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 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, 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, . 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, 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, ~~~> 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, . 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|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, 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, tvr4 u Kjr L. V ^^^^^ - < * M3O6O07 QAio TC THE UNIVERSITY OF CALIFORNIA LIBRARY pel Street, New Haven. ^ Are publishers of the following VALUABLE SCHOOL BOOKS R ; or Introdur I. Olney, An . AritLm'.-: ^ C< OF THE UNITED ;