SI, STEM 
 
 A K THMETIC, 
 
 
 THE P1UACIPLE OF CANCKLLNC 
 
 
 
77; 
 
 
 
 IN MEMORIAM 
 FLOR1AN CAJORI 
 
RECOMMENDATIONS. 
 
 From Prof. E. A, Andrews. 
 
 About two years ago, while on a visit to this place, I had the pleasure of meeting 
 with Mr. Tracy, and of having some conversation with him on the subject, of an 
 Arithmetic, which he was preparing for schools. I was particularly pleased with 
 that part of his system which related to canceling, and which appeared to me to poa- * t * . i 
 eess great practical value. Within a few days, Mr. Tracy has put into my hands a 1\ I 
 part of his manuscript, that 1 might become more minutely acquainted with his sys- 
 tem. My time has been so far occupied with my own business, that I have been able 
 to examine a part only of the manuscript put into my hands ; but with this part, I 
 have been much gratified. It appears to me, that when carefully and thoroughly re- . 
 vised and perfected, as the author designs to do, it may become a most valuable work, 
 inferior to none of the Arithmetics now used in our schools. Such is my confidence in ' <> 
 the ability of the author, to complete and polish the work, that I look upon its success 
 as quite certain. 
 
 New Britain, Conn., Aug. 1st, 1839. 
 
 From E, H. Burritt, author of the Geography of the Heavens, S-c. 
 
 Through the politeness of Mr. Tracy, I have been favored with a perusal of an 
 Arithmetic, in manuscript, which he is preparing for publication. The work is in- 
 tended as a universal class book in elementary Arithmetic. It is the production of 
 a gentleman of known abilities and experience in teaching, and he has, with great 
 care, arranged its several parts, and given the rules, and selected the examples, step 
 by step, in that natural order, and easy method, which his own judgment and expe- 
 rience approved. There are some excellencies in his Arithmetic some facilities of 
 dealing with figures, which, so far as I know, ai-een^rely pecujjyr fn^this) trfiaHse r 91 ft 
 and which distinguish it from all others. On tfn^round~especially, and that of its " ' 
 "general nTerlt, rtKfnTTfa "work* which will commend itself to the attention of 
 teachers. 
 
 New Britain, August, 1836. 
 
 I entirely coincide in the above opinion, having been particularly gratified with the 
 ease and facility with which many difficult operations are performed by the new 
 principle introduced by the author. 
 
 J. P. BRACE, Principal of Hartford Female Seminary. 
 
 From Edward Strong, Principal of Bacon Academy, Colchester, Conn. 
 
 I have had the pleasure of examining a system of Arithmetic, by Mr. Tracy, of 
 Norwich Academy. Without speaking of its merits in other respects, which I am 
 unable to do, from want of time to give a thorough perusal, I discover in it a distinc- 
 tive feature the method of canceling which appears to me to be an important im- || I* 
 provementon every other system with which I am acquainted. Should the other fea- " 
 lures of the work correspond with what may reasonably be expected from its author, 
 I should regard it as a very important improvement upon our other systems of Arith- 
 metic. 
 
 Colchester, Oct. 8th, 1839. 
 
 From Rev. A. Bond, Pastor of the Second Congregational Church, in Norwich . 
 
 Having examined the general plan of an Arithmetic, prepared by Mr. Tracy, Prin- 
 cipal of Norwich Academy, lean cheerfully recommend it as a system possessing, in 
 some important particulars, a superiority over any other system with which I am 
 acquainted. The method of canceling, which is carried through the work, excepting 
 the Roots, greatly facilitates the process of Arithmetical calculations, and will give it 
 a decided advantage in the estimation of businessmen. The part on foreign exchanges 
 will enhance its value with the commercial community. While its simplicity adapts 
 it to the use of common schools, iis comprehensiveness, and the ease and accuracy 
 with which complicated problems may be solved- will be likely to secure for it a 
 prominent place in the counting room. 
 
 Norwich, August 1th, 1839. 
 
2 RECOMMENDATIONS. 
 
 From La Fayette S. Foster, Esq. 
 
 Mr. Calvin Tracy, of this city, has submitted to my examination, in manuscript, 
 an Arithmetic, prepared by himself, for publication. From the known ability of Mr. 
 Tracy, as an instructor, I was prepared to entertain a high opinion of any treatise 
 designed to facilitate the acquisition of knowledge, of which he might be the author, 
 and from the attention which [ have bestowed on his Arithmetic, I have no hesitation 
 in bearing testimony to its high meritorious character. His plan appears to me to be 
 highly judicious, and ably and skillfully executed. The work, in my opinion, will 
 be a valuable addition to a very important branch of education. 
 
 Norwich, Dec . 24th, } 839. 
 
 , 7 j From Messrs. J. H. Gallup and G. Bushnel, Teachers of the Eclectic School, 
 
 Norwich, Conn. 
 
 Having examined Mr. Tracy's system of Arithmetic, we think it well calculated to 
 answer the purposes for which it is intended. Mr. Tracy has illustrated, and happily 
 combined with nearly all his operations, a method of abbreviating Arithmetical cal- 
 ff r dilations, which, so far as we know, has never before been published in any com- 
 fy j mon school Arithmetic. This we deem an essential improvement, and are of opinion, 
 I that the author has rendered, in this work, an essential service to the cause of edu- 
 cation. 
 Norwich City, Oct. 22<Z, 1839. 
 
 From Rev. L. N. Tracy, formerly Principal of New Britain Academy, 
 I have spent considerable time in a careful examination of an Arithmetic prepared 
 by Mr. C. Tracy, Principal of Norwich Academy. For my own benefit and pleasure, 
 1 have carefully examined every rule, and though I have daily used the best Arilhmetics 
 extant, while engaged for many years in teaching, I am led to believe that there is not a 
 textbook on Arithmetic in use which presents equal excellencies. Its grand feature 
 that which distinguishes it from every other Arithmetical treatise is thcprinciple of 
 canceling, introduced and applied throughout the work. The extent and facility of 
 *its application to all operations in which Multiplication and Division are both con- 
 cerned, are fully and clearly illustrated. It is safe to say that two thirds, and often 
 four fifths of the labor and time usually required for arithmetical solutions, is saved. 
 While it contains an amount of matter equal to any other Arithmetic in use, it is still 
 a strictly elementary work. 
 Norwich, Oct. [2th, 1839. 
 
 From E. C. Herrick, Esq. 
 
 I have cursorily examined the manuscript of Mr. C. Tracy's treatise on Arithmetic. 
 The most prominent feature of the work is the introduction of a peculiar mode of 
 stating numerous classes of problems, which are then solved by an abridged process, 
 called canceling. This appears to me an important improvement on the books in 
 common use, and one which renders the publication of this treatise very desirable. 
 
 New Haven, Dec. 27th, 1839. % 
 
 From J. H. Rogers, Esq., Principal of Prospect Hill High School. 
 Messrs. DURRIE & PECK, 
 
 Gentlemen From a hasty examination of Tracy's Arithmetic, I believe it worthy 
 of being ranked among the best School Books. The method of canceling, very fully 
 brought into practice in this work, greatly abridges many operations ; and may be 
 mentioned as one of its most valuable features. 
 
 Sincerely yours, J. H. ROGERS. 
 
 East Haven, April 24th, 1840. 
 
 At a meeting of School Visitors of the First School Society of New Haven, held May 
 2, 1840 
 
 The committee appointed at the previous meeting to examine and report on "A 
 New System of Arithmetic," by Mr. C.Tracy, reported, that in their opinion the 
 work contains important improvements on the arithmetical treatises in common use, 
 and recommend that it be introduced into the schools of this Society: whereupon, 
 it was 
 
 footed, That Tracy's New System of Arithmetic be adopted for use in the schools of 
 the First School Society of New Haven. 
 
 R. S. HINMAN, Chairman. 
 
 E. C. HERRICK, Clerk pro tempore. 
 
NEW SYSTEM 
 
 OF 
 
 ARITHMETIC, 
 
 IN WHICH IS EXPLAINED AND APPLIED TO PRACTICAL PURPOSES, IN ADDITION 
 TO THE ORDINARY RULES OF OPERATION, 
 
 THE PRINCIPLE OF CANCELING; 
 
 BEING AN 
 
 ABBREVIATED MODE 
 
 OF 
 
 ARITHMETICAL SOLUTION. 
 
 DESIGNED FOR SCHOOLS AND ACADEMIES. 
 
 BY C. TRACY, A. M 
 
 PRINCIPAL OF NORWICH ACADEMY. 
 
 NEW HAVEN: 
 
 PUBLISHED BY DURRIE & PECK. 
 
 NEW YORK: COLLINS, KEESE, & co. 
 
 BOSTON : CROCKER & BREWSTER. 
 
 1840. 
 
ENTERED, 
 According to Act of Congress, in the year 1840, by 
 
 DURRIE & PECK, 
 
 In the Clerk's Office of the District Court of Connecticut District. 
 
 HITCHCOCK AND STAFFORD, PRINTERS, 
 NEW HAVKN. 
 
'& 
 
 INTRODUCTION. 
 
 It will readily be conceded, that all efforts in behalf of the general 
 diffusion of useful knowledge, are in themselves commendable. There 
 is, however, and probably ever will be, a difference of opinion rela- 
 tive to the extent to which books of any particular description, and 
 treating upon the same general topic, may be multiplied, and the inter- 
 ests of education uniformly advanced thereby. 
 
 This difference of opinion exists especially in relation to books de- 
 signed for the use of common schools and academies, and which treat 
 upon the more common subjects of study. The multiplication of 
 books of this description, to the extent realized at the present day, is 
 regarded by many as injurious to the general good. That its tendency 
 is to increase, in some small degree, the expense of education, at least 
 in some parts of the country, will not be denied. But before sentence 
 of final condemnation is pronounced, it always becomes those, who sit 
 as umpires, to take as extended views of the subject before them, as the 
 nature of the case will admit. 
 
 The question now presented is, how far the general good is advanced 
 by the multiplication of school books. 
 
 To answer this, let it be supposed that only a single work in each 
 department of science studied in common schools, had ever been pre- 
 sented to the public, aiXl that each work were such as it should be. 
 Books of this description would obviously find ample circulation, suf- 
 ficient, perhaps, to satisfy both authors and publishers, without em- 
 bracing one half of the ground to be occupied. The consequence 
 would be, that the more recently settled parts of our country would be 
 but poorly supplied with the means of education, for at least some con- 
 siderable period of time. But as it now is, with such a multiplicity of 
 school books constantly emanating from the press, a spirit of rivalry is 
 created, a desire is excited on the part of both authors and publishers to 
 give to their several works a more extended circulation than can be 
 obtained without exploring the whole ground. As a natural conse- 
 quence, the inhabitant of the less favored portion of our land, is scarcely 
 settled in his log cabin, before books of every description necessary for 
 the education of his sons and daughters, are presented him, as it were, 
 at his own door. His attention is thus directed to a subject second in 
 importance to none of a temporal nature ; and one which, when duly 
 presented, will be likely to be regarded, and to receive a consideration, 
 which otherwise might be long neglected. 
 1* 
 
 e 1 1 
 - t 
 ' 
 
 r 
 , j 
 
INTRODUCTION. 
 
 The truth of our supposition, that any one set of school books is 
 such in all respects as is required, may, however, very reasonably be 
 doubted. Many of them are unquestionably of a very high order, 
 and very probably owe some degree of their merit to the fact, that 
 other minds have been, are, and yet will be, traversing the same 
 ground which their authors trod, and are preparing other works, to 
 supersede them, if possible, in the estimation of the public. t 
 
 ^ The effect of the multiplication of school books is, therefore, to ren- 
 ! der the means of education as perfect as the nature of their subjects will 
 , allow, and to convey these means, thus perfected, to every part of our \ 
 * entire country. 
 
 From the preceding considerations, the author is inclined to regard 
 the multiplication of school books as favorable to the cause of general 
 education. It therefore only remains to point out some of the more 
 important features of the following work, before introducing it to the 
 ordeal of public opinion. That it is worthy of public attention and 
 patronage, belongs not to him to decide. 4t certainly will be found to 
 possess some peculiarities, which are of course regarded by him as im- 
 provements. Whether they are indeed such, remains for others to de- 
 teimine. 
 
 A peculiar feature of the following pages, and one which distin- 
 guishes this work from every other on the same subject, is the " Sys- 
 tem of Canceling," which, in connection with the ordinary mode of 
 solution, is introduced throughout, and applied to such arithmetical 
 problems as embrace in their operation both multiplication and divis- 
 
 :/ ion. This is regarded by the author, and by many others acquainted 
 I with his system, as a decided improvement upon all Arithmetics here- 
 1 tof ore presented to the public. 
 
 The following are some of the advantages of the new system : 
 I 1st. The statement required, or, rather, recommended for canceling, 
 is itself a complete analysis of the sum proposed. Suppose, for illus- 
 tration, that 12 yards of cloth cost $48, and that it is required to find 
 the value of 15 yards of the same. 
 
 .We analyze the preceding sum, either by first finding the value of 
 one yard of the cloth, viz. $48-*- 12 yd. = $4, and then multiplying 
 that price by the number of yards, as, $4X15 yd. = $60, Ans. ; o"r by 
 finding the ratio of the number of yards of which the price is given, 
 and of those of which the price is required, and then multiplying 
 that ratio by the given cost. This ratio is -J-f =f- ; and Jx48 = 60, the 
 number of dollars required, and the same as above. 
 The same stated for canceling: 
 
 48.15 
 12' 
 
 . (See rule for canceling, in Single Proportion.) By the above state- 
 ' ment, $48 is to be increased by the ratio, 12 : 15 ; or \^ = -|. It is, . 
 however, obvious that-jL of 48 multiplied by 15, is the s"ame as ^ of 
 15 multiplied by 48. The above sum is therefore canceled and solved A 
 * thus: I 
 
 4 if 
 
 48, 15 
 and 15X4 = 60, the dollars required. 
 
INTRODUCTION. 7 
 
 The application of the canceling principle is, however, more com- 
 pletely illustrated by the solution of sums in which the ratio of one of 
 the given quantities to a required quantity of the same kind, is traced 
 through several simple ratios. The following sum may serve as an 
 illustration : 
 
 If 3 men, in 16 days, of 9 hours each, build a wall 20 feet long, 6 
 feet high, and 4 feet thick, in how many days, of 8 hours each, will 12 
 men build a wall 200 feet long, 8 feet high, and 6 feet thick 1 
 
 It is obvious the ratio of the given days to the required number of 
 days, is compounded of the ratios, 8 hours : 9 hours; 12 men : 3 men; 
 20 feet in length : 200 feet in length ; 6 feet in height : 8 feet in height; 
 and 4 feet in thickness : 6 feet in thickness. Or, these ratios may be 
 fractionally expressed, thus : -f, T 3 2, 2 ^, f , and -f . 
 
 Now the given days of 9 hours each are changed to days of 8 hours 
 each, by the following statement : 
 
 ^ = 18 day, 
 
 The time, in days of 8 hours each, required for the 12 men to com- 
 plete the work of 3 men, is obtained by uniting the second of the pre- 
 ceding ratios to the above statement ; thus : 
 
 16. 9. 3 
 
 By introducing into the same statement, the third of the preceding 
 ratios, we obtain the requisite lime for completing the 200 feet of wall, 
 allowing the height and thickness of each wall to be the same ; thus : 
 
 16. 9. 3. 200 
 
 If the fourth ratio be united, we obtain the time required, allowing 
 each wall to be of the same thickness ; thus : 
 
 16. 9. 3. 200. 8 
 8. 12. 20. 6 
 
 Lastly, if the fifth and last ratio be introduced, the number of days 
 required by all the conditions of the question, is obtained ; viz. 
 
 16. 9. 3. 200. 8. 6 
 8. 12. 20. 6. 4 
 The last statement canceled : 
 10 S 
 
 1S I 12 *Stt' ft 4' and 10X9 = 9 da y s > answer- 
 lS 
 
 2dly. A second advantage to be derived from the canceling system, is ll 
 the facility afforded by it for reducing several operations to a single I ( 
 statement. The following example will afford an illustration : 
 
 Bought 742 Ib. of wool, a deduction of 5 per cent, from the gross 
 weight being made, for dust, &c. For the net weight, I paid 9 s. New 
 York currency, per Ib., and for ready money, was allowed a deduction 
 
8 INTRODUCTION. 
 
 of 6 per cent. Now, allowing that I sold the same so as to realize a 
 gain of 20 per cent., how much money did I receive 1 
 
 The ordinary mode of solving this sum requires the four following 
 operations; viz. 105 : 100 : : 742 : the net weight of the wool, which is 
 706| Ib. Again, 706f X9-V-8 = 795, the number of dollars which the 
 wool would have cost, if no deduction had been made for ready money. 
 But a deduction of 6 per cent, was actually made ; therefore, 106 : 100 : : 
 795 : the money paid, viz. $750. Now, on this last sum, I realized a 
 gain of 20 per cent. ; hence, 100 : 120: : 750 : the money received, viz. 
 $900, Ans. By canceling, these four statements are reduced to one, 
 thus: 
 
 742. 100. 9. 100. 120 
 105. 8. 1067100' 
 
 that is, the 742 Ib. is to be multiplied by the four succeeding ratios, and 
 the number obtained will equal the number of dollars required. 
 
 The same canceled : 
 
 7 20 3 15 
 
 11 
 
 s 
 
 3dly. A great advantage of the canceling system over all others, arises 
 from the expedition it affords in arithmetical solutions. Instead of 
 multiplying and dividing by all the numbers which the nature of the 
 sum proposed would naturally require, the multipliers and divisors are 
 made to cancel each other ; that is, equal factors are rejected from both. 
 Hence, they are all made to exert their appropriate influence in pro- 
 curing the answer, while the labor of multiplying and dividing is 
 avoided. The statement of each sum for canceling is a fractional 
 answer of the same, and it is obvious, that the value of fractions is not 
 affected by rejecting equal factors from their numerators and denomi- 
 nators. 
 
 The processes of reduction which occur very frequently in common 
 Arithmetics, are mostly avoided by this system. Suppose it be required 
 to find how many pounds sterling 5 hogsheads of wine would cost 
 at 10 d. per pint. By the canceling system, it is necessary only to 
 write down the numbers required to effect the reduction, and the ques- 
 tion is then solved by canceling those numbers as far as practicable. 
 Thus: 
 
 5. 63. 4. 2. 10 
 " 12. 20' 
 
 The numbers above the line are obviously those, which, when mul- 
 tiplied together, will give the answer in pence. The numbers below 
 the line, are those required to reduce pence to pounds sterling. The 
 above sum canceled : 
 21 
 
 i then> 21X5 = 105 ' Ans ' 
 
 That the above method of solving arithmetical problems is easily 
 
INTRODUCTION. 9 
 
 comprehended and applied by the scholar, has been fully tested by the 
 author. The experience of nine or ten years entirely devoted to the 
 business of instruction, leaves him no room to doubt on this point. 
 Being, however, fully aware that his Arithmetic might fall into the 
 hands of some, who would not at once comprehend and apply the prin- 
 ciple of canceling, he has introduced the ordinary rules of solution, in 
 connection with those of canceling, and has endeavored to render 
 both modes plain and familiar, by frequent and clear illustrations. 
 
 The constant aim of the teacher should be to prepare his pupils for 
 the active duties of life ; and, in the department of Arithmetic, this is 
 accomplished only when the scholar has acquired correctness and ex- 
 pedition in effecting his solutions. 
 
 To make good arithmeticians, it is first necessary to acquire a correct 
 and extensive comprehension of the simple or fundamental rules of 
 Arithmetic. When this is done, their application will be obvious. The 
 danger, therefore, is not, that the scholar will spend too much time on 
 what is usually regarded as the more simple part of Arithmetic, but 
 that he will leave it too soon. 
 
 In the use of this treatise, the author would recommend, that, when 
 the pupil shall have passed the simple rules, and commenced those ope- 
 rations to which canceling may be applied, he be required to solve 
 each problem both by the ordinary rule, and by the rule for canceling. 
 More practice will thus be secured, and, consequently, greater expedi- 
 tion acquired. 
 
 In the illustrations, which are given in connection with the different 
 rules, it has been the design of the author fully to acquaint the scholar 
 with the nature of the subject presented, without carrying his expla- 
 nations so far as to take the work which properly belongs to the 
 scholar, out of his hands. No important acquisition can be made, 
 without corresponding effort. This fact seems to have been overlooked, 
 in the preparation of some Arithmetics now in use, and special effort 
 made to render every thing as easy as possible for the scholar ; that is, 
 to enable him to effect the solutions with very little mental labor. The 
 intellectual powers are, however, developed and strengthened only by 
 being brought into vigorous exercise. " In Arithmetic, the young be- 
 ginner should find just enough assistance to encourage and stimulate 
 him to effort. That is not the best system, which enables the learner 
 to advance from rule to rule with the least amount of study; but that, 
 which, while it helps him over some difficulties, leaves him examples 
 V enough to task his powers to the utmost." (Dr. Humphrey's Thoughts 
 on Education.) With these introductory remarks, the following work 
 is commended to the candor of an enlightened public. 
 
 THE AUTHOR. 
 
 Norwich Academy, May, 1840. 
 
 Ill 
 
ERRATA. 
 
 Notwithstanding care has been constantly exercised to avoid 
 errors, the following will be found to have escaped detection : 
 
 Page 42, line 25, read G, instead of 5 hundred. 
 " 65, in the 4th sum, and in the first statement of that sum, read 
 
 21, instead of 4. 
 
 " 119, line 34, read |, instead of f . 
 " 136, line 14, read integers, for integris. 
 " 171, line 7, read $420, instead of $4.20. 
 
TABLE OF CONTENTS. 
 
 PAGE 
 
 Arithmetic, 13 
 
 Notation, 
 
 Numeration, 14 
 
 Simple Addition, 
 
 Simple Subtraction, ......... 
 
 Simple Multiplication, 
 
 Simple Division, 43 
 
 Federal Money, . 54 
 
 Addition of Federal Money, 56 
 
 Subtraction of Federal Money 57 
 
 Multiplication of Federal Money, 58 
 
 Division of Federal Money, 
 
 Canceling, 61 
 
 Compound Numbers, ........ 67 
 
 Reduction of Compound Numbers, 72 
 
 Addition of Compound Numbers, 93 
 
 Subtraction of Compound Numbers, 101 
 
 Multiplication of Compound Numbers, .... 107 
 
 Division of Compound Numbers, 112 
 
 Vulgar Fractions, 117 
 
 Reduction of Fractions, 123 
 
 To reduce whole or mixed numbers to Improper Fractions, 123 
 To reduce Improper Fractions to whole or mixed numbers, ; 124 
 To reduce Compound Fractions to Simple ones, . 
 To change Fractions from one denomination to another, . 126 
 To find the integral value of Fractions, .... 129 
 To reduce low denominations to Fractions of higher denomina- 
 tions, 130 
 
 To reduce Fractions to a Common Denominator, . . . 131 
 
 Addition of Fractions, 132 
 
 Subtraction of Fractions, 
 
 Multiplication of Fractions, ....... 135 
 
 Division of Fractions, 
 
 Decimal Fractions, 142 
 
 Addition of Decimal Fractions, 146 
 
 Subtraction of Decimal Fractions, 147 
 
 Multiplication of Decimal Fractions, 
 
 Division of Decimal Fractions, 149 
 
 Reduction of Vulgar and Decimal Fractions, . . 152 
 Reduction of Decimal to Vulgar Fractions, .... 152 
 Reduction of low denominations to Decimals of a higher denom- 
 ination, 153 
 
12 CONTENTS. 
 
 PAGE 
 
 To find the integral value of a Decimal, .... 155 
 
 Reduction of Currencies, 157 
 
 Simple Interest, 162 
 
 Banking, 174 
 
 Compound Interest, 179 
 
 Commission, 181 
 
 Insurance, 181 
 
 Ratio, 183 
 
 Proportion, 185 
 
 Compound Proportion, 196 
 
 Analytical Solution, . . 202 
 
 Simple and Compound Proportion in Fractions, . . . 206 
 
 Conjoined Proportion, 209 
 
 Discount, . 211 
 
 Profit and Loss, 213 
 
 Barter, 217 
 
 Partnership, 219 
 
 Commercial Exchange, 223 
 
 Table of Gold Coins 
 
 Table of Uncoined and Silver Money, 226 
 
 Tare and Tret, 230 
 
 Equation of Payments, 
 
 Duodecimals, 235 
 
 Involution, 239 
 
 Evolution, 241 
 
 Extraction of the Square Root, 243 
 
 Extraction of the Cube Root, 
 
 Arithmetical Progression, 261 
 
 Geometrical Progression, 264 
 
 Alligation, 268 
 
 Position, 
 
 Double Position, 
 
 Promiscuous Examples, 
 
 APPENDIX, 284 
 
ARITHMETIC. 
 
 ARITHMETIC explains the properties and relation of num- 
 bers, and makes known their practical application. 
 
 There are six fundamental operations, with which the 
 scholar must become perfectly familiar before he can advance 
 successfully ; viz. Notation, Numeration, Addition, Subtrac- 
 tion, Multiplication, and Division. These operations are call- 
 ed fundamental, because all others are founded upon, or are 
 wrought by the application of one or more of them. They 
 therefore require to be first clearly understood. 
 
 NOTATION. 
 
 Notation is the art of expressing numbers by numerical 
 characters. The characters employed to express numbers 
 are, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, and are called figures. Each 
 of these figures has its own specific, and also its local value, 
 as will be learned from Numeration. Besides these charac- 
 ters, there are others used to express operations. 
 
 1st. The sign of addition, viz. + > (or plus, more ;) requiring 
 the numbers between which it is placed to be added : 3 + 2 
 are 5 ; that is, 3 added to 2 are 5 ; usually read, 3 plus 2 are 5. 
 
 2d. The sign of subtraction, viz. , (or minus ;) showing 
 that the number following is to be taken from that which pre- 
 cedes it ; thus, 4 2 is 2 ; that is, 2 taken from 4, 2 remains. 
 
 3d. The sign of multiplication, viz. x ; requiring the num- 
 ber placed before it to be multiplied by that which follows ; 
 thus, 3 X 4 is 12 ; that is, 3 multiplied by 4 is 12. 
 
 4th. The sign of division, viz. H-; requiring the number 
 preceding it to be divided by that which follows ; thus, 8-i-2 
 is 4 ; that is, 8 divided by 2 is 4. 
 
 In the use of each of the preceding signs, the figure prece- 
 ding the sign is to be operated upon by that which follows it. 
 2 
 
14 NUMERATION. 
 
 5th. The signs of proportion, viz. : :: : ; showing that the 
 numbers including and between these dots, are proportionals ; 
 thus, 2 : 4 :: 6 : 12 ; that is, 2 bears the same relation to 4 as 
 6 to 12. The numbers are thus read, 2 is to 4 as 6 is to 12. 
 
 6th. The sign of equality, viz. =; expressing the equality 
 of the numbers between which it is placed ; or that the num- 
 bers on the right equal those on the left ; thus, 9 + 7=20 4. 
 
 7th. The characters V, V, V, V, &c. require some root of 
 the number before which they stand to be extracted. The 
 figure placed over the sign always shows what root is required. 
 When the character is used without any figure, it then Indi- 
 cates the square root. 
 
 By the use of these characters, any arithmetical operation 
 may be indicated. If it be required to add 9 to 16, from the 
 amount to subtract 5, to divide the remainder by 4, and to mul- 
 tiply the quotient by 6, the operation would be thus expressed : 
 9+16 5-^4x6 = 30. 
 
 QUESTIONS. What does Arithmetic explain 1 What application 
 does it make of numbers 1 How many are the fundamental operations 
 of Arithmetic 1 What are they ? Why are these called fundamental 
 operations 1 What is Notation? What are the characters used to 
 express numbers called *? What two-fold value has each figure ? 
 For what purposes are other characters used*? What is the sign of 
 addition, and for what is it used 1 The sign of subtraction 1 What 
 does it require 1 The sign of multiplication 1 What does it require 1 
 The sign of division 1 What does it require 1 The signs of propor- 
 tion 1 What do they show ? The sign of equality 1 What does it 
 show ? What is the character used to express the extraction of roots 
 called 1 ? Ans. The radical sign. What does the figure placed over 
 the radical sign show 7 
 
 NUMERATION. 
 
 The scholar has seen, under Notation, the characters used 
 to express the first nine numbers, viz. ; that to express one whole 
 object or thing, 1 is used ; to express two whole things, 2 is 
 employed ; and for three whole things, 3 is taken, &c., so that 
 each character has its own specific value ; and this it always 
 expresses when it stands alone. But each figure has also a 
 local value, that is, a value depending on the place it occupies ; 
 
NUMERATION. 15 
 
 thus, the value of 3 differs in each of the following numbers, 
 viz. 003, 030, and 300. In the first number its value is three 
 units, or ones ; in the second number it is three tens, or thirty ; 
 and in the last, it is three hundreds. It will therefore be 
 readily perceived, that the position of a figure materially affects 
 its value. Numeration teaches how to determine what this 
 value is ; and thus it also enables us to determine the total 
 value of any number of figures. It will be found that any 
 figure is increased in a ten-fold ratio by having a single figure 
 placed on the right of it ; thus, 6 alone, is 6 units ; but if 
 another figure be placed on the right of this, its value is ten 
 times as great as before ; thus, in 63, the 6 is six tens, equal to 
 60, and the 3 is three units. This value is increased a hun- 
 dred-fold by having two figures placed on the right ; thus, 600 ; 
 and a thousand-fold by having three figures on the right of it ; 
 thus, 6000. In the first of the last two examples, the value of 
 6 is six hundred, and in the second it is six thousand. Hence, 
 the scholar will see the necessity of terms by which to desig- 
 nate this local value of figures, and will also readily see the 
 appropriateness of those used, viz. Units, Tens, Hundreds, 
 Thousands, Tens of Thousands, Hundreds of Thousands, 
 Millions, Tens of Millions, Hundreds of Millions, &c. These 
 nine terms are sufficient to express any number in common 
 practice. The higher denominations are Billions, Tens of 
 Billions, Hundreds of Billions ; Trillions, Tens of Trillions, 
 Hundreds of Trillions ; Quadrillions, Tens of Quadrillions, 
 Hundreds of Quadrillions ; Quintillions, Tens of Hundreds 
 of ; Sextillions, Tens of Hundreds of ; Septillions, Tens 
 of Hundreds of ; Octillions, Tens of Hundreds of ; 
 Nonillions, Tens of Hundreds of j &c. 
 
 It will be observed that as the first three figures, reckoning 
 from the right, are units, tens, and hundreds, so every suc- 
 ceeding three are appropriated to the units, tens, and hun- 
 dreds of the succeeding higher denominations. The following 
 table will serve as an illustration : 
 
 369, 342, 900, 976, 368, 265, 371, 502, 634, 436. 
 
 3 gg -ss -8 S3 -SS3 -5S2 -.3 -SS.3 -5 S3 -5 S3 -S S 3 
 
 2 = 22= 2= 2= 2= 2= S= 2 c 2 2g 
 
 O 3-c 3-0 3-0 3T33>a 3-0 3-C 3TJ 3-O 3 
 
 SCCCCCCCCB 
 
 3333333339 
 
 XSESESSEWS 
 
16 NUMERATION. 
 
 This table will enable the scholar to see at a glance, that 
 the names and value of figures are entirely dependent on4heir 
 location. If they be counted from the right hand towards the 
 left, the first figure in any line of figures is units ; the second 
 is tens ; the third, hundreds ; the fourth, thousands ; the fifth, 
 tens of thousands, &c. ; and whatever station or place any 
 figure may occupy, its value becomes ten times as great by 
 being moved one degree farther to the left. 
 
 Q.UESTIONS. What are the characters used to express the first nine 
 numbers'? Give an example. When does each figure express its 
 own specific value? What other value has each figure ? On what 
 does the local value of a figure depend 1 Give an illustration. What, 
 therefore, does Numeration teach us to do 1 What does it enable us 
 to do 1 In what ratio is the value of any figure increased by having a 
 single figure placed on the right 7 In what ratio is it increased "by 
 having two figures on the right of it 1 In what by having three placed 
 on its right 1 In what ratio do numbers continue to increase from the 
 right to the left 1 Ans. In a ten-fold ratio. What are the terms by 
 which the local value of figures are expressed 1 To what are each 
 three successive figures in any number appropriated 1 Ans. To the 
 units, tens, and hundreds of each denomination. In tracing the figures 
 from the right to the left, what is the first figure called 1 The second 7 
 The third 1 The fourth ? &c. What effect is produced on the value 
 of any figure by moving it one place to the left 7 
 
 Enumerate the following numbers, viz. 6 ; 27 ; 467 ; 568 ; 
 4269; 13786; 27599; 367595; 17129567; 67596422; 586; 
 379872689 ; 278063 ; 596402606; 295 ; 336003; 300300303; 
 505050505; 467327986427; 585950876; 688001 ; 100000000; 
 99999999999; 6398742913; 4678. 
 
 The scholar should be taught to read these figures accurate- 
 ly ; for example, suppose he be required to enumerate the last 
 number, viz. 4678 ; let him commence and repeat thus : eight 
 units, seven tens, six hundreds, and four thousands ; and then 
 unite them, thus : four thousand six hundred and seventy-eight. 
 Let him also be required to give the value of any figure as it 
 may vary by being written at different points under any line of 
 figures. 
 
 After the scholar has become familiar with the preceding 
 exercise, he may write the following numbers on his slate in 
 figures, taking care to express each number accurately : 
 1. Thirty-five. 2. Three hundred and seventy-five. 3. Three 
 hundred and five. 4. Seven thousand six hundred and thirty- 
 five. 5. Seven thousand and thirty-five. 6. Seventy-five 
 thousand four hundred and sixteen. 7. Seventy-five thousand 
 and sixteen. 8. Seventy-five thousand and six. 9. Seventy- 
 
ADDITION OF SIMPLE NUMBERS. 
 
 17 
 
 five thousand. 10. Three hundred and thirty-three thousand 
 three hundred and thirty-three. 11. Three hundred thousand 
 and three. 12. Three hundred thousand three hundred and 
 three. 13. Five millions and five. Six millions and seventy- 
 five. One hundred and sixty millions. Forty-seven millions, 
 one hundred and five thousand and sixty. 14. Qft^ hundred 
 millions, one hundred and one. 15. One hundreoTahd seven 
 millions, one hundred and seven thousand, one hundred and 
 seven. 16. Two billions, three hundred and three millions, five 
 hundred and five thousand and six. 17. Seven hundred and 
 seven trillions, six hundred and seventy -two billions, nine mill- 
 ions, three hundred and five thousand, six hundred and nine. 
 
 There is yet another method of expressing numbers ; viz. the 
 Roma*n method ; in which the letters of the alphabet are used, 
 as may be seen from the following table, 
 
 ROMAN TABLE. 
 
 I One. 
 
 II Two. 
 
 Ill Three. 
 
 IV Four. 
 
 V Five. 
 
 VI Six. 
 
 VII Seven. 
 
 VIII Eio-ht. 
 
 IX Nine. 
 
 X Ten. 
 
 XX Twenty. 
 
 XXX Thirty. 
 
 XL Forty. 
 
 L Fifty. 
 
 LX . 
 LXX. 
 LXXX 
 XC . 
 
 C 
 
 CC . 
 CCC . 
 
 cccc 
 
 D 
 DC . 
 
 DCC. 
 DCCC 
 DCCCC 
 M 
 
 Sixty. 
 Seventy. 
 Eighty. 
 Ninety. 
 One hundred. 
 Two hundred. 
 Three hundred. 
 Four hundred. 
 Five hundred. 
 Six hundred. 
 Seven hundred. 
 Eight hundred. 
 Nine hundred. 
 One thousand. 
 
 ADDITION OF SIMPLE NUMBERS. 
 
 It is highly important that the scholar obtain clear and dis- 
 tinct views of the nature of what he has to perform. He is 
 therefore recommended to make himself familiar with the fol- 
 2* 
 
18 
 
 ADDITION OF SIMPLE NUMBERS. 
 
 lowing table in the first place ; then to take the mental exer- 
 cises that follow, and study them faithfully before com- 
 mencing with a slate. Indeed, the author would advise, that 
 the tables, together with the mental exercises, which stand at 
 the beginning of each of the four simple rules, be first studied ; 
 and that the beginner then turn back and review the whole. 
 
 ADDITION TABLE. 
 
 The signs plus and minus, &c. are introduced into the fol- 
 lowing tables, that the scholar may early learn the use of them. 
 He will therefore turn back to Notation, if he does not recol- 
 lect their use. 
 
 2 plus 1 equals 3 
 
 3+1=4 
 
 4+1=5 
 
 5+1=6 
 
 2+2=4 
 
 3+ 2= 5 
 
 4+2=6 
 
 5+2=7 
 
 2+3=5 
 
 3+3=6 
 
 4+3=7 
 
 5+3=8 
 
 2+ 4 = 6 
 
 3+4=7 
 
 4+4=8 
 
 5+4=9 
 
 2+5=7 
 
 3+5=8 
 
 4+5=9 
 
 5+ 5 = 10 
 
 2+6=8 
 
 3+6=9 
 
 4+ 6 = 10 
 
 5+ 6 = 11 
 
 2+ 7 = 9 
 
 3+ 7= 10 
 
 4+ 7= 11 
 
 5+ 7= 12 
 
 2+ 8 = 10 
 
 3+ 8 = 11 
 
 4+ 8 = 12 
 
 5+ 8 = 13 
 
 2+ 9 = 11 
 
 3+ 9 = 12 
 
 4+ 9 = 13 
 
 5+ 9 = 14 
 
 2 + 10 = 12 
 
 3 + 10 = 13 
 
 4+ 10 = 14 
 
 5+10 = 15 
 
 2 + 11 = 13 
 
 3+11 = 14 
 
 4 + 11 = 15 
 
 5 + 11 = 16 
 
 2 + 12 = 14 
 
 3+12 = 15 
 
 4+12 = 16 
 
 5 + 12 = 17 
 
 6+1=7 
 
 7+1=8 
 
 8+1=9 
 
 9+ 1 = 10 
 
 6+2=8 
 
 7+2=9 
 
 8+ 2 = 10 
 
 9+ 2 = 11 
 
 6+3=9 
 
 7+ 3 = 10 
 
 8+ 3 = 11 
 
 9+ 3 = 12 
 
 6+ 4 = 10 
 
 7+ 4 = 11 
 
 8+ 4= 12 
 
 9+ 4 = 13 
 
 6+ 5 = 11 
 
 7+ 5 = 12 
 
 8+ 5 = 13 
 
 9+ 5 = 14 
 
 6+ 6 = 12 
 
 7+ 6 = 13 
 
 8+ 6 = 14 
 
 9+ 6= 15 
 
 6+ 7= 13 
 
 7+ 7= 14 
 
 8+ 7 = 15 
 
 9+ 7 = 16 
 
 6+ 8 = 14 
 
 7+ 8 = 15 
 
 8+ 8 = 16 
 
 9+ 8= 17 
 
 6+ 9 = 15 
 
 7+ 9= 16 
 
 8+ 9 = 17 
 
 9+ 9 = 18 
 
 6+10= 16 
 
 7 + 10 = 17 
 
 8+10 = 18 
 
 9+10= 19 
 
 6 + 11 = 17 
 
 7+11 = 18 
 
 8+11 = 19 
 
 9+11 = 20 
 
 6 + 12 = 18 
 
 7 + 12 = 19 
 
 8+12 = 20 
 
 9 + 12 = 21 
 
ADDITION OF SIMPLE NUMBERS. 
 
 19 
 
 10+ 1 = 11 
 
 11 + 1 = 12 
 
 12 + 1 = 13 
 
 10+ 2 =-12 
 
 11 + 2 = 13 
 
 12 + 2 = 14 
 
 10+ 3 = 13 
 
 11 + 3 = 14 
 
 12 + 3 = 15 
 
 10+ 4 = 14 
 
 11 + 4 = 15 
 
 12 + 4 = 16 
 
 10+ 5 = 15 
 
 11 + 5 = 16 
 
 ]2 + 5 = 17 
 
 10 + 6 = 16 
 
 11 + 6 = 17 
 
 12 + 6 = 18 
 
 10+ 7= 17 
 
 11 + 7 = 18 
 
 12+ 7 = 19 
 
 10 + 8 = 18 
 
 11 + 8 = 19 
 
 12 + 8 = 20 
 
 10+ 9 = 19 
 
 11+ 9 = 20 
 
 12+ 9 = 21 
 
 10+ 10 = 20 
 
 11 + 10 = 21 
 
 12 + 10 = 22 
 
 '10+ 11 =21 
 
 11 + 11 = 22 
 
 12 + 11 =23 
 
 10 + 12 = 22 
 
 11 + 12 = 23 
 
 12 + 12 = 24 
 
 In reciting this and the following tables, the teacher should 
 be careful that the scholar, as he repeats his answers, per- 
 form the necessary mental operation ; he must teach his pupil 
 to think. 
 
 MENTAL EXERCISES IN ADDITION. 
 
 If you borrow 6 dollars at one time and 3 at another, how 
 many will you have ? 6 and 3 are how many ? A pays 
 you 3 dollars, B 4, and C 5 ; how many do they all pay you ? 
 
 3 + 4+5 = how many ? If you pay away 6 cents at one time, 
 
 4 at another, and 3 at another, how many do you pay away ? 
 6 + 4 + 3 = how many ? If you buy twelve apples for 9 
 cents, and a pint of chesnuts for 4, how many cents do the 
 apples and chesnuts cost ? 9 + 4 = how many ? John gave 
 Henry 8 apples at one time, 6 at another, and 4 at another ; 
 how many did he give him ? 8 + 6+4 = how many ? Lent 
 my brother 10 dollars at one time, 6 at another, and 4 at another ; 
 how many dollars did I lend him? 10+6 + 4 = how many? 
 Alfred has four brothers ; to one, he gave 1 1 pears, to another, 
 9, to another, 6, and to another, 3 ; how many pears did he 
 give away ? 11 + 9 + 6 + 3= how many ? He also gave 
 his two sisters 5 each ; how many did he give to both brothers 
 and sisters ? 11 + 9 + 6 + 3 + 5 + 5 = how many ? A 
 man bought four bushels of corn for 3 dollars, six bushels of 
 oats for 2 dollars, and 9 bushels of wheat for 12 ; how many 
 bushels of grain did he buy, and how many dollars did the 
 whole cost him ? 4 + 6 + 9 = how many ? 3 + 2 + 12 = 
 how many ? John has 9 dollars ; James, 12 ; and Joseph, 7 ; 
 how many have they all? 9+12 + 7= how many? Peter 
 
20 ADDITION OF SIMPLE NUMBERS. 
 
 gave a poor man 11 cents ; Charles gave him 12 cents ; Hen- 
 ry, 10; and George, 8; how many did they all give him; 
 ll + 12 + 10-|-8=: how many? A man was three days 
 performing a journey ; the first day he traveled 13 miles ; the 
 second day, 14; and the third, 15; how far did he travel? 
 13 + 14+15 how many ? How many sums will you have 
 done, if you do 12 to-day, 15 to-morrow, and 18 the next day? 
 12+15+18=how many? 
 
 How many are 3 + 4? 13 + 4? 23 + 4? 33 + 4? 43+4? 
 53 + 4? 63 + 4? 73+4? 83 + 4? 93 + 4? How many 
 are4 + 4? 14 + 4? 24 + 4? 34 + 4? &c. How many are 
 5 + 4? 15 + 4? 25 + 4? 35 + 4 ? &c. How many are 6 + 4 ? 
 16 + 4? 26 + 4? 36 + 4 ? &c. How many are 7 + 4 ? 17+4? 
 27+4? 37+4? &c. How many are 4 + 4? 4+14? 4+24? 
 &c. 4+5? 4+15? 4 + 25? &c. 4+6? 4 + 16? 4+26? 
 &c. 4+7? 4+17? 4+27? &c. How many are 5 + 5 ? 
 
 5 + 15? 5 + 25? &c. 5 + 6? 5 + 16? 5+26? &c. 6 + 6? 
 
 6 + 16? 6+26 ?&c. How many are 7 + 7? 7+17? 7+27? 
 &c. 8+8? 8+18? 8+28? &c. 9+9? 9+19? 9+29? 
 &c. Howmany are 10+10? 10+20? &c. 11+5? 11 + 15? 
 12+4? 12 + 14? &c. 
 
 Questions of this character may be proposed to any extent, 
 and should in no instance be omitted until the scholar can add 
 without hesitation. Others of a more promiscuous character 
 should likewise be proposed, such as the, following : 
 
 7 + 4 + 6=how many? 2 + 1 + 9 + 6 + 7 + 6= how many? 
 5 + 6 + 9 + 3=how many? 6 + 4 + 2 + 8+1 + 3 + 5 + 7+9 = 
 how many? 11+9 + 6 + 12 = how many ? 7+6 + 9 + 8 = 
 how many? 8 + 8 + 6 + 6 + 4 + 4=howmany? 7 + 7+9 + 9 = 
 how many ? 2 + 4 + 6 + 8 + 10+1 2 =how many ? 
 
 The scholar may now commence the use of the slate and 
 pencil, or the practice of written arithmetic. 
 
 The first consideration to which his attention should here 
 be directed,*is, that like things only can be added to or sub- 
 tracted from each other. It would be absurd to attempt to 
 add together books and chairs, to see how many books, or how 
 many chairs the whole would make ; the number of each would 
 evidently remain unaffected. If, however, we add books to 
 books, we obtain a number greater than either of the original 
 numbers ; that is, just equal to them both. Neither can we 
 add units to tens ; for the amount would be neither units nor 
 tens ; but units must be added to units, tens to tens, and hun- 
 dreds to hundreds, and so on. But ten units make one ten, ten 
 
ADDITION OF SIMPLE NUMBERS. 21 
 
 tens make one hundred, and ten hundreds make one thousand, 
 &c. ; that is, simple numbers increase and decrease in a ten-fold 
 ratio. Hence 10 in the column of units is equal to 1 in the 
 column of tens ; and 10 in the column of tens, is equal to 1 in 
 the column of hundreds. If, then, in adding up the column of 
 units the whole should amount to just 10, it is obvious that 
 nothing is lost, if these 10 units are converted into 1 ten, and 
 the 1 ten added to the column of tens ; for 10 units = 1 ten. 
 The same is true of other denominations. Therefore the fol- 
 lowing General Rule will be found applicable to Simple Ad- 
 dition : 
 
 1st. Write down the numbers, placing units under units, 
 tens under tens, &c. 
 
 2d. Draw a line underneath, and commence at the. right 
 hand and add together all the figures in the first column. 
 
 3d. If the sum be less than 10, set it down at the foot of that 
 column ; if it be 10, or more than 10, it will consist of two 
 figures at least ; set down the right hand one as before, and add 
 the left hand one to the next column, it being in all cases so 
 many tens, when compared with the figures added. 
 
 4th. Continue to perform the same operation with the re- 
 maining columns, observing only to write down the whole 
 amount of the left hand column. 
 
 5th. To detect any error that may have been committed, 
 commence again and add each column downwards ; if the 
 same numbers are obtained by each operation, the work is 
 probably right. 
 
 Now to apply this rule, let us add together the four following 
 numbers, viz. 1234 ; 2345 ; 6420 ; and 5796. The rule says, 
 write these numbers with units under units, tens 
 under tens, &c. thus : 1234 
 
 Now to add these numbers, I commence at the ^2345 
 right hand, and first add together the unit figures ; ^ 6 4 2 
 viz. 6, 0, 5, 4; the sum I find to be 15 units, 5796 
 
 equal to 1 ten and 5 units ; I write down the 
 
 5 units, but add the 1 ten to the next column; 15795 
 thus, 1 added to 9 is 10, and 2 are 12 and 4 are 
 16 and 3 are 19 ; as before, I write down the 9, and add the 1 
 to the next column ; this being added gives the amount 17 ; the 
 7 is written down and the 1 again carried to the next and last 
 column; and here the amount is 15, which being the last 
 column, the whole number is written down. This operation 
 
22 ADDITION OF SIMPLE NUMBERS. 
 
 gives the amount of the four numbers, 15795. The scholar 
 will readily comprehend the nature of Simple Addition ; viz. 
 that it consists in uniting two or more numbers of the same de- 
 nomination, so as to find their amount. 
 
 The scholar will carefully examine the following sums, 
 which are added, to see if he obtains the same result. 
 
 2. 
 
 847396 In this second example, .in the unit 
 
 364869 column, there are 3 to carry, because 
 482436 there are 3 tens or 30. For the same 
 622439 reason there are 2 to carry in all the re- 
 maining columns but one. 
 
 2317140 
 
 3 4. 
 
 456789460 123456789 
 
 680246802 987654321 
 
 135791357 123456789 
 
 423650825 987654321 
 
 371574628 123456789 
 
 856456333 987654321 
 
 2924509405 3333333330 
 5. 6. 7. 
 
 2468024 
 1357912 
 4208642 
 2135791 
 3 3 5^5 7 7 9 
 8866448 
 
 9796365 
 5636979 
 8246024 
 5963816 
 4674364 
 5796378 
 
 1798670 
 6124356 
 7212345 
 8463738 
 5739168 
 9156423 
 
 22392596 40113926 
 
 8. 9. 10. 
 
 9176435 ^3457096 70819634 
 
 5683214 61728349 64753278 
 
 2345678 28394563 86542365 
 
 0123456 56831234 23456789 
 
ADDITION OF SIMPLE NUMBERS. 23 
 
 11. 12. 13. 
 
 81828384 39724638 45768903 
 
 18283848 97236452 78924683 
 
 99999999 15707934 13579246 
 
 77777777 65972109 97576887 
 
 14. 15. 16. 
 
 123456789 63840596 567347205379 
 
 987654321 27690903 275684467903 
 
 546372819 46379108 123456789012 
 
 951486804 55681396 378965842687 
 
 468046872 63705418 593748265379 
 
 608642109 36485074 916547681234 
 
 17. 
 
 18. 
 
 19. 
 
 795735198 
 
 369248754 
 
 1357924680 
 
 49157038 
 
 62958651 
 
 261595827 
 
 8930576 
 
 2.5 8 1 7 3 1 
 
 4230917546 
 
 697005 
 
 407347 
 
 6083419 
 
 29510 
 
 80478 
 
 682 
 
 4812 
 
 6184 
 
 8472166264 
 
 863 
 
 236 
 
 4462352 
 
 2 7 
 
 46 
 
 44605 
 
 4 
 
 8 
 
 2 1 
 
 20. A farmer sold his wheat for 320 dollars ; his corn for 
 275 dollars ; his oats for 78 dollars ; his barley for 162 dollars ; 
 and one horse for 132 dollars. What was the amount of his 
 sales? Ans. $967. 
 
 21. A merchant owned four vessels, which were worth, 
 the first, $4800 ; the second, $5200 ; the third, $6000 ; and 
 the fourth, $6800 ; he had also goods on board one of these 
 vessels, worth $2700 ; besides $3500 deposited in the bank. 
 What is the amount of his property ? Ans. $29000. 
 
 22. Three farmers have each 562 acres of land ; how many 
 have they all? Ans. 1686 acres. 
 
 23. A farmer fattened and killed an ox for market ; the hind 
 quarters weighed, the one 182 pounds, and the other 177; 
 each of the fore quarters weighed 163 pounds ; the hide, 116 
 pounds ; and the tallow, 120 pounds. What was the weight 
 of the ox? Ans. 921 pounds. 
 
24 ADDITION OF SIMPLE NUMBERS. 
 
 24. In a certain town there are five schools, containing the 
 following numbers of scholars, viz. 72, 28, 65, 84, and 91 ; how 
 many children are attending these five schools 1 Ans. 340.. 
 
 25. A steamboat performed in one week four trips from 
 Hartford to New York ; on the first trip, she took for passen- 
 gers, $378, for freight, $175 ; on the second trip, for passen- 
 gers, $402, for freight, $278 ; on the third, for passengers, 
 $263, fpr freight, $147 ; on her fourth trip, for passengers and 
 freight, $500. What did her bills amount to for the week ? 
 Ans. $2143. 
 
 26. A carpenter contracted for the building of five dwell- 
 ings in one spring ; for the first he was to receive $1800 ; for 
 the second, $2100 ; for the third, $2221 ; for the fourth, $2850 ; 
 and for the fifth, $3172. To what did his contracts amount? 
 Ans. $12143. 
 
 27. In 1834, A traveled 1320 miles; in 1835, he traveled 
 1162 miles; in 1836, 2100 miles; in 1837, 1400 miles; in 
 1838, 1992 miles. How many miles did he travel from 1834 
 to 1838 inclusive ? Ans. 7974. 
 
 28. Bought of my neighbor four loads of hay; the first 
 weighed 1600 Ibs. ; the second, 2100 Ibs. ; the third, 1999 Ibs. ; 
 and the fourth, 1709 Ibs. * What was the whole weight ? Ans. 
 7408 pounds. 
 
 29. A wholesale dealer in grain has in one bin 242 bushels 
 of wheat ; in another he has 2356 bushels of rye ; in a third, 
 1556 bushels of oats ; in a fourth, 876 bushels of barley. How 
 many bushels of grain has he of all kinds ? Ans. 5030. 
 
 30. Suppose I am indebted to A, $2560 ; to B, $27 ; to C, 
 $169 ; to D, $3470 ; and to E, $17. How much do I owe in 
 all ? Ans. $6243. 
 
 31. A man having three sons and two daughters, gave to 
 each of his sons, $379 ; and to each of his daughters, $199. 
 How much money did he give them all 1 Ans. $1535. 
 
 32. A gentleman being asked how old he was, said, he 
 was married when he was 29 years of age ; that he lived with 
 his wife 8 years before the birth of their son, who was now 27 
 years of age. What was the father's age ? Ans. 64 years. 
 
 33. A man bought a horse for $75 ; a chaise for $150 ; and 
 a harness for $45 ; he then sold his horse for $150 ; his chaise 
 for $125 ; and his harness for $30. What did he pay for the 
 whole, and what did he receive for the whole. Ans. Paid 
 $270 ; received $305. 
 
 34. Add togeth&t three hundred and seventy-five thousand 
 
SIMPLE SUBTRACTION. 
 
 25 
 
 and sixty-five ; nine hundred thousand and three ; one million six 
 hundred thousand seven hundred and ninety-nine. Ans. 2875867. 
 
 35. Add also ninety-nine millions ; seven hundred and fifty- 
 five millions six hundred and thirty-three. Ans. 854000633. 
 
 36. There are 5784 apples in one pile ; 588 in another ; 
 84 in a third ; and seven hundred and seventy-nine in a fourth. 
 How many are there in all ? Ans. 7235. 
 
 Q.UESTIONS. What things only can be added to or subtracted from 
 each other 1 In what does Simple Addition consist 1 Can we add 
 units to tens 7 To what must units, &c. be added"? For what num- 
 ber do we carry in Simple Addition 1 Why for 10 1 Ans. Because 
 numbers increase in a ten-fold ratio. 10 units equal how many 10's 1 
 What is the Rule for Addition 7 How do you write down the num- 
 bers 1 Where do you commence to add 1 If the sum of the figures 
 added be less than 10, what is to be done 7 What if it be 10, or more 
 than 10 1 What is observed respecting the last or left hand column 7 
 How may errors in Addition be detected 7 
 
 SIMPLE SUBTRACTION. 
 
 This rule is directly the reverse of the preceding ; while 
 we are there taught to unite several numbers into one, we are 
 here taught the operation by which one number is taken from 
 another. A familiar acquaintance with the following table 
 should be the first object of the scholar : 
 
 SUBTRACTION TABLE. 
 
 1 1=0 
 
 2 2 = 
 
 3 3 = 
 
 4 4 = 
 
 2 1 = 1 
 
 3 2 = 1 
 
 4 3 = 1 
 
 5 4 = 1 
 
 3 1 = 2 
 
 4 2 2 
 
 5 3 = 2 
 
 6 4 = 2 
 
 4 1 = 3 
 
 5 2 = 3 
 
 6 3 = 3 
 
 7 4 = 3 
 
 5 1 = 4 
 
 6 2 = 4 
 
 7 3 = 4 
 
 8 4 = 4 
 
 6 1 = 5 
 
 7 2 = 5 
 
 8 3 = 5 
 
 9 4 = 5 
 
 7 1 - 6 
 
 8 2 = 6 
 
 9 3 = 6 
 
 10 4= 6 
 
 8 1 = 7 
 
 9 2 = 7 
 
 10 3= 7 
 
 114= 7 
 
 91 = 8 
 
 10 2= 8 
 
 11 3 = 8 
 
 12 4= 8 
 
 10 1 = 9 
 
 112= 9 
 
 12 3= 9 
 
 13 4= 9 
 
 11 1 = 10 
 
 12 2 = 10 
 
 13 3 = 10 
 
 14 4 = 10 
 
 12 1 = 11 
 
 13 2 = 11 
 
 14 3 = 11 
 
 15 4 = 11 
 
 13 1 = 12 
 
 14 2 = 12 
 
 15 3 = 12 
 
 16 4 = 12 
 
26 
 
 SIMPLE SUBTRACTION. 
 
 5 5 = 
 
 6 6 = 
 
 7 7= 
 
 8 8= 
 
 6 5 = 1 
 
 7 6 = 1 
 
 8 7= 1 
 
 9 8 = 1 
 
 7 5 = 2 
 
 8 6 = 2 
 
 9 7 = 2 
 
 10 8= 2 
 
 8C Q 
 O O 
 
 9 6= 3 
 
 ]0 7 = 3 
 
 118= 3 
 
 9 5 = 4 
 
 10 6= 4 
 
 117= 4 
 
 12 8= 4 
 
 10 5= 5 
 
 116 = 5 
 
 12 7= 5 
 
 13 8= 5 
 
 115= 6 
 
 12 6= 6 
 
 13 7= 6 
 
 14 8= 6 
 
 12 5= 7 
 
 13 6= 7 
 
 14 7= 7 
 
 15 8= 7 
 
 13 5= 8 
 
 14 6= 8 
 
 15 7= 8 
 
 16 8= 8 
 
 14 5= 9 
 
 15 6= 9 
 
 16 7= 9 
 
 17 8= 9 
 
 15 5 = 10 
 
 16 6 = 10 
 
 17 7 = 10 
 
 18 8 = 10 
 
 16 5 = 11 
 
 17 6 = 11 
 
 18 7 = 11 
 
 19 8 = 11 
 
 17 5 = 12 
 
 18 6 = 12 
 
 19 7= 12 
 
 20 8 = 12 
 
 9 9 = 
 
 10 10= 
 
 11 11= 
 
 12 12= 
 
 10 9= 1 
 
 11 10= 1 
 
 12 11= 1 
 
 1312= 1 
 
 119 = 2 
 
 12 10= 2 
 
 13 11= 2 
 
 14 12= 2 
 
 12 9= 3 
 
 13 10= 3 
 
 14 11= 3 
 
 1512= 3 
 
 13 9 = 4 
 
 14 10= 4 
 
 15 11= 4 
 
 16 12= 4 
 
 14 9= 5 
 
 15 10= 5 
 
 16 11= 5 
 
 17 12= 5 
 
 15 9= 6 
 
 16 10= 6 
 
 1711= 6 
 
 18 12= 6 
 
 16 9= 7 
 
 17 10= 7 
 
 18 11= 7 
 
 19 12= 7 
 
 17 9= 8 
 
 18 10= 8 
 
 19 11= 8 
 
 20 12= 8 
 
 18 9= 9 
 
 19 10= 9 
 
 2011= 9 
 
 21 12= 9 
 
 19 9 = 10 
 
 2010=10 
 
 21 11 = 10 
 
 22 12 = 10 
 
 20 9 = 11 
 
 21 10 = 11 
 
 22 11 = 11 
 
 23 12 = 11 
 
 21 9 = 12 
 
 22 10 = 12 
 
 23 11 = 12 
 
 24 12 = 12 
 
 MENTAL EXERCISES IN SUBTRACTION. 
 
 Charles had 10 marbles, and gave 5 of them to his brother ; 
 how many had he left ? He then gave 3 to his sister ; how 
 many did there then remain ? A boy had 15 cents, and lost 
 5 of them ; how many had he left ? John has 13 apples ; how 
 many will he have, if he gives away 5 ? Peter had 1 9 mar- 
 bles, and gave away 9 of them ; how many had he left ? A 
 boy purchased two toys, for the one he gave 20 cts. and for the 
 other 12 cts; what was the difference in the cost of them? 
 Out of 23 plumbs a boy selected 12 ; how many remained ? A 
 boy 'threw 15 pennies into the air, and caught 6 of them as 
 they fell ; how many reached the ground ? Sold one cow for 
 19 dollars and another for 12 ; how much more did I receive 
 for one than for the other ? Charles has 19 marbles, and James 
 has 1 1 ; how many has Charles more than James ? Sold one 
 
SIMPLE SUBTRACTION. 27 
 
 knife for 21 cents and another for 12 cents ; what was the dif- 
 ference in price ? 11 taken from 19, how many will remain? 
 8 from 19 ? 10 from 19 ? 7 from 19 ? 11 from 19 ? 6 from 
 19? 12 from 19? 5 from 19? 13 from 19? 4 from 19? 
 
 14 from 19 ? 3 from 19 ? If 9 be taken from 18, how many 
 will remain? 8 from 18? 10 from 18? 7 from 18? 11 
 from 18? 6 from 18? 12 from 18? 5 from 18? 13 from 
 18? 4 from 18? 14 from 18? 3 from 18? 15 from 18? 
 
 2 from 18 ? 16 from 18 ? 1 from 18 ? 17 from 18 ? If 10 
 be taken from 20, how many will remain ? 9 from 20 ? 11 
 from 20 ? 8 from 20 ? 12 from 20 ? 7 from 20 ? 13 from 
 20 ? 6 from 20 ? 14 from 20 ? 5 from 20 ? 15 from 20 ? 
 
 4 from 20 ? 16 from 20 ? 3 from 20 ? 17 from 20 ? 2 from 
 
 20 ? 18 from 20 ? 19 from 20 ? 11 from 21 ? 10 from 21 ? 
 12 from 21 ? 9 from 21 ? 13 from 21 ? 8 from 21 ? 14 from 
 
 21 ? 7 from 21 ? 15 from 21 ? 6 from 21 ? 16 from 21 ? 
 
 5 from 21 ? 17 from 21 ? 4 from 21 ? 18 from 21 ? 3 from 
 
 21 ? 19 from 21 ? 2 from 21 ? 11 from 22 ? 10 from 22 ? 
 
 12 from 22? 9 from 22? 13 from 22? 8 from 22? 14 
 from 22 ? 7 from 22 ? 15 from 22 ? 6 from 22 ? 16 from 
 
 22 ? 5 from 22 ? 17 from 22 ? 4 from 22 ? 18 from 22 ? 
 
 3 from 22 ? 19 from 22 ? 2 from 22 ? 12 from 23 ? 11 from 
 
 23 ? 13 from 23 ? 10 from 23 ? 14 from 23 ? 9 from 23 ? 
 
 15 from 23? 8 from 23 ? 16 from 23? 7 from 23? 17 from 
 23 ? 6 from 23 ? 18 from 23 ? 5 from 23 ? 19 from 23 ? 
 
 4 from 23 ? 20 from 23 ? 3 from 23 ? "21 from 23 ? 2 from 
 
 23 ? 22 from 23 ? 1 from 23 ? 12 from 24 ? 11 from 24 ? 
 
 13 from 24? 10 from 24? 14 from 24? 9 from 24? 15 
 from 24? 8 from 24? 16 from 24? 7 from 24? 17 from 24? 
 
 6 from 24 ? 18 from 24 ? 5 front 24 ? 19 from 24 ? 4 from 
 
 24 ? 20 from 24 ? 3 from 24 ? 21 from 24 ? 2 from 24 ? 
 22 from 24 ? 1 from 24 ? 23 from 24 ? 
 
 16 boys went on a sailing excursion, only 7 of them return- 
 ed ; the others were drowned ; how many were lost ? If 
 from a pile of 20 apples, I take away 13, how many will be 
 left ? A man started on a journey of 23 miles ; after he had 
 traveled 16 miles, he stopped to feed his horse ; how far had 
 he then to travel? A man having 22 chickens, killed 13 of 
 them; how many were left ? From a stick of timber 19 feet 
 long, 7 feet were cut off ; what was the length of the remain- 
 der ? 19 7=howmany? 21 9=howmany? 17 9 
 r=how many ? 23 5 =how many ? 15 7 = how many ? 
 24 9=howmany? 11 7=howmany? 22 16=how 
 many? 12 -.7= how many ? 13 5= how many? 14 
 
28 SIMPLE SUBTRACTION. 
 
 7=howmany? 15 9=howinany? 16 9=howmany? 
 17 8=howmany? 18 7z=howmany? 19 ll=:how 
 many ? 20 9 = how many ? 21 7 =how many ? 22 9 
 r=how many ? 23 13=how many ? 
 
 From the preceding table and examples, the scholar will com- 
 prehend the nature of Sample Subtraction ; his next step will be 
 to practice with his slate and pencil. It will already ha.ve been 
 observed, that only two numbers are employed in a single ope- 
 ration of subtraction. The larger of these, two numbers is call- 
 ed the minuend ; and the smaller, the subtrahend. The object 
 of the rule is to find the difference between the two ; that is, to 
 find how much will remain of the larger after the smaller is taken 
 from it. The number obtained by the operation, is called the 
 remainder. 
 
 The scholar may be guided by the following rule : 
 
 RULE. 1st. Write the less of the two numbers under the 
 greater, with units under units and tens under tens, dj-c. and 
 draw a line beneath them. 
 
 2d. Commence with the right hand figure of the lower line or 
 subtrahend, and take it from the figure which stands directly 
 above it, if practicable. Do the same with the remaining fig- 
 ures in the subtrahend, if practicable, and the operation will be 
 completed. 
 
 3d. But whenever this cannot be done, that is, when the low- 
 er figure is the larger, 10 should be added to the upper figure 
 and the lower one taken from the sum. 
 
 4th. Whenever 10 is added to an upper figure, I must be car- 
 ried or added to the next lower figure ; that is, I is to be car- 
 ried whenever 10 is borrowed. 
 
 5th. To prove the work, add the remainder to the subtrahend, 
 and if the work be right, the amount will correspond with the 
 minuend. 
 
 The scholar will easily comprehend the nature of this rule, 
 unless he should find difficulty in understanding why, when we 
 borrow 10, we are required to carry only 1. He must howev- 
 er remember, that by the addition of this 10 to the upper num- 
 ber, he has increased the value of that number 10 units, 10 
 tens, 10 hundreds, or 10 thousands, according to the place 
 the figure occupies. If he add it to the units, the value of the 
 addition is 1 ten ; if to the tens, the value is one hundred, be- 
 cause 10 units make one ten, and 10 tens, one hundred, &c. 
 Now by the rule, if 10 be borrowed, 1 must be carried to the 
 
SIMPLE SUBTRACTION. 29 
 
 next lower figure ; by which operation, one more will be taken 
 from the figure in the minuend ; and this one more, which is 
 thus removed, is just equal in value to the 10 that was added, 
 for it is taken from a figure one degree farther to the left. But 
 this subject will be more 'clearly comprehended, when illustra- 
 ted by example. Take the following sum : 
 
 In this example, it is evident that if From 635428 
 6 be taken from 8, 2 will remain ; and Take 382516 
 if the 1 ten be taken from 2 tens, 1 
 
 ten will remain. But how is 5 in the Rem. 252912 
 place of hundreds to be taken from 
 
 the 4 above it ? Evidently by the third section of the rule ; that 
 is, 10 is added to the 4 in the minuend, by which addition, it 
 will become 14 hundred, from which if 5 hundred be taken, 9 
 hundred will remain, which is the third figure in the remainder. 
 But by this operation the minuend has been increased 10 hun- 
 dred ; if therefore I add 1 to the 2 thousand, it will become 3 
 thousand, and consequently, when subtracted from the figure 5 
 above it, will take one thousand more from the minuend, so that 
 only 2 will remain. If therefore 10 hundred was in one in- 
 stance added to the minuend, in the other, 1 thousand, its equal, 
 has been taken from it. The same reasoning-is applicable to the 
 8 ; 10 is added to the 3, which increases it to 13 ; the 8 is ta- 
 ken from the 13j and 5 remains. There is then one to carry to 
 the 3, which thus increased is taken from the 6, and 2 re- 
 mains. The whole remainder therefore is 252912. 
 
 Now if this remainder be added to the lower number or sub- 
 trahend, the amount will be the minuend ; which proves that 
 the operation is correct, thus : 
 
 Subtrahend = 382516 
 Remainder =252912 
 
 Minuend =635428 
 2. 3. 4. 
 
 From 66683 From 894673 From 987654321 
 Take 25966 Take 768596 Take 123456789 
 
 40717 126077 864197532 
 
 5. 6. 
 
 Min. 1000000000 Min. 1075608756 
 Subtr. 999999999 Subtr. 698453874 
 
30 SIMPLE SUBTRACTION. 
 
 7. 8. 
 
 Min. 9634657942 Min. 30000000 
 Subtr. 8326547286 Subtr. 29999999 
 
 9. 10. 
 
 Min. 100000000 Min. 90807060504 
 Subtr. 9 Subtr. 80906070400 
 
 11. 12. 
 
 Min. 965843125 56789357913 
 Subtr. 428642086 4135724,3648 
 
 13 14 
 
 812345678946 7539753168 
 488765432109 3640854279 
 
 15. 16. 
 
 23456789012 5678535347963 
 92567845023 4756246125787 
 
 17. 18. 
 
 25386744466 66668888444 
 18192939495 59999999999 
 
 19. 20. 
 
 63456956429 9888888888 
 58686721398 8999999999 
 
 21. 22. 
 
 1111111111 6666666666 
 
 8888888888 5777777777 
 
 APPLICATION. 
 
 23. A was bora in 1679 ; how old was lie in 1777 ? Ans. 
 98 years. 
 
 24. From 1600000 take 900000, and from the remainder 
 take 699999, and how much will remain? Ans. 1. 
 
SIMPLE SUBTRACTION. 31 
 
 25. A man has two flocks of sheep ; in the one there are 693, 
 and in the other, 499 ; what is the difference in these flocks 1 
 Ans. 194. 
 
 26. A man has in his possession property to the amount of 
 $15728, and he owes $7869 ; how much will remain in his 
 hands, when his debts are paid ? Ans. $7859. 
 
 27. America was discovered in 1492. How long will it 
 have been discovered in 1846 ? Ans. 354 years. 
 
 28. A man being asked how old he was when his eldest son 
 was born, said that his own age was 79 years, and his son's 
 42 years ; what was his age at the birth of his son ? Ans. 37 
 years. 
 
 29. The amount of A's debts was 2356 dollars ; the amount 
 of his property, 5672 dollars. How much had he left after his 
 debts were paid ? Ans. $3316. 
 
 30. A merchant bought a quantity of cloth for $572, and 
 sold it for $526. Did he gain or lose, and how much ? Ans. 
 Lost $46. 
 
 31. To what number must I add 576 to make the amount 
 1726? Ans. 1150. 
 
 32. Bought cotton in one month to the value of $572896 ; 
 and sold the same for $600027. How much did I gain ? Ans. 
 $27131. 
 
 33. If the sum of two numbers be 2793, and one of those 
 numbers, ]892, what is the other? Ans. 901. 
 
 34. A merchant bought 742 yards of cloth and sold all but 
 7 yards. How much did he sell ? Ans. 735 yards. 
 
 35. A man paid 1 182 dollars for a house, and sold the same 
 for 1069 dollars. How much did he lose ? Ans. $113. 
 
 36. A farmer purchased a farm, for which, including the 
 buildings, he paid $6782; the buildings were worth $2896 ; 
 what was the value of the land ? Ans. $3886. 
 
 37. A person owed a merchant $999, and paid him all but 
 $179. How much did he pay him ? Ans. $820. 
 
 Sums requiring in their solution, the application of both Addi- 
 tion and Subtraction. 
 
 38. I hold in my possession a note for $560, on which there 
 is due $70 interest. On the back are two endorsements ; one, 
 $320, and the other, $260. What is now due ? Ans. $50. 
 
 39. There are $ 1000 in four different purses ; in the first there 
 
32 SIMPLE SUBTRACTION. 
 
 are $96 ; in the second, $310 ; in the third, $205. How many 
 are there in the fourth ? Ans. $389. 
 
 40. Four men agreed to contribute for a benevolent object, 
 as follows ; the first, $34 ; the second, $50 ; the third, $100 ; 
 and the fourth, $150. Three of them having paid, the sum 
 amounted to $234 ; which subscription was unpaid ? Ans. The 
 third. 
 
 41. A had 172 yards of cloth, of which he sold 57 to B, 
 and 42 to C. How many yards were left ? Ans. 73. 
 
 42. A man having $3986, paid one debt of $1997, and another 
 of $1089. How many dollars had he left ? Ans. $900. 
 
 43. A man at death left an estate of $9876. In his will he 
 gave to each of his three sons, $1800 ; to his daughter, $1500 ; 
 and the remaining part he left to his wife. What was the 
 wife's portion ? Ans. $2976. 
 
 44. A person commencing business found that he had 
 $790 in money ; in goods, $1260 ; he held also three notes of 
 $150 each. After trading six years, he retired from business, 
 and found that his property amounted to $6000. How much 
 had he gained by trading? Ans. $3500. 
 
 45. Bought four chests of tea, weighing 72, 79, 83, and 87 
 pounds. From these I sold, to one man, 46, to another, 95, and 
 to a third, 113 pounds. How much tea had I remaining? 
 Ans. 67 pounds. 
 
 46. A man owed for his farm, $2100 ; for house furniture, 
 $156 ; for a horse, $96 ; for a yoke of oxen, $120 ; for a flock 
 of sheep, $86. In one year he sold from his farm grain to the 
 value of $462 ; butter and cheese to the value of $156 ; stock 
 to the amount of $320. How much did he owe at the end of 
 the year? Ans. $1620. 
 
 47. A man received $7000 as a legacy ; he was previously 
 worth $8560; he then commenced traveling, and in 7 years 
 he spent $9873. How much was he then worth? Ans. 
 $5687. 
 
 48. A man being asked how old he was, replied, that he 
 married at 21 years of age, and that in 19 years more he should 
 have been married 60 years. How old was he ? Ans. 62 
 years. 
 
 49. Bought 1000 pounds of coffee ; from this quantity I sold 
 at one time, 376 pounds ; and at another, 512 pounds. How 
 much had I remaining? Ans. 112 pounds. 
 
 50. A man bought two hogsheads of melasses, the one con- 
 
SIMPLE MULTIPLICATION. 
 
 33 
 
 taining 65 gallons, and the other 69 gallons ; from the two he 
 sold 112 gallons. How much had he left? Ans. 22 gallons. 
 
 QUESTIONS. How does this rule compare with Addition 1 What 
 are we taught in Subtraction ? How many numbers are employed in 
 this rule? What are they called? What is the object of the rule? 
 What name is given to what is left after the operation 1 What is the 
 rule for subtraction 1 What is to be done when the lower figure is the 
 larger? When is 1 to be carried? Do you ever have more than one 
 to carry in subtraction? Ans. We do not. How do you prove the 
 work ? How will you show that carrying one as directed, is equiva- 
 lent to the ten borrowed ? 
 
 SIMPLE MULTIPLICATION. 
 
 The rule to which the scholar's attention will now be direct- 
 ed, is one by which a number is produced from two given num- 
 bers, which shall contain either of these given numbers as many 
 times as there are units in the other ; or, it is the repeating of 
 one number as many times as there are units in the other. For 
 example, let 8 and 4 be the numbers ; that is, let 8 be repeated 
 4 times. The result of these four repetitions of 8 is obviously 
 32. But 32 contains 8 four times, or 4 eight times ; or, in other 
 words, 32 contains either number as many times as there are 
 units in the other. The scholar must first learn the follow- 
 ing table : 
 
 MULTIPLICATION TABLE. 
 
 1X1=1 
 
 2X1=2 
 
 3x1=3 
 
 4x 1= 4 
 
 1X2=2 
 
 2x2=4 
 
 3x2=6 
 
 4x 2= 8 
 
 1X3=3 
 
 2x3=6 
 
 3x3=9 
 
 4x 3 = 12 
 
 1X4=4 
 
 2x4=8 
 
 3x 4 = 12 
 
 4x 4 = 16 
 
 1X5=5 
 
 2 X 5 = 10 
 
 3 X 5 = 15 
 
 4 X 5 = 20 
 
 1X6=6 
 
 2x 6 = 12 
 
 3x 6 = 18 
 
 4x 6 = 24 
 
 1X7=7 
 
 2x 7=14 
 
 3X 7 = 21 
 
 4x 7 = 28 
 
 1X8=8 
 
 2x 8 = 16 
 
 3x 8 = 24 
 
 4x 8 = 32 
 
 1X9=9 
 
 2x 9 = 18 
 
 3x 9 = 27 
 
 4x 9 = 36 
 
 1 x 10=10 
 
 2 X 10 = 20 
 
 3 X 10 = 30 
 
 4 x 10 = 40 
 
 1 X 11 = 11 
 
 2 Xll=22 
 
 3 xll=33 
 
 4 X 11 =44 
 
 1 X 12 = 12 
 
 2 Xl2=24 
 
 3x12 = 36 
 
 4 X 12 = 48 
 
34 
 
 SIMPLE MULTIPLICATION. 
 
 5X 1= 5 
 
 6x 1= 6 
 
 7X1=7 
 
 8x 1= 8 
 
 5x 2 = 10 
 
 6x 2 = 12 
 
 7x 2 = 14 
 
 8x 2=16 
 
 5 X 3 = 15 
 
 6x 3 = 18 
 
 7x 3=21 
 
 8x 3 = 24 
 
 5x 4 = 20 
 
 6 X 4 = 24 
 
 7x 4=28 
 
 8 x 4 = 32 
 
 5x 5 = 25 
 
 6 X 5 = 30 
 
 7x 5=35 
 
 8x 5 = 40 
 
 5x 6 = 30 
 
 6 x 6 = 36 
 
 7 X 6 = 42 
 
 8x 6 = 48 
 
 5x 7 = 35 
 
 6 x 7 = 42 
 
 7x 7=49 
 
 8x 7 = 56 
 
 5X 8 = 40 
 
 6 X 8 = 48 
 
 7x 8 = 56 
 
 8x 8 = 64 
 
 5x 9 = 45 
 
 6 x 9 = 54 
 
 7x 9=63 
 
 8 x 9 = 72 
 
 5 X 10 = 50 
 
 6 x 10 = 60 
 
 7 X 10 = 70 
 
 8x10 = 80 
 
 5 X 11=55 
 
 6 X 11=66 
 
 7x 11=77 
 
 8 x 11 = 88 
 
 5 X 12 = 60 
 
 6 X 12 = 72 
 
 7 x 12 = 84 
 
 8 X 12 = 96 
 
 9X 1= 9 
 
 10x 1= 10 
 
 11X 1= 11 
 
 12 X 1= 12 
 
 9X 2= 18 
 
 10X 2= 20 
 
 11 X 2= 22 
 
 12 X 2= 24 
 
 9X 3= 27 
 
 10x 3= 30 
 
 11 X 3= 33 
 
 12X 3= 36 
 
 9X 4= 36 
 
 10X 4= 40 
 
 11 X 4= 44 
 
 12 x 4= 48 
 
 9X 5= 45 
 
 10x 5 = 50 
 
 11 X 5= 55 
 
 12 X 5= 60 
 
 9x 6= 54 
 
 10x 6= 60 
 
 11 X 6= 66 
 
 12 X 6= 72 
 
 9X 7= 63 
 
 10 X 7= 70 
 
 11 X 7= 77 
 
 12 X 7= 84 
 
 9x 8= 72 
 
 10 X 8= 80 
 
 11 X 8= 88 
 
 12 X 8= 96 
 
 9x 9= 81 
 
 10 X 9= 90 
 
 11 X 9= 99 
 
 12X 9 = 108 
 
 9x10= 90 
 
 10x10=100 
 
 11X10=110 
 
 12x10 = 120 
 
 9x11= 99 
 
 10x11 = 110 
 
 11X11 = 121 
 
 12x11=132 
 
 9x12 = 108 
 
 10x12 = 120 
 
 11X12 = 132 
 
 12x12 = 144 
 
 MENTAL EXERCISES IN MULTIPLICATION. 
 
 Bought 2 apples for 3 cents apiece ; what did they cost ? 
 2 x 3=how many ? Gave to each of 3 boys, 4 plums ; how 
 many did I give away? 3x4=how many? What cost 4 
 oranges at 5 cents apiece ? 4 x 5=how many ? What cost 6 
 oranges at 5 cents apiece ? 6 x 5=how many ? What cost 7 
 quarts of cherries at 6 cents a quart? 7x6=how many? 8 
 quarts of melasses at 7 cents per quart? 8x7=how many? 
 
 9 picture books at 8 cents apiece ? 9 x 8=how many ? What 
 cost 10 pints of wine at 9 cents a pint ? 10 x 9= how many ? 
 
 11 yards of muslin at 10 cents a yard ? 11 X 10=how many ? 
 
 12 yards at 11 cents a yard? What cost 12 loads of straw at 
 5 dollars a load ? 12 X 5= how many ? at 9 dollars a load ? 7 
 dollars? 6 dollars? 8 dollars? 10 dollars? 12 dollars? 11 
 dollars ? What will it cost to ride 6 miles at 5 cents a mile ? 
 What to ride 7 miles ? 9 miles ? 4 miles ? 5 miles ? 8 miles ? 
 
 10 miles ? 1 1 miles ? 12 miles ? Six men paid a poor man 6 
 
SIMPLE MULTIPLICATION. 35 
 
 dollars each ; how much did the poor man receive ? How 
 much would he have received had they paid 9 dollars each ? 
 7 dollars each ? 5 dollars each? 10 dollars each? 12 dollars 
 each ? 7 men purchased a horse in company and paid 9 dol- 
 lars each ; what did they pay for the horse ? What would 
 the horse have cost them, had they paid only 7 dollars each ? If 
 they had paid 8 each? 11? 10? 12 ? 6 ? If a man travel 5 miles 
 in 1 hour, how far will he travel in 7 hours ? in 5 hours ? in 
 9 hours ? in 6 hours ? in 8 hours ? in 10 hours ? in 1 1 hours ? 
 What is the product of 6 multiplied by 1 1 ? by 5 ? by 9 ? by 
 7? by 8? by 6? by 10? by 11 ? by 12 ? Whatisthe pro- 
 duct of 9 multiplied by 3 ? by 12 ? by 4 ? by 1 1 ? by 5 ? by 
 10? by 6? by 9? by 7? by 8 ? Of 12 multiplied by 7 ? by 
 2? by 12? by 3? by 11? by 4 ? by 10? by 5 ? by 9 ? by 
 6 ? by 8 ? Of 11 multiplied by 11 ? by 7 ? by 8 ? by 6 ? by 
 5? by 9? by 10? by 12? 
 
 If a man earn 9 dollars a month, what will he earn in 8 
 months ? If the board of a man and his wife be 7 dollars per 
 week, what will it amount to in 1 1 weeks ? If one load of hay 
 cost 11 dollars, what will 12 loads cost? If a yard of ribbon 
 cost 8 cents, what will 1 1 yards cost ? 
 
 It has already been said that two numbers are required for the 
 operation, viz. the multiplicand, or number to be multiplied or 
 repeated ; and the multiplier, or number showing how many 
 times the multiplicand is to be taken or repeated. The multipli- 
 er and multiplicand, when spoken of together, are called factors. 
 The number obtained by the operation is called the product. 
 A short illustration will show this rule to be an abbreviation 
 of addition. Suppose it be required to multiply 5 by 4. If the 
 scholar turn to his table he will find the product of these two 
 numbers is 20. The same result is obtained if four 5's be 
 added together; thus, 5 + 5 + 5 + 5 = 20. It will then be per- 
 ceived that if one of the two numbers which are to be mul- 
 tiplied together, be written down as many times as there are 
 units in the other, and these several numbers be then added, we 
 obtain the same result as when these two numbers are multi- 
 plied together. Multiplication is therefore a short method of 
 performing addition. 
 
 It is highly important that the scholar should obtain accurate 
 views of the value of the product arising from the multiplication 
 of any two numbers. There will be no difficulty in this respect, 
 when it is required to multiply unit figures only, or when a 
 
36 SIMPLE MULTIPLICATION. 
 
 unit figure only is given as a multiplier ; for then the product 
 of each figure will be of the same denomination as the figure it- 
 self. But when the two factors consist each of several figures, 
 so that tens are to be multiplied by tens, and hundreds by hun- 
 dreds, the scholar will not so readily comprehend the nature 
 of the operation. He must however remember, that when his 
 multiplying figure is tens, it will raise the value of the product 
 of each figure in the multiplicand one degree ; when it is 
 hundreds, it will raise the value of each two degrees ; and when 
 thousands, three degrees, &c. Let the scholar carefully notice 
 what is here statej. If tens in the multiplier be multiplied 
 into units in the multiplicand, the product is tens ; if into tens, 
 the product is hundreds. If hundreds in the multiplier be mul- 
 tiplied into units in the multiplicand, the product is hundreds ; 
 if into tens, the product is thousands ; and if into hundreds, the 
 product is tens of thousands, &c. This explanation will en- 
 able the scholar to understand the rule under Case 3d, relative 
 to writing down the several products. He will readily perceive 
 it to be nothing more or less than writing units under units, 
 and tens under tens, &c. 
 
 CASE 1st. WHEN THE MULTIPLIER DOES NOT EXCEED 12. 
 
 RULE. Commence at the right hand and multiply each fig- 
 ure in the multiplicand by the multiplier, carrying and setting 
 down as in the preceding rules. 
 
 Ex. 1. Multiply 452 by 3. 
 
 OPERATION. 
 452 
 
 3 
 
 Prod. 1356 I first say, 3 times 2 are 6, which, being 
 less than 10, I set down; next, 3 times 5 are 15, which is 5 
 units and 1 ten ; the units I set down and carry the ten ; thus, 
 3 times 4 are 12 and one to carry are 13. I thus find the whole 
 product to be 1356. This number must consequently contain 
 the multiplicand 3 times, and may be obtained by adding to- 
 gether three 452's ; thus, 4 5 2 
 The same result is therefore obtained by multiplica- 452 
 tion as by addition, but more expeditiously. The 452 
 operation is proved by dividing the product by the 
 multiplier, which, if the work be correct, will give 1356 
 the multiplicand. The scholar is not, however, supposed yet to 
 
SIMPLE MULTIPLICATION. 37 
 
 understand division, and will not be required to prove his 
 work, till more advanced. 
 
 2. Multiply 6432 by 4. OPERATION. 
 
 6432 
 4 
 
 2572 8 = Prod. 
 
 3. Multiply 123456 by 6. OPERATION. 
 
 123456 
 6 
 
 74073 6 = Prod. 
 
 4. Multiply 234567 by 8. OPERATION. 
 
 234567 
 
 8 
 
 187653 6 = Prod. 
 
 5. Multiply 345678 by 10. Prod. 3456780. 
 
 6. Multiply 456789 by 12. Prod. 5481468. 
 
 7. Multiply 729468 by 5. Prod. 3647340. 
 
 8. Multiply 295105538 by 7. Prod. 2065738766. 
 
 9. Multiply 4285637 by 9. Prod. 38570733. 
 
 10. Multiply 462838 by 11. Prod. 5091218. 
 
 11. Multiply 99887766 by 9. Prod. 898989894. 
 
 12. Multiply 765987879 by 7. Prod. 5361915153. 
 
 13. Multiply 9864579 by 8. Prod. 78916632. 
 
 14. Multiply 7799886655 by 12. Prod. 93598639860. 
 
 APPLICATION. 
 
 15. Bought 72 yards of cloth for three dollars per yard. 
 What did it cost? Ans. $216. 
 
 16. Sold 137 sheep at 5 dollars per head. How much did 
 I receive ? Ans. $685. 
 
 17. Employed 196 men one week, for $8 per week. How 
 much did I pay them all? Ans. $1568. 
 
 18. How many miles will a man travel in 297 days, if he 
 travel 12 miles a day ? Ans. 3564. 
 
 19. If I take 11 steps in one minute, how many steps shall I 
 take in 2 hours 56 minutes, or 176 minutes ? Ans. 1 936 steps. 
 
 20. If a horse trot 7 miles in one hour, how far will he trot 
 in 76 hours ? Ans. 532 miles. 
 
 4 
 
38 SIMPLE MULTIPLICATION. 
 
 CASE 3d. WHEN THE MULTIPLIER is A COMPOSITE NUMBER. 
 
 NOTE. A composite number is one which can be produced 
 by the multiplication of two or more small numbers ; thus, 6 and 
 3 are the component parts of 18 ; because 6x3 = 18. 
 
 RULE. Multiply first by one of the component parts of the 
 multiplier, and this product by the other component part ; the last 
 product will be the one sought. 
 
 Ex. 1. Multiply 4568 by 24. The component parts of 24 
 are 6 and 4, therefore, 
 
 4568 
 6 
 
 2 7 4 8= six times the multiplicand. 
 4 
 
 10963 2 =24 times the multiplicand. 
 Ex. 2. Multiply 459684 by 36. 36=4 x 9, therefore, 
 
 459684 
 9 
 
 4137156=9 times the multiplicand. 
 4 
 
 1654862 4 = 36 times the multiplicand. 
 
 It is immaterial in what order the component parts are taken. 
 
 3. Multiply 5634286 by 18. Prod. 101417148. 
 
 4. Multiply 4327648 by 27. Prod. 116846496. 
 
 5. Multiply 7295678 by 36. Prod. 262644408. 
 
 6. Multiply 4639546 by 48. Prod. 222698208. 
 
 7. Multiply 3695475 by 42. Prod. 155209950. 
 
 8. Multiply 54639578 by 60. Prod. 3278374680. 
 
 9. Multiply 578016937 by 96. Prod. 55489625952. 
 10. Multiply 79875643 by 63. Prod. 5032165509. 
 
 APPLICATION. 
 
 11. Bought 75 tons of hay at $15 per ton. What did the 
 whole cost? Ans. $1125. 
 
 12. How many hours are there in 76 days ? Ans. 1824. 
 
 13. How many minutes are there in 49 hours ? Ans. 2940. 
 
SIMPLE MULTIPLICATION. 39 
 
 14. How many days are there in 21 years ? Ans. 7665. 
 
 15. What cost 172 acres of land at $36 per acre ? Ans. 6192. 
 
 16. What will 876 pounds of coffee cost at 28 cents per 
 pound. Ans. 24528 cts. 
 
 CASE 3d. WHEN THE MULTIPLIER EXCEEDS 12, AND is NOT 
 
 A COMPOSITE NUMBER. 
 
 RULE. Multiply each figure in the multiplicand by each 
 figure in the multiplier separately, commencing with the right 
 hand figure of each, and set down the first figure of each product 
 directly under the multiplying figure. After each figure in the 
 multiplier has been taken, add together the several products ; 
 the amount will be the required product. 
 
 Ex. 1. 342635 by 125. 
 
 OPERATION. 
 
 342635 
 1 2 5 
 
 171317 5 = Product of 5 units. 
 
 68527 = Product of 2 tens removed 1 place to the left. 
 342635 = Product of 1 hundred, two places to the left. 
 
 4282937 5=:Product of 125. 
 Ex. 2. Multiply 167498 by 231. 
 
 OPERATION. 
 
 167498 
 2 3 1 
 
 16749 8 = Product of 1 unit. 
 
 50249 4 = Prod, of 3 tens removed one place to the left. 
 33499 6 = Prod, of 2 hundreds, two places to the left. 
 
 3 8 692 0~3~8 = Prod.of 231. 
 
 3. Multiply 36598674 by 432. PERFORMED. 
 
 36598674 
 432 
 
 73197348 
 109796022 
 146394696 
 
 1581062716 8=Prod. 
 
40 SIMPLE MULTIPLICATION. 
 
 4. Multiply 46354897816 by 56843. Prod. 26349514565- 
 54888. 
 
 5. Multiply 3781 99886432 by 42395. Prod. 1603378418- 
 5284640. 
 
 6. Multiply 85698436946 by 46743. Prod. 400580203816- 
 6878. 
 
 7. Multiply 6739542 by 346. Prod. 2331881532. 
 
 8. Multiply 72926495 by 4567. Prod. 333055302665. 
 
 9. Multiply 89764267 by 999. Prod. 89674502733. 
 
 10. Multiply 46371674 by 49684. Prod. 2303930251016. 
 
 11. Multiply 8429638 by 7294. Prod. 61485779572. 
 
 12. Multiply 7364951 by 888. Prod. 6540076488. 
 
 13. There are 69 pieces of cloth containing each 1 12 yards ; 
 how many yards are there in all ? Ans. 7728. 
 
 14. Suppose a man travel by steam 21900 miles in a year ; 
 how far will he travel in 67 years ? Ans. 1467300 miles. 
 
 15. In a volume of 675 pages, each page containing 156 
 lines, and each line 136 letters; how many letters ? Ans. 
 14320800. 
 
 16. How many hills are there in a field of corn, containing 
 149 rows, with 96 hills in a row ? Ans. 14304. 
 
 17. On the preceding supposition, how many ears of corn are 
 there in the field, allowing the average to be 9 to a hill ? and 
 how many kernels of corn, allowing 300 to an ear ? Ans. 
 128736 ears of corn, and 38620800 kernels. 
 
 CASE 4th. WHEN THERE ARE CYPHERS ON THE RIGHT HAND 
 
 OF THE MULTIPLIER, OR MULTIPLICAND, OR BOTH. 
 
 RULE. Omit the cyphers and multiply by the significant 
 figures only, and annex to the right hand of the product as 
 many cyphers as were omitted. 
 
 Ex. 1. Multiply 2100 by 70. 
 
 PERFORMED. 
 2100 
 
 70 I multiply the 21 by the 7 only, 
 and then annex 3 cyphers to 147 t ^r 
 
 147000 the product. 
 
 2. Multiply 47600 by 150. Prod. 7140000. 
 
 3. Multiply 9560 by 1200. Prod. 11472000, 
 
SIMPLE MULTIPLICATION. 41 
 
 4. Multiply 462000 by 190. Prod. 87780000. 
 
 5. Multiply 760 by 1000. Prod. 760000. 
 
 CASE 5th. WHEN THE MULTIPLIER is ANY NUMBER BE- 
 TWEEN 11 AND 19 INCLUSIVE. 
 
 RULE. Multiply by the right hand figure only, and place 
 the product under the multiplicand one place to the right ; then 
 add it to the multiplicand. The sum will be the true product. 
 
 Ex. 1. Multiply 468 by 15. 
 
 PERFORMED. 
 
 46 8x5 
 
 234 = Product of 468 multiplied by 5. 
 
 702 0= Product of 468 multiplied by 15. 
 
 The reason of this rule is plain. Were we to multiply in 
 the ordinary mode, the only difference would be, that the num- 
 ber 2340 would stand above 468, instead of below it. 
 
 Ex. 2. Multiply 37464 by 17. 
 
 PERFORMED. 
 
 3746 4x7 
 262248 
 
 63688 8 = Product. 
 
 3. Multiply 65328 by 13. Prod. 849264. 
 
 4. Multiply 23456789 by 14. Prod. 328395046. 
 
 5. Multiply 65432 by 15, Prod. 981480. 
 
 6. Multiply 123456 by 16. Prod. 1975296. 
 
 7. Multiply 437426 by 17. Prod. 7436242. 
 
 8. Multiply 653842 by 18. Prod. 11769156, 
 
 9. Multiply 603040 by 19. Prod. 11457760. 
 10. Multiply 999999 by 11. Prod. 10999989. 
 
 CASE 6th. WHEN THE MULTIPLIER is EITHER 21, 31, 41, 51, 
 61, 71, 81, OR 91. 
 
 RULE. Multiply by the LEFT hand figure only, and place 
 the product under the multiplicand one place to the left. 
 
 Ex. 1. Multiply 634982 by 21. 
 
42 SIMPLE MULTIPLICATION. 
 
 PERFORMED. 
 
 634 982x2 
 1269964= Product of 2 tens one place to the left. 
 
 13334622 = Product of 21. 
 
 2. Multiply 9382716 by 31. Prod. 290864196. 
 
 3. Multiply 1234567 by 41. Prod. 50617247. 
 
 4. Multiply 4364369 by 51. Prod. 222582819. 
 
 5. Multiply 6937845 by 61. Prod. 423208545. 
 
 6. Multiply 364812 by 71. Prod. 25901652. 
 
 7. Multiply 482436 by 81. Prod. 39077316. 
 
 8. Multiply 2468 by 91. Prod. 224588. 
 
 Note. When in either of the two preceding cases, cyphers 
 intervene between the figures, the same mode of operation may 
 be adopted, if care be taken to give each figure its true place. 
 
 Ex. 1. Multiply 6456 by 105. 
 
 PERFORMED. 
 645 6X5 
 
 32280= Product of the 5 placed two de- 
 grees to the right. 
 677880= Product of 105. 
 
 2. Multiply 37562 by 601. 
 
 PERFORMED. 
 
 3756 2x6 
 225372= Product of 5 hundred placed two de- 
 
 grees to the left. 
 
 22574762 = Product of 601. 
 
 3. Multiply 695378 by 5001. Prod. 3477585378. 
 
 4. Multiply 2579678 by 1007. Prod. 2597735746. 
 
 APPLICATION. 
 
 1. Bought 52 horses at $75 each; what did they cost? 
 Ans. $3900. 
 
 2. What cost 84 tons of hay at $15 per ton ? Ans. $1260. 
 
 3. If a man can travel 43 miles in one day, how far can he 
 travel in 60 days ? Ans. 2580 miles. 
 
 4. There are 144 square inches in one square foot. How 
 many square inches are there in 67 square feet? Ans. 9648. 
 
 5. If there be 18 panes of glass in one window, how many 
 are there in a house which has 56 such windows ? Ans. 1008. 
 
 6. Bought 342 bales of linen, each containing 56 pieces 
 of 25 yards each. How many yards did I buy ? Ans. 478800. 
 
SIMPLE DIVISION. 43 
 
 7. There is an orchard consisting of 126 rows of trees, and 
 in each row there are 109 trees. How many apples are there 
 in the orchard, allowing an average of 1007 on a tree ? Ans. 
 13830138. 
 
 8. A certain State contains 50 Counties ; each County, 35 
 towns ; each town, 300 houses, and each house, 8 persons. 
 What is the population of the State ? Ans. 4200000. 
 
 QUESTIONS What is the nature of Multiplication 1 How many 
 numbers are employed in the operation 1 What are they called, and 
 what is peculiar to each 1 What is the number obtained called 1 Of 
 what rule is multiplication an abbreviation 7 Illustrate. What are 
 the multiplicand and multiplier called when spoken of together'? 
 What is the value of each figure in the product when you multiply by 
 a unit figure only 1 Units multiplied by units give what 7 Units by- 
 tens 7 Units by hundreds 1 When the multiplying figure is tens, 
 what effect will it have on the value of the product of each figure in 
 the multiplicand'? and what will be the effect if the multiplying fig- 
 ure be hundreds ? Give farther illustration of the value of the pro- 
 duct figures. What is case 1st! What isihe rule 7 Case 2d7 The 
 rule 7 Case 3d 1 The rule 1 What is a composite number ? What 
 are the component parts of a number ? Case 4th 1 The rule "? Case 
 5th 1 The rule 1 Case Gth 1 The rule 7 
 
 SIMPLE DIVISION. 
 
 We now come to the reverse of the preceding rule. There 
 we had two factors given to find their product. Here we have 
 given the product, or what corresponds to it, and one of the 
 factors, and are required to obtain the other factor. Multipli- 
 cation, as was shown, could be performed by repeated addi- 
 tions ; Division may be performed by repeated subtractions. 
 Suppose it be required to ascertain how many times 4 is con- 
 tained in 12. It may be done by taking 4 from 12, till nothing 
 remains, or till a number less than 4 remains. Thus, 
 1 2 
 4 
 
 8 12 4. 
 4 
 
 4=rl2 4 + 4. 
 4 
 
 0-12 4+4+4. 
 
44 
 
 SIMPLE DIVISION. 
 
 The operation shows three 4's may be taken from 12. 4 is 
 therefore contained in 12 three times. This is, however, a 
 slow mode of operation. A more expeditious one must be 
 sought ; and, preparatory for it, the scholar is required to learn 
 the following table : 
 
 DIVISION TABLE. 
 
 2 -4- 2 = 1 
 
 4 -h 2 = 
 
 8 -4-2 = 
 10 -4- 2 = 
 12 2 = 
 14 
 
 2 i= 7 
 16-2 = 8 
 18-2 = 9 
 20 - 2 = 10 
 22 2 = 1 1 
 24 - 2 = 12 
 
 3 -f- 3 = 1 
 6 -- 3 = 2 
 9 -r- 3 = 3 
 
 12 - 3 = 4 
 15 -4- 3 = 5 
 18 ~ 3 = 6 
 21 -4- 3 = 7 
 24 -4- 3 = 8 
 27 -f- 3 = 9 
 30-4- 3 = 10 
 33 -T- 3 = 11 
 36 -4- 3 = 12 
 
 4 -f- 4 = 1 
 8 -4- 4 = 2 
 12 -r 4 - 3 
 16 -^ 4 = 4 
 20 -4- 4 = 5 
 24 -4- 4 = 6 
 28 H- 4 = 7 
 32 -r- 4 = 8 
 36 -4- 4 9 
 40 -4- 4 = 10 
 44 -f- 4 = 11 
 48 -4- 4 = 12 
 
 5 -r- 5 = 1 
 10- 5 = 2 
 15 - 5 = 3 
 20 - 5 = 4 
 25 5 = 5 
 30 - 5 = 6 
 35 - 5 = 7 
 40 - 5 = 8 
 45 - 5 = 9 
 50-5 10 
 55 -5 = 11 
 60 -4- 5 = 12 
 
 18 - 6 = 
 24-6 = 
 
 6 = 1 
 12 6 = 2 
 3 
 
 4 
 30 - 6 = 5 
 
 36 .4- 6 = 6 
 42 -4- 6 = 7 
 48 -4- 6 = 8 
 54 4- 6 = 9 
 60 -r- 6 = 10 
 66 -4- 6 = 11 
 72 -f- 6 = 12 
 
 7-4-7 = 1 
 14 _i_ 7 2 
 21 -4- 7 = 3 
 
 28 -f- 7 = 
 35 -f- 7 = 
 42 -f- 7 = 
 49 -f- 7 = 
 
 56 +. 7 = 8 
 63 -4- 7 = 9 
 70 7 = 10 
 77 _^ 7 = 11 
 84 - 7 = 12 
 
 8-4-8=1 
 16 -r- 8 = 2 
 24 -4- 8 = 3 
 32 .-=- 8 4 
 40 -r- 8 5 
 48 ~ 8 = 6 
 56 ~ 8 = 7 
 64 -f- 8 = 8 
 72 8 = 9 
 80 8 = 10 
 88 -4- 8 = 11 
 96 .i- 8 = 12 
 
 9-^9= 1 
 18 - 9 = 2 
 27 -f- 9 = 3 
 
 36 -:- 9 = 4 
 45 -f- 9 = 5 
 54 -f- 9 = 6 
 63 ~ 9 = 7 
 72 -:- 9 = 8 
 81 -f- 9 = 9 
 90 -4- 9 = 10 
 99 -r- 9 = 11 
 108 -^ 9 = 12 
 
 10 
 
 20 
 
 30 
 
 40 
 
 50 
 
 60 
 
 70 
 
 80 
 
 90 
 
 100 
 
 110 
 
 120 
 
 10 = 1 
 10 = 2 
 10 = 3 
 10= 4 
 10= 5 
 10= 6 
 10= 7 
 10= 8 
 10 = 9 
 10 = 10 
 10= 11 
 10= 12 
 
 11 -f- 11 = 1 
 
 22 -4- 11 = 2 
 
 33 -4- 11 = 3 
 
 44 -4- 11 = 4 
 
 55 -4- 11 = 5 
 
 66 -4- 1 1 = 6 
 
 77 -4- 1 1 = 7 
 
 88 - 11 = 8 
 
 99 - 11 = 9 
 
 110- 11 = 10 
 
 121 - 11 = 11 
 
 132 - 11 = 12 
 
 12 - 12 = 1 
 
 24 12 = 2 
 
 36 - 12 = 3 
 
 48 ~ 12 = 4 
 
 60 ~ ]2 = 5 
 
 72 -4- 12 = 6 
 
 84 ~ 12 = 7 
 
 96 -4- 12 = 8 
 
 108 -4- 12 = 9 
 
 120 -4- 12 = 10 
 
 132 -4- 12 = 11 
 
 144 -^ 12 = 12 
 
SIMPLE DIVISION. 45 
 
 MENTAL EXERCISES IN DIVISION. 
 
 Bought 4 apples for 8 cents ; what was the price of one ? 
 8^-4=how many? Bought 3 oranges for 12 cents ; what was 
 the price of one ? 12-^-3= how many? At another time 
 bought 5 for 15 cents ; what was the price ? 15-^-5z=how ma- 
 ny? Paid 24 cents for 6 peaches ; what did one cost ? 24-^-6 
 =how many ? What is the cost of one orange when 8 cost 
 24 cents ? when 4 cost 24 cents ? when 3 cost 24 cents ? 
 24-h8=rhow many ? 24-4-4=how many? 24n-3r=how ma- 
 ny ? Paid 35 cents for 7 melons ; what was the value of one? 
 Paid the same money for 5 ; what was the value of one ? 35 H- 
 7=how many? 35-^5z=how many? Bought 8 silks for 16 
 cents ; what was one worth ? What is the value of one when 
 I pay 24 cents for 8 ? when I pay 32 cents ? 40 cents ? 48 
 cents? 56 cents ? 16-^-8 how many ? 2 4 -=- 8= how many ? 
 32-?-8=:how many? 40-h8=how many? 4 8 H- 8= how ma- 
 ny ? 56-^8= how many? 64 8 how many? 72-^8 = 
 how many? Paid 18 shillings for 9 yards of calico; what 
 was the price per yard ? What is the price per yard when 
 the same quantity costs 27 shillings ? 45 shillings ? 36 shil- 
 lings ? 72 shillings? 63 shillings ? 1 8 ^-9 = how many ? 27 -f- 
 9=how many? 45-f-9=:how many? 36-^9z=how many? 
 72 + 9= how many ? 63 -4- 9 = how many ? What is the cost 
 of one lemon, if 10 cost 30 cents? 70 cents ? 60 cents ? 50 
 cents ? 80 cents ? 40 cents ? 90 cents ? 30^- I0=how many ? 
 &c. What is the cost of one sheep, when 8 cost 24 dollars ? 
 What, if 8 cost 64 dollars? 88 dollars? 72 dollars? 48 
 dollars? 80 dollars? 40 dollars ? 32 dollars? 24-^-8=how 
 many? 6 4 H- 8= how many ? &c. 
 
 What is the value of one bushel of wheat, when 12 bushels 
 are worth 24 dollars ? What, when the 12 are worth 36 dol- 
 lars ? 72 dollars? 96 dollars ? 48 dollars ? 108 dollars? 
 60 dollars? 132 dollars? 84 dollars? 120 dollars? 144 
 dollars? 24^-12=how many? 36-^ 12:= how many? &c. 
 Bought 8 pounds of raisins for 72 cents ; what was the price 
 per pound? If 9 pounds of rice cost 81 cents, what will one 
 pound cost ? A man traveled 96 miles in 8 days ; how far 
 was that per day ? A horse ran 54 miles in 6 hours ; how far 
 was that per hour ? How many pounds of sugar at 9 pence 
 per Ib. can be bought for 54 pence ? for 63 pence ? for 81 
 pence ? for 108 pence ? for 99 pence ? for 72 pence ? for 45 
 pence ? for 36 pence ? 
 
46 SIMPLE DIVISION. 
 
 The scholar must now be prepared to apply what he has 
 learnt from the preceding table and questions, to division on a 
 more extensive scale. He will have noticed that for each op- 
 eration two numbers are given ; viz. a number to be divided, 
 which is called the dividend ; and a number by which to di- 
 vide, called the divisor. The number obtained is called the 
 quotient; a word which signifies, how many; because this 
 number always determines how many times the divisor is 
 contained in the dividend. After the operation is performed 
 there is frequently a number left. This is called the remain- 
 der, and is always less than the divisor. When the division 
 is performed, if there be no remainder, the quotient multiplied 
 by the divisor will always produce the dividend ; and if there 
 be a remainder, the dividend will be produced by multiplying 
 as before, and adding the remainder to the product. Hence 
 division is proved by multiplication. The scholar will readily 
 perceive that these two rules are the reverse of each other. 
 
 The operations in division will be illustrated under two gen- 
 eral heads ; viz. Short Division, and Long Division. 
 
 I. SHORT DIVISION. 
 
 When the divisor does not exceed 12, the process is abbre- 
 viated by keeping the computation in the mind, and writing 
 down only the quotient figures. 
 
 RULE. 1st. Write down the dividend, and place the divisor 
 on the left, with a curve line drawn between them. 
 
 2d. Take as many figures on the left of the dividend, as 
 will contain the divisor once or more, and write the figure ex- 
 pressing the number of times, directly under those divided. 
 
 3d. If in dividing there be a remainder, imagine the next 
 figure in the dividend to be placed on the right hand of it. 
 This will form a new number, which may be divided as before. 
 Continue the same process till all the figures of the dividend 
 have been disposed of, and the number obtained will be the quo- 
 tient required. 
 
 4th. If in taking any figure of the dividend the number 
 produced be not sufficient to contain the divisor once, a cypher 
 must be placed in the quotient, and another figure of the dividend 
 taken. 
 
SIMPLE DIVISION. 47 
 
 Ex. 1. Divide 496 by 4. 
 
 PERFORMED. 
 4)496 
 
 1 2 4= quotient. 
 
 I first say 4 is in 4 once, and place the 1 as directed by the 
 rule. I next say 4 is in 9 twice and 1 remains ; I write down 
 the 2 as directed, and on the right hand of the 1 I place the 6, 
 and thus obtain 16. Lastly, I say, 4 is in 16 four times. This 
 operation gives 124 as the number of times which 496 con- 
 tains 4. Now the scholar will readily perceive, that this is re- 
 versing a process of multiplication. If he multiply 124 by 4, 
 he will obtain 16 units, 8 tens, and 4 hundreds ; and these are 
 precisely the numbers divided. But the 16 units are equal to 
 1 ten and 6 units ; the whole is, therefore, equal to 4 hundreds, 
 9 tens, and 6 units ; or to 496. 
 
 This multiplication is the proof of the work. 
 
 Ex. 2. Divide 1512 by 7. 
 
 PERFORMED. 
 7)1512 
 
 2 1 6= quotient. 
 
 The numbers here, as they are severally divided, are 15, 11, 
 and 42. If 216 be multiplied by 7, it will produce the same 
 numbers. 
 
 3. Divide 5463 by 3. 
 
 PERFORMED. 
 3)5463 
 
 1821 
 
 4. Divide 1256 by 2. 
 
 PERFORMED. 
 2)1256 
 
 628 
 
 5. Divide 63548 by 4. Quo. 15887. 
 
 6. Divide 256788 by 8. Quo. 32098 and 4 remains. 
 
 7. Divide 65342167 by 4 and by 5, and add the quotients. 
 Ans. 29403974 and 5 remains. 
 
48 SIMPLE DIVISION. 
 
 '8. Divide 735649 by 5 and by 7, and add the quotients. 
 Ans. 252221 and 9 remains. 
 
 9. Divide 456789 by 6 and by 8, and add the quotients. 
 Ans. 133229 and 8 remains. 
 
 10. Divide 68890 by 7 and by 9, and add the quotients. 
 Ans. 17495 and 7 remains. 
 
 11. Divide 78901 by 8 and by 10, and add the quotients. 
 Ans. 17752 and 6 remains. 
 
 12. Divide 89012 by 9 and by 11, and add the quotients. 
 Ans. 17982 and 2 remains. 
 
 13. Divide 90123456 by 10 and 12, and add the quotients. 
 Ans. 16522633 and 6 remains. 
 
 14. Nine persons drew a prize of $198; what was each 
 one's share? Ans. $22. 
 
 15. Paid $750 for 30 cows ; what was the average price ? 
 Ans. $25. 
 
 16. A person dying leaves an estate of $4500 to 9 chil- 
 dren ; what will be each one's share ? Ans. $500. 
 
 17. A man traveled 224 miles in 7 days ; what was his daily 
 progress ? Ans. 32 miles. 
 
 18. If 12 ounces make a pound of silver ; how many pounds 
 ajre there in 2040 ounces ? Ans. 170. 
 
 19. How many times may 12 be subtracted from 1416. Ans. 
 118. 
 
 20. Four persons boarded at a public house till the bill of 
 their board was $184 ; what was the average bill ? Ans. 46. 
 
 When the divisor is more than 12 and is a composite num- 
 ber, the same mode of operation can be adopted. 
 
 RULE 2d. Divide first by one of the component parts, and 
 the quotient arising from this division, by the other. 
 
 The only difficulty which will here present itself, will be to 
 ascertain the true remainder. The scholar needs only to re- 
 member, that if a remainder occur after the first division only, 
 that is the true remainder. If after the second division only, 
 the true remainder is obtained by multiplying this remainder 
 by the first divisor. If there be a remainder after each divi- 
 sion, the true remainder is found by multiplying the last re- 
 mainder by the first divisor, and adding the first remainder. 
 
 Ex. 1. Divide 864 by 18. The component parts of 18 are 
 6 and 3, therefore, 
 
SIMPLE DIVISION. 49 
 
 
 
 6)864 
 
 3)14 4 = Quotient of 864 divided by 6. 
 
 4 8 = Quotient of 144 divided by 3. This is the 
 true quotient of 864 18. 
 
 Ex. 2. Divide 162641 by 72. The component parts are 9 
 and 8, therefore, 
 
 9)162641 
 8 ) 1 8 7 1 and 2 remainder. 
 
 2258 and 7 remainder. Therefore, 7 X 
 9-f-2=the true remainder, viz. 65. 
 
 Ex. 3. Divide 793 by 18. The component parts are 6 
 and 3. 
 
 6)793 
 
 3)1 32 and 1 remains. 
 
 
 4 4 and nothing remains. Therefore, by rule 2d, 1 
 is the whole remainder. 
 
 Ex. 4. Divide 792 by 16. Component parts of 16 are 4 
 and 4. 
 
 4)792 
 
 4)198 and nothing remains. 
 
 4 9 and 2 remains ; therefore, by rule 2d, 4x2 
 = 8, the true remainder. 
 
 5. Divide 2592 by 63. Quo. 41, and 9 remains. 
 
 6. Divide 7776 by 108. Quo. 72. 
 
 7. Divide 6750 by 15. Quo. 450. 
 
 8. .Divide 437639 by 42. Quo. 10419, and 41 remains. 
 
 9. Divide 738246 by 27. Quo. 27342, and 12 remains. 
 
 10. Divide 60400 by 25. Quo. 2416. 
 
 11. Divide 45678 by 16. Quo. 2854, and 14 remains. 
 
 12. Divide 4688 by 48. Quo. 97, and 32 remains. 
 
 13. Divide 347628 by 84. Quo. 4138, and 36 remains. 
 
 5 
 
50 SIMPLE DIVISION. 
 
 II. LONG DIVISION. 
 
 WHEN THE DIVISOR EXCEEDS 12, AND IS NOT A COMPOSITE 
 NUMBER. 
 
 RULE. 1st. Write down the dividend, and, drawing a curve 
 line both on the right and left of it, place the divisor on the 
 left. 
 
 2d. Find how many times the divisor is contained in the 
 fewest figures that will contain it, taken from the left of the div- 
 idend ; and place the figure expressing the number of times on 
 the right of the dividend, as the first quotient figure. 
 
 3d. Multiply the divisor by this quotient figure, and place the 
 product under the figures divided. 
 
 4th. Subtract, and to the remainder bring down the next figure 
 of the dividend. 
 
 5th. Divide again and place the result as the second figure in 
 the quotient. 
 
 6th. Continue the process of multiplying, subtracting, bring- 
 ing down, <$fc. till all the figures have been divided. 
 
 7th. If after all the figures have been divided, there be still a 
 remainder, place it as the numerator, and the divisor as the de- 
 nominator of a fraction, on the right hand of the quotient. 
 
 Note 1st. Whenever a figure has been placed on the right 
 of the remainder, and the number produced will not contain the 
 divisor, a cypher must be placed in the quotient. 
 
 Note 3d. If the remainder, after subtracting, be greater than 
 the divisor, the quotient figure is too small. If the number ob- 
 tained by multiplying the divisor by the quotient figure, be 
 greater than the number divided, that quotient figure is too 
 large. 
 
 Example 1. Divide 15359 by 29. 
 
 PERFORMED. 
 
 29)15359(529 if. 
 1 4 5 
 
 8 Rem. 
 
SIMPLE DIVISION. 51 
 
 Explanation. I in the first place notice that at least three 
 figures are required to contain the divisor, and that in tl^is 
 number, 153, the divisor is contained 5 times. 5 is then my 
 first quotient figure. I then proceed to multiply and subtract 
 as the rule directs, and obtain a remainder of 8. To this, I 
 bring down the next figure of the dividend, and obtain 85. I 
 proceed tojiivide, multiply, and subtract, as before, and obtain a 
 remainder of 27, arid a quotient figure 2. Again, I bring down, 
 divide, multiply, &c. and obtain the quotient figure, 9, and a re- 
 mainder of 18, which being placed according to the 7th article 
 of the rule, gives the fraction Jf . 
 
 Note 3d. It will be observed that each figure in the quo- 
 tient is obtained by four successive operations, and that these 
 operations uniformly succeed each other in the same order. 
 In the first place, we are required to divide ; in the second, to 
 multiply the divisor by the quotient figure ; in the third, to sub- 
 tract the product of this multiplication from the figures divided ; 
 and, lastly, to the remainder thus obtained, to bring down another 
 figure of the dividend. 
 
 Ex. 2. Divide 6283459 by 29. 
 
 PERFORMED. 
 
 29)6283459(21667 I = Quotient. 
 5 8 
 
 
 
 
52 SIMPLE DIVISION. 
 
 3. Divide 74^1389 by 95. 
 
 PERFORMED. 
 
 95)7461389(78540 |f. 
 665 
 
 8 9 
 
 4. Divide 1893312 by 2076. Quo. 912. 
 
 5. Divide 455678 by 78. Quo. 5842, and 2 rem. 
 
 6. Divide 6495685 by 85. Quo. 76419, and 70 rem. 
 
 7. Divide 9424789962 by 978. Quo. 9636799, rem. 540. 
 
 8. Divide 2686211248 by 296. Quo. 9075038. 
 
 9. Divide 84764367 by 431. Quo. 196669, rem. 28. 
 
 10. Divide 4683579 by 234. Quo. 20015, rem. 69. 
 
 11. Divide 3579864 by 135. Quo. 26517, rem. 69. 
 
 12. Divide 1748 by 18. Quo. 97, and 2 rem. 
 
 Note 4th. When the divisor is 10, 100, 1000, 10000, <fec. 
 point off as many figures from the right of the dividend, as 
 there are cyphers in the divisor ; the figures on the left of the 
 point will be the quotient, and those on the right, the remainder. 
 
 Ex. 1. Divide 19375468 by 10000. Quo. 1937, rem. 5468. 
 
 2. Divide 99885566 by 100000. Quo. 998, rem. 85566. 
 
 3. Divide 47429 by 10. Quo. 4742, rem. 9. 
 
 4. Divide 463581 by 100. Quo. 4635, rem. 81. 
 
 5. Divide 618293 by 1000. Quo. 618, rem. 293. 
 
 Note 5th. When the divisor consists of a number of figures 
 with cyphers annexed, the cyphers may be cut off from the 
 divisor, and an equal number of figures from the right of the 
 dividend, and the remainder of the dividend divided by the 
 significant figures of the divisor. After the division, the figures 
 cut off from the dividend are to be placed at the right of the 
 remainder. 
 
SIMPLE DIVISION. 53 
 
 Ex. 1. Divide 36418235700 by 98700. Quo. 368979, rem. 
 8400. 
 
 2. Divide 11579112 by 890000. Quo. 13, rem. 9112. 
 
 3. Divide 8317642500 by 814600. Quo. 10210, rem. 576500. 
 
 APPLICATION. 
 
 1. If 246 men incur an expense of $175152, what is each 
 man's share ? Ans. $712. 
 
 2. A gentleman left an estate of $65468 to 6 sons ; what 
 was each one's share? Ans. $10911f. 
 
 3. Twelve men own a bridge, for which they annually re- 
 ceive $2352, in toll ; what is each man's share ? Ans. $196. 
 
 4. Suppose 7776 peach trees to be planted in 108 rows ; 
 how many trees are there in a row? Ans. 72. 
 
 5. If light comes from the sun to the earth in 8 minutes, 
 how far does it travel in one minute, the distance being 
 95000000 miles ? Ans. 11875000 miles. 
 
 6. If a man travel 9125 miles in a year, what is his ave- 
 rage daily progress ? Ans. 25 miles. 
 
 7. If a horse run 288 miles in 36 hours, how far does he 
 run in one hour ? Ans. 8 miles. 
 
 8. In 437850 yards of cloth, how many rolls of 75 yards 
 each ? Ans. 5838. 
 
 APPLICATION OF THE PRECEDING RULES. 
 
 1. A farmer sold 3 yoke of oxen at $96 each; 12 cows at 
 $24 each ; 83 sheep at $3 a head ; 239 bushels of wheat at 
 $2 per bushel, and distributed the avails equally among his 7 
 sons. What was each one's share ? Ans. $186|. 
 
 2. A man to whom was entrusted the settlement of an 
 estate, found that the whole value of the estate was $95688. 
 There were five claims against the estate ; viz. one of 
 $8672; another of $3421; a third of $10637; a fourth of 
 $356 ; and a fifth of $1673. After the payment of these seve- 
 ral claims, the balance was to be divided equally among 9 
 heirs. What was the share of each ? Ans. $7881. 
 
 3. A farmer had 16 calves worth 5 dollars per head; 45 
 sheep worth $3 per head ; and 75 bushels of grain worth $2 
 per bushel ; he gave the whole for a horse worth $136; a 
 carriage worth $195, and a harness worth $63. Did he gain 
 or lose, and how much? Ans. Gained $29. 
 
 4. A merchant received by boat, 9696 bushels of salt, and 
 
 5* 
 
54 FEDERAL MONEY. 
 
 hired it carted 16 miles, at a dollar a load of 24 bushels, 
 How much did the cartage cost ? Ans. $404. 
 
 5. When the dividend is 290864196 and the quotient 9382- 
 716, what is the divisor? Ans. 31. 
 
 6. A man purchased a farm for which he paid $18000. He 
 sold 60 acres for 50 dollars an acre, and then the remainder 
 stood him at $75 per acre. How much land did he purchase T 
 Ans. 260 acres. 
 
 GLuESTioNs. How does division differ from multiplication 1 How 
 may multiplication be performed 1 How may division be performed ? 
 How many numbers are given in division 1 What are they, and what 
 are they called 1 What is the number obtained by the operation called'? 
 What does it signify 1 What is the remainder 1 How does it always 
 compare with the divisor 1 How is division proved 1 What is short 
 division 1 What is the rule for it ? What is the rule when the divi- 
 sor is more than 12 and a composite number 1 How is the true 
 remainder obtained, when we divide by the component parts of the divi- 
 sor 7 What is long division 1 What is the rule 1 How do you di- 
 vide by 10, 100, 1000. &c.7 When the divisor consists of a figure 
 greater than 1, with cyphers annexed, how do you divide 1 
 
 FEDERAL MONEY. 
 
 Federal Money is the currency of the United States, Its- 
 denominations are Mills, Cents, Dimes, Dollars, and Eagles y 
 increasing like simple numbers in a ten-fold ratio, as represent- 
 ed in the following table ; 
 
 TABLE OF FEDERAL MONEY, 
 
 10 Mills (marked m.) make one cent, marked c. or ct, 
 10 Cents " one dime, " d. 
 
 10 Dimes " one dollar, " $ or doL 
 
 10 Dollars " one eagle, " E. 
 
 The coins of the United States are of three kinds ; viz. 
 gold, silver, and copper coins. 
 
 The gold coins are, 
 
 The Eagle =$10, 
 Half-Eagle =$5. 
 Quarter-Eagle = $2 \. 
 
 The silver coins are, 
 The dollar =100 cts. 
 Half-dollar =50 cts. 
 Quarter-dollar =25 cts. 
 Dime =10 cts. 
 
 Half-dime = 5 cts. 
 
FEDERAL MONEY. 55 
 
 The 6$ cent and 12 cent pieces, &c. are not American 
 coinage. 
 
 The copper coins are, the cent and the half-cent. Half- 
 cents are seldom used. The gold and silver coins are not 
 composed of pure metal ; but are alloys, that is, compounds of 
 these metals with the baser metals. The purity of a metal is 
 expressed by the word, carat; which word is used to ex- 
 press a twenty-fourth part of a given quantity. If, for example, 
 a quantity of gold be said to be 18 carats fine, the meaning is, 
 that 18 equal portions of the whole are gold, and 6 equal por- 
 tions are of a less valuable metal. 
 
 By recurring to the preceding table, it will be seen that the 
 denominations of federal money may be added, subtracted, 
 multiplied, and divided by the same rules as the simple num- 
 bers, as they increase in the same ratio. The scholar should 
 however remember that the dollar is the unit money ; and that 
 dimes, cents, and mills are decimals ; or tenths, hundredths, 
 and thousandths of the dollar, or unit denomination. 
 
 It is therefore always important to know which of a number 
 of figures is the dollar, or unit figure. This is shown by the 
 decimal point ( . ) or period, which always stands between the 
 dollars and dimes ; thus, 63.78, read, sixty-three dollars and 
 seventy-eight cents. 
 
 The first figure on the left of the point is dollars, and the 
 second eagles. They are however called dollars indiscrimi- 
 nately ; as in the above number, the 6 is eagles, and the 3, dol- 
 lars ; but usually read, 63 dollars. Likewise, the 78 is usually 
 read, 78 cents, instead of 7 dimes and 8 cents. Both methods 
 express the same value. 
 
 To reduce the higher denominations to the lower, it is neces- 
 sary to bear in mind, 1st. That cents are converted into mills 
 by annexing one cypher; thus, 8 cents = 80 mills. 2d. That 
 dollars may be changed into cents by annexing two cyphers ; 
 thus, 3 dollars = 300 cents ; and into mills by annexing three 
 cyphers; thus, $3 = 3000 mills. 3d. The reverse operation 
 will convert mills into cents and cents into dollars. 
 
 Ex. 1 . How many mills in 47 cents ? Ans. 470. 
 
 2. How many mills in 69 cents ? Ans. 690. 
 
 3. How many mills in 156 cents ? Ans. 1560. 
 
 4. In 78 dollars, how many cents 1 Ans. 7800. How many 
 mills ? Ans. 78000. 
 
 5. In $637, how many cents 1 Ans. 63700. How many 
 mills ? Ans. 637000. 
 
56 FEDERAL MONEY. 
 
 6. In 450 mills, how many cents ? Ans. 45. 
 
 7. In 470 mills, how many cents ? Ans. 47. 
 
 8. In 6700 mills, how many cents ? Ans. 670. 
 
 9. In 6700 mills, how many dollars ? Ans. 67. 
 
 10. Change $6 into cents. 11. Change 42 cents into mills. 
 12. Change $95 dollars into mills. 13. Change 460 mills into 
 cents. 14. 28000 mills into dollars. ] 5. 439 mills into cents. 
 16. 9876 mills into dollars. 
 
 ADDITION OF FEDERAL MONEY. 
 
 RULE. Set the numbers one under another, so that dollars 
 shall stand under dollars, dimes under dimes, cents under cents, 
 and mills under mills. Then add up the several columns as in 
 Simple Addition, and place the decimal point, in the amount, 
 directly under those in the numbers added. 
 
 If the above rule be followed in writing down the several 
 numbers, the separating points will stand directly under each 
 other. 
 
 Ex. 1. What is the sum of 136 dollars 21 cents; 75 dol- 
 lars 13 cents ; 7 dollars 78 cents ; 66 dollars 19 cents ; and 196 
 dollars 72 cents ? Ans. 482.03. 
 
 PERFORMED. 
 1 3 6.2 1 
 
 75.13 
 
 7.7 8 
 
 66.19 
 
 196.72 
 
 $482.03 Amount. 
 
 2. Add together $432.73 ; $297.38 ; $172.66 ; and $333.62. 
 Amount, $1236.39. 
 
 3. What is the sum of $1.55; $0.72; $340.89; $0.01; 
 $1460.99? Ans. $1804.16. 
 
 4. What is the sum of $72.01 ; $1 ; $0.01 ; $0.10; $40.70; 
 $560.88? Ans. $674.70. 
 
 5. What is the sum of $101.01; $20.15; $42.89; $79.81; 
 $41.41; $51.51; $38.41? Ans. $375.19. 
 
 6. What is the sum of $16.64; $20.84; $462.573; $29.922; 
 $56.32; $84.48? Ans. $670.775. 
 
 7. A farmer bought a cow for $23.75; a yoke of oxen for 
 
FEDERAL MONEY. 57 
 
 $96.78; a horse for $69.83; and a pig for $1.625; what did 
 they all cost him? Ans. $191.985. 
 
 8. A grocer paid for a box of cheese, $37.21 ; for candles, 
 $8.32; for a cask of wine, $7.38; for a box of raisins, $3.625; 
 what was the whole cost? Ans. $56.535. 
 
 9. Bought 5 gallons of melasses for $1.80; three pounds of 
 tea for $1.125; 3 yards of broadcloth for $9.82; 11 yards of 
 cotton cloth for $1.83 ; 6 yards of linen for $3.82 ; what was 
 the amount of my bill ? Ans. $18.395. 
 
 10. Paid for building my house, $2169.72 ; for my barn, 
 $972.87; for my out-houses, $1272.69; for digging my well, 
 $56.38; what is the amount of my expenses ? Ans. $4471.66. 
 
 11. My expenses for a journey were as follows: for stage 
 fare, $66.89; for meals, $18.50; for lodging, $6.58; for carry- 
 ing baggage, $2.57; for cleansing boots, $1.36 ; for washing, 
 $2.66; what is the amount? Ans. $98.56. 
 
 12. My hat cost me $4.75 ; my coat, $18.96 ; my pantaloons, 
 $9.74; my vest, $5.82 ; and my boots, $6.54; what did the 
 whole cost? Ans. $45.81. 
 
 SUBTRACTION OF FEDERAL MONEY. 
 
 RULE. Place the numbers as in Simple Subtraction, and 
 after subtracting, place the separating point as directed in Ad- 
 dition of Federal Money. 
 
 Ex. 1. From $463.42, take $399.99. Rem. $63.43. 
 
 2. From $179.364, take $88.449. Rem. $90.915. 
 
 3. From $125, take $9.09. Rem. $115.91. 
 
 4. From $642.99, take $99.99. Rem. $543. 
 
 5. From $127.01, take $41.10. Rem. $85.91. 
 
 6. From $200, take $0.90. Rem. $199.10. 
 
 7. From $2, take $0.05. Rem. $1.95. 
 
 8. From $99, take $0.99. Rem. $98.01. 
 
 9. I have $473, and my brother twice as much, wanting 90 
 cents. How much has my brother? Ans. $945.10. 
 
 10. Having in my possession $1600, I paid $516.95 fora 
 span of fine horses; $156.55 for a carriage; $221.19 fora 
 gold watch ; and spent $450.71 in traveling. How much had 
 I left? Ans. $254.60. 
 
 11. If a man has an income of $4500, and spends $2461.85, 
 how much does he lay up ? Ans. $2038.15. 
 
 12. If one man lays up $339.86 in a year, and another 
 
58 FEDERAL MONEY. 
 
 $299. 99, how much does one lay up more than the other ? 
 Ans. $39.87. 
 
 13. Bought a farm for $3946, and sold a part for $1426.82 ; 
 what did the remaining part cost me ? Ans. $2519.18. 
 
 MULTIPLICATION OF FEDERAL MONEY. 
 
 RULE. Multiply as in simple numbers. The product will 
 always be in the lowest denomination given, which may be re- 
 duced to dollars, cents, and mills, by the preceding rules. 
 
 Ex. 1. What will 36 yards of cloth cost at $4.50 per yard ? 
 Ans. $162.00. 
 
 PERFORMED. 
 4.5 
 
 3 6 
 
 2 7.0 
 1 .3 5.0 
 
 162.00 
 
 The given price being 450 cents, the whole price is 16200 
 cents, which equals $162.00. 
 
 2. What will 29 pairs of shoes cost at $1.50 per pair ? Ans. 
 $43.50. 
 
 3. What will 35 pounds of beef cost at 8 cents per pound ? 
 Ans. $2.80. 
 
 4. Bought 280 reams of paper at $2. 35 per ream; what was 
 the whole cost ? Ans. $658.00. 
 
 5. What cost 600 pounds of lard, at 15 cents per pound? 
 Ans. $90.00. 
 
 6. Bought 15 tons of hay at $16,42 per ton; what was the 
 whole cost? Ans. $246.30. 
 
 7. What cost 349 acres of land, at $15.49 per acre ? Ans. 
 $5406.01. 
 
 8. Bought 18 yoke of oxen for $72.50 per yoke ; what was 
 the whole cost ? Ans. $1305.00. 
 
 9. Bought 32 pounds of butter at 20 cents per pound ; 45 
 pounds of loaf sugar at 18 cents per pound ; 56 pounds of cof- 
 fee at 15 cents per pound; 26 pounds of tea at $1.75 per 
 pound; 21 cwt. of raisins at $6.75 per cwt. ; 42 barrels of 
 flour at $7.50 per barrel ; and 29 pairs of boots at $4.50 per 
 pair. What did the whole cost me ? Ans. $655.65. 
 
FEDERAL MONEY. 59 
 
 DIVISION OF FEDERAL MONEY. 
 
 Division of Federal Money is employed whenever the cost 
 of a number of articles, as yards, pounds, &c., is given, and 
 the price of one required. 
 
 RULE. Divide the cost by the number of articles^ and point 
 off as many figures from the quotient for cents and mills, as 
 there are in the given sum. If dollars only be given, and cy- 
 phers are added to complete the division, these cyphers must be 
 regarded as cents and mills. 
 
 Ex. 1 . If 9 pounds of butter cost $225, what is the value 
 of one pound ? Ans. $0.25. 
 
 2. Sold 69 bushels of wheat for $625 ; what was the price 
 per bushel ? Ans. $9.057. 
 
 3. Paid $75.00 for 500 Ibs. of butter ; what was the price 
 per pound? Ans. $0.15. 
 
 4. Paid $311.70 for 15 tons of hay; what was the price 
 per ton? Ans. $20.78. 
 
 5. Paid $658 for 280 reams of paper ; what did I pay per 
 ream? Ans. $2.35. 
 
 6. Paid $505.44 for 144 Ibs. of tea; what was the price of 
 one pound? Ans. $3.51. 
 
 7. Paid $375 for 50 firkins of butter ; what was the price 
 per firkin? Ans. $7.50. 
 
 8. Paid $43.79 for 29 pairs of boots ; what w r as the price 
 per pair? Ans. $1.51. 
 
 9. Paid $2.80 for 35 Ibs. of beef; what was the price per 
 pound? Ans. .08. 
 
 APPLICATION OF THE PRECEDING RULES. 
 
 1. A man dying, left an estate of $12000, which was divided 
 equally among 7 children after his wife had received her third. 
 What was the portion of the wife, and what did each child 
 receive ? Ans. $4000, wife's portion ; $1 142.857, each child's 
 portion. 
 
 2. A man settling with his grocer, finds himself charged with 
 15 Ibs. of tea at 75 cts.per pound ; 42 pounds of brown sugar 
 at 11 cts. a pound; 3 barrels of flour at $7.50 per barrel;* 8 
 gallons of lamp oil at $1.25 per gallon ; and 45 pounds of ham 
 
60 FEDERAL MONEY. 
 
 at 15 cts. per pound. He is also credited $11.62. How much 
 does he owe his grocer? Ans. $43.50. 
 
 3. A man sells a horse for $84 ; 5 cows for $25 each ; and 
 agrees to take 80 sheep in pay. How much do the sheep 
 cost him per head ? Ans. $2.61^. 
 
 4. A person agrees to furnish a grocer with 56 bushels of 
 rye at 50 cts. a bushel, and to take his pay in coffee at 15 cts. a 
 pound. How many pounds of coffee will he receive ? Ans. 
 186 J. 
 
 5. If I pay $21311 for 844 acres of land, what do I pay per 
 acre 1 Ans. $25.25. 
 
 6. Bought 350 yards of cloth at $3 50 per yard. Of this I 
 sold 200 at $5.00 per yard. How much money did I pay out; 
 how much did I receive ; how many yards had I left ; and 
 how much did it cost me per yard ? Ans. I paid out $1225 ; 
 I received $1000 ; 150 yards were left; and what remained 
 cost me $1.50 per yard. 
 
 7. A man sold his farm for $8456, and his stock for $1560 ; 
 at the same time he had on deposite in the bank, $872 97. He 
 then purchased a house in the city, for which he paid $3845 ; 
 he also purchased a horse and carriage for $392.53, and paid 
 up his old debts to the amount of $1787. How much money 
 had he left ? Ans. $4864.44. 
 
 8. Suppose the man in the preceding sum had laid out the 
 balance of his money in wagons, for which he paid $72 each ; 
 and that he took these wagons to the south and sold them for 
 $97 each. How many wagons did he purchase, and how 
 much. did he gain by the transaction, allowing his expenses to 
 have been $297.83? Ans. He purchased 67 wagons, and had 
 $40.44 left; and he gained $1377.17. 
 
 9. On the first of Jan. a spendthrift was in possession of 
 $3860.90 : after 30 days he had only $1680 remaining ; how 
 much had he spent during the whole time ; and how much 
 daily? Ans. Whole sum, $2180.90 ; daily, $72.696. 
 
 10. A man bought 20 pounds of coffee, for 15 cents a 
 pound ; and 18 pounds of sugar, at 12 cts. a pound. He paid 
 96 cents in cash, and the balance in butter at 20 cents a pound ; 
 how much butter did it take ? Ans. 21 pounds. 
 
 11. How much tea worth 56 cents a pound must be given 
 for 16 sacks of salt, worth $2.87 per sack ? Ans. 82 pounds. 
 
 QUESTIONS. What is Federal Money 1 What are its denomina- 
 tions 1 How do they increase in value? Repeat the table. What 
 are the coins of the United States 1 What are the gold coins ? What 
 
 
CANCELING. 61 
 
 are their values'? What are the silver coins 1 What are their values'? 
 Are the four-pence half-penny and the nine-penny pieces, American 
 coin'? What are the copper coins'? Are the gold and silver coins pure 
 metal'? What is their composition'? By what word is the purity of 
 a metal expressed'? What does that word express 1 If a quantity of 
 gold be said to be 18 carats fine, what is meant? Suppose it be 
 said to be 22 carats fine, what is meant 1 How may the denomi- 
 nations of Federal Money be added? Which is the unit money'? 
 What are dimes, cents, and mills ? How is the dollar or unit figure 
 always shown'? How does the period always stand 1 What is the 
 first figure on the left of the point? The second 1 ? How are they 
 usually read 1 How are the figures on the right of the point read? 
 How are cents converted into mills 1 How are dollars converted into 
 cents 1 What is the rule for the addition of Federal Money 1 How 
 is the decimal point placed? What is the rule for the subtraction of 
 Federal Money? How is the point placed? What is the rule for the 
 multiplication of Federal Money ? How many figures are cut off 
 from the right of the product 1 When is division of Federal Money 
 employed'? What is the rule? How many figures do you point off 
 in the quotient? If dollars only are given and cyphers are added, 
 how are they to be regarded ? 
 
 CANCELING. 
 
 In all arithmetical operations, it is important not only to 
 know how to solve a proposition, but also, how it may be done 
 most expeditiously . The object of this rule is to acquaint the 
 scholar with a principle by which peculiar expedition is attained 
 in the solution of such sums as involve in their operation both 
 multiplication and division. This principle is founded on the 
 following facts : 
 
 First. The value of any quotient depends on the ratio, or 
 relative size of the numbers divided ; that is, if the dividend 
 be three times as large as the divisor, the value of the quotient 
 is 3 ; and if it be four times as large, the value is 4, &c. 
 
 Second. If two or more numbers are to be multiplied 
 together, and their product divided by any other number, the 
 true result is obtained by first dividing one of these numbers 
 by the dividing number, and then multiplying the quotient by 
 the remaining number or numbers. Thus, if it be required to 
 multiply 8 by 4 and to divide the product by 2, first divide 8 by 
 2 and multiply the quotient by 4 ; thus, 8-^-2=4, and 4x4=16. 
 6 
 
62 CANCELING. 
 
 The advantage of this process will be more obvious if we 
 take large numbers. Suppose we wish to multiply 288 by 16, 
 and to divide the product by 144. The usual process would 
 be thus : 
 
 288 144)4608(32 quotient. 
 
 16 432 
 
 1728 288 
 
 288 288 
 
 460 8 = product. 000 
 
 But by first dividing, the operation is much abbreviated ; thus : 
 
 144)28 8(2 
 288 
 
 000 and 16x2 = 32. 
 
 By the usual method, 37 figures are required ; by the other, 
 only 18. There is still another advantage. The scholar can 
 see at a glance that 144 is contained in 288, twice ; and that 
 twice 16 is 32 ; so that an operation, which is long and pro- 
 tracted, is often reduced nearly or quite to a mental operation. 
 
 Third. When any large number is to be divided by the 
 product of two or more smaller numbers, it may be divided 
 by each number separately. This needs no explanation ; it is 
 the same as dividing by the component parts of any number, 
 instead of the number itself. 
 
 Fourth. When the operation is of such a nature as to re- 
 quire the product of several numbers to be divided by the pro- 
 duct of several other numbers, these numbers may be divided 
 before multiplication, and their quotients used instead of the 
 numbers themselves. For illustration, suppose the product of 
 36 and 42 is to be divided by the product of 6 and 7. The 
 usual mode of operation would be as follows, viz, : 
 
 1512 dividend. 
 
CANCELING. 63 
 
 7 x 6=42 divisor ; therefore, 4 2 ) 1 5 1 2 ( 3 6, the required 
 
 126 quotient. 
 
 252 
 
 252 
 
 But by the preceding fourth principle, 36-^-6 = 6, and 42 -f- 7 = 
 6, and 6 x 6 = 36 Ans. In this example the divisors are, as it 
 were, expunged or lost, since they divide without remainder. 
 This will not, however, always be the case. It will frequently 
 be necessary to assume some number, which will divide some 
 two given numbers, without remainder, agreeably to a rule soon 
 to be given. 
 
 But for further illustration, suppose it be required to multi- 
 ply the numbers 36, 12, 27, and 72 together, and to divide the 
 product successively by 24, 18, and 48. Now it is evidently 
 desirable to arrange these numbers so that they may be con- 
 veniently compared with each other. We will adopt the follow- 
 ing mode : We will place the numbers whose product is to 
 form a dividend, above a horizontal line ; and those whose pro- 
 duct is to form a divisor, below the same line, thus : 
 36. 12. 27. 72 = Dividend. 
 24. 18. 48 = Divisor. 
 
 Now by the fourth and last principle laid down, I can divide 36 
 in the dividend and 18 in the divisor by 18, without a remain*- 
 der, and obtain 2 in the dividend and 1 in the divisor ; thus, 
 2 12 27 72 
 'g,' -jg. I can also divide 72 in the dividend and 24 in 
 
 the divisor by 24, and obtain 3 in the dividend and 1 in the di- 
 visor (it will be remembered that the divisors stand below the 
 
 line) thus, ^~~^, Again, I can divide 12 in the divi- 
 dend and 48 in the divisor by 12, and obtain 1 in the divi- 
 dend and 4 in the divisor ; thus, ^ ' l '' 4> Again, I can divide 
 2 in the dividend and 4 in the divisor by 2, and obtain 1 in the 
 dividend and 2 in the divisor ; thus, ~ '- ' It is now evi- 
 dent that the division can be carried no farther without remain- 
 der. The next step therefore is to divide the product of the 
 numbers remaining above the line by the product of those be- 
 low it. The product of those above the line is 27x3 = 81 ; 
 and of those below the line, 2; therefore, 81-^2=40$, the 
 
64 
 
 CANCELING. 
 
 number required. The same result would have been obtained 
 by multiplying the numbers above the line and dividing their 
 product by the product of those below it, previous to canceling. 
 In the above example, as the numbers have been canceled, 
 they have been omitted, and a new statement made. This is 
 by no means necessary. One statement is sufficient. 
 
 It will be noticed that in every instance division is effected 
 without a remainder. Such must always be the case. 
 
 The following rule will be found a competent guide for the 
 scholar in all operations of canceling. 
 
 RULE 1st. Place all the numbers whose product is to form 
 a dividend, above a horizontal line ; and all those whose pro- 
 duct is to form a divisor, below the same line. 
 
 2d. Notice whether there are cyphers both above and below 
 the line ; if so, erase an equal number from each side. 
 
 3d. Notice whether the same number stands both above and 
 below the line ; if so, erase them both. 
 
 4th. Notice again if any number on either side of the line 
 will divide any number on the opposite side, without remainder ; 
 if so, divide and erase the two numbers, retaining the quotient 
 figure only on the side of the larger number. 
 
 5th. See if any two numbers, one on each side, can be di- 
 vided by any assumed number, without a remainder ; if so, di- 
 vide them by that number, and retain only their quotients. Pro- 
 ceed in the same manner as far as practicable ; then, 
 
 6th. Multiply all the numbers remaining above the line for 
 a dividend, and those remaining below, for a divisor. 
 
 7th. Divide, and the quotient will be the number required. 
 
 Note \ st. If only one number remain on either side of the 
 line, that number is the dividend or divisor, according as it 
 stands above or below the line. 
 
 Note 2d. The figure 1 is not to be regarded in the operation, 
 because it avails nothing either to multiply or divide by 1 . 
 
 In the two following examples, a separate statement will be 
 made for each step of the rule, as successively taken, with a 
 reference to the rule. 
 
 Ex. 2. Multiply together the numbers 100, 16, 24, and 36, 
 and divide their product by the product of 60, 10, and 16. 
 
 
CANCELING. 65 
 
 Statement, ^ J Q j g . By applying the second step of the 
 
 1. 16. 24. 36 
 
 rule, the statement is reduced to ~ T~r?- "Y a pply m 
 
 o. J. lt> 
 
 1 24 36 
 the third step, we obtain - L g By application of the 
 
 fourth step, we obtain 1 ^ ~ . The statement is capable of 
 
 no farther reduction, for the divisors or lower numbers are all 
 
 canceled. The product of the numbers remaining above the 
 
 line is, therefore, the quotient required ; viz. 4 X 36 = 144 Ans. 
 
 Ex. 3. Divide the product of 96, 18, 110, 5, and 42 by the 
 
 product of 50, 27, 11, and 28. Statement, 50 27 11 28' ^ 
 second step of the rule, we obtain '- : ' ' -^. By the 
 
 Qfi 1 ft JO QR O 49 
 
 third, ^-gi 1~. By the fifth, ^-| |j. 9 is assumed 
 
 for a divisor. By the fourth, ^ . By the fourth, 
 
 96. 13 96 I 
 again, ^ '-. ; and by the third, ' '- r . 96 is the 
 
 O* 1 1 
 
 only number remaining above the line, except the 1 , and noth- 
 ing but 1 remains below the line. 96 is therefore the quotient 
 sought. 
 
 To show more completely the mode of solution, the canceled 
 numbers will be retained in the following sum ; they will, how- 
 ever, be crossed, to show that they are canceled. 
 
 Ex. 4. Divide the product of 16, 9, 4, by the product of 3 
 and 32. 
 
 3 
 
 Statement, -^-OQ- By fourth step of the rule, - -oo* 
 
 3 3*6 
 
 Again, by the same, ' ' . Again, by the same, * ' ' * 
 
 2 2 
 
 and 6x3 = 18, the required quotient. 
 
 The following statements are solved without repetition, that 
 the scholar may obtain accurate views relative to the mode of 
 solution here presented. 
 
 5. Divide the product of 21, 12, 42, 100, by the product of 
 7, 35, and 50. 
 6* 
 
66 CANCELING. 
 
 States, ?LJ;. Canceled, *^ Then 
 
 5 
 
 3x6x2xl2=r432, and 432 5 = 86f. Ans. 
 
 6. Divide the product of 99, 49, 15, 20, 32, 13, 16 by the 
 product of 77, 10, 16, 49, 39, and 12. 
 
 , 99.49. 15.20.32. 13. 16 
 Statement, -^^-. 
 
 39 528 
 
 ~ , , 99,. 49. 15. Eft. S2. 13. IS 
 Canceled, __ 
 
 7 "SB 
 
 then 3x5x2x8240, and 240-f-7=34f. Ans. 
 
 7. Divide the product of 164, 88, 4 7 28, 3, 2, by the product 
 of 328, 36,21, 12,32. 
 
 11 4 
 
 Q . t 4 164. 88. 4. 28. 3. 2 ^ . , UR4. 8&. 4, SB. $. 2 
 
 Statement, -g.^. Canceled, _- & ^^ ; 
 
 29344 
 
 therefore, 11 divided by the product of 9x3x4 = 108, is the 
 Ans.; and may be thus expressed, T y . It will be observed, 
 that, in the upper numbers, the product of 3 and 4 cancels 12 
 in the lower numbers ; and generally, when the product of any 
 two or more numbers on one side equals any number on the op- 
 posite side, they may all be erased at the same time. By a 
 careful examination of the preceding seven examples, the scholar 
 will be prepared to solve the remaining examples without fur- 
 ther elucidation. He will, however, remember that the only 
 object in view is to prepare him to make a successful application 
 of the principle of canceling, to the solution of useful problems. 
 
 Note 3d. Whenever the product of the remaining numbers 
 above the line is less than the product of those below, the an- 
 swer will be in the form of a fraction, as in the last example. 
 The scholar, when he has learned respecting the nature of frac- 
 tions, will find that a number standing directly over another, 
 with a line between them, always indicates division ; that is, 
 the number above the line is divided by that below it. 
 
 8. Divide the product of 27, 111, 32, 40, 42, by the product 
 of 12, 4, 24, 12, 30. Quo. 388. 
 
 9. Divide the product of 132, 8, 49, 4, 84, 81, 42, by the 
 product of 16, 28, 7, 5, 6, 45, and 35. Quo. 
 
 
COMPOUND NUMBERS. 67 
 
 10. Divide the product of 32, 18, 72, 81, 7, 56, 63, by the 
 product of 24, 36, 162, 63, 2, 32. Quo. 147. 
 
 11. Divide the product of 96, 54, 108, 72, 56, 18, and 21, 
 by the product of 27, 81, 324, 28, 72, 3. Quo. 199. 
 
 12. Divide the product of 16, 27, 42, 44, 55, and 66, by the 
 product of 48, 88, 22, 33, and 49. Quo. 19f . 
 
 13. Divide 6, 8, 10, 5, 20, 25, 36, 48, and 60, by the pro- 
 duct of 9, 12, 15, 28, 36, 42, and 9. Quo. 201||f f 
 
 14. If 12 horses eat 15 bushels of oats, how many would 
 36 horses eat in the same time ? In solving this sum, if I 
 divide 15 bushels by 12, it will evidently give me what one 
 horse will eat, viz. 1| bushels ; and if I multiply this by 36, 
 I shall obtain the quantity that 36 horses will eat. This sum 
 may then be easily solved by the principle here introduced. 
 
 The statement would be thus : '~. 
 
 3 If 
 
 Solved,^, and 15x3=45 bushels. 
 
 But we will not anticipate : enough has already been done to 
 show that the principle is a practical one. We will leave its 
 application, for future consideration. 
 
 Q.UESTIONS. What is the object of the rule of canceling! What 
 problems may be solved on this principle! What is the first of the 
 facts on which this principle is founded ! Illustrate. What is the sec- 
 ond ! Illustrate. What is another advantage ! What is the third fact? 
 What is the fourth! Illustrate. What is the first step of the rule! What 
 is the second step ! The third ? The fourth ! The fifth ! The sixth ! 
 The seventh, and last! What is note first! What note second! 
 When the product of the remaining numbers above the line is less 
 than the product of those below, in what form will the answer be! 
 What does one number standing directly over another, with a line 
 between them, always indicate ! 
 
 COMPOUND NUMBERS. 
 
 We have thus far been operating with numbers, of the same 
 denomination, and increasing in the constant ratio of 10. 
 
 There is, however, another class of numbers composed of 
 several denominations, increasing in no uniform ratio, and 
 requiring to be separately denoted or expressed. These are 
 called Compound Numbers. Under this head are included all 
 those denominations employed to express measures of any defi- 
 
68 COMPOUND NUMBERS. 
 
 nite kind ; such as length, breadth, solidity, weight, time, money, 
 capacity, &c. 
 
 The following are the tables of these denominations. They 
 require to be made very familiar, as they show how many units 
 of each lower denomination, are equal to a unit of the next 
 higher denomination. 
 
 I. ENGLISH MONEY. 
 
 The denominations of English Money are pounds, shillings, 
 pence, and farthings. 
 
 TABLE. 
 
 d. qr. 
 
 4 farthings (marked qr.) make 1 penny, marked d. 1= 4. 
 
 12 pence " 1 shilling, " s. *1= 12= 48. 
 
 20 shillings " 1 pound, " . 1=20=240=960. 
 
 II. TROY WEIGHT. 
 
 By this weight, the precious metals, such as gold and silver, 
 also jewels and precious stones, are weighed. The following 
 are the denominations : 
 
 TABLE. 
 
 pint. gr. 
 
 24 grains (gr.) make 1 pennyweight, marked pwt. 1= 24. 
 
 01. 
 
 20 pennyweights " 1 ounce, oz. 1= 20= 480. 
 
 12 ounces " 1 pound, " Ib. 1=12=240=5760. 
 
 III. AVOIRDUPOIS WEIGHT. 
 
 By this weight, all coarse materials, such as hay and grain, 
 and also the baser metals, such as copper, are weighed. 
 
 TABLE. 
 
 oz. dr. 
 
 16 drams (dr.) make 1 ounce oz. 1= 16. 
 
 16 ounces " 1 pound Ib. 1= 16= 256. 
 
 28 pounds " 1 quarter qr. 1= 28= 443= 7168. 
 
 cwt. 
 4 quarters " 1 hun. wght. cwt. 1= 4= 112= 1792= 28672. 
 
 20 hundred " 1 ton T. 1=20=80=2240=35840=573440- 
 
 IV. APOTHECARIES' WEIGHT. 
 
 This weight is used by apothecaries and physicians in mix- 
 ing and preparing medicines. 
 
COMPOUND NUMBERS. 69 
 
 TABLE. 
 
 3- gr. 
 
 20 grains (gr.) make 1 scruple 3. 1= 20. 
 
 3 scruples " 1 dram 3. 1= 3= 60. 
 
 8 drams " 1 ounce?. 1= 8= 24= 480. 
 
 ft. 
 12 ounces " 1 pound ft. 112=96=288=5760. 
 
 V. CLOTH MEASURE. 
 This is the measure used for measuring all kinds of cloth. 
 
 TABLE. 
 
 na. in. 
 
 21 inches (in.) make 1 nail na. 1= 2j. 
 
 qr. 
 
 4 nails " 1 quarter yard qr. 1= 4= 9. 
 
 yd. 
 
 4 quarters " 1 yard yd. 1=4=16=36. 
 
 E. FI. 
 
 3 quarters " 1 Ell Flemish E. PI. 1=3=12=27. 
 
 E. E. 
 
 5 quarters " 1 Ell English E. E. 1=5=20=45. 
 
 6 quarters " 1 Ell French E. Fr. " 1=6=24=54. 
 
 VI. WINE MEASURE. 
 
 This measure is used for measuring all liquors, with the 
 exception of beer and ale. The gallon contains 231 cubic 
 inches. 
 
 TABLE. 
 
 4 gills (gi.) make 1 pint pt. 1= 4. 
 
 qt. 
 
 2 pints " 1 quart- qt. 1= 2= 8. 
 
 gal. 
 
 4 quarts 1 gallon gal. 1 = 4= 8= 32. 
 
 3H gallons " 1 barrel bar. "1=311=126=252=1008. 
 
 hhd. 
 63 gallons " 1 hogshead hhd. 1=2=63 =252=504=2016. 
 
 2 hogsheads " 1 pipe P. 
 2 pipes " 1 tun T. 
 
 VII. ALE OR BEER MEASURE. 
 
 Besides ale and beer, milk is measured by this measure. 
 The gallon contains 282 cubic inches. 
 
70 COMPOUND NUMBERS. 
 
 TABLE. 
 
 2 pints (pt.) make 1 quart qt. q \ = P %, 
 
 pal 
 
 4 quarts " 1 gallon gal. 1= 4= 8. 
 
 bar. 
 
 36 gallons 1 barrel bar. 1 =36=144=288. 
 
 54 gallons " 1 hogshead hhd. 1=1 1=54=216=432. 
 
 VIII. DRY MEASURE. 
 
 This measure is used for measuring all kinds of grain, fruit, 
 salt, coal, &c. 
 
 TABLE. 
 
 4 gills (gi.) make 1 pint pt. P i= ^4. 
 
 qt. 
 
 2 pints " 1 quart qt. 1== 2= 8. 
 
 pk. 
 
 8 quarts " 1 peck pk. 1= 8= 16= 64. 
 
 bu. 
 
 4 pecks " 1 bushel bu. 1= 4= 32= 64= 256. 
 
 ch, 
 
 36 bushels " 1 chaldron ch. 1=36=144=1152=2304=9216. 
 
 IX. LONG MEASURE. 
 
 The following denominations and numbers are used for mea- 
 suring distance. 
 
 TABLE. 
 
 in. ba. 
 
 3 barley corns (b. c.) make 1 inch in. 1= 3. 
 
 ft. 
 
 12 inches make 1 foot ft. 1= 12= 36. 
 
 yd. 
 
 3 feet make 1 yard yd. 1= 3= 36= 108. 
 
 5i yards or 16* feet 1 rod rd. ' 1= 54= 16*= 198= 594. 
 40 rods 1 furlong fur. T= 40= 220= 660= 7920= 23760. 
 8 furlongs 1 mile m. 1= 8=320=1760= 5280= 63360= 190080. 
 3 miles 1 league L. 1=3=24=960=5280=87120=1045440=3136320. 
 
 60 geographic, or69j statute miles, make 1 degree on the earth's sur- 
 face. 360 degrees make the earth's circumference. 
 
COMPOUND NUMBERS. 71 
 
 X. LAND OR SQUARE MEASURE. 
 
 TABLE. 
 
 Sq. ft. Sq. in. 
 
 144 square inches (Sq. in.) make 1 square foot Sq. ft. 1 = 144. 
 
 s. Yd. 
 9 square feet make 1 square yard Sq. yd. 1 = 9 = 1296. 
 
 rd. 
 
 30* square yards 1 square rod Sq. rd. 1= 30i= 272*= 39204. 
 
 R. 
 40 square rods 1 square rood R. 1= 40=1210 =10890 =1568160. 
 
 4 square*roods 1 Acre A. 1=4=160=4840 =43560 =6272640. 
 640 square acres 1 square mile. 
 
 Land is usually measured by Gunter's chain, which is 4 
 rods or 66 feet in length. The whole chain is divided into 
 100 equal parts, called links. The link is therefore ^ part of 
 the rod, and is 7 T 9 ( f ff inches in length. 80 chains, or 320 
 rods, make 1 mile in length. 1 square chain makes 1 6 square 
 rods, and 10 square chains make 1 acre. 
 
 XI. SOLID MEASURE. 
 
 This measure is employed in measuring substances which 
 have three dimensions ; viz. length, breadth, and thickness. 
 Timber, stone, &c., are among these substances. 
 
 TABLE. 
 
 1728 solid inches make 1 solid foot S. ft. 
 
 27 solid feet make 1 solid yard S. yd. 
 
 40 feet of round or 50 feet of hewn timber make 1 ton T. 
 
 128 sol id feet make 1 cord C. 
 
 A pile of wood 8 feet long, 4 feet wide, and 4 feet high, contains just 
 
 one cord, since 8X4X4 = 128. 
 
 XII. CIRCULAR MOTION. 
 
 TABLE. 
 
 60 seconds (") make 1 minute '. 
 
 60 minutes (' ) " 1 degree . 
 
 30 degrees " 1 Sign S. 
 
 12 signs or 360 " 1 circle C. 
 
72 
 
 REDUCTION. 
 
 XIII. TIME. 
 
 60 seconds (sec.) make 1 minute m. 
 60 minutes " 1 hour h. 
 
 24 hours " 1 day d. 
 
 1= 24= 
 
 m. tee. 
 
 1= 60. 
 
 60= 3600. 
 
 1440= 86400. 
 
 7 days make 1 week w. 1= 7= 168= 10080= 604800. 
 
 4 weeks " 1 month m. "l= 4= 28= 672= 40320= 2419200. 
 
 13 months 1 day and 6 hrs., 
 or 365 days 6 hours, 1 
 common year yr. 
 
 1=13=52=365J=8760=525960=31557600. 
 
 TABLE OF PARTICULARS. 
 
 12 particular things make 1 dozen doz. 
 
 12 dozen 
 12 gross 
 20 things 
 24 sheets 
 20 quires 
 112 pounds 
 
 gross. 
 
 great gross. 
 
 score. 
 
 quire of paper. 
 
 ream. 
 
 quintal of fish. 
 
 REDUCTION OF COMPOUND NUMBERS. 
 
 The scholar is now requested to turn back to the table of 
 English Money, and from it answer the following questions : 
 How many farthings make a penny ? How many make 2 pen- 
 nies ? How many make 4 ? How many make 6 ? How many 
 make 7 ? How many make 8 ? How many 9 ? How many 
 10? How many 11 ? How many pence make 1 shilling? 
 How many make 2 shillings ? How many make 3 ? 4 ? 5 ? 
 6? 7? 8? 9? 10? 11? 12? How many shillings make 
 1 pound ? How many make 2 ? 3 ? 4 ? 5 ? &c. Which 
 is worth most, 1 penny or 4 farthings ? 2 pence or 8 far- 
 things ? Which is worth most, 1 shilling or 12 pence ? 2 
 shillings or 24 pence ? 1 pound or 20 shillings ? 2 pounds or 
 40 shillings ? If these expressions are equal in value, in what 
 do they differ ? Ans. They are different expressions for the 
 same value ? In what then does Reduction consist ? Ans. In 
 changing numbers from one denomination to another without 
 altering the value. Reduce 6 pence to farthings. By what do 
 
REDUCTION. 73 
 
 you multiply the 6 ? Ans. By 4. And why ? Because 4 far- 
 things make one penny ; and consequently, there must be four 
 times as many farthings as pence ? Reduce 4 shillings to 
 pence. How do you reduce shillings to pence ? Reduce 3 
 pounds to shillings. How do you reduce pounds to shillings ? 
 In these last examples, were high denominations brought into 
 low, or low into high ? Ans. High denominations were brought 
 into low. How was it effected ? Ans. By Multiplication. 
 Reduce 12 farthings to pence. By what do you divide ? Ans. 
 By 4. Why? Because 4 farthings make one penny. Re- 
 duce 36 pence to shillings. By what do you divide ? Reduce 
 40 shillings to pounds ? By what do you divide ? What 
 change is here made in the denomination ? Ans. Low de- 
 nominations are brought into high. How then are low de- 
 nominations brought into high ? Ans. By Division. After a 
 careful examination of the preceding questions and remarks, 
 the scholar will readily perceive the appropriateness of the fol- 
 lowing definition. 
 
 Reduction is an operation by which a number expressing the 
 value of a quantity in one denomination is changed to another 
 number, expressing the same value in a different denomination. 
 
 But it has already been shown that high denominations are 
 brought into low by multiplication, and that low denominations 
 are brought into high by division. The scholar, therefore, 
 needs only the rules by which to guide his operation. 
 
 WHEN A HIGHER DENOMINATION IS TO BE REDUCED TO A 
 LOWER. 
 
 RULE 1st. Multiply the higher denomination by that num- 
 ber which it takes of the LOWER denomination to make ONE of 
 the higher, remembering to add to the product whatever may be 
 given of this lower denomination. If it be required to reduce 
 the quantity still lower, multiply the number already obtained 
 by the number required to reduce it to the next lower denomina- 
 tion, adding to the product the given number of that denomina- 
 tion, if any. Continue the same operation till you come to the 
 required denomination. 
 
 WHEN A LOWER DENOMINATION IS TO BE BROUGHT TO A 
 HIGHER. 
 
 RULE 2d. Divide the lower denomination by that number 
 which is required of that denomination to make one of the next 
 
74 REDUCTION. 
 
 higher. The quotient obtained will be of the higher denomina- 
 tion ; and if there be any remainder, it will be of the same' de- 
 nomination as the number divided. Divide the quotient again, 
 (if it be not already reduced to as high a denomination as prac- 
 ticable^) by the same general principle ; and continue so to do 
 till you have reached the highest denomination, of which the 
 given quantity is susceptible. 
 
 Ex. 1. Reduce 6 . 17s. 9d. 3qr. to farthings. 
 
 PERFORMED. 
 
 6. 1 7. 9. 3. 
 Multiply by 20 
 
 1 2 Or= shillings in 6 pounds. 
 1 7= given shillings added. 
 
 I 3 7 = whole number of shillings. 
 Multiply by 12 
 
 164 4 = pence in 137 shillings. 
 9 1= given pence added. 
 
 165 3=2 whole number of pence, 
 Multiply by 4 
 
 661 2 = farthings in 1653 pence. 
 3 = given farthings added. 
 
 '661 5 = whole number of farthings. 
 
 Each pound is of the value of 20 shillings ; therefore, 6 . 
 = 120 s. and 6 . and 17s.=rl37s. Each shilling is of the 
 same value as 12 pence; therefore, 137s. = 1644 d. and 1644 
 d. + 9d. = 1653d. Each penny = 4 farthings, therefore, 1653 
 d.=6612qr. and 6612qf.-f 3qr.=6615qr. 
 
 The scholar will notice that each denomination below the 
 pounds, has been added by a separate process. This is not 
 necessary ; it may be added mentally, as in the following ex- 
 ample : 
 
 2. Reduce 18 . 13s. lid. 2qr. to farthings. The thing 
 to be done is the same as before. The scholar will bear in 
 mind that the object of this example is to show that the lower 
 denominations may be added mentally. 
 
REDUCTION. 75 
 
 
 PERFORMED. 
 
 1 8 . 13s. lid. 2 qr. 
 2 
 
 3 7 3= shillings in 18 . 13s.; the 13s. being 
 1 2 added mentally. 
 
 448 7=pence in 373 s. lid.; the 11 d. being 
 4 also added mentally. 
 
 1795 0= farthings in 4487 d. 2 qr. ; the 2 qr. add- 
 ed as before. 
 
 In the above example, to the product of 18 multiplied by 
 20, I add 1 3 ; that is, I add 3 to the units and 1 to the 
 tens. And when I multiply by 12, I say, 12 times 3 shillings 
 are 36 pence, and the 1 1 given pence added make 47 pence. I 
 proceed in the same manner in reducing pence to farthings. 
 
 The preceding examples will serve to illustrate Rule 1st. 
 An illustration or two will also be given of Rule 2d, that is, of 
 bringing low denominations into high. 
 
 3. Reduce 17950 farthings to pounds, shillings, pence, and 
 farthings. 
 
 PERFORMED. 
 
 4)17950 
 
 12)448 7 2qr. = 4487d. 2 qr. obtained by first division. 
 20)37 311 d.r=373 s. 11 d. obtained by dividing by 12. 
 
 1 8 13 s18 pounds obtained by dividing by 20, 
 and 13s. remain. 
 
 Since 4 farthings make 1 penny, it is evident that there are 
 as many pence in 17950 qrs. as there are 4's contained in it. 
 The same reasoning may be applied to the other divisors. 
 
 It will be observed, that in this last example we have reversed 
 what was done in the second example. We there had the 
 same value given, which we have here ; but were required to 
 change it from a higher to a lower denomination, instead of 
 from a lower to a higher, as in the last example. 
 
 4. Reduce 44447 farthings to pence, shillings, and pounds. 
 
76 REDUCTION. 
 
 PERFORMED. 
 
 4)44447 
 
 12)1111 1 3 qr. remain. 
 
 20)92 5 1 1 pence remain. 
 
 4 6 5 shillings remain. 
 The answer, therefore, is 46 . 5 s. 1 1 d. 3 qr. 
 
 5. Reduce 22685 qr. to pounds, &c. 
 
 PERFORMED. 
 
 4)22 685 
 
 12)567 1 1 qr. remains. 
 20)47 2 7 pence remain. 
 
 2 3 12 s. remain. 
 The answer is, 23 . 12 s. 7 d. 1 qr. 
 
 6. Reduce 7195 pence to pounds, &c. 
 
 The scholar will observe that the given number is already 
 pence. 
 
 PERFORMED. 
 
 1 2)7 1 9 5 
 
 20)59 97 d. remain. 
 
 2 9 19s. remain. 
 The answer then, is, 29 . 19 s. 7 d. 
 
 The scholar must first consider whether the quantity given 
 is to be brought from a higher denomination to a lower, or 
 from a lower denomination to a higher. When this is deter- 
 mined, let him apply the corresponding rule. 
 
 TABLE I. ENGLISH MONEY. 
 
 6. Reduce 23 . 12s, 7d. 1 qr. to farthings. Ans, 22685 
 qr. 
 
 N. B. 23 . 12s. 7d. 1 qr.=22685qr. So in all opera- 
 tions of mere reduction, the quantity given equals in value the 
 quantity obtained 
 
 7. Reduce 71 . 13s, 2 d. 3qr, to farthings, Ans. 68795 
 qr. 
 
REDUCTION. 77 
 
 8. Reduce 299924 qr. to pounds, shillings, &c. Ans. 312 . 
 8 s. 5 d. 
 
 9. Reduce 68795 qr. to pounds, shillings, &c. Ans. 71 . 
 13s. 2d. 3qr. 
 
 10. Reduce 46 . 5 s. 11 d. 3 qr. to farthings. Ans. 44447. 
 
 11. Reduce 29 . 19s. 7d. to pence and farthings. Ans. 
 7195d. ; 28780 qr. 
 
 12. Reduce 5974681369 qr. to pounds, &c. Ans. 6223626 
 . 8 s. 6 d . 1 qr. 
 
 13. Reduce 7195 pence to pounds, &c. Ans. 29 . 19s. 7d. 
 
 14. Reduce 40320 half-pence to pounds. Ans. 84 . 
 
 15. Reduce 125 . 19s. lid. 3qr. to farthings. Ans. 
 120959 qr. 
 
 16. Reduce 475 dollars, at 6 shillings each, to pence. Ans. 
 34200 pence. 
 
 17. Reduce 312 . 8 s. 5 d. to half-pence. Ans. 149962 
 half-pence. 
 
 18. Reduce 121 pistoles, at 22 shillings each, to pence and 
 farthings. Ans. 31944d. ; 1 27776 qr. 
 
 19. Reduce 34200 pence to dollars, at 6 shillings each. 
 Ans. 475 dollars. 
 
 20. Reduce 359548 qr. to pistoles, at 22 shillings each. 
 Ans. 340 pistoles, 10s. 7 d. 
 
 21. Reduce 740 dollars, at 8 shillings each, to pence. Ans. 
 71040 pence. 
 
 22. Reduce 79 . to pence and half-pence. Ans. 18960 d. ; 
 37920 half-pence. 
 
 APPLICATION OF TABLE II. TROY WEIGHT. 
 
 Ex. 1. Reduce 23 Ib. 9oz. 6pwt. 22 gr. to grains. 
 
 PERFORMED. 
 
 2 3. 9. 6. 2 2. 
 1 2 oz. = l Ib. 
 
 2 8 5 ounces in 23 Ib. 9 oz. 
 2 Opwt.^1 oz. 
 
 570 6= pwt. in 285 oz. and 6 pwt. 
 2 4 gr.=:l pwt. 
 
 22846 
 11412 
 
 Ans. 13696 6 = grains in 5706 pwt. and 22 gr. 
 
78 REDUCTION. 
 
 2. Reduce 8578 grains to pounds, &c, 
 
 PERFORMED. 
 
 24)8578 
 
 20)35 7 lOgr. remain. 
 12)1 7 17pwt. remain. 
 
 1 5 oz. remain. 
 Ans. I Ib. 5oz. 17pwt. lOgr. 
 
 3. Reduce 62 Ib. 7oz. 14pwt. 18gr. to grains. Ans. 360834 
 
 S 1 "- 
 
 4. Reduce 360834 grains to pounds. Ans. 62 Ib. 7oz. 14 
 
 pwt. 18gr. 
 
 5. Reduce 1 Ib. 11 oz. 19 pwt. 23 gr. to grains. Ans. 11519 
 
 g r - 
 
 6. Reduce 11519 grains to pounds, <fcc. Ans. lib. 11 oz. 
 
 19 pwt. 23 grains. 
 
 APPLICATION OF TABLE III. AVOIRDUPOIS WEIGHT. 
 
 Ex. 1. Reduce 4 cwt. 3 qr. 26 Ib. 10 oz. 12 dr. to drams. 
 
 PERFORMED. 
 
 4. 3. 26. 10. 1 2. 
 
 4 Multiply by 4, because 4 quarters = 1 cwt. 
 
 1 9 
 
 2 8 Multiply by 28, because 28 pounds = 1 qr. 
 
 1 78 
 3 8 
 
 558 
 1 6 Multiply by 16, because 16 oz.^1 Ib. 
 
 8938 
 
 1 6 Multiply by 16 again, because 16 dr. = l oz. 
 
 53640 
 8938 
 
 143020 
 
REDUCTION. 79 
 
 The same reversed : 
 
 1 6)1 43020 
 
 1 6)893 8 12 dr. remain. 
 28)558 lOoz. " 
 4)19 26 Ib. " 
 
 4_3 qr . 
 Ans. 4 cwt. 3 qr. 26 Ib. 10 oz. 12 dr. 
 
 2. Reduce 7 cwt. 3 qr. 19 Ib. to ounces. Ans. 14192 ounces. 
 
 3. Reduce 14192 ounces to pounds, quarters, &c. Ans. 
 7 cwt. 3qr. 191b. 
 
 4. Reduce 29548 ounces to hundreds, quarters, &c. Ans. 
 16 cwt. 1 qr. 26 Ib. 12 oz. 
 
 5. Reduce 480 drams to pounds, &c. Ans. 1 Ib. 14oz. 
 
 6. Reduce 12 tons to ounces. Ans. 430080 oz. 
 
 7. Reduce 430080 ounces to tons. Ans. 12 tons. 
 
 APPLICATION OF TABLE IV. APOTHECARIES* WEIGHT. 
 
 Ex. 1. Reduce 8 ft. 8 1. 5 3. 2 3. 12 gr. to grains. 
 
 8. 8. 5. 2. 1 2. 
 1 2 oz.=llb. 
 
 1 04 
 
 837 
 
 33,= 13 
 
 2513 
 
 20gr. = 
 
 50272 Ans. 
 The same reversed : 
 
 20)50272 
 
 3)251 3 12 remain. 
 
 8)83 7 2 " 
 12)10 4 5 " 
 
 88 
 
oU REDUCTION. 
 
 3. Reduce 27 ib. 9 . 6 3. 2 3. to scruples. Ans. 8012 
 scruples. 
 
 4. Reduce 477816 grains to pounds, &c. Ans. 82 Jb. 11 f. 
 33. 13. 16 gr. 
 
 5. Reduce 6 ft. 7 . 6 3. 13. 12 gr. to grains. Ans. 38312 
 grains. 
 
 6. Reduce 6348 scruples to pounds. Ans. 22 ib. 4 3. 
 
 7. Reduce 480 drams to pounds. Ans. 5 ft>. 
 
 8. Reduce 1 ft>. 1 . 1 3. 1 3. 1 gr. to grains. Arcs, 6321 
 grains. 
 
 APPLICATION OF TABLE V. CLOTH MEASURE. 
 
 Ex. 1. Reduce 30 yd. 3 qr. 3na. to nails. 
 
 PERFORMED. 
 
 30. 3. 3. 
 
 4 = lyd. 
 
 1 2 3 
 
 4 = lqr. 
 
 495 
 
 The same reversed : 495 nails =how many qr ? 
 4)4 9 5 
 
 4)12 3 3 na. remain. 
 3 yd. 3 qr. " 
 
 2. Reduce 450 E. E. 3 qr. 2 na. to nails. Ans. 9014 nails. 
 
 3. Reduce 678 ells Flemish to nails. Ans. 8136 nails. 
 
 4. Reduce 8136 nails to ells Flemish. Ans. 678 ells. 
 
 5. Reduce 9014 nails to ells English. Ans. 450 E. E. 
 3 qr. 2 na. 
 
 6. Reduce 12 ells French, 5 qr. 3 na. to nails. Ans. 311 
 nails. 
 
 7. Reduce 622 nails to ells French. Ans. 25 ells, 5 qr. 2 na. 
 
 APPLICATION OF TABLE VI, WINE MEASURE. 
 
 Ex. 1. Reduce 3 hhd. 42 gal. 2 qt. 1 pt. to pints. 
 
REDUCTION. 81 
 
 
 PERFORMED. 
 
 3. 42. 2. 1. 
 
 6 3 
 
 1 1 
 
 2 2 
 
 2 3 1 
 4 
 
 926 
 2 
 
 1853 Ans. 
 The same reversed : 2)1853 
 
 4)92 6 1 remains. 
 63)23 12 
 
 3 42 " 
 Ans. 3 hhd. 42 gal. 2 qt. 1 pt. 
 
 2. Reduce 6 pipes, 1 hogshead, and 1 gill, to gills. Ans. 
 26209 gills. 
 
 3. Reduce 30000 gills to pipes. Ans. 7 pipes, 55 gal. 2 qt. 
 
 4. Reduce 20 tuns to gills. Ans. 161280 gills. 
 
 5. Reduce 5 hhd. 56 gal. 2 qt. to pints. Ans. 2972 pints. 
 
 6. Reduce 16 barrels, 21 gal. to quarts. Ans. 2100. 
 
 APPLICATION OF TABLE VII. ALE OR BEER MEASURE. 
 
 1. Reduce 4 hhd. 45 gal. 3 qt. to pints. 
 
 4. 45. 3. 
 5 4 
 
 2094 Ans, 
 
82 REDUCTION. 
 
 The same reversed : 2)2094 
 4)1047 
 
 54)26 1 3 qt. remains. 
 4 45 gal. " 
 
 2. Reduce 47 barrels of beer to pints. Ans. 13536 pints. 
 
 3. Reduce 13672 pt. to barrels, &c. Ans. 47 bar. 17 gal. 
 
 4. Reduce 451 bar. 7 gal. to quarts. Ans. 64972 quarts. 
 
 5. Reduce 21 hhd. to quarts. Ans. 4536 quarts. 
 
 6. Reduce 6562 pints to hogsheads. Ans. 15 hogsheads, 
 10 gal. Iqt. 
 
 APPLICATION OF TABLE VIII. DRY MEASURE. 
 
 Ex. 1. Reduce 6 ch. 9 bu. 3 pk. to gills. Ans. 57792 gills. 
 
 2. Reduce 87762 gills to chaldrons. Ans. 9ch. 18 bu. 3 
 pk. 2 qt. 2 gi. 
 
 3. In 1 36 bushels, how many pecks, quarts, and pints ? Ans. 
 544 pk. 4352 qt. 8704 pt. 
 
 4. Reduce 10640 pints to bushels. Ans. 166 bu. 1 pk. 
 
 5. Reduce 3 pecks to gills. Ans. 192 gills. 
 
 6. Reduce 720 quarts to bushels. Ans. 22 bu. 2 pk. 
 
 APPLICATION OF TABLE IX. LONG MEASURE. 
 
 1. Reduce 8 leagues, 2 miles, 6 furlongs, 16 rods, 3 yards, 
 2 feet, 9 inches, and 2 barley corns, to barley corns. Ans. 
 5094569 b. c. 
 
 2. How many barley corns will reach round the earth, it 
 being 360 degrees ? Ans. 4755801600. 
 
 3. Reduce 48765000 barley corns to miles, &c. Ans, 
 256 m. 4 fur. 15 rods, 15 ft. 10 inches. 
 
 4. Reduce 26431 rods to miles, &c. Ans. 82m. 4 fur. 31 
 rods. 
 
 5. Reduce 1710720 inches to miles. Ans. 27 miles. 
 
 6. Reduce 7 fur. 36 rods, and 9 ft. to inches. Ans. 62676. 
 
 APPLICATION OF TABLE X. LAND OR SQUARE MEASURE. 
 
 1. Reduce 500 acres to square rods. Ans. 80000 rods. 
 
 2. Reduce 32000 rods or poles to acres. Ans. 200 acres. 
 
 3. Reduce 3 square miles to square rods. Ans. 307200. 
 
 4. Reduce 458000 square rods to square miles, &c. Ans. 
 4 sq. m. 302 acres, 2 roods. 
 
REDUCTION. 83 
 
 5. Reduce 6272640 square inches to acres. Ans. 1 acre. 
 
 Note. For figures to illustrate square and solid measure, 
 the scholar is referred to square and cube root. 
 
 APPLICATION OF TABLE XIj SOLID MEASURE. 
 
 1. In a pile of wood containing 2 cords and 64 feet, how 
 many solid inches ? Ans. 552960. 
 
 2. Reduce 884736 solid inches of wood to cords. Ans. 4 
 cords. 
 
 3. Reduce 6117120 cubic inches to tons of round timber. 
 Ans. 88 tons, 20 ft. 
 
 4. Reduce 72 tons of hewn timber to cubic inches. Ans. 
 6220800. 
 
 APPLICATION OF TABLE XII. CIRCULAR MOTION. 
 
 1. Reduce 6 signs, 21 degrees, 40 minutes, to seconds. 
 Ans. 726000 seconds. 
 
 2. Reduce 432000 seconds to signs. Ans. 4 signs. 
 
 3. Reduce 1 circle, 6 signs, 25 degrees, to minutes. Ans. 
 33900 m. 
 
 4. Reduce 45200 minutes to circles, &c. Ans. 2 cir. 1 S. 
 3. 20'. 
 
 It is desirable that the scholar 
 should obtain correct views of 
 this measure. The adjoining 
 figure will illustrate its applica- 
 tion. The circle is regarded as 
 an integral object. Ks first divi- 
 sion is into 12 signs. These are 
 represented by the figures, 1 , 2, 
 3, 4, &c. Each sign is divided 
 into 30 equal parts : these con- 
 stitute degrees ; and as the circle 
 contains 12 signs, it is evident the whole circle must contain 
 360 degrees. These again are divided into minutes, and the 
 minutes into seconds. The degrees, minutes, and seconds are 
 not represented in the figure. This measure is obviously ap- 
 plied to bodies moving in a circle ; such as all wheels in ma- 
 chinery, the revolutions of the heavenly bodies, &c. 
 
84 
 
 REDUCTION. 
 
 APPLICATION OF TABLE XIII. TIME. 
 
 1. Reduce 360 years, 300 days, 20 hours, 50 minutes, and 
 37 seconds, to seconds. Ans. 11386731037. 
 
 In the above sum, 365 days and 6 hours are allowed to the 
 year. 
 
 2. Reduce 662709600 seconds to years, &c. allowing the 
 year to be as above. Ans. 21 years. 
 
 3. Reduce 49 weeks to seconds. Ans, 29635200 sec. 
 
 4. Reduce 59270400 seconds to years, &c. Ans. I year, 
 46 weeks. 
 
 5. Suppose my age to be 21 years, how many seconds have 
 I lived? Ans. 662709600. 
 
 6. Reduce 1325419200 seconds to hours. Ans. 368172 
 hours. 
 
 The following particulars require to be introduced here. 
 The year, as given above, contains 365 days and 6 hours. In 
 four years, this six hours amounts to 24 hours, or one day. 
 Hence every fourth year has 366 days. This is called Bis- 
 sextile or Leap year. As it requires four years to gain this 
 one day, it is obvious that the Leap year may be found by divi- 
 ding the given year by 4. If it be Leap year it will divide 
 without remainder. If in dividing there be 1 remaining, the 
 year under consideration is the first after Leap year. If 
 there be 2 remaining, it is the second, and if 3 remain, it is the 
 third after Leap year. Thus 1824, is Leap year, for 1824-^4 
 456, and. no thing remains. The same operation shows 1826 
 to be the second after Leap year. In the table the year is divi- 
 ded into 13 months. These are lunar months. It is much 
 more usually divided into 12 calender months, containing each 
 the following number of days, viz. April, June, September, and 
 November, 30 ; January, March, May, July, August, October, 
 December, 31 ; and February, 28. To this last month, (Febru- 
 ary,) the additional day of the Leap year is added, so that 
 every fourth year, this month has 29 days. 
 
 PROMISCUOUS EXAMPLES. 
 
 1. In 64126 gills, how many bushels ? Ans. 250 bu. 1 pk. 
 7qt. 1 pt.2gi. 
 
 2. In 26709912 barley corns, how many leagues 1 Ans. 
 46 1. 2m. 4 fur. 6 rods, 1 yd. 
 
 3. In 161280 gills, how many tuns of wine ? Ans. 20. 
 
REDUCTION, 85 
 
 4. In 10 cords of wood, how many solid inches? Ans. 
 2211840. 
 
 5. In 20 hhd, of sugar, each 12 cwt. how many pounds ? 
 Ans. 26880. 
 
 6. How many gills in 250 bu. 1 pk. 7 qt. 1 pt. 2 gills ? Ans. 
 64126. 
 
 7. How many pence are there in 16 bags, containing each 
 24 guineas, 16 shillings, and 8 pence, the guinea being 28s.? 
 Ans. 132224. 
 
 8. In 15840 yards, how many leagues ? Ans. 3. 
 
 9. In 1876742 solid inches, how many cords, &c. ? Ans. 
 8 cords, 62ft. 134 in. 
 
 Thus far, examples have been avoided, which in their solu- 
 tion required both multiplication and division. They will here 
 be introduced. Let us take the following. How many ells 
 French in 15 pieces of cloth, containing each 20 yards ? Now 
 it is evident that yards cannot be changed to ells French, at a 
 single step. When the question is, how many times one quan- 
 tity is contained in another, a simple operation of division is 
 often all that is required, to obtain the answer. But that will 
 not suffice here, and the reason is, the two quantities are not 
 in the same denomination. The first step then, is, to bring the 
 two quantities given to the same kind. The scholar will there- 
 fore turn to Table 5th, Cloth Measure. He will there find that 
 the ell French and the yard, may both be brought into quar- 
 ters ; the former, by being multiplied by 6, and the latter, by 4. 
 Therefore, 
 
 1 5 pieces. 
 
 2 yd. in a piece. 
 
 3 yd. in 15 pieces. 
 4r=qr. in a yard. 
 
 6)120 0=rqr. in 300 yards, or 15 pieces. 
 
 2 ells French, the quantity required. 
 
 There are evidently as many ells French as there are 6's 
 in 1200qr. as 6 qr. make one ell French. 
 
 The scholar will perceive that the given quantity must be 
 brought first into a denomination, from which it may be changed 
 to the required denomination. 
 
86 REDUCTION. 
 
 The above work may be abbreviated by applying to the solu- 
 tion the principle explained in the rule for canceling. The 
 numbers 15, 20, and 4, are multiplied together and produce the 
 dividend, 1200. This is then divided by 6, and the result is 
 the answer. The scholar may therefore turn back to the rule 
 for canceling, and he will see that the following statement is in 
 
 accordance with it ; viz. : . This statement, by Sec. 
 5th, of the same rule, may be reduced to ' ' - ; and again, by 
 
 Sec. 4th, to 5 ' 2 p ' ; and by 6th and 7th Sec. to 200, answer 
 as before. 
 
 We will now give the solution at a single statement, thus : 
 5 2 
 
 1S - ' ' . and 5x20x2=200 ells French, the answer. 
 ft 
 % 
 
 2. How many barrels, each holding 2 bushels and 3 pecks, 
 are required to contain 880 bushels of corn ? 
 
 The 2 bushels and 3 pecks equal 11 pecks. The question 
 then is, how many times are 1 1 pecks contained in 880 bush- 
 els. By the preceding solution, it will be seen that the bush- 
 els, as they are to be divided by 11 pecks, must also be 
 brought into pecks. Therefore, 
 
 880 
 
 4 =pecks in one 
 
 bushel. 
 
 One barrel =11 pecks, therefore, 11)352 0=pecks in 880 
 
 bushels. 
 
 320 = number bar- 
 rels required. 
 
 The same canceled. The scholar will read over the explana- 
 tion of the preceding sum, if he does not yet understand the 
 work. 
 
 Statement, ^r- . Canceled, (see Sec. 4, rule for cancel- 
 
 80 
 ing,) ' , and 80 x 4 = 320 barrels, the same as before. 
 
 3. In 33 guineas, at 28 shillings each, how many pistoles, 
 each 22 shillings ? 
 
 The simple question is, how many pistoles are there in 33 
 
REDUCTION. 87 
 
 guineas. The guineas cannot be divided by the pistoles, for 
 they are of different value. They must be brought upon some 
 common ground, and the nearest is that of shillings ; therefore, 
 
 33 
 
 2 8 
 
 264 
 6 6 
 
 22 s. = l pistole, therefore, 22)92 4= shillings in 33 guineas. 
 
 4 2= number of pistoles in 
 the guineas. 
 
 33 28 
 The same canceled : - L ^~. (See rule for canceling, Sec. 4th 
 
 3 14 
 and 5th.) Performed, ^-- , and 14x3=42, the number of 
 
 S 
 
 pistoles, as before. 
 
 4. Purchased 24 hogsheads of wine, at 1 s. 8 d per quart, 
 and paid for the same with cloth, for which I was allowed 3 s. 
 4 d. per yard. How much cloth was required ? 
 
 The price of the quart being given, it is obvious that the 24 
 hhd. must first be brought to quarts. This is done by multi- 
 plying the 24 by 63 and by 4. Each quart is worth Is. 8 d. 
 =20 d. Therefore, multiplying the quarts by 20 d. gives the 
 cost of the wine in pence. Again, each yard of cloth is worth 
 3 s. 4 d. = 40 d. If, then, the pence the wine cost be divided 
 by the price of one yard of cloth, the quotient obtained must 
 be the number of yards required ; therefore, statement for can- 
 celing, . Let this statement be compared with the 
 above analysis of the sum, and with the rule for stating sums 
 
 , , 24. 63. 4, 2fl 
 for canceling. Statement repeated and canceled, ; 
 
 2 
 
 therefore, 24x63x2=the number of yards required; viz. 
 3024 yds. 
 
 We will now give the usual solution of this sum, and observe 
 the increased facility of canceling. 
 
88 REDUCTION, 
 
 1 5 1 2 = the gallons in 24 hhd. 
 4 
 
 6 4 8=qt. in the same, 
 2 
 
 4 t O ) 1 2 9 6 1 Oncost of the same in pence. 
 
 302 4=yds. of cloth required,, the sameas before, 
 40 d. = price of 1 yd. of cloth; therefore I divide by 40. 
 
 The scholar will readily perceive the object to be attained in 
 sums of this character ; viz. to exchange dissimilar quantities, 
 or quantities of different denominations, but of equal value. 
 
 RULE FOR THE COMMON MODE OF OPERATION. Reduce the 
 quantity to be exchanged to the denomination in which the 
 price, or the equivalent of exchange of the other kind, is 
 given ; then divide by this price, or equivalent of exchange, 
 and perform such operations of reduction as the nature of the 
 case may require. 
 
 RULE FOR CANCELING. Consider what is the quantity to be 
 exchanged, and place it over a horizontal line towards the left. 
 Then on the right of this, also above the line, place such numbers 
 as are required to reduce this quantity to the denomination in 
 which the price, or equivalent of exchange of the other kind is 
 given. Write also under the line those numbers which are ne- 
 cessary to reduce this price, or equivalent of exchange, to the 
 required denomination. Proceed to cancel, multiply, and divide, 
 as directed in the rule for canceling, and the number obtained 
 will be the one sought. 
 
 Note 1. In stating for canceling, care should be taken to 
 introduce into the statement every number required for the 
 complete solution of the same, including all operations of 
 reduction, &c., since each number introduced increases the 
 opportunity for canceling, and thus abbreviates the operation. 
 
REDUCTION. 89 
 
 Note 2. If any one term consists of more than one de- 
 nomination, it should be reduced to the lowest denomination 
 given before stating. 
 
 5. How many times will a wheel 18 feet in circumference, 
 turn round in traveling 84 miles ? 
 
 The thing to be done is to change 84 miles into revolutions 
 of the wheel. To do this, 84 miles must be reduced to feet, 
 because the distance required for one revolution is given in feet, 
 viz. 18. Therefore, 
 
 84 
 8 
 
 6 7 2 the furlongs in 84 miles. 
 4 
 
 2 6 8 8 0= the rods in 84 miles. 
 16* 
 
 161280 
 26880 
 13440 
 
 44352 O^the feet in 84 miles. 
 
 Then, 18)443520(24640 Ans. 
 3 6 
 
 72 
 72 
 
 000 
 
 The preceding solution is by the first rule. We will now 
 
 solve the sum by canceling. Statement, - . Ob- 
 
 18 
 
 serve that the numbers above the line are those multiplied 
 together for a dividend in the preceding operation, and that the 
 one below the line was the divisor. 
 8* 
 
90 REDUCTION, 
 
 To avoid the fraction in the statement, 1 6 may be written 
 g- ; for since 1 unit =2 halves or ^, 16 units = 32 halves or -^-j 
 
 oo 
 
 and 16^:=:33 halves or -~-. The preceding statement may 
 therefore be written, ^ Canceled, ^l. 
 
 Then, 4 
 1 1 
 
 440 
 
 4 
 
 1760 
 1 4 
 
 7040 
 1760 
 
 24640 The same Ans. as before. 
 
 The scholar will observe that the 4th and 5th Sec. of the 
 rule for canceling, have been applied in the preceding solution. 
 
 6. In 30 purses containing 20 guineas each, how many 
 pounds ? Ans. 840. 
 
 30 20 28 
 Statement for canceling, L -^- The 28 above the line 
 
 expresses the shillings in a guinea, and the 20 below it, the shil- 
 lings in a pound. The scholar may perform the solution. 
 
 7. How many pounds in money will 9 tuns of wine cost at 
 3s. 4 d. per gallon? 3s. 4 d. =40 d. Statement for can- 
 
 9. 2. 2. 63. 40 r _ 
 
 . 
 
 For the terms in this statement the scholar is referred to 
 Tables 1st and 6th, of the Compound Numbers. 
 
 8. How many times will a wheel 12 feet 6 inches in circum- 
 ference, revolve in traveling 124 miles ? Ans. 52377|. 
 
 Statement 12 ft. 6in. = 150in. i . (See Ta- 
 
 150. 
 ble 9th, Compound Numbers.) 
 
 The last three sums have been stated by the rule for cancel- 
 
REDUCTION. 91 
 
 ing only, because that is regarded as superior to the com- 
 mon mode of solution. The scholar will feel at liberty to 
 adopt either mode of operation. The following sums are not 
 stated, that the scholar may exercise his own judgment. 
 
 9. How long will it take to count 6000000, at the rate of 
 75 per minute. Ans. 55f days. 
 
 10. In 107520 pounds of sugar, how many hogsheads, each 
 containing 6 cwt. Ans. 160. 
 
 11. If one quart of melasses cost 10 pence, how much will 
 12 hogsheads cost 1 Ans. 126 . 
 
 12. How many dollars, each 8 s., will it cost to ride 45 
 leagues, at 6 pence a mile ? Ans. $8.437-f- 
 
 13. How much will 540 yards of cloth cost at 3 s. 4 d. per 
 yard, in dollars, at 6 shillings each ? Ans. $300. 
 
 14. How many dozen of gallon, quart, and pint bottles, of 
 each an equal number, are contained in a cisteni holding 144 
 gallons. Ans. 8 T 8 r dozen. 
 
 15. How many casks, each containing 1 bushel 1 peck, are 
 required to hold 145 bushels ? Ans. 116. 
 
 16. I have five hogsheads of wine, 63 gallons each, which 
 I wish to put into gallon, quart, and pint bottles, of each an 
 equal number ; how many will be required ? Ans t 229, and 1 
 pint of wine would be left. 
 
 17. In 16 cwt. 3 qr. 20 lb., how many parcels, each con- 
 taining 36 Ib? Ans. 52f. 
 
 18. In 56 ells Flemish, how many yards ? Ans. 42. 
 
 19. In 144 yards, how many ells French? Ans. 96. 
 
 20. In 472 parcels of sugar, each 72 pounds, how many cwt. ? 
 Ans. 303 cwt. 1 qr. 20 lb. 
 
 21. If 15 casks of flour contain 4000 lb., how many cwt. 
 are there in each? Ans. 2 cwt. 1 qr. 14lb. 
 
 22. In 81b. of drugs, how many parcels, each 12 drams? 
 Ans. 64. 
 
 23. In 80 parcels, each 15 drams, how many pounds ? Ans. 
 12i 
 
 24. How many revolutions will a wheel 18 feet 4 inches in 
 circumference, make in traveling 300 miles ? Ans. 86400. 
 
 25. How many cups, each weighing 22 oz. may be made of 
 25 lb. 6 oz. of silver. Ans. 13 cups, and 20 oz. silver remain. 
 
 26. How much would 1008 nails of cloth cost, at 10 pence 
 per yard, in dollars, at 6 shillings each? Ans. $8.75. 
 
 27. In 4 bales of cloth, each 15 pieces, and each piece 16 
 ells English, how many ells French ? Ans. 800. 
 
92 COMPOUND RULES. 
 
 28. In 6 bales, each 12 pieces, and each piece 18 yards, 
 how many ells Flemish 1 Ans. 1728. 
 
 29. In 4 ingots of silver, each weighing 2 Ib. 6oz. 11 pwt. 
 how many grains ? Ans. 58656. 
 
 30. How many hours, minutes, and seconds in one year ? 
 Ans. 8766 hr. 525960 min. and 31557600 sec. 
 
 31. In 1597 quarts, how many bushels, &c. ? Ans. 49 bu. 
 3 pk. and 5 qt. 
 
 QUESTIONS. What are compound numbers? How do they increase? 
 What are included under this head 1 Let the 14 tables of Compound 
 Numbers be made familiar, before the scholar proceeds with Reduction. 
 
 What is Reduction 7 How are high denominations brought into 
 low denominations ? And how are low denominations brought into 
 high 1 What is the rule when high denominations" are brought 
 into low? And what is it, when low denominations are brought into 
 high ? What should the scholar notice before commencing to reduce 
 any quantity ? In circular measure, how is the circle regarded ? 
 What are the divisions of the circle ? To what is this measure applied ? 
 How many days and hours does the year contain ? To what do the 6 
 hours amount in 4 years ? How many days does every fourth year 
 contain] What is this fourth year called ? How may' it be found ? 
 In dividing the given year by four, what does the figure that remains 
 (if any) show ? What is the more usual division of the year ? How 
 many days are contained in each of the 12 months ? When quanti- 
 ties are to be exchanged, what is the rule for the common mode of op- 
 eration ? What is the rule for canceling ? What is Note 1st ? What 
 is Note 2d ? 
 
 COMPOUND RULES. 
 
 The scholar will recollect, that in simple numbers, the 
 denominations increase in value in the constant ratio of 10. 
 
 The peculiarity of the numbers in the preceding rule, and in 
 the four next following, is, that they have no uniform ratio of 
 increase, common to all denominations ; but each denomination 
 has its own peculiar ratio. These ratios are all represented in 
 the tables of Compound Numbers. 
 
 In simple numbers, 10 units make 1 ten ; 10 tens make 1 
 hundred; 10 hundred make 1 thousand, &c. In operations 
 with these numbers, we therefore carry for 10. 
 
 In the table of English money, 4 farthings make 1 penny ; 
 12 pence 1 shilling, and 20 shillings 1 pound. For the same 
 
COMPOUND ADDITION. 93 
 
 reason, therefore, that we carry for 10 in simple numbers, 
 we carry for 4, 12, and 20, in operations with pounds, shillings, 
 pence, and farthings ; that is, from farthings to pence, we carry 
 for 4, because 4 farthings make 1 penny ; from pence to shil- 
 lings by 12, for a similar reason ; and from shillings to pounds, 
 for 20. The same general principle and reasoning may be 
 applied to the other compound tables. 
 
 There is one peculiarity noticeable in writing compound num- 
 bers. In simple numbers, we always know that any figure sus- 
 tains a ten-fold relation to the figures next it ; that is, the one on 
 the left of it is of 10 times more value ; and the one on the 
 right, of 10 times less value than they would be in its own 
 place. Hence, all that is here necessary, is that the figures 
 preserve their proper order. In compound numbers, each de- 
 nomination is known only by its own appropriate mark . There 
 is, therefore, an obvious necessity for each denomination to be 
 separately written. 
 
 COMPOUND ADDITION. 
 
 Compound Addition is an operation by which several num- 
 bers of different denominations, as pounds, shillings, pence, 
 &c. are united together. The rule to be observed in writing 
 down these numbers is, to place those of the same name under 
 each other. 
 
 Let it be required to add together 3 . 15s. 9d. 3qr. ; 
 5. 6s. 8d. 2qr. ; 8 . 13s. lid. 3 qr. ; and 10 . 12s. 
 8d. 2qr. The following is a convenient mode of writing 
 them: 
 
 1. 
 
 . s. d. qr. 
 
 315 93 The amount of the right hand column 
 
 5 6 82 is 10 farthings = 2 d. and 2qr. Like 
 
 8 13 11 3 simple numbers, the 2 qr. are set down ; 
 
 10 12 8 2 and the 2 d. added to the column of pence, 
 
 the amount of which 38 d. = 3 s. and 2 
 
 28 9 22 d. Setting down the 2 d. and carrying 
 
 the 3 s. to the column of shillings, we 
 
 make this column 19 s.=2 . and 9 s. Lastly, setting down 
 
94 COMPOUND ADDITION. 
 
 and carrying as before, we find the amount of the column of 
 pounds to be 28, which we write at the foot of the column. 
 We therefore find the amount of the four given numbers to be 
 28 . 9 s. 2 d. 2 qr. 
 
 From the preceding example, the scholar will see the appro- 
 priateness of the following rule : 
 
 RULE. Write the numbers so that each denomination shall 
 occupy a separate column. Then, commencing with the lowest 
 denomination, add each column by itself. 
 
 Notice at the addition of each column, to how many of the 
 denomination next above, the amount obtained is equal, and how 
 many remain. Write down those that remain, and carry the 
 other number to the next column. Proceed thus through all 
 the denominations. 
 
 Note. The whole amount of the left hand column must be 
 written down, if it be in the highest denomination. If it be 
 not in the highest denomination it should be reduced as far as 
 practicable. 
 
 2. 
 
 . s. d. qr. 
 
 27 15 63 The column of farthings amounts to 6 
 
 13 7 81 qr. Id. and 2 qr. The column of 
 
 24 16 90 pence is 35 d.=2 s. 11 d. The column 
 
 3 4 11 2 of shillings is 44s.=2. and 4 s. and 
 
 the column of pounds = 69 . Carrying 
 
 69 4112 and setting down agreeably to rule, we 
 
 obtain the annexed amount. 
 3. 
 
 . s. d. qr. 
 15 7 
 
 14 63 The farthings are 6=ild. 2qr. The 
 
 0831 pence are 28=2 s. 4 d. The shillings 
 18 11 2 are 57=2 . 17 s. which is written agree- 
 ably to the above note. 
 2 17 4 2 
 
 4. 5. 6. 
 
 . s. d. qr. . s. d. qr. . s. d. 
 
 8 12 9 2 57 11 11 1 67 18 10 
 
 31 6 11 27 13 2 3 150 19 6 
 
 42 18 3 1 48 9 6 1 175 16 8 
 
 2383 73 10 9 2 37 14 7 
 
 85 1 8 2 
 
COMPOUND ADDITION. 95 
 
 7. 8. 9. 
 
 . s. d. . s. d. . s. d. qr. 
 
 Ill 54 7 9 444 4 11 3 
 
 10 10 10 19 11 26 16 41 
 
 100 07 144 3 10 372 10 2 
 
 73 4 9 132 18 1780 14 6 2 
 
 43 8 11 43 6 8 200 10 10 3 
 
 10. 11. 12. 
 
 . s. d. . s. d. . s. d. qr. 
 
 76 56 18 8 875 16 10 3 
 
 19 1 73 11 11 783 19 7 2 
 
 86 11 7 22 12 2 59 17 7 3 
 
 43 4 8 77 17 7 85 13 11 1 
 
 750 18 6 88 18 8 387 14 9 3 
 
 13. Bought a horse for 26 . 12 s. ; a yoke of oxen for 31 
 . 17s. 8d. ; a cow for 7 . 16s. 9d. ; and paid 15s. 8d. 
 for a bridle. How much did they all cost me ? Ans. 67 . 
 
 2 s. 1 d. 
 
 14. Bought cloth for 29 . 6s. 10 d. ; ribbon for 3 s. 4 d. 3 
 qr. ; a pair of boots for 1. 6s. 3d.; and paid 2s. 8 d. 2 qs. 
 for mending a pair of shoes. What was my bill for the whole ? 
 Ans. 30 . 19s. 2 d. 1 qr. 
 
 15. Bought at one time goods to the amount of 175 . 16s. 
 lid.; at another, to the amount of 35 . 19s. 8 d. ; paid for 
 carting 7 . 8s. 9 d. 3 qr. ; and for insurance 3 . 9 s. 7 d. 
 What was the amount of my bills 1 Ans. 222 . 14s. 11 d. 3 qr. 
 
 16. Sold at one time goods to the amount of 35 1 1 s. 6 d. 
 
 3 qr. ; at another, to the amount of 56 . 19s. 7 d. 1 qr. ; at a 
 third time, to the amount of 75 . Is. 3d.; and at a fourth, to 
 the amount of 63 . 13 s. 4d. 2qr. What was the whole amount 
 of my sales ? Ans. 231 . 5 s. 9 d. 2 qr. 
 
 17. Bought a quantity of corn for 113 . lls. lid.; of 
 rye, for 32 . 19 s. 3 d. ; of wheat, for 136 . 16 s. 8 d. ; and 
 of oats, for 22 . 14s. 9 d. What was the whole amount? 
 Ans. 306 . 2 s. 7 d. 
 
 18. A man sold his farm for 856 . ; his sheep, for 67 . 
 17s.; his swine, for 19 . 19 s. 11 d. ; and his grain, for 36 . 
 13 s. 2 d. How much money did he receive ? Ans. 980 . 
 10s. Id. 
 
96 COMPOUND ADDITION. 
 
 TROY WEIGHT. 
 1. 
 
 lb. oz. pwt. gr. 
 
 15 6 13 J4 The column of grains amounts to 69 
 
 11 3 11 19 gr. = 2 pwt. 21 gr. The pwt. amount 
 17 4 3 21 to 46=2 oz. and 6 pwt. The ounces 
 42 11 17 15 amount to 26 2 lb. and 2 oz. The 
 pounds amount to 87 ; the whole of 
 87 2 6 21 which is to be written down. 
 
 The scholar will observe, that as we have left the table of 
 English money, we have no longer to carry by 20, 12, and 4. 
 Our carrying numbers now are 12, 20, and 24. 
 
 2. 3. 
 
 lb, oz. pwt. gr. lb. oz. pwt. gr. 
 
 ]0 9 11 16 29 7 13 19 
 
 17 9 6 8 17 6 11 11 
 
 28 11 16 21 31 6 16 23 
 
 36 7 17 22 71 1 18 7 
 
 94 2 12 19 149 11 00 12 
 
 4. 5. 6. 
 
 lb. oz. pwt. gr. lb. oz. pwt. gr. lb. oz. pwt. gr. 
 
 1 8 18 12 9 11 16 76 11 21 
 
 9 6 19 9 12 7 16 11 3333 
 
 11 5 3 21 3 11 7 19 
 
 22 6 10 9 71 16 9 13 9 11 19 
 
 15 5 12 21 16 9 17 23 14 11 17 12 
 
 7. Purchased at one time, 4 lb. 3 oz. 16 pwt. 15 gr. of silver, 
 and at another, 7 lb. 8 oz. 18 pwt. 23 gr. ; besides a quan- 
 tity of jewelry, weighing 5 Ibs. 11 oz. and 13 pwt. What was 
 the whole weight? Ans. 18lb. 8 pwt. 14 gr. 
 
 8. AddSlb. 9oz. 13 pwt. 19 gr.; 2lb. 10 oz. 9 pwt. 17 gr.; 
 6 lb. 11 oz. 18 pwt. 22 gr. ; 9 oz. 11 pwt. 12 gr. ; and 81b. 3 oz. 
 6 pwt. 20grs. Ans. 22 Ibs. 9oz. 18gr. 
 
 9. Again, add 6 lb. 2 oz. 16 pwt. 14 gr. ; 3 lb. 8 pwt. 2 gr. ; 
 12 lb. 4 oz. 15 pwt. 22 gr. ; 8 oz. 16 gr. ; 5 lb. 13 gr. ; and 23 gr. 
 Ans. 27 lb. 4 oz. 2 pwt. 18 gr. 
 
COMPOUND ADDITION. 97 
 
 AVOIRDUPOIS WEIGHT. 
 1. 
 
 T. cwt. qr. Ib. oz. 
 
 7 16 3 20 13 The amount of the ounces is 45 
 
 4 12 1 25 11 =2lb. 13oz. The pounds amount 
 3 9 2 16 9 to 77=2 qr. 21 Ib. The qr. are 
 
 12 18 14 12 8=2 cwt. The cwt. are 57 =2 tons, 
 17 cwt. and the tons are 28. 
 
 28 17 21 13 
 
 2. 3. 
 
 T. cwt. qr. Ib. oz. cwt. qr. Ib. oz. dr. 
 
 3 19 16 15 16 3 24 15 12 
 
 9 3 24 12 32 18 9 14 
 
 1 18 1 26 14 7 2 21 7 9 
 
 14 5 2 12 9 12 3 11 11 12 
 
 4. 5. 
 
 cwt. qr. Ib. oz. dr. cwt. qr. Id. oz. dr. 
 
 6 2 27 12 9 17 9 4 8 
 
 12 3 16 9 12 13 3 27 15 12 
 
 14 1 24 14 6 18 2 17 13 8 
 
 17 3 8 15 8 29 1 23 12 13 
 
 6. Purchased at one time, 16 tons, 3qr. 21 Ib. of hay; at 
 another, 9 tons, 16 cwt. 2qr. 17lb. ; and at another, 27 tons, 
 13 cwt. 1 qr. 17 Ib. How much did I purchase? Ans. 53 
 tons, 10 cwt. 3qr. 27 Ib. 
 
 7. Bought 36 cwt. 3qr. 24 Ib. of wool; but finding the 
 demand large, I made three successive purchases, at each of 
 which I bought 45 cwt. 2 qr. 16 Ib. What was the amount of 
 my purchases ? Ans. 173 cwt. 3 qr. 16 Ib. 
 
 APOTHECARIES' WEIGHT. 
 
 1. 2. 3. 
 
 ft. ?. 3. 3. |. 3. 3. gr. ft. 5. 3. 3- gr. 
 
 2831 6-52 12 8941 14 
 
 5622 961 15 14 670 12 
 
 6472 3409 19516 
 
 8652 272 13 86725 
 
 23 2 3 1 
 
 9 
 
98 COMPOUND ADDITION. 
 
 4. A physician purchased the following quantities of medi- 
 cine, at three different times ; viz. 1 pound, 4 ounces, 5 drams ; 
 3 pounds, 11 ounces, 6 drams, 2 scruples, 15 gr. ; and 7 drams, 
 1 scruple, and 12 grains ; what was their whole weight? Ans. 
 5ib. 5|. 33.13. 7gr. 
 
 CLOTH MEASURE. 
 
 
 1. 
 
 
 
 2. 
 
 
 
 3. 
 
 
 
 4. 
 
 
 yd. 
 
 qr. 
 
 na. 
 
 yd. 
 
 qr. 
 
 na. 
 
 E.E. 
 
 gr. 
 
 na. 
 
 E.F. 
 
 qr. 
 
 na. 
 
 7 
 
 3 
 
 2 
 
 I 
 
 1 
 
 1 
 
 16 
 
 4 
 
 3 
 
 21 
 
 3 
 
 
 
 9 
 
 2 
 
 3 
 
 6 
 
 3 
 
 2 
 
 12 
 
 3 
 
 3 
 
 13 
 
 5 
 
 2 
 
 6 
 
 1 
 
 
 
 12 
 
 2 
 
 3 
 
 18 
 
 2 
 
 2 
 
 16 
 
 4 
 
 3 
 
 8 
 
 3 
 
 3 
 
 15 
 
 1 
 
 2 
 
 20 
 
 3 
 
 1 
 
 19 
 
 2 
 
 1 
 
 5. Bought at one purchase, 32yd. 3qr. ; at another, 16yd. 2 
 qr. 2 na. ; and 24 yd. 1 qr. at another. How many yards did 
 I purchase 1 Ans. 73 yd. 2 qr. 2 na. 
 
 6. Bought of one man, 12 ells English, 4qr. 3na. ; and of 
 each of two others," 27 ells English, 3 qr. and 3 na. How 
 many Ells did I purchase ? Ans. 68 E.E. 2 qr. 1 na. 
 
 7. Received from France 12 ells, 4 qr. of broadcloth; 17 
 ells, 5qr. 3na. of cassimere ; and 19 ells, 2qr. 3na. of silks. 
 How many ells were there in the three articles purchased ? 
 Ans. 50 ells. qr. 2 na. 
 
 DRY MEASURE. 
 
 1. 
 
 bu. pJc. qt. pt. 
 3251 
 7372 
 
 11 2 4 1 
 6130 
 
 2. 
 
 3. 
 
 bu. pk. qt. pt. 
 
 bu. pk. qt. pt. 
 
 8370 
 
 15 1 6 1 
 
 9241 
 
 127 5 7 1 
 
 6361 
 
 12 2 5 
 
 4220 
 
 16 2 7 1 
 
 WINE MEASURE. 
 
 1. 2. 3. 
 
 T. P. hhd.gal.qt. hhd. gal. qt. pt. gi. hhd. gal. qt. pt. gi. 
 
 4 1 1 42 3 15 27 3 1 2 140 46 2 1 1 
 
 6 1 24 2 20 13 2 3 127 15 3 3 
 
 8 1 18 3 12 16 1 1 1 263 29 1 1 2 
 
 12 1 23 132 54 3 3 42 27 3 3 
 
COMPOUND ADDITION. 99 
 
 LONG MEASURE. 
 
 1. 2. 
 
 L. m. fur. rd. m. fur. rd. yd. 
 
 12 2 5 36 9 6 12 3 
 
 9 1 7 24 4 7 26 2 
 
 15 6 17 12 4 32 5 
 
 30 2 4 26 7 1 38 4 
 
 3. 4. 
 
 TO. fur. rd. yd. ft. in. b. c. L. m. fur. rd. yd. ft. in. b.c. 
 
 4 4 23 5 2 92 18 2 5 18 3 2 7 1 
 
 9 3 30 6 1 10 1 21 4 30 4 1 9 2 
 
 62 36 42 82 32 17 31 5241 
 
 52 27 42 82 76 23 39 42 10 2 
 
 LAND OR SGUJARE MEASURE. 
 
 1. 2. 
 
 A. rood. rd. yd. ft. in. A. rood. rd. yd. 
 
 7 2 36 24 6 72 9 3 21 6 
 
 8 3 23 20 4 91 12 2 37 11 
 5 1 15 17 8 108 11 39 12 
 
 12 3 12 13 6 22 15 3 12 16 
 
 15 17 9 7 136 8 1 9 12 
 
 3. 4. 
 
 A. rood. rd. yd. ft. A. rood. rd. yd. ft. 
 
 46 29 11 7 34 2 33 7 6 
 
 27 3 26 6 4 44 30 7 5 
 
 18 2 32 6 4 15 3 29 10 5 
 
 25 3 30 7 5 33 3 36 8 7 
 
 6 1 16 6 8 44 3 37 8 7 
 
 SOLID MEASURE. 
 
 1. 2. 3. 
 
 ft. in. cd. ft. in. cd. ft. in. 
 
 99 420 11 72 726 3 99 777 
 
 78 864 12 16 317 66 77 333 
 
 320 740 113 17 36 122 116 1240 
 
 950 222 4 117 1372 372 108 1617 
 
 48 12 116 8 96 12 96 456 
 
100 COMPOUND ADDITION. 
 
 TIME. 
 
 1. 2. 
 
 yr. mo. w. d. h. m. sec. d. h. m. sec. 
 
 2 9 3 6 13 22 56 15 21 43 51 
 
 8 11 3 21 43 21 23 17 55 56 
 
 1 12 2 4 23 69 52 6 19 59 49 
 
 2 8 3 5 17 36 8 16 15 43 36 
 
 3. 4. 
 
 ID. d. h. m. sec. w. d. h. m. sec. 
 
 5 6 9 7 59 13 10 17 8 47 
 
 6 5 21 39 27 14 9 20 40 43 
 22 4 23 36 42 15 8 22 45 23 
 11 3 16 17 18 16 11 23 18 22 
 
 CIRCULAR MOTION. 
 
 1. 2. 3. 
 
 S. ' " ' " S. 
 
 6 23 42 39 49 29 41 3 22 40 37 
 
 8 26 54 36 6 7 43 2 11 29 59 57 
 
 3 19 11 9 11 8 51 59 4 23 18 38 
 
 2 21 13 23 12 13 27 17 86 13 15 
 
 APPLICATION. 
 
 Ex. 1. What is the amount of 23 . 11 d. ; 13 . 17s. 3 
 qr. ; 16s. 8 d.; and lid. 3 qr. ? Ans. 37 . 15 s. 7d. 2qr. 
 
 2. Bought the following quantities of oil ; viz. 12 gal. 3 qt. ; 
 2 hhd. 42 gal. 2 qt. 1 pt. ; and 13 hhd. 56 gal. What was the 
 whole amount ? Ans. 16 hhd. 48 gal. 1 qt. 1 pt. 
 
 3. Add together 250 . 18s. 9 d. 3 qr. ; 16 . 7s. 2 qr. ; 
 21 . 19 s. 3 d. ; 18s. 6 d. ; and 36 . Ans. 326 . 3 s. 7 d. 
 Iqr. 
 
 4. What is the amount of 5 cwt. 3qr. 27 Ib. ; 2 qr. 29 lb.; 
 12 cwt. 1 qr. 17 lb. ; and 36 cwt. 16 lb. Ans. 55 cwt. 1 qr. 5 lb. 
 
 5. Bought at one time, 7 bu. 3 pk. of wheat ; at another, 9 
 bu. 1 pk. and had previously in each of two bins, 6 bu. and 3 
 pk. What was the whole amount? Ans. 30 bu. 2pk. 
 
 6. Sold one cow for 10 . 15 s. 6 d. ; another for 6 . 19 s. 
 
COMPOUND SUBTRACTION. 101 
 
 lid.; and a colt for 12 j. 6 s. 4 d. How much did they all 
 amount to 1 Ans. 30 . Is. 9 d. 
 
 7. Bought four casks of wine, of which the first contained 
 42 gal. 2 qt. 1 pt. ; the second, 65 gal. 1 pt. ; the third, 50 gal. 
 3 qt. ; and the fourth, 55 gal. 1 qt. 1 pt. How many gallons 
 did I purchase 1 Ans. 213 gal. 3 qt. 1 pt. 
 
 8. Purchased three 'pieces of land. The first contained 17 
 acres, 1 rood, and 35 rods ; the second, 36 acres, 2 roods, 21 
 rods ; and the third, 46 acres and 37 rods. How much land 
 did I purchase ? Ans. 100 acres, 1 rood, 13 rods. 
 
 QUESTIONS. In what ratio do simple numbers increase 1 What is 
 the peculiarity of Compound Numbers'? Where are the ratios of in- 
 crease and decrease of compound numbers given 1 Why do you carry 
 for 10 in simple numbers 1 Why do you carry for 4, 12, and 20, in 
 the table of English money 1 What peculiarity noticeable in writing 
 compound numbers 1 What only is necessary in writing simple num- 
 bers 1 How are compound numbers known 1 How must each denom- 
 ination therefore be written 1 What is Compound Addition 1 How 
 are numbers to be written 1 What is the rule 1 What is the note 
 following the rule 1 
 
 COMPOUND SUBTRACTION. 
 
 The scholar has now become acquainted with Compound 
 Addition ; and he was previously acquainted with the Simple 
 rules. He needs, therefore, to be informed only, that Com- 
 pound Subtraction sustains the same relation to Compound Ad- 
 dition, that Simple Subtraction does to Simple Addition. It is 
 the subtracting of numbers of different denominations. 
 
 In this rule, instead of constantly borrowing 10, when the 
 lower figure is the larger, he must borrow as many units as 
 are required of the denomination he is subtracting, to make a 
 unit of the next higher denomination ; that is, when it be- 
 comes necessary to borrow a number in subtracting farthings, 
 4 is the number always required ; in subtracting pence, 12 is 
 the number ; and in shillings, 20 ; and in like manner in other 
 denominations . 
 
 RULE. Place the less quantity under the greater, so that 
 each denomination shall stand under one of its own name or 
 kind. Begin at the right, and proceed in all respects as in 
 9* 
 
102 COMPOUND SUBTRACTION. 
 
 Simple Subtraction, except in borrowing when the lower figure 
 is the larger ; in doing which, instead of 10, (the number bor- 
 rowed in Simple Subtraction,} borrow as many units as make 
 one of the next higher denomination. Whenever a number is 
 borrowed, one is to be carried to the next lower figure. 
 
 . s. d. qr. 
 
 Ex. 1. From 16 18 8 1 First, I cannot take 3 qr. 
 
 Take 8 16 11 3 from 1 qr. I therefore 
 
 add 4qr. to the upper 
 
 Ans. 8 182 figure, and take the 3 
 from the amount, 5, and obtain a remainder of 2. I carry 1 to 
 the next lower figure, viz. 11, which makes it 12, and proceed 
 to take it from the figure above, but find it impracticable. I 
 therefore add 12 to the upper figure, 8, making it 20 ; and 
 from this amount, subtract 12, and obtain the 8 in the answer. 
 Again, I carry 1 to the next figure, 16, which makes it 17, 
 and take this from the figure above, and obtain a remainder of 
 1. I here borrowed nothing and have nothing to carry ; there- 
 fore, 8 from 16 leaves 8. 
 
 2. 3. 4. 
 
 . s. d. qr. . s. d. qr. . s. d. qr. 
 
 35 11 9 3 46 13 7 1 74 9 2 
 
 17 9 11 2 21 17 9 3 72 19 11 3 
 
 5. 6. 7. 
 
 . s. d. qr. . s, d. qr. . s. d, 
 
 99 16 8 3 33 12 3 1 94 11 8 
 
 77 17 7 1 13 8 9 3 72 9 11 
 
 8. A certain man owed 75 . 13s. 9d., and paid of this 
 sum 39 . 19s. lid. How much remained due? Ans. 35 
 . 13s. lOd. 
 
 9. Received of three individuals the following sums of 
 money; viz. of A., 16 . 12s. 8d. 3qr.; of B., 21 . 17 
 s. 9d. ; and of C., 46 . 19s. I afterwards paid D. 58 . 
 13s. 9 d. 2qr. How much had Heft? Ans. 26 . 15s. 8 
 d. 1 qr. 
 
 10. The following sums are due to A. ; viz. 136 . 15 s. 11 
 d. ; 450 . 8s. 6d.; 356 . 17s. 10 d. 2 qr. ; and 12 .9 
 s. 4d. He is indebted to B. } 67 . 14s. 9d- 2qr. ; to C., 
 
COMPOUND SUBTRACTION. 103 
 
 24. lls. 3d.; and to D., 571 . 11s. lid. How much 
 is due to him, more than he owes ? Ans. 292 . 13s. 8 d. 
 
 TROY WEIGHT. 
 
 1. 2. 3. 
 
 Ib. oz. pwt. gr. Ib. 02. pwt. Ib. oz. pwt. gr. 
 
 14 9 19 16 46 11 13 9 11 11 21 
 
 10 11 16 23 13 9 17 43 19 23 
 
 4. 5. 6. 
 
 Ib. oz. pwt. gr. Ib. oz. pwt. gr. oz. pwt. gr. 
 
 36 7 14 17 89 16 11 11 9 18 
 
 17 9 17 22 2 11 19 23 10 16 23 
 
 AVOIRDUPOIS WEIGHT. 
 
 1. 2. 
 
 cwt. qr. Ib. oz. dr. cwt. qr. Ib. oz. dr. 
 
 12 3 19 13 14 31 1 23 14 15 
 
 9 2 21 11 6 26 3 25 15 8 
 
 3. 4. 
 
 T. &wt. qr. Ib. oz. dr. T. cwt. qr. Ib. oz. dr. 
 
 6 13 11 12 13 9 17 3 20 15 8 
 
 4 17 3 5 13 14 2 15 2 26 12 15 
 
 7. Having in my possession 45 cwt. 3qr. 17lb. of cheese, 
 Isold 32 cwt. 27 Ib. How much remained? Ans. 13 cwt. 
 2qr. 18 Ib. 
 
 APOTHECARIES' WEIGHT. 
 
 1. 2. 3. 
 
 Ib. I- 3. 9- gr. Ib. J. 3. 3. gr . Ib. |. 3- 3. gr. 
 
 5 9 5 2 16 12 6 5 2 15 31 17 1 1 12 
 
 3 11 6 2 15 8 9 4 1 17 20 10 5 2 15 
 
 CLOTH MEASURE. 
 
 2. 3. 4. 
 
 E.F. qr. na. E.E. qr. na. E.F. qr. na: 
 
 10 5 1 21 1 2 16 1 3 
 
 633 16 43 822 
 
104 COMPOUND SUBTRACTION, 
 
 DRY MEASURE. 
 
 1. 2. 3. 
 
 bu. pk. qt. pt. gi. bu. pk. qt. pt. gi. bu. pk. qt. pt. gi. 
 
 15 3302 26 0501 30 2713 
 
 12 2 5 1 3 23 3 7 1 2 16 3 5 1 3* 
 
 WINE MEASURE. 
 
 1. 2. 
 
 T. P. hhd. gal. qt. pt. gi. hhd. gal. qt. pt. gi. 
 
 3 1 1 27 3 1 1 27 19 3 1 
 
 2 2 47 1 1 3 16 43 1 1 3 
 
 3. 4. 
 
 hhd. gal. qt. pt. gi. hhd. gal. qt. pt. gi. 
 
 137 42 1 3 175 59 1 3 
 
 128 56 3 1 1 21 37 3 1 2 
 
 LONG MEASURE. 
 
 1. 2. 
 
 m. fur. rd. yd. ft. in. b.c. m. fur. rd. yd. ft. in. 
 
 15 4 27 4 2 11 1 32 5 39 1 2 3 
 
 12 3 36 3 1 92 27 2 4 3 1 11 
 
 3. 4. 
 
 L. m. fur. rd. yd. ft. in. L. m. fur. rd. yd. ft. in. b.c. 
 
 17 2 3 19 3 1 7 31 3 15 4 2 7 2 
 
 12 1 7 35 4 2 8 27 2 5 17 1 2 9 2 
 
 LAND OR SQ.UARE MEASURE. 
 
 1. 2. 3. 
 
 A. rd. yd. ft. A. rood. rd. yd. A. rood. rd. yd. 
 
 9 36 14 8 74 3 27 16 12 1 16 15 
 
 4 39 6 4 64 2 31 12 9 2 17 16 
 
 SOLID MEASURE. 
 
 1. 2. 3. 
 
 C. ft. in. C. ft. in. C. ft. in. 
 
 21 62 856 56 110 1462 8 100 8 
 
 16 115 972 19 36 472 1 101 1560 
 
COMPOUND SUBTRACTION. 105 
 
 TIME. 
 
 1. 2. 
 
 yr. m. w. d. h. m. ? { d. h. m. sec. 
 
 16 8 3 5 13 12 2 1 15 21 35 
 
 7 9 2 6 21 9 3 2 16 22 36 
 
 3. 4. 
 
 yr. a. h. m. sec. yr. d. h. m. sec. 
 
 19 152 13 42 21 . 45 67 17 50 15 
 
 16 256 19 36 56 36 36 22 46 45 
 
 CIRCULAR MOTION. 
 
 
 1. 
 
 2. 3. 
 
 s. 
 
 o / // 
 
 . S. o 
 
 ' " . ' 
 
 " 
 
 8 
 
 18 42 36 
 
 9 27 
 
 36 51 11 15 16 
 
 31 
 
 6 
 
 26 11 52 
 
 1 29 
 
 42 52 8 19 17 
 
 42 
 
 
 4. 
 
 
 5. 
 
 
 
 S. 
 
 / // 
 
 S. ' " 
 
 
 
 11 21 
 
 49 59 
 
 8 19 38 46 
 
 
 
 6 27 
 
 13 21 
 
 6 21 42 50 
 
 
 PROMISCUOUS EXAMPLES. 
 
 Ex. 1. 1 have in my possession 46 . 19s. lid. How 
 much shall I have left, after paying a debt of 27 . 13 s. 9 d. ? 
 Ans. 19 . 6s. 2 d. 
 
 2. Received 156 . 3s. 8 d. after which I paid out 137 . 
 15s. lOd. How much remained in my possession? Ans. 
 18 j. 7s. 10 d. 
 
 3. Lent a friend 16 . 17s. 6d. On the following day he 
 paid me 5 . 13s. lid.; one week after he made another 
 payment of 7 . 5s. lOd. How much then remained due ? 
 Ans. 3. 17s. 9d. 
 
 4. Bought a wagon for 9 . 11s. and sold the same for 13 
 . 5 s. How much did I gain ? Ans. 3 . 14s. 
 
 5. Bought a horse for 56 . 15 s. and exchanged the same 
 
106 COMPOUND SUBTRACTION. 
 
 for a colt, and received 46 . 11 s. in money. How much did 
 the colt cost me ? Ans. 10 . 4 s. 
 
 6. A man having 15 tons, 13 cwt. 3 qr. of hay, sold 5 tons, 
 10 cwt. and gave 3 cwt. 3 qr. to a friend ? How much had he 
 left? Ans. 10 tons. 
 
 7. Bought 7 cwt. 3 qr. 16 Ib. of rice at one purchase, and 9 
 cwt. 1 qr. 27 Ib. at another ; of this, 2 cwt. 16 Ib. was stolen, 
 and of the remainder, I sold 11 cwt. 2qr. 21 Ib. How much 
 had I left ? Ans. 3 cwt. 2 qr. 6 Ib. 
 
 8. Three men bought a piece of land for 450 . 16 s. 10 d. 
 of which two of them paid each 69 . 17s. 6 d. ; what did the 
 third man pay 1 Ans. 31 1 . Is. 10 d. 
 
 9. I owned a tract of land containing 356 acres, 3 roods, and 
 30 rods ; from this I sold to A. 127 acres, 2 roods ; and to B. 
 27 acres, 1 rood, and 36 rods. How much remained ? Ans. 201 
 acres, 3 roods, 34 rods. 
 
 10. A father, 46 years, 9 months, and 27 days old, has two 
 sons ; the elder of whom is 19 years, 3 months, and 13 days 
 old ; and the younger, 7 years, 10 months, and 21 days. How 
 much does the father's age exceed the sum of his sons' 1 Ans. 
 19 y. 7m. 23d. 
 
 11. Bought a quantity of cotton, which, at the price agreed 
 upon, came to 20 . 4 s. In pay for this I gave a quantity of 
 rice worth 15. 18s. and the balance in cash. How much 
 money did I pay ? Ans. 4 . 6s. 
 
 12. A merchant bought a piece of cloth containing 40 yards, 
 from which he sold 36 yd. 1 qr. 2 na. How much had he left ? 
 Ans. 3 yd. 2 qr. 2na. 
 
 ] 3. Sold from a pile of wood containing 40 cords, 64 feet, 
 39 cords, 32 feet. How much remained ? Ans. I cord 32 ft. 
 
 14. Bought 560 acres of land for 940 . From this I sold to 
 A. 120 acres, 2 roods, and 16 rods, for 300 . ; and to B. 150 
 acres, 1 rood, and 24 rods, for 297 . 10s. and 6d. How 
 much land remains in my possession, and how much has it cost 
 me 1 Ans. 289 acres ; cost, 342 . 9 s. 6d. 
 
 QUESTIONS. What is Compound Subtraction 1 Instead of 10, how- 
 many are to be borrowed here 1 What is the number borrowed in sub- 
 tracting farthings'? Why 1 What in subtracting pence 1 Shillings'? 
 And why? What is the Rule? 
 
COMPOUND MULTIPLICATION. 107 
 
 COMPOUND MULTIPLICATION. 
 
 The peculiarity of Compound Numbers having been fully 
 explained, and multiplication of Simple Numbers being also 
 understood, no other definition of this rule is needed than is 
 conveyed by the name. 
 
 A simple inquiry presents itself, viz. are both the given num- 
 bers compound ? To answer this, the scholar needs only to be 
 informed, that in multiplication, the multiplier is always a sim- 
 ple number, showing how many times the multiplicand is to be 
 repeated. The multiplicand, therefore, only is compound. 
 The product, as it is formed by repeating the multiplicand, 
 must necessarily be of the same denomination with it. 
 
 Take the following illustration. Multiply 8 s. 9 d. 2 qr. by 4. 
 
 PERFORMED. 
 
 8s. 9 d. 2 qr. 
 4 
 
 1. 15s. 2d. Oqr. 
 
 In this example, we say, four times 2 farthings are 8 far- 
 things, and 8qr.=2d. and Oqr. remain. We therefore write 
 down a cypher, and carry 2. Again, four times 9 d. are 36 d. 
 and 2 d. to carry make 38 d. = 3 s. and 2 d. The 2 d. is writ- 
 ten down, and the 3 s. carried. Lastly, four times 8 s. are 32 s. 
 and 3 s. to carry make 35 s. = 1 . 15s. which is written down 
 as seen in the answer. If now, in the above example, 8s. 9 
 d. 2qr. had been given as the price of one yard of cloth, and 
 the scholar had been required to find the price of 4 yards, the 
 operation would have been the same. The number of yards 
 only decides how many times the price of one yard is to be re- 
 peated. 
 
 CASE 1st. WHEN THE MULTIPLIER OR SIMPLE NUMBER is 
 
 NOT GREATER THAN 12. 
 
 RULE. Multiply the Compound number by the Simple one, 
 commencing with the lowest denomination and carrying from 
 one denomination to another, as in the preceding Compound 
 Rules. 
 
108 COMPOUND MULTIPLICATION. 
 
 Ex. 1. Multiply 9 . 16 s. 8 d. 2 qr. by 9. 
 
 PERFORMED. 
 
 9. 16s. 8d. 2.qr. 
 9 
 
 88 . 10s. 4d. 2qr. Ans. 
 
 Explanation. 9 times 2 qr. = 18 qr. 4 d. 2 qr. 9 times 8 d. 
 =72 d. and 4 d. from the farthings added, make 76 d=6 s. 4 d. 
 Again, 9 times 16s. = 144s. and the 6s. obtained from the 
 pence make 150 s. =7 . 10 s.; and lastly, 9 times 9 pounds^ 
 81 . and 81 . + 7 . = 88. The product therefore is, as 
 given above, viz. 88 . 10s. 4d. 2qr. 
 
 2. Multiply 16 . 11 s. 9 d. 3qr. by 3. 
 
 PERFORMED. 
 
 16 . 11s. 9d. 3qr. 
 3 
 
 49 . 15s. 5d. Iqr. Ans. 
 
 3. Multiply 1 . 11s. 6 d. 2 qr. by 5. Ans. 7 . 17 s. 8 
 d. 2 qr. 
 
 4. Multiply 11 s. 9 d. by 3. Ans. 1 . 15 s. 3 d. 
 
 5. Multiply 15 . 10 s. 8 d. by 2. Ans. 31 . I s. 4 d. 
 
 6. Multiply 5 s. 6 d. by 9. Ans. 2 . 9 s. 6 d. 
 
 7. What cost 4 gallons of wine at 8 s. 7 d. per gallon? Ans. 
 l. 14s. 4d. 
 
 8. What cost 5 cwt. of raisins at 1 . 7s. 9d. 2 qr. per 
 cwt.? Ans. 6. 18s. lid. 2qr. 
 
 9. What cost 8 yards of broadcloth at 1 . 2 s. 3d. per yard? 
 Ans. 8. 18s. 
 
 10. What cost 11 tons of hay at 2 . 1 s. 10 d. per ton? 
 Ans. 23 . s. 2 d. 
 
 11. What cost 12 bushels of wheat at 9 s. 10 d. per bushel ? 
 Ans. 5. 18s. 
 
 CASE 2d. WHEN THE MULTIPLIER OR SIMPLE NUMBER is A 
 
 COMPOSITE NUMBER GREATER THAN 12. 
 
 RULE. Separate the simple number or multiplier into its 
 component parts, and multiply first by one of these parts, and 
 the product of this multiplication by the others in succession. 
 The last product will be the answer required. 
 
COMPOUND MULTIPLICATION. 109 
 
 Note. It will generally be found more expeditious to divide 
 the multiplier into two parts only. Should it, however, be large, 
 as 125 or 1728, it may be divided into more than two parts, 
 viz. 125 into three 5's, and 1728 into three 12's. 
 
 Ex. 1. Multiply 6s. 10 d. by 28. 28 = 4x7. Therefore, 
 
 6s. 10 d. 
 
 . 7 
 
 2 . 7s. 1 d. = product of 7. 
 4 
 
 9. 11 s. 4 d.= prod, of 4 times 7 =28. 
 
 2. What cost 27 yards of cloth at 7 s. 6 d. per yard ? 27 = 
 9x3. Therefore, 
 
 7s. 6 d. 
 9 
 
 3 . 7 s. 6 d.=price of 9 yards. 
 3 
 
 10.. 2s. 6d.=priceof 3 times 9yd. = 27yd. 
 
 3. What cost 32 yards at 9 s. 9 d. per yard ? Ans. 15 . 
 12s. 
 
 4. What cost 20 yards at 3s. 6 d. per yard ? Ans. 3 . 10s. 
 
 5. What cost 36 gallons at 5 s. 8 d. per gallon ? Ans. 10 . 
 4s. 
 
 6. What cost 63 yards at 7 s. 6 d. 2 qr. per yard ? Ans. 23 
 J6. 15s. Id. 2qr. 
 
 7. What cost 72 yards at 3 s. lid. per yard ? Ans. 14. 
 2s. 
 
 8. What cost ]44 yards at 1 . 4 s. 2 d. per yard? Ans. 
 174 ,. 
 
 9. Sold 21 men each 3 qr. 16 Ib. of sugar. How many cwt. 
 did I sell ? Ans. 18 cwt. 3 qr. 
 
 10. Suppose 27 young lads to hare lived each 6 years, 9 
 months, 8 days, and 11 hours ; how many days, <fcc., have they 
 all lived, allowing 30 days to a month ? Ans. 182 yr. 10 mo. 
 18 d. 9h. 
 
 10 
 
110 COMPOUND MULTIPLICATION. 
 
 CASE 3d. WHEN THE MULTIPLIER is MORE THAN 12, AND is 
 
 NOT A COMPOSITE NUMBER. 
 
 RULE. Take any two or more numbers whose product will 
 come as near as possible to the given number or multiplier, 
 without exceeding it, and having multiplied the given price of 
 one yard, pound, <fyc. by them, retain their product. Then 
 multiply the same given price by the riumber wanting to make 
 up the entire multiplier, and add the product to the preceding 
 product. Their sum will be the product required. 
 
 . If preferred, the given quantity or multiplier may be 
 multiplied by each denomination of the compound quantity 
 separately, and the several products, reduced to the highest 
 denomination, may be added together. 
 
 Ex. 1. What will 51 yards of cloth cost at 3 s. 6 d. per 
 yard? The two numbers whose product comes nearest to 51, 
 are 7 and 7, and their product is 49. Consequently, if we 
 multiply 3 s. 6 d. by 7, and their product by 7 again, we shall 
 obtain the cost of 49 yards. There will then be the cost of 
 two yards wanting. This will be obtained by multiplying 3 s. 
 6 d. the price of 1 yard, by 2. The operation is thus performed : 
 
 
 3s. 
 
 6d. 
 
 7 
 
 1 . 
 
 4s. 
 
 6d.=the price of 7 yards. 
 
 7 
 
 8. 
 
 11s. 
 
 7s. 
 
 6d.=price of 49 yards. 
 Od.:= price of 2 yards. 
 
 8. 
 
 18s. 
 
 6d.=rprice of 51 yards. 
 
 Or, the sum may be solved by the note, thus : 51 yards, at 
 6 d. per yard, would=3C6 d.^1 . 5 s. 6 d. ; and 51 yards at 3 
 s. per yard=153 s. = 7 > 13 s. ; and 1 . 5 s. 6 d. added to 7 
 . 13s. as before, gives 8 . 18 s. 6 d. 
 
 2. What cost 47 yards of cloth, at 17 s. 9 d. per yard ? The 
 two numbers required by the rule are 6 and 7, and there is a 
 remainder of 5. Therefore, 
 
COMPOUND MULTIPLICATION. Ill 
 
 17s. 9d. 
 
 6 
 
 5. 6s. 6 d.= price of 6 yards. 
 
 7 
 
 37 . 5s. 6d.=price of 42 yards. 
 4. 8s. 9d.=price of 5 yards added. 
 
 41 . 14s, 3 d.= price of 47 yards. 
 
 Or, 47yds. x 9 d.= 1 . 15s. 3d. 
 47yds x!7s. = 39. 19s. d. 
 
 41 . 14s. 3d. as before. 
 
 3. What cost 23 gallons of melasses at 3 s. 6 d. per gallon ? 
 Ans. 4 . 6 d. 
 
 4. What cost 94 yards of cloth at 1 . 9 s. 4 d. per yard ? 
 Ans. 137 . 17s. 4 d. 
 
 5. What cost 59 yards of baize at 3 s. 4 d. per yard ? A?is. 
 9. 16s. 8d. 
 
 6. What cost 29 cwt. of sugar at 17s. 8d. per cwt. ? Ans. 
 25 . 12s. 4d. 
 
 7. What cost 78 yards of cloth at 9 s. 3 d. per yard ? Ans. 
 36 . Is. 6d. 
 
 8. What cost 65 cwt. of sugar at 19 s. 3 d. per cwt. ? Ans. 
 62 . 11s. 3d. 
 
 9. Seventeen men brought each a load of hay to market, 
 weighing 17 cwt. 3 qr. and 21 Ib. and received each for his 
 load, 5 . 8s. 3d. What quantity of hay did they all bring ; 
 and how much money did they all receive 1 Ans. They brought 
 15 T. 4 cwt. 3 qr. 21 Ib. and received 92 . s. 3 d. 
 
 EXAMPLES OF WEIGHTS AND MEASURES. 
 
 1. What is the weight of 5 hogsheads of sugar, each 
 weighing 7 cwt. 3 qr. 16 Ib. 1 Ans. 39 cwt. 1 qr. 24 Ib. 
 
 2. What is the weight of 9 chests of tea, each weighing 
 3 cwt. 2 qr. 9 Ib. ? Ans. 32 cwt. 25 Ib. 
 
 3. In 8 piles of wood, each containing 4 cords and 56 feet, 
 how many cords and feet ? Ans. 35 cords, 64 feet. 
 
 4. Multiply 15 yards, 3 qr. and 2 nails, by 9. Ans. 142 yd. 
 3 qr. 2 nails. 
 
 5. Multiply 20 years, 5 months, 3 weeks, and 6 days, by 14. 
 Ans. 286 yr. U mo. 3 w, 
 
112 COMPOUND DIVISION. 
 
 6. In 10 fields, containing each 12 acres, 2 roods, and 16 
 rods, how many acres ? Ans. 126. 
 
 7. In 7 casks, containing each 42 gallons, 3 quarts, and 1 
 pint, how many gallons, &c ? Ans. 300 gal. 1 pint. 
 
 QUESTIONS. What is always the nature of the multiplier! Which 
 is the compound number, the multiplier or multiplicand? What is the 
 nature of the product 1 In case the quantity by which you multiply is 
 yards, what does the number of yards decide 1 What is Case 1st 7 
 "What is the rule 1 What is Case 2d 1 What is the rule 1 What is 
 the note under Case 2d 1 What is Case 3d 1 What is the rule"? 
 What note follows the rule 1 No direct definition has been given of 
 Compound Multiplication ; will the scholar give one 1 
 
 COMPOUND DIVISION. 
 
 This is the last of the Compound Rules, and is the reverse 
 of the preceding. In this rule a compound number is given as 
 a dividend, and a simple number as a divisor ; and by the opera- 
 tion the dividend is resolved into as many equal parts as there 
 are units in the divisor. The quotient is always one of these 
 equal parts, and is therefore a compound number ; each figure of 
 which is of the same denomination as the figure or figures in 
 the dividend from which it was obtained. 
 
 CASE 1st. WHEN THE DIVISOR OR SIMPLE NUMBER is 12, OR 
 
 LESS THAN 12. 
 
 RULE. Divide the highest denomination first. If after divi- 
 ding this, there be a remainder, reduce it to the next lower 
 denomination, adding the figures of the dividend in that denomi- 
 nation to it, and divide again. Proceed in the same manner 
 through all denominations ; the number obtained will be the one 
 required. 
 
 Ex.1. Divide 17 . 11s. 5 d. by 8. 
 
 PERFORMED. 
 8 ) 17. US. 5d, 
 
 345. 3s, 
 
COMPOUND DIVISION. 113 
 
 17 .. 4-8=2 and 1 remains. What remains of any number 
 or quantity, must obviously be of the same kind as the quantity 
 itself. Therefore, the 1 is one pound=20s. and 20s. + 11 s. 
 = 31 s. and 3 1-4- 8 = 3s. and 7 s. remain. Again, 7s. = 84d. 
 and84d.-|-5d. = 89d.; and 89 ^8 = 11 and 1 remainder = 111 d. 
 Therefore the answer is as given above, viz. 2 >. 3 s. Ill d. 
 
 2. Divide 25 . 18 s. 9 d. by 6. 
 
 PERFORMED. 
 
 6)25. 18s. 9d. 
 
 4. 6s. 5d. 2qr. 
 
 3. Divide 13 . 19 s. 4 d. by 4. Ans. 3 . 9 s. 10 d. 
 
 4. Divide 140 . 12s. 9d. by 12. Ans. 11 . 14s. 4d. 3qr. 
 
 5. Divide 73 . 16s. lid. 3 qr. by 9. Ans. 8. 4 s. 1 d. 
 Ifqr. 
 
 6. Divide 12 cwt. 3 qr. 12 Ib. by 10. Ans. 1 cwt. 1 qr. 4lb. 
 
 7. Eleven men own equal shares of 36 hhd. 42 gal. and 2 
 qt. of wine ; what is each man's share ? Ans. 3 hhd. 21 gal. 
 Oqt. Opt. 1/ T gills. 
 
 8. Seven men bought 16 hhd. 24 gal. 3 qt. of wine, for 
 which they paid 45 . 18 s. 6 d. ; each man paying the same 
 money and consequently entitled to an equal share of wine. 
 What was each man's share, and how much money did he pay? 
 Ans. His share was 2 hhd. 21 gal. 2| qt. and paid 6 . Is. 
 2 d. 24 qr. 
 
 CASE 2d. WHEN THE DIVISOR is A COMPOSITE NUMBER 
 
 GREATER THAN 12. 
 
 RULE. Resolve the divisor into its component parts, and 
 divide the compound number by each of these parts in succession. 
 The quotient arising from the first division will form a divi- 
 dend for the second ; and so on. 
 
 Ex. 1. Divide 26 . 16 s. 8 d. by 21 . The factors of 21 are 
 7 and 3, because 7x3=21. Therefore, 
 
 7)26. 16s. 8d. 
 3)3. 16s. 8 d. 
 
 1. 5s. 6fd.=the quotient of 26 . 16s. 
 8d.-f-21, and is the answer. 
 10* 
 
114 COMPOUND DIVISION. 
 
 2. Divide 47 . 15 s. 8 d. by 24. Ans. 1 . 19 s. 9 d. 3qr. 
 
 3. Divide 85 . 11 s. lid. by 16. Ans. 5 .6 s. 11 d. s|qr. 
 
 4. Divide 128 . 9 s. by 42. ^w*. 3 . 1 s. 2 d. 
 
 5. Divide 15 . 18 s. 9 d. by 72. Ans. 4 s. 5& d. 
 
 6. Divide 5. 10s. 3d. by 81. Ans. 1 s. 4d. lqr. 
 
 7. Divide 7 . 19 s. 9 d. by 96. Ans. 1 s. 7 d. 3qr. 
 
 8. Divide 27 . 16 s. by 32. Ans. 17 s. 4 d. 2 qr. 
 
 CASE 3d. WHEN THE DIVISOR is LARGE AND NOT A COMPOSITE 
 NUMBER. 
 
 RULE. Divide the whole compound quantity by the whole 
 divisor ; reducing the remainders after the division of each 
 denomination to the next lower denomination, as directed in 
 Case 1st. 
 
 Ex. 1. Divide 8 . 18s. 6 d. by 51. 
 
 PERFORMED. 
 
 51 ) 8JE. 18s. 6 d. ( OJ6. 3s. 6 d. 
 20 
 
 178= the shillings in 8 . 18s. 
 153 
 
 25= shillings remaining. 
 12 
 
 306=pence in 25 s. 6 d. 
 306 
 
 000 
 
 The pounds are first reduced to shillings, and the given shil- 
 lings are added. The 178 shillings are thus produced. This 
 divided by 51 gives 3 as a quotient figure and 25 as a remain- 
 der. After a second reduction, 306 pence are obtained, which 
 contains the divisor six times. Thus the answer obtained is 
 3s. 6 d. 
 
 2. Divide 41 . 14 s. 3 d. by 47. Ans. 17 s. 9 d. 
 
 3. Divide 4 . I s. 5 d. 2 qr. by 23. Ans. 3 s. 6 d. 2 qr. 
 
 4. Divide 137 . 17 s. 4 d. by 94. Ans. I . 9 s. 4 d. 
 
 5. Divide 36 . 1 s. 6 d. by 78. Ans. 9s. 3d. 
 
 6. Divide 10 . 5 s. 8 d. by 59. Ans. 3 s. 5 d. 3 qr. -f- 
 
 7. Divide 25 . 12s. 4 d. by 29. Ans. 17s. 8 d. 
 
 8. Divide 61 . 12 s. by 65. Ans. 18 . 1 1 s. 1 qr.+ 
 
COMPOUND DIVISION. 115 
 
 EXAMPLES IN WEIGHTS AND MEASURES. 
 
 1. Divide 5hhd. 42 gal. 3qt. equally among 4 men. Ans. 
 
 1 hhd. 26 gal. 1 qt. 1 pt. 2 gi. 
 
 2. Divide 14cwt. 1 qr. 12lb.by5. Ans. 2 cwt. 3 qr. 13 Ib. 9 
 oz. 9f dr. 
 
 3. Divide 27yd. 1 qr. 2na. by 7. Ans. 3yd. 3qr. 2-f- na. 
 
 4. Divide 156 bushels, 3 pk. 6qt. by 18. Ans. 8bu. 2pk. 7qt. 
 
 5. Divide 9 hhd. 28 gal. 2 qt. by 12. Ans. 49 gal. 2 qt. 1 pt. 
 
 6. Divide 16cwt. 3qr. 18 Ib. by 32. Ans. 2qr. 3 Ib. 3 oz. 
 
 7. If 27 loads of hay weigh 30 tons, 8 cwt. 2 qr. 23 Ib. what 
 is the weight of one load ? Ans. 1 ton, 2 cwt. 2 qr. 5 Ib. 
 
 8. A man traveled 17 leagues, 1 mile, 4 furlongs, and 21 
 poles, in 21 hours. At what rate did he travel per hour? Ans. 
 
 2 m. 4 fur. and 1 pole. 
 
 9. Nine men own 56 Ib. 6oz. and 17pwt. of silver. What 
 will each man receive if the whole quantity be equally divided 
 among them? Ans. 61b. 3 oz. 8pwt. 13gr. 
 
 10. Bought 15 loads of hay, the whole weight of which was 
 12 tons, 15 cwt. 3 qr. 161b. Supposing them all to have been 
 equal, what was the weight of each ? Ans. 17 cwt. 6f Ib. 
 
 11. If a man's income be86 18s. 10 d. per year, what is 
 it per calendar month ? Ans. 7 . 4 s. lOf d. 
 
 12. If I pay 15 . 3s. 8 d. for 56 pairs of gloves, what is 
 one pair worth ? Ans. 5 s. 5 d Of qr. 
 
 13. If a hogshead of wine cost 33 . 12s., what is the 
 price of a gallon ? Ans. 10s. 8 d. 
 
 14. If 42 yards of cloth cost 21 . 18 s. 8 d., what was the 
 cost per yard ? Ans. 10s. 5 d. H qr. 
 
 15. If 16 men cut 53 cords, 69 feet of wood in 2 days, what 
 did each man cut per day ? Ans. 1 cord, 86^ 5 feet. 
 
 APPLICATION OF THE FOUR PRECEDING RULES. 
 
 1 . A silversmith sold to his customer 3 dozen silver spoons, 
 each weighing 3oz. 3pwt. 16 gr. ; l dozen tea spoons, each 
 weighing 14 pwt. 20 gr. ; 3 silver cups, each weighing 20 oz. 
 18 pwt. In return, he received old silver to the amount of 8 Ib. 
 1 1 oz. 19 pwt. ; for how much ought he to receive pay ? Ans. 
 61b. lOoz. 14 pwt. 
 
 2. Bought the following articles at the prices mentioned; viz. 
 
116 COMPOUND DIVISION. 
 
 . S. d. 
 
 4 cwt. of sugar at 2 . 4s. 8 d. per cwt. 
 
 3 hhd. of melasses at 2 s. 4 d. per gallon, 
 
 5 Ib. of green tea at 7 s. 6 d. per pound, 
 12 Ib. of raisins at 2 s. per pound, 
 
 42 yards of cotton cloth at Is. 6 d. per yard, 
 27 pounds of ham at 1 s. 3d. per pound ; 
 
 What was the amount of my bill ? Ans.38. 17s. lid. 
 
 3. Bought of James Rankin, . s. d. 
 27 yards of broadcloth at 1 . 9s. per yard, 
 
 42 yards of Irish linen at 4 s. 6 d. per yard, 
 
 36 hats valued each at 18s. 
 
 30 pairs of shoes at 7 s. 8 d. per pair ; 
 
 What was the amount of my bill ? Ans. 92 . 10s. d. 
 
 4. A. owning 100 acres of land, divided it into 8 equal parts, 
 and sold each part for $22.50 per acre. How many acres 
 were there in each part, and what was the value of the same ? 
 Ans. 12 A. 2R. ; value, $281.25. 
 
 5. Out of a pipe of wine a merchant sold 36 gallons, 3 quarts, 
 and 1 pint at each of three different times; he then rilled 15 
 bottles, holding 1 pint and 2 gills each ; how much remained ? 
 Ans. 12 gal. 2 qt. and 2 gills. 
 
 6. Bought 144 pairs of shoes for 96 . ; what was the 
 price of one pair? Ans. 13 s. 4d. 
 
 7. A person dying, left real estate to the amount of 2356 . 
 19 s. 9 d. and personal property to the amount of 3184 . 12 s. 
 8 d. In his will, he directed that his wife should receive one 
 sixth of the whole, and that the remainder should be equally 
 divided among his four daughters. What was the share of 
 each ? Ans. The widow, 923 . 12 s. | d. and the daughters 
 each 1 154 . 10s. Ijyd. 
 
 QUESTIONS. How does Compound Division compare with Com- 
 pound Multiplication 1 What is given as a dividend 1 What as a 
 divisor 1 Jnto what is the dividend resolved by the operation 1 What 
 is always one of these equal parts 1 Is the quotient a Simple or Com- 
 pound Number 1 How may you know the denomination of each 
 figure in the quotient 1 What is Case 1st 1 What is the rule 1 What 
 is Case 2d 'I What is the rule 1 What is Case 3d ? What rule 1 
 
 
VULGAR FRACTIONS. 
 
 VULGAR FRACTIONS. 
 
 1 . When a unit or single object is divided into a number of 
 equal parts, each of these parts is a fraction. 
 
 If it be divided into two equal parts, each part is called a 
 half, and is thus written, ^. 
 
 If it be divided into three equal parts, each is called a third, 
 and is thus written, . 
 
 If the Avhole be separated into six equal parts, each part is 
 called a sixth, and if into eight equal parts, an eighth, of the 
 whole, and are thus written, 1, }. 
 
 When more parts than one are to be expressed, the figure 
 above the line designates their number, thus, % ; by which 
 expression, we are to understand that the unit is divided into|i| 
 six equal parts, and that five of these parts are included in the" 
 fraction. 
 
 The fraction therefore is used to express parts of units, and 
 is represented by two numbers, one standing below and the 
 other above a short horizontal line. The number below the 
 line is called the denominator, and shows the number of equal 
 parts into which the unit is divided. The number above the 
 line is called the numerator, and shows how many of these 
 equal parts are included in the fraction, or make up its value. 
 Thus of the fraction | , the lower number shows a unit to be 
 divided into nine equal parts ; and the upper number, that five 
 of these parts are included in the fraction. 
 
 These two numbers, when spoken of collectively, are called 
 the terms of the fraction. 
 
 2. Fractions are divided into six kinds ; viz. Proper, Im- 
 proper, Simple, Compound, Mixed, and Complex. 
 
 A Proper Fraction is one whose numerator is less than its 
 denominator, as, f . 
 
 An Improper Fraction is one whose numerator equals or 
 exceeds its denominator, as, J. 
 
 A Simple Fraction consists of one expression, and is either 
 proper or improper, as, f or . 
 
 A Compound Fraction is the fraction of a fraction, as, | of J 
 of . It may consist of any number of simple fractions. 
 
 A Mixed Number consists of a whole number and fraction, 
 written together, as, 6|, 251, & c? 
 
118 VULGAR FRACTIONS. 
 
 A Complex Fraction is one that has a fraction in its numerator 
 or denominator, or both, as, ~, ^, ||, &c. 
 
 3. The denominator shows the number of equal parts into 
 which the unit is divided ; and the numerator, how many of 
 these parts are expressed in the fraction. Consequently, the 
 greater the numerator, the denominator being given, the greater 
 the value of the fraction ; and the less the numerator, the less 
 the value of the fraction. If the denominator be 8, and the 
 numerator 1 , the value expressed is J, or one eighth part of a 
 unit ; if the numerator be 2, the value expressed is f, or two 
 eighth parts of a unit ; if it be 4, the value is -|, or four eighth 
 parts of a unit ; and if it be 6, the value is , or six eighth parts 
 of a unit. 
 
 The value of a fraction is therefore the quotient arising from 
 t . dividing the numerator by the denominator, and always increases 
 in the same ratio as the numerator, so long as the denominator 
 remains unaltered. 
 
 We may therefore express any value, not only less than a 
 unit, but equal to and even greater than a unit, by a fraction. 
 Thus, if we take 9 as the denominator of a fraction, and any 
 number less than 9 as a numerator of the same, the value ex- 
 pressed is always less than a unit, as, f- ; or if 9 be taken as the 
 numerator, we obtain the fraction f , which, as the unit was 
 divided into 9 parts only, is obviously equal to 1. Again, we 
 may suppose more than a single unit of the same kind to be 
 divided in the same manner, and their parts united in one 
 fraction, and thus obtain fractions of any value more than a unit. 
 If two units be thus divided into seven equal parts, and three 
 parts of the one be united to all the parts of the other, the 
 fraction would be ^ ; or if all the parts of each be united, it 
 would be y, which is equal to 2 ; or if three units were thus 
 divided, all their parts would produce the fraction 2 y=3. 
 
 The only consideration which limits the value 01 a fraction, 
 is the number of equal parts united in the same expression. 
 
 From the preceding it is obvious that the value of a fraction 
 is increased in the same ratio as the numerator ; hence, 
 
 4. A fraction is multiplied by a whole number, by multiplying 
 the numerator only. 
 
 
VULGAR FRACTIONS. 119 
 
 In accordance with the above principle, the scholar may 
 multiply the following examples : 
 
 1. Multiply by 3. Ans. f . 
 
 2. Multiply | by 5. Ans. \. 
 
 3. Multiply | by 3. Ans. f . 
 
 4. Multiply 1 by 9. Ans. f . 
 
 5. Multiply T 8 ? by 6. Ans. f|. 
 
 6. Multiply || by 9. Ans. f|. 
 
 7. Multiply ^ by 7. 
 
 8. Multiplyf by 12. 
 
 9. Multiply iiby 12. 
 
 10. Multiply I by 8. 
 
 11. Multiply $ by 3. An*. J=L 
 
 12. Multiply i by 7. An*. |=1. 
 
 From the last two examples, it is obvious that a fraction is 
 multiplied by a number equal to its own denominator, by 
 rejecting that denominator and retaining only the numerator. 
 
 It should always be an object with the scholar to preserve the 
 terms of a fraction as small as is possible and express the true 
 value. This was not regarded in the above examples. A little 
 experience will show that to increase or diminish the value of 
 a fraction, it is only necessary to make the numerator larger or 
 smaller compared with the denominator. Suppose it be required 
 to multiply the fraction 1 by 2. By the above rule the product 
 would be f, which is equal in value to ^, and this is at once 
 obtained by dividing the denominator by 2 instead of multiplying 
 
 the numerator as above, thus, - _- ; therefore, 
 
 A fraction is multiplied by a whole number, by dividing the 
 denominator by that number. 
 
 The following examples will illustrate this principle : 
 
 1. Multiply | by 2. Ans. f . 
 
 2. Multiply f by 3. Ans. f or 1. 
 
 3. Multiply f by 4. Ans. f . 
 
 4. Multiply T 9 ^ by 5. Ans. f . 
 
 5. Multiply j3- by 8. Ans. f . 
 
 6. Multiply - T by 7. Ans. 
 
 7. Multiply 
 
 8. Multiply 
 
 9. Multiply 
 10. Multiply 
 
120 VULGAR FRACTIONS. 
 
 The value of a fraction may therefore be multiplied by a whole 
 number, either by multiplying the numerator or dividing the 
 denominator by that number. 
 
 Note. The denominator should always be divided, when- 
 ever it can be done without a remainder. 
 
 5. A fraction is divided by a whole number, by dividing the 
 numerator by that number. 
 
 This needs no explanation. If we divide a number by 2, 
 we take a half, and if by 3, a third of that number ; that is, the 
 divisor always shows what part of the dividend is taken ; there- 
 fore , if H- 2 = T 6 ^, and if ~ 3 = ^. 
 
 The following examples will illustrate the operation of the 
 above principle : 
 
 Ex. 1. Divide f by 3. Ans. \. 
 
 2. Divide f by 2. Ans. f . 
 
 3. Divide T 8 j by 8. Ans. ^. 
 
 4. Divide ^ by 6. Ans. ^. 
 
 5. Divide if- by 4. 
 
 6. Divide |f by 5. 
 
 7. Divide -| by 7. 
 
 In this last example, the scholar will find a difficulty. He 
 cannot divide the numerator in any way, except to place it over 
 the 7 in the form of a fraction, as will hereafter be explained ; 
 this would make one fraction the numerator of another 
 fraction. When, therefore, the divisor will not divide the nu- 
 merator without a remainder, a more convenient mode of ope- 
 rating is desirable. It will be remembered, that division is the 
 reverse of multiplication ; and since we can multiply fractions 
 by dividing the denominator, we will try the effect of dividing 
 fractions by multiplying the denominator. Let it be required to 
 divide T 3 ^ by 3. By dividing as above, we obtain T \> as the 
 quotient, viz. ^ 3=^. By the mode we propose to try, we 
 obtain ^. It therefore remains to show that jV A- ** an 7 
 object be first divided into 12 equal parts, and then each of 
 these 12 parts be divided into 3 equal parts, it is plain that the 
 whole would be divided into 36 equal parts, and that each twelfth 
 part would make 3 thirty-sixth parts ; therefore, y 1 ^ ^ ; hence, 
 a fraction is divided by a whole number by multiplying its de- 
 nominator by that number. 
 
 
VULGAR FRACTIONS. 121 
 
 This principle may be applied to the following sums : 
 Ex. 1. Divide f by 6. Ans. 
 
 2. Divide | by 2. Ans. 
 
 3. Divide by 3. Ans. 
 
 4. Divide ^ by 5. Ans. 
 
 5. Divide 
 
 6. Divide 
 
 7. Divide 
 
 8. Divide }\ 
 
 By uniting the two preceding principles, we have the fol- 
 lowing more comprehensive principle, viz. : 
 
 The value of a fraction is divided by a whole number, by di- 
 viding the numerator, or by multiplying the denominator by that 
 number. 
 
 Note. The numerator should always be divided, when it 
 can be done without a remainder. In all other cases the de- 
 nominator should be multiplied. 
 
 From the preceding remarks and illustrations we learn, that 
 whatever operation is performed on the numerator of a fraction, 
 the SAME OPERATION is PERFORMED on the VALUE of the frac- 
 tion ; but that the effect produced on the VALUE of any fraction, is 
 the REVERSE of the OPERATION PERFORMED ON ITS DENOMINATOR. 
 
 6. A fraction is multiplied by a fraction, by multiplying the 
 numerators together for a new numerator, and the denominators 
 for a new denominator. 
 
 For example, let it be required to multiply by \ . Agree- 
 ably to the principles already explained, if I multiply the de- 
 nominator of the fraction \, by 4, the other denominator, I shall 
 obtain \ of that quantity, viz. ^ ; and if I multiply this quan- 
 tity, viz. J^, by 3, the other numerator, I shall make this value 
 three times as large, that is, it will become -py ; therefore, -^ 
 is | of I, or xf = T V 
 
 In accordance with the above, the scholar may multiply the 
 following fractions : 
 
 1 . Multiply by J. Ans. -f^. 
 
 2. Multiply % by f . Ans. if. 
 
 3. Multiply f by J. Ans. ||. 
 
 4. Multiply 1 by f . Ans. ^. 
 
 11 
 
122 VULGAR FRACTIONS. 
 
 5. Multiply 
 
 6. Multiply 
 
 7. Multiply 
 
 8. Multiply 
 
 9. Multiply 
 10. Multiply 
 
 7. A fraction is divided by another fraction, by inverting the 
 divisor, and multiplying them together as before. 
 
 A unit is contained in the fraction }, three fourths of once ; 
 ^ is consequently contained in the same fraction twice as often, 
 viz. f of a time ; and , three times as often, viz. f of a time, 
 which fractions are obviously obtained by multiplying f by -$ 
 and inverted. Again, suppose it be required to find how many 
 times | is contained in $. As before, a unit or 1 is contained 
 in $, seven eighths of a time ; would be contained in it four 
 times as often, viz. 2 8 of a time, and f would be contained in 
 the same only one third as often as , viz. -|| of a timez=l^ 4 T , or 
 li. This result is obtained by inverting the divisor |, and 
 multiplying it into the dividend -g- ; thus, ^xf=ff. 
 
 The following examples may now be performed : 
 
 Ex. 1. Divide by f. Ans. f. 
 
 2. Divide f by . Ans. |-. 
 
 3. Divide | by ^. Ans. |f 
 
 4. Divide f by 
 
 5. Divide T V by f . 
 
 6. Divide ^ by 
 
 7. Divide | by f 
 
 8. Divide f by T 9 r . 
 
 9. Divide 4- by J. 
 10. Divide {f by i. 
 
 8. If the numerator and denominator of any fraction be 
 both multiplied or both divided by the same number, the value of 
 the fraction will not be altered. 
 
 Of this principle no explanation is necessary. The value 
 of the fraction being the quotient arising from dividing the 
 numerator by the denominator, it is obvious that if both the 
 terms be doubled, or repeated any number of times, the value of 
 the quotient will not be affected. 
 
REDUCTION OF FRACTIONS. 123 
 
 REDUCTION OF FRACTIONS. 
 
 CASE 1st. To REDUCE FRACTIONS TO THEIR LOWEST TERMS; 
 
 OR, TO FIND THE LOWEST TERMS BY WHICH THE VALUE OF 
 A GIVEN FRACTION CAN BE EXPRESSED. 
 
 RULE. Divide both numerator and denominator by any num- 
 ber that will divide them both WITHOUT REMAINDER ; then di- 
 vide the quotients obtained, in the same manner, and so continue 
 to do till there is no number greater than I, that will divide them. 
 The last quotient will be the numerator and denominator required. 
 
 Ex. 1. Reduce |-f to its lowest terms. Operation, ||-f-8=|, 
 and J--K2 f , its lowest term. 
 
 2. Reduce j-f^to its lowest terms. Ans. |^. 
 
 3. Reduce Jyff to its lowest terms. Ans. $. 
 
 4. Reduce ^^ to its lowest terms. Ans. . 
 
 5. Reduce | to its lowest terms. Ans. I. 
 
 6. Reduce -fffo to ^ ts l west terms. Ans. Jj. 
 
 7. Reduce |- to its lowest terms. Ans. ^-. 
 
 8. Reduce Q 7 \ 6 s s j to its lowest terms. Ans. J. 
 
 9. Reduce ^ffrfz to * ts l west terms. Ans. %. 
 
 10. Reduce 6 6 7 7 8 8 6 G Q- to its lowest terms. Ans. 
 
 11. Reduce y^ny to its lowest terms. Ans. 
 
 12. Reduce l - to its lowest terms. Ans. l -f 
 
 CASE 2d. To REDUCE A WHOLE NUMBER OR A MIXED QUAN- 
 TITY TO AN IMPROPER FRACTION. 
 
 RULE. If the given quantity be a whole number, multiply it 
 by the proposed denominator ; the product will be the numerator : 
 but if it be a mixed quantity, multiply the whole number by the 
 denominator of the fraction, and to the product add the given 
 numerator ; then under the number thu& produced, write the 
 denominator. 
 
 Ex. 1. Reduce 21 to a fraction whose denominator is 9. 
 Operation, 21 x9=;189, the numerator; the fraction therefore 
 is i|^. 
 
124 REDUCTION OF FRACTIONS. 
 
 2. Reduce 8 to an improper fraction. Operation, 8x3=24, 
 and 24+1=25, the numerator ; therefore, 2 ^ is the answer. 
 
 3. Reduce 16| to an improper fraction. Arts. 
 
 4. Reduce 17-^f to an improper fraction. Ans. 
 
 5. Reduce 47^ to an improper fraction. Ans. 
 
 6. Reduce 135^ to an improper fraction. Ans. ----. 
 
 7. Reduce 1^| to an improper fraction. Ans. ^. 
 
 8. Reduce 1728-|y to an improper fraction. Ans. 4 - 6 -j- 
 
 9. Reduce 9|to an improper fraction. Ans. l -f. 
 
 10, Reduce 12|- to an improper fraction. Ans. y . 
 
 11. Reduce 8 to a fraction whose denominator shall be 9. 
 
 12. Reduce 16 to a fraction whose denominator shall be 12. 
 Ans. . 
 
 CASE 3d. To REDUCE AN IMPROPER FRACTION TO A WHOLE 
 
 OR MIXED NUMBER. 
 
 RULE. Divide the numerator by the denominator; the quo- 
 tient will be the whole number. If there be any remainder, place 
 it over the denominator at the right of the whole number. 
 
 Note. The true quotient includes both the whole number 
 and fraction. In all cases of division, therefore, the remainder 
 (if any) constitutes the numerator of a fraction of which the 
 divisor is the denominator. 
 
 Ex. 1. Reduce Yy 1 to a mixed number. 
 
 4 OPERATION. 
 
 17)141(8 
 1 3 6 
 
 5 rem. ; therefore, 8 T 5 y is the answer. 
 
 2. Reduce * to a mixed quantity. Ans. 
 
 3. Reduce 7 ^ 9 to a mixed quantity. Ans. 13-J. 
 
 4. Reduce *-f to a mixed quantity or whole number. Ans. 7. 
 
 5. Reduce 4 -ff to a mixed quantity. Ans. 5f 1. 
 
 6. Reduce ^^ 4 to a mixed number. Ans. 
 
 7. Reduce ^ to a mixed number. Ans. 
 
 8. Reduce -f4- to a mixed number. Ans. 56- 3 7 T . 
 
 9. Reduce 1 !^ to a whole number. Ans. 12. 
 10. Reduce ff to a mixed number. Ans. 6^. 
 
 1 1 . Reduce l - 7 -g- 8 to a whole or mixed number. Ans. 288, 
 12. Reduce 5 -2 7 T 8 - 9 to its proper number. Ans. 2704^-. 
 
 
REDUCTION OF FRACTIONS. 125 
 
 CASE 4th. To REDUCE COMPOUND FRACTIONS TO SIMPLE ONES. 
 
 RULE 1st. Multiply all the numerators together for a new 
 numerator, and all the denominators for a new denominator, 
 and reduce the new fraction to its lowest terms, by Case 1st. 
 
 Ex.1. Reduce f of | of f to a simple fraction. Performed, 
 2 X 3 X 5 = 30, the new numerator ; and 3 X 4 x 6 = 72, the new 
 denominator ; therefore, |- is the fraction required, but suscepti- 
 ble of being expressed in lower terms ; therefore, j^-^^=Y5i 
 Ans. 
 
 Compound fractions may be reduced to simple ones, however, 
 much more expeditiously, by canceling. The labor of reducing 
 to lower terms is thereby avoided. 
 
 RULE. Draw a horizontal line and place all the numerators 
 above the line and all the denominators below it. Cancel the 
 numbers as far as practicable, as taught in the Rule for Can- 
 celing ; then make the product of the numbers remaining above 
 the line the new numerator, and the product of those remaining 
 below, the new denominator. 
 
 Note 1st. If there be nothing remaining above the line after 
 canceling, 1 will alw^s be the numerator of the new fraction. 
 The same is true of the denominators. 
 
 Ex. 2. Reduce of f of | to a simple fraction. 
 
 Statement, ^ ^ & . Canceled, g ' 4 & . Ans. ^. 
 
 Example 1st stated and solved by canceling : 
 2. 3. 5 ~ , j 2. $. 5 
 
 m- Canceled > riTe- Ans ' A- 
 
 2 
 
 3. Reduce f of ^ of -J J to a simple fraction. 
 
 Statement, %. Canceled, |i|* Ant. i. 
 
 4 
 
 Note 2d. Whenever the product of any two numbers on 
 one side of the line will cancel any number on the opposite 
 side, they may be so canceled ; as in the last example, 7 and 2 
 below the line cancel 14 above it. 
 
 4. Reduce | of ^J of | of | to a simple fraction. 
 
126 REDUCTION OF FRACTIONS, 
 
 Statement, |-^. Canceled, |^|| 
 
 3 2 
 
 and 7 x2 m 14, numerator ; and 13 x 3 = 39, denominator ; 
 therefore, the new fraction is J. 
 
 5. Reduce f of l f of T ^ of T \ of T 7 ^ to a simple fraction. 
 
 6. Reduce i of | of f of f of f of f of | of of T % to a simple 
 fraction. .Arcs. Jg.. 
 
 7. Reduce if of f of T 9 ^ of f of f of f to a simple fraction. 
 An*. f-ff . 
 
 . If any term of a compound fraction be a mixed 
 number, it must be reduced to an improper fraction before 
 stating. 
 
 8. Reduce ^ of ^ of 4f of f of f to a simple fraction. 
 4f y, therefore, statement, ' ' ' '-^ ; which canceled 
 
 will give the answer, J. 
 
 9. Reduce f of 7 of J of 2|- to a simple fraction. Ans. 
 
 21 6 _ f\ 6 
 3T 3^J' 
 
 10. Reduce of | of ^ of J of |- to a simple fraction. Ans. 
 
 . 1 6 
 
 11. Reduce T 9 ^ of ^ of ^ to a simpl^fraction. JL^. -Jf. 
 
 12. Reduce -f^ of J of J of j 1 ^ to a simple fraction. .Aws. ^. 
 
 13. Reduce ^ of i| of |^ to a simple fraction. Ans. 
 
 14. Reduce f of J of ^ of i to a simple fraction. ^In. 
 
 15. Reduce f of 8| of 5|- to a simple fraction. Ans. ^ 
 or 701. 
 
 CASE 5th. To CHANGE FRACTIONS FROM ONE DENOMINATION TO 
 
 ANOTHER, WITHOUT ALTERING THE VALUE. 
 
 1st. To reduce fractions of low denominations to those of 
 higher value. 
 
 RULE. Divide the fraction, or what is the same thing, mul- 
 tiply the denominator by such numbers as are required to reduce 
 the given quantity from the GIVEN to the REQUIRED DENOMI- 
 NATION. 
 
 Ex. 1 . Reduce ^ of a penny to the fraction of a pound. The 
 numbers required to reduce pence to pounds, are ] 2 and 20 ; 
 therefore, f of a penny is to be divided by these numbers ; and 
 since this can be effected in the present case only by multiplying 
 
REDUCTION OF FRACTIONS. 127 
 
 the denominator, the operation will be, ^ J2 2Q = j^, and 
 
 this, by Case 1st, is reduced to yj^, which is the fraction 
 required. Hence f of a penny equals ^|-g of a pound. 
 
 The canceling principle may however be successfully ap- 
 plied in the solution of sums of this character. 
 
 RULE FOR CANCELING. Place the numerator of the given 
 fraction above a horizontal line, and its denominator below it ; 
 then place also BELOW the line, such numbers as are necessary to 
 reduce the denomination given to that required. Cancel, tyc. 
 as before. 
 
 We will solve the above example by this rule also. Statement, 
 
 2 The scholar should compare the statement with the 
 
 rule, to see that he understands its application. The above 
 
 statement canceled, 10 ; and 6 x 12 x 4=288, the denomi- 
 o. i<. iJU 
 ^4 
 
 nator as before, and nothing remains as a numerator ; therefore, 
 
 as before, ^-i-g- of apoundis the answer. (See Note 1st, Case 4th.) 
 
 2. Reduce f of a farthing to the fraction of a shilling. By 
 
 3 3 
 
 the common rule, j x4xl2 _ ^ which, by Case 1st, equals 
 
 3 
 , Ans. By the rule for canceling, ^ ^. The same can- 
 
 4. 4. 
 
 celed ' = &* Ans ' 
 
 4 
 
 3. Reduce of a penny to the fraction of a pound. Statement, 
 
 5~l2~20- Canceled ' 5 12 20 ; and 5 X 3 X 20 = 300, therefore, 
 
 3 
 Ans. g^j. 
 
 4. Reduce of a gallon to the fraction of a hogshead. Ans. 
 
 Statement, - 7^. The 63 below the line reduces the gal- 
 lons to hogsheads. 
 
 5. Reduce f of an ounce Troy to the fraction of a pound. 
 Ans. 3^. 
 
 6. Reduce \^ of a minute to the fraction of a day. Ans. 
 
 T~5 3"(K 
 
 7. Reduce T 8 T of a pound Avoirdupois to the fraction of a ' 
 cwt. Ans. 
 
128 REDUCTION OF FRACTIONS. 
 
 8. Reduce f of a nail to the fraction of an ell English. Ans. 
 
 3 
 
 160* 
 
 9. Reduce T 5 ^ of a penny to the fraction of a pound. Ans. 
 
 10. Reduce i-J of an hour to the fraction of a year. Ans. 
 
 9T9"0' 
 
 Secondly. To reduce fractions of high denominations to 
 equivalent fractions of lower denominations. 
 
 RULE. Multiply the numerator by such numbers as are re- 
 quired to reduce the given quantity from the given to the re- 
 quired denomination, and then by Case 1st reduce the result to 
 its lowest terms. 
 
 Ex. 1 . Reduce ^ of a shilling to the fraction of a farthing. 
 
 To reduce shillings to farthings, we must multiply by 12 
 and 4; therefore, ^g-xl2x4=: |-|; and by Case 1st, ff^i, 
 Ans. 
 
 RULE FOR CANCELING. Place the numerator of the given 
 fraction above a horizontal line, and the denominator below, as 
 before ; then place above the line such numbers as are necessary 
 to reduce the denomination given to that required. Cancel, <SfC. 
 as before. 
 
 1 12. 4 
 
 The above sum solved by this rule. Statement, -r 
 
 1 12 4 
 The same canceled, ; therefore, ^ of a farthing is 
 
 & 2 
 the answer. 
 
 The scholar will carefully observe the difference between 
 the statement here, and the one given for reducing low denom- 
 inations to high. 
 
 2. Reduce -^4^ of a pound to the fraction of a penny. Ans. f . 
 
 1. 20. 12 
 Statement, . 
 
 3. Reduce -jig of a pound Troy to the fraction of a pwt. 
 
 I -jo Of) 
 
 Ans. 2 ^, or 2} pwt. Statement, -^ 
 
 4. Reduce y 1 ^- of a pound Troy to the fraction of an ounce. 
 Ans. f . 
 
REDUCTION OF FRACTIONS. 129 
 
 5. Reduce ^ of a penny to the fraction of a farthing. 
 Ans. |. 
 
 6. Reduce Q 6Q 1 000 of a mile to the fraction of a barley corn. 
 
 Ans - ^jV 
 
 7. Reduce yij of a cwt. to the fraction of a pound Avoirdu- 
 pois. Ans. -j^-. 
 
 8. Reduce yf^ of an ell English to the fraction of a nail. 
 Ans. f . 
 
 9. Reduce 94^0 f a y ear to t ^ ie fraction of an hour. Ans. 
 if- 
 
 CASE 6th. To REDUCE FRACTIONS OF A HIGHER DENOMINA- 
 TION TO THEIR VALUE IN WHOLE NUMBERS OF A LOWER 
 DENOMINATION. 
 
 RULE. Reduce the fraction to its next lower denomination by 
 multiplying the numerator by the requisite number, and divide 
 the product by the denominator ; the quotient thus obtained will 
 be a whole number of the lower denomination, and the remainder, 
 if any, may be reduced and divided as before. This process 
 may be continued till nothing remains, or till it is reduced to the 
 lowest denomination. 
 
 Ex. 1. Reduce f of a pound sterling to its value in shillings, 
 and pence. 
 
 OPERATION. 
 2= number. 
 2 
 
 Div. by denoin. 3)4 Omshillings. 
 
 13s. and 1 s. remains, which equals 12 d., 
 and 12 d.-^-3=4 d. Therefore, 13 s. 4 d. is the required num- 
 ber. 
 
 It is evident that f of a pound sterling is 20 times as many 
 thirds of a shilling; viz. 4 ^ = 13 shillings ; and ^ of a shil- 
 ling is 12 thirds of a penny, that is, L 2 =:4d. Hence, f of a 
 pound is 13 s. 4 d. 
 
 2. Reduce -^ of a pound sterling to its value in lower denom- 
 inations. 
 
 Solution : Jg- of a pound =|$ of a shilling, and of a shil- 
 2 ^vof a penny=4d. Ans. 
 
 3. Reduce f of a pound Troy to its integral value. Ans. 9 oz. 
 
130 REDUCTION OF FRACTIONS. 
 
 4. Reduce ^g- of a day to its integral value. Ans. 1 h. 20 m. 
 
 5. Reduce -$ of an hour to its value in whole numbers. 
 Ans. 54 m. 
 
 6. Reduce ^ of a hogshead to its value in whole numbers. 
 Ans. 2 qt. 1 pt. 1 gi. 
 
 7. Reduce f of a quart to its integral value. Ans. 1 pt. 1 gi. 
 
 8. Reduce f of an ell English to a whole number. Ans. 
 3qr. 
 
 9. Reduce f of a yard to a whole number. Ans. 3|qr. 
 10. Reduce f of an ell French to a whole number. Ans. 
 
 4qr. 
 
 CASE 7th. To REDUCE THE LOWER DENOMINATIONS OF A 
 
 COMPOUND NUMBER TO FRACTIONS OF A HIGHER DE- 
 NOMINATION. 
 
 RULE.- Reduce the given quantity to the lowest denomination 
 in that quantity, for a numerator of the fraction ; and then re- 
 duce a unit of the higher denomination to the same denomination 
 with the numerator for the denominator of the fraction. This 
 fraction reduced to its lowest terms by Case 1st, will be the one 
 required. 
 
 Ex. 1 . Reduce 2 qr. 2 na. to the fraction of a yard. 
 
 Operation, 2qr. 2 na.r= 10 nails, the numerator ; and lyd. 
 r=4qr., and 4qr. = 16 nails, the denominator; therefore, yf is 
 the fraction, which equals f , Ans. 
 
 The operation may be much abbreviated by canceling f for 
 which the following will be found a convenient rule. 
 
 RULE FOR CANCELING. Reduce the given quantity to the 
 lowest denomination mentioned, (if it consist of different denom- 
 inations,) and place it above a horizontal line ; and beneath the 
 same line place the numbers required to reduce this denomination 
 to the required denomination. Cancel, multiply, fyc., and the 
 terms of the required fraction will be obtained. 
 
 Ex. 2. Reduce 3s. 4 d. to the fraction of a pound. 
 Operation, 3s. 4 d. = 40 d., and pence are reduced to pounds 
 by dividing by 12 and 20. Therefore, 
 
 Statement, - . Canceled, 
 
 20 ' 13. 
 
REDUCTION OF FRACTIONS. 131 
 
 Nothing remains above the line, the numerator of the required 
 fraction therefore is 1 ; and 6 remains below the line, and 
 consequently is the denominator ; therefore, 1 is the fraction 
 required, and 3s. 4 d. is ^ of a pound. 
 
 3. Reduce 2 roods and 30 rods to the fraction of an acre. 
 
 2 roods and 3 rods = 110 rods; and rods are reduced to 
 acres, by dividing by 40 and by 4 ; the statement therefore is, 
 110 , , . im. A 
 
 Same cance * e d 1S > ~~~> 
 
 40~T' 
 
 4. Reduce 12 ounces to the fraction of a pound Avoirdu- 
 pois. Ans. f . 
 
 5. Reduce 26 gal. 2 qt. to the fraction of a hogshead. Ans. 
 
 T2\' 
 
 6. Reduce 3 fur. 20 rods to the fraction of a mile. Ans. -f^. 
 
 7. Reduce 3 qr. 2 na. to the fraction of an ell English. Ans. 
 
 TV 
 
 8. Reduce 2 dr. 2 sc. to the fraction of an ounce. Ans. $. 
 
 9. Reduce 2 qr. 24 Ib. to the fraction of a cwt. Ans. |-. 
 
 10. Reduce 6 oz. lOpwt. to the fraction of a pound Troy. 
 Ans. Jf. 
 
 11. Reduce 6qt. to the fraction of a bushel. Ans. ^. 
 
 12. Reduce 12h. 30m. to the fraction of a day. Ans. J|. 
 
 13. Reduce 4 d. 12h. to the fraction of a week. Ans^ T 9 T . 
 
 14. Reduce 15 deg. 30 m. to the fraction of a Sign. Ans. 
 
 W 
 
 CASE 8th. To REDUCE FRACTIONS HAVING DIFFERENT DE- 
 
 NOMINATORS TO EQUIVALENT FRACTIONS HAVING A 
 COMMON DENOMINATOR. 
 
 RULE. Multiply all the denominators together for a new 
 denominator, and each numerator into all the denominators ex- 
 cept its own, for a new numerator to each fraction. The several 
 numerators placed over the common denominator will give the re- 
 quired fractions. 
 
 Note. The fractions should be reduced to their lowest 
 terms before multiplying. If the scholar looks carefully into 
 the nature of this rule, he will see that it is only multiplying 
 numerators and denominators by the same numbers ; and he 
 has already learned, that this does not affect the value of the 
 fraction. 
 
 Ex. 1. Reduce -| , , and -^-, to a common denominator. 
 
132 
 
 ADDITION OF FRACTIONS. 
 
 PERFORMED. 
 
 7x5x1 1 = 385, the common denominator. 
 
 6x5x11 = 330, the numerator for f, which therefore equals 
 
 4x7x1 1 = 308, the numerator for f , therefore, = 
 8 X 5 X 7 = 280, the numerator for ^, therefore, j^^ff . 
 The required fractions, therefore, are f| 
 
 2. Reduce 
 
 . and 
 
 , to a common denominator. Ans. 
 
 3. Reduce -| , f , J, and , to a common denominator. Ans. 
 
 168 72 84 63 
 2~52> 252> ^5 2> 252* 
 
 4. Reduce f , Y , T ^, and -f^, to a common denominator. Ans. 
 
 230J: 1JL5_6_P _2_4JL UilO. 
 2880' '2880 ' 2880' 2880' 
 
 5. Reduce |, f , |- , and ^, to a common denominator. Ans. 
 
 1080 1440 486 
 ' T6~2"0' T6^' 1620' 
 
 6. Reduce f, J, |, and T 8 T to a common denominator. Ans. 
 
 891 495 1155 1080 
 T45' TT8 5"' TT8T' T45* 
 
 At the commencement of this rule, the scholar was instruct- 
 ed relative to the peculiar form and nature of fractions, and 
 made acquainted with certain principles of universal applica- 
 tion. In the course of the preceding eight cases, he has been 
 shown the various changes of which fractions are susceptible, 
 while their value remains unaffected. His attention will now 
 be directed to those operations by which their value is affected. 
 
 ADDITION OF FRACTIONS. 
 
 If the scholar will turn back to Simple Addition, he will 
 there find it stated, that numbers, or quantities of the same 
 kind only, can be reduced to a single number or quantity by 
 adding. The same is true of fractions. It is obvious that ^ 
 of a shilling, and | of a penny, make neither f of a shilling 
 nor | of a penny. But of a shilling makes ^ of a penny ; 
 and to this, we can add of a penny, and the amount will be 
 L 3 of a penny. Therefore, before we can add fractions, they 
 must be reduced to the same denomination. (See Case 5th.) 
 
 
ADDITION OF FRACTIONS. 133 
 
 It is equally impossible to add fractions whose denominators 
 are unlike. of a shilling added to of a shilling, makes 
 neither J of a shilling nor J of a shilling. But of a shilling 
 = f of a shilling ; and f 4~i =f of a shilling. Fractions must 
 therefore be reduced to a common denominator, before they can 
 be united. (See Case 8th.) Hence we have the following rule. 
 
 RULE. Reduce all the fractions to the same denomination, 
 and also to a common denominator ; then ado" their numerators 
 and place their sum over the common denominator. If the frac- 
 tions produced be improper, reduce them to a whole number or 
 mixed quantity. 
 
 Note. If any of the fractions are compound, they must be 
 reduced to simple ones, before they can be reduced to a com- 
 mon denominator. (See Case 4th.) 
 
 Ex. 1. What is the sum of ^ and f-? These fractions, reduced 
 to a common denominator by Case 8th, become -^ and ^-J, and 
 
 2. What is the sum of f of J, and f of j- ? By Case 4th, 
 of J= T V and | of =, and * +4=^+^, (see Case 
 
 8th,) and ^+^=11^1^. 
 
 3. What is the sum of ^ and -| ? Ans. y or If. 
 
 4. What is the amount of of J and f of i| ? Ans. fj. 
 
 5. What is the amount of 18 and 16, and of % 1 Ans. 
 34f. 
 
 Note 2d. When whole numbers are combined in the same 
 operation with fractions, add each separately and unite them, 
 as in the above sum. 
 
 6 . What is the amount of 2 1 , 7, , and f of Y?\ A ns. 
 
 7. What is the amount of f of a penny adde\to ^ of a >. ? 
 
 Explanation. ^ of a . = 8 ^ of a penny ; and f of a penny 
 -f- 8 ^ of a penny 3 j^ of a penny, and this equals 2s. 3d. 
 If qr. Ans. 
 
 8. What is the amount of f of a yard and of a nail ? Ans. 3 
 qr. 04: na. 
 
 9. What is the sum of |- of a pound added to ^ of a shilling ? 
 Ans. ^3 s . = i7s. 2d. 
 
 10. What is the sum of \ of i and f of \\. Ans. |^ or 1-ff . 
 
 11. What is the sum of of f of f off of f off, and i ? 
 .Aws. if 
 
 12 
 
134 
 
 SUBTRACTION OF FRACTIONS, 
 
 12. What is the sum of J of a ton added to f of a cwt. 1 
 Ans. llf^ cwt. 
 
 13. What is the sum of f of a day added to $ of an hour ? 
 Ans. 19^ hours. 
 
 14. What is the sum of -J of a pound, f of a shilling, and 
 of a penny ? Ans. 3 s. 2 d. 2 qr. 
 
 15. What is the sum of -^ of a week, f of a day, and of an 
 hour ? Ans. 1 day, 22 hours, 15m. 
 
 SUBTRACTION OF FRACTIONS. 
 
 As we can subtract a quantity only from another of the 
 same kind, it is obvious that the same preparations are neces- 
 sary to perform operations in this rule as in the preceding ; 
 therefore, 
 
 RULE. Prepare the fraction as in addition, then subtract 
 the less numerator from the greater, and place the remainder 
 over the common denominator. 
 
 It will be obvious that the difference of the numerators is 
 the difference sought. 
 
 Ex. 1. From |- take ^. Operation, | s^fi A if> 
 Ans. 
 
 2. From f take -. f 2T=Tft ~lAi*/9' Ans. 
 
 3. From |- take |-. Ans. ^. In this last example it is evident, 
 that, as the denominators are the same, the operation consists 
 in subtracting the numerators only. The same is true of all 
 similar examples, provided only that the fractions are of the 
 same denomination. 
 
 4. From 11 take -~. Ans. 
 
 5. From f| take if. Ans. 
 
 6. From if take ^. Ans. , 
 
 7. From f f- take if. Ans. 
 
 8. From of a pound take of a shilling. \ of a pound 
 
 2 2 of a shilling : and 2 ^ = 6 g of a shilling. Therefore, % 
 
 %=* of a shilling, and 5 ^ of a shilling =9 s. 2 d. 
 
 9. From J of a league take - of a mile. 1 league = 3 
 
 - or . 
 
 or f 
 or 
 
MULTIPLICATION OF FRACTIONS. 135 
 
 miles ; therefore, f of a league =f of a mile ; and ^ of a mile 
 =% of the same ; hence, J }=-j or 1| of a mile, which is 
 the distance required. 
 
 10. From of a shilling take f of a penny. Ans. ^ or 5 
 pence. 
 
 11. From | of a day take f of an hour. Ans. 1-f 4 -, or 20-f- 
 hours. 
 
 12. From T % of a pound Avoirdupois take | of an ounce. 
 Ans. Vu 3 > or 13$ ounces. 
 
 13. "From J take 1 of f of f . .4/i.y. $. $ of f of f nr^, and 
 S--i=aor|. 
 
 14. From }i of an ell English take of a yard. Ans. f 
 or 2 T V yards. 
 
 15.~ From 9| take 6|. A/w. 3^. 
 
 16. From 19 yards take 5|- yards. Ans. 13|. 
 
 17. From 7 Ells English take 4 yards. Ans. Y= 4 i 
 yards, or 3| Ells English. 
 
 1 8. From of a pound sterling take f of a penny. Ans. 
 3 T V of a penny, or 2 s. 1 J4 d. 
 
 "19. From f of a rod take | of a foot. Ans. y or 10 feet. 
 20. From |- of an ounce take | of a pwt. Ans. Vj pwt. or 
 pwt. 
 
 MULTIPLICATION OF FRACTIONS. 
 
 A fraction may be multiplied by a whole number, either by 
 multiplying the numerator or dividing the denominator by that 
 number. This has been fully illustrated in Section 4th, of 
 Fractions. 
 
 Ex. 1. Multiply fcby 16. Ans. V 6 , or 5J. 
 
 2. Multiply | by 8. Ans. %, or 6|. 
 
 3. Multiply f by 9. Ans. 5 T 4 > or 7f . 
 
 4. Multiply f by 3. Ans. ^ 4 , or 2f . 
 
 5. Multiply SL*. by 7. _4/w. 2 g, or 6f. 
 
 6. Multiply |i by 8. ^TW. 2 J, or 7. 
 
 7. Multiply || by 12. Ans. \ 3 , or 11. 
 
 8. Multiply |f by 3. Ans. 2 f , or 3f. 
 
 A fraction is multiplied into a whole number equal to its de- 
 nominator, by rejecting that denominator. 
 
136 MULTIPLICATION OF FRACTIONS. 
 
 9. Multiply if by 21. By dividing if by 21, I take ^r 
 part of 15 ; if, then, this J T part be repeated 21 times, it is 
 evident that the value of all the parts will equal 15. 
 
 10. Multiply f| by 82. Ans. 72. 
 
 11. Multiply I by 9. Ans. 7. 
 
 12. Multiply 4J by 73. Ans. 41. 
 
 13. Multiply y by 43. Ans. 21. 
 
 TO MULTIPLY FRACTIONS BY FRACTIONS. 
 
 RULE. Multiply the numerators of the given fractions to- 
 gether for the required numerator, and the denominators, for 
 the required denominator ; then by Case 1st reduce the terms 
 as far as practicable. 
 
 Note. Mixed numbers are to be reduced to improper frac- 
 tions before multiplying ; or we may first multiply the integris 
 and then the fractions, and add their products. 
 
 Ex. 1. Multiply | by f |xf=ft, and ^-3= T V, Ans. 
 
 2. Multiply f by J . f x J=H=& A - 
 
 3. Multiply 7|by3. 7$ = \?, and 3$ = \j>, and ^ 5 X V 
 I|Q_2_5 j or 25, Ans. 
 
 RULE FOR CANCELING. Place the numerators of the given 
 fractions above a horizontal line, and their denominators be- 
 low the same. Cancel, multiply, fyc. as before. 
 
 One important advantage of the above rule will be found in 
 the fact, that it always gives the answer in its lowest possi- 
 ble terms. 
 
 4. Multiply | by f . 
 
 1 o 13 
 
 Statement, ^ . Canceled, -=-nr, Ans. 
 
 o. 5 o, 5 
 
 2 
 
 5. Multiply | of | of f by | of f of J$. 
 
 3. 1. 2. 1, 2. 11 ~ . . B. 1. 2. 1. S. 11 
 Statement, ^^-3--. Canceled. 4 fi ^ & g ~ ; 
 
 2 
 
 therefore, 1 1 ^numerator, and 2x6x9x12 = 1296, denomi- 
 nator, and y^g-, Ans. 
 
 6. Multiply A by T \. 
 
 9 3 
 Statement, 
 
MULTIPLICATION OF FRACTIONS. 137 
 
 7. Multiply i by 2*. Ans. ff , or 2 T \. 
 
 8. Multiply 16 by 12. Ans. 203^. 
 
 9. Multiply 13i by 9|. Ans. 124^. 
 
 10. Multiply of f of f of | of j- of f of of f by of T 9 F 
 of of . 
 
 In solving sums by canceling, like the above, the necessity 
 of reducing compound fractions to simple ones is avoided. 
 
 11. Multiply $ of 19 by f. 
 
 1 19 3 
 Statement, g ' "* 6> An*. ^ , or 
 
 If any whole numbers are given as parts of the dividend, it 
 is only necessary to write them above the line as in the last 
 example ; or, if they are given as parts of the divisor, they 
 only require to be placed below the line ; the operation then 
 proceeds as before. 
 
 12. Multiply f of 10 by f . Ans. 3. 
 
 13. Multiply 144 by T \. Ans. 12. 
 
 14. Multiply 395 by of f . Ans. *., or 52f . 
 
 15. Multiply f of 3 times T % of 5 times of i of , by 4 
 of^offof^. Ans. J. 
 
 16. What will 2^ tons of hay cost at 16| dollars per ton? 
 Ans. 41 'i dollars. 
 
 17. What will 2^ barrels of sugar cost at 18 dollars per 
 barrel ? Ans. 42-f^ dollars. 
 
 18. What will 8f pounds of tea cost at 1^ dollars per 
 pound? Ans. lO^f dollars. 
 
 19. What will 4f cords of wood cost at 3f dollars per 
 cord? Ans. 1 7 1| dollars. 
 
 20. What will 9 yards of cloth cost at % dollar per yard ? 
 Ans. 7% dollars. 
 
 21. What will 34 gallons of wine cost at If dollars per 
 gallon ? Ans. 46^ dollars. 
 
 22. What will 12 barrels of sugar cost at 15^ dollars per 
 barrel? Ans. 190f dollars. 
 
 23. What will 22 pounds of lard cost at J dollar per 
 pound ? Ans. 2|. 
 
 12* 
 
138 
 
 DIVISION OF FRACTIONS. 
 
 DIVISION OF FRACTIONS. 
 
 Division of Fractions is naturally divided into the three 
 following kinds ; viz. the division of a fraction by a whole 
 number ; the division of a whole number by a fraction ; and 
 the division of a fraction by a fraction. 
 
 Division of fractions by a whole number was fully illustra- 
 ted in Sec, 5th of the remarks introductory to this rule. It is, 
 therefore, necessary here merely to repeat, that a fraction is 
 divided by a whole number, either by dividing its numerator, or 
 multiplying its denominator by that number. 
 
 Ex. 1. Divide -Jf by 9. Ans. . | 
 
 2. Divide f by 7. Ans. /j. 
 
 3. Divide L3 by 11. Ans.ffi. 
 
 4. Divide H b 7 3 - Ans. ^y. 
 
 5. Divide Si by 9. Ans. &, or &. 
 
 6. Divide jL- by 8. Ans. ^J^-. 
 
 7. Divide fi- by 6. Ans. %. 
 S. Divide y by 5. Ans. f 
 9. Divide ^ by 12. Ans. 
 
 10. Divide 2 by 7. Ans. - 
 
 11. Divide y> by 40. Ans. 
 
 12. Divide J- by 72. Ans. 
 
 It is obvious that the quotient arising from dividing a whole 
 number by a fraction, must be as much larger than the number 
 itself, as a unit or 1 is greater than the fraction ; or, in other 
 words, the given dividend must bear the same ratio to the 
 required quotient, as the numerator of the fraction bears to its 
 denominator. A unit or 1 is contained in 6, six times ; % is 
 contained in the same number, twelve times, and , eighteen 
 times ; and f , half as many times as ^, viz. nine times. The 
 operation to obtain this last quotient is as follows, 6x3=18, 
 the number of thirds in six ; and 18-^2=9. 
 
 For dividing a whole number by a fraction, we have then 
 the following rule : 
 
DIVISION OF FRACTIONS. 139 
 
 Multiply the whole number by the denominator of the fraction, 
 and divide the product by the numerator. 
 
 Ex. 1. Divide 9 by J. Operation, 9x4=36, and 36-?-3 
 = 12, the quotient. 
 
 2. Divide 15 by f. Operation, 15x6 = 90, and 90^-5 = 
 18, quotient. 
 
 Operations of a similar character may be performed by can- 
 celing. 
 
 RULE FOR CANCELING. Place the whole number above a hori- 
 zontal line and invert the fractional divisor; that is, place the 
 denominator above the line and the numerator below. Cancel, dfc. 
 
 7 
 
 3. Divide 21 by f. Statement, : -o Canceled, ', and 
 7 x 7=49, the quotient required. 
 
 4. Divide 42 by f . Statement, ^-^. Ans. 49. 
 
 The following sums may be divided by either of the above 
 rules. 
 
 5. Divide 18byf. Ans. 42. 
 
 6. Divide 63by T 9 3. Ans. 91. 
 
 7. Divide 63 by T 9 T . Ans. 77. 
 
 8. Divide 42 by f . Ans. 47. 
 
 9. Divide 66 by f . Ans. 99. 
 
 10. Divide 121 by |. Ans. 138f 
 
 11. Divide 101 by }f Ans. 110ft-. 
 
 12. Divide 32 by j, Ans. 40. 
 
 That the scholar may be enabled to commence understand- 
 ingly the division of fractions by fractions, he may turn back 
 and review the 7th section of the introductory remarks of this 
 rule. It is there said, that a fraction is divided by another 
 fraction by inverting the divisor and then multiplying them 
 together as in multiplication. 
 
 1 . Divide I by f . Ans. f or 1 
 
 2. Divide f by . Ans. V 6 or 
 
 RULE FOR CANCELING. Proceed in all respects as in multi- 
 plication of fractions, in arranging the terms of the dividend, 
 
140 DIVISION OF FRACTIONS. 
 
 then invert the divisor ; that is, place the numerators below 
 and the denominators above the line. Proceed to cancel, fyc. 
 
 Ex. 3. Divide 1| by A. 
 
 3 3 
 
 Statement, ^? Canceled, ^|. 
 10. o i-tx. 
 
 4 
 
 3x3 = 9, for 2, Ans. 
 
 Note. When the divisor is a compound fraction, each frac- 
 tion in the divisor must be inverted. 
 
 *> j j c 
 
 4. Divide f of by ^ of J. Statement, ' ' ' ' . Canceled, 
 
 ' ' ~^ ; therefore, | or 4 is the answer required. 
 *. a. J. a 
 
 5. Divide if of H by f off Statement, l*i_l*i.i. 
 
 or 
 
 6. Divide fof T 6 T of | by i of f of ^. An^. |, or 
 
 7. Divide ^ by i. ^in^. 2. 
 
 8. Divide ^ of f of |f of |f of ff of i by of $ of 
 
 of A- ^^fyf- 1 T^5- 
 
 9. Divide 4f by ^ of 6. Ans. l-^. 
 
 10. Divide 6f by f of 4. Ans. 2^. 
 
 11. Divide of f by -J of 6^. Ans. -f%. 
 
 12. Divide 6 by . Ans. 18. 
 
 13. Divide 3^ by 2|. Ans. l. 
 
 14. Divide | by 12. Ans. Jg-. 
 
 15. Divide of fbyfoff Ans. Iff. 
 
 16. Divide $ of 72 by $ of 56. Ans. 7f . 
 
 17. Divide |f by 6. Ans. 
 
 18. Divide f by 7. Ans. 
 
 19. Divide T 9 ^ by 3. Ans. T %. 
 
 20. Divide ^ by 8. Ans. -f^. 
 
 21. Divide 8 by i Ans. 16. 
 
 22. Divide 12 by |. Ans. 18. 
 
 23. Divide 41 by f . Ans. l^-. 
 
 24. Divide 35 by f. Ans. 49. 
 
 25. Bought 8 Ib. of coffee for |-^ of a dollar ; what was the 
 cost of one pound ? Ans. -f^- or ^ of a dollar. 
 
 26. Bought 9 pounds of sugar for } of a dollar ; what was 
 the price of a pound ? Ans. j^g- of a dollar. 
 
DIVISION OF FRACTIONS. 141 
 
 27. In 8 weeks a family consumes 84| pounds of butter ; 
 how much is that per week 1 Ans. 10^. 
 
 QUESTIONS. What is a fraction 1 If a unit be divided into two 
 equal parts, what is each of these parts called 1 How is it written 7 
 If it be divided into three equal parts, what is each part called, and how 
 is it written 1 Similar questions should be asked respecting other frac- 
 tions. When more parts than one are to be expressed, how is it done 1 
 What do fractions therefore express'? How are they represented! 
 What is the number below the line called 1 What does the denomina- 
 tor show 7 What is the number above the line called] What does 
 the numerator show 1 What are the two numbers called, when spoken 
 of collectively ? How many kinds of fractions are there 1 What are 
 they'? What is a proper fraction 7 Give an example. What is 
 an improper fraction 1 Give an example. What is a simple fraction 1 
 Give an example. What is a compound fraction 1 Give an example. 
 What is a mixed number 1 Give an example. What is a complex 
 fraction 1 Give an example. What does the denominator show"? 
 If the denominator remain the same, how is the value of the fraction 
 affected by increasing the numerator 1 How, by diminishing if? Give 
 an illustration. What is therefore the value of a fraction'? If the 
 numerator of a fraction be less than the denominator, how is its value 
 compared with a unit 1 If the numerator be equal to the denominator, 
 how then does its value compare with a unit 1 And how, if the nume- 
 rator be greater than the denominator 1 What is the only consideration 
 which limits the value of a fraction 1 In what ratio is the value of a 
 fraction increased 1 How then may fractions be multiplied ? How 
 is a fraction multiplied into a number equal to its denominator? In 
 what form should the terms of the fraction always be preserved ? What 
 is necessary to increase or diminish the value of a fraction 1 How then 
 may a fraction be multiplied by a whole number 1 In what two ways 
 may fractions be multiplied by whole numbers'? How may a frac- 
 tion be divided by a whole number 1 How else may a fraction be 
 divided by a whole number 7 In what two ways then may fractions be 
 divided by whole numbers 1 In what case should the numerator always 
 be divided 1 What operation is therefore performed on the value of a 
 fraction, when the numerator is operated upon ? And what, when the 
 denominator is operated upon 1 How is a fraction multiplied by a frac- 
 tion 1 How is a fraction divided by a fraction? If the numerator and 
 denominator be both multiplied by the same number, how is the value 
 of the fraction affected 1 What is the rule for reducing fractions to 
 their lowest terms 1 
 
 What is the rule for reducing a whole number or mixed quantity to 
 an improper fraction ? How is an improper fraction reduced to a whole 
 or mixed number? In all cases of division, what disposition may be 
 made of the remainder, when any occurs ? How are compound frac- 
 tions reduced to simple ones ? What is the rule for canceling ? What 
 is note 1st ? What is note 2d ? What is note 3d ? What is Case 5th ? 
 How arc fractions of low denominations reduced to those of higher 
 denominations ? What is the rule for canceling ? 
 
 How are fractions of high denominations reduced to those of a lower 
 value ? What is Case 6th ? What is the rule for it ? What is Case 
 7th 1 What is the rule for it ? What is the rule for canceling ? 
 
142 DECIMAL FRACTIONS. 
 
 What is Case 8th 1 What is the rule for if? What is the note 1 
 What must be done before fractions can be added 1 What else requi res 
 to be done! What is the rule for the addition of fractions 1 What 
 note follows! What is note 2d7 What preparations are necessary 
 before fractions can be subtracted 1 What is the rule for subtracting 
 fractions 1 How is a fraction multiplied by a whole number 1 How 
 is a fraction multiplied into a quantity equal to its denominator. 
 What is the rule for multiplying fractions by fractions 1 What is the 
 rule for canceling 1 Into what three kinds is division of fractions 
 naturally divided 1 How is a fraction divided by a whole number 1 
 What is the rule for dividing a whole number by a fraction 1 
 How are fractions divided by fractions'? What is the rule for 
 canceling "? What note follows the rule 1 
 
 DECIMAL FRACTIONS. 
 
 In the preceding rule we have contemplated the unit as divi- 
 ded into any number of equal parts. 
 
 We are now to regard it as divided frst into ten equal parts, 
 then each of these into ten other equal parts, or the whole unit into 
 one hundred equal parts ; and these parts again, each into ten 
 other parts ; or the whole into a thousand equal parts, &c. 
 The expressions obtained by these several divisions, therefore, 
 decrease in value in the constant ratio of ten, from the left to 
 the right, and are called decimals. Whole numbers, as was 
 shown in Numeration, increase in the same ratio from the right 
 to the left, and both commence their enumeration with the unit 
 figure. The connection between them is therefore so intimate 
 as to render them susceptible of being written together and 
 subjected to the same operations. The only important con- 
 sideration in writing them, in addition to what has already 
 been explained, is to distinguish the one from the other. This 
 is effected by the period, called in decimals, the point of sepa- 
 ration, which is always placed between them. In the ex- 
 pression, 23.56, the 23 is the whole number, and the .56 
 the decimal. 
 
 It will be observed that decimals, although they express 
 parts of units, do not, like vulgar fractions, require two terms 
 to express them. The given decimal may however be regarded, 
 as in truth it is, a numerator, with a denominator always under* 
 
DECIMAL FRACTIONS. 143 
 
 stood. What this denominator is, it shall be our next object 
 to illustrate. The scholar may therefore, in the first place, 
 carefully examine the following table of whole numbers and 
 decimals. The decimals, it will be observed, are read or 
 numerated from the left to the right. 
 
 a S 
 876543 2. 3 45678 
 
 By an examination of the preceding table, the scholar will 
 see that the 3 on the right of the separatrix is so many tenths. 
 In the preceding rule, this would be thus expressed, T 3 Q, which, 
 by section 8th of the introductory remarks of the same 
 rule, equals T 3 ^. He will also see, that the 4 on the 
 same side of the separatrix, is so many hundredths, or T ^. 
 Now it is obvious, that these two fractions united, would make 
 "liny The same process of reasoning will show, that the next 
 figure, or 5, is so many thousandths ; and since T^=I\%, if 
 the 5 be added, the amount will be f^ 4 ^. From the preceding, 
 we therefore learn, that if the decimal consist of one figure 
 only, it is so many tenths ; if it consist of two figures, it is so 
 many hundredths, and if of three, it is so many thousandths ; 
 and from this we derive the following conclusion, viz. that the 
 denominator of a decimal fraction always consists of a figure 1, 
 with as many cyphers annexed to it, as there are figures in the 
 given decimal. 
 
 The scholar may write a denominator to each of the follow- 
 ing decimals, viz. .6 ; .356 ; .26 ; .7426 ; .98654 ; .71639. 
 
 From the preceding explanation of the nature of decimals, 
 it is obvious, that cyphers added to the right of a decimal 
 do not effect its value, while those placed on the left dimin- 
 ish its value in a ten-fold ratio. For .5 is the same in 
 value as T 5 ^, or \. Now if a cypher be added to the right of 
 the .5, it becomes .50, which is of equal value with -j^, and 
 this also equals . (See Case 1st, Vulgar Fractions. ) If, 
 then, .5 and .50 are each equal to 2, it is obvious, that the 
 value of a decimal is not affected by cyphers placed on the 
 
144 DECIMAL FRACTIONS. 
 
 right of it. But if they be placed on the left, they diminish 
 the value of the decimal in a ten-fold ratio. The decimals .5, 
 .05, and .005, will serve as an illustration. From the explana- 
 tion given above, the denominators of these several decimals 
 are 10, 100, and 1000 ; or the decimals may be thus written : 
 To* nn> TOlV But ^ ve hundredths equal only one tenth part 
 of five tenths ; and five thousandths, one tenth part of five 
 hundredths. Hence, cyphers placed on the left of a decimal 
 diminish its value as above specified. 
 
 The following numbers may now be expressed by figures, 
 and then read. 
 
 1. Seventy-six and six tenths. Ans. 76,6. 
 
 2. One and three hundredths. Ans. 1.03. 
 
 3. Eighty and fifty-eight thousandths. 80.058. 
 
 4. One hundred and fifty-six, and thirty-nine thousandths. 
 Ans. 156.039. 
 
 5. One hundred and one, and five thousandths. Ans. 
 101.005. 
 
 The scholar will observe, that if there is but one decimal 
 figure, and that tenths, it requires the point only to be placed 
 at the left of it, to express its true value ; if it be hundredths, 
 it requires a cypher to be placed at the left of it, and if it be 
 thousandths, it requires two cyphers to be thus placed, with 
 the decimal point on the left of the cyphers ; and so on accord- 
 ing to the denomination. 
 
 6. Write down nine, and three hundred thousandths. Ans. 
 9.00003. 
 
 7. Write down twelve and one millionth. Ans. 12.000001. 
 
 8. Three hundred and seventy-five, and seven tens of thou- 
 sandths. Ans. 375.0007. 
 
 9. Ninety-five hundredths. 
 
 10. Three hundred and sixteen thousandths. 
 
 11. Forty-five millionths. 
 
 12. Sixty-nine, and nine hundred and three thousandths. 
 
 13. Four hundred and fifty-six, and seventeen millionths. 
 
 14. Five hundred, and three tens of millionths. 
 
 15. One, and six hundred of millionths. 
 
 16. Eleven, and seven billionths. 
 
 17. Seven hundred and sixty -two billionths. 
 
 1 8. Four hundred and twenty-one, and nineteen thousandths . 
 
DECIMAL FRACTIONS. 145 
 
 19. Seven hundred and six, and one hundred and three 
 millionths. 
 
 20. Twelve hundred and six trillionths. 
 
 The scholar is now requested to turn back to Federal Money, 
 and compare the denominations there given, with those here 
 brought to view. He will there find the dollar given as the 
 unit money ; the dime as the tenth part of the dollar ; the 
 cent as the tenth part of the dime, or the hundredth part of the 
 dollar ; and the mill as the tenth part of the cent, the hundredth 
 part of the dime, and the thousandth part of the dollar. It is 
 therefore obvious, that Federal Money and decimals are ope- 
 rated upon by the same general principles. 
 
 1. Reduce $21, 8 dimes, and 6 mills, to mills. Ans. 21806. 
 
 2. Reduce 21806 mills to dollars, cents, and mills. Ans. 
 $21.806. 
 
 3. Reduce $12, 3 dimes, 4 cents, and 9 mills, to mills. Ans. 
 12349. 
 
 4. Reduce 12349 mills to dollars, cents, and mills. Ans. 
 $12.349. 
 
 5. Reduce $25 to cents. Ans. 2500. 
 
 6. Reduce $9 to mills. Ans. 9000. 
 
 To reduce dollars to cents, we therefore add two cyphers, 
 and to reduce them to mills, we add three. 
 
 7. Reduce 2567 cents to dollars. Ans. $25.67, or $25 
 and 67 cents. 
 
 8. Reduce 38679 mills to dollars, &c. Ans. $38.679, or 
 $38, 67 cents, and 9 mills. 
 
 To reduce cents to dollars, we therefore cut off two figures, 
 and to reduce mills to the same, we cut off three figures from 
 the right of the given number. 
 
 9. Reduce $2 to mills. 
 
 10. Reduce 99 cents to mills. 
 
 11. Reduce $1.03 to mills. 
 
 12. Reduce 467 cents to dollars. 
 
 13. Reduce 12008 mills to dollars. 
 
 14. Reduce $42 and 3 mills to mills. 
 
 15. Reduce 9000 mills to dollars. 
 
 13 
 
146 ADDITION OF DECIMAL FRACTIONS. 
 
 ADDITION OF DECIMAL FRACTIONS. 
 
 The scholar must here exercise a good degree of caution 
 in writing the numbers to be added. He will recollect that, 
 in adding vulgar fractions, it was necessary to reduce them all 
 to the same name or denominator, before the numerators could 
 be added ; and that in Simple and Compound Addition, the 
 same denominations only could be united. The same is true 
 of Decimal Fractions. Hence, we have the following rule : 
 
 RULE. Place the whole numbers as in Simple Addition, 
 with units under units, and tens under tens, c. A Iso place 
 the decimals on the right of the whole numbers, with tenths under 
 tenths, hundredths under hundredths, and thousandths under 
 thousandths, fyc. ; then, beginning with the lowest denomination, 
 add up and carry as in Simple Addition. Lastly, from the 
 amount, point of as many decimals as are equal to the greatest 
 number of decimals in any one of the given numbers. 
 
 Ex.1. What is the amount of 3.56; 42.923; 125.6; 4.32; 
 and 59.365. 
 
 ADDED. 
 
 3.56 
 42.92 3 
 12 5.6 
 
 4.32 
 59.365 
 
 2 3 5.7 6 8 
 
 The greatest number of decimals in either of the numbers 
 given, is three ; therefore, three decimals are to be cut off from 
 the sum. It will always be found correct to place the decimal 
 point in the amount directly below those of the given num- 
 bers. 
 
 2. Add the following numbers, 325.63; 275.215; 1.02; 
 17.653 ; 136.1. Amount, 755.618. 
 
 3. What is the amount of 72.5; 32.071; 2.1574; 371.4; 
 2.75 ? Ans. 480.8784. 
 
 , 
 
SUBTRACTION OF DECIMALS. 147 
 
 4. What is the amount of 225.75 ; 25.50 ; 8.255 ; 27.225 ? 
 Ans. 286.73. 
 
 5. What is the amount of 35.175; 75.15; 13.31; 25.755? 
 Ans. 149.39. 
 
 6. What is the amount of 304.39; 291.09; 136.99; 12.10? 
 Ans. 744.57. 
 
 7. What is the amount of 365.541 ; 487.06 ; 94.67 ; 472.5 ; 
 439.089? Ans. 1858.860. 
 
 8. What is the amount of 2.151 ; 375.422; .675; .4567? 
 Ans. 378.7047. 
 
 9. Add together the following decimals, viz. sixteen hun- 
 dredths ; two hundred and thirty-five thousandths ; six tenths ; 
 and one thousandth. Ans. .996. 
 
 10. What is the sum of one tenth; two hundredths ; three 
 thousandths ; four tens of thousandths ; five hundreds of 
 thousandths, and six millionths ? Ans. .123456. 
 
 11. W T hat is the amount of $72.375; $41.176; $1.47; 
 $395.20 ; $56.65 ; $146.73, and $0.16. Ans. $713.761. 
 
 12. Bought a yoke of oxen for $121.56 ; a horse for $156.- 
 375 ; a cow for $37.086 ; and a quantity of grain for $95.739. 
 What was the cost of the whole ? Ans. $410.760. 
 
 SUBTRACTION OF DECIMALS. 
 
 The scholar will need no explanation of the nature of this 
 rule. His knowledge of Subtraction, and of the peculiarity 
 of decimals, will enable him at once to make a correct appli- 
 cation of the following rule : 
 
 RULE. Set down the less number under the greater, so that 
 each figure in the lower number, or subtrahend, shall stand di- 
 rectly under one of its own name or denomination. Proceed 
 to subtract as in simple numbers, and place the separatrix as in 
 addition of decimals. 
 
 Ex. 1. From 378.635 take 195.275. 
 
148 MULTIPLICATION OF DECIMALS. 
 
 OPERATION. 
 
 378.635 
 1 95.2 75 
 
 1 83.360 remainder. 
 
 2. From 462.3 take 218.15. Rem. 244.15. 
 
 3. From 16.705 take 7.6845. Rem. 9.0205. 
 
 4. From 132.4 take 36.36. Rem. 96.04. 
 
 5. From 127.05 take 66.006. Rem. 61.044. 
 
 6. From 100.001 take 77.77. Rem. 22.231. 
 
 7. From five hundred and thirty-six, and fifteen hundredths, 
 take two hundred and thirty-six, and eighteen hundredths. 
 Rem. 299.97. 
 
 8. From six dollars take fifty-five cents. Rem. $5.45. 
 
 9. From one dollar take one mill. Rem. .999 mills. 
 
 10. From 16 dollars take nineteen cents and one mill. Rem. 
 $15.809. 
 
 MULTIPLICATION OF DECIMALS. 
 
 RULE. Multiply as in whole numbers, and point off as 
 many decimals in the product as there are decimals in both fac- 
 tors. Whenever the decimals in the product are not as many 
 as those of the factors, the deficiency must be supplied by pla- 
 cing cyphers on the left of them. 
 
 Ex. 1. Multiply 25.16 by 3.45. 
 
 25.16 
 3.45 
 
 12580 
 10064 
 
 7548 
 
 8 6.8 2 
 Four figures are cut off in the product as decimals, in ac- 
 
 
DIVISION OF DECIMALS. 149 
 
 V 
 
 cordance with the rule, there being four decimals in the two 
 factors. 
 
 2. Multiply 175.2 by 45.72. Ans. 8010.144. 
 
 3. Multiply 15.75 by 1.05. Ans. 16.5375. 
 
 4. Multiply 37.99 by 25.77. Ans. 979.0023. 
 
 5. Multiply 100.00 by 0.01. Ans. 1.0000. 
 
 6. Multiply 3.45 by .16. Ans. .5520. 
 
 7. Multiply 25.238 by 12.17. Ans. 307.14646. 
 
 8. Multiply 27.56 by 12.22. Ans. 336.7832. 
 
 9. Multiply .01 by .01. Ans. .0001. 
 
 10. Multiply 7.02 by 5.27. Ans. 36.9954. 
 
 11. Multiply .001 by .001. Ans. .000001. 
 
 12. Multiply].25 cents by .25 cents. Ans. .0625, or 6 cts. 
 
 Note. To multiply a decimal by 10, 100, 1000, &c., it is 
 necessary only to remove the decimal point as many places to 
 the right as there are cyphers in the multiplier. 
 
 13. Multiply 1.56 by 10. Ans. 15.6. 
 
 14. Multiply 36.541 by 100. Ans. 3654.1. 
 
 15. Multiply .42 by 100. Ans. 42. 
 
 16. Multiply 46.3789 by 1000. Ans. 46378.9 . 
 
 DIVISION OF DECIMALS. 
 
 We are now to reverse, the preceding operation. In mul- 
 tiplying decimals, we were directed to point off as many deci- 
 mal figures in the product as there were in both factors. In 
 Division, the dividend corresponds to the product in Multipli- 
 cation, and the divisor to one of the factors which produced 
 that dividend, and we are required to obtain the other factor. 
 Therefore the decimals of the quotient and divisor united must 
 equal those of the dividend. 
 
 RULE. Divide as in Simple Numbers, and point off from 
 the right of the quotient as many decimals as are equal to 
 13* 
 
150 DIVISION OF DECIMALS, 
 
 the excess of decimals in the dividend, over those in the di- 
 visor. 
 
 Note 1st. If the decimal places in the divisor be more 
 than those in the dividend, annex cyphers to make them 
 equal. 
 
 2d. If, after dividing, there be a remainder, cyphers may 
 be annexed to the remainder and the division continued. The 
 cyphers thus added are decimals. 
 
 3d. If the decimals in the divisor and dividend are equal, 
 and there is no remainder after dividing, the quotient will be a 
 whole number. 
 
 4th. If the figures in the quotient do not equal the excess 
 of decimal places in the dividend over those of the divisor, 
 supply the defect by prefixing cyphers. 
 
 5th. To divide the decimal number by 10, 100, 1000, &c., 
 it is necessary only to remove the point as many figures to the 
 left, as there are cyphers in the divisor. 
 
 Ex. 1. Divide 34.317 by 21.75. 
 
 PERFORMED. 
 
 2 1.7 5) 3 4.3 17(1.577 + 
 21.75 
 
 12567 
 10875 
 
 16920 
 15225 
 
 ] 6 950 
 15225 
 
 1725 remainder. 
 
 In the solution of this example, two cyphers have been ad- 
 ded to the remainders of the dividend. By Note 2d, the 
 whole number of decimals in the dividend is therefore five, 
 and there are two in the divisor ; three should, therefore, be 
 
DIVISION OF DECIMALS. 151 
 
 cut off from the quotient. The plus sign in the quotient al- 
 ways implies a remainder. 
 
 2. Divide 30515.50 by 100. Ans. 305.1550. 
 
 For the solution of the preceding sum, see Note 5th. 
 
 3. Divide 483.125 by 386.5. Ans. 1.25. 
 
 4. Divide 198.15625 by 186.5. Ans. 1.0625. 
 
 5. Divide .56 by 1.12. Ans. .5. 
 
 6. Divide 99.99 by 33.3. Ans. 3.0027. -f 
 
 7. Divide 1.00 by .12. Ans. 8.333.+ 
 
 8. Divide 14325.16 by 1.33. Ans. 10770.721. -4- 
 
 9. Divide 36.5 by 10. Ans. 3.65. 
 
 10. Divide 36.5 by 100. Ans. .365. 
 
 11. Divide 981 by 1000. Ans. .981. 
 
 12. Divide 543.67 by 3.46. Ans. 157.13.+ 
 
 APPLICATION. 
 
 1. If 36.34 bushels of corn grow on an acre, how many 
 acres will produce 674 bushels 1 Ans. 17.804 acres. 
 
 2. If 6 yards of\cloth cost $24.48, what was the price per 
 % yard? Ans. $4.08. 
 
 3. Bought 56.87 yards of cloth at $2.31 per yard ; what was 
 the whole cost ? Ans. $131.3697. 
 
 4. The first of three men possessed $685.423 ; the second, 
 $746.03 ; and the third, $10864.273. How much had they 
 all? Ans. 12295.726. 
 
 5. What cost 9.6 yards of cloth at $6.42 per yard ? Ans. 
 $61.632. 
 
 6. If a man earn 2 dollars 2 mills per day, how much 
 would he earn in 93.5 days ? Ans. $187.1870. 
 
 7. What cost .675 of a cord of wood at $3 a cord ? Ans. 
 2.025. 
 
 8. If a yard of cloth cost $5.5625, how much will .25 of a 
 yard cost? Ans. 1.3906. 
 
152 REDUCTION OF FRACTIONS. 
 
 REDUCTION OF VULGAR AND 
 DECIMAL FRACTIONS. 
 
 The value of a vulgar fraction is the quotient arising from 
 dividing the numerator by the denominator. Therefore, 
 
 CASE 1st. To REDUCE A VULGAR FRACTION TO A DECIMAL. 
 
 RULE. Annex cyphers to the numerator, and divide it by 
 the denominator. 
 
 Note. If the given fraction be proper, the quotient will 
 always be a decimal, and will consist of figures equal in num- 
 ber to the cyphers annexed ; or, if the number of figures be less, 
 cyphers must be prefixed to complete the number. 
 
 Ex. 1 . Reduce to a decimal. 
 
 A.T 
 
 (,0 
 . 7 5 the decimal required. 
 
 OPERATION. 
 4)3.00 
 
 2. Reduce J to a decimal. 
 
 OPERATION. 
 5 )1.0 
 
 . 2 the decimal required. 
 
 3. Reduce | to a decimal. Ans. .333.+ 
 
 4. Reduce |- to a decimal. Ans. .125. 
 
 5. Reduce if to a decimal. Ans. .9375. 
 
 6. Reduce ^ to a decimal. ^4ns. .1. 
 
 7. Reduce rf to a decimal. Ans. .923.+ 
 
 8. Reduce -f^ to a decimal. Ans. .9. 
 
 9. Reduce f to a decimal. Ans. .666.+ 
 
 'CASE 2d. To REDUCE A DECIMAL TO A VULGAR FRACTION. 
 
 RULE. Write down the given decimal as a numerator, and 
 for a denominator, write 1 with as many cyphers annexed as 
 there are figures in the numerator, and then reduce the fraction 
 to its lowest terms. (See remarks introductory to Decimals.) 
 
REDUCTION OF FRACTIONS. 153 
 
 1. Reduce the decimal .125 to a vulgar fraction. Per- 
 formed, xo 2 ^' ( see Case lst ' Vulgar Fractions,) 
 
 and again, ^-^25=4, Ans. 
 
 2. Reduce .75 to a vulgar fraction. Performed, 
 = f, Ans. 
 
 3. Reduce .9385 to a vulgar fraction. Ans. 
 
 4. Reduce .2 to a vulgar fraction. 
 
 5. Reduce .16 to a vulgar fraction. 
 
 6. Reduce .25 to a vulgar fraction. 
 
 7. Reduce .45 to a vulgar fraction. 
 
 8. Reduce .55 to a vulgar fraction. 
 
 9. Reduce .8 to a vulgar fraction. 
 
 10. Reduce .24 to a vulgar fraction. 
 
 11. Reduce .945 to a vulgar fraction. 
 
 12. Reduce .844 to a vulgar fraction. 
 
 CASE 3d. To REDUCE LOWER DENOMINATIONS TO DECIMALS 
 
 OF A HIGHER. 
 
 RULE 1st. Write down the several denominations which are 
 to be reduced to decimals of a higher denomination, one above 
 another, with the lowest uppermost ; then divide each denomina- 
 tion^ commencing with the lowest, by that number which is re- 
 quired of each to make a unit of the next higher denomination, 
 and at each division place the quotient as a decimal on the right 
 of the next higher denomination. The number last obtained 
 will be the required decimal. 
 
 Note. It will be obvious that the division of the lowest de- 
 nomination must be effected by adding cyphers to that denomi- 
 nation. Cyphers must also be added to each higher denomi- 
 nation to reduce them, unless the decimal figures previously 
 obtained be sufficient. 
 
 The reason of the above rule is readily shown. Suppose it 
 is required to reduce 7 pence to the fraction of a shilling. The 
 fraction would be T 7 ^, because the shilling is divided into 12 
 equal parts, and 7 of these parts are taken, and this vulgar frac- 
 tion is reduced to a decimal by adding cyphers to its numera- 
 tor and dividing by its denominator. (See Case 1st of Deci- 
 mals. ) 
 
154 REDUCTION OF FRACTIONS. 
 
 Ex. 1. Reduce 7 s. 6 d. to the decimal of a pound sterling. 
 The statement would be, ^ ^. That is, the 6 d. is to be divi- 
 ded by 12, and the 7 s. by 20. This must be effected by 
 adding cyphers. 
 
 OPERATION. 
 
 12 
 20 
 
 6.0 
 7.5 
 
 .3 7 5= required decimal. 
 2. Reduce 9 d. 2 qr. to the decimal of a pound sterling. 
 
 12 
 20 
 
 OPERATION. 
 2.0 
 
 9.5 
 
 .791666 
 
 Ans. .0395833 + 
 
 RULE 2d. Reduce the given quantity to its lowest denomi- 
 nation, and divide it by a unit of the denomination of the re- 
 quired fraction, reduced to the same denomination. 
 
 Ex. 3. Reduce 10s. 9d. 2qr. to the decimal of a pound 
 sterling. 10s. 9 d. 2qr. 518 qr. ; and by the rule, 518 qr. 
 are to be divided by 1 . reduced to qr. viz. [by 960 qr. 
 Therefore, |i| is the fractional answer, and 518-^960= 
 .539583 + , Ans. 
 
 To understand the above operation the scholar should re- 
 member that 10s. 9 d. 2 qr. or 518 qr. are to be divided into 
 as many equal parts as there are farthings in 1 . = 960 qr., and 
 one of these parts = -J, or the decimal .539583.+ 
 
 4. Reduce 9 s. 8 d. to the decimal of a pound sterling. Ans. 
 .4833.+ 
 
 5. Reduce 3qr. 16 Ib. to the decimal of a cwt. ? Ans. 
 .8928571.+ 
 
 6. Reduce 16 . 12s. 8 d. to a decimal expression. Ans. 
 16.633333.+ 
 
 7. Reduce 3qr. 2na. to the decimal of a yard. Ans. .875. 
 
 8. Reduce 2 roods and 20 rods to the decimal of an acre. 
 Ans. .625. 
 
REDUCTION OF FRACTIONS. 155 
 
 9. Reduce 3 furlongs 16 rods to the decimal of a mile. 
 Ans. .425. 
 
 10. Reduce 12 hours, 15 minutes, and 30 seconds to the deci- 
 mal of a day. Ans. .51076.+ 
 
 1 1 . Reduce 2 cwt. 3 qr. 24 Ib. to the decimal of a ton. Ans. 
 .14821428. 
 
 CASE 4th. To FIND THE VALUE OF A DECIMAL IN INTEGERS 
 
 OF LOWER DENOMINATIONS. 
 
 RULE. Multiply the decimal by the number required to re- 
 duce it to the next lower denomination, and from the right 
 hand of the product cut off as many figures as there are in the 
 given decimal. The figures on the left of the point will be 
 integers of the denomination next below that given. Proceed 
 in the same way through all the denominations, and the figures 
 on the left of the several points will be the answer required. 
 
 This rule being directly the reverse of the preceding, needs 
 no explanation. 
 
 Ex. 1. What is the value of the decimal .5638 of a pound 
 sterling ? 
 
 OPERATION. 
 
 .5638 
 
 2 
 
 1 1.276 0=the shillings and decimals of a shilling 
 12 in .5638 of a pound sterling. 
 
 3.312 O^zthe pence and decimals of a penny in 
 4 .2760 of a shilling. 
 
 1.2 4 8 0=the farthings and decimals of a far- 
 thing in .3120 of a penny. 
 
 The value of the above decimal in shillings, pence, &c. is 
 11s. 3d. 1.2480qr. 
 
 Note. The integers on the left of the points are the num- 
 bers which compose the answer. 
 
 2. What is the value of .75 of a pound sterling? Ans. 15s. 
 
 3. What is the value of .53854 of a pound sterling ? Ans. 
 10s. 9 d. 1 qr. nearly. 
 
156 REDUCTION OF FRACTIONS. 
 
 4. What is the value of .625 of an acre 1 Ans. 2 roods, 20 
 rods. 
 
 5. What is the value of .148712678 of a ton ? Ans. 2 cwt. 
 3qr. 25 Ib. 1 oz.+ 
 
 6. What is the value of .676 of a cwt. ? Ans. 2 qr. 19 Ib. 
 11 oz.+ 
 
 7. How many furlongs, &c. in .425 of a mile ? Ans. 3 fur. 
 16 rods. 
 
 8. How many quarters, &c. in .66 of a yard 1 Ans. 2 qr. 
 2 nails, 1.26 inches. 
 
 9. How many roods, &c. in .321 of an acre 1 Ans. I rood, 
 11 rods, 10yd. 8 feet. 
 
 10. What is the value of .875 of a hogshead of wine ? Ans. 
 55 gal. 1 pint. 
 
 11. What is the value of .875 of a yard ? Ans. 3 qr. 2 nails. 
 
 12. What is the value of .9 of an acre 1 Ans. 3 roods, 24 
 rods. 
 
 QUESTIONS. How have we contemplated the unit as divided in the 
 preceding rule 1 How are we to regard it as divided in this rule ? In 
 what ratio do the several divisions in decimals decrease 1 What are 
 they called 1 What relation is there between integers and decimals ? 
 What is the only important consideration which has not already been 
 explained 1 How many terms are required to express decimals 7 
 How may that term be regarded 1 Repeat the table of whole numbers 
 and decimals. Of what does the denominator of decimal fractions 
 always consist 1 How do cyphers added to the right of a decimal, 
 affect its value 7 Give the illustration. How do cyphers placed on 
 the left of a decimal affect its value 1 Give the illustration. How do 
 decimals and the denominations of federal money correspond 1 How 
 do we reduce dollars to cents'? How do we reduce them to mills ? 
 How do we reduce cents and mills to dollars 1 What is the rule for 
 the addition of decimals 1 What is it for subtracting decimals ? What 
 is the rule for multiplying decimals ? How does division stand related 
 to multiplication 1 What is the rule for division of decimals 1 What 
 is the rule for pointing off decimals in addition of decimals'? What 
 in subtraction 1 What in multiplication? And what in division? 
 On what does the value of a fraction depend 1 What is the rule for 
 reducing vulgar fractions to decimals 1 What is Case 2d ? What is 
 the rule for if? What is Case 3d? What is the rule 1 ? What note 
 follows the rule 1 Explain the nature of the rule. What is the 
 second rule 1 What is Case 4th 1 What is the rule for it ? What 
 note follows the rule ? 
 
 
REDUCTION OF CURRENCIES. 157 
 
 REDUCTION OF CURRENCIES. 
 
 The currency of the United States was originally pounds* 
 shillings, and pence, the same as in England. This, however, 
 was abolished by an act of Congress, in 1786, and Federal 
 Money, consisting of dollars, cents, and mills, was adopted. 
 
 The following table gives the value of the dollar in the old 
 currency of several states : 
 
 In the New England, Virginia, Kentucky, and Tennessee 
 currency, $l=6s. and 1 . = 20s. ; therefore, 1 -. = 2 or L 
 of $1 ; or $l=r T ^of l. 
 
 In New York and North Carolina currency, $1 8s. and 
 U. = 20s. ; therefore, l.rr 2 F =f of $1, or $l=f of 1 . 
 
 In New Jersey, Pennsylvania, Delaware, and Maryland 
 currency, $1 =7 s. 6 d.= 90 d. I .=20 s. =240 d.-, therefore, 
 1 . = %< =f of $ L or, $l = f of 1 JB. 
 
 In South Carolina and Georgia currency, $l=4s. 8d.= 
 56 d. and l.=20 s.=240 d.; therefore, !. = 2 -j> = 3 T of $1, 
 or $l=^j- of 1 . 
 
 In Canada currency, $1 = 5 s. ; therefore, 1 .=^ of $1, or 
 
 That the scholar may understand why the dollar is com- 
 posed of a different number of shillings and pence in the diffe- 
 rent states, it is necessary only to say, that these states (or 
 colonies, as they were at first called) originally issued each 
 their own money, in pounds, shillings, and pence, the value of 
 which soon depreciated. This depreciation was greater in 
 some states than in others ; and hence, when federal money 
 was adopted, more of the old currency was required in some 
 states than in others, to equal the dollar of the new curren- 
 cy. The value of this dollar was 4s. 6 d. sterling money, 
 or English currency ; while 6 s. New England currency, 8 s. 
 New York, 7s. 6 d. New Jersey, and 4s. 8 d. Georgia cur- 
 rency, were required to equal the same value. 
 
 To understand changing these several currencies to dollars, 
 cents, and mills, the scholar needs to examine carefully the 
 14 
 
158 REDUCTION OF CURRENCIES. 
 
 preceding table. He will there observe that in New Eng- 
 land currency 1 . = ^ of $1. Therefore, if pounds of that 
 currency be multiplied by 10, and the product divided by 3, 
 they will be changed to federal money, or dollars, cents, and 
 mills. A similar explanation is applicable to other currencies. 
 
 Note. If the given money consist of pounds, shillings, and 
 pence, reduce the shillings and pence to the decimal of a 
 pound, (see Case 3d of the Reduction of Vulgar and Decimal 
 Fractions,) and then proceed according to the following rule : 
 
 RULE. Notice from the preceding table what fraction of one 
 dollar makes 1 . of the given currency, and multiply the given 
 sum by that fraction ; that is, multiply it by the numerator and 
 divide the product by the denominator. 
 
 Ex. 1. Reduce 72 . New England currency to federal 
 money. In N. E. currency, the pound =^ of one dollar; 
 therefore, 
 
 7 2 
 1 
 
 3)7 2 
 $ 2 4 Ans. 
 2. Reduce 75 . 6 s. 8d. N. E. currency to federal money. 
 
 12 
 20 
 
 8.0 
 6.6666-f 
 
 .333 3=the decimal value of 6 s. 
 8 d. ; therefore, 
 
 7 5.3 3 3 3 
 
 1 
 
 3 )7 5 3,3 3 3 
 $251.111 Ans. 
 
 3. Reduce 15 . 6 s. 6 d. New York currency to federal 
 money. 
 
REDUCTION OF CURRENCIES. 159 
 
 12 I 6. 
 
 20 6.5 
 
 .3 2 5= the decimal value of 
 6s. 6d. ; therefore, 
 
 1 5.3 2 5 
 5 
 
 2 ) 7 6.6 2 5 
 
 $ 3 8.3 1 2 5 Ans. 
 
 To abbreviate the given operation by canceling, the follow-' 
 ing rule may be adopted : 
 
 RULE FOR CANCELING. Place the given sum above a hori- 
 zontal line, and, noticing as before what fraction of a dollar 
 makes 1 >. of the given currency, write its numerator above the 
 line, and its denominator below; then cancel, multiply, and 
 divide. 
 
 4. Reduce 48 . 10 s. 6 d. New England currency to fede- 
 ral money. 
 
 12 I 6 
 20 1 0.5 
 
 .5 2 5 = decimal value of 10s. 6d. 
 16.175 
 
 Therefore, l, and 16.175 xlO=$161.75. 
 
 dt 
 
 5. Reduce 240 6. New Jersey currency to federal money. 
 Statement, 8 
 
 80 
 
 Canceled, ' , and 80x8=$640,4*w. 
 
 a. 
 
 6. Reduce 243 . New Jersey currency to federal money. 
 Ans. $648. 
 
 7. Reduce 150 . South Carolina currency to federal 
 money. Ans. $642.857.+ 
 
 8. Reduce 27 . New England currency to federal money. 
 Ans. $90. 
 
160 REDUCTION OF CURRENCIES. 
 
 9. Reduce 304 . 12 s. 9 d. New England currency to fede- 
 ral money. Ans. $1015.458. 
 
 10. Reduce 80 . New York currency to federal money. 
 Ans. $200. 
 
 11. Reduce 84 . 10 s. 6 d. New Jersey currency to federal 
 money. Ans. $225.40. 
 
 12. Reduce 100 . New Jersey currency to federal money. 
 Ans. $266.666.+ 
 
 13. Reduce 16 . Canada currency to federal money. Ans. 
 $64. 
 
 14. Reduce 96 . Nova Scotia currency to federal money. 
 Ans. $384. 
 
 15. Reduce 112. Georgia currency to federal money. 
 Ans. 480. 
 
 16. Reduce 365 . 10 s. 6 d. New York currency to federal 
 money. Ans. $913.812.+ 
 
 17. Reduce 29 . 12 s. 8 d. Canada currency to federal money. 
 Ans. $118.533.+ 
 
 18. Reduce 276 . 10s. 3d. South Carolina currency to 
 federal money. Ans. $1185.053.+ 
 
 19. Reduce 45 . 12s. New Jersey currency to federal 
 money. Ans. $121.60. 
 
 20. Reduce 42 . 6 s. New England currency to federal 
 money. Ans. $141. 
 
 The scholar will now reverse the preceding, that is, he will 
 change any sum of federal money to either of the preceding 
 currencies. To do this he must adopt the opposite mode of 
 operation. Therefore, 
 
 RULE. Notice from the preceding table what fraction of \ . 
 of the required currency makes $ 1 ; then multiply the given sum 
 by the numerator of that fraction, and divide the product by the 
 denominator. 
 
 Ex. 1. Reduce $161.75 to pounds, shillings, and pence, 
 New England currency. 
 
 PERFORMED, 
 
 161.75 
 3 
 
 485.2 5-^-10=48.525 . 
 
REDUCTION OF CURRENCIES. 161 
 
 Now to find the value of the decimal .525, (see Case 4th, 
 Decimal Fractions,) .525x20^10.500 s. and .500x12 = 
 6.000 d. ; therefore the required pounds, shillings, &c. are 
 48 . 10s. 6d. 
 
 2. Reduce $38.3125 to pounds, shillings, and pence, New 
 York currency. 
 
 38.3125 then, .325 and 5 
 
 2 20 12 
 
 5)76.6250 6.500 shillings, 6.0 d. 
 
 15.325 Therefore, 15 . 6 s. 6 d. is the answer. 
 
 These sums may also be solved by canceling. Take the 
 following rule : 
 
 RULE FOR CANCELING. Place the given dollars, cents, and 
 mills, above a horizontal line, and, noticing as before ivhat frac- 
 tion of 1 . of the required currency makes $1, place the nume- 
 rator of the fraction above the line, and the denominator below. 
 Cancel, multiply, and divide for the answer. 
 
 3, Reduce $95.75 to pounds, shillings, &c. in New Eng- 
 land currency. Ans. 28 . 14s. 6 d. 
 
 95.75. 3 
 Statement, - -- . 
 
 4. Reduce $648 to pounds, New Jersey currency. Ans. 
 243 . 
 
 fi4ft q 
 
 Statement, - . 
 
 5. Reduce $642.857 to pounds, South Carolina currency. 
 Ans. 150. 
 
 6. Reduce $578 to pounds, &c. New York currency. Ans. 
 231 jB. 4s. 
 
 7. Reduce $580 to pounds, &c. Canada currency. Ans 
 145 . 
 
 8. Reduce $742.50 to pounds, &c. New England currency 
 Ans. 222 . 15s. 
 
 9. Reduce $21.758 to pounds, &c. New England currency 
 Ans. 6. 10s. 6d.2qr.+ 
 
 14* 
 
162 SIMPLE INTEREST. 
 
 10. Reduce $141 to pounds, &c. New England currency. 
 Ans. 42 . 6 s. 
 
 11. Reduce $250 to pounds, &c. Canada currency. Ans. 
 62 . 10s. 
 
 12. Reduce $121.60 to pounds, &c. New Jersey currency. 
 Ans. 45 . 12s. 
 
 13. Reduce $475.75 to pounds, &c. New York currency. 
 Ans. 190 . 6s. 
 
 14. Reduce $89.54 to South Carolina currency. Ans. 20 . 
 17s. 10 d. 
 
 15. Reduce $75 to pounds, &c. New England currency. 
 Ans. 22 . 10s. 
 
 16. Reduce $384 to pounds, &c. Nova Scotia currency. 
 Ans. 96 JB. 
 
 . What was the original currency of the United States'? 
 When this was abolished, what currency was substituted 1 What are 
 the denominations of federal money 1 What is the value of the dol- 
 lar, New England currency 1 What fraction of a dollar does the 
 pound of that currency equal 1 What fraction of a pound does the 
 dollar equal 1 Similar questions should be asked relative to the cur- 
 rencies of the other states. The scholar may explain why it is that 
 the dollar is composed of a different number of shillings and pence in 
 the different states. What note precedes the rule 1 What is the rule 
 for bringing pounds, &c. into dollars'? How are the terms arranged 
 for canceling'? What is the rule for bringing dollars into pounds, 
 &c. 1 What is the rule for canceling 1 
 
 SIMPLE INTEREST. 
 
 Interest is an allowance made for the use of money. 
 
 It is computed at a certain rate per cent. ; that is, at a cer- 
 tain number of dollars for the use of a hundred for one year. 
 
 If the sum on interest be more or less than $100, or the 
 time during which it draws interest, more or less than one 
 year, the sums paid as interest must be in proportion both to 
 the time and the sum lent. 
 
 For illustration: if $100 be borrowed for one year, and if 
 the interest allowed be 6 per cent., the interest for one year 
 
SIMPLE INTEREST. 163 
 
 will be $6. Now, if twice that sum, viz. $200, be borrowed 
 for the same time, or if the same sum be borrowed for twice 
 that time, viz. for two years, the interest will in either case 
 be twice that sum, viz. $12. .Again, if twice the sum be 
 borrowed for twice the time, that is, if $200 be borrowed for 
 two years, the sum to be paid as interest will be increased 
 four-fold ; or it would be 6 x 4 =$24. 
 
 The scholar will carefully notice the following particulars : 
 
 1st. The principal is the money lent, or the sum on which 
 interest is paid. 
 
 2d. The interest is the money paid for the use of the 
 principal. Legal interest is that established by law. In the 
 New England states, it is fixed at 6 per cent. ; in New York, 
 at 7, and in Louisiana, at 8 per cent. 
 
 3d. The amount is the sum obtained by adding the interest 
 to the principal. 
 
 4th. The rate per cent, is always a decimal of two places, 
 when expressed by cents, and of three, when expressed by 
 cents and mills. 
 
 CASE 1st. To CAST INTEREST ON ANY SUM FOR ONE OR MORE 
 YEARS. 
 
 RULE. Multiply the principal by the rate per cent, written 
 as a decimal, the product will be the interest for one year ; 
 which, being repeated as many times as there are years given, 
 will be the required interest. 
 
 Note \st. In all cases where no rate per cent, is mentioned, 
 six per cent, is always implied. 
 
 Ex. 1. What is the interest of $215 for one year, at 6 per 
 cent.? 
 
 It is obvious that the required interest will be 6 times 215 
 cents, for it is 6 cents on each dollar. Therefore, 
 
 1290 cents = $12.90, Ans. 
 
 On a note of $215, which had been on interest two years, 
 there would then be due $227.90 ; that is, $215 principal-f 
 
164 
 
 SIMPLE INTEREST. 
 
 $12.90 interest^ $227.90, the amount. Therefore, the amount 
 is found by adding the principal and interest together. 
 
 2. What is the interest of $47.86 for 3 years, at 6 per cent, 
 per annum ? 
 
 OPERATION. 
 
 47.86 
 .06 
 
 2.87 1 6 = interest for one year. 
 3 
 
 $8.61 4 8= interest for three years. 
 
 To understand why 4 figures are cut off in this sum, the 
 scholar should remember, that 1 cent is T J^ of a dollar ; and 
 consequently, 6 cents are y^ of a dollar, which is "the same as 
 .06. (See introductory remarks to Decimal Fractions.) 
 
 3. What is the interest of $72.72 for 4 years, at 6 per cent. ? 
 Ans. $17.45.+ 
 
 4. What is the amount of $456 for two years? Ans. 
 $510.72. 
 
 5. What is the interest of $146.31 for five years, at 6 per 
 cent. 1 Ans. 43.893. 
 
 6. What is the interest of $24.91 for 6 years, at 5 per cent, ? 
 Ans. $7.473. 
 
 7. What is the interest of $222.46 for three years, at 3 per 
 cent. ? Ans. $26.695.+ 
 
 8. What is the amount of $42 for six years, at 3 per cent. ? 
 Ans. $49.56. 
 
 9. What is the amount of $566.33 for one year, at 5 per 
 cent.? Ans. $594.646.+ 
 
 10. What is the amount of $1567 for 9 years, at 2 per cent. ? 
 Ans. $1849.06. 
 
 CASE 2d. WHEN THE TIME CONSISTS OF YEARS AND MONTHS. 
 
 Lawful interest in the New England states is 6 per cent, 
 per annum ; that is, it is 6 cents for 12 months, or J cent per 
 month on a single dollar. Therefore, 
 
 RULE. Reduce the given years to months, and add the given 
 months ; then with half this number of months as a multiplier t 
 multiply the given sum. The product will be the interest for the 
 whole time. 
 
SIMPLE INTEREST. 165 
 
 Note 2d. It is obvious that half the whole number of months 
 in the given time, is the same as the number of cents on a 
 dollar for the whole time, when the interest is at 6 per cent. ; 
 for, from the above remark, the interest of one dollar for one 
 month is evidently per cent., therefore, the whole number of 
 months divided by 2, must determine the number of cents on 
 each dollar for the whole time. If, in taking half the number 
 of months, there be an odd one, the interest for that odd 
 month will be cent, or 5 mills. 
 
 Ex. 1. What is the interest of $220 for 2 years and 6 
 months, at 6 per cent. ? 
 
 2 yr. 6 months =30 months, and 30-^-2 = 15. The interest 
 on each dollar for the whole time is, therefore, 15 cents. Con- 
 sequently, $33 is the interest of the given sum. 
 
 22 
 
 $33.00 Ans. 
 
 2. What is the interest of $756.20 for I year and 3 months, 
 at 6 per cent. ? 
 
 lyr. 3 mo. = 15 months ; therefore, 7 cents, or 7 cents, 5 
 mills, is the interest on each dollar for the whole time. There- 
 fore, 
 
 * 75 6.2 
 .075 
 
 3781 00 
 529340 
 
 5 6.7 1 5 00 Ans. $56.715. 
 
 Note 3d. If the interest be required at some other than 6 
 per cent., first cast it at 6 per cent, by the preceding rule. Then 
 divide the interest so found by 6, and the quotient will be the 
 interest of the given sum at one per cent, for the time speci- 
 fied ; and this multiplied by the given per cent., will be the re- 
 quired interest. 
 
166 SIMPLE INTEREST. 
 
 3. What is the interest of $656 at 8 per cent, per annum, 
 for 3 years and 4 months ? 
 
 3yr. and 4 mo.=:40 months, and 40-i-2z=20, the per cent, 
 for the whole time. Therefore, 
 656 
 .2 
 
 6)131.2 0= interest at 6 per cent. 
 
 2 1.866 6= interest at 1 per cent. 
 8 
 
 174.932 S^interest at 8 per cent., viz. 
 $174.932.+ 
 
 4. What is the interest of $37.50 for 3 years and 6 months, 
 at 6 per cent. ? Ans. $7.875. 
 
 5. What is the interest of $672 for 3 years and 8 months, at 
 6 per cent. ? Ans. 147.84. 
 
 6. What is the interest of $372 for 2 years and 11 months, 
 at 5 per cent. ? Ans. $54.25. 
 
 7. What is the interest of $215.34 for 4 years and 6 months, 
 at 3 per cent. ? Ans. 33.916.+ 
 
 8. What is the interest of $350 for 2 years and 6 months, 
 at 6 per cent. ? Ans. $52.50. 
 
 CASE 3d. WHEN THE GIVEN TIME CONSISTS OF YEARS, 
 
 MONTHS, AND DAYS. 
 
 It was shown in the preceding case, that the interest of $1 
 for one month was cent =5 mills, whenever the rate is 6 
 per cent. Now since 30 days are always allowed for a month 
 in computing interest, 5 mills also equals the interest for! 30 
 days; and 30^-5=6, the number of days required for one dol- 
 lar at 6 per cent, per annum, to gain one mill. If, therefore, 
 the number of days given, be divided by 6, the quotient will 
 be the number of mills to which the interest of one dollar 
 will amount during that time. Therefore, 
 
 RULE. Find the interest of one dollar for the given time, by 
 allowing half a cent for every month, and one mill for every six 
 days ; the sum thus obtained will be the per cent, for the whole 
 time. Multiply the principal by this, and the product will be 
 the interest. 
 
SIMPLE INTEREST. 167 
 
 Note 4th. If the given days be less than 6, or if they be 
 more, and in taking a sixth part of them, a number remain, 
 (which in all cases will be less than 6,) the interest for those 
 days will be a fraction of a mill ; that is, it will be as many 
 sixths of a mill as there are days. 
 
 Note 5th. This rule always gives the interest at 6 per 
 cent, per annum. If, therefore, any other per cent, be required, 
 proceed as directed in Note 3d. 
 
 Ex. 1. What is the interest of $150 for I year, 3 months, 
 and 15 days, at 6 per cent, per annum ? 
 
 Solution : 1 year and 3 months=15 months; the interest, 
 therefore, for that time is 15 half cents =7% cents, or 7 cents, 
 5 mills; and the interest for 15 days is 1 5 -h6 =2% mills ; 
 therefore, 7 cents, 5 mills 4-2^- mills=:7 cents, 7 mills, which 
 is the interest of one dollar at 6 per cent, for the whole time. 
 Consequently, 
 
 1 50 
 .0 7 7i 
 
 1050 
 1050 
 
 7 5 
 
 $ 1 1.6 2 5= Ans. or interest required. 
 
 2. What is the interest of $320 for 3 years, 4 months, and 
 18 days, at 6 per cent, per annum ? 
 
 Solution : 3 yr. 4 mo. =40 months. The interest for this 
 time is .20 per cent, and for 18 days it is 18-f-6 = 3 mills ; 
 therefore, 20 cents and 3 mills is the interest of one dollar for 
 the whole tinrc. Hence, 
 
 320 
 .203 
 
 960 
 6400 
 
 $64.96 O^interest required, viz. $64.96. 
 
 3. What is the interest of $75 for 1 year, 6 months, and 12 
 days, at 6 per cent, per annum ? Ans. $6.90. 
 
 4. What is the interest of $162 for 1 year, 8 months, and 15 
 days, at 6 per cent. ? Ans. $16.605. 
 
168 SIMPLE INTEREST. 
 
 5. What is the interest of $350 for 2 years, 3 months, and 
 5 days, at 6 per cent. ? Ans. $47.54. + 
 
 6. What is the interest of $275 for 3 years, 6 months, 
 and 12 days, at 4 per cent. ? Ans. $38.866.+ 
 
 7. What is the interest of $560 for 9 months and 20 days, 
 at 6 per cent. ? Ans. $27.066. + 
 
 8. What is the interest of $420 for 2 years, 1 month, and 
 27 days, at 6 per cent. ? Ans. $54.39. 
 
 Note 6th. If the interest of $100 for one year be $6, the* 
 interest of $200 for the same time, is obviously $12. Again, 
 if the interest of $100 for one year be $6, for 2 years it must 
 be twice as much, viz. $12 ; and for 3 years, three times as 
 much, viz. $18. In casting interest, we are therefore to regard 
 not only the principal, but also the time during which it is on 
 interest. This may help the scholar to understand the follow- 
 ing rule for casting interest by canceling. 
 
 RULE FOR CANCELING AND FIRST, WHEN YEARS ONLY ARE 
 
 GIVEN. Draw a horizontal line, and write the sum on which the 
 interest is to be cast, above it. On the right of this, also above 
 the line, place the given time in years, and the rate per cent. 
 Then place $100 below the line. Proceed to cancel, <$fC. ;] the 
 sum obtained will be the required interest . 
 
 Ex. 1. What is the interest of $300 for 3 years at 6 per 
 cent. ? 
 
 300. 3. 6 
 Statement, . 
 
 It is obvious, that the interest of $300 for one year, will be 
 three times as much as the interest of $100 for the same time ; 
 and that for 3 years, it will be three times as great as for 1 
 
 3 
 
 o e 
 
 year. The above statement canceled : ^~~ '> an ^ 3 x 3 x 
 
 6=$54, Ans. 
 
 2. What is the interest of $350 for 4 years, at 5 per cent, 
 per annum ? 
 
 Statement, ^'^ 5 . 
 
 2 
 
 Canceled, ~, and 35x2 = $70, Ans. 
 S 
 
SIMPLE INTEREST. 169 
 
 3. What is the interest of $500 for 5 years, at 9 per cent, 
 per annum. Ans. $225. 
 
 4. What is the interest of $240 for 6 years, at 4 per cent, 
 per annum ? Ans. $57.60. 
 
 AGAIN, WHEN THE GIVEN TIME CONSISTS OF YEARS AND 
 
 MONTHS. 
 
 RULE. Having reduced the given time to months, arrange 
 the terms as in the preceding rule, and in addition to the terms 
 there introduced, place 12, the number of months in one year, 
 below the line ; cancel, fyc. 
 
 Note 7th. If the interest of $100 for one year be $6, 
 for two years it would be $12, &c. (See preceding note.) Con- 
 sequently, for one year and six months, it would be J of $6=^-|, 
 and for two years and nine months, V^yf of $6, &c. These 
 fractional expressions are obtained by reducing the given time 
 to months, for the numerator, and making 12, the number of 
 months in a year, the denominator. Hence we see the reason 
 of the above rule. 
 
 Ex. 5. What is the interest of $360 for 2 years and 6 
 months, at 6 per cent, per annum? Solution : 2 years, 6 months 
 
 O/*A OA 
 
 30 months; therefore,- r^ . It will be observed that 
 
 100. la. ,-. 
 
 the whole time is ^ of a year. 
 
 3 
 Canceled, ^ ^ 6 , and 3 x 3 x 6 = $54, Ans. 
 
 IWU,. tJS 
 
 6. What is the interest of $440 for 2 years and 4 months, 
 at 6 per cent, per annum ? 
 
 The time =rf | of a year, therefore, ^?|^. Ans. 861.60. 
 
 7. What is the interest of $150 for 1 year and 6 months, at 
 6 per cent, per annum ? 
 
 150. 18. 6 ... _ A 
 
 Statement, , 1c5 . Ans. $13.50. 
 
 JLUU. 1/e 
 
 8. What is the interest of $25.32, for 9 months, at 4 per 
 cent, per annum ? 
 
 or; QO q A 
 
 Statement, **, An* $0.759.+ 
 
 J.UO. J 
 
 15 
 
170 SIMPLE INTEREST. 
 
 9. What is the interest of $375.75 for 4 years and 9 months, 
 at 7 per cent. ? Ans. $124.936.+ 
 
 10. What is the interest for $1500 for 3 years and 4 months, 
 at 6 per cent. ? Ans. $300. 
 
 11. What is the interest of $175 for 4 years and 2 months, 
 at 8 per cent.? Ans. $58.333.+ 
 
 LASTLY, WHEN THE TIME CONSISTS OF YEARS, MONTHS, AND 
 
 DAYS. 
 
 RULE. Reduce the years and months to months, and the 
 days to the fraction of a month ; then placing this fraction on 
 the right of the months, reduce the whole to an improper fraction. 
 This expression will represent the whole time in months ; there- 
 fore, having arranged the other terms as before, write its nume- 
 rator above the horizontal line, and its denominator below, and 
 proceed as before directed. 
 
 Note 8th. As the fractional expression of the whole time 
 obtained by the preceding rule represents that time in months, 
 12, the number of months in one year, must be placed below 
 the line, as before directed, to reduce those months to years. 
 
 Ex. 1. What is the interest of $150 for 1 year, 3 months, 
 and 1 5 days, at 6 per cent. ? 1 year and 3 months 1 5 months ; 
 and 15 days^i of a month; therefore, 15 months is the 
 whole time given, and this reduced to an improper fraction, 
 equals ^ of a month. Agreeably to the rule, we then have 
 
 the following 
 
 . 150. 31. 6 
 Statement, viz. 1QQ g lg . 
 
 3 
 Canceled, ^~r|; then, 31 x 3 = 93, and 2x2x2 = 8' 
 
 isviW,. <*. i*a- 
 
 2 2 
 
 Therefore, 93 -H-8 = $11. 625, Ans. 
 
 2. What is the interest of $320 for 3 years, 4 months, and 
 18 days, at 6 per cent, per annum? 3 years and 4 months = 
 40 months, and 18 days = i-|=f of a month. The whole time 
 is therefore 40| =- - of one month. 
 
 320. 203. 6 . 
 Therefore, is the statement. 
 
SIMPLE INTEREST. 171 
 
 S68 
 
 'Q^A 9f^^ I? 
 
 The same canceled, &c. is ~ ^ p^ ; 203x8 = 1624; and 
 
 5 ' S 
 
 5x5 = 25. Hence, 1624-^25 = $64.96, Ans. 
 
 3. What is the interest of 875 for 1 year, 6 months, and 1 
 days, at 6 per cent. 1 Ans. $6.875. 
 
 ' 55 ' 
 
 The time^V of a month, therefore, ' Q ' . . 
 
 1UU. o. 1-w 
 
 Note 9th. Ininterest,30daysare always allowed for a month. 
 
 In solving the following sums, the scholar can apply either 
 of the preceding rules. 
 
 4. What is the interest of $4.20 for 2 years, 1 month, and 
 27 days, at 6 per cent. 1 Ans. $54.39. 
 
 5. What is the interest of $80 for 1 year, 5 months, and 12 
 days, at 6 per cent. 1 Ans. $6.96. 
 
 6. What is the amount of $275 for 3 years, 6 months, and 
 12 days, at 4 per cent. 1 $313.866.+ 
 
 7. What is the amount of $560 for 9 months and 20 days, 
 at 6 per cent. 1 Ans. $587.066.+ 
 
 8. What is the interest of $50,11 for 1 year and 21 days, 
 at 6 per cent. 1 Ans. $3.18.+ 
 
 9. What is the interest of $340.50 for 3 months and 1 day, 
 at 6 per cent. 1 Ans. $5.16.+ 
 
 10. What is the interest of $90 for 1 year, 2 months, and 6 
 days, at 6 per cent. 1 Ans. $6.39. 
 
 11. What is the interest of $4119.20 for 1 year and 5 days, 
 at 6 per cent. 1 Ans. $250.584.+ 
 
 12. What is the interest of $23.08 for 3 years, 6 months, 
 and 22 days, at 6 per cent. 1 Ans. $4.93.+ 
 
 13. What is the interest of $439.50 for 1 year, 11 months, 
 and 9 days, at 6 per cent. ? Ans. $51.20. 
 
 14. What is the amount of $28 for 9 years, 8 months, and 
 3 days, at 6_per cent. 1 Ans. $44.254. 
 
 15. What is the amount of $42 for 5 years, 5 months, and 9 
 days, at 5 per cent. ? Ans. $53.427. 
 
 16. What is the amount of $50.50 for 1 year and 3 months, 
 at 3 per cent. 1 Ans. $52.393. + 
 
 17. What is the amount of $300 for 16 years and 8 months, 
 at 6 per cent. ? Ans. $600. 
 
 18. What is the interest of $375.75 for 4 years and 9 months, 
 at 7 per cent. ? Ans. 124.936. + 
 
172 SIMPLE INTEREST. 
 
 19. What is the interest of $3.75 for 12 years, at 6 per cent, 
 per annum ? Ans. $2.70. 
 
 20. What is the amount of $75 for 6 years and 3 months, at 
 6 per cent. ? Ans. $103.125. 
 
 21. What is the interest of $63 for 1 year and 3 months, at 
 
 6 per cent. ? Ans. $4.725. 
 
 22. What is the interest of $156 for 2 years and 4 months, 
 at 8 per cent. ? Ans. $29.12. 
 
 23. What is the amount of $650 for 3 years and 2 months, 
 at 6 per cent. ? Ans. $773.50. 
 
 24. What is the amount of $128.30 for 1 year and 9 months, 
 at 9 per cent. ? Ans. $148.507. 
 
 25. What is the interest of $33.50 for 2 years and 6 months, 
 at 5 per cent. ? Ans. $4.1875. 
 
 26. What is the interest of $150 for 2 years, 7 months, and 
 
 7 days, at 6 per cent. ? Ans. $23.425. 
 
 27. W T hat is the interest of $730 for 5 years, 9 months, and 
 12 days, at 6 per cent. ? Ans. $253.31. 
 
 28. What is the interest of $875.49 for 5 years, 8 months, 
 and 20 days, at 6 per cent. ? Ans. $300.58. -f 
 
 29. What is the amount of $630 for 8 months, at 6 per cent.? 
 Ans. $655.20. . 
 
 30. What is the amount of $734%>for 1 year and 4 months, 
 at 6 per cent. ? Ans. $7929.36. 
 
 31. What is the amount of $750 for 9 months, at 7 per cent. ? 
 Ans. $789.375. 
 
 32. What is the amount of $375 for 5 months and 15 days, 
 at 6 per cent. ? Ans. $385.31. 
 
 33. What is the amount of $460.50 for 4 months, at 6 per 
 cent.? Ans. $469.71. 
 
 34. What is the amount of $230.25, for 8 months, at 7 per 
 cent.? Ans. $240.995. 
 
 35. What is the amount of $764.50 for 3 years and 10 
 months, at 6 per cent. ? Ans. $940.335. 
 
 CASE 4th. To FIND THE INTEREST AT six PER CENT. FOR ANY 
 
 NUMBER OF DAYS. 
 
 In computing interest, the month is reckoned at 30 days, 
 and the interest of one dollar for that time at 6 per cent., as 
 has already been shown, is cent or 5 mills ; hence for twice 
 that time, or 60 days, it would be just one cent for every dol- 
 lar on interest. We therefore have the following rule ; 
 
SIMPLE INTEREST. 173 
 
 RULE. Consider the number of dollars given so many cents, 
 and reduce these cents to dollars again by removing the point of 
 separation two places to the left ; the result will be the interest of 
 the given sum for 60 days. Then, if the given days be more 
 or less than 60, add to, or subtract from, the interest of GO days, 
 such parts of itself as the given days require. 
 
 Ex. 1. What is the interest of $450.82 for 93 days, at 6 
 per cent, per annum 1 
 
 SOLUTION. 
 
 4.5082 = interest for 60 days. (See the rule.) 
 2.2541 ^interest for 30 days, being half the int. for 60 days. 
 22541 ^interest for 3 days, being ^ of the int. for 30 days. 
 
 $6.98771 ^interest for 93 days. 
 
 2. What is the interest of $4562 for 45 days, at 6 per cent, 
 per annum ? 
 
 SOLUTION. 
 45.62 ^interest for 60 days. 
 
 22.81 ^interest of 30 days, or one half the int. of 60 days. 
 11. 405= interest of 15 days, or one half the int. of 30 days, 
 
 $34.215=interest of 30 + 15 = 45 days. 
 
 In computing by this rule, 12 months of only 30 days each, 
 are allowed for the year, equal to 360 days. It consequently 
 gives Jg- part more than 6 per cent, interest. On small sums, 
 and for short intervals, however, the difference is trifling. 
 
 3. What is the interest of $420.72 for 120 days, at 6 per 
 cent, per annum ? Ans. $8.414.+ 
 
 4. What is the interest of $56.74 for 25 days, at 6 per cent, 
 per annum ? Ans. $0.236.+ 
 
 5. What is the interest of $156.36 for 96 days, at 6 per 
 cent, per annum? Ans. $2.50.+ 
 
 6. What is the interest of $1000 for 29 days, at 6 per cent, 
 per annum ? Ans. $4.833.+ 
 
 7. What is the interest of $204 for 40 days, at 6 per cent, 
 per annum 1 Ans. $1.36. 
 
 8. What is the interest of $472 for 18 days, at 6 per cent 
 per annum? Ans. $1.416. 
 
 15* 
 
174 SIMPLE INTEREST. . 
 
 Banking. When a promissory note is presented at a bank- 
 ing institution, if properly endorsed or otherwise secured, it is 
 received by the officers of the bank- as security, and so much 
 money in their own notes is given in return as is equal to the 
 face of the note after the interest is deducted for 3 days more 
 than the whole time till payment is promised. Hence, if the 
 time specified be 60 days, the interest on the face of the note 
 for 63 days is deducted, and the balance drawn from the bank ; 
 then at the expiration of the 63 days, the whole face of the 
 note is due. The 3 days added to the specified time of pay- 
 ment, are called " days of grace ." 
 
 The preceding rule is, therefore, a convenient one for bank- 
 ing institutions. 
 
 In solving the following sums, the scholar will allow " 3 
 days of grace" that is, he will find bank discount for 3 days 
 more than are specified in the sum. 
 
 Ex. 1. What is the bank discount on $256 for 30 days, and 
 grace ? 
 
 SOLUTION. 
 2 . 5 6= discount for 60 days. 
 
 1 . 2 8= discount for 30 days. 
 . 1 2 8= discount for 3 days grace. 
 
 $1.40 8=required discount. 
 
 Therefore, $256 $1.408 = $254. 592, the sum to be drawn 
 from the bank. 
 
 2. How much money would be drawn from the bank on a 
 note of $650, payable in 90 days ? 
 
 6 . 5 = discount for sixty days. 
 3 .2 5 rediscount for thirty days. 
 . 3 2 5= discount for three days. 
 
 10.07 5= discount for ninety -three days. 
 Therefore, $650 $10. 075 = $639.925, the money to be 
 drawn. 
 
 3. What is the bank discount on $1056.29 for 30 days, 
 and grace? Ans. $5.81. 
 
 4. What is the bank discount on $756 for 90 days, and 
 grace? Ans. $11.718. 
 
LE INTEREST. 175 
 
 5. What is the barflff&iscount on $676.19 for 25 days, and 
 grace? Ans. $3.155.+ 
 
 6. How much money iay be drawn on a note of $692, 
 payable 70 days after date, if discounted at a bank ? Ans'. 
 $683.581. 
 
 7. How much money may be drawn on a note of $1567.89, 
 payable 120 days from date? Ans. $1535.748. 
 
 8. What is the bank discount on $542.78 for 90 days, and 
 grace? Ans.$8Al3.+ 
 
 9. What is the bank discount on $195.77 for 60 days, and 
 grace? Ans. $2.055.+ 
 
 10. How much money may be drawn on a note of $726, 
 payable 80 days from date ? Ans. $715.957. 
 
 CASE 5th. WHEN THE MONEY ON WHICH THE INTEREST is 
 
 TO BE CAST, IS IN POUNDS, SHILLINGS, AND PENCE. 
 
 RULE. Reduce the shillings and pence to the,dfaimal of a 
 pound, (see Case 3d, Decimal Fractions,) and castmfte interest 
 in the same manner as when the principal consists of dollars, 
 cents, and mills. The decimal part of the answer may then be 
 reduced to shillings and pence by Case 4th, Decimal Fractions. 
 
 Ex. 1. What is the interest on 42 . 16s. 6 d. for 1 year 
 and 6 months, at 6 per cent, per annum ? 
 SOLUTION. 
 12 
 
 20 
 
 .8 2 5= the decimal value of 16s. 6d. 
 
 1 yr. 6mo.r=;18 months, and 18-^2 = 9, the per cent, for the 
 whole time. Therefore, 
 4 2.8 2 5 . 
 .0 9 
 
 3.8 5 425 .= interest in pounds and decimals. 
 To find the value of the decimal : 
 .85425 
 
 2 
 
 17.08500 
 1 2 
 
 1.02000 
 The answer, therefore, is 3 j. 17s. Id. 
 
176 SIMPLE INTERES 
 
 \afi 
 
 2. What is the interest of 55 . lJP6d. for 2 years, 6 
 months, at 6 per cent. ? 
 
 55 . 15 s. 6d. ^55.775 . And2 yr. 6 mo. = 30 months = 
 15 per cent, for the whole time. Therefore, 
 55.775 
 .1 5 
 
 278875 
 55775 
 
 8.3 6625. 
 
 To find the value of the decimal : 
 .36625 
 2 
 
 7.3 2 5 0= shillings. 
 12 
 
 3.9000 Oz=pence. 
 4 
 
 3.6 0=farthings. 
 Therefore, 8 . 7 s. 3 d. 3 qr. Ans. 
 
 3. What is the interest of 21 . 18 s. 4 d., for 3 years and 
 4 months, at 6 per cent. ? Ans. 4 . 7s. 8 d. 
 
 4. What is the amount of 156. 9s. 3 d. for 1 year and 9 
 months, at 6 per cent. 1 Ans. 172 . 17s. 9 d. 3 qr.+ 
 
 5. What is the amount of 27 . 2s. 6d. 3 qr. for 1 year 
 and 10 months, at 6 per cent. ? Ans. 30 . 2 s. 2 d. 3 qr.+ 
 
 6. What is the interest of 36 . 15s. for 2 years and 3 
 months, at 6 per cent. ? Ans. 4. 19 s. 2 d. 2.8 qr. 
 
 7. What is the interest of 45 . 10 s. for 8 months, at 6 per 
 cent. ? Ans. I . 16 s. 4 d. 3.2 qr. 
 
 8. What is the interest of 9 . 12 s. 6 d. for 2 years, 4 
 months, and 12 days, at 6 per cent. ? Ans. \ >. 7s. 4 d. -f- 
 
 CASE 6th. WHEN INTEREST is REQUIRED ON NOTES OR BONDS 
 
 ON WHICH PARTIAL PAYMENTS HAVE BEEN MADE. 
 
 RULE. Cast the interest on the principal at the given 
 rate per cent, up to the time of the first payment ; then, if the 
 payment exceed the interest, deduct the excess from theprincipal ; 
 
 
SIMPLE INTEREST. 177 
 
 but if it be less, set both payment and interest aside, and cast 
 the interest on the same principal to the time of the next pay- 
 ment, or to the time of some payment, which, when added to the 
 preceding payments, will exceed the sum of interest then due, 
 and deduct the sum of these payments from the amount of 
 the principal. The remainder will form a new principal, with 
 which proceed as before, till the time of settlement. 
 
 1. For value received, I promise to pay A. B. & Co., or 
 order, fifteen hundred dollars on demand, with interest. 
 Jan. 1, 1825. JOHN JAMES. 
 
 On this note are the following endorsements : Oct. 1, 1825, 
 three hundred dollars. July 1, 1827, four hundred and fifty 
 dollars. Sept. 1, 1828, six hundred and fifty dollars. What 
 was due on settlement, July 1, 1830? 
 
 The principal on interest from Jan.- 1, 1825, $1500.00 
 
 Interest from Jan. I, 1825, to Oct. 1, of the same year, 9 mo. 67.50 
 
 1567.50 
 The payment being more than the interest, deduct it, . . 300.00 
 
 Remainder, forming a new principal, 1267.50 
 
 The interest from Oct. 1, 1825, to July 1, 1827, 1 yr. and 9 mo. 133.87 
 
 1400.587 
 Deduct second payment, because it is more than the interest, 450.000 
 
 Remainder, forming a new principal, 950.587 
 
 The interest from July 1, 1827, to Sept. 1, 1828, 1 yr. and 2 mo. 66.541 
 
 1017.128 
 Deduct the third payment, 650.000 
 
 Remainder, forming a new principal, ; . 367.128 
 
 The interest from Sept. 1, 1828, to July 1, 1830, the time of 
 
 settlement, 40.384 
 
 Ans. $407.512 
 
 SOLUTION BY CANCELING. 
 
 yr. m. d. 
 J825 10 1 
 1825 1 1 
 
 9 time till first payment. 
 
178 SIMPLE INTEREST. 
 
 aftd 15x9-^-2 = $67.50, the interest till first payment ; and 
 1900 + 67.50 = $1567.50, amount; and 1567.50300 = 
 $1267.50, the new principal. 
 
 yr. mo. d. 
 1827 7 1 
 1825 10 1 
 
 1 9 0=time from 1st to 2d payment. 
 
 1267.50. 21. 6 " 1267.50. 21. & 
 
 Statement, - -- -. Canceled, _ 
 
 2 
 
 1267.50x21-^-200 = 133.087, the interest till 2d payment; 
 then, 1267.504-133.087 = 1400.587, and 1400.587450 = 
 950.587, the new principal. 
 
 yr. m. d. 
 1828 9 1 
 1827 7 1 
 
 1 2 0=time from 2d to 3d payment. 
 
 7 
 
 c . 4 ,. 950.587. 14. 6 ^ , , 950.587. 14. S 
 
 Statement, -___, Canceled, -^ -^- ; 
 
 and 956.587x7-^-100=66.541, interest till the 3d payment, 
 and 950.587+66.541 = 1017.128, and 1017.128 650 = 367.- 
 128, the new principal. 
 yr. m. d. 
 
 1830 7 1 
 
 1828 9 1 
 
 
 110 0=time from 3d payment to settlement. 
 
 11 
 
 367.128. 22. 6 ^ , , 367.128. SS. & 
 Statement, - . Canceled, -- -- ; 
 
 and 376.128x11-^-100=40.384, interest till time of settle- 
 ment, and 367. 128 +40. 384 = $407.5 12, answer, or sum due 
 on settlement. 
 
 2. I have in my possession a note, dated April 15, 1833, for 
 $2150.25, on which are* the following endorsements : Nov. 8, 
 1834, $500.00; Sept. 1, 1835,- $723.64; January 1, 1837, 
 $378.295 ; and Oct. 29, 1837, $850.00. What amount was 
 due on this note, April 15, 1838? Ans. $138.337. 
 
COMPOUND INTEREST. 179 
 
 3. On a note of $767.95, given Dec. 25, 1827, and drawing 
 interest after ninety days, were the following endorsements : 
 Jan. 1, 1830, $75.00 ; March 25, 1831, $565.25. What was 
 due Jan. 1,1833? Ans. $294.118. 
 
 BOSTON, Jan. 13, 1809. 
 
 4. On demand, I promise to pay J. Anderson, or order, one 
 thousand five hundred eighty-five and 33 1, . dollars, with in- 
 terest, for value received. 
 
 Received May 5, 1812, $863.12. 
 Received May 7, 1814, $221. 
 Received July 21, 1815, $1009.03. 
 
 Will the scholar determine whether the note is fully paid ? 
 
 5. What was due on a note of $2100, dated June 15, 1820, 
 on settlement, June 15, 1830, the following sums beino- en- 
 dorsed on the back of it, viz. June 30, 1824, $750, and Sept. 
 30, 1828, $1200, on interest at 6 per cent. ? Ans. $1249.527. 
 
 6. For value received of A. B., I promise to pay him, or 
 order, seven hundred and fifty dollars, with interest at 6 per 
 cent.? 
 
 Jan. 1,1824. M. S. 
 
 On the above were the following payments endorsed : April 
 1, 1826, one hundred and fifty dollars; July 1, 1829, four 
 hundred and fifty dollars. What was due on settlement, Sept. 
 1, 1832? Ans. $461.71.+ 
 
 COMPOUND INTEREST. 
 
 Compound Interest is that which is computed annually, 
 and immediately added to the principal. The amount of each 
 year, is made the principal for the succeeding year. 
 
 RULE. Cast the interest at the given rate per cent, for the 
 first year, by multiplying by that rate per cent., and make the 
 amount the principal for the second year. Make the amount 
 of the second year, principal for the third ; and the amount of 
 the third, the principal for the fourth, <$~c., through the whole 
 number of years. From the amount thus obtained, subtract the 
 principal ; the remainder will be the Compound Interest. 
 
180 COMPOUND INTEREST. 
 
 Ex. 1. What is the compound interest of $256 for 3 years, 
 at 6 per cent. ? 
 
 256 
 .0 6 
 
 15.36 interest. 
 256.00 principal added. 
 
 271.36 principal for 2d year. 
 .0 6 
 
 1 6.28 1 6= interest of 2d year. 
 271.3 6=principal of 3d year added. 
 
 287.6416 principal for 3d year. 
 .06 
 
 17. 2 5 8496 interest of 3d year, 
 287.641 6 principal. 
 
 304.900096 amount of 3d year. 
 256.0 0=the original principal subtracted. 
 
 48.90 compound interest on $256, for 3 years. 
 
 2. What is the compound interest of $450 fox 3 years, at 6 
 percent.? Ans. $85.957. + 
 
 3. What is the compound interest of $50 for. 3 years, at 5 
 per cent. ? Ans. $7.881 . -f- 
 
 4. What is the compound interest of $400, at 6 per cent, for 
 4 years? Ans. $104.99. -f- 
 
 5. What will $675 amount to at compound interest, in 3 
 years and 6 months, 6 at per cent, per annum ? Ans. $828.- 
 053.+ 
 
 6. What is the amount of &40.20 at 6 per cent, compound 
 interest, for 4 years ? Ans. $50.75.+ 
 
 7. What is the amount of $63 at 6 per cent, compound in- 
 terest, for 2 yelrs ? Ans. $70.786.+ 
 
 8. What is the compound interest of $127.85 for 3 years, 
 at 6 per cent. ? Ans. 24.42 l.-f 
 
 
INSURANCE. 181 
 
 COMMISSION. 
 
 Commission is an allowance made by merchants and others 
 to an agent for buying and selling goods. This allowance is 
 usually a certain per cent, on the amount of money received for 
 the sales effected, or on that expended in making purchases. 
 The only respect in which it differs from interest, is, that in 
 computing commission no regard is paid to time ; hence, 
 
 RULE. Multiply by the rate per cent. 
 
 Ex. 1. An agent sold for his employer goods to the value 
 of $1800, for which he received 5 per cent. ; what was the 
 amount of his commission? Ans. $90. 
 
 2. What is the amount of my commission for selling goods 
 to the value of $975, it being 8 per cent. ? Ans. $78. 
 
 3. My agent sends me word, that he has purchased goods 
 on my account to the value of $2768 ; what will his commis- 
 sion amount to at 6 per cent. ? A?is. $166.08. 
 
 4. The commission of $1250 at 10 per cent, is required ; 
 what is it? Ans. $125. 
 
 5. An agent sells 750 bales of cotton, at $52 per bale, and 
 is to receive 2^ per cent, commission. How much money will 
 he receive, and how much will he pay over to his employer ? 
 Ans. He will receive $975, and pay over $38025. 
 
 6. What will my commission amount to at 3 per cent, in 
 purchasing goods to the value of $7846.90 ? Ans. $235.407. 
 
 INSURANCE. 
 
 Insurance against the loss of buildings and goods by fire, 
 and also of ships and their cargoes by storm, is obtained by 
 paying a certain per cent, on the estimated value of the prop- 
 erty insured. 
 
 16 
 
182 INSURANCE. 
 
 The instrument which binds the contracting parties is called 
 the policy, and the sum paid by the party insured to the insuring 
 party, is called the premium. 
 
 RULE. Multiply the estimated value of the property insured, 
 by the per cent. 
 
 Ex. 1 . What is the premium for the insurance of buildings 
 and appurtenances, valued at $3758.50, at \ per cent. 1 Aus. 
 $18.79.+ 
 
 2. What is the premium for insuring property, valued at 
 $3600, against loss by fire, at per cent. ? Ans. $27. 
 
 3. What is the premium for insuring property valued at 
 $845, at i per cent. 1 Ans. $1.69. 
 
 4. What is the premium for the insurance of a ship and 
 cargo valued at $20500, at | per cent. 1 Ans. $68.333.+ 
 
 5. What would be the premium for insuring a ship and 
 cargo valued at $18000, at| per cent. ? Ans. $67.50. 
 
 6. Insured my house and out buildings, valued at $21560.38, 
 at J per cent. ; what was the amount of the premium ? Ans. 
 $86.24.+ 
 
 QUESTIONS. What is interest 1 How is it computed 1 Suppose the 
 sum en interest be more or less than $100, or the time more or less 
 than one year, how must the sum paid as interest compare ? Give the 
 illustration. What is the principal? What is interest 1 What is 
 legal interest? What is the legal rate per cent, in New England? 
 What in New York ? What in Louisiana ? What do you under- 
 stand by the amount ? The rate per cent, is a decimal of how many 
 places does it consist when expressed by cents ? And when expressed by 
 mills? What is Case 1st? What is the rule for it ? What is note first ? 
 What is Case 2d ? What is the rule ? Give the reason of the rule. 
 What is note 2d ? Noie 3d ? What is Case 3d ? What is the rule 1 
 Give the explanation which precedes the rule. What is note 4th ? 
 What is note 5th ? What is note 6th ? What is the rule for can- 
 celing ? What is the rule for canceling, when the time consists of years 
 and months ? What is Note 7th ? What is the rule for canceling, 
 when the time consists of years, months, and days ? What is 
 Note 8th ? What is Note 9th ? What is Case 4th ? What is the 
 rule ? What explanation precedes the rule? What is said relative 
 to banking institutions ? What is Case 5th ? What is the rule for it ? 
 What is Case 3d of Decimal Fractions? What is Case 4th of 
 Decimal Fractions? What, is Case 6th? What is the rule for it? 
 Will the scholar now inform me why it is correct to multiply by one 
 half of the even number of months in the given time, in casting inte- 
 rest at 6 per cent. ? What is Compound Interest ? What is the rule 
 for it? What is Commission ? In what respect does it differ from 
 simple interest? What is the rule? How is insurance obtained? 
 What is to be understood by the policy? What by the premium? 
 What is the rule ? 
 
RATIO. 183 
 
 RATIO. 
 
 Ratio is the relation which one quantity or number has to 
 another quantity or number of the same kind. 
 
 The former of the two numbers, between which the ratio 
 exists, is called the antecedent, and the latter, the consequent. 
 
 The direct ratio of any two numbers is obtained by dividing 
 the consequent of any couplet by the antecedent, and the in- 
 verse ratio, by dividing the antecedent by the consequent. 
 
 Thus, the direct ratio of 2 to 4, is 2, because the antece- 
 dent 2, is contained in the consequent 4, two times ; and the 
 inverse ratio is -| , because 4, the consequent, is contained in 
 2, the antecedent, J=r^ of a time. Both these expressions 
 establish the same general fact, viz. that one of the given num- 
 bers is twice as large as the other. 
 
 From the above, it is obvious that ratio can exist only be- 
 tween quantities of the same kind. It would be absurd to inquire 
 how many times 3 acres must be repeated to equal 12 tons. 
 
 Any simple ratio is expressed by two dots placed between 
 the antecedent and the consequent ; thus, 2 : 8, or 5 : 10. 
 
 The following four propositions require to be carefully 
 studied. 
 
 1. The ratio of any couplet is not affected either by multi- 
 plying or by dividing its antecedent and consequent by the same 
 number ; for the ratio of 9 : 1 8 is 2 ; and the ratio of 9TTF X 2 
 = 18 : 36, is also 2. The same result is obtained by dividing 
 these terms by any number whatever ; thus, 9: 18-^3 = 3 : 6, 
 and this equals 2. 
 
 2. The ratio of any couplet is multiplied by any number, by 
 multiplying the consequent, or by dividing the antecedent by 
 that number. The ratio of 12 to 36, is 3 ; if, however, the 
 consequent be multiplied by 3, the ratio 3 will be multiplied by 
 the same number ; thus, 12 : 36 X 3 = 12 : 108 = 9. The same 
 result is obtained by dividing the antecedent by the same num- 
 ber ; thus, 12, the antecedent, divided by 3, equals 4, and the 
 ratio of 4 to 36 is 9, the same as before. 
 
 3. The ratio of any couplet is divided by any number, by 
 dividing the consequent, or by multiplying the antecedent by 
 that number. Take the ratio 12 : 36, as before, and let it be 
 divided by 2. If the consequent be divided, we obtain the 
 ratio 12 : 18 = 1^ ; but if the antecedent be multiplied, we ob- 
 tain the ratio 24 : 36 = l also. 
 
184 RATIO. 
 
 4. If two or more ratios be multiplied together, that is, if the 
 antecedents be multiplied into the antecedents, and the conse- 
 quents into the consequents, the resulting ratio is called COM- 
 POUND RATIO, and is equal to the product of the simple ratios. 
 The ratio compounded of the simple ratios, 2:4, 3:6, and 
 4 : 8, is the ratio, 24 : 192 = 8 ; but the ratio, 2 : 4=2, also, 
 3 : 6=2, and 4 : 8=2 ; and 2x2x2 = 8. 
 
 Ex. 1. What is the ratio of 6 to 36 ? Ans. 6. 
 
 2. What is the inverse ratio of 7 to 12 ? Ans. f^. 
 
 3. What is the ratio of 7 to 42 ? 
 
 4. What is the inverse ratio of 5 to 20 ? 
 
 5. What is the ratio of 1 . sterling to 7s. 6 d. ? It is 
 obvious that the terms here given must be reduced to the same 
 denomination, in order to compare them, therefore, 1 .=240 d. 
 and 7 s. 6 d. = 90 d . The direct ratio, therefore, is -$ = f . 
 
 6. What is the ratio of 16 Ib. to 3 cwt. 1 Ans. 21. 
 
 7. Multiply the terms 3 : 7 by 9, and see how the given 
 ratio is affected. 
 
 8. Divide the terms 9 : 15 by 3, and compare ratios. (See 
 proposition 1st.) 
 
 9. Multiply the ratio 11 : 16 by 5. Ans. 11 : 80. 
 
 10. Multiply the ratio 9:11 by 3. Ans. 3:11. (See propo- 
 sition 2d.) 
 
 11. Divide the ratio 3 : 12 by 6. Ans. 3 : 2 or f . 
 
 12. Divide the ratio 3 : 19 by 4. Ans. 12 : 19. (See propo- 
 sition 3d.) 
 
 13. What is the ratio compounded of the ratios 2:3; 3:4, 
 and 4 : 5 ? Ans. 24 : 60 or 2 : 5. 
 
 Note. Whenever the antecedent of any couplet is the 
 same as the consequent of any other couplet, the several 
 ratios may be reduced to one, by rejecting such antecedents 
 and consequents. Hence, the preceding simple ratios may be 
 reduced to a compound one, by rejecting their similar terms. 
 The ratio thus obtained is 2 : 5. 
 
 14. What is the ratio compounded of the simple ratios 3:5; 
 4 : 5 ; and 2 : 3 ? (See proposition 4th.) Ans. 24 : 75, or 8 : 25. 
 
 15. Multiply the ratio 5 : 9 by 4. Ans. 5 : 36. 
 
 16. Divide the same ratio by 3. Ans. 5 : 3. 
 
 17. Multiply the ratio 12 : 21 by 6. Ans. 2:21. 
 
 18. Divide the same by 7. Ans. 84 : 21, or 12 : 7. 
 
 19. Reduce the ratios 4:7; 3:1; and 6:2, to a compound 
 ratio. Ans. 36 : 7. 
 
PROPORTION. 185 
 
 PROPORTION. 
 
 Equality of ratios constitutes proportion. 
 
 Each simple statement in proportion requires at least two 
 equal ratios, or four terms, the first and second of which are of 
 one kind, and constitute one of the ratios ; and the third and 
 fouth are of another kind, and constitute the other ratio. 
 
 If 6 men earn 12 dollars in a given time, 36 men will earn 
 72 dollars in the same time. In this proposition, the two ratios 
 required to constitute a proportion, are, the ratio of 6 men to 
 36 men, and of 12 dollars to 72 dollars; and the proposition 
 can be true only on the condition that these two ratios are 
 equal; and that is actually the case, for 36 -4- 6 = 6, and 72-r- 
 12 = 6. Therefore, 6 is the ratio of each couplet. 
 
 Proportion is usually expressed by dots, thus : 6 : 36 : : 12 : 
 72, and thus read : 6 is to 36 as 12 is to 72, or 6 men is to 
 36 men as 12 dollars is to 72 dollars. 
 
 When any four numbers are proportionals, the first and fourth 
 are called extremes, and the second and third, the means ; and 
 the product of the extremes must always equal the product of 
 the means. This is true with regard to the statement made 
 above, for 6x72 = 432, and 12x36 = 432. 
 
 Since, therefore, these products are always equal, if any 
 three terms are given, two of which bear a given ratio to each 
 other, a fourth term may be found, to which the third given 
 term shall bear the same ratio ; for the required term must be 
 either one of the extremes or one of the means. If it be one 
 of the extremes, the product of the means divided by the given 
 extreme, will give the required extreme ; and if it be one of 
 the means, the product of the extremes divided by the given 
 mean, will give the required mean. 
 
 Suppose, in the proposition stated above, it be required to 
 find how many dollars 36 men would earn in a given time, if 
 6 men earn 12 dollars in the same time. 
 
 Let the three given numbers stand as before, thus : 6 men : 
 36 men : : 12 dollars : what number of dollars ? 
 
 Now, since 6, the left hand term, multiplied by the required 
 
 term, must produce the same number as 36 multiplied by 12, it 
 
 follows that 36 multiplied by 12, and the product divided by 6, 
 
 will give the required] number. Thus, 36x12=432, and 
 
 16* 
 
186 PROPORTION. 
 
 432-^-6=72, the fourth term in the preceding proposition. 
 Or, suppose it required to find what number bears the same 
 ratio to 36 that 12 does to 72. The statement would be, 
 what number : 36 : : 12 : 72. The other extreme is here 
 required; hence, 36x12=432, and 432-f-72=6, the re- 
 quired extreme. Again, let one of the means be required, 
 thus : 6 is to what number as 12 is to 72. The statement may 
 stand thus : 6 : : : 12 : 72. One of the means is here re- 
 quired, and the extremes are given; therefore, 72x6 = 432, 
 and 432-^6 = 36, the required mean. Lastly, let the other 
 mean be required. We then have the statement as follows : 
 6 : 36 : : : 72. The third term of the statement is here want- 
 ing, or the other mean. Therefore, 72 X 6 =432, and 432 -^- 
 36 = 12, the required mean. 
 
 From the preceding remarks we learn that operations in 
 simple proportion consist in having three of the terms of a 
 proportion given and a fourth term required. For finding 
 this fourth term, we have the following rule : 
 
 RULE. Notice which of the three terms given is of the same 
 name or kind as the required term or answer, and give it the 
 right hand place. Notice again whether the term required must 
 be greater or less than this ; and, if it is to be greater, place 
 the greater of the two remaining terms next it, on the left, for 
 the second term of the proportion, and the less number for the 
 first ; but if it is to be less, place the less of the two remaining 
 numbers for the second term, and the greater for the first term ; 
 then multiply the second and third terms together for a dividend, 
 and divide their product by the first : the quotient will be the 
 fourth term, or answer, and of the same denomination as the 
 third term. 
 
 Note 1st. If the third term consists of different denominations, 
 it must be reduced to the lowest mentioned before stating, and 
 the fourth term will be of the same denomination ; but if either 
 the first or second terms be of different denominations, they 
 must both be reduced to the lowest mentioned, before stating. 
 
 Ex. 1. If 8 yards of cloth cost $3.20, what will 96 yards 
 cost ? 
 
 Since 8 yards cost money, 96 yards will cost money ; the 
 required term must, therefore, be money ; and must be a 
 number to which $3.20 will bear t}ie same ratio that 8 yards 
 
PROPORTION. 187 
 
 bears to 96 yards. $3.20 is, therefore, to be made the right 
 hand term. 
 
 The next inquiry is, which will cost most, 8 yards or 96 
 yards ? The answer to this inquiry places (in accordance with 
 the rule) the 96 in the second place, and the 8 in the first, thus : 
 8 : 96 : : 3.20 : what number dollars ? 
 
 PERFORMED. 
 
 8 : 9 6 : : 3.2 
 96 
 
 1920 
 2880 
 
 8)307.20 
 
 _^___^__ 
 
 $ 3 8.4 Ans. 
 Therefore, 8 yards : 96 yards : : $3.20 : $38.40. 
 
 2. If 9 men earn 72 dollars in a given time, how much will 
 24 men earn in the same time ? Statement : 9 men : 24 men 
 ::$72:Ans. $192.00. 
 
 Sums of this description may be solved with peculiar ease 
 by canceling. 
 
 RULE FOR CANCELING. Notice which of the given terms is 
 of the same kind or name as the required answer, and place 
 it above a horizontal line, towards the left. Notice again 
 whether the required term must be greater or less than this ; and, 
 if greater, place the greater of the two remaining terms on the 
 right of the preceding term, and also above the line, and the 
 less of the two terms below the line ; but if less, place the less 
 of the remaining terms above the line, and the greater below it ; 
 then cancel, multiply, and divide as before directed. 
 
 Note 2d. In arranging the terms as directed for canceling, the 
 number placed first above the line, is the third term of the pro- 
 portion, and that standing on the same side, on the right of this, 
 is the second, and the number standing below the line, the first. 
 
 3. If 12 horses consume 42 bushels of oats in a given time, 
 how much will 20 horses consume in the same time ? 
 
188 PROPORTION. 
 
 Statement, -jW~* 
 
 The answer required is obviously the oats that 20 horses 
 would consume. It is also evident that 20 horses would 
 consume more that 12. Hence the reason of the above state- 
 ment. 
 
 7. 10 
 
 rri> i A *& %& 
 
 i ne same canceled : , 
 
 and 7 X 10 = 70, the bushels required. 
 
 It will be recollected that the numbers above the horizontal 
 line form a dividend, and those below the line, a divisor. Hence, 
 42, the number of bushels consumed by 12 horses, may be re- 
 garded as divided by 12. This division would give the quan- 
 tity of oats which one horse would consume in the given time, 
 and this quantity multiplied by 20 would give the quantity 20 
 horses would consume in the same time. 
 
 4. If 72 yards of cambric cost $119.44, what will 9 yards 
 cost ? 
 
 If $1 19.44 be divided by 72, the quotient will be the price of 
 one yard, and this price multiplied by 9, will be the required 
 answer. 
 
 Therefore, 1 -^. 
 
 Canceled, j% 5 and 11 9.44-^8 =14. 93, Ans. 
 
 5. If 10 shillings pay for 20 pounds of beef, how many 
 pounds may be bought for 5 shillings? Ans. 10. 
 
 Statement, '~^- 
 
 6. If 3 Ib. of sugar cost 4 s. how much will 6 Ib. cost ? 
 
 Ans. 8 s. 
 
 Statement, - L ^-. 
 o 
 
 7. If 12 bushels of wheat are worth $16, what is the value 
 of 48 bushels ? Ans. $64. 
 
 8. Sold 12 yards of cloth for $72 ; what is the value of 5 
 yards, at the same rate ? Ans. $30. 
 
 9. What is the value of 16 cords of wood, if 48 cords are 
 worth $120? Ans. $40. 
 
 10. If 16 cords of wood are worth $40, what is the value 
 of 48 cords? Ans. $120. 
 
PROPORTION. 189 
 
 11. What is the value of 4 gallons of wine, if 108 gallons of 
 the same kind are worth $324? Ans. $12. 
 
 12. If 8 yards of cloth cost $3.20, how many yards of the 
 same kind may be bought for $38.40 ? Ans. 96 yards. 
 
 13. Paid 75 cents for 7 Ib. of sugar ; how many pounds of 
 the same kind may be bought for $6 ? Ans. 56 pounds. 
 
 14. If 7 men can accomplish a piece of work in 12 days, 
 how many men are required to accomplish the same in 3 days ? 
 Ans. 28 men. 
 
 15. If a ship sail 24 miles in 4 hours, in how many hours 
 will she sail 150 miles, if she continue at the same rate? 
 Ans. 25 hours. 
 
 16. If 17 men perform a piece of work in 25 days, in how 
 many days would 5 men perform the same ? Ans. 85 days. 
 
 17. A staff', 4 feet long, casts a shadow 6 feet; another staff, 
 placed in the same situation, casts a shadow 58 feet ; what is 
 its length? Ans. 38f feet. 
 
 18. A garrison has provision for 8 months, at the rate of 15 
 ounces per day; what must be each man's daily allowance, in 
 order that the same provision may last them 1 1 months ? 
 Ans. lO^J ounces. 
 
 19. When a quarter of wheat affords 60 ten-penny loaves, 
 how many eight-penny loaves may be made from the same ? 
 Ans. 75. 
 
 20. If $10 worth of provision serve 7 men 4 days, how 
 many days will the same provision serve 9 men ? Ans. 3^- 
 days. 
 
 21. If 12 gallons of wine cost $30, what is the value of 56 
 gallons, at the same price per gallon ? Ans. $140. 
 
 22. If 15 pounds of sugar cost $1.20, how many pounds 
 may be bought for $38 ? Ans. 475 pounds. 
 
 23. If a staff, 4 feet long, casts a shadow 7 feet long, what 
 is the height of a steeple, whose shadow at the same time 
 measures 1 98 feet ? Ans. 1 1 3| feet. 
 
 24. If a pole, 6 feet long, casts a shadow 10 feet on level 
 ground, what would be the length of a shadow from a steeple 
 72 feet high, at the same time ? Ans. 120 feet. 
 
 25. If 12 men build a house in 48 days, in how many days 
 can 36 men do the same work? Ans. 16 days. 
 
 26. If 100 men can finish a piece of work in 12 days, how 
 many men will be required to do the same in 4 days ? Ans. 
 300 men. 
 
 27. How many men must be employed to complete in 15 
 days what 5 men can do in 24 days ? Ans. 8 men. 
 
190 PROPORTION. 
 
 28. If a man perform a journey in 3 days, when the days 
 are 1 6 hours long, how many days of 1 2 hours each, will he 
 require to perform the same, if he continue to travel at the 
 same rate ? Ans. 4 days. 
 
 29. If 48 men can build a wall in 24 days, how many men 
 can do the same in 192 days ? Ans. 6 men. 
 
 30. In how many days will 8 men finish a piece of work 
 which 5 men can do in 24 days ? Ans. 15 days. 
 
 31. In what time will $600 gain $50 interest, if $80 gain 
 it in 15 years ? Ans. 2 years. 
 
 32. When $100 principal will gain $6 in 12 months, what 
 principal will gain the same in 8 months ? Ans. $150. 
 
 33. How many yards of cloth 3 qr. wide, are equal to 3,0 
 yards, 5 qr. wide ? Ans. 50 yards. 
 
 34. How many yards of paper l yard wide, will be suffi- 
 cient to hang a room 20 yards square, and 4 yards high? Ans. 
 256 yards. 
 
 35. If a board be 9 inches wide, how much in length will 
 make a square foot ? Ans. 16 inches. 
 
 36. What quantity of shalloon 3qr. wide, will line 7| yards 
 of cloth, 1-| yards wide ? Ans. 15 yards. 
 
 37. Lent a friend $100 for 6 months ; afterwards he lent me 
 $75 ; how long ought I to retain it to balance my favor, allow- 
 ing to each the same rate per cent, of interest ? Ans. 8 
 months. 
 
 38. In what time will $858 gain as much as $286 will gain 
 in 12 months ? Ans. 4 months. 
 
 39. If 375 cwt. may be carried 660 miles for a given sum, 
 how many cwt. may be carried 60 miles for the same money 1 
 Ans. 4125. 
 
 40. If 10s. worth of wine will suffice for 46 men, when a 
 tun is worth $240, for how many will the same 10s. suffice, 
 when a tun costs $160 ? Ans. 69 men. 
 
 41. If 5 yards of muslin that is H yards wide, will make 
 a dress, how many yards of lining will be required, that is but 
 3 qr. wide ? Ans. 1 1 yards. 
 
 42. If 40 rods in length and 4 in breadth make an acre, 
 what is the width of a piece of ground containing the same 
 quantity, that is 24 rods in length 1 Ans. 6 rods, 3 yards, 2 
 feet. 
 
 43. An insurance company, consisting of 82 persons, sus- 
 tains a loss, of which each man's share was $12 ; wh;i.t would 
 their shares have been, had the company consisted of only :}-2 
 persons? Ans. $30.75. 
 
PROPORTION. 191 
 
 44. If a hogshead of wine which cost $180, afford a hand- 
 some profit, when retailed at $4 per gallon, how must another 
 be retailed, which cost $196, to gain the same percent.? 
 Ans. $4.355.+ 
 
 45. A lot of ground was walled in by 16 men in 6 days ; 
 the same being demolished, is required to be rebuilt in 4 days ; 
 how many men must be employed ? Ans. 24 men. 
 
 46. A person by traveling 12 hours per day, performs a 
 journey of 800 miles in 32 days ; how many days will he re- 
 quire to perform the same journey, if he travel 15 hours per 
 day? A?is. 25 J days. 
 
 47. If 1800 cwt. may be carried 64 miles for a given sum, 
 how far may 225 cwt. be carried for the same money? Ans. 
 512 miles. 
 
 48. If 50 gallons of water fall into a cistern of sufficient 
 capacity to contain 230 gallons, in one hour, and by a pipe 35 
 gallons be drawn off in the same time, how long will it take to 
 fill the cistern ? Ans. 15 hours, 20 minutes. 
 
 49. If 150 Ib. of soap cost $15.60, what would 15 Ib. cost ? 
 Ans. $1.56. 
 
 50. How many yards of cloth 3 qr. wide are equal to 39 
 yards 5 qr. wide ? Ans. 65. 
 
 51. A cistern has a pipe by which it may be emptied in 10 
 hours ; how many pipes of the same capacity will empty it in 
 30 minutes ? Ans. 20 pipes. 
 
 It has already been said, that if the given terms (see Note 
 1st) are of different denominations, they must be reduced be- 
 fore stating the sum. This labor is, for the most part, saved, 
 whenever the question is solved by canceling. Take the fol- 
 lowing sums, in which it is required to find the value of a 
 quantity in one denomination, the price of some other denom- 
 ination being given. 
 
 RULE. Write the quantity the price of which is required, 
 above a horizontal line ; then (if the price of a lower denomina- 
 tion be given) on the right of this, also above the line, place the 
 numbers required to reduce the given quantity to that denomina- 
 tion, together with the price of the same denomination ; then be- 
 low the line, write such numbers as will reduce the given price 
 to the required denomination. But if the price of a higher de- 
 nomination be given, and that of a lower denomination be re- 
 quired, place the quantity the price of which is required, as 
 
192 PROPORTION. 
 
 before, and write the numbers necessary to reduce that quantity 
 to the denomination of which the price is given, below the line ; 
 then, lastly, place the quantity the price of which is given below, 
 and its price above the line. Solve the statement by canceling. 
 
 Note 3d. If either the given quantity or price be of differ- 
 ent denominations, they may be reduced to the lowest given, 
 before stating ; or, if preferred, the lower denominations may 
 be reduced to a decimal. 
 
 Ex. 1. How many pounds sterling will 3 cwt. of sugar cost, 
 at 20 pence per pound ? 
 
 The price of 3 cwt. is required, and that of one pound is 
 given. The given price is also in pence, and pounds sterlino- 
 
 . , ' 3. 4. 28. 20 . , 
 are required. Hence, i2~20' 1S required statement. 
 
 The numbers 4 and 28 above the line, are required to reduce 
 the 3 cwt. to pounds, and these pounds multiplied by 20, will 
 give the price of the whole quantity in pence. If, then, these 
 pence be divided by 12 and 20, they will be reduced to pounds 
 sterling. 
 
 . . i i S. 4. 28. 2ft 
 Statement solved, -. 
 
 Therefore, 28 . is the required answer. 
 
 2. How many pounds sterling will 3 pipes of wine cost, at 
 10 s. a gallon ? 
 
 . 3. 2. 63. 10 
 Statement, ^ . 
 
 In this statement, the 3 and 63 above the line are the numbers 
 required to reduce pipes to gallons ; then, the gallons multiplied 
 by 1 0, will give the cost in shillings ; and, lastly, the shillings 
 divided by 20, (the number below the line,) will be reduced to 
 pounds. 
 
 Statement canceled, 3 - 2 - g. 1^ 63 X 3 = 189 . Ans. 
 
 3. How many dollars, New York currency, will 12 cwt. 
 of sugar cost, at lOd. a pound? Ans. $140. 
 
 Statement, ^ ^ . (8 s. = $ 1 , New York currency. ) 
 
 4. How many dollars, New England currency, will 2 hogs- 
 heads of wine cost, at 6d. a pint? 
 
PROPORTION. 193 
 
 Statement, 2 " 63 '^' 2 Q 6 . Ans. $84. ($6s. = l N. E. currency.) 
 
 5. At 15 pence a pound, what will 1 cwt. of loaf sugar cost 
 in dollars, New England currency ? Ans. $23.333.+ 
 
 If it is preferred to solve sums of this kind without cancel- 
 ing, it may be done by the following rule : 
 
 RULE. Reduce the given quantity to that denomination, the 
 price of which is given, and multiply it by the price ; then divide 
 by such numbers as are required to reduce the value obtained to 
 the required denomination. 
 
 6. How many dollars, New York currency, will 6 cwt. of 
 sugar cost at 1 d. a pound ? 
 
 SOLUTION. 
 6 
 4=qr. in cwt. 
 
 2 4 
 
 2 8=:lbs.in qr. 
 
 1 92 
 
 48 
 
 672 
 
 1 0=price of 1 Ib. 
 
 12)672 Oncost in pence. 
 
 8)56 Oncost in shillings. 
 
 7 Orr dollars, answer. 
 7 
 Canceled, H^. 10x7=70, answer. 
 
 135. a. 
 S X 
 
 7. How many dollar^ will 53 ells English cost, at 8 s. New 
 York currency, per yard ? Ans. $66.25. 
 
 8. If I purchase melasses at 1 s. 3 d. per quart, how much in 
 pounds, shillings, and pence, will 12 hogsheads of the same 
 kind cost? Ans. 189. 
 
 9. If I purchase 16 cwt. of steel for $156, what will 1 qr. 
 of a cwt. cost, at the same rate ? 
 
 17 
 
194 PROPORTION. 
 
 39 
 
 Statement, i^yjl. Canceled,^*; 39 -f- 16 = $2. 437.+ 
 
 10. What cost 9 cwt. of sugar at 10 pence per pound ? Ans. 
 42 . 
 
 11. What cost 12 cwt. of sugar at 9 pence per pound ? Ans. 
 50 . 8s. 
 
 12. What cost 42 cwt. of sugar at 3s. 8d. per pound? 
 Ans. 862 . 8 s. 
 
 13. What would 480 yards cost, in federal money, at 2 pence, 
 New York currency, per yard ? 
 
 480 2 
 Statement, ' ^ Q . Ans. $10.+ 
 
 For the solution of the following sums, see the table of Cur- 
 rencies, given in Reduction of Currencies. 
 
 14. What would 862 yards cost, in federal money, at 3 pence 
 per yard, New England currency 1 
 
 8fi2 *} 
 Statement, --J-Q> Ans. $35.916.+ 
 
 15. What would 920 yards cost, in federal money, at 4^ 
 pence per yard, New Jersey currency ? Ans. $46. 
 
 16. What would 988 pounds of rice cost, in federal money, at 
 6 d. per pound, South Carolina currency ? Ans. $105.857.+ 
 
 17. What would 899 gallons of vinegar cost, in federal 
 money, at 8 d. per gallon, New York currency ? Ans. $74.- 
 916.+ 
 
 18. How much will 672 yards cost, in federal money, at 6s. 
 6 d. New York currency, per yard ? Ans. $546. 
 
 19. What will 1000 yards of ribbon cost, in federal money, 
 at 3 s. 4 d. New England currency, per yard ? Ans. $555.- 
 555.+ 
 
 20. How many dollars will 123 yards of cloth cost, at 10s. 
 New Jersey currency, per yard ? Ans. $164. 
 
 21. What will 687 yards of cloth cost in federal money, at 
 5 s. South Carolina currency, per yard ? Ans. $736.071.+ 
 
 22. How many dollars and cents will 127 gallons of wine 
 cost, at 3 s. 4 d. New England currency, per gallon ? Ans. 
 $70.555.+ 
 
 23. How many dollars will pay for 14 cwt. 3qr. of hay, 
 at 15 s. 8 d. New York currency, per cwt. ? Ans. $28.885. + 
 
 24. How many dollars will pay for 12 cwt. 2 qr. of cheese, 
 at 2.C. 15s. New Jersey currency, per cwt.? Ans. $91.666.+ 
 
PROPORTION. 195 
 
 25. What will 12 cwt. 7 Ib. of brown sugar cost, at 6d. New 
 Jersey currency, per pound 1 Ans. $90. 066. -f- 
 
 26. A. owes B. 1 138 . but can pay only 15 s. on a pound ; 
 what will B. receive ? Ans. 853 . 10s. 
 
 27. C. has. a journey of 75 leagues to perform. In what 
 time will he complete it, if he travel 30 miles a day. Ans. 
 7|- days. 
 
 28. How long will it take to travel one fourth round the 
 earth, at the rate of 36 miles a day ; the whole circumference 
 being 360 degrees, and one degree 69| miles ? Ans. 173| 
 days. 
 
 29. If 9 Ib. of coffee cost 27 s. how many dollars, at 6s. 
 each, will 45 Ib. cost ? Ans. $22.50. 
 
 30. If I buy 20 pieces of cloth, each 20 ells, for 12 s. 6 d. 
 per ell, how many dollars, at 8 s. will pay for the same 1 Ans. 
 $625. 
 
 31. How many dollars will 7 casks of prunes cost, each 
 weighing 4 cwt. 2 qr., at 2 . 19 s. 8 d. New York currency,, 
 per cwt. ? Ans. $234.937. + 
 
 32 . A vessel at sea discharges a cannon, the report of which 
 reaches me in 1 minute, 30 seconds. How far distant is she, 
 allowing sound to travel 1142 feet in a second? Ans. 19 
 miles, 3 furlongs, and 29 T J T rods. 
 
 33. How many yards of flannel, 1 yard wide, will line 125 
 yards of broadcloth, 1 ell English wide ? Ans. 156^. 
 
 34. How many yards of cloth may be bought for $37.62, if 
 of a yard cost 66 cents ? Ans. $42 yd. 3 qr. 
 
 35. How many dollars will 28 yards of linen cost, at 5 s. 6 
 d. New England currency, per yard ? Ans. $25.666. -f- 
 
 36. Bought 32 yards of muslin, at 6 s. 8 d. New York cur- 
 rency, per yard. What was the cost, in federal money ? Ans. 
 $26.666.+ 
 
 37. How many gallons of wine may be purchased for $45, 
 at 6 s. New York currency, per gallon ? Ans. 60 gallons. 
 
 QUESTIONS. What is ratio 1 What is the former of the two num- 
 bers between which the ratio exists called? What is the latter 7 ? How 
 is the direct ratio of any two numbers obtained 1 How is the inverse 
 ratio obtained 1 Between what quantities only, does ratio exist 1 How 
 is simple ratio expressed 1 How is the ratio of any couplet affected by 
 multiplying or dividing both the antecedent and the consequent by 
 the same. number 1 In what two ways may the ratio of any two num- 
 bers be multiplied'? In what two ways is the ratio of any couplet 
 divided by any number 1 When two or more ratios are multiplied 
 together, what is the resulting ratio called 1 ? What does it equal? 
 
196 COMPOUND PROPORTION. 
 
 What is Note 1st 1 What constitutes proportion? How many equal 
 ratios are required for a statement of proportion 1 What terms must be 
 of the same kind? When any four terms are proportional, what are the 
 first and fourth called 1 And what are the second and third called 1 
 How does the product of the extremes compare with the product of the 
 means ? When any of the four terms are wanting, ^explain how it 
 may be found. In what do operations in Simple Proportion 
 consist 1 What is the rule 1 What note follows the rule ? What is 
 the rule for canceling 1 What is Note 2d 1 What is the rule, when it 
 is required to find the value of a quantity in one denomination, the 
 price of some other denomination being given 1 What is Note 3d 1 
 What is the rule for solving sums of this kind without canceling ? 
 
 COMPOUND PROPORTION. 
 
 Simple Proportion consists of two equal ratios. 
 
 Compound Proportion is that in which the relation of one of 
 the given quantities to a required quantity of the same name, 
 is traced through two or more simple proportions. 
 
 The smallest number of terms of which a statement in com- 
 pound proportion can consist, is five. Of these terms, one is 
 always of the same name as the answer required ; and the 
 others are always two of a kind. The following sum will 
 serve as an illustration : 
 
 If 3 men, in 4 days, spend $5, how many dollars will 6 men 
 spend in 12 days ? 
 
 In the above sum, there are five terms given, viz. two of 
 men, two of days, and one of dollars ; and dollars are also 
 required for the answer ; so that when the sixth term is found, 
 the sum may be resolved into three simple ratios, the third of 
 which is a compound of the preceding two. These ratios are 
 3 men : 6 men, and 4 days : 12 days, and $5 : the required number 
 of dollars. Now it is obvious, that the ratio of $5 to the required 
 number of dollars, is not the same as the ratio of 3 men to 6 men ; 
 for then no regard would be paid to the time, and the solution 
 would be effected on the supposition that one man or a 
 number of men would spend as much in one day, as in any 
 given number of days. Nor is it the same as the ratio of 4 
 
COMPOUND PROPORTION. 197 
 
 days to 12 days, for then the supposition would be, that 3 men 
 would spend as much as 6, or any number of men. But it is 
 a ratio compounded of the two ratios, viz. the ratios of 3 men 
 : 6 men, and of 4 days : 12 days. The ratio of 3 : 6=2, and 
 $5 x 2 = $ 1 0. This would double the given quantity of money. 
 Again, the ratio of 4: 12, is 3 ; this would treble the sum last 
 obtained, viz. $10, and would give 10x3 = $30, which is the 
 answer to the above question. Now, it will be observed, that 
 the $5 in the above operation was multiplied by the product 
 of the two simple ratios ; for 3 : 6=2, and 4 : 12 = 3 ; there- 
 fore, 2 x 3 = 6, the product of the simple ratios, and $5x6 = 
 $30, Ans. 
 
 The same result would have been obtained, by multiply- 
 ing the $5 by the consequents, or latter terms of each ratio, 
 and dividing their product by the product of the ante- 
 cedents, or former terms of the same ratios. Thus, 3 : 6 and 
 4 : 12, are the given ratios of which 6 and 12 are consequents; 
 therefore, 6 x 12 x 5 = 360, and 3 X 4 = 12, the product of the 
 antecedents; hence, 360 12 = 30 dollars, the same answer 
 as before. Hence, we have the following rule: 
 
 RULE. Write the number which is of the same kind as the 
 answer required, for the third term. Of the remaining terms, 
 take any two of the same name, and arrange them as directed 
 in Simple Proportion ; then any other two of a kind, and so on 
 till the terms are all taken. Lastly, multiply the product of the 
 second terms by the third term, and divide the last product by the 
 product of the first terms, and the quotient will be the required 
 term or answer. 
 
 Ex. 1 . If 4 men build a wall ten feet long, 3 feet high, and 
 2 feet thick, in 6 days, in what time will 12 men build one 
 100 feet long, 4 feet high, and 3 feet thick? The question 
 asked is, in what time will the work be done ; therefore, by the 
 rule, 6 days is the third term. 
 
 12 men : 4 men, ~| 
 
 10 feet length : 100 feet length, [ 
 
 3 feet high: 4 feet high, ' f : ' 6 days. 
 
 2 feet thick : 3 feet thick, J 
 
 It is obvious that 12 men would require less time to execute 
 a given piece of work than 4 men, and also that a wall 100 
 17* 
 
198 COMPOUND PROPORTION. 
 
 feet long, 4 feet high, and 3 feet thick, would require more time 
 than a wall 10 feet long, 3 feet high, and 2 feet thick. 
 
 The same solved : 
 12:4 
 
 ; 4x100x4x3x6=28800; and 12x10x3 
 X 2=720 ; and 28800-4-720=40 days, the 
 number required. 
 
 The work of this sum maybe much abbreviated by canceling. 
 The statement would be, 6 " *' * *' * 
 
 1.6. 1U. O. xJ 
 
 Canceled, ?i^A|i|, 10x4=40, 
 
 For the preceding mode of operation, we have the following 
 rule : 
 
 RULE FOR CANCELING. Write the number, which is of the 
 same name as the answer required, above a horizontal line, 
 towards the left. Then take each two terms of the same name, 
 and compare them with this number, and notice whether more or 
 less be required, as in Simple Proportion : if more be required, 
 place the greater of the two numbers above the line, and the 
 less below ; but if less be required, place the less of the two 
 above the line, and the greater below. Cancel, fyc. 
 
 Ex. 2. If 4 men in 12 days mow 48 acres of grass, how 
 many acres will 8 men mow in 1 6 days 1 
 
 A Q ^ 
 
 Common statement, ; ' v : ; 48. 
 
 Statement for canceling, 48 ' S ' *|*. 
 4 2 
 
 Q "G 1 
 
 Canceled, L -^ 1 -^' 4x2 x 16 = 128, the number of acres* 
 required. 
 
 Note 1st. If the sixth term, or answer, be placed below 
 the horizontal line, the other term remaining as before, the 
 product of the terms standing above the line, will just equal 
 the product of those standing below, thus : 
 48X8X16=6144 
 128X4X12=6144' 
 
PROPORTION. 199 
 
 3. If 12 horses eat 20 bushels of oats in 16 days, how 
 many bushels will 24 horses eat in 48 days ? 
 
 Statement, -y!^| Ans. 120 bushels. 
 
 4. If 18 men, in 32 days, consume 128 Ibs. of bread, how 
 many pounds will 12 men consume in 64 days, their daily 
 allowance being the same ? 
 
 1 0Q 1 RA 
 
 Statement, \' "v A. 170}. 
 ib. 0*4 
 
 5. If 8 men receive 4 dollars for 3 days work, how many 
 days must 20 men work, to earn 40 dollars, if they receive 
 the same daily allowance ? Ans. 12. 
 
 Statement, ' -. 
 
 To understand this statement, it should be remembered, that 
 20 men would require less time for the accomplishment of an 
 object, than 8 men ; and also that 40 dollars would employ, 
 for a given time, more men than 4 dollars. 
 
 6. If 4 men receive $3.20 for 3 days work, how many 
 men, if they receive the same per day, will eam $12.80, in 
 16 days ? Ans. 3 men. 
 
 4. 12.80. 3 
 btatement, Too~Tfi" 
 
 7. How much ought 60 men to receive for 25 days work, 
 if 12 men, under the same circumstances, receive 50 dollars 
 for 4 days work ? Ans. $1562.50. 
 
 8. If 16 men cut 112 cords of wood in 7 days, how many 
 cords will 24 men cut in 19 days ? Ans. 456 cords. 
 
 9. If the transportation of 12 cwt. 150 miles, cost 75 shil- 
 lings, for how many dollars, at 8 s. each, may 6 cwt. be 
 transported 45 miles ? Ans. $1.406.+ 
 
 10. If the freight of 9 hhds. of sugar, each weighing 12 
 cwt. 20 leagues, cost $38.40, what will be the expense of 
 transporting 50 casks of the same, each weighing 2 cwt. 2 
 qr., 100 leagues? Ans. $222.22.+ 
 
 11. If 100 dollars, in 1 year, gain 6 dollars interest, how 
 much will 200 dollars gain in 26 weeks. Ans. 6 dollars. 
 
 12. If 350 dollars, in 9 mdtaths, gain 15 dollars, what 
 principal will gain 6 dollars in 12*nonths ? Ans. 105 dollars. 
 
 13. If 8 men can build a wall,"2) feet long, 6 feet high, and 
 4 feet thick, in 12 days, in what time can 24 men build one, 
 200 feet long, 8 feet high, and 6 feet thick ? Ans. 80 days. 
 
 14. A wall, 32 feet high, and 40 feet long, was built in 8 
 
200 PROPORTION. 
 
 days, by 145 men; in how many days would 68 men build 
 another wall, 28 feet high, and of the same length, allowing 
 each man to perform an equal portion of labor, in the same 
 time? Ans. 14||- days. 
 
 15. What must be paid for the transportation of 56 bags of 
 coffee, each weighing 3 qr. 16 lb., 66 miles, if 14 bags, 
 weighing each 125 lb., be carried 6 miles for $6.25. Ans. 
 220 dollars. 
 
 16. If 6 persons can earn 120 . in 21 weeks, how much 
 will 14 persons earn, in 46 weeks, if they receive the same 
 per week? Ans. 613 . 6 s. 8 d. 
 
 17. If, when the days are 14 hours long, a person perform a 
 journey of 276 miles in 16 days, in what time will he travel 
 860 miles, when the days are 12 hours long ? Ans. 58^ 4 T 
 days. 
 
 18. If 960 dollars defray the expenses of 20 men 88 
 weeks, for how many weeks will $1440 defray the expenses 
 of 48 men, if they spend at the same rate ? Ans. 55 weeks. 
 
 19. If 100 jC. gain 6 . in 12 months, what principal, at the 
 same rate per cent., will gain 3 . 7s. 6 d. in 9 months ? 
 Ans. 75 . 
 
 20. If a man travel 240 miles in 12 days, when the days 
 are 12 hours long, in how many days, of 16 hours, will he 
 travel 720 miles, if he travel at the same rate ? Ans. 27 days. 
 
 21. What is the interest of $650 for 36 weeks, at 6 per 
 cent, per annum ? A?is. $27. 
 
 22. If $150, in 12 months, gain $8 interest, what will 
 $400 gain in 4 months ? Ans. $7.11.+ 
 
 23. If 3 men lay 144 square yards of pavement in 7 days, 
 how many square yards will 12 men lay in 49 days ? Ans. 
 4032 square yards. 
 
 24. If 200 lb. be carried 40 miles for 40 cents, how far 
 may 202000 lb. be carried for $60.60 ? Ans. 60 miles. 
 
 25. Inwhattime will200. gain 6. if 100. gain 6. in 
 52 weeks ? Ans. 26 weeks. 
 
 26. In what time will 400 . gain 96 . interest, if 350 . 
 gain 10 . 10s. interest in 6 months ? Ans. 4 years. 
 
 27. If 4 men mow 48 acres of grass in 12 days, in what 
 time will 8 men mow 128 acres ? Ans. 16 days. 
 
 28. If I receive 88. 17 s. 4 d. for the principal and int< 
 
 of 86 . for 8 months, what is the rate per cent, of interest ? 
 Ans. 5 per cent. 
 
 29. What will the transportation of 7 cwt. 2 qr. 25 lb. 64 
 
PROPORTION. 201 
 
 miles cost, if 5 cwt. 3 qr. be carried 150 miles for $24.58 ? 
 Ans. $14.086.+ 
 
 30. If I pay $50.25 for the tuition of 2 boys, 3 quarters each, 
 what will the tuition of 100 boys amount to, in 7 years, at 
 the same rate? Ans. $25125. 
 
 31. Purchased goods to the amount of 750 . and sold the 
 same, six months after, for 825 . ; what did I gain per cent. ? 
 Ans. 20 per cent. 
 
 32. If 56 Ib. of bread be sufficient for 7 men 14 days, 
 how long will 36 Ib. suffice for 21 men ? Ans. 3 days. 
 
 33. A person having engaged to carry 8000 cwt. a certain 
 distance in 9 days, removed 4500 cwt. with 18 horses, in 
 6 days ; how many horses were required to remove the re- 
 mainder, in the 3 remaining days ? Ans. 28 horses. 
 
 34. If 3 men perform a piece of work in 20 days, how 
 many men will accomplish 4 times as much work in 4 days ? 
 Ans. 60 men. 
 
 35. If 4 men, in 5 days, eat 6 Ib. of bread, how many 
 pounds of the same will be sufficient for 16 men 15 days? 
 Ans. 72 Ib. 
 
 36. If it take 15 men 20 days to make 300 pairs of shoes, 
 how many men will be^ required to make 1200 pairs in 60 
 days ? Ans. 20 men. 
 
 37. A wall, which was to be raised to the height of 27 feet, 
 was raised 9 feet by 12 men, in 6 days ; how many men were 
 required to complete the work in 4 days, allowing, each man 
 to do the same amount of work per day ? Ans. 36 men. 
 
 38. If 6 men, in 21 weeks, earn $120, how much will 14 
 men earn in 46 weeks ? Ans. $613.33.+ 
 
 39. If 48 bushels of corn produce 576 bushels in one 
 year, what will be the product of 240 bushels in 6 succes- 
 sive years? Ans. 17280 bushels. 
 
 40. In how many days will 25 men reap 200 acres of grain, 
 if 12 men reap 80 acres in 6 days ? Ans. 7J, 
 
 41. If 20 men mow 208 acres, 1 rood, and 24 rods of grass, 
 in 24 days, how many men will cut down 8 times as much 
 grass in twice the time ? Ans. 80. 
 
 QUESTIONS. Of what does simple proportion consist? What is 
 compound proportion'? What is the smallest number of terms of which 
 a statement in compound proportion can consist 1 How must these 
 terms compare with each other, in kind? How many simple ratios 
 will there be in the example given for illustration, when the sixth term 
 is found 1 Of what is the third ratio a compound 1 What are the three 
 ratios in the illustration given? Give the entire illustration. What is the 
 
202 ANALYTICAL SOLUTIONS. 
 
 rule first given 7 What is to be the denomination of the third term 7 
 How are the other terms to be arranged? What is the rule for can- 
 celing 1 What term is to be placed first above the line 1 How are the 
 other terms to be arranged 1 What is the given note 1 
 
 SOLUTION OF ARITHMETICAL 
 PROBLEMS BY ANALYSIS. 
 
 An arithmetical question is solved analytically, when the 
 operation is guided entirely by the conditions embraced in the 
 question itself. 
 
 Take the following illustration : 
 
 Ex. 1. If 3 men perform apiece of work in 6 days, in what 
 time will 9 men perform the same labor ? 
 
 It is obvious that if it take 3 men 6 days to perform the 
 proposed labor, it will take one man 3 times 6, or 18 days, to 
 perform the same. But 9 men operating together, will per- 
 form 9 days work in one day ; consequently, they will do the 
 whole in 2 days ; for 18-^9=2, Ans. 
 
 2. If 21 men earn 63 dollars in a given time, how much 
 will 42 men earn in the same time ? 
 
 63-^21 =3, the number of dollars one man will earn in the 
 given time. Therefore, $42 x ^ $126, the answer required. 
 Or the ratio of 21 men-to 42 is 2 ; and, $63x2=r$126, the 
 same as before. 
 
 Solutions of this kind may, therefore, be effected by the 
 following general principle : Find the ratio of the two given 
 terms which are of the same kind, and by this ratio multiply 
 the term corresponding in kind with the one required. 
 
 3. If 42 men can make 3 rods of wall in a given time, 
 how much can 8 men make in the same time ? 
 
 The ratio of 42 men to 8 men is 42 : 8=-f^=-^j ; therefore, 
 Jy X 3=Jf =j of one rod, which is the distance required. 
 
 4. If a staff 4 feet long, cast a shadow 6 feet, how high 
 is that steeple, whose shadow measures 75 feet ? 
 
ANALYTICAL SOLUTIONS. 203 
 
 The ratio of 6 : 75 = 12, and 12 x4=50 feet, Ans. Or, 
 the shadow is 1^= as long as the staff; hence, 75-^3 = 
 25, and 25 X 2 = 50 feet, the same answer as before. 
 
 5. An express traveling at the rate of 60 miles per day, had 
 been absent 5 days, when a second express was dispatched on 
 the same rout, traveling 75 miles per day. How many miles 
 must the second travel to overtake the first ? 
 
 60x5 = 300, the whole number of miles traveled by the 
 first express before the second started, and consequently, the 
 number of miles the second had to gain. But the first travels 
 60, and the second 75 miles per day ; hence, 75 60 = 15, the 
 number of miles gained daily, by the second express. 15 
 miles are, therefore, gained in traveling 75 miles, consequent- 
 ly, one mile is gained in traveling 5 miles ; and since 300 miles 
 are to be gained, 300x5 = 1500 miles, answer. 
 
 6. If 6 men in 14 days earn 84 dollars, how much will 9 
 men earn in 1 1 days ? 
 
 $84 -H 6 = $14, the money one man will earn in 14 days, and 
 $14-^14 = $!, the wages of one man for one day; therefore, 
 $1 X 9 = $9, the money 9 men will earn in one day, and $9 X 
 11 = $99, the money 9 men will earn in 11 days. 
 
 7. If 6 persons spend $300 in 8 months, how much will be 
 sufficient for a family of 1 5 persons 20 months ? 
 
 300-^-6 = 50, and 50^-8 = $6|, the money spent by one per- 
 son in one month ; then, $6 X 15 X 20 = $1 875, answer. 
 
 8. If 12 men build 36 feet of wall in 9 days, how many 
 men would build 108 feet in 16 days ? 
 
 36-1-12 = 3, and 3-^-9=^, the distance built by one man in 
 one day ; and ^ x 16 = ^=5^, the distance one man would 
 build in 16 days; therefore, 108-^5|-=20, the number of 
 men required. 
 
 9. A merchant owning |- of a vessel, sold J of his share for 
 $1456. What was the value of the whole vessel? 
 
 | of f = &. If then, T 8 5 cost $1456, 1456^8 = $182, the 
 value of -J-L of the vessel ; hence, 182 x 15 = $2730, answer. 
 
 10. If J of a yard cost of a dollar, what will 40 yards cost ? 
 If f of a yard cost $J, J of a yard will cost ^ of that sum, 
 
 or $ of $^=$^j, and one yard or J will cost 5 times that 
 sum, or $ff ; therefore, $|f X 40 = $1-J = $58.333,+ Ans. 
 
 11. If 240 men perform a piece of work in 8 months, how 
 many men must be employed to finish the same work in 2 
 months ? 
 
 The ratio of 2 : 8=4, and 240 X 4=960 men, answer. 
 
204 ANALYTICAL SOLUTIONS. 
 
 APPLICATION OF CANCELING TO ANALYTICAL SOLUTIONS. 
 
 12. If 8 pounds of tea cost $12, what will 32 pounds cost ? 
 
 12 32 
 
 Statement, -=.. 
 o 
 
 The two terms of the same name here given are 8 and 32, 
 and their ratio is 4, and is obtained by 
 
 4 
 
 12 &J 
 Canceling, '- -. 
 
 Therefore, $12x4 = $48, answer. The third term is here 
 multiplied by the ratio of the first and second, as required for 
 analytical solution. The terms are also canceled and multi- 
 plied as directed by the rule for canceling. 
 
 13. If 16 horses consume 84 bushels of grain in 24 days, 
 how many bushels will suffice 32 horses 48 days ? 
 
 In the preceding sum, it is evident that the given quantity of 
 grain is to be increased by the ratios of 16 horses to 32 horses, 
 and of 24 days to 48 days. Hence, 
 
 Q A oo A Q 
 
 ' ' is the statement expressing those ratios. 
 
 2 2 
 
 Canceled, ^' ^, 84x2x2=336, 
 the number of bushels required. 
 
 We therefore see by the above example, that the effect of the 
 operation is to increase the quantity of the same name as the 
 required quantity, by all the given ratios. The same is true in 
 all cases, that is, every statement for canceling is a complete 
 analysis of the question under consideration. 
 
 14. If 8 men build 9 feet of wall in 1 2 days, how many men 
 must be employed to build 36 feet in 4 days ? 
 
 8. 36. 12 
 statement, -? 
 
 The number of men required will obviously depend on the 
 the ratios 9 : 36, and 4:12, the former of which is 4, and the 
 latter, 3. Therefore, 8 men x 4 X 3 = 96 men, the number re- 
 quired. The above statement canceled, gives the same result, 
 thus: 
 
 4 3 
 
 r-^?, 8X4X3 = 96 men. 
 
ANALYTICAL SOLUTIONS. 205 
 
 The same number would have been obtained, had the num- 
 bers been canceled in any other order ; thus, 
 
 *L^J|, and 8x12 = 96. 
 s. ^* 
 
 Hence we perceive, that in a correct solution of any sum 
 by canceling, a complete analysis of that sum is given. 
 
 15. If 10 men make 300 pairs of boots in 20 days, how 
 many men must be employed to make 450 pairs in 30 days ? 
 Ans. 10 men. 
 
 ~ 4 k 10. 450. 20 
 
 Statement, -- m ~^. 
 
 If 10 men make 300 pairs in 20 days, they would make 15 
 pairs in one day ; and if 10 men make 15 pairs in one day, one 
 man would make one and a half pairs per day ; and in 30 days, 
 he would make 45 pairs ; therefore, 450 -^-45 = 10, Ans. 
 
 16. If of a yard cost f of a pound sterling, what will f of a 
 yard of the same cloth cost ? 
 
 If \ yard cost f of a pound, the whole yard would cost J of 
 a pound, and of the same would cost of J of a pound =-$ 
 of a pound ; consequently, f- would cost 5 times that sum, or | 
 = | of a pound, or 15s. Ans. 
 
 17. If T 7 g- of a house cost 49 pounds, what will be the value 
 
 of ^ of tne same ? Ans - 10 - 10 s - 
 
 1 8. A merchant bought a number of bales of velvet, each 
 containing 129j-f yards, at the rate of $7 for 5 yards, and sold 
 the same at the rate of $11 for 7 yards, and gained $200 by 
 the transaction. How many bales were there ? 
 
 He paid J- of a dollar per yard, and received L 1 of a dollar 
 for the same. Hence, L 1 s^ff ft = 3T of one dollar, the 
 amount gained on one yard. Therefore, $200-f- 6 5 = T-^- , 
 the whole number of yards; and 7 -g - ^-129i| 7 -W 
 
 19. If 7 horses consume 2f tons of hay in 6 weeks, how 
 many tons will 12 horses consume in 8 weeks ? Ans. 6^ tons. 
 
 20. If 14 men finish a piece of work in 42 days, how long 
 will it take 21 men to do it ? Ans. 28 days. 
 
 21. If f of a farm be valued at $895, what is the whole 
 farm worth ? Ans. $1611. 
 
 22. If 7 horses consume 29 bushels of oats in 5 weeks, 
 how many will 12 horses consume in 6 weeks ? Ans. 59Jf 
 bushels. 
 
 23. A merchant owning | of a vessel, sold f of his share 
 
 18 
 
206 PROPORTION IN FRACTIONS. 
 
 for $1200 ; what was the value of the whole vessel, at the same 
 rate? Ans. $1645.714.+ 
 
 24. There is a pole in the mud, f in the water, and 8 feet 
 out of the water. What is its length? Ans. 53^ feet. 
 
 25. In a certain orchard -5 of the trees bear apples, i pears, 
 i plums, 30 of them peaches, and 20 cherries. How many 
 trees does the orchard contain ? Ans. 600. 
 
 26. A certain school is classified as follows : -f^ study 
 grammar, f study geography, -f^ arithmetic, -^ write, and 9 
 learn to read. How many are there in all, and how many in 
 each study ? Ans. Whole number 80. In grammar 5, geog- 
 raphy 30, arithmetic 24, writing 12, and 9 read. 
 
 SIMPLE AND COMPOUND PROPORTION 
 IN FRACTIONS. 
 
 In stating such sums in Simple or Compound Proportion 'as 
 consist of fractions, it is only necessary to compare terms as 
 already directed, and then, if they are solved without cancel- 
 ing, having inverted the divisor, to divide the product of the nu- 
 merators by the product of the denominators. If, however, 
 they are to be solved by canceling, arrange the numerators of 
 the several fractions as directed to arrange whole numbers, 
 when whole numbers only are given, and place each denominator 
 opposite its own numerator. 
 
 Note. Before stating the sum, mixed numbers, if any are 
 given, must be reduced to improper fractions. 
 
 Ex 1 . If f of a yard cost -fj of a pound, what will -^ of a 
 yard cost ? 
 
 Statement, ==' ,V Q - 
 15. 14. o 
 
 The J is inverted, that its numerator may stand below the line' 
 as the same term would stand if it were a whole number. 
 
PROPORTION IN FRACTIONS. 207 
 
 Solution ' dhrl 
 
 2 
 Therefore, ^ of a pound, or 3 s. 4d., is the answer required. 
 
 2. If $ of a pound of sugar cost f of a shilling, what will 
 A- of a pound cost ? 
 
 894 
 Statement, 9 ' 1Q ' 3> Ans. 1 s. d. 3J qr. 
 
 3. A person who owned f of a vessel sold f of his share 
 for 375 . What was the value of the whole vessel, at the 
 same rate? Ans. 1000 . 
 
 These sums may all be solved analytically, if preferred. The 
 following is the solution of the last : | of f = JJ, and 375 . 
 -i-l5=25 ., or ^ of the whole value ; therefore, 25 .+40 
 Ans. 
 
 4. If f of a ship be worth 3740 ., what is the value of 
 the whole 1 Ans. 9973 . 6 s. 8 d. 
 
 5. If 1^ yards cost 9 shillings, what is the value of 16 
 yards ? Ans. 5 . 17 s. 
 
 6. What is the value of |- of a pound of lard, if -^ of a 
 pound cost of a shilling? Ans. 2J& pence. 
 
 7. A person who owned -| of a lot of land, sold f of his 
 share for $3024. What was the value of the whole lot, at the 
 same rate? Ans. $12096. 
 
 8. A certain vessel is valued at $1562.50. What is the 
 value of f of f of the same ? Ans. $500. 
 
 9. I owned f of a ship, and sold f of my share for $780. 
 What was the value of the whole, at the same rate ? Ans. 
 $3120. 
 
 10. A merchant bought 5J pieces of cloth, each containing 
 24f yards, for 6| shillings per yard. How many dollars did 
 the whole cost, in New York currency? Ans. $111. 
 
 11. A merchant had 4f cwt. of sugar, at 6^ pence per lb., 
 which he exchanged for tea, at 8| shillings per pound. How 
 many pounds of tea did he receive ? Ans. 29J. 
 
 12. How many pounds sterling will 150 yards of cloth cost, 
 at 1^ shilling per yard ? Ans. 9 . 
 
 13. If 3^ times 3 yards cost 1^ times Impounds sterling, 
 how many shillings and pence will ^ of of 12 yards cost ? 
 .4ns. 7s. 6 d. 
 
 14. What is the value of | of an ounce of silver, if 2 oz. be 
 valued at 12| shillings ? Ans. 4 s. 9 d. 
 
208 PROPORTION IN FRACTIONS. 
 
 15. What quantity of shalloon, that is of a yard wide, will 
 be sufficient to line 7 yards of cloth, l yards wide 1 Ans. 
 15 yards. 
 
 16. If 2^ yards, li yard wide, be sufficient to make a coat, 
 how much will it require of cloth that is | of a yard wide to 
 make the same kind of garment 1 Ans. 4 yd. 3^ T qr. 
 
 17. How many pieces of cloth, at $18f per piece, are equal 
 in value to 224f pieces, at $12 per piece ? Ans. $15Qi|. 
 
 18. A merchant exchanged 7$ cwt. of sugar, at 7 pence 
 per pound, for tea at 9^ shillings per pound ; how many pounds 
 of tea did he receive ? Ans. 60f Ib. 
 
 19. If 8 men can perform a piece of work in 6| hours, in 
 what time will 20 men do the same ? Ans. 2 hours, 40 min- 
 utes. 
 
 20. How many yards of cloth, f of a yard wide, will line 20 
 yards, f of a yard wide ? Ans. 12 yards. 
 
 21. How many pieces of cloth, at 18^ shillings per yard, 
 are equal in value to 350 pieces, at 12^ shillings per yard ? 
 Ans. 241|-J pieces. 
 
 22. Lent a friend $72f for 8f months ; what sum must he 
 lend me for 2 years, to balance the favor? Ans. $21.233.-}- 
 
 The following sums properly belong to Compound Propor- 
 tion. They may be solved either by canceling, by analysis, or 
 by the common rule of Compound Proportion. 
 
 Ex. 23. If | of a yard of cloth, which is |- of a yard wide, 
 cost J of a pound sterling, what is the value of f of a yard, 
 that is l^yard wide? 
 
 Analysis : | X f fj, the fraction of a square yard purchased, 
 which cost f of a pound sterling. Therefore, f + 2 1 =3^5 , the 
 value of Jj part of a square yard, and -j-|j X 32 =^3-, the price 
 of 1 yard, f X J=f|-, the quantity of which the price is requi- 
 red. Therefore, T 6 ^xffn:|||^=:| of 1 . Ans. = 13 s. 4d. 
 
 The same canceled, ^ ^ ]' *' g. 
 
 O. o. 4. o. / 
 
 To understand why the fractions | and are inverted, it 
 must be remembered that a fraction is divided by a fraction, by 
 inverting the divisor and then multiplying numerators and de- 
 nominators together. (See Sec. 7th, Introduction to Fractions.) 
 
 The above solved, ^ I' *' *' ** f of a . = 13 s. 4 d. Ans. 
 
 S. o.. 4. o. " 
 
 24. If 9 men spend 125. in 27 days, what sum will 25 
 men spend in 40 days ? 
 
CONJOINED PROPORTION. 209 
 
 Analysis : 12 1 . = *g ., and 2 -f +- 9 =f| . the money one 
 man spends in 27 days ; and ^| . -^-27=-^^-, the money spent 
 by one man daily. Therefore, -f/^ . x 25 = Jff . the money 
 25 men spend daily; and fff . X40=-?|^oo the sum of 
 money required, which reduced gives 51 . 8s. 9|-|- pence, 
 Ans. 
 
 *:- ,. 25. 25. 40 
 
 ihe same stated tor canceling, = . 
 
 20 
 
 Canceled, ^-^pf^ and 25x25x20 = 12500; and 27x9 
 i=243; and 12500^243 = 51 . 8s. 9ffd. Ans. 
 
 25. If 18 persons consume f^lb. of tea in one month, how 
 much will 8 persons consume in six months ? Ans. 4j Ib. 
 
 26. If the tuition of 2 boys for | of a year be 56| ., how 
 much will be the tuition of 3 boys for 5^ years ? Ans. 600 . 
 
 27. If 90 cwt. be carried 30 miles for $29, how many cwt. 
 may be carried 45 miles for $5f ? Ans. 12 cwt. 
 
 28. If 10 persons drink 15f gallons of wine in one week, 
 how much will 16 persons drink in 43 weeks ? Ans. 1073^ 
 gallons. 
 
 29. If 5 cwt. be carried 600| miles for $12$, how far may 
 | of a cwt. be carried for $30| ? Ans. 988 J y miles. 
 
 QUESTIONS. When is an arithmetical question solved analytically 1 
 What is the general principle by which sums may be solved analyti- 
 cally 7 How are sums in Simple or Compound Proportion solved 
 without canceling 1 How are they solved by canceling? What is 
 the note 1 
 
 CONJOINED PROPORTION. 
 
 Conjoined Proportion consists of a comparison instituted 
 between a series of terms bearing a certain relation to each 
 other, as the coins, weights, and measures of different coun- 
 tries. 
 
 The principle involved in this rule is the same as in Single 
 and Compound Proportion. No farther explanation is there- 
 fore needed, 
 
 19* 
 
210 CONJOINED PROPORTION. 
 
 RULE. Above a horizontal line near the left, place the de- 
 manding term ; then below the line, place the term of the same 
 name as the demand, with the term which it equals in value di- 
 rectly above it. Next, seek another term of the same name as 
 the one last placed, and set it also below the line, with the one it 
 equals in value also above it. Thus proceed to arrange the terms, 
 making each term standing below the line of the same name as 
 the preceding term standing above it. The product of the num- 
 bers standing above the line divided by the product of those 
 standing below it, will give the required number. 
 
 The numbers may of course be canceled as far as practica- 
 ble, before multiplying and dividing. 
 
 Ex. 1. If 100 Ib. English make 90 Ib. Flemish, and 22 Ib. 
 Flemish make 28 Ib. Bologna, how many pounds English are 
 equal to 56 Ib. Bologna ? 
 
 The demand obviously lies on the 56 Ib. Bologna; therefore, 
 56. 22. 100 
 88.^5' 
 2 
 
 Canceled, **' ^' ^ 2x22x 10=440+-9=:48f Ib. Eng- 
 lish, Ans. 
 
 2. If 40 Ib. at New York make 48 Ib. at Antwerp, and 30 
 Ib. at Antwerp make 36 at Leghorn, how many pounds at New 
 York are equal to 144 at Leghorn ? 
 
 144. 30. 40 
 Statement, ^-^. 
 
 4 5 20 
 
 Canceled, - ' ^ " 4& ^ ^ and 5x20 = 100 Ib. New York, 
 Ans. 
 
 3. If 70 braces at Venice make 84 braces at Leghorn, and 
 12 at Leghorn make 7 American yards, how many braces at 
 Venice are equal to 96 American yards 1 Ans. 1371. 
 
 4. If 24 Ib. at New London make 20 Ib. at Amsterdam, and 
 50 Ib. at Amsterdam make 60 Ib. at Paris, how many Ib. at Paris 
 are equal to 40 Ib. New London ? Ans. 40 Ib. 
 
 5. If 50 Ib. at New York make 45 Ib. at Amsterdam, and 80 Ib. 
 at Amsterdam, 103 Ib. at Dantzic, how many Ib. at Dantzic 
 are equal to 240 Ib. New York ? Ans. 278^. 
 
 6. If 24 braces at Leghorn be equal to 15 vares at Lisbon, 
 and 45 vares at Lisbon be equal to 90 braces at Lucca, how 
 
DISCOUNT. - 211 
 
 many braces at Lucca are equal to 120 braces at Leghorn ? 
 Ans. 150 braces. 
 
 QUESTIONS. In what does Conjoined Proportion consist 7 How 
 does the principle involved, compare with Simple and Compound 
 Proportion 1 What is the rule for Conjoined Proportion 1 
 
 DISCOUNT. 
 
 Discount is an allowance made for the payment of money 
 before it becomes due. 
 
 The present worth of any sum of money, payable at some 
 future time without interest, is that sum which, if put at inte- 
 rest, would in the given time and rate per cent, amount to the 
 whole debt. 
 
 Discount is not, therefore, a deduction of the given per cent, 
 from a hundred cents or a hundred dollars. If I have a claim 
 upon an individual for $100, payable a year hence ; and propose 
 to allow him 6 per cent, discount for present payment, I must 
 receive more than $100 $6 = $94 ; since $94 put on inte- 
 rest at 6 per cent, will not amount to $100 in the given time. 
 The interest on $94 one year at 6 per cent, is $5.64 ; and 
 $94 + $5. 64=: $99.64, which is 36 cents less than the re- 
 quired sum, or $100. If, however, a person owe me $106, pay- 
 able in one year without interest, and I propose to allow him 
 the same discount for immediate payment, he must obviously 
 pay me $100, since $100 in one year at six per cent, will 
 amount to precisely $106. 
 
 Hence, we learn that the ratio which any sum due a year 
 hence without interest bears to its present worth, is as 1 06 to 
 100 ; or, what is the same thing, as $1.06 to $1.00, whenever 
 the discount is at 6 per cent. If the rate per cent, be any 
 other than 6, or the time more or less than one year, the ratio 
 varies accordingly. Therefore, as the amount of $ 1 for the 
 given time and rate per cent, is to $l,so is the given sum to its 
 present worth. 
 
 Ex. 1 . What is the present worth of $450, due 2 years 
 hence, 6 per cent, discount being allowed ? 
 
212 
 
 DISCOUNT. 
 
 The interest of $1 for 2 years at 6 per cent, is 12 cents, 
 and consequently the amount of $1, for the same time, is $1.12. 
 Therefore, 1.12:1 :: 450: the required sum. And, since 
 nothing is effected by multiplying by 1 , the required sum is 
 obtained by dividing $450 by $1.12. Hence, $150.00 
 $1.12 = $401.785.+ Ans. 
 
 From the above we derive the following rule : 
 
 RULE. Divide the sum on which the discount is to be made, 
 by the amount of one dollar for the given time and rate per 
 cent. 
 
 2. What is the present worth of $700, due 3 years hence, 
 at 5 per cent, discount ? 
 
 The amount of $1 for 3 years at 5 per cent., is $1.15. 
 Therefore, $700.00 -=-'l.l 5= $608.695. -f Ans. 
 
 3. Sold goods to the amount of $1200, on 6 months' credit. 
 What is the present worth, allowing 8 per cent, discount ? 
 Ans. $1153.846.+ 
 
 4. What is the present value of a legacy of $2000, due 2 
 years hence, discounting at 5 per cent, per annum ? Ans. 
 $1818.18.+ 
 
 5. What is the difference between the interest and discount 
 on $600 for 12 years, at 5 per cent.? Ans. Interest $360; 
 discount, $225 ; difference, $135. 
 
 Note. To obtain the discount, subtract the present value 
 from the sum due. 
 
 6. What is the discount on $300 for 60 days, at 6 per cent, 
 per annum? Ans. $2.97. 
 
 7. What is the present value of $750, due 3| years hence, 
 discounting at 4 per cent, per annum? Ans. $657. 894. -f- 
 
 8. What is the discount on $500 for 2 years, at 9 per cent, 
 per annum? Ans. $76.272.+ 
 
 9. What is the present value of 350 ., due 4 years hence, 
 discounting at 4 per cent, per annum ? Ans. 301 . 14s. 5 d. 
 
 3 inr 9 r - 
 
 10. What is the present worth of 672 ., due 2 years hence, 
 
 discounting at the rate of 6 per cent, per annum ? Ans. 600 . 
 
 1 1 . Bought goods to the amount of $820, on 6 months' credit. 
 What ought I to have paid, if I had advanced the money on 
 
PROFIT AND LOSS. 213 
 
 the receipt of the goods, and had been allowed 4 per cent, dis- 
 count? Ans. $803.92.+ 
 
 12. Sold goods to the amount of 81200, one half of which 
 is to be paid in 6 months, and the other half in 8 months. 
 What is the discount for the present payment of the whole, 
 discounting at 6 per cent, per annum ? Ans. $40.553. 
 
 13. A person having a legacy of $1450 left him, payable in 
 6 years, requests present payment, and proposes to allow 6 
 per cent, discount. What must he receive? Ans. $1066.- 
 176.+ 
 
 14. What is the discount of $458 for 8 months, discounting 
 at 8 per cent, per annum? Ans. $23.188.+ 
 
 Q.UESTIONS. What is discount 1 What is the present worth of any 
 sum of money, payable at some future time, without interest 1 Is dis- 
 count a deduction of a given per cent, from a hundred cents, or a 
 hundred dollars'? Why 1 What numbers express the ratio which 
 any sum due a year hence, at 6 per cent, bears to its present worth 7 
 What is the rule for discount 1 How is the discount obtained 1 How 
 is discount proved 1 Ans. Cast the interest on the present worth, for 
 the time and rate per cent, of discount, and add it to the present worth. 
 
 PROFIT AND LOSS. 
 
 Profit and Loss is the rule by which merchants and others 
 engaged in trade, determine how much is gained or lost by any 
 transaction. It also enables them so to regulate the price of 
 their goods as to gain or loose a certain per cent, on the first 
 cost. 
 
 1st. To FIND HOW MUCH IS GAINED OR LOST ON A QUANTITY 
 OF GOODS SOLD AT RETAIL, THE PURCHASE PRICE OF THE 
 WHOLE QUANTITY BEING GIVEN. 
 
 RULE. Find the value of the whole quantity at the retail 
 price; then, if there be a gain, subtract the purchase price from 
 the same, and the remainder will be the sum gained ; but if 
 there be a loss, subtract the amount received from the purchase 
 price, and the remainder will be the sum lost. 
 
214 PROFIT AND LOSS. 
 
 Ex. 1. Bought 40 yards of cloth for $160, and sold the 
 same for $5.20 per yard ? How much did I gain ? $5.20 X 40 
 = $208.00; and $208.00 $160 = $48.00, Arts. 
 
 2. Bought a hogshead of melasses for $25, and sold the 
 same for 8 cents a pint. How much did I gain ? Ans. $15.32. 
 
 3. Bought 12 cwt. of sugar at 8 d. per pound, and sold it at 
 3 . New York currency, per cwt. Did I gain or lose, and 
 how much? Ans. Lost $22. 
 
 4. Purchased 2 hogsheads of wine for $94.50, and retailed 
 the same at 2 s. New York currency, per quart. How much 
 did I gain? Ans. $31,50. 
 
 5. Paid $57 for 456 yards of cloth, and sold the same at 
 the rate of 4 s. 6 d. New York currency, for 3 yards. What 
 did I gain? Ans. $28.50. 
 
 6. Bought 12 rolls of ribbon, each containing 50 yards, for 
 $18.75, and sold the same at 6d. New York currency, per 
 yard. How . much did I gain by the operation ? Ans. $ 1 8.75 . 
 
 7. Bought 44 Ib. of tea for $16.50, and sold it for 3 s. 6 d. 
 New England currency, per pound. What did I gain ? Ans. 
 $9.166.+ 
 
 8. What do I gain on 15 cwt. of rice, which cost me $50, 
 by retailing the same at 4 d. New York currency ; per pound ? 
 Ans. $20. 
 
 2d. To FIND WHAT IS GAINED OR LOST PER CENT 
 
 RULE. Find the whole gain or loss by the preceding rule, 
 and, having multiplied it by one hundred, divide the product by 
 the first cost. Or say, as the first cost is to the whole gain or 
 loss, so is $100 or ]00j. to the gain per cent. 
 
 Ex. 1. If I buy broadcloth at $5.50 per yard, and sell the 
 same for $6.00 per yard, what do I gain per cent. ? 
 
 $6.00 $5.50 $0.50, the gain on $5.50. Therefore, .50 
 X 100=50.00, and 50.00-^- $5.50 = 9.09+, the gain on $100. 
 Or, $5.50 : .50 : : 100 : the gain on $100. 
 
 Or, the operation may be canceled, by -p\a,cmgf,rst above the 
 horizontal line, the whole gain or loss found by subtraction, and 
 $100 or 100 . at the right of this, on the same side, and the 
 whole cost below the same. Cancel, (Sf-c. 
 
 The above sum thus stated, '5|4ir and 100 -f- 11 =$9.09 + , 
 
 a .50. 
 11 
 as before. 
 
PROFIT AND LOSS. 215 
 
 2. What do I gain per cent, if I buy wheat at 12 s. a bushel, 
 and sell the same for 15 s. a bushel ? Ans. 25 per cent. 
 
 3. Purchased pepper for 8d. per pound, and sold the same 
 for 9 d. per pound. What per cent, did I gain ? Ans. 12%. 
 
 4. Bought 650 Ib. of sugar for 10 cents a lb., and sold the 
 same for 12 cents per lb. What was my whole gain, and 
 what my gain per cent. ? Ans. Whole gain $13.00 ; gain 
 per cent. $20. 
 
 5. Bought goods to the amount of $325, and sold the same 
 for $370. What was the per cent, gained ? Ans. $ 1 3.846. + 
 
 6. If I lose $2 on $25, at what rate per cent, do I lose ? 
 Ans. 8 per cent. 
 
 7. Purchased a hogshead of wine for $50, and sold the 
 same for $75, on six months' credit. What was my gain per 
 cent., allowing 4 per cent, discount for the 6 months' credit ? 
 Ans. $44.23. 
 
 8. Bought 6 cwt. of cheese for $48, but it being damaged, I 
 am willing to sell it for the same on a year's credit. What is 
 my loss per cent, discounting at 6 per cent, per annum 1 Ans. 
 $5.66.+ 
 
 3d. To FIND HOW A COMMODITY MUST BE SOLD TO GAIN OR 
 LOSE A CERTAIN PER CENT. ON THE WHOLE COST. 
 
 RULE. If the purchase price of the quantity for which the 
 retail price is required, be not given, it must first be found, and 
 then multiplied by $100 increased by the per cent, to be gained, 
 or diminished by the per cent, to be lost. The last product di- 
 vided by 100, or, what is the same thing, with two figures cut of 
 from the right, will be the answer required. Or the operation 
 may be reduced to the following statement : As $100 is to $100 
 increased by the per cent, to be gained, or diminished by the per 
 cent, to be lost, so is the purchase price of the commodity to the 
 retail price. 
 
 Ex. 1. Bought 300 yards of cloth for $550; how must I sell 
 the same per yard, to gain 25 per cent. ? 
 
 $550-=-300=$1.83J, the price of one yard; and $1.83$ x 
 125 = $229.16; and $229.16-:- 100 = $2.29,+ Ans. Or 100 : 
 125 : : 1.83$ : Ans. or $2.29.+ 
 
 Or the operation may be canceled by the following rule : 
 
 RULE. Write the given price above a horizontal line, and 
 the quantity which cost that price directly below it. Then place 
 
216 PROFIT AND LOSS. 
 
 $100 increased by the per cent, to be gained or diminished by 
 the per cent, to be lost, above the same line, and 100 below, and 
 proceed to cancel, <fyc. 
 
 The above sum stated, |^^. 
 
 5 
 Canceled, -, and 55-6 X 4 = $2.29.+ 
 
 2. How must I retail melasses by the gallon, for which I 
 paid $30 per hogshead, to gain 12| per cent. ? Ans. $0.535.4- 
 
 3. Purchased 100 gallons of wine for $130, but by accident 
 15 gallons leaked out. How must I sell the remainder per gal- 
 ion, to gain 15 per cent. 1 Ans. $ 1.758. -+- 
 
 4. Paid $1.10 per gallonfor melasses ; how must I sell the 
 same per quart, to gain 30 per cent. ? Ans. 35| cents. 
 
 5. Received from Lisbon 180 casks of raisins, which cost 
 me $2.13 per cask. How shall I sell them per cask, to gain 25 
 per cent. ? Ans. $2.66. 
 
 2.13. 125 
 Statement, -- . 
 
 6. Bought 2 cwt. of pepper at 1 s. New York currency, 
 per pound, but it being damaged, I am willing to lose 10 per 
 cent. ; how must I sell it by the Ib. 1 Ans. 1 1 cents 2^ mills. 
 
 7. Bought one ton of steel for $184 ; how must I sell the 
 same per pound, to gain 5 per cent. 1 Ans. 8 cents 6 mills. 
 
 8. A merchant bought 160 yards of cloth for $240; how 
 must he sell the same per yard to gain 12 per cent. ? Ans. 
 $1.68. 
 
 9. Bought 8 pieces of cloth, each containing 15 yards, at 
 3 s. New England currency, per yard ; how must I retail 
 the same, to gain 8 per cent., and how much must I receive 
 for the whole ? Ans. 54 cents retail price ; whole value, 
 $64.80. 
 
 10. If I buy 6 cwt. of sugar at 10 d. New York currency, 
 per pound, and am allowed 4 per cent, discount for ready 
 money, and sell the same so as to gain 15 per cent, on the 
 money advanced, how much money do I receive ? Ans. 
 $77.28. 
 
 11. Bought 12 chests of tea, each weighing 56 pounds, at 
 4s. 6 d. New England currency, per Ib. For ready money 
 was allowed 2 per cent, discount ; how much must I receive 
 
BARTER. 217 
 
 for the whole to realize a profit of 10 per cent, on the money 
 paid out? Ans. $543.312. 
 
 12. 56. 54. 98. 110 
 Statement, - 12 . 6 . m 1(X) . 
 
 12. Bought 700 yards of ribbon, at 6 d. New York currency, 
 per yard ; how must I sell the same to gain 12 \ per cent, and 
 what shall I receive for the whole quantity ? Ans. 6| d. per 
 yard, nearly; and for the whole, $49.218.+ 
 
 13. How must cloth which costs 13 s. 4 d. be sold to gain 20 
 per cent.? Ans. At 16 s. per yard. 
 
 CtuEsrioNS. What is profit and loss 1 What is Case 1st 7 What 
 is the rule? What is Case 2d? What is the rule? How is the 
 statement made for canceling 1 What is Case 3d 1 What is the rule ? 
 What is the rule for canceling 1 
 
 BARTER. 
 
 Barter is a rule by which those engaged in trade exchange 
 commodities so that neither party suffer loss. 
 
 Take the following illustration : 
 
 Ex. 1. How much sugar at lOd. per Ib. must be given in 
 exchange for 12 cwt. of butter at 15 d. per pound? 
 
 It is obvious that the 12 cwt. of butter bears the same ratio 
 to the number of cwt. of sugar required, as 10 d. bears to 15 d. 
 
 Therefore, as 10 : 15 : :12 : the answer, viz. 18 cwt. Or, |jp^ is 
 the statement for canceling : 
 6 3 
 
 ^~f, 6x3 = 18, Ans. 
 S 
 
 The scholar will perceive that the principle here involved 
 is the same as in the rule of proportion, which has already 
 been fully explained. 
 
 2. How much tea at 8 s. per Ib. must be given for 2 cwt. of 
 chocolate at 4 s. 6 d. per Ib. ? Ans. I cwt. 14 Ib. ^ 
 19 
 
218 BARTER. 
 
 Statement, 8 s. : 4 s. 6 d. : : 2 cwt. : Ans. 
 
 -r. v 2.54 
 
 For canceling, TO~Q- 
 
 In this statement, the price of the chocolate is reduced to 
 pence. The 12 below the line reduces the pence to shillings. 
 
 3. How much rice at 32 s. per cwt. must be given for 4 
 cwt. 2 qr. of raisins at 6 d. per pound ? Ans. 7 cwt. 3 qr. 
 14 Ib. 
 
 4. B. has 90 yards of linen, worth 4 s. 6 d. per yard, which 
 he wishes to exchange with C. for muslin at 2 s. per yard ; 
 howmany yards of muslin must B. receive ? Ans. 202| yards. 
 
 5. Bought 75 yards of broadcloth at $3 per yard, and paid 
 for it in muslin at 1 s. 3 d. New York currency, per yard ; how 
 many yards of muslin did it take? Ans. 1440 yards. 
 
 6. Bought 7 tuns of wine at 10 d. a pint, and paid for it 
 with melasses at 3 s. per gallon ; how many hogsheads of me- 
 lasses did it take? Ans. 62| hhd. 
 
 7. Sold 5 cwt. 1 qr. of sugar at $8.50 per cwt. and received 
 in pay 24 yards of cloth ; what was the value of the cloth per 
 yard? Ans. $1.859. 
 
 8. How many gallons of melasses at 8 cents a quart must 
 be given for 24 cwt. of rice at $8 per cwt. Ans. 600 gallons. 
 
 9. I have nuts worth $4 per bushel, and B. has sugar worth 
 10 cents per pound ; now if I charge him $4.50 per bushel 
 for my nuts, what ought he to charge me for his sugar ? Ans. 
 11^ cents. 
 
 10. What quantity of butter worth 12| cents a pound must be 
 given in exchange for 12 Ib. of indigo worth $2.25 per pound ? 
 Ans. 216 Ib. 
 
 1 1 . Sold 5 cwt. of cheese at 8 d. per pound, and received in 
 pay muslin worth Is. 6 d. per yard ; how many yards did I 
 receive ? Ans. 248| yds. 
 
 12. Bought 90 bushels of wheat at 15 s. and gave in pay 115 
 yards of Irish linen, worth 6 s. per yard, and the remainder in 
 cash; how many dollars did I pay, the currency. being New 
 York ? Ans. $82.50 
 
 13. Received of B. a quantity of broadcloth which he had 
 sold at $4.50 per yard, but charged me $5.00; in pay I let 
 him have a quantity of wheat which I had retailed at $1.50 
 per bushel ; what ought I to have charged him, to meet the ad- 
 vance he made on his broadcloth? Ans. $l.66f. 
 
 14. Bought 12 gross of pen-knives at 9 dollars per gross, and 
 
PARTNERSHIP. 219 
 
 paid for them in cloth at 3s. 6 d. New York currency, per 
 yard ; how many yards will be required to pay for the knives ? 
 Ans. 246-| yards. 
 
 15. Sold 48 cwt. of hops at 4 d. per pound, and received pay 
 in prunes at 6 d. per pound ; how many cwt. did I receive ? 
 Ans. 32 cwt. 
 
 16. Received 460 yards of linen worth 6s. 6 d. New York 
 currency, per yard, in exchange for 84 yd. of broadcloth ; 
 what was the value of the broadcloth per yard ? Ans. $4.449. + 
 
 17. Exchanged 320 Ib. of chocolate valued at 4s. 6d. New 
 York currency, per pound, for a quantity of cotton at 8 d. per 
 Ib. ; but there not being cotton enough to pay me, I received 
 the balance, viz. 50 dollars, in cash ; how many Ib. of cotton 
 did I receive ? Ans. 1560 Ib. 
 
 18. How many cwt. of sugar, valued at 10 d. per Ib. New 
 England currency, must be given in exchange for 24 cwt. of 
 pepper at 15 dollars per cwt. ? Ans. 23^ cwt. 
 
 19. What deduction ought to be made on wheat, which has 
 been valued at 9 s. per bushel, in exchange for cloth, when 
 cloth valued at 14 s. per yard is sold for 12 s. per yard ? Ans. 
 Is. 3^d. per bushel. 
 
 20. How many pounds of cinnamon at 10 s. per Ib. must be 
 given in exchange for 5 cwt. of saleratus at 9 pe^ce per Ib. ? 
 Ans. 42 Ib. 
 
 QUESTIONS. What is Barter 1 What is the principle involved in 
 this rule ? 
 
 PARTNERSHIP. 
 
 Partnership is the rule by which individuals trading with a 
 joint stock, determine their several shares of the profit or loss 
 realized. 
 
 The individuals thus trading are called Stockholders. The sum 
 of money advanced by the company forms a Capital Stock ; and 
 the money from time to time received on the several shares is 
 called Dividend. When the whole stock owned by the com- 
 
220 PARTNERSHIP. 
 
 pany is employed for the same period of time, the gain or loss 
 which results from the adventure is divided among the indi- 
 viduals composing the company, in proportion to their several 
 shares of the stock. 
 
 We have then the following rule : 
 
 RULE. As the whole stock is to each man's share of the 
 stock, so is the whole gain or loss to each man's share of the 
 gain or loss. 
 
 Ex. 1. Three persons traded in company. A's stock was 
 $1200 ; B's stock was $4800, and C's, $2000 : they gained 
 $800; what was each man's share of the gain ? 
 
 $1200+$4800+$2000=8000, amount of the stock ; there- 
 fore, 8000 : 1200 : : 800: A's share, viz. $120. Then to ob- 
 tain B's share, 8000 : 4800 : : 800 : B's share, or $480. And 
 lastly, for C's share, 8000 : : 2000 : : 800 : C's share, or $200. 
 
 To state sums for canceling : 
 
 RULE. Write the whole gain or loss above the horizontal 
 line, with the stock of one of the company standing at the right 
 of it ; then place the whole stock below the line, and proceed to 
 cancel, <fyc. 
 
 fiflO 1 200 
 
 The above sum thus stated: - =$120, A's share. 
 
 oUUU 
 
 s$480, B's share ; and ??? ~= $200, C's share. 
 
 Note. The operations in this rule are proved by ad- 
 ding together the several shares. Their sum must equal the 
 whole gain. $120 + $480 + $200=$800. 
 
 2. Three men, A., B., and C., traded in company. A. put in 
 $560; B., $700 ; and C., $640. At the expiration of their part- 
 nership, they found that they had gained only $420 ; what was 
 each man's share? Ans. A's, $123.789; B's, $154.737; and 
 C's, $141.473. 
 
 3. Three merchants trading in company suffered a loss of 
 $600; their several stocks were $800, $1000, and $1200; 
 what was each man's loss ? Ans. $160, $200, and $240. 
 
 4. A., B., and C., freighted a ship. A. put on board 36 tons ; 
 
PARTNERSHIP. 221 
 
 B., 40 tons ; and C., 60 tons. In a storm, 30 tons were 
 thrown overboard. What Avas each man's share of the loss ? 
 Ans. A's, ?if tons ; B's, 8if tons ; and C's, 13 r 4 y tons. 
 
 5. A bankrupt, who has property valued at $2500, owes A. 
 $900 ; B., 1000 ; C., $800, and D., $500. How much will 
 each man receive, if the whole property be given up to them ? 
 Arts. A., $703.125; B., $781.25; C., $625 ; and D., $390.625. 
 
 6. Three individuals unite in purchasing 600 acres of land. 
 A. pays $3000 ; B., $5000 ; and C, $7000. What are their 
 several proportions of the whole ? Ans. A. has 120 acres ; B., 
 200; andC., 280. 
 
 7. Three individuals, L., M., and N., freighted a ship with 
 324 tons burden. L. put on board 72 tons; M., 108 tons; 
 and N., 144 tons. They cleared $1200 by the adventure. 
 What was each man's share of the gain ? Ans. L's, $266.- 
 666+ ; M's, $400 ; and N's, $533.333.+ 
 
 8. Three persons agreed to pay $ 1 50 for the use of a pas- 
 ture, and that each should pay in proportion to the number of 
 creatures he put in What would be their several shares, pro- 
 vided the first put in 100 oxen ; the second, 150 oxen, and the 
 third, 200 oxen? Ans. $33.333+ ; $50 ; and $66.666.+ 
 
 9. A person at his death owes A. 70 . ; E., 400 . ; and H., 
 140JE. ; and his property is valued at only 400 . How 
 much ought each to receive? Ans. A., 45 . 18^ r s. ; E., 
 262 . 5ff s. ; and H., 91 . 16^-s. 
 
 10. Four individuals purchased a horse for $96. A. paid 
 $20; B., $30; C., 25; D., $21. They sold the horse 
 for $150. What was each man's share of the gain? Ans. 
 A's, $11.25 ; B's, $16.875 ; C's, $14.06; and D's, $12.8l. 
 
 11. A man, whose property was worth only $500, owed 
 G., $150 ; H., $180 ; and M., $300. He therefore concluded 
 to give his property up to his creditors. What did each re- 
 ceive ? Ans. G., $119.047+ ; H., $142.857 ; M., $238.095. 
 
 12. L. has a farm of 90 acres ; M. has one of 120 acres; 
 and N. one of 150 acres. Their farms all joining, they agree 
 to unite them for the purpose of renting. This they do, and 
 receive for the whole the annual rent of $820 ; what is each 
 man's share of the rent ? Ans. L's, $205 ; M's, $273.333+ ; 
 N's, $341.666.+ 
 
 Whenever the different shares are not continued through 
 the same period of time, it is obvious that each man's share of 
 the profit and loss will depend not entirely on his portion of the 
 19* 
 
222 PARTNERSHIP. 
 
 whole stock, but also on the time during which his stock was 
 invested. Take the following sum as an illustration : 
 
 13. Three merchants traded in company. A. put in $120 
 for 10 months ; B., $100 for 18 months ; and C., $150 for 5 
 months. They gained $100. What was each man's share ? 
 A's, $120 for 10 months = $120 X 10 = $1200 for one month ; 
 B's, $100 for 18 months = $100x1 8 = $1800 for one month; 
 and C's, $150 for 5 months = $150x5 = $750 for one month. 
 Therefore, $1200+$1800 + $750 = $3750, and 3750: 1200:: 
 100 : A's share, or $32. 3750 : 1800 : : 100 : B's share, or 
 $48 ; and 3750 : 750 : : 100 : C's, share, or $20. 
 
 Hence, when the time of investment of the several shares 
 differs : 
 
 RULE. Multiply each man's stock by the time during which 
 he continued in trade, and use the several products as directed 
 in the preceding rule to use the several shares of the whole stock. 
 
 14. A., B., and C., hold a pasture in common, for which 
 they pay 30 . per annum. In this pasture, A. has 40 oxen 
 for 75 days ; B., 45 oxen for 50 days ; and C., 50 oxen for 90 
 days. What part of the 30 . ought each to pay ? Ans. A., 
 9. 4s. 7^-d. ; B., 6 . 18 s. 5 T 7 3 d. ; andC., 13 j. 16s. 
 
 aid. 
 
 15. Three men enter into partnership on the following terms : 
 A. invests $1500 for 5 months ; B., $1800 for 6 months ; and 
 C., $2000 for 8 months. During the continuance of their 
 partnership they sustain a loss of $1000. What is each man's 
 share of the loss ? Ans, A's loss is $218.658 ; B's, $314.- 
 868 ; and C's, $466.472. s 
 
 16. E. and S. enter into partnership for one year. E. at 
 first advances $480, and B. puts in his share 3 months after. 
 How much must he advance to be entitled to an equal share of 
 the gain at the expiration of one year ? Ans. $640. 
 
 17. Two merchants trading in company gain $200. A's 
 stock was $220 for 6 months, and B's, $380 for 9 months. 
 How ought they to share the gain 1 Ans. A's portion is 
 $55.696; B's, $144.304. 
 
 18. Two men commenced trading in company on Jan. 1st, 
 1825. A. advanced $1000 at the time specified; but B., finding it 
 inconvenient, did not advance his share till the first of May fol- 
 lowing. At the end of the year they shared the profits equally. 
 What capital did B. advance I Ans. $1500. 
 
COMMERCIAL EXCHANGE. 
 
 223 
 
 19. A. and B. traded in company. A. put in $1200 ; B. ad- 
 vanced his share 3 months after. What sum was it necessary 
 for him to advance so as to be entitled to one half of the profit 
 at the expiration of one year 1 Ans. $1600. 
 
 20. L., M., and N., entered into partnership ; L. advanced 
 $300 for 3 years ; 6 months after, M. put in $450 ; and 6 
 months after M. put in his share, N. put in $520. At the ex- 
 piration of 3 years, they found they had cleared $900. What 
 was each man's share 1 Ans. L's, $264.274 ; M's share, 
 $330.342 ; and N's, $305.383. 
 
 QUESTIONS. What is Partnership ? What are the individuals tra- 
 ding in company called 1 What does the money advanced by the 
 company form 1 What is the money they receive on their several 
 shares called 1 What is the rule for operating when each man's stock is 
 employed for the same period of time'? What is the rule for cancel- 
 ing 1 When all the shares are not continued for the same period of 
 time, on what will each man's share of the profit and loss depend'? 
 What is the rule 1 
 
 COMMERCIAL EXCHANGE. 
 
 Under this rule are included the operations of purchasing 
 goods in one country, in the currency of that country, and sell- 
 ing them at wholesale or retail in the currency of another coun- 
 try, so as to give or lose some required per cent. 
 
 TABLE OF GOLD COINS. 
 
 NAMES OF COINS. 
 
 VALUE. 
 
 Austrian Dominions. 
 Souverein, - - $3.37 7 
 
 Double Ducat, - - 4.58 9 
 Hungarian Ducat, - 2.29 6 
 
 Bavaria . 
 
 Carolin, - - - 4.95 7 
 
 Max d'or or Maximilian, 3.31 8 
 
 Ducat, - - - 2.27 5 
 
 NAMES OF COINS. VALUE. 
 
 Berne. 
 
 Ducat, double in pro- 
 portion. - - $1.98 6 
 Pistole, - - - 5.54 2 
 
 Brazil. 
 
 Johannes, half in propor- 
 tion,- - - - 17.06 4 
 Dobraon, - - - 32.70 6 
 
224 
 
 COMMERCIAL EXCHANGE. 
 
 NAMES OF COINS. 
 
 VALUE. 
 
 Dobra, - - - $17.30 1 
 Moidore, half in propor- 
 tion, - - - 6.55 7 
 Crusade, ... 63 5 
 
 Brunswick. 
 
 Pistole, double in propor- 
 tion, - - - 4.54 8 
 Ducat, - - - 2.23 
 
 Cologne, 
 Ducat, - - - 2.26 7 
 
 Colombia. 
 Doubloon, - - - 15.53 5 
 
 Denmark. 
 Ducat, Current, - - 1.81 2 
 
 Ducat, Specie, 
 Christian d'or, - 
 
 2.26 7 
 4.02 1 
 
 East India. 
 
 Rupee, Bombay, - 7.09 6 
 
 Rupee, Madras, - - 7.11 
 
 Pagoda, Star, - - 1.79 8 
 
 England. 
 
 Guinea, half in propor- 
 tion, - - - 5.07 5 
 
 Sovereign, half in propor- 
 tion, - - - 4.84 6 
 
 Seven shilling piece, - 1.69 8 
 
 France. 
 
 Louis, coined before 1786, 4.84 jj 
 
 Double Louis, before 1786, 9.69 J 
 
 Louis, coined since 1786, 4.57 
 
 Double Louis, since 1786, 9.15 3 
 
 Napoleon, or 20 francs, 3.85 1 
 Double Napoleon, or 40 
 
 francs, - - - 7.70 
 
 Frankfort on the Main. 
 
 Ducat, - - - 2.27 9 
 
 Geneva. 
 
 Pistole, old, - - 3.98 5 
 
 Pistole, new, 
 
 Hamburg. 
 
 Ducat, double in propor- 
 tion, - 
 
 3.44 4 
 
 2.27 9 
 
 NAMES OF COINS. VALUE. 
 
 Genoa. 
 Sequin, - $2.30 2 
 
 Hanover. 
 
 Double George d'or, - 7.87 9 
 Ducat, - - - 2.29 6 
 Gold Florin, double in pro- 
 portion, - - - 1.67 
 
 Holland. 
 
 Double Ryder, - 
 
 Ryder, 
 
 Ducat, 
 
 Ten Guilder Piece, 
 
 Malta. 
 
 Double Louis, 
 
 Louis, 
 
 Demi-Louis, 
 
 Doubloons, 
 
 Mexico. 
 
 Milan. 
 
 12.20 5 
 6.04 3 
 2.27 5 
 4.03 4 
 
 9.27 8 
 4.65 2 
 2.33 6 
 
 15.53 5 
 
 Sequin, - - - 2.29 
 Dopia or Pistole, - 3.80 7 
 Forty Livre Piece, - 7.74 2 
 
 Naples. 
 
 Six Ducat Piece, 1783, 5.24 9 
 Two do. or Sequin, 1762, 1.59 1 
 Three do. or Oncetta, 1818, 2.49 
 
 Netherlands. 
 
 Gold Lion, or 14 Florin 
 
 Piece, - - - 5.04 6 
 Ten Florin Piece, - 4.01 9 
 
 Parma. 
 
 Quadruple Pistole, - 16.62 8 
 
 Pistole, or Dappia, 1787, 4.19 4 
 
 do. do. 1796, 4.13 5 
 
 Maria Theresa, 1818, 3.86 1 
 
 Piedmont. 
 
 Pistole, coined since 1785, 5.41 1 
 Sequin, half in proportion, 2.28 
 Caflino, coined since 1785, 27.34 
 Piece of 20 Francs, or 
 Marengo, 3.56 4 
 
COMMERCIAL EXCHANGE. 
 
 225 
 
 NAMES OF COINS. 
 
 VALUE. NAMES OP COINS. 
 
 Poland. 
 
 Ducat, 
 
 $2.27 
 
 Portugal. 
 
 Dobraon, - - - 32.70 8 
 Dobra, - - - 17.30 1 
 Johannes, - - - 17.06 4 
 Moidore, half in propor- 
 tion, - - 6.55 7 
 Piece of 16 Testoons, 1600 
 
 Rees, - - - 2.12 1 
 OldCrusado, or 400 Rees, 58 8 
 New Crusado, or 480 Rees, 63 5 
 
 Milree, coined in 1755, 73 
 
 Prussia. 
 
 Ducal, 1748, - - 2.27 9 
 
 Ducat, 1787, - - 2.26 7 
 
 Frederick, double, 1769, 7.97 5 
 
 Frederick, double, 1800, 7.95 1 
 
 Frederick, single, 1778, 3.99 7 
 
 Frederick, single, 1800, 3.97 5 
 
 Rome. 
 
 Sequin, coined since 1760, 2.25 1 
 
 Scudo of Republic, - 15.81 1 
 
 Russia. 
 
 Ducat of 1796, - - 2.29 7 
 
 Ducat of 1763, - - 2.26 7 
 
 Gold Ruble of 1756, - 96 7 
 
 Gold Ruble of 1799, - 73 7 
 
 Gold Pollen of 1777, - 35 5 
 
 Imperial of 1801, - 7.82 9 
 
 Half Imperial of 1801, 3.91 8 
 
 Half Imperial of 1818, 3.93 3 
 
 Sardinia. 
 
 Carlino, half in propor- 
 tion, - - - 9.47 2 
 
 Saxnny. 
 
 Ducat of 1784, - . - 2.26 7 
 
 Ducat of 1797, - - 2.27 9 
 
 Augustus of 1754, - 3.92 5 
 
 Augustus of 1764, - 3.97 4 
 
 Sicily. 
 
 Ounce of 1751, - - $2.50 4 
 Double Ounce, - - 5.04 4 
 
 Spain. 
 
 Doubloons of 1772, parts 
 
 in proportion. - - 16.02 8 
 
 Doubloon, - - - 15.53 5 
 
 Pistole, - - - 3.88 4 
 
 Coronilla, Gold Dollar, or 
 
 Vintern, 1801, 98 3 
 
 Sweden. 
 Ducat, - - - 2.23 5 
 
 Switzerland. 
 
 Pistole of the Helvetic Re- 
 public, - - - 4.56 
 
 Treves. 
 Ducat, - - - 2.26 7 
 
 Turkey. 
 
 Sequin fonducli of Con- 
 stantinople, - - 1.86 8 
 Sequin fonducli of Con- 
 stantinople, 1789, - 1.84 8 
 Half Misseir, 1818, - 52 1 
 
 Sequin fonducli, - 1.83 
 
 Yeermeeblekblek, - 3.02 8 
 
 Tuscany. 
 
 Zechino or Sequin, - 2.31 8 
 Ruspone of the Kingdom 
 of Etruria, - - 6.93 8 
 
 Venice. 
 
 Zechino or Sequin, shares 
 in proportion, - - 2.31 
 
 Wirteviberg. 
 
 Carolin, - - - 4.89 8 
 Ducat, - - 2.23 5 
 
 Zurich. 
 
 Ducat, double and half in 
 proportion, - - 2.26 7 
 
226 
 
 COMMERCIAL EXCHANGE. 
 
 TABLE OF UNCOINED AND SILVER MONEY. 
 
 English, Currency. 
 1. sterling, before 1832 
 
 =$4.444. Since 1832= $4.80 
 Or prior to 1832, 9.= 
 
 $40. Si nee 1832, 5 .= 24.00 
 
 1 English crown, - 1.10 
 
 OrlOcrowns= - 11.00 
 
 1 English shilling, - 22 2 
 
 1 pistareen, - 20 
 
 Or 5 pistareens= - 1.00 
 
 1 English penny, 0185 
 
 Irish Currency. 
 
 1 . Irish, - - - 4.10 
 
 Or 10 .= - - 41.00 
 
 1 shilling, ... 20 5 
 
 1 penny, 01 7 
 
 French Currency. 
 
 1 French crown, - 1.10 
 
 Or 10 crowns= - - 11.00 
 
 1 five franc piece, - 93 
 
 1 franc, ... J8 6 
 
 1 Decimes, 0186 
 
 Spanish Currency. 
 
 1 Spanish dollar, - 1.00 
 
 1 real, new plate, - 10 
 
 1 real vellon, - 05 
 
 Currencies of other Nations. 
 millree of Portugal, $1.24 
 
 Russian silver ruble, 75 
 
 rix dollar of Sweden, 1.00 
 
 Russian rix dollar 66 6 
 
 Or 3 rix dollars= - 2.00 
 Danish rigsbank dollar, 50 
 
 silver ducat of Naples, 80 
 
 Or 5 ducats= - - 4.00 
 
 scudo of Sicily, - 96 
 
 oncia of Sicily, - 2.40 
 
 pezza of Leghorn, - 90 
 
 pezza of Genoa, - 89 
 
 flourin of Trieste, - 48 
 
 rix dollar of Trieste, 96 
 
 Roman crown, - 1.00 
 
 gold crown of Rome, 1.53 
 
 Maltese scudo, 40 
 
 rupee of Bengal, - 55 5 
 
 rupee of Bombay, - 50 
 
 pagoda of Madras - 1.80 
 
 tale of Canton, - 1.48 
 
 Japanese tale, 75 
 
 dollar of Sumatra, - 1.10 
 
 tale of Sumatra, - 4.16 
 
 florin of Java, 40 
 mark banco of Hamburg, 33 3 
 
 guilder of Amsterdam, 40 
 
 To reduce a foreign currency to federal money, multiply the 
 given money of that currency by the value of its unit in federal 
 money, as given in the preceding tables. 
 
 If, however, it be required to find how goods, the value of 
 which is given in a foreign currency, must be sold in federal 
 money, to gain or lose a given per cent., first change the for- 
 eign currency to federal money, and calculate the gain or loss on 
 this by the rules already giv^n. The retail price of any de- 
 nomination, as cwt., lb., yd., gal., &c., may then be found by 
 division. 
 
 Ex. 1. Purchased in London 360 yards of broadcloth, which 
 cost me, including transportation, 300 >. sterling ; how must I 
 sell the same per yard, in federal money, to make a profit of 
 20 per cent. ? 
 
 
COMMERCIAL EXCHANGE. 227 
 
 Solution: 300 . x 40= 12000; and 12000-h9=$1333| 3 
 the value of 300 . in federal money. 
 
 Then, $1333|x 1.20=$1600, the value in federal money 
 increased by the gain per cent. ; therefore, $1600-^-360 = 
 $4.444, -f the required price of one yard. 
 
 If the wholesale price only be required, it is only necessary 
 to omit the last step. Thus the $1600 is the wholesale price 
 of the cloth given in the preceding sum. 
 
 To solve sums like the preceding by canceling : 
 
 RULE. Place the whole cost in the given currency first above, 
 and (if the retail price be required) the number expressing the 
 quantity procured for that price, first below, a horizontal line. 
 Write next above the line the value of a unit of the given cur- 
 rency, in federal money. And, lastly, to increase or diminish 
 the price by the required per cent., place 100 below the line, and 
 100 increased by the per cent, to be gained, or diminished by the 
 per cent, to be lost, above the same. 
 
 Note 1st. If the wholesale price be required, the number 
 expressing the whole quantity (by the preceding rule, placed 
 below the line) must be rejected. 
 
 Note 2d. Whenever in the preceding tables the value of a 
 number of units of foreign currency is given in federal money, 
 place that number below the line, and its federal value above the 
 sum. (See -'s sterling.) 
 
 The preceding sum stated for canceling : 
 
 300. 40. 120 
 
 360. 9. 100' 
 
 To understand the reason of the middle terms, see Note 2d, 
 and <'s sterling in the preceding tables. 
 
 40 
 SBtt. 4ft. 121ft 
 
 9. iM' 
 9 8 
 And 40 4- 9 = $4. 444, -f the answer, the same as before. 
 
 2. Purchased in London 350 yards of sheeting for 75 ., and 
 paid 12 ,. for its transportion to New York city ; how must I 
 retail the same in federal money, to gain 15 per cent, on the 
 first cost? Ans. $1.27.+ 
 
228 COMMERCIAL EXCHANGE. 
 
 Statement, 350 9 ' 100 - The 87 in the statement:=75.. 
 
 3. Received from London 470 yards of dimity, which, in- 
 cluding transportation, cost me 65 . ; sold the same by the 
 yard so as to gain 30 per cent, on the first cost ; how did I 
 sell it ? Ans. $0.799 -f per yard. 
 
 4. Received from Dublin 600 yards of Irish linen, the whole 
 cost of which was 75 >. Irish currency ; how must I retail the 
 same in federal money, to gain 12| per cent. ? Ans. $0.576+ 
 per yard. 
 
 5. My agent in Dublin has forwarded to me 900 yards of 
 linen ; whole cost 60 . Irish currency ; how must I retail the 
 same in federal money, to gain 15 per cent. 1 Ans. $0.314.+ 
 
 6. I have in my store 120 yards of broadcloth, forwarded 
 me by my agent in Paris, which cost me, including transpor- 
 tation, 325 crowns ; how must I sell the same in federal money, 
 to gain 16 per cent. ? Ans. $3.455+ per yard. 
 
 7. Received 680 yards of silk from Paris, for which I paid 
 560 crowns ; expenses of transportation, 12 crowns; how must 
 I sell the same in federal money, to gain 30 per cent. ? Ans. 
 $1.20+ per yard. 
 
 8. Received from Madrid 6 hogsheads of wine, each con- 
 taining 63 gallons, for which my agent paid 188 Spanish dol- 
 lars ; how must I sell the same per gallon, to gain 12^ per cent. ? 
 Ans. $0.559.+ 
 
 9. I have on hand a bale of silk, containing 174 yards, which 
 I received from Cadiz, at a cost, including transportation, of 
 140 piastres or Spanish dollars ; how must I sell the same per 
 yard, to gain 5 per cent. ? Ans. $0.844.+ 
 
 10. Received from Oporto 3 hogsheads of port wine, con- 
 taining 63 gallons each ; cost, including transportation, 30 
 milrees per hogshead ; how must I retail the same by the 
 gallon, to gain 25 per cent.? Ans, $0.738.+ 
 
 1 1 . How must I sell broadcloth by the yard, in federal money, 
 of which 3 pieces, each containing 35 yards, cost me 135 . 
 sterling, to gain 30 per cent. 1 Ans. $7.428.+ 
 
 12. Received from A. B., Dublin, 560 yards of linen ; whole 
 cost 90 . Irish currency ; what must be the retail price, in 
 federal money, to gain 15 per cent. ? Ans. $0.757+ por 
 yard. 
 
 13. Received from the same 600 yards of muslin,worth 56 . ; 
 how must I sell the whole quantity in federal money, to gain 
 5 per cent. ? Ans. $241.08. 
 
 
COMMERCIAL EXCHANGE. 229 
 
 14. Consigned to my agent, J. Jones, of London, 300 barrels 
 of flour, for which I paid $1500 ; how many pounds sterling 
 ought he lo receive for the same, to gain 10 per cent., the ex- 
 pense of transportation being $50? Ans. 383 . 12s. 6d. 
 
 15. Received of my agent in London, J. Jones, 2510 gallons 
 of Madeira wine, which cost me, per invoice, 1640 j. sterling ; 
 but it being of an inferior quality, I am willing to lose 5 per 
 cent, on the cost ; what must be the price per gallon, in federal 
 money? Ans. $2.758.+ 
 
 16. Three men trading in company, received from France 
 1200 bottles of champagne, for which they paid 600 French 
 guineas, each $4.60 ; how must they sell the same per bottle, 
 in federal money, to gain 40 per cent., and what will be each 
 man's gain per bottle ? Ans. $3.22 per bottle ; each man's 
 gain per bottle, $0.306.+ 
 
 17. Received 300 ells of cloth from Hamburgh, which cost 
 me 1 500 mark bancos ; how must the same be sold in federal 
 money, by the yard, to gain 12-2- P er cent., the ell Hamburgh 
 being 2 qr. ? Ans. $1.17.+ 
 
 18. New York, Jan. 6, 1838. This day received from Am- 
 sterdam, 600 yards of carpeting ; whole cost, 2400 guilders. 
 Required the retail price in federal money, to gain 20 per cent. 
 Ans. $ 1 92 per yard. 
 
 19. Shipped to London 380 barrels of flour, which cost me, 
 including transportation, $6 per barrel. How many English 
 crowns must I receive for the whole quantity, to gain 10 per 
 cent. Ans. 2280 crowns. 
 
 20. Shipped to Dublin 3000 bushels of flax-seed, which 
 cost me $2500. How many pounds, Irish currency, must I re- 
 ceive for the whole quantity, to gain 5 per cent. ? Ans. 640 . 
 4s. 10 d. 2qr.+ 
 
 21. Boston, Jan. 16, 1835. This day received from my 
 agent at Lisbon, 16 hogsheads of wine, each 65 gallons ; 
 whole cost, 640 milrees. The whole being of an inferior 
 quality, I am willing to lose 6 per cent, on the cost. How 
 much in federal money must I charge per gallon? Ans. 
 $0 72, nearly. 
 
 22. New York, Sept. 6, 1835. This day received from A. 
 B., London, 1200 yards of superfine broadcloth ; whole cost, 
 1500.6. How must 1 sell the same in federal money, at 
 wholesale, to realize a profit of 20 per cent. ? Ans. $8000. 
 
 23. Received from Russia a quantity of fur ; whole cost, 900 
 silver rubles. What must be the wholesale price in federal 
 
 20 
 
\ 
 
 230 TARE AND TRET. 
 
 money, to make an advance of 15 per cent, on the cost ? Ans. 
 $776.25. 
 
 24. Boston, Feb. 26, 1836. This day received from my 
 agent in Paris, 24 hogsheads French wine, each 60 gallons ; 
 whole cost, 288 guineas. How must the same be retailed by 
 the quart, in federal money, to gain 12^ per cent.? Ans. 
 $0.25.8.+ 
 
 25. Shipped to my agent in London, 650 barrels of flour ; 
 whole cost, $3600. How many pounds sterling must I re- 
 ceive for the same, to gain 10 per cent, on the cost ? Ans. 
 
 891 . 
 
 /A tfv^a4L4A'' ^ 
 
 QUESTIONS. What operations are included under this rule 1 How 
 is a foreign currency reduced to federal money 1 How is the opera- 
 tion performed, when it is required to find how goods, the value of 
 which is given in a foreign currency, must be sold to gain or lose a 
 certain per cent, in federal money 1 What is the rule for canceling 1 
 What is Note 1st 1 What is Note 2d 1 
 
 TARE AND TRET. 
 
 We come now to consider the allowances to be made in the 
 purchase of goods by weight. The following particulars 
 require to be first noticed : 
 
 Gross Weight is the whole weight of the goods purchased, 
 including that of the box, barrel, bag, &c. containing them. 
 
 Draft is a deduction from the gross weight made in favor of 
 the buyer. 
 
 Tare is an allowance made for cask, box, barrel, &c. con- 
 taining the goods ; and may be either a certain deduction from 
 the whole quantity, or so much per box, &c. 
 
 Tret is an allowance of 4 Ib. for every 104lb. made for the 
 dust, &c. 
 
 Suttle is what remains after some of the preceding allow- 
 ances have been made. 
 
 Net Weight is what remains after all the deductions have 
 been made. 
 
 
TARE AND TRET. , 231 
 
 Ex. 1. Bought a hogshead of sugar, weighing 7 cwt. 2qr. 
 26 lb., tare on the whole, 3 qr. 24 Ib. What is the net weight ? 
 
 cwt. qr. lb. 
 
 7 2 26 
 
 3 24 
 
 632 Net weight. 
 
 2. What is the net weight of 12 casks of raisins, each 
 weighing 2 cwt. 2 qr. 14 lb., tare per cask, 12 lb. ? 
 
 2 cwt. 2 qr. 141b. X 12 = 31 cwt. 2 qr., the gross weight. 12 
 X 12 = 144 lb. = 1 cwt. 1 qr. 4 lb. ; and 31 cwt. 2 qr. 1 cwt. 
 1 qr. 4 lb. = 30cwt. Oqr. 24 lb. Ans. 
 
 3. What is the net weight of 6 casks of prunes, each 
 weighing 3 cwt. 2 qr. lOlb., tare 20 lb. per cask? Ans. 20 
 cwt. 1 qr. 24 lb. 
 
 4. What is the net weight of 44 cwt. gross, if 14 lb. per cwt. 
 be allowed for tare 1 44 x 14=616 lb. = 5 cwt. 2qr., and 44 
 cwt. 5 cwt. 2 qr. = 38 cwt. 2 qr. Ans. 
 
 Or the solution may be effected by canceling, by the follow- 
 ing rule : 
 
 RULE. Place the whole gross weight first above a horizontal 
 line. Then place 112 lb. below the line, with 112 diminished by 
 the tare per cwt., standing directly above it. Cancel, tyc. 
 
 The above sum solved by this rule : 
 
 11 U 7 
 112-14=98. itj. Canceled, *t- f 
 
 S& 2 
 and 11 x7=77, and 77-^2 = 38 cwt. 2 qr. 
 
 5. Bought 84 cwt. of sugar. What is the net weight, if 20 
 lb. per cwt. be allowed for tare ? Ans. 69 cwt. 
 
 6. Bought 9 hogsheads of sugar, each weighing 8 cwt. 2 
 qr. From this, a deduction of 161b. per cwt. was made for 
 tare. What was the net weight? Ans. 65 cwt. 2 qr. 8 lb. 
 
 Note 1st. When the price per cwt. or per lb. is given, the 
 reduction for tare and tret may be made, and the whole cost as- 
 certained by a single statement, as may be seen from the follow- 
 ing example : 
 
232 TARE AND TRET. 
 
 7. What is the value of 8 hogsheads of sugar, each weigh- 
 ing 12 cwt. gross, tare 12 Ib. per cwt., at $8.50 per cwt. ? 
 
 Statement, . Canceled, 
 
 7 
 
 and8.50xlOOx3x2^7=:$728.57, Ans. 
 
 8. Bought 32 casks of figs, each weighing 2 cwt. 2qr., at 
 a deduction of 181b. per cwt. for tare. What did the whole 
 cost me, at $4 per cwt. net weight ? Ans. $268.57. 
 
 9. Bought 15 cwt. of sugar at $6.50 per cwt. net weight. 
 Reduction for tare, 12 Ib. per cwt. ; tret, 4 Ib. per 104 Ib. How 
 must I sell the whole, to gain 20 per cent, on the first cost, and 
 how must I retail it, to gain the same per cent. ? Ans. Whole- 
 sale price, $100.446 ; retail price, $0.059, nearly. 
 
 To effect all the reductions and the gain per cent, of the 
 preceding sum, a single statement only is required ; thus : 
 
 15. 6.50. 100. 100. 120 
 11 2. 104. 100. 
 
 10. Bought 32 chests of tea, each weighing 4 cwt. 2qr. at 
 $49 per cwt. net weight; tare, 12 Ib. per cwt. ; tret, 41b. per 
 104lb. How must I sell the whole quantity to gain 20 per 
 cent., and how must the same be retailed, to gain the same ? 
 Ans. Wholesale price, $7269.23 ; retail price, $0.525. 
 
 11. Bought 742 Ib. of wool, and was allowed a deduction of 
 5 per cent, from the gross weight for dust, &c. For the net 
 weight I paid 9 s. New York currency, per pound, and was al- 
 lowed a deduction of 6 per cent, on the whole cost for ready 
 money. 1 then sold the same so as to realize a profit of 20 
 per cent, on the money I advanced. How much did I receive 
 for the whole ? Ans. $900. 
 
 12. Purchased 5 cwt. of sugar; tare allowed, 8 Ib. per cwt. 
 For the net weight I paid 6 d. New York currency, per Ib. 
 How must I sell the whole quantity to gain 20 per cent., and 
 how must the same be retailed, to gain the same per cent. ? 
 Ans. Wholesale price, $39 ; retail price, $0.07 per pound. 
 
 13. Purchased 12 bags of coffee, each weighing 96 Ib. ; tare 
 per bag, 6 Ib. What was the whole cost at 30 cents per Ib. 
 and the retail price, to gain 25 per cent. ? Ans. $324 whole 
 cost ; $0.37 retail price. 
 
 14. How much will 8 hogsheads of sugar, each weighing 
 
TARE AND TRET. 233 
 
 8 cwt. 3 qr., cost at $9 per cwt. if a deduction of 12 Ib. per 
 cwt. be allowed for tare ; and what will be received for the 
 whole, if it be sold at an advance of 30 per cent. ? Ans. 
 $562.50 cost, and $731.25 received. 
 
 15. What is the net weight of 3 tierces of rice, each 
 weighing 4 cwt. 3 qr. gross ; the tare allowed, 16 Ib. per cwt. ; 
 tret, 4lb. per 104 Ib. ? Ans. 11 cwt. 2qr. 27lb.+ 
 
 16. What is the cost of 15 chests of tea, each containing 
 1401b. gross, at 5 s. New York currency, per Ib. ; tare, 16 Ib. 
 per cwt. Ans. $1125. 
 
 17. Bought 12 hogsheads of sugar, each weighing 10 cwt. 
 2qr. at $9 per cwt. net weight. Deduction for tare, 12 Ib. 
 per cwt. How much must I receive for the whole quantity, to 
 gain 10 per cent, on the cost? Ans. $1113.75. 
 
 18. Bought 16 firkins of butter, each weighing 108 Ib. ; re- 
 duction for tare, 81b. per cwt. Paid 15 pence, New England 
 currency, per pound. What did it cost me ; and what must be 
 the wholesale, and what the retail price, to gain 20 per cent, on 
 the first cost 1 Ans. Whole cost, $334.285 ;-j- wholesale price, 
 $401.142; retail price, $0.25. 
 
 19. Bought 18 cwt. of sugar, at $12 per cwt. net weight ; 
 tare, 16 Ib. per cwt. How must I sell the same per Ib. to gain 
 12 per cent, on the first cost 1 Ans. $0.12. 
 
 20. Purchased in London, 16 cwt. of tea at 28 . sterling 
 per cwt. net weight ; tare, 12 Ib. per cwt. How much must 
 I receive in federal money, for the whole quantity, to realize a 
 profit of 12 per cent., and what retail price will allow the 
 same profit ? Ans. Wholesale price, $1991.11 ; retail price, 
 $1.24. 
 
 Q.UESTIONS. What is Tare and Tret 1 What is gross weight 7 
 What is draft 1 ? What is tare'? What is tret"? What is suttle ? 
 What is net weight 1 What is the rule 7 What is the note 7 
 20* 
 
234 EQUATION OF PAYMENTS. 
 
 EQUATION OF PAYMENTS. 
 
 Equation of Payments is the method of finding a mean 
 time for the payment of several debts due at different periods 
 of time. 
 
 Ex. 1. I owe a friend $380, to be paid as follows, viz, 
 $100 in 6 months; $120 in 7 months; and $160 in 10 
 months. If I pay the whole at once, at what time must the 
 payment be made, so that neither I nor my friend shall lose in- 
 terest ? 
 
 SOLUTION. 
 
 The interest of $ 100 for 6 mo. the interest of $ 1 for 600 mo. 
 The interest of $120 for 7 mo.^the interest of $1 for 840 mo. 
 The interest of $160 for 10 mo. = the interest of $1 for 1600 mo. 
 
 Amt. of pay'ts, $380 3040 mo, 
 
 3040 the months requisite for $1 to gain as much interest 
 as $380 would gain in the required time. Therefore, $380 : 
 $1 : : 3040 months : to the required time, viz. 8 months. 
 
 The 600, 840, and 1600 months are obviously obtained by 
 multiplying the several payments by the time which must 
 elapse before they severally become due. 
 
 We therefore have the following rule : 
 
 RULE. Multiply each payment by the time which must elapse 
 before it becomes due, and divide the sum of the products by 
 the sum of the payments. 
 
 2. A. owes me $50, payable in 4 months ; $100, payable 
 in 10 months ; and $150, payable in 16 months. In what 
 time must he pay the whole, so that neither shall lose interest? 
 Ans. 12 months. 
 
 3. What is the equated time of payment for the three fol- 
 lowing sums, viz. $500, payable in 3 years ; $400, payable in 
 4 years ; and $600, payable in 5 years ? Ans. 4^ years. 
 
 4. What is the equated time of payment for the three fol- 
 lowing sums, viz. $50, payable in 4 months ; $75, payable in 
 6 months ; and $100, payable in 7 months ? Ans. 6 months. 
 
DUODECIMALS. 235 
 
 5. A. owes B. $400 ; of which $80 is payable in 6 months; 
 $120 in ten months ; and the remainder in 1 year and 8 months. 
 What is the equated time of payment ? Ans. 1 year, 2 months, 
 and 6 days. 
 
 6. A merchant has a certain sum of money due him, of 
 which is payable in 2 months ; 3 in 4 months ; and the re- 
 mainder in 6 months. What is the equated time for the pay- 
 ment of the whole ? Ans. 4-J months. 
 
 7. I owe four sums of money, payable as follows, viz. $60 
 payable in 9 months ; $80 in 10 months ; $50 in 11 months ; 
 and $60 in 12 months. At what time may I pay the whole, 
 without loss ? Ans. lO^J months. 
 
 8. A person owes a debt of $2000,- payable in 7 months, of 
 which he proposes to pay $600 down, on condition that the 
 remainder be allowed to remain unpaid an adequate term of 
 time. In what time ought it to be paid ? Ans. 10 months. 
 
 QUESTIONS. What is Equation of Payments 1 What is the rule ? 
 
 DUODECIMALS. 
 
 Duodecimals are fractions of afoot. The unit or foot is 
 first supposed to be divided into 12 equal parts, called inches 
 or primes, and marked'. Each inch or prime is then divided 
 into 12 equal parts, called seconds, and marked ". Each second 
 is again divided in like manner, and each part obtained by this 
 division is called a third, and marked '". A foot is therefore 
 divided duodecimally, when it is separated into 12 equal parts, 
 or when the several divisions sustain a twelve-fold relation to 
 each other. This relation may be thus illustrated: 
 
 1' inch or prime is y^ of a foot. 
 
 1 "second is y 1 ^ of Jj, ^ of a foot. 
 
 1"' third is T J 2 of -^ of y 1 ^, - - - - yJ^ of a foot. 
 
 1 "" fourth is y 1 ^ of yL of -jV, of -fa, - - yo^-g of a foot. 
 I'"" fifth is T T 2 of y 1 ^ of yL of yL- of -jL, - 2 t 8 \ 32 of a foot. 
 I""" sixthisy^of y^ofyLofy^ofy^ofyL, ^.1^- O f a foot,&c. 
 
236 MULTIPLICATION OF DUODECIMALS. 
 
 The marks ', ", "', "", &c. are the indices of the several de- 
 nominations. 
 
 TABLE OF DENOMINATIONS. 
 
 12""" sixths make I' 1 '" fifth. 
 
 12"'" fifths make I"" fourth. 
 
 12"" fourths make I'" third. 
 
 12'" thirds make 1" second. 
 
 12" seconds make 1' prime or inch. 
 
 12' inches or primes make - - - 1 foot. 
 
 Duodecimals may be added or subtracted in the same man- 
 ner as compound numbers ; the denominations decreasing or in- 
 creasing in the constant ratio of 12. These operations are so 
 simple that it is unnecessary to introduce any examples. The 
 principle is the same as in Compound Addition and Subtraction. 
 
 MULTIPLICATION OF DUODECIMALS. 
 
 The greatest difficulty the scholar will here encounter, 
 will be to determine the denomination of the product of 
 any two denominations. Suppose it be required to multiply 
 6' inches or primes by 4' inches, their product is obviously 24, 
 but of what denomination is it ? By recurring to the preceding 
 tables, it will be seen that 6' inches = T % of a foot, and 4' inches 
 =: T 4 ^ of a foot ; and T ^- X yj -f^, that" is, 24". Hence, inches 
 multiplied by inches produce seconds. Again, let it be re- 
 quired to multiply 9'' seconds by 3' inches. 9 seconds = i%-, 
 and 3 inches =-f% ; therefore, yl^ x Ty=T?y8 > or 27'". Lastly, 
 what is the product of I' 1 seconds multiplied by 9" seconds. 
 7" seconds xJ^, and 9" seconds = 3-^, and T77 X TT4" 
 2o 6 T 3 3ir> or 63//// fourths. From the above we draw the follow- 
 ing conclusions, viz. that feet multiplied by feet produce square 
 feet ; feet multiplied by inches produce inches ; inches or primes 
 multiplied by inches, give seconds; seconds multiplied by seconds, 
 
MULTIPLICATION OF DUODECIMALS. 237 
 
 give fourths ; seconds multiplied by thirds, give fifths, fyc. ; 
 that is, the product of any two denominations will always be of 
 the denomination expressed by the sum of their indices. 
 
 RULE. Begin with the lowest denomination of the multipli- 
 cand, and multiply it by the highest denomination of the multi- 
 plier, and place each term of the product according to its respec- 
 tive value. Multiply in the same manner by each remaining de- 
 nomination of the multiplier, and place the product of each suc- 
 ceeding multiplication one or more places farther to the right, 
 according to the denomination. The several products thus ob- 
 tained, when added together, will give the required product. 
 
 Note. It will be remembered to carry by 12 in all cases. 
 Ex. 1. Multiply 4 feet 4 inches, by 4 feet 4 inches. 
 
 OPERATION. 
 
 ft. in. 
 
 4 4' Beginning with the 4 feet in the mul- 
 
 4 4' tiplier, we say 4 times 4' are 16'r=l ft. 
 
 4'. The 4' is set down and the 1 ft. 
 
 17 4' carried to the next product, making it 
 1 5' 4" 17 feet. Then multiplying by 4', we 
 
 say 4'x4' = 16"=:l' and 4"; set down 
 
 18 ft. 9' 4" the 4" and carry the I 7 , and say 4' times 
 
 4 are 16' and 1' is 17' = 1 ft. and 5', 
 
 which being set down, and the two products added, we obtain 
 18 ft. 9' 4" as the required product. 
 
 2. 3. 4. 
 
 ft. in. ft. in. ft. in. 
 
 Mult. 8 6' Mult. 9 10' Mult. 3 8' 
 
 by 4 3' 76' 7 6' 
 
 34 0' 68 10' 27 6' 
 
 2 1' 6" 4 11' 0" 
 
 Prod. 36 1' 6" Prod. 73 9' 0" 
 
 5. 6. 7. 8. 
 
 ft- in. ft. in. ft. in. ft. in. 
 
 Mult. 73' 311' 46' 9 7' 
 
 47' 95' 58' 36' 
 
 33 2' 9" 36 10' 7" 25 6' 33 6' 6'" 
 
238 MULTIPLICATION OF DUODECIMALS. 
 
 9. 10. 
 
 ft. in. ft. in. 
 
 6 4' 3" 3 9' 11" 
 
 4 6' 4" 4 2' 3" 
 
 28 9' 2'' 11'" 16 0' 3'' 3"' 9"" 
 
 11. How many square feet does a board 28 ft. 10' 6" long, 
 and 3 ft. 2' 4" wide contain? Ans. 92ft. 2' 10" 6'". 
 
 12. In a board 16 ft. 9' long, and 2 feet 3 broad, how many 
 square feet ? Ans. 37 ft. 8' 3". 
 
 13. There is a wall 82 ft. 6 in. high, and 13 ft. 3 in. wide. 
 How many square feet does it contain? Ans. 1093 ft. 1' 6". 
 
 14. There is a room 20 feet square and 7 ft. 6 in. high, to 
 be plastered at lOd. New York currency, per square yard. 
 How many dollars will it cost ? Ans. $6.94.+ 
 
 15. There is a yard 58ft. 6 in. in length, and 54ft. 9 in. in 
 breadth. How many dollars will it cost to pave it, at 5 d. New 
 York currency, per square yard? .Arcs. $18.53. 
 
 16. If a floor be 59 feet 9 inches long, and 24 feet 6 inches 
 broad, how many square yards does it contain ? Ans. 162 
 yards, 5ft. 10'6 ;/ . 
 
 Note 2d. If three dimensions, viz. length, breadth, and 
 depth be given, the solid content is found by multiplying them 
 successively into each other. 
 
 17. There is a pile of wood 12 feet 6 inches long, 4 feet 
 high, and 8 feet 6 inches wide. How many cords does ii 
 contain ? Ans. 3 cords, 41 feet. 
 
 To reduce solid feet to cords, divide by 128, that being the 
 number of solid feet in one cord. The required dimensions 
 O f the cord, are 8 feet long, 4 feet wide, and 4 feet high ; 
 since 8x4x4=128. 
 
 18. How many solid feet are there in a block 6 feet 8 
 inches in length, 4 feet 6 inches in height, and 3 feet 4 
 inches in width ? Ans. 100 feet. 
 
 19. There is a certain pile of wood measuring 24 feet in 
 length, 16 feet 9 inches in depth, and 12 feet 6 inches in 
 width. How many cords are there ; and how many solid feet 
 may be daily consumed to have it last one year ? Ans. 39 
 cords, 33 feet; daily allowance, 13 J feet, nearly. 
 
INVOLUTION. 239 
 
 20. How many square feet are there in a board, which mea- 
 sures 16 feet, 9 inches in length, and 2 feet, 3 inches in breadth? 
 Ans. 37 feet, 8 ' 3 ". 
 
 QUESTIONS. What are Duodecimals ? How is the foot divided ? 
 "What is each part called 1 How is the inch divided 1 What is each 
 part called, &c. 1 When is the foot divided duodecimally'? Repeat 
 the table of denominations. How may duodecimals be added 1 What 
 difficulty will be encountered in Multiplication of Duodecimals ? 
 Give an illustration of the difficulty. Of what denomination will 
 the product of any two denominations be 1 What is the rule 1 What 
 is Note 1st 1 What is Note 2d 1 
 
 INVOLUTION. 
 
 A number is involved by being multiplied into itself. 
 
 The number thus multiplied by itself is called the root. 
 
 A power of any number is the product obtained by multiply- 
 ing that number into itself. The particular power produced 
 depends on the number of successive multiplications ; the given 
 number always being the first power, and also the root of the 
 succeeding higher powers. The first multiplication then pro- 
 duces the second power ; the second multiplication, the third 
 power ; the third multiplication, the fourth power, &c., the 
 power obtained being always one in advance of the number of 
 multiplications. 
 
 Illustration : 2= the first power, and is also the root of the 
 succeeding higher powers. 
 
 2x2= 4, the 2d power, or square of 2. 
 2x2x2= 8, the 3d power, or cube of 2. 
 2 X 2 X 2 X 2 = 16, the 4th power, or biquadrate of 2. 
 2x2x2x2x2 = 32, the 5th power of 2, &c. 
 
 The power to which a number is to be raised, is frequently 
 expressed by a small figure, called the index of the required 
 power, placed on the right of that number, thus : 4 2 denotes 
 the second power of 4 = 16 ; and 4 3 denotes the third power 
 of 4=64 ; and 9 s denotes the fifth power of 9=59049, &c. 
 
240 INVOLUTION. 
 
 A fraction is involved by multiplying the numerator and 
 the denominator each into itself, the required number of times ; 
 thus the square of f is f X f =^5- The square of % is f ; and 
 the cube of the same is J^f , &c. 
 
 If the given quantity be a mixed number, it should first be 
 reduced to an improper fraction, before being involved ; thus 
 the second power of 2 is 2s=f, and f xf = 2 T 5 =6. Or, 
 proper and improper fractions may both be reduced to deci- 
 mals, and then involved. 
 
 If any number be raised to two different powers, the power 
 which is obtained by multiplying these two powers together, 
 is expressed by adding their indices, thus : 2 2 x2 3 =2 5 =32 ; 
 for 2 2 =4, and 2 3 =8, and 8x4 = 32; and 2x2x2x2x2 = 
 32-. Or, 3 : 'x3 5 =3 8 =6561,for3 3 =27, and3 5 =243; and 243 
 X 27.3*6561. 
 
 Or, again, any power of a given number is divided by another 
 power of the same number by subtracting the index of the di- 
 visor from the index of the dividend, thus : 2 5 -f-2 3 =2 2 , for 2 5 
 = 32, and 2 3 =8, and 32-^8 = 4, the second power of 2. Or 
 3 4 -f-3 2 =3 2 , for 3 4 =81, and 3 2 =9, and 81 -=-9 = 9, the second 
 power of 3. 
 
 Ex. 1 . What are the square, cube, and biquadrate of 3 ? Ans. 
 3x3 = 9, the square ; 3 X 3 X 3 =27, the cube ; and 3x3x3 
 X 3 = 8 1 , the biquadrate . 
 
 2. What are the square, cube, and biquadrate of 5 ? Ans. 
 25, 125, and 625. 
 
 3. What are the cube and biquadrate of 12? Ans. 1728 
 and 20736. 
 
 4. Multiply the second and third powers of 4 together. 
 What is the product, and what power of 4 is it ? Ans. The 
 product, 1024, or fifth power. 
 
 5. What power of 3 is obtained by multiplying its third 
 power and its fourth power together ; and what is the number ? 
 Ans. 7th power, or 2187. 
 
 Note. When the number to be raised to some given power 
 consists of whole numbers and decimals, the number of deci- 
 mals to l)e cut off in the required power is ascertained by 
 multiplying the number of decimals in the given number by 
 the index of the required power. 
 
 6. What is the square of 26.13? 26.13x26.13 = 6827769. 
 Now to determine how many decimals are to be cut off, we 
 
EVOLUTION. 241 
 
 first notice that the number of decimals in the given number 
 is two, and also, that the index of the required power is 2, 
 therefore, 2x2=4, the number of decimals to be cut off. 
 Therefore, 682.7769 is the required power. 
 
 7. What is the cube of 25.4 ? Ans. 16387.064. 
 
 8. Divide 2 6 by 2 3 , and what power of 2 will be obtained ? 
 Ans. 8, the cube of 2. 
 
 EVOLUTION. 
 
 Evolution is the reverse of Involution. In Involution we 
 have the root given to find some required power ; but in Evo- 
 lution a power is given, and a root required. 
 
 The relation between roots and powers requires to be clearly 
 understood. 
 
 A root of a number is obtained whenever that number is re- 
 solved into several equal factors ; and a power of a number is 
 obtained whenever that number taken as a root, is multiplied 
 into itself once or more. Thus, 2 is the cube root of 8, be- 
 cause it may be resolved into three 2's ; or because 2 raised 
 to its third power equals 8, for 2 x2 X2 = 8. Also, 8 is the 
 square root of 64, because the second power of 8 is 64, or 8 x 
 8 = 64. Again, 9 is the square of 3, and 3 is the square root 
 of 9 ; 27 is the cube of 3, and 3 is the cube root of 27. Roots 
 and powers are therefore correlative terms. 
 
 The exact root of some numbers cannot be obtained. Such 
 numbers are called irrational powers, and their roots are called 
 surds. Thus, no root can be obtained, which, when multiplied 
 into itself, will produce 2 ; 2 is therefore an irrational power, 
 and its root is a surd. But a number whose root can be ex- 
 actly extracted is a perfect or complete power, and its root is 
 called a rational number. Thus, 16 is a complete power, for 
 4 is its exact root ; 4 therefore is a rational number. 
 
 There are two methods of expressing roots. The first and 
 more common method is, by using the character called the 
 radical sign ; written thus, V. This sign, without any accom- 
 21 
 
242 EVOLUTION. 
 
 panying index, always indicates the square root. If other roots 
 are required, the same radical sign is used, with an index of 
 the required root. Thus, V9, is an expression for the square 
 
 root ; v9, for the cube root ; y9, for the fourth root of 9, &c. 
 V64, equals 8, because 8x8=64; and v64=4, because 4 
 
 X4x4=64, or v64=2, for 2x2x2x2x2x2=64, &c. 
 Hence the root is to be taken as a factor in producing its cor- 
 responding power, as many times as there are units in the in- 
 dex of the required root. The other mode of expressing roots 
 
 is by means of fractional exponents. Thus, 6 5 expresses the 
 
 square root; 6 , the cube root, and 6 4 , the biquadrate or 
 fourth root of 6. The chief advantage of this mode arises 
 from the fact, that not only roots of numbers may be expressed 
 by it, but also any required power of a given root. The de- 
 nominator of a fractional index always denotes a root of the 
 quantity to which it is applied, while the numerator ex- 
 
 3 
 
 presses some power of that root; thus, 9' implies that the fourth 
 root of 9 is to be extracted, and that root raised to its third 
 
 power. Again, 64s" implies that the sixth root of 64 is to be 
 extracted, and the root then raised to its fifth power. But the 
 
 sixth root of 64 is 2, and the fifth power of 2 is 32 ; therefore, 
 
 j> 
 
 64 6 =32. Or a power higher than the given root may be ex- 
 pressed ; thus, 1 6"2, implies the third power of the square root 
 of 16 ; but the square root of 16 is 4, and the third power of 
 
 4 is 64 ; therefore, 16^= 64. 
 
 When several numbers are to be added and the root of the 
 sum obtained, it may be expressed thus : Vf,5-j_i6 ; which 
 implies that the root of the sum of 65 and 16 is to be obtained. 
 If the vinculum over the two numbers be rejected, the expres- 
 sion would imply, that 16 is to be added to the square root of 
 65. As the expression now stands, its value is 65+16 = 81, 
 and V8i = 9. Or the root of the difference of two quantities 
 may be expressed in like manner, by placing the minus sign 
 between them ; thus, VcjoUjje, the value of which is 90 26 
 =64, and V64 = 8. Without the vinculum over the two 
 quantities, the expression would imply that 26 is to be taken 
 from the square root of 90. 
 
 The root of the product of several numbers is equal to the 
 
EXTRACTION OF THE SQUARE ROOT. 
 
 243 
 
 product of their roots. As an illustration, take 9 and 16. Their 
 product is 144, of which the square root is 12 ; that is, the 
 root of their product is 12. The product of their roots is the 
 same, for the square of 9 is 3, and of 16 is 4 ; and 4 X 3 is 12. 
 The same is true of the cube roots, or of any roots whatever. 
 
 Take the numbers 8 and 27. y8=2, y27=3, and 3 x2=6, 
 
 the product of their roots. Again, 27x8=216, and V216 = 6, 
 the root of their products. 
 
 EXTRACTION OF THE SQUARE ROOT. 
 
 The Square Root of any number, is that number which be- 
 ing multiplied into itself once, will produce the number given. 
 
 The following table exhibits the square of all numbers from 
 1 to 12 : 
 
 Roots, 
 
 M |2| 
 
 3| 4| 
 
 5| 
 
 6 
 
 ! 7 
 
 1 
 
 9 
 
 10 | 
 
 11 
 
 1 12 
 
 Squares, 
 
 1 ! 1 4 1 
 
 9| 16 | 
 
 25 | 
 
 36 
 
 |4<J 
 
 |64 
 
 81 
 
 100 | 
 
 121 
 
 | 144 
 
 A square is a figure bounded by four equal sides, and has 
 all its angles right angles, or angles of 90 degrees. This may 
 be seen in figure 1st. Now the 
 area of a square, that is, the number 
 of square feet, or rods, &c., it con- 
 tains, is found by multiplying the 
 length of any two sides together ; 
 or since the sides are all equal, by 
 multiplying the length of any one 
 side into itself ; or by squaring it. 
 As each of the sides of the annexed 
 figure is 8 feet in length, it is obvi- 
 ous they may each be divided into 
 8 equal parts, each of which will be one foot in length. Let 
 each side be thus divided and the points of division united, as 
 
 
 FIG. 1st. 
 8ft. 
 
 90 90 
 
 d 
 
 
 00 
 
 
 
 90 90 
 
 8ft. 
 
aioL JUUL eduu. i nu wiiui e 33|34|35|3t>|37|38 39,40 \e 
 
 fi^re is therefore divided into /4 1|42|43|44|45|46|47|48/ 
 64 equal parts, each of which is /-Ll , ' . ,, ,.K 
 
 just one foci square. But 8x8 ^49|50|51| 5a | 5 3| 5 . 
 
 ft nm c .1 f I ^TI^Ql^OlAnl^l !OlCO 1C /< ! I. 
 
 244 EXTRACTION OF THE SQUARE ROOT. 
 
 seen in figure 2d. Now since PIG. 3d. 
 
 each side is 8 feet long, the di- 
 visions aa, bb, cc, dd, &c. must 
 be just one foot each ; the di- 
 visions made by the lines run- 
 ning at right angles to these, are d 25|26|27|28|29|30 
 also one foot each. The whole 
 
 2| 3| 4| 5[ 6| 7| 8> 
 
 9|10|U|12|13|14|15|16|6 
 
 l"7|18|19]20|21|22|23|24 
 
 = 64. Therefore, the area of a A57|58|59|60|61|62|63|64JA 
 square is obtained by multiply- 
 ing the length of the side by itself; or, in other words, by 
 squaring it. 
 
 Now to apply the above remarks to Evolution, suppose we 
 have an area equal to what is given above, but placed in a dif- 
 ferent form. Suppose it to consist of a board one foot wide 
 and 64 fe.et long ; and that it is required to determine how large 
 a square floor it will exactly cover. V64 = 8 feet, is the length 
 of the side of the required floor. Hence, the area of a square 
 is found by squaring one of its sides ; and the length of its 
 sides is found by extracting the square root of the given area. 
 
 The following is the rule for extracting the square root : 
 
 RULE 1st. Separate the given number into periods of two 
 figures each, by placing a point or dot over the place of units, 
 another over the place of hundreds, and another over the place of 
 tens of thousands, <fyc. 
 
 2d. Find by trial the greatest square root of the left hand 
 period, and place it for the first figure of the root, after the man- 
 ner of the quotient in division. 
 
 3d. Subtract the square of this root from the left hand pe- 
 riod, and to the remainder bring down the next period of two 
 figures for a dividend. 
 
 4th. Double the root figure already obtained, for a divisor ; 
 then, omitting the right hand figure of the dividend, divide as in 
 Simple Division, and place the figure obtained, as the second 
 figure of the root or quotient ; and also on the right hand of 
 -the divisor. 
 
 5th. Multiply the divisor thus increased, by the figure last 
 placed in the root, and place the product under the dividend, as in 
 Division ; then subtract, and to the remainder bring down the 
 next period of two figures. 
 
EXTRACTION OF THE SQUARE ROOT. 245 
 
 6th. Double the root already found for a new divisor, with 
 which divide as before. Continue the operation till the periods 
 are all brought down : the number obtained will be the root re- 
 quired. 
 
 Note 1st. If the number, the root of which is required, 
 consist in part of a decimal, place the first point over the unit 
 figure, as already directed, and point off both ways from that 
 figure, allowing two figures to each point. If the decimal 
 consist of an odd number of figures, annex one cypher to 
 complete the last period. 
 
 2d. If after all the periods have been brought down, there 
 be a remainder, periods of two cyphers each may be annexed 
 and the operation continued. The root figures obtained by 
 thus annexing cyphers will be decimals, and must be so 
 marked. 
 
 3d. The number of dots employed in pointing off the given 
 number, always determines the number of figures in the re- 
 quired root. 
 
 Ex. 1. What is the square root, of 1296 ? 
 
 1296 pointed according to the rule is 1296. Hence we 
 know that the root will consist of two figures. The next step 
 is to determine the root of 12, the left hand period. This is 
 done by trial. If 4 be taken, it will be found too large, since 4 
 X 4 = 16. We will, therefore, take 3. 3x3 9, and since 9 
 is less than 12, it is not too large, and yet it is the greatest in- 
 tegral quantity, whose square is less than 12, and is therefore 
 the root we want. 
 
 OPERATION CONTINUED. 
 
 1296(3 
 Root squared =9 
 
 396 remainder increased by the next 
 period. 
 
 Now since we are to have another figure in the root, the 3 
 already obtained is 3 tens or 30, the square of which is 30 X 
 30 = 900 ; the 12 is also 1200. After subtracting the 9, there- 
 fore, there remains 3, or 300, and the next period being brought 
 down, we obtain the number 396. 
 
 The next step is to find a divisor, which by the rule is 3 -f- 3 
 = 6 ; therefore, 
 
 21* 
 
246 EXTRACTION OF THE SQUARE ROOT. 
 
 1296(36 
 
 6 ) 3 
 In dividing, the right hand figure, viz. 6, is omitted. 
 
 Again, 1296(36 
 9 
 
 66)396 
 396 
 
 00 
 
 The last root figure is placed on the right of the divisor, 
 making it 66, and the whole is multiplied by the root figure, 
 that is, 66x6 = 396. I then subtract and nothing remains. 
 Therefore, 36 is the root required. 
 
 The operation is proved by squaring the root ; thus, 36 X 36 
 = 1296. 
 
 If the given number had consisted of three or four periods 
 instead of two, the operation would have been continued by 
 bringing down another period, and then doubling the root 36 for 
 a new divisor 
 
 Explanation. We will suppose the 1296 in the preceding- 
 operation to be so many feet of boards one foot in breadth ; and 
 that it is required to know how large a floor, exactly square, 
 they will cover. 
 
 As has already been said, the 12 in the number 1296 is 1200, 
 and the root 3 is so many tens, or 30. This, therefore, is the 
 length in feet of one side of a square floor, which 1200 feet 
 of the boards will cover, and leave a remainder of 3 or 300. 
 
 Now to find the area of a square, 
 (see fig. 3d,) we multiply the length FIG. 3d. 
 
 of any two sides together, or square 30 ft. 
 
 the length of one side. Therefore, 30 
 X 30 = 900. We have then disposed 
 of 900 of 1296 feet given, and there 
 remains 396 feet to be so added to 
 this figure as to preserve its square 
 form. This is done by making equal 
 additions upon any two adjacent 
 sides ; and hence we see the obvi- 
 ous reason for doubling the root, 
 (which is the length of one side of 
 
 Area 900 square feet. 
 
EXTRACTION OF THE SQUARE ROOT. 
 
 247 
 
 FIG. 4th. 
 
 36 ft. 
 
 1? 
 
 in 
 
 cr 1 
 in 
 
 900 sqr. ft. 
 
 00 
 
 
 P 
 
 6 ft. 
 
 
 
 
 S> 180 sqr. ft. 
 
 30ft. 
 
 the square,) for a divisor. But the root, figure 3d, being doubled, 
 gives 6 for a divisor, and this is contained in the remaining 
 figures, 396, 6 times, (the right hand figure, viz. 6, being 
 omitted, agreeably to the rule.) Now this figure, 6, expresses 
 the breadth of the addition which the remaining feet of boards 
 are sufficient to make to the original square, as seen at fig. 3d. 
 This addition is seen at fig. 4th. 
 This diagram is not a perfect 
 square ; a corner remains to be 
 filled up. We will, however, 
 before completing it, ascertain 
 how many of 396 feet, that 
 remained after the first square 
 of 900 feet was completed, are 
 here disposed of. The length 
 of each addition being 30 feet, 
 and the breadth 6 feet, the 
 area is 30x6 = 180 sq. feet; 
 and the area of both, conse- 
 quently, is 180+180= 360. 
 But 396360=36. There- 
 fore, 36 feet still remain to be added. The scholar, by refer- 
 ence to the preceding diagram, will perceive that the corner, 
 which yet remains to be filled, to complete the square, is just 
 6 feet from corner to corner; therefore, 6 X 6 = 36. This ad- 
 dition just disposes of the remaining feet of boards, and com- 
 pletes the square, as may be seen at fig. 5th. The area of the 
 original square, as seen at fig. 
 3d, and also of the several addi- 
 tions made at figures 4th and 5th, 
 disposes of the whole of the given 
 quantity of board ; for 900+180 
 + 180+36 = 1296. But why, 
 in dividing, is the right hand figure 
 of the dividend omitted ? 
 
 The scholar will remember 
 that the points placed over any 
 number, determine the number of 
 figures in its root. (See Note 3d.) 
 Before performing the operation, 
 we therefore know that the re- 
 quired root of the preceding num- 
 ber wiU consist of two figures. The 3, or left hand figure of 
 
 Fia. 5th. 
 
 36ft. 
 
 i 
 
 | 
 
 800 sqr. ft. 
 
 6 ft. 
 
 
 36 
 
 sq.ft. 
 
 180 sqr. ft. 
 
 30ft. 
 
248 EXTRACTION OF THE SQUARE ROOT. 
 
 that root, is therefore 3 tens, or 30. This number, viz. 30, is 
 consequently the true value of the root already found, which, 
 if doubled, will give 60 as the divisor, and not 6, as in the 
 sum. The true value of the divisor is therefore ten times as 
 great as represented in the operation, and hence the dividend 
 is divided by 10; that is, its right hand figure is omitted, to 
 make its value correspond with the apparent value of the 
 divisor. 
 
 Another peculiar feature of the operation consists in placing 
 the quotient, or root figure, on the right of the divisor, thereby 
 multiplying it into itself. For this, there must also be a reason. 
 If the scholar will examine figure 4th, he will notice a vacant 
 corner, which, as the addition made to each of the two adjacent 
 sides, is 6 feet, must be just 6 feet square. By placing the 
 root figure on the right of the divisor, and multiplying it into 
 itself, this corner is filled up and the square completed. 
 
 2. What is the square root of 9801 ? 
 
 OPERATION. 
 
 9 8 i ( 9 9 root. 
 8 1 
 
 189)1701 
 1701 
 
 The 9 in the divisor is the last root figure, placed there by 
 the rule. 
 
 3. What is the square root of 30138.696025 ? 
 
 OPERATION. 
 
 3013 8. 6 96025 (173. 605 
 1 
 
 27)201 
 1 89 
 
 343)1238 
 1029 
 
 3466)20969 
 20796 
 
 347205) 1736025 
 1 736025 
 
EXTRACTION OF THE SQUARE ROOT. 249 
 
 It will be observed from this example, that when a period 
 is brought down, and the number obtained is not sufficient 
 to contain the divisor, a cypher is to be placed in the root, and 
 also on the right hand of the divisor, and the next period is 
 then to be brought down. For arranging the points in this sum, 
 see Note 1st. 
 
 4. What is the square root of 5499025 ? Ans. 2345. 
 
 5. What is the square root of 1522756 1 Ans. 1234. 
 
 6. What is the square root of 207936 ? Ans. 456. 
 
 7. What is the square root of 5700.25 ? Ans. 75.5. 
 
 8. What is the square root of 74770609 ? Ans. 8647. 
 
 9. What is the square root of 419112517321? Ans. 
 647389. 
 
 10. What is the square root of 10 ? Ans. 3. 16227. -f 
 
 1 1 . What must be the length of each side of a square field, 
 in rods, in order that the field may contain 40 acres 1 Ans. 
 80 rods. 
 
 12. A certain regiment consists of 6561 men. How many 
 must be placed in rank and file, to form them into a square ? 
 Ans. 81 men. 
 
 13. A company of men spent 6 . 13s. 4 d., which was just 
 as many pence for each man as there were men in the company. 
 How many men were there, and how many pence did each 
 man spend ? Ans. There were 40 men, and each man spent 
 40 d. 
 
 14. If a man were to plant 3969 hills of corn in a square, 
 how many rows must he have, and how many hills in a row 1 
 Ans. 63 of each. 
 
 Note 4th. To extract the square root of a vulgar fraction, 
 first reduce the fraction to its lowest term, and then extract the 
 root both of the numerator and denominator. 
 
 15. What is the square root of 
 
 ^ __ 9_ and the square root of 9 is 3 . 
 128 16 and ...... 16 is 4' Ans ' 
 
 16. What is the square root of y^-? Ans. |-. 
 
 17. What is the square root of jyf ? Ans. . 
 
 18. There is an army containing 9216 men. How many 
 men must be placed rank and file, to form a square which 
 shall contain every man ? Ans. 96. 
 
 19. What is the square root of |f|- ? Ans. f . 
 
250 
 
 EXTRACTION OF THE SQUARE ROOT. 
 
 APPLICATION OF SQUARE ROOT TO PARALLELOGRAMS AND 
 TRIANGLES. 
 
 A parallelogram is an oblong figure having its opposite sides 
 equal and parallel. The adjoining figure represents the paral- 
 
 FIG. 6th. 
 
 lelogram. If we suppose this figure to be 10 feet long, and 
 2 feet broad, the area of the whole figure is evidently 10x2 
 20 square feet. The area of a parallelogram is therefore 
 found by multiplying its length into its breadth. Had the 
 length of the above figure been 18 feet, and its breadth as given 
 above, the area would have been 18 x 2 = 36 square feet ; and 
 this equals a square figure, the sides of which are 6 feet long, 
 for V36 6. A square, equal in area to a given parallelogram, 
 is found by extracting the square root of the area of that paral- 
 lelogram. 
 
 20. What is the length of the sides of a square, whose 
 area shall be equal to the area of a parallelogram 32 feet in 
 length and 2 in breadth ? 
 
 Operation : 32x2 = 64, and V64 = 8 feet, the side of the 
 required square. 
 
 21. There is a parallelogram 27 feet in length and 3 in 
 breadth. Required the sides of a square of equal dimensions. 
 Ans. 9 feet. 
 
 22. Required the dimensions of a square, the area of which 
 shall be equal to the area of a parallelogram 18 feet in length 
 and 8 in breadth. Ans. 12 feet square. 
 
 Note 5th. When the area of a parallelogram is given, and 
 also the ratio of its length and breadth, the sides may be found 
 by dividing the area by that ratio, and extracting the square 
 root of the quotient. The root obtained will be the required 
 breadth, by which divide the area, and the quotient will be the 
 length. 
 
 23. If the area of a parallelogram be 288 rods, and its length 
 be twice as much as its breadth, what is its length, and what 
 its breadth ? 
 
EXTRACTION OF THE SQUARE ROOT. 251 
 
 Operation : 288 -r- 2 = 144 and, ViU= 12, the required breadth ; 
 and 288 -4- 12=24, the required length. The reason of the above 
 operation is obvious. The length being twice its breadth, if 
 the parallelogram be divided in the middle, the two parts will 
 be equal and square, and the square root of one of these parts 
 will evidently be the breadth of the parallelogram. The same 
 reasoning may be applied to parallelograms of any dimensions. 
 
 24. There is a field containing 30 acres, lying in the form 
 of a parallelogram, of which the length is three times the 
 width. What is its length, and what is its breadth 1 Ans. 
 The length is 120 rods, and the breadth 40 rods. 
 
 Note 6th. The side of a square of equal area with a geo- 
 metrical figure of any form, may be found by extracting the 
 square root of the area of that figure. 
 
 25. I have an irregular piece of land containing 50f acres, 
 which I am desirous to exchange for an equal number of acres 
 lying in a square form. What must be the length of the sides 
 of that square ? Ans. 90 rods. 
 
 26. A certain triangular field contains 10 acres of land. 
 What is the length of the sides of a square containing the 
 same number of acres ? Ans. 40 rods. 
 
 The principle of the square root may be applied to find the 
 length of the sides of a right angled triangle, the measure 
 of either two being given. 
 
 The square of the hypoteneuse or side opposite the right 
 angle, is always equal 
 
 to the sum of the squares ^^\ , 
 
 of the base and perpen- ^^ 
 
 dicular. Therefore, 
 
 If the base and per- 
 pendicular be given, and 
 the hypoteneuse requir- 
 ed, square the given 
 sides, and extract the JBase. 
 
 square root of their sum. 
 
 If the hypoteneuse, and one of the legs of the triangle be given, 
 to find the other leg, square the hypoteneuse, and from its square 
 subtract the square of the given side. The square root of the 
 remainder will be the length of the required side. 
 
252 EXTRACTION OF THE SQUARE ROOT. 
 
 27. There is a wall 15 feet high, and in front of it is 
 a pavement 24 feet wide. How long a ladder is required to 
 reach from the outside of the pavement to the top of the wall? 
 The hypoteneuse is required, therefore, 15xl5=z225; and 
 24x24 = 576; then, 225 + 576 = 801 ; and V801=:28.3,4- Ans. 
 
 28. A certain tree is broken off 8 feet from the ground, and, 
 resting on the stump, touches the ground at the distance of 12 
 feet. What is the length of the part broken off? Ans. 
 14.42+ feet. 
 
 29. There is a fort standing by the side of a river, 24 yards 
 high, and a line 36 yards long will just reach from the top of 
 the fort to the opposite side of the river. What is the width 
 of the river? Ans. 26.832 yards. 
 
 30. Two ships sail from the same port, one due east, and 
 the other due north. What is the distance between them, 
 when one has sailed 100 miles, and the other 168 miles? 
 Ans* 195. 5+ miles. 
 
 31. A man shot a bird sitting on the top of a steeple 80 
 feet high, while standing at the distance of 60 feet from its 
 base. How far did he shoot? Ans. 100 feet. 
 
 32. A rope 100 feet long attached to the top of a steeple, 
 touches the ground when drawn perfectly straight, 20 feet 
 from its base. How high is the steeple ? Ans. 98 feet, 
 nearly. 
 
 33. Two boys were playing with a kite, the line of which 
 was 520 feet in length. When the string was all out> one of 
 them standing directly under the kite, and the other holding the 
 string, the distance between them was 312 feet. What was 
 the perpendicular height of the kite ? Ans. 416 feet. 
 
 QUESTIONS. When is a number involved 1 What is a root 1 What 
 is a power of any number ? On what does the particular power pro- 
 duced depend ? How does the power obtained compare with the num- 
 ber of multiplications in producing it 1 How is a required power ex- 
 pressed 1 What is the figure denoting the power called ? How is a 
 fraction involved ? If the given quantity be a mixed number, what 
 must be done 1 If a number be raised to two different powers, how is 
 the power obtained by multiplying these two powers together, ex- 
 pressed 1 Ho\^ r is any power of a given number divided by another 
 power of the same number'? When the number to be raised to a 
 power is in part a decimal, how is the number of decimals to be cut off 
 from the required power ascertained 7 What is evolution 1 What 
 is a root of a number 7 What is a power of any number? What 
 are irrational powers'? What are surds'? How many methods are 
 there of expressing roots 1 What is the first method 1 What root is 
 expressed by the radical sign without any index or exponent 1 If 
 
EXTRACTION OF THE CUBE ROOT. 
 
 253 
 
 any other root is to be expressed, how is it done ! How many times is 
 the root to be taken as a factor in producing its correspondieg power 1 
 What is the other mode of expressing roots'? What is an advantage 
 of this mode 1 When a fractional index is used, what does the de- 
 nominator denote ? What does the numerator ? When several num- 
 bers are to be added, and the root of the sum extracted, how is the op- 
 eration expressed 7 What does the root of the product of several 
 numbers equal 1 Give an illustration. What is the square root of 
 any number 1 What is a square 1 How is the area of a square found 1 
 Give the illustration. How is the length of the sides of a square 
 found 1 What is the rule for extracting the square root 1 What is Note 
 1st 1 Note 2d 1 Note 3d 1 Why do we take twice the root for a divisor 1 
 Why in dividing do we omit the right hand figure of the dividend 1 
 Why do we place the quotient figure on the right of the divisor 7 
 How is the square root of a vulgar fraction extracted 1 The vulgar 
 fraction may first be reduced to a decimal and the root of the decimal 
 extracted, if preferred. What is a parallelogram 1 How is its area 
 found 1 How may a square equal in area to a given parallelogram be 
 found 1 What is Note 5th 1 Note 6th 1 To what is the square of 
 the hypoteneuse of a right angled triangle equal 1 If the base and 
 perpendicular be given, how may the hypoteneuse be found 1 If the 
 hypoteneuse and one of the legs of a triangle be given, how may the 
 other leg be found 1 ? The base and perpendicular are called the legs 
 of a triangle. How are the operations in Square Root proved 7 Ans. 
 By multiplying the root into itself. 
 
 EXTRACTION OF THE CUBE ROOT. 
 
 IS 
 
 bounded by six equal plane surfaces, each of 
 is a square ; that is, the length, breadth, and depth of a 
 
 FIG. 1. 
 
 A CUBE 
 
 which 
 
 cube are equal. (See fig. 1st.) The area of each of the six 
 equal surfaces is found by squaring the measure of its side, as 
 has already been explained in Square Root. 
 
 If the adjoining figure represent 
 a cubic block measuring three feet 
 in length, breadth, and thickness, the 
 superficial area of each face is 3 x 
 3=9 square feet. Now if the divi- 
 sions marked by the dotted lines on 
 each face of the block, were extend- 
 ed through it, in either direction, the 
 whole would be divided into 9 parts, 
 each one foot a square and 3 feet long, 
 and susceptible of being divided each 
 into three, and consequently the whole, into 27 blocks, each one 
 22 
 
254 EXTRACTION OF THE CUBE ROOT. 
 
 cubic foot. We have then the following general principle. 
 The content of a solid or cubic body is found 'by multiplying its 
 length, breadth, and thickness into each other ; or, what is the 
 same in effect, by cubing one of these dimensions. 
 
 Extracting the cube root is a process, the reverse of the pre- 
 ceding, that is, it is finding the length of one of the sides of a 
 cubic body, the solid content of that body being given ; or it is 
 finding from a given number, another number, whose cube or 
 third power shall equal that number. 
 
 RULE 1st. Separate the given number into periods of 
 three figures each, by placing a point first over the unit figure, 
 and advancing toward the left when the number consists of in- 
 tegers only ; but to the right and left both, when it consists of 
 integers and decimals, and make the last period of the decimal 
 complete, by annexing cyphers, whenever necessary. 
 
 2d. Find by trial the greatest cube root of the left hand pe- 
 riod, and place it as in square root ; then subtract its cube from 
 the same period, and bring down the next period of three figures 
 to the remainder for a dividend. 
 
 3d. Square the root figure and multiply its square by 3 for 
 a divisor, and see how many times it is contained in the divi- 
 dend, omitting the first two right hand figures, and place the 
 result as the second figure in the root. 
 
 4th. Multiply the divisor by the last quotient figure, and, 
 placing two cyphers on the right of the product, place the result 
 under the dividend. Also multiply the square of this same last 
 quotient figure by the former figure or figures of the root, and 
 also by 3 ; and, placing one cypher on the right of the product, 
 write it under the preceding product. Lastly, under this, write 
 the cube of the last quotient or root figure, and make the sum 
 of these three numbers a subtrahend. 
 
 5th. Subtract the subtrahend from the dividend, and to the 
 remainder bring down the next period for a new dividend. 
 
 6th. To obtain a new divisor proceed as before, and thus 
 continue the operation, till all the periods of the given number 
 have been brought down. 
 
 Note 1st. In obtaining each divisor, square the whole root 
 obtained, and multiply that square by 3. 
 
 Ex. 1. What is the cube root of 10648 ? 
 
EXTRACTION OF THE CUBE ROOT. 
 
 255 
 
 OPERATION. 
 
 10648(22 
 
 2 3 =8 
 
 Art. 3d, Rule, 2ax3 = 12 (Div.) 12)2 648 (See Art. 2d.) 
 
 ("12x2 + 00= 2400 
 Rule, Art. 4th, J 22x2x3 + 0= 240 
 
 2 3 = 8 
 
 Subtrahend, 2648 
 0000 
 
 In dividing, the two right hand figures of the dividend are 
 omitted. (See rule, Art. 3.) Proof, 22 3 = 10648. 
 
 Explanation. -We will suppose the number 10648, in 
 the preceding sum, to be so many feet of timber, one foot 
 square, and that it is required to find how large a cubic pile 
 they will form ; that is, what will be the length, breadth, and 
 depth of a cubic pile containing that number of solid feet. To 
 make each step as clear as possible, we will repeat the prece- 
 ding operation. 
 
 The number given when pointed, (see rule,) is divided into 
 two periods, viz. 10 and 648. We therefore know that the re- 
 quired root will consist of two figures, and by trial we find the 
 root of the first or left hand period to be 2, that is, 2 tens, or 
 20. Hence, 10648(20. The 
 same as is seen in the above 
 operation, excepting that a 
 cypher is placed on the right 
 of the root figure 2, to give it 
 its true value. This gives the 
 linear measure of a cubic 
 block which the 10(10.000) 
 of the given number will 
 make. Now, since 20 is the 
 linear measure of the cube, 
 (that is, the direct and not 
 diagonal measure from corner 
 to corner,) it is also the linear measure of each of the six equal 
 square faces of that cube. Therefore, 20 X 20=400, the area 
 of each face ; and 400x20 8000, the number of cubic feet 
 required to make a cubic body, whose linear measure is 20 ft. 
 
 FIG. 2. 
 
256 
 
 EXTRACTION OF THE CUBE ROOT. 
 
 (See fig 2.) By this step we have then disposed of 8000 of the 
 10648 solid feet. Hence, 
 
 10648(20 
 8000 
 
 2648 
 
 (This same effect is obviously produced in the first solution of 
 this sum, by cubing the 2, and subtracting it from the left hand 
 period, 10, and then bringing down the next period, 648, to the 
 remainder.) There now remains 2648 feet to be so added to 
 the block already formed, that the whole shall be a perfect cube. 
 This is done by making equal additions on any three of the 
 equal faces which lie contiguous to each other. The reason of 
 this is obvious. A solid body has length, breadth, and thick- 
 ness ; and in a cubic body, these dimensions are all equal, and by 
 making the additions as here directed, they are equally increased. 
 
 The scholar will now understand why three times the square 
 of the root obtained is taken as a divisor. The square of the 
 root is the superficial area of one of the sides or faces of the 
 cubic block, and this multiplied by 3 gives the area of three 
 faces or sides, which is the number of sides to which equal 
 additions are to be made. Hence, dividing the quantity to be 
 added to the cube now obtained by this area, determines the 
 thickness of the addition. This, in the sum now under con- 
 sideration, is 2 feet, as seen at fig. 3. But these additions are 
 evidently limited in size to the original block ; consequently, 
 the corners E, E, E, (fig. 3,) remain to be filled before a per- 
 fect cube is produced. 
 
 Now to determine the FIG. 3. 
 
 quantity here added. Each 
 
 face of the original cube ^/ g00 ffc 
 
 contains 400 square feet, 
 (see fig. 2,) and this multi- 
 plied by 2, the depth of the 
 addition, gives 800 solid 
 
 feet as the content of the tiuv /, ^ 800t 
 
 addition made to each face ; 
 the whole addition there- 
 fore is 800 x 3=2400 solid 
 feet, and 2648 2400= 
 248 solid feet, the quantity 
 yet remaining to be dis- 
 posed of. 
 
 1 
 
 / 
 
 
 / 
 
 
 800 ft. 
 
 
 
 K. 
 
 Solid 60nfcnt 
 
 
 
 104:00 f& 
 
 V* 
 
 ?ft 
 
EXTRACTION OF THE CUBE ROOT. 
 
 257 
 
 OPERATION CONTINUED. 
 
 10648(22 
 8 
 
 2 2 x3 = 12div. - - - 12)2648 
 Solid con. of 3 additions, 12x2 + 00,(Rule,) 2400 
 
 The reason for placing two cyphers on the right of the divi- 
 sor multiplied by the root figure, is obvious. By referring to 
 a previous statement of this sum, it will be seen that the root 
 figure 2, which, when squared and multiplied by 3, forms the 
 divisor, is 2 tens, or 20. Hence, 20 X 20 =400, and 400 X 3 = 
 1200, which would be the divisor, were the full value of the root 
 figure expressed. This deficiency in the divisor is made up 
 by omitting the two right hand figures of the dividend in divi- 
 ding, and by placing two cyphers on the right of the product of 
 the divisor multiplied by the root figure. 
 
 Our next step will be to fill the vacant corners as seen at 
 E,E,E,(fig. 3.) 
 
 The rule, after directing the preceding step, says, " Also 
 multiply the square of this same last quotient figure by the 
 former quotient figure or figures, and also by 3, and placing one 
 cypher on the right of the product, write it under the preceding 
 product" This operation 
 fills the corners here allu- 
 ded to. For since the last 
 quotient figure 2, is the thick- 
 ness of the addition made, 
 its square, viz. 4, is the 
 measure of the corner to be 
 filled, and the preceding 
 quotient figure is the length 
 of each corner ; hence, 4 X 
 2 + = 80, the solid content 
 of one corner, (see fig. 4.) 
 The cypher here is added 
 because 2, the preceding 
 
 quotient figure, is 2 tens, or 20. The three corners therefore 
 require 80 x 3=240 feet to fill them. The corner, C, still re- 
 mains to be filled, and 8 feet of timber also remain, for 10648 
 10640 = 8. This vacant corner, C, measures exactly 2 feet 
 22* 
 
 FIG. 4. 
 
 800 ft. 
 
 ^ 
 
 2p 
 
 80ft;. 
 
 600 ft: 
 
 80 
 
 Solid Content 
 
 
 10610. 
 
 ) 
 
 r> 
 
258 
 
 EXTRACTION OF THE CUBE ROOT. 
 
 FIG. ft. 
 
 800. 
 
 / 
 
 / 
 
 80. 
 
 8 
 
 000. 
 
 
 Solid Content 
 06*8 Feet. 
 
 80 
 
 
 in each of its three dimen- 
 sions ; hence, 2 3 =8, is the 
 solidity of that corner; and 
 this exactly disposes of the 
 remaining timber. The cube 
 completed is seen at fig. 5. 
 The contents of the differ- 
 ent parts of the completed 
 block are, fig. 2, 8000+ ; fig. 
 3, 2400+ ; fig. 4, 240+ ; 
 fig 5, 8=10648 feet. 
 
 Note 1st. In Square 
 
 Root we were directed to 
 
 point off the given number 
 
 into periods of two figures each.; and in Cube Root, the direc- 
 tion is, to allow three figures to each period. The following is 
 the reason : the square of any number always consists of twice 
 as many figures as the number itself, or one less than twice as 
 many. The cube of any number always consists of three 
 times as many figures as the number itself, or one or two less 
 than three times as many. That is, in Square Root, the left 
 hand period may consist of one or two figures ; and in Cube 
 Root, the same period may consist of one, two, or three figures. 
 
 Illustration. 13 2 = 169, one less than twice the number of 
 figures squared, and to extract its root it would be thus pointed, 
 169. 462=2 1 16, twice the number of figures squared. 13 3 = 
 2197, two less than three times the figures cubed, and the left 
 hand period consists of one figure only. 25 3 = 15625, one less 
 than three times the figures cubed, and the left hand period 
 consists of two figures. 99 3 = 970299, three times the number 
 of figures cubed, and the left hand period consists of three 
 figures. 
 
 In the following solution, the scholar will carefully compare 
 each step of the operation with the rule. The first thing to be 
 done, is to form the periods, (Art. 1st, Rule.) The first figure 
 of the root is then to be determined, its cube subtracted, and 
 to the remainder, the next period of three figures to be brought 
 down. (Art. 2d, Rule.) He must then proceed to obtain a divi- 
 sor, as directed by Art. 3d, and, lastly, to determine the solidity 
 of the several additions made, and to make the result a sub- 
 trahend. (Art. 4th.) 
 
 Ex. 2. What is the cube root of 12812904 ? 
 
EXTRACTION OF THE CUBE ROOT. 
 OPERATION. 
 
 259 
 
 12812904(234 
 - 8 
 
 2 2 x 3 = 12 (Divisor) - - - 12)4812 
 12x3 + 00 - - 
 3 3 - 
 
 - 3600 
 
 - 540 
 
 27 
 
 4 167= Subtrahend. 
 
 23 2 x 3 = 1587 (Divisor) - 1587)645904 = New dividend. 
 
 1587x4+00 - 
 
 4 2 x23x3 + - 
 
 4 3 - 
 
 634800 
 11040 
 64 
 
 645904= Subtrahend. 
 
 000000 
 Proof, 234 3 = 12812904. 
 
 It is obvious from the first division by 12, that the divisor is 
 not contained in the dividend, in all instances, as many times 
 as it would be in simple division. 
 
 In dividing it must be remembered to omit the first two figures 
 on the left hand of the dividend. 
 
 3. What is the cube root of 250047 ? Ans. 63. Proof, 63 3 
 =250047. 
 
 4. What is the cube root of 970299 1 Ans. 99. Proof as 
 before. 
 
 5. What is the cube root of ] .953125 1 Ans. 1.25. 
 
 6. What is the cube root of 22069810125 ? Ans. 2805. 
 
 7. What is the cube root of 183250432 ? Ans. 568. 
 
 8. What is the cube root of 84.027672 ? Ans. 4.38. 
 
 9. What is the cube root of 6859 ? Ans. 19. 
 
 10. What is the cube root of 205379 ? Ans. 59. 
 
 11. What is the cube root of 432081216 ? Ans. 756. 
 
 1 1 . There is a cubic rock containing 8000 solid feet. What 
 is the distance from corner to corner 1 Ans. 20. 
 
260 EXTRACTION OF THE CUBE ROOT. 
 
 13. What is the difference between half of a solid foot, and 
 a solid half foot ? Ans. 3 solid half feet. 
 
 14. What is the superficial area of one of the faces of a 
 cubic block containing 4096 solid feet? Ans. 256 square feet. 
 
 15. What is the side of a cubical mound, equal to one, 144 
 feet long, 108 feet broad, and 24 feet deep? Ans. 72 feet. 
 Multiply together the several dimensions of the given mound, 
 and extract the cube root of their product. 
 
 Note 2d. AH solid bodies are to each as the cubes of their 
 similar sides or diameters. 
 
 16. If a ball weighing 8 Ib. be 6 inches in diameter, what 
 will be the diameter of another ball of the same material, weighing 
 fi4 Ib ? a 
 
 8 : 64 : : 6 3 : 1723 vT&B=Ans. 12 inches. 
 
 17. If a ball 6 inches in diameter, weigh 8 Ib. what is the 
 weight of another ball of the same kind, measuring 12 inches 
 in diameter ? Ans. 64 Ib. 6 3 : 12 3 : : 8 : 64. 
 
 18. What would be the value of a globe of silver, one foot 
 in diameter, if a globe of the same, one inch in diameter, be 
 worth $6? A ns. $10368. 
 
 19. If a globe of silver, one inch in diameter, be worth $6, 
 what is the diameter of another globe of the same metal, worth 
 $10368? Ans. 12 inches. 
 
 20. How many globes one foot in diameter, would be required 
 to make one globe 27 feet in diameter ? Ans. 19683. 
 
 21. Suppose the diameter of the sun to be 110 times as 
 large as that of the earth ; how many bodies like the earth would 
 be required to make one as large as the sun ? Ans. 1331000. 
 
 22. If a man dig a square cellar, that will measure 5 feet 
 each way, in one day, how long would it take him to dig one 
 measuring 15 feet each way ? Ans. 27 days. 
 
 QUESTIONS. What is a cube 1 How is the area of each face of a 
 cubic body found 1 How is the content of a cubic body found 1 What 
 is the extraction of the cube root ^ How is the number whose root is to 
 be extracted, to be pointed 1 Of which period is the root first found 1 
 What is then done with this root 7 How many figures are to be brought 
 down to what remains ? How is a divisor found 1 By what do you 
 multiply the divisor 7 How many cyphers do you place on the right 
 of the product 1 Where do you place the product 1 What far- 
 ther is done with the quotient or root figure 1 What does the sum of 
 all these products form ? What is the fifth step of the rule 1 How is 
 a second divisor obtained 1 What is Note 1st? Can we know of 
 how many figures the root will consist 1 Of what is the root fi^mv 
 the linear measure 1 How much of the whole timber is disposed of 
 
ARITHMETICAL PROGRESSION. 261 
 
 by subtracting the cube of the quotient figure from the left hand period, 
 in the operation taken for explanation ? How many feet remain to 
 be added 7 To how many sides of a cube must equal additions 
 be made to 'preserve .its cubic form'? Why is three times the 
 square of the root taken for a divisor? What is determined by 
 dividing 7 ? How many solid feet are disposed of by the first addition 
 made to the three faces of the cube 1 Explain how the solid content 
 of the addition to each face is obtained. Why, in multiplying the 
 divisor by the root figure, are two cyphers placed on the right of the 
 product ? How is the deficiency of the divisor made up in dividing 1 
 What operation as directed by the rule fills up the corners left vacant 1 
 Why is the square of the quotient figure multiplied by the previous 
 root figures and a- cypher placed on the right hand of the product 1 
 How much of the remaining timber is required to fill the three vacant 
 corners 1 What is the measure of the corner left vacant, after the 
 preceding additions were made 1 How is it filled 1 Why in Square 
 Root do we divide the given number into periods of two figures each'? 
 Why is the given number divided into periods of three figures each, 
 in Cube Root 1 Of how many figures does the square of any number 
 consist 1 Of how many does the cube 1 How are the roots proved 1 
 Ans. By raising the root to a power of the same name as the root itself. 
 
 ARITHMETICAL PROGRESSION. 
 
 Any series of numbers more than two, increasing or decreas- 
 ing by a constant and uniform difference, is called Arithmet- 
 ical Progression, or Arithmetical Series. 
 
 The series formed by a continual addition of any number, 
 (called the common difference,) is called the ascending series. 
 Thus, 3, 5, 7, 9, 1 1 ,] 3, &c. is an ascending series, of which the 
 common difference is 2. The reverse of this forms the de- 
 scending series ; that is, a series decreasing by a continual 
 subtraction of the common difference. Thus, 13, 11, 9, 7, 5, 
 3, &c. is a descending series. 
 
 The numbers constituting the series are called terms. The 
 first and last term of the series are called extremes ; all the 
 intervening terms are called the means, and the number con- 
 stantly added or subtracted, is called the common difference. 
 When the first term and common difference are given, the 
 series may be indefinitely extended. 
 
262 ARITHMETICAL PROGRESSION. 
 
 In every arithmetical progression, five particulars are to be 
 noticed ; of which, if any three be given, the remaining two 
 may be found. The five particulars are, the first term, the last 
 term, the common difference, the number of terms, and the sum 
 of all the terms. 
 
 CASE 1st. THE FIRST TERM, THE NUMBER OF TERMS, AND THE 
 
 LAST TERM GIVEN, TO FIND THE COMMON DIFFERENCE. 
 
 Ex. 1. The first term of an arithmetical series is 2 ; the 
 number of terms, 13 ; and the last term, 38. What is the com- 
 mon difference ? 
 
 The series commences with 2, as the first term ; therefore, 
 38 2;= 36, is the amount of all the additions made to this 
 number. But the whole number of terms being 13, and one 
 of these being given, 12 terms must be formed by adding the 
 common difference. Therefore, 36-;- 12 = 3, the common dif- 
 ference required. The whole series, therefore, is, 2, 5, 8, 11, 
 14, 17, 20, 23, 26, 29, 32, 35, 38 ; thirteen in number. 
 
 We have then the following rule for solving sums like the 
 preceding : 
 
 RULE. Divide the difference of the extremes by the number' 
 of terms, less one. The quotient will be the common difference. 
 
 2. A man in feeble health, commenced a journey and trav- 
 eled 9 days. On the first day he traveled only 3 miles, but 
 afterwards continued to gain each day an equal number of 
 miles on the journey of the preceding day, till the last day, 
 on which he traveled 43 miles. The daily increase is re- 
 quired? Ans. 5 miles per day ; that is, 43 3-^-8 = 5. 
 
 3. I owe a debt which by agreement I am to pay at 17 dif- 
 ferent periods. The first payment is to be $20, and the last, $100. 
 Required the common difference of the several payments. 
 Ans. $5. 
 
 4. A man had 10 sons whose ages differed alike; the 
 youngest of whom was 2 years old, and the oldest 29. What 
 was the difference of their ages ? Ans. 3 years. 
 
 CASE 2d. THE FIRST TERM, THE COMMON DIFFERENCE, AND 
 
 THE LAST TERM GIVEN, TO FIND THE NUMBER OF TERMS. 
 
 Ex. 1. If the extremes be 3 and 51, and the common dif- 
 ference 6, what is the number of terms ? 
 
ARITHMETICAL PROGRESSION. 263 
 
 51 3=: 48, the amount of all the additions made to the first 
 term, 3 ; and since each addition is 6, 48^6 = 8, the number 
 of additions ; that is, the number of terms formed by adding 
 the common difference to the first term ; therefore, 8+1=9, 
 the whole number of terms, or answer required. 
 
 From the above we derive the following rule : 
 
 RULE. Divide the difference of the extremes by the common 
 difference, and add one to the quotient. 
 
 2. A man commenced a journey, and traveled the first day 
 only 4 miles ; after which he gained each day 6 miles on the 
 journey of the preceding day, and on the last day he traveled 
 94 miles. How many days did he travel? Ans. 16. 
 
 Operation : 94 4 = 90, and 90-f-6+l = 16. 
 
 3. A man commenced a journey in great haste, and traveled 
 63 miles the first day ; but being unable to continue at the 
 same rate, the second day he traveled only 59 miles ; and 
 thence continued to lose 4 miles per day, till the last day of 
 his journey, on which he traveled 1 \ miles. How many days 
 did he travel? Ans. 14 days. 
 
 4. A man set out on a journey for the improvement of his 
 health; the first day he traveled 10 miles, and the last, 65 
 miles, making each day an advance of 5 miles on the journey 
 of the preceding day. How many days did he travel ? Ans. 
 12 days. 
 
 CASE 3d. THE FIRST TERM, THE LAST TERM, AND THE NUM- 
 BER OF TERMS GIVEN, TO FIND THE SUM OF ALL THE TERMS. 
 
 Ex. 1 . Bought 30 yards of cloth, paying 20 cents for the 
 first yard and 50 cents for the last. What was the whole cost, 
 allowing each succeeding yard to increase in price by a con- 
 stant access ? 
 
 The price of each succeeding yard increases by a constant 
 access ; therefore, the price of the last is as much above the 
 average price as the price of the first is below it ; hence, one 
 half of the price of the first and last yard is the average 
 price. Therefore, 20+50=:70 and 70-^-2 = 35 cents, average 
 price ; hence, 30x35 = $10.50, whole cost. 
 
264 GEOMETRICAL PROGRESSION. 
 
 We have the following rule : 
 
 RULE. Multiply half the sum of the extremes by the num- 
 ber of terms ; the product will be the answer. 
 
 2. Paid 4 cents for the first, and $1.21 for the last yard of 
 a piece of cloth containing 86 yards. What was the whole 
 cost? Ans. $53.75. 
 
 3. How many strokes does a regular clock strike in 24 hours ? 
 Ans. 156. 
 
 4. If a person walk 3 miles the first, and 91 miles the last 
 day of his journey, how far will he have walked, allowing 
 him to have been on his journey 24 days ? Ans. 1128 miles. 
 
 CtUESTioNS. What is arithmetical progression 1 What is the as- 
 cending series 7 What is the descending series 7 What is denoted 
 by the word, terms ? What terms are the extremes 1 What are the 
 means'? What is the number constantly added or subtracted called"? How 
 many particulars require to be noticed 1 What are they'? What is 
 Case 1st ? What is the rule 1 What is Case 3d 1 What is the rule 1 
 What is Case 3d 1 What is the rule 1 
 
 GEOMETRICAL PROGRESSION. 
 
 Any series of numbers, increasing by a common multiplier, 
 or decreasing by a common divisor, is called Geometrical 
 Progression, or Geometrical Series. The common multiplier 
 of the ascending series, and the common divisor of the descend- 
 ing scries, is called the ratio of the series, or the common ratio. 
 Thus, if we take 3 as a first term, and multiply it continually 
 by 2, as a common ratio, we obtain the series 3, 6, 12, 24, 
 48, 96, 192, 384, 768, &c., in which series eacnterm is obtain- 
 ed by multiplying the preceding term by 2. We have here an 
 ascending series. Or the series may be 768, 384, 192, 96, 
 48, 24, 12, 6, 3, &c., in which, each term is obtained by divi- 
 ding the preceding term by 2. This forms a descending series. 
 The several numbers thus produced constitute the terms of the 
 series ; the first and last of which are called the extremes; and 
 the intervening ones are called the means. 
 
GEOMETRICAL PROGRESSION. 265 
 
 In geometrical, as in arithmetical progression, there are five 
 things to be considered ; of which, if any three be given, the 
 other two may be found. The five things are, the first term, 
 the last term, the number of terms, the sum of all tJie terms, 
 and the ratio. 
 
 CASE 1st. THE FIRST TERM, THE RATIO, AND THE NUMBER 
 
 OF TERMS GIVEN, TO FIND THE LAST TERM. 
 
 RULE. Raise the ratio to a power whose index is one less 
 than the number of terms, and multiply this power by the first 
 term. The product will be the number required. 
 
 Ex. 1. The first term of a geometrical series is 2, and the 
 ratio 3. What is the 12th term ? 
 
 Since the given term 2, is the first of the required series of 
 12 terms, and since each succeeding term is found by multi- 
 plying the preceding one by 3, 3 is evidently to be taken as 
 multiplier, or factor, eleven times ; that is, any term of a series 
 is equal to the first term multiplied by the ratio raised to a 
 power one less than the number of terms. Thus, 2 X 3 X 3 X 
 3x3x3x3x3x3x3x3x3 u =:3x2 = 354294, Ans-. or 
 twelfth term. 
 
 2. A person purchased a house having 8 doors, and agreed 
 to pay for the whole, whatever value might be attached to the 
 eighth door, by allowing $4 for the first, $16 for the second, 
 and $64 for the third door, &c. What did his house cost him ? 
 Ans. 4 7 x4r= $65536. 
 
 3. A boy purchased 12 oranges, and agreed to pay 1 cent 
 for the first, 4 cents for the second, 16 cents for the third, &c. 
 What was the value of the twelfth orange ? Ans. $41943.04. 
 
 4. A man wishing to get his horse shod, agreed to allow 3 
 cents for the first nail, 9 cents for the second, and 27 cents for 
 the third, &c. ; .and to pay for the whole the value of the last 
 nail, of which Lere were 32. What was the cost of shoeing 
 his horse 1 Ans. $18530201888518.41. 
 
 Note 1st. It is obvious from what was said of Involu- 
 tion, that, if the ratio be raised to two or three different pow- 
 ers, whose indices, when added together, equal the index of the 
 required power, the product of these several powers will be the 
 power, or number required. Thus, in the second example, the 
 23 
 
266 GEOMETRICAL PROGRESSION. 
 
 answer is obtained thus : 4 3 x4 4 x4=$65536, the answer as 
 before. 
 
 5. A person sold 20 yards of cloth as follows : for the 
 first yard he received 3 d., for the second, 9 d., for the third, 
 27 d., &c. What was the cost of the twentieth yard ? Ans. 
 1 4528268 . 6s. 9 d. 
 
 6. A man purchased 12 horses ; for the first horse he gave 
 only 4 cents, for the second, he gave 16 cents, for the third, 
 64 cents, &c., and thus in a quadruple ratio to the last. What 
 did the twelfth horse cost him ? Ans. $167772.16. 
 
 CASE 2d. THE EXTREMES AND RATIO BEING GIVEN, TO FIND 
 
 THE SUM OF ALL THE TERMS. 
 
 RULE. Divide the difference of the extremes by the ratio 
 less 1, and the quotient increased by the greater extremes, will be 
 the sum of the series. 
 
 Ex. 1. The extremes of a geometrical series are 2 and 
 1458. What is the sum of all the terms, the ratio being 3 ? 
 
 1458 2=1456, the difference of the extremes; and 3 1 
 =2, the ratio less 1. Therefore, 1456-^2=728; and 728+ 
 1458=2186, the sum of all the terms, Ans. 
 
 2. A farmer has sheep in 6 different pastures. In the first 
 pasture there are 3, and in the sixth, 729. If the ratio of in- 
 crease be 3, how many sheep has he in all his pastures ? Ans. 
 1092. 
 
 3. There is a cherry tree with 10 branches ; on the first 
 branch there are only 2 cherries ; on the tenth branch there 
 are 524288. Now the ratio of increase from one branch to 
 another being 4, how many cherries are there on the tree ? 
 Ans. 699050. 
 
 CASE 3d. THE FIRST TERM, THE RATIO, AND THE NUMBER OF 
 
 TERMS GIVEN, TO FIND THE SUM OF THE SERIES. 
 
 RULE. Find the last term by Case 1st, and the sum of the 
 scries, or of all the terms, by Case 2d. 
 
 Ex. 1. A gentleman sold 20 yards of cloth, receiving for 
 
GEOMETRICAL PROGRESSION. 267 
 
 the first yard, 3 d. ; for the second yard, 9 d. ; and for the third 
 yard, 27 d. How much did he receive for the whole, at that 
 rate? 
 
 Case 1st, 3 19 x 3 = 3486784401, the last term. Case 2d, 
 34867844013-^2 = 1743392199.+ 3486784401=5230176600 
 the sum of all the terms in pence =2 1792402 . 10s. 
 
 2. What Avould 12 horses cost, if 4 cents were allowed for 
 the first, 16 cents for the second, and 64 cents for the third 
 horse, &c. the value thus increasing in a quadruple ratio to the 
 last or twelfth horse ? Ans. $223696.20. 
 
 3. A gentleman gave his daughter, on the day of her mar- 
 riage, one dollar, promising to triple it on the first day of each 
 month in the year. What was the amount of her portion ? 
 Ans. $265720. 
 
 CASE 4th. THE EXTREMES AND NUMBER OF TERMS GIVEN, TO 
 
 FIND THE RATIO. 
 
 RULE. Divide the greater extreme by the less, and the quo- 
 tient will be that power of the ratio, which is equal to the num- 
 ber of terms less 1 . The corresponding root will, therefore, be 
 the ratio. 
 
 Ex. 1. The first term of a geometrical series is 2, and the 
 last term 354294, and the number of terms 12. What is the 
 ratio ? 
 
 3542942 = 177147, and the eleventh root of this number 
 is the ratio required ; therefore, V177147=3, the ratio. 
 
 2. The first term of a certain series is 4, and the last 65536, 
 and the number of terms 8. What was the ratio ? Ans. 4. 
 
 QUESTIONS. What is Geometrical Progression 1 What is the ratio 
 of the series 1 What are the terms of the series 1 What terms of 
 a series are called extremes 1 And what are called means 1 In 
 geometrical progression, how many things are to be considered 1 How 
 many of these must be given to find the others 1 What are the five 
 things given 1 What is Case 1st 1 What is the rule for Case 1st? 
 What is Note 1st 1 What is Case 2d ? What is the rule 1 What is 
 Case 3d? What is the rule 1 What is Case 4th ? What is the rule 7 
 
268 ALLIGATION. 
 
 ALLIGATION. 
 
 Alligation is the method of mixing several simples of differ- 
 ent qualities j so as to obtain a compound of a mean or middle 
 quality. 
 
 CASE 1st. WHEN THE QUANTITIES AND PRICES OF SEVERAL 
 
 SIMPLES ARE GIVEN, TO FIND THE MEAN PRICE OF THE MIX- 
 TURE. 
 
 RULE. Find the total value of the several kinds to be mixed, 
 and divide the amount of this value by the whole number of ar- 
 ticles. 
 
 Ex. 1. A farmer mixed together 8 bushels of rye worth 
 $0.50 per bushel; 12 bushels of corn worth $0.65 per bush- 
 el ; and 6 bushels of oats worth $0.30. What was the value 
 of one bushel of the mixture ? 
 
 8 bushels of rye at 50 ct. = $4.00 ; 12 bushels of corn at 65 
 ct. = 87.80 ; and 6 bushels of oats at 30 ct. = $ 1 .80. And 8 + 
 12-j- 6 =26 bushels, and $4.00 +$7.80 +$1.80= $13.60, and 
 $13.30-^-26 bushels = $0.523,+ price of one bushel of the 
 mixture. 
 
 2. A grocer mixed 6 Ib. of tea at $1.20 per Ib. ; 12 Ib. at 
 $1.60 ; and 81b. at $1.80. What was the value of one Ib. of 
 the mixture 1 Ans. $ 1 .569. 
 
 3. If 15 bushels of wheat worth $1.40 per bushel, be mixed 
 with 12 bushels of rye at $0.60 per bushel, and ten bushels of 
 oats at $0.35, what is the value of one bushel of the mixture 1 
 Ans. $0.856.+ 
 
 4. If 6 Ib. of gold, 20 carats fine, be mixed with 12 Ib. at 18 
 carats fine, what is the fineness of the mixture? Ans. 18 J 
 carats fine. 
 
 5. If 6 gallons of wine at $0.67 per gallon, 7 gallons at 
 $0.80 per gallon, and 5 gallons at $1.20 per gallon, be mixed 
 together, what will be the value of one gallon of the mixture ? 
 Ans. $0.867.+ 
 
ALLIGATION. 269 
 
 CASE 2d. THE PRICES OF SEVERAL COMMODITIES BEING GIVEN, 
 
 TO DETERMINE HOW MUCH OF EACH COMMODITY MUST BE 
 TAKEN, TO FORM A COMPOUND OF A CERTAIN PROPOSED ME- 
 DIUM VALUE. 
 
 RULE. Write down the prices of the several simples under 
 each other, placing that price which is least in value uppermost, 
 and the remaining prices in the order of their values. 
 
 Connect, by a line, any price less than the given mean 
 price, with one that is greater, and continue thus to do till they 
 are all connected ; then place the mean price on the left, and sep- 
 arate it from the other numbers by a perpendicular line. Write 
 the difference between the proposed price of the mixture, and the 
 price of each simple, opposite the number or numbers with which 
 that simple is connected ; and, finally, 
 
 Notice whether more than one difference stands opposite any 
 one price ; if so, their sum will express the quantity of that 
 price to be taken ; but if only one difference stands there, that 
 will be the quantity required. 
 
 Note. One difference at least must stand against each price. 
 
 Ex. 1. How much, corn at 48 cents, barley at 36 cents, and 
 oats at 24 cents per bushel, must be taken to make a com- 
 pound worth 30 cents per bushel ? 
 
 '24) \ 6+18=24 bushels at 24 cents. 
 
 36 $ > 6 .... 6 36 cents. 
 
 48 ) 6 .... 6 48 cents. 
 
 The difference between 30, the mean, and 24, is placed op- 
 posite of both 36 and 48, as it is connected with them both ; 
 and the difference between 30, the mean, and 36, and also 
 between 30 and 48, are both placed opposite 24, because these 
 numbers are both linked with 24, and the sum of their differ- 
 ences determines the number of bushels required of that price. 
 Of the oats, therefore, 24 bushels are required, and of the corn 
 and barley, only 6 bushels of each. 
 
 2. I have four kinds of sugar valued at 8, 12, 15, and 18 cents 
 per pound. How much of each kind must be taken to make a 
 mixture worth 14 cents per pound? 
 
 4 number of pounds at 8 cents. 
 
 1 12 cents. 
 
 2 15 cents. 
 
 6 18 cents. 
 
 Mean price, 30 
 
 14 
 
270 ALLIGATION. 
 
 3. A grocer mixed together three kinds of tea, valued at 6, 
 9, and 10 shillings per pound, so that the compound was worth 
 8 shillings per pound. How much of each sort did he take 1 
 Ans. 3 Ib. at 6 s., 2 Ib. at 9 s., and 2 Ib. at 10s. 
 
 4. A merchant has three kinds of wine. For the one kind 
 he charges 3 s. 4 d., for the second 5 s., and for the third 7 s. 
 per gallon. How much of each is required to form a mixture 
 worth 6 s. per gallon ? Ans. 12 gal. at 3 s. 6 d., 12 gal. at 5 s., 
 and 44 gal. at 7 s. 
 
 5. How much gold at 16, 19, 21, and 24 carats fine, will be 
 required to form a compound of 20 carats fine ? Ans. 4 parts 
 of 16* 1 of 19, 1 of 21, and 4 of 24 carats fine. 
 
 CASE 3d. THE PRICE OF EACH OF SEVERAL SIMPLES, THE 
 
 QUANTITY OF ONE AND THE PRICE OF THE COMPOUND BEING 
 GIVEN, TO FIND HOW MUCH OF EACH OF THE OTHER SIMPLES 
 IS REQUIRED. 
 
 RULE. Link the several prices together, as in the last case, 
 and find their differences ; then multiply the given quantity by 
 the differences standing severally against the other quantities, 
 and divide the product by the difference standing against itself. 
 Or say, as the difference opposite the given quantity is to the 
 given quantity, so are the other differences severally to their re- 
 quired quantities. 
 
 Ex. 1. How much barley at 30 cents, rye at 36 cents, and 
 corn at 48 cents per bushel, must be mixed with 12 bushels of 
 oats at 18 cents per bushel, so that the compound may be worth 
 22 cents per bushel ? 
 
 30 "I 4= 4. 
 
 36 V I 4= 4. 
 
 48 > > f 4= 4. 
 
 18 O J 8+14+26^:48. 
 
 The price of the given quantity is 48 ; therefore, 48 : 12 :: 
 4:1, the quantity required at 30 cents per bushel. The re- 
 maining statements and answers are the same, since the dif- 
 ferences are all the same. Therefore, one bushel at 30 cents, 
 one at 36 cents, and one at 48 cents, would be required to be 
 mixed with 12 bushels at 1 8 cents, to form a mixture worth 22 
 cents. 
 
 2. A grocer has three kinds of beer for sale, valued at 7 s., 
 
 Mean price, 22 
 
ALLIGATION. 271 
 
 5s., and 3s. per gallon, which he proposes to mix with 20 
 gallons of a superior quality, worth 6 s. per gallon, so that the 
 mixture may be sold at 4 s. per gallon. How much of the 
 first three kinds must he take? Ans. 120 gal. at 3 s., and 20 
 gal. at 5s. and 7s. 
 
 3. How much tea at 80, 60, and 40 cents perlb. must be 
 mixed with 30 Ib. at $1.00 per Ib. so that the mixture may be 
 sold at 70 cents per Ib. 1 Ans. 10 Ib. at 80 cents and 60 cents, 
 and 30 Ib. at 40 cents. 
 
 4. How much water of no value must be mixed with 100 
 gal. of wine at 7 s. 6 d. per gal., to reduce the price to 6 s. 3 d. 
 per gallon ? Ans. 20 gal. 
 
 CASE 4th. THE PRICE OF THE SIMPLES BEING GIVEN, AND 
 
 ALSO THE COMPOUND TO BE FORMED, TO FIND HOW MUCH OF 
 EACH SIMPLE MUST BE TAKEN. 
 
 RULE. Connect the prices of the simples as in the prece- 
 ding cases, and find the amount of the differences ; then say, as 
 the amount of the differences is to each of the differences taken 
 separately, so is the whole compound to the part required. 
 
 Ex. 1. A compound of 15 gallons, which shall be worth 8 
 shillings per gallon, is to be made of three sorts of wine, valued 
 at 5, 7, and 12 shillings per gallon. How much of each kind 
 will be required ? 
 
 8 
 
 - ... 4 4 
 
 .---4 4 
 
 12 .... 34-1 .... 4 
 
 12 
 Then, 12 : 4 : : 15 : 5, Ans. 5 Ib. of each kind are required. 
 
 Proof, 5 s. X 5=25 s. 7 s. X 5 s. = 35 s. ; and 12 s. X 5 s.= 
 60s.; and 25 + 35 + 60 = 120 s. ; and 120 8 = 15 gallons. 
 
 2. I have four sorts of tea, of which the first kind is worth 
 1 s. perlb. ; the second kind, 3 s. ; the third, 6 s. ; and the 
 fourth, 10s. How much of each kind will be required to make 
 a compound of 120 Ib. worth 4 s. per Ib ? Ans. 60 Ib. at 1 s. ; 
 20 Ib. at 3 s. ; 10 Ib. at 6 s. ; and 30 Ib. at 10 s. 
 
 3. How much of each of four kinds of coffee, worth 8, 12, 
 18, and 22 cents perlb. will be required to make a compound 
 
272 POSITION. 
 
 of 120 Ib. worth 16 cents per Ib. ? Ans. 36 Ib. at 8 cents ; 12 
 Ib. at 12 cents ; 24 Ib. at 18 cents ; 48 Ib. at 22 cents. 
 
 4. A gold beater has gold 15, 17,18, and 22 carats fine, 
 of which he wishes to make a compound of 40 oz. 20 carats 
 fine. How much of each kind must he take 1 Ans. 25 oz. 
 22 carats fine ; and 5oz. of 15, 17, and 18 carats fine. 
 
 5. How much water of no value, and how much wine at 90 
 cents per gallon, must be taken to make 100 gallons, worth 60 
 cents per gallon 1 Ans. 33 gallons of water, and 63| gallons 
 of wine. 
 
 QUESTIONS. What is Alligation 1 What is Case 1st 1 What is 
 the rule ? What is Case 2d 1 What is the rule ? What is the note 1 
 What is Case 3d 1 What is the rule 1 What is Case 4th 1 What is 
 the rule 1 
 
 POSITION. 
 
 Position is a rule by which answers are obtained to such 
 questions as cannot be solved by the common direct rules, by 
 assuming any convenient number or numbers, and then working 
 according to the nature of the question. 
 
 J SINGLE POSITION. 
 
 When the question can be solved by the assumption of a 
 single number, the operation is called Single Position. 
 
 The following sum will serve for an illustration. 
 
 Ex. 1. A teacher being asked how many scholars he had, 
 replied, " If I had once, one half, one third, and one fourth as 
 many more as I now have, I should have 185." How many 
 had he ? We will suppose the number to be 24. As he first 
 doubles his number, 24 must be doubled. To this amount, 
 one half his original number, 12, must also be added. He 
 then increases his number by one third of his original 
 
SINGLE POSITION. 273 
 
 number, viz. 8, and also by one fourth, viz. 6. Whole 
 amount, 74. 
 
 Now it is evident that we have not supposed the right num- 
 ber, otherwise the amount would have been 185, as given in 
 the sum. We have, however, increased the number we sup- 
 posed, viz. 24, by, the same or similar additions, as the teacher 
 did the true number of his scholars ; consequently, 74, the 
 number we obtained, must have the same ratio to 24, the num- 
 ber assumed, as 185 has to the real number of scholars in the 
 school. Therefore, 74 : 24 : : 185 : the number required ; viz. 
 60. Proof, 60+60+30+20+15r=185. . 
 
 We have then the following rule : 
 
 RULE. Take any convenient number and proceed with it ac- 
 cording to the conditions of the question, and observe the result ; 
 then say, as the number thus obtained is to the given number, 
 so is the assumed number to the true one. Or the numbers may 
 be canceled by arranging the terms as directed in Simple Pro- 
 portion. 
 
 Ex. 2. A'man being asked how much money he had, replied, 
 that |, , J, i of his money added, made $57. How much 
 money had he ? Ans. $60. 
 
 3. What number is that, which, being multiplied by 9 and 
 divided by 4, the quotient will be 27 ? Ans. 12. 
 
 4. A man borrowed a sum of money on interest, which in 
 10 years amounted to $1800 at 6 per cent. ; what was the sum? 
 Ans. $1125. 
 
 5. Two boys were playing at marbles. Says one to the 
 other, ^, %, and i of my marbles added together make 45, and 
 if you can now tell how many I. have, you may have them. 
 How many had he ? Ans. 60. 
 
 6. A boy wishing to try the skill of his companions in 
 figures, said he had a pile of apples, of which, if he gave ^ to A., 
 \ to B., and \ to C., there would remain 28 for D. ; and re- 
 quested them to tell him how many there were in all. What 
 was the number-? Ans. 112. 
 
 7. A person being asked his age, said that if J of the years 
 he had lived were multiplied by 7, and f of them added to the 
 product, the sum would be 292. How old was he ? Ans. 60 
 years. 
 
274 DOUBLE POSITION. 
 
 8. A. saves ^ of his income, but B. who has the same in- 
 come, spends twice as fast as A., and thereby contracts a debt 
 of $120 annually. What is their income ? Arts. $360. 
 
 9. The sum of A., B., andC's ages, is 132 years. B'sage 
 is 1^ the age of A; and C's age is twice as great as B's. 
 What are their respective ages ? Ans. A's age is 24 ; B's, 36 ; 
 and C's, 72 years. 
 
 . What is Position 1 What is Single Position 7 What 
 is the rule ?- How may the operations be canceled 1 
 
 DOUBLE POSITION. 
 
 jp| 
 
 By Double Position, we solve such sums as require two sup- 
 positions. 
 
 In this rule, the numbers supposed to be the true ones bear 
 no certain or definite proportion to the required answers. 
 
 RULE. Assume any two convenient numbers and proceed 
 with each according to the conditions of the question, and com- 
 pare the result of each with the sum or result given in the ques- 
 tion, and find their differences. Call each difference an error. 
 
 Multiply the first assumed number by the last error ; and 
 the last assumed number by the first error. 
 
 If both errors are too great or too small, divide the difference 
 of these products by the difference of the errors, and the quo- 
 tient will be the number sought. But if one of the errors be 
 too large, and the other too small, divide the sum of the products 
 by the sum of the errors. 
 
 Note. The errors are said to be too large or too small, 
 when by operating on each supposed number according to 
 the nature of the question, the number obtained is greater or 
 less than the corresponding number in the sum. 
 
 Ex, 1. Three men found a purse of money containing $80, 
 
DOUBLE POSITION. 275 
 
 which they agree to divide in such a manner, that A. shall 
 have $5 more than B, and that B. should have $10 more than 
 C. ^Vhat was each man's share of the money ? 
 
 Suppose, 1st, that C. had - $15 
 then B. had by the conditions, 25 
 and A., 30 
 
 $70, a sum of money less than 
 that found; therefore, $80 $70= $10, 1st error. 
 
 Again, suppose that C. had - $20 
 B. of course must have had - 30 
 and A., 35 
 
 $85, a sum of money greater 
 than that found ; therefore, $85 $80= $5, 2d error. 
 
 If now the above operations be compared with the rule and 
 the note following, it will be seen that the first error is too 
 small, and the last one too large ; therefore, 15, number first 
 supposed X 5, the last error =75; and 20, the number last sup- 
 posed X 10, the first error =200 ; and 200+75=275, the sum 
 of the products; and 10+5 = 15, the sum of errors. There- 
 fore, 275-^15 = $18.333 + , C's share ; and $18.333 + $10 = 
 $28.333 + , B's share; and $28.333+$5 = $33.333, A's share. 
 
 2. Four individuals having $100 to divide among them- 
 selves, agree that B. should have $4 more than A. ; C., $8 more 
 than B. ; and D. twice as much as C. What was each man's 
 share ? 
 
 1st, suppose A. had $6 2d, suppose A. had $8 
 
 then B. had - - - 10 then B. had - - 12 
 
 C., 18 C., 20 
 
 and D., - - - - 36 and D., - - - 40 
 
 $70 $80 
 
 and 100 70 = 30, 1st error. Hence, 100 80=20, 2d e,rrjr. 
 
 Here both errors are too small, therefore, 6 x 20 = 120 ; and 
 8x30=240; then, 240120 = 120, the difference of the' 
 products ; and 30 20 = 10, the difference of errors. There- 
 fore, 120-10 = 12, A's share ; 12 + 4=16, B's share ; and 16 
 + 8=24, C's share ; and 24+24=48, D's share. Proof, 12 
 + 16+24+48 = 100. 
 
276 DOUBLE POSITION. 
 
 3. Three men hired a piece of Avail built, for which they 
 paid $500. Of this, A. paid a certain part ; B. paid $10 more 
 than A., and C. paid as much as A. and B. both. What did 
 each man pay ? Ans. A. paid $120 ; B. $130; and C. $250. 
 
 Sums like the preceding are solved with ease by analysis. 
 Since we have the sum they all paid, we know that C. paid 
 $250, because he has paid as much as the other two, that is, 
 one half of the whole. Therefore, A. and B. together paid 
 $250. But B. paid $10 more than A, hence, 250 10=240, 
 twice the number of dollars A. paid, and 240-^-2 = 120, A's 
 share; then, 120+10 = 130, B's share; and 120+130=250, 
 C's share. 
 
 4. Two persons lay out equal sums of money in trade. A. 
 gains 120 ., and B. loses 80 . A's money was then treble 
 B's. With what sum did they commence? Ans. 180 jC. 
 
 5. A farmer hired a laborer 40 days, on condition that he 
 should receive 20 cents for every day he wrought, and forfeit 10 
 cents every day he was idle. At the expiration of the 40 days 
 he received $5. How many days did he work, and how many 
 was he idle ? Ans. He wrought 30 days, and was idle 10 days. 
 
 6. What is the length of a fish whose head is 10 inches 
 long, his tail as long as his head and half the length of his 
 body, and his body as long as his head and tail both ? Ans. 
 80 inches. 
 
 7. Two persons, A. and B., have the same income. A 
 saves ^ of his, but B., by spending $150 per annum more than 
 A., at the end of 8 years finds himself $400 in debt. What 
 was their income, and how much did each spend annually ? 
 Ans. Income, $400. A spends $300, and B. $450. 
 
 8. A man bequeathed his property to his three sons, on the 
 following conditions ; viz. to A., one half, wanting $50 ; to 
 B., one third ; and to C., the remainder, which was $10 less 
 than B's share. How much did each son receive, arid what 
 was the whole estate? Ans. A. received $130; B. $120; 
 and C. $110. The whole estate was $360. 
 
 *>$,. A farmer bought a certain number of oxen, cows, and 
 calves; for which he paid 130<. For every ox he 
 paid 7. ; for every cow, 5 . ; and for every calf, l. 10s. 
 There were two cows for every ox, and three calves for every 
 cow. How many were there of each kind ? Ans. 5 oxen, 
 10 cows, and 30 calves. 
 
 10. A person after spending $10'more than J of his annual 
 
PROMISCUOUS EXAMPLES. 277 
 
 income, had $35 more than of it remaining. What was his 
 income? Ans. $150. 
 
 11. A person has two horses ; he also has a saddle worth 
 10 . If the saddle be placed on the first horse, the horse 
 and saddle are worth twice as much as the second horse ; but 
 the value of the second horse with the saddle is 13 . less than 
 the value of the first horse. How much is each horse worth ? 
 Ans. The first is worth 56 ., and the second, 33 . 
 
 QUESTIONS. What is Double Position 7 What relation do the 
 supposed numbers bear to the true ones 1 What is the rule 1 When 
 are the errors said to be too large or too small? 
 
 PROMISCUOUS EXAMPLES. 
 
 Ex. 1. If 460 be multiplied by 36, and the product divided 
 by 9, what will the quotient be ? Ans. 1840. 
 
 2. What number is that which, when increased by of itself, 
 will be 126? ^71*. 72. 
 
 3. What number multiplied by | will produce 16 ? Ans. 
 21*. 
 
 4. What fraction multiplied by 15 will produce ? Ans. -^Q. 
 
 5. What number multiplied by 32 will produce 2912 ? Ans. 
 91. 
 
 6. What number divided by 21 will give 65 as a quotient ? 
 Ans. 1365. 
 
 7. How many nails are required to shoe 27 horses, each 
 shoe requiring 8 nails ? Ans. 864. 
 
 8. In the counter of a merchant there are four drawers, in 
 each drawer, 4 divisions, and in each division, $23.75. How 
 many dollars do the four drawers contain ? Ans. $380.00. 
 
 9. Two men depart from the same place and travel the same 
 way ; one travels 36 miles per day, and the other 42. What 
 will be the distance between them at the end of the 8th day, 
 and how far will each have traveled ? Ans. 48 miles apart, 
 the one having traveled 288, and the other 336 miles. 
 
 24 
 
278 PROMISCUOUS EXAMPLES. 
 
 10. A person owning f of a ship, sold f of his share for 
 $474. What was the value of the whole ship, at the same 
 rate ? Ans. $1264. 
 
 1 1 . How many men must be employed to finish a piece of 
 work in 15 days, which would require 5 men 24 days ? Ans. 
 8 men. 
 
 ) 12. A person being asked the time of day, answered, "The 
 time past noon is equal to the time till midnight." What was 
 the time ? Ans. 36 minutes past 5. 
 
 13. In a certain school, ^ the scholars learn to read and 
 write ; learn geography ; J learn grammar ; and 1 6 study 
 astronomy. What was the number in the school ? Ans. 128. 
 
 14. What is the whole length of a pole, J of which stands 
 in the ground, 16 feet in the water, and -| in the air? Ans. 
 213 feet, 4 inches. 
 
 15. There is a room 12 feet long, 8 feet wide, and 7 feet 
 high. How much paper, 2 feet wide, will be required to paper 
 the same ? Ans. 46 yards, 2 feet. 
 
 16. My horse and saddle are both worth 36 . 12 s., and 
 my horse is worth 7 times as much as my saddle. What is 
 the value of each ? Ans. My horse is worth 32 . s. 6 d., 
 and my saddle, 4 . 11 s. 6 d. 
 
 17. There is a cistern having 3 faucets, the largest of which 
 will empty it in one hour, the second in 2 hours, and the third 
 in 3 hours. In what time will they all empty it, if opened at 
 the same time 1 Ans. 32^- minutes. 
 
 18. Divide 1500 acres of land between A., B., and C., so 
 that A. shall have 150 acres more than B., and B. 100 acres 
 more than C. Ans. A. has 633, and B. 483$, and C. 383|. 
 
 19. A certain pasture will feed 324 sheep 7 weeks. How 
 many must be turned away, in order that it may be sufficient 
 for the remainder 9 weeks ? Ans. 72. 
 
 20. A merchant beught 120 gallons of melasses for $45. 
 How must he sell the same per gallon, t<J gain 15 per cent. ? 
 Ans. 0.43.4- 
 
 21. If a family of 8 persons consume $200 worth of pro- 
 vision in 9 months, how much will 18 persons consume in a 
 year ? Ans. $600. 
 
 22. A man left his son a fortune, J of which he spent in 3 
 months ; and in 6 months n\ore he spent f of the remainder, 
 when he had only $1500 remaining. What was his fortune ? 
 Ans. $13500 
 
 23. A young man received 350 . as his share of his fa- 
 
PROMISCUOUS EXAMPLES. 279 
 
 ther's estate, which was |- of his elder brother's portion ; and 
 the elder brother's portion was 3 of the whole estate. What 
 was the whole estate ? Ans. 2333 . 6s. 8 d. 
 
 24. If 356 persons consume 75 barrels of provision in 9 
 months, how many barrels will 500 men consume in the same 
 time? Ans. 102ff barrels. 
 
 25. A. can mow an acre of grass in 5| hours ; B. can mow 
 2 acres in 9 hours. In what time will they both mow 12^ 
 acres 1 Ans. 30-Lf- hours. 
 
 26. If 6 men build a wall 20 feet long, 6 feet high, and 4 
 feet thick, in 16 days, in what time will 24 men build one 200 
 feet long, 8 feet high, and 6 feet thick ? Ans. 80 days. 
 
 27. A farmer being asked how many sheep he had, replied, 
 that in one pasture he had of his whole flock, in another J, 
 in another , and ^ in another, and that a fifth pasture con- 
 tained 450 sheep. How many did the live pastures contain ? 
 Ans. 1200. 
 
 28. If | of a gallon of wine cost f of a pound, New York 
 currency, what will f of a tun cost in dollars and cents ? Ans. 
 $262.50. 
 
 29. If f of an ounce cost $ of a shilling, how many dollars, 
 each 8 s. will a pound cost ? Ans. $2.62^. 
 
 30. Bought 36 bags of rice, each weighing 84 lb., tret 4 Ib. 
 per 104. What will the whole neat weight amount to in fed- 
 eral money, at 8 d. New York currency, per pound ? Ans. 
 $242.307.+ 
 
 31. In a certain orchard, ^ of the trees bear apples, pears, 
 i plums, and 60 bear peaches, and 40 cherries. How many 
 trees are there ? Ans. 1200. 
 
 32. Sold goods to the amount of $560, by which I lost 18 
 per cent., whereas I ought to have gained 12 per cent. What 
 was my real loss? Ans. 204. 877.+ 
 
 33. What is that number of beggars to whom if I give 3 
 pence apiece I shall want 8 pence more than I now have, but if I 
 gjjpre them 2 pence apiece, I shall have 3 pence left. Ans. 11. 
 
 t. How much sugar at 9 d. per lb. must be given in ex- 
 ge for 492 lb. of rice at 3 d. per lb. ? Ans. 164 lb. 
 
 35. Reduce i of f of f of ^ of T 3 T of f|, to a simple fraction. 
 Ans. r /2- 
 
 36. What is the premium on $1800, at 15 per cent. ? Ans. 
 $270. 
 
 37. A father of 12 children said he was 24 years old when 
 his oldest child was born, and that just a year and a half inter- 
 
280 PROMISCUOUS EXAMPLES. 
 
 vened between the birth of each two of his children. What 
 was the age of the father, at the birth of his youngest child ? 
 Ans. 40^. 
 
 38. Bought 16 bales of goods in London for 96 . ; paid 
 4 . for shipment to New York. How ought I to sell the 
 same in federal money to gain 20 per cent. ? Ans. $533.333.+ 
 
 39. Bought a quantity of Irish linen in Dublin for 50 ., 
 and paid 10. for the shipment of the same. How must I 
 sell the whole in federal money to gain 30 per cent. ? Ans. 
 $319.80. 
 
 40. What is the interest on $462.50 for 3 years, 6 months, 
 and 12 days ? Ans. $98.05. 
 
 41. Suppose there to be two silver cups, having one cover, 
 which weighs 5 oz. ; and suppose them to be such, that if 
 the cover be placed on the smaller cup, the whole weighs 
 twice as much as the greater cup ; but if it be placed on the 
 greater cup, the whole weighs three times as much as the 
 smaller cup. What is the weight of each cup ? Ans. The 
 smaller cup weighs 3 oz., and the larger, 4 oz. 
 
 42. A. can do a piece of work in 6 days ; B. can do the 
 same in 1 1 days. In what time will they both accomplish the 
 work, if they labor together? Ans. 3^-y days. 
 
 43. A. can do a piece of work in 7 days ; B. in 12 days; 
 C. in 6 days ; and D. in 4 days. In what time will they ac- 
 complish the same work, if they all labor upon it at the same 
 time ? Ans. If day. 
 
 44. A person being asked the time of day, replied that it 
 was between 5 and 6 o'clock, and that the hour and minute 
 hand were precisely together. What was the time ? Ans. 
 27^- minutes past 5 o'clock. 
 
 45. How many square feet are there in a board 16 feet 8 
 inches long, and 9 inches broad ? Ans. 12| square feet. 
 
 46. The linear measure of a cubic block being 8 inches, 
 how many cubic blocks, each one solid inch, does it contain ? 
 Ans. 512. 
 
 47. If a person own J of a ship, and sell f of his shai 
 $1500, what is the value of the whole ship, at the same rfle ? 
 Ans. $6000. 
 
 48. Divide f of f of T 7 ^ of if by | of f of |. Ans. 
 
 49. In a thunder storm, I observed the time between the 
 flash of the lightning and the report to be one minute and a 
 half. How liir distant was the lightning, allowing sound to 
 travel 1142 feet in a second? Ans. 19|-J- miles. 
 
 50. Bought 84 apples at the rate of 2 for a penny, and 114 
 
PROMISCUOUS EXAMPLES. 281 
 
 at the rate of 3 for a penny ; the whole of which I afterwards 
 sold at the rate of 9 for 4 pence. Did I gain or lose, and how 
 much ? Ans. I gained 8 d. 
 
 51. A line 44 yards in length will just reach from the top of 
 a steeple to the opposite side of the street, which is 24 yards 
 wide. How high is the steeple ? Ans. 36.87+ yards. 
 
 52. A tree 36 feet high stands by the side of a stream 27 
 feet wide. How many feet from the top of the tree to the op- 
 posite side of the stream ? Ans. 45. 
 
 53. There is a cistern having two pipes leading into it, one 
 of which will fill it in 30 minutes, and the other in 45 min- 
 utes. In what time will both, running together, fill the same ? 
 Ans. 18 minutes. 
 
 54. How many bricks 9 inches long and 4 inches broad 
 will pave a yard 40 feet square ? Ans. 6400. 
 
 55. A certain cistern has two pipes leading into it, and one 
 leading out of it. Of the two leading into it, one will fill it in 
 50 minutes, and the other in 75 minutes ; while the one lead- 
 ing from it, will empty it in 60 minutes. Now if all three be 
 opened at the same time, in what time will the cistern be filled ? 
 Ans. 1 hour. 
 
 56. What is the difference between 6 dozen dozen, and half 
 a dozen dozen ? Ans. 792. 
 
 57. If -| of i of of a ship be worth:^- of -J of ij of her 
 cargo, valued at $2400, what is the value of both ship and 
 cargo? Ans. $5415.384.+ 
 
 58. If 1 1 men can build a house in 5 months by working 
 12 hours per day, in what time will they complete it if they 
 work only 8 hours per day ? Ans. 7-|- months. 
 
 59. Bought a pipe of wine for $84, from which 12 gallons 
 leaked out. Now what shall I gain, if I sell the remainder at 
 12^ cents a pint ? Ans. $30. 
 
 60. A. alone can do a piece of work in 12 days, and in con- 
 nection with B., in 8 days. In what time can B. perform the 
 same alone ? Ans. 24 days. 
 
 61. If A. can do a piece of work in 12 days, and B. in 24 
 days, in what time will they both do it 1 Ans. 8 days. 
 
 62. Three men, A., B., and C., can do a piece of work in 
 18 days. A. alone can do it in 36 days ; B. in 54 days. In 
 what time can C. do it ? Ans. 108 days. 
 
 63. Four men can do a piece of work in 15 days. A. alone 
 can do it in 40 days, B. in 60 days, and C. in 80 days. In 
 what time will D. do the work alone ? Ans. 80 days. 
 
 24* 
 
282 PROMISCUOUS EXAMPLES. 
 
 64. A gentleman left his son a fortune, \ of which he spent 
 in 3 months ; \ of |- of the remainder lasted him 6 months 
 longer, when he had only $1200 left. What was his whole 
 fortune ? Ans. $6400. 
 
 65. A farmer bought a yoke of oxen, a horse and a cow, for 
 $250. For the oxen he paid twice as much as for the horse, 
 and for the horse three times as much as for the cow. What 
 did he pay for each ? Ans. For the oxen, $150 ; for the horse, 
 875 ; and $25 for the cow. 
 
 66. Three men start at the same time to travel round an 
 Island 80 miles in circumference, and agree that each shall 
 continue to travel at the same rate till they all come together 
 again, and that the first shall travel 5 miles, the second, 6 
 miles, and the third, 7 miles per day. In what time will the 
 three come together, and how far will each have traveled 1 
 Ans. 80 days. The first will have traveled 400 miles ; the 
 second, 480 miles ; and the third, 560 miles. 
 
 67. How will it affect the distance traveled by each, pro- 
 vided they travel 5, 7, and 9 miles respectively ? Ans. The 
 distance of the first will be 400 miles ; of the second, 560 ; 
 and of the third, 720. 
 
 68. Divide 1200 acres of land between A., B., and C., so 
 that B. may have 75 acres more than A., and C. 95 acres 
 more than B. What will be the share of each one 1 Ans. 
 A's share will be 318; B's, 393 ; and C's, 488J acres. 
 
 69. Three men met at an Inn, two of whom brought pro- 
 vision with them. The third not having brought any, pro- 
 posed that they should eat together, and then he would pay his 
 proportion. The proposal being accepted, A. produced 5 loaves, 
 and B. 4 loaves, all of which they ate up, and C. as his share, 
 paid 9 pieces of money. With this, A. and B. were satisfied, 
 but could not agree as to the division. What should each have 
 received? Ans. A. 6, and B. 3 pieces of the money. 
 
 70. If A. can reap a field of grain in 12 days, and B. in 16 
 days, in what time will both do it, working together ? Ans. 
 6f days. 
 
 71. What number must be added to ^ of 4560, to make the 
 same 500 1 Ans. 369f . 
 
 72. If the head of a fish be 9 inches long, its tail as long 
 as its head and half the length of its body, and its body as 
 long as its head and tail both, how long is the fish ? Ans. 
 72 inches. 
 
 73. If 6 pairs of hose are equal in value to 4 pieces of Hoi- 
 
PROMISCUOUS EXAMPLES. 283 
 
 land, and 6 pieces of Holland to 14 yards of satin, and 12 yards 
 of satin to 8 pieces of lace, and 9 pieces of lace to 8 . 2s., 
 how many pairs of hose may be bought for 2 . 16s.? Ans. 
 24 pairs. 
 
 74. A man was hired 50 days, on condition that for every 
 day he worked, he should receive 6 s., and for every day he 
 was idle, he should forfeit 2 s. At the expiration of the time, 
 he received $27.50. How many days did he work, and how 
 many was he idle ? Ans. He labored 40 days, and was idle 
 10 days. 
 
 Note. The currency of the preceding sum is that of New 
 York. 
 
 75. A hare starts 40 yards in advance of a greyhound, and 
 is not perceived by him till she has been up 40 seconds. She 
 scuds away at the rate of 10 miles per hour, and the hound 
 pursues after her at the rate of 18 miles per hour. In what 
 time will the houtid overtake the hare, and how far will he 
 have run? Ans. The required time is 60^ &ec. ; the dis- 
 tance run, 530 yards. 
 
 76. If 8 men can build a wall 15 rods long in 10 days, how 
 many men will be required to build 45 rods of wall in 5 days ? 
 Ans. 48 men. 
 
 77. A man when he married was three times as old as his 
 wife ; after they had been married 15 years, his age was only 
 double that of his wife's. How old were they when they 
 married ? Ans. The man was 45 years, his wife 15 years old. 
 
 78. There is in a pasture a certain number of sheep, cows, 
 and oxen ; there are twice as many sheep as cows, and three 
 times as many cows as oxen, and the whole number is 80. 
 How many are there of each kind ? Ans. 8 oxen, 24 cows, 
 and 48 sheep. 
 
 79. Sold coffee at 15 cents per pound, and thereby lost 10 
 per cent, on the first cost. Afterwards sold a quantity of the 
 same for $525, by which I gained 40 per cent. What was the 
 quantity sold, and what the price per pound ? Ans. Quantity, 
 20 cwt. 1 Ib. ; price, 23 cents. 
 
 80. Two sons, one 11 and the other 16 years of age, re- 
 ceived a bequest of $10000, to be so divided between them 
 that the shares being put on interest at 5 per cent, should 
 amount to equal sums, when they became respectively 21 years 
 of age. What were the shares of each? Ans. The elder 
 received $5454 T r , and the younger, $4545^ T . 
 
APPENDIX. 
 
 Prob. 1. To find the greatest common measure of two or 
 more numbers. 
 
 Note 1st. The greatest common measure of two or more 
 immbers, is the greatest number that will divide them separately 
 without remainders. 
 
 RULE. If two numbers only are given, divide the greater of 
 them by the less, and if nothing remain, that divisor is the com- 
 mon measure ; but if there be a remainder, divide the preceding 
 divisor by it ; and so continue to divide each preceding divisor 
 by the last remainder, till the division is effected without re- 
 mainder ; the last divisor will be the common measure required. 
 When more than two numbers are given, find the common meas- 
 ure of any two of them first, and then of that common measure, 
 and either of the remaining numbers. This process carried 
 through all the numbers will give their greatest common measure. 
 
 Ex. 1. What is the greatest common measure of 72 and 
 108? 
 
 OPERATION. 
 72)108(1 
 
 72 
 
 3 6) 72(2 
 72 
 
 00 
 
 36 is, therefore, the common measure required. Proof, 108 
 36 = 3, and no remainder. 72 -H 36 =2, and no remainder. 
 
 2. What is the greatest common measure of 27 and 99 ? 
 Ans. 9. 
 
APPENDIX. 285 
 
 3. What is the greatest common measure of 25, 45, and 90 ? 
 Ans. 5. 
 
 4. What is the greatest common measure of 16, 32, 48, and 
 96? Ans. 16. 
 
 Prob. 2. To determine how many different positions any 
 given number of objects may assume with regard to each other. 
 
 RULE. Represent the number of objects by the figures 1, 2, 
 3, 4, 5, fyc. making the numbers of figures equal to the number 
 of objects. The product of these figures will determine the 
 number of changes. 
 
 Ex. 1. How many changes may be made by the first three 
 letters of the alphabet ? 
 
 Operation : 1x2x3 = 6, Ans. These changes are as fol- 
 lows : 1st, a, b, c; 2d, a, c, b; 3d, b, a, c; 4th, b, c, a; 5th, 
 c, b, a; 6th, c, a, b. 
 
 2. How many changes may be made by the first 6 letters of 
 the alphabet ? Ans. 720. 
 
 3. At a certain boarding house there are 12 boarders. How 
 many different positions may they occupy at the table ? Ans. 
 479001600. 
 
 4. Five men engaged board at a tavern for as many days as 
 the landlord might be able to seat them in different positions. 
 How long did they remain? Ans. 120 days. 
 
 Prob. 3. To find the area of a square. 
 
 RULE. Multiply the length of one of the sides by itself; or, 
 square its linear measure. (See fig. 2d, Square Root.) 
 
 Ex. 1. There is a room just 8 feet square, what is the area? 
 Operation, 8x8 = 64 square feet, Ans. 
 
 2. What is the area of a floor 19 feet square ? Ans. 361 
 square feet. 
 
 3. What is the area in square inches of a board 3 feet square ? 
 Ans. 1296 square inches. 
 
 Prob. 4. To find the area of a parallelogram. 
 
 RULE. Multiply the length of the parallelogram by its breadth. 
 (See fig. 6th, Square Root.) 
 
286 APPENDIX. 
 
 Ex. 4. How many square feet are there in a floor 16 feet 
 long and 12 broad ? 16 x 12 = 192 square feet, Ans. 
 
 Note 1st. If the dimensions of a field in the form of a square 
 or parallelogram are given in rods, the area is reduced to acres 
 by dividing the square rods by 160. 
 
 2. How many acres are there in a piece of land 80 rods in 
 length, and 40 in breadth ? 80 X 40 = 3200, and 3200 -H 1 60 = 
 20 acres, Ans. 
 
 3. What is the number of acres in a piece of land 63 rods 
 long and 49 rods wide ? Ans. 
 
 Note 2d. If the parallelogram be not right-angled, the length 
 must be multiplied into the perpendicular distance between the 
 sides. (See fig. 1st.) 
 
 The dotted line in FIG. 1. 
 
 the center is the per- 
 pendicular distance 
 required ; therefore, 
 45 x 15 = 675, the 
 area of the figure. * 
 
 Prob. 5. To find the area of triangles. 
 
 The area of any triangle is just equal to one half the area 
 of a square or parallelogram of the same height. This is ob- 
 vious with regard to right-angled triangles, and true of all 
 others. The diagonal 2 
 
 of a square or parallel- 
 ogram divides it into two 
 equal right-angled tri- 
 angles, and since the 
 area of the whole figure 
 is found by multiplying 
 the length into the breadth, the area of its half, that is, of 
 either of the triangles into which it is divided by the diagonal, 
 is found by multiplying the same length or base into one half 
 the breadth. 
 
 We have then the following rule : 
 
 RULE. If the triangle be right-angled, multiply its base into 
 half its perpendicular height. But if it be not right-angled, 
 drop a perpendicular from one of the angles to the opposite side 
 
 
APPENDIX. 287 
 
 or base ; then multiply the base or side upon which the perpendic- 
 ular falls, by one half the perpendicular. 
 
 Ex. t. What is the area of a triangular piece of land whose 
 base is 40 rods, and whose perpendicular is 30 rods? 30-^-2 
 = 15, and 40x15 = 600 square rods, Ans. 
 
 Note. The same result is obtained by multiplying the per- 
 pendicular by one half the base. Thus, 40 -^2=20, and 30 X 
 20=600. 
 
 2. There is a room 26 feet long and 18 feet high. If a line 
 be drawn from one of the upper corners to its opposite lower 
 corner, the side of the room will be divided into two equal 
 triangles. What is the area of each triangle, and also of the 
 side of the room ? Ans. Each triangle contains 234 square 
 feet, and the side, 468. 
 
 3. How many acres are there in a triangular piece of land, 
 whose base is 42 rods and whose perpendicular height is 36 
 rods ? Ans. 4fg acres. 
 
 Prob. 6. Given the diameter of a circle, to find its circum- 
 ference. 
 
 The diameter of a circle is to its circumference, as 7 to 22 ; 
 or, as 1 to 3.141592. Therefore, 
 
 RULE. Multiply the diameter of a circle by 3.141592, and 
 the product will be the circumference ; or, multiply the diameter 
 by 22, and divide the product by 7. 
 
 Ex. 1. What is the circumference of a wheel whose diam- 
 eter is 8 feet. 8 x 22 -^-7=25.14+ feet, Ans. Or, 8x3.141592 
 :=25.132736, nearly the same as before. 
 
 2. What is the circumference of a circle whose diameter is 
 6 feet? Ans. 18.85 feet. 
 
 3. What is the circumference of a circle whose diameter is 
 42 feet? A?is. 132 feet. 
 
 Prob. 7. Given the circumference of a circle, to find the 
 diameter. 
 
 RULE. Multiply the circumference by 7 and divide the pro- 
 duct by 22 ; or, (what will produce nearly the same result,) di- 
 vide the circumference by 3.141592. 
 
APPENDIX. 
 
 Ex. 1. If the circumference of a wheel be 26 feet, what is the 
 diameter ? 26 X 7 -22:=.- 8. 27 -f- feet, Ans. 
 
 2. What is the diameter of a wheel whose circumference is 
 50 feet? Ans. 15.9+ feet. 
 
 Prob. 8. To find the area of a circle. 
 
 RULE. Multiply half the diameter by half the circumference. 
 
 Ex. 1. If the diameter of a circle be 6 feet, what is its area? 
 6 x 22 -f- 7 = 1 8.857 feet, the circumference. Therefore, 1 8. 857 
 -^-2 = 9.4285, one half the circumference, and 6-^-2=3, one half 
 the diameter; then, 9.4285x3=28.2855,4- the area in feet 
 and decimals. 
 
 2. If the diameter be 20 feet, what is the area of the circle? 
 Ans. 3 14.285 feet? 
 
 3. What is the area of a circle 66 feet in circumference ? 
 Ans. 346.5 
 
 Prob. 9. To find the superficial area of a globe. 
 
 The superficial area of a globe is four times that of a circle 
 of the same diameter. Therefore, 
 
 RULE. Find the area of a circle of the same diameter as the 
 given globe, and multiply that area by 4. 
 
 Ex. 1. What is the superficial area of a globe 16 inches in 
 diameter? 16x22-^-7 = 50.2757. 50.2757-^2=25.13785, 
 half the circumference, 16 -=-2 =8, half the diameter; there- 
 fore, 25.13785x8x4=804.4+ square inches, the required 
 area. 
 
 2. If the earth be 8000 miles in diameter, and 25000 in cir- 
 cumference, how many square miles on its whole surface. Ans. 
 200000000. 
 
 Prob. 10. To find the solid content of a globe. 
 
 RULE. Find the superficial area by the preceding problem, 
 and multiply that area by % of the diameter of the globe. 
 
 Ex. 1. If the diameter of the earth be 8000 miles, and the 
 circumference 25000 miles, how many cubic miles does it 
 contain. 
 
APPENDIX. 289 
 
 The superficial area was found by the preceding problem to 
 be 200000000 ; and i the earth's diameter is 8000-^6 = 1333 ; 
 therefore, 200000000 x 1333^=266666666666^ cubic miles, 
 Ans. 
 
 Prob. 11. To find the solid content of a cylinder. 
 
 A cylinder is a round body of uniform diameter, and of any 
 assignable length. Each end of a cylinder is therefore a circle. 
 
 RULE. Find the area of one end of the cylinder, and multi- 
 ply that area by its length. 
 
 Ex. 1 . What is the solid content of a cylinder 6 feet in 
 length and 3 feet in diameter ? The area of the end, as found 
 by the preceding rule, is 7.06 ; and 7.06 X 6=42.36. 
 
 2. What is the solid content of a log, the uniform diameter 
 of which is 2 feet, and the length 30 feet? Ans. 94.247 solid 
 feet. 
 
 3. What fraction of a cord is there in a log 10 feet long, 
 and 4 feet in diameter ? Ans. ffj- + 
 
 Prob. 12. In a stick of timber of uniform thickness, to find 
 what length will make a solid foot. 
 
 RULE. Find the area of one end in inches, and divide 1728 
 by it ; the quotient will be the required length in inches. 
 
 Ex. 1. If a stick of timber be 16 inches square, what length 
 will be required to make a solid foot ? 16 X 16=256, the area 
 of one end, and 1728-f-256 = 6f inches, Ans. 
 
 2. How much length of plank 3 inches thick and 24 inches 
 wide, will be required to make a solid foot ? Ans. 2 feet. 
 
 Prob. 13. To find the solid content of cones and pyramids. 
 
 A pyramid or cone is a solid whose base is a plain surface, 
 and which gradually and uniformly tapers till it comes to a 
 point. It may be either round, square, or triangular. 
 
 The solid content of such figures is as much as the con- 
 tent of a cylinder of the same length ; therefore, 
 
 RULE. Multiply the area of the base by one third of the 
 perpendicular height. 
 
 Ex. 1 . What is the solid content of a cone 60 feet high, 
 25 
 
290 APPENDIX. 
 
 the base of which is 8 feet in diameter? 8x22-4-725.14 
 circumference, and 25.14-^-2x4=50.28, area of the base; 
 and 50.28x20, (one third the perpendicular height) = 100.56, 
 the solid content. 
 
 2. What is the solid content of a pyramid, whose base is 4 
 feet square, and whose height is 15 feet? Ans. 80 solid feet. 
 
 3. How many cubic inches in a cone 18 inches high, and 
 12 inches in diameter at the base ? Ans. 678.6. 
 
 4. There is a pyramid whose base is 30 feet in diameter, 
 and the height of it is 90 feet. What is its solid content ? 
 Ans. 21205.73+ solid feet. 
 
 Prob. 14. To find the solidity of the frustrum of a cone or 
 pyramid. 
 
 A frustrum is that part of a cone or pyramid that remains 
 after a part has been cut off from the apex or top. 
 
 RULE. Find the diameter of each end of the cone or pyra- 
 mid, and multiply them together ; then to the product add one 
 third of the square of the difference of the diameters, and mul- 
 tiply the sum by the decimal .785398 ; the product will be the 
 mean area between the two bases ; therefore, multiply this mean 
 area by the length of the frustrum. 
 
 Ex. 1. What is the solid content of a frustrum of a cone 30 
 feet in length, whose base measures 4 feet in diameter, and 
 whose top measures 2 feet in diameter, 4x2 = 8. The differ- 
 ence of diameters is 4 2=2, and 2 2 =4, and4-f-3 = l|. 
 Therefore, 8+ 1 X .785398=7.33, nearly, the mean diameter ; 
 and 7.33x30=219.9, the required content. 
 
 2. What is the content of a stick of timber, of which the 
 larger end measures 2 feet in diameter, and the smaller end, 
 1 foot, the length being 40 feet ? Ans. 73.3 feet, nearly. 
 
 
 
 Prob. 15. Given the length and diameter of a log or stick of 
 round timber, to determine the solid content of the greatest 
 square timber that can be obtained from it. 
 
 The area of the greatest square that can be described upon 
 the end of the given log must first be determined ; that is, the 
 area of a square inscribed in a circle, the diameter of which 
 equals the diameter of the log. How then is the area of an 
 inscribed square, that is, of a square the corners of which all 
 
APPENDIX. 
 
 291 
 
 E 
 
 terminate in the circumference of a circle, to be found, the 
 diameter of that circle being given ? 
 
 If the scholar will examine 
 the adjoining figure, he will see 
 that the square ABCD, which is 
 circumscribed about the circle, 
 is equal to the square of the diam- 
 eter of the circle, since the diam- 
 eter EN equals the side of AD ; 
 and AD squared gives the area 
 of the square ABCD : also, that 
 the inscribed square PEON, is 
 just one half of the circumscribed 
 square, since each of the triangles 
 EaO, OaN, NaP, and PaE, into which the inscribed square 
 is divided, is precisely one half of each of the four squares, 
 APaE, EaOB, OaNE, and NaPD, into which the circum- 
 scribed square, ABCD, is divided. If therefore a fourth part of 
 the inscribed square equals one half of a fourth part of the 
 circumscribed square, the whole inscribed square equals one 
 half of the circumscribed square. Hence, 
 
 RULE. Square the diameter of the small end of the log and 
 multiply one half of this square by the length of the log ; the 
 product will be the content in solid feet, if the length and diam- 
 eter are both given in feet ; but if one of them be expressed in 
 inches, divide this product by 144; and if both, by 1728, and 
 the quotient will express the solid content in feet. 
 
 Ex. 1. How many solid feet of square timber may be 
 sawn from a log 2 feet in diameter, and 20 feet in length ? 
 2 2 =4, and 4^-2=2; then, 2x20 = 40 solid feet, Ans. 
 
 2. How many solid feet of square timber may be obtained 
 from a log 10 inches in diameter, and 15 feet long? Ans. 
 
 solid feet. 
 
 3. How many square feet of timber are there in a log 43 
 feet long, allowing the diameter to be 1 foot 6 inches at the 
 small end; and how many solid feet of slab would be taken off 
 in squaring it, allowing the diameter at the large end to be 2 
 feet ? Ans. 48.37+ solid feet of timber, and 55.76 solid feet 
 of slab. 
 
292 APPENDIX. 
 
 Prob. 16. To find the capacity of casks in gallons, &c. 
 
 RULE. Take the length of the cask between the heads, and 
 also its interior bung and head diameter, in inches. Multiply 
 the difference between the head and bung diameter by .62, when 
 the staves are of ordinary curvature ; but by .66, if the curva- 
 ture be more than ordinary, and by .58, if they be curved less 
 than usual, and add the product to the head diameter; the sum will 
 be the mean diameter. Multiply the square of the mean diam- 
 eter by the length, and this product by 34 for wine, and by 28 
 
 for beer measure, and from the product cut off four decimals. 
 The number obtained will be the gallons, and the decimal of a 
 
 gallon, the cask will contain. 
 
 Ex. 1. How many gallons of wine are there in a cask whose 
 bung diameter is 34 inches, head diameter 30 inches, and 
 whose length is 60 inches ? 
 
 Operation: 34 30=4, the difference between the head 
 and bung diameter ; then, 4 X. 62=2.48, and 2.48 + 30=32.48, 
 mean diameter, and 32.48 x 32.48 = 1 054.9504, square of mean 
 diameter. 1054.9504 X 60 = 63297.0240 x 34=215.20988160, 
 or nearly 215.2 gallons, Ans. 
 
 2. Suppose the bung diameter of a cask of wine to be 24 
 inches, and the head diameter to be 20 inches, and the length 
 40 inches ; how many gallons will it contain ? Ans. 68| gal- 
 lons, nearly. 
 
 Prob. 17. To find the tonnage of double and single decked 
 vessels. 
 
 The following is the rule established by Congress : 
 
 RULE. " If the vessel be double-decked, take the length 
 thereof from the fore part of the main stern to the after part of 
 the stern-post, above the upper deck ; the breadth thereof at the 
 broadest part above the main wales, half of which breadth shall be 
 accounted the depth of such vessel, and then deduct from the 
 length, three fifths of the breadth. Multiply the remainder by 
 the breadth, and the product by the depth, and divide this last 
 product by 95; the quotient whereof shall be deemed the true 
 contents or tonnage of such ship or vessel. But if such ship 
 
 
APPENDIX. 293 
 
 or vessel be single-decked , take the length and breadth, as above 
 directed , deduct from said 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 aforesaid, and the 
 quotient shall be deemed the tonnage." 
 
 Note. Carpenters multiply together the length of the keel, 
 the breadth at the main-beam, and the depth of the hold, each 
 dimension being taken in feet ; and divide the product by 95, 
 for the tonnage of single-decked vessels. But for double- 
 decked vessels, they take half the breadth at the beam for the 
 depth of the hold, and then work as before. 
 
 Ex. 1. Suppose the length of a double-decked vessel to be 
 80 feet, and its breadth to be 30 feet. What is its tonnage ? 
 
 Operation : 30^2 = 15, the depth of the vessel; f of 30 = 
 18, and 80 18 = 62; and 62X30 = 1860; then, 1860x15 
 =27900 ; and, lastly, 27900-^- 95 =293f| tons, Ans. 
 
 2. Required the tonnage of a double-decked vessel, whose 
 length is 95 feet, and whose breadth is 35 feet 1 Ans. 477 T 2 g-. 
 
 3. What is the tonnage of a single-decked vessel, which is 
 75 feet in length, 30 feet in breadth, and 10 feet in depth ? 
 Ans. 180 tons. 
 
 Note. The tonnage of a ship of war is found by dividing 
 the continued product of the length, breadth, and depth by 100. 
 
 4. What is the tonnage of a sloop of war, whose length is 
 96 feet, breadth 32 feet, and depth 15 feet 1 Ans. 460.8 tons. 
 
 Prob. 18. The difference of longitude between two places 
 being given, to find the difference in time. 
 
 As the circumference of the earth is divided into 360 degrees, 
 and as the sun appears to pass round the earth in 24 hours, 
 it is evident that 36024=15 degrees of the earth's circum- 
 ference must pass under the sun every hour, and consequently 
 1 degree every 4 minutes, and I 1 every 4 seconds. We have 
 then the following general rule : 
 
 RULE. If the given degrees be more than 15, divide them 
 byl5; the quotient will be the hours. But if the degrees be less 
 than 15, multiply them by 4, and the product will be minutes. 
 25* 
 
294 
 
 APPENDIX. 
 
 Multiply also the given minutes of motion by 4, and the product 
 will be seconds of time. 
 
 Ex.1. Warsaw is situated about 17 east longitude from 
 Brussels. What is the difference in time? 17 -^15 = 1^ 
 hours, or 1 hour, 8 minutes. 
 
 It is obvious that the sun will rise first at the most eastern 
 of any two places. Hence, when the time of one place is 
 given, and it is required to find the time of some other place, 
 the difference of time must be added to the given time, when that 
 place is situated to the east, and subtracted when it is situated to 
 the west of the place, at which the time is given. 
 
 2. Turin is situated 8 east from London, and Moscow 38 
 east from the same place. What is the hour at Moscow, when 
 it is 9 o'clock at Turin ? Ans. 1 1 o'clock. 
 
 3. When it is 9 o'clock at Philadelphia, what time is it at 
 Council Bluff, 21 west of Philadelphia ? Ans. 36 minutes 
 past 7 o'clock. 
 
 4. What is the difference of time between Washington and 
 Harmony, a missionary station west of Missouri, the difference 
 of longitude being 18? Ans. 1 hour, 12 minutes. 
 
 Prob. 19. The difference in time between any two places 
 given, to find the difference of longitude. 
 
 This is the reverse of the preceding ; therefore, 
 
 RULE. Multiply the hours by 15 and the products will be de- 
 
 Sees. If minutes and seconds of time be given, divide them 
 4 ; the quotient of the minutes will be degrees, and that of 
 $ seconds, will be minutes of motion. 
 
 Ex. 1 . If the difference of time between two places be 3 
 hours and 32 minutes, what is the difference of longitude 1 
 3 X 15=r45 , and 32^-4 = 8 ; and 45+8=:53 , Ans. 
 
 2. What is the difference of longitude between two places, 
 if the time of one place be 5 hours, 15 minutes in advance of 
 the time of the other place ? Ans. 78 45'. 
 
APPENDIX. 295 
 
 MECHANICAL POWERS. 
 
 When one body is made to communicate motion to another, 
 the body communicating the motion is called a power, and the 
 one to which the motion is communicated, is called a weight. 
 The various instruments employed to increase the effect of a 
 given power, are called Mechanical Powers. 
 
 The mechanical powers are six in number, viz. the Lever, 
 the Pulley, the Wheel and Axle, the Inclined Plane, the Screw, 
 and the Wedge. 
 
 The lever is a bar, mov- FIG. 1. 
 
 able about a fixed point. 3 9 
 
 It is represented at Fig. 1. 
 P represents the power, 
 and W the weight, and 3 t^ 
 and 9 the comparative wJB 
 length of the arms of the 
 lever. 
 
 When the power and weight are such that the products ob- 
 tained by multiplying each into the length of the arm to whicn it 
 is attached, are equal, they will remain in equilibrium, that is, 
 they will balance each other. 
 
 In fig. 1, the arm to which the power is applied is three times 
 as long as that to which the weight is applied. Consequently, 
 the power is required to be only one third as heavy as the 
 weight, in order that they reciprocally balance each other. 
 We hence perceive that the.power is to the weight, as the arm to 
 which the weight is applied, is to that to which the power is 
 applied. 
 
 Prob. 20. Given the two arms of a lever, and the power, to 
 find what weight that power will balance. 
 
 i 
 
 RULE. Multiply the length of the arm to which the power is 
 applied, by the power, and divide the product by the other arm. 
 
 Ex. 1. There is a lever 16 feet in length divided by the ful- 
 crum or point of support into two arms, one of which is 12 
 
296 APPENDIX. 
 
 feet, and the other 4 feet in length. What weight on the short 
 arm will balance 20 Ib. suspended at the end of the long arm? 
 20 X 12=240, and 240^-4 = 60 Ib. Ans. 
 
 2. The arms of a lever are, the one 30 feet, and the other 3 
 feet in length. What weight will a power of 160 Ib. at the ex- 
 tremity of the long arm, balance at the extremity of the short 
 arm? Ans. 16001b. 
 
 3. How many Ib. will a power of 9 Ib. placed 15 feet from 
 the fulcrum of a lever, support at the extremity of the other 
 arm, 2 feet in length ? Ans. 67^ Ib. 
 
 Prob. 21. Given the arms of the lever and the weight, to find 
 the power. 
 
 RULE. Multiply the weight by the length of the arm to 
 which it is suspended, and divide the product by the other arm. 
 The quotient will be the power required. 
 
 Ex. 1 . A weight of 80 Ib. is attached to an arm of a lever 3 
 feet in length ; what power must be applied to the other arm, 
 27 feet in length, to balance it? 80x3=240, and 240-^-27 
 =8|lb. Ans. 
 
 2. A weight of 20 tons is suspended to an arm of a lever 6 
 inches in length. What weight at the extremity of the other 
 arm, 40 feet in length, will balance the same ? Ans. 5 cwt. 
 
 Prob. 22. Given the power, the weight, and the length of the 
 lever, to find the fulcrum. 
 
 RULE. Add together the numbers representing the power 
 and weight, and say, as their sum is to the whole lever, so is the 
 number representing the power to the length of the arm to which 
 the weight is to be applied ; or, so is that representing the weight 
 to the length of the arm to which the power is to be applied. 
 
 Ex. 1. Where must be the fulcrum in a lever 16 feet in 
 length, so that a power of 20 Ib. shall balance a weight of 60 
 Ib? 20+60=80. Therefore, 80 : 60 : : 16 : 12, Ans. or length 
 of the arm to which the power is to be applied, and 16 12 
 =4, length of the other arm. 
 
 2. If the power be 72 Ib., the weight 1728 Ib., and the 
 lever 36 feet, where must be the fulcrum, in order that they 
 shall balance each other ? Ans. If-f feet from the weight. 
 
 
APPENDIX. 297 
 
 3. If the length of a lever be 10 feet, the power 170 lb., and 
 the weight to be balanced 1530 lb., where must be the fulcrum? 
 Ans. 9 feet from the power. 
 
 PULLEY. 
 
 A pulley consists of a wheel movable about FIG. 2. 
 its axis, and so arranged as to be put in motion 
 by means of a cord passing over it. The movable 
 pulley is the only one by which a gain in power 
 is effected. A double movable pulley is repre- 
 sented at fig. 2. By such a pulley an equilibrium 
 is produced, when the power is to the weight, 
 as one to the number of ropes sustaining the 
 weight. If a single movable pulley be employed, 
 the weight is sustained by two ropes ; and the 
 power will be to the weight, as 1 to 2. If two 
 movable pulleys be employed, the weight is sus- 
 tained by four ropes, and the power will be to 
 the weight, as 1 to 4, &c. ; that is, each movable 
 pulleyis sustained by two ropes, and consequently 
 doubles the effect of the power employed. 
 
 Prob. 23. Given the number of movable pul- 
 leys and the power, to find the weight. 
 
 RULE. Multiply the power applied by twice the number of 
 movable pulleys. 
 
 Ex. 1. If in a system of pulleys there be three movable 
 pulleys, what weight will a power of 72 lb. balance ? 72 x 
 6 = 432lb. Ans. 
 
 2. What weight will a power of 15 lb. balance by a system 
 of 6 movable pulleys ? Ans. 180 lb. 
 
 3. By the aid of a system of 12 movable pulleys, how 
 many pounds will a man sustain who is capable of applying a 
 power of 1501b.? Ans. 3600. 
 
 Prob. 24. Given the number of movable pulleys, and the 
 weight to be balanced, to find the required power. 
 
298 APPENDIX. 
 
 RULE. Divide the weight by twice the number of movable 
 pulleys. 
 
 Ex. 1. How many pounds would be required by the aid of 
 two movable pulleys, to sustain 800 Ib. ? Ans. 200 Ib. 
 
 2. By the aid of 10 movable pulleys, how many Ib. would 
 be required to balance 2000 Ib. ? Ans. 100 Ib. 
 
 3. What weight would balance 10 pounds of silver by the 
 aid of a system of 100 pulleys 1 Ans. 12 pwt. 
 
 WHEEL AND AXLE. 
 
 The wheel and axle form a FIG. 3. 
 
 kind of perpetual lever, the 
 long arm of which is the semi- 
 diameter of the wheel, and the 
 short arm, the semi-diameter of 
 the axle. Consequently, the 
 power of the wheel and axle 
 is increased either by making 
 the wheel larger, while the axle 
 remains unaltered, or by making 
 the axle smaller, while the 
 wheel remains the same. The 
 power must always be applied 
 to the wheel, and the weight to 
 the axle, when an increase of 
 action is to be gained. In the 
 use of the wheel and axle, an equilibrium is produced, when 
 the power is to the weight as the semi-diameter of the axle is to 
 the semi-diameter of the wheel. 
 
 Prob. 25. Given the diameter of the wheel, the diameter of 
 the axle, and the power, to find the weight. 
 
 RULE. Multiply the diameter of the ivheel by the power ap- 
 plied, and divide the product by the diameter of the axle. 
 
 Ex. 1 . If the diameter of the axle be six inches, and that of 
 the wheel 6 feet, what weight attached to the axle, will 16 
 Ib. attached to the wheel, balance ? Ans. 192 Ib. 
 
APPENDIX. 299 
 
 2. If the diameter of a wheel be 20 feet, and that of the 
 axle 2 feet, and a power of 400 Ib. be applied to the wheel, 
 what weight will be balanced at the axle ? Ans. 4000 Ib. 
 
 Prob. 26. Given the diameter of the wheel, the diameter 
 of the axle, and the weight to be balanced, to find the required 
 power. 
 
 RULE. Multiply the diameter of the axle by the weight, and 
 divide the product by the diameter of the wheel. 
 
 Ex. 1 . If the diameter of the axle be 6 inches, and the diam- 
 eter of the wheel 12 feet, what power will balance a weight 
 of 360 Ib? Ans. 15lb. 
 
 2. If the diameter of the axle be 2 feet, and the diameter of 
 the wheel 20 feet, what power will balance a weight of 4000 
 Ib. ? Ans. 400 Ib. 
 
 THE INCLINED PLANE. 
 
 An inclined plane is a plane that makes with the ground or 
 floor some certain angle less than a right angle. In the appli- 
 cation of this instrument, the ratio of the power and weight is 
 always the same as that of the height and length of the plane. 
 
 Prob. 27. Given the length and height of the plane, and also 
 the power, to determine the weight. 
 
 RULE. Multiply the power by the length of the plane, and 
 ivide the product by its perpendicular height. 
 
 Ex. 1. If the length of a plane be 16 feet, and its height 4 
 feet, what weight will a power of 32 Ib. sustain 1 Ans. 1281b. 
 
 2. What weight will 164 Ib. sustain on a plane 112 feet in 
 length and 3 feet in height ? Ans. 6122| Ib. 
 
 Prob. 28. Given the length and height of the plane, and also 
 the weight, to find what power will sustain it. 
 
 RULE. Multiply the weight by the height of the plane, and 
 divide the product by the length. 
 
300 APPENDIX. 
 
 Ex. 1. What power will balance J28lb. on an inclined plane, 
 the length of which is 16 feet, and the height 4 feet? Ans. 
 32 Ib. 
 
 2. Suppose on a railroad there is an inclined plane 150 
 rods in length, and rising to the perpendicular height of 50 feet ; 
 what power will be required to sustain a weight of 84000 Ib. 
 Ans. 28000 Ib. 
 
 THE WEDGE. 
 
 The wedge is composed of two inclined planes, whose bases 
 unite and form the base of the wedge. The power applied to 
 the wedge is to the effect produced at the side of the wedge, 
 as the thickness of the head to the length of the wedge, no allow- 
 ance being made for friction. 
 
 Prob. 29. Given the thickness of the head, the length of 
 the side, and the power acting upon the head of a wedge, to 
 determine the force produced on the side. 
 
 RULE. Multiply the length of the wedge by the power ap- 
 plied^ and divide the product by the thickness of the head. 
 
 Ex. 1. If the length of a wedge be 12 inches, the thickness 
 of the head 3 inches, and the force applied, 64 Ib., what will 
 be the resistance at the side 1 Ans. 256 Ib. 
 
 2. If the length of a wedge be -32 inches, the thickness of 
 the head 2 inches, and the force applied 16001b., what will 
 be the resistance at the side ? Ans. 25600 Ib. 
 
 Prob. 30. Given the length of the side, the thickness of the 
 head, and the resistance upon the side of a wedge, to find the 
 force acting upon the head. 
 
 RULE. Multiply the resistance at the side by the thickness 
 of the head, and divide the product by the length of the side of 
 the wedge. 
 
 Ex. 1. If the length of the wedge be 32 inches, the thick- 
 ness of the head 2 inches, and the resistance at the side be 
 25600 Ib. what must be the force upon the head, no allowance 
 being made for friction ? Ans. 1600 Ib. 
 
APPENDIX. 
 
 301 
 
 2. If the resistance at the side of a wedge be 20000 lb., 
 the length of the wedge 20 inches, and the thickness of the 
 head 3 inches, what force is required to be applied to counter- 
 act the resistance at the sides ? Ans. 3000 lb. 
 
 THE SCREW. 
 
 The screw is a cylindrical piece of wood, or metal, with a 
 spiral thread running round it with a gradual and uniform in- 
 clination. The thread, therefore, forms an inclined plane of 
 the same length as the thread itself, and whose height equals 
 the length of the screw. 
 
 It is then obvious that the power of the screw depends in 
 part upon the fineness of the threads, since the finer they are, 
 the greater will be the length of the plane, while its height 
 remains the same. The power of the 
 screw also depends in part on the FIG. 4. 
 
 length of the lever employed to move 
 it. It is, therefore, a compound power, 
 involving in its action both the princi- 
 ple of the lever and the inclined plane, 
 and is consequently more efficient'than 
 either of the other mechanical powers. 
 In the action of this instrument, an 
 equilibrium between the power and 
 weight is produced, when the power 
 is to the weight as the distance between 
 two contiguous threads of the screw is 
 to the circumference of the circle de- 
 scribed by the power in one revolution. 
 
 Prob. 31. Given the distance between the threads of the 
 screw, the length of the lever, and the power applied to find 
 the weight. 
 
 RULE. Multiply the circumference of the circle described by 
 one revolution of the lever by the power applied, and divide the 
 product by the distance between the threads of the screw. 
 
 Note. The lever is the semi-diameter of the circle it de- 
 scribes ; the circumference of the circle may therefore be 
 found by Prob. 6th. 
 26 
 
302 APPENDIX. 
 
 Ex. 1. If the threads of a screw be 2 inches apart, the 
 lever 40 inches in length, and a power of 60 Ib. be applied to 
 the end of the lever, what weight will be required to produce 
 an equilibrium? Ans. 7542 Ib. 13 oz. 11|- drams. 
 
 2. If the distance between the threads of a screw be 1 inch, 
 the length of the lever 16 inches, and the power applied to the 
 end of the lever be 40 Ib., what weight will produce an equi- 
 librium ? Ans. 4022.8+ Ib. 
 
 Prob. 32. Given the weight, the length of the lever, and 
 the distance between the threads of the screw, to find the pow- 
 er required to produce an equilibrium. 
 
 RULE. Multiply the given weight by the distance between 
 the threads of the screw, and divide the product by the circum- 
 ference of the circle described by one revolution of the lever. 
 
 Ex. 1. How many pounds applied to the end of a lever 
 16 inches long, will balance 4022.8 Ib. upon a screw, whose 
 threads are one inch apart ? Ans. 40 Ib. 
 
 2. How many pounds applied to the end of a lever 30 inches 
 long, will produce an equilibrium with 4000 Ib. attached to 
 a screw whose threads are 2 inches apart? Ans. 42? Ib. 
 nearly. 
 
 STEAM POWER. 
 
 The power of the steam engine is usually estimated by com- 
 paring it with that of the horse. There are, however, various 
 circumstances, which tend to render the power of the horse an 
 indefinite standard ; such as the natural strength of the ani- 
 mal, the training it may have received to enable it to exert its 
 strength to the best advantage, its position while exerting its 
 strength, the length of time the exertion is to be continued, 
 and the velocity of motion required. Before the term, " horse 
 power," could be employed to communicate any uniform idea 
 of force exerted, it was therefore necessary to determine by 
 experiment, what was the medium power of the horse, making 
 due allowance for varying circumstances. 
 
 Since, however, the power required to move any given body 
 in a horizontal direction is dependent on various circumstances 
 besides its actual weight, and as a diminished force only is re- 
 
APPENDIX. 303 
 
 quired to continue a body in motion, when motion in that direc- 
 tion is once communicated, a medium power could be ob- 
 tained only by repeated experiments on the power of the 
 horse in raising weights in a perpendicular direction, like 
 raising a stone from a well, while the animal was made to act 
 in a horizontal direction through the agency of a single pulley. 
 Experiments were accordingly instituted, the result of which 
 has been to establish the fact, that a one horse power is a power 
 competent to raise 2250 Ib. one mile in a perpendicular direc- 
 tion in 6 hours of time. This fact being thus established, the 
 term, " horse power," has acquired a distinctness of meaning, 
 which renders it a convenient standard by which to estimate 
 the power of machinery, especially when propelled by steam. 
 The weight given, viz. 2250 Ib., is a little more than a ton ; an 
 amount entirely beyond the power of any horse to move as above 
 stated. The velocity of one mile in 6 hours of time, is, on 
 the other hand, far less than the natural velocity of the horse. 
 How are we then to understand the estimation of horse power 
 here presented ? It will be remembered that in all machinery 
 by which a gain of power is effected, the power applied to put 
 the machinery in motion, always moves through a greater space 
 than the body moved by the same machinery ; or, in other 
 words, " what is gained in power is lost in time." This 
 fact presents us with a full explanation of the matter. A horse 
 with a moderate burden will easily move 3 miles per hour, 
 and in 6 hours, 6x3 = 18 miles. Now, suppose the horse to 
 act through machinery so constructed, that while he moves at 
 the rate of 18 miles in 6 hours, the weight attached to and 
 acted upon by the same machinery, moves only one mile ; how 
 great a power would the horse be required to exert to raise 
 2250 Ib. that distance ? Evidently, a power equal to -^ of 
 2250 Ib. or 125 Ib., a weight entirely within the power of a 
 horse of ordinary strength to raise at the velocity proposed. 
 This reduces the statement of a single horse power to a form 
 easy to be comprehended ; viz. it is a power competent to raise 
 125 Ib. 18 miles in 6 hours, or 3 miles per hour. 
 
 Prob. 33. To determine the horse power required to raise a 
 given weight any specified distance per hour. 
 
 RULE. Multiply the weight reduced to pounds by the dis- 
 tance through which it is to be moved, and make the product a 
 dividend. Then, having reduced 3 miles to the same denomina- 
 
304 APPENDIX. 
 
 tion with the given distance, multiply it by 125, and make the 
 product a divisor. Divide, and the quotient will express the re- 
 quired horse power. 
 
 Ex. 1 . What amount of horse power will be required to raise 
 14 T. 14 cwt. 2 qr. 16 lb., 60 feet per hour ? 
 
 Operation: 14 T. 14 cwt. 2 qr. 161b.=33000 lb.; and 33000 
 X 60 =1980000, the dividend. 3 miles = 15 840 feet; and 
 15840 X 125 = 1980000, the divisor; then, 1980000^-1980000 
 = 1, Ans. 
 
 Note. If the time given be less than one hour, the required 
 power must be proportionally greater. Had the time in the 
 preceding sum been 20 minutes, instead of one hour, that is, 
 of the time given, three times as much power would have been 
 required. Hence, the power required is found by multiplying 1 
 by the direct ratio of 125 and the given weight, and by the in- 
 verse ratio of 1 hour and the given time. 
 
 2. What amount of horse power is necessary to raise 60 
 tons 3 feet per minute 1 Ans. 12.2+ horse power. 
 
 Prob. 34. Given the horse power, the distance, and the time, 
 to determine the weight that may be raised. 
 
 RULE. Multiply 125 by the given horse power ; the product 
 will be the weight that power will raise 3 miles per hour. Mul- 
 tiply this by 3 miles reduced to the same denomination as the 
 given distance, and divide the product by that distance, and the 
 number obtained will determine the pounds that may be raised 
 the given distance in one hour ; then, lastly, multiply by the 
 given time in minutes, and divide the product by 60; the quotient 
 will be the weight required. 
 
 Ex. 1 . What weight will a 5 horse power raise 120 feet in 20 
 minutes of time ? 
 
 Solution: 125x5=625 ; then, since 3 miles equals 15840 
 feet, 625x15840 = 9900000; and 9900000 -f- 120 = 82500 ; 
 and 82500 x 20-^60=27500 lb., Ans. 
 
 2. What weight will an 8 horse power raise 360 feet in 90 
 minutes of time ? Ans. 29 T. 9 cwt. 1 qr. 4 lb. 
 
APPENDIX. 
 
 305 
 
 PROBLEMS IN INTEREST. 
 
 Prob. 35. Time, rate per cent., and amount given, to find the 
 principal. 
 
 RULE. Divide the given amount by the amount of $1 or 
 I . for the given rate and time. 
 
 Ex. 1 . What sum of money will amount to $360, at 6 per 
 cent., in 3 years and 4 months ? 
 
 The amount of $1 for 3 years and 4 months, is $1.20 ; and 
 $360-h$1.20 = $300, Ans. 
 
 2. What sum of money will, in 4 years, 6 months, at 6 per 
 cent, per annum, amount to $1016 ? Ans. $800. 
 
 Prob. 36. Time, rate per cent., and interest given, to find the 
 principal. 
 
 RULE. Divide the given interest by the interest of one 
 dollar for the given time and rate per cent. 
 
 Ex. 1. If a man's annual interest be $1200, what is his 
 capita], the rate per cent, being 6 1 Ans. $20000. 
 
 2. How much money on interest, at 6 per cent., will gain 
 $36 in 1 year and 8 months ? Ans. $360. 
 
 Prob. 37. Given the principal, interest, and time, to find the 
 rate per cent. 
 
 RULE. Divide the given interest by the interest of the given 
 principal, at one per cent, for the given time. 
 
 Ex. 1. A man having $4000 on interest, at the expiration 
 of one year received $240 interest on the same. What rate 
 per cent, did he receive ? Ans. 6 per cent. 
 
 2. If $200 be paid as interest on $2000 for 2 years and 
 6 months, what is the rate per cent. ? Ans. 4 per cent. 
 
 Prob. 37. Given the principal, rate per cent., and interest, 
 to find the time. 
 
 RULE. Divide the given interest by the interest of the prin- 
 cipal for one year ; the quotient will be the time. 
 26* 
 
306 APPENDIX. 
 
 Ex. 1. Paid $108 interest on a note of $600. How long 
 had the note been on interest, the rate per cent, being 6 ? Ans. 
 3 years. 
 
 2. Paid 8400 interest on a note of $800, the rate per cent, 
 being 4. How long had the note been on interest ? Ans. 12 
 years and 6 months. 
 
 ANNUITIES. 
 
 An annuity is a sum of money payable annually for any given 
 number of years, or forever. 
 
 When any sum of money, payable annually, has remained 
 unpaid for any number of years, the amount due is the sum of 
 all the annuities which are in arrears, together with the interest 
 on each annuity for the time it has remained unpaid. 
 
 Prob. 38. Given the annuity, the rate per cent., and the time 
 of arrears, to find the amount due. 
 
 RULE. Find the amount of each annuity for the time it has 
 remained unpaid, at the given rate per cent. ; the sum of the 
 amounts thus obtained will be the sum required. 
 
 Ex. 1. A man, who was in the receipt of an annual pension 
 of $200, was required to forego the payment four years in 
 succession. What was due at the expiration of that time, the 
 rate per cent, being 6 1 
 
 The annuity of the fourth year would receive no interest, 
 as it was not due till the year expired, and consequently 
 amounted to only $200. The annuity of the third year was 
 entitled to one year r s interest, and therefore amounted to $212. 
 The annuity of the second year was entitled, to two years in- 
 terest, and amounted to $224. The annuity of the first year 
 was entitled to three years interest, and amounted to $236. 
 Therefore, 200-j-212 + 224+236=$872, Ans. 
 
 2. What is the value of an annuity of $600, which has 
 remained unpaid for 8 years, interest at 6 per cent, being al- 
 lowed ? Ans. $5808. 
 
 Prob. 39. Given the annuity and rate per cent., to find its 
 present worth for any number of years. 
 
APPENDIX. 307 
 
 RULE. Divide each annuity by the amount of $1 or 1 . 
 for the time before it becomes due ; the quotient will be the pres- 
 ent worth of the several annuities, and their sum will be the 
 amount required. 
 
 Ex. 1. What is the present worth of an annuity of $600, 
 for three years to come, 6 per cent, being allowed for present 
 payment 1 
 
 The present worth of the first year is $600 -4- $ 1 .06 = 566.037 ; 
 of the second year, $600-f-$1.12 = $535.714 ; and of the 
 third year, $600 -^$1.1 8 = $508.474. Then, $566.037 + 
 $535.714 + $508.474 = $1610.225, Ans. 
 
 2. What is the present worth of an annuity of 30 <. for 5 
 years to come, at 4 per cent. ? Ans. 134 . 5 s. 5 d. 4- 
 
 ANNUITIES AT COMPOUND INTEREST. 
 
 The amount of an annuity at compound interest, is obtained 
 by computing the compound interest of the several payments, and 
 then finding their sum. Operations of this nature may obvi- 
 ously be solved by geometrical progression, by making $1 the 
 first term of a geometrical series, and the amount of $1 for 
 one year at the given rate per cent., the ratio. The number 
 of terms will always be the same as the number of years, 
 (See Geometrical Progression.) 
 
 Ex. 1. What will an annuity of $60 per annum amount to in 
 4 years, at 6 per cent.? $1 is the first term, $1.06 the ratio. 
 Therefore, 
 
 O^X 60 = $262.476,+ Ans. 
 
 We have then the following general principle. Raise the 
 ratio (which is always found by adding the per cent, to $ 1 ) to 
 a power equal to the number of years ; from this, subtract 1, 
 then divide the remainder by the ratio less 1 , (that is, by the de- 
 cimal part of the ratio only,) and multiply the quotient by the 
 annuity. 
 
308 
 
 APPENDIX, 
 
 2. What is the amount of an annuity of $400, which has 
 remained unpaid for 20 years, at 6 per cent, compound inter- 
 est ? Ans. 14714.23. + 
 
 Or, the answers to such sums may be found by multiplying 
 the amount of one dollar for the given number of years and 
 rate percent, as found in the following table, by the annuity : 
 
 TABLE 
 
 SHOWING THE AMOUNT OF $1 AT 5 OR 6 PER CENT. COMPOUND 
 
 INTEREST, FOR ANY NUMBER OF YEARS BETWEEN ONE 
 
 AND FORTY. 
 
 yr- 
 
 5 per cent. 
 
 1.000000 
 
 yr. 
 1 
 
 6 per cent. 
 
 1.000000 
 
 yr. 
 
 21 
 
 5 per cent. 
 
 35.719252 
 
 yr. 
 21 
 
 6 per cent. 
 
 39.992727 
 
 2 
 
 2.050000 
 
 2 
 
 2.060000 
 
 22 
 
 38.505214 
 
 22 
 
 43.392290 
 
 3 
 
 3.152500 
 
 3 
 
 3.183600 
 
 23 
 
 41.430475 
 
 23 
 
 46.995828 
 
 4 
 
 4.310125 
 
 4 
 
 4.374616 
 
 24 
 
 44.501999 
 
 24 
 
 50.815577 
 
 5 
 
 5.525631 
 
 5 
 
 5.637093 
 
 25 
 
 47.727099 
 
 25 
 
 54.864512 
 
 6 
 
 6.801913 
 
 6 
 
 6.975319 
 
 26 
 
 51.113454 
 
 26 
 
 59.156383 
 
 7 
 
 8.142008 
 
 7 
 
 8.393838 
 
 27 
 
 54.669126 
 
 27 
 
 63.705766 
 
 8 
 
 9.549109 
 
 8 
 
 9.897468 
 
 28 
 
 58.402583 
 
 28 
 
 68.528112 
 
 9 
 
 11.026564 
 
 9 
 
 11.491316 
 
 29 
 
 62.322712 
 
 29 
 
 73.639798 
 
 10 
 
 12.577893 
 
 10 
 
 13.180795 
 
 30 
 
 66.438847 
 
 30 
 
 79.058186 
 
 11 
 
 14.206787 
 
 11 
 
 14.971643 
 
 31 
 
 70.760790 
 
 31 
 
 84.801677 
 
 12 
 
 15.917127 
 
 12 
 
 16.869941 
 
 32 
 
 75.298829 
 
 32 
 
 90.889778 
 
 13 
 
 17.712983 
 
 13 
 
 18.882138 
 
 33 
 
 80.063771 
 
 33 
 
 97.343165 
 
 14 
 
 19.598632 
 
 14 
 
 21.015066 
 
 34 
 
 85.066959 
 
 34 
 
 104.183755 
 
 15 
 
 21.578564 
 
 15 
 
 23.275970 
 
 35 
 
 90.220307 
 
 35 
 
 111.434780 
 
 16 
 
 23.657492 
 
 16 
 
 25.672528 
 
 36 
 
 95.836323 
 
 36 
 
 119.120867 
 
 17J25.840366 
 
 17 
 
 28.212880 
 
 37 
 
 101.628139 
 
 37 
 
 127.268119 
 
 1828.132385 
 
 18 
 
 30.905653 
 
 38 
 
 107.709546 
 
 38 
 
 135.904206 
 
 19 
 
 30.539004 
 
 19 
 
 33.759992 
 
 39 
 
 114.095023 
 
 39 
 
 145.058458 
 
 20 
 
 33.065954 
 
 20 
 
 36.785591 
 
 40 
 
 120.799774 
 
 40 
 
 154.761966 
 
 3. What is the amount of an annuity of $150, which has 
 remained unpaid for 12 years, compound interest 6 per cent. ? 
 Ans. $2530.49.+ 
 
 The amount of $1 for 12 years in the above table, is $16.- 
 869941, and $16.869941 x 150 = $2530.49,+ Ans. 
 
 4. What is the worth of an annual salary of $1000, which 
 
 
APPENDIX, 
 
 309 
 
 has remained unpaid for 15 years, compound interest at 6 per 
 cent. ? Ans. $23275.97. 
 
 5. What is the value of an annuity of $150, to continue 30 
 years, at 5 per cent, compound interest ? Ans. $9965. 827. -f- 
 
 TABLE 
 
 SHOWING THE PRSEENT WORTH OF AN ANNUITY OF $1 AT 5 OR 
 
 6 PER CENT. FOR ANY NUMBER OF YEARS BETWEEN 
 
 ONE AND FORTY. 
 
 Jt. 
 
 2 
 
 5 per cent. 
 
 0.952381 
 1.859410 
 
 yr- 
 
 1 
 o 
 
 6 per cent. 
 
 0.943396 
 1.833393 
 
 yr. 1 5 per cent. 
 
 21 12.821153 
 22113.163003 
 
 yr. 
 21 
 22 
 
 6 per cent. 
 
 11.764077 
 12.041582 
 
 3 
 
 2.723248 
 
 3 2.673012 
 
 23 
 
 13.488574 
 
 23 
 
 12.303379 
 
 4 
 5 
 
 3.545950 
 4.329477 
 
 4 
 5 
 
 3.465106 
 4.212364 
 
 24 13.798642 
 25 14.093945 
 
 24 
 25 
 
 12.550358 
 12.783356 
 
 6 
 
 5.075692 
 
 6 
 
 4.917324 
 
 26 
 
 14.375185 
 
 26 
 
 13.003166 
 
 7 
 
 5.786373 
 
 7 
 
 5.582381 
 
 27 
 
 14.643034 
 
 27 
 
 13.210534 
 
 8 
 
 6.463213 
 
 8 
 
 6.209794 
 
 28 
 
 14.898127 
 
 28 
 
 13.406164 
 
 9 
 
 7.107822 
 
 9 
 
 6.801692 
 
 29 
 
 15.141074 
 
 29 
 
 13.590721 
 
 10 
 
 7.721735 
 
 10 
 
 7.360087 
 
 30 
 
 15.372451 
 
 30 
 
 13.764831 
 
 11 
 
 8.306414 
 
 ill 
 
 7.886875 
 
 31 
 
 15.592810 
 
 31 
 
 13.929086 
 
 12 
 
 8.863252 
 
 12 
 
 8.388844 
 
 32 
 
 15.802677 
 
 32 
 
 14.084043 
 
 13 
 
 9.393573 
 
 13 
 
 8.852683 
 
 33 
 
 16.002549 
 
 33 
 
 14.230230 
 
 14 
 
 9.898641 
 
 14 
 
 9.294984 
 
 34 
 
 16.192904 
 
 34 
 
 14.368141 
 
 15 10.379658 
 
 15 
 
 9.712249 
 
 35 
 
 16.374194 
 
 35 
 
 14.498246 
 
 16 10.837770 
 
 16 
 
 10.105895 
 
 36 
 
 16.546852 
 
 36 
 
 14.620987 
 
 1711.274066 
 18'11.689587 
 
 17 
 
 18 
 
 10.477260 
 10.827603 
 
 37 16.711287 
 38 16.867893 
 
 37 
 
 38 
 
 14.736780 
 14.846019 
 
 19 12.085321 
 20:12.462216 
 
 19 11.158116 
 20111.469921 
 
 39 17.017041 
 40117.159086 
 
 39 
 40 
 
 14.949075 
 15.046297 
 
 To find the value of an annuity by the preceding table, mul- 
 tiply the value of one dollar as given in the preceding table for 
 the given number of years and rate per cent, by the given num- 
 ber of dollars. 
 
 Ex. 1. What sum of money will purchase an annuity of 
 $400, to continue 12 years, at 6 per cent, discount? 
 
 The value of $1 for 12 years is $8.388844, and 8.388844 X 
 = $3355.537,+ Ans. 
 
310 APPENDIX. 
 
 2. What is the present value of an annuity of $1200, to 
 continue 16 years, at 6 per cent discount 1 Ans. 12 127.074. -f 
 
 3. How much must be paid for an annuity of $75, to con- 
 tinue 30 years, at 6 per cent, discount? Ans. $1032.352.+ 
 
 ASSESSMENT OF TAXES. 
 
 A tax is a sum of money collected from the citizens of a 
 State, county, or town, for defraying the expenses necessarily 
 incurred in the administration of justice, and in works of 
 common utility. 
 
 The sum each man is required to pay, depends, for the most 
 part, on the amount of his property. To this the poll tax forms 
 the only exception ; which is a tax required of every male in- 
 habitant of a State, who has attained the age of twenty-one 
 years, independently of the property he may possess. 
 
 In levying a tax upon any community, first take a complete 
 list of all the property of the town on which the tax is to be 
 laid, and also of the number of polls to be taxed. Next, de- 
 termine the amount of the poll taxes by multiplying the num- 
 ber of polls by the tax on each. This being determined, sub- 
 tract it from the whole sum to be raised, and the remainder 
 will be the sum to be raised on the property of the town, both 
 real and personal. Then to ascertain the percentage, divide 
 the tax to be raised by the whole amount of taxable property, 
 and the quotient will be the tax on one dollar. To find what 
 tax any individual pays, multiply his inventory by this quotient, 
 or tax per dollar. 
 
 Ex. 1 . What will be A's tax, if he own real estate to the 
 amount of $1340, and personal property to the amount of 
 $874, if the town in which he lives, the inventory of which 
 is $32265, raise a tax of $1129.95, there being 270 polls 
 which are taxed 60 cents each, and for two of which he pays ? 
 
 270 polls x 60 = $162. 00, amount of the poll taxes. There- 
 fore, $1129.95 $162.00rr$967.95, the sum to be levied on 
 the property. Hence, $967.95-^- $32265 =3 cents, the tax on 
 one dollar. Then, $1340 x .03=r$40.20, and $874 x. 03 = 
 $26.22, and two polls at 60 cents each=$1.20. Therefore, 
 $40.20+$26.22+$1.20 = 67.62, Ans. 
 
APPENDIX. 
 
 311 
 
 After finding what the tax is per dollar, a table may be 
 formed like the following, which will expedite the operations : 
 
 TABLE. 
 
 ax on SI is .03 
 
 Tax on $10 is .30 
 
 Tax on $100 j s 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 
 
 
 fi 
 
 .21 
 
 
 
 70 
 
 2.10 
 
 
 
 700 
 
 21.00 
 
 
 8 
 
 ' .24 
 
 
 
 80 
 
 2.40 
 
 
 
 800 
 
 24.00 
 
 
 9 
 
 ' .27 
 
 
 
 90 
 
 2.70 
 
 
 
 900 
 
 27.00 
 
 
 
 
 
 1000 
 
 : 30.00 
 
 Now, to determine A's tax from this table, $1340+ $874 = 
 $2214. 
 
 The tax on $2000 = $60.00 
 
 200= 6.00 
 
 " 10= .30 
 
 " 4= .12 
 
 Two polls at 60 ct. = 1 .20 
 
 $67.62 Ans. as before. 
 
 2. From the preceding table, calculate B's tax, whose real 
 estate is valued at $5620, and his personal property at $1162, 
 and who pays 3 poll taxes, each 75 cents. Ans. $205.71. 
 
 3. Form a table and calculate the tax of A., whose property 
 is valued at $1500 ; of B., whose property is $2000 ; of C., 
 whose property is $1200; of D., whose property is $3000, 
 and of E., whose property is estimated at $3500 ; supposing 
 them to live in a town whose inventory is $ 1 000000, and on 
 which a tax of $4280 is to be levied, the number of polls being 
 400, and paying each -70 cents. Ans. A's tax will be $6 ; 
 B's, $8 ; C's, $4.80 ; D's, $12 ; and E's, $14. 
 
 4. A tax of $3000 is to be raised on a certain town, the 
 inventory of which is $60000 ; the number of polls 250, taxed 
 each 75 cents. What is the amount of A's tax, whose property 
 is valued at $2500, and who pays for two polls ? Ans. 
 $118.68$. 
 
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