# Full text of "The western practical arithmetic : wherein the rules are illustrated, and their principles explained : containing a great variety of exercises, particularly adapted to the currency of the United States : with an appendix containing the canceling system, abbreviations in multiplication, mensuration, and the roots : designed for the use of schools and private students"

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IRLF S CIN ' /. ] i. IN MEMORIAM FLOR1AN CAJOR1 THE WESTERN^, . PRACTICAL ARITHMETIC, WHEBE1N THE RULES ARE ILLUSTRATE, AND THEIR PRINCIPLES EXPLAINED: CONTAINING A GREAT VARIETY OF EXERCISES, PART^CULAJILY-ADAPTED TO THE CURRENCY OF THE UNITED STATES. WITH AN APPENDIX: ""S CONTAINING THE CANCELING SYSTEM, ABBREVIATIONS IN MULTIPLl-f CATION, MENSURATION, AND THE ROOTS. -) DEIQNEO FOB THE USE OF SCHOOLS AND PRIVATE STUDENTa COMPILED BY JOHN L. TALBOTT. PUBLISHED BY E. MORGAN & CO., Ill MAIN STREET. 1 853. District of Ohio, to wit : ? District Clerk's Office. 5 BE IT BEMEMBEBED, that on the seventeenth day of January, An> no Domini eighteen hundred and forty-five, E. MOBGAN & Co., of the said District, have deposited in this office the title of a book, the title of which is in the words following, to wit : u The Western Practical Arithmetic, wherein the rules are illus- trated, and their principles explained, containing a great variety of exercises, particularly adapted to the currency of the United States : with an Appendix, containing the Canceling System, Abbreviations in Multiplication, Mensuration and the roots ; designed for the use of schools and private students; compiled by JOHN L.TALuoTT;"the right whereof they claim as proprietors, in conformity with an act of Congress, entitled "An act to amend the several acts respecting copy- rights." WM. MINER, Clerk of District. PREFACE. V&^T'f-' I.v presenting this work to the public, the author makes no pre- tensions to having discovered any new spring by which to put the youthful mind into action, nor any new method of communicating a knowledge of Arithmetic. He has founded his work on the belief that labor and labor only, can insure success in any pursuit ; and -that labor should always be bestowed upon those objects which pro- duce the greatest useful result. In the selection and arrangement of matter, therefore, those rules that are of the most general use, have been presented first, and their exercises made extensive, that the pupil many early become familiar with their principles, and expert in their application. The explanations accompanying the rules, are designed to facili- tate the progress of private students, and to diminish the labor of teachers, especially in large schools, where they are unable to give to each pupil the necessary explanations. The MESSUUATIOW of Carpenters', Masons', Plasterers' and Pavers' work, &c., will be found an acceptable part of Arithmetic, to every man of business, and a practical knowledge of it will con- tribute much to the security and satisfaction of both workmen and employers, in estimating amounts of work. This has been intro- duced in consequence of numerous applications to the author to measure various kinds of work, and for instruction in particular rules of Mensuration. The system of Book Keeping, is thought to be sufficient for all the purposes of farmers, mechanics and retailers, in that necessary branch of a business education. a* - How far the author has succeeded in his attempts to compile a useful work, particularly adapted to the circumstances of the Western People, remains for them to judge, and for experience to determine. NOTICE. THE favorable reception of this treatise and the increasing demand for it, have induced the publishers to revise, enlarge, and otherwise improve the work. Such alterations and amendments have been made as the experience of the author and of other intelligent and successful teachers has suggested; it is therefore presumed, that the work will be found more useful, and consequently more acceptable than heretofore. Numerous testimonials to the merits of the work, have been re- ceived; but its general adoption without any efforts to force its introduction, and its intrinsic worth, are our main reliance; we have therefore given it a thorough revision, and now submit the result of our labors to a discerning public. Febr ii ajy T J841. P MORGAN & Co. _ -J.elk v5j CONTENTS. PACK. Numeration, ~ 5 Simple Addition, 12 Simple Subtraction, 15 Simple Multiplication, 18 Simple Division, 23 Addition of Federal Money, - 30 Subtraction of Federal Money, 32 Multiplication of Federal Money, 34 Division of Federal Money, 36 Reduction, 38 Compound Addition, 60 Compound Subtraction, 66 Compound Multiplication, 72 Compound Division, 80 Simple Proportion, 88 Compound Proportion, 100 Practice, 103 Tare and Tret, 109 Interest, 112 Compound Interest, 118 Insurance, Commission and Brokage, 121 Discount, 122 Equation of Payments, 124 Barter, 126 Loss and Gain, 128 Fellowship, 132 ar Fractions, 135 Reduction of Vulgar Fractions, 136 Addition of Vulgar Fractions, 144 Subtraction of Vulgar Fractions, 146 Multiplication of Vulgar Fractions,. 147 Division of Vulgar Fractions, 148 Decimal Fractions, . 149. Addition of Decimals, 150 Subtraction of Decimals, ib. Multiplication of Decimals, 151 Division of Decimals, - ib. Reduction of Decimals, 153 Proportion in Decimals, 155 Compound Proportion in Decimals, 156 Mensuration, > ib. Involution, . . 167 Evolution, 169 Square Root, ib. Cube Root, 174 Roots of All Powers, 179 Arithmetical Progression 180 Geometrical Progression 184 Appendix 194 Exchange *.* 189 Promiscuous Exercise . . .... ,190 ARITHMETIC. ARITHMETIC is that part of MATHEMATICS which treats of numbers. It is both a science and an art; the science explains the nature of numbers, and the principles upon which the rules are founded^ while the art relates merely to the application of the various rules. All the operations of arithmetic are conducted by means of FIVE fundamental rates, viz., Numeration, (which includes Notation,} Addition, Subtraction, Multiplication, and Division. NUMERATION AND NOTATION. Numeration is the art of representing figures or num- bers by words ; Notation is the art of representing num- bers by characters called figures. ' All numbers are represented by the following charac- ters, which are called figures or digits. 0, 1, 2, 3, 4, 5, 6, 7, 8, 0. nought, one, two, three, four, five, six, seven, eight, nine. The one is often called a unit, it signifies a whole tb"ig of a kind ; two signifies two units or ones ; three s ^nifies three units or ones, <fcc. The value which the figures have when standing alone is called their simple value ; but in order to denote numbers higher than 9, it is necessary to give them ano- ther value called a local value, which depends entirely on the order or place in which they stand. Thus, when we wish to write the number ten in figures, we do it by combining the characters already known, placing a 1 on the left hand of the 0, thus, 10, which is read ten. This 10 expresses ten of the units denoted by 1, but as it is only a single ten it is called a unit, and the 1 being written in the second order or second place from the right hand to express it, it is called a unit of the second order, the first place being called the place 6 NUMERATION AND NOTATION. of units, and the second, the place of tens ; ten units of the first order making one unit of the second order. When units simply are named, units of the first order are always meant, when units of any other order are intended, the name of the order is always added. Two tens or twenty, are written 20. Three tens or thirty, " " 30. Four tens or forty, " " 40. Five ten* or fifty, " " 50. Six tens or sixty, " " 60. Seven tens or seventy, " " 70. Eight tens or eighty, " " 80. Nine tens or ninety, " " 90. Ten tens or one hundred, " " 100. The numbers between 10 and 20, between 20 and 30, between 30 and 40, <fec. may easily be expressed by considering the tens and units of which they are composed. Thus, eleven being composed of one ten and one unit, is expressed thus, 11, twenty-thcee being composed of two tens, and three units, is expressed thus, 23. &c. Sixteen being 1 ten and 6 units, is written thus, 16. Thirty-nine being 3 tens and 9 units, is written 39. Sixty-four being 6 tens and 4 units, is written 64. Ninety-five being 9 tens and 5 units, is written 95. Ten tens or one hundred forms a unit of the third order ; it is expressed by placing a 1 in the third pin e, and filling the first and second places with cyphei . thus, 100. Two hundred is expressed thus, 200 Three hundred thus, 300, <fcc. With the orders of units, tens, and hundreds, all the numbers between one and one thousand may be readily expressed. For example, in the number four hundred and twenty-seven, there are 4 hundreds, 2 tens, and 7 units, that is, 4 units of the third order, 2 units of the second order, and 7 units of the first order. = I Hence the number is written thus, 427 In the number three hundred and five, there are 3 hundreds, no tens, and 5 units, or 3 units of the third, NUMERATION AND NOTATION. 7 none of the second, and five of the first order, hence the number is written thus, * I 305 Ten units of the order of hundreds, that is ten hun- dreds form a unit of the fourth order, called thousands, written thus, 1000. In the same manner ten units of the fourth order form a unit of the fifth order, called tens of thousands. The following may be regarded as the principles of Notation and Numeration. 1st. Ten units of the first or lowest order, make one unit of the second order; ten units of the second order, make one unit of the third order, and universally ten units of any order make a unit of the next higher order. 2d. Jill numbers are expressed by the nine digits, and the cypher, and this is effected by giving to the same figure different values according to the place it occupies. Thu^, 4 in the first place is 4 units, in the second place 4 tens or forty, and so on. This tenfold increase of value by changing the place of the same figure is usually expressed by saying that figures in- crease from right to left in a tenfold proportion. The names of the orders are to be learned from the NUMERATION TABLE. 3 2 o S | **" 3 **"" JS aa rS "^ | | | | ^ (| 987654321 The orders are likewise divided into periods of ei places each, according to the following table. NUMERATION AND NOTATION. of Billions. ofMilli llioni. of Unit*. f^ I ii II g 2 ' . . s o 'S S .3 S3 X! P o 5 "5 W2 "5 "^ !<= - The periods succeeding those in the table, are Tril- lions, Quadrillions, Quintillions, Sextillions, Sepiil- lions, Octillions, and Nonnillions, and analogical names might be formed for the succeeding higher periods. From the preceding remarks the pupil will readily understand the reason of the following rule for numer- ating or expressing figures by words. RULE. Commence at the right hand, and separate the given number into periods, then beginning at the left hand, read the figures of each period as if they stood alone, and then add the name of the period. Thus, the number 8304000508245, when divided into periods, becomes 8,304000,508245, and is read, Eight billion, three hundred and four thousand mil- lion, Jive hundred and eight thousand two hundred and forty-Jive. The name unit of the right hand pe- riod is commonly omitted in reading. EXERCISES IN NUMERATION. Ex. 1. 35 10. 3700054 2. 204 3. 513 4. 2000 5. 3054 6. 7428 7 10345 8. 40024 X 9. 61304 11. 6130425 12. 2701030 13. 3705423 14. 6803217 15. 2003005 16. 70032004 17. 62003005 18. 91010010 19. 20. 21. 22. 23. 24. 25. 26. 27. 20031025 68723145 901023406 820302008 310275603 600000501 3000400230024 80000102051003 50000021375604 28. 4000012000040250014 29. 1000982000375000482000354000271000032561804 NUMERATION AND NOTATION. 9 From the preceding tables and remarks, the pupil will likewise readily understand the reason of the following rule for notation, or expressing numbers by figures. RULE. Make a sufficient number of cyphers or dots, and divide them into periods, then underneath these dots write each figure in its proper order and fill the vacant orders with cyphers. NOTE. The object of the dots or cyphers, being to guide the learner at first, after a little practice he may dispense with them. Ex. 1. Write down in figures the number twenty millions three hundred and four thousand and forty. Here millions being the highest period named, we write cyphers to correspond with that, and the period of units, and then underneath these place the significant figures in their proper order, and afterwards fill the vacant orders with cyphers. 000000, 000000 20304040 The pupil must recollect that cyphers being of no use except to fill vacant orders, are never to be placed to the left of whole numbers. EXERCISES IN NOTATION. Express the following numbers in figures. EXAMPLES. 2. Seventy-five. 3. Ninety. 4. One hundred and five. 5. Three hundred and twenty. 6. Nine hundred and four. 7. Eight hundred and ninety. 8. Two thousand three hundred and five. 9. Six thousand and forty. 10. Seven thousand and four. 11. Eight thousand and ninety-five. 12. Ten thousand five hundred and fifty-six 13. Forty thousand and forty. 14. Ninety-five thousand two hundred and sixty-seven. 15. Eighty thousand one hundred and nine. j] 10 NUMERATION AND NOTATION. 16. One hundred and thirty-six thousand two hundred and seventy five. )i 17. Three hundred and seven thousand and sixty-four. Ji 18. Five hundred thousand and five. || 19. One million, two hundred and forty-seven thousand, four hundred and twenty-three. |i 20. Ten millions, forty thousand and twenty. 21. Sixty millions, seventeen thousand and two. 22. One hundred and four millions two hundred and four thousand and sixty- five. 23. Five hundred and three millions, one hundred and two thousand and nine. 24. Ninety one thousand and two millions, and four. 25. Sixty billions, three millions and forty-one thousand. 26. One billion, one hundred million, one thousand and one. The Roman method of representing nnmbers, is by means of certain capital letters of the Roman alphabet. Thus: I II III IV V VI VII VIII IX X XI XII XIII XIV XV XVI XVII one two three four five six seven eight nine ten eleven twelve thirteen fourteen fifteen sixteen seventeen xvni XIX XX XXX XL L LX LXX LXXX xc c cc ccc cccc D M MDCCCXXXVIII eighteen nineteen twenty thirty forty fifty sixty seventy eighty ninety one hundred two hundred three hundred four hundred five hundred one thousand 1838 NOTE 1 . As often ag any letter is repeated, so often is its vilue re- peated. NOTE 2. A less character before a greater one, diminishes its value NOTE 3. A les* character after a greater one, increases its value. EXPLANATION OF CHARACTERS. 11 QUESTIONS. What is Arithmetic ? When is it a science ? When is it an art ? What are the fundamental rules of arith- metic ? What is numeration ? What is notation ? What does a unit signify ? What does two signify ? Three, <fec. ? What is meant by the simple value of a unit? What does the local value of a figure depead on ? How do you write the number ten in figures ? Why is the one in this case called a unit of the second order? How many units of the first order does it take to make a unit of the second order ? How many units of the second order does it require to form a unit of the third order? &c. Repeat the principles of notation and numeration. l| Repeat the names of each of the first nine orders as ex- [i pressed in the numeration table. Repeat the name of each of the periods. Repeat the Rule for numeration. Repeat the^ Rule for notation. EXPLANATION OF CHARACTERS. Signs. Significations. = equal; as 20s. = 1. -f- more ; as 6 -f- 2 = 8. ^ less ; as 8 2 = 6. X into, with, or multiplied by ; as 6 X 2 = 12. -J- by (i. e. divided by ;) as 6 -f- 2 = 3 ; or, 2)6(3. : : : : proportionality; as 2 : 4 : : 6 : 12. J or, J Square Root ; as J 64 = 8. J Cube Root; as J 64 = 4. iy Fourth Root ; as J 16 = 2, &c. A vinculum ; denoting the several quantities over which it is drawn, to be considered jointly as a simple quantity. 12 SIMPLE ADDITION. SIMPLE ADDITION. SIMPLE ADDITION is the art of collecting several nuin- jers, of the same name, into one sum. RULE. Place the numbers with units under units, tens under tens, &,c. Begin the addition at the units, or right hand column, and add together all the figures in that column ; then, if the amount be less than ten, set down the whole sum: but if greater than ten, see how many tens there are, and set down the number above the even tens, and carry one for each ten to the next column, and proceed with it as in the first. Proof. Begin the addition at the top of each column, and proceed as before, and if the result be the same, it is presumed to be right, EXAMPLES. (1) (2) (3) (4) 432 231 214 4 213 413 121 2 121 121 312 5 213 132 321 3 979 sum 897 sum 968 sum 1 4 sum (5) Here 4, 2, 1, Sand 6 make 15. In fifteen there 27636 * s one ^ en aiu ' ^ ve lln ' ls * Set down *h e f' ve units 7 Q ft Q 2 un ^ er tne umts c l uinn * an d carry one for the ten ' * to the next or tens column. 38941 67832 Then 1,4, 3, 4, 9 and 3 make 24; in 24 there are 59244 two tens, and four over: set down the foui under the column of tens, and carry two to the next or hundreds column &c., to the last, where thr whole 273545 amount may be set down. SIMPLE ADDITION. 13 (6) (7) (8) 47386 99786 72752 29492 86937 37823 18583 27849 78794 89294 49878 23567 28887 72937 98372 74392 48732 12345 288034 386119 323653 (9) (10) (11) 47823 72683 84736 73714 95892 78928 27834 82783 27849 23925 94973 63782 67883 76892 28637 62734 43987 73862 (12) (13) (14) 73684 9376 7379 7 5 723 7463 473 8 7 729 6893 9 489 7 4 8 7 2 483 937 6 8 9 6 9 2 432 APPLICATION. 1. Add 224 dollars, 365 dollars, 427 dollars, and 784 dollars, together. 224 365 427 784 Answer^ $1800 Dollars 14 SIMPLE ADDITION. 2. Add 3742 bushels, 493 bushels, 927 bushels, 643 bushels, and 953 bushels, together. Answer, 6758 bushels. 3. Add 7346 acres, 9387 acres, 8756 acres, 8394 acres, 32724 acres. Ans. 66607 acres. 4. Henry received at one time 15 apples, at another 115, at another 19. How many did he receive? Ans. 149. 5. A person raised in one year 724 bushels of corn, in another 3498 bushels, in another 9872. How much in all? Ans. 14094 bushels. 6. A rnan on a journey, travelled the first day 37 miles, the second 33 miles, the third 40 miles, the fourth 35 miles. How far did he travel ill the four days? Ans. 145 miles. 7. A has a flock of sheep containing 34. B has a flock of 47, and C of fifty-four. How many sheep are there in the three flocks? Ans. 135. 8. The distance from Philadelphia to Bristol is 20 miles; from Bristol to Trenton, 10 miles; from Trenton to Princeton, 12 miles; from Princeton to Brunswick, 18 miles ; from Brunswick to New York, 30 miles. How many miles from Philadelphia to New York? Ans. 90. 9. A person bought of one merchant, 10 barrels of flour, of another 20 barrels, of another 95 barrels. How many barrels did he buy ? Ans. 125 barrels. 10. A wine-merchant has in one cask 75 gallons, in another 65, in a third 57, in a fourth 83 ; in a fifth 74, and in a sixth 67 gallons. How many gallons has he in all f Ans. 421 gallons. Questions. How many primary rules of Arithmetic are there f What are they called ? What is addition? How do you place numbers to be added? Where do you begin the addition * Why do you carry one for ten, in preference to any other number? Ans. Because it takes ten ones to make one ten, ten tens to make one hundred, &c. (See table, page 9.) SDIPLE SUBTRACTION. 15 | SIMPLE SUBTRACTION. SIMPLE SUBTRACTION is taking a less number from a greater, of the same name, to show the difference be- tween them. The greater number is called the minuend. The less number is called the subtrahend. The difference, or what is left, is called the remainder RULE. Place the less number under the greater, with units under units, tens under tens, &LC. Then draw a line under them; begin at the right hand or units place, and subtract each figure of the sub- trahend from the figure of the minuend that is above it, and set the remainder below. When the figure in the subtrahend is greater than the one above it, borrow one (which is one ten) from the next figure, and add it to the figure of the minuend; then subtract from the sum. Proof. Add the remainder and the subtrahend to- gether, and if the sum equal the minuend, the work is ! presumed to be right. EXAMPLES. (1) (2) 79252743 Minuend 9738476 34120312 Subtrahend 2614253 451324!' 1 Remainder 7124223 (3) Here we cannot take seven from two ; then we 726398 -j must borrow one from the 8: that one is one fen; R A 9 Pi r Q >y tnen len an< ^ two are * vve l ve > now ta ^ e seven from ' twelve, and five remain. One is borrowed from the 8, leaving only 7 ; 838345 then take 3 from 7, and 4 remain : or, suppose 8 to remain untliminished ; and to cancel the one which is borrowed from the 8, add one to the 3 below, making four; then four from eight and four remais-j as before, &c. 16 SIMPLE SUBTRACTION. (4) (5) 9273847 82703682 2641386 27341237 6632461 55362445 (6) (7) 7837286 273683070 3273195 4321725 4564091 269361345 (8) W 68427362 593784283 34613524 54321432 (10) (11) 792836842 92037842 24653128 41372761 APPLICATION. 1. From 78 take 32 and what will remain? Answer, 46. 2. From 478 take 324. What will remain? Ans. 154. 3. Charles had 723 apples, and sold 421. How ma- ny has he left? Ans. 302 4. James had 9768 dollars, and gave for a house and lot 3453 dollars. How many has he left? Ans. 6315, 5. A farmer had 3849 acres of land; he gave to his sons 2135 acres. How many acres has he left for himself? Ans. 1714. 6. There are two piles of bricks, one contains 7SUH, and the other 4389. How many more are there in the oae than in the other? Ans. 3507. SIMPLE SUBTRACTION. 17 7. Bought 100 bags of coffee, weighing 14510 Ibs., and sold thereof 63 bags weighing 6871 pounds; how many bags, and how many pounds remain unsold? Ans. 37 bags, and 7639 Ibs. 8. A man bought a chaise for 175 dollars, and to pay for it gave a wagon worth 37 dollars, and the rest in money. How much money did he pay ? Ans. 138 dollars. 9. A man deposited in bank 8752 dollars, and drew out at one time 4234 dollars, at another 1700 dollars, at another 962 dollars, and at another 49 dollars. How much had he remaining in bank? Ans. 1807 dollars. 10. A merchant bought 4875 bushels of wheat, and sold 2976 bushels. How many bushels remain in his possession? Ans. 1899. 1 1. A grocer bought 25 hogsheads of sugar, containing 250 hundred weight, and sold 9 hogsheads, containing 75 hundred weight. How many hogsheads and how many hundred weight had he left ? Ans. 16 hogsheads, and 175 hundred weight. 12. A traveller who was 1300 miles from home, trav- elled homeward 235 miles in one week; in the next 275 miles; in the next 325 miles; and in the next 290 miles. How far had he still to go, before he would reach home ? Ans. 175 miles. Questions. What is subtraction? What is the greater number called ? What is the less number called? What is the difference called? How do you place numbers for subtraction? Where do you begin the subtraction? When the lower figure is greater than the upper one, how do you proceed? Why is the one you borrow, one ten. Ans. Because ten ones make one ten; and if I borrow one ten it will make ten ones again, &c. How do you prove subtraction' M^MMMMHWBMMMMMHMMHMMMMMBIMMHB^^ Vi* 18 SIMPLE MULTIPLICATION. SIMPLE MULTIPLICATION. I SIMPLE MULTIPLICATION is a short method of perform- ing particular cases of addition. The number to be multiplied, is the multiplicand. The number to be multiplied by, is the multiplier. The number produced is the product. The multiplicand and multiplier are sometimes called factors. MULTIPLICATION TABLE. Twice 3 times 4 times 5 times 6 times 7 times 1 make 2 1 make 3 1 make 4 1 make 5 1 make 6 1 make 7 2 4 2 6 2 8 2 10 2 12 2 14 3 6 3 9 3 12 3 15 3 18 3 21 4 6 4 12 4 16 4 20 4 24 4 28 5 10 5 15 5 20 5 25 5 30 5 35 6 12 6 18 6 24 6 30 e 36 6 42 7 14 7 21 7 28 7 35 7 42 7 49 8 16 8 24 8 32 8 40 8 48 8 56 9 18 9 27 9 36 q 45 9 54 9 63 10 20 10 30 10 40 10 50 10 60 10 70 11 22 LI 33 11 44 11 55 11 66 11 77 12 24 12 36 12 48 J2 60 12 72 12 84 8 times 9 times 10 times 11 times 12 times 1 make \ 1 make 9 1 make 10 1 make 11 1 make 12 2 1( 5 2 18 2 20 2 22 2 24 3 & [ 3 27 3 30 3 33 3 36 4 3S } 4 36 4 40 4 44 4 48 5 4( 5 45 5 50 5 55 5 60 6 ' 46 } 6 54 G 60 6 66 6 72 7 5( 5 7 63 7 70 7 77 7 84 i 8 W [ 8 72 8 80 8 88 [ 3 96 9 7$ I 9 81 9 90 9 99 9 108 10 8( 1 10 90 10 100 10 110 10 120 11 8* j 11 99 11 110 11 121 11 132 12 9( i 12 108 12 120 12 132 12 144 CASE 1. When the multiplier does not exceed 12. RULE. Place the multiplier under the units figure of he multiplicand; and multiply each figure of the multi- I'licand in succession, and set down the amount, and :arry, as in addition. Proof. Multiply the multiplier by the multiplicand. SOIFLE MULTIPLICATION. 19 EXAMPLES. 4231 Multiplicand 34253 7342 2 Multiplier 3 4 8462 Product 102759 29368 36563 8375 4378 9286 567 8 182815 50250 30646 74288 4375 7862 3724 7482 9 10 11 12 39375 78620 40964 897^4 EXERCISES 1 Multiply 4218 by 2 Product 8436 2 7321 by 3 2196 I 3 87692 by 4 3507t>e 4 900078 by 9 8100702 5 826870 by 10 8268700 6 278976 by 11 3068736 7 569769 by 12 6837228 CASE 2. When the multiplier exceeds 12. RULE. Place the multiplier as before, with units un- der units, &c. Then multiply all the figures of the mul- tiplicand by the units figure of the multiplier, setting down the product as before. Proceed with the tens figure in the same manner, ob- serving to set the product of the first figure in the tens place and with the hundreds figure placing the first product in hundreds place, &.C., and add the several pro ducts together. 20 SIMPLE MULTIPLICATION. EXAMPLES. 43752 Here we multiply by the 6 or units figure 436 as before: then by the 3 or tens figure, pla- cing the first product in the second or tens 262512 place, immediately under the three, in the 131256 multiplier. In like manner we use the 4, 175008 placing the first product in the third or Atm- dreds place, immediately under the 4; alter 19075872 which we add the several products together, and the work is done. 73684 427 515788 147368 294736 31463068 1 Mu tiply 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 > 17 18 EXEBCISES. 4736 by 5762 by 6483 by 7368 by 4327 ,y 7382 by 4728 by 7584 by 5678 by 7G83 by 4962 by 7384 by 4376 by 7923 by 6842 by 7648 by 8473 by 9372 by 37462 563 112386 224772 187310 21091106 34 Product 161024 43 54 45 56 67 76 87 78 89 98 87 97 78 89 523 456 567 247766 350082 331560 242312 494594 359328 659808 442884 683787 486276 642408 424472 617994 608938 3999904 !| 3863688 1 5313924 | SIMPLE MULTIPLICATION. 21 NOTE 1. When either or both of the factors have noughts on the right hand, they may be omitted in the operation, and annexed to the product. Thus : 47 I 000 734 I 00 42 00 42 000 94 1468 188 2936 -Product 197400000 Product 3082800000 NOTE 2. When the multiplier is the exact product of any two factors in the multiplication table, the opera- tion may be performed by separating th multiplier into its components, and multiplying first by the one, then its product by the other. Thus : 754 by 36 754 754 754 9 3 6 36 2272 4524 4524 12 6 2262 27144 27144 27144 27144 9 and 4, or 3 and 12, or 6 and 6, multiplied together, produce 36, and by using either pair, according to the above note, the true result is obtained. EXERCISES. 1 Multiply 756 by 42 Product 31752 2 645 by 24 15480 3 876 by 48 42048 4 963 by 56 53028 5 827 by 72 59544 6 946 by 81 76626 7 875 by 84 73500 8 948 by 96 91008 9 795 by 108 85860 Tiie pupil may work these by all the several pairs of components that he can find in the multiplier. 22 SIMPLE MULTIPLICATION. NOTE 3. When the multiplier is not the exact pro duct of any two numbers in the table, use two factors whose product is short of the multiplier, then multiply the sum by the number required to supply the deficiency and add its product to that obtained by the two .factors. 583X3 4 11680 1749 13409 1 Multiply 2 3 4 5 EXAMPLES. 583X2 7 12243 1166 583X5 3 1749 6 10494 2915 13409 13409 13409 EXERCISES. 846 by 26 784 by 29 975 by 34 859 by 43 794 by 59 Product 21996 22736 33150 36937 46846 PROMISCUOUS EXERCISES. 1. Charles has 24 marbles, and John has 13 times as many; how many has John? - Ans. 312. 2. A gentleman owns 17 houses, for each of which he receives 250 dollars rent; how much does he receive for them all? Ans. 4250. 3. A laborer hired himself to a farmer for 1 1 years, at 150 dollars a year; how much did he receive? Ans. 1650 dollars. 4. A person wishes to purchase 26 shares of Bank stock at 75 dollars a share ; what must he pay ? Ans. 1950 dollars. 5. A mason having built a house, found that 98470 bricks were in it; suppose he desires to build 19 such houses, how many bricks must he obtain for the pur- pose Ans. 1870930. SIMPLE DIVISION. 23 SIMPLE DIVISION. SIMPLE DIVISION is a short method of performing sev- eral subtractions. The number to be divided is called the dividend. The number by which it is to be divided is called the divisor. The number of times that the divisor is contained in the dividend, is called the quotient. So many figures of the dividend as are taken to be divided at one time, is called a dividual. If any thing remain when the operation is completed, it is called the remainder. CASE i. SHORT DIVISION. When the divisor does not exceed 12. RULE. Place the divisor on the left hand side of the number to be divided. Consider how often the divisor is contained in the first figure or figures of the dividend, and set down the result below ; observing how many remain, if any. If there be no remainder, consider how often the divisor is con- tained in the next figure : but if there be a remainder, call it so many tens, and add the next figure to it, and divide the sum, placing the result beneath, as before. Proof. Multiply the quotient by the divisor; add the remainder, if any, and the product will equal the divi- dend. EXAMPLES. Dividend. Divisor 2)182 2)648 3)963 4)484 Quotient 241 324 24 SIMPLE DIVISION. 3)741851 Here 3 are contained In 7 two times* and one re- o/iTOQO I Q T*^ n \*A^ mains ; place the two under 247283-H Remainder theseven , and suppose the one that remains to be one ten, and add the next fig- Proof 741851 ure ( 4 ) to *> which make3 fourteen. Now 3 are contained in fourteen 4 times, and 2 remain. Set the 4 down under the 4 in the dividend, and suppose the two that remain to be two fen$, and add the next figure (1) to it, which make twenty- one. Now 3 into 21 go 7 times, and no remainder. Place the 7 un- der the 1 in the dividend, and proceed in the same manner with the other figures. 4)65270167 5)6572686 163175414-3 Rem. 1314537+1 Rem 4 5 Proof C5270167 Proof 6572G86 6)8739627 7)4873692 8)9273684 0)8379286 10)946873 11)893726 12)98796 SIMPLE DIVISION. 25 EXERCISES. Divide 7893762 by 6 Ans. 1315627 9387984 by 7 , 13411404 6928437 by 8 8660545 9276874 by 9 10307637 8672934 by 10 8673934 6873842 by 11 6248948 7369287 by 12 6141073 CASE 2. LONG DIVISION. When the divisor exceeds 12. RULE. Place the divisor to the left hand of the divi- dend, as in case 1. Consider how often the divisor is contained in the least number of figures into which it can be divided; and set down the result at the right hand of the dividend. Multiply the divisor by the quotient figure thus found, and set the product under the dividual or figures supposed to be divided. Subtract the product from the dividual, and set down what remains. Bring down the next figure of the divi- dend, and proceed as before, till all the figures are brought down and divided. EXAMPLES. Divisor. Dividend. Quotient. 27)984376(36458 Twenty-seven into 98 go 3 81 times: multiply the divisor (27) by 3 and set the product under the dividual (98) and subtract. To the remainder (17) bring down the next fig- ure (4) of the dividend. Now 27 into 174 go 6 times. Place the 6 in the quotient and mul- tiply (27) the divisor, by 6, and set the product under 174 and subtract as before, &,c. 226 216 10 Remainder. 26 SIMPLE DIVISION. Divisor. Dividend. Quotient. 42) 98754 (2351 84 42 Divisor 147 4702 126 9404 12 Remainder 215 210 98754 Proof 54 42 12 Remainder 32)789627(24675 65)1827538(28115 64 130 149 527 128 520 216 192 242 224 187 75 65 103 65 388 160 325 27 Rem. 63 Rem. EXEltCIgES. Divide 8769 by 13 Quo. 674 Rem. 7 476 by 15 31 11 958 by 18 53 4 1475 by 28 52 19 4277 by 31 137 30 25757 by 37 696 5 1 63125 by 123 513 26 1 253622 by 422 601 SIMPLE DIVISION. 27 NOTE 1. Cyphers on the rigkt^hand of the divisor may be omitted in the operation, observing to separate as many figures from the right of the cttvidend, which must be annexed to the remainder. EXA3TPLES. 54 00)1463 108 383 378 Rem. 540 40(27 32 | 0)7617 64 121 96 257 256 3(238 Rem. 13 EXERCISES. Jf| Divide 40220 by 1900 Ans. 21 Keai. 320 137000 by 1GOO 85 1000 99607765 by 27000 3689 4765 2304108 by 5800 397 1508 NOTE 2. When the divisor is the exact product of any two numbers in the multiplication table, the opera- tion may be performed by dividing first by one of the component parts, and then the quotient by the other. To get the true remainder, multiply the last remain- der by the first divisor, and add the first remainder. EXAMPLES. 7 98754 42 [o | 14107 5 first remainder 2351 1 last remainder 7 first divisor add |i 5 first remainder 12 true remainder 28 SIMPLE DIVISION. 984376 27 * ' 3281251 36458~-3 X 3 + 1 =10 Rem EXERCISES. 9756 by 35 Quotient 278 Rcm. 26 8491 by 81 104 67 44767 by 18 2487 1 92017 by 56 1643 9 38751 by 48 807 15 7S4071 by 72 10195 31 APPLICATION. 1. SevWn boys have 161 apples, which they divide equally among them. How many does each have? Answer, 23. 2. What is the quotient, if 8736 be divided by 8, and that quotient by 4? Ans. 273. 3. If 350 dollars be equally divided among 7 men, what will be the share of each ? Ans. 50. 4. How many times are 27 contained in 952? Ans. 35 times and 7 over. 5. Suppose 2072 trees planted in 14 rows. How ma- ny trees will there be in each row? Ans. 148. 6. Several boys who went to gather nuts, collected 4741, of which each boy received 431. How many boys were there? Ans. 11. 7. If the expense of erecting a bridge, which is 15036 dollars, be equally defrayed by 179 persons, what must each pay? Ans. 84 dollars. 8. Suppose a man receive in one year 2920 dollars; how much a day is his income at that rate; and sup- pose that his expenses for the year amount to 1769 dol- lars. How much will he save in a year? Ans. His income will be 8 dollars a day; he will save 1151 dollars in a year. SIMPLE DIVISION. 29 Questions. What is division? What do you call the number that is to be divided? What do you call the number you divide by ? What do you call the number obtained by division? What do you call that which is left when the work is done ? When the divisor does not exceed 12, how do you per- form the operation? When the divisor exceed 12, how do you proceed? How do you prove division? How may the operation be performed when there are cyphers at the right hand of the divisor? How may it be performed when the divisor is the exact product of two numbers in the multiplication table? How do you obtain the true remainder in the last case? PROMISCUOUS EXERCISES iN THE PRECEDING RULES. 1. If the contents of five bags of dollars, containing $295, $410, $371, $355, and $520, be divided equally among 25 persons, how much is the share of each? Ans. $78.04 2. A man possessed of an esnue of $30,000, disposed of it in the following manner: to his brother he gave $1500, and the balance to his 5 sons, to be equally di- vided among them. What was each one's share ? Ans. $5700. 3. What number is it, which being added to 9709 will make 110901? Ans. 101192. 4. Add up twice 3?7, three times 794, four times 31196, five times 15S80, six times 95280, and once 33304. Ans. 812,344. 5. Three merchants have a stock of 14876 dollars, of which A owns 4963 dollars, B 5188, and C the re- mainder. How much does C own? Ans. 4725 dolls. 30 FEDERAL MONEY. FEDERAL MONEY, OR MONEY OF THE UNITED STATES. TABLE. 10 mills make 1 cent 10 cents 1 dime 10 dimes 1 dollar 10 dollars 1 eagle These denominations bear the same relation to each other as those of units, tens, hundreds, &c. Federal money is therefore added, subtracted, multiplied, and divided by the same rules as Simple Addition, Subtrac- tion, Multiplication, and Division. ADDITION OP FEDERAL MONEY. Rule. Place the numbers one under another, with mills on the right, cents, dimes, &,c., in succession ; observing to keep mills under mills, cents under cents, &c. Then proceed as in simple addition. When halves or- fourths of a cent occur, find their amount in fourths, and consider how many cents these fourths will make, and carry them to the column of cents. EXAMPLES. Eagles. Dolls. Dimes. Cents Mills. Dolls. Ds. Cts. 789 978 684 637 482 E. 27 3 6 453570 NOTE. In common business transactions, eagles, dimes, and mills are not used: dollars, cents, and frac- tions of a cent, are the only denominations kept in j accounts. 3 7 8 9 5 7 4 9 8 7 2 3 8 7 9 8 8 9 8 6 4 7 8 9 8 FEDERAL MONEY. EXAMPLES. Ds. cts. 34 62 56 27 23 27 31 82 68 42 169 , 85 Ds. cts. 468 , 31 723 , 62 845 , 92 736 , 25 846 , 31 428 , 62 EXERCISES. (2) Ds. cts. 927 , 24 768 , 32 427 , 56 792 , 34 587 , 62 842 , 27 Ds. cts. 427 , 68 342 , 31 427 , 26 793 , 84 273 , 42 2264 , 51 (3) Ds. cts, 273 45 846 283 846 674 273 37 75 91 75 25 Ds. cts. 437 , 62i 386 , 814 243 , 18| 427 , 37* 428 , 12i One halfis two-fourths; and one half more make four fourths, and three fourths more make seven fourths, and one fourth more make eight fourths, and one half (or two fourths) more make ten fourtiis. Four fourths make one cent, then ten fourtiis make two cents, and leave two fourths, or one half cent. Set down the i cent, and carry the two cents to the next column. 1923 , 12* Ds. cts. 274 , 814 362 , 87* 421 , 184 625 , 314 241 , 561 Ds. cts. 27 , 68| 36 , 81* 28 , 62i 37 , 934 24 , 62i Ds. cts. 56 , 064 32 , 12i 36 , 25 42 , 62i 54 , 814 32 FEDERAL MONEY. APPLICATION. 1. Add 48 dollars 20 cents; 14 dollars 58 cents; 100 dollars 25 cents; and 84 dollars 36 cents. Ans. 247 dollars 39 cents. 2. Add $7,62*, $34,314, $72,064, $41,314, $25, 68|, and $87,43| together, and tell the amount. Ans. $268,43|. 3. Bought a hat for $4,25 cents ; a pair of shoes for $2,25; a pair of stockings for $1,25, and a pair of gloves for 75 cents. What is the cost of the whole ? Ans. $8 50 cents. 4. If I buy coffee for $1,18|, tea for $2,50, cloves for 87 ft, mace for 93|, cinnamon for $l,87i, raisins for $2,68|, nutmegs for 37i, candles for 87, and wine for $1,93|, what must I pay for them? Ans. $13,25. Questions. What relation do mills, cents, dimes, tec., bear to each other? How are the addition, subtraction, multiplication, and division of Federal money performed? How do you place the numbers to be added? How do you proceed when halves, fourths, &,c., occur? SUBTRACTION OF FEDERAL MONEY. RULE. Place the less under the greater, with dollars under dollars, and cents under cents; then, if there are no fractions, proceed as in simple subtraction. If there is a fraction in the upper sum and none in the lower, set it down as a part of the remainder, and proceed as before. If there is a fraction in each sum, and the lower be less than the upper, subtract the lower from the upper, and set down the difference. If the lower fraction be greater than the upper one, borrow one cent, and call it four fourths, and add them to the upper fraction, and subtract the lower one from the sum. Proof. As in simple subtraction. FEDERAL MONEY. 33 EXAMPLES. Ds cts. Ds. cte. Ds. cts. 32 ,62 43 , 68| 75 , 68* 21 ,31 21 , 25 24 , 12i $11 , 31 $22 , 43| 51 , 564 NOTE. TJiree fourths cannot be taken from two Ds. cts. fourths : then borrow one cent from the two cents, 271 62* which has four fourths in it: add the/our fourths to , or> ' QQ 3 tne t wo fourths, this makes six fourths ; subtract lo/5 , Jof three fourths from six fourths, and three fourths (|) remain. Set down the and add one to the next 138 , 68| 3, as in simple subtraction. EXERCISES. Ds. cts. Ds. cts. Ds. cts. 65 , 49 520 , 314 436 , 31* 35 , 12i 210 , 12* 243 , 18| Ds. cts. Ds. cts. Ds. cts. 273 , 62* 237 , 564 732 , 314 124 . 37* 142 , 874 261 , 68| APPLICATION. 1. Subtract $432,68| from 1000,93|. Ans. $568,25. 2. Subtraction shows the difference between two numbers; what is the difference between $37,62 * and $93,87*. Ans. $56,25. 3. Bought goods to the amount of $545,95, and paid at the time of purchase $350; How much remains un- paid? Ans. $195,95. 4. A merchant bought a quantity of cotil?e, for which he paid $560. He afterwards sold it foi $610,87* How much did he gain by the transaction ? Ans. $50,87*. n > 34 FEDERAL MONEY. Questions. How do you place the numbers in subtraction of Federal Money ? How do you perform the operation? If a fraction occur in the upper line or minuend, what do you do with it? If a fraction occur in each, how do you proceed ? Suppose the lower fraction is greater than the upper one, how do you proceed ? How do you prove subtraction of Federal Money ? MULTIPLICATION OF FEDERAL MONEY. RULE. Set the multiplier under the multiplicand, and if there be no fractions, proceed as in simple multi- plication; observing to separate the cents from the dollars in the product. If there is a fraction in the sum, multiply it, and see how many cents are in the product; set down the frac- tion that is over, and proceed as before. Or if the multiplier exceeds 12, multiply the sum, omitting the fractions; then multiply the fraction, and add the number of cents contained in the product, to the product of the rest of the sum. EXAMPLES. Ds. cts. 10 , 56* 2 $50 , 00 $21 , 12* $118 , 12* Ds. cts. Ds. cts. 10,87* 125 times 4,18* 24 times * are 125 one half make 24 72 fourths: four 125 halves: 2 fourths are con- 5435 into 125 go 62 1672tained 18 times in 2174 times, leaving 836 72 fourths, rnak- 1087 one; that is, 18 ing 18 cents. 62* one half, mak- -ing 62* cents. $100,50 $1359,37* FEDERAL MONEY. EXERCISES. 1 Multiply 2 3 4 5 G $145,18| 7,874 28,684 42,314 137,62* 79,004 by by by by by by 207 7 47 68 58 67 35 Ans. $1016,314 370,124 1950,75 2454,124 9220,874 16354,034 APPLICATION. 1. What will 8 pounds of cheese come to, at 18 cents a pound? Ans. $1 44 cts. 2. What is the value of 12 yards of linen, at 35 cents ;a yard? Ans. $4 20 cts. 3. What cost 29 yards of cloth at $2 25 cts. a yard? Ans. $65 25 cts. 4. What wMl 213 barrels of flour cost, at $5 25 cents a barrel? Ans. $1118 25 cts. 5. Bought 321 barrels of cider at $1 25 cts. a barrel. What did it amount to? Ans. $401 25 cts. 6. What will 580 bushels of salt cost at $1 124 cts. a bushel.. Ans. $652 50 cts. 7. What is the value of 2 pieces of cloth, one contain- ing 38 yards, and the other 26 yards, at $3 874 cts. a yard? ' Ans. $248. 8. What will be the cost of 132 pieces of linen at $17 374 cts. each? Ans. $2293 50 cts. 9. What will 8 cords of wood amount to, at 4 dollars { 50 cents a cord? Ans. 36 dollars. 10. Sold 213 barrels of flour for 6 dollars 25 cents per barrel. What is the amount? Ans. 1331 dols. 25 cts. 11. Bought )08 pounds of coffee at 21 cents a pound. What is the amount? Ans. 64 dols. 68 cts. 12. Bought 217 gallons of brandy at $1 18| cts. per gallon; and sold it for $1 374 cts. per gallon. What was the amount paid for the whole; the sum it sold for; and the gain? Ans. Prime cost, $257 68|: sold for $298 374; gain, $40,681 36 FEDERAL MONEY DIVISION. RULE. Divide as in simple division. When a re- mainder occurs, multiply it by 4; and add the number of fourths that are in the fraction of the sum (if any) to its product: divide this product by the divisor, and its quotient will be fourths, which annex to the quotient. Proof. As in simple division. Ds. cts. 2)45,22 22,61 Ds. cts. 25)629,68|(25,1S| 50 EXAMPLES. Ds. cts. 3)63.181 21,064 Ds. cts. 2)25,374 12,684 32)78800(24,624 64 129 148 125 128 46 200 25 192 218 80 200 64 18 Here 18 cents remain; multiply 16 4 18 cents by four, brings them to 4 fourths of a cent; add the |, this makes 75 fourths: divide 75 fourths by 25, and are obtained, whicii 3*2/4(2 or 4 25)7 ''(<* place in the quotient. 64 Divid D. cts. 56,15 96,00 156,00 58,14 417,96 494,45 627,38 by by by by by by by EXERCISES. 10 5 4 38 129 341 508 Quotient 5,61 4 - 19,20 - 39,00 - 1,53 -- 3,24 1,234 FEDERAL MONEY. 37 APPLICATION. 1. If 7 pounds of butter cost $1,89 cts., what is the value of 1 pound? Ans. 27 cts. 2. If 8 Ibs. of coffee cost $2,04 cts., what is the price of one pound? Ans. 25i cts. 3. Bought 29 yds. of fine linen for $65,25 cts., what was the price per yard? Ans. $2,25. 4. Paid $58,75 cts. for 235 yds. of muslin, what was it per yard? Ans. 25 cts. 5. A piece of cloth containing 72 yds. cost $450, what was it per yard? Ans. $6,25. Questions. How do you perform division of Federal Money? How do you proceed when a remainder occurs? PROMISCUOUS EXERCISES IN THE PRECEDING RULES. 1. Bought 18 barrels of potatoes, each containing 3 bushels, at 25 cts. a bushel, what did they cost? Ans. $13,50. 2. A farmer sold 30 bushels of rye at 87 cts. a bushel, 30 bushels of corn at 53 cts. a bushel; 8 bushel of beans at $1,25 cts. a bushel; 2 yoke of oxen at $62 a yoke; 10 calves at $4 a piece; 15 barrels of cider at $2,37i a barrel, what was the amount of the whole? Ans. $251,62*. 3. What will be the price of four bales of goods, each bale containing 60 pieces, and each piece 49 yards, at 374 cents a yard? Ans. $4410. 4. Add $324,43* cts. $208,09* cts. and $507,90* cts. together, and divide the sum by 2, and what will be the result? Ans. $520,21|. 5. Divide 400 dollars, equally, among 20 persons. What will be the portion of each person? Ans. $20. 6. Divide 1728 dollars, equally among 12 persons. What does each one of them share? Ans. $144. 7. If 240 bushels cost 420 dollars; what is the cost of one bushel at the same rate? Ans. $1.75. 38 REDUCTION. ' REDUCTION. REDUCTION is the changing o f a sum, or quantity, from one denomination to another, without altering the value. CASE 1. To reduce a sum, or quantity, to a lower denomination than its own. RTTLE. Multiply the sum, or quantity, by that num- ber of the lower denomination which makes one of its own. If there are one or more denominations between the denomination of the given sum, and that to which it is to be changed, first change it to the next lower than its own; then to the next lower, and so on to the deno- mination required. DRY MEASURE. TABLE. 2 pints (pts.) make 1 quart, qt. 8quarts - 1 peck, pc. 4 pecks - 1 bushel, bu. NOTE. This measure is used for measuring grain, salt, fruit, &,c. EXAMPLES. NOTE. 1. To reduce bushels to pecks, multiply by 4, because each bushel has 4 pecks in it. 1. Reduce 23 bushels to pecks. bit. 23 4 Amt. 92 pecks. 2. Reduce 35 bushels to pecks. Amt. 140 peeks. NOTE. 2. To reduce pecks to quarts, multiply by 8, because each peck has 8 quarts in it. REDUCTION. 39 3. Reduce 27 pecks to quarts. 8 Amt. 216 quarts. 4. Reduce 43 pecks to quarts. Amt. 344 quarts. NOTE. 3. To reduce quarts to pints multiply by 2, because each quart has 2 pints in it. 5. Reduce 43 quarts to pints. qt. 43 2 Amt. 86 pints. 6. Reduce 32 quarts to pints. Amt. 64 pints. Reduce 34 bushels to pints. . fctt. 34 4 Multiply the bushels by 4 to bring them to pecks. 136 8 Multiply the pecks by 8 to bring - them to quarts. 1088 2 And multiply the quarts by 2 to bring - them to pints. Amt. 2176 pints. 7. Reduce 56 pecks to pints. Amt. 896 pints- 8. Reduce 47 bushels to quarts. Amt. 1504 qt 9. Reduce 85 bushels to pints. Amt. 5440 pt 10. Reduce 63 pecks to quarts. Amt. 504 qt. 11. Reduce 132 bushels to quarts. Amt. 4224 qt. 12. Reduce 234 bushels to pints. Amt. 14976 pt. NOTE. 4. When several denominations occur, reduce the highest denomination to the next lower one, and this again to the next lower, and so on ; observing to add the amount of each denomination, the number there is of that denomination in the given sum. 40 REDUCTION. EXAMPLES. 1. Reduce 23 bushels, 3 pecks, 5 quarts, 1 pint, to pints. bu. pe. qt. pt. 23-3-5-1 4 Multiply the bushels by 4 to bring them to pecks, and 92 add the 3 pecks to the amount, 3 which makes 95 pecks. Multiply the pecks by 8 to bring them to quarts, and add the 5 quarts, which makes 765 quarts. Multiply the quarts by 2 to bring them to pints, and add the 1 pint which makes 1531 pints. 1531 amt. Or thus: bu. pe. qt. 23 3 - 5 - 4 Multiply by 4 as above; add the 3, and set down the 95 amount, &c. 8 765 2 1531 Amt. as before. EXERCISES. 1. Reduce 13 bushels, 2 pecks, 7 quarts, 1 pint to pints. Amt. 879 pints. REDUCTION. 41 2. Reduce 24 bushels, 3 pecks, 1 quart to quarts. Amt. 793 qt. 3. Reduce 7 bushels, 3 pecks to quarts. Amt. 248 qt. 4. Reduce 3 pecks, 2 quarts to pints. Amt. 52 pt. 5. Reduce 7 quarts, 1 pint, to pints. Amt. 15 pt. 6. Reduce 32 bushels, pecks, 1 quart to pints. Amt. 2050. 7* Reduce 5 bushels, 1 peck, quarts, 1 pint to pints. Amt. 337 pt. 8. Reduce 43 bushels, 1 peck to pints. Amt. 2768 pt. Question*. What is reduction? For what is case first used ? How do you reduce a sum to a lower denomination than its own? How do you reduce bushels to pecks? Why do you multiply by 4 ? How do you reduce pecks to quarts? Why do you multiply by 8? How do you reduce quarts to pints? How do you reduce bushels to pints? AVOIRDUPOIS WEIGHT. TABLE. 16 drams (dr.) make 1 ounce, oz. 16 ounces 1 pound, Ib. 28 pounds 1 quarter of a cwt. qr. 4 quarters, (or 112 lb.)* 1 hundred weight, cwt 20 hundred weight 1 ton, T. NOTE. By this weight are weighed, tea, sugar, cof- fee, flour and other things subject to waste, and all the metals, except silver and gold. * The gross hundred weight of 112 pounds is nearly out cf use: the decimal hundred weight of 100 pounds is taking its place. 42 REDUCTION. EXAMPLES. 1. Reduce 23 tons to hundred weight. tons. 23 20 Amt. 460 cwt. 2. Reduce 34 hundred weight to quarters. cwt. 34 4 Amt. 136 quarters. 3. Reduce 42 quarters to pounds, qrs. 42 28 336 84 Amt. 1176 pounds. 4. Reduce 73 pounds to ounces. Ibs. 73 16 438 73 Amt. 1168 ounces. 5. Reduce 54 ounces to drams. oz. 54 16 t 324 54 Amt. 864 drams. REDUCTION. 43 6. Reduce 35 tons to drams. tons. 35 20 700 cwt. 4 2800 qr. 28 22400 5600 78400 Ib. 16 470400 78400 1254400 oz. 7526400 1254400 Amount. 20070400 drams. EXERCISES. 7 Reduce 24 pounds to drams. Amt. 6144 dr. 8 Reduce 36 hundred weight to pounds. Amt. 4032 Ib. 9. Reduce 73 quarters" to ounces. Amt. 32704 oz. 10. Reduce 2 tons to pounds. Amt. 4480 Ib. 11 Reduce 4 tons to drams. Amt. 2293760 dr. 44 REDUCTION. 12. Reduce 3 tons, 13 cwt., 2 qu., 14 Ibs., to pounds. T. cwt. qr. Ib. 3-13-2-14 20 60 Or thus: 13 T. cwt. qr. Ib. 3-13-2-14 73 20 4 73 292 4 2 294 294 28 28 2366 2352 588 588 8246 pounds 8232 14 8246 pounds. 13. Reduce 2 tons. 15 cwt. 2 qr. to quarters. Amt. 222 qr. 14. Reduce 3 tons. 25 Ib. to pounds, ^nt. 6745 Ib. 15. Reduce 5 cwt. 3 qr. 14 Ib. to ounces. Amt. 10528 oz. 16. Reduce 2 cwt. 2 qr 14 ounces to drams. Amt. 71,904 dr. TROY WEIGHT. TABLE. 24 grains (gr.) make 1 pennyweight, dwt. 20 pennyweights - 1 ounce, oz. 12 ounces - 1 pound, Ib. NOTE. By this weight, jewels, gold, silver, and I liquors, are weighed. REDUCTION. 45 EXASEPLES. 1. Reduce 32 pounds to ounces. Ib. ' 32 12 Amt. 384 ounces. 2. Reduce 23 ounces to pennyweights. oz. 23 20 Amt. 460 dwt. 3. Reduce 43 pennyweights to grains, dwt. 43 24 172 86 Amt 1032 grains. 4. Reduce 53 pounds to grams. Ibs. 53 12 50880 25440 Amt. 305280 grains. EXERCISES. 1. Reduce 24 ounces to grains. Amt. 11520 gr. 2. Reduce 32 pound* to pennyweights. Amt. 7680 dwt. 3. Reduce 132 pounds to ounces. Amt. 1584 oz. 4. Reduce 234 ounce's to grains. Amt. 112320 gr. 46 REDUCTION. 5. Reduce 463 pounds to grains. Amt. 2666880 gr. 6. Reduce 47 pounds, 10 ounces, 15 pennyweights to pennyweights. Amt. 11495 dv/t. 7. Reduce 5 pounds, 6 ounces, 4 pennyweights, 20 grains to grains. Amt. 31796 gr. APOTHECARIES WEIGHT. TABLE. 20 grains (gr.) make 1 scruple, sc. 9 3 scruples - 1 dram-, dr. 3 8 drams - 1 ounce, oz. 3 12 ounces - 1 pound, Ib. NOTE. By this weight apothecaries mix their medi- cines, but they buy and sell by Avoirdupois Weight. EXERCISES. 1. Reduce 32 pounds to ounces. Amt. 384 oz. 2. Reduce 43 ounces to drams. Amt. 344 dr. 3. Reduce 27 drams to scruples. Amt. 81 sc. 4. Reduce 37 scruples to grains. Amt. 740 gr. 5. Reduce 28 pounds to drams. Amt. 2688 dr. 6. Reduce 36 ounces to scruples. Amt. 864 sc. 7. Reduce 27 drams to grains. Amt. 1620 gr. 8. Reduce 23 pounds to grains. Arrt. 132480 gr. 9. Reduce 3 pounds, 5 ounces, 2 scruples to scru- ples. Amt. 986. sc. 10. Reduce 7 ounces, 5 drams, 14 grains to grains. Amt. 3674 gr. 11. Reduce 27 pounds, 7 ounces, 2 drams, 1 scruple, 2 grains, to grains. Amt. 159022 gr. CLOTH MEASURE. TABLE. 4 nails (na.) make 1 quarter of a yard, qr. 4 quarters - 1 yard, yd, 3 quarters - 1 Ell Flemish, E. Fl 5 quarters - 1 Ell English, E. E. 6 quarters - 1 Ell French, E. Fr. NOTE. By this measure cloth, tapes, linen, muslin, &c., are measured. REDUCTION. 47 EXERCISES. 1. Reduce 24 yards to quarters. Amt. 96 qr. 2. Reduce 32 quarters to nails. Amt. 128 na. 3. Reduce 27 yards to nails. Amt. 432 na. 4. Reduce 46 Flemish ells to quarters. Amt. 138 qr. 5. Reduce 27 English ells to quarters. Amt. 135 qr. 6. Reduce 34 French ells to quarters. Amt. 204 qr. 7. Reduce 45 Flemish ells to nails. Amt. 540 na. 8. Reduce 36 English ells to nails. Amt. 720 na. 9. Reduce 54 French ells to nails. Amt. 1296 na. 10. Reduce 13 yards, 3 quarters to quarters. Amt. 55 qr. 11. Reduce 3 quarters, 2 nails, to nails. Amt. 14 na. 12. Reduce 24 yards, 2 nails to nails. Amt. 386 na. 13. Reduce 13 E. ells, 2 qrs., 3 nails to nails. Amt. 271 na. LONG MEASURE. TABLE. 12 inches (in.) make 1 foot, ft. 3 feet 1 yard, yd. 5* yards - 1 Rod, Pole, or Perch, p. 40 poles 1 Furlong. 8 Furlong 1 Mile. 3 Miles 1 League. 60 Geographic, orJ , Hpjrrpp 694 Statute Miles j NOTE. This measure is used for length and di tances. A Hand is a measure of four inches, and is used in measuring the height of horses. A Fathom is 6 feet, and is chiefly used in measuring the depth of water. REDUCTION. EXERCISES. 1. Reduce 23 leagues to miles. Amt. 69 m. 2. Reduce 43 miles to furlongs. Amt. 344 f. 3. Reduce 27 furlongs to poles. Amt. 1080 p. 4. Reduce 56 poles to yards. Amt. 308 yd. 5. Reduce 132 yards to feet. Amt. 396 ft. 6. Reduce 76 feet to inches. Amt. 912 in 7. Reduce 24 miles to poles. Amt. 7680 p. 8. Reduce 32 furlongs to yards. Amt. 7040 yd. 9. Reduce 86 poles to inches. Amt. 1 7028 in. 10. Reduce 26 leagues to yards. Amt. 137280 yd. 11. Reduce 52 miles to feet. Amt. 274560* ft. 12. Reduce 5 leagues to inches. Amt. 950400 in. 13. Reduce 24 degrees to statute miles. Amt. 1668 m. 14. Reduce 12 miles, 3 furlongs, 25 poles to poles. Amt. 3985 po. 15. Reduce 14 leagues, 2 furlongs to poles. Amt. 13520 po. 16. Reduce 3 leagues, 2 miles, 6 furlongs, 18 poles to yards. , Amt. 20779 yds. LAND, OR SQUARE MEASURE. TABLE. 144 square inches make 9 square feet 304 square yards 40 square perches - 4 roods square foot, square yard, square perch, rood, acre, ft. yd. P- r. NOTE. This measure is used to ascertain the quan- tity of lands, and of other things having length and breadth to be estimated. EXERCISES. 1. Reduce 27 acres to roods. . Amt. 108 r. || 2. Reduce 53 roods to perches. ' Amt. 2120 p. |] REDUCTION. 49 3. Reduce 28 perches to square yards. Amt. 847 sq. yds. 4. Reduce 36 square yards to square feet. 324 ft. 5 Reduce 27 square feet to square inches. Amt. 3888 in. 6. Reduce 34 acres to perches. Amt. 5440 p. 7. Reduce 42 roods to square yards. Amt. 50820 sq. yds. 8. Reduce 24 square perches to square feet. 6534 ft. 9. Reduce 32 roods to square feet. Amt. 348480 ft. 10. Reduce 23 acres to square inches. Amt. 144270720 sq. in. 11. Reduce 11 acres, 2 roods, 19 perches to perches. Amt. 1859 p. 12. Reduce 17 acres, 3 roods to perches. Amt. 2840 p. 13. Reduce 12 acres, 12 roods, 12 perches to square yards. Amt. 60863 sq. yd. CUBIC, OR SOLID MEASURE. TABLE. 1728 cubick inches make 1 cubic foot 27 feet 1 cubic yard 40 feet of round timber, or ) , m 50 feet of hewn timber, | l Ton or load 128 solid feet 1 Cord of wood NOTE. This measure is employed in measuring solids, having length, breadth, and thickness to ba esti- mated. EXERCISES. 1 Re Ace 29 cords of wood to cubick feet. Amt. 3712 c. i 2 Reduce 32 cubic yds. to feet. Amt. 864 c. f. 3 Reduce 23 cubic feet to inches. Amt. 39744 c. in. 4 Reduce 32 cubic yds. to inches. Amt. 1492992 c. in. 5 Reduce 2 cords of wood to inches. Amt. 442368 c. in 6 Reduce 3 cords, 10 feet to feet. Amt. 394 f> 7 .Reduce 1 cord. 3 feet, 136 inches fo inches. Amt. 226504 ta- 50 REDUCTION. LIQUID MEASURE. TABLE. 4 gills make 1 pint pt. 2 pints (pts) 1 quart qt. 4 quarts 1 gallon gal. 42 gallons 1 tierce te. 63 gallons 1 hogshead hhd. 2 hogsheads 1 pipe or butt pi. 2 pipes 1 tun. T NOTE. This measure is employed in measuring cider, oil, beer, &c. EXERCISES. 1 Reduce 23 tuns to pipes. Amt. 46 pi. 2 Reduce 43 pipes to hogsheads. Amt. 86 hhd. 3 Reduce 34 hogsheads to gallons. Amt. 2142 gal. 4 Reduce 27 tierces to gallons. Amt. 1134 gal. 5 Reduce 53 gallons to quarts. Amt. 212 qt. 6 Reduce 724 quarts to pints. Amt. 1448 pt. 7 Reduce 37 pints to gills. Amt. 148 g. 8 Reduce 12 pipes to gallons. Amt. 1512 gal. 9 Reduce 4 hogsheads to quarts Amt. 1008 qt. 10 Reduce 32 gallons to gills. Amt. 1024 g. 11 Reduce 2 tuns to gills. Amt. 16128 gills 12 Reduce 32 gals 3 qts. to pints. Amt. 262 pt. 13 Reduce 2 hogsheads, 27 gals. 3 qts to quarts. Amt. 615 qt. 14 Reduce 3 tons, 1 hogshead, 15 gals. 1 qt to pints. Amt. 6674 pt. MOTION, OR CIRCLE MEASURE. TABLE. * 60 seconds ("sec) make 1 minuta ^in. 60 minutes 1 degree deg. 30 degrees 1 sine , sin. 12 sines (or 360 degrees) 1 revolution NOTE. This measure is employed by astronomers, navigators, &c. REDUCTION. 51 EXERCISES. 1 Reduce 5 sines to degrees. Ami. 150 2 Reduce 8 degrees to minutes. Amt. 480 1 3 Reduce 6 minutes to seconds. Amt. 360 sec. 4 Reduce 12 sines to seconds. Amt. 1296000 sec. 5 Reduce 3 sines 15 degrees to minutes. Amt. 6300 min. TIME. TABLE. 60 seconds (sec) make 1 minute min. 60 minutes 1 hour H. 24 hours 1 day 7 days 1 week 12 months (or 365 days) 1 year. NOTE. The true year, according to the latest and most accurate observations, consists of 365 d. 5 h. 48 m. and 58 sec : this amounts to nearly 365$ days. The com- mon year is reckoned 305 days, and every fourth or leap year one day more on account of the fraction omit- ted each year, which being put together, every fourth year is added to it, making leap year 366 days. The year is divided into 12 months as follows. The fourth, eleventh, ninth and sixth, Have thirty days to each affixed, And every other thirty-one, Except the second month alone, Which has but twenty-eight in fine, Till leap year gives it twenty-nine. OR THUS: Thirty days hath September, April, June, and November, February hath twenty-eight alone, And each of the rest has thirty one. When the year can be divided by four, without a re- mainder, it is bissextile, or leap year. 52 REDUCTION. EXERCISES. 1 Reduce 42 years to months. Amt. 504 m. 2 Reduce 23 days to hours. Amt. 552 h. 3 Reduce 36 hours to minutes. Amt. 2160 min. 4 Reduce 25 minutes to seconds. Amt. 1500 sec. 5 Reduce 14 days to minutes. Amt. 20160 min 6 Reduce 52 hours to seconds. Amt. 187200 sec. 7 Reduce 13 weeks to hours. Amt. 2184 h. 8 Reduce 12 weeks to minutes. Amt. 120960 min. 9 Reduce 3 years to minutes, allowing 365 days to each year. Amt. 1576800 min. 10 Reduce 15 years and 6 months to months. Amt. 186 m. 11 Reduce 4 weeks, 3 days, 22 hours, to hours. Amt. 766 h. 12 Reduce 7 years, 24 days, 43 minutes, to seconds. Amt. 222828180 sec. STERLING MONEY. TABLE. 4 farthings (qr) make 1 penny d. 12 pence 1 shilling s. 20 shillings 1 pound Farthings are usually written as fractions of a penny, thus- i one farthing 4 two farthings or a half penny. I three farthings. EXERCISES. 1 Reduce 14 pounds to shillings. Amt. 280 s* 2 Reduce 23 shillings to pence. Amt. 276 d. 3 Reduce 34 pence to farthings. Amt. 136 qr. 4 Reduce 4 pounds to pence. Amt. 960 d. 5 Reduce 13 shillings to farthings. Amt. 624 qr. 6 Reduce 16 pounds to farthings. Amt. 15360 qr. 7 Reduce 13 pounds 14 shillings, to pence. Amt. 3288 d. 8 Reduce 3 pounds 15 shillings 6 pence to farthings. Amt. 3624 qr. REDUCTION. 53 FEDERAL MONEY. TABLE. 10 mills make 1 cent 10 cents 1 dime 10 dimes 1 dollar 10 dollars 1 eagle EXERCISES. 1 Reduce 5 eagles to cents. Amt. 5000 ct. 2 Reduce 3 dollars to mills. Amt. 3000 m. 3 Reduce 15 dimes to cents. Amt. 150 ct. 4 Reduce 3 eagles, 5 dollars to cts. Amt.3500 ct. 5 Reduce 7 dollars, 3 dimes, 6 cents, to mills. Amt. 7360 m. As eagles, dimes and mills are not used in accounts, they will generally be omitted in the subsequent exer- cises of this work. 4 fourths, or 3 thirds, or 2 halves, make 1 cent. 100 cents - 1 dollar. 6 Reduce 125 cents to halves of a cent. Amt. 250 halves. 7 Reduce 32 cents to fourths of a cent. Amt. 128 fourths. 8 Reduce 23 dollars to cents. Amt. 2300 ct. 9 Reduce 25 dollars 15 cents to cents. Amt. 2515 ct. 10 Reduce 15 dollars 374 cents to halves of a cent. Amt. 3075 halves. 11 Reduce 21 dollars 15 cents to thirds of a cent. Amt. 6345 thirds. 12 Reduce 5 dols. 37i cents to fourths of a cent. Amt. 2150 fourths. 13 Reduce 15 dollars 33* cts. o thirds of- a cent. Amt. 4600 thirds NOTE. To reduce dollars to cenrs annex two cyphers : thus 53 dollars are 5300 cents. To reduce dollars and cents to cents, place them to- 54 lllIDUCTiOX. gather without any separating point, and the amount will be cents. Thus 35 dollars 24 cents are 3524 cents. Questions. For what purpose is Dry measure used ? For what is Avoirdupois weight used? For what is Troy weight employed? For what is Apothecaries weight employed? For what is Cloth measure employed? For what is Long measure used? For what is Land or Square measure used? For what is Cubick -measure employed? For what is Liquid measure employed? For wha.t is Sterling currency used? For what is Federal currency used? CASE 2. To reduce a sum or quantity to a HIGHER denomina- tion than its own RULE. Divide the sum or quantity by that number of its own denomination which makes one of the denomina- tion to which it is to be changed. When there are one or more denominations between the denomination of the given sum and that to which it is to be changed; first change it to the next higher than its own, and then to4he next higher, antl so on. Remainders are always of the same denominations as the sums divided. DRY MEASURE. EXAMPLES. 1 Reduce 25 pints to quarts. pts. NOTE. Divide by 2, because every 2 pints 2)25 make one quart. In 25 are 12 two's and 1 over, that is 12 quarts and 1 pint. ql.12 Ipt 2 Reduce 43 quarts to pecks. qt. Divide by 8, because every 8 qts. make 8)43 1 peck. In 43 are 5 eights and 3 over, that is 5 pecks and 3 quarts. pe. 5 3qt ^ REDUCTION. 55 3 Reduce 2G pecks to bushels. bu. Divide by four because every 4 pecks 4)26 make 1 bushel. In 26 are 6 fours and 2 over; that is 6 bushels and 2 pecks. bu. 6 2 pecks 4 Reduce 359 pints to bushels. pi. Divide pints by 2, brings them 2)359 to quarts; divide quarts by 8, brings them to pecks, and divide pecks by 8)179 1 pt. 4 brings them to bushels. 4) 22 3qt 5 b. 2 p. 3 qt. 1 pt. 5 Reduce 81 quarts to bushels. A. 2 bu. 2 pe. 1 qt. 6 Reduce 134 pints to pecks. 8 pe. 3 qt. 7 Reduce 194 pints to bushels, 3 bu, pe. 1 qt. Questions, What is reduction? For what is case second used ? How do you reduce a sum to a higher denomination than its own? When there are one or more denominations between the denomination of the given sum and the one to which you wish to reduce it, how do you proceed ? Of what denomination is the remainder always! How do you bring pints to quarts ? How do you bring quarts to pecks? How do you bring pecks to bushels? How do you bring pints to bushels? AVOIRDUPOIS WEIGHT. 1 Reduce 65 cwt. to tons. Result 3 tons 5 cwt. 2 Reduce 27 quarters to cwt. Res. 6 cwt. 3 qr. 3 Reduce 109 pounds to qr. Res. 3 qr. 25 Ib. 4 Reduce 123 ounces to pounds. Res. 7 Ib. 11 oz. 5 Reduce 234 drams to ounces. Res. 14 oz. 10 dr. 6 Reduce 4274 drams to pounds. Res. 16 Ib. 11 oz. 2 dr. | 7 Reduce 175 quarters to tons. Res. 2 tons 3 cwt. 3 qr. j 8 Reduce 6745 pounds to tons. Res. 3 tens 25 Ib. i 56 REDUCTION. TROY WEIGHT. 1 Reduce 378 ounces to pounds. Result, 31 Ibs. 6 oz. 2 Reduce 235 pennyweights to ounces. Res. 11 oz. 15 dwt. 3 Reduce 748 grains to pennyweights. Res. 31 dwt. 4 grains. 4 Reduce 678 pennyweights to pounds. Res. 2 Ibs. 9 oz 18 dwt. 5 Reduce 732 grains to ounces. -* Res. 1 oz. 10 dwt. 12 grains. 6 Reduce 14752 grains to pounds. Res. 2 Ibs. 6 oz. 14 dwt. 16 gr. APOTHECARIES WEIGHT. 1 Reduce 432 ounces to pounds. Res. 36 Ibs. 2 Reduce 782 drams to ounces. Res. 97 oz. 6 dr. 3 Reduce 91 scruples to drams. Res. 30 dr. 1 scr. 4 Reduce 192 grains to scruples. Res. 9 sc. 12 gr. 5 Reduce 258 scruples to ounces. Res. 10 oz. 5 dr. 1 scr. 6 Reduce 12660 grains to pounds. Res.. 2 lb.2 oz. 3 drs. CLOTH MEASURE. 1 Reduce 60 quarters to yards. Res. 15 yds. 2 Reduce 60 quarters to English ell&. Res. 12 E. ells. 3 Reduce 60 quarters to French ells. Res. 10 Fr. ells. 4 Reduce 60 quarters to Flemish ells. Res. 20 Fl. ells. 5 Reduce 52 nails to quarters. Res. 13 qr. 6 Reduce 123 nails to yards. Res. 7 yds. 2 qr. 3 na. 7 Reduce 543 nails to English ells. Res. 27 ells. qr. 3 nails. LONG MEASURE. 1 Reduce 36 miles to leagues. Res. 12 1. 2 Reduce 75 furlongs to miles. Res. 9 m. 3f. KEDUCTION. 57 3 Reduce 295 poles to furlongs. Res. 7f. 15 p. 4 Reduce 286 yards to poles. Res. 52 p. 5 Reduce 365 feet to yards. Res. 121 yds. 2 ft. 6 Reduce 759 inches to feet. Res. 63 ft. 3 in. 7 Reduce 253 inciies to yards. Res. 7 yds. ft 1 inch. 8 Reduce 2792 poles to leagues. Res. 2 1. 2 m. 5 f. 32 p. SQUARE MEASURE. 1 Reduce 287 roods to acres. Result 71 a. 3 r. 2 Reduce 245 perches to roods. Res. 6 r. 5 p. 3 Reduce 756 square feet to yards. Res. 84 yds. 4 Reduce 4731 square yards to perches. Yds. Res. 156 p. 12 yds. 304 4 121 18924 121 682 605 774 726 156 p. Bring the 30^ yards and the 4731 yards both to fourths, and divide. The remainder 48, is fourths of a yard; divided by four, brings it to yards, the true icmaiader. Rem. 12 yards. 5 Reduce 3575 square inches to feet. Res. 24 feet 119 inches. 6 Reduce 1728 square perches to acres. Res. 10 a. 3 r. 8 p. CUBIC MEASURE. 1 Reduce 789 cubic feet to cords. Result, c. 21 ft. 2 Reduce 343 cubic feet to yards. Res. 12 yds. 19 ft. 3 Reduce 9386 cubic inches to feet. Res. 5 "fi. 746 in. 4 Reduce 70353i> cubic inches to cords. Res. 3 c. 23 ft. 243 in. c 2 58 REDUCTION. LIQUID MEASURE. 1 Reduce 25 pipes tq.tuns. Res. 12 T. 1 P. 2 Reduce 34 hogsheads to pipes. Res. 17 P. 3 Reduce 1575 gallons to hogsheads. Res. 25 hhds. 4 Reduce 163 quarts to gallons. Res. 40 gal. 3 at. 5 Reduce 6048 pints to tuns. Res. 3 tuns. MOTION. 1 Reduce 1440 seconds to minutes. Result, 24 min 2 Reduce 720 minutes to degrees. Res. 12 deg. 3 Reduce 342 degrees to sines. Res. 11 sines 12 deg. 4 Reduce 443907 seconds to sines. Res. 4 sines 3 deg. 18 min. 27 sec. TIME. 1 Reduce 1800 seconds to minutes. Result, 30 m. 2 Reduce 720 minutes to hours. Res. 12h. 3 Reduce 744 months to years.- Res. 62 yrs. 4 Reduce 4649 minutes to days. Res. 3d. 5h. 29 min- 5 Reduce 48888 minutes to weeks. Res. 4w. 5d. 22hrs. 48 min. STERLING MONEY. 1 Reduce 78 shillings to pounds Res.3. 18.9. 2 Reduce 93 pence to shillings. R^p. 7*. 9d. 3 Reduce 39 farthings to pence. Res. JW. 3qr 4 Reduce 656 pence to pounds. Res. 2 14s. Sd. 5 Reduce 781 farthings to shillings. Res. 16s. 3d. Iqr. 6 Reduce 6529 farthings to pounds. Res. 6. 16s. OcZ. 1 qr. FEDERAL MONEY. 1 Reduce 250 halves to cents. Res. 125 cts. 2 Reduce 128 fourths to cents. Res. 32 cts. Reduce 2343 cents to dollars. Res. 23 dol. 43 cts. 4 Reduce 15371 cents to dollars. Res. 15 dol. 374 cts. 5 Reduce 6150 half cents to dollars. Res. 30 dol. 75 cts. NOTE. To reduce cents to dollars, separate two figures 01 1 the right hand for cents; those on the left will be dollars. REDUCTION. 59 PROMISCUOUS EXERCISES. 1 How many bushels in 738 quarts? Ans. 23 bushels 2 quarts. 2 In 7 bushels, how many pints? Ans. 448 pints. 3 How many cwt. in 5356 ounces? . Ans. 2 cwt. 3qr. 2Glb. 12 oz. 4 How many drams in 3 qr. 23 Ib. 14 oz. ? Ans. 27610 drams. 5 How many grains in 9 oz. 14 dwt. 3 gr.? Ans. 4659 gr. 6 How many pounds in 7432 dwt. ?Ans. 301b. lloz. 12 dwt. 7 How many scruples in 15 Ib. 1 oz. 6 drams? Ans. 4362 scr. 8 How many ounces in 218 scruples? Ans. 9 oz. 9 How many furlongs in 2346 yards? Ans. lOf. 26p. 3yd. 10 How many poles in 3 leagues? Ans. 2880 poles. 1 1 How many yards in 84 nails? Ans. 5 yd. 1 qr. 12 How many perches in 4719 square yards Ans. 156 perches. 13 How many square yards in one acre? Ans. 4840 sq. yds. 14 How many hogsheads in 9728 gills? Ans. 4 hhds. 52 gal. 15 How many pints in 2 pipes? Ans. 2010 pints. 16 How many minutes in 3 days 6 hours? Ans. 4680m. 17 How many hours in 2 weeks and 4 days? Ans. 432 hours. 18 How many shillings in 27 four pences? Ans. 9 s. 19 How many cords of wood in 9334 cubic feet? Ans. 73 cords 20 feet. ( .20 How many cubic feet in 9 cords? Ans. 1152 feet.j 21 How many inches round the globe, which is 360 de- grees of 69i miles each? Ans. 1,565,267,200 inches, Enumerate the answer. 6t) CO3IPOUXD ADDITION. COMPOUND ADDITION. COMPOUND ADDITION is the art of collecting several numbers of different denominations into one sum. RULE. i Place the numbers so that those of the same denom- ination may stand directly under each other, observing to set the lowest denomination on the right, the next lowest next, &,c. Then add up the several columns beginning with the lowest denomination : divide the sum by as many of the number of that denomination as it takes to make one of the next; and so on. Proof. As in Simple Addition. DRY MEASURE. EXAMPLES. bu. pe. qt pt. The first column on the right makes 3271 ^ ve P U) ts. F ' ve pi nts make two quarts. and leaves one pint. Set down the one pint under the column of pints and carry the two quarts to the column of quarts. 6321 The column of quarts with the two quarts 2261 added makes twenty four quarts. Twen- ty-four quarts make three pecks and leave T~ ~ _ no quarts. Set down undei the column ^* <i U 1 O f quarts and carry the three pecks to the column of pecks. The column of pecks with the three pecks added makes fourteen pecks. Fourteen pecks make three bushels and leave two pecks. Set down the two pecks under the column of pecks and carry the three to the column of bushels. The column of bushels with the three bushels added makes twenty-four bushels. Here set down the whole amount. COMPOUND ADDITION. 61 bu. pe. qt. pt. bu. pe. qt. bu. pe. qt. pt. 23 371 437 371 34 261 526 261 42 351 615 050 51 141 .704 141 23 231 833 331 14 1 2 1 412 021 11 3 4 1 324 211 202 3 2 1 40 3 7 3270 APPLICATION. 1 Add 2 bu. 3 pe.;7 bu. 3qt.;4 bu. 1 pe. 1 pt.; 6 bu. 4 qt. 1 pt.; and 3 pe. 1 qt. together. Amount 21 bu. pe. 1 qt. pt. 2 Add 3 bu. 2 pe.Sqt. 1 pt.; 7 bu. 7 qt. 1 pt.; 3 pe. 1 pt.; 4 bu. 5 qt.; 4 bu. 3 pe.; 8 bu. 3 pe. 7 qt. 1 pt. together. Amt.29 bu. 2 pe. qt. pt. 3 Add 7 bu. 1 pt.; 3 pe. 7 qt. 1 pt.; 6 qt. 1 pt.; 9 bu. 3 pe. 6 qt. 1 pt.; 3 bu.3 qt.; 4 bu. 1 pe. Amt. 25 bu. 2 pe. 4 In a wagon load of grain contained in seven sacks, viz: in the first 4 bu. 3 pe. 1 qt.; in the second 5 bu. 7 qt. 1 pt. the third 3 bu. 1 pe. 1 pt. fourth, 3 bu. 2 pe. 6 qt. fifth, 5 bu. sixth, 4 bu. 1 pe. 1 pt.: and in the seventh 6 bu. 1 pe. 1 pt. How many bushels? Questions. What is Compound Addition? How do you place the numbers to be added? Do you place the greater or smaller denominations in the right hand column? Where do you begin the addition? When the first column is added, how do you proceed with the sum? When you divide the sum by as many of that denom- ination as make one of the next; which do you set down, the remainder or the quotient? What do you do with the quotient? 62 COMPOUND ADDITION. In what particular does compound addition differ from simple addition? Do you carry one for every ten in compound addition? Since you do not carry one for every, ten, how many do you always carry? A. One fo* as many of any de- nomination as make one of the next. Here the pupil will have something with which to compare simple addii-ion, in which he carries one for every ten. This comparison will improve and correct his understanding of the elementary rules. AVOIRDUPOIS WEIGHT. T. cut. qr. Ib. T. cwt. qr. Ib. oz. dr. (1) 15 3 2 15 (2) 7 11 2 16 4 13 4839 15 7 3 8 16 7 82 19 1 10 13S 19 1 12 8 13 163 8 3 17 42 8 3 19 12 4 34 15 2 24 357 6 2 8 3 3 300 16 1 19 APOTHECARIES' WEIGHT. 339 & 3 3 9 sr- (1) 6 3 L 2 (2) 84 7 t> 12 19 9 5 1 132 5 2 182 7 3 2 16 2 2 2 8 57 6 1 1427 6 7 19 40 5 14 6 1 9 306 7 3 2 TROY WEIGHT. Ib. oz. dwt. Ib. oz.dwt. gr. (1) 47 10 12 (2) 185 2 19 20 38 8 6 56 9 15 6 16 11 4 1472 11 2 17 7 2 16 385 8 5 13 9 11 10 8 7 12 124 6 9 C02IPOUND ADDITION. 63 CLOTH MEASURE. yds. qr. na. E.E. qr. na. EFL qr. na. (1) 75 3 2 (2) 72 3 2 (3) 19 2 3 163 13 536 2 1 728 1 2 245 2 847 1 3 142 1 738 3 1 1453 2 ,816 1785 23 41 2 32 1 2 3009 1 1 LONG MEASURE. L. M.fur. P. yd. ft. iji. (1) 5 2 4 17 (2) 3 2 11 16 1 3 10 1 1 9 72 5 24 2 8 526 3 12 3 1 10 834 2 6 34 2 4 38 3 12 6 2 7 1493 2 2 29 SQUARE MEASURE A. R. P. A. R. P. (1) 39 2 37 (2) 487 2 17 G2 1 17 25 3 28 68 38 67 32 129 3 12 45 1 16 532 1 18 26 29 832 2 2 o< 04 COMPOUND ADDITION. CUBIC MEASURE. yd. ft. in. (I) 75 22 1412 9 26 195 3 19 1091 28 15 1110 49 24 218 18 17 1225 cords, feet. in. (2) 37 119 1015 9 110 159 48 127 1071 8 21 9 111 956 9 27 28 1091 186 18 67 135 122 863 3 In four piles of wood; the first containing 32 feet 149 inches; the second 121 feet 1436 inches; the third 97 feet 498 inches; the fourth 115 feet 1356 inches; how much did the whole amount to? Ans. 2 cords, 110 ft. 1711 in. 4 In six boat-loads of wood: the first containing 22 cords 114 feet, 987 in.; the second 18 cords, 121 feet, 1436 in.; the third 21 cords, 109 feet, 1629 in.; the fourth 15 cords, 82 >et, 1321 in.; the fifth 16 cords, 98 feet, 1111 in.; the sixth 24 cords, 89 feet, 987 in. How much did they contain? Ans. 120 c. 105 ft. 559 in. LIQUID MEASURE. T. hhd. gal lihd. gal. qt. pt. (1) 18 2 54 (2) 385 42 3 1 62 1 39 27 36 2 327 4 132 17 46 1 19 729 25 285 3 28 173 47 2 1 740 1 18 COMPOUND ADD/TION. 65 MOTJOIV i ' STNf o i // (1) 17 55 48 ^) 1 25 49 51 1 37 51 2 4 21 36 28 19 45 4 19 47 18 19 19 37 1 25 25 39 67 13 1 10 15 24 24 3. Add 5sm 10 46' 38'; 11 37' 18"; Isin. 1712' 18"; Isin. 52"; Isin. 15 12' 23"; and 11 57' 29" to- gether. Ans. II sin. 6 46' 58". 4. Add 45'; Isin. 9 18"; 14 21' 34"; 2sin. 8 13' 54"; 4sin. 7 12' 19"; and 47' 32" together. Ans. Sain. 10 20' 37". TIME, we. d. h. H. min. sec. (2) 3 5 20 (3) 20 52 40 2 3 17 k 122 12 35 3 6 22 68 9 17 4 16 135 37 12 3 19 24 35 28 231 7 11 3 22 371 7 12 STERLING MONEY. 8. d. s. d. .?. d. (1) 2 3 4 (2) 7 9 44 (3) 4 6 4 712 13 7 6| 47 19 7 973 452 159 53 5 2 24 10 18 10$ 78 6 11 23 13 114 4 Add 703 Is. 4d.-, 39 4*. 9<Z.; 162 17s. 2d ; 459 15*.; 473 12*. Sd together. Ans. 1898 16*. lid. 66 COMPOUND SUBTRACTION. 5 Add the following sums : viz. 69 18s. 7d.; 175 2s. Qd. ; 1582 19s 4cZ.; 175 13s. 9d.; 143 13s. 8d.-> and 212 Os. 7d Ans. 2359 8s. 5d. COMPOUND SUBTRACTION. COMPOUND SUBTRACTION is 'the art of finding the dif- ference between two numbers consisting of several de- nominations. RULE. Place the numbers, as in compound, addition with the less under the greater: then begin at the right hand denomination and subtract the lower number from the upper, and set down the remainder. If the upper number of any denomination be less than the lower one, add to the upper one as many as it takes of that to make one of the next; subtract the lower num- ber from the amount and set down the remainder as before. Proof. As in simple subtraction. EXAMPLES. bu. pe. qt. pt bu. pe. qt. pt. 7241 42 361* 3120 31 231 4121 11 1 3 bush. pe. qt. 925 We cannot take 7 quarts from 5 quarts ; then 237 borsow 1 from the 2 pecks. One peck has 8 __________^_ quarts in it: 8 quarts added to the 5 quarts, f* n (* make 13 quarts. Take 7 qts. from 13 qts. and ~ .6 qts. remain. Sot down the 6 qt. Because I borrowed 1 from the 2 quarts, I /nust add one to the 3 below it, which makes the lower figure 4. Now [ pecks from 2 pecks we cannot take : then borrow one bushel from the I ; that bushel has 4 pecks in it; 4 p. and 2 p. make 6 p. Now 4 p. "rom 6 p. and 2 pecks remain, which set down. Because I borrowed 1 from the 9, I must add 1 to the figure below t. 1 to 2 make 3. Take 3 from 9 and 6 remain. Set down the 6, aid the work is done. COMPOUND SUBTRACTION. 67 bu. pe. qt. pt. bu. pe. qt. pt. 8271 8130 4361 4251 bu. pe. qt. bu. pe. qt. pt. 95 3 2 28 2 2 22 1 14 3 5 1 APPLICATION. 1. From a granary containing 94 bushels, 2 pecks, 7 quarts, have been taken 43 bush. 3 pe. 5 qr. How much remains? Ans. 50 bush. 3 pe. 2 qt. 2. From a wagon load of corn containing 63 bushels, 3 pecks, 4 qts., have been sold 27 bush. 3 pe. 7 qt. 1 pt. How much remains unsold? Ans. 35 bu. 3 p. 4qt, 1 pt. Questions. What is compound subtraction? In what particular does it differ from simple subtrac- tion? How do you place the numbers in compound subtrac tion? Where do you begin the operation? When the upper number of any denomination is less than the lower one, how do you proceed? Do you borrow one from the next? Do you call the number you borrow, one ten, as in simple subtraction? What do you call it? Ans. I call it one peck, or one yard, or one mile, as the case may be ? What do you do with it then? Ans. I reduce it to quarts, or to feet, or to furlongs, &c. according to the nature of the case; then add these 68 COMPOUND SUBTRACTION. to the upper figure on the right, subtract the lower figure from the sum, and set down the remainder. When you borrow one from the upper figure, why do you add one to the figure below it? NOTE. Upon a clear conception of the principles involved in these questions, depends the pupil's correct knowledge of the science of Arithmetic. AVOIRDUPOIS WEIGHT. tons cwt. qr. tons civt. qr. Ib. cwt. qr. Ib. oz. dr. From 45 11 3 52 12 3 15 17 Take 15 10 2 24 10 26 6 3 21 15 9 Rem. 30 1 1 28 2 2 17 10 6 0. 7 1. Subtract 76 tons, 18 hundred weight, 3 quarters, from 195 tons, 2 hundred weight, 2 quarters. Ans. 118 tons, 3 cwt. 3 qr. 2. Subtract 14 pounds, 6 ounces, 3 drams from 20 pounds, 2 ounces. Ans. 5 Ibs. 11 oz. 13 dr. APOTHECARIES WEIGHT, ft g 3 ft 3 3 & gr. 1090 16 48 9 6 1 4 106 2 7 1 10 2 8 983 10 7 3. From 59ft 1 23 take 53ffi 73 63. Ans. 5fc 5g 5J. 4. Subtract 14fc 1>3 13 from G9. Ans. 54ft x>3 73. TROY WEIGHT. Ib. oz. dwt. gr. Ib. oz. dwt. gr. Ib. oz. dwt. gr. 10 6 18 8 3 2 106 15 4 2 20 2 1 18 6 10 6 2 20 6 6 15 4 4. Subtract I4lb. Goz. \\dwt. from 92lb. I2dwt. 6 gr. Ans. lib. Qoz. Idwt. 6 3 *r. 5. From 16Z&. take I2lb. lloz. iQdwt. \\gr. Ans. 3Z&. Ooz. Qdwt. I3gr. COMPOUND SUBTRACTION. 60 CLOTH MEASURE. yds. qrs. na. E.E. qrs. na. E. FL qrs. na From 71 3 1 42 2 51 2 2 Take 14 2 3 19 2 3 42 2 1 Rem. 57 2 22 2 3 901 4. Subtract 95 yards, 3 quarters, 2 nails, from 156 yards, 2 quarters, 3 nails. Ans. 60 yds. 3 qr. 1 nail. 5. Subtract 14 English ells, 1 quarter, 2 nails, from 52 English ells, 3 quarters, 2 nails. Ans. 38 yds. 2 qr. LONG MEASURE. L. M.fur. L. M. fur. P. yds. ft. in. From 24 I 7 58 1 19 6 2 10 Take 18 2 4 10 7 20 327 Rem. 523 46 00 39 303 4. Subtract 45 miles, 5 furlongs, 20 poles, from 320 miles, 3 furlongs, 36 poles. Ans. 274 M. 6 F. 16 P. 5. Subtract 15 yards, 2 feet, 6 inches, from 36 yards, 1 foot, 11 inches. Ans. 20 yds. 2 ft. 5 inches. LAND, OR SQUARE MEASURE. A. R. P. A. R. P. Yds. ft. in. From 96 3 36 195 22 25 2 72 Take 25 2 39 36 3 1 14 7 10 Rem. Tl 37 158 3 1 10 4 62 4. Subtract 36 acres, 2 roods, from 900 acres, 3 roods, 16 perches. 864 A. 1 R. 16 P. 5. Subtract 72 acres, from 360 acres, 2 roods, 29 perches. 288 A. 2 R. 29 P. CUBICK MEASURE. yds. ft. in. cords, ft. in. 79 11 917 349 97 1250 17 25 1095 192 127 1349 61 12 1550 156 97 1629 70 COMPOUND SUBTRACTION. 1. From a pile of wood containing 432 cords, 27 feet, and 1432 inches, have been hauled 156 cords, 92 feet, 946 inches: how much remains? Ans. 275 cords, 63 feet, 486 in. 2. From a bank of earth containing 2984 yards, 18 feet, have been taken 143G yards, 21 feet, what re- mains? Ans. 1547 yds. 24 feet. LIQUID MEASURE. T. hhd.gal.qt.pt. T. hhd.gal.qt.pt. 2 3 50 1 100 1 19 2 1 1 2 16 3 1 99 1 28 3 1 1 1 33 1 1 3. If I purchase 2hhd. of wine, and to oblige a friend send him 29^aZ., what quantity have I left? Ans. \hlid. 34gal. 4. Bought 1 pipe of wine, 4hhd. of brandy, 2 barrels of beer; I have since sold 93 gallons of wine, 29 of brandy, 1 barrel of beer: how much of each have I remaining? Ans. 33gal. of wine, 223gal. of brandy, and 1 barrel of beer. MOTION. O f n gi n> O , ,/ 79 21 31 6 10 12 48 41 41 52 38 39 2 37 39 39 3 1 33 19 A circle being 12 sines, how far has the hand of ? watch to pass, after having gone through 4 sines, 23' .5' 29" ? Sin. ' " 12 4 23 15 29 COMPOUND SUBTRACTION. 71 2 A person resioing in latitude 27 C 32' 45" north, wishes to visit a place 52 24' 18' north. How many degrees, minutes, and seconds northward must he travel? Ans. 24 51' 33'. TIME Y. M. w. d. ho. min. sec. 69 3 1 3 40 20 16 2 6 2 57 36 H.min.sec. Y. M. 16 29 33 18 11 7 36 44 9 10 53 2 42 44 4. From 900 Y. take 111F. 6m. Ans. 788 Y. 6m. 5. If I take IF. 1M. from 6Y. what space of time will still remain? Ans. 4F. 11 M. NOTE. To ascertain the amount of time passed be- tween two events, set down the year, month, and day of the latter event, and place those of the former below it, and subtract. 6. A bond was given 24th July, (7th month) 1809, and paid off 13th August, 1821. yrs. mo. ds. 1821 8 13 1809 7 24 12 20 7. The declaration of independence of the United States passed Congress, 4th July, (7th month) 1776; and the declaration of the late war with Great Britain, 18th June, (6th mo.) 1812. How many years, &,c. be- tween them? Ans. 35yr. llmo. 14d. STERLING MONEY. . ,. d. s. d. s. d. 146 19 lOi 47 6 71 419 7 6 7 19 9| 139 0| 28 lOi 227 8 94 72 COMPOUND MULTIPLICATION. 4. Subtract 200 9s. from 1000 Us. Ans. 800 2s. 5. I have a purse of money containing 1000 2s. 4id.: it I take out 60 7*. 8|d. what sum will be left? Ans. 939 14*. 7|d. COMPOUND MULTIPLICATION. COMPOUND MULTIPLICATION is the art of multiplying numbers composed of several denominations. CASE 1. When the multiplier does not exceed 12. RULE. Place the number to be multiplied as directed in compound addition,- and set the multiplier undes-the lowest denomination. Multiply as in simple multiplication, and divide the product of each denomination by as many as it takes of that to make one of the next greater; set down the re- mainder (if any) and carry the quotient to the product of the next denomination. Proof. Double the multiplicand and multiply by half the multiplier. EXAMPLES. Bu. pe. qt. pt. 7 times 1 pint make 7 pints ; 2 pt. 7 2 5 1 make 1 qt. ; then 7 pt. make 3 qt. and ~ leave 1 pt. Set down the 1 pt. and carry the 3 qt. to the product of the next figure. 53 261 7 times 5 qt. make 35 qt. to which add the 3 qt. which make 38 qt. ; 8 qt. make one peck ; then 38 qt. make 4 pecks and leave 6 qts. Set down the six quarts and carry the 4 pecks. 7 times 2 pecks make 14 pecks ; add the 4 pecks, makes 18 pe. ; 4 pecks make one bushel ; then 18 pe. make 4 bushels and leave 2 pecks. Set down the 2 pecks and carry the 4 bushels. 7 times 7 bushels make 49 bushels ; add the 4 bushels, makes 53 bushels, which set down, and the work is done. COMPOUND MULTIPLICATION. 73 Bu. pe. qt. pt. Bu. pe. qt. pt. 9361 23 2 5 1 5 8 49 3 1 189 1 4 1. In one vessel are contained 29 bushels 2 pecks and 5 quarts : how many in 9 such vessels ? Ans. 266 bu. 3pe.5qt. 2. If one tub will contain 8 bu. 3 pe. 5 qt. how much will 11 such tubs contain? Ans 97 bu. 3 pe. 7 qt. CASE 2. When the multiplier exceeds 12, and is the exact product of two factors in the multiplication table. RULE. Multiply the given sum by one of the factors, and the product by the other factor. Proof. Change the factors. EXAMPLES. 1. Multiply 3 bushels, 2 pecks, 7 qt. by 24. Product. bu. pe. qt. bu. pe. qt. 327 327 6 4 22 1 2 14 3 4 4 6 89 1 Proof 89 1 OR THUS: bu. pe. qt. bu. pe. qt. 327 327 3 8 11 5 o 3 8 3 89 1 89 1 2. Multiply 7 bushels, 3 pecks, 5 quarts, by 36. Product, 284 bu. 2 pe. 4 qt. 3. Multiply 19 bushels, 2 pecks, 3 quarts, bv 42 D 74 COMPOUND MULTIPLICATION. CASE 3. When the multiplier exceeds 12, and is NOT the product of any two factors in the multiplication table. RULE. Multiply by the two factors whose product is the least short of the given multiplier; then multiply the given sum by the number which supplies the deficiency; and add its product to the sum produced by the two factors. EXAMPLES. 1. Multiply 21 bushels, 1 peck, 7 quarts, by 23. Prod. bu. pe. qt. bu. pe. qt. 21 1 7X3 21 1 '7x2 5 3 OR THUS: 107 1 3 64 1 5 4 7 429 1 4 product of 20 450 3 3 product of 21 64 1 5 product of 3 42 3 6 product of 2 493 3 1 product of 23 493 3 1 product of 23 2. Multiply 19 bushels, 3 pecks, 7 quarts, by 34. Product, 678 bu. 3 pe. 6 qt. 3. Multiply 7 bushels, 3 pecks, 4 quarts, by 59. 4. Multiply 9 bushels, 3 pecks, 2 quarts, by 47. 5. Multiply 15 bushels, 1 peck, 7 quarts, by 78. 6. Multiply 12 bushels, 2 pecks, by 92. 7. Multiply 1? bushels, 3 quarts, by 98. 8. How many bushels in 104 sacks, each containing 7 bushels, 2 pecks, 3 quarts? 9. How many bushels of wheat on 125 acres, con- taining 21 bushels, 3 pecks each? L. COMPOUND MUI/TIPLICATIOX. 75 CASE 4. Wlien the multiplier is greater than the product iff any two numbers in the multiplication table. RULE. Multiply the given number by 10, as many times less one as there are figures in the multiplier. Multiply that product by the left hand figure of the multiplier. Multiply the given sum by the units figure of the multiplier; the product of the first 10 by the tens figure of the multiplier; the hundreds product by the hundreds figure of the multiplier, and so on, till you have multi- plied by all the figures of the multiplier except the left hand one. Add all the products together, and you have the pro- duct required. EXAMPLES. 1. Multiply 3 bushels, 3 pecks, 1 quart, by 45G. bu. pc. qt. Product 1724 bu. 1 pe. 3 3 1X6 10 Because there are 3 figures, multiply . 2 times by 10. Multiply that product by ty- o v E the kft hand figure (4) of the multiplier. ^ X Multiply the given number by the units figure (6) and set the product beneath. Multiply the 10's product by the tens 378 4 figure (5) of the multipliar. Add the several products. 1512 2 22 2 6 189 2 1724 1 Product 76 COMPOUND MULTIPLICATION. 2. Multiply 53 bushels, 2 pecks, 7 quarts, by 2345. bu. pe. qt. 53 2 7X5 10 537 6+4 10 5371 4+3 10 53718 107437 2 161 < 5 2 2148 3 268 2 product of the 2000 4 " " 300 " 40 3 " "5 125970 1 7 Product of the 2345 NOTE. Let the pupil try experiments, by multiplying simple num- bers in this way. 3. Multiply 72 bushels, 1 peck, 2 quarts, by 4723. Product, 341531 bu. 3 pe. 6 qt. 4. Multiply 13 bushels, 2 pecks, 4 quarts, by 5124. Questions. What is compound multiplication? In what does it differ from simple multiplication? When the multiplier does not exceed 12, how do you proceed ? How many do you always carry ? How do you prove compound multiplication? How do you proceed when the multiplier exceeds 12, and is the exact product of two numbers in the multi- plication table? When the multiplier exceeds 12, and is not the exact product of any two numbers in the table, how do you proceed? COMPOUND MULTIPLICATION. 77 How do you proceed when the multiplier is greater than the product of any two numbers in the table ? AVOIRDUPOIS WEIGHT. tons. cwt. qrs. cwt. qr. Ib. oz. dr. 23 12 3 7 3 14 9 6 4 6 94 11 47 1 3 8 tons act. qr. cwt. qr. Ib. oz. dr. 7 15 3 7 3 24 12 14 8 9 5. Multiply 7 tons, 16 cwt. 3 qr. by 24. Product, 188 T. 2 cwt. 6. Multiply 3 cwt. 2 qr.21 Ib. 14 oz. by 30. Product, 110 cwt. 3 qr. 12 Ib. 4 oz. 7. Multiply 3 tons, 7 cwt. 2 qr. by 34. Product, 114 tons 15 cwt. APOTHECARIES WEIGHT. tt g 3 9 fc g 3 9 # fc 3 3 Bgr. 4821 53 10 2 12 17 5 6 1 4 5 9 12 23 5 3 2 TROY WEIGHT. Ib. oz. dwt. Ib. oz. dwt. gr. Ib. oz. diet. gr. 67 5 16 43 8 10 113 6 6 246 134 11 12 4. Multiply 41 Ib. 6 oz. 18 dwt. 2 gr. by 7. Ans. 291 Ib. oz. 6 dwt. 14 gr. 5. Multiply 91 Ib. 4 oz. 14 dwt. 16 gr. Ly 8. Ans. 731 Ib. 1 oz. 17 dwt. 8 gi 7 78 COMPOUND MULTIPLICATION. CLOTH MEASURE. yd. qr. na. E. E. qr. na. E.Fl. qr. na. E.Fr. qr. na. 20 2 3 37 4 2 18 3 14 1 3 6 8 12 9 124 2 5. If 19 yd. 1 qr. 2 na. be multiplied by 5, what num- ber of yards will there be ? Ans. 96 yds. 3 qr. 2 na. 6. Multiply 56 Ells Eng. 3 qr. by 9. Ans. 509 Ells E. 2 qr. LONG MEASURE. deg.m.fur.p. 1. m.fur.p. m. fur. p. yd. ft. in, 8 1 3 36 4 2 2 29 18 3 20 1 2 10 12 7 5 96 17 6 32 4. Multiply 6 deg. 40 m. 7 fur. by 10. Ans. 65 deg. 61 m. 2 fur. 5. Multiply 44 m. 6 fur. 20 p. by 7. Ans. 313 m. 5 fur. 20 D LAND, OR SQUARE MEASURE. a. r. p. a. r. p. a. r. p. 49 2 17 19 3 20 10 33 2 6 9 99 34 4. How many acres will 10 men reap in one day, allowing them 1 acre 3 roods 11 perches each? Ans. 18 A. OR. 30 P. 5. Multiply 63 acres 3 roods 18 perches, by 11. Ans. 702 A. 1 R. 38 P. 6. How many acres in 15 lots, containing 17 acres, 2 roods, and 20 perches each? Ans. 264 A. 1 R. 20P. COMPOUND MULTIPLICATION. 79 CUBIC MEASURE. cords, ft. in. yd. fir in. 7 28 1327 19 23 1421 6 8 43 44 1050 159 1 1000 cords, ft. in. yd. ft. in. 21 56 1432 27 13 1291 7 9 5 In a pile of wood are 14 cords 92 feet; how much in 24 such piles? Ans. 353 c. 32 ft. 6 In a cellar, are contained 42 yards 25 feet ; what are the contents of 23 such cellars? Ans. 987 yd. 8 ft. LIQUID MEASURE. hhd. gal. qt. T.hhd.gal. qt. pt. pi. hhd.gal. qt. pt. 8 43 2 1 2 16 3 1 4 1 19 3 1 4 '10 5 34 48 4 Multiply 3 T. 2 hhd. 50 gal. 2 qt. by 8. Ans. 29 T. 2 hhd. 26 gal. qt. 5 Multiply 4 hhd. 41 gal. 1 pt. by 10. Ans. 46 hhd. 33 gal. 1 qt. pt. MOTION. sin. ' sin. ' " 3 27 48 1 24 48 25 7 9 27 14 36 16 13 15 45 3 If a planet move through 2sin. 15 23' of its orbit in one day ; how far will it advance in 8 days. Ans. 205m. 3 4' 80 COMPOUND DIVISION TIME. weeks rf. h. d. h. min. sec. o 5 23 3 14 25 36 8 9 8 30 5 16 32 9 50 24 4 If a man can perform a piece of work in 2 yr. 3 mo., how long would it take him to perform 5 such? Ans. il yr. 3 mo, 5 If a laborer dig a drain in 2 weeks, 3 days, how long a time would he require to dig 9 such drains? Ans. 21 weeks 6 days. STERLING MONEY. s. d. s. d. s. d. 246 13 3| 14 6 04 111 11 10* 11 9 10 2713 6 5i *. d. . s d. 4 Multiply 37 6 9i by 5 Prod. 186 13 Hi 5 56 8 7| by 9 507 17 9| COMPOUND DIVISION. COMPOUND DIVISION is the art of dividing a sum which consists of several denominations. CASE 1. When the divisor does not exceed 12. RULE. Divide the several denominations of the given sum, one after another, beginning with the highest, and set their respective quotients underneath. When a remainder occurs, reduce it to the neXt lower denomination, and add it to the number of ihe next de- nomination, and divide the sum as before. COMPOUND DIVISION. 81 EXAMPLES, bu. pe. qt. pt. 7) 25 2 6 1 ^ ere ^ * nto ^ ku. ^ times and ' 4 remain. Set down the 3. Reduce the 4 bushels to pecks, 3251 which makes 16 pecks: add 16 pecks to 2 pecks, which make 18 "peeks. Now 7 into 18 jpe. 2 times, and leave 4. Set down the 2. Reduce the 4 pecks to quarts, which makes 32 qts. Add 32 qt. to 6 qt. makes 38 qt. 7 into 38 qt. 5 times, and 3 remain. Set down the 5. Reduce the 3 qt. to pt. makes 6 pt., add 6 pt. to 1 pt. makes 7 pints. 7 into 7, 1 time. Set down the 1, and the work is completed. bu. pe. qt. bu. pe. qt. 2) 8 2 6 3) 9 3 6 4 1 3 4 Divide 34 bu. 3 pe. 6 qt. between 9 persons. Ans. 3 bu. 3 pe. 4 qt. 5 92 bu. 3 pe. belong equally to 7 persons; what is the share of each? Ans. 13 bu. 1 pe. CASE 2. When the divisor exceeds 12, and is the exact product of two numbers in the multiplication table. RULE. Divide the sum by one of the factors, and the quotient by the other. Multiply the last remainder by the first divisor, and add the first remainder for the true remainder, as in simple division, note 2. EXAMPLES. 1 Divide 89 bu. 3 pe. 7 qt. by 28. Quotient 3 bu. 6 qt. 1 pt. 18 .era. 2 82 COMPOUND DIVISION. '4 bu. 89 22 1 73 Rem, 3 6 5 Rem. 4 First divisor 20 3 First rem. True rem. 23 Quarts 2 28)46 pints Pt. 1 and a rem, of 18 pints undivided. 3. Divide 78 bushels, 3 pecks, 4 quarts, among 32 persons; what will be the share of each? Ans. 2 bu. 1 pe. 6 qt. 1 pt. and a remainder of 24 pints undivided. CASE 3. When the divisor is more than 12, and is NOT the exact product of any two numbers in the multiplication table RULE. Divide the highest denomination of the given sum, as in case 2, simple division; and reduce the remainder, if any, to the next lower denomination; add the number of that denomination to the result, and divide as before. EXAMPLES. 1. Divide 77 bushels, 1 peck, 7 quarts, by 23. Quotient, 3 bu. 1 pe. 3 qt. 1 pt.!3rem. COMPOUND DIVISION. 83 EXAMPLES. bu. pe. qt.bu.pe.qt. pt. 23)79 1 7(3 1 6 1+3 pint remaining. 69 10 4 23)41(1 peck 23 18 8 23)151(6 quarts 138 13 2 23)20(1 pint 23 Rem. 3 pints 2. A boat load of corn, containing 4927 bushels, 3 pecks, is owned equally by 29 persons : wnat is the share of each? Ans. 169 bu. 3 pe. 5 qt. 1 pt., and a rem. of 1 pint. Questions. What is compound division? When the divisor does not exceed 12, how do you proceed ? When a remainder occurs, what do you do with it? Where the divisor exceeds 12, but is the product of two numbers in the table, how do you perform the operation ? How do you find the true remainder in the latter case ? When the divisor is more than 12, and is not the pro- duel of any two numbers in the table, how do you per- form the operation? 84 COMPOUND DIVISION. AVOIRDUPOIS WEIGHT. tons cwt. qr. Ib. Ib. oz. dr. 6)37 17 3 27 7)40 12 14 6619-1 rem. 6 10 15-5 R. tons cwt. qr. cwt. qr. Ib. oz. dr. 8)92 3 3 9)75 3 23 14 12 5. A quantity of iron weighing 473 tons, 19 cwt., 3 quarters, is owned equally by 22 persons; what is the share of each? Ans. 21 T. 10 cwt. 3 qr. 16 Ib. 8 oz. 11 dr. Rem. 14 dr. APOTHECARIES WEIGHT. fc 3 3 9 & 3 3 9 gr. 4)23 7 5 1 5)41 6 7 2 14 5 10 7 1 8361 2-4rem. fc 5 3 9 fc 3 3 9 ST. 6)46 912 7)93 7 5 2 14 5. Divide 127fe 3g 63 into 17 equal parcels: how much in each parcel? Ans. 7fe 5g 63 2^ 16gr. 8 rem. TROY WEIGHT. Ib oz. dwt. Ib. oz. dwt. gr. 8)34 10 15 7)45 11 16 22 Ib. oz. dwt. Ib. oz. dwt. gr. 9)78 9 16 8)82 7 14 21 COMPOUND DIVISION. 85 CLOTH MEASUEE. yd. qr. na. Ells E. qr. na. 5)27 3 1 6)37 3 o 1 1+4 rem. yd. qr. na. EllsE. qr. na. 7)45 3 2 8)37 3 1 LONG MEASURE. 1. m. f. m. f. p. yd. 6)37 2 2 7)46 7 17 3 607 6 5 25 2 yd. ft. in. m. f. p. yd. ft. 6)53 2 9 7)87 6 23 4 2 , 5. A traveller has a journey of 946 miles, 6 furlongs, to perform in 26 days; how far must, he travel each day? Ans. 36 m. 3 f. 12 p. 8 rem. LAND, OR SQUARE MEASURE. ; A. R. P. A. R. P. yds. 7)37 3 27 9)423 3 28 2 5 1 26+5 Rem. 47 16 13+5 Rem. 3. A farm containing 746 acres, 3 roods, 29 poles, is to be divided equally between 9 heirs; what is the share of each? Ans. 82 A. 3 R. 38 P. and 7 rem. CUBIC MEASURE. cords ft. yd. ft. in. 8)97 48 9)148 16 493 12 22 16 13 1398+7R. 86 COMPOUND DIVISION. 3. A boat load of wood, containing 92 cords 87 feet, is to be divided between 3 persons; what is the share of each? Ans. 30 c. 114 ft. 1 rem, 4. A quantity of earth, containing 6987 yards, 25 feet, is to be removed by 29 carters; how much must each remove? Ans. 240 yd. 20 \ LIQUID MEASURE. tuns.Jihd. gal. hhd. gal. qt. pt. 5)37 3 45 6)57 36 3 1 7 2 . 21+3 Rem. 9 37 2 l-(-l Rem tuns hhd. gal. hhd. gal. qt. pt. 7)84 2 32 8)93 43 3 1 5. A quantity of liquor owned equally by 27 person.* the whole quantity being 431 hhd. 47 gals; what i the share of each? Ans. 15 hhd. 62 gal. 1 qt.; 17 rem. MOTION. Sin. ' " Sin. 8)9 16 45 36 9)11 23 48 54 1 5 50 42 TIME. yr. mo. we. da. ho. min. sec 11)848 10 12)24 6 20 32 24 77 2 2 13 42 42 yr. mo. da. ho. min. s*,c. 4)375 8 7)37 16 28 32 COMPOUND DIVISION. 87 STERLING MONEY. *. d. *. d. 6)82 14 6 8)143 7 10 13 15 9 17 18 5| s. d. s. d. 7)78 10 11 9)98 17 1 s. d. s. d. 19)36 16 3(1 18 9 6 Divide 113 13s. 4d. by 31. What is the quotient? Ans. 3 135. 4d 7 Divide 189 14s. by 95. Quotient, 1 19s PROMISCUOTIS EXERCISES 1 In 35 dollars how many cents? Ans. 3500. 2 How many miles are there in 98 furlongs ? Ans. 12M. 2fur. 3 How many weeks are there in 365 days? Ans. 52we. 1 da. 4 In 84 half cents how many cents? Ans. 42cts. 5 In 8 tons 15 cwt., how many hundred weight? Ans. 175 cwt. 6 How many perches are there in 63 roods ? Ans. 2520 square per. 7 How many pounds in 157s.? Ans. 7 17s. 8 In 175 pecks how many bushels ? Ans. 43bu. 3pe. 9 In 7642 cents how many dollars ? Ans. $76 42cts 10 In 103 pints how many quarts? Ans. 51qt. Ipt. 11 How many minutes are there in 720 seconds? Ans. 12min. 12 In 7 hogsheads, 33 gallons, how many gallons? Ans. 474 gal. 88 SINGLE RULE OF THREE, PROPORTION ; OR, THE SINGLE RULE OF THREE. PROPORTION is an Equality of RATIOS ;* That is, four numbers are proportional, when the first has the same ratio to the second as the THIRD has to the FOURTH, Thus> as 12 : 4 : : 24 ; 8; or as 4 : 12 : : 8 : 24, The ratio of 12 to 4 is 3 and the ratio of 24 to 8 is 3. Or, the ratio of 4 to 12 is , and the ratio of 8 to 24 is . Then Four numbers are proportional, when the first is as many times the second or the same part of the second, as the third is of the fourth. Or, when the ratio of the first to the second equals the ratio of the third to the fourth. Tha two quantities compared are called the TERMS of the ratio: the first is called the ANTECEDENT, and the second the CONSEQUENT. In any series of four proportionals, the first and fourth terms are called the EXTREMES, and the second and third the MEANS. The product of the Means, equals the product of the Ex- tremes. Thus in either series above, 12X8=96, and 24X4=96. Now suppose we have the three first terms of a series in propor- tion, and we wish to find the fourth. Divide the product of the second and third terms by the first, and the quotient will be the fourth term. In this manner let the fourth term be found in each of the following series. 2 : 4 : : 8 is to what? Am. 16. 3 : 11 : : 9 is to what? tins. 33. 4 : 6 : : 6 is to what ? Jlns. 9. 2 : 9 : : 8 is to what? Jlns. 30. 5 : 7 : : 15 is to" what? Am. 21. RULE. number which is of the name or kind in which the answer is required, in th<? third place: Set that * RATIO is the relation of one thing to another of the same kind in regard to magnitude or quantity. SINGLE RULE OF THREE. 89 And, if the answer must be greater than the third term, set the greater of the remaining two terms in the second, and the less in the first place ; but, if the an- swer must be less than the third term, set the less in the second, and the greater in the first place. When the first and second terms are not of the same denomination, reduce one or both of them titlt they are; and, if the third consist of several denominations, reduce it to the lowest, then Multiply the second and third terms together and divide the product by the FIRST, and the quotient will be the fourth term or answer. NOTE. The answer will be of the same denomination as the third term , and, in many instances, must be reduced to a greater denomination. EXAMPLES. 1. If four pounds of sugar cost 50 cents, what will 24 pounds cost at the same rate? Ans. $3,00. 1st Term. 2d Term. 3d Term. t Lj _^_ lu _j t^^^-^j y_,- ,_j In this question the answer is jj j, required to be money: therefore ** C "' money (the 50 cts.) must be in As 4 t he third place. Because 24 50 pounds will cost more than 4 _ pounds the greater (24 Ibs.) 4)1200 must occupy the second place: ' _ and the remaining term (4 Ibs.) ~ 2. If 24 pounds, cost 300 cents (or 3 dollars ;) how- many pounds may be purchased for 50 cents at the same rate? Ans. 4 Ibs. cts. els. Ibs. In this question the answer is re- As 300 : 50 : : 24 quired to be in pounds,- therefore 50 pound 8 (th 6 24) must be in the third place, Because 50 cts. will purchase less 300)1200 than 300 cts. the lens (50 cts.) - must occupy the second place : and 4 the remaining term (300 cts.) the first. 90 SINGLE RULE OF THREE. 3. Bought a load of corn containing 27 bushels, 3 pecks, at 50 cts. per bushel, what did it cost? Ans. $13,87^. bit. bu. pe. cts. 1 : 27 3 : : 50 4 4 Because pecks occur in the second term, the first and second are reduced 4 111 to pecks. 50 4)5550 $13,87^ 4. What are 42 gallons worth, if 3 gallons 2 quarts cost $1,20? Ans. $14,40. gals. qts. gals. D. cts. 3 2 : 42 : : 1,20 44 As quarts occur in the first term, the first and second are reduced to 14 168 q uarts - 120 14)20160 $14,40 5. If 8 bushels 2 pecks cost $4,25, how many bush- els can I purchase with $38,25 ? Ans. 76 bu. 2 pe. D. cts. D. cts. bu. pe. 4,25 : 38,25 : : 8 2 34 4 As two denominations occur in the third term, it is reduced *A r^ t0 the leSS ; hellCe the reSult ' S d4pe. pecks> whicil must be re< luced 11475 to bushels. 425)130050(306 pe. 1275 pe. 2550 4)306 2550 76 bu. 2 pe. SINGLE RULE OF THREE. 91 G. What will 5 Ib. 6 oz. 5 d\vt. of silver-ware cost at $1,50 per ounce? Ans. $99,37^. oz. 1 20 20 U). oz. dwt. : 565 12 66 20 1325 150 66250 1325 D. ds. 1,50 As dwts. arc in the second term, the first and second must be reduced to dwts. . 20)198750 f99,37| 7. AVhen 3 yards and 8 feet of plastering cost $1,40, what will be the cost of 16 yards? Ans. $5,76. ydv.ff. yds. D.cts. 38 : 16 : : 1,40 9 9 35 144 140 5760 144 35)20160(5,76 cts. 8. How many yards of cloth can be purchased for 95 collars, if 4 yd. 3 qr. cost $9,50 ? Ans. 47 yd. 2 qr. ; or 47| yd. $ ct. $ ct. yd. qr. qr. As? 9,50 : 95,00 : : 4 3 4)190(47 yd. 2 qr. 92 SINGLE RULE OF THREE. NOTE. The operation may, in many instances, be contracted by dividing the second or third term by the first ; or the first by either |! of the others, or by any number that will divide the first and either of the others without a remainder; and, using the quotients instead of the original numbers. 9. If 24 yards cost $96, what will 8 yards cost? Aus. $32. yds. yds. D. yds. yds. D. 24a : *8a : : 96c or 24a : 8 : : 96a 3c Ans. 32. 4 32 10 If 36 bushels cost $72; what will 12 bu. cost? Ans. $24. bu. bu. D. bu. bu. D. 3Ga : I2a : : 72c 12)36 : 12 : : 72a 3c 24 3a 1 24 APPLICATION. 1 When 4 bushels of apples cost $2,25, what must be paid for 20 bushels ? Ans. $11,25. bit. bu. D. cts. 4 : 20 : : 2,25 2 How many yards of cloth can I buy for $60, when 5 yards cost $12? Ans. 25 yds. D. D. yds. 12 : 60 : : 5 3 If 6 horses eat 21 bushels of oats in a given time ; how much will 20 horses eat in the same time ? Ans. 70 bu. ,4 If 20 horses eat 70 bushels of oats in a certain time ; how much will 4 horses eat in the same time? Ans. 14 bu. 5 If a family ofHen persons use 7 bushels 3 pecks of wheat in a month; how much will serve them when there are 30 in the family? Ans. 23 bu. 1 pe. 6 If 14 Ibs. of sugar cost 75 cents, how many pounds can be bought for three dollars? Ans. 56 Ibs. 7 If 4 hats cost 12 dollars, what will 27 feats cost at the same rate? Ans. $81. SINGLE RULE OF THREE. 93 8 If 20 yards of cloth cost $85, what will 324 yards cost at the same rate ? Ans. $1377. 9 If 2 gallons of molasses cost 70 cents, what will 2 hogsheads cost ? Ans. $44,10. 10 If 1 yard of cloth cost $3,25 cts., what will be the cost of 6 pieces, each containing 12 yds. 2 qrs. ? Ans. $243,75 cts. 11 If 3 paces or common steps of a person be equal to 2 yards, how many yards will 160 paces make? Ans. 106 yds. 2 ft. 12 If a person can count 300 in 2 minutes, how many can he count in a day ? Ans. 216000. 13 What quantity of wine at 60 cts. per gallon can be bought for $37,80 cts. Ans. 63 gal. 14 If 8 persons drink a barrel of cider in 10 days, how many persons would it require to drink a barrel in 4 days? Ans. 20. 15 If 8 yards of cloth cost $12, what will 32 yards [cost? Ans. $48. 16 If 3 bushels of corn cost $1,20, what will 13 bush- els cost? Ans. $5,20. 17 If 9 dollars will buy 6 yards of cloth, how many yards will 30 dollars b^y ? Ans. 20. 18 If a man drink 3 gills of spirits in a day, how much will he drink in a year ? Ans. 34 gal. 1 pt. 3 gi. 19 If 12 horses eat 30 bushels of oats in a week, how many bushels will serve 44 horses the same time ? -Ans. 110. 20 If a perpendicular staff 6 feet long, cast a shadow 5 feet 4 inches, how high is that tree whose shadow is 104 feet long at the same time? Ans. 117 feet. EXERCISES. 1 If 12 acres, 2 roods, produce 525 bushels of corn, how many bushels will 62 acres, 2 roods produce ? Ans. 2625 bu. 2 If 7 men plough 6 acres, 3 roods in a certain time, how many acres will 96 men plough in the same time ? Ans. 92 A. 2 R. 11 Per. 12 yd.-f 94 SINGLE RULE OF THREE. 3 Suppose 3 men lay 9 squares* of flooring in 2 day s ; low many men must be employed to lay 45 squares in he same time? Ans. 15 men. 4 If 7 pavers lay 210 yards of pavement in one day; low many pavers would be required to lay 120 yards in he same time? Ans. 4 pavers. 5 If 2 hands saw 360 square feet of oak timber in 2 days; how many feet will 8 hands saw in the same time? Ans. 1440 feet. 6 An engineer having raised a certain work one hun- dred yards in 24 days, with 5 men; how many men must be employed to perform a like quantity in 15 days? Ans. 8 men. 7 If 3 paces or common steps be equal to 2 yards ; how many vards will 160 such paces make? Ans. 108 yd. 2 ft. 8 If a carriage wheel in turning twice round, advance 33 feet 10 inches; how far would it go in turning round 63360 times? Ans. 203 miles. 9 Sound flies at the rate of 1142 feet in 1 second of time; how far off may the report of a gun be heard in 1 minute and 3 seconds? Ans. 13 miles, 5 fur., poles, 2 yd. 10 If a carter haul 100 bushelst>f coal at every 3 loads; how many days will it require for him to load a boat with 3600 bushels,* suppose he haul 9 loads a day? Ans. 12 days. bu, lu. da. da. As 300 : 3600 : : 1 : 12. 11 If 8 men can reap a field of wheat in 4 days; how many days will it require for 16 men to do it? Ans. 2 days. 12 -Sold 10 yards of linen at 5 dollars 50 cents; what was it a yard? Ans. 55 cents. *A SQUARE is 10 feet long and 10 feet wide, or 100 square feet. This measure is employed in estimating the quantity of flooring, roof- ing, weather-boarding, &c. SINGLE RULE OF THREE. 95 13 If 7 pounds of cheese cost 87 i cents; what must I pay for 122 pounds? Ans. 15 dol. 25 ct. 14 If 1 ounce of silver cost 72 cents; what will 3 pounds 5 ounces come to? Ans. 29 dol . 52 ct. Why do you multiply the second and third terms to- gether and divide by the first? What will 24 pounds of bacon cost at 50 ct. for every 4 pounds? Ib. Ib. ct. ct. As 4 : 24 ::50 : 300 cts. 4)50 12} the price of lib. 24 50 25 If 4 pounds cost 50 cents, divide the 50 cents by 4, gives the price of 1 pound: thus 4 into 50 12*. If 1 pound cost 12 cents, 2 pounds will cost twice that; three pounds three times, and 24 pounds will cost 24 times 12! ct. ; that is 300 cts. or 3 dol. If the second and third terms be multiplied together, and their product divided by the first, the result will be the Bame as it is when the third is di- vided by the first, and the quotient multiplied by the second. 300 ct. the price of 24 Ib. 15 If 15 yards of broad cloth cost 80 dollars; what will 75 cost. Ans. 400 dollars. 16 A man bought li yards linen for $2 50 cts.; what is the worth of 1 qr. 2 na. at the same rate. Ans. 62* ct. 17 If 321 bushels of salt cost $240 75 cents; what was it per bushel? Ans. 75 cents. 18 If the moon move 13 deg. 10 min. 35 sec. in one day; in what time does it perform one revolution? Ans. 27 da. 7 hrs. 43 min. deg. min. sec. deg. da. As 13 10 35:360 ::1 96 SINGLE RULE OF THREE. 19 If a staff 4 feet long, cast a shade 7 feet on level ground ; how high is a steeple whose shade is at the same time 198 feet. Ans. H3\ 20 If a man's annual income be 1333 dollars, and he spend 2 dollars 14 cents a day ; what will he save at the end of one year? Ans. $551 90 ct. 21 Suppose A. owes B. 791 dols. 60 ct., and can pay only 374 cts. on the dollar; how much must B. receive? Ans. $296 85 ct. 22 Bought 3 casks of raisins, each containing 3 cwt 1 qr. 14 lb.; how much did they cost at $6 21 ct. pei cwt.? Ans. $62,874 PROPORTION DIRECT AND INVERSE. Hitherto, proportion has been treated in general terms ; it now remains to consider the two kinds, DIRECT and INVERSE. DIRECT PROPORTION is that in which more requires more, or LESS requires LESS. Thus: yd. yd. dol. If 2 yd. cost 4 dol., 124 yd. being As 2 * I'M: * 4 more than 2 yd., will cost more than 4 dol. yd. yd. dol. And, if 124 yd. cost 248 dol.; 2yd. As 124 2 248 being leis wil1 cost less " That is, more yards require more money, and less yards cost less money INVERSE PROPORTION is that in which more requires less; and less requires more. Thus: If 12 men built a wall It is supposed that 12 men perform in 4 davs; how many ft . P iec J of * ork I" * da r s: w a , like "'i o j n P 18ce work is to be done in o days; men can do it in 8 days i this will require a kss uumne i of men : that is. more days require less men. STATED. Here it is supposed that 12 men da. dct. m. m. performed a piece of work in 8 days: As 4 : 8 inversely : : 12 . 6 nke . iece I s to be done in ? d *y. s J this will require more men. 1 hat is, more days require less men, and less da. da. m. m. days require more men. As 8: 4 directly:: 12.6 SINGLE RULE OP THREE. 97 All the past exercises in proportion are Direct the "ollowing will be INVERSE PROPORTION. Questions in Inverse Proportion, may be stated and solved by the same rule that is given for Direct Propor- tion. EXERCISES. 1 If 6 mowers mow a meadow in 12 days; in what time will 24 mowers do it? Ans, 3 da. 2 If a man perform a journey in 6 days, when the days are 8 hours long; in what time will he do it when they are 12 hours long? Ans. 4 da. 3 If, when wheat is 83 cents a bushel, the cent loaf weighs 9 ounces ; what ought it to weigh when wheat is $ 1 244 cts. a bushel? Ans. 6 oz. 4 If 100 dollars principal in 12 months gain 6 dollars interest; what principal will gain the same interest in 8 months? Ans. $150. 5. If 12 inches long and 12 inches wide, make 1 square foot; how long must a board be that is 9 inches wide, to make 12 square feet? Ans. 16 ft. 6 A. lent B. 500 dollars for 6 months ; how long must B. lend A. 220 dollars to be equivalent? Ans. 13 months, 19 days.-f 7 There is a cistern having a pipe that will empty it in 12 hours; how many pipes of the same capacity will empty it in 15 minutes ? Ans. 48 pipes. 8 A certain building was raised in 8 months by 120 workmen, but the same being demolished, it is required to be rebuilt in 2 months; how many workmen must be employed? Ans. 480 men> 9 If for 48 dollars 225 cwt. be carried 512 miles; how many hundred weight may be carried 64 miles for the same money? Ans. 1800 cwt. *A month is estimated at 30 days, unless a particular month C . referred to. _^_^____ 98 SINGLE RULE OF THREE. 10 If 48 men can build a wall in 24 days; how many men can do it in 192 days ? Ans. 6 men, 11 How many yards of carpeting 2 ft. 6 in breadth, will cover a floor that is 27 feet long and 20 feet wide I Ans. 72. 12 What quantity of shalloon that is 3 quarters wide, will line 7i yards of cloth that is U yd. wide? Ans. 15 yd. 13 How many yards of matting that is 3 quarters wide, will cover a floor that is 18 feet w r ide and 60 feet long? Ans. 160. 14 In what time will $600 gain the same interest that $80 will gain in 15 years? Ans. 2 years. Questions* What is proportion? When is the proportion direct? When is it inverse? Why is the proportion inverse in. the last question? A. because it is more money requiring less time. Why is the proportion inverse in the llth question? A. because the shalloon is narrower than the cloth; that is less width requiring more length. Why is the 10th question inverse? PROMISCUOUS EXERCISES. 1 A certain steeple standing upon level ground, casts a shadow to the distance of 633 feet 4 inches, when a staff 3 feet long, perpendicularly erected, casts a shadow of 6 feet 4 inches; what is the height of the steeple? Ans. 300 ft. 2 A ship's company of 15 persons is supposed to have bread enough for a voyage, allowing each person 8 oun- ces a day, when they take up a crew of 5 persons, with whom they are willing to share; what wi" be the daily allowance of each person now? Ans. 6 oz. 3 Bought 215 yards of broad cloth at 6 dollars a yard; SINGLE RULE OF THREE. 99 what was the prime cost, and how must I sell it per yard to gain $135 on the whole. Aiis. prime cost $1290,00; to be sold for $G,62f per yard. 4 If 100 men can complete a piece of work in 12 days; how many can do it in 3 days? Ans. 400 men. 5 If a board be 4| inches wide; how long a piece will it take to make 1 square foot? Ans. 32 in. G A pole, whose height is 25 feet, at noon casts a shadow to the distance of 33 feet 10 inches ; what is the breadth of a river which runs due East at the bottom of a tower 250 feet high, whose shadow extends just to the opposite edge of the water? Ans. 338 ft. 4 in. 7 A plain of a certain extent having supplied a body of 3000 horses with forage for 18 days ; how long would it have supplied 2000 horses ? Ans. 27 da. 8 A piece of ground 1 rod wide and 160 rods long, makes 1 acre*; how wide a piece must I have across the end of a farm 32 rods wide to make an acre ? Ans. 5 rods. 9 I have a floor 24 feet long, and 15 feet wide, which I wish to cover with carpeting that is 3 quarters of a yard wide; how many yards must I buy. Ans. 53 yards, 1 foot. 10 How much land at $2,50 an acre must be given in exchange for3GO acres worth $3,75 an acre? Ans. 540 acres. 11 What is the weight of a pea to a steel-yard, which is 39 inches from the centre of motion, will balance a weight of 208 Ibs., suspended at the draught end 3 quar- ters of an inch? Ans. 4 Ib. 12 If $28 will pay for the carriage of 6 cwt. 150 miles ; how far should 24 cwt. be carried for the same money? Ans. 37 i miles. 100 DOUBLE RULE OF THREE. COMPOUND PROPORTION; OR THE DOUBLE RULE OF THREE. DIRECT AND INVERSE. COMPOUND PROPORTION is two or more series of pro- portionals combined. Five, seven, nine, or other odd number of terms, is always given to find a sixth, eighth, or tenth, &c*, or answer. Rule for the Statement. Place the numbers that is of the denomination in which the answer is required to be, in the third place. Then: Consider separately each pair of similar terms and place them agreeably to the rule for SIMPLE PROPORTION. An OR, Work by two separate statements in simple propor- tion.* Rule for the Solution. Reduce the several pairs of terms to similar denom- inations as in single proportion, and the last to the lowest denomination given : Then Multiply the two initials, or left hand terms togethei for a DIVISOR, and the other three for a DIVIDEND. Divide the latter by thejformer, and the quotient wil< be the answer, in that denomination to which the third term was reduced. EXAMPLES. 1. If 6 men in 8 days build 40 rods of wall, how rnucl will 18 men build in 20 days? Aiis. 300 rods * It would be well for the pupil to work each sum both ways. DOUBLE RULE OP THREE. 101 The answer is required to be rods gi ven in rods : then rods must be Afi the third term. If 6 men build 40 rods, 18 men will build more,; then more (18 men) must occupy the second, and less (6 men) the 48 360 first P lace - If 8 days produce 40 rods, 20 days will produce more; then more (20 days) must occupy the 48)14400(300 *eomd, and few (8 days) the>j* 144 place. N. B. The first pair, or two upper terms must be alike. Also the lower pair must be alike. That is, both must be men or both days, both hours or both bushels, &c. 2. If 6 men in eight days eat lOlb. of bread, how much will 12 men eat in 24 days? Ans 60. men 6 : 12) in IK days 8 : 24 : Contracted. ^ 6 : 12 2j 288 8 : 24 3 ' 10 6 48)2880(60 Ans. 10 288 60 Ans. 3. Suppose 4 men in 12 days mow 48 acres, how many acres can 8 men mow in 16 days? Ans. 128A. 4." If 10 bushels of oats be sufficient for 18 horses 20 days, how many bushels will serve 60 horses 36 days, at that rate? Ans. 60bu. 5. Suppose the wages of six persons for 21 weeks be 288 dollars, what must 14 persons receive for 46 weeks? Ans. $1472. 6. If the carriage of 8cwt. 128 miles cost f 12.80, what must be paid for the carriage of 4cwt. 32 miles? Ans. $1.60. 7. If 371b. of beef be sufficient for 12 persons 4 days, how many Ib. will suffice 38 men 16 days? Ans. 4681b. lOi oz. 102 DOUBLE RULE OF THREE. 8. If a man can travel 305 miles m 30 days, when the days are 14 hours long, in how many days can he travel 1056* miles, when the days are 12^ hours long? Ans. 116days.-f2540. 9. If the carriage of 24cwt. for 45 miles be 18 dol- lars, how much will it cost to convey 76cwt. 121 miles? Ans/$153 26 cts.+720. 10. A person having engaged to remove 000cwt. in 9 days; removed 4500cwt. in 6 days, with 18 horses: how many horses will be required to remove the balance in the remaining 3 days? Ans. 28 horses. 11. If 3 men reap 12 acres 3 roods in 4 days 3 hours, how many acres can 9 men reap in 17 days? Ans. 153 acres. Analysis. If 3 m. reap 12 a. 3 r. 1 m. rea 4 a. 1 r. and 9 m. reap 38 a. 1 r. If 4 d. 3h. reap 38 a. 1 r. Id. reap 9 a. ami 17 d. reap 153 a. Ans. 1836 9180 153)93636(612 roods, or 153 acrea 918 183 153 306 306 *The day is here estimated at twelve hours. PRACTICE. 103 12. If 40 men build 32 rods of wall in 8 days, work- ing 10 hours each day,- in how many days will 60 men build 48 rods, working 12 hours a day ? Ans. 6 days, 8 hours. Men men Multiply all the initial 60 : 40 terms (or 60, 32, and 12) rods rods days together for a divisor: and 32 : 48 : : 8 the other four for a dim- hours hours dend. 12 : 10 ,. 13. If 36 men dig a cellar 60 feet long, 24 feet wide, and 8 feet deep, in 16 days, working 16 hours per day, ho\v many men can dig a cellar 80 feet long, 40 feet wide, and 12 feet deep, in 20 days, working 12 hours per day? Ans. 128. Questions. What is ccfnpound proportion? How do you state questions in compound proportion? Which terms do you multiply together for a divisor? Which for a dividend? What other method is there? PRACTICE. PRACTICE is a short and expeditious method of per- forming various calculations in business. CASE 1. When the given price is LESS than one dollar. RULE. Set down the given number as one dollar, and take such aliquot part* or parts of that number, as the price is of one dollar, for the answer. *An aliquot part of a number is any number that will divide A without a remainder; thus 4 is an aliquot part of 20; and 8 of 40: and 25 cents is an aliquot part of 100 cents, (or $1.) because 25 c are contained in 100 cts-, an even number of times, without a remain- der. 104 PRACTICE. TABLE OF ALIQUOT PARTS. CTS. CWT. ROODS. 50 1 -v 1 10 r roods 33i 2 i ""3 o 11 > 25 4 J 4 P 20 r f -*j 2 iV p P 121 10 o ? p 1 D- qr. perches 20 i o 3 V iff p" 2 i 10 ,\] 5* 1 TO 1 o 20 |" o 4. 1 2J Ibs. 10 ^ p 2 I 5T- 16 i o 8 i 14 t- i 5 i < o 8 r* 4 ^ 7 l 7 T6" 2 it EXAMPLES . 1. What will 826 bushels of wheat come to at 25 cts. a bushel? Ans. $206 50 cts. cts. $ 25 3 r |8 26 82(5 bushels, at one dollar a bushel, will cost 826 dollars : at 25 cents, or ^ of a dh 1 kc rcn dollar, it will cost one fourth as much. $ J <6UO,OU 2. What will 934 gallons of molasses cost, at 50 cts. a gallon? Ans. $467. cts. $ 50 \ r 934 $ 467 3. What will 1832 bushels of salt cost, at 75 cents a bushel? Ans. $1374. cts. <fc At 50 cents the cost will be 50 \ _ _ sP r 1832 i as much as at one dollar. At 25 cents the cost will be 25 ] r - i as much as at 50 cts. 916 cost at 50 c. 458 cost at 25 c. $1374 cost at 75 c. 50 25 PRACTICE. 105 <J As before ; the cost at 50 cts. will 1H^2 ke i as much as at 1 dollar. At 25 cts. it will be i as much 916 ct. at 50. 458 ct. at 25. as at 1 dollar. $1374 ct. at 75. 4. What will 680 pounds of sugar cost, at 1 cents a pound? $68. 5. What will 742 pounds of pork cost, at 61- cts. a pound? Ans. $46 37* cts. 6. What must I pay for 371 pounds of bacon, at 12|- cts. a pound? Ans. $46 37i cts. 7. How much will 8750 bushels of rye cost, at 62| cts. a bushel? Ans. $5468 75 cts. 8. How much must be paid for 4360 square feet of marble, at 87 i cents a foot? Ans. $3815. 9. What will 468 square yards of plastering cost, at 18| cents a yard? Ans. $87 75 cts. 10. How much will be the cost of laying 856 perches of stone, at 93| cents a perch? Ans. $802 50 cts. 11. What will the digging of a cellar, containing 180 cubic yards, cost, at 20 cents a yard ? Ans. $36. 12. What will be the cost of hauling 248 cords of wood, at 3H cents a cord? Ans. $77 50 cts. 13. What must be paid for 432 perches of stone, at 37i cts. a perch ? Ans. $162. 14. How much must be given for 724 days labor, at 56* cents a day? Ans. $407 25. 15. What will 742 bushels cost at 10 cts. Ans. $71 20 16. 732 15 10980 17. 732 20 14640 18. 475 25 11875 19. 684 30 20520 20. 756 35 26460 21. 927 40 37080 22. 824 50 41200 23. 682 55 375 10 24. 341 60 20460 25. 784 70 54880 E 2 '~ 106 PRACTICE. 28. What will 352 busnels cost at 64 cts.? Ans.$22 00. 29. 436 124 54 50 30. 724 18| 135 75 31. 956 314 298 75 32. 742 37* 278 25 33. 274 43| 119 87* 34. 732 56* 411 75 35. 845 624 528 124 36. 684 68| 470 25 37. 274 814 222 624 38. 386 93| 361 874 CASE 2. When the given price is MORE than one dollar. RULE. Multiply the given sum by the number of dollars, and take the aliquot part or parts for the cents. ! as in Case 1. EXAMPLES. 1. What will 342 cords of wood cost, at 3 dollars 75 cents a cord? Ans. $1282 50. cts. $ 50 r 342 342 cords at $1, will cost $342; at $3 3 it will cost 3 times $342; at 50 cts. it will cost i as much as it will at $1. ; and at 25 cents, 4 as much as it will at 50 cts. ; 1026 which added together, will be the cost at 25 J r 171 $375. 85 50 1282 50 2. What will 250 acr. cost at $4 624 AJIS. $1 156 25 3. 435 5 874 2555 624 4. 273 6 124 1672 124 5. 942 7 374 6947 25 6. 846 368| 3119624 7. 957 5 75 5502 75 8. 236 6 93| 1637 25 9. 754 3 564 2686 124 10. 932 27 25 25397 00 PRACTICE. CASE 3. 107 When the given quantity consists of several denom- inations. RULE. Multiply the given price by the number of hundred weight, acres, yards, or pounds, &/c. and take the aliquot parts for the quarters, roods, feet, or ounces, &c. EXAMPLES. 1. What will 240 acres, 2 roods, 10 perches, cost at $ 15 25 cents an acre ? Ans. $3668 57* cts. 2 r. 10 p. 1525 240 61000 3050 762i 951-J-2 rem. 3668574 2. What will 29 yards, 4 feet, of stone pavement cost, at $2 25 cents a yard? 3 square feet Ans. $66 25 cts. 29 2025 450 ,75 25 6625 108 PRACTICE. 3. What will 32 pounds 8 ounces of silver cost, at $15,62i a pound? Ans. $510 41 i. 6 oz. Troy 2oz. $1562i 32 16 3124 4686 78H 26CH+2Rem. 510 4H 4. What will 27 cwt. 3 qrs. cost, at $23 50 cts. a cwt.? Ans. $652 12i. 5. What will be the cost of 47 Ib. 10 oz. (Troy) at $1 25 cts. Ans. $59 79. 6. What will 64 yds. 3 qrs. cost, at $2 25 a yard? Ans. $145 68| . 7 Sold 83 yards 2 qrs. of cloth at $10 50 a yard; what does it amount to? Ans. $876 75. 8. What will the laying of 28 squares, 75 feet of floor- ing cost, at $2 25 cts. a square? Ans. $64 68|. 9. What is the cost of 27 cords, 96 feet of fire wood, at $3 75 a cord. Ans. $104 064 10. What is the value of 428 gals. 3 qts. at $1 40 cts. a gallon ? Ans. $600 25 cts. 11. What is the value of 765 gals. 3 qt. 1 pt. at $2 18| cents a gallon? Ans. $1675 34| cts. 12. What is the value of 5 hhds. 3H gals, at $47 12 cts. a hogshead? Ans. $259 16 cts. 13. What is the value of 17 hhds. 15 gals. 3 qts. at $64 75 cts. per hogshead? Ans. $1116 93 cts. 7m. 14. What is the value of 120 bu. 2 pecks, at 35 cents a bushel? Ans. $42 17 cts. 5 m. 15. What is the value of 780 bu. 2 pecks, 2 qts. at $1 17 cts. a bushel? Ans. $913 25 cts.+ 16. What is the value of 1354 bu. 1 peck, 5 qts. 1 pt. at 25 cts. a bushel? Ans. $338 60 cts. 5m.+ 17. What is the value of 35 acres 2 roods 18 perches, at 51 dollars 35 cts. an acre? Ans. $1935 53 cts. 9m. TARE AND TEET. 109 Questions What is practice? What is the rule for the solution of questions in prac- tice? What is an aliquot part? Are 50 cts. an aliquot part of 100 cents? What part of $1 is fifty cents? What part of $1 is 33i cents? What part of $1 is 25 cents? What part of $1 is 12i cents? What part of $1 is 10 cents? What part of $1 is 20 cents? What part of $1 is 5 cents? What part of $1 is 4 cents? What part of $1 is 6* cents? TARE AND TRET. TARE AND TRET are allowances made on the weight of some particular commodities. Gross weight is the weight of the goods, together with the vessel that contains them. Tare is an allowance for the weight of the vessel. Tret* is an allowance of 4 Ib. for every 104, for waste, &,c. Neat weight is the weight of the goods, after all allow- ances are made. BULB. Subtract the tare from the gross, and the remainder is the neat weight. EXAMPLES. 1 Bought a chest of tea, weighing gross 63 Ib., tare 8 Ib. what are the neat weight and value, at 85 cents per Ib? * As tret is never regularly allowed in this country ; no account -of it is taken in this work. To 110 TARE AND TRET. lb. 85 ct. 63 gross or, weight of the chest and tea 55 lb. 8 tare or, weight of the chest 425 55 neat or, weight of the tea itself 425 $40,75 value, 2 Bought 5 bags of coffee, weighing each 97 lb. gross, tare of the whole 7 lb. what are the neat weight arid value, at 25 cents per lb.? Ans. 478 lb. neat $119,50. 3 The gross weight of a hogshead of sulphur is 1344 lb.; the tare 138 lb. what are the neat weight and its value, at $4,75 per 100 lb.? Ans. 1206 lb. neat $57,28i 4 Bought 3 barrels of sugar, weighing as follows, viz 236 lb. gross, 23 lb. tare 217 lb. gross, 22 lb. tare ^25 lb. gross, 23 lb. tare what are the neat weight and value, at $8 per 100 lb.? Ans. 610 lb. neat $48,80. 5 Sold 3 hogsheads of sugar, weighing each 12 cwt. 2 qrs. 14 lb. gross; tare 2 cwt. 1 qr. 27 lb. what are the neat weight and value, at $11,50 per cwt.? Ans. 35 cwt. 1 qr. 15 lb. neat $406 91 i cts. 6 What is the neat weight of 15 tierces of rice, weigh- ing 48 cwt. 3 qrs. 12 lb. gross; tare 6 cwt. 12 lb., and what is the value, at $5,25 per cwt. ? Ans. 42 cwt. 3 qrs. neat $224,43$. 7 What is the neat weight of 28 hogsheads of tobacco, weighing 201 cwt. 3 qrs. 12 ib. gross; tare 3140 lb.; and what does it come to at $5 per cwt. ? Ans. 173 cwt. 3 qrs. 8 lb $869 10| cts. 8 Bought 17 bags of grain, weighing 3561 lb. gross; tare 2 lb. per bag what is the neat? Ans. 3527 lb. 9 What is the neat weight of 16 bags of pepper, each weighing 65 lb. gross; tare 1ft lb. per bag and what is the amount at 30 cents per lb.? Ans. 1016 lb. neat $304,80. TARE AND TRET. Ill 10 In 14 hogsheads of sugar, weighing 89 cwt. 3 qrs.i 17 Ib. gross; tare 100 Ib. per hhd. how much ne&t weight, and what is its value, at $9 per cwt.? Ans. 77 cwt. 1 qr. 17 Ib. neat $696,611. 11 What are the neat weight and value of 16 hhds. of tobacco, each weighing 5 cwt. 1 qr. 19 Ib. gross; tare 101 Ib. per hhd., at 2 6s. lOd. per cwt.? Ans. 72 cwt. 1 qr. 4 Ib. neat 169. 5s. 4id. 12 Bought 6 hhds. of sugar, each 1126 Ib." gross; tare 117 Ib. per hhd. what are the neat weight and value at $8,75 per cwt.? Ans. 6054 Ibs. $529,72*. 13 What are the neat weight and cost of a hogshead of sugar weighing gross 986 Ib. ; tare 12 per cent, (or 12 Ib. for every WO Ib.) at $8 per neat hundred pounds? Ib. Ib. Ib. Ib. Ib. Ib. Ib. Ib. AslOO:986:: 12:118, Ib. 986 gross. 118, tare. 868, neat weight. Or as 100: 88:: 986: 868, Ib. 883 8 dol. $69,44 the value. 14 What are the neat weight and value of 4 hhds. of sugar weighing gross 45001b. tare 12 Ib. per cent, at $8, 75 percent.? Ans. 3960 Ibs. neat $346,50. 15 Bought 10 hhds. of sugar, each 920 Ib. gross; tare 10 Ib. percent. what arelhe neat weight and value at $-9,25 per cwt.? Ans. 8280 Ib. neat- $765,90. 16 Sold 3 casks of alum, each 675 Ib. gross; tare 13 Ib. percent. what are the neat weight and value at $4, 25 per cent. Ans. 1762 Ib. neat $74,87.4375. Or, 1762 Ib. neat $74,88. 5nearly. 17 What is the neat weight of48001b.gross: tare 12 Ib. per cent.? Ans. 4224 Ib. 18 What are the neat weight and value of 4 hhds. of sugar, each 12 cwt. 1 qr. 14 Ib. gross; tare 12 Ib. per cwt. at $12,20 per cwt.? Ans. 44 cwt. 22 Il> neat $539 19i cts. 112 INTEREST. 19 Bought 17 hhds. of sugar, weighing 201 cwt. 2 qrs. 13 Ib. gross; tare 10 Ib. per cwt. what are the neat weight and value at $14 per cwt.? Ans. 183 cwt. 2 qrs. 13 Ib. neat- $2570 62 i cts. INTEREST. INTEREST is an allowance made for the use of money. Principal is the sum for which interest is to be com- puted. Rate per cent, per annum is the interest of 100 dol- lars for one year. Amount is the principal and interest added together. CASE 1. When the time is one year and the rate per cent, is any number of dollars. RULE. Multiply the principal by the rate per cent., and divide by 100 j the quotient will be the interest for one year. EXAMPLES. 1. What is the interest of 500 dollars for 1 year, at 6 per cent, per annum? $500 6 100-f-$30100 Ans. 2. What is the interest of 225 dollars for 1 year, at Y dollars per cent, per annum? Ans. $15 75. 3. What is the interest of 384 dollars 50 cents, for 1 year, at 5 dollars per cent, per annum? Ans. $19 22i. 4. What is the amount of $275 for 1 year, at 6 per cent, per annum? Ans- $291 50. $275 6 16,50 interest 275,00 principal $291,50 amount i:\-TEKEST. 113 5. What is the interest of 1654 dollars 81 cents for 1 year, at 5 dollars per cent, per annum? Ans. $82 74-)-. 6. What is the interest of 1500 dollars, for 1 year, at i dollar per cent, per annum? Ans. $7 50. 7. What is the amount of $736, at 6 per cent, per annum, for 1 year. 780 dols. 16. 8. What is the interest of 524 dollars, for 1 year, at 5J dollars per cent, per annum? Ans. $27 51. 9. What would be the interest of 842 dollars, for 1 year, at 54 dollars per cent, per annum? Ans. $46 31. CASE 2. When the interest is required for several years. RULE. Find the interest for one year, and multiply the interest for one year by the number of years. EXAMPLES. 1. What is the interest of 500 dollars, for 4 years, at 6 dollars per cent, per annum? $500 6 < 3000 4 $12000 Ans, 2. What will be the interest of 540 dollars, for 2 years, at 5 dollars per cent, per annum?* Ans. $54 00. 3. What would be the interest of 482 dollars, for 7 years, at 6 dollars per cent, per annum? Ans. $202 44. 4 What is the amount of $736 81i with 7 years, nine months interest due on it, at 6 per cent, per annum? Ans. $1079 43|. Note. If the interest is required for years and months, multiply the interest for 1 year by the number of years, and take the aliquot parts of the interest for 1 year, for the months. 1 10* 114 Omo. 3 mo. INTEREST. $736 814 6 4420,874 interest 1 year 7 30946,12i interest 7 years 2210,431 interest 6 months 1105,21!-)- 3 34261,78 interest for 7 yr. 9 mo. 73681,25 principal -$107943 03 amount 5. What is the amount of $362 25 for 4 years 6 mo. at 6 per cent, per annum? Ans. $460 ()5|, CASE 3v When the interest is required for any number of months, weeks or days, less or more than one year. RULE. Find the interest of the given sum for one year Then, by proportion, As 1 year Is to the given time, So is the interest of the given sum (for 1 year) To the interest for the time required. Or take the aliquot parts of the interest for one year, for the given time, as in note, Case 2. EXAMPLES. 1. What is the interest of $560 for 2 years and 6 mo. at 5 per ct. per annum? Ans $70 560 5 6 mo. 2800 interest for 1 year 2 years 5600 1400 $70 00 interest for 2 years 6 months. INTEREST. 115 2. What is the interest of 325 dollars, for 4 years and 2 months, at 4 dollars per cent, per annum? Ans. $54 16 cts. 6m. 3. What is the interest of 840 dollars for 5 years and 3 months, at 4 dollars per cent, per annum? Ans. $176 40. 4. WTiat is the interest of 840 dollars, for 5 years and 4 months, at 7 dollars per cent, per annum? Ans. $313 60. 5. What is the interest of 5GO dollars, for 4 months, at 6 dollars per cent, per annum? 560 6 m. m. $ cts. $ cts. 100)33 60 As 12 : 4 : : 33 60 : 11 20 Ans. 6. What is the interest of 1200 dollars, for 15 weeks, at 5 dollars per cent, per annum? Ans. $17 30. 7. What will be the interest of 240 dollars, for 61 days, at 4| dollars per cent, per annum? Ans. $1 90.-J- 8. What is the interest 'of $1000, for 14 months, at 7 per cent, per annum? Ans. $81 60S. 9. What is the interest of 450 dollars, for 6 months and 20 davs, at 5i dollars per cent, per annum? Ans. $1375. 10. What is the interest of 375 dollars 25 cents, for 3 years 2 months 3 weeks and 5 days, at 6 dollars per ct. per annum? Ans. $72 92.-J- 11. What is the amount of $736 for 28 weeks, at 10 per cent, per annum? Ans. $775 63. CASE 4. To fad the interest of any sum for any number of days^ as computed at banks. RULE. Multiply the dollars by the number of days, and divide by 6; the quotient will be the answer in mills. The interest of any number of dollars for 60 days, at 6 per cent, will be exactly the number of cents; and if any other rate per cent, is required, take aliquot parts, and add or subtract according as the rate per cent, is more or less than 6. 116 INTEREST. EXAMPLES. 1. What is the interest of 563 dollars, for 60 days, at 6 per cent, per annum and likewise at 7 per ct. per an.? Ans. $5,63 at 6 per cent. 60 $6,56.8 at 7 per cent. 6)33780 in. at) 6per[ 5630 mills, cent.) $5630 938 Interest at 7 per cent. 6568 mills. 2. What is the interest of 854 dollars, for 30 days, at 6 per cent, per annum? Ans. $4 27. 3. What is the interest of 1100 dollars, for 48 days, at 6 per cent, per annum? Ans. $8 80. 4. What is the interest of 3459 dollars, for 75 days, at 6 per cent, per annum? Ans. $43 23 cts. 7 m.-|- 5. What is the interest of 1500 dollars, for 60 days, at 5 per cent, per annum? Ans. $12 50. CASE 5. TJie amount, time, and rate per cent, given, to Jlnd the principal. RULE. Find the amount of 100 dollars for the time required, at the given rate per cent. Then, by proportion, as the amount of 100 dollars for the time required, (at the given rate per cent.) is to the amount given, so is 100 dollars to the principal required. EXAMPLES. 1. What principal, at interest for 8 years, at 5 per ct. per annum, will amount to 840 dollars? Ans. $600. 5 dollars 8 years 40 Int. of $100 for Syr. 100 $ $ $ $ 140 : 840 ::100 : 600 140 Amt. of $100 for 8 yr. INTEREST. 117 2. What principal, at interest, for 6 years, at 4 per cent, per annum, will amount to $,1240. Ans. $1000 3. What principal, at interest for 5 years, at 6 per ct. per annum, will amount to 2470 dollars? Ans. $1900. CASE 6. The principal, amount, and time given, to find the rate per cent. RULE. Find the interest for the whole time given, by subtracting the principal from the amount. Then, as the principal is to 100 dollars, so is the in- terest of the principal for the given time, to the interest of 100 dollars for the same time. Divide the interest last found by the time, and the quotient will be the rate per cent, per annum, Or by compound proportion. EXAMPLES. 1. At what rate per cent, per annum, will COO dollars amount to 744 dollars, in 4 years? Ans. 6 per cent. $744 amount As 600 : 100 : : 144 : 24. 600 principal yr. $ 4) 24 (6 rate per cent. 144 interest Or by compound proportion : $ $ As 600 : 100 $ $ yr. yr. : : 144 : 6 rate per cent. 4 : 1 2. At what rate per cent, per annum, will $1200 amount to $1476, in 5 years and 9 months? Ans. 4 per cent. 3. If 834 dollars, at interest 2 years and 6 months, amount to 927 dollars 82* cents, what was the rate per cent, per annum? Ans. 44 per cent. 118 COMPOUND INTEREST. CASE 7. To Jlnd the time, when the principal, amount, and rate per cent, are given. RULE. Divide the whole interest by the interest of the principal for one year, and the quotient will be the time required, or by proportion. EPAMPLES. 1. In what time will 400 dollars amount to 520 dol- lars, at 5 per cent, per annum? Ans.6 years. * 9 400 520 5 400 <& A Y Y ~ as jm i. JL 20JOO 20)120(6 20 : 120 : : 1 : 6 Ans. 2. In what time will 1600 amount to 2048, at 4 per cent, per annum? Ans. 7 years. 3. Suppose 1000 dollars, at 4t per cent, per annum, amount to 1281 dollars 25 cents, how long was it at in- terest? Ans.6Y. 3rao. COMPOUND INTEREST. Compound interest is that in which the interest for one year is added to the principal, and that amount is the principal for the second year; and so on for any number of years. RULE. Find the a mount of the given sum for the first year by simple interest, which will be the principal for the second year; then find the amount of the principal for the second year for the principal for the third year; and so on for \ any number of years. Subtract the first principal from the amount, and the remainder will be the compound interest required. . EXAMPLES. 1. What is the compound interest of 150 dollars for 5 years, at 4 per cent, per annum? Ans. $32,49 COMPOUND INTEREST. 119 $150 $150 t 4 6 inst 1st year 6|00 int. 1 yr. 156 amount 1st year 6,24 int. 2d year $156 102,24 amount 2d year 4 6,48.9 ink 3d year 6|24 168,72,9 amount 3d year 6,74.9 ink 4th year. 175,47.8 amount 4th year 7,01.9 int. 5th year 6|48.96 182,49.7 amount 5th year 150,00.0 principal 32,49.7 compound int. for 5 years. 2. What is the compound interest of 760 dollars, for 3 years, at 6 dollars per cent, per annum? Ans. $145 17 cts. 2 m.-f- 3. What is the compound inierest of $242 50 els., for 4 years, at 6 per cent, per annum? An.s. $63 65 cents. 4. What is the amount of 1300 dollars, for 3 years, at 5 dollars per cent, per annum, compound interest? Ans. $1504 91 cts. 2 m.-f 5. How much is tha amount of 3127 dollars, for 4 years, at 4i dollars per cent, per annum, compound in- terest? Ans. $3729 OOcts. 5m. Questions. What is interest? What is the principal? What is the rate per cent. What is the amount? How do you proceed when the interest for several years is required? 120 COMPOUND INTEREST. What is to be noted f the interest is required for years and months? When the interest is required for any number of weeks or days, less or more than one year, how do you perform the operation? How do you proceed to find the interest, at 6 per cent, for any number of daj's, as computed at banks ? What is to be observed when the interest is at any other rate than 6 per cent.? How do you proceed, when the principal, amount, and time are given, to find the rate per cent.? How do you find the time, when the principal, amount, and rate per cent, are given ? What is compound interest ? How is compound interest computed? PROMISCUOUS EXERCISES. 1. What is the interest of 620 dollars 25 cents for 5 years, at 5i per cent, per annum? Ans. $170 56 8 m -f- 2 What is the interest of $420, for 1 year, at 7 per cent, per annum? Ans. $29 40. 3 What is the interest of 1450 dollars, for 60 days, at 6 per cent, per annum? Ans. $14 50 cts. 4 What is the compound interest of $626 25, for 3 years, at 5i per cent, per annum? Ans. $103 91.+ 5 What is the interest of $1659 for 3 weeks, at 4 per cent, per annum? Ans. $3 82i.-f- 6 In what time will 500 dollars amount to 1000 dol- lars ut 8 per cent, per annum, simple interest? Ans. 12 years, 6 months. 7 What principal, at interest for 6 years and 6 months, at 2 per cent, per annum, will amount to 250 dollars? Ans. $221 23 cts. 9 m. 8 At what rate per cent, per annum, will $300 amount to $450, in 5 years? Ans. 10 per cent. INSURANCE, COMMISSION, AND BROKAGE. 121 INSURANCE, COMMISSION, AND BROKAGE. INSURANCE, Commission and Brokage,are allowances made to insurers, factors, and brokers, at such rate per cent, as may be agreed on between the parties. RULE. Proceed in the same manner as though you were re- quired to find the interest of the given sum for one year. EXAMPLES. 1 What is the commission on 625 dollars, at 4 dollars per cent? $625 4 Ans. $25,00 2 What is the commission on $1320, at 5 per cent.? Ans. $66. 3 What is the commission on 3450 dollars, at 44 dol- lars percent.? Ans. $155,25. 4 The sales of certain goods amount to 1680 dollars: what sum is to be received for them, allowing 2| dollars per cent, for commission? Ans. $1633,80. 5 What is the insurance of $760, at 6i per cent. ? Ans. $49,40. 6 What is the insurance of 5630 dollars, at 7f dollars per cent.? Ans. $436 32 cts. 5 m 7 A merchant sent a ship and cargo to sea, valued at 17654 dollars: what would be the amount of insurance, at 18| dollars per cent.? Ans. $3310 12i cts 8 What is the brokage on 2150 dollars at 2 per cent.? Ans. $43 9 When a broker sells goods to the amount of 984 dollars 50 cents, what is his commission, at 14 dollar per cent.? Ans. $12 30i cts.-f- 10 If a broker buys goods for me, amounting to 1050 h 122 DISCOUNT. dollars 75 cents, what sum must I pay him, allowing him 14 per cent.? Ans. $24 76 cts. 1 m.+ Questions. What are Insurance, Commission, and Brokage? How do you proceed to find the Insurance, Commis- sion, or Brokage? In what does this rule differ from interest? It takes no account of time. DISCOUNT. DISCOUNT is an abatement of so much money from any sum to be received before it is due, as the remainder would gain, put to interest for the given time and rate per cent. RULE. Find the interest of 100 dollars for the given time at the given rate per cent. Add the interest so found to 100 dollars, then by pro- portion, As the amount of 100 dollars for the given time, Is to the given sum, So is 100 dollars, To the present worth. If the discount be required, subtract the present worth from the given sum, and the remainder will be the dis- count. NOTE. When discount is made without regard to time, it is fouuti precisely like the interest for one year. EXAMPLES. 1 What is the present worth of 420 dollars, due in 2 years, discount at 6 per cent, per annum? Ans. $375. DISCOUNT. 123 $ $ $ $ r 6 112 :420:: UK) .375 2 12 100 112 2 What is the present worth of 850 dollars, due in 3 months, at 6 per cent, per annum? Ans. $837 43| cts.-f 3 What is the discount of 645 dollars, for 9 months, at 6 per cent, per annum? Ans. $27 77i cts. 4 What is the present worth of 775 dollars 50 cents, due in 4 years, at 5 per cent, per annum? Ans. $646,25. 5 What is the present worth of 580 dollars, due in 8 months, at 6 per cent, per annum? Ans. $557,69.-|- 6 What is the present worth of 954 dollars, due in 3 years, at 4i per cent, per annum? Ans. $840 52 cts. 8 m.+ 7 What is the discount of 205 dollars, due in 15 months, at 7 per cent per annum? Ans. $16 49 cts. 5 m.+ 8 Bought goods amounting to 775 dollars, at 9 months' credit: how much ready money must be paid, allowing a discount of 5 per cent, per annum? Ans. $746 98 cts. 7 m. 9 I owe A. to the value of 1005 dollars, to pay as fol- io^: viz. 475 dollars in 10 months, and the remainder in To months; what is the present worth, allowing dis- count at 6 per cent, per annum? Ans. $945 40 cts. 4 m. 10 What is the difference between the interest of 2260 dollars, at 6 per cent, per annum, for 5 years, and the discount of the same sum for the same time and rate percent.? Ans. $156 46 cts. 2m.-4- 124 EQUATION OF PAYMENTS. 11 What is the discount of 520 dollars, at 5 per cent.? $520 5 $26,00 Ans. % 12 How much is the discount of $782, at 4 per cent.? Ans. $31, 28 13 What is the discount of 476 dollars, at 3 per cent.? Ans. $14,28. 14 Bought goods on credit, amounting to 1385 dol- lars : how much ready money must he paid for them, if a discount of 6 per cent, be allowed? Ans. $1301,90. 15 I hold A.'s note for 650 jdollars ; but I agree to al- low him a discount of 4 per cent, for present payment: what sum mustl receive? Ans. $620,75. Questions. What is discount? What is first to be done ? After having found the interest of 100 dollars, at tht given time and rate per cent., what is next to be done ? After having added the interest so found to 100 dol- lars or pounds, by what rule do you work to find the dis- count? When discount is made without regard to time, how is it found? EQUATION OF PAYMENTS. EQUATION is a method of reducing several stated times, at which money is payable, to one mean, or equa- ted time, when the whole sum shall be paid. RULE. Multiply each payment by its time, and divide the sum of all the products by the whole debt, the quotient will be the equated time. EQUATION OF PAYMENTS. 125 Proof. The interest of the sum payable at the equa- ted time, at any given rate, will equal the interest of the several payments for their respective times. EXAMPLES. 1 C. owes D. 100 dollars, of which the sum of 50 dollars is to be paid at 2 months, and 50 at 4 months ; but they agree to reduce them to one payment; when must the whole be paid? Ans. 3 months. 50X2=100 50x4=200 100)300(3 months 2 A merchant hath owing to him 300 dollars, to be paid as follows : 50 dollars at 2 months, 100 dollars at 5 months, and the rest at 8 months ; and it is agreed to make one payment of the whole; when must that time be? Ans. 6 months. 3 F. owes H. 2400 dollars of which 480 dollars are to be paid present, 900 dollars at 5 months, and the rest at 10 months; but they agree to make one payment of the whole, and wish to know the time? Ans. 6 months. 4 K. is indebted to L. 480 dollars which is to be dis- charged at 4 several payments, that is i at 2 months, i at 4 months, i at 6 months, and i at 8 months; but they agreeing to make one payment of the whole, the equa- ted time is therefore demanded? Ans. 5 months. 5 P. owes Q. 420 dollars, which will be due 6 months hence, but P. is willing to pay him 60 dollars now, pro- vided he can have the rest forborn a longer time : it is agreed on; the time of forbearance therefore is required? Ans. 7 months. 6 A merchant bought goods to the amount of 2000 dollars and agreed to pay 400 dollars at the time of pur- chase, 800 dollars at 5 months, and the rest at 10 months; but it is agreed to imike one payment of the whole; what is the mean or equated time? Ans 6 months. 11* 126 BARTER. BARTER. BARTER is the exchanging of one kind of goods for another, duly proportioning their values, &c. RULE. The questions that come under this head, may be done by the compound rules, the Rule of Three, or Practice, as may be most convenient. EXAMPLES. 1 A country storekeeper bought 150 bushels of salt, at 56 cents per bushel ; and is to pay for it in corn, at 33* cents per bushel ; how much corn will pay for the salt? ct. ct. lu. bu. As 33* : 56 : : 150 : 252 OR Ct. 56 334)8400 150 3 3 1|00)252K>0 252 bushels of corn. Cost of the salt. 8400 cts. 2 How much wheat, at 1 dollar 25 cents per bushel, will pay for 35 sheep, at 2 dollars 25 cents a piece? Ans.63bush. 3 How much sugar, at 9 cents per Ib. will pay for i dozen pair of shoes, at 1 dollar 75 cents per pair? Ans. 233* Ibs. 4 Row much tea, at 80 cents per Ib. will pay for 560 Ibs. of pork, at 5 cents per Ib.? Ans. 35 Ibs. 5 Bought 4 hats for 3 dollars 50 cents; 4 dollars; 4 dollars 50 cents ; and 5 dollars how much corn at 32 cents will pay for them? Ans. 53 bush. 4 qts. 6 A. has 420 bushels of corn, which he barters with B. for cats, and is to receive 4 bushels of oats for 3 of corn how many bushels of oats must A receive? Ans. 560 bush. BARTER. 127 7 A boy bartered 735 pears for marbles, giving 5 pears for 2 marbles how many marbles ought he to have received? Ans. 294 marbles. 8 A boy exchanges marbles for pears, and gives 2 marbles for 5 pears how many pears should he receive for 294 marbles? Ans. 735 pears. 9 A farmer bartered 3 barrels of flour, at 5 dollars 25 cents per barrel, for sugar and coffee, to receive an equal quantity of each ho\v much of each must he receive, admitting the sugar to be valued at 9 cents per Ib. and the coffee at 14 cents? Ans. 68i Ib. nearly. 10 A bartered 42 hat?, at 1 dollar 25 cents per hat, with B. for 50 pair of shoes, at 1 dollar 12i cents per pair who must receive money, and how much? Ans. B. $3,75. 11 Sold 75 barrels of herrings, at 2 dollars 75 cents per barrel, for which I am to receive 75 bushels of wheat, at 1 dollar 8 cents per bushel, and the residue in money how much money must I receive? Ans. 125 dolls. 25 cts. 12 Sold 35 yards of domestic, at 20 cents* per yard, and am to receive the amount in apples, at 25 cents per bushel how many bushels must I have? Ans. 28 bush. 13 Gave 35 yards of domestic for 28 bushels of ap- ples, at 25 cents per bushel what was the domestic rated at per yard? Ans. 20 cts. 14 What is rice per Ib. when 340 Ib. are given for 4 yards of cloth, at 4 dollars 25 cents per yard? Ans. 5 cts. 15 Gave in barter 65 Ibs. of tea for 156 gallons of rum, at 33i per gallon what was the tea rated at? Ans. 80 cts. per Ib. 16 Q. has coffee worth 16 cents per pound, but in barter raised it to 18 cts.; B. has broad cloth worth 4 dollars 64 cents per yard what must B. raise his cloth to, so as to make a fair barter with Q? Ans. $5,23. 128 LOSS AND GAIN. 17 B. had 45 hats, at 4 dollars per hat, for which A. gives him 81 dollars 25 cents in cash, and the rest in pork, at 5 cents per Ib ; how much pork will be required ? Ans. 1975 Ib. 18 Two merchants barter; A. receives 20 cwt. of cheese, at 2 dollars 87 cents per cwt.; B. 8 pieces of linen, at 9 dollars 78 cents per piece; which of them must receive money, and how much? Ans. A. $20,84. 1>9 If 24 yards of cloth be given for 5 cwt. 1 qr. of tobacco, at 5 dollars 7 cents per hundred; what is the cloth rated at per yard? Ans. $1. 109. 20 A. barters 40 yards of cloth, at 98 cents per yard, with B. for 284 Ibs. of tea, at 1 dollar 53 cents per Ib. ; which must pay balance, and how much ? Ans. A. $4,405. 21 A has 74 cwt. of sugar, at 8 cents per Ib., for which B. gave him 124 cwt. of cheese , what was the cheese rated at per Ib.'? Ans. $. 048. 22 What quantity of sugar, at 8 cts. per Ib. must be given in barter for 20 cwt. of tobacco, at 8 dollars per cwt.? Ans. 17 cwt. 3 qrs. 12 Ib. 23 P. has coffee, which he barters with Q. at 11 cts. per Ib. more than it cost him, against tea, which stands Q. in 1 dollar 33 cents the Ib., but he puts it at 1 dollar 66 cents ; query, the prime cost of the coffee ? Ans. $. 443+ LOSS AND GAIN. By Loss AND GAIN, merchants and dealers compute their gains or losses. RULE. Work by the Compound Rules, by Proportion, or in Practice, as may be most convenient. LOSS AND GAIN. '129 EXAMPLES. 1 Bought 1234 Ibs. of coffee, at 12j cts. per lb., and sold the whole for 160 dollars ; did I lose or gain by it, and how much? Ans. gained $5,75. 2 Bought 120 dozen knives, at 2 dollars 50 cents per dozen, and sold them at 18| cents a piece; did I gain or lose, and how much? Ans. lost $30. 3 Bought 1234 yards of muslin, for 17i cents, and sold it at 20 cents per yard; what was the gain? Ans. $30,85. 4 Bought 10 chests of tea, each 63 Ibs. neat, for 600 dollars, and retailed it at 87i cents per lb.; did I gain or lose, and how much ? Ans. lost 48 dol. 75 ct. 5 Gave 285 dollars 25 cents for 4564 Ibs. of bacon, and sold it for 365 dollars 12 cents ; what was the gain per lb? Ans. . 1| cts. 6 Bought 1234 yards of muslin, for 246 dollars 80 cents, and sold it for 215 dollars 95 cents; what did I lose per yard? Ans. $. 2i cts. 7 Gave 25 cts. per bushel for corn, and sold it at 28 cents; what is the gain per cent.? Ans. 12 dolls, per 100 dolls. 8 Sold corn at 25 cts. per bushel, and 4 cts. loss; what was the loss per cent.? Ans. $13,79. 9 Bought 13 cwt. 25 Ibs. of sugar, for 106 dollars, and sold it at 9i cts. per lb. ; what did I gain per cent.? Ans. 32 dolls. 73 cts. 10 Bought 128 gallons of wine for 150 dollars, and retailed it at 20 cts. per pint; what was the gain per cent.? Ans. 34 dolls. 40 cts. 11 Sold a quantity of goods, for 748 dollars 66 cents, and gained 10 per cent; what did I give for them.? Ans. 680 dols. 60 cts. F 2 130 LOSS AND GAIN. dols. dols. dols. 100 110 : 100: : 748,66 10 100 110 110)74866,00($680,60 12 Sold goods to the amount of $'1234, and gained at the rate of 20 per cent.; what was the prime cost? Ans. $1028,33* 13 Soldji quantity of goods, for $475, and at a loss of 12 per cent.; what did 1 give for them? dols. dols. dols. 100 88 : 100 : : 475 12 100 88 88)47500(539,77+Ans. 14 Sold hats to the amount of $136, at 20 per cent, loss; \vhat was the first cost? Ans. $170. 15 Laid out $755 in salt; how much must I sell it for, so as to gain 12 per cent.? 12 100 As 100 : : 112 : 755 : : 845,60 Ans. 16 Bought 32 yards of mole skin for 128 dollars; what must I sell it for per yard, so as to gain 20 per cent.? Ans. 4 dols. 80 cts.-j- 17 Bought 17 yards of silk for 21 dollars; how much per yard /mist I Detail it for, and gain 25 per cent. ? Ans. 1 dol. 54 cts.-f- 18 Bought 64 ya.rds of muslin for 1C dollars 50 cents, )>ut proving a bad bargain, I am willing to lose 8 per cent; what must I sell it at per yard? Ans.19cts.4m.-f- 19 When hats are bought at 48 cents, and sold at 5 i cents; what is the gain per cent.? Ans. 12 i LOSS AND GAIN. 131 20 If, when cloth is sold for 84 cents per yard, there is gained 10 per cent.; what will be the gain percent, when it is sold for 1 dollar 2 cents per yard? Ans. 33 dols. 68 cts.+ 21 Bought a chest of tea, weighing 490 Ibs. for $122 50 ct. and sold it for $137 20 cents; what was the profit on each lb.? Ans. 3 cts. 22 Bought 12 pieces of white cloth, for 16 dollars 50 cents per piece; paid 2 dollars 87 cents a piece for dying; for how much must I sell them each, to gain 20 per cent. ? Ans. 23 dols. 244. 23 If 28 pieces of stuff be purchased at 9 dollars 00 cents per piece, and 10 of them sold at 14 dollars 40 cents, and 8 at 12 dollars per piece; at what rate must the rest be disposed of, to gam 10 per cent. by the whole? Ans. 5 dols. 568. 24 Sold a yard of cloth for 1 dollar 55 cents, by which was gained at the rate of 15 per cent.; but if it had been sold for 1 dollar 72 cents ; what would have been the gain per cent.? Ans. 27 dols. 69-j- 25 If, when cloth is sold at $. 935 a yard, the gain is 10 dollars per cent. ; what is the gain or loss per cent., when it is sold at 80 cents per yard ? *Ans. 5 dollars 88+loss. 26 A draper bought 100 yards of broad cloth, for which he gave $56 I desire to know how he must sell it per yard, to gain $19 in the whole? Ans. 75 ct. per yard. 27 A draper bought 100 yards of broad cloth for $56; I demand how he must sell it per yard, to gain $15 in laying out $100? Ans. 64 ct. 4 in. 28 Bought knives at 11 cents, and sold them at 12 cents; what will I gain by laying out 100 dollars in knives? . Ans. 9 dols. 09+ 29 Bought knives at 11 cents, and sold them at 12 cents; what did I gain by selling to the amount of 100 j dollars? . Ans. 8 dols. 333+ jj 132 FELLOWSHIP. 11 - 30 If by selling 1 Ib. of peppejr for 10i cents, there are 2 cents- lost; how much is the loss per cent.? Ans. 16 dols, 31 A merchant receives from Lisbon, 180 casks of raisins, which stands him in here 2 dollars 13 cents each, and by selling them at 3 dollars 68 cents per cwt., he gains 25 per cent.; required the weight of each cask, one with another? Ans. 81 Ib. * FELLOWSHIP. FELLOWSHIP is a method by which merchants and others adjust the division of property, loss, or gain, &,c., in proportion to their several claims. CASE 1. SIMPLE FELLOWSHIP. When the claims are in proportion to the amount of stock, labor, &c., without regard to time. RULE. (By Proportion.) As the whole amount of stock or labor, Is to each man's portion, So is the whole property, loss, or gain, To each man's share of it. Proof. The sum of all the shares must equal the whole gain, &,c. EXAMPLES. 1 Two men bought a stock of goods for 480 dollars, of which A. paid 320, and B. 160. They gained 128 dollars by the transaction ; what was the share of each ? Ans. A. received 85 dols. 33J cts. and B. 42 dollars 661 cts. $ $ $ $ ct. Proof A's. stock $320 As 480 : 320 : : 128 : 85,33* $85,331 H'p. stock 160 42,601 .^ . & (ft (JN (ft s%4- _ Whole st'k 480 As480:160::128 : 42,661 128,00 FELLOWSHIP. 133 2 Three workmen having undertaken to do a piece of work for 275 dollars, agreed to divide their profits in proportion to the amount of labor each one performed. M. labored 50 days, N. 65 days, and O. 85 days : what was the share of each? Ans. M. received 68 dols. 75 cts. ; N. 89 dols. 37 i cts. ; aridO. 116 dols. 87 i cts. 3 A merchant being deceased, worth 1800 dollars, is found to owe the following sums: to A. 1200 dollars, to B. 500 dollars, to C. 700 dollars : how much is each to have in proportion to the debt? Ans. A. 900 dols., B. 375 dols, and C. 525 dols. 4 Three drovers pay among them 60 dollars for pas- ture, into which they put 200 cattle, of which A. had 50, B. 80, and C. 70 : I would know how much each had to pay? Ans. A. 15 do!?., B. 24 dols., C. 21 dols. 5 A man failing, owes the following sums: to A. 120 dollars, to B. 250 dollars 75 cents, to C. 800 dollars, to D.208 dollars 25 cents j and his whole effects were found to amount to but 650 dollars : what will each one receive in proportion to his demand ? Ans. A. $ 88.73.+ C. $221.84.+ B. $185.42.+ D. $153.99+ 6 A bankrupt is indebted to A. 500 dollars 37 i cents to B. 228 dollars to C. 1291 dollars 23 cents to D. 709 dollars 40 cents ; and his estate is worth 2046 dol- lars 75 cents: how much does he pay per cent., and what does each creditor receive? Ans. He pays 75 per cent., and A. receives 375 dollars 27! cts.; B. 171 dols.,- C. G68 dols. 42| cts.; and D. 532 dols. 5 cts. 7 If a man is indebted to A. 250 dollars 50 cents, to B. 500 dollars, to C. 349 dollars 50 cents, but when he comes to make a settlement, it is found he is worth but 960 dollars, how much will each one receive, if it be in proportion to their respective claims? (A. $218 61 cts. 8 m.+ Ans. {B. $436 36 cts. 3 m.+ (C. $305 01 ct. 8 m.+ 12 134 FELLOWSHIP. CASE 2. COMPOUND FELLOWSHIP. When the respective stocks are considered with rela- tion to time. RULE. (By Proportion.) Multiply each man's stock by its time; add the several products together; then: As the sum of the products Is to each particular product, So is the whole gain or loss - To each man's share of the gain or loss. EXAMPLES. 1 Three merchants traded together; A put in 120 dollars for 9 months, B. 100 dollars for 16 months, and C. 100 dollars for 14 months, and they gained 100 dol- lars; what is each man's share? $ mo. A's. stock 120 X 9 = 1080 B's. stock 100 X 16 = 1600 C's. stock 100 X 14 = 1400 Sum 4080 Sum. Prod. $ $ As 4080 : 1080:: 100 : 26,47+ A's. share. As 4080 : 1600 ::100 : 39,214 f B's. share. As 4080 : 1400 ::100 : 34,31+ C's, share. 2 Three men traded together; L. put in 88 dollars for 3 months, M. 120 dollars for 4 months, and N. 300 dol- lars for 6 months ; they gained 184 dollars: what will each man receive of the gain? L. $ 19 09 cts. 4 m Ans. M. $ 34 71 cts. 6 m. N. $130 18 cts. 8 m. VULGAR FRACTIONS. 135 3 Two merchants entered into partnership for 16 months : A. put in at first $600, and at the end of 9 nonths put in $100 more; B. put in at first $750, and at lie end of 6 months took out $250, at the close of the time their gain was $386, what was the share of each? Ans. A's. share was $200,794; B's. share was $185,20. 4 A., B., and C., made a stock for 12 months; A. put in at first $873,60, and 4 months after he put in $96,00 more ; B. put in at first $979,20, and at the end of 7 months he took out $206,40; C. put in at first $355,20, and 3 months after he put in $206,40, and 5 months after that he put in $240,00 more. At the end of 12 mouths, their gain is found to be $3446,40 ; what is each man's share of the gain? (A's. share is $1.334,821 Ans. ?B's. - - $1271,61i-f (CTs. - - $839,96 Questions. What is Fellowship? By what rule are its operations performed? When is Fellowship simple? When is it compound? In what respect is Fellowship compound? Ans. The proportion is compound : that is, the divi- sion of property, gain, &c., is founded on the compound proportion of the stock and time. VULGAR FRACTIONS. A VULGAR FRACTION is a part, or parts of a unit ex* pressed by two numbers placed one above the other with a line between them. As ^, *,&c. The number below the line is the denominator, the number above the line is the numerator. The denominator denotes the number of parts into which the unit is divided. L36 VULGAR FRACTIONS. The numerator shows how many of those parts are to taken. Fractions are either proper, improper, or compound. A proper fraction is one whose numerator is less than its denominator, as f- or y. An improper fraction is one whose numerator is greater than its denominator, as | or |. A compound fraction is a fraction of a fraction, as \ of 3-, or | of J. A mixed number is a whole number and a fraction. REDUCTION OF VULGAR FRACTIONS. CASE 1. To reduce a fraction to its lowest terms. RULE. Divide the terms by any number that will divide both without a remainder, and divide the quotient in the same manner, and so on till no number greater than one will divide them : the fraction is then at its lowest terms. EXAMPLES. 1. Reduce T \ 4 T to its lowest terms. == result. 2. Reduce - to its lowest terms. Res. \ 3. Reduce ~~ to its lowest terms. Res. 4. Reduce JJ to its lowest terms. Res. |* NOTE. When a divisor cannot readily be found, divide the denominator by the numerator, and that divisor by the remainder, and so on, till nothing remain: the last divisor is the common measure of the two numbers j with which proceed as before. 5. Reduce -f/ 7 to its lowest terms. Res. f- VULGAR FRACTIONS. 137 5 Reduce ~~j to its lowest terms. Rss.*|. 85 85)136(1 Here 17 being the last divisor, 85 is the common measure of 85 - and 136. 51)85(1 51 34)51(1 j 85 ) (5 34 17 - = J- I 136) (8 17)34(2 34 6. Reduce j to its lowest terms. Res. . 7. Reduce \\\ to its lowest terms. Res. f . 8. Reduce |f \\ to its lowest terms. Res. i-^. ~ CASE 2. To reduce a mixed number to an improper fraction. RULE. Multiply the whole number by the denominator, and add the numerator to the product for the numerator of the improper fraction, and place the denominator under it. EXAMPLES. 1. Reduce 12 J to an improper fraction. 12 Res. '. 112 Nine 12's are 108; add - 4 makes 112 ninths. 9 2. Reduce 17 ^to an improper fraction. Res. ! f 2 3. Reduce 45 | to an improper fraction. Res. I | 7 4. Reduce 24 ~ to an imp roper fraction. Res. 4 y T 12* 138 . VULGAR FRACTIONS. CASE 3. To reduce an improper fraction to its proper value. RULE. Divide the numerator by the denominator, and the quotient will be the whole number; the remainder, if any, will be the numerator of tho fraction. EXAMPLES. 1 Reduce Y to its proper value. Res. 3 J. 5 2 Reduce l J 2 to its proper terms. Res. 12 J. 3 Reduce l ~ 2 to its proper terms. Res. 17 |. 4 Reduce 4 T y to its proper terms. Res. 24 ~. CASE 4. To reduce several fractions to other fractions having a common denominator, and retaining their value. RULE. Multiply each numerator into all the denominators but its own, for the respective numerators; and all the denominators together, for a common^ denominator. EXAMPLES. 1 Reduce f- J and f to a common denominator. Res. 4| f|, and ?}. 2X4X6=48) 3X3X6=54V Numerators 5X3X4=60) 3X4x6=72, common denominator. Then we have If for f ; || for J, and 4| for f . Reduce each new fraction to its lowest terms, and the result will prove the work to be right. 2. Reduce J, f , and ^, to a common denominator 7> p<3 216 240 J 1 6 8 1168 2JJJ ana 3. Reduce J, |, f , and T \, to a common denominator. T> Ae 216 288 360 nn ,1 252 , ^ 1X6S - 43 2> 432? 43 2> attd 4 3; 2' 4. Keduce |, ^, |, and , to a common denominator. [_>, lv , R3 480 432 Mn J 50 \ ixes. - f .,0, y-oo, -7-30-5 ^nci y^-^. VULGAK TRACTIONS. 139 NOTE. It is often convenient to use the least possible common denominator; to find which, divide the denomi- nators by any number that will divide two or more of them without a remainder, setting down those that would have remainders; then multiply all the divisors and all the quotients together. 5 6 8 1 1 51713 4X3X2X5X7X3=2520 common denom. Which may be divided separately by 2, 3, 4, 5, 6, 7, 8, and 9, without a remainder. 1 1 EXAMPLES. 5 Find the least common denominator for J, J, T 5 j, and ~, and compute their equivalent fractions. Res --- -i 2 - --- --*. 240 com. denom. 00 X 3=180 30X 7=210 20X11=220 15 X 5= 75 12 X 9=108 6 Reduce J, |, |, T \, and ~, to their least common denominator. Res. T 9 ~* V^U r^l- i4^, and 144. 2(7 7 Reduce f , T \, T 9 , |xjto their least common denom- natnr Poo 1 5 16 8" 35 on J 110 nator. Kes. 240, 24^, ^ T o and ?T . 8 Reduce the above fractions to a common denomina- tor, by the general rule, Case 4. 19200 21504 3-OT20J - 14080 3*120' 5X10X16X24=19200 7X 8x16x24=21504 9X 8X10X24=17280 11 X 8X10X16=14080 140 VULGAR FRACTIONS. CASE 5. To reduce a compound fraction to a simple one. RULE. Multiply the numerators together for a new numera- tor; and the denominators together for a new denomina- tor. EXAMPLES. 1 Reduce of of to a single fraction. Res. T . 3X5X9 135 9 4X6X10 240 16 2 Reduce of ~ of ~ to a single fraction. Res. T y T . 3 Reduce j of of J to a single fraction. Res. -/^. 4 Reduce ~J of f- of i- to a single fraction. Res. T S T . CASE 6. To reduce a fraction of one denomination to the fraction of another denomination, but greater, retaining the same value. RULE. Multiply the denominator of the fraction by the num- ber of that denomination which it takes to make one of the next, and so on to the denomination required, and place the numerator of the given fraction over it. EXAMPLES. 1 Reduce f of a quart to the fraction of a bushel.* qt. 2 2 1 = Result, 1 T of a bushel. 3X8X4=9(5 48 2 Reduce J of an ounce, Troy, to the fraction of a pound. Res. T 3 , or ~ of a pound. 3 Reduce j of a nail to the fraction of a yard. Res. \, or - 2 ~ of a yard. 3 Reduce | of a perch to the fraction of an acre. Res. * That is, what part of a bushel are two-thirds of a quart ? VULGAR FRACTIONS. 141 4 Reduce ~j of a pint to the fraction of a hogshead Res. T ^ T of a hhd. CASE 7. To reduce ike fraction of one denomination to the frac- tion of another, but less, retaining the same value. RULE. Multiply the given numerator by the parts of that be- tween it and that to which it is to be reduced, and place the product over the given denominator for the fraction required. EXAMPLES. 1 Reduce j 7 of a bushel to the fraction of a quart. Res. of a quart. 2 Reduce -^ of a yard to the fraction of a nail. Res. J of a nail. 3 Reduce y/j ^ f an acre to tne fraction of a perch. Res. I of a perch. 4 Reduce 7 of a hogshead to the fraction of a pint. Res. ~ of a pint. 5 Reduce yJg-^ of a day to the fraction of a minute. Res. TT of a minute. CASE 8. To reduce a fraction to its proper value or quantity, in whole numbers. RULE. Multiply the numerator by the parts of the integer, and divide by the denominator. EXAMPLES. 1 Reduce J of a yard to its proper quantity. Res. 3 qr. 2 na. 7 eighths of a yd. 4 eighths of a qr. 4 4 8)28 8)16 3 i quarters 2 nails 142 VULGAR FRACTIONS. 2 Reduce of a pound, avoirdupois, to its proper quantity. Res. 8 oz. 14 dr. 3 Reduce J of a pound, Troy, to its proper quantity. Res. 9oz. 4 Reduce - of a mile to its proper quantity. Res. 4 fur. 125 yd. 2 fLlfinch, 5 Reduce -f-g of an acre to its proper quantity. Res. 1 rood, 30 pf ich. 6 Reduce J of a dollar to its proper quantity. Res. GO cents. 7 Reduce ~ of a pound to its proper value. Res. 6s. 8d. 8 Reduce j~ of a year (365 days) to its proper quan- tity. Res. 225 days. 9 Reduce J of a tun to its proper quantity. Res. 3 hhd. 7 gal. 10 Reduce J of a ton to its proper quantity. Res. 15 cwt. 2 qr. 6 Ib. 3 oz. 8| dr CASE 9. To reduce a given quantity to a fraction of any greate-i denomination of the same kind, RULE. Reduce the given quantity to the lowest denomination mentioned for a numerator; and th^ integer to the same denomination, for a denominator. EXAMPLES 1 Reduce 3 qr. 2 na. to the fraction of a yard. Res. J of a yard, qr. na. 3 2 4 yd. 141 (7 1X4X4=16) (8 2 Reduce 2 roods 20 perches to the fraction of an acre. Res. | of an acre. VULGAR FRACTIONS. 143 3 Reduce 6 furlongs 16 poles to the fraction of a mile. Res. j of a mile. 4 Reduce 9 ounces, Troy, to the fraction of a pound. Res. f of a pound. 5 Reduce 7 hours 12 minutes to the fraction of a day. Res. -^g- of a day. ADDITION OF VULGAR FRACTIONS. RULE. Reduce the given fractions, if necessary, to single ones, or to a common denominator; add all the numerators together, and place the sum over the com- mon denominator. EXAMPLES. f- 4 W - 7 ft 3 A ry 9 " To 3 I 7 t- 8 19 or 2 f 26 T 8 o= 8 28 NOTE 1. When the fractions are of different denom- inators, reduce them to a common denominator, and proceed as above. (See Note, page 139.) 4 Add J, , T i, T 5 and- A- together. Result 3^\. 240 60X 3=180 30X 7=210 20X11=22C 15X 5= 75 12X 9=108 IT 73 2T7 ^40" - 2T7 5 Add f , f , J, -& and Vi together. 6 Add |, T \, T V, and II together, 7 Add |, T 3 , and T 4 together. 8 Add f and f together. 9 Add -j^, ii and | together Res. 4ii Res. 2/-V i||= r Vo Res. U 8 T Res. 2 f VV 144 VULGAR FRACTIONS. NOTS 2. When mixed numbers occur, place them as in examples 2 and 3; proceed with the fractions as di- rected in Note 1 ; and if they amount to one or more integers, carry them to the integers, and proceed as in simple addition. 10 Add 5|, 61, and 41 together. Res. l7, 24 8x2=16 3x7=21 12X1=12 49) V =2 24 \ Ti- 11 Add 21 and 3J together. 12 Add 74 and 5 .together. 13 Add 171 and | together. 14 Add 4, 6, 9, and { 15 Add 5, 7|, | and T together. together. Result 61. Res. 12|J Res. 181. Res. 2011. Res. 13vY ' ' 2" NOTE 3. When compound fractions are given, re- duce them to single fractions, and proceed as before. 16 Add T V of 11, T C T of T \, and T 7 j of f together. Res. j" 9240 common denominator 77X99=7623 60X^8=-2880 77X35=2695 131981 9240) 17 Add | of 1 and * of 1 together. 18 Add 1|, J of 1, and 9^ together. Res Res. 11 . - 19 Add If -, 6J, of 1, and 71 together. Res. 16 T 7 2\. NOTE 4. When the given fractions are of several denominations, reduce them to their proper values or quantities, and add as in the following example. 13 G VULGAR FRACTIONS. 145 20 Add of a pound, to ~ of a shilling. Result 15s. 10 r Vl. 15 s. d. } of a 15 6$ ft of a *. 3| 5x2=10 3X3= 9 15 10 T 4 T 19 V =1 15 21 Add 2 of a pound, to J of a shilling. Result 18s. 3d. 22 Add J of a penny, to J- of a pound. Res. 2s. 3d. Iqr. \, 23 Add i Ib. troy, to ^ of an ounce, Res. 6oz. lldwt. 16grs. 24 Add J of a mile, to f$ of a furlong. Res. Gfu. 28p. 25 Add j of a yard, to of a foot. Res. 2ft. 2 in. 26 Add | of a day, to |. of an hour. Res. 8h. SOmin. 27 Add ~ of a week, \ of a day, and \ of an hour together. Res. 2 days, 14 hours, 30min. SUBTRACTION OF VULGAR FRACTIONS. KT7LE. Prepare the given fractions as in Addition; then sub- tract the less from the greater, and place the difference over the common denominator. EXAMPLES. 1 Take | from J. Rem. ^. 2 Take T ^ from f^. Rem. {. 3 Take T \ from T V Rem. |. 4 Take f from f . Rem. s y 35 5x3=15 7X^=14 t 3T 146 VULGAR FRACTIONS. I tf tf ft V . Tf 3 V T T2" 2 IT* 3" 5 12 com. cknom 60 com. denom. 24 "t TOO From J of a pound take ~ of a shilling. 15 com denom. s. d. | of a pound =15 6| [ 5x2=10 !\ of a shilling = 3| | 3x3=9 s. 15 3 T r j Ans. jj Fron* | of a take | of a shilling. Res. 14s. 3d. From I of a Ib. troy, take 1 of an ounce. Res. 8oz. IGdwt. 16grs. From 1 of a yard take I of an inch. Res. 5in. 4. From f of a j take S of | of a shilling. Res. 10s. 7d. Iqr. * MULTIPLICATION OF VULGAR FRACTIONS. RULE. Prepare the given fractions, if necessary; then mill tiply the numerators together for a new numerator, and the denominators together for a new denominator. EXAMPLES. 1 Multiply f by T V Res. T 1 T 12 1 TT7 TT VULGAR FRACTIONS. 147 2 Multiply T V by J. Res. ,Y 3 Multiply f by V- Res - 2 Jf> or 2 T- 4 Multiply 12J by 7f . Res. 96|. 12| = V and 7f=Y; then V X V =' U 9=96 l- 5 Multiply 71 by 8|. Res. 61|. 6 Multiply 41 by 1. Res. T 8 T . 7 Multiply |- "by 13 T V Res. 12i. 8 Multiply I of | by T \ of ft. Res. y T . 9 Multiply 4| by f of J. Res. 2f . 10 Multiply \ of 7 by |. Res. If. 11 Multiply 21 by 1*, and multiply the product by of | of f . Res. | DIVISION OF VULGAR FRACTIONS. RULE. Prepare the given fractions, if necessary, then invert the divisor, and proceed as in Multiplication. EXAMPLES. 1 Divide J by 1. Res. ||. 8X4=32 7X9=63 2 Divide 4 by f . Res. I 3 Divide II by |. Res. Iff 4 Divide H by 4 T 8 . Res. ^ 5 Divide 3JJ- by 9. Res. 1 6 Divide J by 4. Res. y \ 7 Divide 4 by |. Res. 4| 8 Divide 1 of f by f of J. Res. f 9 Divide | of 19 by f of |. Res. 7| 10 Divide 4| by | of 4. Res. 2-^ 11 Divide | of 1 'by f of 7|. Res. T | T 12 Divide 5205^ by J of 91. Res. 71| 148 VULGAR FRACTIONS. PROPORTION IN VULGAR FRACTIONS. RULE. State the question, (as in page 89) reduce each term to its simplest form, invert the first term or terms, and proceed as in Multiplication of Vulgar Fractions EXERCISES. 1 If J yd. cost $f ; what will | yd. cost? Ans. 50c yd. yd. D. 3 . 3 . . 5 r PI-i/in 4 v 3 y 5 - " 'I 1 - ^A // i J j* - I1( :I1 3- A j A -g- Ta'o" *n? *Jv Ci9i 2 If 1 bu. cost $1 ; what will f bu. cost? Ans. $2,80. bu. bu. D. 1 : i :: 1. Then <X JX J-=y 1 ?=$2f J=$2,80. 3 If A owned | of a toll-bridge, and sold f of his share for $681 ; what is the whole value? Ans. $1520. I of | : V : : 6 f 4 ' that is, -/ : 1 : : "f 4 . Then 4 If I barter 5| cwt. of sugar at 6| cts. per lb., for indigo at $4 T 5 per lb. ; how much indigo must I re- ceive? Ans. 10 lb. 5 oz. 2j dr.-f D. cts. cwt. cts. cts. lb. 4f- : 6| : : 5-; thatis 6 ^ : y : : 5 V 6 - Then 16 4 9' i| T X V X 5 V 3 6 = YAW = 10 ^. 5 oz. 2 dr. J^J. f* V 5 If the cent roll weighs 6| oz., when wheat is 68| cents per bu. ; what is the cost of wheat per bu. when it weighs 4 oz? Ans. $1,03}. oz. oz. cts. 41 : 61 : : 68|; then ^X V X a j s = 4 }|| = 103i. 64 4 2_5 2_5 2 ^T 5 6 How many men will reap 417| acres in 121 dayb, if 5 men reap 521 i n 61 days? Ans. 20. a. a. men. 1O1 . l 5 V 2 V 2 8 8 y 2 5 V 5 1^1 . bj 2T X 25X J X TXy NOTE. Tn multiplying, omit the numbers that 420 *"i both the upper and lower series. DECIMAL FRACTIONS. 149 DECIMAL FRACTIONS. A decimal fraction is a fraction whose denominator is 1, w^th as many cyphers annexed as there are figures in I the numerator, and is usually expressed by writing the numerator only with a point prefixed to it: thus T \, T 7 ^, T 6 o 2 <ro> are decimal fractions, and are expressed by .5, .75, .625. A mixed number, consisting of a whole number and a decimal, as 25 T \, is written thus, 25.5. As in numeration of whole numbers the values of the figures increase in a tenfold proportion, from the right hand to the left; so in decimals, their values decrease in the same proportion, from the left hand to the right, which is exemplified in the following TABLE. Whole numbers. Decimals. NOTE. Cyphers annexed to Decimals, neither in- crease nor decrease their value; thus, .5, .50, .500, be- m g T 5 o> T S OO> T\Yo> are f tne same value: but cyphers prefixed to decimals, decrease them in a tenfold propor- I tion; thus .5, .05, .005; being T \, T f^, T \^, are of dif- ferent values. 150 DECIMAL FRACTIONS. ADDITION OF DECIMALS. RULE. Place the given numbers according to their values, viz. units under units, tenths under tenths, &c., and add as in addition of whole numbers; observing to sot the point in the sum exactly under those of the given num- bers. EXAMPLES. .12 2.16 .14 .1 .15 .134 3.45 .24 4.12 .75 .21 40.02 .122 15.4 .92 743 35.4 .36 76.36 63.25 345 36.1 .141 120.16 25. .002 125.32 .567 425.04 4. 1.554 242.45 6 Add .5, .75, .125, 496, and .750 together. 7 Add .15, 126.5, 650.17, 940.113, and 722.2560 together. 8 Add 420., 372.45, .270, 965.02, and 1.1756 to- gether. SUBTRACTION OF DECIMALS. RULE. Place the numbers as in addition, with the less under the greater, and subtract as in whole numbers; setting the point in the remainder under those in the given numbers. EXAMPLES. .4562 56.12 .4314 5672.1 32.456 .316 1.242 .312 321.12 1.33 .1402 54.878 6 From 100.17 take 1.146. 7 From 146.265 take 45.3278. 8 From 4560. take .720. DECIMAL FRACTIONS. 151 MULTIPLICATION OF DECIMALS. RULE. Multiply as in whole numbers, and from the right hand of the product, separate as many figures for deci- mals, as there are decimal figures in both the factors. EXAMPLES. 1 Multiply .612 by 4,12 2 Multiply 1.007 by .041. .612 1.007 4.12 .041 2.52144 3 Multiply 37.9 by 46.5 4 36.5 by 7.27 5 29.831 by .952 6 3.92 by 196. 7 .285 by .003 8 4.001 by .004 9 .00071 by .121 1007 4028 .041287 Product 1762.35 265.355 28.399112 768.32 _ .000855 .016004 .00008591 DIVISION OF DECIMALS. RULE. Divide as in whole numbers, and from the right hand of the quotient, separate as many figures for decimals as the decimal figures of the dividend exceed those of the divisor. If there are not so many figures as the rule requires, supply the defect by prefixing cyphers. 152 DECIMAL FRACTION*. EXAMPLES. 1 Divide .863972 by .92 2 Divide 4.13 by 572.4, .92).863972(.9391 572.4)4.130000(.00721+ 828 . 40068 359 12320 276 11448 837 8720 828 5724 92 2996 92 3 Divide 19.25 by 38.5 Quotient .5 4 234.70525 by 64,25 3.653 5 1.0012 by .075 13.34+ 6 .1606 by .44 .365 7 .1606 by 4.4 .0365 8 .1606 by 44. .00365 9 9. by .9 10. 10 .9 by 9. .1 11 186.9 by 7.476 25. NOTE 1. When a whole number is to be divided by a greater whole number, cyphers must be affixed to the dividend, as decimal figures. 12 Divide 3 by 4 Quotient .75 13 275 by 3842 .071577+ 14 210 by* 240 .875 NOTE 2. When any whole number is divided by ano- ther, if there be a remainder, cyphers may be affixed to the dividend, and the quotient continued. 15 Divide 382 by 25 Quotient 15.28 16 13689 by 75 182.52 17 315 by 124 2.5403+ DECIMAL FRACTIONS. 153 REDUCTION OF DECIMALS. CASE 1. To reduce a vulgar fraction to a decimal. RULE. Place cyphers to the right of the numerator, until you can divide it by the denominator, and continue to divide until there is no remainder left; or if it be a number which will never come out without a remainder, until it is carried out to a convenient number of decimal places. EXAMPLES. 1 Reduce j to a decimal. 5)40 .8 Ans. 2 Reduce | to a decimal. Ans. .875. 3 Reduce JJ to a decimal. Ans. .70833.-(- 4 Reduce T 3 T y* to a decimal. Ans. .1762.-f- 5 Reduce ^f to a decimal. Ans. .4566.-J- CASE 2. To reduce any given sum or quantity to the decimal of any higher given denomination. RULE. Reduce the given sum or quantity to the lowest de- nomination mentioned in it. Reduce onet>f that denomination of which you wish to make it a decimal, to the same denomination with the given sum. Divide the given quantity so reduced by one of the denomination of which you wish to make it a decimal ; the quotient will be the decimal required. 154 DECIMAL FRACTIONS. EXAMPLES. 1 Reduce 3s. 6d. to the decimal of a pound. 3a. 6d.= 42 240)42.000(.175 decimals. Ans, 1. =240 240 1800 1680 1200 1200 2 Reduce 2R. 4P. to the decimal of an acre. Answer, .525. 3 Reduce 2 qr. 2 nails to the decimal of a yard. Ans. .625. 4 Reduce 5 minutes to the decimal of an hour. Ans. .08333. 5 Reduce 10 graias to the decimal of an ounce, apothecaries' weight. Ans. .02083.-)- 6 Reduce 2 quarts 1 pint to the decimal of a hogs- head. Ans. .00992.+ CASE 3. To reduce a decimal fraction to its proper value. RULE. Multiply the given fraction continually by the denom- ination next lower than that of which it is a decimal, for the proper value. EXAMPLES. 1 What is the value of .375 of a dollar? Ans. 37cts. .375 100 37.500 10 5.000 2 What is the value of .1361 of a .? Ans. 2s. 8Jd v DECIMAL TRACTIONS. 155 3 What is the value of .235 of a day? Ans. 5 hours, 38 min. 24 sec. I 4 What is the value of .42 of a gallon? Ans. 1 quart, 1.36 pt. 5 What is the value of .253 of a shilling? Ans. 3.036d. G What is the value of .436 of a yard? Ans. 1 qr. 2.976 na. 7 What is the value of .9 of an acre? Ans. 3R. 24P. PROPORTION IN DECIMALS. RULE. State the question as the rule of three, in whole num- bers, only ohserve, when you multiply and divide, to place the decimal points according to the rules of multi- plication and division of decimals. EXA3IPLES. 1 If 4.21b of coffee cost 8s. 2.3J., what cost 639.2511).? Ib. Ib. s. d. s. d. 4.2 : 639.25 : : 8 2.3 : 62 6 9.49 Ans. 2 When 1.4 yard cost 13s. what, will 15 yards come to at the same price? Ans. 6 19s. 3d. 1.71 qr. 3 If I sell 1 qr. of cloth for 2 dollars 34.5 cents, what is it per yard? Ans. $9 38 cts. 4 A merchant sold 10.5 cwt. of sugar, for 108.30 dol- lars, for which he paid 84 dollars 39.12 cents; what did he gain per cwt. by the sale? Ans. $2 27 cts. 7m.-f- 5 How many pieces of cloth, at .20.8 dollars per piece, are equal in value to 240 pieces, at 12.6 dollars per piece? Ans. 145.38-J- pieces. 6 If, when the price of wheat is 74.6 cents per bush- el, the penny roll weighs 5.2 oz., what should it be per bushel when the penny roll weighs 3.5 oz.? Ans. $1 10 cts. 8m.+ QueMion. How do you perform operations in the rule of three in decimals? 156 MENSURATION. COMPOUND PROPORTION, IN DECIMALS. Questions in this rule are wrought as in whole num- bers, placing the points agreeably to former directions. EXAMPLES. 1 If 3 men receive 8.9j for 19.5 days labor, how much must 20 men have for 100.25 days? Ans. 305. Os. 8.2d. i If 19.5: oO.25 days ' - Os ' 2 If 2 persons receive 1.625s. for 1 day's labor, how much should 4 persons have for 10.5 days? Ans. 4. 17s. lid. 3 If the interest of 76.5for 9.5 months, be 15.24. what sum will gain 6 in 12.75 months ? Ans. 22 8s. 9|d. 4 How many men will reap 417.6 acres in 12 days, if 5 men reap 52.2 acres in 6 days? Ans. 20 men. 5 If a cellar 22.5 feet long, 17.3 feet wide, and 10.25 feet deep, be dug in 2.5 days, by 6 men, working 12.3 hours a day, how many days of 8.2 hours, should 9 men take to dig another, measuring 45 feet long, 34.6 wide, and 12.3 deep? Ans. 12 (lays. MENSURATION. MENSURATION is employed in measuring masons' and carpenters' work, plastering, painting and paving; also, for measuring timber in all its forms, and for estimating quantity in length, superfices, and solids, whenever yards, feet, inches, &.C., are employed. The denominations are, foot, inch, second, third, and fourth, 12 Fourths ,' one 1 Third'" 12 Thirds one 1 Second" 12 Seconds one 1 Inch. /. 12 Inches one 1 Foot. Ft. MENSURATION. 157 ADDITION. RULE. Proceed as in Compound Addition. EXAMPLES. Ft. I. " Ft. I. " Ft. I. " '" " 25 G 3 7'2 4 G 17 9 2 3 11 14 2 9 54 3 2 18 11 10 8 9 35 11 10 14 8 22 11 5 4 9 45 10 11 2G 32 14 10 11 10 8 600 19 4 12 4 10 490 14 00 10 2840 132 4 9 4 Four floors in a certain building contain each 1084 feet, 9in. 8"j how many feet are there in ail? Ans. 4339ft. 2 in. 8". 5 There are six mahogany boards, the first measures 27 ft. 3in., the second 25 ft. llin., the third, 23ft. lOin., the fourth 20ft. 9in., the fifth 20ft. Gin., and the sixth 18 feet 5 in.; how many feet do they contain? Ans. 13tift. Sin. SUBTRACTION. RULE. Proceed as in Compound Subtraction. EXAMPLES. Ft. I. " Ft. I. " Ft. I. " '" "" 75 9 9 84 6 4 100 10 8 10 11 14 6 11 72 9 8 97 2 4 6 8 61 2 10 4 If 19ft. lOin. be cut from a board which contains 41ft. 7in. how much will be left? Ans. 21ft. 9in. 5 Bought a raft of boards containing 59621ft. 8in., of which are since sold 3 parcels, each 14905ft. 5in.; how many feet remain? Ans. 14905ft. 5in 14 158 MENSURATION. MULTIPLICATION. CASE 1. When the feet of the multiplier do not exceed 12. RULE. Set the feet of the multiplier under the lowest denom- ination of the multiplicand, as in the following example ; then multiply as in Compound Multiplication, by each denomination of the multiplier separately, observing to place the right hand figure, or number, of each product, under that denomination of the multiplier by which it is produced. EXAMPLES. 1 Multiply 10 feet G inches by 4 feet 6 inches. Product 47 feet, 3 in Ft. I. " 10 6 A table 10 feet 6 inches long, 4 6 and 1 foot wide, will make 10 feet 6 inches, or 10& feet, square meas- 530 are. 42 And 4 feet 6 inches, or 4i feet wide, will make 4d times 10i, or 47 3 47i feet, or 47 feet 3 inches. OR THUS: ft. in. 10 6 44 ft. 5 3 42 47 3 NOTE 1. If there are no feet in the multiplier, sup- ply their place with a cypher; and in like manner sup- ply the place of any other denomination between the highest and lowest. MENSURATION. 159 10ft. Gin. or 101 feet long. 1 i i 1 1 I 1 1 1 1 1 1 1 1 4fimoa 1 OA TYinlfA 4^fi ! | 1 1 1 1 1 1 tlllJtxb J.V/S illrtlVt/ TK^li* and itime 104 make 5i ft. 1 ' 1 1 1 1 1 1 Added, make 474. CD | 1 1 1 1 1 1 1 i i i i i i I Ft. 7. " Ft. I. Ft. I. " " \ 2 Multiply 9 7 by 3 6 Res. 33 6 6 3 3 11 bv 9 5 36 10 7 4 8 6 9 by 7 3 8 62 6 7 9 5 28 10 6 by 3 2 4 92 2 10 6 CASE 2. When the feet of the multiplier exceed 12. RULE. Multiply by the feet of the multiplier as in Compound Multiplication, and take parts for the inches, &,c. EXAMPLES. 1 Multiply 112ft. 3in. 5" by 42ft. 4in. 6" Ft. I. " ' 112 3 5 6X7=42 G73 8 6 J 7 4 I 4715 11 6 '" "" 37 5 1 i 8 6 4 8 ] L 8 6 4758 9 4 G Ft. I. " Ft. 7. " Ft. I. " "/ 2 Multiply 76 7 by 19 10 Res. 1518 10 10 3 127 6 by 184 8 23545 4 71 2 6 by 81 1 8 5777 922 160 MENSURATION APPLICATION. 1 A certain board is 28ft. lOin. 6" long, and 3ft. 2 in 4" wide; how many square feet does it contain? Ans. 92ft. Sin. 10" 6'-. 2 If a board be 23ft. 3in. long, and 3ft. Gin. wide, how many square feet does it contain? Ans. 81ft. 4in. 6 3 A certain partition is 82ft. Gin. by 13ft. 3in.; how many square feet does it contain? Ans. 1093ft. lin. G". 4 If a floor be 79ft. Sin. by 38ft. llin., how many square feet are therein? Ans. 3100ft. 4in. 4". NOTE. Divide the square feet by 9, and the quotient will be square yards. 5 If a ceiling be 59ft. 9in. long, and 24ft. Gin. broad, how many square yards does it contain? Ans. lG2yd. 5ft.-4- Ft. I. 6 in. * 59 9 3 179 3 8 1434 29 10 6 9)14G3 10 6 162yd. 5ft. 6 How many yards are contained in a pavement 50 feet 9 inches long, and 18 feet 4 inches wide? Ans. 115yd. 5ft 5in. 7 How many yards in a ceiling 92ft. 4in. long, 22ft. 8in. wide? Ans. 232yd. 4ft. 10in.+ 8 How many squares in a floor 37ft. Gin. long, and 21ft. 9in. wide? Ans. 8 squares, 15 feet.-f- A square is 10 feet long and 10 feet wide, or 100 square feet. It is used in estimating flooring, roofing, weather-boarding, &-c. 9 How many squares of weather-boarding on the side of a house 43 feet 6 in. long, and 18ft. Sin. high? Ans. 8 squares, 12ft. MENSURATION. 161 10 How many squares in a roof 36ft. 4in. long, 15ft. 9in. wide? Ans. 5 sq. 72ft.-f NOTE 2. To measure a triangle. Multiply the base by one half the perpendicular height, and the product will be its superficial content. 11 Let C, H, and G, represent a triangle, whose base is 40 feet, and perpendicular height 28 feet; how many feet does it contain ? Ans. 560 feet. 40 feet 40 feet feet. 40 14 half the perpendicular 160 40 560 12 How manv square feet in a triangle 80 feet long and 36 feet high? Ans. 1440ft. 13 In a triangular pavement 46 feet long, and 24 feet at the place of its greatest width, how many yards ; and how many bricks, allowing 41 to every square yard? Ans. 61yd. 3ft., and 2514 bricks 14 In the gable ends of a house, which is 63 feet long and 22 feet high, from the "square of the building" to the top, how many squares? Ans. 6sq. 93in NOTE 3, To find the circumference of a circle, when the diameter is given: Say, As 7 are to 22, so is the diameter to the circumfer- ence ; or the contrary, As 22 are to 7, so is the circumference to the diam- eter. U* 162 MENSURATION. The diameter of a circle is 14 feet; what is the circumference? Ans. 44. As 7 : 22 : : 14 : 44 The circumference of a circle is 44ft.; what is the diameter? Ans. 14. NOTE 4. To find the superficial contents of a circle. Multiply half the circumference by half the diameter 15 How many square feet in a circle whose diameter is 14 feet, and circumference 44? Ans 154 ft. half circumference, 22 half diameter, x 7 154 feet 16 How many square feet in a circle whose circum- ference is 16 feet? Ans. 20 sq. ft. halfcir. 8 As 22 : 7:: 16 : 5 halfdiam.2i 20 17 How many square feet in a circle whose diameter is 21 feet? Ans. 346* ft. NOTE 5. To find the superficial contents of a globe. Multiply the circumference by the diameter. 18 What are the superficial contents of a globe whose diameter is 70 feet, and circumference 220 feet? Ans. 15400 sq. ft. 19 How many square feet of cloth would be re- quired to cover a globe, whose diameter is 28 feet, and circumference 88? Ans. 2464 ft. 20 How many yards of canvass would be required to make a balloon of a globular form, 20yardsin diameter? Ans. 1257 sq. yds. NOTE 6. To find the solid contents of a cube,* or of a square stick of timber, or a .pile of wood, fyc. Multiply the length by the breadth, and that product by the thickness. * A cube is a solid body, contained by 6 equal sides, ail of which are exact squares. 3UENSUKATION 163 21 What are the solid contents of a cube whose di- ameter is 4 feet? Ans. 64 feet. 4 feet 4 16 4 64 22 What is the solid contents of a stick of timber 2 feet thick, 3 feet wide, and 36 feet long? Ans. 216 solid feet. 36 feet 3 108 2 216 23 How many solid feet in a block of marble 3 feet thick, 7 feet wide, and 13 feet long? Ans. 273 sol. ft. 24 In a cube whose diameter is 7 feet, how many solid feet? Ans. 343 feet. 25 How many solid feet in a pile of wood 28 feet long, 8 feet wide, and 10 feet high; and how many cords does it contain*? Ans. 2240 feet; 17 cords 64 ft. or, 17 i cords. 26 In a cellar 36 feet long, 27 feet wide, and 4i feet deep, how 'many solid yards? Ans. 162 yards. 27 How many perches* of stone in a wall 42 feet long, 84 feet high, and 2 feet thick? Ans. 28,8 per. feet 24,75)714.00(28,8 28 In a 12 inch brick wall, 52 feet long and 36 feet high, how many bricks, allowing 21 to every square foot of wall? Ans. * A perch is 16* feet long, 1 ft. wide, and 1 foot hig! solid feet. 164 MENSURATION. 29 In an 8 inch brick wall, 82 feet long and 16 feet high; how many bricks, allowing 14 bricks for every square foot of wall? Ans. 18368. 30 In a 16 inch brick wall, 148 feet long and 42 feet high, how many bricks, allowing 28 bricks to the square foot? Ans. 174048. 31 How many bricks in 3 walls, the first 68 feet long. 18 feet 6 inches high, 16 inches thick; the second 72 ft. 6in. long, 19ft. 4in. high, 12in. thick; the third 43ft. 4in. long, 12ft. Sin. high, 8in. thick? Ans. 72343.+ NOTE 7. To find the solid contents of a cylinder.* Find the contents of one end by Note 4, and multiply that product by the length. 32 What are the solid contents of a cylinder whose diameter is 14 feet, arid length 16 feet? Ans. 2464ft. half circumference 22 half diameter 7 2464 feet 33 What are the contents of u circular well, 7 feet in diameter, and 62 feet deep? Ans. 2387 ft. 34 What are the solid contents of a tub whose diam- eter is 6 feet and height 7 feet? Ans. 198 ft. 35 How many feet in a circular well, 10 feet diame- ter and 20 feet deep ? Ans. 157y feet. NOTE 8. To find the solid contents of the f rust rum of a cone, To the sum of the squares of the two diame- ters, in inches, add their product ; multiply this sum by one-third of the depth, and this last product by the deci- mal .754. Eu. El. The result will be the contents in cubic inches, which ***' s desired. * A c /Under is a long round body, whose diameter is every where t k . e same,. MENSURATION. 165 The liquid gallon of Ohio contains 231 cubic inches. The dry gallon contains 268i cubic inches. The bushel, of grain, contains 2150| cubic inches. The bushel, of coal, lime, &c. 2688 cubic inches. EXAMPLES. 1 In a circular vessel, whose greater diameter is 80 inches, the less 71, and the depth 34, what is the contents in liquid gallons ; and also in bushels of grain ? A C 659.72-1- gallons. 3 * I 70.86 bushels. 80X80=6400= square of 80. 71X71 = 5041= square of 71. 11441 80X71=5680= product of 80 and 71. 17121 in depth = lU=j of the depth. 194038X. 7854=152397.4452 inches. Divide by 231 for gallons, and 2150| for bushels. 2 The greater diameter of a tub is 38 inches, the less 20.2, and the depth 21 ; what is the content in gallons ? Ans. 62. 34+ gallons. 3 The top diameter of a tube is 22 inches, the bot- tom 40, and the height 60 ; what is its contents in gal- lons, also in bushels of grain ? . C 20 1.55-1- gals. *'l 21.64+bu. 4 How many barrels, of 32 gallons each, in a cistern, whose greater diameter is 8 feet 6 inches, the less 8 feet, and depth 7 feet 9 inches ? Ans. 96 barrels 28-j-gals. 5 How many bushels of grain in i bin that is 8 feet long, 4 feet wide, and 6 feet high ? Ans. 123-{-bu. 6 How many bushels of coal in a" boat 60 feet long, 16 feet wide, and 4% feet deep ? Ans 2777+bushels. 1 66 MENSURATION. NOTE 9. To find the solid contents of a round stick of timber of a taper from one end to the other. Find I the circumference a little nearer the larger than the smaller end; from this, by Note 3, find the diameter: multiply half the diameter by half the circumference, and the product by the length.* EXAMPLES. 1 What are the solid contents of a round stick of tim- ber 10 feet long, and 2.61 feet circumference? Ans, 5.4 feet.+ As 22 : 7 : : 2.61 : .83 diameter 1,305 half circumference .415 half diameter 6525 1305 5220 .541575 10 length 5.415750 2 How many solid feet in a log 40 feet long, which girts 66 inches? Ans. 96.25 ft. As 22 : 7 : : 66 : 21 in diameter 33 xlOiX40= 13860. 144)13860(96.25 NOTE 10. To find the solid contents of a globe. Multiply the cube of the diameter by .5236. EXAMPLES. 1 What are the solid contents of a globe whose diame- ter is 14 inches? Ans. 1436.75in.+ 14 X 14 X 14=2744. 2744 X .5236= 1436.7584. 2 What are the contents of a balloon of a globular form, 42 feet in diameter? Ans. 38792.4 ft.-{- 3 How many solid miles are contained in the earth, or globe, which we inhabit? * This method, thoitgli not quite accurate, is sufficiently near the truth for the purpose of measuring timber. INVOLUTION. 167 Suppose the diameter to be 7954 mhes: then, 7954 X 7954x7954=503218686664 the cube of the earth's axis, or diameter; then, 50321 8686664 X .5236=2634853(34337 cubick miles. Ans. NOTE. The solidity of a globe may be found by the circumference, thus Multiply the cube of the circum- ference by .016887 the product will be the contents. INVOLUTION, OR THE RAISING OF POWERS. The product arising from any number multiplied by itself, any number of times, is called its power, as fol- lows t 2x2= 4 the square, or 2d power of 2. 2x2x2= 8 3d power or cube of 2. 2X2X2X2=16 4th power of 2. The number which denotes a power is called its index. NOTE. When any power of a vulgar fraction is re- quired, first raise the numerator to the required "power, and then the denominator to the required power, and place the numerator over the denominator as before : thus, the 4th power of | Questions. What is the product, arising from the multiplication of any figure by itself a given number of times, called? What is the number which denotes a power, called ? How do you proceed to find any required power of a vulgar fraction? 168 INVOLUTION. i Table of the first nine Powers. \ 8s ] | 6 II 1 1 e, i i 1 1 1 1 1 1 1 1 1 1 i 2 4 8 16 32 64 128 256 512 3 9 416 525 63fi 749 864 981 27 64 125 216 343 512 729 81 256 625 1296 2401 4096 6561 243 1024 3125 77-76 16807 32768 59049 729 4096 15625 46656 117649 262144 531441 2187 16384 78125 279936 823543 2097152 4782969 6561 65536 390625 1679616 5764801 16777216 43046721 19683 - 262144 1953125 10077696 40353607 134217728 387420489 EXAMPLES. 1 What is the square of 32? 32 64 96 1024 Ans 2 What is the cube of 14? 14 14 Ans. 2744. 56 196 14 784 196 2744 3 What is the sixth power of 2.8? Ans 481.890304 4 What is the third power of .263? Ans. .018191447. TUB SQUARE ROOT. 169 EVOLUTION, OR THE EXTRACTING OF ROOTS. The root of a number, or power, is such a number, as being multiplied into itself a certain number of times, will produce that power, Thus 2 is the square root of 4, because 2x2=4; and 4 is the cube root of 64, be- cause 4x4X4=64, and soon. THE SQUARE ROOT. The square of a number is the product arising from that number multiplied into itself. Extraction of the square root is the finding of such a number, as being multiplied by itself, will produce the number proposed. Or, it is finding the length of one side of a square. RULE. 1 Separate the given number into periods of two fig- ures, each, beginning at the units place. 2 Find the greatest square contained in the left hand period, and set its root on the right of the given number: subtract said square from the left hand period, and to the remainder bring down the next period for a dividual. 3 Double the root for a divisor, and try how often this divisor (with the figure used in the trial thereto annexed) is contained in the dividual: set the number of times in the root; then, multiply and subtract as in division, and bring down the next period to the remainder for a new dividual. 4 Double the ascertained root for a new divisor, and proceed as before, till all the periods are brought down. NOTE. If, when all the periods are brought down, there be a re- mainder, annex cyphers to the given number, for decimals, and pro- ceed till the root is obtained with a sufficient degree of exactness. Observe that the decimal periods are to be pointed off from the de- cimal point toward the right hand : and that there must be as many whole number figures in the root, as there are periods of whole num- bers, and as many decimal figures as there are periods of decimals. 15 170 THE SQUARE ROOT. PROOF. Square the root, adding in the remainder, (if any,) and the result will equal the given number. EXAMPLES. 1 What is the square root of 5499025? 5,49,90,25(2345 Ans. 4 2345 2345 43)149 129 11725 9380 464)2090 7035 1856 4690 4685)23425 5499025 Proof. 23425 2 What is the square root of 106929? Ans. 327. 3 What is the square root of 451584? Ans. 672. 4 What is the square root of 36372961? Ans. 6031. 5 What is the square root of 7596796? Ans. 2756.2S+ 6 What is the square root of 3271.4007? Ans. 57.19+ 7 What is the square root of 4.372594? Ans. 2.091+ 8 What is the square root of 10.4976? Ans. 3.24 9 What is the square root of .00032754? Ans. .01809+ 10 What is the square root of 10? Ans. 3.16224 To extract the Square Root of a Vulgar Fraction. RULE. Reduce the fraction to its lowest terms, then extract the square root of the numerator for a new numerator, and the square root of the denominator for a new deno- minator. NOTE. If the fraction be a surd, that k, one whose root can never be exactly found, reduce it tD a decimal, and extract the root therefrom. THE SQUARE ROOT. 171 EXAMPLES. 1 What is the square root of |54f J ^ ns - 1- 2 What is the square root of f 14-j-? Ans. J. 3 What is the square root of flf ? Ans. .93309-[- To extract the Square Root of a Mixed Number. RULE. Reduce the mixed number to an improper fraction, and procee'd as in the foregoing examples : or, Reduce the fractional part to a decimal, annex it to the whole number, and extract the square root there- from. EXAMPLES. 1 What is the square root of 37J|? Ans. 61. 2 What is the square root of 27^? Ans. 51. 3 What is the square root of Soli? Ans. 9.27-|- APPLICATION. 1 The square of a certain number is 105G25: what is that number? Ans. 325. 2 A certain square pavement contains 20736 square stones, all of the same size; what number is contained in one of its sides? Ans. 144. 3 If 484 trees be planted at an equal distance from each ether, so as to form a square orchard, how many will be in a row each way? Ans. 22. 4 A certain number of men gave 30s Id. for a chari- table purpose; each man gave as many pence as there were men: how many men were there) Ans. 19. 5 The wall of a certain fortress is 17 feet high, which is surrounded by a ditch 20 feet in breadth; how long must a ladder be to reach from the outside of the ditch to the top of the wall? Ans. 26.24-J-feet. NOTE. The square of the longest side of a right angled triangle is equal to the sum of the squares of the othef two sides ; and consequently, the difference of the square of the longest, and either of the other, Ditch, is the square of the remaining one. 172 THE SQUARE ROOT. 6 A certain castle which is 45 yards high, is surroun- ded by a ditch 60 yards broad ; what length must a ladder be to reach from the outside of the Hitch to the top of the castle ? Ans. 75 yards. 7 A line 27 yards long, will exactly reach from the top of a fort to the opposite bank of a river, which is known to be 23 yards broad ; what is the height of the fort? Ans. 14.1.42-|-yards. 8 Suppose a ladder 40 feet long be so planted as to reach a window 33 feet from the ground, on one side of the street, and without moving it at the foot, will reach a window on the other side 21 feet high; what is the breadth of the street? Ans. 56.64+feet. 9 Two ships depart from the same port; one of them sails due west 50 leagues, the other due south 84 leagues; how far are thev asunder?? Ans. 97.75+ Or, 97|+leagues. Questions. What is a square? A square- is a surface whose length and breadth are equal, and whose angles' (or cor- ners) ore right angles, (or square.) What is its square root? The square root is the length of the side of a square. If the square be sixteen, what is the root? Why is the root four? If the root be three, what is the SQUARE? What is the square root of twenty-five? What is the square of five ? What is the square root of thirty-six ? What is the square of six? How do you point off a number whose square root is to be extracted? What is the next step? What do you subtract from the period? What do you annex* to the remainder? THE SQUARE ROOT. 173 Illustration of the Rule for extracting the Square Root. The reason for pointing off the given number into periods of two figures each, is, that the product of any whole number contains just as many figures as are in both the multiplier and the multiplicand, or but one less ; consequently, the square contains just double as many figures as the root, or one less. A E B Suppose the figure ABCD contains 1849 square feet, and that the number consists of two periods; then there must be two figures in the D 120 11 1600 9 120 root. The largest root whose square can be taken out of the left hand period, is 4, (or as it will stand in ten's place in the root, it is 40,) and the square of this is 16 (of 1600.) This taken from the whole C square ABCD, or 1849, leaves 249. 18,49(43 16 83)249 249 GIRD AEIIG HFCI EBFH ABCD Now double GH or HI, which is 40, for a divisor, omitting the cypher to leave place for the next quotient figure, to complete the divi- sor. 80 into 249 are contained 3 times ; this 3 is the width of the oblong ALHG, or HFCI. But the square is imperfect without EBFH; then annex the three to the divisor. Now multiply this perfect divisor by the last figure of the root, to get the = 1849 Quantity m tne two oblong figures, and the small square which comprises the great square ABCD. 15* 174 THE CUBE ROOT. How do you find the divisor? Why do you place the new quotient figure in the units place of the divisor? How do you prove the square root? THE CUBE ROOT. The cube of a number is the product of that number multiplied into its square ? Extraction of the cube root is finding such a number as, being multiplied into its square, will produce the number whose cube root is extracted. RULE. Separate the given number into periods of three fig ures each, beginning at the units place. Find the great- est cube in the left hand period, and set its root in the quotient; subtract said cube from the period, and to the remainder bring down the next period for a dividual. Square the root, and multiply the square by three hundred for a divisor. See how often the divisor is contained in the dividual, and place the result in the quotient. Multiply the divisor by the last found quotient figure ; square the last found figure multiply the square by the preceding figure or figures of the quotient, and this pro- duct by thirty; and cube the last figure. Add these three products together, and subtract their amount from the dividual. To the remainder add the next period, and proceed as before, until the periods are all brought down. When a remainder occurs, annex periods of cyphers to obtain decimals, which may be carried to any conve- nient number. NOTE 1. The cube root of a vulgar fraction is found by reducing it to its lowest terms, and extracting the root rf the numerator for a numerator, and of the denomica- THE CUBE ROOT. 175 tor for a denominator. If it be a surd,* extract the root of its equivalent decimal. EXAMPLES. 1 What is the cube root of 99252847? 99,252,847(463 Ans. 463. 4X4X4=64 4X4X300=4880 Div. 4800X6= 6X6X4X30= 35252 463 463 28800 4320 1389 0X6X6= 216 2778 1852 Subtrahend 33336 214369 46X46X300=634800! 1916847 463 Div. 634800x3=1904400 643107 3X3X46X30= 12420 ' 1286214 3X3X3= 27 857476 Subtrahend 1916847 Proof 99252847 2 What is the cube root of 84604519? -Ans. 439. 3 What is the cube root of 259694072? Ans. 638. 4 What is the cube root of 32461759? Ans. 319. 5 What is the cube root of 5735339? Ans. 179. 6 What is the cube root of 48228544? Ans. 364. 7 What is the cube root of 673373097125? Ans. 87C5. 8 What is the cube root of 7532641? Ans. 196.02-f- 9 What is the cube root of 5382674. Ans. 175.2-j- 10 What is the cube root of 15926.972504? Ans. 25.16+ When decimals occur, point the periods both ways, beginning at the decimal point, and if the last period of the decimal be not complete, add one or more cyphers. A mixed number may be reduced to an improper fraction, or a decimal, and the root thereof extracted. * A surd is a quantity whose root cannot exactly be formed, quantity whose root can be found, is called a rational quantity. 176 THE CUBE ROOT. 1 What is the cube root of ^VV ? Ans. . 2 What is the cube root of |J? Ans 5-. 3 What is the cube root of ? Ans . 4 What is the cube root of 12UJ Ans. 2J. 5 What is the cube root of Sl^s ? Ans. 31. SURDS. 6 What is the cube root of 7^? Ans. 1.93-(- 7 What is the cube root of 91? Ans. 2.092+ APrLICATION. 1 The cube of a certain number is 103823; what is that number? Ans. 47 2 The cube of a certain number is 1728 j what num- ber is it? Ans. 12. 4 There is a cistern or vat of a cubical form, which contains 1331 cubical feet: what are the length, breadth and depth of it? Ans. each 11 feet. 4 A certain stone of a cubical form contains 474552 solid inches j what is the superficial content of one of its sides? * Ans. 6084 inches. Questions. What is a cube? A cube is a solid body contained by six equal square sides. What is the cube root? It is the length of one side of a cube. What is the square of the cube root? It is the su- perficial contents of one side of a cube. How do you point off a number whose cube root is tc be extracted? What is the first figure of the root? It is the root of the greatest cube in the first period. When you subtract the cube from the first period, what do you do? How do you find the divisor? What is the first step towards finding the subtrahend ? What is the second? What is the third? When a remainder occurs, how do you proceed? How do you prove the cube root? THE CUBE ROOT. 177 Illustration of the Rule for extracting the Cube Root. The reason for pointing off the number into periods of three figures eacli, is similar to the one given in the Square Root; for the number of figures in any cube will never exceed three times the figures in the root, and will never be more than two figures less. OPERATION. 15,625 I 25 8 2X2X2= 2X2X300=1200 5X5X2X30= 5X5X5= 7625 6000 1500 125 7625 Fig. 1. In Ois number there are two pel ->ods : of course there will be two figures in the root. "The greatest cube in the left hand period (15) is 8, the root of which is 2;" therefore, % is the first figure of the root, and as we shall have another figure in the root, the 2 stands for 2 tens, or 20. But the cube root is the length of one of the sides of the cube, whose length, breadth and thickness are equal : then the cube whose root is 20, contahw 20X20 X 20=8000. "Subtract the cube thus found (8) from said period, and to the remainder bring down the next period," or, subtract the 8000 from the whole given number (15625) and 7625 will remain. Thus 8000 feet are disposed of in the cube, Fit 1. 20ft .ong, 80 ft wide, and 20 ft. nigh. The cube is to be enlarged by the addition of 7625 feet which remain. In doing this, the figure must be enlarged on three sides, to make it longer^ and wider, and higher, to maintain the complete cubic form. The next step is, to find a divisor; and this must oe the number of square feet contained in the three sides to which the addition must be made. Hence we ^multiply the square of the quotient Jigure by 300."* That is, 2 X 2 X 300=1200 : or 20 X -0 X 3= 1 200 feet, which is the superficial content of the three sides, A, B, and C. H 2 178 THE CUBE ROOT. Fig. 2. Fig. 4. Proof. 20X20X20= 20X20X3X5= 5X5X20X3= 5X5X5= 8000 6000 1500 125 25X25X25= 15625 This "divisor (1200) is contained in, ike dividual" (7625) 5 times : then 5 is the second quotient figure ; that is, the addition to each of the three sides is 5 feet thick ; if 1200 feet cover the three sides one foot thick, 5 feet thick will require 5 times as many; that is 1200X5= 6000. But when the additions are made to the three squares there will be a deficienc)- along the whole length ol the sides of the squares be- tween the additions, which must be supplied before the cube will be complete. These deficiencies will be three, as may be seen at NNN in Fig. 2, therefore it is that we "multiply the square of the last figure by the prece- ding figure, and by 30," (that is, 5X5X^X30,) or 5 X 5 X 20 X 3=1500vvhich is the quantity required to supply the three deficiencies. Figure 3, represents the solid with these deficiencies supplied, and discovers an other deficiency, where they approach each other at ooo. Lastly, "cube the last fig- ure;' 1 '' this is done to fill the deficiency left at the comer, in filling up the other defi- ciencies. This corner is limited by the three portions applied to fill the former va . cancies, which were 5 feet in breadth ; consequently the cube of 5 will be the solid contents of the corner. Fig. 4 represents this deficiency (eee) supplied, and the cube complete. ROOTS OP ALL POWERS. 179 The illustration is much better made by means of 8 blocks 'of the ollowing description: One cube of about 3 inches diameter; three lieces each 3 inches square, inch thick; three pieces each & inch square, 3 inches long ; and one cube % inch. A set of these should >elong to the apparatus of every Professional Teacher. A GENERAL RULE FOR EXTRACTING THE ROOTS OF ALL POWERS. 1 Point the given number into periods, agreeably to the required root. 2 Find the first figure of the root by the table of pow- ers, or by trial ; subtract its power from the left hand seriod, and to the remainder bring down the first figure in the next period for a dividend. 3 Involve the root to the next inferior power to that which is given, and multiply it by the nuhiber denoting the given power, for a divisor; by which find a second figure of the root. 4 Involve the whole ascertained root to the given power, and subtract it from the first and second periods Bring down the first figure of the next period to the re- mainder, for a new dividend; to which, find a new divi- sor, as before; and so proceed. Note. The roots of the 4th, 6th, 8th, 9th, and 12th powers, may be obtained more readily thus : For the 4th root take the square root of the square root. For the 6th, take the square root of the cube root. For the 8th, take the square root of the 4th root. For the 9th, take the cube root of the cube root. For the 12th, take the cube root of the 4lh root. EXAMPLES. 1 What is the 5th root of 916132832? 9161,32832(62 Ans. 7776 6X6X6X6X6=7776 6X6X6X6X5=6480 div 6480)13853 916132832 62x62x62x62x62=916132832 916132832 180 ARITHMETICAL PROGRESSION. 2 What is the fourth root of 140283207936? Ans. 612. 3 What is the sixth root of 782757789696 ? Ans. 96. 4 What is the seventh root of 194754273881 ? Ans. 41. 5 What is the ninth root of 1352605460594688 ? Ans. 48. ARITHMETICAL PROGRESSION. A SERIES of numbers, increasing or decreasing by a common difference, is called an Arithmetical Progres- sion. Thus 3, 5, 7, 9, 11, 13, 15, &c., is an ascending se- ries, whose common difference is 2. And 16, 13, 10, 7, 4, 1, is a descending series, whose common difference is 3. The three most important properties of an arithmeti- cal series are the following : I. The sum of the two extremes is equal to twice the mean, or to the sum of any two terms equidistant from the mean. In the above series 3 -{-15= twice 9, which is the mean or middle term; and 5-f-13 which are equidistant from the mean. II. The difference of the extremes, is equal to the common difference multiplied by the number of terms, less one. In the above, the number of terms 7 1=6; then the common difference 2X6 = 12, which is equal to 15 3. Or the number of terms in the other series 61=5 ; then 5X3 = 15, which is equal to 161. III. The sum of all the terms is equal to the product of the mean, or of half the sum of the extremes, multi- plied by the number of terms. As above, the mean is 9 ; which, multiplied by 7, the number of terms, gives 63= 15-j-13-|- 11 4-9-f 7+5-1-3. Or, 15+3 = 18; half of which is 9. Then, 9X7 =63, as before. ARITHMETICAL PROGRESSION. 181 And 16-fl = 17; half of which is 8. This multi- tiplied by 6, the number of terms, =51 = 16-J-13-{-10 +7+4+1. NOTE 1. To find the last term, multiply the common difference by the number of terms, less one, and add the product to the first term in an increasing series ; or, sub- tract the product from the first term in a decreasing series. EXAMPLES. 1 If the first term is 3, the common difference 2, and the number of terms 7, what is the last term ? Ans. 15. 7 1=6 ; the 2X6+3=15, the last term. 2 The first term being 16, the common difference 3, and the number of terms 6, what is the last term ? Ans. 1. 61=5; then 3X5 = 15. Then 16 15=1, the last term. 3 What is the last term in a series, whose first term is 5, the common difference 4, and the number of terms 25? Ans. 10K 4 Suppose, in the above, the first term is 3 ; what is the last term ? Ans. 99. 5 A man bought 50 yards of calico at 6 cents for the first yard, 9 for the second, 12 for the third, &c.; what did he pay for the last? Ans. SI. 53. NOTE 2. To find the mean term, take half the sum of the extremes. EXAMPLES. 1 The first term is 3, and the last 15 ; what is the arith- metical mean ? Ans. 9. 3+15=18 ; the 18-^-2=9, the mean term. 2 The weight of 5 packages of goods is, severally, 180, 150, 120, 90, 60 pounds ; what is the mean or average weight ? Ans. 1201bs. NOTE 3. To find the sum of all the terras, multiply the mean term by the number of terms ; or, the mean 182 ARITHMETICAL PROGRESSION. by the sum of the two extremes, and take half the pro- duct. EXAMPLES. 1 The mean term is 11, and the number of terms 9; what is the sum of the series ? Ans. 99. 2 The first term is 5, the last 32, and the number of terms 10 ; what is the sum of the series ? Ans. 185. 3 How many strokes does the hammer of a common clock strike in 12 hours ? Ans. 78. 4 What debt can be discharged in one year, by week- ly payments in arithmetical progression, the first being $12, and the last, or fifty-second, payment $1236 ? Ans. 32448. NOTE 4. To find the common difference, divide the difference of the extremes by the number of terms, less one. EXAMPLES. 1 The ages of 8 boys form an arithmetical series the youngest is 4 years old and the oldest is 18 pwhat is the common difference ? Ans. 2. 2 A debt can be discharged in one year, by weekly payments in arithmetical progression the first is $12, and the last $1236 ; what is the common difference ? Ans. $24. NOTE 5. To find the number of terms, divide the difference of the extremes by the common difference, and add 1 to the quotient. EXAMPLES. 1 In a series, whose extremes are 4 and 1000, and the common difference 12, what is the number of terms ? Ans. 84. 2 If a man, on a journey, travels 18 miles the fir^t day, increasing the distance 2 miles each day, and on the last day goes 48 miles, how many days did he travel ? / Ans. 16. ARITHMETICAL PROGRESSION. 183 PROMISCUOUS EXERCISES. 1 24 persons bestowed charity to a beggar the first gave him 12 cents, the second 18, &c., in an aiithmeti- cal series ; what sum did he receive ? Ans. $19.44. 2 Suppose 100 apples were placed in a right line, 2 yards apart, and a basket 2 yards from the first ; how far would a boy travel to gather them up singly, and re- turn with each separately to the basket ? Ans. 20200 yards. 3 In a drove of 400 hogs, 5 of the largest weighs, on an average, 280 pounds a-piece, and 5 of the smallest 180 ; what is the mean Or average weight of them all, and what is the whole weight ? . C 2301bs. mean weight. S * I 92000 Ibs. whole wt. 4 How many acres in a piece of land 80 rods wide at one end and 60 at the other, and 1 20 rods long ? Ans. 52. It may be observed, that the natural numbers 1, 2, 3, 4, 5, 6, 7, &c., is an arithmetical series, whose first term is 1, and common difference 1 ; and that the last term is equal to the number of terms. From this series, we may form another by adding to each figure the sum of all the preceding, and we shall have 1, 3, 6, 10, 15,21,28, &c. These are called triangular numbers, because they may be represented by points, forming equilateral trian- gles, thus : Hence we perceive, that the sum of the natural numbers, to any degree, expresses the triangular number of the same degree. In the same manner, the square numbers 1, 4, 9, 16, 184 GEOMETRICAL PROGRESSION. 25, 36, &c., may be expressed by points, arranged in squares, thus : Natural numbers -1, 2, 3, 4, 5, 6, 7, 8, 9, &c. Triangular numbers 1, 3, 6, 10, 15, 21, 28, 36, &c. Square numbers 1, 4, 9, 16, 25, 36, 49, 64, 81, <fcc. Cube numbers 1, 8, 27, 64, 125, 216, 343, 512, 729, &c. GEOMETRICAL PROGRESSION. A SERIES of numbers, increasing or decreasing by a common ratio, is called a Geometrical Progression. Thus 2, 4, 8, 16, 32, 64, 128, is an increasing series whose common ratio is 2 ; And 729, 243, 81, 27, 9, 3, is a decreasing series, whose common ratio is . The most important properties of a geometrical series are the following : I. The product of the extremes is equal to the square of the mean ; or, to the product of any two terms equi distant from the mean. In the above, 128X2 = 16X 16, or 32X8, &c. Also 729X3=243X9, or 81X27, between which the mean falls. Hence, the mean term, in a geometrical series, is th square root of the product of the extremes, or of an) two terms equidistant from the mean. II. The last term of an increasing series, is the pro duct of the first term, multiplied by the ratio involved t( the power which is one less than the number of terms in a decreasing series, it is the quotient of the first term divided by the power. In an increasing series, whose first term is 2, ratio 3 GEOMETRICAL PROGRESSION. 185 and the number of terras 5, we have, by the natural method, 2, 6, 18, 54, 162, which gives 162 for the last term. Or, by the artificial method, the ratio involved in the 4th power, which is one less than the number of terms, 3X3X3X3=81; these, multiplied by the first term, 2X81, gives 162, the last term, as before. Again, let the first term be 4 ; then we have 4, 12, 36, 108, 324, or 3X3X3X3 = 81. Then multiply the first term, 4, by 81=324, the last term. In a decreasing series, with the first term 243, ratio , and the number of terms 5, we have 243, 81, 27, 9, 3 ; then 3X3X3X3=81, and 243-f-81 gives 3, which is the last term, as before. NOTE 1. To find the last term, involve the ratio to the power which is one less than the number of terms, and multiply the first term by the power. EXAMPLES. 1 What is the eighth term of an increasing geometri- cal series, whose first term is 4, and ratio 2 ? Ans. 512. 2X2X2X2X2X2X2 = 128; then 4X128 = 512, which is the eighth term. Or, 4, 8, 16, 32, 64, 128, 256, 512, the eighth term, as before. 2 Required the last term of an increasing series, whose first term is 15, ratio 2, and number of terms 10. Ans. 7680. 3 A boy purchased 18 oranges, at 1 cent for the first, 4 for the second, 16 for the third, &c.; what was the price of the last? Ans. $171798691.84. NOTE 2. To find the sum of all the terms, multiply I the last term by the ratio, and from the product subtract the first term ; then divide the remainder by the ratio, less one. EXAMPLES. 1 What is the sum of a series, whose first term is 2, ratio 3, and number of terms 5 ? Ans. 242. 186 GEOMETRICAL PROGRESSION. Find (he last term by Note 1 . 162 last term. 2 3 ratio. 486 2 first term. 31=2)484 242 242 sum of the sums. 2 What is the sum of the series, whose first term ii 4, ratio 5, and number of terms 7 ? Ans. 15624. Find the last term by Note 1. 12500 last term. 4 5 ratio. 20 100 62500 500 4 first term. 2500 12500 51=4)62496 15624 15624 sum of the series. 3 What is the sum of a series, whose first term is 3 ratio 4, and number of terms 7 ? Ans. 16383. Last term, Ratio, First term, . 41=3)49149 Sum of the series 16383 Write down the series, multiply it by the ratio, an< subtract the first series from the second, thus : 3 12 192 768 3072 12288 12 192 768 3072 12288 49152 GEOMETRICAL PROGRESSION. 187 Hero the terms all cancel, but the first of the upper and last of the lower series. Then we have 49152 3 Divide by ratio, less 1 : 4 1=3)49149 The sum of the series, 16383 Now, as we multiplied the given series by the ratio, which is 4, and subtracted once the series from the pro- duct, the remainder is three times the given series. We, therefore, divide by 3, which is the ratio, less 1 : the quo- tient is the sum of the series. 4 At 2 cents for the first *ounce, 6 for the second, 18 for the third, &c., what would a pound of gold cost ? Ans. $5314.40. 5 Sold 10 yard3 of velvet, at 4 mills for the first yard, 20 for the second, 100 for the third, &c.; what did the piece cost ? Ans. $9765.62,4. 6 What is the cost of a coat with 14 buttons, at 5 mills for the first, 15 for the second, 45 for the third, &c.? Ans. $11957.42. 7 What is the cost of 16 yards of cloth, at 3 cents for the first yard, 12 for the second, 48 for the third, &c.? Ans. $42949672.95.- NOTE 3. The sum of a geometrical series is found by the extremes and the ratio, independent of the num- ber of terms ; hence, whether the number of terms be many or few, there is no variation in the rule. We may, therefore, require the sum of the seres, 6, 3, 1, 1, -*-, &c., to infinity, provided we can determine the value of the other extreme. Now, we see the terms decrease as the series advances ; and the hundredth term, for example, would be exceedingly small, the thousandth too small to be estimated, the millionth still less, and the infinite term would be nothing : not, as some tell us, " extremely small," or, " too little to be considered," &c., but abso- lutely nothing. 188 GEOMETRICAL PROGRESSION. Now let us consider the series as inverted ; then 6 will be the last term, and 3 the ratio. By the rule, multiply the last term by the ratio, subtract the first term, and divide by the ratio, less 1. Here we have the sum c 6X30 the series, =9, the answer. EXAMPLES. 1 What is the sum of the infinite series, 1-j-i-f-j -f-yV' &c.? Invert the series : is the last term, and 2 X 100 the ratio; hence, - =1, the answer. 2 What is the sum of the infinite series y\, T f^, _3_xiO y^, &c.? -- - - = |, the answer. 9 3 What is the value of \, ~, T V j, &c., to infinity ? Ans. |. 4 What is the value of |, 5, i, f, &c., to infinity? Ans. f. 5 What is the value of 1, |, T 9 g, &c., to infinity ? Ans. 4. Here the ratio is |. 6 What is the value of f , 5 4 T , T j, &c., to infinity ' Ans. |. 7 What is the value of .777, &c., to infinity ? This may be expressed by T \, T J T , T \o &c - 8 What is the sum of .6666, &c., to infinity ? Ans. . 9 What is the value of .232323, &c., infinitely ex- tended ? Ans. ff This may be expressed by T 2 /o T f lo & c - EXCHANGE. . 189 \ EXCHANGE. THE object of exchange is to find how much of the money of one country is equivalent to a given sum of the money of another. By the par of exchange between two countries, is meant the intrinsic value of the one, compared with the other ; it is estimated by the weight and fineness of the coins. The course of exchange, at any time, is the sum of the money of one country which, at that time, is given for a certain sum of money of another country. The course of exchange varies according to the circumstances of trade. All the calculations in exchange can be per- formed by the Rule of Three. EXAMPLES. 1 A sovereign, of England, is worth $4.84,6. A mer- chant of New York is indebted to a merchant of London $7520 ; how many sovereigns will it require to discharge the debt ? 2 A six-ducat piece of Naples, is wprth $5.25. A merchant of Naples is indebted to a merchant in Boston $6940 ; how many such pieces shall he remit ? 3 The current rupee of Calcutta is 44.4 cents. A merchant of Philadelphia has a claim against a mer- chant of Calcutta of $437 ; how many rupees shall he draw ? 4 The piaster is 20 cents ; how many piasters in $1128.22? 5 Reduce 7218 rupees to Federal money, at 46 cents per rupee. 6 The dollar of Bencoolen is reckoned at $1.10, Fed- eral money ; how much Federal money in $2740 of Bencoolen ? 7 The Prussian rix-dollar is worth 66| cents. Re- duce $7348.32 into Prussian money. 190 PROMISCUOUS EXERCISES. Ex. 1. If l sterling is worth $4,44 cts. 4 ms. ; what is 65^6 sterling worth ? Ans. $288,86. 2. What is the value of $500 in English money, at $4,44,4, per sterling? Ans. 112 10s. 2<L 3. What is the value of 125^ 7s. at $4,44,4, per sterling ? Ans. $557,05,5. 4. What is the value of $1000, in English money, at $4,44,4, per sterling Ans. 225^ 5*d. PROMISCUOUS EXERCISES. 1. A merchant had 1000 dollars in bank ; he drew out at one time $237.50, at another time, $116.09, and at another, $241.061 : after which he deposited at one time 1500 dollars, and at another time $750.50 ; how much had he in bank after making the last deposite ? Ans. $2655.84. 2 Sold 8 bales of linen, 4 of which contained 9 pieces each, and in each piece was 35 yards ; the other 4 bales contained 12 pieces each, and in each piece was 27 yards ; how many pieces and how many yards were in all ? Ans. 84 pieces, 2556 yards. 3 If a man leave 6509 dollars to his wife and two sons, thus to his wife f , to his elder son | of the remainder, and to his other son the rest ; what is the share of each ? ("Wife's share $2440.87 1. Ans. 1 Elder son's $2440.87^. [Other son's $1627.25. 4 What is the commission on $2176.50, at 2 per cent? Ans. $54.411. 5 If a tower is 384 feet high from the foundation, a sixth part of which is under the earth, and an eighth part under water, how much is visible above the water ? Ans. 272 feet. 6 How many bricks 9 inches long and 4 inches wide, will pave a yard that is 20 feet square ? Ans. 1600. 7 What is the value of a slab of marble, the length of which is 5 feet 7 inches, and the breadth 1 foot 10 inches, at 1 dollar per foot? Ans. $10.232. 8 A certain stone measures 4 feet 6 inches in length, PROMISCUOUS EXERCISES. 191 2 feet 9 inches in breadth, and 3 feet 4 inches in depth; how many solid feet does it contain? Ans. 41 ft. 3 in. 9 A line 35 yards long will exactly reach from the top of a fort, standing on the brink of a river, to the op- posite bank, known to be 27 yards from the foot of the wall; what is the height of the wall? Ans. 22 yards 3\ inches.-f- 10 The account of a certain school is as follows, viz. ~ of the boys learn geometry, f learn grammar, ~ learn arithmetic, o learn to write, and 9 learn to read: what number is there of each? A (5 who learn geometry, 30 grammar, 24 I arithmetic, 12 writing, and 9 reading. 11 A merchant, in bartering with a farmer for wood at $5 per cord, rated his molasses at $25 per hhd., which was worth no more than $'20 ; what price ought the far- mer to have asked for his wood to be equal to the mer- chant's bartering price? Ans. $6,25. 12 A and B dissolve partnership, and equally divide their gain : A's share, which was $332 50 cts., lay for 21 months; B's for 9 months only: the adventure of B. is required. Ans. $775 834 cts. 13 If a water-hogshead holds 110 gals, and the pipe which fills the hogshead discharges 15 gal. in 3 minutes, and the tap will discharge 20 gal. in 5 minutes, and these were both left running one hour, how many gallons would the hogshead then contain; and if the tap was then stopped, in what time would the hogshead be rilled? Ans. 60 gal., and filled in 10 min. 14 A has B's note for $500 75 cts., with 9 months in- terest, at 6 per cent., v^e on it, for which B gave him 5064 feet of boards, a> <\ ,*,ts per foot, with 140 pounds of tallow, at 13 cts. pei n^Mid, and is to pay the rest in flax seed, at 92$ cts. per b.Vnljhow many bushels of flax seed must A receive, to ba,A\ce the note? Ans. 409Jf . bushels. 15 A, B, an3 C,in company, had put in $5762: A's money was in 5 months, B's 7, and C's 9 months: they gained $780, which was so divided, that 1 of A's was J of B'SJ and j of B's was 1 of C's : but B, having Deceived 192 PROMISCUOUS EXERCISES. $2087, absconded: what did each gain, and put in; and what did A and C gain or lose hv B's misconduct? f A's stock $2494,887 gain 260 A J B's do $2227,577 clo 325 " ] C's do $1039,536 do 195 I^A and C would gain $465,577 16 When 100 boxes of prunes cost 2 dollars 10 cents each, and by selling them at 3 dollars 50 cents per cwt. the gain is 25 percent., the weight of each box, one with another, is required. Ans. 84 Ib. 17 There are two columns, in the ruins of Persepo- iis, left standing upright; one is 64 feet above the plain, the other 50. Between these, in a right line stands an ancient statue, the head whereof is 97 feet from the sum- mit of the higher, and 86 feet from the top of the lower column, and the distance between the lower column and the centre of the statue's base, is 76 feet; the distance between the top of the columns is required. Ans. 157-J-ft 18 If I see the flash of a cannon, fired from a fort on the other side of a river, and hear the report 47 seconds afterwards, what distance was the fort from where I stood? Ans. 53674 feet. NOTE. Sound, if not interrupted, will move at the rate of about 1142 feet in a second of time. 19 What is the difference between the interest of $1000 at 6 per cent, for 8 years, and the discount of the same sum at the same rate, and for the same time ? Ans. The interest exceeds the discount by $155 67 cts. 5 m. 20 If a tower be built in the following manner, 7 \\ of its height of stone, 27 feet of brick, and i of its height of wood, what was the height of the tower? Ans. 113 feet 4 inches. 21 A captain, 2 lieutenants, and 30 seamen, take a prize worth $7002, which they divide into 100 shares, of which the captain takes 12, the two lieutenants each 5, and the remainder is to be divided equally among the sailors; how much will each man receive? Ans. Cap- tain's share, $840.24, each lieutenant's, $350,10, and each seaman's, $182,05,2. PREFACE TO APPENDIX. THE author of the foregoing work, has long contemplated its ex- Jgnsion in an Ajjoejidix, which is now offered to the public in hopes | the usefulness of the whole may be extended, and the science of; arithmetic advanced. The view of numbers, and the abridged modes of operation herein presented, it is believed, will be found ac- ceptable alike to the business-man and to the scholar. The Appendix recognizes the scientific character of numbers, and gives bold and enlarged views of_arithmetical operations. The method of cancelling is not new^ but_for a long period it has scarcely been known! ITls, however, coming into general use; and it is carried much farther in this part of the work than in any we have hitherto known. In Europe, this system has been very generally adopted in the higher schools, and in this country it is fast becoming known and as far as it is known, it supercedes the usual modes of operation. To the method of stating problems in Proportion by comparing cause and effect, we invite special attention. On critical examina- tion it will be found more easy and more rational than any other method. Other methods of statement sometimes require the number of men to be multiplied into feet of wall days into acres of grass, &c., all of which, though correct as abstract proportion in numbers, is unnatural and void of strict philosophical expression not so witt this method. The peculiarities which the student will here find in the Extrac- tion of Roots, and in Mensuration, are not all new indeed there an be nothing new in principle but as far as the author's know- ledge extends, he is not aware that these abbreviations have ever been collected in any arithmetical work. The impression seems to have been, that the people could not comprehend arithmetical breV' ity, nor appreciate mathematical beauty"; but the author thinking otherwise, presents this brief, yet comprehensive, Appendix to the public, in the full assurance that whoever will pay due attention to the subject will be highly gratified and abundantly rewarded. I 193 APPENDIX. THE pupil having passed over the common routine of Arithmetic, and supposed to be able to perform all its ne^ cessary operations in the usual way, we now present nim with some new modes of practical operations, by which the labors will be much abridged and the science pre- sented with more of its roses and less of its thorns. We commence with a systematic study of numbers. The following are called prime numbers ; because no one can be divided by any number less than itself with- out producing a fraction. 123. 5. 7. ..11. 13... 17 . 19 ... 23 .. . . 29 . 31 37 ... 41 . 43 . . . 47 53 . . . 57 . 59 . 61 . . . . 67 . . . 71 . 73 79 ... 83 ... 87 . 89 97 ... 101 The points represent the composite numbers; and here it can be observed that there are 29 prime numbers, and of course 71 composite numbers in the first hun- dred the prime numbers becoming fewer as the num- bers rise higher. Observe the following series : 5 10 15 20 25 30 35 40 45 50, and so on. Every body knows that our Geometric scale of numbers i* 1 10 100 1000, &c. Now we wish the student to observe the numbers, 5 20 25 50 75 125 500, as being not only in the preceding series, but aliquot parts of some number in our Geometric scale. For ex- ample, 25 is I of 100 ; 125 is 1- of 1000, &c. We now charge the student to make his eye familiar with all the preceding series the prime numbers as be- ing unmanageable and inconvenient, and the others the very reverse ; but the full importance of such a study can only appear in the sequel. 194 APPENDIX. 195 By a little attention to the relation of numbers, we may often contract operations in multiplication. A dead unifomity of operation in all cases indicates a mechani- cal and not a scientific knowledge of numbers. As a uniform principle, it is much easier to multiply by the small numbers 2, 3, 4, 5, than by 7, 8, 9. EXAMPLES. ^ Multiply 4532 by 39 Commence with the 3 tens. Mul- 13596 tiply this 13596 by 3, because 3X3=9, 40788 and place the product in the place of units. 176748 Multiply 576 Multiply this last number, by. ... 186 3456, (which is 6 times 576) by 3, and place the product (6X3=18.) 3456 in the place of tens, and we 10368 have 180 times 576. Ob- serve the same principle in 107136 the following examples. Multiply .... 576 Multiply 40788 by 618 by .... 497 Commence with 6. 3456 (7X7=49.) 285516 (6X3 = 18.) 10368 1998612 355968 20271636 Observe, that in this last example 497 is 3 less than 500 ; 500 is \ of 1000, thefore 2)40788*000 20394 000 Subtract (3X40788=122364.) 122 364 20271 636 196 APPENDIX. Multiply ........... 785460 by ............. 14412 First multiply by 12, then that 9425520 product by "2. 113106240 11320049520 Multiply 86416 by 135. Observe, 135 is. 125+10 - 125 is | of 1000, therefore 8)8 64 1 6'0 10802 000 864 160 Product 11666160 tg are from Ray's Arithmetic, page 34 : Multiply 1646 by 365. As the first factor is even, and the last ends in 5, we mentally half the one and double the other, which will not affect the product, but very much contract the operation, we then have 823 730 24690 5761 600790 We can do the same in all such cases. Multiply 999 by 777. 777000 Subtract 777 Product 776223 APPENDIX. 197 Multiply. 61524 by. ... 7209 5537 1 6 Multiply this product of 9 by 4429728 8, because 9 times 8 are 72, and place the product in the place Product 443646516 of 100, because it is 7200. Multiply 1243 by 636 7458 First by 600. 44748 Multiply 7458 by 6. Product 790548 Multiply 624 by 85. The product will be the same as the half of 624 by the double of 85 that is, 312 by 170, or 3120 by 17=53040. We do not say that these changes give any advantage in this particular example; we only make them to call out thought and attention from the pupil. Multiply by- This may be done by commencing with the 2 ; then that product by 2 and 3. Or we may commence with the 6 units, and then that product by 4 ; because 4 times 6 are 24. Multiply 4386, or any other number, by 49. We may do this by taking the number 50 times, or ^ of 100 times, and subtracting once the number. Multiply 87742 by 65. This may be done by taking the number 1 (100,) +10+! C( 10 ) times - That is 4387100 877420 438710 Product ..5703230 77* 198 APPENDIX. Multiply 92636 by 150. It will be much more easy to multiply the half of it by 300, which will give the same result. Multiply 679 by 279. Multiply first by 9, and that product by 3, put in the place of 10. Multiply 87603 by 9865. By the common formal rule this would be a tedious operation ; but let us observe that 9865 is 10000 135, 135 is 125-f-10 ; 125 is } of 1000. Therefore, 876030000 Subtract I, 8)87603000 11826405 : 10950375 864203595 Product. 10 times, 776030 11826405 Multiply 818327 by 9874. But 9874 is 10000 less 126. Therefore 8183270000 Less 1 of 818327000 Subtract 103108202 102290875 8080161798 Product. 818327 103109202 Multiply 188 by 135, and that product by 15. In- stead of doing this literally and mechanically according to rule, we may half 188 twice, and double each of the fac- tors that end in 5, and we shall have 47 . 270 . . 30 ; or, 47 ' 8100 4700 376 380700 How far will a ship sail in 365 days, at the rate APPENDIX. 199 of 8 miles per hour? Here 365X24X8730X96; or, 73000 less 730X4=2920 2920 Product 70080 EXERCISES FOR PRACTICE. 299 X 299 X299=what number? Ans. 26730899 999X999X999= what number? Ans. 997002999 4962 X 98 = what number ? Ans. 487276 Multiply 8340745 by 64324. Observe, that 32 is 8 times 4, and 64 is 2 times 32. Multiply 8340745 by 64432. In this last example, commence at the 4 in the place of hundreds. Multiply 24 X 25 X 12 X 5, together. Here it would be no index of even a decent knowledge of numbers, to obey literally and in the order the numbers are given ; yet how many even at the present day would do so ! Take the factor 4, out of the 24, and multiply it into the 25 ; also, the factor 2, out of 12, and put it with 5, which can be done without effort. We then have 6X 100 X 10 X 6=36000, the answer. SECTION II. WE shall say nothing of division in whole numbers, as nothing new or interesting can be offered on that topic ; but we cannot forbear making a few comments on division in decimals. To divide, to cut into parts, will not at all times give a clear understanding of the operation, and confusion frequently arises from taking this view of the subject; we better consider it as one number measuring another. For example : how often will .5 of a foot measure 12 feet ? In other words, divide 12 by .5 ; or, divide 12 by 200 APPENDIX. t. Here, if the student should imagine that 12 must be cut into parts, he would make a great error. He must divide 120 tenths into parts ; in this case, into 5 parts because the 5 is .5 : or he may consider that \ of a foot may be laid down in 12 feet; that is, measure 12 feet 24 times. Or he may reduce the 12 feet to half feet, and then divide by 1. In all cases, the divisor and dividend must be of the same denomination before the division can be effected. But in decimals, these reductions are made so easily, that a thoughtless operator rarely per- ceives them ; hence the difficulty in ascertaining the value of the quotient. We now give a few examples, for the purpose of teaching the pupil how to use his judgment; he will then have learned a rule more valuable than all others. EXAMPLES. Divide 15.34 by 2.7. Here we consider the whole number, 15, is to be divided by less than 3 ; the quo- tient must, therefore, be a little over 5. One figure then, in the quotient, will be whole numbers, the rest decimals. Divide 15.34 by .27. Here we perceive that 15 is to be divided, or rather measured, by less then % of 1 ; therefore the quotient must be more than 3 times 15. Or we may multiply both dividend and divisor by 100, which will not effect the quotient, and then we shall have 1534, to ba divided by 27. Now no one can mis- take how much of the quotient will be whole numbers : the rest, of course, decimals. Divide 45.30 by .015. Conceive both numbers to be multiplied by 1000 ; then the requirement will-be to di- vide 45300 by 15, a common example in whole num- bers. By attention to this operation, the student will have no difficulty in any case where the divisor is less than the dividend. Here is one of the most difficult cases : Divide .003753 by 625.5. In all such examples as this, we insist on the formality of placing a cipher in APPENDIX. 201 the dividend, to represent the place of whole numbers, thus : 625.5)0.0037530( We now consider whether the whole number in the divisor will be contained in the whole number in the div- idend, and we find it will not ; wr, therefore, write a cipher in the quotient to represent the place of whole numbers, and make the decimal on its right, thus, 0. We now consider, that 625 will not go in the 10's, nor in the 100's, nor in the 3, nor in the 37, nor in the 375, but it will go in the 3753. We must make a trial at every step, that is, every time we take in view another place ; and we must take but one at a time. In this case, then, we shall have 0.000006, the quotient. Divide 3 by 30. 30 will not go in 3 ; we, therefore, write for place of whole numbers, and then say 30 in 30 tenths, 1 tenth times ; or, 0.1. Divide .55 by 11. 11)0.55(0.05 11 in 0, no times ; 11 in 5 tenths, times ; 11 in 55 hundredths, 5 hundredths times. It will be observed, that we make the decimal point in the quotient as soon as we ascertain it ; not wait, and then find where it should be by counting, &c. a rule that we regard as unworthy of being followed by all those who can use their reason. EXAMPLES. 1 Divide 9 by 450. Ans. 0.02. V Divide 2.39015 by .007. Anss. 341.45. 3 Divide 100 by .25. Ans. 400.00. 4 If 350 pounds of beef cost $12.25, what is the cost af one pound ? Ans. .035. 5 At $5.75 per yard, how much cloth can be pur- chased with $19.50625 ? Ans. 3.375 yards. i 2 202 APPENDIX. 6 At .07 per cent, per annum, how much capital must be invested to yield $602 ? Ans. $8600. 7 A benevolent individual gave away $600 per annum to charitable objects, which was .12 of his income. What was his income ? SECTION III. Multiplication and Division Combined. WHEN it becomes necessary to multiply two or more numbers together, and divide by a third, or by a product of a third and fourth, it must be literally done, if the numbers are prime. For example : Multiply 19 by 13, and divide that pro- duct by 7. This must be done at full length, because the numbers are prime ; and in all such cases there will result a frac- tion. But when two or more of the numbers are composite numbers, the work can always be contracted. Example : Multiply 375 by 7, and divide that pro- duct by 21. To obtain the answer, it is sufficient to di- vide 376 by 3, which gives 125. The 7 divides the 21, and the factor 3 remains for a divisor. Here it becomes necessary to lay down apian of oper- ation. Draw a perpendicular line, and place all numbers that are to be multiplied together under each other, on the right hand side, and all numbers that are divisors under each other, on the left hand side. EXAMPLES. 1 Multiply 140 by 36, and divide that product by 84. We place the numbers thus : 84 14 84 36 APPENDIX. 203 We may cast out equal factors from each side of the line without affecting the result. In this case, 12 will divide 84 and 36. Then the numbers will stand thus : 140 3 But 7 divides 140, and gives 20, which, multiplied by 3, gives 60 for the result. 2 Multiply 4783 by 39, and divide that product by 13: 4783 A* ** 3. Three times 4783 must be the result. 3. Multiply 80 by 9, that product by 21, and divide the whole by' the product of 60X6X 14. 9 to 3 In the above, divide 60 and 80 by 20, and 14 and 21 by 7, and those numbers will stand cancelled as above, with 3 and 4, 2 and 3 at their sides. Now the product 3X6X2, on the divisor side, is equal to 4 times 9 on the other, and the remaining 3 is the result. Hoping the reader now understands our forms, and comprehends the true philosophical principle, we will give no more abstract examples ; but we will give many practical examples, such as might occur in business, and we prefer taking them from books, that it may not be said we made them expressly for this occasion. Again, it may be observed, that this method of opera- tion may serve for only a few problems. We answer, it will serve for 71 out of 100, according to the theory of numbers, as we have seen there are 71 composite num- bers in the first hundred,, and more as they rise higher. But the prime numbers, 2, 3, and 5, are so small and manageable ami are factors in FO many other numbers, 204 APPENDIX. that they may be considered in as favorable a light as the composite numbers and for this reason we may say, 75 out of every hundred problems can be abridged. But, in actual business, the problems are almost all reduceable by short operations ; as the prices of articles, or amount called for, always corresponds with some aliquot part of our scale of computation. This method may not work a great many problems as they are found in some books, but it will work 90 out of every 100 that ought to be found in books. *In a book, we might find a problem like *his: - What is the cost of 21b. 7oz. 13dwt. of tea at 7s. 5d. per pound ? But the person who should go to a store and call for 31b. 7oz. and 13dwt. of tea, would be a fit subject for a mad-house. The above problem requires downright drudgery, which every one ought to be able to perform, but such drudgery never occurs in business. The following examples are extracted from books in common use, and we mark them in order that any one may find the original. For instance, T. 42, means Tal- bott's Arithmetic, page 42 ; R. 93, means Ray's Arith- metic, page 93 ; E. 123, means Emerson's Arithmetic, page 123, &c. EXAMPLES. 4 How many bushels of apples can be bought for $3, at 15 cts. a peck? (R. 92.) Ans. 5. $ Explanation 3 in 15, 5 times ; 4 5 times 5 are 20, and 20 in 100, 5 times. 5 A 'farmer has 91 bushels of wheat, and he wishes to put it into bags, each of which holds 3 bushels 2 pecks ; how many bags will it take ? (R. 92.) Ans. 26. 3bu. 2 pe. = 14pe. 13 6 What is the value of a piece of gold weighing lib. 3dwt., at 12|cts. per grain ? (R. 92.) Ans. $729. .1243 ? \&L 3 APPENDIX. 205 7 At Sets, a pound what will 6cwt. Iqr. cost? (R. 83.) Ans. $21. 8 At $2.25 a qr. what will 1 ton Icwt. cost? (R. 93.) Ans. $189. 21 cwt. 4 9 9 At 5cts. per oz., what will 7 Ibs. 8oz. cost? (R. 93.) Ans. $6. 10 In this example, we must reduce 7lbs. 8oz. to ounces: 7X16+8 is the same as 14X8+8, or 15X8. 100 15 8 ^=100 5 30 2 10 1 1 A grocer bought a lot of cheese, each weighing 91bs. 15oz., the weight of the whole amounted to 39cwt. 3qr.; how many cheeses were there ? (R. 93.) Ans. 448. qr. 28 16 12 How many casks, each holding 841bs., can be fill- ed out of a hogshead of sugar weighing 15cwt. 3qr? (R. 93.) Ans. 17. 13 A bell of Moscow weighs 288000lbs.; how many tons? (R. 93 part of Ex. 23.) Ans. 128 tons. 11 cwt. Iqr. 20lbs. 14 "VVe prefer dividing (mentally) the pounds into their obvious factors. 7)900 . , 1000 1281 15 How many times will a wheel, which is 9ft. 2in. 18 206 APPENDIX. in circumference, turn round in going 65 miles ? (R. 94, Ex. 32.) Ans. 37440. 65 X10 320 12 16 What will 2 square yards 2 square feet of ground come to at Sets, a square inch ? (R. 94.) Ans. $144 2 square yards 2 square feet=20 square feet. 12 12 17 What will one square yard of gilding cost at 12.5 cents a square inch? (R. 94.) Ans. $162. ft 8 9 144 18 What will 5 yards 2qrs. of cloth cost at 12|cts. a nail? (R. 96.) Ans. $11. 8 I 22 qrs. 19 How many coat patterns, each containing 3 yards 2qrs., can be cut out of a piece of cloth containing 70 Ells Flemish? (R. 95.) Ans. 15. H 7 3 20 What will one hhd. of wine cost at 6lcts. a gill? (R. 95.) Ans. $126. Observe, that 6| = 2 T S ; and, as 4 is a divisor to 25, it must be put on the opposite side of the line. 100 4 63 4 2 4 25 or, 16 63 4 2 4 21 If a person write 10 minutes each day, how APPENDIX. 207 much time will that amount to in 4 years ? (R. 96.) Ans. 10 days 85 hours. 60 24 365 4 10 22 How many yards of carpeting, 2 feet 6 inches in breadth, will cover a floor 27 feet long and 20 feet wide ? (T. 98.) Ans. 72. 2=f. 2 is to divide the 5 ; it must, 5 therefore, go over the line. 3 20 27 2 23 What quantity of shalloon, 3 quarters wide, will line 71 yards of cloth that is 1 yards wide? (T. 98.) Ans. 15. or, 15 N. B. Mixed numbers are reduced to improper frac- tions, and the denominators thrown over the line. 24 How much land, at $2.50 per acre, must be given in exchange for 360 acres, worth $3.75. (T. 99.) Ans. $540. 360 31 or, 90 3 25. What will a bnshel of clover-seed come to at 12; cts. a pint? (Wilson, 41.) Ans. $8. 12 cts.=| of a dollar. The 8 on one side cancels 8 on the other, and leaves 4X2 for the answer. 4 pecks. 26 Suppose a hogshead of molasses, which cost $23, be retailed at 125Cts. a quart; what is the profit on it? (W. 41.) Ans. $8.50. 208 APPENDIX. 63 gallons. Sale Cost 31.50 23 Profit 8.50 27 What will 5 barrels of flour cost at 3|cts. per pound? Ans. $34.30. 28 How many times is of a pint contained in i of a gallon ? (W. 65.) Ans. 6|. or, f 20 We have already remarked, that denominators of fractions must go over the line from the term to which they belong. 29 How many times can a vessel, holding T 9 ^ of a quart, be filled from ~ of a barrel containing 31 gal- lons ? (W. 66.) Ans. 46f . 81* 5 == 10 30 If one acre and 20 rods of ground produce 45 bushels of wheat ; at that rate, how much will nine acres produce ? (W. 90.) Ans. 360. la. 20r. = 180. 45X8=360. 45 N. B. We shall plan no more problems in this sec- tion ; but the following require no real labor, save cor- rect reasoning. When the numbers are properly arrang- ed, a few clips with the pencil, and perhaps a trifling multiplication, will suffice. 31 At l|cts. a gill, how many gallons of cider can be bought for $12 ? (R. 95.) . Ans. 25. 32 How many casks, each containing 12 gallons, can be filled out of a ton of wine ? (R. 95.) Ans. 21. APPENDIX, 209 33 A man retailed 9 barrels of ale, and received for it $129.60 ; at what price did he sell it a pint ? (R. 96.) Ans, Sets. 34 How much butter, at 9cts, per pound, will pay for 12 yards of cloth, at $2.19 per yard ? (W. 79.) Ans. 292 35 At 45| dollars per acre, what will 32 rods of land come to? (W. 79.) Ans, $9.10. 36 How long must a laborer work, at 62;|cts. a day, to earn $25 ? (W. 78.) Ans. 40 days. 37 A merchant sold 275 pounds of iron, at 5|cts, a pound, and took his pay in oats, at 50cts. a bushel; how many bushels did he receive ? (Adams, 53.) Ans. 38. 38 How many yards of ctoth, at $4.66 a yard, must be given for 18 barrels of flour, at $9.32 a barrel? (A. 53.) Ans. 36. 39 How long will it require one of the heavenly bo- dies to move through a quadrant, at the rate of 43' 12" per minute ? (R. 97.) Ans. 2~ hours. 40 If a comet move through an arc of 7 12- per day, how long would it be in passing an arc of 180 ? (R. 97.) Ans. 25 days. 41. What is the cost of 8hhds. of wine, at 5cts. per pint? Ans, $201.60. 42 There are 30| square yards in one perch of land ; how many perches are there in 363 square yards ? Ans. 12. 43 What will 18 1 yards cost, at 75cts. per yard ? Ans. $14.06|. 44 If 16 persons receive $516 for 43 days' work, how much does each man earn per day ? Ans. 75cts. 45 How many times will a wheel, which is 12 feet in circumference, turn round in going a mile ? Ans. 440. 46 An auctioneer sold 45 bags of cotton, each contain- 210 APPENDIX. ng 400 pounds, at 1 mill a pound ; what did the whole come to? (R. 61.) Ans. $18. 47 A mechanic receives $90 for 40 days T work he ivorked 12 hours each day; how much was it per hour? (R. 73.) Ans, ISfcts. 48. A laborer worked 26 days for a farmer, at 87|cts. per day, and took his pay in wheat, at 65cts. per bushel ; how many bushels did he receive ? Ans. 35. SECTION IV. IT is an axiom in philosophy, that equal causes pro- duce equal effects ; and that effects are always propor- tionate to their causes. Now, causes and effects that admit of computation, that is, involve the idea of quantity, may be represented by numbers, which will have the same relation to each other as the things they represent. Keeping these premises in view, then, we have a uni- versal rule, applicable to all cases which can arise under Proportion, simple or compound, direct or inverse namely : RULE. As any given cause is to its effect, so is any required cause (of the same kind] to its effect ; or, so is another given cause, of the same kind, to its requirea effect. The only difficulty the pupil can experience in thi system of proportion, is readily to determine what is cause, and what is effect. But this difficulty is soon over eome, when we consider, that all action, of whatsoevei nature, must be cause and whatever is accomplished by 3 that action, or follows such action, must be effect. y EXAMPLES. > If 10 horses, in 50 days, consume 128 bushels oi APPENDIX. 211 oats, how many bushels will 5 horses consume in 90 days? (W. 113.) Ans. 72. Here it is evident, that the consumption of oats spoken of, in both the supposition and demand, are the true ef- fects ; and the action of the horses, multiplied by the days, must express the amount of cause. We shall therefore state it thus : Cause. Effect. Cause. Effect. 16 : 128 : : 5 : [ ] 50 90 We write the factors, one under another, as above ; their multiplication is understood, but rarely or never actually accomplished. The second effect is an un- known term, or answer, required a bracket, or blank, or point, is left to represent it. When found, the four terms above would be a perfect Geometrical proportion, and the product of the extremes equal to the product of the means. In this example, the product of the means is perfect ; which product, divided by the factors in the extremes, will give what is wanting in the extremes, nainly the answer. Therefore, agreeably to one mode of performing mul- tiplication and division, we draw a line thus, and cancel down : ifi 128 16 - If $480, in 30 months, produce $84 interest, what cap- ital, in 15 months, will produce $21 ? Now capital will not produce interest without time ; and, whatever be the rate per cent., the same capital in a double time, will produce a double interest. Therefore, Cause. Effect. Cause. Effect. 480 : 84 : : [ ] : 21 30 15 212 APPENDIX. Here one element of the second cause is wanting ; that is, the answer to the question. In this case, the extremes are complete; we will, therefore, divide the product of^the extremes by the fac- tors in the mean, and the quotient will give the definite factor, or answer, namely $240. 3 If 7 men, in 12 days, dig a ditch 60 feet long, 8 feet wide, and 6 feet deep, in how many days can 21 men dig a ditch 80 feet long, 3 feet wide, and 8 feet deep? . It is almost too plain for comment, that 7 men, multi- plied by 12 days, must be the first cause, and the con- tents of the ditch they dig, the effect. Therefore, Cause. Effect. Cause. Effect. 7 : 60 : : 21 : 80 12 8 [ ] 3 6 8 Here, as in the preceding example, one of the elements of the second cause is wanting ; or, rather say a factor in the means of a perfect proportion, and can be found as above. Ans. 2 days. 5 If 6 men build a wall in 12 days, how long will it require 20 men to build it ? Ans. 3| days. Questions of this kind are usually classed under the single rule of three inverse ; they do in fact, however, belong to compound proportion : but, as one of the terms is the same in the supposition as in the demand, it may be omitted. The term, in the present example, is, one wall. If we make the number different in the two branches of the question, or connect any conditions with it, such as lengths, breadths, &c., it at once falls under compound proportion of necessity, and may be stated thus : Cause. Effect. Cause. Effect. 6:1 : : 20 1 12 [] 5 If 4 men, in 2^ days, mow 6| acres of gra^c by - APPENDIX. 213 working 8| hours a day, how many acres will 15 men mow in 3 1 days, by working 9 hours a day? Ans. 40 y acres. Cause. Effect. Cause. Effect. 2' ' 6| : : l l '' C ] 81 9 Let the pupil observe, that when a correct statement is made, there will be the same number of elements, or factors, under the same letters, as in the above example under each Cause. We have men, days, and hours, to be multiplied together. When there are fractions in any of the terms, their denominators are to be placed over the line from where the term belongs ; mixed numbers being previously reduced to improper fractions. 6 If 12 oz. of wool make 1 yards of cloth, | of a yard wide, how many yards, 1| wide, will 16 pounds of wool make ? Ans 22 1 yards. With this example, some might hesitate as to arrang- ing it under cause and effect, as the actors, those wh^> made the cloth, whether many or few, have nothing to do with the question. But we take the phrase of the example and say, The wool makes the cloth. Cause. Effect. Cause. Effect. 12 : U : : 16 I 16 7 If the transportation of 12icwt. 206 miles cost $25.75, how far, at the same rate, may 3 tons and 3qrs. be carried for $243 ? Ans. 402f miles. In this example, it is indifferent which we take for the cause and which for the effect. We may say, that the money, $25.75, is the cause of having the weight car- ried ; or, we may say, that carrying the weight is the cause of purchasing the money. There are many ques- tions where it is indifferent which we take for cause, and 214 APPENDIX. which for effect. The above example may be stated thus : Cause. 25| : Or thus : Cause. 12| 206 Or thus : Cause. 206 T Or thus : Effect. 243 Effect. 12icwt. 206 miles. Effect. Cause. 60| C3 Effect. Cause. 243 Effect. 60|cwt, Cause. Effect 60| : 243 CD Effect. Effect. OK3 . OJ.Q -O.T . x-o Cause. 60! Cause. 121 206 These changes show, conclusively, that this method of statement is strictly scientific and philosophical ; and, in all these different arrangements of the terms, the same terms are multiplied together. The most that can be said for the common modes of statements, in the Double Rule of Three, is, that the products, when the terms are multiplied out, are pro- portional. But the first and second terms, taken as a whole, express no particular idea or thing; whereas, in this mode of statement, the thing the philosophical idea is the only sure guide. Nor is this all ; it is very extensive and easy in its application ; it will cover every case that can arise under interest to find time, rate per cent. &c., and thus do away, or suspend, live or six special rules, which encumber every arithmetic. We give a few examp'es to apply this rule. 8 What is the interest of $240 for 3 years at 6 per cent.? Cause. Effect. Cause. Effect. 100 : 6 : 240 :] APPENDIX. 215 To obtain the answer from this statement, we perceive that we must multiply the means together i. e. 240X6, the rate, and that by the 3 years, the time and divide by 100 ; and this is the common rule. 9 The interest of a certain sum of money, at 6 per cent., for 15 months, was $60 ; what was the sum? Ans. $800. Cause. Effect.^ Cause. Effect. 100 : 6 " : : [ ] : 60 12 15 10 If $800, in 15 months, should gain $60, what would be the rate per cent.? Ans. 6. Cause. Effect. Cause. Effect. 800 : 60 : : 100 :[] 15 12 11 Eight hundred dollars was put out at interest, at 6 per cent., and the interest received was $60; how long was it out? Ans. 15 months. Cause. Effect. Cause. Effect. 100 : 6 : : 800 : 60 12 [] 12 If 12 men, working 9 hours a day for 15| days, were able to execute f of a job, how many men may be withdrawn and the residue be finished in 15 days more, f the laborers are employed only 7 hours a day ? (W. 109.) Ans. 4 men. 13 The amount of a note, on interest for 2 years and 6 months, at 6 per cent., is $690; required the principal. (R. 172.) Ans. 600. 14 What is the interest of $1248 for 16 days? (R 162.) Ans. $3.28. Cause. Effect. Cause. Effect. 100 : 6 : : 1248 : [ ] 216 APPENDIX, This can be cancelled down and made very brief. 15 What is the interest of $1200, for 15 days, at 6 per cent.? (R. 162.) Ans. $3.00. 16 How many men will reap 417.6 acres in 12 days, if 5 men reap 52.2 acres in 6 days ? (T. 156.) Ans. 20. 17 If a cellar 22.5 feet long, 17.3 feet wide, and 10.25 feet deep, be dug in 2.5 days, by 6 men working 12.3 hours a day, how many days, of 8.2 hours, should 9 men take to dig another 45 feet long, 44.6 wide, and 12.3 deep ? (T. 156.) Ans. 12. 18 What is the interest of 160 dollars for 36 days, at 7 per cent.? Stated by cause and effect. Ans. $1.12. Cause. Effect. Cause. Effect. 100 : 7 : : 160 : [ ] 12 1.2 19 The interest of a certain sum, for 36 days, is $1.12 the rate per cent, is 7 ; what is the sum ? Ans. $160. 20 The interest of $160 for 36 days, is $1.12 ; wHt was the rate per cent.? Ans. 7. 21 The interest of $160, at 7 per cent., was $1.1$ ; what was the time ? Ans. 36 days. 22 In what time will any sum double itself at 6 per cent.? At any per cent.? Ans. At 6 per cent., in 16J years; at any per cent., divide 100 by the per cent 23 If 2k yards of cloth, 1J wide, cost $3.37, how much will 36 yards cost, 1% yards wide? Ans. $52.79. 24 If 4 men spend | of of f of i-J of ^30, in T 7 T of T \ of 4 of y of 9 days, how many dollars, at 6 shillings each, will 21 men spend in ? of yf of -f of T 4 j of 45 days? (Burnham, 142.) Ans. $630. 25 I lend a friend $200 for 6 months ; how long ought APPENDIX. 217 he to lend me $1000, to requite the favor allowing 30 days to a month ? Ans. 36 days. 26 If 1000 men, besieged in a town, with provisions for 5 weeks, allowing each man 16 ounces per day, be reinforced with 500 men more and supposing that they cannot be relieved until the end of eight weeks how many ounces a day must each man have, that the provis- ions may last them that time ? (W. 183.) Ans. 6J ounces The advocates of this system are of opinion, that there are far too many rules in our common arithmetics ; and to reduce them, and thereby simplify the science, they recommend that all the problems, generally arrang- ed under Profit and Loss, Equation of Payments, &c., should be solved by proportion, and arranged under that head. In this light, they are more simple and in- telligible than they can be made by any special rules. EXERCISES FOR PRACTICE. 1 If I buy cotton-cloth at 2s. per yard, and sell it at 2s. 8d,, what do I gain per cent? (W. 134.) Ans. 33. Statement. If 24 pence gain 8 pence, what will 100 pence gain? Or, 24 : 8 : : 100 : Ans. Or, 3 : 1 : : 100 : Ans. 2 A merchant bought broadcloth, at $5.50 per yard, and sold it for $6.60 ; what did he gain per cent? (W. 135.) Ans. 20. 3 If I buy Irish linen at 2s. 3d. per yard, how must I sell it to gain 25 per cent? Ans. 2s. 9d. 3h. Statement. If 100 pence return 125 pence, how much must 27 pence return ? Or, 100 : 125 : : 27 : Ans. Or, 4 : 5 : : 27 : Ans. 4 If, by selling cloth at $6.50 per yard, I lose 20 per cent., what was the prime cost of it ? (W. 137.) Ans. $8.12. 218 APPENDIX. That is, if 80 I now receive originally cost me 100, what did 6.50 originally cost? 5 By selling calico at 37^ cents a yard, 50 per cent was gained; what was the first cost? (R. 185.) Ans. 25cts. 150 : 100 : : 37| : Ans. Or, 3 : 2 : : 37 : Ans. 6 Sold wine at $1.36 per gallon and lost 15 per cent.; what per cent, would have been gained had the wine been sold for $1.856 per gallon ? (R. 187.) Ans. 16. 7 Bought 126 gallons of wine for 150 dollars, and re- tailed it at 20cts. per pint ; what was the gain per cent.? (T. 129.) Ans. 34f. 8 If $126.50 are paid for llcwt. Iqr. 25lbs. of su- gar, how must it be sold a pound to make 30 per cent, profit? (W. 135.) Ans. 12|cts. 9 If I buy 124cwts. of sugar for $140, at how much must I sell it per pound to make 25 per cent.? Ans. 12jcts. 10 If a firkin of butter, containing 561bs. cost $7, at how much must it be sold per pound to yield 30 per cent, profit? Ans. 16|cts. 11 What is the whole loss, and what is the loss per cent., in laying out $70 for hats, at $1.75 each, and sell- ing them for 25cts. a-piece less than cost? (Burnham, 173.) Ans. Whole loss $10; loss per 100, 14f. 12 A merchant purchases 180 casks of raisins, at 16 shillings per cask, and sells the same at 28 shillings per owl., and gains 25 per cent.; what is the weight of each cask? (B. 174.) Ans. 80lb. We multiply 180 by 16, and to 180 add for 25 per cent., we multiply 16 by 5 and divide by 4. Then di- 4 5 vide by 28, and it gives cwt.; multiply 28 1J2 by 112, and we have pounds; then di- 180 vide by 180, and we have pounds in each cask That is, arrange the numbers as above, and cancel down. APPENDIX. 219 For other examples, the student is referred to the body of the work. SECTION V. Square and Cube Roots. To work the square and cube roots with ease and fa- cility, the pupil must be familiar with the following pro- perties of numbers. I. A square number, multiplied by a square number, the product will be a square number. II. A square number, divided by a square number, the quotient is a square. III. A cube number, multiplied by a cube, the product is a cube. IV. A cube number, divided by a cube, the quotient will be a cube. If the square root of a number is a composite number, the square itself may be divided into integer square fac- tors ; but if the root is a prime number, the square can- not be separated into square factors without fractions. N. B. Substitute the word cube, for square, in the preceding sentence, and tiie same remarks apply to cube numbers. x No person can extract roots with any tolerable degree of skill, without being able to recognize the squares and cubes of the nine digets as soon as seen. Numbers, 1 2 3 4 5 6 N 8 9 i] Squares, 1 A 9 | 35 36 H 64 31 ,100 Cubes, j 1 8 27 64 125 216 J343| 512 729 luool We here wish to remind the reader, that the pupil is supposed to understand the extraction of the roots in 220 APPENDIX. the common way, and we request them not to forget that this is merely an appendix. EXERCISES FOR PRACTICE. 1 What is the square root of 625 ? (R. 220.) Ans 25. If the root is an integer number, we may knpw, by the inspection of the above table, that it must be 25, as the greatest square in 6 is 2, and 5 is the only figure whose square is 5 in its unit place. Again, take 625 Multiply by 4 4 being a square. 2500 The square root of this product is obviously 50 ; but this must be divided by 2, the square root of 4, which gives 25, the root. 2 What is the square root of 6561 ? (R. 220.) Ans. 81. As the unit figure, in this example, is 1, and in the line of squares in the above table, we find 1 only at 1 and 81, we will, therefore, divide 6561 by 81, and we find the quotient 81 ; 81 IS, therefore, the square root. 3 What is the square root of 106729? (T. 170.) Ans. 327. As the unit figure, in this example, is 9, if the number is a square, it must divide by either 9, or 49. After di- viding by 9 we have 11881 for the other factor, a prime number, therefore its root is a prime number= 109. 1 09, multiplied by 3, the root of 9, gives 327 for the answer. 4 What is the root of 451584 ? (T. 179.) Ans. 672. As the unit figure is 4, and in the line of squares we find 4 only at 4 and 64, the above number, if a square, must divide by 4, or 64, or by both. APPENDIX. 221 We will divide it by 4, and we have the factors 4 and 112896. This last factor closes in 6; therefore, by looking at the table, we see it must divide by 16, or 36, <fec. &c. We divide by 36, and we have the factors 36 and 3136 ; divide this last by 16, and we have 16 and 196 ; divide this last fraction by 4, and we have 4 and 49. Take now our divisors, and last factor, 49, and we have for the original number the product of 4X36X 16 X4X49; the roots of which are 2X 6X4X2X7, the products of which are 672, the answer. 5 Extract the square root of 2025. (E. 163.) Ans. 45. Divide by 25, and we have its square factors, 25 and 81. Roots of these factors are 5X9=45, the answer; Again, multiply by the square number 4, when a num- ber ends in 25, and we have 8100, root 90, half of which, because we multiplied by 4, the square of 2, is 45, the answer. 6 What is the square root of 390625 ? (R. 220.) Ans. 625. 390625 Multiply by 4, ... 4 1562500 Multiply by 4 again, 4 6250000 As the number, independent of the ciphers, still ends in 25, we multiply again by 4, and we have 25000000. The root of this is, obviously, 5000. Divide by 2 three times, or by 8, and we have 625, the answer. So far, some may think this more curious than useful. However this may be, there are problems where much labor may be saved by attending to the foregoing princi- ples. The following are some of them : Find a mean proportional between 4 and 256. Ans. 32. 222 APPENDIX. Find a mean proportional between 4 and 196. Ans. 28. Find a mean proportional between 25 and 81. Ans. 45. As the above are square numbers, multiply their square roots together for the answers. EXAMPLES. 1 If 484 trees be planted at an equal distance from each other, so as to form a square, how many will be in a row each way. (T. 171.) Ans. 22. Factors 4 and 121 2X11 roots=22. 2 A section of land, in the Western states, is a square, consisting of 640 acres ; what is the length in rods of one of its sides? (W. 147.) Ans. 320. Nine out of ten of our teachers would actually reduce the acres to square rods, by multiplying by 160, and ex- tract the square root of the product but this would show too little attention to numbers. Remove one of the ciphers from one number to the other, and we have 64 to be multiplied by 1600, both square numbers, whose roots are 8 and 40 product 320, the answer. 3 What must be the side of a square field, that shall contain an area equal to another field of rectangular shape, the two adjacent sides of which are 18 by 72 rods. (W. 147.) Ans. 36 rods. 18 by 72 is the same as the half of 18 by the double of 72, or 9 by 144, square numbers, roots 3X12=36, the answer. 4 A has two fields, one containing 10 acres and the other 12| ; what will be the length of the side of a field containing as many acres as both of them ? (R. 220.) Ans. 60 rods. 22 ,5 XI 60 is the same as 225X16; roots 15X4=60, the answer. 5 What is the mean proportional between 24 and 96 ? Ans. 48. APPENDIX. 223 6 What is the mean proportional between 18 and 32? Ans. 24. Problems on the Right-angled Triangle. 1 The top of a castle is 45 yards high, and is sur- rounded with a ditch 60 yards wide ; required the length of a ladder that will reach from the outside of the ditch to the top of the castle. Ans. 75 yards. This is almost invariably done by squaring 45 and 60, adding them together, and extracting the square root ; but so much labor is never necessary when the numbers have a common divisor, or when the side sought is ex- pressed by a composite number. Take 45 and 60 ; both may be divided by 15, and they will be reduced to 3 and 4. Square these, 9+16 =25. The square root of 25 is 5, which, multiplied by 15, gives 75, the answer. 2 Two brothers left their father's house, and went, one 64 miles clue west, the other 48 miles due north, and purchased farms ; how far are they from each other? (E. 171.) Ans. 80 miles. Divide by. - 16)64 48 4 3 16+9=25,5X16=80. 3 The hypothenuse of a right-angled triangle is 520 feet, the base 312 feet; what is the perpendicular? Ans. 416. Divide by .... 52)520 312 2)10 6 5 3 25 9=16, root 4. Multiply by 104 Answer . . . .^ . . 416 4 Required the height of a May-pole, whose top be- ing broken off, struck the ground at the distance of 15 feet from the foot, and measured 39 feet. Ans. 75 feet 224 APPENDIX. 5 A hawk, perched on the top of a perpendicular tree, 77 feet high, was brought down by a sportsman, standing off 14 rods, on a level with its base ; what dis- ance, in yards, did he shoot? (W. 149.) Ans. 81.154-yards If this problem is worked with skill, it will be requi- site to extract the root of 10 only. 6 If the diagonal of a rectangular field is 40 rods, and one of the sides 32, what is the other ? (W. 150.) Ans. 24. Cubes and Cube Root. Cubes, whose roots are composite numbers, may be divided by cube factors. Cube numbers, whose unit figure is 5, may be multiplied by the cube number 8, and that period reduced to ciphers. 1 What is the cube root of 91125 ? Ans. 45. Multiply by 8 729000 Now 729 being the cube of 9, the root of 729000 is 90 ; divide this by 2, the cube root of 8, and we have 45, the answer. 2 The contents of a cubical cellar are 1953.125 cubic feet ; what is the length of one of its sides ? (R. 225.) Ans. 12.5 feet. 1953.125 Multiply by 8, - - .. 8 15625.000 Multiply by 8 again, 8 125.000 The cube root of this is 50 ; divide by 4, because we multiplied by 8 twice, and we have 12.5 the answer. 3 The number 195112 is a cube; what is its root? Ans. 58. APPENDIX. 225 The cube numbers are 8, 27, 64, 125, 216, 343, 512, 729. Comparing these numbers with 195112, and we observe, that the root, in the place of tens, cannot be more than 5, and the root, in the place of units, must be some num- ber which, when cubed, give 2 for its unit figure and 8 js the only figure possible ; the root of the whole is, therefore, 58. 4 The number 912673 is a cube ; what is its root ? Ans. 97. Observe, the root of the superior period must be 9, and the root of the unit period must be some number which will give 3 for its unit figure when cubed, and 7 is the only figure that will answer. In this manner, we can speedily and easily obtain the cube roots of all cube numbers containing not more than two periods, or determine whether they are cubes or surds. The following numbers are cubes ; required their roots. 1 What is the cube root of 59319 ? Ans. 39. 2 What is the cube root of 79507 ? Ans. 43. 3 What is the cube root of 117649? Ans. 49. 4 What is the cube root of 110592 ? Ans. 48. 5. What is the cube root of 357911 ? Ans. 71. 6 What is the cube root of 389017 ? Ans. 73. 8 What is the cube root of 571787 ? Ans. 83. When a cube has more than two periods, it can gener- erally be reduced to two by dividing by some one or more ^f the cube numbers, unless the root is a prime number. The number 4741632, is a cube; required its root. He re we observe, that the unit figure is 2 ; the unit fig- i re of the root must, therefore, be the root of 512, as that is the only cube of the 9 digits whose unit figure is 2. The cube root of 512 is 8 ; therefore, 8 is the unit figure in the root, and the root is an even number, and 226 APPENDIX. can be divided by 2 and, of course, the cube itself can be divided by 8, the cube of 2. 8)4741632 592704 Now, as the first number was a cube, and being di- vided by a cube, the number 592704 must be a cube, and, by inspection, as previously explained, its root must be 84, which, multiplied by 2, gives 168, the root required. The number 13312053, is a cube ; what is its root ? Ans. 237. As there are three periods, there must be three figures, units, tens, and hundreds, in the root ; the hundreds must be 2, the units must be 7. Let us then divide the 2d figure, or the tens, in the usual way, and we have 237 for the root. Again, divide 13312053 by 27, and we have 493039 for another factor. The root of this last number must be 79, which, multiplied by 3, the cube root of 27, gives 237, as before. The number 18609625 is a cube ; what is its root? As this cube ends with 5, we will multiply it by 8 : 18609625 8 148877000 As the first is a cube, this product must be a cube ; and, as far as labor is concerned, it is the same as reduced to two periods, and the root, we perceive at once, must be 530, which, divided by 2, gives 265 for the root re- quired. N. B. It a number, whose unit figure is 5, be mult plied by 8, -md does not result in three ciphers on the right, the number is not a cube. To find the Approximate Cube Root of Surds. The usual way of direct extraction, is too tedious to APPENDIX. 227 be much practiced, if any shorter method can possibly be obtained. By the invention of logarithms, a very short method has been found; but, before that event, several eminent mathematicians bestowed much time and labor to obtain short practical rules and some of their rules are too ingenious and useful to be lost, notwith- standing the invention of logarithms has nearly super- ceded their absolute value in practice. There is no exact and constant relation between pow- ers and their roots ; for this reason, all rules (save by logarithms, and the direct and tedious one,) must be more or less approximate; but, nevertheless, with common judgment and care, we can arrive at results as near as by the direct methdd. In order to obtain a rule, let us take two cube numbers, as near in value to each other as practicable, and compare them with their roots. 216000 and 226981 are cubes ; their roots are 60 and 61. Now 216000 is not to 226981 as 60 to 61. But let us double the first and add it to the second, and double the second and add it to the first, and we shall have 658981 and 669962, which are to each other very nearly as 60 to 61. Or, by the principles of proportion, the first is to the difference between the first and second, as is the third to the difference between the third and fourth. That is, 658981 : 10981 : : 60 : 1 very nearly. Now one is the difference between the two roots, and if the last root, or 61, was unknown, this proportion would give it very nearly. EXAMPLES. 1. Required the cube root of 66. The cube root of 64 is 4. Now it is manifest, that the cube root of 66 is a little more than 4, and by taking a similar proportion to the preceding, we have 228 . APPENDIX. 64X2 = 128 2X66=132 66 64 194 : 196 : : 4 : to root of 06. Or, 194 : 2 : : 4 : to a correction. 194)8.0000(0.04124 176 240 194 460 388 720 Therefore, the cube root of 66 is 4.04124 2 Required the cube root of 123. Suppose it 5 ; cube it, and we have 125. Now we perceive, that the cube of 5 being greater than 123, the correction for 5 must be subtracted 2X125=250 246 Add 123 125 As 373 : 371 : : 5 : root of 123. Or, 373 : 2 : : 5 : correction for 5. 373)10.0000(0.02681 746 2 540 From 5.00000 2238 take 0.02681 3020 Ans. 4.97319 2984 360 APPENDIX. 229 From what precedes, we may draw the following RULE. Take the nearest rational cube to the given number, and call it the assumed cube; or, assume a root to the given number and cube it. Double the as- sumed cube and add the number to it ; also, double, the number and add the assumed cube to it. Take the difference of these sums, then say, J?s double of the as- sumed cube, added to the number, is to this difference^ so is the assumed root to a correction. This correction, added to or subtracted from the assum- ed root, as the case may require, will give the cube root very nearly. By repeating the operation with the root last found as an assumed root, we may obtain results to any degree of exactness ; one operation, however, is generally suf- ficient. 3 What is the cube root of 28"? Ans. 3.03658-|-. 4 What is die cube root of 26? Ans. 2.96249-}-. 5 What is the cube root of 214 ? Ans. 5.98142-h 6 What is the cube root of 346 ? Ans. 7.02034-f-. The above being very near integral cubes that is, 28 and 26 are both near the cube number 27, 214 is near 216, &c., all numbers, very near cube, numbers are easy of solution. We now give other examples, more distant from inte- gral cubes, to show that the labor must be more lengthy and tedious, though the operation is the same. EXAMPLES. 1 What is the cube root of 3214? Ans. 14.75758. Suppose tho root is 15 its cube is 3375, which, being greater than 3214, shows that 15 is too great; the cor- rection will, therefore, be subtractive. 230 APPENDIX. By the rule, 9964 : 161 : : 15 : 0.2423, the cor- rection. Assumed root, 15.0000 Less 2423 Root nearly ...-.. 14.7577 Now assume 14.7 for the root, and go over the opera- tion again, and you will have the true root to 8 or 10 places of decimals. 2 What is the cube root of 14760213677 1 Ans. 2453. Suppose the root 2400, &c. Take the correction to the nearest unit, and you will find it 53. 3 What is the cube root of 980922617856? Ans. 9936. Suppose the root to be 10000. 4 What is the cube root of 9 ? Ans. 2.08008. 5 What is the cube root of 9^? Ans. 2.092-f 6 What is the cube root of 41 ? Ans. 3.44S2-J-. When it is requisite to multiply several numbers to- gether and extract the cube root, try to change them into cube factors, and extract the root before the multiplica- tion. EXAMPLES. 1 What is the side of a cubical mound equal to one 288 feet long, 216 feet broad, and 48 feet high? (R. 225.) The common way of doing this, is to multiply these numbers together and extract the root, a lengthy opera- tion. But, observe, that 216 is a cube number, and 288 =2X12X12, and 48=4X12; therefore, the whole product is 216X8X12X12X12. Now the cube root of 216 is 6, of 8 is 2, and of 12 3 is 12, and the product of 6X2 X 12 = 144, the answer. 2 Required tlie cube root of the product of 448 X392, in a brief manner. Ans. 56. 3 If you have a pile of wood 32 feet long, 4 feet APPENDIX. 231 wide, and 4 feet high, what v/ould be the side of a cubic pile containing the same quantity ? Ans. 8 feet. Proposition The solid contents of cubes or spheres are to each other as the cubes of their like dimensions. (See Geometry.) EXAMPLES. 1 Mercury is about 2000 miles in diameter, and the earth about 8000; what is their relative magnitudes ? Ans. As 1 to 64. 2 Mars is about 4000 miles in diameter, the earth 8000 ; what is their relative magnitudes ? Ans. As 1 to 8. In the preceding examples we, of course, do not cube the numbers given, but their smallest integral relations. 3 The diameter of the earth, to that of the sun, is nearly as 1 to Ills ; what is their relative magnitudes, or bulks? Ans. As 1 to 1384472 nearly. 4 If a ball, 6 inches in diameter, weigh 32lbs., what will be the weight of a ball, of the same metal, whose diameter is 3 inches ? (R. 225) Ans. 41bs. 6:3:: 2:1 8 : 1 : : 32 : 4 5 Suppose an iron ball, of 4 inches in diameter, to weigh 9 pounds ; required the weight of a spherical shell of 9 inches in diameter, and 1 inch thick. Ans. 541bs. 4oz. 6 If a cable, 12 inches about, require an anchor of 18cwt., of what weight must an anchor be for a 15 inch cable? (Pike, 211.) Ans 35cwt. 15lbs. SECTION VI. Mensuration, Gauging, fyc. N. B. EVERY problem that follows, can be done with much less labor than they are usually done. 232 APPENDIX. EXERCISES FOR PRACTICE 1 What is the difference of area between a square of 40 rods on a side, and an equilateral rhombus of 40 rods to a side, but of a perpendicular altitude of only 34 rods ? (W. 175.) Ans. 240 rods. 2 There is a barn, 50 feel "by 36, and 20 feet high to the eaves ; how many boards will it take to cover the body, if the boards were all 15 inches wide and 10 feet long ? Ans. 275-j- boards. 3 On a base of 120 rods in length, a surveyor wished to lay off a rectangular lot of land, to contain 60 acres ; what distance in rods must he run out from his base line ? (W. 175.) Ans. 80 rods. 4 How many square yards in a triangle, whose base is 48 feet, and perpendicular height 254 feet? (W. 175.) Ans. 67s yards. 5 A man bought a farm 198 rods long, and 150 rods wide, and agreed to give $32 per acre ; what did the farm come ta? Ans. $5940. N. B. Make no attempt to compute the number of acres definitely. 6. If the forward wheels of a coach are 4 feet, and the hind ones 5 feet in diameter, how many more times will the former revolve than the latter in going a mile ? (W. 176.) Ans 84. N. B. In this problem use 7 to 22. 7 How many square feet in a board, 2 feet wide at the larger, 1 foot 8 inches at the smaller end, and 14 feet long ? (W. 176.) Ans. 25| feet. 8 The plate supporting the rafters of a house, being 40 feet long, 14 inches wide, and 8 inches thick, how many solid feet does it contain? (W. 177.) Ans. 31| feet. 12 12 12 40 12 14 8 inches APPENDIX. 233 Cancel down, and this is the form for all solids. 9 If a pile of wood be 60 feet long, 12 feet high, and 6 feet wide, how many cords does it contain ? Ans 33? cords. 10 The bin of a granary is 10 feet long, 5 feet wide, and 4 feet high ; allowing the cubical contents of a dry gallon to be 268| cubic inches, how many bushels of grain will it contain ? Ans. 160*. 1 1 If you wanted a bin to contain twice as much as mentioned in the last problem, with a length of 12 feet, and a breadth of 6 feet, of what height must it be ? (W. 177.) Ans. 5| feet. 12 A canal contractor engaged to excavate 2 miles of canal across a plane, at Sots per cubic yard the canal to be 54 feet wide at top, 40 at bottom, and 4$ feet deep ; what did it amount to ? Ans. $6617.60. 13 There is a circular cistern, of uniform diameter, whose depth is 8 feet, and diameter 5 feet; what is its capacity, allowing 231 inches to the gallon, and how much would its capacity be increased by adding 6 inches to its diameter ? . C 1175.04 gallons its capacity, S> I 246| gallons increase. 14 What would be the produce of a kernel of wheat in 11 years, at 20 fold, the produce of each year being sowed the next-? allowing 5000 kernels to a quart? (W. 166.) Ans. 64000000 bushels. N. B. If we blindly perform all the labor indicated by set rules, the above would be a tedious operation ; but it is extremely brief in the hands of a skillful operator. 15 The length of a room being 20 feet, its breadth 14 feet 6 inches, and its height 10 feet 4 inches ; how much will the coloring come to at 27cts. per square yard, de ducting a fire-place of 4 feet by 4 feet 4 inches, and tw. windows, each 6 feet by 3 feet 2 inches? (R. 232.) Ans. $19.73. 1 6 What will the paving of a foot-path come to, at 1 8 cents per square yard, the length being 35 feet 4 inches, and the breath 8 feet 3 inches ? (R. 233.) Ans. $5.83. 234 APPENDIX. 17 What will it cost to roof a building 40 feet long, the rafters on each side being 18 feet 6 inches long, at $3.50 per 100 square feet? (R. 233.) Ans. $51.80. 18 There is a block of marble, in the form of a paral- lelepiped, whose length is 3 feet 2 inches, breadth 2 feel 8 inches, and depth 2 feet 6 Inches ; what will it cost at Slcts. per cubic foot? (R. 234.) Ans. $17.10. 19 What will it cost to build a wall 320 feet long, 6 feet high, and 15 inches thick, allowing 20 bricks to the solid foot, at $5 T 8 ^ per thousand bricks ? (R. 234.) Ans. $282. 20 How many bricks 8 inches long, 4 inches wide, and 2^ inches thick, will it take to build a wall 120 feet long, 8 feet high, and 1 foot 6 inches thick ? (R. 234.) Ans. 34560. 21 What will it cost to build a brick wall 240 feet long, 6 feet high, and 3 feet thick, at $3.25 per 1000 bricks each brick being 9 inches long, 4 inches wide, and 2 inches thick ? (R. 234.) Ans. $336.96. 22 A ship's hold is 75 feet long, 18 wide, and 7| deep; how many bales of goods 3 feet long, 2| deep, and 2| wide, may be stowed therein, leaving a gangway, the whole length, of 3? feet wide ? (Pike, 470.) Ans. 385,4-f-. N. B. Do this by one operation after taking out the gangway mentally, by subtraction. 23. A stick of timber is 16 inches broad and 8 inches thick ; how many feet in length must be taken to make 20 solid feet? Ans. 22&. 24 There is a square pyramid, each side of whose base is 30 inches, and whose perpendicular height is 120 inches, to be divided by sections, parallel to its base, into three equal parts ; required the perpendicular height of each part. (P. 371.) Ans. The height of the lower section is 15.2 inches ; the height of the middle section is 21.6 inches ; the height of the top section is 83.2 inches. N. B. In solving this problem, remember that solids, APPENDIX. 235 of the same shape, are to each other as the cubes of their like sides. 25 A man wishes to make a cistern of 8 feet in diameter, to contain 60 barrels, at 32 gallons each, and 231 cubic inches to a gallon ; what shall be the depth of the cis- tern ? 60X32X231 gives the cubic inches the cistern is to contain. This divided by the circular end, expressed in in- ches, will give the depth in inches. 8X12X22 .JNow, = the circumference in inches. But the circumference of any circle, multiplied by | of its diameter, giver its area. 8X12X22X24 Then, .... = the area. Hence, $ 7X7X 5=245 ; which, divided by 4, gives 6U inches for the answer. SECTION VII. Miscellaneous Examples. 1 One-half, one-third, and one-fourth of a certain num ber, added together, make 130 ; what is the number? Ans. 120. To solve this by arithmetic, we must consider the number as the whole of a thing, or a unit. Then 5 -f -r-? = T 6 2+T 4 2"i"T 3 2 or VI- Then, by proportion, if \ make 130, what will 1, or 4-5, make ? That is, 130 T* -' "V 236 APPENDIX. By multiplying the first and last terms by 12, the pro- portion becomes 13 : 130 : : 12 Or, .1 : 10 : : 12 2 One-fourth of a certain number exceeds one-sixth of the same number by 20 ; what is the number ? Here, again, the number must be considered as a sin- gle thing ; then $ of it is not, strictly, one-fourth of a unit, but | of that number, or that thing. In algebra, this thing, or number, would be represented by some let- ter, as x or y and the question is strictly an algebraical one. But all such questions in algebra, can be solved by fractions and proportion in arithmetic ; and, indeed, all questions, that involve simple equations only, can be resolved by arithmetic but by algebra they are much easier. In days that we re, we always found a rule in arithme- tic called Position, which included such problems as the preceding, and some few others^ but could not take in any questions involving powers, or roots as. powers and roots are not in arithmetical proportion to each other. For example, 16 and 64 are square numbers, and their square roots are 4 and 8, or as 1 to 2 ; but the numbers themselves, 16 and 64, are to each other as 1 to 4, a dif- ferent relation from the roots. Example : A man, having a purse of money, being asked how much was in it, answered, The square root of it, added to the half of it, made 220 dollars ; how much was in the purse.? It is evident, that this question must be excluded from any proportional operation ; for, unless we first suppose the right number, the result of the supposition will not be to the given result as the supposed number to the true number and when this proportion fails, supposition, that is, "Position fails ;" and if we suppose the true number, it is then an operation of chance, and is, in fact, no problem at all. True it is, we can give a rule, or rather give orders which, followed, will reduce the problem, but it will not APPENDIX. 237 be arithmetic; and, to put into an arithmetical work what is not arithmetic, we hold to be deceptive and im- proper. We have remarked, that the solution of these prob- lems are much easier by algebra than by arithmetic; but we would remind the pupil, that the solution of a problem is a small object compared with a principle in- volved in a solution, or with a knowledge of the science of numbers. As one step to secure this latter object, we give a few more such problems. 3 A post is J in the earth, | in the water, and 13 feet above the water ; what is the length of the post ? Ans. 35 feet. Add ^ and f ; not ^ and ~ of the number 1, but i and f of the whole post. These parts, added together, make || ; the remaining ij must be 13 feet. Then, by pro- portion, 35 35 Or, ..... 13 : 13 : : 35 : 35 the answer. 4 A and B have the same income. A contracts an an- nual debt amounting to ~ of it, B lives upon | of it ; at the end of ten years, B lends to A money enough to pay off his debts, and has 160 dollars to spare; what is the income of each ? Ans. $280. If B lives on |, he saves ^ ; out of this he pays A's debts, 4. Hence, from jr subtract 1 7 , and there remains ? 2 j. This, in 10 years, is worth 160 dollars; therefore, in 1 year it is worth 16 dollars. Now, by proportion, / T : 16 : : || : the answer ; Or, 2 : 16 : : 35 : the answer; Or, . . 1 : 8 : : 35 : 280, the answer. 5 Of the trees in an orchard, are apple trees, -i pear trees, and the remainder peach trees which are 20 more than | of the whole ; what is the whole number in the orchard ? Ans. 800. The apple, the pear, and the peach trees, make the whole; therefore, add + '-=+-=- 238 APPENDIX. This wants /$, or ^L, of being the whole ; therefore, we must conclude, that the 20, not taken into the account, is V of the whole. Hence, 20X40=800 ; or, by pro- portion, T V : 20 : : }{. : answer. That is, 1 : 20 : : 40 : 800, the answer. 6 A, B, and C, would divide $200 between them, so that B may have $6 more than A, and C $8 more than B ; how many dollars for each ? (R. 229.) Ans. A 60, B 66, C 74. Observe, that A has the least sum, B 86 more than A, and C $14 more than A ; hence, $20 is to be reserved, and the remaining 180 to be divided equally among them; which, of course, gives $60 to A, and $60-}-6 to B, and 60+6+8 to C. 7 A gentleman bought a chaise, horse, and harness for $378 the horse came to twice the price of the harness, and the chaise to twice the price of both the horse and harness ; what did he give for each ? (R. 229.) This problem is generally given under position, but it is really one in simple division. Take the idea of shares. Divide the money into shares : it will take 1 share to purchase the harness, 2 shares to purchase the horse, 6 shares to purchase the chaise ; therefore, divide the whole into 9 shares, and we shall have $42 for one share, i. e. $42 for the harness, $84 for the horse, $252 for the chaise. In conclusion, we would caution the young arithme- tician against imbibing the idea, that he understands arith- metic, from works on arithmetic alone. We must rise above a plane, to have a fair view of the objects on its surface, and we must rise above arithmetic before we can understand all its scientific relations. Nor must we conclude that we are perfect in num- bers, because we may excel our comrades in disentang- ling knotty questions. Science is a different thing from acuteness at solving intricacies ; and all men. of true science, more or less despise all things intended to puz- zle, for there are enough objects of useful inquiry and investigation, on which to expend all our powers of mind. APPENDIX. 239 Algebra is but a continuation of arithmetic; and as soon as we acquire a good practical knowledge of arith- metic, so as to have a clear comprehension of fractions and general proportion, we may, yea, we should, com- mence algebra. But when we advance in algebra, we should be careful not to look down with the spirit of con- tempt on arithmetic we may study the one in order to understand the other. The following properties of num- bers, the student must take as facts, unless he is an alge- braist. They cannot be demonstrated without the aid of that science.* 1 If from any number, the sum of its digits be sub- tracted, the remainder is divisible by 9. Take, for example, 34 ; subtract 7, and we have 27, which is divisible by 9. Again, take 438 ; subtract the sum of 4+3+8=15, and we have 423=9X47, and so with any other number. 2 If the sum of the digits of any number be divisi- ble by 9, the number itself is divisible by 9. Thus the numbers 72, 81, 99, 171, 387, 51489, &c., the sum of whose digits is divisible by 9, are themselves divisible by 9. All numbers divisible by 9, are, of course, divisible by 3. 3 If the sum of the digits of any number be divisi- ble by 3, then the number itself is divisible by 3. Thus 18, 27, 54, 75, 111, 123, 258, &c. &c., are all divisible by 3. 4 If from any number the sum of the digits stand- ing in the ODD places be subtracted, and the sum of the digits standing in the EVEN places be added, then the result is divisible by 11. Take any number, say 785432 ; then subtract the sum of 2+4+8 = 14, and add 3+5+7=15, or, in this ex- ample, add 1, and we have 785433 = 11X71403. * The demonstrations of these properties may be found in Bridge's Algebra, section XLII. 240 APPENDIX. 5 If the sum of the digits standing in the EVEN pla- ces, be equal to the sum of the digits standing in the ODD places, then the number is divisible by 11. Thus the numbers 121, 363, 12133, 48422, &c., are all divisible by 11. Our numbers increase in a ten-fold proportion, from the right to the left, which is called, the root of the scale ; but, if the scale was 7, in lieu of ten, then, what is now true for 9 and 11, would be then true for 6 and 8. 6 We may change numbers from, the scale of ten to any other scale, by dividing the number by the number denoting the scale, continually saving the remainders and forming a new number by them. Example Change 63 into an equivalent number, wherein the value of the digit shall increase in a five-fold proportion ; in other words, where the root of the scale shall be 5. 5)63 5)12(3= first remainder. 5)2(2= second remainder. 0(2= third remainder. Now 223 is the value of 63, in a scale where the num- bers increase in a five-fold proportion. Change 3714 to its equivalent value on a scale of 4. Ans. 322002. Numbers may also be considered as arising from the continued multiplication of certain factors. Ji perfect number, is one which is equal to the nun of all its divisors. They are not numerous, and may be expressed thus : 2 (2 2 1)=2X3=6. 2 2 (2 3 1) =4X7=28. 2 4 (2 5 l) = 16X 31=496. 2 6 (2 7 1) = 64X 127=8128. 2 10 (2 U *) = 1024X 2047=2096128. A PRACTICAL SYSTEM OF BOOK-KEEPING, FOR MECHANICS AND RETAILERS. BOOK-KEEPING is the method of recording a system- atic account of business transactions. It is of two kinds Single and Double Entry. The former, only, will be noticed in this work. On account of the simplicity of Single Entry, it is, perhaps, the best wliich can be recommended to farmers, mechanics, and retailers. It consists of two principal books the Day Book, or Wast* Book, and the Lcger, and one auxiliary book, the Cas/trBook. TIIE DAY BOOK. This book is ruled with a column on the left hand for the date, and three columns on the right, the first, for the folio or page of the Leger, to which the account is transferred; and tiie last two for dollars and cents. This book exhibits a minute history of business trans- actions in the order of time in which they occ ur, with every circumstance, necessary to render the tra asaction plain and intelligible. 17 " * CINCINNATI, 1833. Jan. 4 6 8 11 a 14 14 15 David Judkins, Dr. To 10 Ibs. coffee at 17cts.$l 70 " 25 Ibs. sugar, at LO cts. 2 50 Timothy W. Coolidge, Dr. To 1 bl. sugar, weighing 135 Ibs. neat, at 8* cts. $11 47i 1 bag coffee, 98 Ibs. at 15 cents, 14 70 Geo. H. Eaton, Dr. To 1 bl. flour, $3 874 1 Ib. Y. H. Tea, I i2i " 1 keg lard, neat weight 60 Ibs., at 6i cts. 375 David Judkins, By cash on account, Cr. James Wilson, Dr. To 3 weeks boarding, at 2 dollars per week, Hiram Ames, Dr. To 12 yards broadcloth at 6 dol- lars per yard, $72 00 " 30 yds. muslin at 14 cts. 4 20 Cr. By an order on J. Jones, for Gro- ceries, $51 00 " Cash, 20 00 Timothy W. Coolidge, Cr. By a bill of carpenter work, James Wilson, Dr. To 1 bl. vinegar, $3 25 " 1 keg lOd. nails, weight "111 Ibs. at 8 cts. 888 76 71 25 12 CINCINNATI, 1833. Jan. 19 21 25 George Ha rail ton, By 1 set of Fancy chairs, George H. Eaten, By cash, 10 balance account, R >bert Young, Dr. To cash on account, $3 00 " 10 Ibs. N. O. sugar, at 10 cents, 1 00 12 Ibs. coffee at 163 cts. 2 00 29 T( " 31 Feb. 13 Cr. Cr. Alb.Y.H. Tea, 62* Jackson Moore, Dr. 21 Ibs. Ham, at 10 cts. $2 10 I box soap, 30 Ibs. at 5 cents, 150 David Judkins, Dr. 30 To 12 bis. apples at 1 5 cents, By 47 bush, corn, at cents, Cr. James Wilson, By cash on account, Cr Thomas Hilton, Dr 3 To cash, $5 00 1 Ib. Y. H. Tea, 1 06 16 Ibs. Rice, at 64 cts. 1 00 Cr. By 13 days labor, at 87* cts. George Hamilton, Dr 10 To 1 canister Imperial Tea, 14 Ibs. at $1 87 i per Ib. Hiram Ames, Cr. By order on E. Disney for goods, 15 11 76 iO CINCINNATI, 1833. .Feb. 15 21 27 March 6 James Wilson, Dr. To 10 gais. molasses, at 40 cents, $4 00 " 4 Ibs. Old Hyson Tea, at 93 cents, 372 Robert Young, Dr. To 1 box sperm candles, "25 Ibs., at 30 cts. perlb. $7 50 " 2 bu. dried fruit, at $1 " 25 per bushel, 2 50 Robert Young, \ Cr. By 12 cords of wood at $2 25 Ames &, Smith, Dr To 18 Ibs. sole leather, at " 25 cents, $4 50 " 1 side upper leather, 2 75 "3 calf skins, $125, 375 Hiram Ames, Dr To 500 feet whits pine boards, ai " $12 50 per M. $6 25 " 25 bu. potatoes, at 50 cts. 12 50 1 ton of Hay, 10 00 David Judkins, Dr To 200 Ibs. flour, at $175 15 Thomas Hilton, Dr To cash paid his order to William Cuolirlge, George Hamilton, Dr To 1 copy Whelpley's Com- pend, $1 2i " 1 ream letter-paper, 4 51 1 doz. Spelling books, 1 0( 27 11 18 cts. 00 72 00 00 75 50 75 CINCINNATI, 1833. Mar. 20 23 Ames and Smith, , Cr 25 By 1 hhd. sugar, weight 1317 Ib neat, at 74 cts. 28 30 31 Theodore P. Letton, Dr. To sharpening his plough,*$ 1 ,00 " Shoeing his horse, 1,6 " Repairing chain, 25 .Timothy W. Coolidge, Dr. To 2 qrs. tuition of himself at evening school, at $3 per qr. Thomas Hilton, Cr By the hire of his horse 10 days at 62 i cts. per day, T. P. Letton, Dr. To 1 ream wrapping paper, $1,62* " 1 beaver hat, 5,00 " 1 set silver tea-spoons, 6,00 Henry C. Sanxay, Cr. By my order on him in favor of Jno. Torrence for stationary, Ames and Smith, Dr. To 2000ft. clear pine boards, at $20 per M. $40,00 500 common do. at $8, 4,00 5000 shingles, at $2,25 1 1,25 " Cash to balance account, 32,52 Jackson Moore, By painting my house, Cr. 98 cts. 874 00 25 774 624 874 77 00 6 CINCINNATI, 1833. March 31 31 31 31 31 Thomas Hilton, Dr. To 4 bu. wheat, at $1 25, $5 00 1 bl. mess pork, 9 00 " 2 bu. salt, at 50 cts. 1 00 " 8 Ibs. brown sugar, 11 cts. 88 2 4 3 2 f- $ 15 14 I) 14 11 cts 88 00 00 00 50 George Hamilton, Dr. To 12 cedar posts, at 25c $300 1 plough, 937* " 1 scythe, 1 62d Jackson Moore, Dr. To repairing his wagon and plough, Thomas Hilton, Cr. By 1 pair shoes, $1 50 " 1 mahogany table, 12 50 George Hamilton, Cr. By an order on J. Hulse, $5 00 cash, 650 END OF THE DAY BOOK. r THE LEGER. THIS book is used to collect, the scattered accounts of the Day Book, and toarrunpe all tliat relates to each individual, into one separate statement. The i-usiness of transferring the accounts from the Day Book to the Leger, is called posting. The Leger is ruled with a double line in the middle of the page, to sepa- rate the debits from the credits. Each side has two columns for dollar* and cents, one for the page of the Day Book, from which the particular item ia brought, and a column for the date. When an account is posted, the page of the Leger on which this account is kept, is written in the column for that purpose in the Day Book, and also the page of the Day Book from which the account was posted, is written in the 2d column of the Leger. Tn posting, begin with the first account in the Day Book, which you will perceive is the name of David Judkins. Enter his name in he first page of the Leger, in a large, fair hand, with Dr. on the left and Cr. on the right. As there are several articles charged to D. judkins on the 4th of January, instead of specifying each article in the Leger, we merely say, For Sundries, and enter the amount in the proper columns see Leger, pane 1. The Leger has an index or alphabet, in which the narn3 of persons are arranged under their initial letters, with the page in the Leger where the account may be found. ALPHABET TO THE LEGER. A Ames, Hiram - 2 Auies & Smith - 4 I J Judkins, David 1' R B Balance 5 K s Sanxay, H. C. - 4 C Coolidge, T. W. - 1 L Letton, T. P. - 4 T D M Moore, Jackson 3 U E Eaton, George H- 1 N V F W Wilson, Jas. - 2 G P X Y Young, Robt. - 3 H Hilton, Thomas - 2 Hamilton, Geo. - 3 a Z I 1 Dr. David Judkins, Cr. 183& Jan. 4 30 Mar.7 Apl.l NOTE.- tno sums and for w that the person in Fo sundries 2 Apples 3 Flour 4 To balance of account bro't down, -The Dr. on the left entered on that side 'liich they owe you. sums entered on that ider whose account t 4 9 3 16 = 4 Land ofth TheC side c hey st 20 )0 30 70 30 sideo epage /r. on >f tlie and, a Jan. 6 " 30 r the page are those he right 1 page are id for wl By cash 2 t Corn 3 I " Bal ' ^ 11 signifies debtor, for articles sold t )and side signifies c "or articles receive irh you owe him. J|00 3 40 1 30 3 70 and that o others, redit, or 1 of the Timothy W. Coolidge. Jan. ^ Mar. 20 Apl. 1 NOTE.- idge has It appea which dt and tlier To sundries "2 " Tuition 5 To balance -The Dr. aide of this received of Die, and a that the total aino duct the $25 00 (whi 3 will remain a balan R 32 7 aero the C unt o - ti ata ceof 17* 00 17* 174 unt ah r. side f my? ndsto S717i Jan 14 Apl. 4 nws the a shows w iccountaf his credit due me. By work 2f< " Balance \ \ mount of articles ! lat F have receive.^ 'ainst liim if $32 1 on the ri;jht of the ,500 7 17* g.17* Hr. Cool- of 1 hn. 7J, from account) George H. Eaton. Jan. 6 NOT*.- ttellcHi 1* fully c To sundries ,$ -Thte account prese e J. H Eaton noihii loaed. 1 nts e g< an 75 iua) si 1 tlrat Jan21 ma on ho he owes By cash |3 th sides; hence H 1 me nothing. Tlu 875" s evident ; sccount Dr. Hiram Ames. Cr. 2 Jan 1 i r Mar.O Fo sund.' 2 " Corn 4 713 28 2U 75 Jan. 11 Feb. 13 Apl. 1 By sund. " Order " Bal 2 3 7) 10 23 00 00 95 104 95 104 95 Apl. 1 To Bal ,23 95 Thomas Hilton. iFeb.3 Mar 8 " 21 To sund. S Cash. 4 do. G 7 18 15 Ot> 75 88 Feb. 3 Mar23 31 Apl. 1 By labor, " hire of horse, Sund. Bal 4 5 6 11 : OS 14 ( 10( m >5 )0 )6i 41 69 41 e >9 Apl. 1 To Bal 10 06* James Wilson. Jan. 11 11 Feb. 21 To boar- ding, > " Sunds. L " do. 2 6 2 12 4 7 00 13 72 Jan. 31 April 1 : By cash " Bal 3 15 10 00 85 25 S5 25 85 April ] I To bal- ance bro't down, 10 85 s=s NOTE. When an account is settled only, and sot fully paid, as in the above, arid several preceding accounts, the balance, whether it he in your favor or against you, is brought down and placed distinctly by itself, and serves for the beginning of a new account, as you perceive has been done in the above exam- ple, the balance being f 1085. 3 Dr. George Hamilton. Cr. Feb. 10 Mar.15 To Tea, " Sund. 3 20 25 75 Jan. T Mar 3 J By chairs ft 1 Sunds. 6 11 00 50 " 31 " do. 14 00 Apl. L Balance, 10 50 47 00 4? 00 April 1 To Bal 10 50 Robert Young. Jan. 27 Feb. 15 To sund- ries, [ " do. * * 6< I 1.0 ( 52* )0 Feb.27 By wood, 4 tf '00 Apl. 1 Bal. 101 m 27 ( )0 2" ^00 Apl.l By J?aZ. It 1374 NOTE. I n the above ace cnmt th ediffi jrence bel ween the Dr. an d C r. side is $10 37^, by which I perceive that the balance against me, in favor of Robert Voting, is S 10 37*. Jackson Moore, Jan. 29 To Sun- Mar.31 J Jy paint- dries, c 3 e ing, a 21 00 Mar.31 Sunds. t 6C Apl. 1 Bal. 11 4 210 21 00 ~ Apl. 1 1 }y bal- a ince bro't -, own, 11 40 Dr. Ames 4* Smith. Cr. 4 Mar. 6 30 To Sundries " do. 4 5 1 i 87 00 77 By sugar, 1 fa = 77 77 98 77 NV*TK. amount or This account, like i the Dr. side Mng . the ust on equ 3 on al to the first p that on the age, is fully c Cr. osed, the T. P. Lctton. Mar. 20 " 20 To Sundries do. 5 5 '2 87^ ti'2i Ap. 1 By Bal. 15 50 15 50 15 50 April 1 To balance 15 50 NOTE.- ence is, th tail, hi the In the above accoirr at T. P. Letton ow< Day Book, page 5. itt >a n iere ie $ is ni 155(1 i sum on tl for sundrj e Cr side, and articles ezpret the! eedii ifer- ide- Henry C. Sanxay. A P i i: Ho balance, 1 s 1C ;p| Mar.28 By my order, 5 12 1 10 f, 371 r = Apl. 1 By bal 121 ^7i NOTE. Tn f his account it will he perceived that as there in no amount char- ged to B.C. Sanxay, on the Dr. side, I owe hint $12 t!7|. 5 Dr. Balance. Cr. April 1 ToD. Judkins, 11 " T. W. Coolidge, 1 1 H.Ames, 2 " T. Hilton, 2 " J. Wilson, 2 " G. Hamilton, 3 " T. P. Letton, 4 4 2: lO 10 15 W 7* )5 ! So }.'i >o April 1 By R. Youn, " J. Moore, " II. Sanxay, ! 101 i ) 12 7J 40 -7| 82 54 34 G5 NOTK. This account exhibits the exact state of your hooks. It is made from the preceding accounts in the Leser. The Dr. side is an exhibit of the amounts due to you by others, arid the Cr. side the amounts due by you to oth- ers. It is noi strictly necessary that this account 'should he introduced in tne Le<jer in single entry : it will he found convenient, however, to balance the book at stated intervals, and transfer the balances to the new accounts 1 elow, 1 as in the preceding Leger, and when that is done, a balance account like the " above will be found convenient, as presenting, at one view, the exact state of your Leger. FORM OF A BILL FROM THE PRECEDING. January 4 30 March 8 January 8 30 Mr. David Judkins, To Edward Thomson, Dr. To lOlbs. coffee at 17 cts. - - - - 1 7C 25 Its. sugar at 10 cts. .... 2 5( 4 3 1(5 12 20 00 50 TO 40 12 bbls. apples at 75 200 IDS. flour at $1 cts. . . . . 00 40 Cr. 3 47 bushels corn at 20 els. .... 9 Errors excepted. Balance due Rec'd payment in full, EDWARD THOMSON. 4 30 THE CASH BOOK. The Cash Book is used to record the daily receipts and payments of mo- ney. It ia ruled nearly the same as the Leger ; the Dr. side exhibits the amount of money received, and the Cr. side, the amount paid out. Subtract the sum of the Or. from that of the Dr. and the balance will always be equal to the amount of cash on hand. FORM OF A CASH BOOK. Dr. Cash. Cr. 1833 Jan. 1 " 1 1 " 1 To cash on ha:.d, Cash rec'd oC J. Young. " '' H. Sanxay. " " I). Judkins 7; H "25 1C 81 40 GO 00 J833 Jan. 1 " 1 " 1 By rent of house paid T. P. Letton, Paid note to R. Hand, Family expenses, By cash on hand, 18 5(J 4 52 UO 00 37 84 Li>5 21 12o21 Jan. 2 " 2 " 2 To cash on hand, of T. Coolidge, Cash found on Main St. 59 2: 29 81 16 00 Jan. 2 " 2 By cash paid Ames & Smith, By cask an hand, 20 85 00 00 'Or, 00 105 00 ! Jan. 3 To cash on hand, 65 00 Talbott, J.;, THE UNIVERSITY OF CALIFORNIA LIBRARY YA 02436 LHABLE SCHOOL BOOKS PUBU 1ED BY E. MORGAN & ; ' . ill JHativ-st : -v \'. f J : ?,oyED MATHEMATICAL WGMS. '] ;;g works foriK a complete Elementary COU V ,-K ol JTatliott's I*iifle Arithmetic tie -ginnerg; or, the first steps of a course of p; < : , B - Practical Arithmetic, ; ; ! ' h added ^he PRicrsf*-:.i-/ nn;^^ :- ; ;: ,,: Deviations, ali h ^,; \he most competeet teachers, and other men lgetora. AH Elementary Trea ?3, designed to facilitate the comprehension, demonstrs id applicuuon ofche leading prvacipies of Algebrs v luchl, A.,, M,, Prufes&OT o*" Mathematics aati' Kstura! fhi 3j)hy, Cincinnati s Iflitehel'9;- ', < . , ; ' - / , in w hi cb rhe reasoning ii analysis, s-uti the proportions according to tlieir immediate deper -'* ; ;, By O, M~ Mitcliel, A. M. Professor 02 iV id ,-jaturai Philosophy, Cincinnati college. A Unirersal Kef to the , Fa which the numencal- the Appendix of this Arithme'; !abo? of algebru;-' ? one foarth the us: .; ^>iHtions> to . -!e highly rcc.' <;;8sentiai math Bictisnary remium His! cry Oi tlie IJniied Church -.- 1 - >a. ch Ea-tion,