce .4245 rods to feet, &c, 
 
 10. Reduce .1824 hhds.to gallons, &c. 
 
 II. Reduce .4826 gal. to qts., &c. 
 
 12. Reduce .4258 day to hours, &c. 
 
 13. Reduce .845 hr. to minutes and seconds. 
 
 QnsT. 3-18. How are decimal compound numbers reduced to whole ones of a lowar 
 denomination! 
 
ARTS. 347-353.] CIRCULATING DECIMALS. 217 
 
 SECTION X. 
 PERIODICAL OR CIRCULATING DECIMALS.* 
 
 ART. 349. Decimals which consist of the same figures or set 
 of figures repeated, are called PERIODICAL, OR CIRCULATING DECI- 
 IIJLS. (Art. ,339.) 
 
 35O. The repeating figures are called periods, or repetends. 
 If one figure only repeats, it is called a single period, or repetend ; 
 as .11111, &c. ; .33333, &c. 
 
 When the same set of figures recurs at equal intervals, it is 
 called a compound period, or repetend; as .01010101, &c. ; 
 .123123123, &c. 
 
 351* If other figures arise before the period commences, the 
 decimal is said to be a mixed periodical ; all others are called 
 pure, or simple periodicals. Thus .42C31G31, &c., is a mixed 
 periodical ; and .33333, &c., is a pure periodical decimal. 
 
 OBS. 1. When a pure circulating decimal contains as many figures as there 
 are units in the denominator less one, it is sometimes called a perfect period, 
 or repetend, (Art. 344.) Thus, i=.14'2857, &c., and is a perfect period. 
 
 2. The deciiral figures which precede the period, are often called the ter- 
 minate part of the fraction. 
 
 352. C ; rculating decimals are expressed by writing the period 
 once with a dot over its first and last figure when compound ; and 
 when singl* by writing the repeating figure only once with a dot 
 over it. Thus .40135135, &c., is written .40135 and .33, etc., .3. 
 
 353^. Similar periods are such as begin at the same place or 
 distance from the decimal point ; as .1 and .3, or 2.34 and 3.70, &c. 
 
 Dissimilar periods are such as be^in at different places ; as 
 .123 and .42325 
 
 Similar and conterminous periods are such as begin and end in 
 the same places; as .2321 and 1034. 
 
 * Should Periodical Decimals be deemed too intricate for younger classes, they can be 
 omitted till review 
 
218 PERIODICAL OR [SECT. X 
 
 REDUCTION OP CIRCULATING DECIMA1 S. 
 
 CASE I. To reduce pure circulating decimals to common frac- 
 tions. 
 
 354* Tc investigate this problem, let us recur to the origin 
 of circulating decimals, or the manner of obtaining them. For 
 example, -=.11111, <fec., or .1 ; therefore the true value of .11111, 
 &c., or .1, must be i from which it arose. For the same reason 
 .22222, &c., or .2, must be twice as much or f ; (Art. 186;) 
 .33333, (fee., or .3=f; .4=f; .5=%, (fee. 
 
 Again, -gV=.010101, &c., or .61 ; consequently .010101, &c., 01 
 .oi^-g-V; .020202, <fec., or .02=-&; .030303, <fec., or .03=tV 
 .070707, &c., or .67=VV> &c. So also -9-^=. 00 100 1001, &c. 
 or .6oi ; therefore .001001, &c., or .6oi=-9-h- ; .602=-9-f-9 ; &c. 
 
 In like manner =.142857 ; (Art. 337;) and i42857=iiltf|; 
 for, multiplying the numerator and denominator of by 142857, 
 we have iff Iff . (Art. 191.) So f is twice as much as ; f, three 
 times as much, &c. Thus it will be seen that the value of a pure 
 periodical decimal is expressed by the common fraction whose 
 numerator is the given period, and whose denominator is as many 
 9s as there are figures in the period. Hence, 
 
 355. To reduce a pure circulating decimal to a common 
 fraction. 
 
 Make the given period the numerator, and the denominator will 
 be as many 9s as there are figures in the period. 
 
 Ex. 1. Reduce .3 to a common fraction. Ans. f, or f. 
 
 2. Reduce .6 to a common fraction. Ans. -f, or |. 
 
 3. Reduce .18 to a common fraction. 
 
 4. Reduce .123 to a common fraction. 
 
 5. Reduce .297 to a common fraction. 
 
 6. Reduce .72 to a common fraction. 
 
 7. Reduce .09 to a common fraction. 
 
 8. Reduce .045 to a common fraction. 
 
 9. Reduce .142857 to a common fraction. 
 10. Reduce .076923toa common fraction 
 
ARTS. 354 357. J CIRCULATING DECIMALS. 219 
 
 CASE II. To i educe mixed circulating decimals to common 
 fractions. 
 
 3 50. 11. Reduce .16 to a common fraction. 
 
 Analysis. Separating the mixed decimal into its terminate and 
 periodical part, we have .16 = .! +.06. (Art. 320.) Now .1 =fa ; 
 (Art. 312 ;) and .06=-&; for, the pure period .6=f, (Art. 351,) 
 and since the mixed period .06, begins in hundred ths* place, its 
 value is evidently only fa as much ; but f-MO=VV- (Art. 227.) 
 Therefore .16=-tV+"9 I o- Now -fa and - 9 \, reduced to a common 
 denominator and added together, make $, or . Ans. 
 
 OBS. In mixed circulating decimals, if the period begins in hundredths' place 
 it is evident from the preceding analysis that the value of the periodical part is 
 only -fa as much as it would be, if the period were pure or begun in tenths' 
 place ; when the period begins in thousandths' place, its value is only -j-J-g- part 
 as much, &c. Thus .6=f ; .06=f-s-10=-^- : .00(i=f-j-lOO= r f ir , &c. 
 
 357. Hence, the denominator of the periodical part of a 
 mixed circulating decimal, is always as many Os as there are fig- 
 ures in the period with as many cipher* annexed as there are deci- 
 mals in the terminate part. 
 
 12. Reduce .8567923 to a commop fraction. 
 
 Solution. Reasoning as before .8567923=-^ftr+ 9 ?nufl' Re- 
 ducing these two fractions to the least common denominator, 
 (Art. 261.) t\*oX99909H^-ilHru whose denominator is the same 
 as that of the other. Now 
 
 Contraction. To multiply by 99999, annex as many 
 
 8500000 ciphers to the multiplicand as there are 
 
 85 9s in the multiplier, &c. (Art. 105.) This 
 
 8499915 1st Nu. gives the numerator of the first fraction 
 
 67923 2d Nu. or terminate part, to which add the nu- 
 
 8567838 merator of the second or periodical part, 
 
 9999900 Ans. and the sum will be the numerator of the 
 
 answer. The denominator is the same as 
 
 that of the second or periodical part. 
 
 10* 
 
220 PERIODICAL OR [SECT. X. 
 
 Second Method. 
 
 8567923 the given circulating decimal. 
 
 85 the terminate part which is subtracted 
 8567838 the numerator of the answer. 
 
 Note. 1. The reason of this operation may be shown thus: 8567923ra 
 8500000-J-G7923 Now 8500000 85+67923 is equal to 8567923- -85. 
 
 2. Jt is evident that the required denominator is the same as that of the 
 periodical part; (Art. 357;) for, the denominator of the periodical part is the 
 least common multiple of the two denominators. Hence, 
 
 358. To reduce a mixed circulating decimal to a common 
 fraction. 
 
 Change Loth tlie terminate and periodical part to common frac- 
 tions 'separately, and their sum will be the answer required. 
 
 Or, from the given mixed periodical, subtract the terminate part, 
 and tlie remainder will be the numerator required. The denominator 
 is always as many 9s as there are figures in the period with as 
 many ciphers annexed as there are decimals in the terminate part. 
 
 PROOF. Change the common fraction back to a decimal, and if the 
 result is the same as the given circulating decimal, the work is right. 
 
 13. Reduce .138 to a common fraction. Ans. -J-JHJ-, or -^-. 
 
 14. Reduce .53 to a common fraction. 
 
 15. Reduce .5925 to a common fraction. 
 
 16. Reduce .583 to a common fraction. 
 
 17. Reduce .0227 to a commmon fraction. 
 
 18. Reduce .4745 to a common fraction. 
 
 19. Reduce .5925 to a common fraction. 
 
 20. Reduce .008497133 to a common fraction. 
 
 CASE III. Dissimilar periodicals reduced to similar and confer* 
 minous ones. 
 
 359* In changing dissimilar periods, or repetends, to similar 
 and conterminous ones, the following particulars require attention. 
 
 1, Any terminate decimal may be considered as interminate 
 by annexing ciphers continually to the numeratoj . Thus .46= 
 .460000, &c.=.460 
 
ARTS. 358-361. J CIRCULATING DECIMALS. 221 
 
 2. Any pure periodical may be considered as mixed, by taking 
 the given period for the terminate part, and making the given 
 period the intcrminate part. Thus .4G=.4G + .004G, (fee. 
 
 3. A single period may be regarded as a compound periodical. 
 Thus .3 may become .33, or .333 ; so .63 may be made .0333, 
 or .03033, &c. 
 
 4. A single period may also be made to begin at a lower order, 
 regarding its higher orders as terminate decimals. Thus .3 may 
 be made .33, or .3333, (fee. 
 
 5. Compound periods may also be made to begin at a lower 
 order. Thus .30 may be changed to .303, or .30303, (fee. ; or by 
 extending the number of .places .479 may be made .47979, or 
 .4797979, <fec. ; or making both changes at once .532 may be 
 changed to .5325325, (fee. Hence, 
 
 3 GO. To make any number of dissimilar periodical decimals 
 similar. 
 
 Move the points, so that each period shall begin at the same order 
 a r> the period which has the most figures in its terminate part. 
 
 21. Change 0.814, 3.20, and .083 to similar and conterminous 
 periods. 
 
 Operation. Having made the given periods simi- 
 
 0. 814 = 0. 81481481 lar, the next step is to make them con- 
 
 3.20 = 3.20202026 terminous. Now as one of the given 
 
 .083 = 0.08333333 periods contains 3 figures, another 2, 
 
 and the other 1, it is evident the new 
 
 periodical must contain a number of figures which is some multi- 
 ple of the number of figures in the different periods; viz: 3, 2, 
 and 1. But the least common multiple of 3, 2, and 1 is 6 ; there- 
 fore the new periods must at least contain figures. Hence, 
 
 36 ! To make any number of dissimilar periodical decimals, 
 similar and conterminous. 
 
 Firm make the periods similar ; (Art. 300;) then extend the fig- 
 ure;> of each to as many places, as there are units in the le^st com- 
 mon multiple of the NUMBER of periodical fiyures contained in each 
 \>f the given decimals. (Art. 176.) 
 
222 PERIODICAL OR T^ECT. X. 
 
 L' 
 
 22. Change 46.162, 5.26, 63.423, .486, and 12.5, to similar 
 and conterminous periodicals. 
 
 Operation. 
 
 46.162 = 46.16216216 The numbers of periodical figures in 
 
 5.26 = 5.26262626 the given decimals are 3, 2, 3, and 1 ; 
 
 03.423 = 63.42342342 and the least common multiple of 
 
 .486= 0.48666666 them is 6. Therefore the new periods 
 
 125 =12.50600006 must each have 6 figures; 
 
 23 Make .27, .3, and .045 similar and conterminous. 
 
 24 Make 4.321, 6.4263, and .6 similar and conterminous. 
 
 ADDITION OF CIRCULATING DECIMALS. 
 
 Ex. 1. What is the sum of 17. 23 +41. 2476 + 8. 61+1.5 + 
 35.423 ? 
 
 Operation. 
 
 Dissimilar. Sim. &. Conterminous. 
 
 17.23 =17.2323232 First make the given decimals sim- 
 
 41.2476 = 41.2476476 ilar and conterminous. (Art. 361.) 
 
 % 8.61 = 8.6161616 Then add the periodical parts as in 
 
 1.5 = 1.5000006 simple addition, and since there are 
 
 35.423 =35.4232323 6 figures in the period, divide their 
 
 Ans. 104.0i93648 sum D 7 999999; for this would be 
 
 its denominator, if the sum of the 
 
 periodicals were expressed by a common fraction. (Art. 355.) 
 Setting down the remainder for the repeating decimals, carry the 
 quotient 1 to the next column, and proceed as in addition of 
 whole numbers.. Hence, 
 
 362* We derive the following general 
 
 RULE FOR ADDING CIRCULATING DECIMALS. 
 
 first make the periods similar and conterminous, and find iheij 
 ^um as in Simple Addition. Divide this sum by as me ny 9s as 
 there are figures in the period, set the remainder under the figures 
 added for the period of the sum, carry the quotient to the 
 column, and proceed with the rest as in Simple Addition. 
 
ARTS. 362, 363.] CIRCULATING DECIMALS. 223 
 
 OBS. If the remainder has not so many figures as the period, ciphers must 
 be prefixed to make up the deficiency. 
 
 2. What is the sum of 24.132 + 2. 23+85.24 + 67.6 ? 
 
 3. What is the sum of 328.126+81.23+5.624+61.6*? 
 
 4. What is the sum of 31.62 + 7.824 + 8.392+.027 ? 
 
 5. What is the sum of 462.34 + 60.82 + 71.164 + .35 ? 
 
 6. What is the sum of 60.25+.84+6.435+.45+45.24 ? 
 
 7. What is the sum of 9.814+1.5+87.26+0,83+124.09? 
 
 8. What is the sum of 3. 6 + 78.3476 + 735. 3+.37S + . 27 + 
 187.4? 
 
 9. What is the sum of 5391.357+72.38+187.21+4.2965 + 
 217.8496+42.176+.523+58.30048? 
 
 10. What is the sum of . 162 + 134.09+2. 93 + 97. 26 + 3.769-SO 
 +99.083+1.5+.814? 
 
 SUBTRACTION OP CIRCULATING DECIMALS. 
 Ex. 1. From 52.86 take 8.37235. 
 
 Operation. We first make the given decimals sn i- 
 
 52.86=52.86868 lar and conterminous, then subtract as n 
 
 8.37235= 8.37235 whole numbers. But since the period tn 
 
 44.49632 tne lower line is larger than that above 
 
 it, we must borrow 1 from the next higher 
 
 order. This will make the right hand figure of the remainder one 
 less than if it was a terminate decimal. Hence, 
 
 363* We derive the following general 
 
 RULE FOR SUBTRACTING CIRCULATING DECIMALS. 
 
 Make the periods similar and conterminous, and subtrttt as in, 
 ivhole numbers. If the period in the lower line is larger than that 
 above it, diminish the right hand figure of the remainder ^y 1. 
 
 OBS. The reason for diminishing the right hand figure of the remainder by 
 1, if -he period in the lower line is larger than that above it, may be explained 
 thus: 
 
 When the period in the lower l ; ne is larger than that above it, we must evi- 
 dently borrow 1 from the next higher order. Now if the given decimals were 
 extended to a second period, in this period the lower number would also bo 
 
224 PERIODICAL OR [SECT. X. 
 
 larger than that above it, and therefore we must borrow 1. But having bor- 
 rowed 1 in the seoDnd period, we must also carry one to the next figure in the 
 lower line, or, what is the same in effect, diminish the right hand figure of the 
 remainder by 1. 
 
 2. From 85.62 take 13.76432. Ans. 71.86193. 
 
 3. From 476.32 take 84.7697. 
 
 4. From 3.8564 take .0382. 
 
 5. From 46.123 take 41.3. 
 
 6. From 801.6 take 400.75. 
 
 7. From 4.7824 take .87. 
 
 8. From 1419.6 take 1200.9. 
 
 9. From .634852 take .021. 
 
 10. From 8482.421 take 6031.035. 
 
 MULTIPLICATION OF CIRCULATING DECIMALS 
 Ex. 1. What is the product of .36 into .25 ? 
 
 Operation. We first reduce the given periodi- 
 
 .36=ff=-^ f cals to common fractions ; (Art. 357 ;) 
 
 .25 = -[\+9 5 u=-ff. then multiply them together. (Art 
 
 Now T 4 rXli=-9 9 9^ 219.) Finally, we reduce the product 
 
 But ^j^=.092 Ans. to a periodical decimal. Hence, 
 
 364* We derive the following general 
 
 RULE FOR MULTIPLYING CIRCULATING DECIMALS. 
 
 First reduce the given periodicals to common fractions, and mul- 
 tiply them together as usual. (Art. 219.) Finally, reduce the prod- 
 uct to decimals and it will be the answer required. 
 
 OBS. If the numerators and denominators have common factors, the opera- 
 tion may be contracted by canceling those factors before the multiplication ifl 
 psrformet:. (Art. 221.) 
 
 2. What is the product of 37.23 into .26 ? Ans. 9.928. 
 
 3. What is the product of .123 into 6 ? 
 
 4. What is the product of .245 into 7.3 ? 
 
 5. What is the product of 24.6 into 15.7 ? 
 
 6. What is the product of 48.23 into 16.13 ? 
 
ARTS. 364. 365.J CIRCULATING DECIMALS. 
 
 7. What is the product of 8574.3 into 87.5 ? 
 
 8. What is the product of 3.973 into .8 ? 
 
 9. What is the product of 49640.54 into .70503 ? 
 10. What is the product of 7.72 into .297 ? 
 
 DIVISION OF CIRCULATING DECIMALS. 
 
 Ex. 1. Divide 234.6 by .7. 
 
 Operation. We first reduce the divisor 
 
 234.6 = 234-f=- 2L | A and dividend to common frac- 
 
 .7=|- tions; (Art. 357 ;) and divide 
 
 Now -H^i^-H^Xf^-ft 3 - one by the other; (Art. 229 ;) 
 
 And -^^=301.714285 Ans. then reduce the quotient to a 
 
 decimal. (Art. 337.) Hence, 
 
 365* We derive the following general 
 
 RULE FOR DIVIDING CIRCULATING DECIMALS. 
 
 Reduce the divisor and dividend to common fractions ; divide 
 one fraction by the other, and reduce the quotient to decimals. 
 
 OBS. After the divisor is inverted, if the numerators and denominator* have 
 factors common to both, the operation may be contracted by canceling those , 
 fectors. (Art. 232.) 
 
 2. Divide 319.28007112 by 764.5. Ans. 0.4176325. 
 
 3. Divide 18.56 by .3. 
 
 4. Divide .6 by .123. 
 
 5. Divide 2.297 by .297. 
 
 6. Divide 750730.518 by 87.5. 
 
 7. Divide 42630.6 by 28421.3. 
 
 8. Divide 80000.27 by 20000.36. 
 
 9. Divide 24.081 by .386. 
 10. Divide .36 by .25. 
 
226 REDUCTION OP fSfiCT. X] 
 
 SECTION XI. 
 FEDERAL MONEY. 
 
 ART. 366* FEDERAL MONEY is the currency of the United 
 States. Its denominations, we have seen, are Eayles, Dollars, 
 .Dimes, Cents, and Mills. (Art. 244.) 
 
 36 7 All accounts in the United States are required by law 
 to be kept in dollars, cents, and mills. Eagles are expressed in 
 dollars, and dimes in cents. Thus, instead of 8 eagles, we say, 
 80 dollars ; instead of 6 dimes and 7 cents, we say, 67 cents, &c. 
 
 368. Federal Money is based upon the Decimal system of 
 Notation. Its denominations increase and decrease from right to 
 left and left to right in a tenfold ratio, like whole numbers and 
 decimals. (Art. 244. Obs. 1.) 
 
 369* The dollar is regarded as the unit ; cents and mills are 
 fractional parts of the dollar, and are distinguished from it by 
 a decimal point or separatrix (.) in the same manner as common 
 decimals are distinguished from whole numbers. (Art. 311.) 
 Dollars therefore occupy units' place of simple numbers ; eayles, 
 or tens of dollars, tens' place, &c. Dimes, or tenths of a dollar, 
 occupy the place of tentlis in decimals ; cents, or hundred ths of a 
 dollar, the place of hundredths ; mills, or thousandths of a dollar, 
 the place of thousandths ; tenths of a mill, or ten thousandths of 
 a dollar, the place of ten thousandths, &c. 
 
 OBS. 1. Since dimes in business transactions are expressed in cents, two places 
 of decimals are assigned to cents. If therefore the number of cents is less than 
 10, a cipher must, always be placed on the left hand of them ; for cents are hun- 
 dredths of a dollar, and hundredths occupy the second decimal place. (Art. 313.) 
 For example, 4 cents are written thus .04 ; 7 cents thus .07; &c. 
 
 2. Mills occupy the third place of decimals ; for they are thousandths of e 
 dollar. Consequently, when there are no cents in the given sum, two cipher 
 must be placed before the mills. Hence, 
 
 QUEST. 366. What is Federal Money? 367. In what are accounts kept in the U. S. 
 How would you express 8 eagles ? How express 6 dimes and " cents ? 3f>8. Upon what i.. 
 Federal Money based 1 369. What is regarded as the unit ir Federal Money ? What aiv 
 jents and mills ? How are they distinguished from dollars ? 
 
ARTS. 386-371. J FEDERAL MONEY. 227 
 
 37O. To read any sum of Federal Money. 
 
 Call all the figures on the left of the decimal point dollars ; the 
 first two figures after the point, are cents ; the third figure denotes 
 mills ; the other places on the right are decimals of a mill. Thus, 
 $3.25232 is read, 3 dollars, 25 cents, 2 mills, and 32 hundredth^ 
 of a mill. 
 
 OBS. Sometimes all the figures after the point are read as decimals of a :ol- 
 ar, Thus, $5.356 is read, " 5 and 356 thousandths dollars." 
 
 Write the following sums in Federal money : 
 
 1. 70 dollars, and 8 cents. Ans. $70.08. 
 
 2. 150 dollars, 3 cents, and 5 mills. 
 
 3. 409 dollars, 40 cents, and 3 mills. 
 
 4. 200 dollars, 5 cents, and 2 mills. 
 
 5. 4050 dollars, 65 cents, and 3 mills. 
 
 Note. In business transactions, when dollars and cents are expressed to- 
 gether, the cents are frequently written in the form of a common fraction. 
 Thus, the sum of $75.45, is written 75-^^ dollars. 
 
 REDUCTION OP FEDERAL MONEY. 
 
 CASE I. 
 Ex. 1. How many cents are there in 95 dollars? 
 
 Solution. Since in 1 dollar there are 100 cents, in 95 dollars 
 there are 95 times as many. And 95 X 100=9500. 
 
 Ans. 9500 cents. 
 2. In 20 cents how many mills ? Ans. 200 mills. 
 
 Note. To multiply by 10, 100, &c., we simply annex as many ciphers to 
 Ihe multiplicand, as there are ciphers in the multiplier. (Art. 99.) Hence, 
 
 3 7 1 To reduce dollars to cents, annex two ciphers. 
 To reduce dollars to mills, annex three ciphers. 
 To reduce cents to mills, annex one cipher. 
 
 OBS. To reduce dollars, cents, and mills, to mills, erase the sign of dollars 
 and the scparatrix. Thus, $25,36 reduced to cents, becomes 2536 cents. 
 
 QUEST. 353. How do you read Federal Money 1 Obs. What other mode of reading 
 Federal Money is mentioned 1 354. How are dollars reduced to cents ? Dollars to mills? 
 Cents to mills? Obs. Dollars, cents, and mills, to mills? 
 
228 ADDITION op [SECT. XI 
 
 3. In $12 l:ow man)- cents? Ans. 1200 cents. 
 
 4. In $460 how -many cents? 
 
 5. In $95 how many mills ? 
 
 6. In 90 cents how many mills ? 
 
 7. Reduce $25.15 to cents. 
 
 8. Reduce $864.08 to cents. 
 
 9. Reduce $1265.05 to mills. 
 
 10. Reduce $4580.10 to mills. 
 
 11. Reduce $6886.258 to mills. 
 
 12. Reduce $85625.40 to mills. 
 
 CASE II. 
 
 13. In 6400 cents, how many dollars? 
 
 Suggestion. Since 100 cents make 1 dollar, 6400 cents will 
 make as many dollars as 100 is contained times in 6400. And 
 6400-MOO = 64. Am. $64. 
 
 14. In 260 mills, how many cents? Ans. 26 cents. 
 
 Note. To divide by 10, 100, &c., we simply cut off as many figures from the 
 right of the dividend as there are ciphers in the divisor. (Art. 131.) Hence, 
 
 372. To reduce cents to dollars, cut off two figures on the 
 right. 
 
 To reduce mills to dollars, cut off three figures on the riyht. 
 To reduce mills to cents, cut off one figure on the right. 
 OBS. The figures cut off are cents and mills. 
 
 15. In 626 cents, how many dollars? Ans. $6.26. 
 
 16. In 1516 cents, how many dollars? 
 
 17. In 162 mills, how many cents? 
 
 18. In 1000 mills, how many dollars? 
 
 19. In 2360 mills, how many cents ? 
 
 20. In 3280 mills, how many dollars? 
 21 Reduce 8500 cents to dollars. 
 
 22, Reduce 2345 cents to dollars, &c. 
 
 23. Reduce 92355 mills to dollars, &c. 
 
 QCST. 355. How are cents reduced to dollars? Mills to dollars! Mills to cental 
 Ob*. What are the figures cut off 1 
 
ARTS. 372-374.] FEDERAL MONEY. 229 
 
 24. Reduce 150233 mills to dollars, &c. 
 
 25. Reduce 450341 cents to dollars, &c. 
 
 373. Since Federal Money is expressed according to the de- 
 cimal system of notation, it is evident that it may be subjected to 
 the same operations and treated in the same manner as Decimal 
 fractions. 
 
 ADDITION OF FEDERAL MONEY. 
 
 Ex. 1. A man bought a cloak for $35.375, a hat for $4.875, \ 
 pair of boots for $6.50, and a coat for $23.625 : what did he pay 
 for all? 
 
 Operation. We write the dollars under dollars, cents 
 
 $35.375 under cents, &c. Then add each column sepa- 
 
 4.875 rately, and point off as many figures for cents 
 
 6.50 and mills, in the amount, as there are places 
 
 23.625 of cents and mills in either of the given num~ 
 
 $70.375 Am. bers. Hence, 
 
 374. We derive the following general 
 
 RULE FOR ADDING FEDERAL MONEY. 
 
 Write dollars under dollars, cents under cents, &c., so that the 
 same orders or denominations may stand under each other. Add 
 each column separately, and poini off tlie amount as in addition of 
 decimal fractions. (Art. 320.) 
 
 OES. If either of the given numbers have no cents expressed, it is customary 
 to supply their place by ciphers. 
 
 2. What is the sum of $48.25, $95.60, $40.09, and $81.10 ? 
 
 3. What is the sum of $103.40, $68.253, $89.455, $140 02, 
 and $180? 
 
 4. What is the sum of $136.255, $10.30, $248.50, $65.33, 
 and $100.125? 
 
 QUEST. 357. How is Federal Money added 1 How point off the amount Obi. When 
 ny of the given numbers have no cents expressed, how is their place suppMed ? 
 
230 SUBTRACTION OP [SECT. XL 
 
 5. What is the sum of $170, $400.02, 8130, $250.10, ana 
 $845.22? 
 
 6. What is the sum of $268,45, $800.05, $192.125,- $80.625, 
 and $90.25 ? 
 
 7. What is the sura of $1500.20, $1050.07, $100.70, $95.025, 
 $360.437, and $425? 
 
 8. What is the sum of $2600, $1927.404, $1603.40, $3304.17 
 $165.47, and $2600.08? 
 
 9. A man bought a load of hay for $19.675, a horse for $73 25, 
 a yoke of oxen for $69.56, a cow for $17, and a calf for $5.80: 
 what did he pay for all ? 
 
 10. A lady gave $21.50 for a dress, $9.25 for a bonnet, $28.33 
 for a shawl, and $15.25 for a muff: what was her bill? 
 
 11. A jockey bought a span of horses for $276.87, and sold 
 them so as to gain $73.45 : how much did he sell them for? 
 
 12. A man gave $4925.68 for a farm, and sold it so as to gain 
 $1565.37 : how much did he sell it for? 
 
 13. A man sold a sloop for $7623.87, which was $1141.25 less 
 than cost : how much did it cost ? 
 
 14. A man bought a block of stores for $15268, which was 
 $1721 less than cost: what was the cost? 
 
 15. What is the sum of 134 dolls. 3 cts. 7 mills, 108 dolls. 
 6 cts. 8 mills, 90 dolls. 9 cts. 4 mills, and 46 dolls. 18 cts. 4 mills ? 
 
 ' 16. What is the sum of 61 dolls. 1 ct. 2 mills, 19 dolls. 11 cts. 
 4 mills, 140 dolls, and 80 dolls. 4 cts.? 
 
 17. What is the sum of 140 dolls. 10 cts., 69 dolls. 3 cts. 
 8 mills, 18 dolls. 7 cts., and 29 dolls. 5 mills? 
 
 18. What is the sum of 860 dolls. 8 cts., 298 dolls. 4 cts. 8 mills, 
 416 dolls., 280 dolls. 13 cts., and 91 dolls. ? 
 
 19. What is the sum of 14209 dolls., 65241 dolls., 1050 dolls., 
 610 dolls. 7 cts., and 1000 dolls. 10 cts. ? 
 
 20. What is the sum of 1625 dolls., 4025 dolls., 1863 dolls. 
 75 cts., 16000 dolls., and 48261 dolls.? 
 
 21. What is the sum of 8 thousand dolls., 2 hundred and 60 
 dolls. 5 cts., 19 thousand dolls. 60 cts., 6 hundred dolls. 9 cts.? 
 
 22. What is the sum of 19 thousand dolls. 50 cts., 61 thou- 
 sand dolls. 10 cts., 18 hundred dolls. 3 ctf . ? 
 
ART. 375.] FEDERAL MONEY. 
 
 SUBTRACTION OF FEDERAL MONEY. 
 
 Ex. 1. A merchant bought a quantity of molasses for $75.40 ; 
 and a box of sugar for $42.63 : how much more did he pay for 
 one Uian the other ? 
 
 Operation. We write the less number under the greater, 
 
 $75.40 placing dollars under dollars, &c., then subtract 
 
 42.63 and point off the answer, as in subtraction of 
 
 $32.77 decimals. Hence, 
 
 375* We derive the following general 
 
 RULE FOR SUBTRACTING FEDERAL MONEY. 
 
 Write the less number under the greater, with dollars under dol- 
 lars, cents under cents, &c. ; then subtract and point off the remain- 
 der as in subtraction of decimal fractions. (Art. 322.) 
 
 OBS If either of the given numbers have no cents expressed, it is customary 
 to supply their place by ciphers. 
 
 2. A man bought a horse for 175.50, and sold it for $87.63* 
 how much did he make by his bargain? 
 
 3. If a man deposits $204.65 in a bank, and afterwards checks 
 out $119.83, how much will he have left ? 
 
 4. A man owing $682.40, paid $435.25 : how much does ha 
 still owe ? 
 
 5'. A man owing $982.68, paid all but $64.20 : how much did 
 he pay ? 
 
 6. A merchant bought a quantity of goods for $833.63, and re- 
 tailed them for $1016.85 : how much did he make by the bargain ? 
 
 7. A merchant bought a lot of goods for $1265.82, and sold 
 them for $942.35 : how much did he lose ? 
 
 8. A grocer sold a lot of sugar for $635.20, and made thereby 
 $261 ,38 : how much did he pay for the sugar ? 
 
 9. A man sold his farm for $12250.62, which was $1379.87 
 more than it cost : how much did it cost ? 
 
 QUEST. 358. How is Federal Money subtracted ? How point off the remainder 1 
 Obs When either of the given numbers have no cents, how is their place supplied 1 
 
232 MULTIPLICATION OP [SECT. XJ 
 
 10. From $10600.75 take $8901.26. 
 
 11. From $20206.85 take $10261.062. 
 
 12. From $61219.40 take $100.036. 
 
 13. From $19 take 1 cent and 9 mills. 
 
 14. From 89 dollars take 89 cents. 
 
 15. From 506 dolls, take 316 dolls, and i cts. 
 
 1 6. From 5 dolls. 7 mills take 2 dolls. 7 els. 
 
 17. From 61 dolls. 6 cts. take 29 dolls. 4 mills. 
 
 18 From 11000 dolls. 10 cts. take 110 dolls. 3 cts. 
 19. From 1 00 100 dolls, take 10110 dolls. 10 cts. 
 
 MULTIPLICATION O^ FEDERAL MONEY. 
 
 376. In multiplication of Federal Money, as well as in simple 
 numbers, the multiplier must always be considered an abstract 
 number. (Art. 82. Obs. 2.) 
 
 Ex. 1. What will 8 bbls. of flour cost, at $5.62 per bbl. ? 
 
 Analysis. Since 1 bbl. costs $5.62, 8 bbls. will cost 8 times as 
 much; and $5.62 X8=$44.96 Am. 
 
 2. What cost 21.7 bushels of apples, at 15 cts. per bushel? 
 
 Operation. Reasoning as before, 21.7 bushels will cost 
 
 21.7 21.7 times 15 cents. But in performing the 
 
 .15 multiplication, it is more convenient to make 
 
 1085 the .15 the multiplier, and the result will be 
 
 217 the same as if it was placed for the multipli- 
 
 $3.255 Ans. cand. (Art. 83.) Point off the product as be- 
 
 fore. Hence, 
 
 377. When the price of one article, one pound, one yard, &c., is 
 given, to find the cost of any number of articles, pounds, yards, tfec. 
 
 Multiply the price of one article and the number of articles to- 
 gether, and point off the product as in multiplication of decimal*, 
 (Art. 324.) 
 
 In Multiplication of Federal Money, what must one of the given factor* 
 be considered 1 377. When the price of one article, one pound, &c., is given, how is the 
 cost of any number of articles found 1 
 
ARTS. 376-378.] FEDERAL MONLY. 233 
 
 3. What cost 17.6 yards of cloth, at $4.75 per yard? 
 
 4. Multiply $25.625 by 20.2. 
 
 37 8 From the preceding illustrations we derive the following 
 general 
 
 RULE FOR MULTIPLYING FEDERAL MONEY. 
 
 Multiply as in simple numbers, and point Ojf the product as in 
 multiplication of decimal fractions. (Art. 324.) 
 
 OB3. 1 . When the price, or the quantity contains a common fraction, the 
 fraction may be changed to a decimal. (Art. 337.) 
 
 2. In business operations, when the mills in the answer are 5, or over, it is 
 customary to call them a cent ; when under 5, they are disregarded. 
 
 5. What cost 12 yards of cotton, at 9-J- cts. per yard? 
 
 Solution. 12f yards=12.5, and 9i cts. = .0925 ; now .0925 X 
 125=$1. 15625. Ans. 
 
 6. What cost 45-J- yards of satin, at 87^ cts. per yard ? 
 
 7. What cost 169 bbls. of pork, at $8| per barrel ? 
 
 8. What cost 324-f- Ibs. of sugar, at 12^ cts. a pound? 
 
 9. What cost 97 gals, of oil, at 87- cts. per gallon? 
 
 10. What cost 310 Ibs. of tea, at 62i cts. a pound? 
 
 11. What cost 23i tons of hay, at $8$ per ton ? 
 
 12. What cost 45 bbls. of flour, at $7| per barrel? 
 
 13. At 15i- cts. per doz., what cost 13 dozen of eggs ? 
 
 14. At 8f cts. per pound, what will 32 Ibs. of pork come to? 
 
 15. At $6-}- per bbl, what will 145| bbls. of flour cost? 
 
 16. At 22- cts. per doz., what will a gross of buttons cost? 
 
 17. At 31^- cts. per doz., what cost 45 doz. skeins of silk? 
 
 18. At I7i cts. per yard, what cost 91- yards of calico? 
 
 19. What cost 45 doz. plates, at 62- cts. per doz. ? 
 
 20. What cost 63 doz. pen-knives, at $3 per doz. ? 
 
 21. What cost 19 doz. silver spoons, at $7 per dozen? 
 
 22. What cost 1865i bushels of wheat, at $lf per bushel? 
 
 23. What cost 2560^- yds. of broadcloth, at $5| per yard? 
 
 QUEST. 378. What is the rule for Multiplication of Federal Money? Obs. When tJM 
 wrice or quantity contains a common fraction, what should be done with it ? 
 
234 DIVISION OF [SECT X.L 
 
 DIVISION OP FEDERAL MONEY. 
 
 Ex. 1. A man bought 8 sheep for $42.24 : what did n give 
 apiece ? 
 
 Analysis. If 8 sheep cost $42.24, 1 sheep will cost \ of 4^2.24 ; 
 and $42.24H-8 = $5.28. Am. $5.28. 
 
 PROOF. If 1 sheep costs $5.25, 8 sheep will cost 8 times as 
 much, and $5.28 X8=$42. 24. Hence, 
 
 370. When the number of articles, pounds, yards, <fec., a. 
 the cost of the whole are given, to find the price of one article, ont 
 pound, &c. 
 
 Divide the whole cost by the whole number of articles, and poin 1 
 off the quotient as in division of decimal fractions. (Art. 330.) 
 
 2. A shoemaker sold 15 pair of boots for $67.50: how much 
 did he get a pair ? 
 
 3. A merchant sold 65 Ibs. of sugar for $3.93: how much 
 was that a pound ? 
 
 4. A man bought 6.5 yards of cloth for $20.345 : how much 
 was that per yard ? 
 
 5. How many bbls. of flour, at $5.38 per bbl., can be bought 
 for $34.97 ? 
 
 Analysis. Since $5.38 will buy 1 bbl., Operation. 
 
 $34.97 will buy as many bbls. as $5.38 5.38)34.97(6.5 Ans 
 are contained times in $34.97. We divide 32 28 
 
 as in simple numbers, and point off one de- 2 699 
 
 cimal figure in the quotient. 2 690 
 
 PROOF. $5. 38X6. 5 =$34.97, the given amount. 
 
 38O. Hence, when the price of one article, pound, yard, <fec., 
 and the cost of the whole are given, to find the number of arti- 
 cles, &c. 
 
 Divide the whole cost by the price of one, and point off the quo* 
 tient as in division of decimals. 
 
 GUEST.- 379. When the number of articles, pounds, &c., and the cost of the whole are 
 given, how is the cost of one article found ? 380. When the price of one article, one pound* 
 &.C., and the cost of the whole are given, how is the number of articles found 1 
 
ARTS. 379-381.] FEDERAL MONEY. 235 
 
 6. How many coats, at $12.56, can be bought for $103.085 ? 
 
 7. How many times is $11.13 contained in 87.606 ? 
 
 8. A gentleman distributed $68 equally among 32 poor per- 
 sons : how much did each receive ? 
 
 Operation. 
 
 32)68(2.125 Ans. After dividing the $68 by 32, there is 
 
 64 a remainder of 4 dollars, which should be 
 
 4000 reduced to cents and mills, and then be 
 
 32 divided as before. (Art. 354.) The ciphers 
 
 80 thus annexed must be regarded as deci- 
 
 64 mals ; consequently there will be three 
 
 160 decimal figures in the quotient. 
 160 
 
 
 381. From the preceding illustrations we derive the following 
 general 
 
 RULE FOR DIVIDING FEDERAL MONEY. 
 
 Divide as in simple numbers, and point off the quotient as in 
 division of decimal fractions. (Art. 330.) 
 
 OBS. 1. In dividing Federal Money, if the number of decimals in the divisor 
 is the same as that in the dividend, the quotient will be a whole number. 
 (Art. 330. Obs. 1.) 
 
 2. When there are mare decimals in the divisor than in the dividend, annex 
 as many ciphers to the dividend as are necessary to make its decimal places 
 equal to those in the divisor. The quotient thence arising will be a whole 
 number. (Obs. 1.) 
 
 3. After all the figures of the dividend are divided, if there is a remainder, 
 ciphers may be annexed to it, and the operation may be continued as in divi- 
 won of decimals. (Art. 330. Obs. 3.) The ciphers thus annexed must be re- 
 garded as decimal places of the dividend. 
 
 9. How many gallons of molasses, at 28 cts. per gallon, can 
 you buy for 886.25 ? 
 
 QUEST. 381. What is the rule for Division of Federal Money ? Obs When there is a 
 remainder after a 1 the figures of the dividend are divided, how proceed ? When there are 
 more decimals in the divisor than in the dividend, how proceed! 
 T.H. 
 
236 DIVISION OF [SECT. XL 
 
 10. How many yards of calico, at 13 cts. per yard, can b 
 bought for $73.3 7? 
 
 11. How many doz. of eggs, at 9 cts. per doz., can be bought 
 for $94.185? 
 
 12. At 18f cts. per doz., how many skeins of sewing silk can 
 be bought for $67.50 ? 
 
 13. A man paid $72.25 for 20.5 yards of cloth : how much did 
 he pay per yard ? 
 
 14. A man paid $76.50 for 51 sheep : what was the price per 
 nead ? 
 
 15. A man paid $150 for 24 pair of boots: how much was 
 that a pair ? 
 
 16. If you give $56.25 for 28| bbls. of flour, how much do you 
 pay per barrel ? 
 
 17. If a man gives $316.375 for 87^ yards of cloth, what is 
 that per yard ? 
 
 18. A grocer sold 965- Ibs. of sugar for $81.25 : what did he 
 get a pound ? 
 
 19. The fare from Albany to Buffalo, a distance of 326 miles, 
 is $13.20 : how much is it per mile ? 
 
 20. The fare from Boston to Albany, a distance of 203 miles 
 is $5.50 : how much is it per mile ? 
 
 21. If a clerk's salary is $650 per annum, how much does he 
 receive per day ? 
 
 22. If a man spends $563.38 a year, how much are his average 
 expenses per day ? 
 
 23. At 87 cts. per bushel, how many bushels of wheat can 
 you buy for $1500? 
 
 24. IIow many tons of coal, at $6.625 per ton, can you buy for 
 $752.36 ? 
 
 25. If a man's income is $100 per week, how much is it per 
 hour ? 
 
 26. At $14. 50 per acre, how many acres of land can you buy 
 foi $3560 ? 
 
 27. At $15^ apiece, how many cows can you buy for $7750 ? 
 
 28. At $375.75 apiece, how many carriages cm be bought for 
 $56362.50 ? 
 
ART. 382. 1 FEDERAL MONET. 237 
 
 COUNTING ROOM EXERCISES. 
 
 Ex. 1. What cost 320 yards of satinet, at $1.12 per yard? 
 
 Analysis. If the price were $1 per yard, the cloth would evi- 
 dently cost as many dollars as there are yards. But $1.12 is 
 equal to 1 and dollars ; hence, the cloth will cost i more dollars 
 than there are yards ; consequently, if we add to the number of 
 yards i of itself, it will give the cost. Now i of 320=40, and 
 320+40=360. Ans. $360. 
 
 PROOF. $1.12X320=$360, the same as before. Hence, 
 
 382. When the price of 1 article, 1 pound, &c., is $1.12 
 $1.25 ; $1.37; &c., to find the cost of any number of articles. 
 
 To the given number of articles, add \, \, -f , &c., of itself, as 
 the case may be, and the sum will be the cost required. 
 
 OBS. When the price of 1 article, &c., is $2.12$, $2.25, $3.37j, &<;., the 
 operation may be contracted by multiplying the given number of articles by 
 ^s, 2}, 3f , &c., as the case may be. 
 
 2. What cost 640 bushels of wheat, at $1.25 per bushel? 
 
 3. What cost 372 pair of shoes, at $1.37^ a pair? 
 
 4. What cost 480 bbls. of. cider, at $1.62 a barrel? 
 
 5. What cost 520 yards of silk, at $1.50 per yard? 
 
 6. What cost 720 drums of figs, at $1.87- per drum? 
 
 7. At $2.12- apiece, what will 480 sheep cost? 
 
 8. _At $2.37- apiece, what will 364 vests cost? 
 
 9. At $3.25 per yard, what cost 744 yards of cloth? 
 
 10. At $4.62i apiece, what cost 960 hats? 
 
 11. At $5.12 a pair, what cost 278 pair of boots? 
 
 12 At $7.37^ per lb., what will 365 Ibs. of opium cost? 
 
 13. A collier sold 856 tons of coal, at $6.87i per ton: how 
 much did it amount to ? 
 
 14. At 19. 62^ per acre, what will 537 acres of land cost? 
 lo. What cost 72 Ibs. of flax, at $8.25 per hundred? 
 
 Analysis. 72 pounds are iW of 100 pounds; th;reiOre 72 
 
 Q 2 { Sv72 
 pounds will cost flft of $8.25 ; and -fifo of $8.25= ~-. 
 
238 APPLICATIONS OF [SECT. XI. 
 
 Operation. We multiply the price of 100 Ibs. (88.25) 
 
 $8.25 by 72, the given number of pounds, and the 
 
 72 product $594.00, is the cost of 72 Ibs. at 
 
 1050 $8.25 per pound. But the price is $8.25 per 
 
 5775 hundred; consequently the product $504.00 
 
 $5.9400 Ans. is 100 times too large, and must therefore be 
 
 divided by 100, to give the true answer. But 
 
 to divide by 100, we simply remove the decimal point two places 
 
 towards the left. (Art. 331.) 
 
 16. What cost 367 bricks, at $4.45 per 1000? 
 
 Operation. Reasoning as before, 367 bricks will cost 
 
 4.45 T^Vs- of $4.45. We multiply the price of 1000 
 
 3 67 bricks by the given number of bricks, and di- 
 
 $1.63315 Ans. vide the product by 1000. (Art. 331.) Hence, 
 
 383. To find the cost of articles sold by. the 100, or 1000. 
 
 Multiply the given price by the given number of articles ; then 
 if the price is for 100, divide the product by 100 ; but if the price 
 is for 1000, divide it by 1000. (Art. 331.) 
 
 17. A farmer sold 563 Ibs. of hay, at $1.12 per hundred : how 
 much did it come to ? 
 
 18. What cost 1640 Ibs. of beef, at $6.37i per hundred? 
 
 19. What cost 2719 Ibs. of fish, at $4.20 per hundred? 
 
 20. What is the freight on 3568 Ibs. from New York to Buffalo, 
 ai$1.67 per hundred? 
 
 21. What cost 6521 Ibs. of cheese, at 7f cts. per hundred? 
 
 22. What cost 15214 Ibs. of butter, at 12| cts. per hundred? 
 
 23. At $6.25 per 1000, what cost 865 feet of spruce boards? 
 
 24. At $19.45 per 1000, what cost 2680 feet of pine boards? 
 25 At $67.33 per 1000, what cost 6500 feet of mahogany? 
 26. When ginger is 16.53 per cwt., what is it per pound? 
 
 Analysis. Since 100 Ibs. cost $16.53, 1 Ib. will cost -rbr of' 
 I '6.53. But to divide by 100, we remove the decimal point twa 
 places to the left. (Art. 331.) Ans. $0 1653. 
 
 Q,OBS r. 383. How do you find the cost of articles sold, by the 100, or 1000 7 
 
ARTS. 383-385.] FEDERAL MONEY 239 
 
 27. When pine boards are $21.63 per 1000, what are they per 
 foot? 
 
 Solution. Reasoning as before, 1 foot will cost -reVv of $21.63. 
 Now to divide by 1000, we remove the decimal point three places 
 to the left. (Art. 331.) Ans. $.02163. Hence, 
 
 384. "When the cost of 100, or 1000 articles, pounds, &c., 4 
 given, the price of one is found by simply removing the decimal 
 point in the given cost or dividend, as many places to the left as 
 there are ciphers in the divisor. (Art. 331.) 
 
 28. Bought 1000 bricks for $7.20 : what is that apiece? 
 
 29. If 1000 feet of hemlock boards cost $6.40, what' will one 
 foot cost ? 
 
 30. Bought 42 cwt. of tobacco for $565.82 : what is that per 
 cwt. ; and what per pound ? 
 
 31. Bought 75 cwt. of butter for $966.38: what is that per 
 cwt. ; and what per pound ? 
 
 BILLS, ACCOUNTS, <feC. 
 
 385* A Bill, in mercantile operations, is a paper containing 
 a written statement of the items, and the price or amount of goods 
 sold. 
 
 32. What is the cost of the several articles, and what the amount, 
 of the following bill ? 
 
 NEW YORK, May 21st, 1847. 
 G. B. Grannis, Esq., 
 
 Bought of Mark H. Newman & Co. 9 
 75 Thomson's Mental Arithmetic, at $ .12 
 50 " Practical Arithmetic, " .31 1 
 
 36 Porter's Rhetorical Reader, ' .62 
 
 25 Willson's School History, " .46 - - ' 
 
 80 M'Elligott's Young Analyzer, " .31-J- 
 
 75 Thomson's Day's Algebra, " .50 - 
 
 60 " Lcgendre's Geometry, .47 - 
 
 Received Payment, 
 
 Mark fl. Newman <& Co. 
 
240 BILLS, ACCOUNTS, ETC. [SECT. XI. 
 
 (33.) 
 
 PHILADELPHIA, June 10th, 1847. 
 Hon. Horace Binney, 
 
 Sought of Leverette & Griggs t 
 1G3 Ibs. Butter, at $ .14 
 
 235 Ibs. Coffee, " .08* 
 
 86 Ibs. Chocolate, " .11 
 
 685 Ibs. Sugar, " ,10|- 
 
 21 doz. Eggs, " .13 
 
 860 Ibs. Lard, " .09|- 
 
 What was the cost of the several articles, and what the amount 
 of his bill? 
 
 (84.) 
 
 ALBANY, July 1st, 1847. 
 Messrs. Collins & Brotliers, 
 
 To G. W. Bunker, Dr. 
 For 320 yds. Silk, at $1.12 
 
 " 256 " Broadcloth, " 3.62 
 
 " 175 pair Cotton Hose, ". 0.12 
 
 " 100 " Silk " " 0.871 
 
 " 15 doz. Gloves, " 0.621 
 
 " 120 Straw Hats, " 1.871 
 
 What was the cost of the several articles, and what the amount 
 of his bill? 
 
 (35.) 
 
 ST. Louis, Aug. 25th, 1847. 
 James Henry, Esq. 
 
 To J. L. Hoffman & Co., Dr. 
 
 For 15260 Ibs. Pork, at $0.051 
 
 " 7265 Ibs. Cheese, " 0.081 
 
 " 11521 bu. Corn, " 0.50 
 
 " 1560 bbls. Flour, " 6.12^ 
 
 CREDIT. 
 
 Bj 115C Ibs, Cotton, at 80.06| 
 
 " 8256 Ibs. Sugar, 0.07 
 
 M 6450 gals. Molasses, " 0.3 7i 
 
 " Cash to balance account, - - 
 
 What is the amount of cash requisite to balance the account ? 
 
ARTS. 386, 387. j PERCENTAGE. 241 
 
 SECTION XII. 
 PERCENTAGE. 
 
 ART 386* The terms Percentage and Per Cent, signify a cer- 
 tain allowance on a hundred; that is, a certain part of a hundred, 
 or simply hundred tks. Thus, the expression G per cent, signifies 
 6 hundredths, (ruir,) V per cent., 7 hundredths, (-rfcr,) &c., of the " 
 number, or sum of money under consideration. 
 
 Note. -The terms Percentage and Per Cent, are derived from the Latin per 
 and centum, signifying by I/M hundred. 
 
 387. We have seen that hundredth* are decimal expressions, 
 occupying the first two places of figures on the right of the deci- 
 mal point. (Arts. 311, 314.) Now, since percentage and. per cent. 
 signify hundredths, it is manifest that they can be expressed by 
 decimals, as in the following 
 
 PERCENTAGE TABLE. 
 
 1 per cent. . . . is written thus : .01 
 
 2 per cent. . " " " .02 
 
 3 per cent. ... " " " .03 
 G per cent. . . . " " " .00 
 7 per cent ... u u u .07 
 
 10 per cent. . . . JO 
 
 12 per cent. . * . " " ,13 
 
 50 per cent. . . . .50 
 
 100 per cent. . . " " " 1.00 
 
 103 per cent. . . . 1.03 
 
 1-25 per cent., &c. . . 1/25 
 
 J per cent., that is, $ of 1 per cent " " " .005 
 
 \ per cent., that is, \ of 1 per cent. " " " 0025 
 
 | per cent., that is, $ of 1 per cent. " " .0075 
 
 13 1 per cent. ... " " " .13125 jjt 
 
 251 per cent. . . . ,c .25375 
 
 OBS. I. It will be seen from the preceding Table, that when the given per 
 cent, is /(.'.s.s than 10, a cipher must be prefixed to the ligure ex pressing it, in the 
 same manner as when the number of cents is less than 10. (Art. 3GU. Obs. 1.) 
 
 QPKST. 386. Wh;vt do the terms percentage and per cent, signify 1 What is meant IT 
 6 oer cent., 7 per cent., &.C., of any number, or sum 1 
 
242 PERCENTAGE. [SECT. Xll. 
 
 When the given per cent, is more than 100, it must plainly require a mixed 
 number to express it. (Art. 315. Obs. 2.) 
 
 2. Parts of 1 per cent, may be expressed either by a common fraction, or by 
 decimals Thus, the expression 17f per cent., is equivalent to .17G25 percent. 
 
 3. The first two decimal figures properly denote the per cent., for they are 
 kundredtks ; the other decimals being parts of hundrcdths, express parts of 
 1 per cent. 
 
 EXAMPLES. 
 
 1 Write 1 per cent., 2 per cent., 4 per cent., 6 per cent., 7 per 
 cent., 8 per cent., in decimals. 
 
 2 Write 11 per cent. ; 12; 14; 15; 16; 23; 65; 93. 
 
 3 Write i per ct. ; i;i;|; f; *; i; i; f; f ; 1; *; }; *. 
 
 4. Write 4i per ct. ; 6|; ?i; 9i; 12|; 16-J-; 115; 400|. 
 
 5. An agent collected $700 for a merchant, and received 5 per 
 cent, for his services : how much did he receive ? 
 
 Analysis. Since 5 per cent, is the same as -p^, the agent must 
 have received -Hhr of $700. Now rb- of $700 is $f JHK which is 
 equal to $7 ; and 5 hundredths is 5. times $7, or $35. 
 
 Operation. Since TOTT -05, we multiply the given num- 
 
 $700 ber of dollars by .05, and it gives the answer in 
 
 .05 cents, which we reduce to dollars by pointing 
 
 $35TOO Ans. off 2 decimals. (Art. 372.) Hence, 
 
 388. To calculate percentage on any number, or sum of 
 money. ^ 
 
 Multiply the given number or sum by the given per cent, expressed 
 decimally ; and point off ike product as in multiplication of deci- 
 mal fractions. (Art. 324.) 
 
 OBS. 1. It is important for the learner to observe, that the amount of money 
 collected, is made the basis upon which the percentage is computed. That is, 
 the agent is entitled to 3 dollars, as often as he collects 100 dollars, and not as 
 n as he pays over 100 dollars, as is frequently supposed. For in the latter 
 case he would receive only -f-f 3, instead of -,- -- of the sum in question. This 
 distinction is important, especially in calculating percentage on large sums. 
 
 QUEST. 387. How may per centage or per rent, be expressed? Obs. When the given 
 per cent, is less than 10, how is it written? When more than 100, how? 388. Hew is 
 percentage calculated ? Obs. In collecting money, upon what basis is the per cent, cal- 
 culated ? If the per cent, contains a common fraction which cannot le expressed deci- 
 mally, how proceed 1 
 
ART. 388 ] PERCENTAGE. 243 
 
 2. Hence, if the per cent, contains a common fraction which cannot be ex- 
 pressed decimally, first multiply by the decimal, then by the common fraction 
 of the given per cent., and point off the sum of their products as above. 
 
 6. What is 4| per cent, of $300 ? 
 
 Solution. Expressed decimally, 4 per cent.=.042 ; (Art. 387 
 Obs. 2 ;) and $300 X .042-$! 2.60. 4ns. 
 
 7. What is 3 per cent, of $256.25 ? 
 
 8. What is 2 per cent, of $437.63 ? 
 
 9. What is 2\ per cent, of $138.432 ? 
 10 What is 6 per cent, of $145.13 ? 
 
 11. What is 7 per cent, of $1630.10? 
 
 12. A man borrowed $150, and paid 7 per cent, for the use or 
 it : how much did he pay ? 
 
 13. A merchant bought goods amounting to $1825, and sold 
 them so that he gained 12 per cent. : how much did he gain? 
 
 14. A constable collected $862.56, and charged 5 per cent', for 
 his services : how much did he receive ; and how much did he 
 pay over ? 
 
 15. What is 10 per cent, of $4020.50 ? 
 
 16. What is 8 per cent, of $1675 ? 
 
 17. What is 4 per cent, of $725 ? 
 
 18. What is 5| per cent, of $648.30? 
 
 19. What is 6| per cent, of $1000 ? 
 
 20. W T hat is 7i per cent, of $2000 ? 
 
 21. What is 8| per cent, of $100.25 ? 
 
 22. A farmer having 1500 sheep, lost 25 per cent, of them: 
 how many did he lose ? 
 
 23. A merchant having $1960 on deposit, drew out 20 per cent, 
 of it : how much had he left in the bank ? 
 
 24. A merchant imported 1500 boxes of oranges, and 12 per 
 cent, of them decayed : how many boxes did he lose ; and how 
 many had he left ? 
 
 25.. What is per cent of $1625 ? 
 
 26. What is i per cent, of $2526.40 ? 
 
 27. What is i per cent, of $42260.08 ? 
 
 28. What is i per cent, of $75000 ? 
 
 29. What is -f pvr cent, of $100000? 
 
 11* 
 
244 PERCENTAGE. [SECT. XII. 
 
 30. What is i per cent, of $45241.20 ? 
 
 31. What is i per cent, of $675264 ? 
 
 32. A merchant bought a stock of goods amounting to $4565, 
 and paid 3 per cent, for freight : what was the whole cost of 
 his goods ? 
 
 33. A man's salary is $2000 a year, and he lays up 37 pel 
 cent, of it : how much does he spend ? 
 
 34. A youth who inherited $20000, spent 40 per cent, of it in 
 dissipation : how much had he left ? 
 
 35. Two merchants embarked in business with $18250 capital 
 apiece ; one gained 20 per cent, and the other lost 20 per cent, 
 the first year : what was then the amount of each man's property ? 
 
 36. Two men invested -Si 0000 apiece in stocks; one lost 8 per 
 cent., the other 6 per cent. : what was the difference of their loss ? 
 
 37. What is the difference between 6 per cent, of $1040, and 
 7 per cent, of $905 ? 
 
 . APPLICATIONS OP PERCENTAGE. 
 
 389 PERCENTAGE, or the method of reckoning by hundred ths, 
 is applied to various calculations in the practical concerns of life. 
 Among the most important of these are Commission, Brokerage, 
 the Rise and Fall of Stocks, Interest, Discount, Insurance,^ Profit 
 and Loss, Duties, and Taxes. Its principles, therefore, should be 
 thoroughly understood by every scholar. 
 
 COMMISSION, BROKERAGE, AND STOCKS. 
 
 39O Commission is the per cent, or sum charged by agents 
 for their services in buying and selling goods, or transacting other 
 business. 
 
 OBS. An Agent who buys and sells goods for another, is called a C(nnmi9* 
 sion Merchant, a Factor, or Correspondent. 
 
 391. Brokerage is the per cent, or sum charged by money deal- 
 ers, called Brokers, for negotiating Bills of Exchange, and other 
 monetary operations, and is of the same nature as Commission. 
 
 Q.UST. 390. What is commission 1 Obs. What is au agent who buys and sells goad* 
 for another usually called ? 
 
ARTS. 389-395.] COMMISSION. 245 
 
 392. By the term Stocks, is meant the Capital of moneyed 
 institutions, as incorporated Banks, Manufactories, Railroad and 
 Insurance Compiinies ; also, Government and State Bonds, &c. 
 
 OBS. 1. Stocks are usually divided into portions of $100 each, called shares; 
 anil the owners of these shares are culled Stockholders. 
 
 2. The association or company thus formed, is called a corporations the in- 
 strument specifying the powers, rights, and privileges invested in the corpora- 
 tion / is called a charter. 
 
 393. The original cost or valuation of a share is called its 
 nominal, or par value / the sum for which it can be sold, is its 
 real value. 
 
 OES. 1. The rise or fall of Stocks is reckoned at a certain per cent, of its 
 par value. The term par is a Latin word, which signifies equal, or a stale of 
 equality. 
 
 2. When Stocks sell for their original cost or valuation, they are said to be 
 at par ; when they sell for more than cost, they are said to be above par, at a 
 premium, or an advance ; when they do not sell at cost, they are said to be 
 below par, or at a discount. 
 
 3. Persons who deal in Stocks are usually called Stock Brokers, or Stock 
 Jobbers. 
 
 394. The commission or allowance made to factors and brokers, 
 also the rise and fall of stocks, are usually reckoned at a certain 
 vercentaye on the amount of money employed in the transaction, 
 or on the par value of the given shares. Hence, 
 
 395* To compute commission, brokerage, and the premium or 
 discount on stocks. 
 
 Multiply the given sum l>y the given per cent, expressed in deci- 
 mal*, and point off the product as in Percentage. (Art. 388.) 
 
 OBS. The commission for the collection of bills, taxes, &c., also for the sale 
 or purchase of goods, varies from *2i to 12 or 15 per cent., and should always 
 be reckoned on the amount of money collected, or paid out. or employed in the. 
 transaction. 
 
 The brokerage for the sale or purchase of stocks, varies from J tr f pe 
 cent., reckoned on the par value of the stock. 
 
 QUEST. :<91. What is hrnkernpe 1 :?92. Whnt is meant hy the term storks 1 Ob.- How 
 are stocks nnially fivi<1pd 1 .'!<):{. What is the par value of stocks ? What the real value ? 
 Obs. What is the meaning of the term par ? When are stocks at par ? When above par! 
 When below 1 395. How do you compute commission, brokerage, &.c. t 
 
246 COMMISSION. JSECT. XII 
 
 EXAMPLES. 
 
 1. An auctioneer sold goods amounting to $463, at 3 per cent 
 commission: how much did he receive ? Am. $13.89. 
 
 2. An agent bought goods amounting to $625.375 : what is 
 his commission, at 2 per cent. ? 
 
 3. What is the commission on $1682.25, at 3- per cent. ? 
 
 4. What is the commission on $1463.18, at 5 per cent. ? 
 
 5. What is the commission on $2560.07, at 4 per cent. ? 
 
 6. What is the commission on $10250, at 6 per cent. ? 
 
 7. What is the commission on $8340.60, at 7 per cent. ? 
 
 8. What is the commission on $960.625, at 5i per cent. ? 
 
 9. A commission merchant sold goods to the amount of $6235, 
 at 2 per cent. : what was his commission ? 
 
 10. An attorney collected a debt of $8265.17, and charged 7 
 per cent, for his services : how much did he receive ? 
 
 11. Bought $1108 worth of books, at 4 per cent, commission: 
 what was the amount of commission ? 
 
 12. A tax-gatherer collected $12250, for which he was entitled 
 to 5i per cent, commission : how much did he receive ? 
 
 13. Sold goods amounting to $1432.26: how much was the 
 commission, at 4 per cent. ? 
 
 1 4. A commission merchant sold a quantity of hardware amount- 
 ing to $9240.71 : how much would he receive, allowing 2^ per 
 cent, for selling, and 2 per cent, more for guaranteeing the pay- 
 ment? 
 
 15. An auctioneer sold carpeting amounting to $2136.63, and 
 charged 2^ per cent, for selling, and 2f per cent, for guaranteeing 
 the payment: how much did the auctioneer receive; and how 
 much did he remit the owner ? 
 
 396. Commission merchants, agents, &c., generally keep an 
 account with their employers, and as they make ii, vestments or 
 sales of goods, charge their commission on the amount invested, 
 or the sum employed in the transaction. 
 
 Sometimes, however, a specific amount is sent to an agent or 
 broker, requesting him, after deducting his comnrSsion, to lay out 
 the balance in a certain manner. 
 
ARTS. 396, 397.] BROKERAGE. 247 
 
 16. A gentleman sent his 'agent $1500 to purchase a library: 
 how much had he to lay out after deducting his commission at 5 
 per cent. ; and what was his commission ? 
 
 Note. The money actually laid out by the agent in books, is manifestly the 
 proper basis on which to calculate his commission ; for it would be unjust tt 
 charge commission on the sum he retains. (Art. 395. Obs.) 
 
 Analysis. The money laid out is }$$ of itself, and the commis- 
 sion is -poTT f this sum ; consequently the money laid out added 
 to the commission, must be -HHj* the whole amount. The question 
 therefore resolves itself into this: $1500 is ^-{J-ft of what sum? 
 If $1500 is -HHK rh- must be 1500-j-lOo^W, and 1^=-^ 
 X 100 $1428.57, the sum laid out. Now $1500 $1428.57 
 $71.43, the commission. 
 
 PROOF.- $1428.57 X. 05=171.43; and $1428.57+$7l.43 = 
 $1500, the amount sent. Hence, 
 
 397. To compute commission when it is to be deducted in 
 advance from a given amount, and the balance is to be invested. 
 
 Divide the given amount by $1 increased by the per cent, commis- 
 sion, and the quotient will be the part to be invested. Subtract the 
 part invested from the given amount, and the remainder will be the 
 commission. 
 
 OBS. The commission may also be found by multiplying the sum invested by 
 the given per cent, according to the preceding rule. (Art. 395.) 
 
 17. An 8q;ent received $21500 to lay out in provisions, after 
 deducting 2 A )er cent, commission : Avhat sum did he lay out ? 
 
 18. A country merchant sent $3560 to his agent in the city, to 
 purchase, goods : after taking out his commission, at 3 per cent., 
 how much remained to lay out ? 
 
 19. Baring, Brothers & Co. sent their agents $800000 to buy 
 flour : after deducting 5 per cent, commission, how much would 
 be left to invest? 
 
 20. A broker negotiated a bill of exchange of $82531, at 5 per 
 cent. : how much did he receive for his services ? 
 
 21. What is the brokerage on $94265, at 1| per cent.? 
 
 22. What is the brokerage on $6200, at per cent. ? 
 
248 STOCKS. [SECT. XII 
 
 23. What is the brokerage on $8845.50, at per cent. ? 
 
 24. What is the brokerage on $2500, at per cent. ? 
 
 25. A broker made an investment of $21265, and charged 1^ 
 per cent. : what was the amount of his brokerage ? 
 
 20. If you buy 20 shares of Western Railroad stock, at 7 per 
 cent, advance, how much will your stock cost you? Ans. $2140. 
 
 Note. The stock evidently cost its par value, which is $'2000 and 7 per cent. 
 tesides. Now $2000X.07=$140.00; and $2000 -f-$140=$2140. 
 
 27. What cost 20 shares of bank stock, at 7 per cent, discount? 
 
 Ans. $2000 $140 = $] 860. 
 
 28. What cost 35 shares of New York and Erie Railroad stock, 
 at 5- per cent, premium ? 
 
 29. A merchant bought 45 shares of Commercial Bank stock, 
 at par, and afterwards sold them, at 50 per cent, discount : how 
 much did he lose ? 
 
 30. A man invested $8460 in the New England Manufacturing 
 Co., and afterwards sold out at 4 per cent, advance : how much 
 did he sell his stock for ? 
 
 31. Sold 64 shares of Hudson River Railroad stock, at 10 per 
 cent, premium : how much did they come to ? 
 
 32. A man bought 35 shares of Utica and Syracuse Railroad 
 stock, at par, and afterwards sold them at l per cent, advance : 
 how much did he get for them ? 
 
 33. A man bought 15 shares of Albany and Schenectady Rail- 
 road stock, at 2 per cent, advance, and sold them at 10 per cent, 
 disc. : how much did he sell them for ; and how much did he lose ? 
 
 34. Bought 71 shares in the Albany Gas Co. at 5- per cent. 
 ' premium : how much did they amount to ? 
 
 35. A broker bought 48 shares of Michigan Railroad stock, 
 at. 14 per cent, discount, and sold them at 6 per cent, advance: 
 how much did he make by the operation ? 
 
 36. If I employ a broker to buy me 55 shares of Railroad stock, 
 rhich is 20 per cent, below par, and pay him per cent, broker 
 age, how much will my stock cost me ? 
 
 37. If my agent buys 78 shares of New York and PhihiJel- 
 phia Railroad stock, at 15 per cent, advance, and charges me -f 
 per cent, brokerage, how much will rnj stock cost? 
 
ARTS. 398-40 l.J INTEREST. 249 
 
 INTEREST. 
 
 398. INTEREST is the sum paid for the use of money by the 
 borrower to the lender. It is reckoned at a given per cent, per 
 annum ; that is, so many dollars are paid for the use of $100 
 for one year ; so many cents for 100 cents; so many pounds for 
 100; <fec. 
 
 OBS. The student should be careful to notice the distinction between Com- 
 mission, and Interest. The former is reckoned at a certain per cent, without 
 regard to time; (Art. 395;) the latter is reckoned at a certain percent, for one 
 year ; consequently, for longer or shorter periods than one year, like proportions 
 of the percentage for one year are taken. 
 
 The term per annum, signifies for a year. 
 
 399. The money lent, or that for which interest is paid, is 
 called the principal. 
 
 The per cent, paid per annum, is called the rate. 
 
 The sum of the principal and interest, is called the amount. 
 Thus, if 1 borrow $100 for 1 year, and agree to pay 5 per cent, 
 for the use of it, at the end of the year I must pay the lender 
 $100, the sum which I borrowed, and $5 interest, making $105. 
 The principal in this case, is $100; the interest $5; the rate 5 
 per cent. ; and the amount $105. 
 
 OBS. The term per annum, is seldom expressed in connection with the rale 
 per u'.'if., but it is always understood ; for the rate is the per cent, paid per 
 annum. (Art. 399.) 
 
 400. The rate of interest is usually established by law. It va- 
 nes in different countries and in different parts of our own country. 
 
 OBS. When no rate is mentioned, the rate established by the laws of the 
 State in which the transaction takes place, is always understood to be the one 
 intended by the parties. 
 
 40 1 . Any rate of interest higher than the legal rate, is called 
 usury, and the person exacting it is liable to a heavy penalty. 
 
 Any rate less than the legal rate may be taken, if tht parties 
 concerned so agree. 
 
 QUKST. 398. What is Imerest? How is it reckoned? Obs. What is the difference be- 
 tween Commission and Interest ? What is meant by the term per annum 7 3!>9. What is 
 nteant by the principal? The rate ? The amount ? 409. How is the rate usually deter- 
 mined ? Is it the same everywhere ? Obs. When no rate is mentioned, what rate is un- 
 derstood 7 401. What is anv rate higher than the legal rate called ? 
 

 250 
 
 INTEREST. 
 
 [SECT. XII. 
 
 4O2 The legal rates of interest, and the penalty for usury in 
 the several States of the Union, are as follows : 
 
 Penalty for Usury. 
 Forfeit of the whole debt. 
 Forfeit of three times the usury. 
 Recovery in action with costs. 
 Forfeit of three times the usury. 
 Forfeit of the usury and int. on the debt. 
 Forfeit of the whole debt. 
 Forfeit of the whole debt. 
 Forfeit of the whole debt. 
 Forfeit of the whole debt. 
 Forfeit of the whole debt. 
 Usurious contracts void. 
 Forfeit of double the usury. 
 Forfeit of double the usury. 
 Forfeit of interest and usury with costs. 
 Forfeit of three times the usury. 
 Forfeit of interest and usury. 
 Forfeit of usury and costs. 
 Usurious contracts void. 
 Usurious contracts void. 
 Usury may be recovered with costs. 
 Usuiious contracts void. 
 Forfeit of double the excess. 
 Forfeit of three times the usury, and int. due. 
 Forfeit of the usury, and the interest due. 
 Forfeit of the usury, and one fourth the debt. 
 Forfeit of usury. 
 Forfeit of interest and usury. 
 Forfeit of three times the usury. 
 Forfeit of three times the usury. 
 Usurious contracts void. 
 Usurious contracts void. 
 
 OBS. 1. On debts and judgments in favor of the United States, interest M 
 computed at 6 per cent. 
 
 2. In Canada and Nova Scotia, the legal rate of interest is 6 per cent. In 
 England and France it is 5 per cent. ; in Ireland 6 per cent. In Italy, about 
 the commencement of the 13th century, it varied from 20 to 30 per cent. 
 
 States. 
 
 Legal rates. 
 
 Maine, 
 
 6 per cent. 
 
 N. Hampshire, 
 
 6 per cent. 
 
 Veimont, 
 
 6 per cent. 
 
 M assachusetts, 
 
 6 per cent. 
 
 Rhode Island, 
 
 6 per cent. 
 
 Connecticu , 
 
 6 per cent. 
 
 New York, 
 
 7 per cent. 
 
 New Jersey, 
 
 6 per cent. 
 
 Pennsylvania, 
 
 6 per cent. 
 
 Delaware, 
 
 6 per cent. 
 
 Maryland, 
 
 6 per cent, a 
 
 Virginia, 
 
 6 per cent. 
 
 N. Carolina, 
 
 6 per cent. 
 
 S. Carolina, 
 
 7 per cent. 
 
 Georgia, 
 
 8 per cent. 
 
 Alabama, 
 
 8 per cent. 
 
 Mississippi, 
 
 8 per cent, b 
 
 Louisiana, 
 
 5 per cent, c 
 
 Tennessee, 
 
 6 per cent. 
 
 Kentucky, 
 
 6 per cent. 
 
 Ohio, 
 
 6 per cent. 
 
 Indiana, 
 
 6 per cent. 
 
 Illinois, 
 
 6 per cent, d 
 
 Missouri, 
 
 6 per cent, e 
 
 Michigan, 
 
 7 per cent. 
 
 Arkansas, 
 
 6 per cent. / 
 
 Florida, 
 
 8 per cent. 
 
 Wisconsin, 
 
 7 per cent, g 
 
 Iowa, 
 
 7 per cent, h 
 
 Texas, 
 
 10 per cent. 
 
 Dist. Columbia, 
 
 6 per cent. 
 
 a On tobacco contracts 8 per cent, b By contract as high as 10 per cent, c Bank inter- 
 est G per cent. ; conventional as high as 10 per cent, d By agreement as high as 12 pet 
 cent, e By agreement as high as 10 per cent. / By agreement, any rate not exceeding 10 
 per cent, g By contract as high as 12 per cent, h By agreement as high as 12 per cent. 
 
ARIS. 402, 403 ] INTEREST. 251 
 
 4O3. Ex. l.-What is the interest of $30 for 1 year, at 6 
 per cent. ? 
 
 Analysis. We have seen that 6 per cent, is r^g- ; that is, $6 
 for $100, 6 cents for 100 cents, &c. (Art. 386.) Since therefore 
 the interest of $1 (100 cents) for 1 year is 6 cents, the interest )f 
 $30 for the same time must be 30 times as much; and $30X-06 
 =$1.80. Ans. 
 
 Operation. We first multiply the principal by the 
 
 $30 Prin. given rate per cent, expressed in decimals, 
 
 .06 Rate. as in percentage, and point off as many de- 
 
 $1.80 Int. 1 yr. cimals in the product as there are decimal 
 places in both factors. 
 
 Ex. 2. What is the interest of $140.25 for 1 year, 1 month, and 
 10 days, at 7 per cent. ? What is the amount? 
 
 Operation. 
 
 $140.25 Prin. 1 month is ^ of a year; there- 
 
 .07 Rate. fore the interest for 1 month is -fa 
 
 12)$9.8175 Int. 1 yr. of 1 year's interest. 10 days are % 
 
 3) 8181 " 1 mo. of 1 month, consequently the interest 
 
 2727 " 10 d. for 10 days, is -J- of 1 month's inter- 
 
 $10.9083 Interest. est. The amount is found by add- 
 
 $140.25 Prin. added, ing the principal and interest to- 
 $151.1583 Amount. gether. 
 
 Note. I. In adding the principal and interest, care must be taken to add 
 dollars to dollars, cents to cents, &c. (Art. 374.) 
 
 2. When the rate per cent, is Jess than 10, a cipher must always.be prefixed 
 to the figure denoting it. (Art. 387. Obs. 1.) It is highly important that the 
 principal and the rate should both be written correctly, in order to prevent mis- 
 takes in pointing off the product. 
 
 Ex. 3. What is the interest of $250.80 for 4 years, at 5 per 
 cent. 9 What is the amount ? 
 
 Szlution. $250.80X.05=$12.54, the interest for 1 year. 
 Now $12.54X 4=$50.16, " " 4 years. 
 
 And $250.80+$50.16=$300.96, the amount required. 
 
252 INTEREST. [SECT. X.I1. 
 
 4O J ; . From the foregoing illustrations and, principles we de- 
 duce the following general 
 
 RULE FOR COMPUTING INTEREST. 
 
 I. FOR ONE YEAR. Multiply the principal by the given rate, and 
 frwi tlie product point off as many figures for decimals, as tluro, 
 are decimal places in both factors. (Art. 324.) 
 
 II. FOR TWO OR MORE YEARS. Multiply the interest of 1 yew 
 "by tlie given number of years. 
 
 III. FOR MONTHS. Take such a fractional part of 1 year's in- 
 terest, as is denoted by the (jiven number of months. 
 
 IV. FOR DAYS. Take such a fractional part of one month's in- 
 terest, as is denoted by the (jiven number of days. 
 
 The amount is found by adding the principal and interest together. 
 
 OBS. 1. The reason of this rule is evident from the consideration that the 
 given rate per cent, per annum denotes kundrcdths. (Arts. 386. 39S.) Now 
 when the rate is 6 per cent. \ve multiply by .OH, when 7 per cent, by .07, &c., 
 and point off two figures in the product ; consequently the result will be the 
 same as to multiply by -y^p -j-fo, &c. 
 
 2. In calculating interest, a month, whether it contains 30 or 31 days, or 
 even but 28 or 29, as in the case of February, is assumed to be one twelfth, of a 
 year. Therefore, for 1 month we take -fa of 1 year's interest; for 2 months, --; 
 for 3 months, \ ; for 4 months, -^ ; for G months, ; for 8 months, -f , &c. 
 
 Again, 30 days are commonly considered a month; consequently the interest 
 for 1 day, or any number of days under 30, is so many thirLieUis of a month's 
 interest. (Art. 303. Obs. 2.) Therefore, for 1 day we take -^ of 1 month's 
 interest ; for 2 days, -fa ; for 3 days, -fa ; for 5 days, ; for 10 days, --, &c. 
 
 This practice seems to have been originally adopted on account of its 1 , con- 
 venience. Though not strictly accurate, it is sanctioned by general usage. 
 
 3. Allowing 30 days to a month, and 12 months to a year, a year would con- 
 tain only 300 days, which in point of fact is -3-^3-, or -fa less than an ordinary 
 year. H:nce, 
 
 To find the interest for any number of days with entire accuracy, we must 
 take so many 3G5ths of 1 year's interest, as is demoted by the given number 
 of days ; or. find the interest for the days as above from this subtract -fa of 
 
 QI-EST. 404. How is interest computed for a year? I 1 , w for any number of years J 
 How far months? How for days? How find the amoun' 1 Obs. In reckoning interest, 
 what part of a year is a month considered ? How many* days are commonly considered a 
 Bumth ? Ls this practice accurate ? 
 
Allf. 404.] INTEREST. 253 
 
 itself, and the remainder will be the exact interest. The laws of New York, 
 and several other states, require this deduction to be made. 
 
 In business, when the mills in the result are 5, or o\er, it is customary to 
 add I to the cents ; if under 5, to disregard them. 
 
 EXAMPLES. 
 
 ] What is the interest of $423 for 1 yr., at 7 per cent. ? 
 
 2. What is the interest of "240.31 for 3 yrs., at 6 per cent ? 
 
 3. What is the interest of $403.67 for 2 yrs., at per cent ? 
 
 4. What is the interest of $040 for 1 yr., at 8 per cent ? 
 
 5. What is the interest of $430.45 for 2 yrs.. at 7 per cent. ? 
 
 6. What is the interest of $185.00 for 4 yrs., at per cent. ? 
 
 7. What is the interest of $804.80 for 5 yrs., at 4| per cent. ? 
 
 8. What is the interest of $703 for 4 months, at 7 per cent.? 
 
 9. What is the interest of $940.20 for mo., at per cent. ? 
 
 10. What is the interest of $243.10 for 5 mo., at 8 per cent. ? 
 
 11. What is the interest of $195.82 for 7 mo., at per cent. ? 
 
 12. What is the interest of $425.35 for 9 mo., at per cent. ? 
 
 13. At 7 per cent., what is the int. of $738 for 1 yr. and 2 mo. ? 
 
 14. At per cent., what is the int. of $894 for 1 yr. and 8 mo. ? 
 
 15. At 7 per cent., what is the amount of $920 for mo. ? 
 10. At 7 per cent., what is the amt. of $048 for 2 mo. 15 d. ? 
 
 17. At per cent., what is the amt. of $1000 for 1 mo. lid.? 
 
 18. At 5 per cent., what is the amt. of $1505.45 for 3 mo. ? 
 
 19. At per cent., what is the amt. of $872 for 4 mo. ? 
 
 20. What is the int. of $081 for 10 days, at per cent. ? 
 
 21. What is the int. of $483.20 for 15 d., at 7 per cent. ? 
 
 22. What is the int. of $509.40 for 20 d., at per cent. ? 
 
 23. What is the amt. of $95 for 1 yr. and mo., at 5 percent. ? 
 
 24. What is the amt. of $148 for 8 mo. ]2 d., at per cent ? 
 
 25. What is the amt. of $700 for 30 d., at 7 per cent. ? 
 20. What is the int. of $340 for 00 d., at 5i per cent, ? 
 
 27. What is the int. of $4(585 for 90 d., at 0^ per cent. ? 
 
 28. What is the amt. of $3293 for 30 d., at 7 per cent. ? 
 
 29. What is the amt. of $5205 for 15 d., at (: per cent. ? 
 
 30. What is the int. of $8310 for 10 d., at 7 per cent. ? 
 
 31. What is the int. of $50025 for 21 d., at 7 per cent. ? 
 82. What is ihe amt. of $05256 for 4 mo., at 7 per cent. ? 
 
254 INTEREST. [SECT. XII 
 
 SECOND METHOD OF COMPUTING INTEREST. 
 
 405. There is another method of computing interest, which 
 is very simple and convenient in its application, particularly when 
 the interest is required for months and days, at 6 per cent. 
 
 406. We have seen that for 1 year, the interest of 81 at 6 
 per cent, is 6 cents., or $.06 ; (Art. 404 ;) therefore, 
 
 For 1 month, the interest of $1 is -^ of 6 cents, which is $.005 ; 
 
 " 2 months, " is ^, or -f of 6 cents, " " 001; 
 
 " 3 months, is -j^-, or | of 6 cents, " ' .015; 
 
 " 4 months, " " is -^, or -^ of 6 cents, " .02 1 
 
 " 5 months, is -j\, of G cents, " " .025; 
 
 " 6 months, " " is -$j, or \ of 6 cents, " " .03; 
 
 Hence, The interest of $1 for 1 month, at 6 per cent., is 5 mills; 
 for every 2 months, it is 1 cent ; and for any number of months, 
 it is as many cents, or hundredths of a dollar, as 2 is contained 
 times in the given number of montJis. 
 
 407. Since the interest of $1 for 1 month (30 days) is 5 mills, 
 or $.005, (Art. 406,) 
 
 For 6 days ( of 30 days) the interest of SI is of 5 mills, or $.001 ; 
 
 12 days (| of 30 days) " is -f of 5 mills, or .002; 
 
 18 days (f of 30 days) " is -f of 5 mills, or .003 ; 
 
 " 3 days (-fL of 30 days) " is fj of 5 mills, or .0005 ; 
 
 That is, the interest of $1 for every 6 days, is 1 mill, or $.001 ; 
 and for any number of days, it is as many mills, or thousandth 
 of a dollar, as 6 is contained times in the given number of days. 
 
 4O8 Hence, to find the interest of $1 for any number of 
 days, at 6 per cent. 
 
 Divide the given number of days by 6, and set the first quotient 
 figure in thousandths' place, when the days are 6, or more than 6 ; 
 but in ten thousandths' place, when they are less than 6. 
 
 OBS. For 60 days (2 mo.) the interest of $1 is 1 cent; (Art. 406,; v* nen, 
 therefore, the number of days is 60 or over, the first quotient figure must 
 occupy hundredth^ place. 
 
 QUEST. 408. How find the interest of $1 for any number of days, at 6 per cent. 1 
 
ARTS. 405 -409. J INTEREST. 255 
 
 Ex. 1. What is the interest of $185 for 1 year, 6 months and 
 18 days, at 6 per cent.? 
 
 Analysis. The interest of $1 for 1 year is , Operation. 
 6 cents ; for 6 months it is 3 cents; and for $185 Prill. 
 
 18 days it is 3 mills. (Arts. 406, 407.) Now .093 Int.$l. 
 
 .06 + .03+.003=$.093. Since therefore the 555 
 
 interest of $1 for the given time is $.093, the 1665 
 
 interest of $185 must be 185 times as much. $17.205 Ans. 
 
 4O9 Fr< m these principles we may derive a 
 
 SECOND RULE FOR COMPUTING INTEREST. 
 
 1. To compute the interest on any sum, at 6 per cent. 
 Multiply the principal by the interest of $1 for the given time, 
 
 at 6 per cent., and point off the product as in multiplication of 
 decimals. (Art. 324.) 
 
 II. . To compute int. at any rate, greater or less than 6 per cent. 
 
 First find the interest on the given sum at 6 per cent, j then, 
 add to this interest, or subtract from it, such a fractional part of 
 itself, as the required rate exceeds or falls short of 6 per cent. 
 
 The amount is found by adding the principal and interest to- 
 gether as in the former method. (Art. 404.) 
 
 Ops. ' . The amount may also be found by multiplying the given principal by 
 the amount of one dollar for the time. 
 
 2. The. rental of the first part of this rule, is manifest from the principle that 
 ihe interest of 2 dollars for any given time and rate, must be twice as much as 
 the interest of 1 dollar for the same time arid rate ; the interest of 50 dollars, 
 50 times as much as that of 1 dollar, &c. 
 
 3. When the required rate is 7 per cent., we first find the interest at 6 per cent., 
 then add - of it to itself; if 5 per cent., subtract 4- of it from itself, &c., for the 
 obvious reason, that 7 per cent, is once and 1 sixth, or -^ of G per cent. ; 5 
 per cent, is only A of 6 per cent., &c. 
 
 \. When the decimal denoting the int. of $'1 for the days, is long, or is a repa- 
 tend, it is more accurate to retain the common fraction. (Art. 387. Obs. 2.) 
 
 2. What is the interest of $746 for 4 months and 18 days, at 
 6 pei ont.? Ans. $17.153. 
 
 Quasi. 409. What is the second method of computing interest, at 6 per cent. ? Whea 
 the ti.te uei cent, is greater or less than 6 per cent., how proceed ? 
 
2,56 INTEREST. [SECT. XII. 
 
 3. What is the interest of $240 for G months and 12 days, at 
 7 pei cent. ? 
 
 Operation. 
 
 240 Pnn. The interest of $1 for 6 mo. at 6 
 
 .032 Int. of $1. per ct., is .03 ; for 12 d. it is .002 ; 
 
 480 . and .03 + .002=8.032. 
 
 720 The required rate is 1 per cent. 
 
 6)$7.680=Int. at 6 perct. more than 6 per cent. ; we there- 
 
 1.280= of 6 per ct. fore find the interest at 6 per cent., 
 
 Ans. $8.960 Int. at 7 per ct. and add | of it to itself. 
 
 4. What is the interest of $t>80 for 3 mo., at 5 per cent. ? 
 
 5. What is the interest of $213.08 ibr 1 mo., at 6 per cent. ? 
 
 6. What is the interest of $859 for 1 yr. 2 mo., at 7 per cent.*? 
 
 7. What is the interest of $708 for 1 yr. 7 mo., at 8 per cent.? 
 
 8. What is the interest of $684 for 9 mo., at 6 per cent. ? 
 
 9. At 7 per cent., what is the amount of $387 for 5 mo. ? 
 
 10. At 4 per cent., what is the amt. of $1125 for 1 yr. 2 mo. ? 
 
 11. At 6 per cent., what is the amt. of $1056 for 10 mo. 24 d. ? 
 
 12. At 6 per cent., what is the int. of $1340 for 1 mo. 15 d. ? 
 
 13. At 6 per cent., what is the int. of $815 for 2 mo. 21 d. ? 
 
 14. At 8 per cent., what is the amt. of $961 for 4 mo. 10 d. ? 
 
 15. What is the int. of $2345.10 for 6 mo., at 7 per cent. ? 
 
 16. What is the int. of $1567.18 for 4 mo., at 7 per cent. ? 
 
 17. What is the int. of $3500 for 11 mo., at 10 per cent. ? 
 
 18. What is the int. of $39.375 for 2 yrs., at 12 per cent. ? 
 
 19. What is the int. of $113.61 for 5 yrs., at 15 per cent.? 
 
 20. What is the int. of $1000 for 2 yrs., at 20 per cent. ? 
 
 21. What is the int. of $1260.34 for 10 yrs., at 13 per cent. ? 
 
 22. At 16 per cent,, what is the int. of $150 for 6 years. ? 
 
 23. At 30 per cent., what is the int. of $300 for 1 year. ? 
 
 24. What is the amt. of $12645 for 10 i, at 6 per cent. ? 
 
 25. What is the amt. of $16285 for 24 1, at 7 per cent. ? 
 
 26. At 4^ per cent., what is the int. of $10255 for 8 months ? 
 
 27. At 5| per cent., what is the int. of $17371 for 3 months ? 
 
 28. What is the amt. of $1 for 100 yrs., at 7 per cent. ? 
 
 29. What is the amt. of 1 cent for 100 yrs., at 6 per cent. ? 
 
ARTS. 410, 411. J INTEREST. 257 
 
 41 0. Since the interest of $1 at G per cent, for 12 rno. is 6 
 cents, (Art. 406,) for G mo. it must be 3 cents ; for 3 mo., !- cents ; 
 for 2 mo., 1 cent; for 1 mo. or 30 d. \ cent; for 15 d., | cent; 
 for 20 d. i- cent, &c. That is, the interest of $1 at G per cent. 
 is as many cents as are equal to half the given number of months. 
 
 411. Hence, to compute interest at 6 per cent, by montlis. 
 Multiply the principal by half the number of months, and point 
 
 '}Jf' two more figures for decimals in tJie product than there are deci- 
 mal places in the multiplicand. 
 
 OBS. 1. When there are years and days, reduce the years to months, and 
 the days to a common fraction of a month. 
 
 Or, divide the days by 3, and annex the quotient to the months considered 
 as hundredth*; half of the number thus produced will be the decimal multiplier 
 
 2. The latter method is the same as dividing the days by 6, and setting the 
 fii st quotient figure in thousandth's place ; for, we divide the days by 3 and 
 2, and 3X2=6. (Arts. 407, 408.) 
 
 30. What is the int. of $460.384 for 8 mos. and 15 d., at 6 per ct. ? 
 
 Operation. 
 
 $460.384 We multiply by 4|, for, 8 months-}- 15 
 
 4f days=8i months, and 8^-f-2 = 4|. And 
 
 1841536 since there are three decimals in the mul- 
 
 115096 tiplicand, we point off 5 in the product. 
 $19.56632 Am. 
 
 31. What is the interest of $780 for 4 months, at 6 per cent. ? 
 
 32. What is the interest of $1406 for 3 mo., at 6 per cent. ? 
 
 33. What is the interest of $109 for 2 mo., at 7 per cent. ? 
 
 34. What is the interest of $119.45 for 8 mo., at 6 per cent. ? 
 
 35. What is the interest of $618 for 1 yr. 3 mo., at 6 percent. ? 
 
 36. Wha', is the interest of $861 for 2 yrs. 6 mo., at 6 per cent. ? 
 
 37. What is the interest of $936.40 for 3 yrs., at 6 per cent. ? 
 
 38. What is the interest of $4526 for 6 mo. 2 d., at 6 per cent. ? 
 
 39. What is the interest of $8246 for 10 mo., at 7 per cent. ? 
 
 40. What is the interest of $31285 for 3 mo., at 5 per cent. ? 
 
 41. What is the interest of $17500 for 1 yr. 3 mo., at 7 per ct. ? 
 
 42. What is the amount of $3286 for 8 mo. 15 d., at 6 per ct. t 
 
 43. What is the amount of $15876 for 5 mo. 18 d., at 6 per ct. ? 
 
258 APPLICATIONS OF [SECT. XIl. 
 
 412. We have seen that the interest of $1 at 6 per cent, for 
 any number of days is equal to as many mills, as 6 is contained 
 times in the given days. (Art. 407.) Hence, 
 
 413. To compute interest at 6 per cent, by days. 
 Multiply the principal by one sixth of the given number of days, 
 
 and point off three more figures for decimals in the product t/tan 
 tlicre are decimal places in the principal. (Art. 411. Obs. 2.) 
 
 Or, mdltiply the principal by the given number of days, divid 
 the product by 6, and point off the quotient as above. 
 
 OBS. The product is in mills and parts of a mill. The object, therefore, of 
 pointing off three more places for decimals in the product than there are deci- 
 mals in. the principal, is to reduce it to dollars. (Art. 372.) 
 
 44. What is the interest of 1976.22 for 33 days, at 6 per cent. ? 
 Solution. | of 33 d. = 5i-; and $976.22X5-^=5369.21 mills. 
 
 Pointing off 3 more decimals, we have $5.36921. Ans. 
 
 45. What is the interest of $536.30 for 24 days, at 6 per cent. ? 
 
 46. What is the interest of $7085 for 63 d., at 6 per cent. ? 
 
 47. What is the interest of $8126.21 for 8 d., at 6 per cent. ? 
 
 48. What is the interest of $25681 for 93 d., at 6 per cent. ? 
 
 49. What is the interest of $764.85 for 114 d., at 6 per cent. ? 
 
 APPLICATIONS OP INTEREST. 
 
 414. In the application of interest to business transactions, 
 the following particulars deserve attention. 
 
 1. A promissory note is a writing which contains a promise of the payment 
 of money or other property to another, at or before a time specified, in consid- 
 eration of value received by the promiser or maker of the note. 
 
 Unless a note contains the words " value received," by some authorities it 
 is deemed invalid ; consequently these words should always be inserted. 
 
 2. The person who signs a note is called the r.iaker, drawer, or giver of the 
 note. The person to whom a note is made payable, is called the payee; the 
 person who has the legal possession of a note, is called the Jwlder of it. 
 
 3 A note which is made payable " to order '," " or bearer'* is said to be ncgo 
 tiab.'e ; that is, the holder may sell or transfer it to whom he pleases, and it can 
 be collected by anyone who has lawful possession of it. Notes without these 
 words are not negotiable. (See Nos. 1, 2.) 
 
 4. If the holder of a negotiable note which is made payable to order wishes 
 to sell or transfer it, the law requires him to endorse it, or write his name on 
 the back of it. The person to whom it is transferred, or-the holder of it, ia 
 
A.UTS. 4 J 2-4 15.] INTEREST. 259 
 
 then empowered to collect it of the drawer; if the drawer is unable, or icfuset 
 to pay it, then the endorser is responsible for its payment. (See No. 1.) 
 
 5. When a note is made payable to the bearer, the holder can sell or trans- 
 fer it without endorsing it, or incurring the liability lor its payment. Bank 
 notes or bills are of this description. (See No. 2.) 
 
 6. When a note is made payable to any particular person without the words 
 order or bearer it is nob negotiable; for, it cannot be, collected or sued except in 
 the name^j^he person to whom it is made payable. (See No. 3.) 
 
 7. A note should always specify the time at which it is to be paid; but if no 
 time is mentioned, the presumption is that it is intended to be paid on demand, 
 and the giver must pay it when demanded. 
 
 8. According to custom and the statutes of most of the States, a note or 
 draft is not presented for collection until ikrcc days after the time specified for 
 its payment. These three days are called dai/s of grace. Interest is therefore 
 reckoned for three days more than the time specified in the note. When the 
 last day of grace comes on Sunday, or a national holiday, as the 4th of July, 
 &c.. it is customary to pay a note on the day previous. 
 
 9. If a note is not paid at maturity or the time specified, it is necessary for 
 the holder to notify the endorser of the fact in a legal manner, as soon as cir- 
 cumstances will admit ; otherwise the responsibility of the endorser ceases. 
 
 10. Notes do not draw interest unless they contain the words " with inter- 
 est." But if a note is not paid when it becomes due, it then draws legal in- 
 cerest till paid, though no mention is made of interest. (Art, 400. Obs.) 
 
 11. Notes which contain the words " with interest" though the rate is not 
 mentioned, are entitled to the legal rate established by the State in which the 
 note is made. In writing notes therefore it is unnecessary to specify the rate, 
 unless by agreement it is to be less than the legal rate. 
 
 12. When a note is made payable on a given day, and in a specified article 
 of merchandise, as grain, stock, &c., if the article specified is not tendered at 
 the given time and place, the holder can demand payment in money. Such 
 notes, are not negotiable ; nor is the drawer entitled to the days of grace. 
 
 13. When two or more persons jointly and severally give their note, it may 
 be collected of either of them. (See No. 4.) 
 
 14. The sum for which a note is given, is called the principal, or face oft\it 
 note; and should always be written out in words. 
 
 415* When it is required to compute the interest on a note, 
 we must first find the time for which the note has been on inter 
 est, by subtracting the earlier from the later date; (Art. 303; 
 then cast the interest on the face of the note for the time, by 
 either of the preceding methods. (Arts. 404, 409.) 
 
 OBS. In determining the time, the day on which a note is dated, and tlu't 
 on which it becomes due should not both be reckoned ; it is customary to ex 
 elude the former. 
 
 T.H. 12 
 

 260 APPLICATIONS OF [SECT. XII. 
 
 Ex. 1. What is the interest due on a note j>f $625 frcm FeL 2d, 
 1846, to June 20th, 1847, at 6 per cent.? 
 
 Ojxration. $625 Prin. 
 
 Yrs. mo. ds. 
 
 1847 " 6 " 20 
 
 1846 " 2 " 2 
 1 " 4 " 18 
 
 Int of $1 . 
 
 5000 
 
 $5L875 Ans. 
 Compute the interest on the following notes : 
 
 . _ (No. 1.) 
 
 $450. NEW YORK, June 3d, 1847. 
 
 2. Sixty days after date, I promise to pay George Baker, or 
 order, Four Hundred and Fifty Dollars, with interest, value re- 
 ceived. ALEXANDER HAMILTON. 
 
 _ (No. 2.) 
 
 $630. BOSTON, Aug. 5th, 1847. 
 
 . 3. Thirty days after date, I promise to pay Messrs. Holmes & 
 Homer, or bearer, Six Hundred and Thirty Dollars, with interest, 
 value received. JAMES UNDERWOOD. 
 
 (No. 3.) 
 
 $850. PHILADELPHIA, Sept. 16th, 1847. 
 
 4. Four months after date, I promise to pay Horace Williams, 
 Eight Hundred and Fifty Dollars, with interest, value received. 
 
 JOHN C. ALLEN. 
 
 (No. 4.) 
 $1000. CINCINNATI, Oct. 3d, 1847. 
 
 5. For value received, we jointly and severally promise to pay 
 to the order of Wm. D. Moore & Co., One Thousand Dollars, in 
 one year from date, with interest. JOSEPH HENRY, 
 
 SANDFORD ATWATER. 
 
 6. What is the interest on a note of $634 from Jan. 1st, 1 846, 
 to March 7th, 1847, at 6 per cent. ? 
 
ARTS. 415, 416.] INTEREST. 201 
 
 7. What is the interest on a note of $820 from April 16th, 
 1846, to Jan. 10th, 1847, at 6 per cent. ? 
 
 8. What is the interest on a note of $615.44 from Oct. 1st, 
 1836, to June 13th, 1840, at 4 per cent. ? 
 
 9. What is the interest on a note of $1830.63 from Aug. 16th, 
 1841, to June 19th, 1842, at 7 per cent. ? 
 
 10. What is the amount due on a note of $520 from Sept. 2d, 
 1846, to March 14th, 1847, at 5 per cent. ? 
 
 1 1. What is the amount due on a note of $25000 from Aug. 
 17th, 1845, to Jan. 17th, 1846, at 7 percent? 
 
 12. What is the amount due on a note of $6200 from Feb. 3d, 
 1846. to Jan. 9th, 1847, at 6 per cent.? 
 
 PARTIAL PAYMENTS. 
 
 416 When partial payments are made and endorsed upon 
 Notes and Bonds, the rule for computing the interest adopted by 
 the Supreme Court of the United States, is the following. 
 
 I. " The rule for casting interest, when partial payments have 
 been made, is to apply the payment, in tJie first place, to the dis- 
 charge of the interest then due. 
 
 II.. " If the payment exceeds the interest, the surplus goes towards 
 discharging the principal, and the subsequent interest is to be com- 
 puted on the balance of principal remaining due. 
 
 III. " If the payment be less than the interest, the surplus of 
 interest must not be taken to augment the principal / but interest 
 continues on the former principal until the period when the pay- 
 ments, taken together, exceed tlie interest due, and then the surplus 
 is to be applied towards discharging tlie principal ; and interest is 
 to be computed on the balance as aforesaid" 
 
 Note. The above rule is adopted by New York, MassachiLsetts, and most 
 if the other States of the Union. It is given in the language of the distin- 
 grdshed Chancellor Kent. Johnson's Chancery Reports, Vol. I. p. 17. 
 
 UUHST. 416. What is the genera\ method of casting interest on Notes and Bonds, Then 
 partial pay merits have been made ? 
 
 
262 APPLICATIONS OF [SECT. XII 
 
 $965. NEW YORK, March 8th, 1843.. 
 
 13 For value received, I promise to pay George B. Granniss, 
 or order, Nine Hundred and Sixty-five Dollars, on demand, with 
 interest at 7 per cent. HENRY BROWN. 
 
 The following payments were endorsed on this note : 
 
 Sept. 8th, 1843, received $75.30. 
 
 June 18th, 1844, received $20.38. 
 
 March 24th, 1845, received $80. 
 What was due on taking up the note, Feb. 9th, 1846 ? 
 
 Operation. 
 
 Principal, - 1*965.00 
 
 Interest to first payment, Sept. 8th. (6 months,) 33.775 
 
 Amount due on note Sept. 8th, $998.775 
 
 1st payment, (to be deducted from amount,) - 75.30 
 
 Balance due after 1st pay't., Sept. 8th, 1843, - $923.475 
 Interest on Balance to 2d pay't., June ) ^ Kf . n * 
 
 18th, (9 mo. 10 d.,) 1 
 
 2d pay't., (being less than int. then due,) 20.38 
 Surplus int. unpaid June 18th, 1844, $29.898 
 
 Int. continued on Bal. from June 18th, ) 
 
 to March 24th, 1845, (9 mo. 6 d.,) $ 49 ' 559 79 ' 457 
 
 Amount due March 24th, 1845, - $1002.932 
 3d pay't., (being greater than the int. now due,) > 
 
 is to be deducted from the amount, > ' 
 
 Balance due March 24th, 1845, $922.932 
 
 . Int. on Bal. to Feb. 9th, (10 mo. 15 d.,) - 56.529 
 
 Bal. due on taking up the note, Feb. 9th, 1846, $979.461 
 
 $650. BOSTON, Jan. 1st, 1842. 
 
 14. 1 or value received, I promise to pay John Lincoln, or 
 orler, Six Hundred and Fifty Dollars on demand, with interest 
 at 8 per cent. GEORGE LEWIS. 
 
 Endorsed, Aug. 13th, 1842, $100. 
 Endorsed, April 13th, 1843, $120. 
 What was due on the note, Jan. 20th, 1844? 
 
ART. 417.] INTEREST. 263 
 
 $2460, PHILADELPHIA, April 10th, 1844. 
 
 45. Four months after date, I promise to pay James Buchanan, 
 or order, Two Thousand Four Hundred and Sixty Dollars, with in- 
 terest, at 6 per cent., value received. 
 
 GEORGE WILLIAMS. 
 
 Endorsed, Aug. 20th, 1845, $840. 
 Dec. 26th, 1845, $400. 
 May 2d, 1846, $1000. 
 How much was due Aug. 20th, 1846 ? 
 
 $500a NEW ORLEANS, May 1st, 1845. 
 
 16. Six months after date, I promise to pay John Fairfield, or 
 order, Five Thousand Dollars, with interest at 5 per cent., value 
 received. WILLIAM ADAMS. 
 
 Endorsed, Oct. 1st, 1845, $700. 
 " Feb. 7th, 1846, $45. 
 Sept. 13th, 1846, $480. 
 What was the balance due Jan. 1st, 1S47 ? 
 
 CONNECTICUT RULE. 
 
 417. I. " Compute the interest on the principal to the time of the first pay- 
 ment ; 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 pay- 
 ment, and then deduct the payment as above ; and in like manner, from one 
 payment to another, till all the payments are absorbed; provided the time be- 
 tween one payment and another be one year or more." 
 
 II. " If any payments be made before one year's interest has accrued, then 
 compute the interest on the principal sum due on the obligation, for 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 as above." 
 
 III. " If a year extends beyond the time of payment, then find the tun tint 
 of the principal remaining unpaid up to the time of settlement, likewise the 
 amount of the endorsements from the time they were paid to the time of settle- 
 ment, and deduct the sum of these several amounts from the amount of the 
 principal." 
 
 " If any payments be made of a 'ess sum than the interest arisen at the time 
 of such payment, no interest is to e computed, but only on the principal sum 
 for any period." Kirby's Reports. 
 
264 APPLICATIONS OF [SECT. XIL 
 
 THIRD RULE. 
 
 4 1 . First find the amount of the given principal for the whole time ; then 
 find live amount of each payment from the time it was endorsed to ttvc time of 
 settlement. Finally, subtract the amount of the several payments from iht 
 amount of the principal, and the remainder will be Uic sum due. 
 
 Note.\i will be an excellent exercise for the pupil to cast the interest on 
 the preceding notes by each of the above rules. 
 
 419. To compute Interest on Sterling Money. 
 
 17. What is the interest of 241, 10s. Gd. for 1 year, at 6 pei 
 cent. ? 
 
 Operation. 
 
 241.525 Prin. We first reduce the 10s. 6d. to the 
 
 .06 Rate. decimal of a pound, (Art. 346,) then 
 
 14.49150 Int. 1 yr. multiply the principal by the rate, 
 
 20s.=l. and point off the product as in Art. 
 
 s. 9.83000 404. The 14 on the left of the deci- 
 
 12d.=ls. mal point, denotes pounds; the fig- 
 
 d. 9.96000 ures on the right are decimals of a 
 
 4f.=ld. pound, and must be reduced to shil- 
 
 far. 3.84000 lings, pence, and farthings. (Art. 348.) 
 
 Ans. 14, 9s. 9fd. Hence, 
 
 419. a. To compute the interest on pounds, shillings, pence, 
 and farthings. 
 
 Reduce the given shillings, pence, and farthings to the decimal 
 of a pound ; (Art. 346 ;) then find tlie interest as on dollars and 
 cents ; finally, reduce the decimal figures in the answer to shillings, 
 pence, and farthings. (Art. 348.) 
 
 18. What is the amount of 156, 15s. for 1 year and 4 months, 
 at 5 per cent. ? Ans. 167, 4s. 
 
 19. What is the int. of 275, 12s. 6d. for 1 yr., at 7 per cent. 
 
 20. What is the int. of 89, 7s. 6id. for 2 yrs., at 5 percent. 
 
 21. What is the int. of 500 for 6 mo., at 5 per cent. ? 
 
 22. What is the amt. of 1825, 10s. for 8 mo., at 6 per cent. ? 
 
 23. What is the amt. of 2000 for 10 yrs., at 4| per cent. ? 
 
 QUEST. 419. How is interest computed on pounds, shillings, and peace 1 
 
ARTS. 418-421. J INTEREST. 265 
 
 PROBLEMS IN INTEREST. 
 
 4 2O. It will be observed that there are four parts or terms 
 connected with each of the preceding operations, viz : t/ie prind~ 
 pal, the rate per cent., the time, and the interest, or the amount. 
 These parts or terms have such a relation to each other, that if 
 any three of them are given, the other may be found. The ques- 
 tions, therefore, which may arise in interest^ are numerous ; but 
 tiny may be "reduced to a few general principles, or Problems. 
 
 OE? A number or quantity is said to be given, when its value is stated ; or 
 may bo eaily inferred from the Conditions of the question under consideration. 
 Thus, when the principal and interest are known, the amount may be said to 
 be given, because it is merely the sum of the principal and interest. So, if the 
 principal ind the amount are known, the interest, may be said to be given, be- 
 cause it is the difference between the amount and the principal. 
 
 PROBLEM I. 
 
 421. To find tJte INTEREST, the principal, rate per cent., and 
 the time being given. 
 
 This problem embraces all the preceding examples pertaining 
 to Interest, and has already been illustrated. 
 
 PROBLEM II. 
 
 To find the RATE PER CENT., t/ie principal, the interest, and the 
 time being given. 
 
 Ex. 1. A man borrowed $80 for 5 years, and paid $3G for the 
 use of it : what was the rate per cent. ? 
 
 Analysis. The interest of $80 at 1 per cent, for 1 year is 80 
 cents ; (Art. 404 ;) consequently for 5 years it is 5 times as much, 
 and $.80 X 5 $4. Now since $4 is 1 per cent, on the principal for 
 the given time, $3G must be ^ of 1 per cent., which is equal to 
 9 per cent. (Art. 190.) 
 
 Or, we may reason thus : Since $4 is 1 per cent, on the princi- 
 pal for the given time, $36 must be as many per cent, as $4 is 
 contained times in $36 ; and $36-r-$4=9. Ans. 9 per cent. 
 
 QUKST 420. How many terms are connected with each of the preceding examples 1 
 What are they 1 When three are given, can the fourth be found ? Obs. When is a nun* 
 her or auantity said to be given ? 
 
266 APPLICATIONS OF [SECT. XIL 
 
 PROOF $80X-09=$7.20, the interest of $80 for 1 year at 
 
 9 per cent., and $7.20X5=$36.00, the interest for 5 years, which 
 is equal to the sum paid. Hence, 
 
 422* To find the rate per cent, when the principal, interest,' 
 and time are given. 
 
 Divide the given interest by the interest of the principal at 1 
 per cent, for tJie given time, and the quotient will be the required 
 per cent. 
 
 Or, find the interest of the principal at 1 per cent for the 
 given time ; then make the interest thus found the denominator and 
 the given interest tlie numerator of a common fraction ; reduce this 
 fraction to a whole or mixed number, and the result wilt be the per 
 cent, required. (Art. 196.) 
 
 2. If I loan $500 for 2 years, and receive $50 interest, what is 
 the rate per cent. ? Ans. 5 per cent. 
 
 3. A man borrowed $620 for 8 months, and paid $24.80 for 
 the use of it : what per cent, interest did he pay ? 
 
 4. At what per cent, interest must $2350 be loaned, to gain 
 $7 in 4 months ? 
 
 '5. At what per cent, interest must $1925 be loaned, to gain 
 $154 in 1 year? 
 
 6. A man has $12000 from which he receives $900 interest 
 annually : what per cent, is that ? 
 
 7. A man deposited $2600 in a savings bank, and received $143 
 interest annually : what per cent, was that ? 
 
 8. A man invested $4500 in the Bank of New York, and re- 
 ceived a semi-annual dividend of $lf>7.50 : what per cent, was tht 
 dividend ? 
 
 9. A man paid $16250 for a house, and rented it for $975 a 
 year : what per cent, did it pay ? 
 
 10. A hotel which cost $250000, was rented for $125'X) a year: 
 what per cent, did it pay on the cost? 
 
 11. A capitalist invested $500000 in manufacturing, and re- 
 ceived a semi-annual dividend of $12500 : what per cent, was his 
 dividend ? 
 
 ttoisT. 422. When the principal, interest and time are given, how is flie rite per ct. found f 
 
ARTS. 422, 423.] INTEREST. 267 
 
 PROBLEM III. 
 
 To find the PRINCIPAL, t/ie interest, the rate per cent., and the 
 time being given. 
 
 
 
 12. What sum must be put at interest, at 6 per cent., to gam 
 $75 in 2 years ? 
 
 Analysis. The interest of $1 for 2 years at 6 per cent., (the 
 g ~en time and rate,) is 12 cents. Now 12 cents interest is -^fc 
 ot its principal $1 ; consequently, $75, the given interest, must be 
 ffi; of the principal required. The question therefore resolves 
 itself into this : $75 is iVo" of what number of dollars ? If 75 is 
 iW, T-Jr is A of 175, which is $6|; and +=461x100, which 
 is $625, the principal required. 
 
 Or, we may reason thus : Since 1 2 cents is the interest of 1 
 dollar for the given time and rate, 75 dollars must be the interest 
 of as many dollars for the same time and rate, as 12 cents is con 
 tained times in 75 dollars. And $75 ^-.12 = 625. Ans. $625. 
 
 PROOF. $625X-06=$37.50, the interest for 1 year at the 
 given per cent., and $37.50X2=$75, the given interest. Hence, 
 
 423* To find the principal, when the interest, rate per cent., 
 and time are given. 
 
 .Divide the given interest by the interest of $1 for the given 
 time and rate, expressed in decimals ; and the quotient will be the 
 principal required. 
 
 Or, make the interest of $1 for the given time and rate, the numer- 
 ator, and 100 the denominator of a common fraction ; tlien divide 
 ttte given interest by this fraction, and the quotient will be tlie prin- 
 cipal required. (Art. 234.) 
 
 13. What sum must be put at 7 per cent, interest, to gain $63 
 in 6 months? 
 
 14. What sum must be put at 5 per cent, interest, to gain $90 
 in 4 months ? 
 
 15. What sum must be invested in 6 per cent, stock, to gain 
 $300 in 6 months ? 
 
 QnsT. 423. When the interest, rate per cent., and time arc given, how is the jrinci 
 pal found ? 
 
 12* 
 
268 APPLICATIONS OF [SECT. XII. 
 
 16. What sum must be invested in 7 per cent, stock, to gain 
 $560 in one year? 
 
 17. A man founded a professorship with a salary of Si 000 a 
 year : what sum must be invested at 7 per cent, to produce it ? 
 
 18. What sum must be put at 6 per cent, interest to pay a 
 salary of $1200 a year ? 
 
 19. What sum must be invested in 5 per cent, stock to make a 
 simi-annual dividend of $7 50 ? 
 
 20. A man bequeathed his wife $1250 a year: what sum must 
 be invested at 6 per cent, interest to pay it ? 
 
 PROBLEM IV. 
 
 To find the TIME, tlie principal, the interest, and the rate per 
 cent, being given. 
 
 21. A man loaned $200 at 6 per cent., and received $42 inter- 
 est : how long was it loaned ? 
 
 Analysis. The interest of $200 at 6 per cent, for 1 year is $12. 
 (Art. 404.) Now, since $12 interest requires the principal 1 year 
 at the given per cent., $42 interest will require the same princi^ 
 pal ^-f of 1 year, which is equal to 3 years. (Art. 196.) 
 
 Or, we may reason thus : If $12 interest requires the use of the 
 given principal 1 year, $42 interest will require the same prin- 
 cipal as many years as $12 is contained times in $42, And 
 $42 -i- $12=3.5. Ans. 3.5 years. Hence, , 
 
 424. To find the time, when the principal, interest, and rate 
 per cent, are given. 
 
 Divide the given interest by the interest of the principal at the 
 given rate for 1 year, and the quotient will be the time required. 
 
 Or, make the given interest the numerator, and the interest of the 
 principal for 1 year at tlie given rate the denominator of a common 
 fraction ; reduce this fraction to a whole or mixed number, and it 
 vill be the time required. 
 
 OBS. If the quotient contains a decimal of a year, it should be reduced to 
 months and days. (Art. 348.) 
 
 QUKST. 424. When the principal, interest, and rate per cent, are given, how is the time 
 found ? Obs. When the quotient contains a decimal of a year, what should be done with it 1 
 
ART. 424 ] 
 
 INTEREST 
 
 269 
 
 22. A man loaned $765.50, at 6 per cent., and received $183.72 
 interest : how long was it loaned ? 
 
 23. In what time will 8850 gain $29.75, at 7 per cent, per 
 annum ? 
 
 24. A man received 8136.75 for the use of $1820, which 
 was C per cent, interest for the time : what was the time ? 
 
 25. Tn what time will $6280 gain $471, at 5 per cent, interest? 
 20. li >w long will it take $100, at 5 per cent., to gain $100 
 
 interest; that is, to double itself? 
 
 Operation. The. interest of $100 for 1 year, at 5 per cent., 
 
 $5)^100 is $5. (Art. 404.) 
 
 20 Ans. 20 years. 
 
 PROOF. $100 X .05 X 20=$100, the given principal. (Art. 404.) 
 
 TABLE, 
 
 Showing in what time any given principal will double itself at any rate, 
 from 1 to 20 per cent. Simple Interest. 
 
 Per cent. 
 
 Years. 
 
 Per cent. 
 
 Years. 
 
 Per cent. 
 
 Years. 
 
 Per cent.' 
 
 Years. 
 
 1 
 
 100 
 
 6 
 
 101 
 
 11 
 
 9-rV 
 
 16 
 
 6| 
 
 2 
 
 50 
 
 7 
 
 14? 
 
 12 
 
 8* 
 
 17 
 
 Mf 
 
 3 
 
 33i 
 
 8 
 
 12* 
 
 13 
 
 VA 
 
 18 
 
 5f 
 
 4 
 
 25 
 
 9 
 
 iii 
 
 14 
 
 H 
 
 19 
 
 5A 
 
 5 
 
 20 
 
 10 
 
 10 
 
 15 
 
 ci 
 
 20 
 
 5 
 
 27. How long will it take $365 to double itself, at 6 per cent. ? 
 
 28. How long will it take $1181 to double itself, at 7 per cent. ? 
 
 29. In what time will $2365.24 double itself at 7 per cent.? 
 
 30. In what time will $5640 double itself, at 10 per cent. ? 
 
 31. How long will it take $10000 to gain $5000, at 6 per cent. 
 Interest ? 
 
 32. A man hired $15000, at 7 per cent., and retained it till the 
 principal and interest amounted to $25000 : how long did he 
 have it ? 
 
 33. A man loaned his clerk $25000 to go into business, and 
 agreed to let him have it, at 5 per ct., till it amounted tc $60000 : 
 how long did he have it ? 
 
270 COMPOUND [SECT. XIL 
 
 COMPOUND INTEREST. 
 
 425* Compound Interest is the interest arising not only from 
 the principal, but also from the interest itself, after it becomes 
 due. 
 
 OBS. Compound Interest is often called interest upon interest. When inter- 
 est is paid on the principal only, it is called Simple Interest. 
 
 Ex. 1. What is the compound interest of $842 for 4 years, at 
 6 per cent. ? 
 
 Operation. 
 $842.00 Principal. 
 
 $842X.06 = 50.52 Int. for 1st year. 
 
 ~892.52 Amt. for 1 year. 
 
 $802.52 X. 00= 53.55 Int. for 2d year. 
 
 946.07 Amt. for 2 years. 
 
 $946.07X.06 = 56.76 Int. for 3d year. 
 
 1002.83 Amt. for 3 years. 
 
 $1002.83 X- 06= 00.17 Int. for 4th year. 
 
 1063.00 Amt. for 4 years. 
 
 842.00 Prin. deducted. 
 Ans. $221.00 Compound int. for 4 years. 
 
 426. Hence, to calculate compound interest. 
 
 Cast the interest on the given principal for 1 year, or the specified 
 time, and add it to the principal ; then cast th# interest on this 
 amount for the next year, or specified time, and add it to the prin- 
 cipal as before. Proceed in this manner with each successive year 
 of the proposed time. Finally, subtract the given principal from 
 the last amount, and the remainder will be the compound interest. 
 
 2. What is the compound interest of $600 for 5 years, at 7 per 
 cent.? Ans. $241.53. 
 
 3. What is the compound int. of $1260 for 5 yrs., at 7per cent. ? 
 
 4. What is the amount of $1535 for 6 yrs., at 6 percent, com- 
 pound interest ? 
 
 5. What is the amount of $4000 for 2 yrs., at 7 per cent., paya 
 ble semi-annually ? 
 
 QUEST. 426. How is compound interest calculated? 
 
ARTS. 425,426.] 
 
 INTEREST . 
 
 271 
 
 TABLE, 
 
 Sfwviins; the amount of $1, or 1, at 3, 4, 5, 6, and 7 per cent, compound 
 interest, for any number of years, from 1 to 40. 
 
 j Yrs. 
 
 M prr cent. 
 
 4 per cent. 1 5 per cent. 
 
 6 per cent. 
 
 7 per cent. 
 
 1. 
 
 1.030,000 
 
 1.040,000 
 
 1.050,000 
 
 1.060,000 
 
 1.07,000 
 
 2. 
 
 1.060,900 
 
 1.081,600 
 
 1.102,500 
 
 1.123,600 
 
 1.14,490 
 
 3. 
 
 1.092,727 
 
 1.124,864 
 
 1.157,625 
 
 1.191,016 
 
 1.22,504 
 
 4. 
 
 1.125,509 
 
 1.169,859 
 
 1.215,506 
 
 1.262,477 
 
 1.31,079 
 
 5. 
 
 1.159,274 
 
 1.216,653 
 
 1.276,282 
 
 1.338,226 
 
 1.40,255 
 
 6. 
 
 1.194,052 
 
 1.265,319 
 
 1.340,096 
 
 1.418,519 
 
 1.50,073 
 
 7. 
 
 1.229,874 
 
 1.315,932 
 
 1.407,100 
 
 1.503,630 
 
 1.60,578 
 
 8. 
 
 1.266,770 
 
 1.368,569 1 1.477,455 
 
 1.593,848 
 
 1.71,818 
 
 9. 
 
 1.304,773 
 
 1.423,312 
 
 1.551,328 
 
 1.689,479 
 
 1.83,845 
 
 10. 
 
 1.343,916 
 
 1.480,244 
 
 1.628,895 
 
 1.790*848 
 
 1.96,715 
 
 11. 
 
 1.384,234 
 
 1.539,454 
 
 1.710,339 
 
 1.898,299 
 
 2.10,485 
 
 12. 
 
 1.425,761 
 
 1.601,032 
 
 1.795,856 
 
 2.012,196 
 
 2.25,219 
 
 13. 
 
 1.468,534 
 
 1.665,074 
 
 1.885,649 
 
 2.132,928, 
 
 2.40,984 
 
 14. 
 
 1.512,590 
 
 1.731,676 
 
 1.979.932 
 
 2.260,904 
 
 2.57,853 
 
 15. 
 
 1.557,967 
 
 1.800,944 
 
 2.078,928 
 
 2.396,558 
 
 2.75,903 
 
 16. 
 
 1.604,706 
 
 1.872,981 
 
 2.182,875 
 
 2.540,352 
 
 2.95,216 
 
 17. 
 
 1.652,848 
 
 1.947,900 
 
 2.292,018 
 
 2.692,773 
 
 3.15,881 
 
 18. 
 
 1.702,433 
 
 2.025,817 
 
 2.406,619 
 
 2.854,339 
 
 3.37,293 
 
 19. 
 
 1.753,506 
 
 2.106,849 
 
 2.526,950 
 
 3.025,600 
 
 3.(> 1,652 
 
 20. 
 
 1.806,111 
 
 2.191,123 
 
 2.653,298 
 
 3.207,135 
 
 3.86,968 
 
 21. 
 
 1.860,295 
 
 2.278,768 
 
 2.785,963 
 
 3.399,564 
 
 4.14.056 
 
 22. 
 
 1.916,103 
 
 2.3(.'9,919 
 
 2.925,261 
 
 3.603,537 
 
 4.43,040 
 
 23. 
 
 1.973,587 
 
 2.464,716 
 
 3.071,524 
 
 3.819,750 
 
 4.74,052 
 
 24. 
 
 2.032,794 
 
 2.563,304 
 
 3.225,100 
 
 4.048,935 
 
 6.07,236 
 
 25. 
 
 2.093*,778 
 
 2.665.836 
 
 3.386,355 
 
 4.291,871 
 
 . 42,743 
 
 26. 
 
 2.156,592 
 
 2.772,470 
 
 3.555,673 
 
 4.549,383 
 
 S.80,735 
 
 27. 
 
 2.221,289 
 
 2.883,369 
 
 3.733,456 
 
 4.822,346 
 
 o.21,386 
 
 28. 
 
 2.287,928 
 
 2.998,703 i 3.920,129 
 
 5.111,687 
 
 o.64,883 
 
 29. 
 
 2.356,566 
 
 3.118,651 ! 4.116,136 
 
 5.418,388 
 
 7.11,425 
 
 30. j 427,262 
 
 3.243,398 
 
 4.321.942 
 
 5.743,491 
 
 7.61,225 
 
 31. I ^.500,080 ' 3.373,133 
 
 4.538,039 
 
 6.088,101 
 
 8.14,571 
 
 32. 
 
 2.575,083 3.508,059 4.764,941 
 
 6.453,386 
 
 8.71.527 
 
 33. 
 
 2.652,335 3.648,381 5.003,189 6.840,590 
 
 9.32.533 
 
 34. 
 
 2.731,905 j 3.794,316 5.253,348 7.251,025 
 
 9.97,811 
 
 35. 
 
 2.813,862 3.946,089 5.516,015 7.686.087 
 
 10.6,765 
 
 36. 
 
 2.898,278 4.103,933' 5.791,816 8.147,252 
 
 11.4,239 
 
 37. 
 
 2.985,227 
 
 4.268,090 6.081,407 8.636,087 
 
 12.2,236 
 
 38. 
 
 3.074,783 
 
 4.438,813 6.385,477 9.154,252 
 
 13 0,792 
 
 39. 
 
 3.167,027 I 4.616,366 6.704,751 ! 9.703,507 I'S 9,948 
 
 40. 
 
 3.262,038 4.801,021 j 7.039,989 ; 10.285,72 ^.9*4 
 
DISCOUNT [SECT. XIL 
 
 427. To calculate compound interest by the preceding table. 
 
 Find the amount of $1 or l for the given number of years by 
 the table, multiply it by the given principal, and the product will 
 be the amount required. Subtract the principal from the amount 
 thus found, and the remainder will be the compound interest. 
 
 6. What is the compound interest of $500 for 15 years, at 6 
 per cent. ? What is the amount ? 
 
 Operation. 
 $2.396558 Amt. of $1 for 15 yrs. by Table. 
 
 500 The given principal. 
 $1198.279000 Amt. required. 
 $500 Principal to be subtracted. 
 
 $698.279 Interest required. 
 
 7. What is the amount of $960 for 10 yrs., at 7 per ct. ? 
 
 8. What is the amount of $1000 for 9 yrs., at 5 per ct. ? 
 
 9. What is the compound int. of $1460 for 12 yrs., at 4 per ct. ? 
 
 10. What is the compound int. of $2500 for 15 yrs., at 6 per ct. ? 
 
 11. What is the amount of $5000 for 20 yrs., at 6 per ct. ? 
 
 12. What is the amount of $10000 for 40 yrs., at 7 per ct. ? 
 
 DISCOUNT. 
 
 428* DISCOUNT is the abatement or deduction made for the 
 payment of money before it is due. For example, if I owe a man 
 $100, payable in one year without interest, the present worth of 
 the note is less than $100 ; for, if $100 were put at interest for 
 1 year, at 6 per cent., it would amount to $106 ; at 7 per cent., 
 to $107, &c. In consideration, therefore, of the preseni payment 
 of the note, justice requires that he should make some ciiatement 
 from it. This abatement is called Discount. 
 
 429* The present worth of a debt payable at some future time 
 without interest, is that sum which, being put at legal interest, 
 will amount to the debt, at the time it becomes due. 
 
 QUEST. 428. What is discount 1 429 What '* the present worth of a debt, payable at 
 some future time, without interest 7 
 
ARTS. 427-430.] DISCOUNT. 273 
 
 Ex. 1. What is the present worth of $756, payable in 1 year 
 and 4 months, without interest, when money is worth 6 per cent, 
 per annum ? 
 
 Analysis. The amount, we have seen, is the sum of the prin- 
 cipal and interest. (Art. 399.) Now the amount of $1 for 1 year 
 and 4 months, at 6 per cent., is 81.08; (Art. 404;) that is, the 
 amount is +$% of the principal $1. The question then resolves 
 itself into this : $756 is Ig-g- of what principal ? If $756 is -f ft-g 
 of a certain sum, T-J-g- is TUT of $756 ; now $756-Hl08=$7, and 
 -^=$7X100, which is $700. 
 
 Or, we may reason thus : Since $1.08 (amount) requires $1 
 principal for the given time, $756 (amount) will require as many 
 dollars as 81.08 is contained times in $756 ; and $756-i-$1.08 = 
 $700, the same as before. 
 
 PROOF. $700 X .08=$56, interest for 1 year and 4 months ; and 
 $7004-56 $756, the sum whose present worth is required. Hence, 
 
 43 O. To find the present worth, of any sum, payable at a future 
 time without interest. 
 
 first find the amount of $1 for the time, at the given rate, as 
 in simple interest j then divide the given sum by this amount, and 
 the quotient will be the present worth. (Art. 404.) 
 
 The present worth subtracted from the debt, will give the true 
 discount. 
 
 OBS. This process is often classed among the Problems of Interest, in which 
 the amount, (which answers to the given sum or debt,) the rate per cent., and 
 the time are given, to find the principal, which answers to the present worth. 
 
 2. What is the present worth of $424.83, payable in 4 months, 
 when money is worth 6 per cent. ? What is the discount ? 
 Solution. $424.83-^$1.02=$416.50, Present worth. 
 And $424.83 $416.50=$8.33, Discount. 
 
 3 What is the present worth of $1000, payable in 1 year, 
 when the rate of interest is 7 per cent. ? 
 
 4. What is the present worth of $1645, payable in 1 year and 
 6 months, when the rate of interest is 7 per cent. ? 
 
 UUKST. 430. How do you find the present worth of a det 1 HoW find the discount? 
 
274 BANK DISCOUNT. [SECT. All. 
 
 5. What is the discount on a note for 82300, payable in 6 
 months, when the rate of interest is 8 per cent. ? 
 
 6. What is the discount, at 6 per cent., on $4260, payable in 
 4 months ? 
 
 7. What is the present worth of a note for $4800, due in 3 
 months, when the rate of interest is per cent. ? . 
 
 8. What is the present worth of a draft for $6240, payable ia 
 1 month, when the rate of interest is 6 per cent. ? 
 
 9. A man sold his farm for $3915, payable in 2- years : what 
 is the present worth of the debt, at 6 per cent, discount ? 
 
 10. What is the present worth of a draft of $10000, payable at 
 30 days sight, when interest is 6 per cent, per annum ? 
 
 11. What is the difference between the discount of $8000 for 
 1 year, and the interest of $8000 for 1 year, at 7 per cent. ? 
 
 BANK DISCOUNT. 
 
 431* A Bank, in commerce, is an institution established for 
 the safe keeping and issue of money, for discounting notes, deal- 
 ing in exchange, &c. 
 
 OBS. 1. There are three kinds of banks, viz: banks of deposit, discount, and 
 circulation. 
 
 A bank of deposit receives money to keep, subject to the order of the de- 
 positor. This was the primary object of these institutions. 
 
 A bank of discmmt is one which loans money, or discounts notes, drafts, 
 and bills of exchange. 
 
 A bank of circulation issues bills, or notes of its own, which are redeem- 
 able in specie, at its place of business, and thus become a circulating medium 
 of exchange. Banks of this country generally perform the three-fold office of 
 deposit, discount, and circulation. 
 
 2. The affairs of a bank are managed by a board of directors, chosen annu- 
 ally by the stockholders. (Art. 392. Obs.) The directors appoint a president and 
 cashier, who sign the bills, and transact the ordinary business of the bank. 
 
 A teller is a clerk in a bank, who receives and pays the money on checks. 
 A check is an order for money, drawn on a banker, or t le casl icr, by a de- 
 positor, payable to the bearer. 
 
 3. Banks originated in Italy. The first one was established m Venice, m 
 1171, called the Bank of Venice. 
 
 QUEST. 431. What is a bank ? Obs. Of how many kinds are banks ? 
 
ARTS. 431-433.] BANK DISCOUNT. 275 
 
 432. It is customary for Banks, in discounting a note or 
 draft, to deduct in advance the legal interest on the given sum 
 from the time" it is discounted to the time it becomes due. Hence, 
 
 Bank discount is the same as simple interest paid in advance. 
 Thus, the lank discount on a note of $106, payable in 1 year, at .8 
 per cent., is $6.36, while the true discount is but $6. (Art. 430.) 
 
 OBS. 1 The difference between bank discount and true discount, is the inter- 
 est of tht true discount for the given time. On small sums for a short period 
 his difference is trifling, but when the sum is large, and the time for vhich it 
 is discounted is long, the difference is worthy of notice. 
 
 2. Taking legal interest in advance, according to the general rule of law, a 
 'usury. An exception is generally allowed, however, in favor of notes, drafts, 
 &c., which are payable in less than a year. 
 
 The Safety Fund Banks of the State of New York, though the legal rate 
 of interest is 7 per cent., are not allowed by their charters to take over 6 per 
 cent, discount in advance on notes and drafts which mature within 63 days 
 from the time they are discounted.* 
 
 Banks charge interest for the three days grace. 
 
 CASE I. 
 
 12. What is the bank discount on a note for $850.20 for 6 
 months, at 6 per cent. ? What is the present worth of the note ? 
 
 Operation. 
 $850.20 Principal. 
 
 .03 05 Int. $1 for 6 mo. 3 ds. grace. 
 425100 
 25 5060 
 
 $25.9311 00 Bank discount. 
 And $850.20 $25.93 =$824.27, Present worth. Hence, 
 
 433* To find the bank discount on a note or draft. 
 
 Cast tlie interest on the face of tlie note r draft for three days 
 more than the specified time, and the result will le tlie discount. 
 
 The discount subtracted from the face of the note, ivill give th 
 present worth o* proceeds of a note discounted at a bank. 
 
 QUEST. 432. How do banks usually reckon discount? What then is bank discount! 
 Oft*. What is the difference between bank discount and true discount? Is this difference 
 worth noticing ? How is taking interest in advance generally regarded in law 1 What 
 exception to this rule is allowed ? 
 
 Revised Statutes of New York. (3d edition,) Vol. I. p. 741. 
 
276 BANK DISCOUNT. [SECT. XII. 
 
 Note. Intel est should be computed for the three days grace in each of the 
 following examples. 
 
 1 . What is the bank discount on a note for $^65, payable in 
 6 months, at 6 per cent. ? 
 
 15. What is the bank discount on a note for $972, payable in 
 4 months, at 5 per cent. ? 
 
 16. What is the bank discount on a note for $1492, payable in 
 3 months, at 7 per cent. ? 
 
 17. What is the bank discount on a draft of $028, payable at 
 60 days sight, at 5 per cent. ? 
 
 18. What is the present worth of $2135, payable in 8 months, 
 at 7 per cent. ? 
 
 19. What is the present worth of a note for $2790, payable in 
 1 month, discounted at C per cent, at a bank ? 
 
 20. What is the bank discount, at 5i per cent., on a draft of 
 $1747, payable at 90 days sight? 
 
 21. What is the bank discount, at 4 per cent., on a draft of 
 $3143, payable in 4 months? 
 
 22. What is the bank discount on $5126.63, payable in 30 days, 
 at 8 per cent. ? 
 
 23. What is the bank discount on $3841.27, payable hi 60 days, 
 at 6 per cent. ? 
 
 24. What is the present worth of a note for $6721, payable in 
 10 months, discounted at 6 per cent, at a bank? 
 
 25. What is the present worth of a note for $1500, payable in 
 1 2 days, at 7 per cent, discount ? 
 
 26. What is the bank discount on $10000, payable in 45 days, 
 at 6 per cent. ? 
 
 27. What is the bank discount on $25260, payable in 90 days, 
 at 7 per cent. ? 
 
 28. What is the difference between the true discount and bank 
 discount on $5000 for 10 years, at 6 per cent. ? 
 
 CASE II 
 
 29. A man wishes to make a note payable in 1 year, at 6 
 per cent., the present worth of which, if discounted at a bank, 
 shall be just $200 : for what sum must the note be made ? 
 
A.RT. 434.] BANK DISCOUNT. 277 
 
 Analysis. The present wortli of 8l, payable in 1 year, at 6 
 per cent, discount, is 100 cts. G cts. =94 cts. ; that is, the present 
 worth is -^jfV of the principal or sum discounted. The question 
 then resolves itself into this : $5200 (present worth) is -f^ f . of 
 what sum? Now, if $200 is -ftfa of a certain sum, -rJnr is -fa 
 of 1200; and 8200-^94=82.12766, and iH 82.12766X 100, 
 which is 8212.766. Ans. 
 
 Or, we may reason thus: Since 94 cents present worth requires 
 $1, (100 cents) principal, or sum to be discounted for the ghen 
 time, 8200 present worth will require as many dollars, as 94 cents 
 is contained times in $200 ; and 8200^-8-94=8212.766, 
 
 PROOF. $212.766X-06=8l2.7659, the bank discount for 1 
 year; and 8212.766812.7659=8200, the given sum. Hence, 
 
 434. To find what sum, payable in a specified time, will 
 produce a given amount, when discounted at a bank, at a giveA 
 per cent. 
 
 Divide the given amount to be raised by the present worth of 8l, 
 for the time, at the given rate of bank discount, and the quotient 
 will be the sum required to be discounted. 
 
 30. How large must I make a note payable in 6 months, to raise 
 $400, when discounted at 7 per cent, bank discount? 
 
 31. What sum payable in 4 months must be discounted at a 
 bank, at 5 per cent., to produce 8950 ? 
 
 32. What sum payable in 60 days, will produce $1236, if dis- 
 counted at a bank, at 8 per cent. ? 
 
 33. For what sum must a note be drawn, payable in 34 days, 
 the avails of which, at 6 per cent., bank discount, will be 82500 ? 
 
 34. For what sum must a note be drawn, payable in 90 days, 
 so that the avails, at 7 per cent, bank cfiscount, shall be 83755 ? 
 
 35. A man bought a farm for $4208 cash : how large a note 
 payable in 4 months, must he take to bank to raise the money at 
 6 per cent, discount ? 
 
 QrusT. 134. How find what sum, payable in a given time, will roduce .'i given amount* 
 at a given per cent., bank discount ? 
 
278 INSURANCE. [SECT. XII. 
 
 36 A man wishes to obtain $63240 from a bank at 6 per cent, 
 discount : how large must he make his note, payable in 1 month 
 and 15 days? 
 
 37. What sum payable in 8 months, if discounted at a bank, at 
 6 per sent., will produce $10000? 
 
 38. What sum payable in 4 months, will produce $50000, if 
 discounted at 7 per cent, at a bank ? 
 
 39. A man received $46250 as the avails of a note, payable in 
 60 days, discounted at a bank at 5 per cent. : what was the face 
 of the note ? 
 
 40. A merchant wished to pay a debt of $8246 at a bank, by 
 getting a note payable in 30 days discounted, at 8 per cent. : how 
 large must he make the note ? 
 
 INSURANCE. 
 
 435. INSURANCE is security against loss or damage of prop- 
 erty by fire, storms at sea, and other casualties. This security 
 is usually effected by contract with Insurance Companies, who, 
 for a stipulated sum, agree to restore to the owners the amount 
 insured on their houses, ships, and other property, if destroyed 
 or injured during the specified time of insurance. 
 
 OBS. 1. Insurance on ships and other property at sea is sometimes effected 
 by contract with individuals. It is then called out-door insurance. 
 
 2. The insurers, whether an incorporated company or individuals, are often 
 termed Underwriters. 
 
 436* The written instrument or contract is called the Policy. 
 
 The^s-wm paid for insurance is called the Premium. 
 
 The premium paid is a certain per cent, on the amount of prop- 
 erty insured for 1 year, or during a voyage at sea, or other spe- 
 cified time of risk. 
 
 OBS. 1. Rates of insurance on dwelling-house? and furniture, stores and 
 goods, shops, manufactories, &c., vary from -f to 2 per cent, per annum on 
 the sum insured, according to the exposure of the property and the difficulty 
 of moving the goods in case of casualty. It is a rule with most Insurance 
 
 QUEST. 435. What is Insurance? Obs. When insurance is effected with individuals, 
 what is it called ? What are the insurers sometimes called ? 436. What is meant by the 
 policy? The premium? 
 
ARTS. 435-437.] INSURANCE. 279 
 
 Companies not to insure more than two thirds of the valuo of a building, or 
 goods on land. 
 
 2. Coasting vessels are commonly insured by the season or year. In time 
 of peace, the rate varies from 4 to 7 per cent, per annum; it. time of war it is 
 much higher. Whale ships are generally insured for the voyage, ttt a rate 
 varying from 5 to 8 per cent, on the sum insured. 
 
 3. When the general average of loss is less than 5 per cent., the underwriter* 
 are not liable for its payment. 
 
 CASE I. 
 
 437. To compute Insurance for 1 year, or a specified time. 
 Multiply the sum insured by the given rate per cent., as in inter- 
 est. (Art. 404.) 
 
 Ex. 1. A man effected an insurance on his house for $500, at 
 \\ per cent, per annum : how much premium did he pay ? 
 
 Solution. $1500X.0125 (the rate)=$l 8.75. Ans. 
 
 2. What is the premium for insuring a store to the amount of 
 $2760, at | per cent. ? 
 
 3. What premium must I pay for insuring a quantity of goods, 
 worth '156280, from New York to Liverpool, at 1 per cent. ? 
 
 4. What is the annual premium for insuring a stock of goods, 
 worth $10200, at f per cent. ? 
 
 5. What is the annual premium for insuring a coasting vessel, 
 worth $1600, at 6| per cent.? 
 
 6. A bookseller shipped a quantity of books, valued at $4700, 
 /rom Boston to New Orleans, at 1^ per cent, insurance : what 
 amount of premium did he pay ? 
 
 7. A merchant shipped a cargo of flour, worth $45000, from 
 New York to Liverpool, at 2 per cent. : how much premium did 
 he pay ? 
 
 8. What is the insurance on a cargo of teas, worth $75000, 
 from Canton to Philadelphia, at 2 per cent. ? 
 
 9. What is the annual insurance on a factory, worth $65000, 
 at -f per cent. ? 
 
 10. A powder mill was insured for $1945, at 12^ per cent.: 
 what was the annual premium ? 
 
 QXJKST. 437. How is insurance computed for 1 year or a specified time ? 
 
280 INSURANCE. [SECT. XII 
 
 11. A ship embarking on an exploring expedition, was insured 
 for $45360, at 8 per cent, per annum : what did the insurance 
 amount to in 5 years ? 
 
 12. A policy of insuiance for $45000 was obtained on a. whale 
 ship, at 7 per cent, for the voyage : what was the amount paid 
 for insurance ? 
 
 CASE II. 
 
 1 " If a man pays $16 annually for insuring $800 on his shop, 
 What pei cent, does he pay ? 
 
 Analysis. If $800, the amount insured, costs $16 premium, 
 #1 will cost rJ-o of $16 ; and $16-^800=.02 ; which is 2 per cent. 
 PROOF. $800 X- 02 =$16, the premium paid. Hence, 
 
 43 8, To find the rate per cent, when the sum insured and the 
 annual premium are given. 
 
 Divide the given premium by the sum insured^ and the quotient 
 will be the rate per cent, required. 
 
 Note. This case is similar in principle to Problem II. in Interest. 
 
 14. If a man pays $60 annually for insuring $2400 on his 
 house, what per cent, does it cost him ? 
 
 15. A merchant pays $200 per annum for insuring $8000 on his 
 goods : what per cent, does he pay ? 
 
 16. A grocer paid $122.50 premium on a cargo of flour, 
 worth $12250, from Charleston to Portland: what per cent, did 
 he pay ? 
 
 17. An importer paid $350 insurance on a quantity of cloths, 
 worth $28000, from Havre to New York : what per cent, did he 
 pay? 
 
 CAS E III. 
 
 18. A man pays $45 annually for insuring his library, which is 
 8 per cent, on the amount of his policy : what is the sum insured ? 
 
 Analysis. Since 3 cents will insure $1 at the given rate, for a 
 year, $45 will insure as many dollars as 3 cents are contained 
 times in $45 ; and $45 -f-. 03 =$1500. Ans. 
 
 PROOF. $1500X.03=$45, the given premium. Hence, 
 
ARTS. 438, 439.] INSURANCE. 281 
 
 439*. To find the sum insured when the premium and thfc 
 rute per cent, are given. 
 
 Divide the given premium by the rate per cent., expressed in deci- 
 mals, and the quotient will be the stim insured. 
 
 Note. This case is similar in principle to Problem III. in Interest. 
 
 19. An inr porter paid $650 prenmim on goods from Hamburgh 
 to New York, which was 1-J- per cent, on the amount insured : 
 li >w much did he insure ? 
 
 20. A merchant paid $1640 premium on goods from Philadel- 
 phia to Constantinople, which was 2 per cent, on the worth of 
 the goods insured : how much did he insure ? 
 
 21. A premium of $48*7.50 was paid on a cargo of cotton from 
 New Orleans to Liverpool, which was $ per cet. on its value : 
 what amount was insured on the cargo ? 
 
 22. When the rate of insurance if !- per cent., what sum can 
 you get insured for $860 premium ? 
 
 23. At per cent, per annum, what amount can a man. get in- 
 sured on his house and furniture for $20.50 per annum? 
 
 CASE IV. 
 
 To find what sum must be insured on any given property, so 
 that, if destroyed, its value and the premium may both be recov- 
 ered. 
 
 24. If a man owns a vessel worth $1920, what sum must he get 
 insured on it, at 4 per cent., so that if wrecked, he may recover 
 both the value of the vessel and the premium ? 
 
 Analysis. It is plain, when the rate of insurance is 4 per cent, 
 on a policy of $1, or 100 cents, the owner would receive but 96 
 cents towards his loss ; for, he has paid 4 cents for insurance. 
 Since therefore the recovery of 96 cents requires $1 to be insured, 
 the recovery of $1920 will require as many dollars to be insured 
 as 96 cents is contained times in $1920; and $1920~.96 = 
 $2000. Ans. 
 
 PROOF. $2000X-04=$80, the premium paid, and $2000 
 $80 =$19 20, the value of the vessel. 
 
. 
 
 282 LIFE INSURANCE. [SuCT. XII 
 
 44O. Hence, to find what sum must be insured on a given 
 amount of property, so that if destroyed, both the value of the 
 property and the premium may be recovered. 
 
 Subtract the rate per cent, from $1, then divide the value of the 
 property insured by the remainder, and the quotient will be the sum 
 to be insured. 
 
 25. What sum must be uisured on property worth &8240, at 
 1J per cent., so that the owner may suffer no loss if the property 
 ft) destroyed ? 
 
 26. What sum must be insured on $13460, at 3 per cent., in 
 order to cover both the premium and property insured ? 
 
 27. If I send an adventure to the Sandwich Islands worth 
 $25000, what sum must I get insured, at 7 per cent., that I may 
 sustain no loss in case of a total wreck ? 
 
 LIFE INSURANCE. 
 
 44 1 A LIFE INSURANCE is a contract for the payment of a 
 certain sum of money on the death of an individual, in considera- 
 tion of a stipulated sum paid down, or, more commonly, of an 
 annual premium, to be continued during the life of the assured. 
 
 The average duration of human life is often called the Expecta- 
 tion of Life. This is different in different countries, but it may be 
 determined with great accuracy in any given country, by calcula- 
 tions founded on the register of births and deaths in that country. 
 
 OBS. At birth, the expectation of life, according to the Carlisle Table, is 
 38.72 y. ; at 5, it is 51.25 y. ; at 10, it is 48.82 y. ; at 15, it is 45 y. ; at 20, it 
 is 41.46 y. ; at 25, it is 37.86 y. ; at 30, it is 34.34 y. ; at 35, it is 31 y. ; at 40, 
 it is 27.61 y. ; at 45, it is 24.46 y. ; at 50, it is 21.11 y. ; at 55, it is 17.58 y. ; 
 at 60. it is 14.34 y. ; at 65, it is 11.79 y. ; at 70, it is 9.19 y. ; at 75, it is 7.01 y. ; 
 at 80^ it is 5.51 y. ; at 85, it is 4.12 y. ; at 90, it is 3.28 y. ; at 100, it is 2.28 y. 
 
 442* The premium paid for life insurance, like that for other 
 insurance, is calculated at a certain per cent, on the amount in- 
 sured. The per cent, varies according to the age and employmen 
 of the assured, and the time embraced in the policy. 
 
 UCBST. 441. What is Life Insurance? What is meant by the expectation cf lift t 
 442. How is Life insurance calculated 1 
 
 * See Registers of London, Breslau, Northampton, &c. 
 
ARTS. 440-443.] PROFIT AND LOSS. 283 
 
 OBS. 1. At the age of 21 years, the per cent, on a policy for life is from 1-^ 
 to S2-jp per cent, per annum on the sum insured ; tor 7 years, it is from -- to 
 !- per cent, per annum ; for 1 year, from -fc to 1-f per cent. 
 
 At 30, on a policy for life, it is from 2if^- to 2-,^ per cent, per annum ; for 
 7 years, from \.-^~ to l-^ per cent. ; for 1 year, from l-^ to 1-^ per cent. 
 
 At 40, on a policy for life, it is from 3fV to 3^ per cent.; for 7 years, from 
 1 fc to 2-j 2 - per cent. ; for 1 year, from 1-f to 2^- per cent. 
 
 At 50, on a policy for life, it is from 4-^ to 4-,^ per cent. ; for 7 years, from 
 J^ to 3-,^ per cent. ; for 1 year, from 1-^ to 2-^ per cent. 
 
 At 60, on a policy for life, it is from 6-fc to 7 per cent. ; for 7 years, from 
 IY^ to 5 per cent. ; for I year, from 3 ft - to 4-fo per cent. 
 
 28. A -young man, at the age of 21 years, effected an insurance 
 for $1500 for life, at 2-fV per cent. : what was the annual premium ? 
 
 Am. $31.50. 
 
 29. A man, at the age of 30, effected a life insurance for $2700, 
 for 7 years, at 1W per cent. : what was the premium ? 
 
 30. At 60 years of age, a man effected a life insurance for 1 yea/ 
 for $5750, at 6 per cent. : how much premium did he pay ? 
 
 31. At 40 years of age, a man effected an insurance for $10000 
 for life, at 3- per cent, per annum ; he lived till he was 75 years 
 old : which was the larger, the sum paid for insurance, or the sum 
 insured ? 
 
 PROFIT AND LOSS. 
 
 44*1. PROFIT and Loss in commerce, signify the sum gained 
 or lost in ordinary business transactions. They are reckoned at 
 a certain per cent, on the purchase price, or sum paid for the arti- 
 cles under consideration. 
 
 CASE I . 
 
 To find the AMOUNT of profit or loss, the purchase price and 
 rate per cent, being given. 
 
 Ex. 1. A grocer bought a lot of flour for $84, and sold it for 
 7 per cent, profit : how much did he make by his bargain ? 
 
 QUEST. 443. What is meant by profit and loss ? How are they reckoned ? 444. Ham 
 h the amount of profit or loss found, when the cost and rate per cent, are given ? 
 T.H. 
 
284 PROFIT AND LOSS. [SECT. XII. 
 
 Analysis. Since lie gained 7 per cent, on the cost of the flour, 
 he must have gained -j-fo- of $84. Now -rh> of $S4 is -fa, and 
 j-for is 7 times as much, which is .-f-^f =$5.88. Ans. 
 
 Or thus : If $1 (100 cents) gain 7 cents, $84 will gain 84 times 
 as much ; and $84x.07=$5.88, the same as before. Hence, 
 
 444* To find the amount of profit or loss, when the purchase 
 price and rate per cent, are given. 
 
 Multiply the purchase price by the given per cent, as in percent- 
 aye j and the product will be the amount gained or lost by the trans- 
 action. (Art. 388.) 
 
 OES. In order to obtain the exact profit and loss in mercantile ope/ations, it 
 is manifest that the interest on the cost or purchase price of the goods, during 
 the time they have been on ha-nd, also for the time before payment is received 
 should be taken into consideration. 
 
 2. If I buy a piece of broadcloth for $120, and after keeping it 
 6 months, sell it at 8 per cent, advance on 6 months credit, how 
 much shall I gain if I pay 7 per cent, for the money invested ? 
 
 Ans. $1.20. 
 
 3. If I buy a farm for $1740, and sell it 8 per cent, less than 
 cost, how much do I lose? Ans. $139.20. 
 
 4. If you buy a house for $2180, and sell it at 10 per cent, 
 advance, how much will you gain by your bargain ? 
 
 5. A merchant bought goods amounting to $3400, and retailed 
 them at 20 per cent, profit : how much did he make ? 
 
 6. A grocer bought a lot of flour for $6235, and sold it 15 per 
 cent, less than cost: what was his loss? 
 
 7. A speculator bought a quantity of cotton for $24850, and 
 sold it at 5- per cent, advance : how much did he make by the 
 operation ? 
 
 8. A man bought a block of stores for $58246, and sold them 
 at 118 percent, advance : how much did he gain ? 
 
 9. A man bought wild land amounting to $125000, and after 
 keeping it 10 years, sold it at 50 per cent, advance: allowing 
 money to be worth 6 per cent., did he make rr lose by the oper 
 ation ; and how much ? 
 
ARIS. 444, i45.] PROFIT AND LOSS. 285 
 
 CASE II. 
 
 To find how an article must be sold to gain or lose a specified 
 per cent., the cost being given. 
 
 10. A man bought a building lot for $625, and afterwards sold 
 it so as to gain 10 per cent. : how much did he sell it for ? 
 
 - Operation. Since he gained 10 per cent., it ia 
 
 $625 purchase price. obvious he sold it for the purchase 
 
 .10 percent, profit. price together with 10 per cent, of 
 
 $62.50 profit. that price. We therefore find 10 
 
 $687.50 selling price. per cent, on the cost, and add it to 
 
 itself. (Art. 388.) 
 
 11. A man bought a small house for $840, and afterwards sold 
 it so as to lose 10 per cent. : IIOAV much did he get for it? 
 
 Operation. 
 
 $840 purchase price. Having found the sum lost, (Art. 
 
 .10 per cent. loss. 388,) subtract it from the cost, and 
 
 $84.00 sum lost. the remainder is obviously the sell- 
 
 $756.00 selling price. ing price. Hence, 
 
 445* To find how any article must be sold, in order to gain 
 or lose a given rate per cent, when the cost is given. 
 
 First find the amount of profit or loss on the purchase price at 
 the given rate, as in the last Case ; then the amount thus found 
 added to, or subtracted from the. purchase price, as the case may be, 
 will give the selling price required. 
 
 12. A grocer bought a quantity of cheese for $130.67: for 
 how much must he sell it, 'to gain 20 per cent. ? 
 
 13. Bought a stock of goods for $3460 : for how much must 
 they be sold, to gain 22 per cent. ? 
 
 14. Bought a quantity of flour for $5245 : for how much must 
 it be sold, to gain 13 per cent. ? 
 
 15. Bought 2500 biles of cotton for $30575, which were sold 
 at a less of 3 per cent. : what did they fetch ? 
 
 Q.UEST. 445. What is the method of finding how any article must be sold, in order to 
 gain or lose a given per cent. 1 446. How is the rate per cent, of profit or loss found, when 
 the cost and selling price are given 7 
 
286 PROFIT AND LOSS. [SECT. XII 
 
 CASE III. 
 
 To find the RATE PER CENT, of profit or loss, the cost and sell- 
 ing price being given. 
 
 16. If a merchant buys a quantity of butter for $75, and sells 
 it for $90, what per cent, profit will he make? 
 
 Analysis. Subtracting the cost from the selling price, shows 
 that he gained $15. Now 15 dollars are -ff- of 75 dollars ; there- 
 fore he gained ff of his outlay, or the purchase price of the goods. 
 And -ft reduced to a decimal, is equal to 20 hundred ths, or 20 per 
 cent. (Art. 387. Obs. 3.) 
 
 Or, we may reason- thus: If $75 (outlay) gain 15 dollars, $1 
 will gain ^V of $15. And $15 -r- 75 = .20, the same as before. 
 
 446. Hence, to find the rate per cent, of profit or loss, when 
 the cost and selling prices are given. 
 
 First find the amount gained or lost by subtraction ; then make 
 the gain or loss tJie numerator and the purchase price the denomina- 
 tor of a common fraction ; reduce this fraction to a decimal, and 
 the result will be the per cent, required. (Art. 337.) 
 
 Or, simply annex ciphers to the profit or loss, and divide it by 
 the cost j the quotient will be the per cent. 
 
 OBS. 1. As per cent,, signifies hundredths, the first two decimal figures which 
 occupy the place of hundredths, are properly the per cent.; the other decimals 
 are -parts of 1 per cent. After obtaining two decimal figures, there is some- 
 times an advantage in placing the remainder over the divisor, and annexing 
 it to the decimals thus obtained. (Art. 387. Obs. 3 ) 
 
 2. It should be remembered that the percentage which is gained or last, is 
 always calculated on the purchase price, or the sum paid for the article, ind 
 not on the selling price, or sum received, as it is often supposed. 
 
 17. Bought a quantity of cotton at 6 cents per yard, and sold 
 it alt 8 cents : what per cent, was the profit ? 
 
 18. Bought a quantity of calico, at 12 cents per yard, and sold 
 it at ]^ cents : what per cent, was the profit? 
 
 19. Bought a lot of corn, at- 45 cents per bushel, and sold it at 
 88 cents : what per cent, was the loss ? 
 
 QUKST, -Obs. What figures properly signify the per cent. ? What do the other decimal 
 figures on the lipbt of hundredths denote ? On what is the per cen i. gained ot lost calc 
 
ART. 446.] PROFIT AND LOSS. 287 
 
 20. A grocer bought a pipe of wine for $252, and retailed it 
 at 12- cents per gill : what per cent, did he make? 
 
 21. A man bought a house for $4325, and sold it for $5216: 
 what per cent, did he make ? 
 
 22. A speculator invested $75000 in stocks, which he sold for 
 877225 : what per cent, did he make by the operation? 
 
 CASE IV. 
 
 To find the COST, the selling price and per cent, gained or lost 
 being given. 
 
 23. A man sold a lot of salt for $360, which was 20 per cent, 
 more than cost : what did he pay for the salt ? 
 
 Analysis. The cost is -HHi of itself, and the gain is -iVV of the 
 cost. (Art. 386.) Now iU+Yu Q jr=ioi ; hence, the selling price 
 is -HHI- of the cost. The question then is this: $360 is -Hfr of 
 what sum ? If $360 is -j-fg- of a certain sum, -^ of that sum is 
 -rirr of $360. Now $360~120=$3, and -U$=$3X100, which 
 is $300. Ans. 
 
 Or, if we divide $360, the selling price, by the fraction -HHK 
 the quotient $300, will be the cost. (Art. 234.) 
 
 PROOF. $300X.20=$60.00 the gain; (Art. 388;) 
 and $300+$60=$360, the selling price. 
 
 24. A miller sold a lot of flour for $170, which was 15 per 
 cent, less than cost : how much did the flour cost him ? 
 
 Analysis. Reasoning as before, the cost is -f{H|- of itself, and 
 the loss is tW of the cost. Now -HHJ -iW^i'VV ; consequently 
 the selling price is -fifo Oi tne cost. The question therefore ia 
 this : $170 is -ffa of what sum ? If $170 is T^JT of a certain sum, 
 rfo- is -fa of $170. Now $!70-7-85=$2, and i=$2xlOO, 
 which is $200. Ans. 
 
 Or, thus: Since lie lost 15 per cent., he realized only 85 cents 
 on $1 outlay. Therefore, if 85 cents, selling price, requires $1 
 outlay, $170, selling price, will require as many dollars outlay aa 
 85 cents are contained times in $170 ; and $170^.85 = $200. 
 
 PROOF. $200 X-15=$30.00, the loss ; (Art. 388 ;) 
 and $200 $30=$170, the selling price. Hence, 
 
288 PROFIT AND LOSS. [SECT. Xll. 
 
 447. To find the cost when the selling price and the per cent. 
 gained or lost are given. 
 
 Divide the selling price by $1, increased or diminished by the per 
 cent, gained or lost, as the case may be, and tJie quotient will be lh* 
 cost required. 
 
 Or, make the given per cent, added to or subtracted from 100, as 
 tlte case may be, the numerator, and 100 the denominator of a com- 
 mon fraction ; then divide the selling price by this fraction, and 
 tlie quotient will be the cost. 
 
 OBS. 1. It is not unfrequently supposed that if we find the percentage on 
 the selling price at the given rate, and add the percentage thus found to, or 
 subtract- it from, the selling price, as the case may be, the sum or remainder 
 will be the cost. This is a mistake, and leads to serious errors in the result. 
 It will easily be avoided by remembering, that the basis on which profit and 
 loss are calculated, is always the purchase price or sum paid for the articles 
 under consideration. (Art. 446. Obs. 2.) 
 
 25. A grocer sold a quantity of cheese for $530, which was 15 
 per cent, more than cost : what was the cost ? 
 
 26. A man sold a carriage for $175, which was 15 per cent, 
 less than cost : what was the cost ? 
 
 27. A man sold a farm for $2360, which was 10 per cent, less 
 than cost : what did he give for it ? 
 
 28. An importer sold a library for $3078, which was 12 per 
 cent, advance on the cost : how much did it cost him ? 
 
 29. A merchant sold a cargo of crockery for $12000, which was 
 8 per cent, less than cost : what was the cost ? 
 
 30. A commission merchant sold a lot of cloths for $7265, which 
 was 15 per cent, more than cost: how much did they cost? 
 
 31. A builder sold a house for $17450, which was 2 per cent. 
 less than cost : what was the cost ? 
 
 32. A broker sold stocks to the amount of $45000, which was 
 5 per cent, advance : what was the cost ? 
 
 33. A manufacturer sold a quantity of carpeting for $63240, 
 which was 50 per cent, more than the cost of the materials : what 
 did the materials cost ? 
 
 QUEST. 147. How is the cost found, when the selling price and the rate per cent, 
 gained or lost, are given? Obs. What mistake is sometimes made in finding the os'1 
 How may it be avoided 1 
 
ARTS. 447-450. | DUTIES. 289 
 
 DUTIES. 
 
 448 DUTIES, in commerce, signify a sum of money required 
 by Government to be paid on imported goods. 
 
 OBS. 1. In every port of entry in the United States, the Government has an 
 establishment, called a Custom House, at which the duties on all foreign goods 
 entered at that port, are to be paid. 
 
 2. The persons appointed to inspect the cargoes of vessels engaged in foreign 
 commerce, to examine the invoices of goods, collect the duties, &c., are called 
 custom louse officers. 
 
 449, Duties are of two kinds, specific and ad valorem. A spe- 
 cific duty is a certain sum imposed on a ton, hundred weight, 
 hogshead, gallon, square yard, foot, <fec., without regard to the 
 value of the article. 
 
 Ad valorem duties are those which are imposed on goods, at a 
 certain per cent, on their value or purchase price. 
 
 Note. The term ad valorem is a Latin phrase, signifying according to, or 
 upon the value. 
 
 45O Before specific duties are imposed, it is customary to 
 make certain deductions called tare, draft or tret, leakage, &c. 
 
 Tare is an allowance of a certain number of pounds made for 
 the box, cask, &c., which contains the article under consideration. 
 
 Draft or Tret is an allowance of a certain per cent, (usually 4 
 per cent.) on the weight of goods for waste, or refuse matter. 
 
 Leakage is an allowance of a certain per cent, (usually 2 per 
 cent.) for the waste of liquors contained in casks, &c. 
 
 OBS. 1. AH duties, both specific and ad valorem, are regulated by the Gov- 
 ernment, and have been different at different times and in different countries. 
 
 2. The allowances or deductions for draft, tare, leakage, &c., are different on 
 different articles, and are also regulated by law. 
 
 3. In buying and selling groceries in large quantities, allowances are some- 
 times made for draft, tare, leakage, &c., similar to those in reckoning duties. 
 
 QUEST. 448. What are duties in commerce? 449. Of how many kinds ate ..hey? 
 What arc specific duties ? Ad valorem duties ? Note. What is the meaning of the term 
 ad valorem ? 450. What deductions are made before specific duties are imposed 1 What 
 Is tare? Draft or tret? Leakage? Obs. How are duties regulated? Are allowance* 
 for draft, &c., ever made in buying and selling groceries ! 
 
290 DUTIES. [SECT. XII 
 
 CASE I. Calculation of Specific Duties. 
 
 Ex. 1. What is the specific duty on 15 hhds. of molasses, at 
 10 cents per gallon, allowing 2 per cent, for leakage? 
 
 Analysis, Since there are 63 gallons in one hhd., in 15 hhds. 
 there are 15 times as many, and 63 gals. X 15 = 945 gals. But 2 per 
 cent, of 945 gals, is equal to 945 X. 02, or 18.9 gals. ; (Art. 388 ;) 
 and 945 gals. 18.9 gals. = 926.1 gals., the net gallons. Now if 
 he duty on 1 gallon is 10 cents, on 926.1 gals.it is 926.1 X-10 
 $92.61, the duty required. Hence, 
 
 451. To find the specific duty on any given merchandise. 
 
 First deduct the legal draft, tare, leakage, &c., from the given 
 quantity of goods ; then multiply the remainder by the given duty 
 per gallon, pound, yard, &c., and the product will be the duty re~ 
 
 2. If the specific duty on tea is 12 cents a pound, how much 
 will it be on 30 chests, each weighing 115 Ibs., allowing 12 Ibs. 
 per chest for draft ? 
 
 3. At 4 cents a pound, what is the specific duty on 160 drums 
 of figs, weighing 28 Ibs. apiece, allowing 2 Ibs. a drum for tare? 
 
 4. At 15 cents a pound, what is the specific duty on 63 chests of 
 opium, each weighing 150-lbs., allowing 10 Ibs. per chest for draft? 
 
 5. At 3 cents a pound, what is the specific duty on 250 bags 
 of coffee, weighing 65 Ibs. apiece, allowing 4 per cent, for tret ? 
 
 6. What is the specific duty, at 6 cents a pound, on 173 kegs of 
 tobacco, each weighing 125 Ibs,, allowing 6 Ibs. per keg for tare? 
 
 7. At 5i cents a pound, what is the specific duty on 430 boxes 
 of paints, weighing 175 Ibs. a box, reckoning' the tare at 15 Ibs. 
 per box ? 
 
 8. At 8 cents per gallon, what is the specific duty on 140 Ibs. 
 of olive oil, allowing 2 per cent, for leakage ? 
 
 9. At 22 cents per gallon, what is the specific duty on 50 hhds. 
 of wine, allowing 2 per cent, for leakage ? 
 
 10. At 7 cents per pound, what is the duty on 345 sacks of 
 almonds, weighing 75 Ibs. apiece, allowing 3 per cent, for tare ? 
 
 QUEST. 451. How are specific duties calculated" 
 
ARTS. 151-453.] DUTIES. 291 
 
 CASE II. Calculation of Ad Valorem Duties. 
 
 452. When duties are imposed upon the actual cost of mer- 
 chandise, there are of course no deductions to be made ; conse- 
 quently we have only to find the legal per cent, on the amount of 
 the given invoice, or cost of the goods, and it will be the duty 
 required. 
 
 Ex. 11. What is the ad valorem duty, at 25 per cent., on a 
 $ase of bombazines, invoiced at $450 ? 
 
 Solution. $450 X-25=$l 12.50, the ad valorem duty. Hence, 
 
 453. To find the ad valorem duty on any given merchandise. 
 Multiply the amount of the given invoice by the legal per cent., 
 
 and the product will be the duty required. (Art. 324.) 
 
 OBS. 1. An invoice is a written statement of merchandise, with the value or 
 prices of the articles annexed. 
 
 2. The law requires that the invoice shall be verified by the owner, or one 
 of the owners of the goods, certifying that the invoice annexed contains a true 
 and fail kf nl account, of the actual costs, thereof, and of all charges thereon, arftl 
 no other different discount, bounty, or drawback, but such as has been actually 
 allowed on the same; which oath shall be administered by a consul, or com- 
 mercial agent of the United States, or by some public officer duly authorized 
 to administer oaths in the country where the goods were purchased, and the 
 same shall be duly certified by the said consul, &c. Fraud on the part of the 
 owners, or the consul, &c., who administers the oath, is visited with a heavy 
 penalty. Lajos of the United Stales. 
 
 12. What is the ad valorem duty, at 20 per cent., on an invoice 
 of broadcloths which cost $1240 in Manchester? 
 
 ] 3. What is the ad valorem duty, at 34 per cent., on an invoice 
 of silks, which cost $2110 in Italy? 
 
 14. What is the duty, at 25 per cent., on a quantity of indigo, 
 the invoice of which is $1968 ? 
 
 15. What is the duty on a bale of Irish linens, which cost 
 $3187, at 33 per cent.? 
 
 16. At 25 per cent., what is the duty on an invoice of hosiery, 
 amounting to $2863 ? 
 
 QUEST 453. How are ad valorem duties calculated ? Obs. What is an invoice? What 
 does the law require respecting the invoice of imported goods 1 
 
 13* 
 
292 ASSESSMENT [SECT. XII. 
 
 17. At 33- per cent., what is the duty on an invoice of mousse- 
 line de laines, amounting to $3690 ? 
 
 18. At 35 per cent., what is the duty on an invoice of watches, 
 amounting to $45385 ? 
 
 19. What is the duty, at 20 per cent., on an invoice of boots 
 and shoes, amounting to $63212 ? 
 
 20. What is the duty, at 15 per cent., on a quantity of ready- 
 made clothing, worth $18714? 
 
 21 . What is the duty on $37241 worth of spices, at 30 per ct. ? 
 
 22. What is the duty on $46210 worth of liquor, at 37 per 
 cent. ? 
 
 23. At 22 per cent., what is the duty on $71685 worth of 
 crockery ? 
 
 ASSESSMENT OF TAXES. 
 
 454* A Tax is a sum of money assessed on individuals for 
 the support of Government, Corporations, Parishes, Districts, <fcc. 
 Taxes levied by the Government, are assessed either on the person 
 or property of the citizens. When assessed on the person, they 
 are called poll taxes, and are usually a specific sum. Those as- 
 sessed on the property are usually apportioned at a certain per 
 cent, on the amount of real estate and personal property of each 
 citizen or taxable individual. 
 
 OBS. Property is divided into two kinds, viz : real estate and personal prop- 
 erty. The former denotes possessions that are fixed ; as houses, lands, &c. 
 The latter comprehends all other property ; as money, stocks, notes, mortgages, 
 ships, furniture, carriages, cattle, tools, &c. 
 
 455. When a tax of any given amount is to be assessed, the 
 first thing to be done is to obtain an inventory of the amount of 
 taxable property, both personal and real, in the State, County, 
 Corporation, or District, by which the tax is to be paid ; also the 
 amount of property of every citizen who is to be taxed, together 
 with the number of Polls. 
 
 QUKST. 454. What are taxes ? Upon what are they assessed ? When assessed upon 
 the person, what are they called ? When assessed upon the property, how are they ap- 
 portioned 1 Obs. How is property divided ? What does real estate denote ? What I* 
 personal property 7 455. When a tax is to be assessed, wltat is the first step ? 
 
ARTS. 454-456.] OF TAXES. 293 
 
 OBS. I. By the number of polls is meant the number of taxable individuals , 
 which usually includes every native or naturalized freeman over the age of 21, 
 and under 70 years. In Massachusetts poll taxes are assessot! -jpon every 
 male inhabitant of the state, between the ages of 16 ajid 70 years, whether a 
 citizen or an alien.* 
 
 2. When any part or the whole of a tax is assessed upon the polls, each 
 citizen is taxed a specific sum, without regard to the amount of property he 
 possesses. 
 
 Ex. 1. The tax assessed by a certain town is $990; its prop- 
 erty both personal and real, is valued at $28000, and it contains 
 300 polls, which are assessed 50 cents apiece. What per cent, is 
 the tax ; that is, how much is the tax on a dollar ; and how much 
 is a man's tax who pays for 3 polls, and whose property is valued 
 at $1500? * 
 
 Solution. Since 1 poll pays 50 cents, 300 polls must pay 300 
 times 50 cents, which is $150. Now $990 $150=$840, the 
 sum to be assessed on the property. Now if $28000 is to pay 
 $840, $1 must pay^rsforr of $840; and $840-r$28000=$.03, 
 or 3 per cent. Finally, the tax on $1500, the amount of the 
 man's property, at 3 percent., is $1500X-03=$45 ; and $45 + 
 $1.50 (3 polls)=$46.50, the man's tax. Hence, 
 
 456. To assess a State, County, or other tax. 
 
 I. First find the amount of tax on all the j>olls, if any, at the 
 given rate, and subtract this sum from the whole tax to be assessed. 
 Then dividing the remainder by the whole amount of taxable prop- 
 erty in the State, County, d'c., the quotient will be the per cent, or 
 tax on one dollar. 
 
 II. Multiply the amount of each marts property by the tax on 
 one dollar, and the product will be the tax on Ids property. 
 
 III. Add each mans poll tax to 'the tax he pays on his property^ 
 and the amount will be his whole tax. 
 
 PROCF. When a tax bill is made out, add together the taxes 
 of all the individuals in the town, district, &c., and if the amount 
 is equal to the whole tax assessed, the work is right. 
 
 CinicsT. Obs. What is meant by the number of polls ? 456. How are taws asseu*tf * 
 When a tax bill is made pat, how is its correctness proved ? 
 
 * Revised Statutes of Massachusetts. 
 

 294 ASSESSMENT [SECT. XII. 
 
 2. A certain corporation is taxed $537.50 ; the whole propeHy 
 of the corporation is valued at $35000, and there are 50 polls 
 which are assessed 25 cents apiece. What per cent, is the tax ; 
 and how much is a man's tax, who pays for 2 polls, and whose 
 property is valued at $4240. 
 
 Operation. 
 
 Multiply $.25 the tax on 1 poll, 
 By 50 the number of polls. 
 
 $12.50 Amount on polls. 
 
 But $537.50 12.50=$525, the sum assessed on the corpora- 
 tion ; and $525-r$35000=.015, the per cent, or tax on $1. 
 Now $4240X.015=$63 > 60, the tax on the man's property, 
 And .25X2= .50, the tax for polls. 
 
 Am. $64.10, whole tax. 
 
 3. What is B's tax, who pays for 3 polls, and whose property is 
 valued at $3560 ? 
 
 4. What is C's tax, who is assessed for 1 poll and $5350 ? 
 
 5. The city of New York levied a tax of $1945600; its tax- 
 able property was rated at $243200000 : what per cent, was the 
 tax? 
 
 6. What was A's tax, whose property was valued at $10000 ? 
 
 7. What was B's tax, who was assessed for $15240 ? 
 
 8. What was C's tax, who was assessed for $35460 ? 
 
 457. Having ascertained the expenditures of a State, County, 
 Town, &c., it is necessary in assessing the tax, to take into con 
 sideration the expense of collecting it. Collectors are paid a certain 
 per cent, commission on the amount collected ; (Art. 388. Obs. i ;) 
 consequently, in determining the exact sum to be assessed, allow- 
 ance must be made not only for the commission on the net amount 
 to be raised, but also on the commission itself; for the commis- 
 ion is to be paid out of the money collected. 
 
 9. If the expenses of a town are $950, what sum must be 
 assessed to raise this amount, with 5 per cent, commission for 
 collecting it ? 
 
ARTS. i57-459.] OF TAXES. 295 
 
 Analysis. Since the commission is 5 per cent, the net value 
 of $1 assessment is 95 cents. Therefore, if 95 cents net, require 
 $1 assessment, 8950 net, will require as many dollars assessment, 
 as 95 cents are contained times in $950; and $950-f-$.95=1000. 
 
 Ans. $1000. 
 
 PROOF. $1000X-05=$50, the commission; 
 
 and $1000 $50=8950, the net sum required. Hence, 
 
 458. -To find what sum must be assessed, to raise a given net 
 amount. 
 
 Subtract the given per cent, commission from $1, and the re~ 
 mainder will be the net value of $1 assessment. 
 
 Divide the net amount to be raised by the neb value of $1 assess 
 ment, and the quotient will be the sum to be assessed. 
 
 OBS. To meet the expense of collecting a tax, assessors not unfrequently cal- 
 cujate the commission at the given per cent, on the net amount to be raised, 
 and add it to the tax bill. This method is wrong, and leads to erroneous re- 
 sults. Thus, on a tax of $1000, at 5 per cent, commission, the net amount i 
 $2.50 too small ; on $100000. the error is $250 ; on $1000000, it is $"2500. 
 
 10. What sum must be assessed to raise a net amount of $8500., 
 with 4 per cent, commission for collection ? 
 
 11. What sum must be assessed to raise $15400 net, allowing 
 4 per cent, commission for collection ? 
 
 12. Allowing 5 per cent, for collection, what sum must be 
 assessed to raise $16475 net? 
 
 13. Allowing 3 per cent, for collection, what sum must h> 
 assessed to raise $32860 net? 
 
 FORMATION OF TAX BILLS. 
 
 459. In making out a tax bill for a Town, District, &c., hav- 
 ing found the tax on $1, it is advisable to make a table, show- 
 ing the amount of tax on any number of dollars from 1 to $10 ; 
 then from $10 to $100 ; and from $100 to $1000. 
 
 14. A township composed of 16 citizens, levies a tax of $5700 ; 
 the tywn contains 30 polls, which are assessed 50 cents each, and 
 
 QUEST. 458. How find what sum must be assessed to raise a tax of a given amount? 
 
290 
 
 ASSESSMENT 
 
 [SECT. XII. 
 
 its taxable property is inventoried at $199500. What amount of 
 tax must be raised to pay the debt and 5 per cent, commission 
 for collection ; and what is the tax on a dollar ? 
 
 Solution. The sum to be raised is $6000 ; (Art. 458 ;) and the 
 tax is 3 cents on a dollar. (Art. 456.) Now, since the tax on $1 
 is $.03, it is obvious that multiplying $.03 by 2 will be the tax on 
 $2 ; multiplying it by 3, will be the tax on $3, &c., as seen in th 
 following 
 
 TABLE. 
 
 $1 pays $.03 
 
 $10 pay $.30 
 
 $100 pay $3.00 
 
 2 " .06 
 
 20 " .60 
 
 200 " 6.00 
 
 3 " .09 
 
 30 " .90 
 
 300 " 9.00 
 
 4 " .12 
 
 40 " 1.20 
 
 400 " 12.00 
 
 5 " .15 
 
 50 " 1.50 
 
 500 " 15.00 
 
 6 " .18 
 
 60 " 1.80 
 
 600 " 18.00 
 
 7 " .21 
 
 70 " 2.10 
 
 700 " 21.00 
 
 8 " .24 
 
 80 " 2.40 
 
 800 " 24.00 
 
 9 " .27 
 
 90 " 2.70 
 
 900 " 27.00 
 
 10 " .30 
 
 100 " 3.00 
 
 1000 " 30.00 
 
 15. In the above assessment, what is A. B.'s tax, who is rated 
 at $2256, and pays for 3 polls ? 
 
 Operation. 
 
 $2000 pay $60.00 
 
 200 " 6.00 
 
 50 " 1.50 
 
 6 " .18 
 
 3 polls " 1.50 
 
 Amount, $69.18 
 
 $2256 = 2000 + 200+50 + 6 dollars. 
 Now if we add together the tax paid 
 on each of these sums, as found in the 
 table above, the amount will be the tax 
 on $2256. 
 A. B.'s tax therefore is $69.18. 
 
 16. What is G. A.'s tax, who is assessed for 2 polls, and $2400 
 
 17. What is H. B.'s tax, who is assessed for 1 poll, and $3850 ? 
 
 18. What is W. C.'s tax, who is assessed for 3 polls and $1 5000? 
 
 19. E. D. is assessed for $16024, and 1 poll: what is his tax? 
 
 20. J. F. is assessed for $10450, and 2 polls: what is his tax? 
 
 21. T. G. is assessed for $20680, and 3 polls : what is his tax? 
 
 22. W. H. is assessed for $17530, and 1 poll : what is his tax ? 
 
ART. 4GO.] OF TAXES. &"* 
 
 23. L. J. is assessed for $8760, and 1 poll : what is his tax? 
 
 24. W. L. is assessed for $21000, and 2 polls : what is his tax? 
 
 25. J. K. is assessed for $6530, and 2 polls: what is his tax? 
 
 26. G. L. is assessed for $13480, and 1 poll : what is his tax ? 
 
 27. F. M. is assessed for $12300, and 3 polls : what is his tax? 
 26. C. P. is assessed for $15240, and 2 polls : what is his tax': 
 
 29, J. S. is assessed for $16000, and 1 poll : what is his tax? 
 
 30. R. W. is assessed for $18000, and 2 polls : what is his tax? 
 
 Note. Rate Dills for schools are generally apportioned according to the 
 number of days each scholar has attended. Hence, 
 
 46 O To make out Rate Bills for schools. 
 
 First find the, number of days attendance of all the scholars, 
 and the 'whole amount of expenses, including teacher's salary, fuel, 
 repairs, &c. From the amount of expenses deduct the public 
 money, if any, then divide the remainder by the whole number of 
 days attendance, and the quotient will be the rate per day. Finally, 
 multiply the rate per day by the number of days attendance of each 
 mans children, and the 'product will be his tax. 
 
 OPS. In New York and some other states, the general principle is to include 
 only the Teacher's Salary in the Rate Bill. (Revised Statutes. N. Y.) 
 
 ?l. A certain district paid $130 for teacher's salary, $34 for 
 bocvd, $19.42 for fuel, and $2.58 for repairs ; the district drew 
 $30 public money, and the whole number of days attendance was 
 2400 : what was the rate per day ; and how much was A's tax, 
 wh<- sent 115 days? 
 
 F'vlution. Amount of expenses, $186 $30=$156 ; and $156 
 7-2*00=4.065, the rate per day. Now $.065 X 115=$7.475, 
 A's tax is therefore $7.475. 
 
 3 l \ If the expenses of a district are $313.20, and the whole 
 attendance 3915 days, what is B's tax, who sends 167 days? 
 
 33. A district paid their teacher $115, and their fuel cost 
 21 10 ; it drew $38.50 public money, and the number of days 
 attendance was 1954 : what was C's tax, who sent 69 days? 
 
 34. The expenses of a district were $215.20, and the number 
 of days attendance 2150 : what was D's tax, who ser.t 134 days? 
 
298 ANALYSIS. [SECT. Xlil 
 
 SECTION XIII. 
 ANALYSI.S. 
 
 ART. 46 1 The term Analysis, in physical science, signifies the 
 resolving of a compound body into its elements, or component parts. 
 
 ANALYSIS, in arithmetic, signifies the resolving of numbers into 
 he factors of which they are composed, and the tracing of the 
 relations which they bear to each other. (Art. 95. Obs. 2.) 
 
 OBS. In the preceding sections the studen.t has become acquainted with the 
 method of analyzing particular examples and combinations of numbers, and 
 thence deducing general 'principles and rules. But analysis may be applied 
 with advantage not only to the development of mathematical truths, but also to 
 the solution of a great variety of problems, both in arithmetic and practical 
 life. Indeed, it is the method by which business men generally solve prac- 
 tical questions. A little practice will give the student great facility in it* 
 application. 
 
 462* No specific directions can be given for solving examples 
 by analysis. None in fact are requisite. The judgment, from 
 the conditions of the question, will suggest tlie process. Hence, 
 Analysis may, with propriety, be called the COMMON SENSE RULE. 
 
 OBS. In solving questions analytically, it may be remarked in general, that 
 we reason from the given number to 1, then from 1 to the number required. 
 
 Ex. 1. If 60 yards of cloth cost $240, what will 85 yards cost? 
 
 Analytic solution. Since 60 yds. cost $240, 1 yd. will cost - 6 -V 
 of $240 ; and -gV of $240 is $4. Now if 1 yd. costs $4, 85 yds. 
 will cost 85 times as much; and $4X85=$340. Ans. 
 
 Or, we may reason thus : 85 yds. are -f-fr of 60 yds. ; therefore 
 85 yds. will cost of $240, (the cost of 60 yds.) and f-g- of $240 
 is $240XfS=$340, the same as before. (Arts. 210, 212.) 
 
 OBS. 1. Other solutions of this example might be given; but our preseat c!>- 
 jsct is to show how this and similar questions may be solved by analysis. The 
 
 QUEST. 461. What is meant by analysis in physical science? What in arithmetic 1 
 To what may analysis be advantageously applied ? 462. Can any particular rules be pre-> 
 scribed for solving questions by analysis ? How then will you know how t > proceed 
 Obs. What is the operation of solving questions by analysis called 7 
 
ARTS. 461, 462.] ANALYSIS. 29!) 
 
 former method is the simplest and most strictly analytic, though not so short 
 as the latter. It contains two steps : 
 
 First, we separate the given price of GO yds. ($240) into 60 equal parts, to 
 find the value of one part, or the cost of 1 yd., which is $4. 
 
 Second, we multiply the price of 1 yd. ($4) by 85, the number of yds. whose 
 cost is required, and the product is the answer sought. 
 
 2. This and similar questions are usually placed under the rule of Simple 
 Proportion, or the Ride of Three. 
 
 3. The operation of solving a question by analysis, is called an analytic 
 oluiion. - In reciting the following examples, each one should be analyzed, 
 i)d the reason for every stej given in full. 
 
 2. A man bought a horse, and paid $45 down, which was $ of 
 the price of it : what did he give for the horse ? 
 
 Analysis. Since $45 is -f- of the price, the question resolves 
 itself into this : $45 is f of what sum ? If $45 is f of a certain 
 sum, -f is of $45 ; and of $45 is $9. Now if $9 is 1 seventh, 
 7 sevenths are 7 times as much; and $9xV=$63. Ans. $63. 
 
 PROOF. $ of $6 3 =$9, and 5 sevenths are 5 times as much, 
 which is $45, the sum he paid down for the horse. 
 
 Note. In solving examples of this kind, the learner is often perplexed in 
 finding the value of -f , &c- This difficulty arises from supposing that if & 
 of a certain number is 45, \ of it must be \ of 45. This mistake will be 
 easily avoided by substituting in his mind the word parts for the given de- 
 nominator. Thus, if 5 parts cost $45, \ p ar t w ill cost of $45, which is $9. 
 But this part is a seventh. Now if 1 seventh cost $9, then 7 sevenths will 
 cost 7 times as much. 
 
 3. If 40 cords of wood cost $120, how much will 100 cords 
 sost? 
 
 . 4. Bought 35 tons of hay for $700: how much will 16 tens 
 cost? 
 
 5. What cost 37 gallons of molasses, at $21 a hogshead? 
 
 6. What cost 1500 pounds of hay, at $14 per ton? 
 
 7. What cost 18 quarts of chestnuts, at $3 a bushel? 
 
 8. If 55 tons of hemp cost $660, what will 220 tons cost at the 
 same rate ? 
 
 9. If 165 bushels of apples cost $132, how much will 31 bushelg 
 cost? 
 
300 ANALYSIS. [SECT. XIII. 
 
 10. It 72 bushels of peanuts cost $253.44, what will a pint 
 cost at tne same rate ? 
 
 11. If 150 acres of land cost $7000, what will a square rod 
 cost? 
 
 12. If 2 pipes of wine cost $315, what is that per gill? 
 
 13. A farmer bought a yoke of oxen, and paid $40 in work, 
 whbh was -f- of the cost : what did they cost ? 
 
 14. Bought a house, and paid $030 in goods, which was -fo of 
 the price of it : what was the cost of the house ? 
 
 15 A young man lost $256 by gambling, which was -fg of all 
 he was worth : how much was he worth ? 
 
 16. A man having $1500, paid f of it for 112 acres of land: 
 how much did his land cost per acre ? 
 
 17. If a stack of .hay will keep 350 sheep 90 days, how long 
 will it keep 525 sheep ? 
 
 18. If 440 bbls. of flour will last 15 men 55 months, how long 
 will the same quantity last 28 men? 
 
 19. If 136 men can build a block of stores in 120 days, how 
 long will it take 15 men to build it? 
 
 20. If of a pound of tea cost 40 cents, what will - of a pound 
 cost ? 
 
 21. If -f of a yard of broadcloth cost $2.50, how much will $ 
 of a yard cost ? 
 
 22. Bought -ft of a ton of hay for $3.42 : how much will -ft of 
 a ton cost ? 
 
 23. Bought $ of a hogshead of molasses for $38.19: how 
 much will ^o of a hogshead cost ? 
 
 24. If f of an acre of land cost $108, how much will J- of an 
 acre cost ? 
 
 25. If f of a barrel of beef cost $6.48, how much will f of a 
 barrel cost ? 
 
 26. Paid $4200 for f of a sloop : how much' c?an I afford to 
 iell -^ of the sloop for ? 
 
 2*. Sold 18 baskets of peaches for $34 : how much would 65-J- 
 baskets come to ? 
 
 28. If I pay $60.50 for building 20 Jr rods of wall, how much 
 must I pay for <215| rods? 
 
AKT. 463.J ANALYSIS. 301 
 
 29. A man can hoe a field of corn in 6 days, and a boy can hoe 
 H in 9 days : how long will it take them both together to hoe it ? 
 
 Analysis. Since the man can hoe the field in 6 days, in 1 day 
 he can hoe of it ; and since the boy can hoe it in 9 days, in 1 
 day he can hoe \ of it ; consequently in 1 day they can both hoe 
 1 4. 1=-^ of the field. (Art. 202.) Now if -fa of the field requires 
 them both 1 day, -fa of it will require them of a day, and \\ 
 full require them 18 times as long, or -^ of a day, which is equal 
 to 3f days. Ans. 
 
 30. If A can chop a cord of wood in 4 hours, and B in G hours, 
 how long will it take them both to chop a cord ? 
 
 31. A can dig a cellar in G days, B in 9 days, and C in 12 
 days : how long will it take all of them together to dig it ? 
 
 32. A man bought 25 pounds of tea at 6s. a pound, and paid 
 for it in corn at 4s. a bushel : how many bushels did it take ? 
 
 Analysis. -If 1 Ib. of tea costs 6s., 25 Ibs. will cost 25 times 
 as much, which is 150s. Again, if 4s. will buy 1 bushel of corn, 
 150s. will buy as many bushels as 4s. is contained times in 150s. ; 
 and 150s.-f-4=37 times. Ans. 37i bushels. 
 
 463. The last and similar examples are frequently arranged 
 under the rule of Barter. 
 
 Barter signifies an exchange of articles of commerce at prices 
 agreed upon by the parties. 
 
 OBS. Such examples are so easily solved by Analysis that a specific ride fct 
 them is unnecessary. 
 
 33. A farmer bought 110 Ibs. of sugar at 18 cents a pound, 
 and paid for it in lard at 5 cents a pound : how much lard did 
 it take ? 
 
 34. How much butter, at 12^ cents a pound, must be given for 
 250 Ibs. of tea, at 75 cents a pound ? 
 
 35. How many cords of wood, at $2 per cord, must be given 
 for 56 yds. of cloth, at &4-J- per yard ? 
 
 36. How many pair of boots, at $4.50 a pair, must be given 
 for 50 tons of coal at $9 per ton ? 
 
302 ANALYSIS. [SECT. XIII. 
 
 37. A, B, and C, united in business ; A pit in $250 ; B, $270 ; 
 and C, $340 ; they gained $258 : what was each man's share of 
 the gain ? 
 
 Analysis. The whole sum invested is $250-r-$270+$340 = 
 $860. Now since $860 gain $258, it is plain $1 will gain -g^ of 
 $258, which is 30 cents. And 
 
 If $1 gains 30 cts. $250 will gain $250X-30=$75, A's share, 
 " $1 " " $270 " $270 X. 30= 81, B's share, 
 " $1 " " $340 " $340X-30 = 102, C's share. 
 
 Or, we may reason thus : Since the sum invested is $860, 
 A's part of the investment is equal to $, or -ff- ; 
 B's " ^ffcortt; 
 
 C's " " " , or if. - Consequently, 
 
 A must receive |f of the whole gain $2 5 8 =$7 5 ; 
 B "H " " 258= 81; 
 
 C "if " " 258 = 102; 
 
 PROOF. The whole gain is $258. (Ax. 11.) 
 
 464. When two or more individuals associate themselves to- 
 gether for the purpose of carrying on a joint business, the union 
 is called a partnership or copartnership. 
 
 OBS. The process by which examples like the last one are solved, is often 
 called Fellowship. 
 
 38. A and B join in a speculation ; A advances $1500 and B 
 $2500 ; they gain $1200 : what was each one's share of the gain ? 
 
 39. A, B, and C, entered into partnership ; A furnished $3000, 
 B $4000, and C $5000; they lost $1800: what was each one's 
 share of the loss ? 
 
 40. A's stock is $4200 ; B's $3600 ; and C's $5400 ; the whole 
 gain is $2400 : what is the gain of each ? 
 
 41. A's stock is $7560; B's $8240; C's $9300; and D's 
 $6200 ; the whole gain is $625 : what is the share of each ? 
 
 42. A bankrupt owes one of his creditors $400 ; another $500 \ 
 and a third $600; his property amounts to $1000: how much 
 can he pay on a dollar ; and how much will each of his creditors 
 receive ? 
 
 OBS. The solution of this example is the same in principle as that of Ex. 37 
 
ARTS. 464-4G6.] ANALYSIS. 303 
 
 46 5 Examples like the preceding are commonly arranged 
 under tlw rule of Bankruptcy. 
 
 Note.. A bankrupt, is a person who is insolvent, or unable to pay his just 
 debts. 
 
 43. A bankrupt owes $5000, and his property is worth $3500 : 
 how much can he pay on a dollar ? 
 
 4 t. A man di(d owing $16400, and his effects were sold foi 
 f-4 i 0( : what per cent, did his estate pay ? 
 
 45. If a man owes A $6240, B $8760, and C $9000, and has 
 but $lloOO, how much will each creditor receive? 
 
 46. If I owe $48000, and have property to the amount of 
 $32000, what per cent, can I pay ? 
 
 47 What per cent, can a man pay, whose liabilities are 
 $120000, and whose assets are $45000? 
 
 48. What per cent, can a man pay, whose liabilities are 
 $1500000, and whose assets are $150000? 
 
 4G6. It often happens in storms and other casualties at sea, 
 that masters of vessels are obliged to throw portions of their 
 cargo overboard, or sacrifice the ship and their crew. In such 
 cases, the law requires that the loss shall be divided among the 
 owners of the vessel and cargo, in proportion to the amount of 
 each one's property at stake. 
 
 The process of finding each man's loss, in such instances, is 
 called General Average. 
 
 OBS. The operation is the same as that in solving questions in bankruptcy 
 and partnership. 
 
 49. A, B and C, freighted a ship from New York to Liverpool ; 
 A had on board 100 tons of iron, B 200 tons, and C 300 tons, 
 in a storm 240 tons were thrown overboard : what was the loss 
 of each ? 
 
 50. A packet worth $36000 was loaded with a cargo valued at 
 $65000. In a tempest the master threw overboard $25250 worth 
 of goods ; what per cent, was the general average ? 
 
 51. A steam ship being in distress, the master threw \ of 
 the cargo overboard ; finding she still labored, he afterwards 
 threw overboard ^ of what remained. The steamer was worth 
 
304 ANALYSIS. [SECT. XIII. 
 
 $120000, and the cargo $240000 : what per cent, was the general 
 average, and what would be a man's loss who owned -J- of tl?.? ship 
 and cargo? 
 
 52. A man mixed 25 bushels of peas worth 6s. a bushel, with 
 15 bushels of corn worth 4s. a bushel, and 20 bushels of oats 
 worth 3s. a bushel : what was the mixtuie worth per bushel .' 
 
 Analysis. 25 bu. peas at 6s. = 150s., value of the peas; 
 15 bu. corn at 4s. = 60s., " " corn; 
 and 20 bu. oats at 3s. = 60s., " " oats. 
 The mixture=60 bu. and 270s., value of whole mixture. 
 
 Now if 60 bu. mixture are worth 270s., 1 bu. mixture is worth 
 jV of 270s. ; and 270s.-H60^4i-s. Ans. 
 
 PROOF. 60 bu. at 4s.=270s., the value of the whole mixture. 
 
 467. The process of finding the value of a compound or mix- 
 tare of articles of different values, or of forming a compound 
 which shall have a given value, is called Alligation. Alligation is 
 usually divided into two kinds, Medial and Alternate. 
 
 OBS. 1. When the prices of the several articles and the number or quantity 
 of each are given, the process of finding the value of the mixture, as in the last 
 example, is called Alligation Medial. 
 
 2. When the price of the mixture is given, together with the price of each 
 article, the process of finding how much 6T the several articles must be taken 
 to form the required mixture, is called Alligation Alternate Alligation Alter- 
 nate embraces three varieties of examples, which are pointed out in the follow- 
 ing notes. 
 
 53. If you mix 40 gallons of sperm oil worth 8s. per gallon, 
 with 60 gallons of whale oil worth 3s. per gallon, what will the 
 mixture be worth per gallon ? 
 
 54. At what price per pound can a grocer afford to sell a mix- 
 ture of 30 Ibs. of tea worth 4s. a pound, and 40 Ibs. worth 7s. a 
 pound ? 
 
 55. If 120 Ibs. of butter at 10 cts. a pound are mixed with 24 
 Ibs. at 8 cts. and 24 Ibs. at 5 cts. a pound, what is the mixture 
 worth ? 
 
 56. A tobacconist had three kinds of tobacco, worth 15, 18, 
 and 25 cents a pound: what is a mixture of 100 Ibs of each 
 worth per pound ? 
 
ART 467.] ANALYSIS/ 305 
 
 57. A liquor dealer mixed 200 gallons of alcohol worth 50 cts. 
 a gallon, with 100 gallons of brandy worth $1.75 a gallon: what 
 was the value of the mixture per gallon ? 
 
 58. A grocer sells imperial tea at 10s. a pound, and hyson at 
 4s. : what part of each must he take to form a mixture which he 
 can affc rd to sell at 6s. a pound ? 
 
 Note \ . It will be observed in this example that the price of the mixture 
 W* J . ilsc the price of the several ar ticks or ingredients are given, o find what 
 ]~w jf each the mixture must contain. 
 
 Analysis. Since the imperial is worth 10s. and the required 
 mixture 6s., it is plain he would lose 4s. on every pound of impe- 
 rial which he puts in. And since the hyson is worth 4s .a pound 
 and the mixture 6s., he would gain 2s. on every pound of hyson 
 he puts in. The question then is this : How much hyson must 
 he put in to, make up for the loss on 1 Ib. of imperial ? If 2s. 
 profit require 1 Ib. of hyson, 4s. profit will require twice as much, 
 or 2 Ibs. He must therefore put in 2 Ibs. of hyson to 1 Ib. of im- 
 perial. 
 
 PROOF 2 Ibs. of hyson, at 4s. a pound, are worth 8s., and 
 1 Ib. of imperial is worth 10s. Now 8s.+10s.=18s. And if 
 3 Ibs. mixture are worth 18s., 1 Ib. is worth | of 18s., which is 
 6s., the price of the mixture required. 
 
 59. A farmer has oats which are worth 20 cts. a bushel, rye 
 .55 cts., and barley 60 cts., of which he wishes to make a mixture 
 worth 50 cts. per bushel : what part of each must the mixture 
 contain ? 
 
 Analysis. The prices of the rye and barley must each be com- 
 pared with the price of the oats. If 1 bu. oats gains 30 cts. in 
 the mixture, it will take as many bu. of rye to balance it, as 5 cts. 
 (the loss per bu.) are contained times in 30 cts., viz : 6 bu. Again, 
 since 1 bu. oats gains 30 cts., it will take as many bushels of bar- 
 Jey to balance it, as 10 cts. (the loss per bu.) are contained times 
 in 30 cts., viz: 3 bu. Hence, the mixture must contain 2 paita 
 of oats, 6 parts rye, and 3 parts barley. 
 
 60. Tf a man have four kinds o f sugar worth, 8, 9, 11, and 12 
 
306 ANALYSIS. [SECT. XIII. 
 
 cents a pound respectively, how much of each kind must he take 
 to form a mixture worth 10 cents a pound ? 
 
 NoLe 2. In examples like the preceding, we compare two kinds together, 
 one of a higher and the other of a lower price than the required mixture ; then 
 compare the other two kinds in the same manner. In selecting the pairs to 
 be compared together, it is necessary that the price of one article shall be 
 above, and the other below the price of the mixture. Hence, when there are 
 aeveral articles to be mixed, some cheaper and others dearer than the mixture, 
 a variety of answers may be obtained. Thus, if we compare the highest and 
 lowest, then the other two, the mixture will contain 1 part at 8 cts. ; 1 part at 
 
 cts. ; 1 part at 11 cts.; and 1 part at 12 cts. Again, by comparing those 
 at 8 and 1 1 cts., and those at 9 and 12 cts. together, we obtain for the mixture 
 
 1 part at 8 els. ; 2 parts at 1 1 cts. ; 2 parts at 9 cts. ; and 1 part at 12 cts. 
 Other answers may be found by comparing the first with the third and 
 
 fourth ; and the second with the fourth, &c. 
 
 61. A goldsmith having gold 16, 18, 23, and 24 carats fine, 
 wished to make a mixture 21 carats fine : what part^of each must 
 the mixture contain ? 
 
 62. A farmer had 30 bu. of corn worth 6s. a bu., which he 
 wished to mix with oats worth 3s. a bu., so that the mixture may 
 be worth 4s. per bu. : how many bushels of oats must he use ? 
 
 Note 3. In this example, it will be perceived, that the price of the mix- 
 ture, with the prices of the several articles and the quantity of one of them 
 are given, to find how much of the other article the mixture must contain. 
 
 Analysis. Reasoning as above, we find that the mixture (with 
 out regard to the specified quantity of corn) in order to be worth 
 4s. per bu., must contain 2 bu. of oats to 1 bu. of corn. Hence, 
 if 1 bu. of corn requires 2 bu. of oats to make a mixture of the 
 required value, 30 bu. of corn will require 30 times as much ; and 
 
 2 bu.X 30 = 60 bu., the quantity of oats required. 
 
 63. A merchant wished to mix 100 gallons of oil worth 80 cts. 
 per gallon, with two other kinds worth 30 cts. and 40 cts. per gal- 
 lon, so that the mixture may be worth 60 cts. per gallon : how 
 many gallons of each must it contain ? 
 
 64. A merchant has Havana coffee at 12 cts. and Java at 
 18 cts. per pound, of which he wishes to make a mixture of 150 
 Ibs., which he can sell at 16 cts. a pound: how much of each 
 must he use ? 
 
A.RT. 468.] ANALYSIS. 307 
 
 Note 4. In this example, the whole quantity to be mixed, the jnce of the 
 aixture, and the prices of the several articles are given, to fi.id Juw much of 
 ach must be taken. 
 
 Analysis. On 1 Ib. of the Havana it is obvious he will gain 
 4 cts., and on 1 Ib of the Java he will lose 2 cts. ; therefore to 
 balance the 4 cts. gain he must put in 2 Ibs. of Java; that is, the 
 mixture must contain 1 part of Havana to 2 parts of Java. Now 
 if 3 Ibs. mixture require 1 Ib. Havana, 150 Ibs. mixture, (the quan- 
 tity required,) will require as many pounds of Havana as 3 is con- 
 tained times in 150, viz : 50 Ibs. But the mixture contains twice 
 as much Java as Havana, and 50 Ibs X 2 = 100 Ibs. 
 
 Ans. 50 Ibs. Havana, and 100 Ibs. Java. 
 
 65. It is required to mix 240 Ibs. of different kinds of raisins, 
 worth 8d., 12d., 18d., and 22d. a pound, so that the mixture may 
 be worth lOd. a pound : how much of each must be taken ? 
 
 66. If 10 horses consume 720 quarts of oats in 6 days, how 
 long will it take 30 horses to consume 1*728 quarts? 
 
 Analysis. Since 10 horses will consume 720 qts. in 6 days, 
 1 horse will consume -fa of 720 qts. in the same time ; and -fa of 
 720 qts. is 72 qts. And if 1 horse will consume 72 qts. in 6 days, 
 in 1 day he will consume of 72 qts., which is 12 qts. Again, 
 if 12 qts. last 1 horse 1 day, 1728 qts. will last him as many 
 days as 12 qts. are contained times in 1728 qts., viz: 144 days. 
 Now if 1 horse will consume 1728 qts. in 144 days, 30 horses 
 will consume them in -fa of the time; and 144 d.^-30=4f. 
 
 Ans. 30 horses will consume 1728 qts. in 4f days. 
 
 468. This and similar examples are usually placed under the 
 rule of Compound Proportion, or Double Rule of Three. 
 
 67. If 15 horses consume 40 tons of hay in 30 weeks, how many- 
 horses will it require to consume 56 tons in 70 weeks ? 
 
 68. If 8 men can make 9 rods of wall in 12 days, how long wil 
 \\ take 10 men to make 36 rods? 
 
 69. If 35 bbls. of water will last 950 men 7 months, how many 
 men will 1464 bbls. of water last 1 month? 
 
 T.H. 14 
 
308 ANALYSIS. L SEC 
 
 70. If 13908 men consume 732 bbis. of flour in 2 months, in 
 how long time will 425 men consume 175 bbls. ? 
 
 71. If the interest of $30 for 12 months is $2.10, how much ia 
 the interest of $1560 for 6 months? 
 
 72. If the interest of $750 for 8 months is $28, how much is 
 the interest of $16425 for 6 months ? 
 
 73. A man being asked how much money he had, replied that 
 f, -, and f of it made $980 : what amount did he have ? 
 
 Analysis. It is plain that -f + f -f -f=f. (Art. 202.) The 
 question then resolves itself into this : $980 are ff of what sum ? 
 Now if $980 are ff- of a certafn sum, -gV is -fc of $980 ; and $980 
 i-49=$20, and -ft is $20X24=$480. Ans. 
 
 PROOF. | of $480=$320 ; ? of $480=$360 ; and i of $480 
 =$300. Now $320+$360+$300=$980, according to the con- 
 ditions of the question. 
 
 469. This and similar examples are placed under the rule of 
 Position. The shortest and easiest method of solving them is by 
 Analysis. 
 
 74. A sailor having spent $ of his money for his outfit, depos- 
 ited -f of it in a savings bank, and had $50 left: how much had 
 he at first ? 
 
 75. A man laid out -^ of his money for a house, i for furniture, 
 and had $1500 left; how much had he at first? 
 
 76. A man lost % f his money in gambling, -J- in betting, and | 
 spent -f in drinking ; he had $259 left : how much had he at first? 
 
 77. What number is that -f and of which is 102 ? 
 
 78. What number is that i, |, i, and | of which is 450 ? 
 
 79. What number is that - and of which being added to 
 itself, the sum will be 164 ? 
 
 80. What number is that of which exceeds f- of it by 18 ? 
 
 81. A post stands 40 feet above water, -J- in the water, and J- 
 in the ground : what is the length of the post ? 
 
 82. What will 376 yds. of muslin cost, at 2s. and 6d. per yd.? 
 
 Analysis. 2s. 6d.=i-. Now if 1 yd, costs -, 376 yds. will 
 cost 376 times as much; and iX376=47. Ans. 
 
ARTS. 469, 470. 
 
 ANALYSIS. 
 
 309 
 
 83. If 1 yard of silk costs 50 cents, what will 256 yards 
 cost? 
 
 Analysis. 50 cts.=$. Now if 1 yd. costs $i, 256 yds. will 
 cost 256 times as much; and $iX256=$128. Ans. 
 
 47O. Examples like the preceding, in which the price of a 
 single article is an aliquot part of a dollar, &c., are usually classed 
 under the rule of Practice. 
 
 Practice is denned by a late English author to be " an abridged 
 method of performing operations in the rule of proportion by means 
 of aliquot parts j and it is chiefly employed in computing the 
 prices of commodities." 
 
 OBS. After giving several tables of aliquot parts in money, weight, and 
 measure, the same author proceeds to divide his subject into twelve subdivi- 
 sions or cases, and gives a specific ruk for each case, to be committed to mem- 
 ory by the pupil. It is believed, however, that so many specific rules are worse 
 than useless. They have a tendency to prevent the exercise of thought and 
 reason, while they tax the time and memory of the student with a multiplicity 
 of particular directions for the solution of a class of examples, which his com- 
 mon sense, if permitted to be exercised, will solve more expeditiously by 
 Analysis. 
 
 TABLE OF ALIQUOT PARTS OF $1, l, AND Is. 
 
 Parts of a Dollar. 
 
 Parts of a Pound sterling. 
 
 Parts of a Shilling sterling. 
 
 50 cts.=$i 
 33| cts.=*j 
 
 20 cts. -^ 
 
 10 cts- 
 
 5 cts. %- 
 
 10s. = 
 6s. 8d.= 
 
 5s. = 
 4s. = 
 3s. 4d.= 
 2s. 6d.= 
 2s. = 
 Is. 8d.= 
 Is. = 
 
 = shil. 
 
 6 pence 
 4 pence = 
 3 pence = 
 2 pence : 
 l pence : 
 1 penny: 
 penny: 
 
 7 pence : 
 
 8 pence: . . , . 
 
 9 pence =s.+-s. 
 
 shil. 
 =i shil. 
 =i shil. 
 =i shil. 
 =T"* shil. 
 =^~f shil. 
 
 Note. If the price itself is not an aliquot part of SI, or 1, &c., it may 
 sotaetimes be divided into such parts as will be aliquot parts of $1, 1, &c., 
 or which v r ill be aliquot parts of each other. Thus, 87 cts. is not an a^quot 
 part of $1, but 87 cts. =50+254-12$ cts. Now 50cts.=$J ; 25cts.=$| ; and 
 12$ cts.=$J. Or thus : 50 cte. =-.$$, 25 cts.=$ of 50 cts., and 12$ cts.=:$ of 
 25 cts. 
 
310 ANALYSIS. [SttCT. XIII. 
 
 84. What will 680 bu. of wheat cost, at 87 cts. per bushel? 
 Analysis. It is plain, if the price were $1 per bu., the cost of 
 
 680 bu. would be $680. Hence, 
 
 Were the price 50 cts. the cost would be of $680, which is $340 
 25 cts. " " i-of $680, which is $170 
 
 12 cts. " " -i- of $680, which is $ 85 
 
 But since the price is 50-f-25-hl2 cents, the cost must be $595 
 'Or, thus: $1X680=$680, the cost at $1 per bushel. 
 At 50 cts., or $, it will be of $680, or $340 
 
 " 25 cts., i of 50 cts., " " i of $340, or $170 
 " 12i cts., -^ of 25 cts., " " i of $170, or $ 85 
 Therefore the whole cost is $595. Ans. 
 
 85. What cost 478 yards of cashmere, at 50 cts. per yard? 
 
 86. What cost 1560 Ibs. of tea, at 75 cts. per pound? 
 
 87. What cost 2400 gals, of molasses, at 37 cts. per gal. ? 
 
 88. What cost 1800' yds. of satinet, at 62 cts. per yard? 
 
 89. At 25 cts. per bushel, what cost 1470 bu. of oats? 
 
 90. At 33^ cts. a pound, what cost 1326 Ibs. of ginger? 
 
 91. At 6| cts. per roll, what cost 3216 rolls of tape? 
 
 92. At 8| cts. per pound, what cost 4200 Ibs. of lard ? 
 
 93 At 12 cts. per dozen, what cost 1920 doz. of eggs? 
 
 94 At 16f cts. a pound, what cost 4524 Ibs. of figs ? 
 
 9f At 66-f cts. per yard, what cost 1620 yds. of sarcenet? 
 
 9f What cost 840 bu. of rye, at $f per bushel ? 
 
 9V What cost 690 yds. of cloth, at 6s. 8d. per yard ? 
 
 A vtlysis.^At 1 per yard the cost would be 690. But 
 6s,. Vd. is ^-; therefore the cost must be of 690, which is 
 2?<\ Ans. 
 
 98. What cost 360 gals, of wine, at 16s. per gallon? 
 Analysis. 16s. = 10s. + 5s.-f Is. Now 10s. = i; 5s.=|; 
 U.=i of 5s. 
 
 1 f the price were l per gal., the cost of 360 gals, would be 360. 
 At 10s., i, it will be i of 360, or 180 
 " 5s., i of I Os., " i of 180, or 90 
 " Is., i of 5s., " i of 00, or 18 
 Therefore the whole cost is 288. Ans. 
 
ART. 471." 
 
 ANALYSIS. 
 
 311 
 
 99. What cost 1240 yds. of flannel, at 3s. 4d. per yard? 
 
 100. What cost 2128 Ibs. of spice, at 2s. 6d. ptr pound? 
 
 101. What cost 5250 yds. of lace, at 6d. per yard ? 
 102 What cost 56480 yds. of tape, at l|d. per yard? 
 
 471. Notwithstanding the law requires accounts to be kept 
 in Federal Money, goods are frequently sold at prices stated in 
 the denominations of the old state currencies. 
 
 When the price per yard, pound, &c., stated in those currencies, 
 is an aliquot part of a dollar, the answer may be easily obtained 
 in Federal Money. 
 
 TABLE OF ALIQUOT PARTS IN DIFFERENT STATE CURRENCIES. 
 
 Parts of a Dollar, 
 New York Currency. 
 
 Parts of a Dollar, 
 New K/igland Currency. 
 
 * Parts of a Shilling, 
 N. E. and N. Y. Currency. 
 
 4 shil. =$ 
 
 3 shil. =$ 
 
 6 pence =% shil. 
 
 2s. 8d.=$i 
 
 2 shil. $^- 
 
 4 pence =$ shil. 
 
 2 shil. =$| 
 
 Is. 6d.=t4 
 
 3 pence =} shil. 
 
 Is. 4d.=$ 
 
 1 shil. =$i 
 
 2 pence = shil. 
 
 1 shil. =$1 
 
 4s.=$i-j-$ 
 
 1-^- pence =i- shil. 
 
 5s.=4i+$i 
 
 5s. = $i+$i 
 
 1 penny =-fV shil. 
 
 Note. 1 . In N. Y. currency 8s. make SI ; in N. E. currency 6s. make $1. 
 From example 103 to 119 inclusive, the prices are given in N. Y. currency; 
 from example 120 to 132 inclusive, they are given in N. E. currency. For 
 the mode of reducing the different State currencies to each other and to Federal 
 Money, see Section XVII. 
 
 103. At Is. 4d. per yard, what cost 726 yds. of cambric ? 
 Analysis. If the J3rice were $1 per yard, the cost would be 
 
 $1X726=4726. But Is. 4d.=$i-; therefore the cost must be 
 i of $726, which is $121. Am. 
 
 104. What cost 896 bu. of wheat at 6s. per bushel? 
 Analysis. 6s.=4s.+2s. Now 4s. $; and 2s.= of 4s. 
 At $1 a bushel the cost would be 1896. 
 
 At 4s., $i, it will be of $896, or $448 
 " 2s,, i of 4s., " " i of $448, or $224 
 Therefore the whole cost is $672. Ans. 
 
 Or, thus : 6s=$| ; therefore the number of bu. minus | of itself 
 will be the cost, ind 896224 (| of 896)=672. Ans. $672. 
 
312 ANALYSIS. [SECT. XIII. 
 
 105. What cost 752 yds. of balzorine, at 2s. 8d. per yard? 
 
 106. What cost 1232 yds. of calico, at Is. 6d. per yard? 
 
 107. What cost 763 Ibs. of pepper, at Is. 3d. a pound? 
 
 108. What cost 1116 bu. of apples, at Is. 4d. per bushel? 
 
 109. What cost 1920 yds. of shirting, at Is. 2d. per yard? 
 
 110. At 6s. a basket, what will 1560 baskets of peaches cost? 
 
 111. At 5s. 4d. a pound, what will 1200 Ibs. of tea come to? 
 
 Note. 2. Since 5s. 4<1. is \ less than SI, it is plain 1200 400=$800. Ans 
 
 112. At 7s. per yard, what will 432 yds. of crape cost? 
 
 113. At 6s. 8d. a pound, what cost 972 Ibs. of nutmegs? 
 
 114. At 2s. 8d. a pair, what cost 864 pair of cotton hose? 
 
 115. At 1-^d. a yard, how much will 2800 yds. of tape come to ? 
 
 116. What cost 16^8 yds. of flannel, at 4s. per yard? 
 
 117. What cost 2560 bu. of oats, at '2s. per bushel? 
 
 118. What cost 9600 Ibs. of wool, at 2s. 6d. a pound ? 
 
 119. What cost 3200 Ibs. of sugar, at 6d. per pound? 
 
 120. What cost 600 yds. of damask, at 5s., N. E. cur., per yard ? 
 Note. 3. 5s. N. E. cur. is | less than SI ; hence, GOO 100=S500. Ans 
 
 121. What cost 2500 bu. of potatoes, at Is. 6d. per bushel? 
 
 122. What cost 1440 yds. of gingham, at 2s. per yard? ' 
 
 123. How much will 4848 chickens cost, at Is. apiece? 
 
 124. How much will 1680 slates cost, at Is. 6d. apiece? 
 
 125. How much will 920 turkeys cost, at 4s. 6d. apiece? 
 
 126. What cost 4860 Ibs. of butter, at Is. Id. per pound? 
 
 127. What cost 1260 melons, at 8d. apiece? 
 
 128. What cost 2340 Ibs. of tea, at 4s. a pound? 
 
 129. What cost 240 bu. of peas, at 4s. 6d. per bushel? 
 
 130. What cost 720 pair of gloves, at 5s. 3d. a pair? 
 
 131. What cost 360 bushels of corn, at 3s. per bushel? 
 
 132. What cost 7686 Ibs. of butter, at Is. per pound? 
 133 What cost 960 yds. of silk, at 5s. per yard? 
 
 134. What will 75 Ibs. of butter cost, at $16.80 per cwt. ? 
 
 135. What will 125 Ibs. of wool cost, at $36 per hundred ? 
 
 136. What will 15 ewt. of hemp cost at $60 per ton ? 
 
 137. What will 2500 Ibs. of iron cost, at $72 per ton? 
 
 138. What cost H acre of land, at $120 per acre ? 
 
ARTS. 472-475.J RATIO. 313 
 
 SECTION XIV. 
 RATIO AND PROPORTION 
 
 ART. 472* In comparing numbers or quantities with each 
 other, we may inquire, either how much greater one of the num 
 bers or quantities is than the other ; or how many times one of 
 them contains the other. In finding the answer to either of these 
 inquiries, we discover what is called the relation between the two 
 numbers or quantities. 
 
 473.- The relation between the two quantities thus compared, 
 is of two kinds : 
 
 First, that which is expressed by their difference. 
 
 Second, that which is expressed by the quotient of the one di- 
 vided by the other. 
 
 47 4. RATIO is that relation between two numbers or quanti- 
 . ties, which is expressed by the quotient of the one divided by the 
 other. Thus, the ratio of 6 to 2 is 6-^-2, or 3 ; for 3 is the quo- 
 tient of 6 divided by 2. 
 
 OBS. The relation between two numbers or quantities denoted by their dif- 
 ference, is sometimes called arithmetical ratio ; while that denoted by the quo- 
 tient of the one divided by the other, is called geometrical ratio. Thus 4 is the 
 arithmetical ratio of 8 to 4 ; and 2 is the geometrical ratio of 8 to 4. 
 
 But as the. term arithmetical ratio is merely a substitute for the word differ- 
 ence, the term di (Terence, in the succeeding pages, is used in its stead ; and when 
 the word ratio simply is used, it signifies that which is denoted by the qiwiient 
 of the one divided by the other, as in the article above, 
 
 475. The two given numbers thus compared, when sptken 
 of together, are called a couplet ; when spoken of separately, they 
 are called the terms of the ratio. 
 
 The first term is the antecedent ; and the last, the consequent. 
 
 72. In how many ways are numbers or quantities compared? 474. What Is 
 ratio? 475. What are the two given numbers called when spoken of together? When 
 poke* of separately ? 
 
314 RATIO. [SECT. ^ 
 
 47 6 Ratio is expressed in two ways : 
 
 First, in the form of a fraction, making the antecedent the 
 numerator, and the consequent the denominator. Thus, the ratio 
 of 8 to 4 is written -f ; the ratio of 12 to 3, V> &c. 
 
 Second, by placing two points or a colon ( : ) between the num- 
 bers compared. Thus, the ratio of 8 to 4 is written 8:4; the 
 ratio of 12 to 3, is 12 : 3, &c. The expressions -f-, and 8 : 4, are 
 of the same import, and one may be exchanged for the other, at 
 pleasure. 
 
 OBS 1. The sign ( : ) used to denote ratio , is derived from the sign of divi- 
 sion, -F- ) the horizontal line being omitted. The Engfish mathematicians 
 put the antecedent for the numerator, and the consequent for the denomina- 
 tor as above ; but the French put the consequent for the numerator and the 
 antecedent for the denominator. The English method appears to be equally 
 simple, while it is confessedly the most in accordance with reason. 
 
 2. In order that concrete numbers may have a ratio to each other, they must 
 necessarily express objects so far of the same nature, that one can be properly 
 said to be equal to, or greater, or less than the other. (Art. 294.) Thus a foot 
 has a ratio to a yard ; for one is three times as long as the other ; but a foot 
 has not properly a ratio to an hour, for one cannot be said to be longer or 
 shorter than the other. 
 
 477. A direct ratio is that which arises from dividing the 
 antecedent by the consequent; as 6-:- 2. (Art. 474.) 
 
 478. An inverse, or reciprocal ratio, is the ratio of the recip- 
 rocals of two numbers. (Art. 160. Def. 10.) Thus, the direct 
 ratio of 9 to 3, is 9 : 3, or f ; the reciprocal ratio is \ : ^, or -J-f- 
 | = f ; (Art. 229;) that is, the consequent 3, is divided by the 
 antecedent 9. 
 
 Note. The term inverse, signifies inverted. Hence, 
 
 An inverse, or reciprocal ratio is expressed by inverting the frac- 
 tion which expresses the direct ratio ; or when the notation is by 
 points, by inverting the order of the terms. Thus, 8 is to 4, in- 
 versely, as 4 to 8. 
 
 QUEST. 476. In how many ways is ratio expressed ? The first? The second? Oft*. 
 W 7 hich of the terms do the English mathematicians put for the numerator ? Which do 
 the French ? In order that concrete numbers may have a ratio to each other, what kind 
 of objects must they express ? 477. What is a direct ratio ? 478. What is an inverse at 
 reciprocal ratio ? How is a reciprocal ratio expressed by a fraction ? How ty points ? 
 
ARTS. 476-481.] RATIO. 315 
 
 479* A simple ratio is a ratio which has but one antecedent 
 and one consequent, and may be either direct or inverse ; as 9 : 3, 
 
 48O A compound ratio is the ratio of the products of th<* 
 corresponding terms of two or more simple ratios. Thus, 
 
 The simple ratio of 9 : 3 is 3 ; 
 
 4 And " " of 8 : 4 is 2 ; 
 
 The ratio compounded of these is 72 : 12 = 6. 
 
 OBS 1. A compound ratio is of the same nature as any other ratio. Th 
 term is used to denote the origin of the ratio in particular cases. 
 2. The compound ratio is equal to the product of the simple ratios. 
 
 Ex. 1. What is the ratio of 27 to 9 ? Ans. 3. 
 
 2. What is the ratio of 8 to 32 ? Ans. -J-. 
 
 Required the ratio of the following numbers : 
 
 3. 14 to 7. 13. 324 to 81. 23. 63 Ibs. to 9 oz. 
 
 4. 36 to 9. 14. 802 to 99. 24. 68 yds. to 17 yds. 
 
 5. 54 to 6. 15. 9 to 45. 25. 40 yds. to 20 ft. 
 
 6. 108 to 18. 16. 17 to 68. 26. 60 miles to 4 fur. 
 
 7. 144 to 24. 17. 13 to 52. 27. 45 bu. to 3 pks. 
 
 8. 156 to 17. 18. 27 to 135. ' 28. 6 gals, to 1 hhd. 
 
 9. 261 to 29. 19. 53 to 212. 29. 3 qts. to 20 gal. 
 
 10. 567 to 63. 20. 47 to 329. 30. l to 15s. 
 
 11. '405 to 45. 21. 18 Ibs. to 6 Ibs. 31. 15s. to 3. 
 
 12. 576 to 64. 22. 28 Ibs. to 4 Ibs. 32. 10 to lOd. 
 
 48 1. From the definition of ratio and the mode of expressing 
 it in the form of a fraction, it is obvious that the ratio of two num- 
 bers is the same as the value of a fraction whose numerator and 
 denominator are respectively equal to the antecedent and conse- 
 quent of the given couplet ; for, each is the quotient of the numer- 
 ator divided by the denominator. (Arts. 474, 185.) 
 
 OBS. From the principles of fractions already established, we may, there- 
 fore, deduce the following truths respecting ratios. 
 
 QUEST. 479. What is a simple ratio 1 480. What is a compound ratio ? Obs Doei it 
 differ i/ 1 its nature from other ratios 1 What is the term used to denote ? 
 
 14* 
 
316 RATIO. [SECT. XIV. 
 
 482* To multiply ike antecedent of a couplet by any number, 
 multiplies tlie ratio by that number ; and to divide the antecedent, 
 divides the ratio : for, multiplying the numerator, multiplies the 
 value of the fraction by that number, and dividing the numerator, 
 divides the value. (Arts. 186, 187.) 
 
 Thus, the ratio of 16 : 4 is 4 ; 
 
 The ratio of 16 X 2 : 4 is 8, which equals 4X2; 
 
 And " 16-f-2 : 4 is 2, which equals 4-7-2. 
 
 OBS. With a given consequent the greater the antecedent,, the greater the 
 ratio ; and on the other hand, the greater the ratio, the greater the antece- 
 dent. (Art. 187. Obs.) 
 
 48 3 To multiply the consequent of a couplet by any number, 
 divides the ratio by that number y and to divide the consequent, 
 multiplies the ratio: for, multiplying the denominator, divides the 
 value of the fraction by that number, and dividing the denomina- 
 tor, multiplies the value. (Arts. 188, 189.) 
 
 Thus, the ratio of 16 : 4 is 4 ; 
 
 The " 16 : 4X2 is 2, which equals 4-^2 ; 
 
 And " 16 : 4-i-2 is 8, which equals 4X2. 
 
 OBS. With a given antecedent, the greater the consequent, the less the ratio ; 
 and the greater the ratio, the less the consequent. (Art. 189. Obs.) 
 
 484 To multiply or divide both the antecedent and consequent 
 of a couplet by the same number, does not alter the- ratio : for, mul- 
 tiplying or dividing both the numerator and denominator by the 
 same number, does not alter the value of the fraction. (Art. 191.) 
 
 Thus, the ratio of 12:4 is 3 ; 
 
 The " 12X2: 4X2 is 3; 
 
 And " 12-7-2: 4-;- 2 is 3. 
 
 485* If the two numbers compared are equal, the ratic is 
 unit or 1, and is called a ratio of equality. Thus, the ratio of 
 6X2 : 12 is 1 ; for the value of H= 1. (Art. 196.) 
 
 QUKST. 482. What is the effect of multiplying the antecedent of a couplet by any num 
 her? Of dividing the antecedent ? 483. What is the effect of multiplying the consequent 
 by any number ? Of dividing the consequent 1 Why 1 484. What is the effect of mul- 
 tiplying or dividing both the antecedent and consequent by the same number? Why? 
 485. When the two numbers compared are equal, what is the ratio 7 What is it called ? 
 
ARTS. 482-488.] RATIO. 317 
 
 486. If the antecedent of a couplet is greater than the conse- 
 quent, the ratio is greater than a unit, and is called a ratio of 
 greater inequality. Thus, the ratio of 12 : 4 is 3 ; for the value 
 of Y=3. (Art. 196.) 
 
 487. If the antecedent is less than the consequent, the ratio 
 is less than a unit, and is called a ratio of less inequality. Thus, 
 (he ratio of 3 : G is , or ; for =$. (Art. 195.) 
 
 OBS. 1. The direct ratio of two fractions which have a common numerator, 
 i* the same as the reciprocal ratio of their denominators. Thus, the ratio of 
 ^ : -f is the same as -J- ; -, or 8 : 4. 
 
 2. The ratio of two fractions which have a common denominator, is the 
 same as the ratio of their numerators. Thus, the ratio of |- : -|- is the same 
 as thai of 8 : 4, viz : 2. Hence, 
 
 488. The ratio of any two fractions may be expressed in 
 whole numbers, by reducing them to a common denominator, and 
 then using the numerators for the terms of the ratio. (Art. 484.) 
 Thus, the ratio of % to 4- is the same as -^ : -j\, or 6:2. 
 
 33. What is the direct ratio of 4 to 12, expressed in the lowest 
 terms? Ans. \. 
 
 34. What is the inverse ratio of 4 to 12 ? Ans. -L-j-T J 5=3. 
 
 35. What is the direct ratio of 64 to 8 ? Of 9 to 63 ? 
 
 36. What is the direct ratio of 84 to 21 ? Of 256 to 32 ? 
 
 37. What is the inverse ratio of 4 to 16 ? Of 28 to 7 ? 
 
 38. What is the inverse ratio of 42 to 6 ? Of 8 to 72 ? 
 
 39. Which is the greater, the ratio of 63 to 9, or that of 72 to 8 ? 
 
 40. Which is the greater, the ratio of 86 to 240, or that of 45 
 to 72? 
 
 41. Which is the greater, the ratio of 120 to 85, or that of 
 240 to 170? 
 
 42. Which is the greater, the ratio of 624 to 416, or that of 
 936 to 560? 
 
 43. Is the ratio of 5X6 to 24, a ratio of greater, or less in 
 equality? 
 
 QUZST. 486. When the antecedent is greater than the consequent, whit is the ratio 
 called ? 437. If the antecedent is less than the consequent, what is the ratio called 1 
 488 How mav the ratio of two fractions be expressed in whole numbers 7 
 
318 RATIO. \SECT. XIV 
 
 44. Is the ratio of 6 X 9 to 7 X 8, a ratio of greater, or less in- 
 equality ? 
 
 45. Is the ratio of 2X^X16 to 4X32 a ratio of greater, or less 
 inequality ? 
 
 46. What is the ratio compounded of the ratios of 5 to 3, and 
 12 to 4 ? 
 
 47. What is the ratio compounded of 8 : 10, and 20 : 16 ? 
 
 48. What is the ratio compounded of 3 : 8, and 10:5? 
 
 49. What is the ratio compounded of 18 : 20, and 30 : 40 ? 
 
 50. What is the ratio compounded of 35 : 40, and 60 : 75, and 
 21 tc 19? 
 
 51. What is the ratio compounded of 60 : 40, and 12 : 24, and 
 25:30? 
 
 48 O In a series of ratios, if the consequent of each preced- 
 ing couplet is the antecedent of the following one, the ratio of the 
 first antecedent to the last consequent, is equal to that com- 
 pounded of all the intervening ratios. 
 Thus, in the series of ratios 3 : 4 
 
 4:7 
 7: 16 
 
 the ratio of 3 to 16, is equal to that which is compounded 
 of the ratios of 3 : 4, of 4 : 7, and 7 : 16; for, the compound 
 
 .. . 3X4X7 
 ratio is 
 
 40O If to or from the terms of any couplet, two other num 
 lers having the same ratio be added or subtracted, the sums or re 
 mainders will also have the same ratio. (Thomson's Legendre, 
 B. III., Prop. 1, 2.) Thus, the ratio of 12 : 3 is the same as that of 
 20 : 5. And the ratio of the sum of the antecedents 12 + 20 to the 
 sum of the consequents 3+5, is the same as the ratio cf either 
 couplet. That is, 
 
 12 +20: 3+5:: 12: 3=20: 5, or 
 
 So alsc the ratio of the difference of the antecedents, to the dif- 
 ference of the consequents, is the same. That is, 
 
 2012 12 20 
 20 12 : 5 3 : : 12 : 3=20 5, or 5373-=^= ~^~- 4 - 
 
ARTS. 189-494.] PROPORTION. 319 
 
 49 1 If in several couplets the ratios are equal, the sum of 
 all tht antecedents has the same ratio to the sum of all the conse- 
 quents, which any one of the antecedents has to its consequent. 
 
 ( 12:4 = 3 
 
 Thus, the ratio of] 15:5 = 3 
 ( 18:6 = 3 
 Therefore the ratio of (12 + 15 + 18) : (4+5 + 6)=3. 
 
 OBS. 1. A ratio of greater inequality is diminished by adding the same nurit- 
 for to both terms. Thus, the ratio of 8 : 2, is 4 ; and the ratio of 8+4 : 2+ 
 4 is 2. 
 
 2. A ratio of less inequality is increased by adding the same number to both 
 the terms. Thus, the ratio of 2 : 8 is i, and the ratio of 2+16 : 8+16 is \. 
 
 PROPORTION. 
 
 492* PROPORTION is an equality of ratios. Thus, the two 
 ratios 6 : 3 and 4 : 2 form a proportion ; for f =f , the ratio of each 
 being 2. 
 
 OBS. The terms of the two couplets, that is, the numbers of which the pro- 
 portion is composed, are called proportionals. 
 
 493. Proportion may be expressed in two ways. 
 
 First, by the sign of equality (=) placed between the two 
 ratios. 
 
 Second, by four points (: :) placed between the two ratios. 
 
 Thus, each of the expressions, 12:6 = 4:2, and 12 : 6 : : 4 : 2, 
 is a proportion, one being equivalent to the other. The latter ex- 
 pression is read, " the ratio of 12 to 6 equals the ratio of 4 to 2,'* 
 or simply, " 12 is to 6 as 4 is to 2." 
 
 OBS. The sign (: :) is said to be derived from the sign of equality, thejow 
 points being merely the extremities of the lines. 
 
 49 4 The number of terms in a proportion must at least be 
 four, for the equality is between the ratios of two couplets, and 
 each couplet must have an antecedent and a consequent. (Art. 476.) 
 
 There may, however, be a proportion formed from three num- 
 bers, for one of the numbers may be repeated so as to form two 
 
 QUEST. 492. What is Proportion ? 493. How many ways is proportion expressed T 
 What is the first ? The second ? 494. How many terms must there be in a proportion 1 
 Why 1 Can a proportion be formed of three numbers 7 How ? 
 
320 PROPORTION. [SECT. XIV 
 
 terms. Thus, the numbers 8, 4, and 2, are proportional ; for the 
 ratio of 8 : 4=4 : 2. It will be seen that 4 is the consequent ia 
 the first couplet, and the antecedent in the last. It is therefore 
 a mean proportional between 8 and 2. 
 
 OBS. 1. In this case, the number repeated is called the middle term or mean 
 proportional between the other two numbers. 
 
 The last term is called a third proportional to the other two numbers. Thua 
 2 is a third proportional to 8 and 4. 
 
 2 Care must be taken not to confound proportion with ratio. (Arts. 474, 43*2.) 
 In a simple ratio there are but two terms, an antecedent and a consequent , 
 whereas in a proportion there must at least be four terms, or two couplets. 
 
 Again, one ratio may be greater or less than another; the ratio of 9 to 3 ia 
 greater than the ratio of 8 to 4, and less than that of 18 to 2. QHC proportion, 
 on the other, hand, cannot be greater or less than another ; for equality doea 
 not admit of degrees. 
 
 49 5 The first and last terms of a proportion are called the 
 extremes ; the other two, the means. 
 
 OBS. Homologous terms are either the two antecedents, or the two conse- 
 quents. Analogous terms are the antecedent and consequent of the same 
 couplet. 
 
 496. Direct proportion is an equality between two direct 
 ratios. Thus, 12 : 4 : : 9 : 3 is a direct proportion. 
 
 OBS. In a direct proportion, the first term has the same ratio to the second, 
 as the third has to the fourth. 
 
 497. Inverse or reciprocal proportion is an equality between 
 a direct and a reciprocal ratio. Thus, 8 : 4 :::-; or 8 is to 4, 
 reciprocally, as 3 is to 6. 
 
 OBS. In a reciprocal or inverse proportion, the first term has the same ratio 
 to the second, as the fourth has to the third. 
 
 498. If four numbers are proportional, the product of the ex- 
 tremes is equal to the product of the means. Thus, 8 : 4 : : 6 : 3 ' 
 a proportion; for f=f, (Art. 492,) and 8X3=4X6. 
 
 . Obs. What is the number repeated called ? What is the last term called ji 
 such a case ? What is the difference between proportion and ratio ? 495. Which terms 
 are the extremes ? Which the means 1 Obs- What are homologous terms? Analrgous 
 terms ? 496. What is direct proportion ? Obs. In direct proportion what ratio has the 
 first term to the second ? 497. What is inverse proportion 1 ? Oba. What ratio has the 
 first term to the second in this case 7 498. If four numbers are proportional, what is the 
 puduct of the extremes equW to? 
 
ARTS. 495-501.] PROPORTION. 321 
 
 Again, 12 : 6 : : i : is a proportion. (Art. 496.) 
 And 12xi=6xi 
 
 OBS. 1. The truth of this proposition may also be illustrated in. the following 
 manner : 
 
 The numbers 2 : 3 : : 6 : 9 are obviously proportional. (Art. 492.) 
 
 For, f =f. (Art. 195.) Now, 
 
 Multiplying each ratio by 27, (the product of the denominators,) 
 
 The proportion becomes?^?? =~-- (Art.2t. Ax. 6. 
 
 Dividing both the numerator and the denominator of the first couplet by 3, 
 (Art. 191,) or canceling the denominator 3 and the same factor in 27, (Art. 221,) 
 also canceling the 9, and the same factor in 27, we have 2x9=6x3. But 2 
 and 9 are the extremes of the given proportion, and 3 and 6 are the means ;, 
 hence, the product of the extremes is equal to the product of the means. 
 
 2. Conversely, if the product of the extremes is equal to the product of the 
 means, the four numbers are proportional ; and if the products are not equal, 
 the numbers are not proportional. 
 
 499, Proportion, ii* arithmetic, is usually divided into Simple 
 and Compound. 
 
 SIMPLE PROPORTION. 
 
 500, SIMPLE PROPORTION is an equality between two simple 
 ratios. It may be either direct or inverse. (Arts. 479, 496, 497.) 
 
 The most important application of simple proportion is the 
 solution of that class of examples in which three terms are given to 
 find a fourth. 
 
 501, We have seen that, if four numbers are in proportion, 
 the product of the extremes is equal to the product of the means. 
 (Art. 498.) Hence, 
 
 If the product of the means is divided by one of the extremes, 
 the quotient will be the other extreme ; and if the product of the 
 extremes is divided by one of the means, the quotient will be the 
 
 QUEST. Obs. If the product of the extremes is equal to the product of the means, what 
 le true of the four numbers 1 If the products are not equal, what Is true of them ? 499 
 How is proportion usually divided 1 500. What is simple proportion 7 What is the most 
 Important application of it? 501. If the product of the means is divided by one of th 
 extremes, what will the quotient be ? If the product of the extremes is divided by one of 
 the means, what will the quotient be ? 
 
322 PROPORTION [SECT. XIV 
 
 other mean. For, if the product of two factors is divided by one 
 of them, the quotient will be the other factor. (Art. 156.) 
 
 Take the proportion 8 : 4 : : 6 : 3. 
 
 Now the product 8X3 4 = 6, one of the means ; 
 
 So the product 8x3 6=4, the other mean. 
 
 Again, the product 4x68=3, one of the extremes; 
 
 And the product 4X63 = 8, the other extreme. 
 
 5 O2 If, tJierefore, any three terms of a proportion are given, 
 the fourth may be found by dividing the product of two of tJiem 
 by the other term. 
 
 OBS. Simple Proportion is often called the Rule of TTirce, from the circum- 
 stance that three terms are given to find a fourth. In the older arithmetics, it 
 is also called the Golden Ride. But the fact that these names convey no idea 
 of the nature or object of the rule, seems to be a strong objection to their use, 
 not to say a sufficient reason for discarding them. 
 
 Ex. 1. If the product of the means is. 84, and one of the ex- 
 tremes is 7, what is the other extreme, or term of the proportion ? 
 
 2. If the product of the means is 54, and one of the extremes 
 is 18, what is the other extreme ? 
 
 3. If the product of the means is 720, and one of the extremes 
 is 45, what is the other extreme ? 
 
 4. If the product of the means is 639, and one of the extremes 
 is 213, what is the other extreme ? 
 
 5. If the first three terms of a proportion are 8, 12, and 16, 
 is the fourth term ? 
 
 Solution. 12X16 = 192, and 192-r8 = 24, the fourth term, or 
 number required ; that is, 8 : 1 2 : : 16 : 24. 
 
 6. It is required to find the fourth term of the proportion, the 
 first three terms of which are 36, 30, and 24. 
 
 7. Required the fourth term of the proportion, the first three 
 terms of which are 15, 27, and 31. 
 
 8. Required the fourth term of the proportion whose first three 
 terms are 45, 60, and 90. 
 
 QUEST. Obs. What is simple proportion often called ? Do these terms convey an /deft 
 of the nature or object of the rule 1 
 
ARTS. 502, 503.J PROPORTION. 323 
 
 9. If 8 yds. of broadcloth cost $96, how much will 20 yds. cost 
 at the same rate ? 
 
 Solution. It is plain that 8 yds. has the same ratio to 20 yds. 
 as the cost of 8 yds., viz : $96, has to the cost of 20 yds. That is, 
 
 8 yds. : 20 yds. : : $96 : to the cost of 20 yds. 
 Now $96X20 $1920; and $1920-r8=$240. Ans. 
 
 10. If 35 men will consume a certain quantity of flour in 20 
 lays, how long will it take 50 men to consume it ? 
 
 Note. Since the answer is days, we put the given days for the third term. 
 Then, as the flour will not last 50 men so long as it will 35 men, we put the 
 smaller number of men for the second term, and the larger for the first. 
 
 Operation. 
 
 Men. Men. Days. 
 
 50 : 35 : : 20 : to the number of days required. 
 
 20 Multiply the second and third terms to- 
 
 50 ) 700 gether, and divide the product by the first 
 
 ] 4 days. Ans. term, as in the last example. 
 PROOF. 50X14-35X20. (Art. 498.) 
 
 5O3. From the preceding illustrations and principles, we de- 
 duce the following general 
 
 RULE FOR SIMPLE PROPORTION. 
 
 I. Place that number for the third term, which is of the same 
 kind as the answer or number required. 
 
 11. Then, if by the nature of the question the answer must be 
 greater than the third term, place the greater of the other two num- 
 bers for the second term ; but if it is to be less, place the less of the 
 other two numbers for the second term, and the other for the first. 
 
 III. Finally, multiply the second and third terms together, divide 
 the product by the first, and the quotient will be the answer in the 
 same denomination as the third term. 
 
 PROOF. Multiply the first term and the answer together, and 
 if the product is equal to the product of the second and third terms, 
 the work is right. (Art. 500.) 
 
 QUEST. 503. In arranging the terms in simple proportion, which number is pnt for the 
 third term ? How arrange the other two numbers ? Having stated the question how is the 
 ans wer found 1 Of what U ^nomination is the answer ? How is simple proportion proved I 
 
324 SIMPLE [SECT. XIV. 
 
 Demonstration, -If four numbers are proportional, we have seen that the 
 product of the means is equal to the product of the extremes; (Art. 498;) there- 
 fore the poduct of the second and third terms must be equal to that of the first 
 and fourth. But if the product of two factors is divided by one of them, the 
 quotient will be the other; (Art. 15G ;) consequently, when the first three 
 lerms of a proportion are given, the product of the second and third terms di- 
 vided by the first, must give the fourth term or answer. 
 
 The object of placing that number, which is of the same kind as the answer, 
 for the third term, instead of the second, as is sometimes done, is twofold : 1st, 
 it avoids the necessity of the Rale of Three Inverse ; 2d, the third term, in 
 many cases, has no ratio to the first ; consequently it is inconsistent with tfaa 
 principles of proportion to put it for the second term. Thus, in the ninth ex- 
 ample, if we put $96 for the second term, it would read, 8 yds. : $9G : : 20 
 yds. : $240, the answer. But a yard can have no ratio to a dollar ; for one 
 Cannot be said to be greater nor less than the other. (Art. 476. Obs. 2.) 
 
 OBS. 1. If the first and second terms are compound numbers, reduce them 
 to the lowest denomination mentioned in either, before the multiplication or 
 division is performed. 
 
 When the third term contains different denominations, it must also be re- 
 duced to the lowest denomination mentioned in it. 
 
 2. The process of arranging the terms of a question for solution, or put- 
 ting it into the form of a proportion, is called stating the question. 
 
 3. Questions in Simple Proportion, we have seen, may be solved by 
 Analysis. After solving the following examples by proportion, it will be an ex- 
 ellent exercise for the student to solve them by analysis. (Art. 4G2. Obs. 2.) 
 
 11. If 16 barrels of flour cost $112, what will 129 barrels cost? 
 
 12. If 40 acres of land cost $540, what will 97 acres cost? 
 
 13. If 641 sheep cost $1923, what will 75 sheep cost? 
 
 14. At the rate of 155 miles in 12 days, how far can a man 
 travel in 60 days ? 
 
 15. How much hay, at $17.50 per ton, can you buy for $350 ? 
 
 16. If $45 buy 63 Ibs. of tea, how much will $1540 buy? 
 
 17. If 90 Ibs. of pepper are worth 72 Ibs. of ginger, how many 
 Ibs. of ginger are 64 Ibs. of pepper worth ? 
 
 18. A bankrupt compromised with his creditors, at 64 cts. on a 
 dollar ; how much will be received on a debt of $2563.50 ? 
 
 19. An emigrant has a draft for 1460 sterling: how much is 
 it worth, allowing $4.84 to a pound ? 
 
 27ST. Obs. [f the first and second terms contain different denominate ns, how pro- 
 ceed 7 When the it ird term contains different denominations, what is to be >nne 1 What 
 b meant oy stating a question 1 
 
ART. 504. J PROPORTION. 325 
 
 SIMPLE PROPORTION BY CANCELATION. 
 20. If 72 tons of coal cost $648, how much will 9 tons cost? 
 Operation. Having stated the question as be 
 
 T -?? ' - T <T S - D llS A < * re> we P erce ^ ve tlie factor 9 is com - 
 mon to the first two terms, and 
 
 therefore may be canceled. (Art. 
 
 Now $648-|-8=$81. Ans. _ K1 v J V 
 
 iDi.j 
 
 Or thus, -^ =the answer. (Art. 503.) 
 
 9Vfi4-8 d) y 648 
 But - =~TA =$81, the same as before. Hence, 
 
 72 *>8 
 
 5O4. When the first term has factors common to either of 
 the other two terms. 
 
 Cancel the factors which are common, then proceed according to 
 the rule above. (Arts. 151, 221.) 
 
 PROOF. Place the answer in the denominator, or on the left of 
 the perpendicular line, as the case may be, and if the factors of the 
 divisor exactly cancel those of the dividend, the work is right. 
 
 OBS. 1. The question should be stated, before attempting to cancel the com- 
 mon factors. When the terms are of different denominations, the reduction 
 of them may sometimes be shortened by Cancelation. 
 
 2. Instead of points, it may sometimes be more convenient fo place a per- 
 pendicular line between the first and second terms, as in division of fractions. 
 (Art. 231.) In this case the third term should be placed under the second, 
 with the sign of proportion ( : : ) before it to denote its origin, and its relation 
 n the fourth term or the answer. 
 
 3. It will be perceived that cancelation is applicable in Simple Proportion to 
 all those examples; whose first term has one or more factors common to either 
 of the other terms. 
 
 21 If 24 yds. of cloth cost $63, what will 32 yds. cost ? 
 Operation. 
 
 yds. 
 
 20 yds., 40 When arranged in this way, th 
 
 ::$$$, 21 question is read, 24 yds. is to 320 
 
 Ans. |$21 X 40=$840. yds., as $63 is to the answer required. 
 22. If 20 bu. of oats cost l, how much will 2 quarts cost ? 
 
326 SIMPLE [SECT. XIV 
 
 23. If 12 bbls. of flour cost $88, what will 108 barrels cost 9 
 
 24. If 30 cows cost $480, what will 173 cows cost ? 
 
 25. If a man can travel 240 miles in 16 days, how far can he 
 travel in 29 days? 
 
 26. If 48 men can build a ship in 84 days, how long would it 
 take 1 6 men to build it ? 
 
 27. If ^ of a ton of hay costs -, what will | of a ton cost ? 
 
 ton. ton. 
 
 Solution. J- : -I : : f : Ans. Now,f XlXi=2 Ans. Hence, 
 
 5O5. If the terms in a proportion are fractional, the question 
 is stated, and the answer obtained in the same manner as if they 
 were whole numbers. 
 
 OBS. When the first and second terms are fractions, we may reduce them to 
 a common denominator, and then employ the numerators only ; for the ratio 
 of two fractions which have a common denominator, is the same as the ratio 
 of their numerators. (Art. 487. Obs. 2.) 
 
 28. If f of a cord of wood cost $1.35, what will f of a cord 
 cost? 
 
 29. If of a yard of berege cost 6 shillings, what will $ of a 
 yard cost ? 
 
 30. If f of a yard of sarcenet cost f of a dollar, what will 3| 
 yds. cost? 
 
 31. If | of a pound of chocolate cost | of a dollar, what will 
 25f pounds cost ? 
 
 32. What will 165 melons cost, at $ of a dollar for 5 melons? 
 
 33. A man had 420 acres of land which he wished to divide 
 among his three sons A, B, and C, in proportion to the numbers 
 7, 5, and 3 : how much land would each receive ? 
 
 Solution. Since the several parts are to be proportional to the 
 numbers 7, 5, and 3, the sum of which is 15, it is evident that the 
 Bum of all the given numbers is to any one of them, as the whole 
 quantity to be divided to the part corresponding to the number 
 used as the second term. 
 
 That is 15 : 7 : : 420 A. to A's share, which is 196 acres; 
 
 Also 15:5:: 420A. to B's " " " 140 acres ; 
 
 And 15 : 3 : : 420A. to C's " " " 84 acres. 
 
 PROOF. 196+140+84=420A. the given number. (Ax. 11.) 
 
ARTS 505, 506.] PROPORTION. 327 
 
 5O6. Hence, to divide a given number or quantity into parts 
 which shall be proportional to any given numbers. 
 
 Place the whole number or quantity to be divided for the third 
 term, the sum of the given numbers for the first term, and each of 
 the given numbers respectively for the second; then multiply ard 
 divide as before. (Art. 503.) 
 
 31 A farmer wishes t: mix 100 bushels of provender of oats 
 and corn in the ratio of 3 to 7 : how many bushels of each must 
 he put in ? 
 
 35. Bell metal is composed of 3 parts of copper, and 1 of tin : 
 how much of each ingredient will be used in making a bell which 
 weighs 2567 pounds ? 
 
 36. Gunpowder is composed of 76 parts of nitre, 14 of char- 
 coal, and 10 of sulphur : how much of each of these ingredients 
 will it take to make a ton of powder ? 
 
 37. If 40.12 Ibs. of sugar are worth $5.13, how much can be 
 bought for $125.375 ? 
 
 .38. The Vice-President's salary is $5000 a year: if his daily 
 expenses are $10, how much can he lay up ? 
 
 39. If f Ib. of snuff cost fc, what will 150 Ibs. cost? 
 
 40. If -f of f of f of a sloop cost $1500, what will the whole cost? 
 
 41. If - of -f- of an acre of land on Broadway is worth $8200, 
 how much is - of -f*of an acre worth ? 
 
 42. A man bought of a vessel and sold f of what he bought 
 for $8240, which was just the cost of it : what was the whole 
 vessel worth ? 
 
 43. How many times will the fore wheel of a carriage which is 
 7 ft. 6 in. in circumference turn round in going 100 miles ? 
 
 44. How many times will the hind wheel of a carriage 9 ft. 
 2 in. in circumference, turn round in going the same distance ? 
 
 45. There are two numbers which are to each other as 12 to 
 84, the smaller of which is 75 : what is the larger? 
 
 46. What two numbers are those which are to each other as 
 6 to 6, the greater of which is 240 ? 
 
 47. If two numbers are as 8 to 12, and the less is 320, what 
 is the greater ? 
 
378 PROPORTION. [SECT. XIV. 
 
 48. There are two flocks of sheep which are to each other as 15 
 to 20, and the greater contains 500 : how many does the less con- 
 tain ? 
 
 49. An express traveling 60 miles, a day had been dispatched 
 5 days, when a second was sent after him traveling 75 miles a 
 day : how long will it take the latter to overtake ,ne former ? 
 
 50. A foi has 150 rods the start of a hound, but the hound 
 tuns 6 rods while the fox runs 5 rods : how far must the hound 
 
 un before he catches the fox ? 
 
 51. A stack of hay will keep a cow 20 weeks, and a horse 15 
 weeks : how long will it keep them both ? 
 
 52. A traveler divided 80s. among 4 beggars in such a man- 
 ner, that as often as the first received 10s., the second received 
 f>s., the third 8s., and the fourth 7s. : what did each receive ? 
 
 53. Pure water is composed of oxygen and hydrogen in the ratic 
 of 8 to 1 by weight : what is the weight of each in a cubic foot 
 of tvater, or 1000 ounces avoirdupois? 
 
 COMPOUND PROPORTION. 
 
 5O7 COMPOUND PROPORTION is an equality between a com- 
 pound ratio and a simple one. (Arts. 479, 480.) 
 
 us ' * > : : 12 : 3, is a compound proportion. 
 Into 4:2$ 
 
 That is, 6X4 : 3X2 : : 12 : 3 ; for, 6X4X3 = 3X2X12. 
 
 OBS. Compound proportion is chiefly applied to the solution of exe.mj let 
 which would require two or more statements in simple proportion. It is some- 
 times called Double Ruk of Three. 
 
 Ex. 1. If 8 men can reap 32 acres in 6 days, how many acres 
 can 12 men reap in 15 days? 
 
 Suggestion. When stated in the form of a compound propor- 
 tion, the question will stand thus : 
 
 8m. : 12m. i 
 
 . , , K S : : 32 A. : to the answer. 
 
 6d. : 15a. > 
 
 That is, the product of the antecedents 8X6, has the same 
 
 QUEST. 507. What is compound proportion 1 Ob*. To what Is it cAiefl- rpplled f 
 What is it sometimes called 1 
 
ARTS. 507, 508.] PROPORTION. 329 
 
 ratio tc the product of the consequents 12x15, as 32 has to the 
 answer ; o, simply, 8 into 6:12 into 15 : : 32 : to the answer. 
 
 Operation. The product of the numbers 
 
 32X12X15 = 5760, standing in the 2d and 3d places 
 
 And 8X 6 = 48. divided by the product of those 
 
 Now 5760^-48=120. standing in the first place, will 
 
 Ans. 12,0 acres. give the answer. 
 
 
 
 Not.. The learner will observe that it is not the ratio of 8 to 12 alone, 
 nor that of 6 to 15, which is equal to the ratio of 32 to the answer, as it it 
 sometimes stated ; but it is the ratio compounded of 8 to 12, and 6 to 15, which 
 is equal to the ratio of 32 to the answer. Thus, 8X6 : 12X15 : : 32 : 120, tha 
 answer. A compound proportion when stated as above, is read, " the ratio of 
 8X6 is to 12X15 as 32 is to the answer." 
 
 2 If 6 men can earn 42 in 60 days working 8 hours per day, 
 how much can 10 men earn in 84 days working 12 hours a day? 
 
 Operation. 
 
 State the question, then 
 6m. : 10m. \ 5 ,. . , 
 
 60d 84d ( : : 42 : to Ans. mult 'P'y and dlvlde > as 
 
 V before. 
 
 8hrs. : 12hrs. ) 
 
 10X84X12X42=423360; and 6X60X8=2880. 
 Now 423360H-2880=147. Ans. 147. 
 
 5O8. From the foregoing illustrations we derive the follow- 
 ing general 
 
 RULE FOR COMPOUND PROPORTION. 
 
 I. Place that number which is of the same kind as the answer 
 required for the third term. 
 
 II. Then take the other numbers in pairs, or two of a kind, and 
 arrange them as in simple proportion. (Art. 503.) 
 
 III. Finally, multiply together all the second and third terms t 
 divide the result by the product of the first terms, and the quo- 
 tient will be the fourth term or answer required. 
 
 QUEST. 508. In stating a question in compound proportion, which number do you put fot 
 the third term 1 How arrange the other numbers ? Having stated the question, how i| 
 the answer found"? 
 
330 COMPOUND [SECT. XIV. 
 
 PROOF. Multiply the answer into all of the first terms or ante- 
 cedents of the first couplets, and if the product is equal to the con- 
 tinued product of all the second and third terms, the work is right. 
 (Art. 498.) 
 
 OBS. 1. Among the given numbers there is but one which is of the same 
 kind as the answer. This is sometimes called the odd terrn^ and is always to 
 be placed for the third term. 
 
 2. If the antecedent and conseqiienl of any couplet are compound numbers, 
 they must be reduced to the lowest denomination mentioned in either, before 
 Ihe multiplication is performed. When the third term contains different de- 
 nominations, it must also be reduced to the lowest mentioned in it. 
 
 3. Questions in Compound Proportion may be solved by Analysis ; also by 
 Simple Proportion, by making two or more separate statements. 
 
 3. If 12 horses can plough 11 acres in 5 days, how many 
 horses can plough 33 acres in 18 days? 
 
 4. If a man walking 12 hours a day, can travel 250 miles in 10 
 days, how long will it take him to travel 400 miles, if he walks but 
 10 hours a day ? 
 
 5. If 40 gallons of water will last 20 persons 5 days, how 
 many gallons will 9 persons drink in a year ? 
 
 6. If 16 laborers can earn 15, 12s. in 18 days, how many 
 laborers will it take to earn 35, 2s. in 24 days ? 
 
 COMPOUND PROPORTION BY CANCELATION. 
 
 7. If a person can make 60 rods of wall in 45 days, working 12 
 hours a day, how many rods can he make in 72 days, working 8 
 hours a day ? 
 
 Statement. 
 
 45d. : 72d. ) Rods - 
 12hrs. : 8hrs. [:' 60 : to the answer. That is, 
 
 0,2 4 
 
 = 64 rods. Ans. Hence, 
 
 45X12 
 
 QUEST. How are questions in compound proportion proved ? Obs. Among the given 
 numbers, how many are of the same kind as the answer 1 Can questions in compound 
 proportion be solved in any other way ? 
 
ART. 509.] PROPORTION. 331 
 
 5OO. When the first terms have factors common to the sec- 
 ond or third terms. 
 
 Cancel the factors which are common, then div de the product of 
 those remaining in the second and third terms by the product of 
 those remaining in the first, and the quotient will be the answer. 
 
 PROOF. Place the answer in the denominator, or on the left of 
 the perpendicular line, and if the factors of the divisor and dividend 
 exactly cancel each other, the work is right. 
 
 OBS. 1. Instead of placing points between the antecedents and consequents 
 of the left hand couplets of the proportion, it is sometimes more convenient to 
 put a perpendicular line between them, as in division of fractions. (Art. 23<J.J 
 T.bis will bring all the terms whose product is to be divided on the right of the 
 line, and those whose product is to form the divisor, on the left. In this case 
 the third term should be placed below the second terms, with the sign of pro- 
 portion (: :) before it, to show its origin, and its relation to the answer. 
 
 2. It will be observed that Cancelation can be applied in Compound Pro- 
 portion to all those examples whose^rs^ terms have factors "common to the 
 second terms, or to the third term. 
 
 8. If 24 men can saw 90 cords of wood in 6 days, when the 
 days are 9 hours long, how many cords can 8 men saw in 36 
 days, when they are 1 2 hours long ? 
 Operation. 
 
 fa. 
 
 ,. >d., 2 
 0hrs. 12hrs. 
 
 : 00c., 10 
 
 Ans. 2X12X10 = 240 cords. 
 
 9. If 6 men can make 120 pair of boots in 20 days, working 
 8 hours a day, how long will it take 12 men to make 360 pair, 
 working 10 hours a day? 
 
 10. If 12 men can build a wall 30 ft. long, 6 ft. high, and ' 
 ft. thick, in 18 days, how long will it take 36 men to build on. 
 360 ft. long, 8 ft. high, and 6 ft. thick. 
 
 11. If a horse can travel 120 miles in 4 days when the day 
 are 8 hours long, how far can he travel in 30 days when the days 
 are 10 hours loner ? 
 
 . 509. When the first terms have factors conuncn to the second or third terma, 
 foow proceed ? 
 
 15 
 
 >w pr 
 T.1J 
 
332 CONJOINED [SECT. XIV, 
 
 12. If $250 gam $30 in 2 years, what will be the interest of 
 $750 'for 5 years? 
 
 13. What will be the interest of $500 for 4 years, if $600 will 
 gain $42 in 1 year ? 
 
 14. If $360 gain $14.40 in 8 months, what will $4800 gain in 
 32 months? 
 
 15. If a family of 8 persons spend $200 in 9 months., how 
 much will 18 persons spend in 12 months? 
 
 16. If 15 men, working 12 hours a day, can hoe 60 acres in 20 
 days, how long will it take 30 boys, working 10 hours a day, to 
 hoe 96 acres, 6 men being equal to 10 boys? 
 
 CONJOINED PROPORTION. 
 
 5 1 When each antecedent of a compound ratio is equal in 
 value to its consequent, the proportion is called Conjoined Propor- 
 tion. 
 
 OBS. Conjoined Proportion is often called the chain rule. It is chiefly used 
 in comparing the coins, weights and measures of two countries, through the 
 medium of those of other countries, and in the higher operations of ex- 
 change. The odd term is sometimes called the demand. 
 
 17. If 20 Ibs. United States make 12 Ibs. in Spain ; and 15 Ibs. 
 Spain 20 Ibs. in Denmark; and 40 Ibs. Denmark 60 Ibs. in Russia: 
 how many pounds in Russia are equal to 100 Ibs. U. S. ? 
 
 Operation. Arrange the given terms in 
 
 20 Ibs. U. S. = 12 Ibs. Spain pairs, making the first term the 
 
 15 Ibs. Spain 20 Ibs. Den. antecedent, and its equal the 
 
 40 Ibs. Den. =60 Ibs. Rus. consequent; then since it is 
 
 How many Ibs. R. = 100 Ibs. U. S. required to find how many of 
 
 the last kind are equal to a 
 
 given number (100 Ibs.) of the first, place the odd term or de- 
 mand under the consequents. 
 
 Then, 20X15X40 : 12X20X60 : : 100 : Ans. 
 That is #vt| 12 Cancel the factors common 
 
 20 to both sides, and the prodv ft 
 
 10, 
 
 Ans. 
 
 00, JL of those remaining on the right. 
 
 ::100, 10 divided by the product of those 
 
 1 2 X 1 = 1 2 Ibs. on the left, is t he answer. 
 
AllTrf. 510, 511.] PROPORTION. 333 
 
 511* From these illustrations we derive the following 
 RULE FOR CONJOINED PROPORTION. 
 
 I. Taking the terms in pairs, place the first term on the left of 
 the sign of equality or a perpendicular line for the antecedent, and 
 its equal on the right for the consequent, and so on. Then, if the 
 answer is to be of the same kind as the first term, place tlie odd 
 term under the antecedents j but if not, place it under the conse- 
 quents. 
 
 II. Cancel the factors common to both sides, and if the odd term 
 fells under the consequents, divide the product of the factors re- 
 maining on the right by the product of those on the left, and the quo- 
 tient will be the answer ; but if the odd term falls under tlie ante- 
 cedents, divide the product of tlie factors remaining on the left by 
 the product of those on the right, and the quotient will be the 
 answer. 
 
 PROOF. Reverse the operation, taking the consequents for the 
 antecedents, and the answer for the odd term, and if the result thus 
 obtained is the same as the odd term in the given question, the work 
 is right. 
 
 OBS. In arranging the terms, it should be observed that the Jifst antecedent 
 and the Last consequent will always be of the same kind. 
 
 18. If 100 Ibs. United States, make 95 Ibs. Italian; and 19 Ibw, 
 Italian, 25 Ibs. in Persia ; how many pounds in the U. S. are 
 equal to 50 Ibs. in Persia ? Ans. 40 Ibs. 
 
 19. If 10 yds. at New York make 9 yds. at Athens; and 90 
 yds. at Athens, 112 yds. at Canton; how many yds. at Canton 
 are equal to 50 yds. at New York ? 
 
 20. If 50 yds. of cloth in Boston are worth 45 bbls. of flour in 
 Philadelphia; and 90 bbls. of flour in Philadelphia 127 bales of 
 cotton in New Orleans ; how many bales of cotton at New Or- 
 leans are worth 100 yds. of cloth in Boston? 
 
 21. If $18 U. S. are worth 8 ducats at Frankfort; 12 ducat? 
 at Frankfort 9 pistoles at Geneva ; and 50 pistoles at Geneva, t? 
 rupees at Bombay : how many rupees at Bombay are equal to 
 $100 United States? 
 
334 DUODECIMALS. [SECT. XV. 
 
 SECTION XV 
 DUODECIMALS. 
 
 ART. 512* DUODECIMALS are a species of compound numbers, 
 the denominations 'of which increase and decrease uniformly in a 
 twelvefold ratio. The denominations are feet, inches or primes, 
 seconds, thirds, fourths, fiftlis, tfec. 
 
 Note. The term duodecimal is derived from the Latin numeral du^)decim i 
 which signifies twelve. 
 
 TABLE. 
 
 12 fourths ("") make 1 third, marked "' 
 
 12 thirds " 1 second, " " 
 
 12 seconds " 1 inch or prime, " in. or ' 
 
 1 2 inches or primes " 1 foot, " //. 
 
 Hence 1' =-,V of 1 foot. 
 
 1" =^ of 1 in., or -^ of -f\ of 1 H.=TT of 1 ft. 
 l'"=-iV of 1", or A of -i of tV of 1 ft. =7^ of 1 ft. 
 
 OBS. The accents used to distinguish the different denominations below feet, 
 are called Indices. 
 
 513* Duodecimals may be added and subtracted in the same 
 manner as the other compound numbers. (Arts., 300, 302.) 
 
 MULTIPLICATION OF DUODECIMALS. 
 
 514. Duodecimals are principally applied to the measurement 
 of surfaces and solids. (Arts. 285, 286.) 
 
 Ex. 1. How many square feet are there in aboard 12 ft. 7 in. 
 long, and 4 ft. 3 in. wide ? 
 
 QUEST. 512. What are duodecimals ? What are the denominations? Note. What is 
 the meaning of the term duodecimal ? Repeat the Table. Obs. What are the accent* 
 called, which are used to distinguish the different denominations 1 513. How ar< duodeci- 
 mals added and subtracted ? 514. To what are duodecimals chiefly appliod ? 
 
ARTS. 512-515.] DUODECIMALS 335 
 
 Operation. ^ e ^ rs * multiply each denomination of the 
 
 12 ft 7' multiplicand by the feet in the multiplier, be- 
 
 4 ft 3/ ginning at the right hand. Thus, 4 times 7' 
 
 o~ft ^ - are 28', equal to 2 ft. and 4'. Set the 4 
 
 3 ft 1' 9" un der inches, and carry the 2 feet to the next 
 
 - 53 t ' ^ ^77 product. 4 times 12 ft. are 48 ft. and 2 to 
 
 carry make 50 ft. Again, since 3'=-f\ of a 
 
 ft. and 7'=^r of a ft -> 3/ into 1' is iVt of a ft. = 21", or 1' and 9 y 
 
 Write the 9" one place to the right of inches, and carry the 1' to 
 
 the next product. Then 3' or -^ of a ft. multiplied into 12 ft.= 
 
 ff of a ft., or 36', and 1' to carry make 37' ; but 37'=3 ft. and 1'. 
 
 Now adding the partial products, the sum is 53 ft. 5' 9''. 
 
 Otis. It will be seen from this operation, that feet multiplied into feet, pro- 
 duce feet ; feet into inches, produce inches ; inches into inches, produce 
 seconds, &c. That is, the product of any two factors has as many accents as 
 the factors themselves have. Hence, 
 
 515* To find the denomination of the product of any two 
 factors in duodecimals. 
 
 Add the indices of the two factors together, and the sum will be 
 the index of their product. 
 
 Thus, feet into feet, produce feet; feet into inches, produce 
 inches ; feet into seconds, produce seconds ; feet into thirds, pro- 
 duce thirds ; c. 
 
 Inches into inches, produce seconds ; inches into seconds, pro- 
 duce thirds j inches into fourths, produce fifths, (fee. 
 
 Seconds into seconds, produce fourths ; seconds into thirds, pro- 
 duce fifths j seconds into sixths, produce eiyhtks, &c. 
 
 Thirds into thirds, produce sixths / thirds into fifths, produce 
 eighths ; thirds into sevenths, produce tenths, &c. 
 
 Fourths into fourths, produce eighths ; fourths into eighths, pro- 
 duce twelfths, &c. 
 
 Note. The fooMs considered the unit and has no index. 
 
 Q.TTEST. 515. Hflw find the denomination of the product in duodecimals ? What do feet 
 Into feet produce ? Feet into inches? Feet into seconds ? What do inches into inchefl 
 produce 1 Inches into thirds ? Inches into fourths ? Seconds into seconds ? Seconds 
 Into thirds ? Seconds into eighths 7 Thirds inl i thirds ? Thirds into siiths ? 
 
336 DUODECIMALS. [SECT. XV. 
 
 516* From these illustrations we derive the following 
 
 RULE FOR MULTIPLICATION OF DUODECIMALS. 
 
 1. Place the several terms of tlie multiplier under the corres2>cnd~ 
 ing terms of the multiplicand. 
 
 II. Multiply each term of the multiplicand by each term of the 
 multiplier separately, beginning with the lowest denomination in the 
 multiplicand, and the highest in tlie multiplier, and write the first 
 figure of each partial product one place to the right of that of the 
 pr seeding product, under its corresponding denomination. (Art. 515.) 
 
 III. Finally, add the several partial products toy ether, carrying 
 I for every 12 both in multiplying and adding, and tJie sum will 
 be the answer required. 
 
 UBS. 1. It is sometimes asked whether the inches in duodecimals, are linear, 
 square, or cubic. The answer is, they are neither. An inch is I twelfth of a 
 foot. Hence, in measuring surfaces an inch is -fa of a square foot ; that is, a 
 surface 1 foot wide and 1 inch long. In measuring solids, an inch denotes -fV 
 of a cubic foot. In measuring lumber, these inches are commonly called car- 
 penter's inches. 
 
 2. Mechanics, also surveyors of wood and lumber, in taking dimensions of 
 their work, lumber, &c., often call the inches a fractional part of a foot, and 
 then find the contents in feet and a. fraction of a foot. Sometimes inches are 
 regarded as decimals of a foot. 
 
 3. We have seen that one of the factors in multiplication, is always to be 
 considered an abstract, number. (Art. 82. Obs. 2.) How then, can feet be 
 multiplied by feet, inches by inches, &c. 
 
 It should be observed, that when one geometrical quantity is multiplied by 
 another, some particular extent is to be considered the unit. It is immaterial 
 what this extent is, provided it remains the same in the different parts of the 
 same calculation. Thus, if one of the factors is one fool and the other half & 
 foot, the former being 12 in., and the latter 6 in., the product is 72 in. Though 
 it would be nonsense to say that a given length is repeated as often as another 
 is long, yet there is no impropriety in saying that one is repeated as many times 
 as th.ere are feet or inches in another. 
 
 4. On the principles of duodecimals, it has been supposed that pounds 
 shillings, pence, and farthings can be multiplied by pounds, shillings, pence, 
 and farthings. But it may be asked, what ilcnominalion shillings multiplied 
 by pence, or pence by farthings, will produce? It is absurd to say that 
 2s. and 6d. is repeated 2s. and 6d. times. 
 
 QUEST. 516. What is the rule for multiplication of duodecimals ? Obs. What kind of 
 Inches are those spoken of in me curing surfaces by duodecimals ? In measuring solids 1 
 
ART. 51 6. J DUODECIMALS. 337 
 
 Ex. 2. How many square feet are there in a piece of marble 
 ft. 7 in. 2" long, and 3 ft. 4 in. 7" wide ? 
 
 Note. It is not absolutely necessary to begin to multiply by the highest de- 
 nomination of the multiplier, or to place the lower denomination to the right of 
 the multiplicand. The result will be the same if we begin with the lowest de- 
 nomination of the multiplier, and place the first figure of each partial product 
 nnder the figure by which we multiply. 
 
 Common Method. Second Method. 
 
 9 ft. 7' 2' 9 ft. 7' 2'' 
 
 3 ft. 4' V 3 ft. 4' 7-' 
 
 28 ft. 9' 6" 5' 7" 2'" 
 
 3 ft. 2' 4" 8'" 3 ft. 2' 4" 8'" 
 
 5' 7" 2'" 2'"' 28 ft. 9' 6" 
 
 Ans. 32 ft. 5' 5" 10'" 2"". Ans. 32 ft. 5' 5'' 10'" 2"" 
 
 3. How many square feet are there in aboard 15 ft. 7 in. long, 
 and 1 ft. 10 in. wide ? 
 
 4. How many cubic feet in a stick of timber 15 ft. 3 in. long, 
 2 ft. 4 in. wide, and 1 ft. 8 in. thick? 
 
 6. How many cubic feet in a block of granite 18 ft. 5 in. long, 
 4 ft. 2. in. wide, and 3 ft. 6 in. thick ? 
 
 6. How many square feet in a stock of 10 boards, 15 ft. 8 in. 
 long, and 1 ft. 6 in. wide ? 
 
 7. How many square feet in a stock of 15 boards, 20 ft 3 in. 
 long, and 2 ft. 5 in. wide ? 
 
 8. Multiply 16 ft. 3' 4" by 6 ft. 5' 8" 10"'. 
 
 9. Multiply 20 ft. 4' 8" 5'" by 7 ft. 6' 9 ' 4" . 
 
 10. Multiply 18 ft. 0' 5'- 10'" by 4 ft. 8' 7" 9'". 
 
 11. Multiply 50 ft. 6' 0' 2'" 6"" by 3 ft. 10' 5". 
 
 12. How many cords in a pile of wood 50 ft. 6 m. long, 8 ft. 
 8 in. wide, and 7 ft. 4 in. high ? 
 
 13. If a cistern is 30 ft. 10 in. long, 12 ft. 3 in. wide, and 10 ft. 
 2 in, deep, how many cubic feet will it contain ? 
 
 14. What will it cost to plaster a room 20 ft. 6 in. long, 18 ft. 
 wide, and 10 ft. high, at 12 cts. per square yard ? 
 
 15. How many bricks 8 in. long, 4 in. wide, and 2 in. thick, 
 will make a wall 50 ft long, 10 ft. high, and 2 ft. 6 hi. thick 2 
 
338 EQUATION [SECT. XVI 
 
 SECTION XVI. 
 EQUATION OF PAYMENTS. 
 
 ART 517 EQUATION OF PAYMENTS is the process of finding 
 he equalized or average time when two or more payments due at 
 lifferent times, may be made at once, without loss to either party. 
 
 OBS. The equalized or average time for the payment of several debts, due at 
 different times, is often called the mean time. 
 
 518* From principles already explained, it is manifest, when 
 the rate is fixed, the interest depends both upon the principal and 
 the time. (Art. 404.) Thus, if a given principal produces a cer- 
 tain interest in a given time, 
 
 Double that principal will produce twice that interest ; 
 Half that principal will produce half that interest ; &c. 
 In double that time the same principal will produce twice that 
 interest ; 
 
 In lialf that time, half that interest ; &c. 
 
 519. Hence, it is evident that any given principal will pro- 
 duce the same interest in any given time, as 
 
 One half that principal will produce in double that time ; 
 One third that principal will " " thrice that time ; 
 Twice that principal will " " half that time ; 
 
 Thrice that principal will " " a third of that time ; &c. 
 
 For example, at any given per cent. 
 
 The int. of $2 for 1 year, is the same as the int of $1 for 2 yrs. ; 
 The int. of $3 for 1 year, " " $1 for 3 yrs. ; (fee, 
 
 The int. of $4 for 1 mo. " " " $1 for 4 mos. ; 
 
 The int. of $5 for 1 mo. " " " $1 for 5 mos. ; <fec 
 
 QUEST. 517. What is equation of payments ? Obs. What is the average time for tlw 
 payment of several debts sometimes called? 518. When the ra' is fixed, upon wha 
 does the interest depend t 
 
ARTS. 517-521.] OF PAYMENTS. 339 
 
 5 2O. The interest, therefore, of any given principal for 1 yeai, 
 or 1 month, d'C., is the same, as the interest of 1 dollar for as many 
 years, or months, as there are dollars in the given principal. 
 
 Ex. 1. Suppose you owe a man $15, and are to pay him $5 in 
 10 months, and $10 in 4 months, at what time may both pay- 
 ments be made without loss to either party ? 
 
 Analysis. Since the interest of $5 for 1 month is the same as 
 the interest of $1 for 5 months, (Art. 519,) the interest of 85 for 
 10 months must be equal to the interest of $1 for 10 times 5 
 months. And 5 mo. X 10 50 mo. In like manner the interest 
 of $10 for 4 months is equal to the interest of $1 for 4 times 10 
 months; and 10 mo. X 4 = 40 months. Now 50 months added to 
 40 months make 90 months ; that is, you are entitled to the use 
 of $1 for 90 months. But $1 is -fa of $15", consequently you aro 
 entitled to the use of $15, -^ of 90 months, and 90-f-15 = 6. 
 
 Ans. 6 months 
 Proof. 
 
 The interest of $5 at 6 per cent, for 10 mo. is $5 X-05 = $.25 
 The interest of $10 " " " 4 mo. is $10X.02= .20 
 
 Sum of both $^45~ 
 
 The interest of $15 at 6 per cent, for 6 mo. is 15X-03=$.45. 
 
 521. From these principles we derive the following general 
 
 RULE FOR EQUATION OF PAYMENTS. 
 
 First multiply each debt by the time before it becomes due y then 
 divide the sum of the products thus obtained by the, sum of the debts,, 
 and tJie quotient will be the average time required. 
 
 OBS. 1. If one of the debts is paid down, its product will be nothing ; biu 
 in finding the sum of the debts, this payment must be added with the others. 
 
 2. When there are months a^d days, the months must be reduced to days, 
 or the days to the fractional part of a month. 
 
 3. This rule is based upon the supposition that discount and interest paid in 
 Advance are equal. But this is not exactly true; consequently, the rule, 
 though in general use, is not strictly accurate. (Art. 432. Obs. 1.) 
 
 GUEST. 521. What is the rule for equation of payments ? 
 15* 
 
340 PARTNERSHIP. [SECT. XV 7 !. 
 
 2. If you owe a man $60, payable in 4 months, 8120 payable 
 in 6 months, and $180 payable in 3 months, at what time may 
 you justly pay the whole at once ? 
 
 Operation. 
 
 $ 60X4 $240, the same as $1 for 240 months. (Art. 520.) 
 $120X6 = $720, " " " $1 for 720 
 $180X3 = $540, " " " $1 for 540 
 $360 debts. $1500, sum of products. 
 Now 1 500-7- 360=4i months. Ans. 
 
 3. A merchant bought one lot of goods for $1000 on 5 months ; 
 another for $1000 on 4 months ; another for $1500 on 8 months; 
 what is the average time of all the payments ? 
 
 4. If a man has one debt of $150, due in 3 months ; anothei 
 of $200, due in 4 months ; another of $500, due in 7 months : 
 what is the average time of the whole ? 
 
 O 
 
 5. A man bought a house for $3500, and agreed to pay $500 
 down, and the balance in 6 equal annual instalments : at what 
 time may he pay the whole ? 
 
 6. If you owe one bill of $175, due in 30 days ; another of $81, 
 due in 60 days ; another of $120, due in 65 days, and another of 
 $200, due in 90 days : when may you pay the whole at once ? 
 
 PARTNERSHIP. 
 
 522. PARTNERSHIP is the associating of two or more individ- 
 uals together for the transaction of business. (Art. 464.) The per- 
 sons thus associated are called partners; and the association 
 itself, a company or firm. 
 
 The money employed is called the capital or stock ; and the 
 profit or loss to be shared among the partners, the dividend. 
 
 CASE I. When stock is employed an equal length of time. 
 
 Ex. 1. A and B formed a partnership ; A furnished $600 cap- 
 ital, and B $900 , they gained $300 : what was each partner's 
 share of the gain ? 
 
 QCEST. ">22. What Is partnership ? What are the persons thus associated called t 
 What is the association itself called 7 What is the money employed called ? What th 
 profit or loss ? 
 
ARTS. 522, 523.] PARTNERSHIP. 341 
 
 Analysis. Since the whole stock is $600-{-$900 $1500, A'a 
 part of it was -&uV > and B's part was iWo=f- Now since 
 A put in -f of the stock, he must have f of the gain ; and $300 
 Xf=$120. For the same reason B must have of the gain; 
 and $300Xf=$180. 
 
 Or, we may reason thus : As the whole stock is to the whole 
 gain or loss, so is each man's particular stock to his share of 
 the gain or loss. 
 
 That is, $1500 : $300 : : $600 : A's gain ; or $120, 
 And $1500 : $300 : : $900 : B's gain ; or $180. 
 
 PROOF. $120+ $180=$300, the whole gain. (Art. 21. Ax. 11.) 
 
 523. Hence, to find each partner's share of the gain or loss, 
 when the stock of each is employed for the same time. 
 
 Multiply each man's stock by the whole gain or loss ; divide the 
 product by the whole stock, and the quotient will be his sJiare of the 
 gain or loss. 
 
 Or, make each mans stock the numerator, and the whole stock 
 the denominator of a common fraction ; multiply the gain or loss 
 by the fraction which expresses each mans share of the stock, and 
 the product will be his share of the gain or loss. 
 
 PROOF. Add the several shares of the gain or loss together, and 
 if the sum is equal to the whole gain or loss, tJie work is right. 
 (Art. 21. Ax. 11.) 
 
 OBS. 1. The preceding case is often called Singh Fellowship. But since a 
 pnrf.nerskfp is necessarily composed of (wo or more individuals, it is somewhat 
 difficult to see the propriety of calling it single. 
 
 2. This rule is applicable to questions in Bankruptcy, and all other opera- 
 tions in which there is to be a division of property in specified proportions. 
 (Arts, 405, 466.) 
 
 2. A, B, and C formed a partnership ; A put in $1200 of the 
 capital, B $1600, and C $2000 ; they gained $960 : what was 
 each man's share of the gain ? 
 
 QrnsT. 523. How is each man's share of the gain or loss found, when the stock <rf 
 Bach is employed for the same time 1 How is the opera tio a proved? Obs. VVhstt is U 
 sometimes called ? To what is this rule applicable 1 
 
342 GENERAL [SECT XVI 
 
 3. A, B, and C entered into partnership ; A furnished $2350, 
 B $3200, and C $1820 ; they lost $860 : what ^ras eacl man's 
 share of the loss ? 
 
 4. A bankrupt owes A $2400, B $4600, C $6800, and D $9000 ; 
 his whole effects are worth $11200 : how much will each creditor 
 receive ? 
 
 5. A, B, C, and D, engaged in an adventure ; A put in $170, 
 B $160, $140, and D $130 ; they made $3000 : what was each 
 man's share ? 
 
 CASE II. When the stocks are employed unequal lengths of time. 
 
 6. A and B formed a partnership ; A put in $000 for 4 months, 
 and B put in $400 for 12 months; they gained $763 : what was 
 each man's share of the gain? 
 
 Note. It is obvious that the gain of each depends both upon the Capital he 
 furnished, and the lime it was employed. (Art. 518.) 
 
 Analysis. Since A's capital $900, was employed 4 months, 
 his share of the gain is the same as if he had put in $3600 for 1 
 month; (Art. 519;) for $900X4=$3600. Also B's capital 
 $400, being employed 12 months, his share of the gain is the 
 same as if he had put in $4800 for 1 month; for $400X12 = 
 64800. The sum of $3600 and $4800 is $8400. Therefore, 
 A's share of the gain must be f|^=f . 
 B's " " " " if=4-. 
 Now $763 Xf =$327, A's share. 
 And $7.63 xf=$436, B's share. Hence, 
 
 524* To find each partner's share of the gain or loss, when 
 vae stock of each is employed 'unequal lengths of time. 
 
 Multiply each partner's stock by the time it is employed ; make 
 each mans product the numerator, and the sum of the products the 
 denominator of a common fraction ; then multiply the whole gain 
 or loss by each man's fractional share of the stock, and the product 
 vill be his share of the gain or loss. 
 
 OBS. This case is often called Compound or Double Fellowship. 
 
 QUEST. 524. When the stock of each partner is employed uneqnr I lengths f time, 
 how is each man's share found 1 06*. What is this case sometimes ctl'.ed ? 
 
ARTS. 524-527.] AVERAGE. 343 
 
 7. The firm of X, Y, and Z lost $4500 ; X had 13200 em- 
 ployed for 6 months, Y -$2400 for 7 months, and Z $1800 for 9 
 months : what was each partner's loss ? 
 
 8. A, B, and C hired a pasture for $60; A put in 15 oxen 
 for 20 days, B 17 oxen for 16^ days, and C 22 oxen for 10 days : 
 what r?nt ought each man to pay ? 
 
 9. In a certain adventure A put in $12000 for 4 months, then 
 adding $8000 he continued the whole 2 months longer ; B put 
 
 n $25000, and after 3 months took out $10000, and continued the 
 rest for 3 months longer ; C put in $35000 for 2 months, then 
 withdrawing f of his stock, continued the remainder 4 months 
 longer ; they gained $15000 : what was the share of each ? 
 
 GENERAL AVERAGE. 
 
 525* The term General Average, in commerce, signifies the 
 apportionment of certain losses among the different interests con- 
 cerned, when a part of the cargo, furniture, &c., of a ship has 
 been voluntarily sacrificed to preserve the rest. (Art. 466.) 
 
 The property thus sacrificed is called the jettison. 
 
 520. Losses thus incurred are charged to the ship, the cajffo r 
 and the freir/ht, pro rata j or according to the value of each. 
 
 The contributory interests are to be freed from all charges upon 
 them before the average is made. 
 
 OBS. 1. In estimating the freight, in New York, one-half, but in most ports 
 one-third is deducted from the gross amount, for seamen' 's wages, pilotage, and 
 other small charges. 
 
 2. In the valuation of masts, spars, cables, rigging, &c., of the ship, it is cus- 
 tomary to deduct a third from the cost of replacing them; thus calling the 
 old, two-thirds the value of the new, in making the average. 
 
 3. The cargo is valued at the price it would bring at its destined port, after 
 the storage and other necessary charges are deducted. The property sacri- 
 ficed must be taken into the account as well as that which is saved. 
 
 527* General Average may be calculated both by Analysis 
 and Partnership. (Arts. 464, 522.) 
 
 OBS. 1. Losses arising from the ordinary wear and tear, or from a sacrifice 
 made for the safety of the ship only, or a particular part of the cargo, must 
 be borne by tfr 3 individuals who own the property lost, and hot by general 
 Average 
 
344 
 
 EXCHANGE OP 
 
 [SECT. XVI. 
 
 2. General average is not allowed, unless the peril was imminent $, and the 
 sacrifice indispensable for the safety of the ship and crew. 
 
 10. The ship Minerva from London to New York, had on board 
 a cargo valued at $75000, of which A owned $30000 ; B $27000 ; 
 and C "$18000; the gross amount of freight and passage money 
 was $11040. The ship was worth $40000, and the owner paid 
 $520 for insurance on her. Being overtaken by a severe tempest, 
 the master threw $18000 worth of A's goods overboard, and cut 
 away her mainmast and anchors ; finally, he brought her into 
 port, where it cost $2796.75 to repair the injury: what was the 
 loss of each owner of the ship and cargo ? 
 
 Operation. 
 Ship valued at ... $40000.00 
 Less premium for insurance . 520.00 
 
 $39480.00 
 
 Cargo worth .... 
 Freight and passage money . $11040.00 
 Less one-half for wages of crew 5520.00 
 
 75000.00 
 5520.00 
 
 Amount of contributory interests 
 
 Goods thrown overboard valued at 
 Cost new masts, spars, &c. . $2796.75 
 Less one third for wear of old . 932.25 
 Commissions on repairs 
 Port duties and incidentals 
 
 $120000.00 
 $18000.00 
 
 1864.50 
 15.13 
 120.37 
 
 Amount of loss 
 
 Now $20000X30000-$120000=$5000, loss 
 $20000X27000-$120000=$450 
 $20000X18000 $120000=$3000 
 $20000X39480 $120000=^6580 
 $20000X 5520 $120000 = 8 920 
 
 $20000.00 
 
 of A. 
 B. 
 C. 
 
 Ship. 
 Freight 
 
 PROOF. Whole loss (Ax. 11.) $20000, the 
 
 same as above. 
 
 Note. We may also find what percent, the loss is; then multiply each 
 contributory interest by the per cent. Thus, since $120000 lose &.0000, $1 will 
 | osc T ^_JL__ O f $20000; and 20000-?-$! 20000 =. 1 6f ; that is, the loss is Ibf 
 per cent Now $30000 X-16f =$5000, A's share of the loss. The loss of the 
 others may be found in a similar manner. 
 
ARTS. 528-532.] CURRENCIES. 345 
 
 EXCHANGE OP CURRENCIES. 
 
 528. The term currency, signifies money, or the dnmlating 
 medium of trade. 
 
 5 2O. The intrinsic value of the coins of different nations, de- 
 pends upon their weight and the purity of pie metal of which they 
 are made. (Art. 245. Obs.) 
 
 OBS. For the present standard weight and purity of the coins of the United 
 States, see Arts. 245, 246. For that of British coin, see Art. 248. Obs. 
 
 53O The relative value of foreign coins is determined by the 
 laws of the country and commercial usage. 
 
 OBS, The legal value of a pound sterling in this country has been different 
 at different times. By act of Congress, 1799. it was fixed at $4.44 f-. In 1832 
 its value was raised by the same authority to $4.80; and in 1842, to $4.84. 
 
 531* The process of changing money from the denominations 
 of one country to its equivalent value in the denominations of an- 
 other country, is called Exchange of Currencies. 
 
 CASE I. Reduction of Sterling to Federal Money. 
 Ex. 1. Change 60 sterling to Federal money. 
 Solution. Since l is worth $4.84, 60 are worth 60 times as 
 much, and $4.84X60=$290.40. Ans. 
 
 2 Change 8, 7s. 6d. to Federal money. 
 
 Operation. We first reduce the 7s. 6d. to the decimal 
 
 $4.84 of a pound ; (Art. 346 ;) then multiply $4.84, 
 
 8.375 and 8.375 together, and point off the prod- 
 
 $40.535 Ans. uct as in multiplication of decimals. Hence, 
 
 53.2* To reduce Sterling to Federal Money. 
 
 Multiply the legal value of one pound, $4.84, by the given num- 
 ber of pounds, point off the product as in multiplication of deci- 
 mals, and it will be the answer required. (Art. 324.) 
 
 If the example contains shillings, pence, and farthincs, they must 
 be reduced to the decimal of a pound. 
 
 QUEST. 528. What is meant by currency ? 529. On what does the ir trlnsic val:ie of 
 the coins of different countries depend ? 530. How is the relative value of foreign coiiu 
 determined? Obs. What is the value of a pound sterling? 531. What is me tnt by ei 
 change of currencies 1 532. How is Sterling money reduced to Federal ? 
 
346 EXCHANGE OP [SECT. XVI. 
 
 OBS. 1. The reason of this rule is obvious from the principle that 5 are 
 worth 5 times as much as 1, &c. 
 
 2, The rule usually given for reducing Sterling to Federal Money, is to re- 
 duce the shillings, pence, and farthings to the decimal of a pound, and placing 
 it on the right of the given pounds, divide the whole sn-m by -fa. This rule is 
 based on the law of 1799, which fixed the value of a pound at $4.44$-, and 
 that rf a dollar at 4s. 6d. * But $4.44^ is 9 per cent, of itself, or 40 cents less 
 than $4.84, which is the present legal value -of a pound ; consequently, the 
 result or answer obtained by it, must be 9 per cent, too small. A dollar is now 
 qual to 49.6d. very nearly, instead of 54d. as formerly. 
 
 533* From the preceding rule it is plain that Guineas, Francs 
 Doubloons, and sin foreign coins, may be reduced to Federal money 
 by multiplying the legal value of one by the given number. 
 
 Change the following sums of Sterling to Federal money: 
 
 3. 850, 10s. 8. 1000, 4s. 6d. 13. 50173, 12s. 6ld. 
 
 4. 175, 15s. 9. 1600, 8s. 7|d. 14. 53262, 13s. 8d. 
 6. 85, 13s. 6d. 10. 12531, 10s. 4d. 15. 76387, 15s. 7|d. 
 
 6. 200, 7s. 6d. 11. 43116, 9s. lOd. 16. 58762, 18s. 9|d. 
 
 7. 421, 16s. 4d. 12. 68318, 10s. 3|d. 17. 1000000. 
 
 CASE II. Reduction of Federal to Sterling Money. 
 
 18. Change $40.535 to sterling money. 
 
 Solution. Since $4.84 are worth l, $40.535 are worth as 
 many pounds as $4.84 are contained times in $40.535 ; and 
 $40.535-h$4. 84=8. 375 ; that is 8.375. Now reducing the de- 
 
 7 O 
 
 cimal .375 to shillings and pence, (Art. 348,) we have 8, 7s. 6d. 
 for the answer. Hence, 
 
 534* To reduce Federal to Sterling money. 
 
 Divide the given sum by $4.84, (the value of l,) and point off 
 t\e quotient as in division of decimals. Tlie figures on tlie left 
 Jiand of the decimal point will be pounds ; those on the right, deci- 
 mals of a pound, which must be reduced to shillings, pence, and 
 farthings. (Art. 348.) 
 
 OBS. Federal money may be reduced to Guineas, Francs, or any foreign 
 coin, by dividing the given sum by the value of one guinea, one franc, &c, 
 
 QUEST Obs. How may foreign coins be reduced lo Federal inunei S34. Uow it Fed 
 
 eral money reduced to Sterling? 
 
ARTS. 533-536.] CURRENCIES. 347 
 
 Change the following sums of Federal to Sterling money : 
 
 19. $'396.88. 23. $2160.50. 27. $25265, 
 
 20. 435.60. 24. 975.66. 28. 41470. 
 
 21. 876.25. 25. 4275.10. 29. 50263. 
 
 22. 1265.33, 26. 5300.75. 30. 100000. 
 
 535* Previous to the adoption of Federal money in 17 86, 
 accounts in the United States were kept in pounds, shillings, 
 K'lice, and farthings. 
 
 OBS. At the time Ft b:al money was adopted, the colonial currency or bills 
 of credit issued by the colonies, had more or less depreciated in value : that is, 
 a colonial pound was worth less than a pound Sterling ; a colonial shilling, 
 than a shilling Sterling, &c. This depreciation being greater in some col- 
 onies than in others, gave rise to the different values of the State currencies. 
 
 In N. K. cur., Va., Ky., Tenn., la., 111., Miss., Missou., 6s. or -^.=$1. 
 
 In N. Y. cur., N. C., Ohio, and Mich., - 8s. or -=$!. 
 
 In Penn. cur., New Jer., Del., and Md., 7s. 6d. (7s.) or f=$l. 
 
 In Georgia cur., and South Carolina, 4s. 8d. (4|s.) or -j^ =$1. 
 
 In Canada cur., and Nova Scotia, - - 5s. or ^-=$1. 
 
 Ala., La., Ark., and Florida use Federal Money exclusively. 
 
 CASE III. Reduction of federal Money to State currencies. 
 
 31. Reduce $63.25 to New England currency. 
 
 Solution. Since $1 contains 6s. N. E. cur., $63.25 contains 
 63.25 times as many; and 6s.X63.25 = 379.50s. Now 379-r20 
 =18, 19s., and .5s.Xl2 = 6d. (Art. 348.) Ans. 18. 19s. 6d. 
 
 536* Hence, to reduce Federal money to State currencies. 
 
 Multiply the given sum by the number of shillings which, in the 
 required currency, make $1, and the product will be the answer in 
 shillings, and decimals of a shilling. The shillings s/tould be re~ 
 duced to pounds, and the decimals to pence and farthings, (Art. 348.) 
 
 32. Reduce $450 to New England currency. 
 
 33. Reduce $567.5C to New York currency. 
 
 34. Reduce $840.10 to Pennsylvania currency. 
 
 35. Reduce $1500 to Canada currency. 
 
 QUEST 535. Previous to the adoption of Federal money, in what were accounts kept) 
 136. How is Fede/M monev reduced to the State currencies ? 
 
348 FOREIGN MONEYS [SECT. XVI 
 
 CASE IV. Reduction of State currencies to Federal Money. 
 
 36. Reduce 23, 12s. 6d. N. E. currency, to Federal money. 
 Solution. 23, 12s. 6d. = 472.5s. (Art. 348.) Now since 6s. 
 
 N. E. cur. make $1, 472.5s. will make as many dollars as 6s. is con- 
 tained times in 472.5s. ; and 472.5s.-H6s. = 78.75. Ans. $78.75. 
 
 537. Hence, to reduce State currencies to Federal money. 
 
 Reduce the pounds to shillings, and the given pence and farthing* 
 to the decimal of a shilling ; then divide this sum by the number of 
 shillings which, in the given currency, make $1, and tlie quotient 
 will be the answer in dollars and cents. 
 
 OBS. One state currency may be reduced to another by first reducing the 
 given currency to Federal money, then to the currency required. 
 
 37. Reduce 160, 5s. N. E. currency, to Federal money. 
 
 38. Reduce 245, 13s. 6d. N. Y. currency, to Federal money. 
 
 39. Reduce 369, 15s, 7d. Penn. currency, to Federal money. 
 
 40. Reduce 1800, Georgia currency, to Federal money. 
 
 41. Reduce 5000, Canada currency, to Federal money. 
 
 FOREIGN COINS AND MONEYS OF ACCOUNT. 
 
 538* The denominations of money, in which the laws of a 
 country require accounts to be kept, are called Moneys of account. 
 They are generally represented by a coin of the same name ; 
 sometimes, however, they are merely nominal, like mills in Fede- 
 ral money. (Art. 245.) 
 
 539 Foreign Moneys of Account, with the par value of the 
 unit established by commercial usage, expressed in Federal Money* 
 
 Austria. 60 kreutzers=:l florin ; 1 florin, (silver) is equal to $0.485 
 
 Belgium. 100 cents=l guilder or florin; I guilder, (silver) .40 
 
 The coinage of Belgium in 1832, was made similar to that of France, 
 
 Bencoolen. 8 satellers I soocoo; 4 soocoos=l dollar or rial, - 1.10 
 
 Brazil. 1000 rees=l milree=&828. The silver coin, 1200 rees 394 
 
 Bremen. 5 schwares=l grote; 72 grotes=l rix dollar, (silver) .787 
 
 British India. 12 pice 1 anna; 10 annasr=l Co. rupee, (silver) .445 
 
 The current (silver) rupee of Bengal, Bombay and Madras, is worth .444 
 
 QUEST. 537. How are the several State currencies reduced to Federal Money? 
 * M'CulIoch's Commercial Dictionary ; Kelly's Universal Cambist. 
 
ARTS. 537-539. J OF ACCOUNT. 349 
 
 Buenos Ayres. 8 rials 1 dollar currency, (fluctm ting) - - $0.93 
 
 Canton, 10 cash 1 candarine; 10 can. = 1 mace; 10 mace=l tael 1.48 
 The casn, which is made of copper and lead, is said to be the only 
 
 money coined in China. 
 
 Cape of Good Hope. G stivers 1 schilling; Sschillings^l rix dollar .313 
 Ceylon. 4 pice=l fanam; 12 fan ams=l rix dollar - - .40 
 
 Cuba. 8 rials plate=l dollar; 1 dollar - - 1.00 
 
 Colombia* 8 rials=l dollar; 1 dollar, (variable) mean value - 1.00 
 Chili.^ rials^l dollar; 1 dollar, (silver) - 1.00 
 
 Denmark. 12 pfenings 1 skilling, 16 skillings:=l marc; 6marcs= 
 
 1 r.gsbank or rix dollar, (silver) - - .52 
 
 Egypt. 3 aspers=l para ; 40 paras 1 piastre, (silver) - - .048 
 
 Fiance and Great Britain. See Tables. (Arts. -247, 272.) 
 Greece. 100 lepta=l drachme ; 1 drachma, (silver) - .166 
 
 Holland. 100 cents=l florin or guilder; 1 florin, (silver) .40 
 
 Hamburg. 12 pfenings^l schilling or sol; 16 schillings=l marc 
 Lubs; 3 rnarcs^l rix dollar. The current marc, (silver) $-28; 
 
 marc banco ------- .35 
 
 The term Lubs, signifies money of Lubec. The marc currency 
 
 is the common coin ; the marc banco is based upon certificates 
 
 of deposit of bullion and jewelry in the bank of Hamburg. 
 Invoices and accounts are sometimes made out in pounds, schillings, 
 
 and pence, Flemish, whose subdivisions are like sterling money ; 
 
 the pound Flemish=7 marcs banco. 
 
 Japan. 10 candarines=il mace; 10 mace = l tael - .75 
 
 Java. 100 cents 1 florin; 1 florin, as in Netherlands - - .40 
 
 Also 5 doits=l stiver; 2 stivers^ 1 dubbel; 3 dub. = l schilling; 
 
 4 schillings^ 1 florin - - - - .40 
 
 Malta. 20 granit=l taro; 12tari 1 scudo; 2j scudi^l pezza 1.00 
 
 Mauritius. In public accounts 100 cents^rl dollar - - .968 
 
 In mercantile accounts 20 sols=l livre; 10 livres=l dollar. 
 Manilla. 34 maravedis 1 rial; 8 rials =1 dollar, (Spanish) - 1.00 
 
 Milan. 12denari I soldo; 20 soldi=l lirat - - .20 
 
 Mexico. 8 rials 1 dollar; 1 dollar - 1.00 
 
 Monte Video. 100 centesimos=l rial; 8rials=l dollar - .833 
 
 Naples. 10 grani=:l carlino; 10 carlinir=l ducat, (silver) .80 
 
 Netherlands. Accounts are kept throughout the kingdorr in florins or 
 
 guilders, and cents, as adopted in 1815. See Holland. 
 New South, Wales. Accounts are kept in sterling money. 
 Norway. $ 120 skillings;=l rix dollar specie, (silver) - 1.06 
 
 Papal Slates. 10 bajocchi = l paolo; 10 paoli^l scudc or crown 1.00 
 
 Pern. 8 rialsr=l dollar, (silver) - - - - - 1.00 
 
 * Venezuela, New Grenada, and Ecuador. 
 
 t Grani is the plural of grano, tari of taro, scudi of scudr lire of lira, pezze of pezza. 
 
 ( Norway has no national gold coin 
 
850 
 
 FOREIGN COINS. 
 
 [SECT. XVI. 
 
 Portugal. 400 rees=l cruzado; 1000rees=l milree or crown - $1.12 
 Prussia. 12 pfenings=l grosch, (silver) 30 groschen^l thaler ordol. .69 
 Russia.* 100 copecksr=l rouble, (silver) - - .78 
 
 Sardinia, 100 centesimal lira; 1 lira=l franc, French .186 
 
 Sweden. 12 rundstycks = l skilling; 48 skillings:=l rix dol., specie 1.06 
 Sicit.y.SQ grani=l taro; 30 tari 1 oncia, (gold) - 2.40 
 
 Spain. 2 maravedis=l quinto; 16 quintos=rl rial of oW plate - .10 
 20 rials vellon=l Spanish dollar - 1.00 
 
 The rial of old plate is not a coin, but it is the denomination in 
 
 which invoices and exchanges are generally computed. 
 
 SI. Domingo. 100 centimes =1 dollar; I dollar - 33| 
 
 Tuscany. 12 denari di pezza^l soldo di pezza; 2 soldi di pezza=l 
 
 pezza of 8 rials ; 1 pezza, (silver) .90 
 
 Turkey. -3 aspers 1 para; 40 paras- 1 piastre, (fluctuating) - .05 
 
 Venice. 100 centesimal lira; 1 lira=l franc, French - .186 
 
 Formerly accounts were kept in ducats, lire, &c. 12 denari=l 
 soldo; 20 soldi=llira piccola; 6^ lire piccole=l ducat current; 
 8 lire pic.=l ducat effective. The value of the lira piccola is ,09f 
 
 West. Indies, British. Accounts are kept in pounds, shillings, pence 
 
 and farthings, of the same relative value as in England. The 
 
 value of the pound varies very much in the different islands, and 
 
 is in all cases less than the pound sterling. 
 
 5 4O. The following coins and moneys of account have been 
 made current in the United States, by act of Congress, at the rates 
 annexed. \ 
 
 Pound sterling of Gt. Britain, $4.84 
 Pound of Canada, Nova Scotia, 
 Do. New Brunswick and New- 
 foundland, . . . 4.00 
 Franc of France and Belgium, .186 
 Livre Tournois of France, . .185 
 Florin of Netherlands, . .40 
 Do. Southern States Germany, .40 
 Guilder of Netherlands, . 40 
 Real Vellon of Spain, . .05 
 Do. Plate of Spain, . . .10 
 Milree of Portugal, . . 1.12 
 Do. Azores, . . . .83^ 
 Marc Banco of Hamburg. . .35 
 Thaler or Rix Dollar, Prussia, 
 and North. States Germany, .69 
 
 Rix Dollar of Bremen, . $0.78f 
 
 Specie Dollar of Denmark, 1.05 
 
 Do. Sweden and Norway, . 1.06 
 
 Rouble, silver, of Russia, . .75 
 
 Florin of Austria, . . .485 
 Lira of Lombardo, Venetian 
 
 kingdom, . . . .16 
 
 Lira of Tuscany, . . .16 
 
 Do. of Sardinia, . , ,186 
 
 Ducat of Naples, . . ,80 
 
 Ounce of Sicily, . , 2.40 
 
 Leghorn Livres, . , .16 
 
 Tael of China, . . .1.48 
 
 Rupee, Company, . . .445 
 
 Do. of British India, . . .445 
 
 Pagoda of India, . . 1.84 
 
 * Previous to 1840, accounts were kept in paper roubles, 3 of which made a silver 
 rouble. t Laws of United States. 
 
Amv. 540-543.] EXCHANGE. 351 
 
 541. Foreign gold and silver coins, at the rates established by 
 the Custom Houses and commercial usage* 
 
 Guinea, English. (gold) $5.00 Leghorn Dollar, (s.) $0.90 
 
 Crown, " (silver) 1.12 j Scuda of Malta, (s.) ,40 
 
 S hilling piece, " (s.) .23 
 
 Bunk token, " (S.) .25 
 
 Florin of Basle, (s.) .41 
 
 Moiclore, Brazil, (.) 4.80 
 
 Doubloon, Mexico, (#.) 15.60 
 
 Livre of Neufchatel, (>.; .26* 
 
 Half Joe, Portugal, (g.) 8.53 
 
 Florin, Prussia, (s.) .22f 
 
 Livre of Catalonia. (s.) ,53j I Imperial, Russia, (.) 7.83 
 
 Florence Livre, (s.) .15 
 
 Louis d'or, French, (g.) 4.56 
 Crown, " (s.) 1.05 
 
 Rix Dollar, Rhenish, (.s.) .60$ 
 
 Rix Dollar of Saxony, (s.) .69 
 
 Pistole, Spanish, (ST.) 3.97 
 
 40 Francs, " (if.) 7.66 ! Rial " (s.) .12J 
 
 5 Francs, " (s.) 93 | Cross Pistareen, (s.) .16 
 
 Geneva Livre, (s.) .21 | Other Pistareens, (*.) .18 
 
 10 Thalers, German (g.) 7.80 ! Swiss Livre, (s.) .27 
 
 10 Pauls, Italy, (s.) .97 \ Crown of Tuscany, (*.) 1.05 
 
 Jamaica Pound, nominal, 3.00 | Turkish Piastre, (s.) .05 
 
 Note. The true method of estimating the value of foreign coins, is by their 
 
 weight and purity. 
 
 EXCHANGE. 
 
 542* EXCHANGE, in commerce, signifies the receiving or pay- 
 ing of money in one place, for an equal sum in another, by draft 
 or bill of Exchange. 
 
 OBS. 1. A Bill of Exchange is a written order, addressed to a person, di- 
 recting him to pay at a specified time, a certain sum of money to another per- 
 son, or to his order. 
 
 2. The person who signs the bill is called the drawer or maker ; the person 
 in whose favor it is drawn, the buyer or remitter ; the person on whom it is 
 drawn, the drawee, and after he has accepted it, the accepter ; the person to 
 whom the money is directed to be paid, the payee; and the person who has 
 legal possession of it, the holder. 
 
 3. On the reception of a bill of exchange, it should be immediately pre- 
 sented to the drawee ibr his acceptance. 
 
 543. The acceptance of a bill or draft is a promise to pay it 
 at maturity or the specified time. The common method of accept- 
 
 CirEST. 542. What is meant by exchange ? Obs. What is a hill of exchange! Wha 
 Is the drawer of a bill ? The drawee 1 The payee ? The holder 1 543. What is meant 
 by the acceptance of a bill ? What is the common method of accepting a bill ? 
 
 * See Manual of Gold and Silver Coins by Eckfeldt & Du Bois Ogden on the Tariff of 
 1846 ; Taylor's Gold and Silver Coin Examiner 
 
352 EXCHANGE. [SECT. XVI, 
 
 ing a bill, is for the drawee to write liis name under the word 
 accepted, across the bill, either on its face or back. The drawee 
 is not responsible for its payment, until he has accepted it. 
 
 OBS. L. If the payee wishes to sell or transfer a bill of exchange, it 'is neces- 
 sary for him to endorse it, or write his name on the back of it. 
 
 2. If the endorser directs the bill to be paid to a particular person, it is 
 called a special endorsement, and the person named, is called the endorsee. 
 I' the endorser simply writes his name upon the back of the bill, the endorse- 
 ment is said to be blank. When the endorsement is blank, or when a bill is 
 drawn payable to the bearer, it may be transferred from one to another at 
 pleasure, and the drawee is bound to pay it to the holder at maturity. If the 
 drawee or accepter of a bill fail to pay it. the endorsers are responsible for it. 
 
 544. When acceptance or payment of a bill is refused, the 
 holder should duly notify the endorsers and drawer of the fact 
 by a legal protest, otherwise they will not be responsible for its 
 payment. 
 
 OBS. 1. A protest, is a formal declaration in writing, made by a civil officer 
 termed a notary public, at the request of the holder of a bill, for its non-accept- 
 ance, or non-payment. 
 
 2. When a bill is returned protested for non-acceptance, the drawer must 
 pay it immediately, though the specified time has not arrived, otherwise he is 
 liable to prosecution. 
 
 3. The time specified for the payment of a bill is a matter of agreement be- 
 tween the parties at the time it is negotiated. Some are payable at sight , 
 others in a certain number of days or months after sight, or after date. When 
 payable after sight or date, the day on which they are presented is not reck- 
 oned. W T hen the time is expressed in months, they are always understood to 
 mean calendar months. Hence, if a bill payable in one month is dated the 
 25th of January, it will be due on the 25th of February. And if it is dated 
 
 he 28th, 29th, 30th, or 3lst of January, it will be due on the last day of Feb- 
 ruary. It is customary to allow three days grace on bills ef exchange. 
 
 545. Bills of exchange are usually divided into inland and 
 foreign bills. When the drawer and drawee both reside in the 
 same country, they are termed inland bills or drafts ; when they 
 reside in different countries, foreign bills. 
 
 OBS. In negotiating foreign bills, it is customary to draw fh/ -ee of the sa-mt 
 date and amount, which are called the First, Second and Third of Exchange} 
 and collectively, a Set of Exchange. These are sent by different ship* 01 
 
 Q.UKST. 544.- When the acceptance or payment of a bill is refused, what should be 
 d we 1 Obs. What is a protest? 545. How are bills of exchange divided 1 Obs Whatia 
 meant by a set of exchange ? 
 
ARTS. 544-547.] EXCHANGE. 353 
 
 conveyances, and when the first that arrives, is accepted or paid, the olaer$ 
 become void. The object of this arrangement is to avoid delays, which might 
 arise from accidents, miscarriage, &c. 
 
 FORM OF A FOREIGN BILL OF EXCHANGE. 
 
 Exchange 1000. BOSTON, Oct. 3d, 18-17. 
 
 At ninety days sight of this first of Exchange? (the second and third of 
 the same date and tenor unpaid,) pay George Lewis, Esq., or order, One Thou- 
 sand Pounds sterling, with or without farther advice. 
 
 JOHN W. ADAMB. 
 To Messrs. ROTHSCHILD & Co. 
 Hrokers, London. 
 
 FORM OF AN INLAND BILL OR DRAFT. 
 
 $2500. NEW YORK, Sept. 27th, 1847. 
 
 Thirty days after sight, pay to the order of Messrs. Newman & Co., 
 Twenty-five Hundred Dollars, value received, and charge the same to 
 
 MACY & WOODBURY. 
 To Messrs. D. BAKER & Co. 
 
 Merchants, New Orleans. 
 
 546. The term par of exchange, denotes the standard by 
 which the comparative worth of the money of different countries 
 is estimated. It is either intrinsic or commercial. 
 
 The intrinsic par is the real value of the money of different 
 countries, determined by the weight and purity of their coin. 
 
 The commercial par is a nominal value, fixed by law or commer- 
 cial usage, by which the worth of the money of different countries 
 is estimated. 
 
 OBS. 1. The intrinsic par remains the same, so long as the standard coins 
 of each country are of the same metals, and of the same weight and purity; 
 but, in case the standard coins are of different metals, the intrinsic par must 
 vary, as the comparative values of the metals vary. 
 
 2. The commercial par is conventional, and may at any time be changed 
 by law or custom. 
 
 547 By the term course of exchange is meant the current 
 price which is paid in one place for bills of a given amount drawn 
 on another place. 
 
 OBS. 1. The course of exchange is seldom statimary or at par. It varies 
 
 QCKST. 546. What is meant by par of exchange ? Inlr.nsic par 7 Cuipmerc lal par 1 
 
354 
 
 EXCHANGE. 
 
 . XVI 
 
 according to the circumstances of trade. When the balance of trade is against 
 a country, that is when the exports are less than the imports, bills on the 
 foreign country will be above par, for the reason that there will be a greater 
 demand tor them to pay the balance due abroad. On the other hand, when 
 the balance of trade is in favor of a country, foreign bills will be below pa?', 
 for the reason that fewer will be required. 
 
 2. It should be remarked that the course of exchange can never exceed 
 very much the intrinsic pafvalue; for it is plain that coin or bullion instead of 
 bills will be remitted, whenever the course of exchange is such that the ex 
 ftnse of insuring and transporting it from the debtor to the creditor country 
 & less than the premium for bills, and the exchange will soon sink to par. 
 
 548. Rates of exchange on Great Britain are commonly reck- 
 oned at a certain per cent., on the old commercial par, instead of 
 the new par, 
 
 OBS. 1. According to the old par, the value of a pound sterling is $4.44-|-f 
 as fixed by act of Congress, in 1799. According to the -new par it is $4.84. 
 The intrinsic value of a ster.,or sovereign, according to assays at the U. S. 
 nint, is $4.81)1. The new par is the value fixed by the government in 1842. 
 ind is used in calculating duties, when the invoice is in sterling money. 
 
 2. The old par is nine per cent, less than the new par or legal value ; conse- 
 quently the rate of exchange must reach the nominal premium of 9 per cent 
 before it is at par according to the new standard. 
 
 Table of Exchange showing the. value of l Sterling from 1 to 
 to Yl\ per cent, premium on the old par of $4.44-$-. 
 
 Old Par 
 
 |>4.444 ! 5i per ct. 
 
 $4.689:8 per ct. $4.800 
 
 9i- per ct. $4.878 
 
 1 per ct. 
 
 4.489J6 
 
 4.7llJ8i " 4.811 
 
 10 " 4.889 
 
 2 " 
 
 4.5336i " 
 
 4.733:8i " 4.822 
 
 10i " 4.911 
 
 3 " 
 
 4.578,7 
 
 4.756 8$ " 4.833 
 
 11 " 4.933 
 
 4 " 
 
 4.622 7^ " 
 
 4.767(9 New Par 4.844 
 
 Hi " 4956 
 
 4i " 
 
 4.644 7i " 
 
 4.77819iperct. 4.856 
 
 12 " 4.978 
 
 5 " 
 
 4.667 7f " 
 
 4.789J9i " 4.807 
 
 12i " 5.000 
 
 Note. 1. When exchange is 10 per cent, advance or over, on the old par, 
 it will cause a shipment of specie to England ; for the freight, interest and in- 
 surance will not amount to so much as the premium. When the premium is 
 less than 9 per cent. English funds are, in reality, below their intrinsic par. 
 
 2. The practice of quoting rates of exchange at the old par, is calculated to 
 lead, persons unacquainted with the subject into serious mercantile mistakes, 
 and to degrade our national currency by making it appear to foreign nations 
 to be so much below par. 
 
A.RTS. 548, 549.] EXCHANGE. 355 
 
 * 
 Ex. 1. A merchant negotiated a bill of exchange on London for 
 
 500, 10s., at 8 per cent, premium on the old par : how much 
 did he pay for the bill ? 
 
 Solution. 500, 10s. = 500. 5. (Art. 346.) 
 
 Now $4. 44^X500. 5 = $2224. 44^ at the old par value. 
 Then $2224.44jX .08= 177.95:}- the premium. 
 
 The sum paid $2402.40 -4w$. 
 
 Or, the val. of l by table, $4.80X500.5 =$2402.40. Ans. 
 
 2. A merchant negotiated a bill on Liverpool for 1000, at 
 
 1 per cent, discount from the new par : what did he pay for it ? 
 
 3. What will a bill cost, on England, for 5265, 13s. 6d., at 
 8-J- per cent, advance on the old par ? 
 
 4. How much is a bill worth on France for 1500 francs, at 
 
 2 per cent, above par, which is $.186 per franc? 
 
 5. What will a bill cost on Paris for 56245 francs, exchange 
 being 5 francs and 54 centimes to the dollar ? 
 
 6. What cost a bill of exchange on Hamburg for 2000 marcs 
 banco, at 1 per cent, above par, which is 35 cts. per marc ? 
 
 7. What cost a bill of exchange on St. Petersburg for 2660 
 roubles, at 2 per cent, discount, the par being 75 cts. per rouble ? 
 
 8. What cost an inland bill of exchange at Boston, on New 
 Orleans, for $15265.85, at 1 per cent, advance? 
 
 9. What cost a draft at Albany, on Mobile, for $20260, at 
 2 per cent, premium ? 
 
 10. What cost a draft at St. Louis, on New York, for $35678, 
 at 2-^ per cent, premium ? 
 
 ARBITRATION .OF EXCHANGE. 
 
 549* Arbitration of Exchange is the method of finding the 
 exchange between two countries through the medium of that of 
 other countries. 
 
 OBS, 1. When there is but one intervening country, the operation is terme 
 simple arbitration, when more than one, it is termed compound arbitration. 
 
 2. Problems in Arbitration of exchange are usually solved by conjoined pro- 
 portion. (Art. 511.) Care must be taken to reduce all the quantities which 
 are of the same kind, to the same denomination. 
 T.H. 
 
356 ALLIGATION. [SECT. XVI. 
 
 Ex. 1 . If the exchange of New York on London is 8 per cent, 
 advance on old par, or $4.80 for 1 sterling, and that of Amster- 
 dam on London is 12 florins for l, what is the arbitrated ex- 
 Change of New York on Amsterdam ; that is, how many florins 
 are equal to $1 U. S. ? Ans. $1 = 2 florins. 
 
 2. A merchant in Baltimore wishes to remit 1200 marcs banco 
 to Hamburg, and the exchange of Baltimore on Hamburg is 35 
 cents for 1 marc. He finds the exchange of Baltimore on Paris 
 is 18 cents for 1 franc; that of Paris on London, is 25 francs for 
 l sterling; that of London on Lisbon, is 180 pence for 3 mil- 
 rees ; that of Lisbon on Hamburg, is 5 milrees for 18 marcs banco. 
 How much will he gain by the circuitous exchange ? 
 
 Ans. Direct Ex. $420 ; circuitous Ex. $375 : Gain $45. 
 
 3. A man in England owes a man in Portugal 420 ; the di- 
 rect exchange from London to Lisbon is*70d. for 1 milree ; but 
 the exchange between London and Amsterdam is 48 florins for 
 l sterling; between Amsterdam and Paris it is 16 florins for 
 3 francs ; and between Paris and Lisbon it is 6 francs for 2 mil 
 rees. Is it better for the man in Portugal to have a direct remit- 
 tance from London to Lisbon, or a circuitous one through Amster- 
 dam and Paris ? 
 
 ALLIGATION. 
 
 55O. Alligation is the method of finding the value of a com- 
 pound or mixture of articles of different values, or of forming a 
 compound which shall have a given value. (Art. 467.) 
 
 OBS. 1. The term alligation is derived from the Latin alligo, which signifies 
 to bind or tie together. It had its origin in the manner of connecting the nura- 
 oers together by a curve line in the solution of a certain class of examples. 
 2. Alligation is usually divided into Medial and Alternate. (Art. 467. Ob&) 
 Note. For a new method of Alligation Alternate, see Key, p. 72. 
 
 MEDIAL ALLIGATION. 
 
 551* Medial Alligation is the process of finding the mean 
 price of a mixture of two or more ingredients or articles of dif- 
 ferent values. 
 
 Note. The term medial is derived from the Latin met iius, signifying a mean 
 or average. 
 
ARTS. 550-554] ALLIGATION. 357 
 
 552* To find the mean value of a mixture, when the quantity 
 and the price of each of the ingredients are given. 
 
 Divide the whole cost of the ingredients by the whole quantity 
 mixed, and the quotient will be the mean price of the mixture. 
 
 PROOF. Multiply the whole mixture by the mean price, and if 
 the product is equal to the whole cost, the work is right. 
 
 Ex. 1. A grocer mixed 10 Ibs. of tea worth 5s. a pound, with 
 18 Ibs. worth 3s. a pound, and 20 Ibs. worth 2s. a pound: what 
 is the mixture worth per pound ? 
 
 Solution. 10 Ibs. at 5s. = 50s. 
 
 18 Ibs. at 3s. = 54s. 
 
 20 Ibs. at 2s.=:40s. 
 
 Whole quantity 48 Ibs. and 144s. whole cost of mixture. 
 Now 144s.-i-48 = 3. Am. 3s. a pound. 
 
 2. A drover bought 870 lambs at 75 cts. apiece, and 290 sheep 
 at $1.25 apiece : what is the mean price of the lot per head? 
 
 3. A grocer mixed 12 gals, of wine at 4s. lOd. per gal., with 
 21 gals, at 5s. 3d., and 29-J- gals, at 5s. 8d. : what is a gallon of 
 the mixture worth ? 
 
 ALTERNATE \LLIGATION. 
 
 553* Alternate Alligation is the process of finding what quan- 
 tity of any number of ingredients, whose prices are given, will 
 form a mixture of a given mean price. 
 
 Note The term alternate is derived from the Latin aUernatus, signifying 
 by turns, and in its present application, refers to the connection of the prices 
 which are less than the mean price, with those which are greater. Alternate 
 alligation embraces three varieties of examples 
 
 CASE I. 
 
 554. To find the quantity of each ingredient, when its price 
 and that of the required mixture are given. 
 
 I, Write the prices of tJie ingredients under each other, beginning 
 with tlie least ; tl>n connect, with a curve line, each price which is less 
 than that of the mixture with one or more of those that are greater , 
 and each greater price with one or more af those that are less. 
 
358 ALLIGATION. [SbCT. XVI, 
 
 II. Write the difference between tlie price of the mixture and tliat 
 of each of the ingredients opposite tlie price with which they are 
 connected. If only one difference stands against any price, it will 
 denote the quantity to be taken of that price j but if^ there are more 
 tlutn one, their sum will be the quantity. 
 
 OBS. It is immaterial in what manner we select the pairs of ingredients, 
 provided the price of one of the ingredients is less and the other greater than 
 fchs mean price of the mixture required. 
 
 PROOF. Find tlie value of all the ingredients at their given 
 prices ; if this is equal to the value of the whole mixture at tlie 
 given price, the work is right. 
 
 4. A man mixed four kinds of oil, worth 8s., 9s., 11s., and 12s. 
 per gal. ; the mixture was worth 10s. per gal. : required, the 
 quantity of each. 
 
 10 
 
 
 
 OBS. 1. It is manifest that other answers maybe obtained by connecting 
 the prices in a different manner. 
 
 2. It is also manifest, if we multiply or divide the answers already obtained 
 oy any number, the results will fulfil the conditions of the question ; conse- 
 quently the number of answers is unlimited. 
 
 5. A goldsmith has gold of 18, 20, 22, and 24 carats fine : how 
 much may be taken of each to form a mixture 21 carats fine? 
 
 CASE Tl. 
 
 555* When the quantity of one of the ingredients and the 
 mean price of the mixture are given. 
 
 Fir,d the difference between the price of each ingredient and the 
 mean price of tlie required mixture, as before ; then by proportion, 
 
 As the difference of that ingredient whose quantity is given, is to 
 each particular difference, so is the quantity given to tlie quantity 
 required of each ingredient. 
 

 ARTS. 555, 556.J ALLIGATION. 359 
 
 6. How many pounds of sugar at 10, and 15 cents a pound, 
 must be mixed with 20 Ibs. at 9 cents, so that the mixture may 
 be worth 1 2 cents a pound ? 
 
 Solution. Connecting the prices as directed, the differences 
 between them and the mean, are 3 cts., 3 cts. and 5 cts. 
 Then 3 cts. : 3 cts. : : 20 Ibs. : to the Ibs. at 9 cts. 
 Also 3 cts. : 5 cts. : : 20 Ibs. : " " 10 cts. 
 
 Ans. 20 Ibs. at 9 cts., and 33i Ibs. at 10 cts. 
 
 7. How much gold of 16, 18, and 22 carats fine must be mixed 
 with 10 oz. 24 carats fine, that the mixture maybe 20 carats fine? 
 
 8. How much wool at 20, 30, and 24 cts. a pound must be mixed 
 with 95 Ibs. at 50 cts. to form a mixture worth 40 cts. a pound ? 
 
 CASE III. 
 
 556. When the quantity to be mixed and the mean price of 
 the required mixture are given. 
 
 Find the difference between the price of each ingredient and the 
 mean price of the required mixture, as before ; then by proportion, 
 
 As the sum of tJie differences is to each particular difference, so 
 is the whole quantity to be mixed, to the quantity required of each 
 ingredient. 
 
 9. A grocer has raisins worth 8, 10, and 16 cents a pound : 
 how many of each kind may be taken to form a mixture of 112 
 Ibs. worth 12 cents a pound? 
 
 Solution. The sum of the differences between the prices of 
 
 the ingredients, and the mean price, 6 cts. +4 cts. +4 cts.=14 cts. 
 
 Then 14 cts. : 6 cts. : : 112 Ibs. : to the Ibs. at 16 cts. 
 
 And 14 cts. : 4 cts. : : 112 Ibs. : to the Ibs. at 8 and 10 cts. 
 
 Ans. 48 Ibs. at 16 cts., 32 Ibs. at 10 cts., and 32 Ibs. at 8 cts. 
 
 10. How much wine at 15, 17, 18, and 22 shillings per gallon, 
 may be mixed to form a mixture of 320 gals, worth 20 shillings 
 per gallon ? 
 
 11. How much water must be mixed with wine worth 9s. per 
 gal. to fill a pipe, so that the mixture may be worth Vs. per gal. ? 
 
360 
 
 INVOLUTION. 
 
 [SECT. XVII, 
 
 SECTION XVII. 
 
 INVOLUTION. 
 
 Airr 557* When any number or quantity is multiplied into 
 itself, the product is called a power. Thus, 5X5 = 25; 3X3X3=- 
 27; 2X2X 2X2 = 16; the products 25, 27, and 16 are powers. 
 
 The original number, that is, the number which being multi- 
 plied into itself, produces a power, is called the root of all the 
 powers of that number, because they are derived from it. 
 
 558. Powers are divided into different orders ; as the first, 
 second, third, fourth, fifth power, &c. They take their name from 
 the number of times the given number is used as a factor, in pro- 
 ducing the given power. 
 
 OBS. 1. The first, power of a number is said to be the number itself. Strictly 
 speaking, it is not a power, but a root. (Art. 557.) 
 
 3 yards. 
 
 2. The second power of a number is also called 
 the square; (Art. "257. Obs. 1 ;) for, if the side of a 
 square is 3 yards, then the product of 3x3=9 yards, . 
 will be the area of the given square. (Art. 285.) -^ 
 But 3X3=9 is also the second power of 3 ; hence, it ^ 
 b called the square. M 
 
 3. The diagonal of a square is a line connecting 
 two of its opposite corners. 
 
 3. The third power of a number is also called 
 the cube; (Art. 258. Obs. 1 ;) for, if the side of a 
 cube <s 3 feet, then the product of 3x3x3=27 
 feex, will be the solidity of the given cube. (Art. 
 28G.) But 3X3X3=27 is also the 'third power 
 of 3; hence it is called the cube. Leg. VII. 11. Sen. 
 
 4. The fourth power of a number is called the 
 
 3X3=9 yards. 
 3 feet. 
 
 I 
 
 / 
 
 3X3X3=27 feet. 
 
 QUEST. 557. What is a power ? 558. How are powers divided ? From what do they 
 take their name 7 Obs. What is said to be the first power ? What is the second powtl 
 called 1 The third? The fourth 1 
 
ARTS. 557-56 l.j INVOLUTION. 3C1 
 
 559. Powers are denoted by a small figure placed above the 
 given number at the right hand. 
 
 This figure is called the index or exponent. It shows how many 
 times the given number is employed as a factor to produce the 
 required power. Thus, 
 
 The index of the first power is 1 ; but this is commonly omitted ; 
 for, (2)'==2. 
 
 The index of the second power is 2 ; 
 
 The index of the third power is 3 ; 
 
 The index of the fifth power is 5 ; &c. That is, 
 2 1 2, the first power of 2 ; 
 
 2 2 = 2X2, the square, or second power of 2 ; 
 
 2 3 = 2 X 2 X 2, the cube, or third power of 2 ; 
 
 2< 2X2X2X2, the biquadrate, or fourth power of 2; 
 
 2 5 =2 X 2X2X2X2, the fifth power of 2 ; 
 
 2 1 =2X2X2X2X2X2, the sixth power of 2 ; &c. 
 
 Ex. 1. Express the square of 17, and the cube of 19. 
 
 Ans. 17', 19. 
 
 Express the given powers of the following numbers : 
 
 2. The square of 54. 7. The 2d power of 299. 
 
 3. The cube of 43. 8. The 4th power of 785. 
 
 4. The square of 87. 9. The 5th power of 228. 
 K. The biquadrate of 91. 10. The 8th power of 693. 
 6. The 3d power of 41G. 11. The 32d power of 999. 
 
 5 GO. The process of finding a power of a given number by 
 multiplying it into itself, is called INVOLUTION. 
 
 5G1. Hence, to involve a number to any required power. 
 
 Multiply the given number into itself, till it is taken as a factor, 
 as many times as there are units in the index of the power to which 
 the number is to be raised. (Art. 558.) 
 
 OBS. 1. The number of multiplications in raising a number to any rriveri 
 power, is one less than the imlex of the required power. Thus, 323x3; th 
 3 is taken twice as a factor, but there is but one multiplication. 
 
 UCKST. .V>". How are powers denoted ? What is this figure called 1 What does it 
 show ? What is the index of the first power! Of the second? The third? Fifth 1 
 SoO. What is involution 7 561. How is a number involved to any required power 1 
 
362 INVOLUTION. [SECT. XVD 
 
 2. A Fraction is raised to a power by multiplying it into itself. Thus, th 
 square of | is -f xl=f 
 
 Mixed numbers should be reduced to improper fractions, or the common 
 fraction to a decimal. They may however be involved without reducing them. 
 (Art. 2-20. Obs.) 
 
 3. The process of raising a number to a high power, may often be contracted 
 by multiplying together powers already found. The index of the power thus 
 found, is equal to the sum, of the indices of the powers multiplied together. 
 Thus, 2X2=4; and 4x4=2x2X2x2, or 2*. 
 
 i 35; and 
 
 12. What is the square of 23 ? 
 
 Common Operation. Analytic Operation. 
 
 23 23 = 2 tens or 20+3 units. 
 
 23 23 = 2 tens or 20 + 3 units. 
 69 60+9 
 
 46^ 400+ 60 __ 
 
 "629 A *s. And 400 + 120+9 = 529. Ans. 
 
 It will be seen from this operation that the square of 20 + 3 
 contains the square of the first part, viz : 20X20 = 400, added to 
 twice the product of the two parts, viz: 20X3 + 20X3 = 120, 
 added to the square of the last part, viz : 3X3 = 9. Hence, 
 
 562. The square of the sum of two numbers is equal to the 
 square of the first part, added to twice the product of the two 
 parts, and the square of the last part. 
 
 OBS. 1. The product of any two factors cannot have more figures than both 
 factors, nor but one less than both. For exampfe, take 9, the greatest num- 
 ber which can be expressed by one figure. (Art. 34.) And (9)2, or 9X9=81, 
 has two figures, the same number which both factors have. 99 is the greatest 
 number which can be expressed by two figures ; (Art, 34 ;) and (99) 2 , or 99X 
 99=9eOl, has four figures, the same as both factors have. 
 
 Again, 1 is the smallest number expressed by one figure, and (J) 2 , or 1X1 
 = 1, has but one figure less than both factors. 10 is the smallest number 
 which can be expressed by two figures; and (10) 2 , or 10X10=100, has one 
 figure less than both factors. Hence, 
 
 QUEST. Obs. How many multiplications are there in raising a number to a given 
 power 1 How is a fraction involved ? A mixed number 7 562. What is the square o/ 
 the sum of two numbers equal to? Obs. How many figures are there in the product of 
 any two factors 1 How many figures will the square of a number contain ? The cube ? 
 
ARTS. 562-564'.] EVOLUTION. 363 
 
 2. A sqv^are cannot have more figures than double the number of the root of 
 first power, nor bid one less. 
 
 3. A cube cannot have more figures than triple the number of the root or Ji v st 
 pnver, nor but two less. 
 
 4. All powers of 1 are the same, viz: 1 ; for, 1X1X1X1, &c.=l. 
 
 13. What is the square or second power of 123 ? 
 
 14 The cube of 135 ? 23. The cube of .012 ? 
 
 15. The square of 2880 ? 24. The square of .00125 ? 
 
 16. The 4th power of 10 ? 25. The square of -f ? 
 
 17. The 5th power of 5 ? 26. The cube of i? 
 
 18. The 7th power of 6 ? 27. The square of ? 
 
 19. The 6th power of 7 ? 28. The cube of fW? 
 
 20. The 8th power of 4 ? 29. The square of 4 ? 
 
 21. The 9th power of 9? 30. The square of 7|? 
 
 22. The souare of 2.5 ? 31. The square of 38-JI? 
 
 EVOLUTION. 
 
 503. If we resolve 25 into two equal factors, viz: 5 and 5, 
 each of these equal factors is called a root of 25. So if we re- 
 solve 27 into three equal factors, viz : 3, 3, and 3, each factor is 
 called a root of 27 ; if we resolve 16 into four equal factors, viz : 
 2, 2, 2, and 2, each factor is called a root of 16. And, universally, 
 when a number is resolved into any number of equal factors, each 
 of those factors is said to be a root of that number. Hence, 
 
 564. A root of a number is a factor, which, being multiplied 
 into itself a certain number of times, will produce that number. 
 
 OBS. Roots, as well as powers, are divided into different orders. Thus, 
 when a number is resolved >nto two equal factors, each of these factors is 
 called the second or square root ; when resolved into three equal factors, each 
 of these factors is called the third or cube root, &c. Hence, 
 
 The name of the root expresses the number of ejiual factors into which the given 
 number is to be resolved. 
 
 Roots. | 
 
 1| 2 
 
 1 3| 4| 
 
 5 ' 6 
 
 25] 36 
 
 1 7 
 
 8 | 9 
 
 1 10 
 
 1 11 
 
 12 
 
 Squares. ' 
 
 1 1 4 
 
 1 9 1 16 | 
 
 | 49 
 
 64 | 81 
 
 | 100 
 
 | 121 
 
 144 
 ~i 1728 
 
 Cubes. | 
 
 1| 8 
 
 | 27 | 64 ' 
 
 125 
 
 | 216 
 
 | 343 
 
 512 | 729 
 
 | 1000 
 
 | 1331 
 
 QUEST. Ofo. What are all powers of J 1 564. What is a root if a number 1 Obj. Wha/ 
 does the name of the root express ? 
 
 16* 
 
364 EVOLUTION. [SECT. XVII 
 
 565* The process of resolving numbers into equal factors is 
 called EVOLUTION, or tlie Extraction of Roots. 
 
 OBS. 1. Evolution is the opposite of involution. (Art. 500.) One is finding 
 a power of a number by multiplying it into itself; the other is finding a root by 
 resolving a number into equal factors. Powers and roots are therefore correla- 
 tive terms. If one number is a power of another, the latter is a root of the for- 
 mer. Thus, 27 is the cube of 3 ; and 3 is the cube root of 27. 
 
 2. The learner will be careful to observe, that 
 
 In subtraction, a. number is resolved into two parts; 
 
 In division, a number is resolved into two factors; 
 
 In evolution, a number is resolved into equal factors. 
 
 566* Roots are expressed in two ways ; one by the radical 
 sign (y) placed before a number ; the other by & fractional index 
 
 placed above the number on the right hand. Thus, ^/4, or 4 2 
 denotes the square or 2d root of 4 ; V27, or 27 3 denotes the cube 
 .or 3d root of 27 ; VI 6, or 16 4 denotes the 4th root of 16. 
 
 OBS. 1. The figure placed over the radical sign, denotes the root, or the num- 
 ber of equal factors into which the given number is to be resolved. The figure 
 for the square root is usually omitted, and simply the radical sign -^ is placed 
 before the given number. Thus the square root of 25 is written ^/~25. 
 
 2. When a root is expressed by a fractional index, the denominator, like the 
 figure over the radical sign, denotes the root of the given number. Thus, 
 ;25)* denotes the square root of 25 ; (27)* denotes the cube root of 27. 
 
 3. A fractional index whose numerator is greater than 1, is sometimes used. 
 In such cases the denominator denotes the root, and the numerator the power 
 
 of the given number. Thus, 8' denotes the square of the cube root of 8, or 
 the cube root of the square of 8, each of which is 4. 
 
 4. The radical sign ^ , is derived from the letter r, the initial of the Latin 
 radix, a root. 
 
 1. Express the cube root of 74. Ans. \/74, or 74 3 . 
 
 2. The square root of 119. 5. The square root of . 
 
 3. The 4th root of 231* 6. The cube root of f. 
 
 4. The 9th root of 685. 7. The 4th root of \\. 
 
 6, Express the 3d power of th* 4th root of 6. Ans. 6*. 
 9. Express the 2d power of the 3d root of 81. 
 
 QUEST. 565. What is evolution? Obs. Of what is it the opposite? Into what are 
 nambers resolved in subtraction? In division? In evolution? 566. How many ways 
 are roots expressed ? What are they ? Obs. What does the figure over the radical sign 
 denote ? What the denominator of the fractional index ? 
 
ARTS. 565-570.] SQUARES AND CUBES. 365 
 
 5G7. A number which can be resolved into equal factors, or 
 whose root can be exactly extracted, is called a perfect power, and 
 its root is called a rational number. Thus, 1C, 25, 27, &c., are 
 perfect powers, and their roots 4, 5, 3, are rational numbers. 
 
 568. A number, which cannot be resolved into equal factors, 
 or whose root cannot be exactly extracted, is called an imperfect 
 power ; and its root is called a Surd, or irrational number. Thus, 
 15, 17, 45, <fec., are imperfect powers, and their roots 3.8 + ', 4.1 +; 
 6.7+, &c., are surds, for their roots cannot be exactly extracted. 
 
 OBS. A number may be a perfect power of one degree and an imperfect 
 power of another degree. Thus, 10 is a perfect power of the second degree, 
 but an imperfect power of the third degree ; that is, it is a perfect square but 
 not a perfect cube. Indeed numbers are seldom perfect powers of more than 
 one degree. 1G is a perfect power of the 2d and 4th degrees; 64 is a perfect 
 power of the 2d, 3d and Gth degrees. 
 
 569. Every rooi, as well as power of l,is 1. (Art. 5G2. Obs. 4.) 
 Thus, (I) 2 , (I) 3 , (I) 6 , and -/I, ^1, V% &c., are all equal. 
 
 PROPERTIES OF SQUARES AND CUBES. 
 
 5 TO. The properties of numbers in general, have already been 
 given. The following pertain to square and cubic numbers. 
 
 1. The product of any two or more square numbers, is a square; and the 
 
 2 -2 
 
 product of any two or more cubic numbers, is a cube. Thus 2 X3 =3G, the 
 square of G; and 2 X3 =21G, is the cube of 6. 
 
 2. If a square number is divided by a square, the quotient will be a square. 
 Thus, 144-5-9 = Hi, which is the square of 4. 
 
 3. If a square number is either multiplied or divided by a number that is not 
 a square, neither the product nor quotient will be a square. 
 
 4. If you double the number of times a number is taken as a factor, it will 
 not produce double the product, but the square of it. Thus, 3x3^9, and 3x3 
 X3,<3=81, and not 18. 
 
 5. The product of two different prime numbers cannot be a square. 
 
 G. The product of no two different numbers, which are prime to each otler, 
 will make a square, unless each of those numbers is a square. 
 
 7. The square and cube of an even number are eren. ; and the square am. 
 cube of an odd number are odd. (Art. IGl. Prop. G, 10.) Hence, 
 
 QI-KST. 507. What is a perfect power? What is a rational number 1 508. An imper- 
 . feet jM>wer 7 A surd 1 Obs. Are numbers ever perfect powers of one degree und imperfee 
 powers of another degree 7 5(59. What are all roots and powers of 1 1 
 
366 SQUARES AND CUBES. [SECT. XV11, 
 
 8. The square or cube root of an even number, is even ; and the square or 
 ^ubc root of an odd number, is odd. 
 
 9. Every square number necessarily ends with one of these figures, 1, 4, 5, 
 69; or with an croi number of ciphers preceded by one of these figures. 
 
 10. No number is a square that ends in 2, 3, 7, or 8. 
 
 11. A cubic number may end in any of the natural numbers, 1, 2, 3, 4, 5, 6 
 7, 8, 9, or 0. 
 
 12. All the powers of any number, ending in 5. will also end in 5; and if 
 a number ends in 6, all its powers will end in 6. 
 
 13. E/;ry square number is divisible by 3, and also by 4, or becomes so 
 when diminished by unity. Thus, 4, 9, ib', 25, &c., are all divisible by 3, and 
 by 4, or become so when diminished by 1. 
 
 14. Every square number is divisible also by 5, or becomes so when increased 
 o diminished by unity. Thus, 3b' 1, and 49-f I, are divisible by 5. 
 
 15. Any even square number is divisible by 4. 
 
 16. An odd square number, divided by 4, leaves a remainder of 1. 
 
 17. Every odd square number, decreased by unity, is divisible by 8. 
 
 18. Evciy number is either a square, or is divisible into two, or three, or Jon* 
 squares. Thus 30 is equal to 25-f-4+l ; 33=16-j-lO-|-l ; 63=49-f-9-j-4-j-l. 
 
 19. The product of the sum and difference of two numbers, is equal to the 
 difference of their squares. Thus, (5+3) X (5 3) = 16 ; also 5 3*= 10. 
 
 20. If two numbers are such, that their squares, when added together, form 
 a square, the product of these two numbers is divisible by 6. Thus, 3 and 4, 
 the sum of whose squares, 9-j-16=25, is a square number, and their product 
 12, is divisible by 6. Hence, 
 
 21. To find two numbers, the sum of whose squares shall be a square number. 
 
 Take any two numbers and multiply Uu:m*togclhcr ; Hue double of their prod- 
 uct will be one of the numbers sought, and the difference of their squares will be 
 the other. Thus, take any two numbers, as 2 and 3 ; the double of their prod- 
 uct is 12, and he difference of their squares is 5; now 12 2 -J-5 2 =169, the 
 square of 13. 
 
 22. When two numbers are such, that the difference of their squares is a 
 square number ; the sum and difference of these numbers are themselves square 
 numbers, or the double of square numbers. Thus, 8 and 10 give for the dif- 
 ference of their squares 30 ; and 18, the sum of these numbers, is the double 
 of 9, which is a square number, and 2, their difference, is the double of 1, which 
 is also a square number. 
 
 23. If two numbers, the difference of which is 2, be multiplied together, their 
 product increased by unity, will be the square of the intermediate number. 
 
 21. The sum 01 difference of two numbers, will measure the difference of 
 heir squares. 
 
 25. The sum of two numbers, differing by unity, is equal to the difference 
 of their squares. 
 
 26. The sum of two numbers will measure the sum of their cubes; and th 
 difference of two numbers will measure the difference of their cubes. 
 
ART. 571.] SUUARE ROOT. 37 
 
 27. If a square measures a square, or a cube a cube, the root will al"o -.neas* 
 urc the root. 
 
 28. If one number is prime to another, its square, cube, &c., will also be 
 prime to it. 
 
 29. The difference between an integral cube and its root, is always divisi 
 ble by 6. 
 
 30. If any series of numbers beginning from I, be in continued geometrical 
 proportion, the 3d, 5th, 7th, &c., will be squares; the 4th, 7th, 10th, &c., 
 cubes ; and the 7th will be both a square and a cube. Thus, in the series, 
 1, 2, 4, 8, 16, 32, G4, &c., the 3d, 5th, and 7th terms are squares; the 4th am/ 
 7th are cubes ; and the 7th is both a square and a cube. 
 
 EXTRACTION OF THE SQUARE ROOT. 
 
 571. To extract tJie SQUARE ROOT, is to resolve a given number 
 into two equal factors ; or, to find a number which being multiplied 
 into itself, will produce the given number. (Art. 564. Obs.) 
 
 Ex. 1. What is the square root of 36 ? 
 
 Solution. Resolving the given number into two equal factors, 
 we have 36 = 6X6. Ans. The square root of 36 is 6. 
 
 2. What is the length of one side of a square field whi^h con- 
 tains 529 square rods? 
 
 Operation. Since we may not see what the root of 529 
 
 529(23 is at once, we will separate it into two periods 
 4 by placing a point over the 9 and another over 
 
 43)129 the 5. Now the greatest square of 5, the left 
 
 129 hand period, is 4, the root of which is 2. Plac- 
 
 ing the 2 on the right of the number, we sub- 
 tract its square from the period 5, and to the right of the re- 
 mainder bring down the next period. We then double the 2, the 
 part of the root already found, and, placing it on the left of the 
 dividend for a partial divisor, we perceive it is contained in the 
 dividend, omitting its right hand figure, 3 times. Placing the 3 
 on the right of the root, also on the right of the partial divisor, 
 we multiply the divisor thus completed by 3, and subtract the 
 product from the dividend. The answer is 23 rods/ 
 
 QUEST. 571. What is it to extract the square root of a number 7 
 
368 saUARE ROOT. [SECT. XVII 
 
 bate. Since the root is to contain 2 figures, the 2 stands in tens place, 
 hence the first part of the root found is properly 20 ; which being doubled, 
 gives 40 for the divisor. For convenience we omit the cipher on the right ; 
 and to compensate for this, we omit the right hand figure of the dividend. 
 This is the same as dividing both the divisor and dividend by 10, and therefore 
 does not alter the quotient. (Art. 146.) 
 
 572. Hence, we derive the following general 
 
 RULE FOR EXTRACTING THE SQUARE ROOT. 
 
 I. Separate the given number into periods of two figures each, by 
 placing a point over the units figures, then over every second fig- 
 ure towards the left in whole numbers, and over every second figure 
 towards the right in decimals. 
 
 II. Find the greatest square number in the first or left hand 
 period, and place its root on the right of the number for the first 
 figure in the root. Subtract the square of this figure of the root 
 from the period under consideration ; and to the right of the re- 
 mainder bring down the next period for a dividend. 
 
 III. Double the root just found and place it on the left of the 
 dividend for a partial divisor ; find how many times it is contained 
 in. the dividend, omitting its right hand figure ; place the quotient 
 on the right of the root, also on the right of the partial divisor ; 
 multiply the divisor thus completed by the figure last placed in the 
 root ; subtract the product from the dividend, and to the remainder 
 bring down the next period for a new dividend. 
 
 IV. Double the root already found for a new partial divisor, di- 
 vide, &c., as before, and thus continue tlie operation till the root of 
 all the periods is extracted. 
 
 If there is a remainder after all the periods are brought down, 
 the operation may be continued by annexing periods of ciphers. 
 
 PROOF. Multiply the root into itself ; and if the product it 
 equal to the given number, the work is right. (Art. 5G4.) 
 
 573 Demonstration. Take any number as that in the last example ; thf n 
 separating it into parts, 529=500-|-29. Now the greatest square in 500 is -100 
 the root of which is 20, with a remainder of 100 ; consequently, the first part ol 
 
 QI-KST. 572. What is the first step in extr.-ic )g the sqtnre root ? The second ? Thinlt 
 Fourth 3 When there is a remainder, how proceed 1 How is the square root proved 1 
 
ARTS. 572-574. [ SUI/ARE ROOT. 309 
 
 the root must be 20, and the true remainder is lOO-f-29, or 129. And since 
 there are three figures in the given number, there must be two figures in the 
 root; (.Art. 562. Obs. 2;) but the square of the sum of two numbers, is equal 
 to the square of the first part ad 'ed to twice the product of the two parts and 
 the square of the last part; it follows therefore that the remainder 129, must 
 be t vice the product of 20 into the part of the root still to be found, together 
 with the square of that part. (Art. 562.) Now dividing 129 by 40 the double 
 of 20, the quotient is 3, which being added to 40 makes 43 ; finally, multiply- 
 ing 43 by 3, the product is 129, which is manifestly twice the product of 20 
 into 3, together with the square of 3. In the same manner the operation may 
 be proved in every case. (For illustration of this rule by geometrical figures, 
 *ce Practical Arithmetic, . 318.) 
 
 1. The reason for separating the given numbers into periods of two figures 
 each, is that a square number can not have more figures than double the num- 
 ber oi figures in the root, nor but one less. It also shows how many figures the 
 root will contain, and thus enables us to find part of it at a time. (Art. 562. 
 Obs. 2.) 
 
 2. The reason for doubling that part of the root already found for a divisor, 
 is because the remainder is double the product of the first part of the root into 
 the second part, together with the square of the second part. 
 
 3. In dividing, the right hand figure of the dividend is omitted, because the 
 cipher on the right of the divisor being omitted, the quotient would be 10 
 times too large for the next figure in the root. (Arts. 130, 146.) 
 
 4. The last figure of the root is placed on the right of the divisor simply fot 
 convenience in multiplying it into itself. 
 
 OBS. 1. The product of the divisor completed into the figure last placed in 
 the root, cannot exceed the dividend. Hence, in finding the figure to be placed 
 in the root, some allowance must be made for carrying, when the product of 
 this figure into itself exceeds 9. 
 
 2. If the right hand period of decimals is deficient, it must be completed by 
 annexing a cipher to it. 
 
 3. There will always be as many decimal figures in the root, as there are 
 periods of decimals in the given number. 
 
 574* The square root of a common fraction is found by ex- 
 tracting the root of the numerator and denominator. 
 
 A mixed number should be reduced to an improper fraction. 
 When either the numerator or denominator of a common fraction 
 is not a perfect square, the fraction may be reduced to a decimal, 
 and the approximate root be found as above. 
 
 QUEST. 573. Dem. Why do we separate the given number into periods of two figures 
 each 1 Why double the root thus found for a divtaor ? Why omit the right hand figure 
 of the dividend ? Why place the last figure of the root on the right of the divisor? Obs 
 How many decimal figures will there be in the root? 574. How is the square roc* of a 
 common fraction found ? Of a mixed number ? 
 
370 
 
 SQUARE ROOT. 
 
 XVII. 
 
 "Required the square root of the following numbers : 
 
 3. 
 
 2601. 
 
 10. 
 
 27889 
 
 17. 
 
 566.44. 
 
 24. 
 
 If. 
 
 4. 
 
 5329. 
 
 11. 
 
 961. 
 
 18. 
 
 7.3441. 
 
 25. 
 
 iff 
 
 5. 
 
 784. 
 
 12. 
 
 97. 
 
 19. 
 
 .81796. 
 
 26. 
 
 f. 
 
 6, 
 
 87. 
 
 13. 
 
 7. 
 
 20. 
 
 1169.64. 
 
 27. 
 
 m. 
 
 7 
 
 4761. 
 
 14. 
 
 190. 
 
 21. 
 
 627264. 
 
 28. 
 
 794. 
 
 8. 
 
 7056. 
 
 15. 
 
 43681. 
 
 22. 
 
 3.172181. 
 
 29. 
 
 2071* 
 
 9. 
 
 .9801. 
 
 16. 
 
 47089. 
 
 23. 
 
 10342656. 
 
 30. 
 
 34967 jV. 
 
 31. What is the square root of 152399025 ? 
 
 32. What is the square root of 119550669121 ? 
 
 33. What is the square root of 964.5192360241 ? 
 
 575. When the root is to be extracted to many figures, tha 
 operation may be contracted in the following manner. 1 
 
 First find half, or one more than half the number of figures re' 
 quired in the root ; then having found tlie next true divisor, cut off 
 its right hand figure, and divide the remainder by it ; place Hi* 
 quotient in the root, and continue the operation as in contraction of 
 division of decimals. (Art. 333.) 
 
 34. Required the square root of 365 to eleven figures in the 
 root. Ans. 19.104973174. 
 
 35. Required the square root of 2 to twelve figures. 
 
 36. Required the square root of 3 to seventeen figures. 
 
 APPLICATIONS OF THE SQUARE ROOT. 
 
 576. A triangle is a figure which has three sides and three 
 angles. When one of the sides of a triangle is perpendicular to 
 another side, the angle between them is called a right-angle. 
 
 C 
 
 577. A right-angled triangle is a 
 triangle which has a right-angle. 
 
 The side opposite the right-angle is 
 called the hypothenuse, and the other 
 two sides, the base and perpendicular. 
 The triangle ABC is right-angled at B, 
 and the side AC is the hypothenuse. 
 
ART?. 575-580.J SQUARE ROOT 
 
 371 
 
 578. The square described on the hypothenuse of a right- 
 angled triangle, is equal to the sum of the squares described on 
 the other two sides. (Thomsons Legendre, B. IV. 11, Euc. I. 47.) 
 
 The truth of this principle may be seen from the following geometrical illus- 
 tration. Thus, 
 
 Let the base AB of the right- 
 angled triangle ABC be 4 feet, 
 he perpendicular AC, be 3 feet ; 
 
 hen will the squares described 
 
 n the base AB, and the per- 
 pendicular AC, contain as many 
 square feet as the square de- 
 scribed on the hypothenuse BC. 
 Now (4)-|-(3)a=25 sq. ft. ; and 
 the square described on BC also 
 contains 25 sq. ft. Hence, the 
 square described on the hypothe- 
 nuse of any right-angled trian- 
 gle, is equal to the sum of the 
 squares described on the other two 
 sides. 
 
 OBS. Since the square of the hypothenuse BC, is 25, it follows that the 
 or 5, must be the hypothenuse itself. Hence, 
 
 579, When the base and perpendicular are given, to find the 
 hypothenuse. 
 
 Add the square of the base to the square of ike perpendicular ', 
 the square root of the sum will be the hypothenuse. 
 
 Thus, in the Yight-angled triangle ABC, if the base is 4 and 
 the perpendicular 3, then (4) 2 +(3) 2 =25, and \/25 = 5, the hypo- 
 thenuse. 
 
 580. When the hypothenuse and base are given, to find the 
 perpendicular. 
 
 From the square of the hypothenuse subtract the square of the 
 base, and the square root of the remainder will be the perpendicular. 
 
 QUKST. 576. What is a triangle ? What is a right-angle ? 577. What is a right 
 angled triangle ? What is the side opposite the right-angle called 7 What are the other 
 two sides called ? 578. What is the square described on the hypothenusf -ual tc1 
 
 579. When the base and perpendicular are given, how is the hypothenuse found \ 
 
 580. Whei the hypothenuse and base are given, how is the perpendicular found 7 
 
372 sauARE ROOT. [SECT. XV11. 
 
 Thus, if the Iwpothenuse is 5 and the base 4, then (o) 2 (4)* 
 = 9, and y/Q = S, the perpendicular. 
 
 581. When the hypothenuse and the perpendicular are given, 
 to find the base. 
 
 from the square of the hypothenuse subtract the square of the per- 
 pendicular, and the square root of the remainder will be the base. 
 
 Tl us, if the hypothenuse is 5 and the perpendicular 3, then 
 (o) 2 (3) 2 = 16, and Vl6=4, the base. 
 
 OBS. 1. From the preceding principles it is manifest that the area >f a square 
 may be found by dividing the square of its hypothenuse by 2. (Arts. '285, 578.) 
 
 2. The areas of all similar figures are to each other as tHe squares of their 
 homologous sides, or their like dimensions. (Leg. IV. 25, 27. V. 10.) Hence, 
 
 The s: .n of the areas of equilateral or other similar triangles, also of similar 
 polygons, circles and semicircles described on the base and perpendicular of a 
 right-angled triangle, is equal to the area of a similar figure described on the 
 hypothenuse. 
 
 3. The square of a simple ratio is called a duplicate ratio ; the cube of a sim- 
 ple ratio, a triplicate ratio. * 
 
 The ratio of the square roots of two numbers is called a sub-duplicate ratio; 
 that of the cube roots, a sub-triplicate ratio. 
 
 Ex. 1. If a street is 28 feet wide, and the height of a tower is 
 96 feet, how long must a rope be to reach from the top of the 
 to wer to the opposite side of the street ? 
 
 Solution. (96) 2 + (28) 2 =:10000, and VlOOOO=100 ft. 
 
 2. A ladder 40 feet long being placed at the opposite side of a 
 street 24 feet wide, just reached the top of a house : how high 
 was the house ? 
 
 3. Two ships, one sailing 7 miles, the other 12 miles an hour, 
 spoke each other at sea ; one was going due east, the other due 
 south : how far apart were they at the expiration of 12 hours ? 
 
 4. What is the length of the side of a square farm which con- 
 tains 300 acres; and how far apart are its opposite corners? 
 
 582. A mean proportional between two numbers is equal io 
 the square root of their product. (Arts. 494, 498. Obs. 2.) 
 
 QrtsT. .W When the hypothenuse and perpendicular are given, how is the base 
 found? 
 
ARTS. 581-584.] SCIUARE ROOT 373 
 
 5. Find a mean proportional between 2 ind 8. 
 Solution. 8X2 = 16; and >/16=4. Ans. 
 
 Find a mean proportional between the following numbers: 
 
 6. 4 and 25. 10. 121 and 36. 14. \ and if. 
 
 7. 9 and 36. 11. 196 and 144. 15. |f and -fa. 
 
 8. 16 and 81. 12. 2.56 and 49. 16. H and iff. 
 
 9. 64 and 25. 13. 6.25 and 729. 17, -ff and iff-. 
 
 583* To find the side of a square equal in area to any given 
 surface. 
 
 Extract the square root of the given area, and it will be the side 
 of the square sought. 
 
 OBS. When it is required to find the dimensions of a rectangular field, equal 
 in area to a given surface, and whose length is double, triple, or quadruple, 
 &c., of its breadth, the square root of , , {, of the given surface, will be the 
 width ; and this being doubled, tripled, or quadrupled, as the case may be, 
 will be the length. 
 
 18. What is the side of a square equal in area to a rectangular 
 field 81 rods long, and 49 rods wide? 
 
 19. What is the^side of a square equal in area to a triangular 
 field which contains 160 acres? 
 
 20. What is the side of a square equal in area to a circular 
 eld which contains 640 acres ? 
 
 21. What are the length and 'breadth of a rectangular field 
 which contains 480 acres, and whose length is triple its breadth ? 
 
 22. A general arranged 10952 soldiers,'so that the number in 
 rank was double the file : how many were there in each ? 
 
 584. When the sum of two numbers and the difference of 
 their squares are given, to find the numbers. 
 
 Divide the difference of their squares by the sum of the numbers^ 
 and the quotient will be their difference; then proceed as in 
 Art. 155.. 
 
 23. The sum of two numbers is 42, and the difference of their 
 squares is 756 : what are the numbers? Ans 12 and 30. 
 
 24. The -sum of two numbers is 65, and the diffeierce of their 
 squares is 975 : what are the numbers ? 
 
374 CUBE ROOT. [SECT. XVII 
 
 585. When the difference of two numbers and the difference 
 of their squares are given, to find the numbers. 
 
 .Divide the difference of the squares by the difference of the 
 numbers, and the quotient will be their sum; then proceed as 
 in Art. 155. 
 
 24. The difference of two numbers is 29, and the difference rf 
 their squares is 1885 : what are the numbers ? 
 
 EXTRACTION OP THE CUBE ROOT. 
 
 580. To extract the cube root, is to resolve a given number into 
 three equal factors ; or, to find a number which being multiplied 
 into itself twice, will produce, the given number. (Art. 564.) 
 
 Ex. 1. What is the cube root of 64 ? 
 
 Solution. Resolving the given numbers into three equal fac- 
 tors, we have 64 = 4X4X4. Ans. 4. 
 
 2. What is the cube root of 12167 ? 
 
 Operation. We first separate the 
 
 Col. i. Col. ii. 12167(23 given number into two pe- 
 
 1st term 2_ 4X2= 8 riods, by placing a point 
 
 2d " 4~~ 1200 divisor,) 4 167 over the units' figure, then 
 
 3d " 63 1389X3= 4167 over thousands. This shows 
 
 us that the root must have 
 
 two figures, (Art. 562. Obs. 3,) and thus enables us to find part 
 of it at a time. 
 
 Beginning with the left hand period, we find the greatest cube 
 of 12 is 8, the root of which is 2. Place the 2 on the right of 
 the given number for the first part of the root, and also in Col. I. 
 on the left of the number. Multiplying the 2 into itself, write the 
 product 4 in Col. II. ; and multiplying 4 by 2, subtract its product 
 from the period, and to the right of the remainder bring down the 
 next period for a dividend. Then adding 2, the first figure of the 
 root, to the first term of Col. I., and multiplying the sum by 2, 
 we add the product 8 to the 1st term of Col. II., and to this sum 
 
 Q JEST. 586. Whit is it to extract the cube root 1 
 
ARTS. 585-587.] CUBE ROOT. 375 
 
 annex two ciphers, for a divisor ; also add 2, the first figure of 
 the root, to the 2d term of Col. I. Finding the divisor is con- 
 tained in the dividend 3 times, we place the 3 in the root, also on 
 the right of the 3d term of Col. I. Then multiply the 3d terra 
 thus increased, by 3, the figure last placed in the root, and add 
 the . product to the divisor. Finally, we multiply this sum by 3, 
 ant subtiact the product from the dividend. Ans. 23. 
 
 587, Hence, we derive the following general 
 
 RULE FOR EXTRACTING THE CUBE ROOT. 
 
 I. Separate the given number into periods of three figures each, 
 placing a point over units, then over every third figure towards the 
 left in whole numbers, and over every third figure towards the 
 right in decimals. 
 
 II. Find the greatest cube in the first period on the left hand ; 
 place its root on the right of the number for the first figure of the 
 root, and also in Col. I. on the left of the number. Then multi- 
 plying this figure into itself, set the product for the first term in 
 Col. II. ; and multiplying this term by the same figure again, sub- 
 tract this product from the period, and to the remainder bringdown 
 the next period for a dividend. 
 
 III. Adding the figure placed in the root to the first term in 
 Col. I., multiply the sum by the same figure, add the product to the 
 first term, in Col. II., and to this sum annex two ciphers, for a di- 
 visor ; also add the figure of the root to the second term of Col. I. 
 
 IV. Find how many times the divisor is contained in the divi- 
 dend, and place the result in the root, and also on the right of the 
 third t</ m of Col. I. Next multiply the third term thus increased by 
 the figure last placed in the root, and add the product to the divisor ; 
 then multiply this sum by the same figure, and subtract the product 
 from the dividend. To the remainder bring down the next period 
 for a new dividend. 
 
 V. Find a new divisor in the same manner that the last divisor 
 was found, then divide, &c., as before ; thus continue the operation 
 till the root of all the periods is found. 
 
 QUEST. 587. What is the first step in extracting the cube root? The second' Third t 
 Fourth 1 Fifth 1 How is the cube root proved 1 
 
376 CUBE ROOT. [SECT. XVII 
 
 PROOF Multiply the root into itself twice, and if the last prod- 
 uct is equal to the given number, the work is right. 
 
 OBS. 1. When there is a remainder, periods of ciphers may be added, as in 
 square root. 
 
 2. If the right hand period of decimals is deficient, this defi/jency must be 
 supplied by ciphers. The root must contain as many decimals as there are 
 periods of decimals in the given number. 
 
 & Demonstraiion. r H\m rule depends upon the principle that the cube 
 
 f the sum of two numbers is equal to the cube of the first part, added to * 
 
 times the square of the first part into the last part, added to 3 times the first 
 
 part into the square of the last, added to the cube of the last part. Take any 
 
 number, as 23 ; we have 23=20-j-3. 
 
 Or, (23)3=8000+3b-004-540-}-27=12l67. 
 
 .After subtracting the greatest cube from the left hand period, it is plain the 
 remainder must contain 3 times the square of the first part of the root into the 
 last part, &c. Hence, if we divide the remainder by 3 times the square of the 
 first part of the root, the quotient will be the last part. But it will be seen 
 that the divisor is 3 times the square of the first part of the root, consequently 
 the quotient must be the last part of the root required. 
 
 1. The reason for separating the given number into periods of three figures, 
 is that the cube of a number can not have more figures than triple the number 
 of figures in the root, nor but two less. It also shows how many figures the root 
 will contain, and thus enables us to find part of it at a time. (Art. 562. Obs. 3.) 
 
 2. The reason for annexing 2 ciphers to the divisor is (manifestly) because 
 the first figure of the root, of which the divisor is times the squape, stands in 
 tens' place 
 
 3. Placing the last figure of the root on the right of the 3d term in Col. I., 
 then multiplying it by this figure, and adding the product to the divisor, and 
 this sum being multiplied by the figure last placed in the root, the product will 
 evidently be 3 times the square of the first part of the root into the last part, 
 together with 3 times the first into the square of the last part, and the cube 
 of the last part. In a similar manner the operation may be illustrated in all 
 other cases. 
 
 Note. The preceding method of extracting the cube root was discovered by 
 the late Mr. Homer of Bath, England, and is often called Hornet's Method, 
 (For the common method, and its demonstration by cubical blocks see Piicli- 
 cal Arithmetic, p. 325). 
 
 QUEST. Obs. When there is a remainder, how proceed ? When the right han.l period 
 of decimals is deficient, what must be done 1 How many decimals must the root contain ? 
 588. Why separate the given number into periods of three figures ? Why annex two 
 ciphers to the right of the divisor ? 
 
ARTS. 588-590.] CUBE ROOT. 377 
 
 589. The cube root of a common fraction is found by ex- 
 tracting the root of its numerator and denominator, or by first 
 reducing it to a decimal. 
 
 A mixed number should be reduced to an improper fraction, or 
 the fractional part to a decimal. 
 
 3, Required the cube root of 78314.6. 
 Operation. 
 
 78314.600(42.784-. 
 
 Col. I. 
 
 1st term 4 
 
 Col. II. 78314.1 
 
 16X4 = 64 
 
 2d " 8 
 
 4800, 1st divisor ) 14314 
 
 3d " 122 
 
 5044X2 = 10088 
 
 4th " 124 529200, 2d divisor) 4226600 
 
 5th " 1267 538069X7 = 3766483 
 
 6th " 1274 54698700, 3d divisor) 460117000 
 7th " 12818 54801244X8 = 438409952 
 
 59O* When the root is required to many places of decimals, 
 the operation may be contracted in the following manner. 
 
 First find one more than half the number of decimal figures re- 
 quired. For a new divisor, take as many figures plus one on the 
 left of the last term in Col. II. as remain to be found in the root ; 
 and for a dividend retain one more figure on tJie left of the re- 
 mainder than the divisor has ; then proceed as in the contraction 
 of division of decimals. (Art. 333.) 
 
 Required the cube root of the following numbers : 
 
 4. 91125. 8. 10218313. 12. 37. 16. iff. 
 
 5. 140608. 9. 11543.176. 13. 6. 17. f^-fr. 
 
 6. 571787. 10. 20.570824. 14. 376.- 18. 44f. 
 
 7. 2515456. 11. .241804367. 15. 575. 19. 49-^. 
 
 20. What is the cube root of 2 to eight decimals ? 
 
 21. What is the cube root of -^g- to eleven decimals ? 
 
 22. What is the side of a cubical mound which contains 314432 
 solid feet ? 
 
 Note. -Similar solids are to each other as the cubes of their homologous 
 ides, or like dimensions. (Leg. VII. 20, VIII. 11. Cor.) Hence, 
 
 QUEST. 589. How find the cube root of a common fraction ? Of a mixed number ? 
 
378 HIGHER ROOTS. [SECT. XVIL 
 
 591* To find the side of a cube whose solidity shall be dou- 
 ble, triple, &c., that of a cube whose side is given. 
 
 Cube tlw given side, multiply it by the given proportion, ant* the 
 cube root of the product will be the side of the cube required. 
 
 23. Wk-;t is the side of a cubical bin which contains 8 times as 
 many solid feet as one whose side is 4 feet ? Ans. 8 ft. 
 
 24. What is the side of a cubical block which contains 4 times 
 &s many solid yards as one whose side is 6 feet ? 
 
 25. If a ball 6 inches in diameter weighs 32 Ibs., what is the 
 weight of a ball whose diameter is 3 inches ? 
 
 26. If a globe 4 ft. in diameter weighs 900 Ibs., what is the 
 weight of a globe 3 ft. in diameter? 
 
 592* To find two mean proportionals between two given 
 numbers. 
 
 Divide the greater number by the less, and extract the cube root 
 of the quotient. Multiply the root thus found by the least of the 
 given numbers, and the product will be the least proportional sought ; 
 then multiply the least mean proportional by the same root, and this 
 product will be the greater mean proportional required. 
 
 Find two mean proportionals between the following numbers : 
 
 27. 8 and 216. 29. 12 and 1500. 31. 71 and 15396. 
 
 28. 64 and 512. 30. 40 and 2560. 32. 83 and 60507. 
 
 EXTRACTION OP ROOTS OF HIGHER ORDERS. 
 
 593. When the index denoting the root to be extracted is a 
 composite number. 
 
 First extract the root denoted by one of the prime factors of the 
 y*4)en index / then of this root extract the root denoted by anotlwr 
 f 'ime factor, and so on. Thus, 
 
 For the 4th root, extract the square root twice. 
 For the 6th root, extract the cube root of the square root* 
 For the 8th root, extract the square root three times. 
 For the. 27th root, extract the cube root three times. 
 1. What is the 4th root of 81 ? Ans. 3. 
 
ARTS. 591-594.] HIGHER ROOTS. 379 
 
 2. What is the 8th root of 256 ? 
 
 3. The 4th root of 65530 ? 
 
 4. The 4th root of 19987173376 
 
 5. The 6th root of 46656 ? 
 
 6. The 6th root of 308915776 ? 
 7 The 8th root of 390625 ? 
 
 S The 9th root of 40353607 ? 
 9 The 18th root of 387420489 ? 
 
 10. The 27th root of 134217728? 
 
 594. When the index denoting the root is not a composite 
 number, we have the following general 
 
 RULE FOR EXTRACTING ALL ROOTS. 
 
 1. Point off the number into periods of as many figures each, as 
 there are units in the given index, commencing with the units figure. 
 
 11. Find the first f,gure of the root, and subtract its power from 
 the left hand period ; then to the right of the remainder bring down 
 the first figure in the next period for a dividend. 
 
 III. Involve the root to the power next inferior to that of the index 
 of the required root, and multiply it by the index itself, for a divisor. 
 
 IV. Find how many times the divisor is contained in the divi- 
 dend, and the quotient will be the next figure of the root. 
 
 V. Involve the whole root to the power denoted by the index of 
 the required root, and subtract it from the two left hand periods of 
 f ke given number. 
 
 VI. Finally, bring down the fi.rst figure of the next period to the 
 remainder, for a new dividend, and find a new divisor as before 
 Thus proceed till the whole root is extracted. 
 
 OBS. I. The reason of this rule may be illustrated in the same manner a 
 thaf, for the extraction of the Square and Cube Roots. 
 
 fc. The proof of all roots is by involution. 
 
 3. Any root whatever may be extracted by an extension of the principle ap- 
 plied to the extraction of the cube root. In this general application of the 
 principle, the given number must be divided into periods, each consisting of as 
 many figures as there are units in the index of the required root, and the num- 
 ber of columns employed will be one ess than there are units in the given in- 
 ilex. The operation then proceeds exactly as in the extraction of the cube 
 root ; and if there be a remainder, a like contraction is admissible. 
 T.H. 
 
880 HIGHER ROOTS. [SECT. XV11 
 
 4 
 
 11. Required the 5th root of 35184372088632. 
 
 Operation. 
 
 35184372088832(512 Ans 
 3125 
 = 3125) 3934 
 
 = 345025251 
 
 51 4 X 5 = 33826005) 
 
 512 s =351 84372088832 
 
 12. Required the 5th root of 
 
 13.*' Required the 7th root of 2103580000000000000. 
 
 Note. The preceding method in most of the practical cases, gives ]erhapi 
 as easy solutions, as the nature of the case will admit. But when roots of a 
 very high order are required, the process may be shortened by the following * 
 
 APPROXIMATE RULE. 
 
 595* Call the index of the given power n; and find by trial 
 a number nearly equal to the required root, and call it the assumed 
 root. Raise the assumed root to the power whose index is n. Then, 
 
 As n + 1 time* this power, added to n 1 times the given number, 
 is to n 1 times the same power added to n-j-1 times the yivennum- 
 ber, so is the assumed root to the true root nearly. 
 
 The number thus found may le employed as a new assumed root, 
 and tJie operation repeated to find a result still nearer the true root. 
 
 14. Required the 365th root of 1.06. 
 
 Solution. Take 1 for the assumed root, the 365th power of 
 which is 1 ; and n being 365, we have w + l = 366, and n 1 = 
 864. Then proceed in the following manner : 
 
 1X366 = 366 1X364 = 364 
 
 1.06X364 = 385.84 1.06X366 = 387.96 
 
 As 751.84 : 75L9(F : : 1 : Ans. 
 
 Ans 1.000159b. 
 
 15. The 7th root of 2 ? 17. The 1 2th root of 1.05 ? 
 
 16. The 9th root of 2 ? 18. The 100th root of 100 ? 
 
 * Button's Mathematical Tracts ; also Bonnycastle's Arithmetic* 
 
ARTS. 595-600.] PROGRESSION. 381 
 
 SECTION XVIII. 
 PROGRESSION. 
 
 ART. 596* When there is a series of numbers such, that the 
 ~atios of the first to the second, of the second to the third, &c., are 
 all equal, the numbers are said to be in Continued Proportion, or 
 Progression. Progression is commonly divided into arithmetical 
 and geometrical. 
 
 Note. The terms arithmetical and geometrical are used simply to distin- 
 guish the different kinds of progression. They both belong equally to arith- 
 metic and geometry. 
 
 ARITHMETICAL PROGRESSION. 
 
 5 9 7 Numbers which increase or decrease by a common differ- 
 ence, are in arithmetical progression. (Art. 474. Obs.) 
 
 OBS. 1. Arithmetical progression is sometimes called progression by difference, 
 or equidijjerent scries. 
 
 2. When the numbers increase, the series is called ascending; as, 3, 5, 7, 9, 
 11, &c. When they decrease, the series is called descending ; as, 11, 9, 7, 5, &c. 
 
 598. When four numbers are in arithmetical progression the 
 sum of the extremes is equal to the sum of the means. 
 
 Thus, if 5 3 = 97, then will 5 + 7 = 3 + 9. 
 Again, if three numbers are in arithmetical progression, the sum 
 of the extremes is double the mean. 
 
 Thus, if 9 6=6 3, then will 9 + 3 = 6 + 6. 
 
 599. In any arithmetical progression, the sum of the two ex- 
 tremes is equal to the sum of any other two terms equally distant 
 from the extremes, or equal to double t/ie middle term, when the 
 number of terms is odd. Thus, in the series 1, 3, 5, 7, 9, it is 
 obvious that 1 + 9 = 
 
 GOO. In an ascending series, each succeeding term is found 
 by adding the common difference to the preceding term. Tims, 
 if the first term is 3, and the common difference 2, the series ia 
 3. 5, 7, 9, 11, 13, 15, 19, 17, 21, &c. 
 
382 AKTTHMKTTOnu SECT. XVIII 
 
 In a descending series, each succeeding term is found by sub- 
 tracting the common difference from the preceding term. Thus, 
 if the first term is 15, and the common difference 2, the series u 
 15, 13, 11, 9, 7, &c. 
 
 601. In arithmetical progression there are five parts to be 
 considered, viz : the first term, the last term, the number of terms, 
 the common difference, and the sum of all tlie terms. These parts 
 have such a r relation to each other, that if any three of them are 
 given, the other two may be easily found. 
 
 602. If the sum of the two extremes of an arithmetical pro- 
 gression is multiplied by the number of the terms, the product 
 will be double the sum of all the terms in the series. 
 
 Take the series 2, 4, 6, 8, 10, 12. 
 
 The same inverted 12, 10, 8, 6, 4, 2. 
 
 The sums of the terms are 14, 14, 14, 14, 14, 14. 
 
 Thus, the sum of all the terms in the double series, is equal to 
 the sum of the extremes repeated as many times as there are terms ; 
 that is, the sum of the double series is equal to 12 + 2 multiplied 
 by 6. But this is twice the sum of the single series. Hence, 
 
 603. To find the sum of all the terms, when the extremes 
 and the number of terms are given. 
 
 Multiply half the sum of the extremes by the number of terms, 
 and the product will be tJie sum of the given series. 
 
 OBS. The reason of this process is manifest from the preceding illustration. 
 
 Ex. 1. The extremes of a series are 3 and 25, and the number 
 of terms is 12 : what is the sum of all the terms ? Ans. 168. 
 
 2. What is the sum of the natural series of numbers, 1, 2, 3, 
 4, 5, &c., up to 100? 
 
 3. Hnw many strokes does a common clock strike in 1'2 hours 
 
 GO4 To find the common difference, when the extremes and 
 the number of terms are given. 
 
 Divide the difference of the extremes by the number of terms lea 
 1, and the quotient will be the common difference required. 
 
 OBS. The truth, of this rule is manifest from Art. 603. 
 
ARTS. 601-007. J PROGRESSION 383 
 
 4. The extremes are 5 and 56, and the number of terms is 18: 
 what is the common difference ? Ans. 3. 
 
 5. If the extremes are 3 and 300, and the number of terms 10, 
 what is the common difference ? 
 
 GO5. To find* the number of terms, when the extremes and 
 common difference are given. 
 
 Divide the difference of the extremes by the common difference, and 
 the quotient increased by 1 will be the number of terms. 
 
 OBS. The truth of this principle is manifest from the manner in which the 
 successive terms of a series are formed. (Art. GOO.) 
 
 6. If the extremes are 6 and 470, and the common difference 
 is 8, what is the number of terms ? Ans. 59. 
 
 7. If the extremes are 500 and 70, and the common difference 
 is 10, what is the number of terms? 
 
 GOG. When the sum of the series, the number of terms, and 
 one of the extremes are given, to find the other extreme. 
 
 Divide twice the sum of the series by the number of terms, and 
 from the quotient take the given extreme. 
 
 OBS. The reason of this rule is manifest from Art. 602. 
 
 8. If the sum of a series is 576, the number of terms 24, and 
 the first term 1, what is the last term? Ans. 47. 
 
 9. If the sum of a series is 1275, the number of terms 50, and 
 <he greater extreme 47-J-, what is the less extreme? 
 
 6O7. To find any given term, when the first term and the 
 common difference are given. 
 
 Multiply the common difference by one less than the number of 
 terms required ; then if the series be ascending, add the product to 
 the first term ; but if it be descending, subtract it. 
 
 OBS. The reason of this rule may be seen from the manner in which th 
 succeeding terms of a series are formed. (Art. GOO.) 
 
 10. If the first term of an ascending series is 7, and the common 
 difference 3, what is the 41st term? Ans. 127. 
 
 11. If the first term of a descending series is 100, and the com- 
 mon difference 1-J-, what is til e 54th term ? 
 
384 GEOMETRICAL [SECT. X VIII. 
 
 12. If tli.! first term of an ascending series is Y, and the com- 
 mon difference 5, what is the 100th term? 
 
 608* To find any given number of arithmetical means, when 
 the extremes are given. 
 
 Subtract tlie less extreme from the greater, and divide the re- 
 mainder by 1 more than the number of means required ; the quo- 
 tient will be the common difference, which being continually added 
 to tlie less extreme, or subtracted from the greater extreme, will give 
 the mean terms required. One mean term may be found by taking 
 half the sum of tlie extremes. (Art. 598.) 
 
 OBS. This rule depends upon the same principle as that in Art. 604. 
 
 13. Required 3 arithmetical means between 7 and 35. 
 
 14. Required 6 arithmetical means between 1 and 99. 
 
 GEOMETRICAL PROGRESSION. 
 
 6O9 Numbers which increase by a common multiplier, 01 
 decrease by a common divisor, are in Geometrical Progression. 
 
 The numbers 4, 8, 16, 32, 64, <fec., are in geometrical progres- 
 sion ; and if each preceding term is multiplied by 2, the product 
 will be the succeeding term; thus, 4x2 = 8 ; 8X2 = 16, etc. 
 
 Again, if the order of this series be inverted, the proportion 
 will still be preserved and the common multiplier become a com- 
 mon divisor. Thus; in the series 64, 32, 16, 8, &c., 64-1-2 = 32; 
 32-:-2=16, &c. 
 
 Note. If the first term and ratio are the same, the progression is simply 
 a series of powers; as '2; 2x2; 2x2x2; 2x2X2X2, &c. 
 
 OBS. 1. Geometrical Progression is geometrical proportion continued. It is 
 therefore sometimes called continual proportionals, or progression by quotients. 
 
 If the series increases it is called ascending; if it decreases, descending. 
 
 2. The numbers which form the series, are called the terms of the progres- 
 sion. The common multiplier, or dirisor, is called the ral.io. For most pur- 
 poses, however, it will .be more simple to consider the ratio as always a multi- 
 plier, either integral or fractional. Thus, in the series 64, 32, 16, &c., the 
 ratio is either 2 considered as a divisor, or J considered as a multiplier. 
 
 3. In Geometrical as well as in Arithmetical progression, there are five parts 
 to be considered, viz: the first term, the last term, the number of terms, the ratio, 
 and the sum of alt the terms. These parts have such a relation to each other, 
 that if any three of them are given, the other two may be easily found. 
 
ARTS. 608-011.] PROGRESSION. 385 
 
 6 1 O. To find the last term, when the first term, the ratio, 
 and the number of terms are given. 
 
 Multiply the first term into that power of ike rath ichow index 
 is 1 less than the number of terms, and the product will be Ute 
 last terrr- required. 
 
 OKS. ] The reason of this process may be seen by adverting to the mannet 
 m -wh'cK each successive term is formed. (Art. GOD.) Thus, in the series 4, 8, 
 K; 32, &c. the &l term 8=4X- ) : ll>=4X^X2, or 4X2-; fr2=4X2 3 , &c. 
 
 2. It will be seen that the several Amounts in compound interest, form a 
 gtomctncal series of which the principal is the 1st term ; the amount of $'l for 
 1 year the ratio; and the number of years-j-l the number of terms. Henco 
 the required amount, of compound interest may be found in the same way as 
 the last term of a geometrical series. 
 
 1. If the first term of a geometrical progression is 2, and the 
 ratio 4, wh.it is the 5th term? Art*. 512. 
 
 2. The first term is 04, and the ratio : what is the 5th term? 
 
 3. The first term is 2, and the ratio 3 : what is the 8th term? 
 
 4. The first term is 7, and the ratio 5 : what is the 10th term? 
 
 5. A farmer hired a man for a year, agreeing to give him &1 for 
 the 1st month, 82 for the 2d, &4 for the 3d, and so on, doubling 
 his wages each month : how much did he give the last month? 
 
 G. What is the amount of 8250, at 6 per cent., for 5 years com- 
 pound int. ? Of $500, at 7 per ct., for G years? Of $1000, at 
 5 per ct., for 10 years? 
 
 Oil. To find the sum of the series, when the ratio and the 
 extremes are given. 
 
 Multiply the greatest term into the ratio, from the product sub- 
 tract the least term, and divide the remainder by the ratio less 1. 
 
 OBS. 1. When the first term, the ratio, and the number of terms are given, to 
 find the sum of the series we must first find the last term, then proceed us above. 
 
 2. The sum of an infinite series whose terms -decrease by a common divisor, 
 may be found bit mitUif^uina' Die greatest term info the ratio, and dii't'lri" Iht 
 proffuct ti tJie ratio less 1 The hast, term being infinitely small, is of no com- 
 parative value, and is therefore neglected. $ 
 
 7. What is the sum of the series, whose extremes are 5 and 
 1215, and the ratio 3 ? Ans. 1820. 
 
 8. The extremes of a series are 1 and 512, and the latio 2* 
 what is the sum of the series ? 
 
386 ANNUITIES. [SECT XVIII. 
 
 9. The extremes of a series are 1024 and 15274^, and the ratio 
 is 1 : what is the sum of the series ? 
 
 10. A merchant hired a clerk for a year, and agreed to pay him 
 1 mill the 1st month ; 1 cent the 2d ; 10 cents the 3d, and so on, 
 increasing in a tenfold ratio for each successive month : what was 
 the amount of his wages ? 
 
 11. What is the sum of the infinite series 1 +i+-J--f i, &c. ; that 
 8, the descending series whose first term is 1 and the ratio 2 V 
 
 Ans. 2. 
 
 12. What is the sum of the infinite series 1 +i+i+ izV+'sV, &c. 
 
 612* To find the ratio, when the extremes and number of terms 
 are given. 
 
 Divide the greater extreme by the less, and extract that root of the 
 quotient whose index is 1 less tlmn the number of terms. 
 
 13. The extremes of a series are 3 and 192, and the number 
 of terms 7 : what is the ratio ? Ans. 2. 
 
 14. What is the ratio of a series of 5 terms, whose extremes are 
 7 and 567 ? 
 
 Note. Other formulas in arithmetical and geometrical progression : might be 
 added, but they involve principles with which the student is supposed as yet 
 to be unacquainted. For a fuller discussion of the subject, see Thomson' 
 Day's Algebra. 
 
 ANNUITIES. 
 
 (513. The term annuity properly signifies a sum of money 
 payable annually, for a certain length of time, or forever. 
 
 OBS. 1. Payments made semi-annually, quarterly, monthly, &c., are also 
 called annuities. Annuities therefore embrace pensions, salaries, rents, &c. 
 
 2. When annuities remain unpaid after they are due, they are said to be 
 forborne, or in fir rears. The sum of the annuities in arrears, added to the in- 
 terest due on each, is called the amount. 
 
 The present worth of an annuity is the sum, which being put at interest, 
 will exactly pay the annuity. 
 
 3. When an annuity does not commence till a given time has elapsed, it is 
 called an annuity in reversion; when it continues/0raw, a perpetuity. 
 
 4. In finding the amount of annuities in arrears, it is customary to reckon 
 compound interest on each annuity from the time it is due to the time of pay- 
 ment. The process therefore is the same as find ng th sum of an ascending 
 geometrical series. (Art. 611.) Hence, 
 
ARTS G12-615.J 
 
 ANNUITIES. 
 
 387 
 
 G 1 4. To find the amount of an annuity in arrears. 
 
 Make the annuity t/te first term of a geometrical series, the 
 amount of $1 for 1 year the ratio, and the given number of years 
 t/te number of terms ; then find the sum of the series, and it will be 
 the amount required. (Arts. 610, 611.) 
 
 GBS. When the payments are not yearly, for the amount of $1 for 1 year, use 
 its amount for the time between the payments ; and instead of the number of 
 yut s, use the number of payments that have been omitted, and proceed as before. 
 
 I. What is the amount of an annuity of $100 which has not been paid for 
 3 years, at G per cent, compound interest 1 
 
 . 06)2=112,36; and (112.36X1-06) 100-*-.06= $318.36. 
 
 TABLE, showing the amount of annuity of $1, or 1, at 5, 6, and 7 per cent, 
 for any number of years from I to 20. 
 
 Yrs. 
 
 5 |>er ct. 
 
 6 per ct. i 7 per ct 
 
 Yrs. 
 
 5 per ct. 
 
 6 per ct^ 
 
 7 per ct. 
 
 1 
 
 1.00000 
 
 1.00000 
 
 1.00000 
 
 11 
 
 14.20G78 
 
 14.97164 
 
 15.7836 
 
 2 
 
 2.05000 
 
 2.0GOOO 
 
 2.07000 
 
 12 
 
 15.91712 
 
 16.86994 
 
 17.8884 
 
 3 
 
 3.15250 
 
 3. 18360 1 3.21490 
 
 13 
 
 17.71298 
 
 18.88213 
 
 20.1406 
 
 4 
 
 4.31012 
 
 4.37461 1443994: 
 
 14 
 
 19.598G3 
 
 2 1.0 1506 122.5504 
 
 5 
 
 5.52563 
 
 5.G3709 | 5.75073 ; 
 
 15 
 
 21.5785G 
 
 23.27596 1 25. 1290 
 
 G 
 
 6.80191 
 
 G.97532 
 
 7.15329 
 
 16 
 
 23.65741) 
 
 25.67252 i 27.8880 
 
 7 
 
 8.14201 
 
 8.39383 8.65402 
 
 17 
 
 25.84036 
 
 28.21287 
 
 30.8402 
 
 8 
 
 9.541)11 
 
 9.8974G 
 
 10.2598 
 
 18 
 
 28.13238 
 
 30.90565 
 
 33.9990 
 
 1) 
 
 1 1 .02656 
 
 11.49131 
 
 11.9799 
 
 19 
 
 30.53900 
 
 33.75999 
 
 37.3789 
 
 10 
 
 12.57789 
 
 13.18079 
 
 13.8164 
 
 20 ; 33.0G595 j 3G.78559 j 40.9954 
 
 Note. Multiply the given annuity by the amt. of $1, for the given number 
 of years found in the Table, and the product will be the amount required. 
 
 3. What will an annual rent of $75 amount to in 9 years, at 5 per cent. 1 
 
 4. What is the amount of $200 forborne for 9 years, at 6 per cent. 1 
 
 5. What is the amount of $350 forborne for 10 years, at 7 per cent. ? 
 
 6. What is the amount of $1000 forborne for 20 years, at 6 per cent. ? 
 
 615* To find the present worth of an annuity. 
 
 Find the amount of $1 annuity for the given time as before ; 
 then divide this amount by the amount of $1 at compound, interest 
 for the same time, multiply the quotient by the given annuity, and 
 the product will be the present ivorth. If the annuity is a perpetuity, 
 or to continue forever, multiply it by 100, divide the product by 
 the given rate, and the quotient ivill be the present value required. 
 
 OBS. For the amount of $1 at compound interest, see Table, p. 271. 
 
 7. What is the present worth of an annuity of $40 to continue 5 years, at 
 5 per cent, compound interest"? Ans. $173.178. 
 
 17* 
 
388 PERMUTATIONS. [SECT. XVIII. 
 
 8. What is the present worth of an annuity of $80 to continue forever, at 
 6 per cent. 1 
 
 6 1 G To find the present worth of an annuity in reversion. 
 
 Find the present worth of the annuity from the present time till 
 its termination ; also find its present worth for the time before it 
 commences ; the difference between these two results will be tlie pres- 
 ent worth required. 
 
 9. What is the present worth of $79.625 at 5 per cent., to commence in 
 years tJid continue 6 years 1 Ans. $332.50. 
 
 PERMUTATIONS AND COMBINATIONS. 
 
 617. By Permutations is meant the changes which may be 
 made in the arrangement of any given number of things. 
 
 The tern, combinations, denotes the taking of a less number of 
 things out of a greater, without regard to their order or position. 
 
 618. To find how many permutations or changes may be made 
 in the arrangement of any given number of things. 
 
 Multiply together all the terms of the natural series of numbers 
 from 1 up to the given number, and the product will be the answer. 
 
 1. How many changes may be rung on 5 bells! Ans. 120. 
 
 2. How many diff&rent ways may a class of 8 pupils be arranged 1 
 
 3. How many different ways may a family of 9 children be seated 1 
 
 4. How many ways may the letters in the word arithmetic, be arranged? 
 
 5. A club of 12 persons agreed to dine with a landlord as long as he could 
 seat them differently at the table : how long did their engagement last? 
 
 6 1 9 To find how many combinations may be made out of 
 any given number of different things by taking a given number 
 of them at a time. 
 
 Tak$ the series of numbers, beginning at the number of things 
 given, and decreasing by 1 till the number of terms is equal to the 
 number of things taken at a time ; the product of all the terms 
 \4>iil be the answer required. 
 
 6. How many different words can be formed of 9 letters, taking 3 at a time 1 
 Solution. 9X8X7=504. Ans. 504 words. 
 
 7- Fow many numbers can be expressed by the 9 digits, taking 5 at a time 1 
 Q. Mow many words of fi letters each can be formed out of the 2b' letters of 
 lie alphabet, on the supposition t\at consonants will form a word 1 
 
ARTS. 610-025.] MENSURATION. 389 
 
 ^- 
 
 SECTION XIX. 
 APPLICATION OF ARITHMETIC TO GEOMETRY. 
 
 G2O. In the preceding sections abstract numbers have been 
 applied to concrete substances, or to objects in general, considered 
 arithmetically. On the same principle, geometrical magnitudes 
 aiay be ccmpared or measured by means of the numbers repre- 
 senting their dimensions. (Arts. 7, 510. Oba. 3.) 
 
 Ous The measurement of magnitudes is commonly called mensuration. 
 
 MENSURATION OF SURFACES. 
 
 G2 I In the measurement of surfaces, it is customary to assume 
 a square as the measuring unit, whose side is a linear unit of 
 the name name. (Leg. IV. 4. Sch. Art. 257. Obs. 2.) 
 
 Note. For the demonstration of the following principles, see references. 
 
 G22. To find the area of a parallelogram, also of a square." 
 Multiply the Icnyth l)>j the breadth. (Art. 285, Leg. IV. 5.) 
 
 OBS. When the area and one side of a rectangle are given, the other side it 
 Rrnnd by dividing the area by the given side. (Art. 15G.) 
 
 1. How many acres in a field '240 rods long, and 180 rods wide! 
 
 *J. How many acres in a square field the length of whose side is 3-10 rods! 
 
 3. If the diagonal of a square is 100 rods, what is its urea 7 
 
 4. A rectangular farm of 320 acres, is a mile wide: what is its length! 
 
 G23. To find the area of a rhombus. (Leg. I. Def. 18. IV. 5 ) 
 Multiply the Icnyth by the altitude or perpendicular height. 
 
 5. Find the area of a rhombus whose length is 20 ft., and its altitude 18 ft 
 G24. To find the area of a trapezium. (Leg. IV. 7.) 
 Multiply half the sum of the parallel sides by the altitude. 
 
 6. Find the area of a trapezium the lengths of whose parallel sides art 
 27 ft. and 31 it, and whose altitude is 15 it. 
 
 G25. To find the area of a triangle. . (Leg. IV. C.) 
 Multiply the base by half the altitude or perpendicular hejyht. 
 
 7. Find the area of a triangle whose base is 50 't. } end its altitude 44 ft. 
 
390 MENSURATION. | SECT. XIX. 
 
 626* To find the area of a triangle, the thrpe sides being given. 
 
 From half the sum of the three sides subtract each side respec- 
 tively ; then multiply together half the sum and 'he three remain- 
 ders, and extract the square root of the product. 
 
 9. What is the area of a triangle whose sides are 20, 30, and 10 ft. I 
 
 10. How many acres in a triangle whose sides are each 40 rods 1 
 
 627. To find the circumference of a circle from its diameter. 
 Multiply the diameter by 3.14159. (Leg. V. 11. Sch.) 
 
 Nole. The circumference of a circle is a curve line, all the points of which 
 are equally distant from a point within, called the centre. The diameter cf a 
 circle is a straight line which passes through the centre, and is terminated on 
 both sides by the circumference. The radius or semi-diameter is a straight 
 line drawn from the centre to the circumference, 
 
 11. What is the circumference of a circle, whose diameter is 20 ft. 7 
 
 12. What is the circumference of a circle, whose diameter is 45 rods 1 
 
 G28. To find the diameter of a circle from its circumference. 
 Divide the circumference by 3.14159. 
 
 OBS. The diameter of a circle may also be found by dividing the area by 
 7854, and extracting the square root of the quotient. 
 
 13. What is the diameter of a circle, whose circumference is 314.159 ft. 1 
 
 629. To find the area of a circle. (Leg. V. 11.) 
 
 Multiply half the circumference by half the diameter ; or, mul- 
 tiply the square of the diameter by the decimal .7854. 
 
 15. What is the area of a circle, whose diameter is 50 rods ? 
 
 l(j. Find the area of a circle 200 ft. in diameter, and 628.318 ft. in circura. 
 
 630. To find the side of the greatest square that can be in- 
 scribed in a circle of a given diameter. 
 
 Divide the square of the given diameter by 2, and extract the 
 square root of the quotient. (Art. 581. Obs. 1.) 
 
 17. The diameter of a round table is 4 ft. ; what is the side of-the greatest 
 square table which can be made from it 7 
 
 63 I. To find the side of the greatest equilateral triangle that 
 Can be inscribed in a circle of a given diameter. 
 
 Multiply $ the given diameter by 1.73205. (Leg. V. 4. Sch.) 
 
 18. Required the side of an equilateral triangle inscribed in a circle of 20 ft, 
 <liametcr, 
 
ARTS. 626-637.] MENSURATION. 391 
 
 MEASUREMENT OF SOLIDS. 
 
 632. In the measurement of solids it is customary to assume 
 a cube as the measuring unit, whose sides are squares of the same 
 name. (Art. 258. Obs. 2.) 
 
 633. To find the solidity of bodies whose sides are perpen- 
 dicular to each other. 
 
 Multiph the length, breadth, and thickness together. (Art. 286.) 
 
 OBS. When the contents of a solid body and two of its sides are given, the 
 tftker side is found by dividing the contents by the product of the two given 
 sides. (Art. 159.) 
 
 1. What are the contents of a stick of timber 4 ft. square, and 85J ft. long! 
 
 2. What is the capacity of a cubical vessel, 14 ft. 8 in. deep 1 
 
 634. To find the solidity of a prism. 
 
 Multiply the area of the base by the height. (Leg. VII. 12.) 
 
 OBS. This rule is applicable to all prisms, triangular, quadrangular, pentag- 
 onal, &c., also to all parallelopipedons^ whether rectangular or oblique. 
 
 3. Find the solidity of a prism 4G ft. high, whose base is 7 ft. square 1 
 
 635. To find the lateral surface of a right prism. 
 Multiply the length by the 'perimeter of its base. (Leg. VII. 5.) 
 
 OBS. If we add the areas of both ends to the lateral surface, the sum will be 
 the whole surface of the prism. 
 
 4. What is the surface of a triangular prism, whose sides are each 3 ft., and 
 its length 12 ft. ? 
 
 636. To find the solidity of a pyramid and cone. 
 Multiply the area of the base by \ of the height. (Leg. VII. 18.) 
 
 5. What is the solidity of a pyramid 100 ft. high, whose base is 40 ft. square 1 
 
 6. What is the solidity of a cone 150 ft. high, whose base is 15 ft. in diameter 1 
 
 63 7. To find the lateral or convex surface of a regular pyra- 
 mid, or cone. (Leg. VII. 1G, VIII. 3.) 
 
 Multiply the perimeter of the base by % the slant-height. 
 
 7. What is the lateral surface of a regular pyramid, whose slant-height is, 
 15 ft., and base is 30 ft. square 1 
 
 8. What is the convex surface of a right cr ne, whose slant-height is 94 ft. 
 and the perimeter of its base 37 ft. '{ 
 
393 MENSURATION. [SECT. XIX 
 
 638. To find the solidity of a frustum of a pyramid and cone. 
 To the sum of t/ie areas of the tivo enda, add the square root of 
 
 the product of these areas ; then multiply thin sum by % of ike per- 
 pendicular heiyht. (Leg. VII. 19. Sch., Vlll.G.) 
 
 9. If the two ends of the frustum of a pyramid are 3 ft and 2 ft. square, and 
 the height is 12 ft., what is its solidity 1 
 
 639. The convex surface of a frustum of a pyramid and cone 
 is found by multiplying half the sum of the circumferences of tto 
 to;> ends by the slant-height. (Leg. VII. 17, V11I. 5.) 
 
 10. If the circumferences of the two ends of the frustum of a cone are 18 ft 
 and 14 ft., and its slant-height 11 ft., what is its convex surface 1 
 
 640. To find the solidity of a cylinder. 
 
 Multiply the area of the base by the heir/hi. (Leg. VIII. 2.) 
 
 11. Find the solidity of a cylinder 10 ft. in diameter, and 35 ft. high. 
 
 12. Find the solidity of a cylinder 100 ft. in circumference, and 150 ft. high- 
 
 641. To find the convex surface of a cylinder. 
 
 Multiply the circumference of the base by the heiyht. (Leg. VIII. 1.) 
 
 13. Find the convex surface of a cylinder 5 yds. in diameter, and 5 yds. long. 
 
 642. To find the convex surface of a sphere or globe. 
 Multiply the circumference by the diameter. (Leg. VIII. 9.) 
 
 14. What is the surface of a globe 18 inches in diameter ? 
 
 15. If the diameter of the moon is 2162 miles, what is its surface 1 ? 
 
 643. To find the solidity of a sphere or globe. 
 Multiply the surface by of the diameter. (Leg. VIII. 11.) 
 
 Hi. Find the solidity of a globe 15 inches in diameter. 
 
 17. The diameter of the moon is 21(52 miles: what is its solidity 7 
 
 MEASUREMENT OF LUMBER. 
 
 C44. The area of a board is found by multiplying the length into the men* 
 tread A. (Arts. 622,023.) 
 
 Tr ? solid contents of hewn or square timber are found by multiplying lh 
 wnsih into the mean breadth and depth. 
 
 The solid contents of round timber are found by multiplying the length 
 by \ Uie mean girt or circumference. 
 
 Obs. 1. The mean breadth of a tapering board is found by measuring i' in 
 the middle, >r by taking | the sum of the breadths of the two ends. 
 
ARTS. 638-647.] MENSURATION. 3113 
 
 2. The mean dimensions of square and round timber are found in a similar 
 manner. 
 
 3. The method for finding the solidity of round timber makes an allowance 
 of about - tor waste in hewing. (Arts. 640, 258. Obs. 3.) 
 
 18 Find the area of a board 12 ft. long, and the ends 14 in. and 12 in. wide. 
 
 19 Find the solidity of a joist 10 tl. long, the ends being 8 in. and 4 in. sq. 
 2(1. Find the solidity of a log 50 ft. long, the circumferences of the ends 
 
 oeing 6 ft. and 4 ft. 
 
 GAUGING OF CASKS. 
 
 645. The process of finding the contents or capacities of casks 
 and other vessels is called GAUGING. 
 
 646. r rh& contents of casks are found by multiplying the square of the mean 
 diameter into the length; then this product multiplied bi/ .0034 will give the wine 
 gallons, and multiplied by .0028 will give the beer gallons. 
 
 OES. The mean diameter of a cask is found by adding to the head diameter 
 .7 of the difference between the head and bung diameters when the staves are 
 very much curved ; or by adding .5 when very little curved ; and by adding .55 
 when they are of a medium curve. 
 
 21. How many wine gallons in a cask but little curved, whose length is 4.5 
 in., its bung diameter 40 in., and its head diameter 3G in. 1 
 
 22. How many beer gallons in a cask much curved, whose length is G4 in., 
 its bung diameter 52 in., and head diameter 40 in. 1 
 
 TONNAGE OF VESSELS. 
 
 647. Government Rule. I. If the vessel be double-decked, take the lengln 
 from the fore part of the main stern to the after part of the stern-post, above the 
 upper deck ; then the breadth at the broadest part above the main wales, half 
 of which breadth shall be accounted the depth of such vessel; from the length 
 deduct throe-filths of the breadth, multiply the remainder by the breadth and 
 the product by the depth ; divide the last product by 95, and the quotient shall 
 be deemed the true tonnage of the vessel. 
 
 II. If the vessel be single-decked, take the length and breadth as above di- 
 rected, deduct from the length three-fifths of the breadth, and take the depth 
 from the under side of the deck plank to the ceiling in the hold, then multiply 
 and divide as before, and the quotient shall be deemed the tonnage. 
 
 Carpenter's Rule. The continued product of the length of the keel, the 
 brea 1th at the main beam, and the depth of the hold in feet, divided by 95 will 
 give the tonnage of a single-decked vessel. For a doabl- .decker^ instead of 
 the depth of the hold, take half the breadth at the beam. 
 
 23. What is the government tonnage of a dr uble-decker whose engtl k 
 150 ft., the breadth 35 ft., and the depth 25 ft. 1 
 
 24. What is the carpenter's tonnage of the same vessell 
 
394 MECHANICAL POWERS. [SECT. XIX. 
 
 MECHANICAL POWERS. 
 
 G48 The Mechanical powers are six, viz: the lever, the wlieel 
 and axle, the pulley, the inclined plane, the screw, and the wedge. 
 
 649. When the power and weight act perpendicularly to the arms of a 
 straight fcccr, the power is to the weight, as the distance from die fulcrum to 
 the wcig/U is to the distance from the fulcrum to the power. 
 
 1. If the power is 100 Ibs., the long arm 10 ft., and the short arm 2 ft., what 
 weight can be ra'.sed 1 
 
 2. The arms of a lever are 15 ft. and 4 ft., and the weight raised 500 Ibs.: 
 rhat is the lower 7 
 
 650. When a weight is sustained by a lever resting on two props, 
 
 Tke long arm : the short arm : : the weight, supported by the long arm : the 
 weight supported by the short arm. Hence, 
 
 The whole length : short arm : : whole weight : weight on s. a. (Leg. III. 16.) 
 
 3. A and B carry 25G Ibs. suspended upon a pole 5 ft. from A and 3 ft. from 
 B : how many pounds does each carry 1 
 
 4. A and B carry 90 Ibs. upon a lever 1*2 ft. long : where must it be placed, 
 that B may carry of it 1 
 
 651. The wheel and axle operate on the same principle as the lercr; the 
 semi-diameter of the wheel answers to the long arm, and the semi-diameter of 
 the axle to the short arm. 
 
 5. If the diameter of a wheel is 6 ft., and that of the axle 1 ft., what weight 
 will 100 Ibs. raise 1 
 
 6. A wheel is 8 ft. diameter, an axle 1 ft. : what weight will 200 Ibs. raise 1 
 
 652. In the application of movable pulleys, 
 
 The POWER : the WEIGHT : : 1 : twice l/ie NUMBER of pulleys. 
 
 7. What weight can a power of 200 Ibs. raise with 4 movable pulleys? 
 
 8. What power with 8 pulleys will raise a pillar of granite weighing 10 tono 
 
 653. The perpendicular height of an inclined plane is to its length, as the 
 power to the weight. 
 
 9. What power will draw a train of cars weighing 100000 pounds up an in- 
 clined plane which rises GO ft. to a mile 1 
 
 654. The scrsw acts upon the principle of the inclined plane Henc* , 
 The distance between the threads is to the circumference of a circle described toy 
 
 the pmver, as the power is to the weight. 
 
 10. What weight can be raised by a pr wer of 1000 Ibs. applied to a screw 
 whose threads are 1 inch apart, at the end of a lever 12 ft. long 7 
 
 655. The powr- applied to the head of a wedge is to the weight, as half t.he 
 thickness of the uead is to the length jf its side. In the use of the w;dge, cot 
 less than half the power is lost by fr.ction against the si les. 
 
MISCELLANEOUS SAMPLES. 395 
 
 MISCELLANEOUS EXAMPLES. 
 
 1. The sum of two numbers is 980, and their difference 62: what are the 
 numbers 1 
 
 2. The product of two numbers is 4410, and one is 63 : what is the other 1 
 
 3. What number multiplied by 28^-, will produce 1451 
 
 4. What number multiplied by 6, will be equal to 7| multiplied by 54 } 
 
 5. If an armyof 24000 men have 520000 Ibs. of bread, how long will it . *tt 
 them, allowing each man 1 Ibs. per day 1 
 
 6. What is the interest of $5256 for 60 days, at 7 per cent. 7 
 
 7. What is the amount of $16230 for 4 months, at 6 per cent. ? 
 
 8. What is the bank discount on $1200 for 90 days, at 6 per cent 1 
 
 9. For what sum must a note be made, payable in 4 months, the proceed* 
 of which shall be $1800, discounted at a bank at 7 per cent. 1 
 
 10. A capitalist sent a broker $25000 to invest in cotton, after deducting hi* 
 commission of 2 per cent. : what amount of cotton ought he to receive 1 
 
 11. A merchant bought 500 yards of cloth for $1800: how must he retail it 
 by the yard to gain 25 per cent. 1 
 
 12. A man bought 640 bbls. of beef for $5000, and sold it at a loss of 12 
 per cent. : how much did he get a barrel 1 
 
 13. If a man buys 1000 geographies, at 37 J cents apiece, and retails them 
 at 50 cents, what per cent, will he make 1 
 
 14. A grocer bought 180 boxes of lemons for $360, and sold them at 10 per 
 cent, less than cost : what did he lose 1 
 
 15. How many dollars, each weighing 412 grs., can be made from 16 Ibs. 
 5 oz. of silver 1 
 
 16. How many eagles, weighing 258 grs. apiece, will 21 Ibs. 10 oz. make! 
 
 17. How long a thread can be spun from 1 ton of flax, allowing 5 oz. will 
 make 100 rods of thread 1 
 
 18. How many revolutions will the hind wheel of a carriage 5 ft. 6 in. in 
 circumference, make in 2 miles 4 furlongs? 
 
 19. How many revolutions will the fore wheel of a carriage 4 ft. 7 in. in 
 circumference, make in the same distance 1 
 
 20. 'Bought 1500 doz. buttons for $187.50: what was that per gross 1 
 
 21. A man paid $132 for 40 bbls. of cider: what is that a quart 1 
 
 22. A man paid $150 for 10 rods of land, what was that per acre 1 
 
 23. A man having $2500, laid out - of it in flour, at $5 per barrel : how 
 many barrels did he buy 1 
 
 24. The commander of an exploring expedition found that ^ of his provision* 
 were exhausted in 28 months : how much longer would they last 1 
 
 25 What cost 15-f Ibs. of cheese, at $8 per hundred 7 
 
 26. How many yards of carpeting \ yd. wide will it take to cover a floor 
 18 ft. long and 15 ft. wide 1 
 
 27. If ^ yard of calico cost -f^s., what will of an ell English cost? 
 
 28. How long will 468256 Ibs. of beef last an army of *^45 soldiers, allow- 
 ing each man 1J lb. per day 1 
 
398 MISCELLANEOUS EXAMPLES. 
 
 29. How long would the same quantity of beef last Jhe army, if reinforced 
 jy 2500 men, allowing each man 1 J Ib. per Jay '{ 
 
 30. Bought f of a pipe of wine for $120: what was that per gallon 7 
 
 31. If a man can walk 17 miles in 5 hours, 12 minutes, 31 seconds, how 
 far <*an he walk in 3 hours, 40 minutes, 3u' seconds'? 
 
 32. If a man traveling 1 1 hours per day, performs half his journey in 9 
 days, how long will it take him to go the other half traveling 10 hours a day! 
 
 33. li you lend a man $700 f or 99 days, how long ought he to lend you 
 ftll'OO to requite the favor 1 
 
 34. A milkman's measure was deficient half a gill to a gallon : how much 
 did he cheat his customers in selling 87*20 gallons 7 
 
 35. If 85 yds. of calico cost $10/20, what will 1500 yds. cost 1 
 3G. If 1G50 Ibs. of sugar cost $20:125, what will 87 Ibs. cost! 
 
 37. If 14-24 gals, of oil cost $10b'2, what will '210 gals, cost 7 
 
 38. If wind moves 2$ miles per hour, how long is it in moving from the pole 
 to the equator, a distance of G214 miles'? 
 
 3D. If -j^- of a barrel of flour costs f of a dollar, what will of a bbl. cost 1 
 
 40. If of a ton of chalk cost ^f, what will - of a ton cost 1 
 
 41. If |- of a bushel of wheat cost $|-, how much will f- of a bushel cost 1 
 4'2. If of a ship cost $10000, what will -fa of her cost 1 
 
 43. If IGi bbls. of mackerel cost $b'5f, what will 48^ bbls. cost 7 
 
 44. If 28i gals, of oil cost $31.25, how much will $250 buy 7 
 
 45. At $& for 40-doz. eggs, what will 4b'0-- doz. cost 7 
 
 4G. The President's salary is $25000 per annum : how much can he spend 
 per day, and lay up $10000 of it 7 
 
 47. If one person lies in bed 9 hours per day, and another G hours, how 
 much time will the one gain over the other in '20 years 7 
 
 48. A cistern has three faucets ; the 1st will empty it in 10 min., the 2d in 
 20 min., the 3d in 30 min. : in what time will they all empty it 7 
 
 4!). A man and his wife drink a barrel of beer in 30 days, and the man alone 
 can drink it in 40 days : how long will it last the wife 7 
 
 50. A teacher being asked how many scholars he had, replied \ study Arith- 
 metic, -i- study Latin, -fo study Algebra, -^ study Geometry, and 24 study 
 French : how many scholars had he 7 
 
 51. A man having spent i and of of his money, had 48j left: how much 
 had he at first 7 
 
 52. A man bequeathed -^ -f his property to his wife, to his son, -J- to hig 
 daughter, and the remaindc/, which was $1500, to the Bible Society: what 
 did bis whole property amount to 7 
 
 53 What is that number of which exceeds -|- of it by 45 7 
 
 54. Pewter is composed of 1 12 parts of tin. 15 of lead, and G of brass: hew 
 much will it take of each ingredient to make 5(>50 pounds of pewter 7 
 
 55. Two travelers start at thn same time from Boston and Washington to 
 meet each other; one goes 5 miles an hour, the other 7 miles; the whole dis- 
 tance is 436 miles : how far will each travel 7 
 
MISCELLANEOUS EXAMPLES. 397 
 
 56. A grocer divided ,a barrel of flour into 2 part?, so that the smaller con 
 taincd I as much as the other: how many pounds were there in each 7 
 
 57. A, B, and C, build a ship together; A advanced $1000. B $12000, and 
 C $13000; they gain $5000: what is the gain of each? 
 
 58. A, H, and (J, entered into partnership ; A furnished $000, R and C to- 
 gether $lb<10 ; they gained $%0, of which B took $280 ; how much did A and 
 C gain ; and B and C put in respectively I 
 
 51). The liabilities of a bankrupt are $03240, and his assets $12018: wha 
 per cent, can he pay 7 
 
 00. A bankrupt compromises with his creditors for 37 per cent.: h< w mucB 
 will he pay on a claim of $30507 
 
 01. How much will he pay on a debt of SI 2080 .375 7 
 
 02. A owns f and B -fa of a ship ; A's share is worth $10000 more thai. 
 B's: what is the value of the ship 7 
 
 03. A man gave his oldest son - of hi* t ropcrty less $50; to the second, he 
 gave ; and to the youngest he gave tl > remainder, which was less $10 : 
 what was the amount of his property! 
 
 01. A man and boy together can frame a house in 9 days; the man can 
 frame it alone in 1*2 days: how long will it take the boy to frame it 7 
 
 05. A cistern has a receiving and a discharging pipe ; when both are run- 
 ning it takes 18 hours to fill it; if the latter is closed it requires 15 hours to 
 fill it : if the former is closed, how long will it take the latter to empty it 7 
 
 00. Four men, A, B. C, and 1), spent '255, and agreed that A should pay 
 | ; B ; C J ; and D } : how much must each pay 7 
 
 07. A, B, and C, formed a joint stock of 820, and gained 040. in the 
 division of which A received 5 as often as B did 7, and C 8: how much 
 did each put in and receive 7 
 
 08. A, B, and C, gained a certain sum, of which A and B received $040, 
 B and C $K80, and A and C $800: what was the gain of each 7 
 
 09. What number is that -^ and -J of which being multiplied together, will 
 produce the number itself? 
 
 70. A club spent 2, 12s. Id. ; on settling, each paid as many pence as there 
 were individuals in the party : how many were there in the party 7 
 
 71. The sum of two numbers is 120, and the difference of their squares ii 
 4800 : what are the numbers 7 
 
 72. The di (Terence of two numbers is 53. and the difference of the square* 
 is 10759: what are the numbers 7 
 
 73. The diagonal of a square is 80 ft. : what is its side 7 
 
 74. The diagonal of a square field is 120 rods: what is its are? 7 
 75. Find the side of the greatest square beam which can be hewn..Srora 
 
 log 5 ft. in diameter 7 
 
 70. The mainmast of a ship is 95 ft. long, the diameter of the base Js 3j ft., 
 
 that of the top 2 ft. : what is its solidity 7 
 
 77 A man wished to tie his horse by a rope so that he could feed on just an 
 
 acre of ground now long must the rope be 7 
 
308 MISCELLANEOUS EXAMPLES. 
 
 78. What is the area of a circle 1 mile in circumference "? 
 
 79. If the diameter of the sun is 887000 miles, what is its surface 1 
 
 80. If the diameter of Jupiter is 86255 miles, what is its solidity 1 
 
 81. A conical stack of hay is 20 ft. high, and its base 15 ft. in diametei . 
 what is its weight, allowing 5 Ibs. to a cubic foot 1 
 
 82. How many bushels will a cubical bin contain whose side is 9 ft. 1 
 
 83 How many hogsheads will a cylindrical cistern 10 ft. deep and C^- ft. 
 Uameter contain 1 
 
 84. How far from the end of a stick of timber 30 ft. long, of equal size from 
 end to end, must a lever be placed, so that 3 men, 2 at the lever, and I at the 
 end of the stick, may each carry i of its weight 1 
 
 85. How many different ways may a class of 26 scholars be arranged? 
 
 86. If 100 eggs are placed in a straight line a rod apart, how many miles 
 must a person travel to bring them one by one to a basket placed a rod from 
 me first egg 1 
 
 87. What is the sum of the series 1, !, 2, 2|, 3, &c., to 50 terms'? 
 
 88. A blacksmith agreed to shoe a horse for 1 mill for the first nail in his 
 shoe, 2 mills for the second nail, and so on : the shoes contained 32 nails: how 
 much did he receive 1 
 
 89. Said a mule to an ass, if I take one of your bags, I shall have twice as 
 many as you, and if I give you one of mine, we shall have an equal number: 
 with how many bags was each loaded 1 
 
 90. What number taken from the square of 48 will leave 16 times 541 
 
 91. Divide $1000 between A, B, and C, and give A $120 more than C, and 
 C $95 more than B. 
 
 92. A person being asked the hour of the day, said, that the time past noon 
 was f- of the time till midnight : what was the hour ] 
 
 93. A, B, and C, can trench a meadow in 12 days ; B, C, and D, in 14 days ; 
 C, D, and A, in 15 days; and D, B, and A, in 18 days. In what time would 
 it be done by all of them together, and by each of them singly 1 
 
 94. Suppose A, B, and C, to start from the same point, and to travel in the 
 same direction, round an island 73 miles in compass, A at the rate of 6, B of 
 10, and C of 16 miles per day : in what time will they be next together 1 
 
 95. At what time between 12 and I o'clock do the hour and minute hands 
 of a common clock or watch point in directions exactly 'opposite 1 
 
 96. In how many years will the error of the Julian Calendar involve the 
 loss of a day 1 
 
 97. A man's desk was robbed 3 nights in succession ; the first night half th 
 number of dollars were taken and half a dollar metre ; the second, half the re- 
 mainder was taken and half a dollar more; the third night, half of what was 
 then left and half a dolfar more, when he found he had $50 left, how much 
 had he at first 1 
 
 THE END. 
 
ANSWERS TO EXAMPLES. 
 
 NOTE. At the urgent request of several distinguished Teachers, who hava 
 received Thomson's Higher Arithmetic with favor, the publishers have issued 
 an edition of it, containing the answers in the end of the book. It is hopea 
 that pupile, who may use this edition, will have suificient regard to their own 
 improvement, never to consult the answer till they have made a strenuous and 
 persevering effort to solve the problem themselves. 
 
 N. B. The work without the answers is published as heretofoie. 
 
 ADDITION. ARTS. 59-61. 
 
 Ex. ANS. 
 
 Ex. ANS. 
 
 Ex. ANS. 
 
 1. $5445 
 
 17. 288011295. 
 
 35. 9429190. 
 
 2. 41757 bushels. 
 
 18. 14303433. 
 
 36. 11178170 
 
 3. 11 5 90 pounds. 
 
 19. 100611775. 
 
 37. 10306156. 
 
 4. 831551. 
 
 20. 1805851434. 
 
 38. 10662291. 
 
 5. $5583. 
 
 21. 337351. 
 
 39, 40. Given. 
 
 6. 65440 sq. miles. 
 
 22. 7221. J41. 214. 
 
 7. 10-2451 sq. m. 
 
 23. 4251988. 
 
 42. 253. . 
 
 8. 528524 sq. m. 
 
 24. 3795. 
 
 43. 276. 
 
 9. 650327 sq. m. 
 
 25. 73464390. 
 
 44. 19443. 
 
 10. 1362742 sq. m 
 
 26. 33604444. 
 
 45. 20714. 
 
 11. 233890. 
 
 27. 15821984. 
 
 46. 2476372. 
 
 12. 828463. 
 
 28. 97059404. 
 
 47. $132085946. 
 
 13. 990240. 
 
 29. 1038220930. 
 
 48. $107109740. 
 
 U. 96181521. 
 
 32. $570805. 49. 2069857 tons. 
 
 15. 127215713. 
 
 33. 6460458 yards. 50. $57981492. 
 
 .6. 869754587. 
 
 34. 6657039 pounds 51. Given. 
 
 SUBTRACTION. ART. 76. 
 
 1. $7095. 
 
 10. $12280043. 
 
 19. 5313439. 
 
 2. 28984 bushels. 
 
 11. $23563746. 
 
 20. 543679. 
 
 3. $30954. 
 
 12. 430143 tons. 
 
 21. 2007984. 
 
 4. $46025. 
 
 13. 149237. 
 
 22. 45103074. 
 
 5. 58000000 miles. 
 
 14. 3393329. 
 
 23. 66729549. 
 
 6. $6327597. 
 
 15. 54399581. 
 
 24. 72820280. 
 
 7. $26176670. 
 
 16. 8825431. 
 
 25. 55301760. 
 
 8. $1644737. 
 
 17. 4001722. 
 
 26. 80200180. 
 
 9. $7977899. 
 
 18. 2601900. 
 
 27. 95658143. 
 
400 
 
 ANSWERS. 
 
 [PAGES 46 60. 
 
 SUBTRACTION CONTINUED. A AT. 76 
 
 Ex. ANS. 
 
 Ex. ANS 
 
 Ex. ANS. 
 
 28. 9000001. 
 
 39. 85807625. 
 
 50. 925. 
 
 29. 99899999. 
 
 40. 1598. 
 
 51. 1511. 
 
 30 83128433. 41. 4004. 
 
 52. 41845. 
 
 31. 40592424. 
 
 42. 1384. 
 
 ) $46900, W. 
 
 32. 55352005. 
 
 43. 14061. 
 
 5 ) $69450, 1! 
 
 33, 19957466. 
 
 44. 12494. 
 
 54. $2410 lost. 
 
 81 77919201. 
 
 45. 11547. 
 
 55. 171825. 
 
 35. 70051563. 46. 3295. |56. $1674737. 
 
 30. 53201371. j47. 1006. 
 
 57. $97. 
 
 37. 25311703. 
 
 48. 3707. 
 
 58. $3893. 
 
 38. 86282745. 
 
 49. 2004. 
 
 59. Given. 
 
 MULTIPLICATION. ART. 93. 
 
 1. $24795. 
 
 15. 3931476. 
 
 29. 239968374861. 
 
 2. $36099. 
 
 16. 415143630. 
 
 30. 449148410434. 
 
 3. $56700. 
 
 17. 31884470. 
 
 31. 289975559744. 
 
 4. 90520 miles. 
 
 18. 8468670. 
 
 32. 294144537440. 
 
 .5. 74175 pounds. 
 
 19. 43506216. 
 
 33. 335834314400. 
 
 6. 372500 days. 
 
 20. 11847672. J34. 18834782688. 
 
 7. 960000 rods. 
 
 21. 57380625. \S5. 109588282050. 
 
 8. 20835. 
 
 22. 11050155200. 
 
 36. 654638320927. 
 
 9. 21576. 
 
 23. 12810000. 
 
 37. 396890151372. 
 
 10. 68198. 
 
 24. 48288058. 
 
 38. 554270292192. 
 
 11. 176400. 
 
 25. 3473567604. 
 
 39. 2985984. 
 
 12. 1554768. 
 
 26. 88789980848. 
 
 40. 57111104051. 
 
 13. 5497800. 
 
 27. 9313702853. 
 
 41. 60435595442394 
 
 14. 1674918. 
 
 28. 67226401140. 
 
 42. 87112343040000 
 
 CONTRACTIONS IN MULTIPLICATION. ARTS. 97 1O8. 
 
 8. $1776. 
 
 20. 312046700000. '29. 96000 pounds. 
 
 9. $5760. 
 
 21. 52690078000000 30. 359400000. 
 
 10. $8100. 
 
 22. 6890634570000- 31. 143759940000. 
 
 11. 5782 s. 
 
 000. 
 
 32. 28708635000)00 
 
 12. 23808 miles. 
 
 23. 494603050600- 
 
 34. 123240000. 
 
 13. $11736. 
 
 000000. J35. 2309760000. 
 
 14. 19845 s. 
 
 24. 87831206507- 
 
 36. 26366200000. 
 
 15. $32256. 
 
 000000000. 
 
 37. 144447000000. 
 
 17. 46500 bushels. 
 
 25. 678560051090- 
 
 39. 31276000000. 
 
 18. 365000 days. 
 
 000000000. 
 
 40. 3747600000000- 
 
 19. 1534860000. 
 
 28. 18750 pounds. 
 
 41. 18054680000000 
 
PAGES 61 74.] 
 
 ANSWERS. 
 
 401 
 
 PON TRACTIONS IN MULTIPLICATION CONTINUED. 
 
 Ex. ANS. 
 
 Ex ANS. 
 
 Ex. ANS. 
 
 42. 664726500000- 
 
 74. 4140. 
 
 99. 180600000. 
 
 000. 
 
 75. 27936. 
 
 100. 2722946304. 
 
 43. 1075635900000- 
 
 76. 154250. 
 
 101. 2172069918. 
 
 000. 
 
 77. 11348400. 
 
 102. 7225. 
 
 4 45514. 
 
 78. 34639552. 
 
 103. 65536. 
 
 4F 68476. 
 
 79. 2685942. 
 
 104. 104650. 
 
 47. 400624. 
 
 80. 2801960. 
 
 105. 127447'JO. 
 
 48. 907002. 
 
 81. 72156000. 
 
 106. 31049291000. 
 
 50. 132525. 
 
 82. 1680000000. 
 
 107. 2732116062240 
 
 51. 307664. 
 
 83. 2000000000. 
 
 108. 222310980000 
 
 52. 2333616. 
 
 84. 43644865, 
 
 109. 20066857745- 
 
 53. 5691627. 
 
 85. 81708550. 
 
 896. 
 
 55. 474309. 
 
 86. 401939564. 
 
 110. 1256700743298 
 
 66. 6027966. 
 
 87. 476413195. 
 
 111. 37968807755. 
 
 57. 7293699. 
 
 88. 62220780. 
 
 112. 39073U8478. 
 
 58. 4629537. 
 
 89. 637049231. 
 
 113. 102128*493520 
 
 63. 54530. 
 
 90. 406101366. 
 
 114. 1421400000000 
 
 64. 72819. 
 
 91. 42261696. 
 
 115. 60302400000- 
 
 65. 346896. 
 
 92. 504159579. 
 
 000. 
 
 66. 6624403632. 
 
 93. 6724232757. 
 
 116. 9130020:1000- 
 
 67. 17651712450. < 
 
 <94. 7306359. 
 
 000. 
 
 68. 21983532672. 
 
 95. 21760506. 
 
 117. 6800400^.0000- 
 
 71. 625. 
 
 96. 39429936. 
 
 000. 
 
 72. 2916. 
 
 97. 2283344802. 
 
 118. 4000000000- 
 
 73. 5184. 
 
 98. 650633256. 
 
 000000 
 
 DIVISION. ART. 127. 
 
 1. 45 bu. 13. 
 
 2. 85 bbls. 14. 
 
 3. &68lfi- 15. 
 
 4. $3. 16. 
 
 5. I68ff. 17. 
 
 6. $73972|i. 18. 
 
 7. 20iff days. 19. 
 
 8. m-iWds. 20. 
 
 9. 2773||. 21. 
 
 10. 1139V*. 22 - 
 
 11. 1443-s-V .23. 
 
 12. 1489ff. 124. 
 
 5697f. 
 
 25. 3679. 
 
 35. 826^51- 
 
 3823-f-f. 
 
 26. 4500. 
 
 -iV^j'W' 
 
 4166if. 
 
 27. 50830TW- 
 
 36. 1387805- 
 
 21276|-f. 
 
 28. 630. 
 
 ww- 
 
 12152^. 
 
 29. 235. 
 
 37. 900^00900- 
 
 191-2*^. 
 
 30. 648. 
 
 90^9-j-^T. 
 
 873. 
 
 31. 267l01ff. 
 
 38. IJ009GOO- 
 
 48. 
 
 32. 563. 
 
 oooo-KH. 
 
 48HH- 
 
 33. 8826211- 
 
 39. 900H0900- 
 
 87-iVs- 
 
 Wf- 
 
 009 rhr- 
 
 108. 
 
 34. 23434402- 
 
 
 45- 
 
 aV^"' 
 
 
402 
 
 ANSWERS. 
 
 [PAGES 75 96. 
 
 CONTRACTIONS IN DIVISION. ARTS. 129139. 
 
 fix. AN^ 
 
 Ex. ANS. 
 
 Ex ANS 
 
 Ex ANS 
 
 1, 2. Given. 
 
 18. 36. 
 
 37. 15. 
 
 55. 228378HI 
 
 3. 132f| a . 
 
 19. 68. 
 
 38. 16-ftV. 
 
 56. 941501ff. 
 
 4. 672. 
 
 20. 36ff. 
 
 39. 17. 
 
 57. 478676if. 
 
 5. 460. 
 
 21. 75. 
 
 40. 30. 
 
 58. 59207. 
 
 6. 205. 
 
 23. 1207. 
 
 41. 250 days. 
 
 59. 1826896. 
 
 7 1265. 
 
 24. 1690. 
 
 42. 950 years. 
 
 60. 13S791-M 
 
 8. 20; 34; 56d. 
 
 25. 6512. 
 
 43. $10-iV- 
 
 61. 65964||i 
 
 9. 1650; 7650; 
 
 20. 8654. 
 
 44. $285f* 8 . 
 
 62. 6162^ 5 -. 
 
 $43200. 
 
 27. 83||. 
 
 45. $39rrV 
 
 63. 1583 If-H. 
 
 10. 267, and 
 
 28. 70if. 
 
 46. $2iff. 
 
 64. 21i^ff.' 
 
 50000 R. 
 
 29. 77ff. 
 
 47. $ll- 3 Yr. 
 
 65. 4134i|t. 
 
 11. 144, and 
 
 30. 142if. 
 
 48. $54-3^. 
 
 66. 3966|ff-. 
 
 360791 R. 
 
 31. 94. 
 
 49. $219ifi- 
 
 67. 16581ff. 
 
 12. 5823, and 
 
 32. 194|f. 
 
 50. 18if. 
 
 68. 7405-,^. 
 
 67180309R 
 
 33. 1693if. 
 
 51. 13529||. 
 
 69. 4362-iVs-. 
 
 14. 105 b. 
 
 34. 3795- 2 V 
 
 52. 12466H- 
 
 70. 3186/7V 
 
 15. 184 bbls. 
 
 35. 67. 
 
 53. 12454|f. '71. 9Vf$W. 
 
 16. 197+f. 
 
 36. 203-iVs- 
 
 54. 13446913f 72. 920 T fJHHh r 
 
 CANCELATION. ARTS. 15O, 151. 
 
 2. 45. 
 
 4. 65. 
 
 6, 7. Giten. 
 
 9. 3i. 
 
 3. 63. 
 
 5. 73. 
 
 8. 6. 
 
 10. 3. 
 
 APPLICATIONS OP THE FUNDAMENTAL RULES. 
 
 ARTS. 152159. 
 
 1. Given. 
 
 11. 79.years; 
 
 19. Given. 
 
 27. 632. 
 
 2. 255 acres. 
 
 94 yrs. 
 
 20. 48 beggars. 
 
 28. 974. 
 
 3. 925 bu. 
 
 12. $510| car. 
 
 21. 20 flocks. 
 
 29. 7124; 5516 
 
 4. Given. 
 
 $345i hor. 
 
 22. Given. 
 
 30. 13000; 
 
 5. $190. 
 
 14. 65 years. 
 
 23. 20 years. 
 
 12264. 
 
 6. 11 25 sheep. 
 
 15. 175 rods. 
 
 24. 10 months. 
 
 31. 21151 
 
 8. $2240. 
 
 17. 187825. 
 
 25. 1842. 
 
 20975. 
 
 9. $3436. 
 
 18. 1033062. 
 
 26. 1062. 
 
 32. 786. 
 
 PROPERTIES OF NUMBERS. ARTS. 162, 163. 
 
 19. Given. 
 
 13. 2024122. 
 
 18. 707961. 
 
 22. 1614386. 
 
 10, 20212331. 
 
 14. 1522365. 
 
 19. 1036993. 
 
 23. 118620366. 
 
 11. 2350147. 
 
 15. Given. 
 
 20. 9753020. 
 
 24. 3879090- 
 
 12. 1331124. 
 
 16, 17. Given. 21. 360913096. 
 
 582. 
 
?AGES 97, 98.J 
 
 ANSWERS. 
 
 403 
 
 ANALYSIS OP COMPOSITE NUMBERS. ART. 165. 
 
 Ex. ANS. 
 
 Ex. ANS. 
 
 Ex. ANS. 
 
 4. 9 3X3. 
 
 28. 2, 3, and 7. 
 
 52. 2, and 37. 
 
 5. 2, and 5. 
 
 29. 2, 2, and 11. 
 
 53. 3, 5, and 5. 
 
 6. 2, 2, and 3. 
 
 30. 3, 3, and 5. 
 
 54. 2, 2, and 19. 
 
 7. 2, and 7. 
 
 31. 2, and 23. 
 
 55. 7, and 11. 
 
 8. 3, and 5. 
 
 32. 2, 2, 2, 2, and 3. 
 
 56. 2, 3, and 13. 
 
 9. 2, 2, 2, and 2. 
 
 33. 7, and 7. 
 
 57. 2, 2, 2, 2, and 6 
 
 10. 2, 3, and 3. 
 
 34. 2, 5, and 5. 
 
 58. 3, 3, 3, and 3. 
 
 11 2, 2, and 5. 
 
 35. 3, and 17. 
 
 59. 2, and 41. 
 
 12 3, and 7. 
 
 36. 2, 2, and 13. 
 
 60. 2, 2, 3, and 7. 
 
 13. 2, and 11. 
 
 37. 2, 3, 3, and 3. 
 
 61. 5, and 17. 
 
 14. 2, 2, 2, and 3. 
 
 38. 5, and 11. 
 
 62. 2, and 43. 
 
 15. 5, and 5. 
 
 39. 2, 2, 2, and 7. 
 
 63. 3, and 29. 
 
 16. 2, and 13. 
 
 40. 3, and 19. 
 
 64. 2, 2, 2, and 11. 
 
 17. 3, 3, and 3. 
 
 41. 2, and 29. 
 
 65. 2, 3, 3, and 5. 
 
 18. 2, 2, and 7. 
 
 42. 2, 2, 3, and 5. 
 
 66. 7, and 13. 
 
 19. 2, 3, and 5. 
 
 43. 2, and 31. 
 
 67. 2, 2, and 23. 
 
 20. 2, 2, 2, 2, and 2. 
 
 44. 3, 3, and 7. 
 
 68. 3, and 31. 
 
 21. 3, and 11. 
 
 45. 2, 2, 2, 2, 2, and 2 
 
 69. 2, and 47. 
 
 22. 2, and 17. 
 
 46. 5, and 13. 
 
 70. 5, and 19. 
 
 23. 5, and 7. 
 
 47. 2, 3, and 11. 
 
 71. 2, 2, 2, 2, 2, and 3 
 
 24. 2, 2, 3, and 3. 
 
 48. 2, 2, and 17. 
 
 72. 2, 7, and 7. 
 
 25. 2, and 19. 
 
 49. 3, and 23. 
 
 73. 3, 3, and 11. 
 
 26. 3, and 13. 
 
 50. 2, 5, and 7. 
 
 74. 2, 2, 5 and 5. 
 
 27. 2, 2, 2, and 5. 
 
 51. 2, 2, 2, 3, and 3. 
 
 75. 2, 2, 3, 3, and 3. 
 
 76. 120=2X2X2X3X5. 
 144=2X2X2X2X3X3 
 
 77. 180 = 2X2X3X3X5. 
 420=2X2X3X5X7. 
 
 78. 714=2X3X7X17. 
 836 = 2X2X11X19. 
 
 79 574 = 2X7X41. 
 
 2898 = 2X3X3X7X23. 
 
 80. 11492 = 2X2X13X13X17 
 
 980=2X2X5X7X7. 
 
 81. 650 = 2X5X5X13. 
 1728 = 2X2X2X2X2X2 
 
 X3X3X3. 
 
 82. 1492 = 2X2X373. 
 
 83. 
 84. 
 
 85. 
 86. 
 
 251. 
 
 4604=2X2X1151. 
 16806 = 2X3X2801. 
 71640 = 2X2X2X3X3X5 
 
 X199. 
 
 20780=2X2X5X1039. 
 84570 = 2X3X5X2819. 
 65480 = 2X2X2X5X1637 
 92352 = 2X2X2X2X2X2 
 
 X3X13X37. 
 81660=2X2X3X5X1361, 
 
 T.H. 
 
 18 
 
404 
 
 ANSWERS. 
 
 [PAGES 99 118 
 
 GREATEST COMMON DIVISOR. ARTS. 168171. 
 
 Ex. ANS. 
 
 Ex. ANS. 
 
 Ex. ANS. 
 
 Ex. AN. 
 
 1. Given. 
 2. 3. 
 3. 7. 
 4. 5. 
 6. 2. 
 
 6, 7. Given. 
 8. 15. 
 9. 14. 
 10. 111. 
 11. 39. 
 
 12. 1. 
 13. Given. 
 14. 3. 
 15. 16. 
 17. 15. 
 
 18. 12. 
 
 19. 18. 
 20. 35. 
 21. 6. 
 22. 28. 
 
 LEAST COMMON MULTIPLE. ARTS. 176, 177. 
 
 13. Given. 
 
 4. 90. 
 
 5. 144. 
 
 6. 180. 
 
 7. 360 
 
 8. 720. 
 
 9. 12600. 
 
 10. 504. 
 
 11. 1134. 
 
 12. 15015. 
 
 14. 144. 
 
 15. 600. 
 
 16. 2520. 
 
 17. 252. 
 
 18. 1134. 
 
 19. 360. 
 
 21. 600. 
 
 22. 1440. 
 
 23. 13824. 
 
 24. 51000. 
 
 REDUCTION OF FRACTIONS. ARTS. 195-2O1 
 
 1, 2. Given. 
 
 3. i. 
 
 4. f 
 
 5. --. 
 
 6. f. 
 
 7. f 
 
 8. |f. 
 
 9. f*. 
 
 10. T^. 
 
 11. ii- 
 
 12. i. 
 
 13. li 
 
 14. i. 
 
 15. |. 
 
 16. f. 
 17. 
 18. 
 
 21. 9. 
 
 22. 5. 
 
 23. 3f. 
 
 24. 9|. 
 
 25. 1. 
 
 26. 60. 
 
 27. 21. 
 
 28. 52. 
 
 29. 60f. 
 
 W; * 
 ifil-f; 
 
 63. 
 64. 
 65. 
 66. 
 67. 
 68. 
 69. 
 70. 
 VI. 
 
 72. -mm; 
 
 73, 74, Given. 
 
 ifH; iffX; 
 ; fm* ; ii 
 : HffS; 
 
 30. 
 
 34. 
 35. 
 36. 
 37. 
 38. 
 39. 
 40. 
 41. 
 42. 
 43. 
 44. 
 
 75. 
 
 76. 
 77. 
 78. 
 79. 
 80. 
 81. 
 82. 
 83. 
 84. 
 85. 
 
 23 6 g6 
 
 ni*.' 
 
 J5JLJJL 
 
 47. 
 48. 
 49. 
 52. 
 53. 
 54. 
 55. 
 56. 
 57. 
 58. 
 
 59. f 
 
 60. -ft. 
 
 61. 62 Given 
 
 il; 
 
 i; 
 
 JL& 
 If 4 > 
 
 if; 
 
 *; If; 
 
 a J2-L 
 4 > 2 4 
 
 -ii; -A-; 
 H ; tt ; 
 
 U J CU> GU> 60* 
 
 ft; ft; n; ft- 
 t&** ; ^ ; f "' 
 
 1 o o s > TSnJe > Too* t i oo f 
 
PAGES 118127.] 
 
 ANSWERS. 
 
 405 
 
 REDUCTION OF FRACTIONS CONTINUED. ART. 201. 
 
 El. 
 
 80. 
 
 87. 
 88. 
 
 ANS. 
 
 ttftt; ttftt; nm; 
 fHft- 
 
 -23d.tl -2U.QJL' _J> 8 Q_ ^.lAO. 
 C 3 II > 63 00 > 63UO 6 3 U 
 
 Ex. 
 
 ANS. 
 
 89. 
 90. 
 
 ft; -Aft; ft; & 
 
 ; ttf i tH; ill 
 
 ADDITION OF FRACTIONS. ARTS. 202-2O i. 
 
 1 8. Given. 
 4 3^ a 
 5 . 2 2 T. 
 0. 
 
 2 ^. 
 
 mt. 
 
 9. 
 10. 
 11. 
 12. 
 
 23. & ; 
 
 24. -f^; 
 
 25. iif ; 
 20. 14-f; 
 27. IvVV 
 
 30. Ml* 
 3 1 ^ A & a a 
 
 5. gV- 
 
 0. 5 V=i. 
 
 7. MV- 
 
 8. sWi- 
 
 9. i,VA. 
 
 13. 1-fr. 
 
 14. / 4 V 
 
 15. 2^. 
 10. 214* 
 
 17. 114-f 
 
 18. G|. 
 
 19. 01 f. 
 
 20. 15^ 
 
 017 _6.A_ nl 
 
 SUBTRACTION OF FRACTIONS. AUTS. 2O6 208. 
 
 10. iiJ. 17. 121f. 25. 291 
 
 11. ii. 18. 278|i 2. 003 4 ^. 
 
 12. 23. 21. 
 
 13. -LSA- oo 
 
 15. 
 
 10. 
 
 23. 137f. 
 
 24. 4GG* 
 
 28. -f. 
 
 29. 8203 a . 
 
 30. 7 H|. 
 
 MULTIPLICATION OF FRACTIONS. CASE I. ARTS. 21 1 IT. 
 
 13. Given. 
 
 17. G33|. 
 
 32. 514^. 
 
 48. 0897. 
 
 4. ^=3^. 
 
 18. 3715*. 
 
 33. 3 05 |f i 
 
 49. 15282. 
 
 5. ^ s = 12i-. 
 
 19. 444 Sf. 
 
 35. 10390!-. 
 
 50. 29318. 
 
 D . 18 \ i . 
 
 20. 22G4,V 
 
 30. 174GOH. 
 
 51. 1280. 
 
 7. 37?. 
 
 21. 12519*. 
 
 37. SCGVaV 
 
 52. 279. 
 
 8. 48H- 
 
 24. 108. 
 
 38. 20G73fy. 
 
 53. 4490. 
 
 9. 80- 3 V. 
 
 25. 127. 
 
 39'. 35650|-f 
 
 54. 8113. 
 
 10. GGf?f. 
 
 20. 240. 
 
 40. 23555 HI. 
 
 55. 10413-!*,. 
 
 11. 1 G i T i "$" 
 
 27. 435. 
 
 43. 375. 
 
 50. 5072 ff. 
 
 12. 770- 6 V 
 
 28. 500. 
 
 44. 738. 
 
 57. 5080!*,. 
 
 13. G02ifi. 
 
 29. 02 H. 
 
 45. 1178. 
 
 58. 43452. 
 
 14, 15. Given. 
 
 30. 701|. 
 
 40. 3450. 
 
 59. 74290iV 
 
 16. 735?. 
 
 31. 76A- 
 
 47. 0795. 
 
 CO. 92280ff. 
 
 1 
 
406 
 
 ANSWERS. [PAGES . 28 142. 
 
 MUOIPLICATION OF FRACTIONS CONTINUED. CASE II. 
 
 ARTS. 219, 22O. 
 
 ill. ANS. 
 
 Ex. ANS. 
 
 Ex. ANS. 
 
 Ex. ANS. 
 
 1, 2. Given. 
 
 11. 17. 
 
 20. 3232f&. 
 
 29. 99|. 
 
 3. If = tV- 
 
 12. 53|. 
 
 21. 5143-iWT- 
 
 30. 78f. 
 
 ^& == ^fe t 
 
 13. 242ff. 
 
 22. 6998^-|4l^. 
 
 31. 47tV. 
 
 5. -iVr- 
 
 14. 329. 
 
 23. 167if|. 
 
 32. 86. 
 
 6. i. 
 
 15. 1362f. 
 
 24. 53TVs. 
 
 33. 401f. 
 
 >j f ^ff. 
 
 16. 3198if. 
 
 25. 24091|. 
 
 34. 1474^V|, 
 
 8.LJL& 
 . "| 7 5 . 
 
 17. 451^i. 
 
 26. 466-M-. 
 
 35. $972f. 
 
 9. ff . 
 
 18. 834|^f. 
 
 27. 300000. 
 
 36. $6323^. 
 
 10. Given. 
 
 19. 2001 1 a i 1 r. 
 
 28. 700. 
 
 37. 5501-!^ na. 
 
 CONTRACTIONS IN MULTIPLICATION OP FRACTIONS. 
 ARTS. 221-225. 
 
 3. i. . 
 
 11. f. 
 
 21. 57600. 
 
 32 4762^-. 
 
 4. A. 
 
 12. - 4 V 
 
 22. 99000. 
 
 33 J5937 
 
 6. i. 
 
 13. i. 
 
 23. 187l|. 
 
 34 40187^ 
 
 6. --=2f. 
 
 1'4. A- 
 
 24. 14220. 
 
 35 65450. 
 
 * - f. 
 
 15. A. 
 
 26. 2133^. 
 
 37 1278i. 
 
 8. A- 
 
 16. *VV 
 
 27. 18466|. 
 
 38 4083^. 
 
 9. i. 
 
 19. 493. 
 
 28. 5580. 
 
 39. 8150. 
 
 10. 26. 
 
 20. 8533^ 
 
 29. 430000. 
 
 40. 93833|. 
 
 DIVISION OP FRACTIONS. ARTS. 226241. 
 
 13. Given. 
 
 18, 19. Given. 
 
 33. 9-tV*. 
 
 51. A- 
 
 4. A- 
 
 20. H. 
 
 34. 13iif. 
 
 52. f. 
 
 5- ** 
 
 21. A- 
 
 37. 13-tWr. 
 
 53. 3H. 
 
 64 g 
 TO^^S' 
 
 22. 23|. 
 
 38. STWir- 
 
 54. fr. 
 
 7. A- 
 
 23. 40. 
 
 40. 220^. 
 
 55. 1?. 
 
 8. A=*. 
 
 24. i. 
 
 41. 30fi. 
 
 56. 1. 
 
 o. vw- 
 
 25. i. 
 
 42. 50H- 
 
 58. Hf. 
 
 1013. Given. 
 
 26. 3ff. 
 
 43. 6^. 
 
 59. A- 
 
 14. 1AV- 
 
 29. 135. 
 
 46. 383-ftV- 
 
 60. f;. 
 
 15. 2if. 
 
 30. 168. 
 
 47. 54VVVV. 
 
 61. 5|. 
 
 10. 3H. 
 
 31. 11A- 
 
 49. 2fj. 
 
 63. ^ 
 
 17. m. 
 
 32. llf|. 
 
 50. -H. 
 
 64. i 
 
 APPLICATION OF FRACTIONS. ART. 242. 
 
 1. 88if yds. 
 
 4. $14if. 
 
 7. $1548|. 
 
 10. 5606^ s. 
 
 2. 162tVlbs. 
 
 5. $62H. 
 
 8. $55 16. 
 
 11. $483f. 
 
 3. $54f. 
 
 0. $635^-. 
 
 9. $1515". 
 
 12. $100. 
 
PAGES 143 170.] 
 
 ANSWERS. 
 
 407' 
 
 APPLICATION OF FRACTIONS CONTINUED. ART. 242. 
 
 Ex ANS 
 
 Ex. ANP. 
 
 K.\ ANS 
 
 Ex AN* 
 
 13. &16111-J-. 
 14. 404552fflb 
 15. -Si 8061. 
 16. 36121 bu. 
 17. $30968$-. 
 18. G939-rVm. 
 19. 5229 m. 
 20. $9l7oi. 
 
 21. lOoi yds. 
 22. 218| Ibs. 
 23. 626| gals. 
 24. 20-rVlbs. 
 25. 53-f yds. 
 26. 27 boxes. 
 27. 153f bbls. 
 28. 2911 suits. 
 
 29. 3 8 A rods. 
 30. $3f-g. 
 31. $6f. 
 32. 584f bu. 
 33. 4| doz. 
 34. 6- 6 \ cts. 
 35. 9|fi s. 
 36. $13-,V*. 
 
 37. &5 Hi- 
 38. 42-,-fr tons. 
 39. SlYVVr. 
 40. t51+i*. 
 41. $3fg. 
 42. 2664-Hr d 
 43. $33798J|f. 
 
 REDUCTION. ART. 282. 
 
 2. 68810 far. 
 
 28. 8553600 in. 
 
 52.- 28992 pts. 
 
 3. 86768 far. 
 
 29. 5280000 yds. 
 
 53. 1427 bu. 1 pk. 
 
 4. 284079 far. 
 
 30. 54 m. 7 fur. 38 r. 
 
 54. 130 100 qts. 
 
 5. 966 15 far. 
 
 2 yds. 2 ft. 
 
 55. 36360 inin. 
 
 6. '25, 13s. 6d. 3 far. 
 
 31. 91.2m. 4 fur. 31 r. 
 
 56. 3 1557600 sec. 
 
 7. 433, Is. 2d. 3 far. 
 
 1 yds. 2 ft, 7 in. 
 
 57. 84 wks.6hrs. 45 mm 
 
 8. '266 guin. 18s. 8d. 
 
 32. 5031 rods. 
 
 58. 65 d. 2h. 4m. 40 sec 
 
 9. 1448 sixpences. 
 
 33. 17 m. 20 r. 
 
 59. 3 1556928 sec. 
 
 10. 6050 threepences. 
 
 34. 132105600ft. 
 
 60. 946728000 sec. 
 
 11. 170472 grs. 
 
 35. 2560 na. 
 
 61. 10 yrs. 
 
 12. 9000 pwts. 
 
 36. 5000 qrs. 
 
 62. 397200". 
 
 13. 1010047 grs. 
 
 37. 6396 yds. 2 qrs. Ina 63. 1350000." 
 
 14. 2 Ibs. 1 oz. 10 pwts. 
 
 38. 9302F.e.4qrs.3na 
 
 64. 2126, IT, 54". 
 
 16 grs. 
 
 39. 10156 na. 
 
 65. 555555s. 16o,40'. 
 
 15. 1 77 Ibs. 9 oz. 12 pwts 
 
 40. 7116 qts. 
 
 66. 470660 sq.ft. 
 
 16. 1596 Ibs. 
 
 41. 693 gals. 
 
 67. 43660734 sq. ft. 
 
 17. 564000 oz. 
 
 42. 26528 gi. 
 
 68. 32640858360 sq. in. 
 
 18. 104300 Ibs. 
 
 43. 48 bar. 20 gals. 
 
 69. 582 A. 1 R. 3 r. 
 
 19. 71 680000 drs. 
 
 44. 117 pi. 1 hhd. 46 g. 
 
 269} sq. ft. 
 
 20. lOcwt. 16 Ibs. 
 
 3 qts. 1 pt. 2 gi. 
 
 70. 259200 cu. in. 
 
 21. 133T. 12cwt.351bs 
 
 45. 102 128 gi. 
 
 71. 4551 552 cu. in. 
 
 22. 1 T. 202 Ibs. 1 oz. 
 
 46. 12960 pts. 
 
 72. 10877760 cu. in. 
 
 23. 9 120 drs. 
 
 47. 87 bar. 26 gals. 
 
 73. 49 cu. ft. 1 cu. in. 
 
 24. 37440 sc. 
 
 48. 630 hhds. 44 gals 
 
 74. 306 C. 48 cu. ft. 
 
 25. 64 Ibs. 11 oz. 5 drs. 
 
 49. 19520 pts. 
 
 75 4492800 cu. in. 
 
 26. 881bs.4oz.7drs.2sc 
 
 50. 488 qts. 
 
 76. 52 T. 40 cu. ft. 
 
 27. 142560ft. 
 
 51. 24440 qts. 
 
 180 cu. in. 
 
 APPLICATIONS OF REDUCTION. ARTS. 283 2i. 
 
 1. Given. 
 
 4. 177-flbs. Troy, or 
 
 7. 5 8 Ibs. 4 oz. Troy. 
 
 2. 576 Ibs. avoir. 
 
 145-^f-f Ibs. avoir. 
 
 8. 21f^ 2 - Ibs. Troy. 
 
 3. 691 Ibs. 10 oz. 
 
 5. 265f Ibs. Troy, or 
 
 9. 271 Ibs. 3 oz. 
 
 5-rVr drams. 
 
 218ifilbs.avoir. 
 
 10. Given. 
 
408 
 
 ANSWERS. [PAGES 171- -180 
 
 APPLICATIONS OF REDUCTION CONTINUED. ARTS. 285293. 
 
 F.. Ass. 
 
 E.x. AN*. 
 
 K.\. ANS. 
 
 !1. 3 00 sq.ft. 
 
 30. 3450 wine gals. 
 
 49. 14 hhds. 4SK g* 
 
 12. 14 A. 10 *q. rds.|3l. 8040f w. gals. 
 
 51. 51^f beer gals. 
 
 13. lOSsq. v. 8 sq. ft. 
 
 32. 5184 beer gals. 
 
 52. 45 'ji wine gals. 
 
 14. 440 A. 1 It. 
 
 33. 09 1 2 b. gals. 2 qts 
 
 53. 24f ? w. gals. 
 
 15. 40 A. 
 
 34. 80 i 6 }- l>u. 
 
 54, 1598 w. gals. 
 
 10 30 sq. yds. 
 
 35. 100 bu. 
 
 55. 2207|-f w. gals. 
 
 17 60 sq. yds. 
 
 36. 800 bu. 
 
 56. 3125H qts. 
 
 13. lll|sq. yds. 
 
 37. 897-H- w. gals. 
 
 57. 2734*? gals. 
 
 20. 56 1 cu. ft. 
 
 38. 7l2fff bar. 
 
 59. 8 min. 36 sec. 
 
 21. 120 cu. ft. 
 
 39. 902807H1 hhds. 
 
 60. 39 min. 
 
 22. 80 0. 2 cu. ft. 
 
 41. 622f cu. ft. 
 
 61. 1 hr. 8 m. 40 sec. 
 
 23. 748 cu. ft. 142. 1244f cu. ft. 
 
 62. 33 min. 48 sec. 
 
 24. 750 cu. ft. 
 
 43. 8*1 cu. ft. 
 
 63. 12h.28m. 12s. 
 
 25. 72 cu. yds. 
 
 44. 210^ cu. ft. 
 
 64. Given. 
 
 26. 160 cu. ft. 
 
 45. 842-ft cu. ft. 
 
 65. 4 45'. 
 
 27. 1800 cu. ft. 
 
 47. 42-/A bu. 
 
 66. 12 46'. 
 
 29. 17280 bu. !48. 40-ft- gals. 
 
 67. 13 23'. 
 
 COMPOUND NUMBERS REDUCED TO FRACTIONS. ART. 290. 
 
 1-4. Given. 
 5. 
 
 0. 
 
 8. -ft lb. Troy. 
 
 9. eWlb.Troy 
 10. if lb. avoir. 
 
 HJidi T 
 . 4 U * 
 
 1-2. -f yd. 
 13. -ffjm. 
 14. 
 15 
 
 16. 
 17. 
 18. 
 19. 
 
 A. 
 r/r sq. r. 
 
 igal. 
 i hhd. 
 
 d. 
 
 hr. 
 
 20. TTTsYinr. 
 22. VV- 
 23 --0.- 
 
 24! I' 
 
 25. -gV 
 
 26. fy. 
 
 27. / \. 
 28. 
 
 29. 
 30. 
 31. 
 32. 
 33. 
 34. 
 35. 
 36. 
 
 FRACTIONAL COMPOUND NUMBERS 
 'REDUCED TO WHOLE NUMBERS OF LOWER DENOMINATIONS. ARTS. 297, 298t 
 
 3. 
 
 17s. Gd. 
 
 
 13. 
 
 2 qts. 1 pt 
 
 Hgi- 
 
 25. 
 
 6f hrs. 
 
 4. 
 
 7d. i far. 
 
 
 14. 
 
 55 gals. 1 
 
 pt. 
 
 26. 
 
 2688 min. 
 
 5. 
 
 5 oz. 2 p. 2 
 
 7 g- 
 
 16. 
 
 3 pks. 1 qt. 
 
 Hpts. 
 
 27. 
 
 8- 6 \ na. 
 
 G, 
 
 12 pwts. 1 
 
 2 grs. 
 
 17. 
 
 46 min. 40 
 
 sec. 
 
 28. 
 
 17^-?- qts. 
 
 7. 
 
 10 oz. 10| 
 
 drs. 
 
 18. 
 
 21 hrs. 36 
 
 min. 
 
 29. 
 
 174H? qts, 
 
 S 
 
 57 Ibs. 2oz. 
 
 4|drs. 
 
 19. 
 
 22 sec. 
 
 
 30. 
 
 4|? oz. 
 
 9 
 
 1250lbs. 
 
 
 20. 
 
 17' Sf. 
 
 
 31. 
 
 60 pwts. 
 
 10 
 
 2 ft. 4f in. 
 
 
 22. 
 
 Wd. 
 
 
 32. 
 
 7-j^- r. 
 
 11. 
 
 6 ft. 2| in. 
 
 
 23. 
 
 -aVy- oz. 
 
 
 33. 
 
 A sq. ft. 
 
 12. 
 
 177 r. 12 ft 10 in. 
 
 24. 
 
 Hr. 
 
 
 34. 
 
 70'' 
 
PAGES 182 -190. J ANSWERS. 
 
 409 
 
 COMPOUND ADDITION. ART. 3OO. 
 
 Ex. ANS 
 
 Ex. ANS 
 
 Ex. ANS. 
 
 3. 106, 3s. Id. 
 4. 188, 13s. id. 
 5. 9 T. 8 cwt. 17 Ibs. 
 6 45T. 4 cwt. 57 Ibs. 
 2 oz. 
 7 1 07 Ibs. 7 0.8. p. Ig. 
 8 330 Ibs. 2 o. 3 p. 5 g. 
 9. 4 fur. 1 3 r. 13 ft. Sin. 
 
 10. 1091. 2m. 6 fur. 1ft. 
 11. 114 yds. 3 qrs. 
 1 2. 387 yds. 1 qr. 
 13. 138 A. 114 sq. r. 
 80 sq. ft. 
 14. 468 A. 1R. 33sq.r. 
 15. 43 sq. yds. 5 sq. ft. 
 125 sq. in. 
 
 16. 240 gals. 
 17. 181 hhds. 59 gals 
 1 pt. 1 gi. 
 18. 115 w. 15 h. 25m. 
 19. 322 bu. 1 pk. 5 qts. 
 20. 135 qrs. 3 bu. 3 pka 
 2 qts. 
 
 COMPOUND SUBTRACTION. ARTS. 3O2, 303. 
 
 1. Given. 
 
 2. 9. 2s. 8d. 3 qrs. 
 
 3. 60, 4s. 7d. 3 qrs. 
 
 4. 499, 1 3s. 4d. 2 qrs. 
 
 5. 8 cwt. 1 qr. 6 Ibs. 
 10 oz. 
 
 6. 24 T. 1 cwt. 71 !b a . 
 
 7. 19m. 289 . 2ft. 
 
 8. 1 1. 1 m. 7 fur. lOr. 
 
 ft. 
 
 9. 35 bu. 2 pks. 6 qts. 
 
 10. 19 qrs. 6 bu. 2 pks. 
 
 11. 55 yds. 2 qrs. 3 na. 
 
 12. 44yds. 1 qr. 3 na. 
 
 13. 6 gals. 2 qts. 1 pt. 
 
 14. 48 hhd. 46 g. 2 qts. 
 
 15. 85 A. 119r. 
 
 16. 235 A. 48 r. 
 
 17. 56 C. 90 cu. ft. 
 
 18. 339 cu. ft. 26 in. 
 
 19. 25 3' 15". 
 
 20. 35 3' 30" 
 
 21. 10 26'. 
 
 22. 54 yrs. 2 mos. 2 wka 
 6d.2hrs.45min.6s 
 
 23. Given. 
 
 24. 67 yrs. 9 mos. 22 d 
 
 26. 1 yr. 5 mos. lid. 
 
 27. 3 yrs. 9 mos. 22 d. 
 
 COMPOUND MULTITPLICATION. ART. 305. 
 
 1, 2. Giv*en. 
 
 3. 247, 6s. Id. 
 
 4. 24, 9d. 
 
 5. 17T. 55 Ibs. 
 
 6. 403 T. 17cw f .551bs 
 
 7. 689 Ibs. 8 oz. 16pwts 
 
 8. 6 Ibs. lOoz. lOpwts 
 
 9. 3039 hhds. 39 gals. 
 1 f|t. 1 pt. 
 
 10. 5668 pi. 32 gals. 
 
 11. 2358 yds. 
 
 12. 5375 yds. 
 
 13. 14778 m. 1 fur. 32 r. 
 
 14. 
 
 15. 
 
 ifi. 
 
 17. 
 18. 
 19. 
 
 20. 
 21. 
 
 22. 
 
 2044 1. 1 m. 4 fur. 23. 
 30 r. |24. 
 
 8962 bu. 16 qts. J25. 
 2968 qrs. 5 bu. 2 pks. 26. 
 6 qts. 27. 
 
 7821 A. 20 r. 28. 
 25172 A. 1 R. 3r 29. 
 24645 cu. ft. 930 30. 
 en. in. 31. 
 96350 C. 50 cu. ft. ' 
 
 12783d. lib. 28m. 
 1199 yrs. 9 mos. 
 3 wks. Id. 
 
 15891 13' 30" 
 204 10'. 
 4581 bu. 8 qts. 
 2453 bu. 4 qts. 
 5, 16s. 10 id. 
 679, 3s. 4d. 
 297. 
 
 507, 16s. 3d. 
 36 C. 74 cu. ft. 
 944 in. 
 
 865 Ibs. 12 oz. 
 25418 Ibs. 12 07. 
 8662 gals. 2 qts. 
 
 COMPOUND DIVISION. ART. 307. 
 
 8. 4, 17s. 3d. i qr. 
 
 1-3. Given. 
 
 4. 51 Ibs. 3 oz. 10 pwts. 
 15-f grs. 
 
 5. 31 bu. 14|- qts. 
 
 6. 25 bu. I 1 pts. 
 7 20. Is. 6d. 
 
 9. 10 yds. 3 qrs. If na. 
 
 10. 9 yds. 1 qr. i| na. 
 
 11. 83 m. 2 fur. 26 r. 11 ft v 
 
 12. 214 m. 2 fur. 27 r. 
 4 ft. 4 in. 
 
410 
 
 ANSWERS. [PAGES 196 201 
 
 COMPOUND DIVISION CONTINUED. ART. 307. 
 
 Ex. ANS. 
 
 Ex. ANS 
 
 13. 1 gal. 2 qts. 1 pt. lH gi. 
 
 18. Is. 17 52' 21|f". 
 
 14. 44 hhds. 29 gals. 1 pt. lgi. 
 
 19. 90. 84 ft. 101 6W in. 
 
 15. 24 d. 8 hrs. 42 min. 40 sec. 
 
 20. 6 C. 92 ft. 850|i in. 
 
 16; 10 yrs. 35 d. 1 hr. 13 min. 
 
 21. 6s. 10d. 
 
 llT*j-sec. 
 
 22. 7s. lid. 3^qrs. 
 
 17. 1 48' 41i". 
 
 23. 10s. lid. 2- 7 4qrs. 
 
 ADDITION OP DECIMALS. ART 32O. 
 
 ], 2. Given. 
 
 9. 857.005. |16. 2.471092. 
 
 3. 4-28.1739. 
 
 10. 1097.84143. 
 
 17. 0.0711824. 
 
 4. 103.85-23. 
 
 11. 1408.25559. 
 
 18. 0.3532637. 
 
 5. 14.747274. 
 
 12. 127.05034. 
 
 19. 0.807711. 
 
 6. 60.149. 
 
 13. 33.3182746. 
 
 20. 0.1627165. 
 
 7. 332.1249. 
 
 14. 15674.1613. 
 
 21. 0.996052. 
 
 8. 501.15998. 
 
 15. 1.807. 
 
 
 22. 0.329773. 
 
 SUBTRACTION OF DECIMALS. ART. 3J2. 
 
 1, 2. Given. 
 
 13. 2.291. 
 
 
 24. 0.000999. 
 
 3. 1427.633782. 
 
 14. 9.9999999. 
 
 25. 699.93. 
 
 4. 20.987651. 
 
 15. 8.000001. 
 
 26. 28999.908. 
 
 5. 72.5193401. 
 
 16. 4635.5346. 
 
 27. 255999909.744. 
 
 6. 81.16877. 
 
 17. 541.787. 
 
 28. 0.414. 
 
 7. 0.066721522. 
 
 18. 46.43606. 
 
 29. 0.0041. 
 
 8. 0.01. 
 
 19. 0.0000999. 
 
 30. 0.000000000999. 
 
 9. 9.999999. 
 
 20. 0.0000396. 
 
 31. 0.002873789. 
 
 10. 64.0317753. 
 
 21. 31.99968. 
 
 32. 0.062156. 
 
 11. 24680.12377. 
 
 22. 44.99955. 
 
 33. 0.71699. 
 
 12. 24.75. 
 
 23. 98.99999901. 
 
 34. 0.0000843174. 
 
 MULTIPLICATION OF DECIMALS ART. 324. 
 
 1. 681.45ft. 
 
 13. 36.740232. 
 
 25. 0.00164389993. 
 
 2. 25020 miles. 
 
 14. 919.82036. 
 
 26. 160.86701632806. 
 
 3. 2055.375 gals. 
 
 15. 0.000000072. 
 
 27. 0.06288405909156. 
 
 4. 136.125 nails. 
 
 16. 0.00105175. 
 
 28. 2.5067823. 
 
 5 788.0125 sq. yds. 
 
 17. 390.657556. 
 
 29. 64.327106105314. 
 
 6 43560 sq. ft. 
 
 18. 275.230594. 
 
 30. 0.0000118260069. 
 
 7. 2465.375 sq. rods. 
 
 19. 148.64244532. 
 
 31. 11027.10199543710 
 
 8. 0.250325, 
 
 20. 73.25771882. 
 
 32. 94167471.869654- 
 
 9. 18.93978. 
 
 21. 52.17977576. 
 
 039. 
 
 10. 14.78091. 
 
 22. 0.0306002448. 
 
 33. .OOOOf'006676542- 
 
 11. 0.613836. 
 
 23. 4701.169144360. 
 
 672. 
 
 12. 0.0320016. 
 
 24. 536.660(175952. 
 
 
PAGES 201 213.] ANSWERS. 
 
 411 
 
 CONTRACTIONS IN MULTIPLICATION OF DECIMALS. 
 ARTS. 325-32T. 
 
 Ex. ANS. 
 
 Ex. ANS. 
 
 Ex. ANS. 
 
 1. Given. 
 2. 4:29302.13401. 
 3. 1067123.50123. 
 4. 6)8340.17. 
 5 304672.14067. 
 6. 44632140.32. 
 7. 2134567.82106. 
 8 500. 
 
 9. 75000. 
 10. 6.5. 
 11. 48. 
 12. 2480. 
 13. 381. 
 14, 65.04. 
 15. 834000. 
 16. 10. 
 
 17-20. Given. 
 21. 0.09484. 
 22. 1.262643. 
 23. 0.0769. 
 24. 0.0389254. 
 25. 0.00876. 
 26. 0.002516. 
 27. 0.001789. 
 
 DIVISION OF DECIMALS. ART. 33O. 
 
 1-3. Given. 
 
 4. 13 boxes. 
 
 5. 8 suits. 
 
 6. 4.98347+days. 
 
 7. 82.9997-Hoads. 
 
 8. 27.7173-Hays. 
 
 9. 150.25 bales. 
 
 10. 5.929 1-f. 
 
 11. 6.632. 
 
 12. 79098.8235-f 
 
 13. 0.6344-h 
 
 14. 1210.2344+. 
 
 15. 0.03. 
 
 16. 1 34.8805+. 
 
 17. 59.4060+. . 
 
 18. 24.8266-f. 
 
 19. 4320.67. 
 
 20. 0.02. 
 
 21. 83671000. 
 
 22. 255.1210-J-. 
 
 23. 0.000005. 
 
 24. 60.2589. 
 
 25. 211.076. 
 
 26. 400000. 
 
 27. 60000000. 
 
 28. 4000000. 
 
 29. 3 11. 487360+. 
 
 CONTRACTIONS IN DIVISION OF DECIMALS. ARTS. 331-33 
 
 1, 2. Given. 
 
 3. 67234.567. 
 
 4. 103.42306. 
 
 5. 0.42643621. 
 
 6. 6.72300045. 
 
 7. 0.000012300456. 
 
 8. 0.0000020076346. 
 
 9. Given. 
 
 10. 0.1274. 
 
 11. 0.09471. 
 
 12. 1.611. 
 
 13. 0.04026. 
 
 14. 0.0954776. 
 
 15. 2.0208. 
 
 16. 0.980439. 
 
 DECIMALS REDUCED TO COMMON FRACTIONS. ART. 335. 
 
 1, 2. Given. 
 
 3. i. 
 
 4. if 
 
 5. V. 
 
 10. 
 11. 
 12. 
 13. 
 
 '14. 
 15. 
 16. 
 
 COMMON FRACTIONS REDUCED TO DECIMALS. 
 ARTS. 337344. 
 
 A -3. Given. 
 
 10. 
 
 0.6. 
 
 4. 0.5. 
 
 
 11. 
 
 0.8. 
 
 5. 025. 
 
 
 12. 
 
 0.5. 
 
 6. 0.5. 
 
 
 13. 
 
 0.125. 
 
 7. 075. 
 
 
 14. 
 
 0.25. 
 
 8. 02. 
 
 
 15. 
 
 0.375. 
 
 9. 0.4. 
 
 
 16. 
 
 0.5. 
 
 
 18* 
 
 
 17. 0.625. 
 18. 0.75. 
 19. 0.875. 
 20, 21. Given. 
 
 25. Terminate 
 26. Terminate. 
 27. Interminate, 
 28. Terminate. 
 
 22. Terminate. 
 
 31. 03. 
 
 23. Terminate. 
 
 32. 0.6. 
 
 24. [nterminato. 
 
 33. 0.16. 
 
ANSWERS 
 
 [PAGES 214 224 
 
 COMMON FRACTIONS REDUCED TO DECIMALS CONTINUED. 
 
 Ex. ANS. 
 
 Ex. ANS. 
 
 Ex. ANS. 
 
 Ex. ANS. 
 
 34. 0.3. 
 
 41. 0.714285. 
 
 48. 0.6. 
 
 56. 0.0048828- 
 
 35. 0.6. 
 
 42. 0.857142. 
 
 49. 0.7. 
 
 125. 
 
 36. 0.83. 
 
 43. 0.1. 
 
 50. 0.8. 
 
 57. 0.583. 
 
 37. 0.142857. 
 
 44. 0.2. 
 
 51. 0.1875. 
 
 58. 0.076923. 
 
 38. 0.285714. 
 
 45. 0.3. 
 
 52. 0.076923. 
 
 59. 0.0104895. 
 
 39. 0.428571. 
 
 46. 0.4. 
 
 53. 0.024. 
 
 60. 0.46835443- 
 
 40 0.571428. 
 
 47. 0.5. 
 
 54. 0.0112. 
 55. 0.275. 
 
 03797. 
 
 COMPOUND NUMBERS REDUCED TO DECIMALS. ART. 3 16. 
 
 1, 2. Given. 
 
 3. 0.5375. 
 
 4. 0.825. 
 6. 0.87916. 
 
 6. 0.416 s. 
 
 7. 0.5416 s. 
 
 8. 0.115625 m. 
 
 9. 0.25625m. 
 
 10. 0.2583 hr. 
 
 11. 0.127083 d. 
 
 12. 0.0525 cwt. 
 
 13. 0.46875 Ib. 
 
 14. 0.875 bu. 
 
 15. 0.5625 pk. 
 
 16. 1.125 gals. 
 
 DECIMAL COMPOUND NUMBERS REDUCED TO WHOLE 
 ONES. ART. 348. 
 
 2. 14s. 6d. 
 
 3. 2s. 7d. 3.2 qrs. 
 
 4. Id. 2 qrs. 
 
 5. 9d. 3.6 qrs. 
 
 6. 12 Ibs. 8 oz. 
 
 7. 6 oz. 15.36 drs. 
 
 8. 88 rods. 
 
 9. 7 ft 0.51 in. 
 
 10. 11 gals. 1 qt. 1 pt. 
 3.7184 gills. 
 
 11. 1 qt. Ipt. 3.4432 gi, 
 
 12. 10 h. 13m. 9. 12 sec. 
 
 13. 50 min. 42 sec. 
 
 REDUCTION OP CIRCULATING DECIMALS. ARTS. 35561. 
 
 1, 2. Given. 
 3. if, or -ft. 
 4. 
 
 5. 
 
 9. 
 10. 
 14. 
 15. 
 16. 
 
 - 9 V9,or-rh;.l7. 
 18. 
 19. 
 
 20. T | 
 or 
 23. 0.277. 
 
 0.333. 
 0.045. 
 24. 4.3213. 
 6.4263. 
 0.0000. 
 
 7. Vkor-ji 
 
 ADDITION OF CIRCULATING DECIMALS. ART. 362. 
 
 2. 179.2745563. 
 
 3. 476.65129. 
 
 4. 47.86083. 
 
 5. 594.691 
 
 6. 112,7': 24. 
 
 7. 223.M07744. 
 
 8. 1380.0648193. 
 
 9. 5974.10371. 
 10. 339.626177443 
 
 SUBTRACTION OF CIRCULATING DECIMALS. ART. 363. 
 
 1, 2. Given. 
 
 3. 391.5524. 
 
 4. 3.81824. 
 
 5. 4.789. 
 C. 400.915. 
 7. 3.9046. 
 
 8. 218.60. 
 
 9. 0.013640731. 
 30. 2451.386. 
 
PAGES 225 235.] 
 
 ANSWERS. 
 
 413 
 
 MULTIPLICATION OF CIRCULATING DECIMALS. ART. 3C I. 
 
 Ev ANS. 
 
 Ex. Ass 
 
 Ex. ANS. 
 
 1, 2. Given. 
 3. 0.082. 
 4. 1.8. 
 
 5. 380.185. 
 C. 778.14. 
 7. 750730.518. 
 
 S. 31.701. 
 0. 34008.4199003. 
 10. 2.297. 
 
 DIVISION OF CIRCULATING DECIMALS. ART. 3G5. 
 
 1, 2. Given. 
 
 3. 55. GO. 
 
 4. 5.41403. 
 
 5. 7.72. 
 
 C. 8574.3. 
 
 7. 3.50G493. 
 
 8. 3.145. 
 
 0. 62.323834196- 
 
 soi. 
 
 10. 1.4229240011- 
 85770750088. 
 
 ADDITION OF FEDERAL MONEY. ART. 
 
 1. Given. 
 2. 8205.04. 
 3. 581.128. 
 4. 500.50. 
 5. 1705.34. 
 G. 1431.50. 
 
 7. 3531.432. 
 8. 12200.524 
 0. 185.285. 
 10. 74.33. 
 11. 350.32. 
 12. 0401.05. 
 
 13. 8705.12. 
 14. 10080. 
 15. 378.383. 
 10. 300.100. 
 17. 250.213. 
 
 18. 1045.258 
 10. 821 10.17. 
 20. 71774.75. 
 21. 27800.74. 
 22. 81800.03. 
 
 SUBTRACTION OF FEDERAL MONEY. ART. 375. 
 
 1. Given. 
 2. 12.13. 
 3. 84.82. 
 4. 247.15. 
 5. 018.48. 
 
 0. 183.22. 
 7. 323.47. 
 8. 373.82. 
 0. 10870.75. 
 10. 1000.49. 
 
 11. $0047.788. 
 12. 011 10 304 
 13. 18.081. 
 14. 88.11. 
 15. 180.02. 
 
 10. 2.037. 
 17. 3 2. M50. 
 18. 10800.07. 
 10. 80080.90. 
 
 MULTIPLICATION OF FEDERAL MONEY. ARTS. 377, 378. 
 
 3. 83.00. 
 4. 517.025. 
 6. 30.50375. 
 7. 1440.75. 
 8. 40.50375. 
 
 0. 84.875. 
 10. 103.75. 
 1 I. 205.025. 
 12. 320.25. 
 13. 2.0025. 
 
 14. 2.84375. 
 15. 000.375. 
 10. 2.70. 
 17. 14.0025. 
 18. 15.78375. 
 
 19. 28.125. 
 20. 220.50. 
 21. 142.50. 
 22. 2331.875. 
 23. 14084. 12f 
 
 DIVISION OF FEDERAL MONEY. ARTS. 379-3I. 
 
 1. Given. 5. Given. 10. 543.518 + yds. 
 2. 4.50. 6. 8.207 coats. 11. 001.421+doz. 
 3. 0.00. 7. 7.871+ times. 12. 300 skeins. 
 4. $3.13. 9. 308.035+ gals. 13. $3.524 + . 
 
114 
 
 ANSWERS. [PAGES 230 248 
 
 DIVISION OF FEDERAL MONEY CONTINUED. ART. 31. 
 
 Ex. AN* 
 
 fi\. A MS. 
 
 Ex. Ass. 
 
 14. 81.50. 
 15. 80.25. 
 IG. 81.073 + . 
 17. 83.015+. 
 18. 80.084 + . 
 
 10. 80.04040 + . 
 '20. 80.02700 + . 
 21. 8J.78008 + . 
 22. 81.5435 + . 
 23. 1714.285+ bu. 
 
 24. 113.50377+ tons 
 25. 80.505238 + . 
 20. 245.517+ acres. 
 27. 500 cows. 
 28. 150 carriages. 
 
 APPLICATIONS OF FEDERAL MONEY. ARTS. 32-5. 
 
 1. Given. 
 
 13. $5885. 
 
 28. $0.0072. 
 
 2. 8800. 
 
 14. $10538.625. 
 
 20. $0.0064. 
 
 3. 8511.50. 
 
 15, 16. Given. 
 
 30. $13.4710 + 
 
 4. 8780. 
 
 17. 86.33375. 
 
 per cwt. 1 ; 
 
 5. 8780. 
 
 18. $104.55. 
 
 $0.134710 + 
 
 6, $1350. 
 
 10. $114.108. 
 
 per Ib. 
 
 7. $1020. 
 
 20. $50.5856. \3l. $12.88506 cwt. 
 
 8. $864.50. 
 
 21. $505.3775. 
 
 $0.1288506 Ib. 
 
 9. 82418. 
 
 22. $1001.75. 
 
 32. $120.625. 
 
 10. $4440. 
 
 23. $5.40625. 
 
 33. $208.838. 
 
 11. $1424.75. 
 
 24. $52.126. 
 
 34. $1734.875. 
 
 12. $2691.875. 
 
 25. $437.645. 
 
 35. $13703.78. 
 
 PERCENTAGE. ART. 388. 
 
 5, 6. Given. 
 
 7. $7.6875. 
 8. $8.7526. 
 9. $3.4608. 
 10. $8.7078. 
 11. $114.1070. 
 12. $10.50. 
 13. $210. 
 
 14. $43.13 rec'd. 
 $810.43 paid. 
 15. $402.05. 
 16. $134. 
 17. $32.625. 
 18. $34.03575. 
 19. $62.50. 
 20. $146.666+. 
 21. $8.771875. 
 
 22. 375 sheep. 
 23. $1568. 
 24. 187.5 lost ; 
 1312.5 saved. 
 25. $8.125. 
 26. $6.316. 
 27. $84.52016. 
 28. $250. 
 29. $750. 
 
 30. $90.4824. 
 31. $844.08. 
 32. $4724.775. 
 33. $1250. 
 34. $12000. 
 35. $21900, 1st; 
 $14600, 2d 
 36. $200. 
 37. $0.95. 
 
 APPLICATIONS OF PERCENTAGE. ARTS. 395-97. 
 
 1. Given. 
 2. $12.507. 
 3. $58878. 
 4. $73.159. 
 5. $115.203. 
 6. $615. 
 7. $583.842. 
 8. $52.834. 
 9. $155.875. 
 
 10. $619.887. 
 11. $44.32. 
 12. $673.75. 
 13. $57.29. 
 14. $415.831. 
 15. $106.831 A. 
 $2029.7080. 
 17. $21078.431. 
 18. $3439.613. 
 
 19. $761904.761. 
 20. $4126.55. 
 21. $1413.975. 
 22. $46.50. 
 23. $22.113. 
 24. $9.375. 
 25. $318.975. 
 28. $3692.50. 
 29. $2250. 
 
 30. $8840.70. 
 31. $7072. 
 32. $3552.50. 
 33. $1350 rec'd 
 $180 lost. 
 34. $7490.50. 
 35. $960. 
 36. $4427.50. 
 37. $9028.50, 
 
PAGES 251 270.] ANSWERS. 
 
 415 
 
 INTEREST. ART. 4O4. 
 
 Ex. ANS. 
 
 Kx. A S3 
 
 Kx ANS. 
 
 Ex A us 
 
 1. S'-29.B 1. 
 2. $43.255. 
 3. $40.367. 
 4. $51.20. 
 5. $60.263. 
 6. $44.414. 
 7. $194.58. 
 8. $17.803. 
 9. $28.206. 
 
 10. $8.103. 
 11. $6.853. 
 12. $19.14. 
 13. $60.27. 
 14. $89.40. 
 15. $958.41. 
 16. $657.45. 
 17. $1006.833. 
 18. $1585.018. 
 
 19. $889.44. 
 20. $1.135. 
 21. $1.409. 
 22. $1.898. 
 23. $102.125. 
 24. $154.216. 
 25. $704.083. 
 26. $2.975. 
 27. $76.131. 
 
 28. $3312.209. 
 29. $5278.162. 
 30. $16.158. 
 31. $206.718, at 
 360 days ; 
 $203.886, at 
 365 days. 
 32. $66778.64 
 
 SECOND METHOD. ARTS. 409413. 
 
 4. $8.50. 
 
 15. $82.078. 
 
 26. $307.65. 
 
 38. $137.288. 
 
 5. $1.065. 
 
 16. $39.179. 
 
 27. $227.994. 
 
 39. $481.016. 
 
 6. $70.151. 
 
 17. $320.833. 
 
 28. $8. 
 
 40. $391.062. 
 
 7. $97.28. 
 
 18. $9.8437. 
 
 29. $0.07. 
 
 41. $1531.25. 
 
 8. $30.78. 
 
 19. $85.207. 
 
 31. $15.60. 
 
 42. $3425.655. 
 
 9. $398.287. 
 
 20. $400. 
 
 32. $21.09. 
 
 43. $16320.528. 
 
 10. $1177.50. 
 
 21. $1638.442. 
 
 33. $1.272. 
 
 45. $2.145. 
 
 11. $1113.024. 
 
 22. $144. 
 
 34. $4.778. 
 
 46. $74.392. 
 
 12. $10.05. 
 
 23. $90. 
 
 35. $46.35. 
 
 47. $10.835. 
 
 13. $11.0025. 
 
 24. $12666.075. 
 
 36. $129.15. 
 
 48. $398.055. 
 
 14. $988.761. 
 
 25. $16360.996. 
 
 37. $168.552. 
 
 49. $14.532. 
 
 APPLICATIONS OF INTEREST. ARTS. 415410. 
 
 2. $5.25. 7. $36.08. 
 
 12. $6547.20. 
 
 20. 8, 18s. 9 d. 
 
 3. $3.15. 
 
 8. $91.085. 
 
 14. $499.034. 
 
 21. 12, 10s. 
 
 4. $17. 
 
 9. $107.854+. 
 
 15. $498.595. 
 
 22. 1898, 10s. 
 
 5. $60. 
 
 10. $533.867. 
 
 16. $4149.689. 
 
 4|d. 
 
 6. $45.014. 
 
 11. $25729.166+ 
 
 19. 19, 5s. 10d23. 2900. 
 
 PROBLEMS IN INTEREST. ARTS. 422-424. 
 
 1, 2. Given. 
 
 10. 5 per cent. 
 
 19. $30000. 
 
 28. 14 y. 3 mo. 
 
 3. 6 per cent. 
 
 11. 5" per cent. 
 
 20. $20833^. 
 
 13d. nearly. 
 
 4. 6 per cent. 
 
 13. $1800. 
 
 22. 4 years. 
 
 29. 14 y. 3 mo. 
 
 5. 8 per cent. 
 
 14. $5400. 
 
 23. 6 months. 
 
 13d. nearly. 
 
 6. 7 per cent 
 
 15. $10000. 
 
 24. 1 y. 3 mos. 
 
 30. 10 years. 
 
 7. 5$ per cent. 
 
 16. $8000. 
 
 1 d. nearly. 
 
 31. 8 y. 4 mo. 
 
 8. 7 per cent. 
 
 17. $14285.7143. 
 
 25. 1 y. 6 mo. 
 
 32 9 y. 6 mo. 8 d 
 
 9. 6 per cent. 
 
 18. $20000. 
 
 27. 16 y. 8 mo. S3. 28 years. 
 
 COMPOUND INTEREST. ARTS. 426, t27. 
 
 1, 2. Given. 
 
 5. $4590.09. 
 
 8. $1551.328. 
 
 11. $16035.675. 
 
 3. $507.213. 
 
 6. Given. 
 
 9. $877.506. 
 
 12 $149744. 
 
 4. $2177.426. 
 
 7. $1888.464. 
 
 10. $3491.395. 
 
 
416 
 
 ANSWERS. 
 
 P AG ES 272 - 295 
 
 DISCOUNT. ART. 43O. 
 
 Ex. A\s 
 
 Ex. ANS. 
 
 Kx. ANS. 
 
 Ex. AN* 
 
 '1, "2. Given. 
 ?,. $934.579-f. 
 4. $1488.087+. 
 
 5. $88.461+. 
 6. $83.52+. 
 7. $4729.064+. 
 
 8. $6208.955+. 
 9. $3404.347-f - 
 
 10. $Wo.2l8+. 
 
 11. $36.t'3t>. 
 
 BANK DISCOUNT. ARTS. 433, 431. 
 
 12, 13. Given. 
 
 20. $24.822. 
 
 27. $456.785. 
 
 34. $3821.883. 
 
 14. $14.18-25. 
 
 21. $48.3237. 
 
 28. $1126.523. 
 
 35. $4355.102. 
 
 15. $1(5.605. 
 
 22. $37.595. 
 
 29. Given. 
 
 36. $63717.884. 
 
 16. $26.98. 
 
 23. $43.694. 
 
 30. $414.507. 
 
 37. $10416.666. 
 
 17. $5.495. 
 
 24. $6381.59. 
 
 31. $966.101. 
 
 38. $51194.539. 
 
 18. $2034.1213. 
 
 25. $1495.625. 
 
 32. $1252.70. 
 
 39. $46638.655. 
 
 19. $2774.655. 
 
 26. $80. 
 
 33. $2514.247. 
 
 40. $8301.342. 
 
 INSURANCE. ARTS. 437442. 
 
 1. Given. 
 
 8. $1875. 
 
 16. 1 per cent. 
 
 25. $8365.482. 
 
 2. $20.70. 
 
 9. $487.50. 
 
 17. H per cent. 
 
 26. $13876.288. 
 
 3. $94.20. 
 
 10. $243.125. 
 
 19. $52000. 
 
 27. $27027.027 
 
 4. $63.75. 
 
 11. $192.78. 
 
 20. $65600. 
 
 29. $48.60. 
 
 6. $104. 
 
 12. $3375. 
 
 21. $65000. 
 
 30. $373.75. 
 
 6. $70.50. 
 
 14. 2i per cent. 
 
 22. $573331. 
 
 31. $1UOOO, ins. 
 
 7. $900. 
 
 15. 2 per cent. 
 
 23. $3416s. 
 
 $12250,prem 
 
 PROFIT AND LOSS. ARTS. 444447. 
 
 1-3. Given. 
 
 10, 11. Given. 
 
 19. 15| per cent 
 
 27. $2622.222. 
 
 4. $218. 
 
 12. $156.804. 
 
 20. 100 per cent. 
 
 28. $2-736. 
 
 5. $680. 
 
 13. $4238.50. 
 
 21. 20|H perct 
 
 .29. $13043.478. 
 
 6. $935.25. 
 
 14. $5926.85. 
 
 22. 2f per cent 
 
 30. $6317.391. 
 
 7. $1366.75. 
 
 15. $29504.875. 
 
 23, 24. Given. 
 
 31. $17806.122, 
 
 8. $68730.28. 
 
 17. 23,-^- per ct 
 
 25. $460.869. 
 
 32. $42654.028. 
 
 9. $12500 lost. 
 
 18. 4 per cent. 
 
 26. $205.882. 
 
 33. $42160. 
 
 DUTIES. ARTS. 451-453. 
 
 1. Given. 
 
 7. $3784. 
 
 13. $717.40. 
 
 19. $12642.40. 
 
 2. $370.80, 
 
 8. $345.744. 
 
 14. $492. 
 
 20. $2807.10. 
 
 3. $163.20. 
 
 9. $679.14. 
 
 15. $1051.71. 
 
 21. $11172.30 
 
 4. $1323. 
 
 10. $1882.406. 
 
 16. $715.75. 
 
 22. $17328.75. 
 
 5 $546. 
 
 11. Given. 
 
 17. $1230. 
 
 23. $15770.70. 
 
 6. $1235.22. 
 
 12. $248. 
 
 18. $15884.75. 
 
 
 ASSESSMENT OF TAXES. ARTS. 456, 457 . 
 
 1, 2. Given. 5. f of 1 per cent., or 7. $121. PL, ITs (ax. 
 
 3. $54.15, B's tax. . 8 mills on $1. 8. $283.68, C's tax. 
 
 4. $80.50, C's tax. 6. $80, A's tax. 10. $8854.166. 
 
PAGES 296 310.J ANSWERS. 
 
 417 
 
 ASSESSMENT OF TAXES CONTINUED. ARTS. 459, 460. 
 
 Ex ANS 
 
 Ex. Ass. | Kx. ANS 
 
 11. $ 161 25.654. 
 1-2. Si 7342. 105. 
 13. .$34051.815. 
 16. $73, G. A's. 
 17. $116, H. B's. 
 18. $451.50, W. C's. 
 19. $481 22, E. D's. 
 
 20. $314.50, J. F's. 
 21. $621.9J, T. G's. 
 22. $526.40, W. H's. 
 23. $263.30, L. J's. 
 24. $631.00, W. L's. 
 25. $196.90, J. K's. 
 26. $404.90, G. L's. 
 
 27. $370.50, F. M'a. 
 28. $458.20, C. P's. 
 29. $480.50, J. S'a. 
 30. $541, R. W's. 
 32. $13.36. 
 33. $3.45. 
 34. $13.40. 
 
 ANALYSIS. ARTS. 462 47O 
 
 1, 
 
 2. Given. 
 
 12. 
 
 $0.039-tV. 
 
 22. $7.98. 
 
 
 32. 
 
 Given 
 
 
 3. 
 
 $300. 
 
 13. 
 
 $04. 
 
 
 23. $6.03. 
 
 
 33. 
 
 300 Ibs. 
 
 4. 
 
 $320. 
 
 14. 
 
 $1080. 
 
 24. $160. 
 
 
 34. 
 
 1500 Ibs. 
 
 5. 
 
 $12.33!. 
 
 15. 
 
 $480. 
 
 
 25. $3.70?. 
 
 
 35. 
 
 95.2 cords. 
 
 6. 
 
 $10.50. 
 
 10. 
 
 $8. 
 
 
 26. $3430. 
 
 
 30. 
 
 100 pair. 
 
 7. 
 
 $1.08*. 
 
 17. 
 
 GO days. 
 
 27. $119.918. 
 
 
 38. 
 
 $450, 
 
 A s. 
 
 8. 
 
 $2640. 
 
 18. 
 
 29!f 
 
 mos. 
 
 28. $636.479. 
 
 
 
 $750, 
 
 B's. 
 
 9. 
 
 $24.80. 
 
 19. 
 
 1088 
 
 days. 
 
 29. Given. 
 
 
 39. 
 
 $450, 
 
 A's. 
 
 10. 
 
 $0.055. 
 
 20. 
 
 $0.50 
 
 . 
 
 30. 2 hours. 
 
 
 
 $000, 
 
 B's. 
 
 11. 
 
 $0.29|. 
 
 21. 
 
 $3. 
 
 
 31. 2! days. 
 
 
 
 $750, 
 
 C's. 
 
 40. 
 
 $763. 03-^-, 
 
 A's. 
 
 
 49. 
 
 40 tons. A's. 
 
 65. 
 
 188! 
 
 Ibs. at 8d. 
 
 
 $054.54-^1-, 
 
 B's. 
 
 
 
 80 tons, B's. 
 
 
 
 17! 
 
 Ibs. " 
 
 12d 
 
 
 $981.81-^, 
 
 C's. 
 
 
 
 120 tons, C's. 
 
 
 
 in 
 
 Ibs. 
 
 18d, 
 
 41. 
 
 $150. 95f ( f-f, 
 
 A's 
 
 
 50. 
 
 25 per cent. 
 
 
 
 17! 
 
 Ibs. " 
 
 22d, 
 
 
 $164.53fH 
 
 B's 
 
 
 51. 
 
 33! P er cent. 
 
 67. 
 
 9' horses. 
 
 
 $185.70/1^ 
 
 C's 
 
 
 
 $30000, loss. 
 
 68. 
 
 38| days. 
 
 
 $123.80 3 V 3 
 
 D's 
 
 
 53. 
 
 5s. per gal. 
 
 69. 
 
 278160 men: 
 
 42. 
 
 6G| cts. on $1. 
 
 54. 
 
 5fs. per Ib. 
 
 70. 
 
 15! 7 months 
 
 $260. GOf, 1st. 
 
 55. 
 
 9 cts. 
 
 per Ib. 
 
 71. 
 
 $54.60. 
 
 $333.33|, 2d. 
 
 50. 
 
 19! cts. per Ib. 
 
 72. 
 
 $459. 
 
 90, 
 
 
 $400.000, 3'd. 
 
 57. 
 
 91^ cts. per gal. 
 
 74. 
 
 $600. 
 
 43. 
 
 70 cts. on $1. 
 
 58- 
 
 60. Given. 
 
 75. 
 
 $3600 
 
 44. 
 
 25 per cent. 
 
 
 
 01. 
 
 1 part 
 
 16 car. 
 
 76. 
 
 $030. 
 
 45. 
 
 $2990.00, A's. 
 
 
 1 " 
 
 18 car. 
 
 77. 
 
 72. 
 
 
 $4197.50, B 
 
 's. 
 
 
 
 2! " 
 
 23 car. 
 
 78. 
 
 300. 
 
 $4312.50, C's. 
 
 
 r " 
 
 24 car. 
 
 79. 
 
 120. 
 
 46 
 
 OGf per cent. 
 
 03. 
 
 lOOgals. at80cts. 
 
 80. 
 
 240. 
 
 47 
 
 37! P er cent. 
 
 
 40 
 
 " 30cts. 
 
 81. 
 
 G 
 
 8!f 
 
 
 
 48 
 
 10 per cent. 
 
 
 
 
 40 
 
 " 40cls. 
 
 85. 
 
 $230. 
 
418 
 
 ANSWERS. [PAGES 311 328 
 
 ANALYSIS CONTINUED. ART. 471. 
 
 Ex. ANS. 
 
 Ex. ANS. 
 
 Ex. ANS. 
 
 Ex. APS. 
 
 86. $1170. 
 
 100. 266. 
 
 114. $288. 
 
 127. $140. 
 
 87. $900. 
 
 101. 131| 
 
 115. $43|. 
 
 128. $1560. 
 
 88. $1125. 
 
 102. 353. 
 
 116. $814. 
 
 129. $180. 
 
 89. 1367.50 
 
 105. $2501. 
 
 117. $640. 
 
 130. $630. 
 
 90. $442. 
 
 106. $231. 
 
 118. $3000. 
 
 131. $180. 
 
 91. $201. 
 
 107. S119-&. 
 
 119. $200. 
 
 132. $1281. 
 
 92. $350. 
 
 108. $186. 
 
 121. $625. 
 
 133. $800. 
 
 93. $240. 
 
 109. $280. i!22. $480. 
 
 134. $12.60. 
 
 94. $754. 
 
 110. $1170. ;123. $808. 
 
 135. $45. 
 
 95. $1080. 
 
 111. Given. 
 
 124. $420. 
 
 136. $45. 
 
 96. $630. 
 
 112. $378. 
 
 125. $690. 
 
 137. $90. 
 
 <*9. 206$. 
 
 113. $810. 
 
 126. $877. 
 
 138. $150. 
 
 RATIO. ARTS. 48O 48. 
 
 1 2. Given. 
 
 14. 8-H. 
 
 26. 120. 
 
 40. 45 to 72. 
 
 3. 2. 
 
 15- i. 
 
 27. 60. 
 
 41. Equal. 
 
 4. 4. 
 
 16. i. 
 
 28. -fr. 
 
 42. 936 to 56C. 
 
 5. 9. 
 
 17. *. 
 
 29. V\i> 
 
 43. G. inequality 
 
 6. 6. 
 7. 6. 
 
 18. i. 
 19. 4. 
 
 30. 1J. 
 31. i. 
 
 44. L. inequality 
 45. Equality. 
 
 8. 8. 
 
 20. |. 
 
 32. 240. 
 
 46. 60 : 12=5. 
 
 9. 9. 
 
 21. 3. 
 
 35. 8; f 
 
 47. 1. 
 
 10. 9. 
 
 22. 7. 
 
 36. 4; 8. 
 
 48. f. 
 
 11. 9. 
 
 23. 112 avoir. 
 
 37. 44 \. 
 
 49. H. 
 
 12. 9. 
 
 24. 4. 
 
 38. *; 9. 
 
 50. ttfr- 
 
 13. 4 
 
 25. 6. 
 
 39. 72 to 8. 
 
 51. f. 
 
 SIMPLE PROPORTION. ARTS. 5O2 5O6. 
 
 1. 12. 
 
 17. 51ilbs. 
 
 35. 1925}lbs.cop.j47. 480. 
 
 2. 3. 
 
 18. $1640.64. 
 
 641 |lbs. tin. 
 
 48. 375 sheep. 
 
 3. 16. 
 
 19. $7066.40. 
 
 36. 1520 Ibs.n. 
 
 49. 20 days. 
 
 4. 3. 
 
 22. 3 far. 
 
 280 Ibs. c. - 
 
 50. 400 rods. 
 
 5. Given, 
 
 23. $792. 
 
 200 Ibs. sul. 
 
 51. 8-f weeks. 
 
 6. 20. 
 
 24. $2768. 
 
 37 980.5155 Ibs. 
 
 52. 1, 3s. 6d 
 
 7. 55*. 
 
 25. 435 miles. 
 
 38. $1350. 
 
 1 iV for, 1st 
 
 8 120. 
 
 26. 252 days. 
 
 39. 45. 
 
 1, Is. 2d 
 
 9, 10. Given 
 
 28. $2.70. 
 
 40. $3375. 
 
 -ft- far. 2nd. 
 
 11. $903. 
 
 29. 3s. 3d. 2fq. 
 
 41. $2562.50. 
 
 0, 18.' 9d. 
 
 12. $1309.50 
 
 30. $3.15. 
 
 42. $16480. 
 
 3r\fai.3rd. 
 
 13. $-225. 
 
 31. $8.555. 
 
 43. 70400 times. 
 
 0, 16s. 5d. 
 
 14. 775 miles. 
 
 32. $26.40. 
 
 44. 57600 imes. 
 
 2U far. 4th. 
 
 15. 20 tons. 
 
 34. 30 bu. oats; 
 
 45. 170. 
 
 53. 888 4 oz. ox. 
 
 16. 2156 Ibs. 
 
 70 bu. corn. 
 
 46. 200 -1 llli oz. hy. 
 
PAGES 329 355.] ANSWERS. 
 
 419 
 
 COMPOUND PROPORTION. ARTS. 5O 51 1. 
 
 Ex. AN*. 
 
 Ex. ANS. 
 
 Ex. ANS. 
 
 Ex. ANS. 
 
 3. 10 horses. 
 4. 19*- days. 
 5 1314 gals. 
 6 27 laborers. 
 
 9. 24 days. 
 10. 144 days. 
 11. 1125 miles. 
 12. $225. 
 
 13. $140. 
 14. $768. 
 15. $600. 
 16. 32 days. 
 
 17, 18. Given. 
 19. 56 yds. Can. 
 20. 127 b. N. O. 
 21. 16 rupees. 
 
 1, 2. Given. 
 
 3 28 sq. ft. 6' 10". 
 
 4. 59 cu. ft. 3' 8". 
 
 5. 268 cu. ft. 6' 11" 
 
 6. 235 sq. ft. 
 
 7. 734 sq. ft. 0' 9". 
 
 DUODECIMALS. ART. 516. 
 
 8. 105ft. 5' 4" 5'" 5"" 1 11. 195 ft. 4' 1" 3" 3" 
 
 4 6""". 
 
 9. 154 ft. 3' 1" 5'" 4"" 1 12. 23 C. Ill ft. 3' 
 
 6'"" 8""". !13. 3840 ft. 0' 5". 
 
 10. 85ft. I'll" 0'" 5"" 14. $15.819-|-. 
 
 2'"" 6""". 
 
 15. 33750 bricks 
 
 EQUATION OF PAYMENTS. ART. 521. 
 3. 6 months. * | 4. 6 months. | 5. 3 years. | 6. 62 days. 
 
 PARTNERSHIP. ART. 
 
 1. Given. 
 
 2. $240, A's gain. 
 $320, B's gain. 
 $400, C's gain. 
 
 3. $274.21 f^, A's. 
 &373.40W, B's. 
 $212.37-Hi, C's. 
 
 4. $1178.947, A's. 
 
 $2259.649. B's. 
 
 $3340.351, C's. 
 
 $4421.053, D's. 
 5. $850, A's. 
 
 $800, B's. 
 
 $700, C's. 
 
 $650, D's. 
 7. $1655.172, X's. 
 
 523. 
 
 $1448.276, Y's. 
 $1396.552, Z's. 
 
 8. $22.486, A's. 
 $21.024, B's. 
 $16.490, C's. 
 
 9. $3492.06, A's. 
 $4761.91, B's. 
 $6746.03, C's. 
 
 EXCHANGE OF CURRENCIES. ARTS. 533537. 
 
 3. $4116.42. 
 
 4. $850.63. 
 
 5. $414.667. 
 
 6. $969.815. 
 
 7. $2041.59-)-. 
 
 8. $4841.089-|-. 
 
 9. $7746.082+. 
 
 10. $60652.55-}-. 
 
 11. $208683.8 19+ 
 32 $330661.6054-. 
 
 13. $242840.3694- 
 
 14. $257791.3971-. 
 
 2. $4791.60. 
 
 3 $25391.084+. 
 
 4 $284.58. 
 
 15. $369716.864+. 
 
 16. $284412.622+. 
 
 17. $4840000. 
 
 19. 82. 
 
 20. 90. 
 
 21. 181. lOid. 
 
 22. 261. 8s. 7d. 
 
 23. 446, 7s. 8i,d. 
 
 24. 201, 11s. 7 |d. 
 
 25. 883, 5s. 8Jd. 
 
 26. 1095, 3s. ll|d. 
 
 27. 5220, 9|d. 
 
 28. 8568, 3s. 7}d. 
 
 29. 10384, 18s. 4d. 
 
 30. 20661, 3s. lid. 
 
 32. 135. 
 
 33. 227. 
 
 34. 315, 9d. 
 
 35. 375. 
 
 37. $534.166. 
 
 38. $614.1875. 
 
 39. $986.083. 
 
 40. $7714.285. 
 
 41. $20000. 
 
 EXCHANGE. ART. 548. 
 
 5. $10152-.527+. 
 
 6. $707. 
 
 7. $1881.60. 
 
 8. $15418.509. 
 
 9. $20665.20. 
 10. $36.480.755. 
 
420 
 
 ANSWERS. [PAGES 356 378 
 
 ARBITRATION OF EXCHANGE. ART. 519. 
 
 Ex. Ana. 
 
 Ex. 
 
 ANS. 
 
 Ex. ANS. 
 
 1. 21 florins. 
 
 2. $ 
 
 45 gain. 
 
 3. 180 milrees, circu 
 
 ALLIGATION. ARTS. 552556. 
 
 2. $0.87i. 
 
 3. 5s. 4d. l- qr. 
 
 5 3 grs. at 1 8 car. fine. 
 1 gr. " 20 " 
 1 gr. " 22 " 
 & grs." 24 " 
 
 1. 10 oz. 16 car. fine. 
 5 oz. 18 
 5 oz. 22 " 
 
 8. 133 Ibs. at 20 cts. 
 95 Ibs. at 30 cts. 
 190 Ibs. at 54 cts. 
 
 10. 40 gals, at 15s. 
 40 gals, at 17s. 
 40 gals, at 18s. 
 200 gals, at 22 
 28 gals, water ; 
 98 gals. wine. 
 
 11 
 
 INVOLUTION. ART. 562. 
 
 17. 3125. 
 
 22. 6.25. 
 
 27. VVW- 
 
 18. 279936. 
 
 23. .000001728. 
 
 28. -rfjj-JH 
 
 19. 117649. 
 
 24. .0000015625. 
 
 29. 20}. 
 
 20. 65536. 
 
 25. t-. 
 
 30. 54ff-. 
 
 21. 387420489. 
 
 26. T^. 
 
 31. 1480-f 
 
 IS. Given. 
 13. 15129, 
 14 2460375. 
 
 15. 8294400. 
 
 16. 10000. 
 
 3. 51. 
 
 4. 73. 
 
 5. 28. 
 
 6. 9.327-J-. 
 
 7. 69. 
 
 8. 84. 
 
 9. 99. 
 
 10. 167. 
 
 11. 31. 
 
 APPLICATIONS OF THE SQUARE ROOT. ARTS. 581-585. 
 
 SQUARE ROOT. ARTS. 574, 575. 
 
 12. 9.848+. 
 
 21. 792. 
 
 30. 186.9951-f 
 
 13. 2.6457+. 
 
 22. 1.7810+ 
 
 31. 12345. 
 
 J4. 13.78404+. 
 
 23. 3216. 
 
 32. 345761. 
 
 15. 209. 
 
 24. $-. 
 
 33. 31.05671. 
 
 16. 217. 
 
 25. f-V. 
 
 34. 19.104973174 
 
 17. 23.8. 
 
 26. .79056+. 
 
 35. 1.41421356- 
 
 18. 2.71. 
 
 27. 4.1 683+. 
 
 237 + 
 
 19. .9044+. 
 
 28. 28.181. 
 
 36. 1.732050807 
 
 20. 34.2. 
 
 29. 14.4116+. 
 
 5688772. 
 
 1. Given. 
 2. 32 feet. 
 3. 166.709+m. 
 4. 240rds. side. 
 339.41 12 r.d. 
 5. Given. 
 6. 10. 
 
 7. 18. 
 8. 36. 
 9. 40. 
 10. 66. 
 11. 168. 
 12. 11.2. 
 13. 67.5. 
 
 14. ^ 
 
 15. n- 
 
 16. -ft**- 
 
 17. -Mr- 
 
 18. 63 rods. 
 
 19. 160 rods. 
 
 20. 320 rods. 
 
 21. 480, length; 
 160, breadth. 
 148 in rank; 
 
 74 in file. 
 25 and 40. 
 18 and 47. 
 
 22. 
 
 EXTRACTION OF THE CUBE ROOT. ARTS. 590-92. 
 
 4 45. 
 
 5 52. 
 
 6. 83. 
 
 7. 136. 
 
 8. 217. 
 9 22.6 
 
 10. 2.74. 
 
 11. 0.623. 
 12. 3.332222-4-. 
 13. 1.817121+. 
 14. 7. 21 7652+. 
 15. 8.315517+. x 
 16. Jf-. 
 17. -U. 
 
 18. 3.5463+, 
 19 3|. 
 20. 1.25992104. 
 21. .64:5(55958974. 
 22. 68 ft. 
 24. 3.1748+yds. 
 25. 4 Ibs. 
 
 27. 24 and 72. 
 28. 128 and 256. 
 29. 60 and cOO. 
 30. 160 and 640, 
 31. 426 and 2556. 
 32. 74 7 and 6723. 
 
f AGES 379393.] 
 
 A N S WE RS. 
 
 421 
 
 ROOTS OF HIGHER ORDERS. ARTS. 593-i>5. 
 
 E. ANrf. 
 
 Ex. Ass. 
 
 Ex. AN*. 
 
 Ex. Ass. 
 
 Cx. 
 
 ANS. 
 
 2. 2. 
 3, 16. 
 4 378. 
 
 5. 6. 
 6. 26. 
 7. 5. 
 
 8. 7. 
 
 9. 3. 
 10. 2. 
 
 12. 2.4872+. 
 13. 41 4.5+. 
 15. 1.104089. 
 
 16. 
 17. 
 18. 
 
 1.080059. 
 1.004074. 
 1.047128. 
 
 ARITHMETICAL PROGRESSION. ARTS. 6O3 60S. 
 
 1. Given. 
 
 2. 5050. 
 
 3. 78 strokes. 
 
 4. Given. 
 
 5. 33. 
 7. 44. 
 
 9. 3f. 
 
 11. 33}. 
 
 12. 502. 
 
 13. 14, 21. & 28 
 
 14. 15,29,43,57, 
 
 71, & 85, 
 
 GEOMETRICAL PROGRESSION. ARTS, G1O-12. 
 
 2. 4. 
 
 3. 4374. 
 
 4. 13671875. 
 
 5. $2048. 
 
 6. $334.5563944, amt. 
 
 of $250. 
 
 $750.3651759245, 
 amt. of $500. 
 
 $1628.894614622- 
 37890625, amt. 
 of $1000. 
 
 8. 1023 
 
 9. 43774}. 
 
 10. $111111111.111, 
 12. li. 
 14. 3. 
 
 1, 2. Given. 
 3, $826.992. 
 
 ANNUITIES. ARTS. 614, 615. 
 
 4. $2298.262. I 6. 36785.59. I 8. $1333.333 
 
 5. $4835.74. 7. Given. 9. Gi/-?n. 
 
 PERMUTATIONS AND COMBINATIONS. ARTS. 61 , 619. 
 
 2. 40320 ways. | 4. 3628800 ways. | 7. 15120 numbers. 
 
 3. 362880 ways. 
 
 5. 479001600 days. | 8. 165765600 words. 
 
 MENSURATION OF SURFACES. ARTS. 622631. 
 
 1. 270 acres. 
 
 2. 72:2 acres. 
 
 3. 3H acres. 
 
 4. 320 rods, or 1 m. 
 6. 360 sq. ft. 
 
 6. 435 sq. ft. 
 
 7. 1100 sq.,ft. 
 
 9. 290.4737 sq. ft. 
 
 10. 4 A. 52.82 rods. 
 
 11. 62.8318 ft. 
 
 12. 141.37155 rods. 
 
 13. 100 ft. 
 
 15. 12 A. 43.49375 r. 
 
 16. 31415.9 sq. ft. 
 
 17. 2 ft. 9.94 in. 
 
 18. 17.3205 ft. 
 
 MENSURATION OF SOLIDS. ARTS. 633647. 
 
 1. 1364 cu. ft. 
 2. 3154 ft. 11' 6" 8"'. 
 3. 2615 cu.ft. 1080 in. 
 4. 115 ft. 114.368 in. 
 5 53333 J cu. ft. 
 C. 8835.75. cu. ft. 
 7. 900 sq. ft. 
 8. 1739 sq. ft. 
 9. 76 cu. ft. 
 10. 176 sq. ft. 
 
 11. 2748.891 25 cu.ft. 
 12. 11 9366.25 cu. ft. 
 13. 78 vds. 4 ft. 123- 
 .11 28 in. 
 14. 7 sq. ft. 9.87516 in. 
 15. 14684558.20796 
 sq. miles. 
 16. 1767.14437 cu. in. 
 17. 5291335807.60158 
 cu. in. 
 
 18. 13 sq. ft. 
 19. 4 cu. ft. 
 20. 624 cu. ft. 
 21. 220 gals. 3qts. 1 pt 
 
 22. 451 gals. 2 qta. 
 0.729344 pt. 
 23. 31.71f26-f tons. 
 24. 967.105214 ton*. 
 
422 
 
 ANSWERS. [PAGES 394 398 
 
 MECHANICAL POWERS. ARTS. 618655. 
 
 Ex. ANS. 
 
 Ex. ANS. 
 
 Ex. ANS. 
 
 Ex. ANS. 
 
 1. 500 Ibs. 
 2 133^ Ibs. 
 *. 96 Ibs. A.; 
 
 160 Ibs. B. 
 4. 4 ft. from A. ; 
 8 ft. from B. 
 
 5. 600 Ibs. 
 6. 10661 Ibs. 
 7. 1600 Ibs. 
 
 8. 1250 Ibs. 
 9. 11 36.3636 Ib. 
 10. 904777.92 Ib 
 
 MISCELLANEOUS EXAMPLES. 
 
 459 less. 
 
 34. 136 g. 1 q. 
 
 $440, C's g. 
 
 79, 247170562- 
 
 521 greater 
 
 35. $180. 
 
 $700, B's s. 
 
 2710s. m. 
 
 70. 
 
 36. $10.875. 
 
 $1100,C'ss. 
 
 80. 33600914- 
 
 5^r. 
 
 7. "$156.615. 
 
 59. 20 per cent. 
 
 2264006.2- 
 
 Nfc 
 
 38. 94 d. 3 h. 
 
 60. $1371. 
 
 3104 c. m. 
 
 20 days. 
 
 38m. 10-Hs. 
 
 61. $4755.141. 
 
 81. 5890.5 Ibs. 
 
 $61.32. 
 
 39. $2. 
 
 62. $32000. 
 
 82. 585.80357b 
 
 $16581.65. 
 
 40. 1. 
 
 63. $360. 
 
 83. 39.401 hhd. 
 
 $18.60. 
 
 41. *|f. 
 
 64. 36 days. 
 
 84. 7ift. 
 
 $1843.003. 
 
 42. $4800 
 
 65. 90 hours. 
 
 85. 403291461- 
 
 $24390.243 
 
 43. $197.759. 
 
 66. 51, A's. 
 
 126605635- 
 
 $4.50. 
 
 44. 228 gals. 
 
 34. B's 
 
 584000000. 
 
 $6.875. 
 
 45. $40.29$. 
 
 68, C's 
 
 86. 31m. 180 r 
 
 33i per ct. 
 
 46. $41.095. 
 
 102, D's. 
 
 87. 662^. 
 
 $36. 
 
 47. 2y. 182-J-d. 
 
 67. 160, A's. 
 
 88. $4294967- 
 
 $229|f. 
 
 48. 5-fr min. 
 
 224, B's. 
 
 .295. 
 
 4987i|E. 
 
 49. 120 days 
 
 256, C's. 
 
 89. 5 bags, A. 
 
 2000 miles. 
 
 50. 120 schol. 
 
 205, A's. 
 
 7 bags, M. 
 
 240*0 times. 
 
 51. 292. 
 
 287, B's. 
 
 90. 1440. 
 
 2880 times. 
 
 52. $6000. 
 
 328, C's. 
 
 91. $230, B's. 
 
 $1.50perg. 
 
 53. 600. 
 
 68. $520, D's. 
 
 $325, C's. 
 
 2i| cts. 
 
 54. 5600 Ibs. t. 
 
 $280, A's. 
 
 $445, A's. 
 
 $2400. 
 
 750 Ibs. 1. 
 
 $360, B's. 
 
 92. 5 o'clock, 
 
 437| bbls. 
 
 300 Ibs. b. 
 
 69. 20. 
 
 20 min. 
 
 21 months. 
 
 55. 254| miles. 
 
 70. 25 persons. 
 
 93. lOffi d. all 
 
 $1.328. 
 
 56. 78f Ibs. 
 
 71. 40 and 80. 
 
 47fi d. A 
 
 40 yds. 
 
 Il7f Ibs. 
 
 72. 75 and 128. 
 
 38ff- d. B. 
 
 Is. 3 gV q r s. 
 
 57. $192.307 1 a 3 - 
 
 73. 56.5685 ft. 
 
 27-ftVd.C. 
 
 37ffHfd< 
 
 A's gain. 
 
 74. 7200 rods. 
 
 111^ d. D. 
 
 34*W*d. 
 
 $2307.692 
 
 75. 3.535519ft. 
 
 94. 36 days. 
 
 $1.60.. 
 
 i^B'sg. 
 
 76. 677.-73475f. 
 
 95. 12 o'clock. 
 
 12 miles. 
 
 $2500.000, 
 
 77. 7.13645 r. 
 
 32-^ min. 
 
 12f days. 
 
 C'sg. 
 
 78. 50 A. 3 R. 
 
 96. 128tyrs. 
 
 52 days. 
 
 58. $240,A'sg. 
 
 28.7399+r. 
 
 97. $407. 
 




 r 
 

,'