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 PE AOTICAL ANT» *'Tr, OliETTCAL 
 
 A E I T H M E T I C , 
 
 IX ADBITION TO TUS USUAL ilOPP^ OF ^ PKRATTON, THE SCI- 
 OF NUMBEllS, THE PKCh', 'x SYSTEM, 
 
 AND OTHER TilTOi: ?, 
 
 HCLP A p:; 
 
 BY HORATIO N. ROBiNSOxV, A M., 
 
 AUTHOR OF V C;)l;u-[; Of >rATHE:M.>.av::3. 
 
 JOUS COLLEGES, AXD PKIYATJl STUDFXTS. 
 
 JACOB 
 
 [N STREET. 
 
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 Y Of I 
 
 NIA I 
 
 f i?SITY Of 
 LIK>RH1A 
 
 «X(HIUnB. 
 
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ROBINSON'S 
 
 MATHEMATICAL, 
 
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A 
 
 A NEW 
 
 PKACTICAL AND TIIEOllETIC AL 
 ARITHMETIC, 
 
 ^ IN WHICH, 
 
 IN ADDITION TO THE USUAL MODES OF OPERATION, THE SCIENCE 
 
 OF NUMBERS, THE PRUSSIAN CANCELING SYSTEM, AND 
 
 OTHER IMPORTANT ABBREVIATIONS, HOLD 
 
 A PROMINENT PLACE. 
 
 BY HORATIO N. ROBINSON, A. M. 
 
 AUTHOR OF A COURSE OF MATHP:mATICS. 
 
 DESIGNED FOR SCHOOLS, COLLEGES, AND PRIVATE STUDENTS. \ 
 
 JOHM S. PV^ELL 
 
 Civil ( . ^Ineer, 
 
 8AN FRANCl^Cg, GaL 
 
 CINCINNATI* 
 JACOB ERNST, 112 MAIN STREET 
 
 1854. 
 
woaaia ua. 
 
 Entered according to Act of Congress, in the year 1845, 
 
 BY E. MORGAN & CO., 
 
 In the Clerk's olfice for the District Court of Ohio. 
 
 ^U^Ck 1 
 
 Stereotyped by J. A. Jamee, 
 Cincinnati. 
 
h 
 
 ED'JC. 
 UBRARY 
 
 Civil & Mechanical Engineer. 
 
 SAN FRANCISCO, CAL. 
 PREFACE. 
 
 The public may very properly inquire, what <roocl can be 
 accomplished by addino; another Arithmetic to tlie long list 
 now in existence] Both teachers and the taufjht are already 
 sufficiently perplexed, in making a selection. 
 
 True, there are many arithmetics before us, claiming our 
 attention; but more than half of them are mere collections of 
 questions, ordered to be solved by rules, given in a spiritless 
 manner, having no connected system, and not recognizing the 
 general and universal scientific character of numbers. 
 
 They place arithmetic before the mind of the pupil, as an 
 art, not as a science — they give him directions ichat to do, not 
 what to think, and how to reason; and, in short, many of 
 these books, now pressing for patronage, are already con- 
 demned by all good judges and scientific teachers. 
 
 Within the last ten or fifteen years, the science of arith- 
 metic has undergone many changes, and received many im- 
 portant improvements ; and such improvements have appeared, 
 little by little, and from time to time — some in one book, and 
 some in another — nowhere presenting one grand whole; and 
 the object of this work is, to give unity and system to all the 
 modern improvements \\\\\c\\ preseyii practical utility^ and in- 
 terweave them in the common and general system, or such 
 parts of it as are necessarily retained; and, how far we have 
 accomplished this object, let the competent, the unprejudiced, 
 and the uninterested^ answer. 
 
 These modern improvements generally pass under the name 
 of the canceling system ; but canceling by no means includes 
 them all, nor are they all included in numerical operations- 
 principles, explanations, and forms, have been simplified and 
 improved. 
 
 Durin-g the progress of investigation, there needs must be 
 many fruitless attempts at improvement, and much useless 
 matter must accumulate on the hands of original inquirers; 
 and some such matter occasionally appears to the vv orld, rath- 
 er because it is singular, or new, than because it is practical 
 
 010 
 
VI PREFACE. 
 
 and useful; but we have carefully avoided all that we did not 
 deem of practical utilit}^, either in teaching principles, or in 
 facilitating numerical computations. Several small books 
 have appeared (evidently not written by men of science), pro- 
 fessing- to teach arithmetic in a few lessons, on modern im- 
 proved plans, &c.; but they are mere outlines of rules and 
 forms, unconnected with any general system ; and, of course, 
 they must fail to teach arithmetic as a whole, or long hold 
 public attention. 
 
 Some other authors, more fond of quantity than of quality, 
 not discriminating between the singular and the useful, have 
 collected together, in one great mass, a//that is new, and giv- 
 en it to the world as improvements, and claim merit and pat- 
 ronage, because they have given all. As well might the far- 
 mer claim merit and compensation, for bringing straw and 
 chaff to market with his wheat. 
 
 To make ourselves thoroughly understood on this point, 
 we remark that we have given many useful and practical 
 methods of multiplication; but few or no corresponding meth- 
 ods of division, as the corresponding methods of division, we 
 found, would not be practically useful. For example, there 
 are many practical, obvious, and easy methods of multiplying 
 any niLmber : say 12468 by 9, 99, 98, 97, or any number near 
 10, 100, 1000, &c. And there are corresponding methods of 
 dividing any number by 9, 99, 98, &c.; but the operations are 
 not obvious and easy, as in multiplication, we do not regard 
 them as improvements; for the common way is full as easy, 
 and we will not tax the temper and patience of the pupil, by 
 taking him off into speculations of questionable utility. 
 
 For those fond of speculation, however, we will give the 
 method of dividing by 9, or 99, 98, &c.; and those who please, 
 may extend the principle ; but it is rather too speculative for 
 the common pupil. Divide 1836 by 9. Divide by 10, and 
 the quotient will be too small, because the divisor is then too 
 large; but divide the quotient by 10, continually, as far as 
 tenths: thus. 
 
 Divide by 10, 183, 6 
 
 Again, by 10, 18, 3 
 
 Again, by 10, 1,8 
 
 Again, by 10, 1 
 
 Add, and we have, . . Quo//en^, 202,18, remainder. 
 
 This remainder, 18, will divide by 9, giving 2 to be added 
 
 to 202, making 204 for the actual quotient; but how much 
 
 I more readily 1836 could have been divided by 9, in the usual 
 
PREFACE. 
 
 way 1 There is now and then a case where this method 
 might be preferable. This is one : Divide 3436 by 99. 
 
 "^ Divide by 100, 34, 3G 
 
 Divide this by 100, 34 
 
 Quotient, 34, 70, Remainder. 
 Here is a more difficult case. Divide 49999 by 98. 
 
 Quo. Re. 
 
 Divide by 100, 499, 99 
 
 Double this quotient, and divide by 100, gives 9, 98 
 Double 9, and divide by 100, 18 
 
 Add up, as in decimals. Quotient, 510, 15 
 
 As the divisor was 98, each 100 must have a remainder of 
 2, and the remainders summed up to more than 200, our re- 
 mainder, 15, must be increased by 4 ; then the quotient is 
 510 — remainder 19. This may be curious, but the common 
 way is much better. 
 
 This may be new, but it is not improvement, and by throw- 
 ing all things together, without discrimination, some have 
 brought real excellence into disrepute; and, therefore, we 
 have" taken much care not to introduce any matter that we 
 could not, for real utility, recommend. Clearness and per- 
 spicuity have been our constant aim ; and, although we set a 
 very high price on brevity, it has never been purchased at the 
 expense of perspicuity. 
 
 The reader will observe, that we have laid great stress on 
 the modern improvements. True, we have so; but there are 
 really no new principles — there may be new combinations of 
 them — there are no new mechanical powers, but there are 
 new machines every day; and it is not too much to say, that 
 no arithmetician, however great his knowledge, can be con- 
 sidered very skillful, unless he has paid particular attention 
 to this branch of study within a very few years. Yet, in 
 many respects, the common and ancient modes of operation 
 must ever be employed ; and, in truth, are the best. It must 
 not be understood that we have abandoned them, because we 
 have brought improvements to them. 
 
 For peculiarity and brevity in numerical operation, the 
 reader is referred to the " Supplement to Multiplication and 
 Division, Reduction, Proportion, Interest, Exchange, Mensu- 
 ration, and the roots." 
 
 To the method of stating problems in proportion, by com- 
 paring cause and effect^ we invite special attention. On criti- 
 cal examination, it will be found more easy and more ration- 
 al than any other method. 
 
PREFACE. 
 
 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 propor- 
 tion in numbers^ is unnatural and void of strict philosophical 
 expression ; not so with this method. 
 
 The peculiarities which the student will here find in the 
 extraction of roots, and in mensuration, are not all new — in- 
 deed, there can be nothing- new in principlt — but, as far as the 
 author's knowledg-e extends, he is not aware that these abbre- 
 viations have ever been collected in any arithmetical work, 
 except in the Appendix to Talbott's Arithmetic, and that very 
 recently. 
 
 We place much importance on the canceling or factor sys- 
 tem, in performing operations requiring multiplications and 
 divisions, for their great utility in algebra, when the pupil 
 arrives at that science ; and that teacher who looks no fur- 
 ther than to solve a few problems in arithmetic, has not a 
 sufficiently extended view to judge of the merits of the sys- 
 tem ; therefore, let none such condemn. 
 
 In preparing this treatise, care has been taken to avoid all 
 puzzles, as being unworthy of science; and, all unusual 
 or difficult problems : the object has been to teach the science 
 of numbers — not to try the powers of arithmeticians. We 
 deem it no recommendation to any author, that his questions 
 are very simple, or very severe ; we dislike extremes, on eith- 
 er hand. In choice of problems, we have aimed at sound 
 practical utility, bringing them as near to the real business 
 of active life as possible. To carry out these views, we 
 have added a short Appendix, containing problems under the 
 mechanical powers — though this is not, technically, arith- 
 metic. A brief practical system of book-keeping is subjoin- 
 ed, sufficient for the farmer and mechanic, and for all tliose 
 whose operations do not require the more complicated system 
 of double entry. 
 
 In conclusion, we would remind the reader, that we do not 
 claim to have presented a perfect work ; perfection is, per- 
 haps, impossible ; but it has been our design to present a con- 
 nected, simple, and scientific system, sufficient for the busi- 
 ness man and scholar, without being encumbered witii unne- 
 cessary theories; yet, uniting and systematizing all real im- 
 provements, taking only the practical and useful; and, how 
 far we have attained these objects, an impartial public, aided 
 by experience, must determine. 
 
CONTENTS 
 
 Numeration, Page 12 
 
 Simple Addition, 16 
 
 Simple Addition, applied to Federal Money, .... 20 
 
 Simple Subtraction, 21 
 
 Subtraction in Federal Money, 23 
 
 Simple Multiplication, 24 
 
 Simple Division, 32 
 
 SECTION II. 
 
 Supplement to Multiplication and Division, .... 39 
 
 Compound Numbers, 48 
 
 Reduction, 54 
 
 Compound Addition, 60 
 
 Application of Compound Addition, 64 
 
 Compound Subtraction, 65 
 
 Compound Multiplication, 68 
 
 Compound Division, 71 
 
 SECTION III. 
 
 Vulgar Fractions, 76 
 
 Complex Fractions, 83 
 
 Reduction of Fractions, 85 
 
 Addition of Fractions, 92 
 
 Subtraction of Fractions, 94 
 
 Multiplication of Fractions, 96 
 
 Division of Fractions, 97 
 
 Decimal Fractions, 100 
 
 Addition of Decimals, • 102 
 
 Subtraction of Decimals, 103 
 
 Multiplication of Decimals, 104 
 
 Division of Decimals, , 106 
 
 Reduction of Decimals, ... * 110 
 
 SECTION lY. 
 
 Comparison of Numbers, ... 116 
 
 Proportion, or Rule of Three, 119 
 
X COIS'TE^Tg. 
 
 Application of Proportion, • 122 
 
 Compound Proportion, 134 
 
 Practice, . • 143 
 
 Simple Interest, 151 
 
 Interest on Notes, Bonds, Slc, 159 
 
 Compound Interest, 166 
 
 Discount and Banking, 171 
 
 Per-Centage, 176 
 
 Stocks, 179 
 
 Equation of Payments, 181 
 
 Profit and Loss per cent., 184 
 
 Reduction of Currencies, 188 
 
 Exchange, 192 
 
 Exchange and Per-Centage combined, 195 
 
 Conjoined Proportion, ' 199 
 
 Fellowship, 203 
 
 Taxes, 205 
 
 Compound Fellowship, • 207 
 
 Barter, 210 
 
 SECTION V. 
 
 Mensuration, 212 
 
 Mensuration of Solids, 219 
 
 Involution of Powers, 226 
 
 Square Root, 230 
 
 Application of Square Root, 235 
 
 Problems on the right-angled Triangle, 238 
 
 Cube Root, 241 
 
 Application of Cube Root, 247 
 
 Supplement to Square and Cube Roots, 249 
 
 Abbreviations in Cube Root, 251 
 
 Approximate Cube Roots of Surds, ........ 254 
 
 Roots of all Pov/ers, 258 
 
 Arithmetical Progression, 259 
 
 Geometrical Progression, 262 
 
 Alligation, 266 
 
 Position, 269 
 
 Permutation, 272 
 
 Combination, 273 
 
 Miscellaneous Examples, 274 
 
 Appendix, 280 
 
ARITHMETIC 
 
 Arithmetic is the science of numbers. It would be 
 evidently impossible to express each number by a sepa- 
 rate character ; for in that case we must have an infinite 
 number of characters. 
 
 For example, to express every number, from one up 
 to one million, we must have a million of different signs 
 or marks, unless we can accomplish the end by changes 
 and combinations of a few simple characters. 
 
 (Art. 1.) Numbers were at first written separately, 
 either in words at length, or, as among the Romans, in 
 characters, commonly the letters of the alphabet ; and for 
 larcje numbers, such letters were used as would make up, 
 by the addition of their separate values, the number re- 
 quired. Thus, the number four hundred and fifly-eight 
 was written CCCCLVIIL, the value of C being ©ne hun- 
 dred, L fifty, V five, and I one. 
 
 The system of notation now in use, and which origin- 
 ated in Arabia, is so contrived as to express all numbers 
 with ten characters only. Nine of these are significant, 
 and represent numbers, and the other is used to denote 
 nothing, or the absence of quantity. 
 
 To express numbers greater than nine, recourse is had 
 to a law which assigns different values to the figures, ac- 
 cording to the position which they occupy. According 
 to this law, uniis of the first order occupy the first place 
 on the right of the written expression, units of the second 
 order, the second place, and so of the others. 
 
 The numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, are each, as we 
 perceive, separate characters ; but when we would write 
 
 n 
 
12 ARITHMETIC. 
 
 ten, which is a unit of the second order, M^e place unity in 
 the second place, and a (cypher) in the first, thus, 10, 
 which may either be considered as one unit of tlie second 
 order, or ten units of the first. Proceeding with units of 
 the second orner in the same manner as with those of the 
 first, we shall have 10, 20, 30, 40, 50, 60, 70, 80, 90, and 
 next a unit of the third order, which is written 100, the 
 significant ligure being written in the third place. In like 
 manner, we pass from the third to the fourth order. 
 
 (Art. 2.) To form an adequate and correct conception 
 of large numbers, M-e must have names for the difierent 
 orders of places from the right hand, which will enable us 
 to read off" the numbers in words. This naming and 
 reading is called 
 
 NUMERATION. 
 
 The first order of numbers is called Units. 
 
 The second order, Tens. 
 
 The third order, Hundreds. 
 
 The fourth order, Thousands. 
 
 The fifth order, Tens of Thousands. 
 
 The sixth order, Hundreds of Thousands. 
 
 The seventh order, Millions. 
 
 The eighth order, Tens of Millions. 
 
 The ninth order, Hundreds of Millions. 
 
 The tenth order, Billions. 
 
 The eleventh order, Tens of Billions. 
 
 The twelfth order, Hundreds of Billions. 
 
 The thirteenth order, Trillions. 
 
 And we may thus go on, and repeat the tens and hun- 
 dreds of trillions, and then take another name ; and again 
 and again repeat the tens and hundreds belonging to the 
 ne^v name. 
 
 Hence v/e observe, that we can divide these orders in- 
 to periods of three figures each, and give to each period 
 its distinctive name, reading ofl^ the hundreds, tens cmd 
 uni's, in each period. 
 
NUMERATION. 13 
 
 w c c 
 
 § -^ -I 
 
 KXAMPLE. 
 
 3 S "S r=: 
 
 80.652.941.600.807.362.546.278.009.650.208. 
 
 Which is read, eighty nonillions, six hundred fifty-two 
 octiUions, and so on to the last period, to which the name 
 (units) is not added. 
 
 Divide off by points, and read the following numbers. 
 
 864321 
 
 76247523 
 
 9210043747829 
 
 100210031002478298 
 
 To write numbers from words, we begin on the left 
 hand, taking care to fill up those periods or places that 
 are omitted in the question. 
 
 Write the following numbers : 
 
 Nine millions, seventy-two thousand, and two hun- 
 dred. 
 
 Eight hundred millions, forty-four thousand, and fifty- 
 five. 
 
 Eight billions, sixty-five millions, three hundred and 
 four thousand, and seven. 
 
 Fifty-four sextillions, three hundred trillions, sixty-se- 
 ven millions, four hundred and twenty. 
 
 Seventy decillions, two hundred and thirty-one octil- 
 lions, one billion, one hundred thousand, and three hun- 
 dred. 
 
 Six hundred and forty thousand, four hundred and 
 eighty-one. 
 
 Three millions, two hundred sixty thousand, one hun- 
 dred and six. 
 
 From the system of notation already explained, it is 
 evident that figures have a simple value and a local value. 
 
 When a figure stands in the place of units, it has a 
 simple value only. 
 
 B 
 
14 ARITHMETIC. 
 
 THE FOUR RULES OF ARITHMETIC. 
 
 (Art. 3.) As quantity admits of no other changes than 
 increase and diminution, it is evident that all the operations 
 of arithmetic must be based upon these only. 
 
 Under tlie first are comprehended addition and multi- 
 plication, and under the second, subtraction and division. 
 
 Addition consists in finding the sum of two or more 
 numbers. 
 
 Multiplication is the successive addition of a number 
 to itself a given number of times. 
 
 Subtraction consists in finding the difference between 
 two numbers, or diminishing one given number by an- 
 other. 
 
 Division is the same as subtracting one number succes- 
 sively from another, in order to find how many times the 
 smaller number is contained in the larger. 
 
 Addition, subtraction, multiplication and division, are 
 called i\\efour rules of arithmetic ; all arithmetical oper- 
 ations whatever are combinations of these. 
 
 (Art. 4.) In arithmetical, and in all subsequent math- 
 ematical operations, characters and signs are used to de- 
 note operations and conditions. 
 
 The perpendicular cross, thus +» denotes addition. 
 
 The horizontal line, thus — , denotes subtraction. 
 
 The diamond cross, or a point, thus, X » (either one), 
 indicates multiplication. 
 
 A horizontal line, with a point above and below, thus 
 -^, indicates division. Points, thus, : :; or, :=: pro- 
 portion. Double horizontal lines, thus =, equality. 
 
 The following character represents square root, J. 
 
 The same character, M'ith a small figure annexed, as 3, 
 4, 5, &;c., thus ^J, "^J, ^J, indicates third, fourth and 
 fifth roots. 
 
 (Art. 5.) Numbers are referable to things ; but when 
 no reference is made, the number is said to be abstract ; 
 for instance, tlie number 3 is abstract, and can be applied 
 
THE FOUR RULES OF ARITHMETIC. 15 
 
 to any object whatever, as three apples, three horses, 
 three dollars, &c.; and when it is so applied, it is no long- 
 er abstract. 
 
 Abstra(;t numbers can be added together, as they are 
 then understood to refer to the same thing or scale of 
 measure, but numbers referring to different things, or 
 different scales or standards of measure, cannot be put 
 into one sum. 
 
 Let the pupil early imbibe this important idea, that 
 numbers referring to different things, cannot be added to- 
 gether, under any rule, nor in any part of arithmetic. 
 For example : 3 apples and 5 apples are 8 apples ; but 
 3 apples and 5 dollars are not 8 of the one or the other, 
 and it is manifest that they cannot be added together. 
 
 As an objection to this, a person might say, that in a 
 certain orchard there are 23 apple-trees, 5 pear-trees, and 
 7 cherry-trees ; how many in all ? 
 
 We certainly cannot add them together as apple-trees, 
 because ihey are not all apple-trees, nor are they all pear- 
 trees or cherry-trees ; but we can add them together un- 
 der the general name of trees. 
 
 In the same way, in estimating a farmer's stock, we 
 may add together horses, oxen, cows, sheep, &c., under 
 the general name of animals. 
 
 Indeed we may go further, and say, In a certain field 
 I saw 6 trees, 3 hay-stacks, and 2 oxen ; how many ob- 
 jects were before my vision 1 
 
 Here we may add together trees, hay-stacks, and oxen, 
 or throw them into one sum, under the general name of 
 objects. 
 
 Yet the first assertion holds good, that we cannot add 
 numbers together unless they refer to one common stand- 
 ard. In the first observation above cited, we may put the 
 numbers together as trees, in the second, as animals, in 
 the third, as objects. 
 
 With these preliminary remarks, we commence with 
 tbe first arithmetical operation. 
 
16 ARITHMETIC. 
 
 ADDITION. 
 
 (Art. 6.) Addition is the art of collecting several quantities into 
 one sum. We illustrate this definition by the few following men- 
 tal exercises. 
 
 1. A man bought a load of wood for 4 dollars, a barrel of flour for 
 5 dollars, and a barrel of pork for 13 dollars; how many dollars did 
 he spend ? 
 
 2. A man gave 9 apples to one boy, to another 8, and to another 
 10 ; how many apples did he give away 1 
 
 3. One boy has 11 cents, another has 7 cents, and another has 10 
 cents ; how many cents have the 3 boys ? 
 
 In this last example, shall 11, 7, 10 and 3 be added together; and 
 if not, why ] 
 
 But all numbers cannot be added mentally, because of their mag- 
 nitude ; therefore we must have a more artificial rule ; and as unlike 
 things cannot be added together, preparatory to addition we must write 
 down the dilferent sums, so that units shall stand under units, tens 
 under tens, hundreds under hundreds, <S^c., ^-c, then the student can 
 recognise the rationale of the following 
 
 Rule. Commence with the right hand column, mentally Jin d its 
 sum, and ivrite the unit figure of that sum directly under the col- 
 umn ; add up the next column in the same manner, and increase it 
 by the tens from the last column, the unit of this sum write under 
 the second column, and carry its tens to the next column ,- and so 
 continue, from column to column, to the last one, under which its 
 whole sum mu^t be put doicn. 
 
 The reason for carrying one for every ten, from column to column, 
 is because we have ten characters, and hence, if a sum is more than 
 ten, it cannot be expressed by any single character, and a ten of any 
 order of numbers makes orie of the next higher order. 
 
 We illustrate this by the following example, done out in full, set- 
 ting down the sum of each column in its proper place : 
 
 Here the reason of adding 
 the one of the 14 to the next 
 column, is clearly seen, by in- 
 spection. 
 
 Thus 
 
 4372 
 
 
 5487 
 
 
 3445 
 
 Sum of the units, 
 
 14 
 
 " " " tens, 
 
 19 
 
 " " " hundreds. 
 
 11 
 
 " " " thousands, 
 
 12 
 
 Whole sum, 13304 
 
 After the principle is understood, we do not spread the work out, as 
 above, but mentally add the tens of one column into the next. 
 
 If this mental process is troublesome, and there is danger of losing 
 the number to be carried, the pupil can write units under the first 
 
ADDITION. 17 
 
 column, and the tens over the next column, which will become a part 
 of it : and so on. 
 Tims, 121 2 2 111 
 
 4373 (478 
 
 .5487 Add < 389 Add 
 
 3445 (473 
 
 Sum, 13304 1340 17776 
 
 (I) (2) (3) (4) 
 432 9376 27636 84736 
 213 810 79892 78928 
 121 57 38941 146469 27890 
 213 973 67832 33217 
 92 59244 127076 47263 
 
 Sum 979 
 
 Sum 11308 273545 273545 
 
 N. 15. — There is no simple proof of addition * unless we call a different 
 orfler or manner of putting the sums together proof, in case the final sums 
 
 * We say, no simple proof of addition. There is a proof by casting out 
 the 9's, but its principle is too refined and scientific to be comprehended by 
 persons just learning the science of numbers; we therefore give it in a note. 
 The rule is as follows: 
 
 Add together the figures in the upper horizontal line, reject- Exa7nple. 
 ing the 9's and writing down the excess of 9 a little to the Excesses. 
 
 right. Do the sam^ with every horizontal column. Add up 3497 5 
 these excesses of 9, thus found, and reject the^9 from their 6512 5 
 sum; and if the work is right, this last excess of 9's is the 8-.295 6 
 
 same as the excess of 9's from the sum total. Thus, in our 
 
 example the sum of 5, 5, 6. is 16 ; excess over 9 is 7, and the 18304 7 
 excess of 9's in 18304 is also 7. There is a possibility of the work proving 
 right, and still being wrong; but it is barely a possibility. Thus, if we had 
 made the sum total 18403, it would be wrong, but the excess over 9's, taken 
 horizontally, is still 7. 
 
 This rule is founded on the following property of numbers, namely : .Any 
 number divided by 9 icill leave the same remainder as the sum of its digits di- 
 vided by 9. Thus, 2S divided by 9 leaves a remainder of 1, and 2 added to 
 8 gives 10, which, divided by 9, also leaves 1 : and so with any other num- 
 ber. But, to prove this generally, we may take any number, say 3476, and 
 decompose it thus: 
 
 / 3000=3(1000)=3(999)-f3 
 
 ) 400=4( 100)=4( 99)-f 4 
 
 ^^^^ \ 70=7( 10)=7( 9)-l-7 
 
 ( 6= 6 = 6 
 
 Here, then, we perceive by inspection that 3000 has a certain number of 
 9's, and a remainder of 3 ; 400 has a certain number of 9's, and a remainder 
 of 4; 70 has a certain number of 9's, and a remainder of 7; and 6, being un- 
 der 9, is called a remainder. But these remainders, written horizontal- 
 — 
 
18 
 
 ARITHMETIC. 
 
 a^ree; but tliis gives no demonstraiion lliat an «-rror h;is not hcen made. 
 Thus, in the (3) example we added up ihe eiiiire cohunns in the usual man- 
 ner. Tlieu we took the sum of several parts, and the .sum of these parts, as 
 may be seen by inspeclioii. and tiie final snm.s aijree ; hut if an error was 
 made in the first instance in adding the two upper or two lower sums, 
 tiie same error might be made in the secoiul instance. 
 
 (•^) 
 
 (6) 
 
 (7) 
 
 7298104.3 
 
 1278976 
 
 416785413 
 
 67126459 
 
 7654301 
 
 915123460 
 
 39412767 
 
 876120 
 
 31810213 
 
 7891234 
 
 723456 
 
 7367985 
 
 109126 
 
 31309 
 
 654321 
 
 84172 
 
 4871 
 
 37853 
 
 72120 
 
 978 
 
 2685 
 
 187676921 
 
 10570011 
 
 1371781930 
 
 8. Add 8635, 2190, 7421, 5063, 2196, and 1245 together. 
 
 A71S. 26750. 
 
 9. Add 246034, 298765, 47321, 58653, 64210, 5376, 9821, and 
 340 to<?ether. .4/?.?. 730520. 
 
 10. Add 27104,32540, 10758,6256,704321,730491. 2787316, 
 and 2749104 together. Ans. 7047890. 
 
 11. Add 1, 37, 29504, 6790312, 18757421, and 265 together. 
 
 Ans. 25577540. 
 
 12. Add 562163, 21964, 56321, 18536, 4340, 279, and 73 toge- 
 ther. Am. 663676. 
 
 13. What is the sum of the following^ numbers? viz: Seventy- 
 five ; one thousand and ninety-five ; six thousand four hundred and 
 thirty-five; two hundred and sixty-seven thousand; one thousand 
 four hundred and fifty-fiye ; twenty-seven millions and eighteen ; two 
 hundred and seventy millions and twenty-seven thousand. 
 
 Ans. 297303078. 
 
 APPLICATION. 
 
 1 . Add 3742 bushels, 493 bushels, 927 bushels, 643 bushels, and 
 953 bushels together. Ans. 6758 bushels. 
 
 2. Add 7346 acres, ' 9387 acres, 8756 acres, 8394 acres, and 
 32724 acres. Ans. 66607. 
 
 3. Henry received at <me time 15 apples, at another 115, at ano- 
 ther 19; how many did he receive] Ans. 149. 
 
 4. A person raised in one year 724 bushels of corn, in another 
 3498 bushels, in another 9872; how much in all] 
 
 A)is. 14094 bushels. 
 
 ly 3476, give the original number ; therefore, Uie ssnm of the digits of any 
 number, divided by 9, will give the same reir.'.under as the lumiber divided 
 by 9, which was to be demonstrated. 
 
 Now, tliis property bemg demonstrated, the reason of the rule is obvious ; 
 for the excess of 9's in the several sums added together, and their excess 
 over 9'3 being taken, must be equal to the excess of 9's in the sum total, 
 as the sum of all the parts of a number must equal the whole. 
 
ADDITION. 
 
 5. A man on a journey traveled the first day 37 miles, the second 
 33 miles, the third 40 miles, the fourth 35 miles ; how far did ho travel 
 in the four days ] Atjs. 145 miies. 
 
 6. A has a flock of sheep containing 34, B has a flock of 47, and 
 U a flock of 54 ; how many sheep are there in the three floclcs ? 
 
 A77S. 135 sliecp. 
 
 7. 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 IVevv 
 York 30 miles ; how many miles from Philadelphia to New York ? 
 
 A)}s. 90 miles. 
 
 8. A person bought of one merchant 10 barrels of flour, of ano- 
 ther 20 barrels, of another 95 barrels; how many barrels did he buy ? 
 
 Ans. 125 barrels. 
 
 9. 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? A?is. 421. 
 
 10. An estate is to be shared equally by four heirs, and the portion 
 to each heir is to be 3754 dollars; what is the amount of the estate? 
 
 A>is. 15016 dollars. 
 
 11. A certain regiment of soldiers consists of eight companies, and 
 each company consists of 80 men ; how many soldiers are there in 
 the regiment? Ans. 640. 
 
 12. How many men are there in an army consisting of 52714 in- 
 fantry, 5110 cavalry, 6250 dragoons, 3927 light horse, 928 artillery, 
 250 sappers, and 406 miners ? Ans. 69585. 
 
 13. A merchant deposited 56 dollars in a bank on Monday, 74 on 
 Tuesday, 120 on Wednesday, 96 on Thursday, 170 on Friday, and 
 50 on Saturday ; how much did he deposit during the week ] 
 
 Arts. 466 dollars. 
 
 1 4. A merchant bought at public sale 746 yards of broadcloth, 050 
 yards of muslin, 2100 yards of flannel, and 250 yards of silk ; how 
 many yards in all ? Ans. 3746 yards. 
 
 FEDERAL MONEY. 
 
 (Art. 8.) This is the name given to the established currency of 
 the United States ; and it is purposely adjusted to run on the same 
 scale of computation as abstract numbers : that is, fen of one order 
 making one of the next superior order. Therefore, this money 
 must be added by the same rule as whole or abstract numbers. Its 
 denominations are Eagles, Dollars, Dimes, CeJits, and Mills.* 
 Eagles are the highest, and mills the lowest. Ten mills make a cent, 
 ten cents make a dime, ten dimes make a dollar, and ten dollars 
 make an eagle. In practice, mills, dimes, and eagles are disregarded 
 
 *The coins of the United States are eagles, half-eagles, and quarter-eagles, 
 of gold ; the doUnr, hall'-dollar. quarter-dollar, dime, an half-dime, of silver ; 
 the cent and halt'-cent of copper. 
 
20 ARITHMETIC. 
 
 in naming money, and we have dollars and cents only. The eagles 
 are read ^s dollars, the dimes as cents; and the mills are quite ne- 
 glected in aggregate sums. The dollar is considered as unity ; its 
 mark is $, and a point., called i\\e decimal point, is placed at the right 
 hand of the dollars to separate them from the inferior denominations. 
 'J''hus, twelve dollars and seventy-five cents is written, $12.75. In 
 reading, if all the denomination? were named, we should say one eagle, 
 two dollars, seven dimes, and five cents; but this would he too formal 
 and ijiconvenient. We may change the unit by removing the deci- 
 mal point; thus, in the above sum, twelve dollars and seventy-five 
 cents may be considered 1275 cents, the point only is removed, and a 
 cent is now unity in place of a dollar. 
 
 The following rule for addition in federal money corresponds to the 
 dollar being unity : d stands over dimes, c over cents, and m over 
 mills; but in practice these are not written, for the localily from the 
 decimal point determines the value and denomination of the figure. 
 
 RuLK. rhtce dollars under dollars, cents under cents, SfC, and 
 add as in whole numbers ,- and put the separatrix in the sum to- 
 taL under the separating points above. 
 
 If the cents are less than 10, a cypher must be put in the place 
 of dimes. 
 
 (0 
 
 (2) 
 
 (3) 
 
 (4) 
 
 $ d c m 
 
 $ d c 
 
 d m 
 
 $ d c m 
 
 139.8 6 7 
 
 4319.8 9 
 
 .9 7 6 
 
 81.0 5 3 
 
 1273.5 9 4 
 
 3287.8 
 
 .4 5 8 
 
 67.4 I 2 
 
 9807.(i 
 
 1829.1 6 
 
 .6 2 9 
 
 37.4 5 3 
 
 7695.2 4 
 
 130.0 1 
 
 .5 8 3 
 
 21.6 5 3 
 
 $18916.3 1 
 
 
 
 207.5 7 1 
 
 5. What is the sum of 140 dollars 9 cents, 24 dollars 16 cents, 5 
 dollars 25 cents, 5 dollars 5 cents, and 304 dollars 8 cents and 9 mills. 
 
 Ans. $478,639. 
 
 6. A man left his estate to his two children, who, after payment of 
 the debts, amounting to 645 dollars 75 cents, received each 3842 dol- 
 lars ; how much was the whole estate ? Ans. $8329.75. 
 
 7. If I owe to A 50 dollars and 8 cents, to B 12 dollars more than 
 that sum, and to C 75 cents more than to both the other.s, how much 
 do I owe them all ? Ans. $265.07. 
 
 8. A farmer purchased a cow for $14.75, a horse for $75.14, a wag- 
 on for $31,375, and a harness for $41,625; what was the whole 
 amount? ^/7.s. $161.89. 
 
 9. My grocer sends me the following bill : To 2 pounds of coflfee at 
 12 cents 5 mills per pound; a roll of butter for 87| cents; and 
 a box of soap for 2 dollars and 65 cents; what is the amount of 
 the bill ? Ans. $3,775. 
 
 10. What is the sum of 25 eagles, 62 dollars, 8 dimes, 75 cents, 
 and 5 mills, properly expressed in dollars and cents ? 
 
 Ans. $313.55^. 
 
ADDITION. • 21 
 
 Quesf/oris. 
 
 How many primary rules of arithmetic are there? 
 
 W'liat are they called ? 
 
 What is addition ? 
 
 How do you place nunihers to be added? 
 
 Whore do you begin the addition? 
 
 Why do you carry one for ten, in preference to any other number ? 
 
 W'hy, at the last column, do you write the whole number? 
 
 Why is federal money added as natural or abstract numbers ? 
 
 SUBTRACTION. 
 
 (AnT. 9.) "SiMPLK SuKTnACTiON is taking a less number from 
 a greater, and noting the difference or remainder; thus, 6 taken from 
 10, theditierence is 4; and 5 dollars taken from 7 dollars, leaves a re- 
 mainder of 2 dollars. By signs, these examples may be expressed 
 thus: 10—6=4; 7—5=2. 
 
 " The larger of the two numbers is called the minuend, the lesser 
 is called the subtrahend, and the difference is called their remainder.''^ 
 
 JIEiXTAL EXEnciSES. 
 
 1. Charles had 25 cents, and paid 10 of them for a book ; how 
 many had he left ? 
 
 2. A merchant, 35 years old, had been in trade 13 years ; at what 
 age did he commence business 1 
 
 3. A father who was 37 years of age, had a son 25 years younger; 
 how old was the boy ? 
 
 4. A horse and harness cost 183 dollars. The harness cost 84 
 dollars ; what was the price of the horse 1 
 
 5. Dr. Franklin died in 1790, at the age of 84 ; in what year was 
 he born ? 
 
 6. A man paid 84 dollars for a watch, but was obliged to sell it for 
 07 dollars ; how many dollars did he lose] 
 
 NoTK. — All operations in both addition and sublraction g^o back to the 
 mental proce.ts ; hut when results are written down as last as they are at- 
 tained, it is called written arithmetic. And the art of writing comes in to 
 help ihe mental process, and enables the mind to concentrate its powers on 
 a sing^le point at atime. 
 
 liCt ns write the last problem, and submit it to a more artificial process; 
 but when written, if we consider only one figure at a time, we cannot sub- 
 tract 7 from 4. We must, therefore, decompose (he numbers, thus : 
 
 84=70-1-14 
 67=60-f- 7 
 
 Difference, I0-|- 7, or 17. 
 
 Again, let us subtract 765 from 1234. To analyze this, 1234 must be so 
 
22 
 
 ARITHMETIC. 
 
 separated tliat it would be possible to subtract 7 liundreds, 6 tens, and 5 
 uuils from it, thus: 
 
 1234=1100-1-1204-14 
 
 765= 700-}- 60-f- 5 
 
 Dlff rence, 4tj9= 400-|- GO-j- 9 
 
 "\Vc decompose the numbers to enable the pupil to perceive the raiionale 
 of llie vviiole operation. Thus, it will be observed that we must take 5 uiiiis 
 from 14 uiii:s; but, to make the 4, 14. we must, (as it is called), borrow a ten, 
 or one iVom the tens. Then it will be G I'rom 2. or 12, (but in place of 6 from 
 12. we mav say 7 from 13. the difference is the same). Tlien, 7 fioiri 11, or 
 wiiicli gives the same difference, 8 I>om 12. 4. Hence, when we subtract a 
 figure irom a less one above, we conceive tlie upper figure increased by 10, 
 and the next figure in the lower line increased by one to compensate it. 
 
 From these observations we draw the following 
 
 liuLK, To operate on large numbers, set doivn the less niunher 
 under the greater, units under units, tens under tens, S(C., and draw 
 a line htntath them. 
 
 Then, beginning at the right hand, subtract each figure from the 
 one directly over it, and set down the remainder. 
 
 Bat if the upper figure be the least, suppose it to be increased bij 
 ten ; then make the subtraction, set down the rernainder, and carry 
 one to the next figure of the subtrahend. 
 
 Pkoof. Add the remainder to the subtrahend. If their sum is 
 equal to the minuend, the work may be regarded as right. 
 
 EXAMPLES. 
 
 Minuends, 
 Subtrahends, 
 
 Remainders, 
 
 Proofs, 
 
 859267 
 107895 
 
 751372 
 
 859267 
 
 (4) 
 10000 
 4 
 
 (5) 
 
 3000 
 
 999 
 
 (2) 
 
 6794213 
 
 975678 
 
 5818535 
 
 6794213 
 
 (6) 
 6798 
 4000 
 
 (3) 
 
 19067803 
 
 04202196 
 
 148G5607 
 
 19067803 
 
 (7) (8) 
 10000 87000 
 1 1009 
 
 Remainders, 9996 2001 2798 
 
 9999 
 
 85991 
 
 9. Ft 
 
 10. From 
 
 11. — 
 
 12. — 
 
 13. — 
 
 14. — 
 
 15. — 
 
 16. — 
 
 17. — 
 
 78213609 take 27821890. 
 
 Ans. 50391719. 
 
 196 
 
 487 
 
 875 
 
 967 
 
 1001 
 
 9765 
 
 87696 
 
 455692 
 
 take 
 
 37. 
 
 96. 
 302. 
 351. 
 
 487. 
 
 1307. 
 
 10091. 
 
 300120. 
 
 Resul 
 
 159 
 391 
 573 
 
 616 
 
 514 
 
 8'158 
 
 77605 
 
 155572 
 
SUBTRACTION. 23 
 
 18. From eleven thousand and eleven Ijumlred take. 1521). 
 
 liciii. 10580. 
 
 19. From thirty thousand and ninely-scvi-n, take one thousand six 
 hundred and lilty-four. Ktm. 2841:3. 
 
 20. From one hundred million two hundred and lorty-.seven thou- 
 sand, take one million four hundred and nine. Kern. 9924G591. 
 
 21. Subtract one from one million. Rem. 999999. 
 
 APPLICATION. 
 
 1. Sir Isaac Newton was born in the year 1642, and died in 1727 ; 
 to what age did he live? Ana. 85 years. 
 
 2. A merchant gave his note for 5200 dollars. He paid at one 
 time 2500 dollars, at another time 175 dollars ; what remained due ? 
 
 Ans: 2525 dollars, 
 
 3. -In 1800 there were 903 post-offices in the United States, and in 
 1828 there were 7530 ; how much had their number increased ? 
 
 Aus. 6627. 
 
 4. A farmer owned 360 acres of land, and sold 175 acres; how- 
 many acres had he left! ^1"^. 185. 
 
 5. Massachusetts contains 7800 square miles, and New Hampshire 
 9491 square miles; which is the largest, and by how many square 
 miles ? 
 
 6. New York contains 46085 square miles ; how much larger is it 
 than Massachusetts ? than New Hampshire ? than both these States 
 together ] 
 
 7. The Declaration of Independence was published July 4th, 1776 ; 
 how many years to July 4th, 1850? 
 
 8. Gunpowder was invented in the year 1330 after Christ, and the 
 art of printing in the year 1441 ; how long from each of those events 
 to the year 1850? 
 
 9. The mariner's compass was invented in Europe in the year 
 1302 ; how long since that event to the present year ? 
 
 10. 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 1 
 
 Ans. 1807 dollars. 
 
 11. A merchant bought 5875 bushels of wheat, and sold 2976 
 bushels', how many bushels remain in his possession ? 
 
 Ans. 1899 bushels. 
 
 12. A grocer bought 25 hogsheads of sugar, containing 250 hun- 
 dred 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. 
 
 13. A traveler who was 1300 miles from home, traveled homeward 
 235 miles in one week, in the next 275 miles, in the next 325 miles, 
 and in the next 280 miles; how far had he still to go before he would 
 reach home? Ans. 175 miles. 
 
24 ARITHMETIC. 
 
 SUBTRACTION OF FEDERAL MONEY. 
 
 (Art. 10.) Rule. Place dollars under dollars, cents under 
 cents, Ac- ; proceed as in whole numbers, and put the separatrix in 
 the remainder, under the decimal points above. 
 
 (I) (2) (3) 
 
 $ c $cm $cm 
 
 From 75.48 246 05 6 6327.86 5 
 
 Take 58.76 218.60 7 4961. 
 
 Ans. §16.72 
 
 (2) 
 
 
 $ c 
 
 m 
 
 246 05 
 
 6 
 
 218.60 
 
 7 
 
 
 4. From 36 dollars and 5 cents, take 28 dollars and 60 cents. 
 
 Rem. $7A5. 
 
 5. From two dollars, take 5 dimes and 5 mills. Rem. §1.49^. 
 
 6. From 2 dollars, take 7 cents. Rem. $1.93. 
 
 7. From 4 dollars, take 75 cents. Rerii. $3.25. 
 
 8. Owing my neighbor 60 dollars, I paid him at one time §9.50 ; 
 at another 5il5; at another $30.75; and at another 62^ cents ; how 
 much was then due him? Ans. §4.12^. 
 
 9. A man paid for two sheep 7 dollars and 50 cents, and for a cow 
 twice as much, lacking 75 cents; how much did she cost him ? 
 
 An>: $14.25. 
 
 10. A farmer's bill at a store amounted to $37,625, and he paid 
 $19 and 5 cents in oats ; how much did he still owe ? 
 
 Ans. $18,575. 
 
 1 1. A man owes §100 : he has §47.375 ; how much must he bor- 
 row to pay the debt? Ans. §52.625. 
 
 12. A man sold his house and lot for §2070, which was §62.875 
 more than it cost him ; what did it cost him ] Ans. §2007.125. 
 
 Questioiis. 
 
 What is subtraction? 
 
 What is the greater number called ? 
 
 What is the less number called 1 
 
 What is the difference called ? 
 
 How do you place numbers for subtraction? 
 
 When the lower figure is greater than the upper one, how do you 
 proceed ? 
 
 Why is the 07ie 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 ? 
 
 MULTIPLICATION. 
 
 (Art. 11.) Multiplication is the repetition of the same num- 
 ber, hence, it can be performed by addition. 
 
MULTIPLICATION. 
 
 25 
 
 Tlie number to be repeated is called the multipHcancf, the number 
 of times it is to be repeated is called the multiplier, and the result is 
 called the product. 'I'he multiplier and multiplicand are also called 
 factors of the product. 
 
 EXAMPLE. 
 
 Let 8 be repeated three times, thus: 8-}-8-f-8=24. The short 
 mental operation of saying and comprehending tiiat 3 times 8 are 24, 
 is called multiplication ; 8 is called the multiplicand, 3 the multiplier, 
 and 24 the product. But this is arbitrary : we may call 3 the multi- 
 plicand, and 8 the multiplier ; and it will then be required to repeat 
 3, 8 times : hence, in point of fact, it is indifferent which of the two 
 factors is called the multiplier, and which the multiplicand. 
 
 Any two or more numbers which, multiplied together, produce a 
 third number, are cMed factors of the number. A number may have 
 many different factors ; for example, 36 may have for factors 3 and 
 12, 2 and 18, 4 and 9, 6 and 6, or 2, 3 and 6, because — 
 
 3X12=36 
 2X18=36 
 4X 9=36 
 6X 6=36 
 2X3X 6=36. 
 
 MULTIPLICATION TABLE. 
 
 2X 1= 2 
 
 iX 1= 4 
 
 6X 1= 6 
 
 SX 1 = 
 
 8 
 
 lOX 1= 10 
 
 2X 2= 4 
 
 4X 2= 8 
 
 6X 2=12 
 
 8X 2= 
 
 16 
 
 lOX 2= 20 
 
 2X 3= 6 
 
 iX 3=12 
 
 6X 3=18 
 
 8X 3= 
 
 24 
 
 10 X 3= 30 
 
 2X 4= 8 
 
 4X 4=16 
 
 6X 4=24 
 
 8X 4= 
 
 32 
 
 lOX 4= 40 
 
 2X 5=10 
 
 IX 5=20 
 
 6X 5=30 
 
 SX 5= 
 
 40 
 
 lOX 5= 50 
 
 2X 6=12 
 
 iX 6=24 
 
 6X 6=36 
 
 8X 6= 
 
 48 
 
 lOX 6= 60 
 
 2X 7=14 
 
 4X 7=28 
 
 6X 7=42 
 
 SX 7= 
 
 56 
 
 lOX 7= 70 
 
 2X 8=16 
 
 4X 8=32 
 
 GX 8=48 
 
 8X 8= 
 
 64 
 
 lOX 8= 80 
 
 2X 9=18 
 
 IX 9=36 
 
 .^X 9=54- 
 
 SX 9= 
 
 72 
 
 10 X 9= 90 
 
 2X10=20 
 
 4X10=40 
 
 0X10=60 
 
 8X10= 
 
 80 
 
 10X10=100 
 
 2X11=22 
 
 4X11—11 
 
 SX 11=66 
 
 8X11= 
 
 88 
 
 10X11 = 110 
 
 2X12=24 
 
 4X12=48 
 
 6X12=72 
 
 8X12= 
 
 96 
 
 10X12=120 
 
 3X 1= 3 
 
 5X 1= 5 
 
 7X 1= 7 
 
 9X 1= 
 
 9 
 
 IIX 1= 11 
 
 3X 2= 6 
 
 5X 2=10 
 
 7X 2=14 
 
 9X 2= 
 
 18 
 
 11 X 2= 22 
 
 3X 3= 9 
 
 5X 3=15 
 
 7X 3=21 
 
 9X 3= 
 
 27 
 
 I IX 3= 33 
 
 3X 4=12 
 
 5X 4=20 
 
 7X 4=28 
 
 9X 4= 
 
 36 
 
 11 X 4= 44 
 
 3X 5=15 
 
 3X 5=25 
 
 7X 5=35 
 
 9X 5= 
 
 45 
 
 1 1 X 5= 55 
 
 3X 6=18 
 
 5X 6=30 
 
 7X 6=42 
 
 9X 6= 
 
 54 
 
 IIX 6= 66 
 
 3X 7=21 
 
 ox 7=35 
 
 7X 7=49 
 
 9X 7= 
 
 63 
 
 11 X 7= /7 
 
 3X 8=24 
 
 5x 8=40 
 
 7X 8=56 
 
 9X 8= 
 
 72 
 
 11 X 8= 88 
 
 3X 9=27 
 
 5x 9=45 
 
 7X 9=63 
 
 9X 9= 
 
 81 
 
 IIX 9= 99 
 
 3X10=30 
 
 5X10=50 
 
 7X10=70 
 
 9X10= 
 
 90 
 
 11X10=110 
 
 3X11=33 
 
 5X11=55 
 
 7X11=77 
 
 9X11 = 
 
 99 
 
 11X11=121 
 
 3X12=36 
 
 5X12=60 
 
 7X12=84 
 
 9X12= 
 
 108 
 
 11X12=133 
 
26 ARITHMETIC. 
 
 As multiplication is repeating a sum a given number of times, it is 
 evident that the units, and tens, and hundreds, and every superior or- 
 der of the proposed sum, must be repeated the given number of times ; 
 and in repeating the units, if they amount to any number of tens, and 
 units, the tens must be added into the column of tens, when we re- 
 peat the tens, for the same reason that we carry by tens in addi- 
 tion. 
 
 To analyze the process, let it be required to multiply 468 by 8 ; that 
 is, repeat 468, 8 times. 
 
 468=400-1- 60-|- 8 
 
 Repeat each order 8 times, 8 
 
 3200-|-480-f64 
 
 Or, in a more condensed form, it stands thus 
 Multiplicand, 468 
 Multiplier, 8 
 
 Product, 3744 
 
 From this we perceive, that in all cases the tens which arise from 
 repeating the units must be added in w4th the tens that arise from re- 
 peating the tens; and the hundreds which arise from repeating the 
 tens, must be added to the column of hundreds, &c. ; hence, when 
 the multiplier is a single Jigure, we have the following 
 
 RuLT.. Place ihe multiplier under the units figure of the multi- 
 plicand, and mtiltiply each figure cf the multiplicand in succession, 
 and set down the amount, and carry as in addition, 
 
 (AuT. 12.) Pnoor. The proof of multiplication is division, for 
 they are reverse operations ; but this method of proof cannot be used 
 until the pupil has learned division. But if we vary the method of 
 multiplying, and find the same result, we may be quite conlident that 
 the result is correct; and some persons would regard it as proof and 
 others would not. For instance, if we have any number, say 231, to 
 be multiplied by 8. We may multiply it by 8 directly, and then for 
 proof, divide 8 into any two parts, say 6 and 2, or 5 and 3, and mul- 
 tiply 231 by 6; then multiply 231 by 2, and the two products added 
 together will be the same as the first result, if no mistakes have been 
 made. Hence, to prove multiplication, we have the following 
 
 Rult:. Separate the multiplier into any two parts, and multi- 
 ply by such part separately, and add the two results together, and 
 their sum will be the same as by direct multiplication, provided no 
 mistakes have been made. 
 
MTLTIPLICATION. 27 
 
 EXAMVLES. 
 
 (1) (2) (3) (4) 
 
 3G563 8375 4378 9286 
 
 5 6 7 8 
 
 182315 50250 ;30G46 74288 
 
 APPLICATION-. 
 
 1. If I barrel of flour sells for 555 cents, what will 6 barrels come 
 to "^ Ans>. 33:30 cents. 
 
 2. If 1 load of hay costs 875 cents, what will 7 loails come to? 
 
 A?}s. 6125 cents. 
 
 3. If 1 cord of wood costs 486 cents, what will 8 cords come to? 
 
 Ans. 3888 cents. 
 
 4. If I yard of broadcloth costs 765 cents, what will 9 yards come 
 to ? Ans. 6885 cents. 
 
 5. A lady purchased 54 yards of muslin at 12 cents a yard ; what 
 was the cost of the whole 1 Am: 648 cents. 
 
 6. A farmer sold 49 acres of land at U dollars an acre ; what did 
 the whole amount to? Ans. 5.i9 dollnrs. 
 
 7. A Carpenter worked ten days at 146 cents a day ; ho.v much did 
 his wages amount to? Ans. 1460 cents. 
 
 8. VVliat will 764 pounds of pork come to at 8 cents a pound ? 
 
 Ans. 61 12 cents, 
 
 9. A person spends 9 dollars a week ; how much will he spend in 
 4G week« ? Ans. 4 14 dollars. 
 
 10. If ynu save 12 cents a week, how much will you save in 52 
 weeks, or 1 year] Ans. 624 cents. 
 
 11. James spends for nuts and cakes 8 cents a day ; how much 
 will he spend in a year, or 365 days? Ans. 2920 cents. 
 
 12. .lane spends for cakes, nuts, and apples, 6 cer-ts a dav ; how 
 much will she spend in 216 days? Ans. 1296 cents. 
 
 (Aht. 13.) Before we proceed any further, it is important to re- 
 mark, that multipliers must always be considered as abstract num- 
 bers. Mu!tiplicaii(m is a mere repetition of the Same number, or the 
 same //tin<{;s, a given numl)er of times. 'Vhe number expressing the 
 times anything is to be taken, must be mere count, and cannot repre- 
 sent things. 
 
 In the first example, if 1 barrel of flour sells for 555 cents, vi'hat 
 will 6 barrels come to? 
 
 Here the 555 cents cannot be mulfipl'ud by barrels, but it can and 
 must be repeated as many times as there are barrels.* 'i'he same 
 reason applies to all other questions. 
 
 The quantity which has the same name as the required result is 
 
 *Tliis maybe considered a nice dislinciion ; hut it is essential, as we find 
 so much carelessness on this poini arnon,£r some of our teachers and in some 
 of our hooks. It is not unfr.-ciuenltlia! we find such demands as the'^e : Mul- 
 
28 ARITHMETIC. 
 
 always, properly speaking, the multiplicand, whether we practically 
 make it so or not. Thus, in the 12th example: Jane spends for 
 cakes, nuts, and apples, 6 cents a day ; how much will she spend in 
 216 days ? Here, 6 cents must be taken 216 times, but in practice 
 wc write the operation thus ; 
 
 216 
 6 
 
 1296 
 
 The largest number is taken for the multiplicand, for it makes no 
 difference : the product of two factors is the same, whichever be re- 
 garded as the multiplicand. 
 
 When the multiplier consists of more than one figure, the product 
 of the superior figure will be a supeiior product ; for instance, the 
 product of 6 tens, or 60, must be 10 times as much as the pro;luct 
 of 6 ; though we may multiply by any superior figure, in the same 
 manner as we do by the unit figure, only taking care to give the 
 product its true place in the 07'der of numbers. From these consid- 
 erations arises the following 
 
 Rule. When the multiplier contains more than one JJgure, 
 multiply the multiplicand by each figure of the multiplier, placing 
 the right-hand fig^-ire of each product directly under that figure 
 (f the multiplier Ijy luhic'i it is produced, and take the .sum of all 
 the products. 
 
 EXAMPLES. 
 
 37462 
 .563 
 
 Product of the units, lll^SSo 
 
 Product of the tens, 224773 
 
 Product of the hundreds, 187310 
 
 Entire product of 563, tilD9ll06 
 
 2. Multiply 4962 . . by . . . 9S Ans. 486276. 
 
 3 7331 ..!... 87 642108. 
 
 4 4376 97 424472. 
 
 5 7923 73 617994. 
 
 6 6842 89 608938. 
 
 7 7648 523 3999904. 
 
 P 8473 ..... 456 3S63688, 
 
 9 9372 567 5313924. 
 
 10 75649 579 43800771. 
 
 liply 25 cents by 25 cents; multiply 2 shillings and 6 pence by 2 sliiM^ng-? and 
 6 pence. &c. Now. we can repeal 25 cents, or any other ^^mn of money, as 
 viany times, or an many parts of one time as we pll'a^;o : lun tliere is no 
 such thing as money times money, or imrreis times barrels, or barrels times 
 money. If it were possible to take money times money, what would 
 it produce? 
 
MULTIPLICATION. 29 
 
 11. .Multiply 29S31 . .by. . 952. . . .^77,^.28399112. 
 
 12 252:J8 . . . . 12170 30714G4G0. 
 
 13 991)9 6tJ(iO 66593.340. 
 
 14 87603 9865 ..... 861203o95. 
 
 (AuT. 14.) To multiply by 10, we place a cipher at the right; 
 'i'hus, 5 multiplietl by 10, gives 50; that is, the cipher is put at ihe 
 right of the 5. In the same way, to multiply by 100, we put two 
 ciphers thus— 500; 6 multiplied by 1000, wc have 6000, &c. 
 
 
 
 kxa: 
 
 IPI.E^-. 
 
 
 10. 
 
 Multiply 47654 . 
 
 .by 
 
 . 10 . . . 
 
 . . Ans. 476540. 
 
 11. 
 
 ..... 75478 . 
 
 
 . 100 . . 
 
 . . . 7547600 
 
 12. 
 
 86427 . 
 
 
 1000 . . . 
 
 . . . 8G427000. 
 
 13. 
 
 17846 . 
 
 
 10000 . . 
 
 . . 178160000 
 
 14. 
 
 lOOCO . 
 
 . . 
 
 17S46 . . 
 
 . . 178160000 
 
 It will be observed that examples 13 an.l 14 are tlic same, and pro- 
 duce the sajne results, ft i.s immaterial whether the ciphers are in 
 one factor or the other, or in hotk ; the result will have the same num- 
 ber of ciphers on the right as are contained in both factors. Henct, 
 rvheji ficlors have ciphers on their right, the ciphers may be omit- 
 ted in the operation, and afterwards annexed to the product. Thus : 
 
 15. Multiply 47000 by 4200 Operation .• 
 
 47 
 42 
 
 9i 
 
 188 
 
 Kesuit, 197400000 
 
 16. MuUiplv 73400 by 72000 Result, 3082SOOOOO. 
 
 17 '. 62300 bv 12100 751410000. 
 
 18 785400 by 2000 1570800000. 
 
 When ciphers are between numbers in the multiplier there 
 will be no product corresponding to then), because they signify 
 nothin<T or 770 times. The other partial products exist the same as 
 before. 
 
 EXAMPLES. 
 
 Operation. 
 1. Multiply 672 by 602. 672 
 
 ^ -^ 603 
 
 1344 
 4032 
 
 404544 
 
 c 2 
 
30 ARITHMETIC. 
 
 Here are two partial products ; the product for the units, and the 
 product for the hundreds. 'J'here is no product for the tens, because 
 the multiplier contains no tens. 
 
 2. Multiply 96038 . . by . 6007. . . . Ai^s. 576900266. 
 
 3 40307 .... 30508 1229685956. 
 
 4 3000024 309 1170007416. 
 
 5 83200 904 75212800. 
 
 GENERAL APPLICATION OF MULTIPLICATION. 
 
 1. A gentleman owes 25 laborers 16 dollars each ; how many dol- 
 lars will be required to pay them ? Ans. 400 dollars. 
 
 2. A carpenter owes a journeyman for 43 days' work, at 125 cents 
 per day ; what did the whole amount to ? An.s. 5375 cents. 
 
 3. A merchant buys 440 yards of muslin at 17 cents per yard ; 
 what was the amount of the bill 1 Am. 7480 cents, or $74.80. 
 
 4. A farmer sold 60 bushels of wheat at 125 cents per bushel, 
 40 bushels of rye at 85 cents per bushel, and 34 bushels of corn at 
 50 cents per bushel ; how much money did he receive ? 
 
 ,' . Ans. 126 dollars. 
 
 5. A hogshead contains 63 gallons : what will 3 hogsheads of 
 molasses cost, at 27 cents per gallon ? Ans. 'Si-51.03. 
 
 6. A bushel contains 32 quarts : what will 4 bushels of chestnuts 
 come to, at 5 cents per quart ? Ans. 640 cents, or $6.40. 
 
 7. VVliat will 396 bushels of potatoes come to at 24 cents per 
 bushel ^ Ans. §95.04. 
 
 8. How many dollars will be required to pav for 29 fat oxen, at 
 43 dollars each 1 "' Ans. $1247. 
 
 9. A certain wagon wheel will turn round 350 tinjcs in running 
 a mile : how many times will it turn round in running 25 miles ? 
 
 Ajis. 8750 times. 
 
 10. The semi-diameter of the earth is 3956 miles, and it is 60 
 times that distance to the moon; and the distance to the sun is 400 
 times the distance between the earth and moon ; how many miles is 
 it 10 the sun? /i«s. 94944000" miles. 
 
 1 1 . There are 360 degrees in the circumference of the earth ; and 
 if each degree were 60 miles, what would be the number of miles 
 round the earth ? If 70 miles to a degree, what would be the num- 
 ber of miles ? a S^^ 60— 21 600 miles. 
 
 ^"^' 1 at 70— 25200 miles. 
 N. B. — The circumference is 24856 miles, 
 
 12. If a ship should sail 8 miles an hour for 18 days, how far 
 would it sail? Ans. 3456 miles. 
 
 13. How many panes of glass are there in a house that has 20 
 windows, each containing 24 panes, and 12 windows, each contain- 
 ing 18 panes? Ans. 696 panes. 
 
 14. There are 60 minutes in an hour; how many minutes are 
 there in 24 hours, or one day ? , Ans. 1440. 
 
 15. How many minutes are there in 365 days? 
 
 Ans. 525600. 
 
MULTIPLICATION. 31 
 
 16. A house contains 26 windows, each containing 24 panes of 
 glass; how many panes are there in the whole ? Aus. 6:^4. 
 
 17. There are 20 shillings in a pound ; how manv arc there in 456 
 pounds] Ans.'dl20 shillings. 
 
 18. There are 12 pence in 1 shilling; how many pence are in 254 
 shillings ? yl??5-. 3048. 
 
 19. There are 24 hours in 1 day; how many hours are in 156 
 days? Alls. 3024:. 
 
 20. There are 28 pounds in 1 quarter ot" a cwt. ; how many pounds 
 are in 124 quarters';' yl/^^■. 3472. 
 
 21. There are 20 pieces of cloth, each containing 37 yards, and 49 
 other pieces, each containing 75 yards ; how many yards of cloth 
 are there in all the pieces? Ans. 4415. 
 
 22. Thoro are 24 hours in a day, and 7 days in a week ; how 
 many hours in a week ? Ans. 163. 
 
 23. A merchant buys a piece of cloth containing 97 yards, at 3 
 dollars a yard ; what does the piece cost him 1 An^. 291 dollars. 
 
 (Art. 15.) A composite number is such a number as may be pro- 
 duced by the product of two or more factors multiplied together, 
 'i'hus, 24 is a composite number, because 4 times 6 make 24, or 2 
 times 12 make 24, or 3 times 8 make 24, or 2 X 2 X 2 X ;^="-i4 ; also, 
 56 is a coniposHe number, because it can be composed of the product 
 of 7 times 8. or 2X^X7, &c. 
 
 Obsrvation. — When a multiplier is a composite numuer, we may 
 mitltiply the multiplicand by one factor, and that product by an- 
 other, and so on, until all the factors are used. The lad product 
 will be the result required. 
 
 EXAMPLES, 
 
 1. Multiply 476 by 24 
 
 As 24=4X6, multiply by 4 ' 
 
 1904 
 Now, this product by 6 
 
 Result, 11424 
 
 2. Multiply 425 by 16. (16=4X4) Ans. mOQ. 
 
 3. Multiply 2873 by 54. (54=9X6) Ans. 155142. 
 
 4. Multiply 108 by 28. (28=4X7) An.^. 3024. 
 
 5. Multiply 2042 by 484. (484=12 IX 4) 
 
 (121= IIXH) (4=2X2) 
 2042X11 X 1 1 X2X2=988324, 
 Asa pure mental operation for one who is blind, or has no means 
 at hand of writing figures, this may be preferable to the common 
 method ; but, under ordinary circumstances, the conmion way is the 
 best, for the factors are co!i:imonly larger figures than their aggregate 
 product. For in.stance, if we were required to multi[)ly by 63. the 
 factors of this product are 7 and 9, larger numbers than the mere 
 figures 6 and 3. 
 
32 ARITHMETIC. 
 
 There are many other ways of performing- niuUiplication, but we 
 cannot explain many of them before we acquire the art of dividing. 
 
 Tl)is principie. as last explained, however, is not strictly confined 
 to composite numbers, if the pupi! will consider the natuie of niulti- 
 plicaiion. the real object sought fur. It is, to obiaia Hit amount 
 i:J' l/ic multiplicand as many times as there are units in the 
 multiplier. 
 
 Now, suppose we wish to multiply any number by 37, which is 
 not a composite number, we may consider that 37=36-1-1, and 
 36=6X6. Therefore, wo may obtain the multiplicand 36 times, and 
 then adil it once to f.)rm the product. In the same manner, to multi- 
 ply by 29, we may first multiply by 30, and subtract once the multi- 
 plicand, &c. But these subjects will come up again in the supple- 
 ment to multiplication and division. 
 
 Questions. 
 
 Does it make any difference in the product, which of the two 
 given numbers you make the multiplier? 
 
 Why is the largest number usually made the multiplicand, and 
 the smallest the multiplier ? 
 
 Multiplication is but another method of performing addition : is 
 it, then, indispensable in practice? 
 
 Why do we place the product of the second figure in the multi- 
 plier one place to the left of the product of the first figure ? 
 
 Why, in multiplying, do we begin with the lowest figure in the 
 multi] lier, rather than the highest? 
 
 Provided we place the products for addition in the order of their 
 respective values, does it make any dilTerence in the result which 
 figure of the multiplier we commence with 1 
 
 If you annex three cii>her3 to the right hand of a number, how 
 much is its value increased ? 
 
 DIVISION. 
 
 (AnT. 16.) 1. DivisTov is the art of finding how na any times 
 one number is contained in another. 
 
 2. Division is a short method of finding how many times one 
 number can be subtracted from another. 
 
 3. Division is the converse of multiplication, and the same princi- 
 ples are to be preserved in both multiplication and division. 
 
 4. Division is a concise method of finding how many times one 
 number is greater than another. 
 
 'J'hese four definitions are but different expressions to explain the 
 same priiici[)le. 
 
 The three following technicalities are to be particularly noticed : 
 
 1. The dividend, or number to be divided. 
 
 2. The divisor, or number by which it is divided. 
 
DIVISION. 33 
 
 ;5. The quoiienf, or result of the division, showing how many 
 times the ilivisor is contained in the dividend. 
 
 When tlie divisor is not contained in the dividend an exact num- 
 ber of times, what is over or reinaininir after tlic division, is called 
 the remtdnder ; it must, of course, he less than the divisor. 
 
 The sign comrnordy used to denote division, is -i-: tiius. r2H-3=4. 
 Division is also indicated by placing the divisor under the dividend, 
 
 n 
 
 with a line between them: thus, — =4, which is read, VZ divided bv 
 
 3 
 ;3 equals 4, * 
 
 'I'he following illustration will better chicidate the subject : 3 can 
 be subtracted from 12 four times; thus, 
 12 
 3 one time. 
 
 9 
 
 3 two times. 
 
 6 
 
 % 3 three times. 
 
 3 
 
 3 four times. 
 
 remains. 
 
 How many hats, at 3 dollars apiece, can be bought for 12 dollars ? 
 
 How many times can 3 be subtracted from 12 1 
 
 How many times can 4 be subtracted from 24 ] 
 
 How many times can 8 be subtracted from 27 ? 
 
 At 5 miles per hour, how many hours will it require to travel 40 
 miles ? 
 
 At 4 dollars a yard for cloth, how many yards can be bought for 
 42 dollars? 
 
 How many times can 4 be subtracted from 42, and how many 
 will be left after the last subtraction ? 
 
 The foregoing questions illustrate the general principles of divi- 
 sion ; and the pupil will at once perceive its connection with subtrac- 
 tion ; but we can neither add nor subtract unlike quantities (se(! 
 Art. 5). So in division: the divisor, dividend^ and qiwlient, wiW 
 have a philosophical relation with each other in regard to things. 
 
 Thus : If the dividend be a concrete number, such as dollars, 
 yards, miles, &c., and the divisor an abstract number, or a mere nu- 
 meral, the quotient will be of the same name as the dividend. 
 
 If both the divisor and the dividend are concrete munbers of the 
 same kind, the quotient will be an abstract number, or a mere nume- 
 ral. One or the other, the divisor or the quotient, must be a mere 
 
34 
 
 ARITHMETIC. 
 
 numeral. When hoth the divisor and the quotient are abstract, tlie 
 dividend must also be abstract. 
 
 As division is the converse of multiplication, we now proceed 
 to fexplain it, by taking an example in multiplication to trace its con- 
 verse operation : thus, 
 
 Multiply 704 by 4 ; that is- 
 
 Multiply 
 
 700-1-60-1-4 by 4, 
 4 
 
 2800-1-240-1- lG=r.3056 
 
 The converse operation is, to divide SDfjG Jjy 4 ; and we must 
 make the same division of this number as atK'V?, or. rather, the ope- 
 ration itself will thus separate the number. Au^ as division must be 
 in all respects the converse of multiplication, we rtt ist commence di- 
 vision at the left, or at the highest order of the n«u«ioer. 'J bus, 4 is 
 not contained in 3, (8000,) and the greatest numlxTT ^\i:vkv .30, vvhith 
 is exactly divisible by 4. is 23, or in this example, 20^0 ^-r same 
 number that we perceive in the multiplication — and tlas e)oaM>ntary 
 operation will stand as follows : * 
 
 4)3056(700 700-^ 
 
 28 I 
 
 60 ^Partial quotient* 
 
 4)256(60 
 240 
 
 4)16(4 4 J 
 
 16 764 whole quotient. 
 
 Divide 3496 by 16, 
 
 Here we must consider that 16 is contained in 34 2 times; ^W as 
 34 is 34 hundred, the 2 is 2 hundred, and the operation is thus. 
 
 16)3496(200 
 3200 
 
 200^ 
 
 16)296(10 
 160 
 
 10 
 
 S-Partial quotients. 
 
 10)136(8 
 
 8 218 whole quo. and rem. of 8. 
 
 But, in place of making three problems of this, we can make one, 
 
DIVISION. 
 
 35 
 
 16)3496(218 
 32 
 
 29 
 18 
 
 136 
 128 
 
 From the foregoing, we draw the following 
 
 RcLK. [•"ind how many times the divisor is contained in as many 
 of the kfl-hand figures of the dividend as are ju^t sufficient to con- 
 tain it, and set the result on the right of the dividend fur the first 
 quotient fgure ,• multiply the divisor by it, and place the product 
 under the part of the dividend taken; subtract it therefrom, and 
 to the right of the remainder bring down the next figure in the 
 dividend: divide this number as before, and thus proceed until 
 every figure in the dividend is brought down. 
 
 [NoTK. Remainders may either be carried out to the riglit of the 
 quotient, with a hue between, or be placed above the hue, and the 
 divisor below, in the form of a fraction.] 
 
 When the divisor does not exceed 12, the multiplication and sub- 
 iraciion may be made in the mind, without the formality offsetting 
 down the whole process. In this case, the quvtient is placed under 
 the dividend, with a line drawn between. 
 
 
 EXAMPLES. 
 
 
 (1) 
 
 Divisor 4)654324 
 
 (2) 
 6)8574684 
 
 !3) 
 8)146237456 
 
 Quotient 163581 
 
 1429114 
 
 18'^ 79682 
 
 (4> 
 Divisor 10)1786940 
 
 '5) 
 9)7468542 
 
 11)564521705 
 
 Quotient 178694 
 
 
 
 (7) 
 12)9476521764 
 
 (8) 
 11)546215747 
 
 (9) 
 12)462164684 
 
 1 
 
36 ARITHMETIC. 
 
 The operations may be proved by multiplying the quotients by the 
 divisor, and adding the remainder, if there is any, to the product. If 
 the v»^ork is correct, it will equal the dividend. 
 
 10. What will be tlie quotient of 4562, divided by 3 ] 
 
 Quo. 1520, Rem. 2. 
 
 11. What will be the quotient of 44631, divided by 4? 
 
 Quo. 11157, Teem. 3. 
 
 12. What will be the quotient of 62070, divided by 4? 
 
 Quo. 155)7, Rem. 2. 
 
 13. What will be the quotient of 5374, divided by 5 ? 
 
 Quo. 1074, Rem. 4. 
 
 14. What will be the quotient of 7856, divided by 6 1 
 
 Quo. 1309, Rem. 2. 
 
 15. What will be the quotient of 85362, divided by 7 ? 
 
 Quo. 12194, Rem. 4:. 
 
 16. Divide .... 958 by . 18. . . . Quo. 53 Rem. 4. 
 
 17. Divide. . . 1475 by. 28 .52 . . . 19. 
 
 18. Divide . . . 4277 by . 31 137 .. . 30. 
 
 19. Divide . . . 25757 by . 37 696 .. . 5. 
 
 20. Divide. . .63125 by. 123.. • . . . 513 . . . 26. 
 
 21. Divide . . 253622 by . 422 601 .. . 
 
 22. Divide . . 4049160 by . 328 12345 . . . 
 
 23. Divide . 48905952 by 9876 4952 . . . 
 
 24. Divide. 25312 by . 226 112 .. . 
 
 25. Dividj . . . 43956 by . 666 66 . . . 
 
 26. Divid- . 101442075 by 4025 25203. . . 
 
 27. Divide' . 367376610 by 158694 2315. . . 
 
 (Art. 17) As division is the converse of multiplication, and as 
 we niiiltinly by 10, 100, 1000, &c., by annexing the ciphers, (Anx. 
 14.) thcrctore, when we cut off ciphers, we divide by 10, 100, &c., 
 according to the number of ciphers cut off. But, if there are no 
 ciphers to be cut off. and we wish to divide by 10, 100, 1000, &;c., 
 we may divide by the following 
 
 RiTLE. Cut off as many figures from the right of the dividend 
 as there are ciphers in the divisor. The figures on the left-hand of 
 the point will be the quotient, and those on the right hand 
 the remainder. 
 
 r 10 =785.4 . Ans. 785_4_ 
 
 7854^^ 100 =78.54 78^_1 
 
 1 10 
 
 / 1000=7.854 7J_5_4_ 
 
 ^ 10 
 
 Multiplying a number by 10 increases its value tenfold ; and di- 
 viding it by 10 ditninishes it in the same ratio. The former is effect- 
 ed by removing every figure in it one place farther from the unit's 
 place ; the latter, by bringing them one place nearer. 
 
 When there are ciphers on the right hand of the divisor : 
 
DIVISION. 37 
 
 Cut off lite ciphers, and an equal number of places from the 
 rii^ht of the dividend: in dividinii;, omit the fih;urcs cut off, and 
 annex' tJicm afterwards to the remainder, if any ; othtrwiac, the 
 fiirin-cs cut off from the dividend are the remainder. 
 
 Divide 3704196 by 20. Divide 369183 by 7100. 
 
 (1; (2) 
 2,0)3704196 7i;00)3r)91|83(51T o.|3 ^,4ns. 
 ' 355 
 
 Ana. 18520916 
 
 2 
 
 141 
 71 
 
 70 
 
 3. Divide 2976435 by 2800. 
 
 4. Divide 9400639 by 4700. 
 .5. Divide 6749802 by 9000. 
 6. Divide 4872036 by 1200. 
 
 Wben the dividend is federal money, the operation is the same as 
 in division of whole numbers, and the only difficulty is to decide the 
 value of the quotient ; but this will be treated more at large when we 
 come to division in decimal fractions. 
 
 EXAMPLES. 
 
 !. Divide $375.50 equally among 45 persons. 
 
 Call the whole sum 37550 cents, and divide by 45. 
 
 cents, cents. 
 45)37550(8342 <L; or, 8 dollars 34^ cents. 
 
 360 *^ 
 
 155 
 135 
 
 200 
 180 
 
 20 
 
 e c. 
 
 2. Divide 56 15 by 10 Quotient, $ 5.61^ 
 
 3. Divide 96.00 by 5 19.20 
 
 4. Divide 156 00 by 4 39.00 
 
 5. Divi(fe 58.14 by 38 1.53 
 
 6. Divide 417.96 by 129 3.34 
 
 7. Divide 494.45 by 341 J.45 
 
 8. Divide 627.38 by 508 1.25^ 
 
 D 
 
38 ARITHMETIC. 
 
 APPLICATION. 
 
 1. Suppose 2072 trees planted in 14 rows, how many' trees will 
 there be in each row? Ans. 148. 
 
 2. Several boys, who wont to gather nuts, collected 4741, of which 
 each boy received 431 ; how many boys were there 1 Aufi. 1 1. 
 
 3. If the ex})ense of erecting a bridge, which is 15036 dollars, be 
 equally defrayed by 179 persons, what must each pay ? 
 
 Ans. 84 dollars. 
 
 4. Tt is computed that the distance to the sun is 95,000,000 miles, 
 and that light is 8 minutes traveling from the sun to the earth ; how 
 many miles does it travel per minute? Ans. 11875000. 
 
 5.1 A manufacturer paid 289 dollars a week lo his laborers, giving 
 on an average, 62i cents per day ; how many laborers had he ? 
 
 A71S. 25. 
 
 6. A person bought 27 yards of cloth, for which he paid 6750 
 cents; how much did he pay per yard? Ans. 250 cents. 
 
 7. There are 7 days in a week ; how many weeks in a year of 365 
 days ? Ans. 52 weeks, and 1 day over. 
 
 8. There are 24 hours in a day ; how many days in 2040 hours ? 
 
 Ans. 85 days. 
 
 9. Twenty-three persons dined together : their bill was 46 dollars ; 
 how much had each one to pay ] Ans. 2 dollars. 
 
 10. The sum of 5500 dollars was distributed in shares of 44 dollars 
 each; how many shares did the sum contain? Ans. 125, 
 
 11. A person spent 2880 dollars for land, paying 24 dollars per 
 acre; how many acres did he purchase ? Ans. 120. 
 
 N. B. In resolving this question, we divide dollars by dollars, 
 which some have said could not be done, because we cannot multiply 
 money by money, as explained in Art. 12 ; but the difficulty vanishes 
 when we consider the quotient, 120, as an abstract number, or a mere 
 numeral, as it reallij is. In respect to the aritiiinetical operation, this 
 120 is no more acres of land than it is yards of cloth, or barrels of 
 flour, or anything else. The 120 is merely the number of times that 
 24 dollars can be subtracted from 2SS0 dollars ; and in respect to this 
 question we know, (not by arithmetic, hut by logic.) that the number 
 of acres miist lie the same as this abstract number. We repeat once 
 more that, in division, one or the other, the divisor or the quofieTif, 
 must be abstract; and this is sufficient to correspond to Art. 12, in 
 multiplication. 
 
 12. If we have 24 baskets to pack 1560 eggs, how many must be 
 put into a basket ? Ans. 65. 
 
 13. If 459S0 pounds of bread are to be distributed among 2120 
 soldiers, how many pounds shall each one receive? Ans. 19. 
 
 14. How many times can a three-pint measure be filled from a 
 barrel of beer, containing 36 gallons ? J^ns. 96 times. 
 
 We^iroduce problem 14, (which rather belongs to reduction.) to 
 enforce one important principle in relation to division, which is, that 
 
DIVISION. 39 
 
 whan the tiividcnd and divisor are holh coxchki-k numbers, tlu v must 
 both refer to the name thing, and Oe of the same drivinniia- 
 tion, because division is but. a concise method of performing certain 
 subtractions, and we cannot subtract unlike things. For instance, they 
 must both be dollars, (as in example I !,) or both yards, &c. They 
 must both 1)0 of the same denomination ; one cannot be dollars and 
 the other cents ; one cannot be yards and the other miles, &c. So in 
 example 14 : we cannot divide (ofi) gallons by (3) pints, as, at first 
 view, the question seems to require ; and before we can divide, we 
 must find the number of pints in 30 gallons, and divide that number 
 by 3. That is: When the divisor and dividend are both concrete 
 numbers they must he of the same dcnotnination.'* 
 
 The example is solved thus : One gallon contains 8 pintf?, therefore, 
 multiply 36 l)y 8. and the dividend will be pints, which divide by 3 
 pints, and we have 06 for the times the small measure can be filled. 
 
 15. Into how many parts must I divide the number 8164, so that 
 each part shall be 27, and leave a remainder of 10 7 Ans. 302. 
 
 !&.<. A, B, and C, engaged to do a job of work for 228 dollars, and, 
 together, they accomplished it in 40 days. Tf ow it was agreed that 
 A should have 10 cents a day more than B, and B 10 cents a day 
 more than C ; what was each man's share ? 
 
 Ans. A $80, B $76, and C $73. 
 
 N. B. There will be no difficulty in this problem, provided the 
 extra portions to A and B are taken from the whole sum, before 
 division . 
 
 Questions. 
 
 What is division? 
 
 What do you call the number that is to be divided 1 
 
 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 perform 
 the operation ? 
 
 When the divisor exceeds 12, how do you proceed 1 
 
 How do you prove division ? 
 
 How may the operation be performed when there are ciphers at the 
 right hand of the divisor ? 
 
 When the dividend and quotient are concrete numbers, that is, re- 
 fer to yards, pounds, or to any other special thing, what will be the 
 character of the divisor ? 
 
 Can we divide dollars by dollars'! and if so, what will the quo- 
 tient be ] 
 
 V.AW we divide dollars by cents? if so, why ? if not, why? 
 
 When the dividend is abstract, what must be the character of the 
 divisor anJ quotient ? 
 
 Can we divide bushels by yards? if not, why? 
 
 * We are thus particular, in hopes to induce per«on5 really to study aritli- 
 melic, and not waste so much time in working over it, apparently ex- 
 pecting the fciiovvledge to come to thera. 
 
 y 
 
40 ARITHMETIC. 
 
 SECTION II. 
 
 Orskuvatiox. We have now gone over the ground-rule- of arith- 
 metic, as far as the mere operations on numbers are concerned ; and 
 those persons who are quick in what precedes, may very pro})er!y be 
 said to be quick at figures, though some such persons may be very 
 unsuccessful in future progress; as that will depend upon a pJdluao- 
 pkical, rather than a numerical turn of mind.* A good reasoner can 
 always be a good arithmetician ; on the contrary, one may add, sub- 
 tract, multiply, and divide, with the rapidity of intuition, and, if not 
 a quick and sound reasoner, quickness of operation will only make 
 weak logic the more glaring. It is a mistake to suppose that long 
 practice is most essential to make a good arithmetician. 
 
 Obskuvatiox Secoxd. To apply the knowledge we have already 
 acquired to the common-place business of every-day life, it is neces- 
 sary that we should know something about fractions, although, as yet, 
 we are not prepared to go into the subject of fractions in full. 
 
 We only propose to make use of a few common fractions, such as 
 if 4> h ■§» <^^'3 i'^ ^^^6 following 
 
 SUPPLEMENT TO 
 
 MULTIPLICATION AND DIVISION. 
 
 (AnT. 18.) To multiply by one-half, is to take the multiplicand 
 one-half of one iime ,- that is, take one half of it, or divitle it by 2. 
 
 To multiply by j, take a third of the multiplicand, that is, divide 
 it by 3. 
 
 To multiply by -f , take § first, and multiply that by 2 ; or, multiply 
 by 2 first, and divide the product by 3.-{- 
 
 EXAMPLES. 
 
 1. What will 360 barrels of flour come to, at 5:J- dollars a barrel. 
 At 1 dollar a barrel it would be 360 dollars ; at 5;^ dollars, it would 
 be b\ times as much. 
 
 *It IS not generally understood that the power of reasoning and the power 
 of coinpaiatiou are separate and distinct faculties; and. to be a good mathe- 
 matician, a person must possess both; but it is most important iliat the rea- 
 soning power should be the most iirominent. Indeed, to become a skillfi.ii 
 mathematician, it is almost necessary that the individual sliould make nu- 
 merical compulations with difficulty ; otherwise, he may not be sufficiently 
 prompted to seek lor fibbreviatiuus and artifices, wiiich, more than anything 
 eisi.'. give beauty and polish to science. 
 
 t Sometimes one operation is preferable, and sometimes the other ; good 
 judgment alone can decide, when the case is before us. 
 
MULTIPLICATION AND DIVISION. 41 
 
 360 
 
 5 times, . . . 1800 
 i of a time, . 90 
 
 Ans $1890 
 
 How much at 5| dollars per barrel ? 
 
 360 4)360 
 5| 
 
 90=i of a time. 
 
 5 times 1800 3 
 
 270 
 
 270=1 of 1 time. 
 
 §2070 
 
 2. Forty-eight men were to receive $5| a-piece ; how many dollars 
 were paiiftlieni ? 
 
 3. What will 15 tons of hay come to, at $7^ a ton 1 
 
 4. A farm of 156 acres was sold for $34f an acre; how much did 
 it come to ? 
 
 5. At 16-§ cents a bushel, what will 75 bushels of potatoes cost ? 
 
 Ans. 1250 cents. 
 
 6. What will 27| yards of muslin cost, at 9 cents a yard ? 
 
 A71S. 249| cents. 
 
 7. What will 1| yards of cloth cost at 225 cents per yard ? 
 
 Aiis. 393^ cents. 
 
 8. What is 611 times 35 ? Ans. 242_'_. 
 
 9. What is 350 "times 15i ? Ans. 5337^. 
 
 (Art. 19.) Before we attempt to divide by a mixed number, such 
 as 2^, 3i, 5§, &c., we must explain, or rather observe the principle 
 of division, namely : That, the quofient iviU be the same if we mul- 
 iiplif the dividend and divisor by the same number. Thus, 24 di- 
 vided by 8 gives 3 for a quotient. Now, if we double 24 and 8, or 
 multiply them by any number whatever, and then divide, we shall 
 still have 3 for a quotient. 16)48(3; .32)96(3, &c. 
 
 Now, suppose we have 22 to be divided by 5^ ; we may double 
 both these numbers, and thus be clear of the fraction, and have the 
 same quotient. 5^)22(4 is the same as 11)44(4. 
 
 How many times is 1^ contained in 12 ? Ans. Just as many times 
 as 5 is contained in 48. The 5 is 4 times 1|-, and 48 is 4 times 12. 
 From these observations, we draw the following rule for dividing by 
 a mixed number : 
 
 Rule. Multiply the whole number by the lower term of the 
 fraction ,- add the upper term to the product for a divisor ,- then 
 multiply the dividend by the lower term of the fraction, and then 
 divide. 
 
 S5 ~" 
 
42 ARITHMETIC. 
 
 EXAMPLES. 
 
 1 . There being 5^ yards in a rod, how many rods in 682 yards ? 
 
 .5i)682( Double both, 11)1364(124 ^/js. 
 
 2. How many barrels of flour, at §5| per barrel, can be bought 
 with $500? Ans. Se^, 
 
 3. Thirty-one and a half gallons to the barrel, how many barrels 
 in 485 gallons? Ans. \.5'll, 
 
 4. How many times is li contained in 36? A?is. 30 times. 
 
 N. B. If we multiply both these numbers by 5, they will have 
 the same relation as before, and a quotient is nothing but a relation 
 between two numbers. After multiplication, tlie numbers may be con- 
 sidered as having the denominaiion of fiflhs. 
 
 5. How many times is ^ contained in 12? Ans. 48 times. 
 
 One-fourth, multiplied by 4, gives 1 ; 12, multiplied by 4* gives 48. 
 Now, I in 48 is contained 48 times. 
 
 6. Divide 132 by 2|. Ans. 48. 
 
 7. Divide 121 by 15-^-. Ans. 8. 
 
 8. How many times is ^ contained in 3? Ans. 4 times. 
 
 ( Art. 20.) Operalions in arithmetic may be considerably short- 
 ened, by a little attention to the relation of numbers. A few exam- 
 ples are adduced of contractions in multiplication. 
 
 Suppose it were required to multiply a number — say 746382, by 
 999. (1) 
 
 Multiply by 1000 (Art. 13.) 746382000 
 
 Subtract the multiplicand, 746382 
 
 Product 745635618 
 
 Here, it is evident that the product of 1000 exceeds the product of 
 999, by once the multiplicand. 
 
 2. Multiply 4532 3. Multiply 4532 
 
 bv ... 639 by . . . 963 
 
 63=9X7 40788 40788 
 
 285516 285516 
 
 Product 2895948 Product 4364316 
 
 In both the foregoing examples we multiply the product of 9 by 7, 
 because 7 times 9 are equal to 63. 
 
MULTIPLICATION AND DIVISION. 
 
 43 
 
 Because 9 is in the place of hundreds in example 3, the product for 
 the other two figures is stt two places towards the right. 
 
 In this last example we may commence with the 3 units in the usual 
 way ; then that product hy 2, because 2 times 3 arc 6 ; then the pro- 
 duct of 3 hy 3, which will give the same as the multiplicand by 9. 
 The appearmice of the work, would then be the same as by the usual 
 method, but would be easier, as we actually multiply by smaller 
 numbers. 
 
 4. Multiply 
 by. . . 
 
 . . 40788 
 497 
 
 285516 
 1993612 
 
 20271636 
 
 Product of the 7 units. 
 As 7X7=49, multiply the pro- 
 duct of 7 by 7. 
 
 Observe, that in this last example, 497 is 3 less than 500, and 500 
 is ^ of 1000. Hence, to obtain the product, we take 40788 one 
 thousand times, which is 40788000, 
 
 2)40788000 
 
 500 times is 20394000 
 
 3 times is 122364 
 
 Product, 20271636 
 
 5. Multiply 4962, or any other number, by 98 ; that is, take it 98 
 times. If we take it 100 times, we have 
 
 496200 
 Subtract 2 times 4962, which is 9924 
 
 Product, 486276 
 
 6. Multiply 299X299X299. 
 Take 299 300 times, and subtract 299, &c. 
 
 7. Multiply 999X999X999. 
 
 Product 26730899. 
 
 Prod. 997002999. 
 
 ;Art. 21.) To multiply by 25, take :J- of 100 times, 
 ro multiply by 50, take ^ of 100 times. 
 
 '- by 75, take | of 100 times. 
 
 bv 125, take g of 1000 times. 
 
 by 135, take g of 1000 times-f-10 times. 
 
 by 150, take 100 times-j-^ of 100 times, &c., &c., for 
 
 any other aliquot part of 10, 100, or 1000. 
 
44 ARITHMETIC. 
 
 8. Multiply 86416 by 135. 
 
 125 times is 
 
 8)86416000 
 . 10802000 
 
 10 times is 
 
 8641G0 
 
 Product, 
 
 . 11666160 
 
 9. Multiply any number, say 2636, by 248 
 
 2636 
 248 
 
 Commence with the hundreds, . . . 
 
 Double this for the lens, 
 
 Double this last for the units, . . . . 
 
 . . 5272 
 . . 10544 
 . . 210S8 
 
 
 
 
 Product, 
 
 . . 653728 
 
 Take the same example. 
 
 Multiply by the units: 
 3 times 8=24 ; hence 
 21088 
 3 
 
 Pro 
 
 2636 
 248 
 
 21088 
 63264 
 
 63264 
 
 duct, 653728 
 
 Again, if we add 2 to 248, we h 
 hence, 
 
 250 times is 
 
 ave 250 
 
 : 250 is i of 1000; 
 4)2636000 
 659000 
 
 2 times subtracted 
 
 5272 
 
 Product, , - - - 
 
 
 . . 653728 
 
 10. Multiply 87603 by 9865. B. 
 would be a tedious operation ; but, le 
 —135; 135isl25-[-10; 125 is ^ o 
 
 876030000 Subt 
 1182G405 
 
 ^ the con 
 t us obs€ 
 f 1000. 
 
 ract i : 
 10 tim 
 
 But 987 
 
 imon formal rule, this 
 rve that 9865 is 10000 
 Therefore, 
 
 8)87603000 
 
 10950375 
 eg, 876030 
 
 864203595 Product. 
 
 
 11826405 
 
 11. Multiply 818327 by 9874. 
 Therefore, 
 
 4 is 10000, less 126. 
 
MULTIPLICATION AND DIVISION. 45 
 
 8183270000 Less ^ of 818327000 
 
 Subtract. . J 03 109202 
 
 102290S75 
 
 8080160798 Product. 818327 
 
 103109202 
 
 12. Multiply 188 by 135, and that product by 15. Instead of do- 
 ing this literally and mechanically, according to rule, we may half 188 
 twice, and double each of the factors that end in 5, and we shall have 
 47 . 270 . . 30; or, 
 
 47 
 8 100 
 
 4 70 
 3 7 6 
 
 Product, 3 8 7 
 
 U i 
 
 13. Multiply anv numl-er, say 468232, by 126"> g 
 
 14. by 246 I -^ 
 
 15. by 426 ' 
 
 16. by 486 
 
 17. by 612 
 
 18. Multiply 61524 by 273 and by 327. 
 
 19. Multiply 342516 by 7209 and by 9072. 
 
 20. Multiply 3764 by 199. 
 
 Take 3764 200 times, and from that product subtract 3764. 
 
 21. Muhiply 764 by 498^. 
 
 Take 764 500 times, and from that product subtract 1^ times 764. 
 
 22. Multiply 396 by 21^, or, (which is the same,) 99X87=8700 
 —86=8613. 
 
 N. B. Ninety-nine is i of 396, and 87 is 4 times 2 If. 
 
 23. A man spends 99 cents a day for 19 years, each year consisting 
 of 365 days ; what is the whole amount 1 
 
 We may divide by 5, 25, 50, 125, and other aliquot parts of 10, 
 100, 1000, &c., in the following manner: 
 
 How many times is 5 contained in 20 ! Ans. The same number 
 of times as the double of 5, or 10, is contained in the double of 20, 
 or 40. 
 
 How many times is 50 contained in 762? Ans. The same as 100 
 in 1524. 
 
46 ARITHMETIC. 
 
 How many times is 125 contained in 21251 
 Same as 250 in 4250; 
 Same as 25 in 425 ; 
 Same as 50 in 850; 
 Same as 5 in 85 ; 
 Same as 10 in 170; that is, 17 times. 
 The object of these changes is, to give the learner an accurate and 
 complete knowledge of numbers, and of division ; and the result is 
 not the only object sought for, as many young learners suppose. 
 How many times is 75 contained in 575 ? or, divide 575 by 75. 
 
 Am. 7^. 
 
 Divide 800 by 12^ Quotient, 64. 
 
 Divide 27 by \^ Quo. 1//-, or l^i. 
 
 A person spent 6 dollars for oranges, at 6^ cents apiece ; how many 
 did he purchase ? Ans. 96. 
 
 (Art. 22.) When two or more numbers are to be multiplied to- 
 gether, and one or more of them having a cipher on the right, as 24 
 by 20, we may take the cipher from one number, and annex it to 
 the other, without affecting the product; thus, 24X20, is the same as 
 240X2; 286X1300=28600X13; and 350X70X40=35X7X4 
 XIOOO, &c. 
 
 Every fact of this kind, though extremely simple, will be very use- 
 ful to those who wish to be skillful in operation. 
 
 (Art. 23.) Before we combine division with multiplication, let 
 us take a more systematic view of numbers. 
 
 The following are called prime numbers, because no one can be 
 divided by any number less than itself without producing a fraction : 
 1 2 3.5.7. . .11.13. . . 17 . 19 . . .23 29 
 
 31 .... 37 .. .41 . 43 ... 47 53 ... . 
 
 * 59 . 61 67 . . . 71 . 73 79 . . . 83 . . 
 
 ... 89 97 . . . 101. 
 
 The points represent the composite numbers ; and here it can be 
 observed, that there are 27 prime numbers, and. of course, 73 com- 
 posite numbers in the first hundred, the prime numbers becoming fewer 
 as the numbers rise higher. Observe the following series : 
 
 5 10 15 20 25 30 35 40 45 60 55 60, 
 
 and so on. Every body knows that our arithmetical scale of numbers 
 is 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 arithmetical scale. For example, 25 is i of 100 ; 125 
 is i of 1000, &c. 
 
 We now charge the student to make his eye familiar with all the 
 preceding series — the prime numbers as being unmanageable and in- 
 
MULTIPLICATION AND DIVISION. 47 
 
 convenient, and the others the very reverse ; hut the full importance 
 of such a study cun only appear in the .sequel. 
 
 (Art. 24.) When it liecomes necessary to multiply two or more 
 numbers together, and divide hy a third, or i)y a product of a third and 
 fourth, it must he lifenilli/ done, if the numbers are prime. 
 
 For example: Multiply 19 by 13, and divide that product by 7. 
 
 'I'his must be done at full length, because the numbers are prime : 
 and in all such cases there will result a fraction. 
 
 But, when two or more of the numl)ers are composite numbers, 
 the work can uhcays be contracted. 
 
 Example : Multiply 375 by 7, and divide that product by 21. To 
 obtain the answer, it is sufficient to divide 375 by 3, which gives 125. 
 
 'i he 7 divides the 21, and the foctorS remains for a divisor. Here 
 it becomes necessary to lay down a plan (f operation. 
 
 Dravy 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 ; 
 
 We may cast out equal factors from each side of the line without 
 affecting the result. In this case, 13 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 
 ^'^ 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 60X6X14. 
 
 ^^ 
 
 3 ^0 
 
 6 
 
 2^^ 
 
 ^0 4 
 9 
 # 3 
 
 In the above, divide 60 and 80 by 20, and 14 and 21 by 7, and 
 those numbers will stand canceled as above, with 3 and 4, 2 and 3 at 
 their sides. 
 
48 ARITHMETIC. 
 
 Now, the product 3X6X2, on the divisor side, is equal to 4 limes 
 9 on the other, and the remaining 3 is the result. 
 
 Hoping now that the pupil understands our forms, and compre- 
 hends the true philosopliical principles, we give a few unwrought ex- 
 amples for exercises. 
 
 4. Multiply 84 by 56, and divide the product by 14 ; what is the 
 result ? 
 
 5. What is the result of 75X21, divided by 7 1 
 
 6. What is the result of 126X72, divided by 48 1 
 
 7. What is the result of 5728X49, divided by 56? 
 
 8. What is the result of 64X 18X48, divided by 16X9X13? 
 
 9. What is the result of 125X8X2, divided by 100X24 ? 
 
 10. What is the result of 39X41X360, divided by 82X30? 
 
 1 1. What is the result of 22AX 13X37^, divided by 75X45? 
 
 12. What is the result of 71X19X7, divided by 38X211 
 
 13. What is the result of 221X635, divided by 35X5 ? 
 
 APPLICATION. 
 
 1. A farmer sold 28 bushels of wheat at 85 cents per bushel, and 
 took his pay in cotton cloth at 14 cents per yard ; how many yards 
 did he receive? -4ns, 170. 
 
 2. A person bought 12 yards of cloth at 219 cents per yard, and 
 paid in butter at 9 cents per pound ; how many pounds did it require ? 
 
 Ans. 292. 
 
 3. How many yards of cloth, at $4.66 a yard, must be given for 18 
 barrels of flour at §9.32 a barrel ? A/is. 36. 
 
 4. The children in a Sunday-school contributed 5 dollars to a chari- 
 table object ; each giving 6i cents ; how many children were there ? 
 
 Ans. 80. 
 
 5. A laborer vvrorked 26 days at 87^ cents per day, and look his 
 pay in wheat at 65 cents per bushel ; how many bushels did he re- 
 ceive ? Ans. 35, 
 
 6. A person pays 3^ dollars a week for board ; how many dollars 
 must he pay for 26 weeks? Ans. 13 times 7: $91. 
 
 7. A merchant bought 526 barrels' of flour at $4 50 per barrel, and 
 paid in cloth at $2.25 per yard ; how many yards did it require ? 
 
 Ans. 1052. 
 
 8. How much land, at §2.50 per acre, must be given in exchange 
 for 360 acres at §3.75 per acre ? An^s. 540. 
 
 9. What will 28 pounds of sugar cost, at 9| cents per pound? 
 
 Ans. 7 times 39 cents : or, §2,75. 
 
 10. An auctioneer sold 55 bags of cotton, each containing 400 
 pounds, receiving 1 mill commission on a pound ; how many dollars 
 did his commission come to ? Ans- §22. 
 
 11. How many casks, each containing 1 bushel I peck, are required 
 to hold 145 bushels] Ans. 116. 
 
 12. How much will 540 yards of cloth cost at 3 shillings 4 pence 
 per yard, in dollars, at 6 shillings each 1 Ans. §300. 
 
MULTIPLICATION AND DIVISION. 49 
 
 13. If 1 quart cost 10 pence, how many pounds will IJi hogsheads 
 cost] ybr. £126. 
 
 14. How many pounds of butter at 9^ cents per pound, will pay 
 for 19 yards of muslin at II cents per yard ? Aus. 22. 
 
 15. How many bushels of oats at 22^ cents a bushel, will pay for 
 75 pounds of sugar at 5 cents per pound ? An.". 16-f. 
 
 16. How many bushels of wheat at 75 cents a bushel, will pay for 
 6 yards of cloth at 3^ dollars a yard ? Ans. 28. 
 
 17. How many bushels of corn at 45 cents per bushel, will pur- 
 chase 24 yards of carpeting at 65 cents per yard ] Ans. 34^. 
 
 18. How many tons of hay at 5^ dollars per ton, will pay for 11 
 acres of land at 19 dollars per acre? Ans. ;38. 
 
 19. If a man travel, on an average, 4 miles an hour and 9 hours a 
 day, how many days will be required to pass over 1260 miles ? 
 
 Ans. 35. 
 
 20. How many bushels of barley at 75 cents per bushel, will be re- 
 quired to [);iy a debt of ^45.75? Ans. 61. 
 
 21. How many days' work at 125 cents per day, will pay for 80 
 acres of land at 225 cents per acre ? Ans. 144, 
 
 22. What number, multiplied by 23, will give the same product as 
 75 multipUed by 691 " Ans. 225. 
 
 23. What number, multiplied by 16-^, will give the same product 
 as 300 multiplied by 451 ? Ans. 8200. 
 
 COMPOUND NUMBERS. 
 
 (AuT. 25.) Compound Numbeus are such as express quantities 
 consisting of different denominations, but of the same general kind, 
 such as bushels, pecks, quarts, &c. ; yards, feet, inches, &C. The 
 most proper appellation for these quantities is Denominnte Numbers, 
 because they simply consist of dijftrcnt, denoniinutions, and are not 
 comfiound ; but the name compound numbers has been so long at- 
 tached to them that it would be dilncult to cliange it. 
 
 The following Tables of the denominations of compound numbers 
 are to lie committed to memory before entering upon reduction. 
 
 ENGLISH MONEY. 
 
 The denominations of English Money are, guineas, pounds, shil- 
 lings, pence, and farthings. 
 
 4 firthings, marked far., make 1 penny, marked d. 
 
 12 pence 1 shilliiig, . . . s, 
 
 20 shillings 1 pound, . . . £. 
 
 21 shillings 1 guinea. 
 
50 ARITHMETIC. 
 
 MENTAL EXERCISES. 
 
 In £2 3s., how many shillings ? 
 In 2.S-. 2d., how many pence ? 
 In 35. 2f/., how many pence 1 
 In £Z 2s., how many shillings ? 
 
 It is evident, from the inspection of the table, that to change pounds 
 to shillings we must multiply the£l by 20, and to change shillings to 
 pence we must multiply the shillings by 12, &c. ; and this changing 
 of a quantity from one denomination to another is called Reduction ,- 
 for it is reducing. 
 
 TROY WEIGHT. 
 
 Gold, silver, jewels, and liquors, are weighed by this weight. Its 
 denominations are pounds, ounces, pennyweights, and grains. 
 
 TABLE. 
 
 24 grains, gr., . . . make 1 pennyweight, marked dwt. 
 
 20 pennyweights .... I ounce, oz. 
 
 12 ounces 1 pound, lb. 
 
 MENTAL EXERCISES. 
 
 In 2 pounds 1 ounce, how many ounces ? 
 
 In 2 ounces 3 pennyweights, how many pennyweights? 
 
 In 2 pennyweights and 2 grains, how many grains? 
 
 In 1 pound, how many pennyweights ? 
 
 In 1 ounce, how many grains ? 
 
 APOTHECARIES' WEIGHT. 
 
 This weight is used by apothecaries and physicians in mixing their 
 medicines. Its denominations are pounds, ounces, drams, scruples, and 
 grains. The pound and ounce are the same as the pound and ounce 
 in the Troy weight; the difference belween the two weights consists 
 in the different divisions and subdivisions of the ounce. 
 
 TABLE. 
 
 20 grains, gr., make 1 scruple, marked g 
 
 3 scruples, 1 dram 3 
 
 8 drams, 1 ounce . . . . g 
 
 12 ounces, 1 pound .... lb 
 
 AVOIRDUPOIS WEIGHT. 
 
 By this weight are weighed all coarse articles, such as hay, grain, 
 chandlers' wares, and all the metals, excepting gold and silver. Its 
 denominations are tons, hundreds, quarters, pounds, ounces, and 
 drams. 
 
 'i'he hundred weight is 1 12 pounds, as appears from the table; but 
 at the present time the merchants in our principal cities buy and sell 
 by the 100 pounds, and 20 hundreds a ton. 
 
TABLES. 51 
 
 TABI-E. 
 
 16 drams, dr., make 1 ounce, marked oz. 
 
 1 () ounces 1 pound, . . . . lb. 
 
 23 pounds, 1 quarter, . . , cp: 
 
 4 quarters, 1 Jmndred weight, cict. 
 
 20 hundred. , 1 ton, T. 
 
 MENTAL EXERCISES. 
 
 In 10 pounds, how many ounces ? 
 \n 2 pounds 8 ounces, how many ounces ? 
 In 3 pounds 2 ounces, how many ounces? 
 Ill 10 ounces 10 drams, how many drams 1 
 In 100 pounds, how many ounces'? 
 
 LONG MEASURE. 
 
 Tii>i!sr measure is used when length only is considered. Its denomi- 
 nations are degrees, leagues, miles, furlongs, rods, yards, teet, inches, 
 
 and barley-corns. 
 
 TABLE. 
 
 3 harlev-corns, bar., . . . make 1 inch, . . . marked in. 
 
 12 inche's, 1 foot, ff, 
 
 3 feet 1 yard, yd. 
 
 5h yards, or 16^ feet, 1 rod, perch, or pole, rd. 
 
 40 rods, 1 furlong, fur. 
 
 8 furlongs, or 320 rods, .... 1 mile, 7ni. 
 
 3 miles, 1 league, L. 
 
 60 geoi^raphical or 69A statute "> , , j o 
 
 ° ° ^ ., ■'■ 5-1 decree, di:s^.ox° 
 
 miles, 5 •••"•& 
 
 orn 1 C a great circle, or circumference 
 
 360 degrees, < ° .. ,. 
 
 => ' C ol the earth. 
 
 [Note. A fathom is six feet, and is generally used to measure the 
 depth of water. 
 
 A hand is four inches, and is used to measure the height of horses.] 
 
 MENTAL EXERCISES. 
 
 How many inches in one yard ? 
 
 How many inches in 2 feet 2 inches ? 
 
 How many feet in 2 rods? 
 
 How many feet in 2 rods and two feet? 
 
 In 3 leagues and I mile, how many miles ] 
 
 CLOTH MEASURE. 
 
 Cloth measure is used for measuring all kinds of cloth. Its de- 
 nominations are ells French, ells English, ells Flemish, yards, quar- 
 ters, nails, and inches. 
 
52 ARITHMETIC. 
 
 TABLE. 
 
 2^ inches, in., make 1 nail, . . . marked na. 
 
 4 nails, 1 quarter of a yard, . qj: 
 
 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. 
 
 MEXTAL EXERCISES. 
 
 In 4^ inches, how many nails ? 
 In 2 quarters 2 nails, how many nails? 
 In 3 yards 3 quarters, how many quarters ? 
 In 2 yards 2 quarters, how many quarters? 
 In i yard 2 nails, how many nails] 
 
 LAND OR SQUARE MEASURE. 
 
 Land or square measure is used in measuring land, or anything in 
 which length and breadtli are both considered. 
 
 TABLE. 
 
 144 square inches, sg. in., . . . make 1 square foot, . sq. ft. 
 
 9 square feet I square yard,., .sv^.j/q'. 
 
 30;^ square yards, 1 square pole, . F. 
 
 40 square poles, 1 rood, . . . A'. 
 
 4 roods, 1 acre, , . . . A. 
 
 640 acres, 1 square mile, . M. 
 
 The surveyor's or Gunter's chain is generally used in surveying 
 land. It is four poles, or 66 feet, in length, and is divided into 100 
 links. 
 
 TABLE. 
 
 7jW inches make 1 link, .... marked /. 
 
 4 rods, or 66 feet, .... 1 chain, c. 
 
 80 chains 1 mile, mi, 
 
 I squ ire chain, 16 square poles, . . . . P. 
 
 10 square chains, 1 acre, A. 
 
 Land is generally estimated in square miles, acres, roods, and square 
 poles or perches.* 
 
 MtXTAL EXERCISES. 
 
 In 2 feet square, how many square feet ? Ans. 4. i , r 
 
 A surface 3 feet in length and 2 feet in width— how I j 
 
 many square feet ? ' I i 
 A surface 6 feet in length and 3 in width, how many i ' I 
 
 square feet ? 
 
 ♦More recently in mileB, acres, and decimals of an acre. 
 
TARLK 
 
 53 
 
 A towiishii) 6 miles long and 4 i 
 miles ciiX's it coiitain '.' 
 
 How many square miles in 10 miles square ] 
 
 bruat!, how many square 
 
 SOLID OR CUBIC MEASURE. 
 
 8olid or cubic measure is used in measuring such things as have 
 three (limensions, of length, breadth, and thickness. Its denomina- 
 tions are tons, cords, yards, feet, and inches. 
 
 1728 solid inches, s. in., . . 
 
 '27 solid feet, • 
 
 40 feet of round, or ') 
 
 50 feet of hewn timber, 5 
 
 make 1 solid fool, marked s. fl, 
 
 s- yd. 
 
 1 solid yard, . 
 I ton, . 
 
 Ton. 
 
 12S solid feet=8X4X't, that is, ~) 
 a pile 8 feet in length, 4 feel in > 1 cord of wood, . . C. 
 width, and 4 feet in height, ^ 
 
 [NoTK. A cord foot is 1 foot in length of the pile which makes a 
 cord. It contains 1 6 solid feet.] 
 
 MENTAL EXERCISES. 
 
 A block 1 foot long, 1 wide, and 1 high, is a cubic foot. A block 
 1 foot square at the end and M feet long, how many cubic feet? 
 
 A block containing 4 s([uare inches at the end and 10 inches bng, 
 how many cubic inches? 
 
 WINE MEASURE. 
 
 Wine iriejisure is used in measuring ail liquors, excepting beer and 
 ale. Its denominations are tuns, pipes, hogsheads, barrels, gallons, 
 quarts, pints, and gills. 
 
 TABM-. 
 
 4 trills, o-,-'., 
 
 2 pint.s,\ . 
 
 4 quarts, . 
 
 3 4 gallons, . 
 
 63 gallons, . 
 
 make 1 pint, 
 . . . 1 quar 
 
 ; gallon, . 
 1 barrel, . 
 I hogshead, 
 
 2 hogsheads, 1 pq;e. 
 
 2 pipes, or 4 hc)gsheads, . . . 
 
 tun, 
 
 marked pt. 
 . . . gt 
 . . . gal. 
 .... oar. 
 
 . Md. 
 
 . pi. 
 tun. 
 
 A wine gallon contains 23 1 cubic inches. 
 
 MENTAL EXERCISES. 
 
 In 2 pints, how many gills ? 
 
 In 3 pints 1 gill, how many gills? 
 
 In 3 gallons, how many quarts? 
 
 In 3 gallons, 2 quarts, I pint, how many pints? 
 
 In 1 gallon 1 pint, how many pints ? 
 
54 ARITHMETIC. 
 
 ALE OR BEER MEASURE. 
 
 Ale or beer measure is used in measuring ale, heer, and milk. Its 
 denominations are hogsheads, barrels, gallons, quarts, and pints. 
 
 TABLE, 
 
 2 pints, pt., make 1 quart, . . marked qt. 
 
 4 quarts, 1 gallon, gal 
 
 36 gallons, i barrel, bar. 
 
 54 gallons, 1 hogshead, . . . hhd. 
 
 A gallon, beer measure, contains 282 cubic inches. 
 DRY MEASURE. 
 
 Dry measure is used in measuring all dry articles, such as grain, 
 fruits, roots, salt, coal, &c. Its denominations are caldrons, bushels, 
 pecks, quarts, and pints.] 
 
 TABLE. 
 
 2 pints, pt make I quart, . . . marked qt. 
 
 8 quarts, 1 peck, pk. 
 
 4 pecks, 1 bushel, hu. 
 
 36 bushels, .... • 1 caldron, cul. 
 
 [Note. A gallon; dry measure, contains 2684 cubic inches. A 
 Winchester bushel is 18^ inches in diameter, 8 inches deep, and con- 
 tains 21501 cubic inches. 
 
 5 
 
 A bushel of coal, lime, &c., is 2688 cubic inches. 
 TIME. 
 
 TABLE. 
 
 60 seconds, sec, make 1 minute, . . marked mm. 
 
 60 minutes, 1 hour, h. 
 
 24 hours, 1 day, d. 
 
 7 days 1 week, w. 
 
 12 months, (or 365 days,) ... 1 year, y, 
 
 [Note. The true year, according to the latest and most accurate 
 observations, consists of 365 days, 5 hours. 48 minutes, and 58 seconds ; 
 this amounts to nearly 365^ days. The common year is reckoned 
 365 days, and every fourth or leap year one day more. 
 
 Those years which are divisible by 4, without remainders, are leap 
 years. By estimating 5 hours, 48 minutes, 58 second:=, as 6 hours, 
 accumulated an error, which amounted to more than 1 1 days in 1752 ; 
 and the II days were dropped to correct the style : hence, before that 
 period is called Old Style, subsequently New Style. 
 
 A further correction of one day took place in' the year 1800 ; still 
 a further correction will take place in the year 1900, after which there 
 will be no correction for 200 years.] 
 
REDUCTION. 
 
 55 
 
 MENTAL EXERCISES. 
 
 In 2 hours 10 minutes, how many minutes? 
 
 In 3 minutes 10 seconds, how many seconds? 
 
 In 10 hours, how manyminutesl 
 
 In three hours, how many minutes? how many seconds? 
 
 In 2 hour.--, how many minutes? 
 
 In 13 minutes 3 seconds, how many seconds 1 
 
 CIRCULAR MEASURE, OR MOTION. 
 Circular measure is used in calculating latitude and longitude, and 
 also in measuring the motions of the heavenly bodies. Every circle 
 is supposed to be divided into 360 equal parts, called degrees. 
 
 TABLE. 
 
 60 seconds (") make 1 minute, , . marked 
 
 60 minutes, 1 degree, " 
 
 30 degrees 1 sign, s. 
 
 12signs, or3()0° 1 circle c. 
 
 The signs are becoming obsolete. 'l"he best astronomers of the 
 present day do not use them. 
 
 MENTAL EXERCISES. 
 
 In 1 sign 10 degrees, how many degrees? 
 
 In 2 degrees, how many minutes? 
 
 In 3 degrees 3 minutes, how many minutes ? 
 
 In 2 signs 2 degrees, how many degrees ? 
 
 How many minutes in a quadrant, or 90 degrees? 
 
 PARTICULARS. 
 
 12 units make a dozen. 
 
 12 dozen, a gross. 
 
 144 dozen, a great gross. 
 
 20 units, a score. 
 
 24 sheets of paper, a quire. 
 
 20 quires, a ream. 
 
 BOOKS. 
 
 A sheet folded in two leaves is called a folio. 
 
 folded in four leaves .... a quarto, or 4to. 
 '< folded in eight leaves ... an octavo, or 8vo. 
 " folded in twelve leaves ...ad lodecimo, or 12mo. 
 •* folded in eighteen leaves . . an 18mo. 
 
 REDUCTION". 
 
 (Art. 26.) Redcctiov is reducing or changino; one dcnomina. 
 tion of any quantity to another denomination, without changing its 
 
56 ARITHMETIC. 
 
 real value; and it invoivos no other principles than those called out 
 ill the mental exercises under the tal-.les. 
 
 From these exercises we );erceive that quantities may be reduced 
 from a higher to a lower denomination, l»y mnltiphcation. 
 
 Thus, 2 yards are 6 feet: these, reduced to inches, give 72. 
 
 Conversely, lower denominations may be brought to higher, by 
 division, as pounds to qu irlcrs, hundreds weight, or tons. In all these 
 cases, the quantity remains unchanged, although it is expressed in a 
 diflrrcnt denomination. 
 
 When higher denominations are to be reduced to lower, the expla- 
 nation gives the following 
 
 r?iTT,K. Mnlliply iJie successive denominafions, comincndns; iciih 
 Ihe hig.'ie-f given, by the number iliai ivill make if one of the next 
 hncer, adding, in their proper placea, ihe several inferior denomina- 
 tions expressed in the given number. 
 
 When lower denominations are to be reduced to higher, the expla- 
 nation gives the following 
 
 Rule. Divide the given denomination by the number that will 
 make one of the next higher,- and the quotient thence arising by as 
 manyas7nake one of ike denomination next above that, and soon 
 to ihe required one. 
 
 The farmer of these rules is reduction descending, the latter 
 ascending. Being reverse operations, they pnjve each other. 
 
 As the principles of the operation are the same under each table of 
 weight, measure, or motion, we shall give the examples promiscuously, 
 to exercise quickness of thought. 
 
 1, Reduce 17 lbs. W oz. 15 diets., to pennyweights,* 
 
 nibs. 
 : 1^ cz. 
 
 :o4 
 
 1 1 oz. added. 
 
 In this example, we first multiply by | 
 the number of ounces in a pound, and 215 
 
 then add the ounces, &c,, «Scc. 
 
 20 dwi. 
 
 4300 
 
 15 dwts. added. 
 
 4315 dwts. 
 
 * A practical man would at once perceive that this is 5 pennyweights less 
 than 18 pounds : therefore, 18 X 12X20— 5=the result. 
 
REDUCTION 
 
 57 
 
 2. In 2.') l/js. 9 oz. f/a'/i^ 10 gr., how many grains? 
 
 Ans. 148330. 
 
 3. RediJCfi 7461 fi farthings to pounds. 
 
 4) 71016 
 
 12)18654 pence. 
 
 20)1554 shillings and 6 pence. 
 
 77 and 14 shillings over. 
 
 Am. £77, 145. 6d=746 16 farthings. 
 
 4. In 3278 nails, how many yards? 
 
 4)3278 
 
 We first divide by 4, which brings 
 the number to quarters, and then again 
 by 4, which brings it to yards. 
 
 4)819 . 2 n«. 
 204 . 3 qrs. 
 
 A)ts. 204 yds. 3 qrs. 2 7ia. 
 5. In £1234 155. 7c?., how many farthings? 
 £ s. d. 
 '234 15 7 
 20 
 
 24695 shillings, 15 being added in mentally. 
 12 
 
 296317 pence, 7 being added in mentally. 
 
 .4/15. 1185388 farthings. 
 
 6. Reduce 3060288 cubic inches to tons of round timber. 
 
 1728)3060288(1771 
 
 We first divide by 1728, the num- 
 ber of solid inches in a solid foot, and 
 next by 40, the number of solid feet 
 in a ton 
 
 410)177(1 
 
 44 1 1 
 
 1728 
 
 13322 
 12096 
 
 12268 
 12096 
 
 1728 
 1728 
 
 Ans. 44 tons 11 feet. 
 
58 
 
 ARITHMETIC. 
 
 7. Reduce 43 gallons 3 quarts 1 pint, to pints. 
 43 
 4 
 
 175 
 
 Ans. 351 pints. 
 
 (AiiT. 27.) 8. Reduce 35 tons to drams. 
 
 When there are no inferior denominations, as quarters, pounds, &c., 
 to add in as the operation advances, we may arrange all our multi- 
 pliers before the eye, and take the product of several of them men- 
 tally, to shorten the operation. 
 
 In the present example we multiply 35 tons 
 
 by 20 
 by 4 
 by 28 
 by 16 
 by 16 
 
 Then, 78400 
 
 by 256 gives the answer. 
 
 35X20=700 
 
 4X28=112 
 
 16X16=256 
 
 470400 
 3920 
 1568 
 
 20070400 
 
 9. Reduce 23 bushels 3 pecks 5 quarts 1 pint, to pints. 
 
 N. B. This is 5 pints less than 24 bushels ; therefore, if we reduce 
 24 bushels to pints, deducting 5, we have the answer. 
 
 Multiply 24 
 by ■' 4 
 
 by 8 
 
 by 2 
 
 10. Reduce H acres 3 roods 15 rods, to rods. Ans. 1895. 
 
 1 1. Reduce 24 square rods to square feet. Ans. 6534. 
 
 12. Reduce 10 pipes 1 hogshead and 2 quarts of wine, to pints. 
 
 Ans. 10588. 
 
 13. Reduce 3 hogsheads to gills. Ans. 6048. 
 
 That is . . . 
 
 .... 64 
 
 by 
 
 . . . . 24 
 
 
 128 
 
 
 256 
 
 
 1531 Ans. 
 
REDUCTION. 59 
 
 14. In 1746880 ounces, how many tons, 
 
 Ans. 48 tons, and 14 cwt. 3 qrs. 8 lbs. over. 
 
 15. In 17645 grains, apothecaries' weif^ht, iiow many pounds ? 
 
 Ans. 3 Ib.s., and (i dr. scru. 5 grains over. 
 
 16. In 168474 feet, how many miles'! 
 
 An'\ 31 n)iles, and 7 fur. 58 yds. over. 
 
 17. In 2419200 seconds, how many daysl '" Ajis. 28. 
 
 18. In 547325 fartliings, how many pence, shillings, and pounds? 
 
 Ans. £579 2s. Gii. 1 qr. 
 
 19. In 9173 nails, how massy yards? Ans. 573 yds. 1 qr. 1 n. 
 
 20. How many barleycorns will reach rovnid the earth, supposing 
 it, according to the best calculations, to be 24877 miles ? 
 
 A}is: 4728G20160. 
 
 21. How many seconds are in a solar year, or 365 days 5 hours 48 
 minutes 58 seconds? Ans. .j1556938. 
 
 22. In a lunar month, or 29 days 12 hours 44 minutes 3 seconds, 
 how many seconds] A7is. 2551443. 
 
 23. In an ingot of silver weighing 6 pounds 3 ounces 10 gnins, 
 how many grains? Ans. 36010. 
 
 24. In 5 hundreds 2 quarters 16 pounds, how many pounds ? 
 
 Ans. G32. 
 
 25. In 4 miles 6 furlongs 22 rods, how many rods ? Ans. 1542. 
 
 26. In 11456 geographic miles, how many degrees ? 
 
 Ans. 190 deg. 56 min. 
 
 27. In 164736 inches, how many miles'? 
 
 Ans. 2 miles 4 fur. 176 yds. 
 
 28. In 14764 nails of cloth, how many yards ? 
 
 Ans. 922 yds. 3 qr. 
 
 29. In 2746 quarters, how many ells English ? 
 
 A71S. 549 ells 1 qr. 
 
 30. Reduce 2 tuns to gills. Ans. IG\28. 
 
 31. Reduce 32 gallons 3 quarts to pints. Ans. 262. 
 
 32. Reduce 2 hogsheads 27 gallons 3 quarts, to quarts. 
 
 Ans. 615. 
 
 33. Reduce 3 tuns I hogshead 15 gallons 1 quart, to pints. 
 
 Ans. CG74. 
 
 34. Reduce 1484 yiints to gallons. Ans. 185 gal. 2 qts. 
 
 35. Reduce 167000 gills to barrels of 32 gallons each. 
 
 Ans. 163 barrels 2 gals. 3 qfs. 
 
 36. Reduce 6272640 square inches to acres. Ans. 1. 
 
 37. A grocer purchased 16 hundred weight 3 quarters 20 pounds 
 of sugar ; how many pounds are there in the whole? Ans. 1896. 
 
 33. A tobacconist bought 116 hundred weight 3 quarters of tobac- 
 co; how many pounds were in the whole ? Ans. 13076, 
 
 39. A silversmith has 7 pounds 10 ounces of silver; how many 
 grains are in the whole ? Ans. 45120. 
 
 40. An apothecary has several sorts of drugs, weighing toj^ether 47 
 pounds 6 ounces 4 drams; how many scruples and grains arc in the 
 whole? Ans. 13692 scru., 273810 grs. 
 
60 ARITHMETIC. 
 
 41. A merchant wishes to ship 300 bushels of flaxseeJ in casks con- 
 taining 7 bushels 2 pecks each ; what number of casks are required 1 
 
 Atis. 40. 
 
 42. A certain barn is 64 by 36 feet, and 20 feet high ; another bam 
 is 32 by 24 feet, and 16 feet high: how much larger is the former 
 than the latter ? (8ee solid measure.) Ans. 3^ times larger. 
 
 43. How many times will a wheel 16 feet 6 inches in circumfer- 
 ence, turn round in running 42 miles ? A/is. 13440 times. 
 
 44. There are two places on the equator, one in longitude 40 deg. 
 east of the meridian of Greenwich, tlie other 30 deg. west; what is 
 the number of geographical miles between them ? An.^. 4200. 
 
 45. A ship, during 3 days of storm, had changed her longitude 273 
 geographical miles ; how many degrees and minutes of a degree did 
 she change ] Ans. 4 deg. 33 min. 
 
 46. How many quires and reams of paper are in 500520 sheets ? 
 
 Ans. 20855 quires, 1042 reams 15 quires. 
 
 47. A printer calls for 4 reams 10 quires and 10 sheets of paper to 
 print a book ; how many sheets does he call for ? Ans. 2170. 
 
 48. How many solid inches are in a solid 7 inches long, 7 inches 
 wide, and 7 inches high 1 Ans. 343. 
 
 49. How miny solid inches are in a solid 2 feet 1 inches long, 6 
 inches wide, and 6 inches high 1 Ans. 27X30. 
 
 50. How many solid inches are in a solid 3 .feet 2 mches long by 2 
 feet 2 inches wide and 1 foot 8 inches high? 
 
 A?is. 38X26X20=19760. 
 
 51. How !n;»ny yards of carpeting, 1 yard wide, will be required to 
 carpet a room 18 feet wide and 20 feet long? Ans. 40. 
 
 -^COMPOUND ADDITION. 
 
 (Art. 28.) Compound Addition is simple addition and reduc- 
 tion combined ; but we shall be more clearly comprehended through 
 the mi'dium of an example. 
 
 What is the sum of £19 15s. 8d., £27 18s. 6d., £43 9s. 4d., and 
 £7 6s. 6;]. 
 
 As unlike things cannot be added together, (.\rt. 5.) we must write 
 pounds under pounds, shillings under shillings, and pence under pence, 
 thus: 
 
 £ s. d. 
 
 19 15 8 
 
 27 18 6 
 
 7 6 6 
 
 55 8 
 
UKDUCTION. 01 
 
 Tn'ow, each column, by itself, is a sum in t-implc !i<lcUtion ; and, as 
 luTO are three coliiiniis, we have tliree sums in adililion. But when 
 we make the reduced sum of one column run inlo the next column 
 we unite or caitipuund Hit additions. 
 
 We must commence with tlie lowest column, (the pence column) ; 
 its sufn is 20 pence, which, reduced to shillin.^^s. give 1 .shilling, and H 
 pence over. 'i'he 8 pence must, of course, be [)ut down in the penc<> 
 roluuu^. 'J'he I shilling, added in with the other shillings, all make 
 40 shillings; which, reduced, make 2 pounds and shillings over. 
 The 2 pounds, added into the column of pounds, make the whole 
 number of pounds 55; and the several sums of money make the total 
 sum of £55 Os. 8d. 
 
 If the pupil retains the principle, that compound addition is simple 
 addition and reduction combined, no rule for addition will be required ; 
 but for the sake of conciseness, we give the following 
 
 Rule. 1. Flace the numbers so, that those of the same denomi- 
 nation will stand directly under each other, and in distinct and sepa- 
 rate columns. 
 
 2. Add up the Jis;iires in the lowest denomination, and find, by 
 reduction, how many units or ones of the next higher denomination 
 are contained in their sum, 
 
 3. Set down the remainder below its proper column, and carry 
 those units or ones to the next denomination, which add up in the 
 same manner as before. 
 
 4. Proceed thus through all the denominations, to the highest, 
 whose sum, together with the several remainders, luill give the 
 answer sought. 
 
 N. B. The method of proof, rather a test, is to vary the order of 
 addition : that is. break the sum into several sums, and then take the 
 partial sums for the sum total ; and if two methods give the same sum 
 total, that sum may be regarded as correct. 
 
 ENGLISH MONEY. 
 
 £ 
 
 s. 
 
 d. 
 
 48 
 
 13 
 
 8 
 
 51 
 
 6 
 
 4 
 
 67 
 
 11 
 
 3 
 
 76 
 
 18 
 
 10 
 
 244 
 
 10 
 
 1 
 
 EX A 
 
 MPLE 
 
 s. 
 
 £ 
 
 .<?. 
 
 d. 
 
 876 
 
 12 
 
 8 
 
 542 
 
 11 
 
 9 
 
 213 
 
 6 
 
 4 
 
 457 
 
 16 
 
 7 
 
 
 £ s. d. 
 
 124 15 5 
 
 314 12 11 
 
 145 13 2 
 
 278 17 8 
 
62 
 
 ARITHMETIC. 
 
 
 1 
 
 LON 
 L. mi. fur. rd. 
 316 2 7 29 
 127 1 2 20 
 187 1 15 
 11 1 1 1 
 
 G MEASURE. 
 
 TH MEASUR] 
 7/d. qr. na. 
 *14 3 2 
 25 4 1 
 16 1 
 25 2 2 
 
 yd. ft. iti. bar. 
 90 2 11 2 
 55 1 9 1 
 27 7 
 50 2 2 2 
 
 642 2 4 
 
 25 
 
 
 
 E. Fl. qr. na. 
 
 26 2 4 
 165 2 1 
 172 1 3 
 
 57 2 2 
 
 CLO 
 
 OR 
 
 q. in. 
 104 
 
 27 
 
 2 
 
 128 
 
 1 
 
 \SU 
 A. 
 700 
 375 
 450 
 30 
 
 ^. En. qr. na. 
 
 28 5 1 
 
 120 3 1 
 
 219 1 4 
 
 115 3 1 
 
 423 2 
 
 SQU 
 . CI 
 'E I\ 
 
 ARE ME. 
 
 JBIC ME-^ 
 C. s.ff. 
 116 127 
 317 12 
 418 119 
 737 104 
 
 
 LAND 
 
 sq. yd. sq. ft. . 
 
 197 4 
 
 122 3 
 
 5 8 
 
 237 7 
 
 RE. 
 R. P. 
 
 2 37 
 
 3 25 
 1 31 
 25 
 
 563 5 
 
 117 
 
 
 
 SOLII 
 5. yc?. S.ff. s. in. 
 l'65 25 1129 
 237 26 132 
 350 1 1064 
 222 19 17 
 
 ) OE 
 
 pt. 
 2 
 3 
 1 
 3 
 2 
 
 ^SU 
 
 pi. b 
 2 
 
 1 
 2 
 
 
 1 
 
 RE, 
 
 C. cord ft. 
 287 9 
 126 7 
 316 6 
 419 5 
 
 976 18 614 
 
 lEASURE 
 tun. 
 34 
 25 
 24 
 35 
 77 
 
 
 hhd. gal. qt. 
 
 27 65 3 
 112 60 2 
 
 50 29 
 421 2 
 
 14 39 1 
 
 hd. gul. qt. 
 
 1 27 3 
 
 2 25 2 
 1 27 1 
 
 1 62 3 
 
 2 21 2 
 
 627 7 1 
 
 1 
 
 
 
 

 
 COMPOUND ADDITION. 
 
 
 
 03 
 
 hu. pk. 
 
 qt. pi. 
 
 
 bu. pk. qt. 
 
 
 
 bu. 
 
 pk. qt. pf. 
 
 2:} 3 
 
 7 1 
 
 
 4 3 7 
 
 
 
 
 3 7 1 
 
 3i 2 
 
 6 1 
 
 
 5 2 6 
 
 
 
 
 2 6 1 
 
 42 3 
 
 5 1 
 
 
 6 1 5 
 
 
 
 
 5 
 
 51 1 
 
 4 1 
 
 
 7 4 
 
 
 
 
 1 4 1 
 
 23 2 
 
 3 1 
 
 
 8 3 3 
 
 
 
 
 3 3 1 
 
 14 1 
 
 2 1 
 
 
 4 1 2 
 
 
 
 
 2 1 
 
 11 3 
 
 4 I 
 
 
 3 2 4 
 
 
 
 
 2 1 1 
 
 202 3 
 
 2 1 
 
 
 40 3 7 
 TIME. 
 
 
 
 3 
 
 270 
 
 
 
 
 
 y. m. 
 6 7 
 
 d. 
 
 d. 
 
 h. m. sec. 
 
 
 V' 
 
 m. 
 
 d. h. 
 
 16 
 
 127 
 
 21 46 55 
 
 
 
 
 10 
 
 21 18 
 
 2 1 
 
 24 
 
 24 
 
 16 30 32 
 
 
 
 
 9 
 
 17 9 
 
 9 9 
 
 21 
 
 17 
 
 12 18 00 
 
 
 
 
 11 
 
 23 23 
 
 4 3 
 
 22 
 
 41 
 
 15 19 12 
 
 
 
 
 2 
 
 9 10 
 
 6 7 
 
 25 
 
 14 
 
 9 9 9 
 
 
 
 
 1 
 
 11 11 
 
 
 
 
 
 
 
 
 d. h 
 
 . m. 
 
 sec. 
 
 d. h. m. 
 
 sec. 
 
 
 h. 
 
 m, sec. 
 
 6 4 20 
 
 13 
 
 7 13 
 
 20 
 
 
 
 
 5 12 
 
 4 2 30 
 
 17 
 
 1 2 9 
 
 12 
 
 
 
 
 6 2 
 
 3 7 8 
 
 20 
 
 2 3 4 
 
 6 
 
 
 
 
 17 45 
 
 7 2 42 
 
 8 
 
 3 4 5 
 
 7 
 
 
 
 
 18 12 
 
 
 
 
 
 
 
 
 
 CIRCULAR MEASURE, OR MOTION. 
 
 114 
 
 46 
 
 25 
 
 221 
 
 34 
 
 25 
 
 216 
 
 27 
 
 33 
 
 227 
 
 54 
 
 56 
 
 332 
 
 16 
 
 24 
 
 112 
 
 24 
 
 43 
 
 417 
 
 27 
 
 25 
 
 217 
 
 42 
 
 37 
 
 109 
 
 17 
 
 26 
 
 111 
 
 12 
 
 14 
 
 135 
 
 16 
 
 21 
 
 29 
 
 42 
 
 24 
 
 321 
 
 42 
 
 24 
 
 115 
 
 13 
 
 12 
 
 220 
 
 16 
 
 15 
 
 19 
 
 29 
 
 30 
 
 111 
 
 11 
 
 17 
 
 78 
 
 21 
 
 32 
 
 
 
 
 
 
 
 
 
 
 N. n. In practical astronomy, when the object is to find the posi- 
 tion of a body in a circle, we reject every 360 degrees in addition. 
 
 APPLICATIOX. 
 
 1. A merchant bought 4 barrels of sugar; the first weighed 2 hun- 
 dred weight 3 quarters 20 pounds, the second 1 hundred weight 2 
 quarters and 15 pounds, the third 2 hundred weight 1 quarter 27 
 
64 ARITHMETIC. 
 
 pounds, and the fourth 2 hundred weight 3 quarters and 10 pounds; 
 what was the entire weight of the wholr? 
 
 Alls. 9 cwt. 3 qrs. IG lbs. 
 
 2. A merchant shipped 3 boxes of freight to New Orleans; the 
 first weighed 7 hundred weight 56 pounds, the second 4 hundred 
 weight 80 pounds, and the third 5 hundred weight and 20 pounds ; 
 what was the weight of the whole ] 
 
 This is done Uke simple addition ; modern reform is happily doing 
 away with denominate numbers in many cases. 
 
 3. There are 4 fields ; the first contains 12 acres 2 roods and 3S 
 perches, the second 4 acres 1 rood 26 perches, the third 85 acres 
 roods 19 perches, and the fourth 57 acres 1 rood 2 perches ; how 
 many acres in the four fields? Ans. 159a. 2r. 5p. 
 
 4. In 4 piles of wood, the first containing 32 feot 149 inches, the 
 second 121 feet 1436 inches, the third 97 feet 498 inches, the fourth 
 1 15 feet 1356 inches ; how much did the whole amount to? 
 
 Ans. 2 C. 110 ft. 1711 in. 
 
 5. In 6 boat-loads of wood, the first containing 22 cords 114 feet 
 9S7 inches, the second 18 cords 121 feet 1436 inches, the third 21 
 cords 109 feet 1629 inches, the fourth 15 cords 82 feet 1321 inches, 
 the fifth 16 cords 98 feet 1111 inches, the sixth 2 1 cords 89 feet 987 
 inches ; how much did they all contain ? 
 
 Ans. 120 C. 105 ft. 559 in. 
 
 6. A tailor requires 1 yard 3 quarters 3 nails of cloth for a father's 
 coat, and 1 yard 1 quarter and 2 nails for each of 2 sons ; how much 
 in all? Ans. 4 yds. 2 qrs. 3 na. 
 
 7. How much land is in a farm which consists of 50 acres 2 roods 
 33 rods of wood-land, 25 acres 14 rods meadow-land. 30 acres tillage, 
 20 acres rough pasture land, 12 acres I rood covered with water, and 
 10 acres 25 rods swamp? Ans. 148 a. 32 r. 
 
 8. A farmer sowed 3 fields of wheat ; the first yielded 45 bushels 
 3 pecks, the second 36 bushels 1 peck 7 quarts, and the third 30 
 bushels 2 pecks 1 quart; how much wheat did ho raise? 
 
 Ans. 112 bu. 3pks. 
 
 COMPOUND SUBTRACTION. 
 
 (Akt. 29.) SuBTRACTiov of compound or denominate numbers is 
 essentially the sarne as in simple numbers, except the care that must 
 be taken to borrow or reduce from the tables, in place of borrowing 
 or reducing from 10. For instance, we cannot take 9 pence from 7 
 penre: we must first add, or conceive to be added. 12 pence more, (1 
 shilling.) to 7 pence, borrowing or taking a shilling from the next supe- 
 rior column, &c. From such necessary operations, we deduce the 
 folio win": 
 
COMPOUND SUBTRACTION. 
 
 65 
 
 Kci.K. Place the numbers as hi cnrnpotwd addition. Begin 
 iv:f/i the lowest denominat/oJi in the auhtrahcnd, and take it from 
 the nutnljer above it ; but, if that above be the least, add as many to 
 it as make one of the next higher denomination ,- then subtract, and 
 carry one to the next denomination in the subtrahend. 
 
 The method of proof is the same as in simple subtraction. 
 
 ENGLISH MONEY. 
 
 £ 
 Borrowed 349 
 Paid . . 195 
 
 EXAMPLES. 
 
 s. d. qr. 
 15 6 1 
 
 Lent 
 
 £ 
 791 
 
 s. d. qr. 
 9 8 1 
 
 11 8 1 
 
 Received 197 16 4 2 
 
 Remain;^, 154 3 10 
 
 Due to me 
 
 AVOIRDUPOIS WEIGHT. 
 
 T. cwt. qr. 
 From 45 11 3 
 Take 15 10 2 
 
 Rem. 33 I I 
 
 T. cwt. qr. lbs. 
 52 12 3 15 
 24 10 24 
 
 qr. lbs. oz. dr. 
 3 4 4 
 7 9 8 
 
 CIRCULAR MEASURE, OR MOTION. 
 
 From 79° 21' 
 Take 41° 41' 
 
 31" 
 
 52'' 
 
 From 147«> 19' 42" 
 Take 121^ 37' 50" 
 
 Rem. 37° 39' 39" 
 
 A person residing in latitude 27" 32' 45" north, wishes to visit a 
 place 52° 24' 18" north ; hou' many degrees, minutes, and seconds 
 northward must he travel? Ans. 24° 51' 33". 
 
 y. m 
 
 6 9 
 
 1 6 
 
 5 3 
 
 TIME. 
 
 w. d. h. m. sec. 
 3 I 2;? 40 20 
 2 6 12 57 36 
 
 y. m. d. h. 
 9 8 24 47 
 6 4 19 39 
 
 From 900 years take 1 1 1 years and 6 months. 
 
 Ans. 788 yrs. 6 m. 
 If T take 1 year 1 month from 6 years, what space of time will still 
 remain ? Ans. 4 yrs. 1 1 m. 
 
 _ 
 
6G ARITHMETIC. 
 
 CLOTH MEASURE. 
 
 yd. qr. na. E. Fl. qr. na. E. En. qr. na. 
 
 35 1 2 467 3 1 765 1 3 
 
 19 1 3 291 3 2 149 2 1 
 
 N. B. We deem it unnecessary to give formal examples corres- 
 ponding to all the tables of weights and measures, as the principles 
 of one apply to all. 
 
 ArPLICATIOX. 
 
 1. From 3 pounds 8 shillings and 3 pence, take I pound 7 shillings 
 and 9 pence. Ans. £2 Os. 6d. 
 
 2. From 4 pounds 4 pence, take 19 shillings 5 pence. 
 
 yl;js. £3 0s. Ud. 
 
 3. From 7 pounds, take 20 grains Troy. 
 
 Alls. 6 ll)s. II oz. 19 dwts. 4 grs. 
 
 4. From 9 tons 9 hundred weight, lake 2 tons 3 hundred weight 3 
 quarters and 3 pounds. Ans. 7 T. 5 cwt. qrs. 27 lbs. 
 
 5. From 3 years 4 months 20 days 16 hours, take 1 year 5 months 
 21 days 14 hours. ' Ans. 1 yr. 10 mo. 28 d. 2 h. 
 
 6. What interval of time elapsed between July 12th, 1827, and 
 November 17tli, 1831? Ans. 4 yrs. 4 mo. 5 d. 
 
 7. What is the interval of time between December 23d, 1798, and 
 February 21st. 1811 ? Ans. 12 yrs. 2 mo. 28 d. 
 
 8. 'J'he apparent revolution of the sun is 365 days 5 hours 48 min- 
 utes 58 seconds, that of the moon is 27 days 7 hours 43 minutes 3 
 seconds ; what is the difference ? Ans. 337 d. 2L h. 5 m. .'J4 sec. 
 
 9. What is the difference of longitude between two places, one 75° 
 21' 30" west, and another 71° 20' 3.5" west? Ans. 3° 51' .55". 
 
 10. What is the difference of longitude between two places, one 
 situated 3° 4' '20" east of an assumed meridian, and the other 1° 20' 
 2" west of the same 1 Ans. 10° 24' 22". 
 
 11. What is the difference of latitude between two places, one 
 latitude 5"^ south, the other 5° north ? Ans. 10°. 
 
 12. What is the difference of cold between 3° below zero, and 10° 
 above? Ans. 13°. 
 
 N. B. The last three problems have the algebraic peculiarity of 
 subtraction, the line of demarcation being between them. 
 
 COMPOUND MULTIPLICATION. 
 
 (^Art. 30.) CoMPOuxD MuLTiPLTCATToy IS thc same in relation 
 to simple multiplication, as compound addition is to simple addition : 
 that is, each denomination of the multiplicand is a sum in simple 
 
COMPOUND MULTIPLICATION. 67 
 
 multiplication, and its product may require reduction. Il is, Ihtre- 
 fure, simple mull ipUcation and reduction combined. 
 
 From the above principle, the pupil will recognise the rationale of 
 the following 
 
 Rule. 1. Set down the coinpaund sum to be miilliplied, and un- 
 der its lowest denomination at tlie right hand set the multiplier. 
 
 2. Multiply the number in the lowest denomination by the mul- 
 tiplier, and find how many units of the next higher denomination 
 are contained in the product, setting down what remains. 
 
 3. In like manner multiply the number in the next denomina- 
 tion, and to the product carry or add the units, before found, and 
 find how many xmits of the next higher denomination are in this 
 amount, which carry in like manner to the next product, setting 
 doiun the overplus. 
 
 4. Proceed thus to the highest denominatio7i proposed : so shall 
 the last product, ivilh the srceral remainders, taken as one common 
 number, be the whole anh/unt required. 
 
 The method of procf i^ the same as in simple multiplication. 
 
 EXAMPLES. 
 
 bu. pk. qf. pt. 
 7 2 5 11 7 times I pint make 7 pints ; 2 pints make 
 1 quart ; then, 7 pints make 3 quarts, and 
 leaves 1 pint. Set down the 1 pint, and carry 
 53 2 6 1 the 3 quarts to the product of the next figure. 
 
 7 times 5 quarts make 35 quarts, to which 
 add the 3 quarts, which make 38 quarts; 8 quarts make 1 peck; then 
 38 quarts make 4 pecks, and leaves 6 quarts. Set down the 6 quarts, 
 and carry the 4 pecks. 
 
 7 times 2 pecks make 14 pecks ; add the 4 pecks, and it makes 18 
 pecks. 4 pecks make 1 bushel ; then, 18 pecks make 4 bushels, and 
 leaves 2 pecks. Set down the 2 pecks, and carry the 4 bushels. 
 
 7 times 7 bushels make 49 bushels ; add the 4 bushels, and it 
 makes 53 bushels, which set down, and the work is done. 
 
 bu. pk. qt. pt. bu. pk. qf. pt. 
 
 y 3 6 1 23 2 5 1 
 
 5 8 
 
 49 3 1 189 1 4 
 
 In 1 vessel are contained 29 bushels 2 pecks and 5 quarts ; how 
 many in 9 such vessels ? Ans. 266 bu. 3 pk. 5 qt. 
 
 Observation. When the multiplier is more than 12, and a com- 
 posite number, we may multiply first by one factor, and that product 
 by the other; or, if the multiplier has more than two simple factors, 
 such as 72, which is the product of 2X4X9, we may multiply 
 
68 ARITIliMETIC. 
 
 by the (^ilT^•rpnt factors in succrssion, no matler which we take first; 
 though it is generally more expedient to take the largest first. 
 
 Orseiivatiox SKcoxn. When the multiplier is not a composite 
 number, such as 23, we may take the sum 24 times, by multipiyitig 
 by (i, and that product by 4, and then subtracting once the sum, which 
 will leave 23. 
 
 If the multiplier be 31, we may multiply by 10 and 3, and after- 
 wards add once the number, &c. 
 
 APrLICATlON. 
 
 1. What is the cost of 9 hundred weight of chee-se, at 1 pound 11 
 shillings and 5 pence per hundred weight ? Aris. £14 2s. 9d. 
 
 2. A merchant has .3 chests of tea, each weighing 3 hundred 
 weight 2 quarters and 9 pounds ; what is the weight of the whole? 
 
 Ans. 10 cwt. 2 qrs. 27 lbs. 
 
 3. If a man drink I pint 2 giljs of ale per day for 29 successive 
 days, what quantity will he have drank in all? 
 
 Alls. 5 gal. 1 qt. 1^ pt. 
 
 4. If a soldier's ration of bread be 5 pounds 6 ounces 8 drams per 
 week, what will it amount to in 52 weeks? 
 
 Ans. 2 cwt. 2 qrs. 1 lb. 2 oz. 
 
 5. A merchant bought 9.5 pairs of shoes, at 4 shillings 6 pence and 
 I quarter a [)air ; how much did he pay for the whole ] 
 
 £21 9s. 5Jd. 
 
 6. A gentleman bought 43 silver spoons, each weighing 2 ounces 
 14 pennyweights and 6 grains; what was the weight of the whole? 
 
 Ans. 9 lbs. 8 oz. 12 dwts. 18 gr. 
 
 7. How many yards of cloth in 35 pieces, each containing 27 yards 
 3 quarters 2 nails ? Ans. 975 yds. 2 qrs. 2 na. 
 
 8. A silversmith has 7 tankards, each weighing 3 pounds 4 ounces 
 pennyweights and 22 grains; what is the w^eight of the whole? 
 
 Ans. 23 lbs. 4 oz. 6 dwts. 10 gr. 
 
 9. If a man can perform a piece of work in 2 years and 3 months, 
 how long would it take him to perform 5 such pieces ? 
 
 Ans. 11 yrs. 3 mos. 
 
 10. If a laborer dig a drain in 2 weeks and 3 days, how long a 
 time would he require to dig 9 such drains? Ans. 21 w. 6 d. 
 
 41. It is found by oKservation that the sun, on an average, changes 
 his longitude 0° 59' 8" 33 per day ; how much will he change in 10 
 days? how much in 30? how much in 300? how much in 365? 
 
 ■^ 9 51 23 3 
 
 . I 29 34 9 9 
 ^^^' f295 41 39 
 J 359 45 40 45 
 
COMPOUND DIVISION. 69 
 
 N. B. In this example, 8'' 33 is 8 seconds and 33 hundreds, the 
 33 being decimals. But we have ni)t yet exercised in decimals ; and 
 as thev increase and decrease on iht same ncaU as abstract numbers, 
 we do not tiesilate to introduce them thus incidentally. 
 
 In the table of circular motion, the pupil will lind signs, degrees, 
 and minutes; but the stilus arc no longer rdained in pruc/icul use ,- 
 however, through superstition and old custom, signs will long retain 
 their places in common almanacs. 
 
 12. The moon moves 13*^ 10' 35" in one day, how many degrees, 
 minutes, and seconds, will she move in 17 days ? 
 4X4-1-1=17 
 
 13° 10' 35 
 
 In 4 days it moves 52 42 20 
 
 4 
 
 In 16 days it moves 210 49 20 
 
 In 1 day it moves 13 10 35 
 
 la 17 days it moves 2-i3^ 59' 55'' Ans. 
 
 1:3. The planet Jupiter changes its longitude 4' 59" 2 in one day 
 how much will it change its longitude in 59 days? 
 
 Ans. 4° 54' 18". 
 
 14. The planet Saturn changes its longitude 2' 0" 91 in one day 
 how much will it change in 3ti5 days ? Ans. 12° 15' 37". 
 
 / 
 
 COMPOUND DIVISION.. 
 
 (AnT. 31.) CoMPOUxVD Division — the reverse operation to com- 
 pound multiplication, and, in its general principle, the same as simple 
 division. 
 
 As in simple division, we must divide the highest number of the 
 dividend as a simple number, by the divisor ; and the remainder must 
 be reduced to the next lower denomination, and the number standing 
 in that denomination being added in, we have a new dividend to be 
 divided asa simple number ; and so on through all the denominations. 
 
 This obvious operation gives us the following 
 
 RuLK. Place the divisor on the left of the dividend, as in simple 
 division. 
 
 1. Begin at the left hand, and divide the number of the highest 
 denomination by the divisor^ setting down the quotient in its proper 
 place. 
 
 2. If there be any remainder after this division, reduce it to the 
 next lower denominal/on, which, add to the number, if any, belong- 
 ing to that denomination, and divide the sum by the devisor. 
 
70 ARITHMETIC. 
 
 3. Set down cs;uin this quotient, reduce its remainder to the next 
 h)wer dtnominalion again, and so on through all the denominations 
 to the last. 
 
 N. B. Compound division and compound multiplication prove 
 each other. 
 
 EXAMPLES. 
 
 1. Divide 19 pounds 17 shillings and 6 pence equally among 6 
 men. 
 
 Here we say, 6 in 19 is contained 3 times, and 
 1 pound over, which, reduced to shillings, and 17 
 added, make 37 shillings, which, divided by 6, is 
 contained 6 times, and 1 shilling over, which is 12 
 pence, and the 6 added is 18, which, divided by 
 6, gives 3 pence, 
 bushels 2 pecks 6 quarts 1 pint equally among 
 
 Here 7 into 25 bushels 3 times, and 4 re- 
 mains. Set down the 3. 
 
 Reduce the 4 bushels to pecks, which makes 
 16 pecks: add 16 pecks to 2 pecks, it makes 
 18 pecks. Now, 7 into 18 pecks 2 times, and 
 leaves 4. Set down the 2, &c. 
 
 3. Divide 821 pounds 17 shillings and 9^ pence by 4. 
 
 Ans. £205 9s. 5d. 3 far. 
 
 4. Divide 55 pounds 14 shillings and | pence by 7. 
 
 Ans. £7 1 9s. Id. 32 far. 
 
 5. A farm containing 746 acres 3 roods 29 poles, is to be divided 
 equally between 9 heirs; what is the share of each 1 ^ 
 
 A7is.82 A. 3R.$8^ P. 
 
 6. Divide 3 acres of land into 8 village lots ; what number of poles 
 will each lot contain? Am^. 60. 
 
 7. Divide 3 acres of land into lots, each lot containing 3 roods ; 
 bow many lots will there be ? Ans. 4. 
 
 8. Divide 1 acre 1 rood of land into 10 equal parts; how many 
 poles will each part contain ? Ans. 20. 
 
 9. Put 23 bushels 1 peck of wheat into 10 bags ; how much must 
 each bag hold ? Afis. 2 bu. IJL pk. 
 
 10. Divide 10 gallons and 3 quarts into 6 equal portions: what 
 will each portion be] Ans. 1 gal. 3 quarts ^ pt. 
 
 (Anr. 32.) When the divisor is over 12 and is a composite number, 
 that is, one consisting of two or more simple factors, we may divide 
 as in short division, first by one factor, and that quotient by another, 
 &c. But when the number is prime, having no simple factors, we 
 divide by the whole number at once, after the manner of long division. 
 Thus. 
 
 £ 
 
 s. 
 
 d. 
 
 6)19 
 
 17 
 
 6 
 
 3 
 
 6 
 
 3 
 
 2. Divide 
 
 25 I 
 
 persons. 
 
 
 bu. 
 
 pk. 
 
 r^t.pf 
 
 7)25 
 
 2 
 
 6 1 
 
 3 
 
 2 
 
 5 1 
 
COMPOUND DIVISION. 71 
 
 6. Divide 79 buslicls 1 peck 7 quarts by 23. 
 
 hu. pk.qt. bu. ph. qt.pt. 
 23)79 17(3171 
 69 
 
 — 4X10=40, and 1 peck aJded makes 41, &c. 
 10 
 4 
 
 23)41(1 peck. 
 23 
 
 18 
 
 23)151(6 quarts. 
 138 
 
 13 
 2 
 
 23)26(1 pint. 
 23 
 
 Rem. 3 pints. 
 
 7. A boat load of corn, containing 4927 bushels 3 pecks, is owned 
 equally by 29 persons ; what is the share of each? 
 
 Ans. 169 bu. 3 pk. 5 qt. 1 pt., and I pt. rem. 
 
 8. Divide 542 pounds 7 shillings and 10 pence by 97. 
 
 Ans. £b Us. lOd. 
 
 9. Divide 123 pounds II shillings 2^ pence by 127. 
 
 Ans. £0 19s. 5^d. 
 
 10. Divide 330 hundred weight 3 quarters by 14. 
 
 A>u.2ri cwi. 3 qrs. 18 lbs. 
 
 11. If 35 pieces of cloth, of equal quality, contain 971 yards and 
 
 1 quarter, how many yards in a piece ? Ann. 27 yd. 3 qr. 
 
 12. If 259 acres 1 rood 10 rods of land be divided into 36 equal 
 lots, how much land will be contained in a lot ? 
 
 Ans.t A.OTi. 32^ r. 
 
 13. If 56 pounds of butter cost 4 pounds 18 shillings, what is it 
 per pound? Ans. Is. 9d. 
 
 14. Divide 124 pounds 5 shillings and 4 pence into 32 equal parts. 
 
 Ans. £3 17s. 8d. 
 
 15. Divide 336 bushels 3 pecks 4 quarts by 70. 
 
 Ans. 4 bu. 3 pk. 2 qt. 
 
 16. Divide 336 bushels 3 pecks 4 quarts, by 4 bushels 3 pocks and 
 
 2 quarts. Ans.lQ. 
 
 Reverse the preceding problems, taking the answers for divisors, 
 
72 ARITHMETIC. 
 
 and the former divisors will be quotients; but to effect the division, 
 reduce all to the lowest denomination mentioned, and divide as in sim- 
 ple division. 
 
 If more clear, we may enunciate the 16th example thus: A farmer 
 has 336 bushels 3 pecks 4 quarts of wheat in his granary, which he 
 wishes to put in casks to send away, each cask containing 4 bushels 3 
 pecks and 3 quarts ; how many casks will be required ? 
 
 Ans. 70 casks. 
 
 The operation is thus: 
 
 hu.pk. qt. hu. ph. qi. 
 4 a 2 ) 336 3 4 ( 
 4 4 
 
 19 1347 
 
 8 8 
 
 154 ) 10780^70 Ans. 
 10780' 
 
 17. A certain secret association in Ireland paid a bill of j£4 18s., by 
 assessing its members Is. 9d. apiece ; how many members were there ? 
 
 Alls. 56. 
 
 N. B. This is but a variation of example 13 ; and in like manner 
 every example, up to 16, may be varied. 
 
 18. I have a pitcher which holds 2 quarts and 1 pint; how many 
 times can it be filled from a barrel of cider, which holds 31 gallons 2 
 quarts ? Ans. 502^ 
 
 19. How many times will 13 bushels fill a vessel which holds 1 
 quart 1 pint and 1 gilH Ans. 256 times. 
 
 20. How many suits of clothes, each requiring 3 yards 1 quarter 
 and 2 nails, can be made from a roll of cloth containing 27 yards ? 
 
 Ans. 8. 
 
 MISCELLANEOUS EXAMPLES, 
 
 Applicable to the preceding principles. 
 
 1. At 5 cents a quart, what will 5 bushels cost? Ans. §8. 
 
 2. At 3 cents a pint, what will 3 pecks and 6 quarts cost '\ 
 
 Ans. 90 cents. 
 
 3. At 5 cents a pint, how many gallons can be bought for 
 10 dollars? Ans. 25. 
 
 N. B. As the pupil has been instructed from the very first of this 
 work, from simple addition, that 10 mills make a cent, 10 rents make 
 a dime, and 10 dimes make a dollar, and that these denominations are 
 the same in order as the order of simple numbers; therefore, the re- 
 duction from dollars to cents is to multiply by 100; dollars into mills. 
 
COMPOUND DIVISION. 73 
 
 multiply by 1000; and reduction the other way is to divide by these 
 numbers. 
 
 Some arithmetical writers have treated Federal Monei/ as conipmtnd 
 numbers, ami have gone throui^h all the formality of reduction — addi- 
 tion, subtraction, multiplication, and division of federal money — the 
 same as they do the really compound numbers, pounds, shillings, and 
 pence. 
 
 But, federal money was purposely adjusted to the rcuk nf simple 
 numbers ,- and if it is now proper to treat these denominations as 
 compound, we must suppose the design not accomplished. Federal 
 money belongs to whole numbers and decimal fractions ; and the sub- 
 ject must be incomplete until we pass decimal fractioris. 
 
 Example 3, and most of the examples here inserted, should be done 
 by canceling, as explained in article 21. 
 
 4. At 8 cents a gill, how many gallons will 12 dollars purchase? 
 
 Aris. 4l;. 
 .5. What will 10 ounces 10 pennyweights cost, at 15 cents a pen- 
 nyweight ? * Ans.$-M.50. 
 
 6. At 20 cents a square rod, what will 2 acres 3 roods of land 
 cost? Ans.$fi8. 
 
 7. At $2 a square rod for land, what must be paid for a village lot 
 12 rods long and 5-| rods wide ] Ans. S132. 
 
 8. In 11 bars of gold, each containing 5 pounds 3 ounrcs 2^ pen- 
 nyweights, how many grains ? Ans. 23:]!^8\ 
 
 9. At 5 cents an ounce, what will 10 p.v.inds i ounces of copper 
 cost ? A71S. iS.2(). 
 
 10. IIov.- many kegs, each holding 4 traiions 2 .juarts, can be filled 
 from a liogshead containing G3 gallons ] Aus. 14. 
 
 1 1. .'\.t 6^ cents a quart, what will 6 bushels atul 1 peck cost ( 
 
 Anf,: $12.r)0. 
 
 12. How much will 4 barrels of molasses cost, at 4 cents a pint. 
 
 Any. $10.08. 
 
 13. Divide 2 hours 10 minutes by 5 minutes 5 seconds. 
 
 Ans. 2^^. 
 
 14. How many times is 13° 20' contained in 360° ? 
 
 Ans. 27 times. 
 
 15. If the moon moved 13° 20' in 1 day, how many days would it 
 require to make a revolution nf 360° ] An.^. 27 days. 
 
 leXlf Jupiter changed its longitude 5 minutes of a degree, as seen 
 from the sun, bow many years, of 3^60 days each, would it require to 
 make a revolution ? - O " Ans. 12. 
 
 17. At 8 cents a pint for wine, how many gallons can be bought 
 for 40 dollars? Ans. 6-2^. 
 
 18. At 10 cents a nail, how many yards of cloth can be buu^du for 
 16 dollars? ' J/?.v."]0. 
 
 19. At 12^ cents a quart, how many gallons can be bought for 12 
 dollars? Ans. 24. 
 
74 
 
 ARITHMETIC. 
 
 20. How many little squares, 3 inches long and 3 inches wide, can 
 be cut from a square yard of paper? Ans. 144. 
 
 21. How many bottles, each containing 1 quart 1 gill, will be re- 
 quired to draw off a barrel of cider containing 3U gallons! 
 
 'Ans. 112. 
 
 22. Divide 421 pounds 14 shillings and 8 pence among 3 men, 5 
 women, and 7 boys, and give each man double of the sum given to a 
 woman, and each woman 3 lanes the sum given to a boy ; how much 
 
 IS tne share ct 
 
 ^ 
 
 each ? 
 
 Ans. Each boy must have £10 10s. 
 Each woman, . ... 31 12s. 
 Each m.an, 63 53. 
 
 I0|d. 
 7id. 
 2|d. 
 
 The art of working the preceding problem consists in obtaining the 
 divisor, which is 40. 
 
 23. A man bought a chaise, horse, and harness, for 70 pounds. He 
 gave twice as much for the horse as for the harness, and twice as much 
 for the chaise as for the horse ; what did he give for each ] 
 
 A}is. Harness £10. 
 
 24. A farmer sold some calves and some sheep for 108 dollars; the 
 calves at 5 dollars, the sheep at 8 dollars apiece ; there were twice as 
 many calves as sheep, — what was the num.ber of each sort? 
 
 Aus. 6 sheep and 12 calves. 
 
 25. The planet Jupiter changes its mean longitude 4° 54' 18'^ in 
 59 days; how far will it change in one day ? Aus. 4' 59" 2. 
 
 26. The moon is observed to move over 197° 38' 45" in 15 days; 
 how far will it move in one day ? Ans. 13° 10' 35", 
 
 27. In 365 days, the planet Saturn will change longitude 12° 15' 
 
 37" ; how much will it change in one day ? 
 
 A^7is. 
 
 0" 91. 
 
 28. If the apparent. motion of the sun be 59' 8" in one dav, how 
 many days will it require to make a revolution of 360°? 
 
 Ans. 365|||. 
 
 29. 'I'he moon changes her longitude 13° 10' 3.5" in one day ; how 
 many days, then, will be required to make one revolution ? 
 
 Ans. 27 d. 7 h. 43 m. 
 
 30. If Venus changed her longitude, (as seen from the sun,) 1° 
 36' per day, what, then, would be the time of her revolution 1 
 
 Ans. 225 days. 
 
FRACTIONS, 
 
 SECTION III. 
 
 FRACTIONS. 
 
 (Art. 33.) A part of any one thing is called ;\ fracjion. 
 
 If an api.li', f)r instance, be divided into 3 equal parts, each part 
 will he ont-thlrd, written thus, 5. 
 
 If it be divided into 4 equal parts, each part will be onc-fuurlh, 
 wriiten \. 
 
 !f divided into 5 equal parts, each part will bo \ ; two of these 
 parts must be written 1. 
 
 i bu.-, generally, a fraction must be expressed by two numbers one 
 a!)ove another. The lower number denotes the number of equal 
 part;^ into which the unit is divided. 
 
 The upper number shows how many of these equal parts 
 are taken. 
 
 ileiu-e, as the lower number of a fraction denotes or decides the 
 den:»ii'iiat;()n, whether it be thirds, fourt/iy, ffths, or any other 
 number, it is called the denominator. 
 
 'i'l'.e ur)per number is called the nunjf/Y//or, because it shows the 
 number 1 if parts taken. 
 
 (A;tT. 31.) A fraction may be considered as the result of an im- 
 possible division. Thus, 1, one thing, or ujiity, divided into 5 equal 
 parts, we write I above and ^ under it, or 1. 'J'hree times this is | ; 
 or, we may consider 3 divided by 5, the quotient is 1. 
 
 Hence, the denominator of a fraction may he considered as a divisor, 
 and the numerator as a dividend, and the fraction itself as the result, 
 or quotient, to an example in division. 
 
 When the dividend and divisor are equal, the quotient is 1 ; that 
 i.s, whiMi the numerator and denominator are equal, we have the whole 
 of the thinij;, the whole as an aggregate or unit. 
 
 (Art. 35.) When any fraction is before us, as f, we judge of 
 its value by comparing its numerator with its denominator ; if the nu- 
 merator is e^iual to the denominator, as we have just observed, tlie 
 value of the fraction is 1 ; if the numerator is nearly equal to the de- 
 nominator, the value of the fraction is nearly I ; if the numerator 
 is one-half of the denominator, the value of the fraction i.s one-half. 
 Hence, 1 is the same as ^. 
 
 6 
 
 (Art. 3fi.) As the value of a fraction depends upon the relation 
 of the numerator to the denominator, and as tliis relation is not 
 changed bv dividing both numbefs by the same divisor, we may, there- 
 
76 ARITH3IETIC. 
 
 fore, divide both numerator and denominator by any number that 
 will divide them without a remainder, and the value of the fraction 
 will not be changed.* 
 
 EXAMPLES. 
 
 1. Reduce Li to its lowest terras. Ans.^. 
 
 - 4 
 
 Here, it is evident that we can divide both numerator and denomi- 
 nator by 6, making the fraction |, which is equal to li. 
 
 2. Reduce il to its lowest terms. Ans. f. 
 
 5 6 
 
 3. Reduce _3JL to its lowest terms. An^. ^. 
 
 When the terms of the fraction are large, we may not bring it to its 
 lowest terms by one division only, but we may divide the quotients ob- 
 tained continually until no number greater than 1 will divide both 
 of them \Nithout a remainder. 
 
 4. Reduce Ll^ to its lowest terms. 
 
 7 3 
 
 5. Reduce 1?.^ to its lowest terras. 
 
 2 4 
 
 6. Reduce i?. 1 to its lowest terms. 
 
 6 3 
 
 7. Reduce J^^- to its lowest terms. 
 
 8. Reduce 119. to its lowest terms. 
 
 8 
 
 9. Reduce iii to its lowest terras. 
 
 4 9 2 
 
 10. Reduce 15?- to its lowest terms. 
 
 11, Reduce -1^- to its lowest terms. 
 
 * As a guide to the studem to find suitable divisors for these reduclionSjIet 
 him observe, 
 
 1st. That any number ending with an even number or a cipher, can be 
 divided by -2. 
 
 2d. Any number ending virith 5 or 0. is divisible by 5. 
 
 3J, If tiie right hand place of any number be 0. the whole is div;s'l>le by 
 10; if there be two ciphers, it is divisible by 100; if three ciphers, by lOCO. 
 and so on. which is only cutting ofl" tho?e ciphers. 
 
 4th. If the two right hand figures of any number be divisible liy 4, the 
 whole is divisible by 4: and if the three right hand figures be divisible by 
 8, the whole is divisible by 8. and so on. 
 
 5th. If the sum of the dig ts in any number be divisible by 3 or by 9, the 
 whole is divisible by 3 or by 9. 
 
 6th If the ri?ht hand digit be even, and the sum of all the d.g;is be divisi- 
 ble by 6, then the whole is divisible by 6. 
 
 7th. A number is divisible by 11, when the sum of the 1st. 3d. 5ih. kc. or 
 all the odd places, is equal to' the sum of the 2d, 4th, 6th, &c., or of all the 
 even places of digits. 
 
 Sih. If a number cannot be divided by some cpiantity less than the square 
 root of the same, that number is a prime, or cannot be divided by any 
 number whatever. , „ ^ ^ ■ , 
 
 9th. All prim-.' numbers, except 2 and 5, have either 1. 3. /. or 9. in the 
 Dlace of units; and all other numbers 're composite, or can be divided. 
 
 Ans 
 
 i. 
 
 Ans 
 
 i 
 
 Ans. 
 
 .5, 
 
 6* 
 
 Ans. 2 
 
 J ^ 
 
 Ans. ^ 
 
 o' 
 
 Ans. 
 
 h 
 
 Ans.). 
 
 'J ^ 
 i' 
 
 Ans. ^%_ 
 
 r 
 
FRACTIONS. 77 
 
 I 12. Reduce 3.2 i P. to its lowest tcriii:^. ,-l//.v. J,. 
 
 tj 4 8 II - 
 
 13. Reduce J.JJ'J'J^- to ils lowest leriiis. Ajis. 1. 
 
 6 7 18 5 1) « 
 
 14. Reduce -JJJ'J'jyL to its lowest terms. A7is. }. 
 
 15. Reduce _6_7_8AP_ to its lowest tciins. Ans. J_ 
 
 7 8 6 in' 
 
 16. Reduce -A<L'^j'- to its lowest terms. Ans. 'II. 
 
 10 S C 5 
 
 (AuT. 37.) In place of dividing by small common divisors, in 
 succession, we may lind the greatest common divisor at once, and 
 divide by it. 
 
 RuLK. /Tc) Jhid the greatest common divisor bctivcen two 
 numbers .• 
 
 Divide the greater nutnhcr by the smaller, and this divisor hi/ the 
 remainder, and thus continue dividing the last divisor hi/ the last 
 remainder, till nothing remains. The divisor last used will be the 
 number required. 
 
 What is the greatest common measure of 918 and 1944? 
 
 918)1944(2 
 1836 
 
 108)918(8 
 SG4 
 
 64)108(2 
 108 
 
 The truth of the rule in this problem will he discovered by retracing 
 the above operation, as follows: feince 54 (the last divisor) measures 
 108, it also measures 8X l08-|-64, or 918. Again, since .54 measures 
 108 and 918, it also measures 2X^18-1-108. or 1944. Therefore, 54 
 measures both 918 and 1944. It is, also, \\\e greatei-t common mea- 
 sure ; for, suppose there be a greater — then, since the gre:iter mea- 
 sures 918 and 1944, it also measures the remainder, 108 ; and, since 
 it measures 108 and 918, it also measures the rcmaiiuier, 54, that is, 
 the greater measures the less, wliich is absurd. 
 
 EXAMPLES. 
 
 1. V\'hat is the greatest common measure of the numbers 3;::? and 
 425 1 Ans. 17. 
 
 2. What is the greatest common measure of 2310 and 462G 1 
 
 Ans. G. 
 
 3. Reduce 2.?.i to its lowest terms. 
 
 4 2 .s 
 
 4. Reduce r^'llH. to its lowest terms. 
 
 4 2 6 
 
 g2 
 
78 ARITHMETIC. 
 
 5. Reduce ^6_; 3 to its lowest terms. 
 
 5 9 4 
 
 
 An.. 1». 
 
 6. Reduce 2.5. co to its lowest terais. 
 
 5 ■•) 4 
 
 7. Reduce 2 7 5^ to its lowest terms. 
 
 440 
 
 
 Ans. f . 
 
 8. Reduce -Yy g- to its lowest terms. 
 
 
 Ans. ^=^.. 
 
 9. Reduce JJ>J^ to its lowest terms. 
 1057 
 
 
 Ans. 4. 
 
 (AuT. c58.) We have thus far been s 
 in their simplest forms ; but we have also 
 pound Fractions, Mixed Numbers, and 
 
 peaking ol 
 Improper 
 Complex 
 
 ' proper fractions, 
 Fractions, Corn- 
 Fractions. 
 
 An improper fraction has a numerator greater than its 
 denominator, as t^^ ^^ Sec. 
 
 A compound fraction is a fraction of a 
 
 fraction, as ^ of f . 
 
 A mixed number is a whole number and a fraction standing in one 
 sum, as7|-, 11^, 19|, &c. 
 
 A complex fraction is, where one or both of the ie7'ms of the frac- 
 tion are fractional, as 
 
 2 71 2| 
 
 
 
 3L, 9, 51. 
 
 
 
 By an improper fraction, we understand that the value of the ex- 
 pression is more than a unit in value: thus, 1. As 3. make 1, 1 
 
 must equal 2^. 
 
 Hence, we may reduce improper fractions to mixed numbers, by 
 dividing the numerator by the denominator, and taking the remain- 
 der for^the numerator of the proper fraction. 
 
 EXAMPLES. 
 
 
 
 1. Reduce '_?-7 to its proper terms. 
 
 
 Ans. 11_5_. 
 
 2. Reduce -i 1 1 to its proper terms. 
 
 
 Ans. 153f 
 
 .3. Reduce ^J/ to its proper terms. 
 
 
 Ans. 2^-. 
 
 4. Reduce 1 £1 to its proper terms. 
 
 
 Ans. 301. 
 
 5. Reduce ^_6_i to its proper terms. 
 
 
 Ans. 56 ^!L, 
 
 6. Reduce ' |5 s to its proper terms. 
 
 
 Ans. 54^1 . 
 
 7. Reduce '-P to its proper terms. 
 
 
 Ans. 1|. 
 
 8. Reduce > 1." to its proper terms. 
 
 
 Ans. 23|. 
 
 9. Reduce '^J^-? to its proper terms. 
 
 
 An-. 41. 
 
 10. Reduce "isi to its proper terras. 
 
 
 Ans. 430^. 
 
VULGAR FRACTIONS. 79 
 
 (AiiT. 39.) Whole numbers may be put under a fractional form, 
 by placing unity under them. Thus, i. is manifestly 4, ^ is 9, Y 
 
 is 44, &c. 
 
 Now, by Article 36, the value of any fraction is not changed by 
 dividing both numerator and denominator by the same number. In- 
 versely, then, we shall not alter the value of any fraction by multiply- 
 ing both its terms by the same number. Hence, we can reduce any 
 number into halves, thirds, fourths, or any other number of equal 
 parts, by first placing unity under it, and then multiplying the nume- 
 rator and denominator by the parts required, thus : Reduce (5 units to 
 fifths. First, 1 ; then multiply both terms by 5, and we have 3_o 
 Answer. Again : Reduce 62 to fifths. The result is obviously 3_2 
 — and from this we may draw a rule to reduce mixed numbers to im- 
 proper fractions. 
 
 Rule. Multiply the whole number by the denominator of the 
 fraction, and add in the numerator for a new numerator, and set 
 the denomitiator under it. 
 
 EXAMPLES. 
 
 1. Reduce 4^ to an improper fraction. Ans. 1, 
 
 2. Reduce 711 to an improper fraction. Ans. •'^.a. 
 
 3. Reduce 111 to an improper fraction. Ajis. i|. 
 
 4. Reduce 37§ to an improper fraction. Ans. "-^ . 
 
 53 
 
 5. Reduce 5i to sixteenths: thus, --X i-f = f f , ^«*- 
 
 6. Reduce 21 to sevenths. Ans. 
 
 14 7) 
 
 7 
 
 3_6 7 
 4 
 
 7. Reduce 12J_ to an improper fraction. An 
 
 8. Reduce 341| to an improper fraction. Ans. 
 
 9. Reduce 241 to tenths. Aiis. 2_y • 
 
 72^ 
 
 10. Reduce 24 1, to thirds. Atis. -g- 
 
 11. Reduce 67-^- to an improper fraction. Ans. 7_4_' . 
 
 12. Reduce 131:| to an improper fraction. Ans. 3|5 ^ 
 
 13. Reduce 211 to an improper fraction. Ans. 1 1^ 
 
 If more examples are desired, the pupil can reverse those under 
 Article 38. 
 
80 ARITHMETIC. 
 
 CO.MrOUND FRACTIONS, 
 
 (Art. 40.) We have before dofined Conipountl Fractions to be 
 fractions of other fractions, such as ^ of ^, £. of ^, &c. 
 
 When a compound fraction is reduced to its simple value, it is evi- 
 dent that that simple value must he less than either fraction mentioned, 
 because it must be part of a per/. 
 
 CompoHiid, in arithmetic, always means a product ; and, to reduce 
 compound fractions to single fractions, we must multiply them toge- 
 ther. Hmce, (he same problems may appear under multiplicaiiun 
 of fractions and under ccmpound fraclioJis. 
 
 To find the value of a comriound fraction, let us observe that ^ of -f 
 is certainly 5 ; and we can obtain the same result by nuiltipiying l)oth 
 numerators together for a new numerator, and the denominators for 
 a new denominator, and then reducing down ; or we may cancel the 
 twos in the numerator and denominator. 
 
 Again : what is the value of ^ of -f of ?. ? 
 
 One-half of f is certainly ^ ; then f of ?. is certainly less than 2, 
 and a fraction is made less by increasing its denominator. If we 
 double the denominator only, the fraction is one-half its former value ; 
 if we increase the denominator by 3 times its former value, the frac- 
 tion will he one-third of its former value, &c. Hence, J^^ is the value 
 of 5 of -f of ^. From these observations we draw the following rule 
 for the reduction of compound fractions to single ones. 
 
 Rule. Reduce mixed numbers In improper fractions, and 
 place a unit under whole members to put them in a frac- 
 tional form. 
 
 Then, mxiliiplii all the numerators together for a new numera- 
 tor, and all the denominators for a new denominator, and reduce to 
 Us lowest terms. But, skillful operators reduce by canceling, while 
 performing the general operation. 
 
 EXAMPLKS. 
 
 1. Reduce ^ of 'i of § lo a simple fraction. Ans. _^_. 
 
 2. Reduce -J of 1 of l*. of t to a simple fraction. Ans. 2L, 
 
 3. Reduce 3 of 1 of _?_ of ^ to its simple value. ^4^5. _s_ 
 A. Reduce 2^ of I5 of £ of | to its simple value. Ans. 2. 
 
 5. Reduce ^ of i* of J of ll of 8 to its simple value. 
 
 Ans. 6_5_ 
 
 6. Reduce ^ of ^ of J of i of A of il of 10 to its simple value. 
 
 Ans. \l 
 
VULGAR FRACTIONS. 81 
 
 (AiiT. 41.) To invcstig;itc and ionn a rule to multiply a fraction 
 by a whole jiuniber, we can take some simple fraction, as ^. and if it 
 he miiltipliml by 2, the result must evidently be ? ; and if it be mul- 
 tiplied by 3, the result must be | ; if by 5, the result must be s, &c. 
 Also, take ^, and multiply it by 4 ; the result will be 1, or ^. Also, 
 f multiplied by 2 will give i ; 2 multiplied by 3 is _^, or ■§ ; and from 
 these observations, to multiply a fraction by a whole number, we de- 
 duce the foilnwing 
 
 RuLK. Mulfi'p/i/ the numerator by the whole number ,■ or, when 
 you can, divide ihe denominator by the whole number. 
 
 EXAMPLES, 
 
 1. Multiply ii by 5. Ans. y=2^- 
 
 2. Multiply ^ by 3. Ans. ~=^h- 
 
 3. Multiuly Li by 4, Ans. «l=3ll. 
 
 4. Multiply _!'_ by 100. Ans. 9-^=642. 
 
 5. Multiply ^1 by IS. Ans. 1\. 
 
 6. Multiply 1^3 by 19. Ans. 5^ = . 
 
 7. Multiply 1 by 24. Ans. 56. 
 
 8. Multiply 11 by 105. Ans. 85. 
 
 9. Multiply 3 by 63. Ans. 27. 
 
 10. Multiply 1 by 40. Ans. 25. 
 
 11. Multiply _"- by 11. Ans. 7. 
 
 To multiply a whole number by a fraction is the same as to multi- 
 ply a fraction by a whole number ; for, when two numbers are multi- 
 plied tot^ether, it is indifferent which we call the n)ultiplier, or which 
 the multiplicand. (See Art. 11.) That is, in the last example, ^- 
 multiplied by U, is the same as U multiplied by _'_ ; and so of the 
 other examples. 
 
 (Art. 42.) If we mulliply a fi-adion by its denominator, it 
 luill produce the iiumerator for a product. 
 
 EXAMPLES. 
 
 1. Multiply 1 by 7. By the above rule, the product must be ^, 
 
 or 3. . -, A 
 
 2. Multiply 11 by 19. Ans. 14. 
 
 3. Multiply 5 by 9. Ans. 5. 
 
ARITHMETIC. 
 
 4. Multiply 3i by 3. Ans. 10. 
 
 5. Multiply 7i by 4. Ans. 29. 
 
 6. Multiply 6 1 by 5. ^?is. 33. 
 
 7. Multiply 1^ by 17. Aris. 11. 
 
 8. Multiply ^^ by 29. ^ns. 19. 
 
 9. Multiply 33 by 7. Ans. 24. 
 
 10. Multiply 41.^ by 17. A7is. 79. 
 
 11. Multiply 15- by 21. Ans. 13. 
 
 N. B. Let the pupil remember this article, when he comes to 
 clearing equations of fractions in Algebra. 
 
 (Akt. 43.) We have already defined Complex Fractions to be 
 such as have fractions in the numerator or denominator, or in both, as 
 
 — '-. Now, if we multiply both numerator and denominator of this 
 
 fraction by 2, we shall have J , a simple fraction. 
 
 But, why did we multiply by 2, in preference to any other number ? 
 Ans. Because it was the denominator in the fractional part of the 
 numerator. 
 
 3 94 
 
 Again : — is the same as 1, or -f. Also, — is a complex fraction, 
 4]^ ^ 8g 
 
 and we can clear its numerator of the fraction by rauhiplying it by 3, 
 and we can clear its denominator by multiplying it by 8. Hence, 
 we can banish fractions from both numerators b}' multiplying by 3 
 and then by 8, or multiplying both numerator and denominator by 24 
 at once. 
 
 Thus: 9-1X24=232; 8|X24=20I. Therefore, 232 is the 
 equivalent simple fraction. 
 
 It is now apparent that we can change complex fractions to simple 
 fractions, by the following 
 
 Rule. Multiply both numerator and denominator by the denomi- 
 nators of the partial fractions, or by their product, or by their least 
 common multiple. 
 
 EXAMPLES. 
 2i 
 
 1. Reduce — ^ to a simple fraction. Ans. ll. 
 
 41 *8 
 
 7 
 
 5 
 
 2. Reduce ^rr to a simple fraction. Ans. 1 = 11, 
 
 .0 o 5 5 
 
VULGAR FRACTIONS. 83 
 
 3. Reduce -? to a simple fraction. Ans. 2.1. 
 
 i „. ^^ „, 4| 
 
 4, Reduce - of o, of - ^ (o a simple fraction. 
 
 1 3 **3 4 
 
 3 4 
 
 A71S. 4_8^. 
 
 5. Reduce ~~ to a simiile fraction. A)is. ,"/_. 
 
 1 9 _ I y o 
 
 1 24 3^ 
 
 6. Reduce — of -- of — to a simple fraction. Ans. tl. 
 
 03. 2 2 ^ ^ 
 
 7. Reduce -^ of ^, to a simple fraction. Ans. -k. 
 
 8. Reduce ;;:rj of •§ to a simple fraction. Atis. _«_. 
 
 If we divide i by 2, that is, one-half cut into 2 equal parts, each 
 part will be i. If we divide ^ by 3, or in other words, cut ^ into 3 
 equal parts, each part must be i. But, we can obtain these results 
 mechanically, by multiplying the denominator of the fraciiun by 
 the divisor. 
 
 If we were required to divide 2. into 3 equal parts, each part will be 
 ;.. In this case, then, we can divide the fraction by dividing the 
 numerator, because it is susceptible of being divided into the parts 
 required. 
 
 Therefore, to divide a fraction by a whole number, multiply the 
 denominator of the fraction by the whole number, and place the 
 numerator over the product. Or, when you can, divide the nume- 
 rator by the whole number, and let the denominator remain unchanged. 
 
 EXAMPLES. 
 
 1. Divide I by 3. Ans. i. 
 
 2. Divide 2 by 9. Atis. _2_.. 
 
 3. Divide |. by 5. Ans. ^. 
 
 4. Divide 11 by 13. Ans. J 
 
 1 9 
 
 .5. Divide _»_ by 8. Ajis. _"_. 
 
 6. Divide LL by 7. Ans. JJ^. 
 
 2 1- 14 7 
 
8 4 ARITHMETIC. 
 
 7. Divide LI by 34. Ans. J„. 
 
 2 1-' 42 
 
 8. Divide ^1 by 15. . / Ans. ^. 
 Tiiis is the reverse operation of Art. 41. 
 
 [N. B. Let pupils omit the following Article the first time going 
 I h rough.] .'- 
 
 CONTINUED FRACTIONS. 
 
 (Art. 44.) We have a clearer conception of the value of a frac- 
 tion when its numerator is unity, than when it is in any other form. 
 For in-stance, we have a clearer understanding of the fraction i than of 
 -?^J-, although they differ in value but veri/ little. 
 
 As the value of a fraction is not changed by dividing both nu- 
 morator and denominator by the same number, (Art. 36,) we can 
 always change the numerator to unity, by dividing both terms of the 
 fraction by the numerator; but this will make the denominator a 
 mixed number,* and the fraction a complex fraction. 
 
 If we take the fraction -||^, and divide both numerator and deno- 
 minator by 287, we have — , which shows that the value of the 
 
 fraction is between \ and \, and 5 is its first approximate value. 
 If we take -J.|j, and divide its numerator and denominator by 131, 
 
 we then have ;— -— , and the original fraction becomes 
 1 
 
 If we continue to operate on the fraction ^J^ , until we find the 
 last fraction with unity for a numerator, the original fraction, ||^, 
 will take the following form, which is a continued fraction. 
 
 1 
 
 2-t-l_ 
 5-t-l__ 
 
 6 
 
 *That is: \n case the numerator and denominator are prime to each 
 other, as il is supposed ihev are. If they are not, the fraction must be re- 
 <luced bv Article 36. 
 
FRACTIONS. S;> 
 
 If we noglect tlie fractional paij of the 2ad denominator, ihc second 
 apjToximatc fraction will be 
 
 The first approximate value of the original fraction, 2 "t, is, as 
 
 we have just observed, ^. The second apjiroxiinatt- value is — 
 
 2 
 I 
 The third approximate value is 3T71 
 
 5 
 The fourth approximate value is 
 
 3-fI_ 
 
 2-fl__ 
 5-1-1 
 4 
 And the fifth and last approximate value is the fraction itself 
 
 Let it be observed, that the principle here involved is all encom- 
 passed in Article 36, and that we reiluce fractions to this shape by 
 division ; and, of course, to reduce thorn back again, all we would 
 have to do is to inspect the work, and take the reverse operation. 
 For example, what simjile fraction expresses the value of the third 
 a[)proximate value of the traction under consideration] 
 
 We must commence with the last denominator, 1, and multiply 
 the numerator and denominator of the fraction —7—: hy 5, which gives 
 
 5 
 
 1 
 
 _5_; then, our approximate fraction becomes = ?-l. 
 
 11' ' ^'^ 3_5_ 3 8* 
 
 In this manner we shall find the approximate values of the fraction 
 
 fnhpi 2 11 4fi 287 
 
 By inspection, we perceive that a continued fraction may be defined 
 
 thus : 
 
 A continued fraction is one that has a whole number and a frac- 
 tion for it'^ denominator, which fraction has ahio a whole number 
 and a fraction for its denominatur, aiid so on, as far as the frac- 
 tion may extend; ai}d each fraction must have unity fur its 
 numerator. 
 
86 
 
 ARITHMETIC. 
 
 The partial fractions which compose any continued fraction, are 
 called hit egrut fractious, because each of their numerators must con- 
 sist of a single integer. The integral fractions which compose the 
 continued fractions, which we now have under inspection, are 5, 5, 
 L, 1, 1. But, we have just seen that the approximating fractions 
 are \, 2 n 4_6_ 28.7 
 
 By investigation, we find that the approximating fractions can be 
 drawn from the integral fractions, by the following 
 
 Rule. 1. The first approximating fraction is the same as the 
 first integral fraction. 
 
 2. The second approximating fraction has the denominator of 
 the second integral fraction fur its numerator, and the product of 
 the first two integral denominators, plus 1, for its denominator. 
 
 3. The third approximating fraction may be fanned by multi- 
 plying the terms of the second approximating fraction by the de- 
 nominator of the third integral fraction, and to the products add 
 the terms of the first approximating fraction. 
 
 And thus, in general terms, to form any approximating fraction 
 after the second, 
 
 Multiply the terms of any approximating fraction by the deno- 
 minator of the following integral fraction, and to the products 
 add the corresponding terms of the preceding approximating frac- 
 tion, and we shall have the terms of the next succeeding approxi- 
 mating fraction. 
 
 1. Reduce I2 i_ to a continued fraction : find its integral fractions, 
 and its approximating fractions. 
 
 Ans. ^, I, \, 1, L. integral fractions. 
 
 5 ' G ' 
 
 i 3^ Ll^ --^-^ -r^^i approximating fractioiis. ^ 
 
 2. Reduce the fraction _Y_ into a continued fraction, and find its 
 integral and its approximating fractions. 
 
 Ans. ^, 5, ^, 5, 1, I, integral fractions. 
 
 APPLICATION. 
 
 3. The circumference of any circle is three times its dinnip*'r, 
 plus the fraction -.yj-^JL. Reduce this fraction to a continued frac- 
 tion, and find its first three integral fractions, and its first three ap- 
 proximating fractions. 
 
 Alls. |, _?_, 1., integral fractions. 
 
 Y' rVe' TtV' approximating fractions. 
 
VULGAR FRACTIONS. 87 
 
 By putting the integral 3 to these fractions, and reducing to 
 improper fractions, we have 2_2^ y|-, liA, as the approximating 
 ratios between the circumference and the diameter of a circle. 
 
 4. The solar year is 5 hours 48 minutes 49 seconds above 365 
 da3's: wiuil part of 24 hours, is 5 hours 48 minutes and 49 seconds ? 
 
 To answer this question, we llrst reduce both of these quantities to 
 seconds: fi hours 48 minutes 49 seconds =20929 seconds; 24 
 hours =8(5100 seconds. Then, the proportional part that n hours 
 48 minutes and 49 seconds is of 24 hours is expressed by the fraction 
 
 2 3 L' n 
 8 6 4 00* 
 
 Reduce this fraction to a continued fraction, and find its integral 
 fractions and its approximating fractions. 
 
 j„r X 1 1. 1. J_. 1. L. J_. are its integral fractions. 
 
 And its first six approximating fractions are 
 
 5i 29' 33' 128' I61'2704* 
 
 These fractions show that in every four years we must have one 
 leap year, to make the sun cross the equator on the same day of the 
 month, and this will be only an approximate correction. Seven leap 
 years in 29 years would be better, or 8 in 33, or 31 leap years in 128 
 consecutive years, would be a very near correction, &c. 
 
 5. The planet Mercury performs a revolution round the sun in 87 
 days 23 hours and 15 minutes, nearly, or 126676 minutes. The 
 earth revolves round the sun in 365 days 5 hours and 49 minutes, 
 nearly, or in 525949 minutes. 
 
 Hence, 126676 revolutions of the earth correspond in time to 
 525949 revolutions of Mercury ; for, 525949 X 1-6676 gives the same 
 product as 126676X525949. From this, it appears that if the earth 
 and Mercury are in any particular position in respect to longitude, as 
 seen from the sun, they cannot both be in the same position again, at 
 the same time, until after an elapse of 126676 years; but they can 
 approximate to the same position at different intervals, which can be 
 determined hy making a fraction of the numbers 126676 and 525949, 
 thus 1-116 7^ Reduce this fraction to a continued fraction, and find 
 
 '525949 . . ^ . 
 
 its integral fractions and its approxnnatmg tractions. 
 
 Am. ' I i L t 1 1 i. J-. -'-. 1, are the integral frac- 
 
 -^4' IT' T' T' 2' 1' 1' 6» 6 0' 1' 3' " 
 
 tions, and its approximate fractions are 
 
 1 n 7 13 3 3 ^6 T_9_ J5_2_0_ J}J_2jJ}__ J^JJTLPJL 
 
 4' IjI' l^y"' 51' T3T' T;i I' 328' 2159» 129808' 13 2027' 
 
 1 2 Ifi 76 
 
 52 5 i* 4 9 * 
 
 The first fraction shows that 1 revolution of the earth has an ap- 
 proximate value to 4 revolutions of Mercury. The next fraction shows 
 that 6 revolutions of the earth, (or 6 years,) has a nearer approximate 
 value to 25 revolutions of Mercury, and so on, as the reader can see 
 by inspection. 
 
88 ARITHMETIC. 
 
 N. B. We are aware that ihe applicaiioii of contnuied fraciioiis does 
 not belong to arithmetic, for it involves associations of ideas entirely too 
 h;gh ; but we know from experience thai pupils in general will not take 
 sufficient interest in abstract continued //actions to ui'.(h-;rstand them, unless 
 they are convinced of their practical utility; and this is an apology lor giv- 
 ing the mailer iubsequenl to example 2. 
 
 REDUCTION OF FRACTIONS. 
 
 (AnT. 45.) We have, thus far, treated fracuon.s in the abstract. 
 We shall now begin to apply them to particular things, and change 
 them from »>ne scale of weight or measure to another. But, we shall 
 be best understood by 
 
 EXAMPLES. 
 
 1. Reduce | hundred weight to the fraction of a pound. This is 
 ihe same as asking the value of i of » 12 pounds, which, by Article 
 40, must be A)is. ^±^. 
 
 2. What is the value of | of 1 pound Troy ? that is, what is the 
 value of I of »_2 ounces ? Ans. 7 oz. 4 dwts. 
 
 From the foregoing observations we deduce the following 
 
 Rule. To find the value of a fraction in parts of its integer of 
 lower denomination. 
 
 Multiply the numerator by the parts in the next inferior denom- 
 ination, and divide the product by the denominator, and multiply 
 the remainder, if ariy, by the parts in the next denomination, arid 
 so on. The quotients placed in. cn-der, will express the value of the 
 fraction. 
 
 EXAMPLES. 
 
 1. What is the value of |. of an Ell English, in integers of lower 
 order? Ans. 3 qr. 2^ na. 
 
 2. What is the value of | of an acre, in integers of lower' order? 
 
 Ans. 3 R. 131 P. 
 
 3. What is the value of I of a hogshead ? Ans. 49 gals. 
 
 4. What is the value of 2 of 3 of one hour? Ans. 30 min. 
 
 5. What is the value of -?^ of a day in integers? 
 
 Ans. 3 h. 12 min. 
 
 6. Reduce j of a dollar, or 100 cents, to its integer value. 
 
 Ans. 12^ cts. 
 
 7. Reduce |o of a hogshead to its integer value. Ans. 50 gals. 
 
 8. Reduce ' |. of a day to its integer value. 
 
 Ans. 16 h. 36 m. 55J5_ sec. 
 
 9. Reduce || of a hogshead to its proper quantity. 
 
 Ans. 33 gal. 3 qt. I pt. 11. 
 
REDUCTION OF FRACTIONS. 89 
 
 10. Reduce 1^,- of a £ to its proper quantit}-. Ans. 18 s. 4 d. 
 
 A fraction may be so small, that is, the denominator so large, in 
 relation to the numerator, that we can obtain no integer of the wfe- 
 rior decree. Observe the following 
 
 EXAMPLES. 
 
 11. Reduce ----- of a hogshead to the fraction of a pint. 
 
 Ans. 12. 
 
 8 S 
 
 12. Reduce - J__ of a day to the fraction of a minute. 
 
 15 8 4 •' 
 
 Ans. I'L. 
 
 13. Reduce ." of an acre to the fraction of a rood. 
 
 14 4 
 
 Ans. ^1-. 
 
 ^8 
 
 14. Reduce — 1 of a mile to the fraction of a rod. 
 
 10 5 
 
 Ans. _5_, 
 
 3 3 
 
 15. Reduce ^2_ of a £ to the fraction of a shilling. Atis. ^ s. 
 
 16. Reduce j^^-^ of a hundred weight to the fraction of a pound. 
 
 Ans. A. 
 
 7 
 
 (Art. 46.) The subject-matter of this article is to perform the 
 reverse operations of Art. 45 ; i. e. To find what part one quantity 
 is of another. 
 
 EXAMPLES. 
 
 1. What part of a pound is 3 shiUings and 4 pence ? 
 
 3 s. 4 d.= 40 pence; 
 £1=20 s. =240 pence. 
 
 Now 40 pence is obviously _4j)_ of a £, or 1 of a £. 
 
 What part of a yard is 3 inches'! Manifestly JL=J_. From 
 the above we draw the following 
 
 Rule. Reduce the given quantity to the lowest denomination 
 mentioned^ for the numerator ,- then reduce the integer to the same 
 denomination, for the denominator of the required fraction. 
 
 2. What part of a yard is 3 quarters 2^ nails'? 3 quarters 2^ nails= 
 29 half nails; 1 yard=32 half nails. Ans. |9. 
 
 3. What part of 1 pound Troy is 7 ounces 4 penny-weights ? 
 
 Ans. 3 . 
 
 4. Reduce 3 quarters 7 pounds, to the fraction of 1 hundred weight. 
 
 Ans. ^«^.,.. 
 
 5. What part of 2 rods is 4 yards 1^ feet? Ans. ^^. 
 
90 ARITHMETIC. 
 
 6. What part of 1 bushel is l!l pecks? Anf:. %. 
 
 7. Reduce l:J hundred weight 3 quarters 20 pounds to the fraction 
 of a ton. Am. 12. 
 
 56 
 
 When the proposed quantity is not to be compared to, or measured 
 by, an integer, ike same principle is observed. The following exam- 
 ples will illustrate : 
 
 8. What part of 4 hundred weight 1 quarter 24 pounds, is 3 hun- 
 dred weight 3 quarters 17 pounds 8 ounces ? Ans. g. 
 
 9. A man bought a piece of cloth containing 8 yards 3 quarters; 
 it took 2 yards 2 quarters to make a coat ; what part of the whole 
 piece did it take 1 Ans. ^. 
 
 10. A man has a journey to perform of 35 miles 7 furlongs ; after 
 
 having traveled 15 miles 3 furlongs, what part of the journey has he 
 
 performed? Ans. ill. 
 
 * 2 8 i • 
 
 (AuT. 47.) To reduce fractions from a lower denomination to an 
 equivalent fraction of a higher denomination, we need only to observe 
 the following process : 
 
 1 . V/hit part of a foot is h of an inch ? 
 
 Now, to reduce inches to feet, we have only to divide by 12, and 
 this we mu!-t do, whatever be the number of inches, or parts of an 
 inch. Hence, we must divide 5 by 12, which, by Article 44, must be 
 
 J_, Ans. 
 
 3 6 
 
 2. Reduce A of a cent to the fraction of a dollar. To reduce cents 
 to dollars, we must divide by 100. Hence, _ j._ = _l_ of a dollar, 
 
 •' 6 50120 ' 
 
 the answer. 
 
 The pupil will observe from this, that this fraction must be dvided 
 at once, or in succession, by the number of units between the higher 
 and lower denomination mentioned. 
 
 In the following operations, considerable canceling can be used. 
 
 EXAMPLES. 
 
 1. Reduce j of an ounce Troy, to the fraction of a pound. 
 
 Ans. J_. 
 3 6 
 
 2. Reduce ' • of a minute to the fraction of a day. 
 
 Ans. _^_. 
 
 3. Reduce ^-i_ of a pound avoirdupois, to the fraction of a 
 
 ns. ?_^, 
 
 1. 
 
 Ans. _3_ 
 
 hundred weight. Ans. ?_^. 
 
 4. Reduce f of a nail to the fraction of an ell English. 
 
 5. Reduce _V. of a penny to the fraction of a pound. 
 
 Ans. ^--. 
 576 
 
VULGAR FRACTIONS. 91 
 
 6. Reduce i of a pint to the fraction of a bushel. 
 
 Ans. -i-. 
 
 3 2 
 
 7. Reduce .?_ of an ounce to the fraction of a hundred weight. 
 
 A)is. 
 
 9 8 se 
 
 8. Reduce 1 of an inch to the fraction of an ell English. 
 
 Ans. _L. 
 
 7 5 
 
 (Art. 48.) To reduce a fraction from one denomination to an- 
 other, on the same scale, 
 
 Reduce 7 to fourths. We write I, and then multiply numerator 
 and denominator by 4, making 2_8j ^^'hich is still 7 in value. (See 
 Art. 39.) 
 
 In the same way, let us reduce 4 to fourteenths ; or, in other 
 words, change |. to fourteenths^. 
 
 5 
 6" 
 
 We write '{ and multiply numerator and denominator by 14 
 
 14 14 14 14 
 
 2. How many fifths are (here in i of anything? Ans. 4|. 
 
 3. How many _L are there in J- of any number, or any thing? 
 
 '"* Ans. 6f- 
 
 This is, anything divided into 24 equal parts, 6| such parts is the 
 same portion of the whole ais -|-''j.is of 1. 
 
 4. Reduce 1 to a fraction whose denominator shall be 4. 
 
 3^ 
 Ans. p' 
 
 4 
 
 5. Reduce 11 to a fraction whose denominator shall be 20. 
 
 Ans. -— f- 
 20 
 
 6. How many -V of a shilling are in 4 of a shilling ' 
 
 66 
 
 Alls. _T 
 
 12 
 7. How many pence are there in i. of a shilling 1 Ans. 6£. 
 
 N. B. Observe, that examples 6 and 7 are substantially the same, 
 except that one calls for the answer in parts of a shilling, the other 
 calls for it in pence. 
 
 (AnT. 49.) When several fractions have different denominators. 
 
92 ARITHMETIC. 
 
 it is important, many times, to bring them under one denomination, 
 as we can neither add nor subtract them until this is done. (See 
 Art. 5.) For example, reduce ^ and § to a common denominator, 
 without charginj^ their values. Eighths cannot be reduced to halves 
 in this case, but halves can be reduced to eighths, by multiplying nu- 
 merator and denominator by 4, and then wc have ± and f , 8 being 
 the common denominator. 
 
 Thus, when the deiiominators are muliiples of each other, and the 
 pupil has a clear understanding of the nature of a fraction, he can 
 reduce them all to one, or a common denominator, by multiplying or 
 dividing numerators and denominators, as judgment may dictate. 
 
 EXAMPLES. 
 
 1. Reduce |, _s_, J'_, to equivalent fractions, having a common 
 denominator. Ayis. ^s_^ _6_^ _5_, 
 
 2. Reduce 1 _?_, J'-, to their equivalents, having a common de- 
 nominator. " Ans. _2_, _5_^ _8_. 
 
 All rules for reducing fractions to a common denominator must be 
 based on the principle that, numerators and denominators multiplied 
 or divided by tlie same number, does not change the value of the 
 fraction. On this principle, we may understand the following rule, 
 which is universal in its application : 
 
 Rule. Multiply each numerator irito all the denominators ex- 
 cept its 0W71, for a new numerator. Then multiply alt the deno- 
 minators together for a new denominator, and place it under each 
 new 7iumerator. 
 
 1. Reduce -f, 2, and ^, to a common denominator. 
 
 2X2X4^=16 the numerator for -f. 
 1X3X4=12 - do. - |. 
 3X3X2=18 - do. - |. 
 3X2X4=24 the common denominator. 
 
 The equivalent fractions are 
 
 {1 6-_2 
 24— 3* 
 1 2-_l 
 24 2* 
 2 4 4* 
 
 It is evident that the values of the fractions are not changed, be- 
 cause both the terms of each are multiplied by the same number. 
 
 2. Reduce 1 and 1 to a common denominator. 
 
 Ans. J 63 7' 
 
VULGAR FRACTIONS. 93 
 
 (AiiT. 50.) To follow the above rule in all cases might often 
 require more niechauical labor than necessary ; and to abridge the 
 operation we must lind, not merely the common, but tlie least com- 
 mon denominator. 
 
 Preparatory to this, we must be able to find the least commnn mxil- 
 tiple of tw!) or more numbers ,- that is, the least number divisible 
 bij them icUlwut remainders. 
 
 RfLK. Write the numbers one after the other, and draw a line 
 beneath them ; then, take any prime nund)er which will divide 
 two or more of them tvithout remainder, and divide all the num- 
 bers tiiat will so divide — ivriting the quotients beneath, and all the 
 numbers that are not divisible by it. Find a prime number that 
 will divide two or more numbers in this second line, and proceed as 
 bpfore. Continue the operation tmtil there are no two numbers left 
 bavins; a common divisor; then, multiply all the divisors and re- 
 maining numhers together, and their product will be the least com- 
 mon multiple sought. 
 
 1. Let it be required to find the least common multiple of 12, 15, 
 7, 18, 3, 5, and 35. 
 
 7 
 
 5 
 
 3 
 
 2 
 
 7X5X3X3X2X2=1260. 
 
 First, dividing by the prime number 7, we find that it divides two 
 of the given numbers, 7 and 35 — the rest being written in the line 
 below. We take 5 for the next divisor, which di'^des 15 and 5. 
 Continuing the operation in the same manner, we divide successively 
 by 3 and 2, when the division can be carried no further. Then, mul- 
 tiplying all the divisors together, and that product by the undivided 
 prime numbers left, and the result in this case is 1260. It is evident 
 that this must be the least common multiple, because it contains 07ily 
 the prime factors necessary to make up the given numbers. The 
 same must be true of any other numbers. 
 
 2. Find the least number that is divisible by 9, 12, 16, 24, 36, 
 without remainders. Ans. 144. 
 
 12, 
 
 15, 
 
 7, 
 
 18, 
 
 3, 
 
 5, 
 
 35, 
 
 12, 
 
 15, 
 
 1, 
 
 18, 
 
 3, 
 
 5, 
 
 5, 
 
 12, 
 
 3, 
 
 1, 
 
 18, 
 
 3, 
 
 1, 
 
 1, 
 
 4, 
 
 1, 
 
 I, 
 
 6, 
 
 1 
 
 1, 
 
 1, 
 
 2, 
 
 1, 
 
 1, 
 
 3, 
 
 1, 
 
 1, 
 
 1. 
 
94 ARlTHMliTlC. 
 
 N. B, In the opeiHtion, 9 anri 12 may I'e leO f^nt, as any num- 
 ber divislibe by 36 will be divisible by 9 and 12, 
 
 3. Find the least number divisible by each of the nine digits 
 
 A71S. 2520. 
 
 4. Find the least number divisible by 75, 50, 15, 20, 30, and 45. 
 
 Ans. 900. 
 
 (Aut. 51.) We are now prepared to give the following rule for 
 finding the least common denominator to any set or group of 
 fractions. 
 
 Rule. Find the least common multiple of all the denominators. 
 This will he the common denominator. For the numerators, divide 
 the common denominator by each particular denominator, and mul- 
 tiply the quotient by its numerator. 
 
 EXAMPLES. 
 
 1. Reduce ^, 4, y-, and ^-^ to the least common denominator. 
 7 
 2 
 
 2, 
 
 7, 
 
 16, 
 
 21, 
 
 2, 
 
 1, 
 
 16, 
 
 3. 
 
 1, 
 
 1, 
 
 8, 
 
 3. 
 
 7X2X8X3=336, least common denominator, or denominator to 
 all the new fractions. To obtain the new numerators, divide this by 
 the former dtnominators, and multiply the quotient by the former 
 numerators, thus: 
 
 336X1 
 
 — =168= 1st numerator. 
 
 2^L>lf=192=2d numerator. 
 7 
 
 336X3 „ 
 
 — — — =63=3d numerator. 
 16 
 
 336X3 
 
 — _ - — =32=4th numerator. 
 
 The new fractions equivalent to the given fractions are l^lt i-^, 
 
 J5 3_ Ji2_ 
 3 36» 3 3 6 
 
 2. Reduce _"_. 1. 1. and i, to their least common denominator. 
 
 An^. AHl 231 3.12 1 5 1 
 616' 616' 6Ifi» 6 Tfi' 
 
 3. Reduce _2_, ^^^ :|7^ and J'-^ to their least common denomi- 
 nator. " Ans. J^_ JJ_ Lll 8 . 
 
 150' I50» 150' 150 
 
VULGAR FRACTIONS. UT) 
 
 ^4. Keduce -^ .»_^ ^, and 6, to their least coihuidi) denoiniriator. 
 
 Ans. -s_fi_, JJ, 1-L'l. LJJ?. 
 
 •J32'252'25i;» 252 
 
 5. Reduce S^, 2^, and If, to their least common denominator. 
 
 AflS. 4J l_8 IJ. 
 A ^, 8 » 8 ' S 
 
 4 2-^ 
 
 6. Reduce ^ of ^, 7 andji to their least common denominator. 
 
 s ^ 
 
 AnS. -8^ 4 80 4 5 . 
 2 4' 2 4 » 2 4 
 
 7. Reduce -- of -^, -^ of -^ and 7, to their least common deno- 
 
 minator. Ans.^-?^J 111 2_2_o_5 
 
 3Ts ' 3 1 5' 315' 
 
 ADDITION OF FRACTIONS. 
 
 (Art. 52.) We cannot add unlike things together, (Art. .5,) 
 therefore, betbre fractions can be add(ul, we must be positive that they 
 not only refer to the same integer, (lielong to the same scale of mea- 
 sure.) but have the same denomination of that scale. For illustra- 
 tion, we cannot immedintely add together f of a foot and -J of an 
 inch, for tlie s.uiie reason that we cannot put feet and inches into the 
 same .sum. Equally impossible is it to add fourths and thirds toge- 
 ther. Hence, all the difiiculty a pupil will meet with will be in mak- 
 ing the preparations to add, and not in adding. 
 
 When all the preparations are made, the fractions are added by 
 adding together the numerators, and setting the common denominator 
 under the sum ; for, the sum of ^, £, and g, is ccrtaiiily il. 
 
 Take the second example of the last article (51), and require the 
 in of _?_^ |, ^, and i- 
 
 The equivalent fractions having a common denominator, we see by 
 "erring back to article 51, are A?l. 'l'±l., 252 \_b_i^, 
 
 ° (>16'016'616»G16 
 
 Their sum is 504-|-23I-|-252-4- 154=1 141. 
 
 Am. »-ij'_i 
 
 6 16' 
 
 From the foregoing we deduce the following general 
 
 Rule. Reduce ihe fractions, if necessary, to the same standard 
 or integer of measure ,- compound fractions to single ones, and ail 
 to a common denominator,- then, add the numerators, and place 
 their sum over the common denominator. 
 
 When several fractions have denominators that are multiples of 
 each other, as i, i, |, &c., we can reduce them to a common denomi- 
 nator without any formality of rule, by simply multiplying the nume- 
 rators and denominators of those fractions having the smaller denomi- 
 nators, by such numbers as will bring the denominators alike. 'J'hus, 
 in the above, ^=i> 4=|, and | stands: and their sum is -1=1^. 
 
96 ~ ARITHMETIC. 
 
 In the same manner we find the sum of the fractions ^1, |, | 
 
 and ^1, to be Il^Ve* 
 
 A similar operation will give the sum of the following fractions : 
 L-J-l-Ul-i--'^— It is evident that these denominators measure 24; 
 
 2 T^ 6 1^ n n^ 1 2 
 
 that is, all of them will divide 24 ; hence, we can change all the frac- 
 tions to twenty-fourths, by mere inspection. Thus, 1 = 12 i =_4 
 
 3 = _9_ _"_ = !,£. Sum, 3 9._1|. 
 
 8 24' 12 24 24 " 
 
 When we have mixed numbers, add the whole numbers separately, 
 and then add the fractions. The two sums put together will be the 
 
 sum required. 
 
 EXAMPLE. 
 
 Add 3i, Ih 21, ry^, 9tV, together. Integers, 3-l-7-}-2-i-74-9 
 =29. Fractions, « |+_n_-l-_«_-l--3_+/-=||=|. ' Whole 
 sum, 298. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Add I and 2. Ans. 3 5. 
 
 2. Add I, I ^ and 5^ together. .4ns. li.||. 
 
 3. Add I, i, 2^ and ^ of |, together. Ans. \^^-. 
 
 4. Add together ^^^^J^-^- Ans. U t p. 
 
 5. Add I of i of |, to 6| of i, and f of je , together. 
 
 ylns. 4JL. 
 
 I 8 
 
 6. Add I of a rod to | of a foot. Ans. 13^ feet. 
 
 7. Add i of a mile, ■§ of a furlong, and |- of a rod together. 
 
 Ans. 307|. rods. 
 
 8. Add ^- of 3^, _?_. of 2^, -?_ of U, _i_. of 2|, and !)_ of 3^, 
 together. ^W5. I. 
 
 [This last example is very simple and brief.] 
 
 9. What is the sum of ^ of a pound and |^ of a shilling ? 
 
 ^7W. 'i|5=13s. lOd. 2|q. 
 
 10. What is the sum of § of a yard and ■§ of a quarter ? 
 
 Ans. I_i_ yards. 
 
 11. What is the sum of 2. of a ton, added to ^ of a hundred 
 
 3 / 
 
 weight? A'^is. 123 cwt. 
 
SUBTRACTION OF FRACTIONS. 9' 
 
 12. What is the sum of g of a day added to ^ of an hour ! 
 
 Ana. 9t, hours. 
 
 13. What is the sum of f oi a pound and ^ of a shilling] 
 
 Ans. 3s. 2d. 
 
 14. What is the sum of 1 of a week, | of a day, and i of an 
 hour? Ans. 1 d. 22 h. 15 m. 
 
 SUBTRACTION OF VULGAR FRACTIONS. 
 
 (Art. 53.) We would remind the pupil that, in addition, we took 
 the sum of the numerators, after the fractions were reduced to a com- 
 mon denominator. Hence, the difference of the two fractions must 
 1)3 found by taking the difference of their numerators, when the deno- 
 minators are ahke. For example, the difference between _"_ and _?_, 
 must be -? _ i and the difference between f and f niusl be f, 6lc. 
 These observations must give us the followmg 
 
 RuLK. Prepare ihe fractions as for addition, and place the dif- 
 ference of the numerators over ihe common denominator, fir the dif- 
 ference tf the fractions. 
 
 EXAMPLES. 
 
 1. Find the difference of t and J- . Ans. s 
 
 12 I -i 
 
 2. From # take 1. Ans. _«_. 
 
 * 7 £ 8 
 
 3. From 6| yards take 3| yards. 
 
 Rules would reduce these numbers to improper fractions, and then 
 to a conjmon denominator, thus: ^J and 3J*, difference -J=z'2,^. 
 But, a little tact, which should always be cultivated, will do it more 
 elegantly, thus: 6|=5'Jj and 34=31; hence, 2^ is their dilier- 
 ence. In like manner perform the two following : 
 
 4. From 9L take Q^_. Ans. 2J_. 
 
 5 10 10 
 
 5. From I9i vards take 12_?_ yards. Am. 6L?.. 
 
 6. What is the difference between f of b\ and | of 4i. 
 
 Ans. 1-2. 
 
 ■) 
 
 7. What is the difference betv/een ^ of a rod and | of a foot ? 
 
 Ans. 1^ feet. 
 
 8. From ^ of a yard take f of a quarter. Ans. ^ quarter. 
 
 9. From i. of a pound take -f of ^ of a shilling. 
 
 " Ans. £i_i=10s. 7d. Uq. 
 
 3 6 ^ ^ 
 
 10. From f of a yard take J of a quarter. 
 
 Ans. 'J'=l qr. 3 na. 
 
98 ARITHMETIC. 
 
 11. From 7 ells English take 4J yards. 
 
 Ans. 32 ells English. 
 
 12. From 1 of a pound sterling take | of a penny, 
 
 / * Ans. 2s. iLld. 
 
 13. From 1 of a rod take -f of a foot. Aiis. 3J nr 10^ feet. 
 
 14. From 1 of an ounce take -| of a pennyweight. 
 
 A7if!. ''_5j» dwt., or 16LL dwt. 
 
 2 3 ' 2 S 
 
 MULTIPLICATION OF FRACTIONS. 
 
 (Aht. 54.) Multiplication is the same in its nature, whatever may 
 be the value of the numbers multiplied together. 
 
 When we consider two terms or factors to be multiplied together, 
 a multiplicand and a mulliplier, the multiplicand must be taken as 
 many times as there are units in the multiplier. When there is not 
 a unit in the multiplier, then we are to multiply by a fraction, and it 
 is taking the multij)licand part of one time, and, of course, the product 
 will be less than the multiplicand. 
 
 For example : Let it be required to multiply 2. by 2. Here 1 
 is to be considered the multiplier, which is twice i. Now, we must, 
 obviously, take ^^ 4, of 1 time, — that is, divide it by 7 ; which is, by 
 Article 44, J-. But, we want |, twice as much as 1 : hence, JL. 
 must be the required product. And this we should have obtained by 
 the following 
 
 Rule. To multiply one fraction by another. Multiply the nu- 
 merators together for the numerator of the required fraction, and 
 the denominators together for the denominator ; always canceling, 
 when possible. 
 
 N. B. Mixed numbers must be py-eviously reduced to improper 
 fractions. 
 
 Another explanation : 
 
 1. Multiply I by |. Ans. /^==y\. 
 
 3X2 6 
 Here '-— = — • But, the multiplier is only the l part of 2 ; con- 
 4 4 * 
 
 sequently, i must be diminished corre-spondently, by multiplying its 
 
 denominator by 5. 
 
 2. Multiply 2^ by I of I of 8. 
 
 5X3X2_X8^X2_e^^ 
 2X4X5X1 1 
 
MULTIPLICATION OF FUACTIONS. 09 
 
 
 An.s. 4^. 
 
 f by 7i by ^^„ 
 
 by 7. 
 
 
 Atis. 23. 
 
 , and 4_''_. 
 
 ^n.5. 2J_. 
 
 J 
 
 1. Multiply 5 by -f of ^. ./1/J5. 1. 
 
 2. Multiply J*-, by ^, and that product by 7^-. 
 
 3. Required the product of 21 by 
 
 4. Required the product of 1 
 
 5. Multiply together 5, f, 1 of li, and 41. yl/?.s. sj' =71. 
 G. Multiply fi of 3 times _9_ of 5 times ^ of ^ of i, by 1 of i 
 
 of A of J . 7lrt5. g^. 
 
 7. What will 8^ pounds of tea cost as 1^^ dollars per pound 1 
 
 A71S. 1011 dollars. 
 1 6 
 
 8. What will 4| cords of wood cost, at 3| dollars per cord ? 
 
 Ans. 1713 dollar?. 
 
 1 6 
 
 9. If a horseman travel at the rate of 7^ miles an hour, for 6_"_ 
 hours, how far will he have gone? Ans. 51 -U miles. 
 
 10. What is the cost of 9^ yards of linen, at $lf per yard ? 
 
 Ans. $12.15. 
 
 ^ 6 4 
 
 DIVISION OF FRACTIONS. 
 
 (AuT. .5.5.) The essential nature of division is the same, whether 
 the dividend and divisor are whole numbers or fractions. It is, sim- 
 ply, finding how many times, or what part of a time, one num- 
 ber is contained in another. 
 
 For example: We may demand that 6 shall be divided by j>, and 
 an inattentive pupil might suppose that we required tlie half of 6 ; 
 but this is not true. We demand how many ti'iies ^ is contained in 
 6: or, we might say. Divide 6 into equal parts, and have each part 
 equal to {j ; how many parts will there be ] Ans. 12. Divide 6 by 
 ^. Am. 18. Divide 6 by J_. Ans. 60. Divide 6 by |. Ans. 42. 
 
 Divide 6 by 2. Ans. As 1 is contained in 6 42 times, 2 will be con- 
 
 •'7 7 7 
 
 tained 21 times. That is. We multiply the dividend by the denomi- 
 nator of the divisor, and divide the product by the numerator. 
 
 When the dividend is a fraction, the result may be obtained by the 
 following 
 
 Rule. To divide by a fraction, invert the divisor, and proceed 
 as in muUiplicutidn. 
 
 N. B. When the divisor is a compound fraction^ invert every 
 ierrii ,• and, in all cases, caticel as much as possible. 
 
100 ARITHMETIC. 
 
 EXAMPLES. 
 
 
 
 
 3 
 
 
 21,3 
 Form, — ^— 
 
 Canceled 
 
 #,3 
 
 9. 
 
 
 
 
 
 ^??s. 
 
 48. 
 
 
 
 
 
 u47lS. 
 
 117. 
 
 
 
 
 
 -4/25 
 
 49. 
 
 
 
 
 
 ^ns 
 
 .99. 
 
 I. Form -^ 
 5 4. 
 
 4. 
 
 4. 
 
 5. 
 
 '3: 
 
 
 
 1. Divide 21 by |. 
 
 2. Divide 18 by f. 
 
 3. Divide 63 by _7_. 
 
 4. Divide 42 by ^. 
 
 5. Divide 66 by |. 
 
 6. Divide f of ± by i of I 
 
 Canceled ■ ^' ^' — ; therefore, 4 is the answer required. 
 ^. A- 1- ^• 
 
 7. Divide 4 of _«_ of ^, by i of 6 of -"_. yl"5. 4. 
 
 8. Divide ]^^ of I of 1.^ of 1-f of |8 of i, by ^ of i of _P_ of 4- 
 
 Ans. m=l2J-. 
 4 1 3 4 7 o 
 
 9. Divide 4^ by i of 64. Ans. _'/-,.. 
 10. Divide 6| by | of 42. A«5. -jP_. 
 
 .11. Divide f of I by 4 of 6^. Ans. ^^-. 
 
 (Art. 56.) Fractions must refer to the same scale of weight or 
 measure, before we can divide one by the other. For instance, we 
 cannot divide the fraction of a gallon by the fraction of a pint, or by 
 the fraction of a gill, before we reduce one to the denomination of the 
 other, for the same reason that we cannot add and subtract unlike 
 quantities, (Art. 5.) 
 
 EXAMPLES. 
 
 1. How many times is •§ of a pint contained in 1 of a gallon 1 
 
 Ans. 6^. 
 
 2. How many times can a vessel holding _P_. of a quart, be tilled 
 from 5 of a barrel, containing 31^ gallons ? Ans. 4G|. 
 
 3. How many rods are there in 60^ yards ? Ans. 1 1. 
 * 4. How many times is | of an inch contained in 1 of a yard ? 
 
 Ans. 3i2. 
 5. How many times is 1 of a rod contained in ^ of a foot ? 
 
 A71S. J-. 
 
 » 9 
 
PRACTICAL APPLICATION OF FRACTIONS. 101 
 
 PRACTICAL APPLICATION OF FRACTIOxNS. 
 
 1. What will 51 yards of calico cost, at 31^ cents per yard ? 
 
 Alt.'. '^pl.CA-{- 
 
 2. What will S^ yards of linen cost, at G2^ cents per yard ? 
 
 3. What will 5-} yards of linen cost, at Sl:^- cents per yard ? 
 
 An.v. S4.67-I- 
 
 4. Bought a yard of cloth for ^ of a dollar ; what would be the 
 cost of 8 yards, at the same rate ? Ans. $6^. 
 
 5. If I pay I of a dollar for a gallon of oil, what will I85 gallons 
 cost? ' Ans.^i^^. 
 
 6. What would Jl of an acre of land cost, if ^ of an acre cost 5^27. 
 
 Ans. $48. 
 
 7. Bought f of a fifteen-acre lot, and sold | of what I pur- 
 chased; how much did I sell? Ans. 4^ acres. 
 
 8. What would 395 pounds of sugar cost, at ? of ^ of a dollar 
 per pound? " Ans. $52^. 
 
 9. Add i of a pound sterling to ^ of ^ of a shilling. 
 
 Ans. 2s. 10|d. 
 
 10. Add I of a mile, ^ of a furlong, and J5_ of a rod together. 
 
 Ans. 233^ roJs. 
 IL Add 1 of a hogshead to ^ of a gallon. 
 
 Ans. 54-^- gallons. 
 
 12. What is the value of » of a day? Aiis. 10 h. 17^ m. 
 
 13. What is the value of 1 of a dollar? Ans. 421 cts. 
 
 14. Reduce | of |. of a pound, avoirdupois, to integers. 
 
 Ans. 4 oz. 11|3 dr. 
 
 15. From J- of a ton take f of a hundred weight. 
 
 A71S. -\ cwt. 
 
 16. If ^ of a dollar pay for 3| pounds of sugar, how many cents 
 is it per pound ? Ans. 10. 
 
 17. If $20J pay for 16^ barrels of apples, what is the cost per 
 barrel ? Ans. $1 ,26, nearly. 
 
 18. If ^ of 2 of a farm cost 1200 dollars, what would the whole 
 cost 1 Ans. §7200. 
 
 19. If 6 pounds of tea cost 9^ dollars, what is it per pound ? 
 
 Ans. Sl|. 
 
 20. If 5^ yards of cotton cloth cost § of a dollar, how many cents 
 is it per yard ? Ans. 15_5_. 
 
 21. If ^ of a dollar pay for 1| yards of cloth, what is the price 
 per ell English. Ans. 79_.fi_. 
 
 22. If a piece of cloth measuring 13f yards, cost 27^ dollars, what 
 is the value of 1 yard 1 Ans. |2_2_. 
 
102 arithmp:tic. 
 
 23. If 7 horses consume 2| tons of hay, how much does each con- 
 sume ? Ans. 11. 
 
 24. What part of a rod is 2^ yards 1 A?is. _5_. 
 
 25. Find the difference between 3 of an ell English, and t of 3 
 feet? Ans. IJ_. 
 
 5 6 
 
 Questions. 
 
 How do you judge of the value of a fraction ? 
 
 What numbers do you compare in making up your judgment? 
 
 ]f you double a denominator, what effect does it have upon the 
 fraction ? 
 
 If you double a numerator, what effect does it have? 
 
 If you double both numerator and denominator, what effect ? 
 
 If you add the same number to both numerator and denominator of 
 a fraction, does it change its value? 
 
 How do you reduce complex fractions to simple ones ] 
 
 DECIMAL FRACTIONSr 
 
 (Art. 57.) The common or vulgar division of things into halves, 
 quarters, thirds, &c., &c., so convenient and proper in the every-day 
 little concerns of business life, becomes troublesome and perplexing 
 when we have a variety of these different denominations to add or 
 subtract in large mathematical calculations; and, to surmount this 
 difficulty, we have a more artificial division of things into tenths, those 
 parts again into tenths, and again, if need be, these svibdivisions into 
 tenths, &c., Sec. ; thus having a decreasing order of fractions on the 
 same graded scale as natural numbers, ten of one.order making one 
 of another ; and, of course, we may call them decimals', from deci- 
 mils, the Latin word meaning tenths. From this it is evident tliat, if 
 we divide a tenth into ten equal subdivisions, one of them will be one- 
 hundredth of the unit, &c., &c. ; hence, decimal fractions can have 
 no other denominators than 10, 100, 1000, &c., &c. This being 
 systematized and properly understood, the denominators need not be 
 written. If the fraction is tenths, one place from the unit wdl ex- 
 press it, with a point between ; if hundreds, or two orders from the 
 unit, two places of figures must be taken : thus, _i_ is expressed .05, 
 _^_-_ thus, .005, always as majiy places as the understood denomi- 
 naior contains ciphers. 
 
 Thus, 7_s'_ is written in decimals, .... 7.5. 
 
 4tVo' 4.24. 
 
 I^Vt/WV' 141-1356. 
 
DECIMAL FRACTIONS. 108 
 
 Ciphers written niter a decimal neither increase nor decrease it, for 
 this does not change the place of the figures. 'I'hus, 2^'^ is 2.6, and 
 2.60000 is the same value — the 6 still being in the same orc/e?- of 
 numbers; but, ciphers on the other side, as 2.06, changes the orc?er, 
 and diminishes the value ten-fold for every remove. 
 
 The following shows the scale : 
 
 ■±3 « -r: 
 
 P 
 
 J3 '^ 
 
 C 
 
 llllllllJllia 
 
 333333 3, 333333 
 
 Express the following in figures. • 
 
 1. Sixty -six and seven-tenths. Ans. 66.7. 
 
 2. One and thirty-five-hundredths. Ank: 1.35. 
 
 3. Eighty-one and 1 ten-thousandth. ^ws. 81.0001. 
 
 4. One hundred and 67 ten-thousandths. Ans. 100 0067. 
 
 5. Three and three-hundredihs. Ans. .S.03. 
 
 6. Seventy-five and 75 ten-thousandths. Atis. 7f».C075. 
 Let it be observed, that these fractions vary in value on the same 
 
 scale as whole numbers ; and the denominations of Federal Money 
 are nothing more than whole numbers and decimals, the (lollars being 
 units, and the cents, mills, &c., being decimals of a dollar, as men- 
 tioned in Article 8. 
 
 To familiarize decimals to Federal Money, let the pupil make the 
 following reductions : 
 
 1. Reduce §41.7 dimes and 6 mills, to mills. ^1;?.*. 4I70G. 
 
 2. Reduce 31806 mills to dollars, cents, and mills. 
 
 Am. 31.806. 
 
 3. Reduce $10, 3 dimes 4 cents and 9 mills to mills. 
 
 A)is. 10349. 
 
 4. Reduce 12349 mills to dollars, cents and mills. 
 
 J»5. $12,349. 
 
 5. Reduce $25 to cents Ans. 2500. 
 
 ADDITION OF DECIMALS. 
 
 (Art. 58.) We again call to mind the fact mentioned in Article 
 5, that only like things can be added together. In whole numbers, 
 
104 ARITHMETIC. 
 
 units can only be added to units, tens to tens, &c. So in decimals : 
 tenths can only be added to tenths, &c. Hence, we have the follow- 
 ing rule for addition in decimals: 
 
 Rule. Set down the numbers to be added so that tenths shall 
 fall under tenths, hundredths under hundredths, 6(c. Tliis will 
 brins; all the decimal points directly under each other ,- add up as in 
 wh(jle numbers, and place a separatrix between the units and 
 decimal. 
 
 We shall give but very few examples under this rule, as the opera- 
 tion, (except the case in placing the decimal points under each other,) 
 is the same as in addition of whole numbers. 
 
 EXAMPLES. 
 
 Add 37.03, 621.57, .521, 20.007. 
 
 (2) 
 
 Operation. Operation. 
 
 37.03 .199 
 
 621.57 2.7569 
 
 .521 ,25 
 
 20.007 .654 
 
 Sum, 679.128 Sum, 3.8599 
 
 3. What is the sum of two-tenths, two-hundredths, and two-thou- 
 sandlhs ? Ans. .222. 
 
 4. What is the sum of 14, and 16 ten-lhousandths? 
 
 Ans. 14.0016. 
 
 5. What is the sum of twenty-six and twenty-six hundredths, sev- 
 en-tenths, and six and seventy-three hundredths, four and four thou- 
 sandths ] Ans. 37.694. 
 
 6. What is the sum of 9 dollars 3 mills, 14 dollars 3 dimes 9 cents 
 1 mill, 104 dollars 9 dimes 9 cents 9 mills, 999 dollars 9 dimes 1 mill, 
 4 mills, 6 mills, and I mill 1 ylns. §1 1"28..^05. 
 
 7. What is the sum of 4 dollars 21 cents, 87^ cents, 3 dollars 6^ 
 cents, 2 dollars 2 cents, 5 dollars 372 cents, and 37 dollars 7 cents 1 
 
 Ans. $52.16i. 
 
 SUBTRACTION OF DECIMALS. 
 
 ( Aut. 59.) From the very nature of these numbers, as previously 
 explained, we must have the following 
 
 Rule. Place the numbers as in addition of decimals, subtract as 
 in whole numbers, and place the decimal point in the remainder, 
 under the decimal points above. 
 
DECIMAL FRACTIONS. 105 
 
 EXAMPLES. 
 
 1. From 91.73 take 2.138. 
 
 91.73 (2) 2.73 
 
 2,133 1.9185 
 
 /l/i5. 89.592 Ajis. .8115 
 
 3. Find the (Utrerence between 714 and .916. Ans. 713.0S4. 
 
 4. From .145 take .09GS4. Ans. .04816. 
 
 5. How much greater is 2 than ,298? Ans. 1.702. 
 
 6. What is the ditierence between 75 dollars and 75 cents ? 
 
 7. From 21.004 take 75 hundredths. Ans. 20.254. 
 
 8. From 260.4709 take 154.0578. yln.s6,4131. 
 
 9. From 10.0302 take 2 ten-thousandths. Ans. 10.03. 
 10. From 2.01 take 99 hundredths. Ans. 1.02. 
 
 MULTIPLICATION OF DECIMALS. 
 
 (.Art. 60.) Vie propose now to investigate a rule for the multi- 
 plication of decimals; and as these numbers vary as simple num- 
 bers, ten of one order making one of a superior order, the actual mul- 
 tiplication of the figures must be the same as taught under simple 
 multiplication : it only remains to point out the value of the product. 
 
 Let us multiply .3 by .5. To understand the operation, make vul- 
 gar fractions of them, and we have J^- to multiply by ^-. The 
 product (Art, 54) is JJL ; or, by the definition of decimals, .15. 
 
 Again: Multiply .oVby .13, or _^^_ by _y_. Product, __s_2__, 
 or, by the definition of decimals, .0052. 
 
 Psow, we perceive that every place in a decimal has a cipher cor- 
 responding in a denominator understood ; and the multiplication of 
 these denominators make as many [)laces as there are ciphers (Art. 
 15) ; that is, as many as there are places of decimals in both the fac- 
 tors. Hence, we have the following 
 
 RuLF.. Multiply as in simple niimhers, and point off in the pro- 
 duct, from the right hand, as many figures fur decimals as there 
 are decimal places in both factors ; and if there be not so many in 
 the product, supply the deficiency by prefixing ciphers. 
 
 Multiply ,02345 by .00163. 
 
 .02345 
 .00163 
 
 7035 
 14070 
 23 i 5 
 
 .0000382235 Product. 
 
106 ARITHMETIC. 
 
 2. Multiply 54.25 by 2.2S02. Ans. 123.70536. 
 
 3. Multiply 15.278 by .7854. Ans. 12, iienilv. 
 
 4. Multiply 79.317 by 23.15, Ajis. 1836.88305. 
 
 5. Multiply 350 by .7853. Ans. 274.855. 
 fi. Multiply 20 by .55. Ans. \l. 
 
 7. Multiply one-tenth by one-tenth. A7is. ,01. 
 
 8. Multiply 25 by twenty-live hundredths, Ajis. 6.25. 
 
 9. At 4^ mills apiece, what will 400 quills cost ? Aiis. ^1.80. 
 
 To multiply decimals by 10, 100, 1000, Sfc, remove the decimal 
 paint as many places to ike right as there are ciphers in the multi- 
 plier, agreeably to Article 13. Removing the decimal point one fig- 
 ure to the right increases every figure ten-fold, &c. 
 
 10. Multiply 4.56 by 10. Ans. 45.6. 
 
 11. Multiply §1.75 by 100. Ans. $175. 
 
 12. Multiply .006 by 1000. Afis. 6. 
 
 13. A benevolent person, whose income was $6000, gave .12 of it 
 for charitable purposes ; what did he give away ? Ans. $720. 
 
 14. The capital stock of a bank was | of a million of dollars, and 
 _3_ of it was owned equally by 4 individuals ; how much was owned 
 by each ? Ans. $22500. 
 
 DIVISION OF DECIMALS. 
 
 (AiiT. 61.) Division being the reverse of multiplication, and 
 it having been shown, in multiplication of decimals, that the product 
 of any two factors must contain as many decimal places as the num- 
 ber of decimals in both factors : now, the divisor and quotient, in divi- 
 sion, may be considered as factors of the dividend ; hence, the number of 
 decimal places in the quotient must be equal to the difference of the 
 number of places in the divisor, and the number made use of in the 
 dividend. 
 
 If the dividend really have no decimals, make a decimal point at 
 the right of the unit, and fill up the blank decimal places with ciphers, 
 as many as you wish. In dividing, the decimal places may run out ; 
 we may then annex ciphers and count them as decimals. If the dividend 
 has decimals, but not so many as the divisor, fill up blank places, as 
 before, which will not affect the value of the decimal, (Art. 57). 
 From the above explanations, the reason of the following rule 
 must be evident: 
 
 Rule. Divide as in whole numbers, and poiyit off from the right 
 of the quotient as many places for decimals as the decimal places in 
 the dividend exceed those in the divisor. 
 

 DECIMAL FRACTIONS. 
 
 
 107 1 
 
 1. 
 
 H 
 
 2de( 
 more 
 made 
 two 
 
 2 
 
 3. 
 
 4. 
 
 5. 
 
 fi. 
 
 7. 
 
 EXAMPLES. 
 
 Divide 42.42 by 2.113. 
 
 Operation. 
 2.113)42.4200(20.07 
 42.26 
 
 ICOOO 
 14791 
 
 ^ere were, originally, 
 
 oration we used three 
 
 s with ciphers. This 
 
 taken from it, leaves 
 
 Quo^.O.llI. 
 
 Quot. 11 1. 
 
 Quot. 5.4232. 
 
 Quot. 54.232. 
 
 Quot. 16.278. 
 
 30 by .03. 
 
 ere we rest, and count off the quotient. T 
 :imal places in the dividend, and in the op( 
 ; that is, filled up three blank decimal place 
 } five, and three, the number in the divisor 
 decimal places in the quotient. 
 
 Divide 2.3421 by 21.1. 
 
 Divide 2.3421 by .211. 
 
 Divide .8297592 by .153. 
 
 Divide 8.297592 by .153. 
 
 Divide 12 by .7854. 
 
 Divide 3 by 3 ; divide 3 by .3 ; 3 by .03 ; 
 
 (Art. 62.) The preceding rule for division of decimals is as clear 
 and as explicit as any mere ruJe can be : but, so careless are pupils 
 in general, about the decimal point, that little or no reliance can be 
 placed on the value of their quotients, until they are taught to use 
 iheir judgments in each and every case, independent of any rule. 
 To call into exercise this individual reliance, is the object of the fol- 
 lowing remarks and explanations. 
 
 To divide — to cut into parts — will not, at all times, give a clear 
 understanding of division, and confusion frequently arises from taking 
 this view of the subject; we better consider it as one number mea- 
 suring another. For example : How often will .5 of a foot measure 
 12 feet? In other words, divide 12 by .5, or divide 12 by ^. 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 
 cai^e, into 5 parts, because the 5 is .5 : or, he may consider that ^ of 
 a font 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 
 fill cases, the divisor and dividend must be of the same denomination, 
 before the division can be effected. But, in decimals, those rcductitms 
 are made so easily that a thoughtless operator rarely perceives 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. 
 
108 ARITHMETIC. 
 
 EXAMPLES. 
 
 DiviJe 15..34 by 2.7. Here, we consider the whole number, 1.5, is 
 to be divided by less than 3 : the quotient must, therefore, be a Httle 
 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 than | of 1 ; therefore, the quotient must be 
 more than 3 times 15. Or, we may multiply both dividend and di- 
 visor by 100, which will not aflect the quotient, and then we shall 
 have 1534 to be divided by 27. Now, no one can mistake 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 divide 45300 by 15, a common 
 example in whole numbers. 
 
 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 
 upon the formality of placing a cipher in the dividend, to represent 
 tlie place of whole numbers, thus : 
 
 625.5)0.0037530( 
 
 V/e now consider whether the whole number in the divisor will be 
 contained in the whole number in the dividend, and we find it will 
 not : we, therefore, write a cipher in the quotient, to represent the 
 place of whole numbers, and make the decimal on the right, thus : 0. 
 
 We now consider that 625 will not go in the ID'S, nor in the lOO'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 
 vieiv 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. 
 
 1 1 in 0, no times ; 1 1 in 5 tenths, no times ; 1 1 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 
 
DIVISION OF DKCIMALS. 109 
 
 then find where it should be, by counting, &c., — ;i rule 
 against which we have notliing to say; but good jutig- 
 ment, at ready command, is far superior to any rule that 
 can be formed. 
 
 EXAMPLES* 
 
 1. Divide 9 by 450. ^ns. 0,02. 
 
 2. Divide 2,39015 by ,007. J9ns. 341,45. 
 
 3. Divide 100 by ,25. ^ns. 400,00. 
 
 4. If 350 pounds of beef cost $12,25, what is the cost 
 of one pound ? *^ns. ,035. 
 
 5. At $5,75 per yard, how much cloth can be pur- 
 chased with $19,40625? ^ns. 3,375 yards. 
 ^- 6. At 7 per cent., how much capital must be invested 
 to yield 602 dollars ? ^ns. $8600. 
 
 7. If a contribution, amounting to $36,72, be made by 
 a congregation consisting of 918 persons, how much is it 
 a-piece? ^ns. ,04, or 4 cents. 
 
 8. If 275 lemons cost $2,475, how much is it a-piece? 
 
 ^ns. 9 mills. 
 
 9. A benevolent individual gave away $600 per annum 
 to charitable objects, which was ,12 of his income ; what 
 was his income ? 
 
 (Art. 63.) By paying some litde attention to the rela- 
 tion of numlDcrs, we may abbreviate certain divisions, in 
 a similar manner as we abbreviated multiplication, m 
 Art. 21. 
 
 To divide by ,25 is the same as to multiply by 4. 
 " " by ,5 is the same as to multiply by 2. 
 " " by ,75 is the same as to multiply by 4, and 
 divide by 3. 
 
 To divide by ,125 is the same as to multiply by 8. 
 " " by ,333|- is the same as to multiply by 3. 
 " '« by ,666| is the same as to multiply by 3, 
 and divide by 2, Slc. &c., for any aliquot part of unity. 
 The same is true of aliquot parts of 10, 100, 1000, &c. 
 
 EXAMPLES. 
 
 1. Divide 6 by ,25. ^ns. 24. 
 
 2. Divide 6 by 25. - Jns. ,24. 
 
 3. Divide 15 by 16|,=^ of 100. ^ns. ,9. 
 
110 ARITHMETIC. 
 
 4. Divide 15 by ,16|. ^ns. 90. 
 
 5. Divide 4,27 by ,33^. .^ns. 12,81. 
 
 REDUCTIOIN OF DECIMALS. -.-_ 
 
 (Art. 64.) To reduce a vulgar fraction to its equiva- 
 lent decimal : the value of any fraction is found by di- 
 viding the numerator by the denominator; thus, |=2. 
 But in a proper fraction, we cannot divide the numerator 
 without first reducing it. Annexing one cipher, makes it 
 tenths ; two ciphers, hundredths ; and so on. The quo- 
 tient, therefore, will be decimal parts. Thus : reduce | 
 to a decimal. 
 
 8)1,000 
 
 ,125 
 
 We may take this view of the case : Multiply both 
 numerator and denominator, by 10, 100, 1000, &c. ; 
 which will not change the value of the fraction, (Art. 35.) 
 We then have |^^^. Now divide both numerator and 
 denominator by 8, and we have yVVo' which is a deci- 
 mal fraction, by the definition of decimals. Hence, 
 whichever view we take, we have the following 
 
 Rule. Place a decimal point at the right of the nu- 
 merator, annex ciphers, and divide by the denominator. 
 Compound fractions must first be reduced to simple 
 fractions. 
 
 EXAMPLES. 
 
 1. Reduce \, \, and |, to equivalent decimals. 
 
 Jins. ,25, ,5, ,75. 
 
 2. What decimal is equivalent to |? Ans. ,625. 
 
 3. What decimal is equivalent to | of | ? Ans. ,4. 
 
 4. Reduce |f to a decimal. Ans. ,9375. 
 
 5. Reduce y^'jij to a decimal. Ans. ,0008. 
 
 6. Reduce jf to a decimal. Ans. ,7058 + 
 
 There are many vulgar fractions that cannot be exactly 
 expressed in decimals. The following are some of them : 
 
 2 1 2 4. A _1_ _2_ Stjf, 
 
REDUCTION OF DECIMALS. HI 
 
 Reduce I to a decimal. Ans. ,3333, <fec. This is 
 called a repeating, or infinite decimal. 
 
 Reduce /y to a decimal. Ans. 181818, &c. This is 
 called a circulating decimal, and the repeated figures are 
 called a compoinid rej)etend. Compound repetends may- 
 consist of three or more figures — as, ,123 123, &c. ; or, 
 ,134 91349, &c. 
 
 Reduce ^"3 to a decimal. Ans. ,121212 -|- 
 
 (Art. 65.) To reduce a decimal to a vulgar fraction. 
 
 Rule. Write the dejiominator under the decimal, 
 and reduce to its lowest terms. 
 This rule requires no comment. 
 
 EXAMPLES. 
 
 1. Reduce ,75 to a vulgar fraction. Ans. //^ = |. 
 
 2. Reduce ,125 to a vulgar fraction. Ans. \. 
 
 3. Reduce ,9375 to a vulgar fraction. Ans. i-f. 
 
 4. Reduce ,16 to a vulgar fraction. Ans. ~j. 
 
 5. Reduce ,15 to a vulgar fraction. Ans. t,\. 
 
 6. Reduce ,124 to a vulgar fraction. Ans. ^\. 
 
 7. Reduce ,655 to a vulgar fraction. Aiis. ^fi- 
 
 (Art. 66.) To reduce a repetend, or circulating deci- 
 mal, into its equivalent vulgar fraction. By a little in- 
 spection of the fractions expressed in Art. 64, we per- 
 ceive that the denominators of fractions which make re- 
 petends, are 9, 11, or factors of 9, 11. Hence this 
 
 Rule. Make the repetend the num,erator, and for the 
 denominator take as many 9's as there are figures in 
 the repetend. 
 
 EXAMPLES. 
 
 1. Reduce ,6 to a vulgar fraction. Ans. |. 
 
 2. Reduce 12.3 to a vulgar fraction. Ans. ~^^. 
 
 3. Reduce ,18, &c. to a vulgar fraction. 
 
 Ans. U--=j\. 
 
 We shall resume this subject again, when we treat of 
 geometrical progression. 
 
112 ARITHMETIC. 
 
 (Art. 67.) To find the value of a decimal in known 
 parts, of integers of inferior denominations. 
 
 Rule. To reduce a decimal from a higher to a lower 
 denominaiion : we reason and mnlliply just as we 
 would if the sum to be reduced was a whole number, in 
 place of a decimal^ only observing the decimal point. 
 
 EXAMPLES. 
 
 What is the value, in shillings, pence, &:c., of ^66? or 
 any other number of pounds ? We should certainly mul- 
 tiply by 20, 12. &c.; and whatever we do to one or more 
 pounds, we should do to apart of a pound. 
 
 1. What is the value of ,6237 of a pound sterling, ex- 
 pressed in lower denominations ? 
 ,6237 
 20 
 
 12,4740= the shillings and decimals of a shilling, in the 
 12 decimal ,6237 of a pound. 
 
 5,6880 =the pence and decimals of a penny, in the de- 
 4 cimal ,4740 of a shilling. 
 
 2,7520 =the farthings and decimals of a farthing in the 
 decimal ,6880 of a penny. 
 The integers on the left of the decimal points, are the 
 numbers that compose the answer. 
 
 ^ns. 12 s. 5 d. 2} farthings. 
 
 2. What is the value of ,725 of a £ ? 
 
 ^ns. 15 s. 6d. 
 
 3. What is the value of ,8285 of an acre ? 
 
 ^^ns. 3 roods, 12,56 poles. 
 
 4. What is the value of ,432 of a hundred weight ? 
 
 .^ns. Iqr. 20 lb. 6oz. + 
 
 5. What is the value of ,234 of a day ? 
 
 ^7is. 5 h. 36 m. 57y\ sec. 
 
 6. How many furlongs, &:c., in ,784 of a mile ? 
 
 .^ns. 6 fur. 10 rods, 4 yds. 2 ft. -|- 
 
 7. How many quarters, &.C., in ,936 of an ell Eng- 
 lish ? ^7is. 4 qr. 2 n. + 
 
REDUCTION OF DECIMALS. 113 
 
 (Art. 08.) The reverse of Art. 67. ' To reduce low- 
 er denominations to decimals of a higher. As higher 
 is reduced to lower by midtip/icafiou, lower must be re- 
 duced to higher by division, beginning with the lowest 
 denomination. These observations explain the following 
 
 Rule. Divide each denomination, beginning with 
 the lowei^t, by such a number as will bring it to the next 
 superior denomination, and set the quotient on the right 
 of that denomination, with a decimal point betiveen 
 them. Then divide the next higher, together with its 
 decimal, by the number that will bring it to its next 
 higher denomination, in the same manner as before^ 
 and so on as far as necessary. 
 
 EXAMPLES. 
 
 1. What part of a hundred weight is 3 qr. 14 lb. ? 
 
 Jlns. ,875. 
 3Iore circuitously. 281 14, 
 
 First, (by Art. 46) reduce to a vulgar fraction, 4 1 3,5 
 
 0,875 
 Then (by Art. 64) the vulgar to a decimal, thus : 
 
 3 qr. 14 lb. =98 pounds. Fraction -' « - 875 
 
 1 cwt. = 112 pounds. i-iaction, tt2-»«75. 
 
 2. Reduce 2 quarters 3 nails to the decimal of a yard. 
 
 Ans. ,6875 yd. 
 
 3. What part of a bushel is 3 pecks 1,12 quarts ? 
 
 Ans. ,785 bu. 
 
 4. What part of 1 hogshead is 1,2 pints? 
 
 Ans. ,00238 hhd. 
 
 5. Reduce 5 furlongs, 12 poles, to the decimal of a mile. 
 
 Ans. ,6625. 
 
 6. Reduce 55 minutes, 37 seconds, to the decimal of 
 a day. Ans. ,03862 -|- 
 
 7. Reduce ,32 of a pint to the decimal of a bushel. 
 
 Ans. 0,005. 
 
 8. Reduce 10 ounces, 13 pennyweights, 9 grains, to the 
 decimal of a pound Troy. Ans. ,8890625. 
 
 If more examples are desired, reverse those under 
 Art. 67. 
 
 k2 
 
114 ARITHMETIC. 
 
 Astronomical Problems. 
 The mean length of a tropical year is 365 days, 5 hours, 
 48 minutes, 49 seconds ; what decimal of a day over 365 ? 
 
 Ans. ,24223 + 
 The tropical revolution of the moon is 27 days, 7 hours, 
 43 minutes, 3 seconds; how many days, and decimals 
 of a day? e^ns. 27,321566+ 
 
 MISCELLANEOUS EXAMPLES, 
 
 Applicable to the various rules infractions, both vulgar 
 and decimal. 
 
 1. Which is greatest, V|, or f | ? 
 
 2. Which is most, | of a rod, or | of 12 feet? 
 
 3. If the dividend be ^, the quotient 320, what is the 
 divisor ? 
 
 4. The divisor being \, the quotient y\ ; what is the 
 dividend ? 
 
 5. The divisor being 4520, the dividend | ; what is 
 the quotient? 
 
 6. The Ci\ isor to a number is ,02, the quotient is 420; 
 what is the number ? 
 
 7. What is the difference between f of £T, and i- of 
 a shilling? 
 
 8. What is the sum of 19 and 5 hundredths ; 120 and 
 9 thousandths ; 36 ten-thousandths ; and 406 and eight- 
 millionths ? 
 
 9. From yf^^ take y^'/o u' and divide the difference 
 by ,002. ^'ins. 37,7. 
 
 10. The sum of $120 36 was to be paid by a certain 
 number of persons, and it amounted to 3 dollars 9 mills 
 a-piece ; how many were there of them ? Ans. 40. 
 
 11. Add I of a yard to -^ of a foot. Ans. 1 yd. 
 
 12. Add f of a day to | of an hour. 
 
 Ans. 16 h. 40 m. 
 
 13. Add I of a week, ^ of a day, and \ of an hour. 
 
 Ans. 1 d. 16 h. 36 m. 
 
 14. At 31 i cents per bushel, how many bushels can be 
 purchased for 9 dollars ? Ans. 28, 8^ 
 
 15. At 62^^ cents per basket, how many baskets can be 
 bought for 16 dollars ? Ans. 25,6. 
 
REDUCTION OF DECIMALS. 
 
 115 
 
 141 
 
 16. How many yards of cloth in 4 remnants contain 
 ing, severally, 3 yards 3 quarters, 2 yards 3 nails, Ij 
 yards, and 2^ yards I .^ins. 10,4375. 
 
 47, Trom 1 hundred weight, subtract 3 quarters 
 pounds. ..^?2s. ,11830 cwt. 
 
 18. What is the cost of 3f yards of cloth, at 5| dollars 
 per yard? c^?i5. $19,406 + 
 
 19. At 5| dollars per yard, how much cloth can be pur- 
 chased with 19,40625 dollars? .tljis. 3,375 yd. 
 
 2Q-. At 11,76 dollars per hundred weight, what will i 
 quarter of sugar come to ? ^ns. $1,47. 
 
 21. Traveling at the rate of 4^- miles an hour, in how 
 many hours will a foot-man go 34| miles ? 
 
 *-%is. 7,5 hours. 
 
 22. If 85 yards of cloth be bought for 191,25 dollars, 
 and sold at 2 dollars 87 ^^ cents per yard, how much is 
 the whole profit ? " c^??.s. $53,12'^. 
 
 23. A person bought 3^ pounds of tea, at 1,125 dollars 
 per pound; 8 pounds of coffee, at 12^ cents; 4 gallons 
 of molasses, at 31i cents per gallon; 20 pounds of su- 
 gar, at 8V cents per pound; what is the amount of the 
 bill? " ./^nv. $7,886 -f- 
 
 24. If 4i- gallons of wine cost 5-' dollars, what was it 
 per gallon? ~j9ns. $1,27, nearly. 
 
 25/' Two men start on a journey at the same time; one 
 travels at the rate of 41^ miles per hour, the other, 5| ; 
 how many hours after their departure will they be 34 
 miles apart, and how far will the most advanced have 
 traveled? Jins. In 24 hrs. ; dis. 136 miles. 
 
116 ARITHMETIC. 
 
 SECTION IV 
 
 COMPARISON OF NUMBERS. 
 
 (Art. 69.) A number may be considered as part of 
 another number; for instance, 3 is one-half of 6, and 2 
 is one-fourth of 8 ; and we may mechanicaUy find the 
 V and i, by dividhig the first number by the second ; 
 thus, f = 4-: f = !r' Or, the comparison remains the 
 same, if we divide both numbers, (the numbers we com- 
 pare,) by the smallest number; thus, 3 compares with 6, 
 as 1 to 2 ; 2 to 8, as 1 to 4. 
 
 What part of 3 is 2? Certainly f. What part of 2 
 is 3 ? Certainly |. 
 
 Now, wherever this phraseology applies, (JJliat pari 
 
 of is,) the number after the word is, must be 
 
 taken for a numerator, and the other for a denominator; 
 and reduce the fraction to its most simple form, and we 
 shall have the most simple relation of one number to the 
 other. (See Art. 46.) 
 
 EXAMPLES. 
 
 1. What part of 7 is 2 ? Ans. f . That is, 7 is to 2, 
 as 1 to i. 
 
 2. What part of 12 is 4 ? 
 to 4, as 1 to |. 
 
 3. What part of 19 is 7? 
 
 JJns. |. ( 
 
 3r, we may say 
 
 4. What part of 7 is 19 ? 
 
 
 ./Ins. y. 
 
 5. What part of f is ^ ? 
 
 
 Ans. 1. 
 
 6. AVhat part of f is f ? 
 
 
 Ans. |i. 
 
 Make complex fractions of the two last examples, and 
 reduce them. 
 
 (Art. 70.) Before numbers can be compared in this 
 way, they must be of the same denomination. If, for 
 example, we would compare 5 feet with 2 yards, and 
 asked what part of five is two we could not say ^ ; for 
 we nil know that 2 yards is more than five feet. We 
 
COMPARISON OF NUMBERS. 117 
 
 niiist first reduce the yards to feet; and then the question 
 is, What part of 5 feet is 6 feet? Jlns. |. 
 
 7. What part of 3 hundred weight is 1 hundred weight 
 3 quarters ? Reduce both to quarters ; then compare. 
 
 .fins. -j-\. 
 
 8. What part of 6 shillings is 8 shillings 6 pence ? 
 
 Jins. f 1. 
 
 9. WHiat part of 4,6 is ,46? Ans. j\. 
 
 10. Wliat part of 3 gallons is 2 quarts 1 pint? 
 
 1 1 . Wh;it part of 1 yard is 2 feet 6 inches ? 
 
 ^718, f . 
 
 12. What part of 5 dollars is 35 cents ? Ans, y|-^. 
 Wliatpartof 12 is 101? 
 
 What part of 3 is f,-?" 
 
 AVhat partof 6\ is 6^ ? 
 
 The relation of numbers and things compared in this 
 ■vvay, may be applied to the solution of a great variety of 
 practical problems. 
 
 Example. If 3 men can build 7 rods of wall in a day, 
 how many rods can 9 men build ? Here we compare 9 
 men with 3 men ; and we say, what part of 3 is 9 ? A7is. 
 3 times. Now, the rods of wall must compare, as the 
 men that build them. Hence 7X3=21 rods of wall, 
 for the answer to the question. 
 
 Example 2. H three horses eat 8 bushels of oats in 2 
 weeks, how long will it take them to eat 40 bushels ? 8 
 compares to 40 bushels, as 1 to 5; hence 2X5=10 
 weeks. 
 
 Example 3. If 4'^ tons of hay keep 3 cattle over the 
 w^inter, how many tons will it require to winter 25 cattle? 
 3 cattle compare with 25, as 1 to Y; hence, V x| = y 
 =375 tons, Jlns. 
 
 Example 4. If 19 gallons of molasses cost 12 dollars, 
 what w^ill 3^ quarts cost? Here we cannot directly 
 compare gallons with quarts; one or the other must be 
 reduced. The quarts are ^-^ . Rut to reduce quarts to 
 gallons, we divide by 4; therefore, our 3 J quarts is |^, 
 expressed in gallons. Now, 19 compares with \'l, as 1 
 to ^^5 hence 12XaV=A dollars, or 60 cents the Aiis. 
 
118 ARITHMETIC. 
 
 Example 5. If 16 busliels of oats cost $6,75, what 
 will 320 bushels cost? Oats compares with oats, in the 
 supposition and demand ; as, 16 to 320 ; or, as 1 to 20. 
 Hence, 6,75X20 = 135 dollars. Am. 
 
 Example 6. If 1 acre and 20 rods of ground produce 
 45 bushels of wheat; at that rate, how much will 9 acres 
 produce ? One acre and 20 rods is 180 rods. Nine 
 acres is 9X 160 rods. 180 compares with 9X 160. As 
 20 compares with 160, by dividing both by 9 ; or, as 1 
 to 8, by dividing both by 20. Hence, 45 X 8=360 bush- 
 els, the Ans. 
 
 Example 7. If f of a yard of cloth be worth 4 dollars, 
 what will be the worth of ^ of a yard ? Observe, f =|. 
 Compare | to ^, which is the same as 6 to 7. But 6 is 
 to 7 as 1 to \. Hence ^X4=4f dollars, Ans. 
 
 Example 8. If 3 yards of cloth cost 12| dollars, how 
 many yards may be' bought for 102 dollars? Observe, 
 12| = \'; Y is to '^2 as 1. to |. But i is to 2 as 1 to 
 8. ' Hence 3 X 8=24 yards, the Ans. 
 
 9. Boarding at 12 shillings 6 pence per week, how 
 long will 32 pounds 10 shillings last me ? 
 
 Ans. 52 weeks. 
 
 10. If a barrel of beef last 10 men 95 days, how long 
 will it last 25 men ? Ans. 38 days. 
 
 11. If I lend a man $200 for 60 days, how long ought 
 he to lend me $275 to requite the favor ? 
 
 Ans. 43yy days. 
 
 12. How many years win it require for 5 cents to gain 
 the same interest that $100 does in 1 year ? 
 
 Ans. 2000 years. 
 
 13. If it be required to line cloth i of a yard wide, 
 with lining | wide, what must be the relative quantity of 
 the lining? Ans. ^r. 
 
 14. If 22 cents buy 3 pounds of coffee, how many 
 pounds will 44 dollars buy ? Ans. 600. 
 
 15. If 3 pounds are purchased for 25 cents, what will 
 135 pounds cost? Ans. $11,25. 
 
 16. At 41 dollars per ton, how many ton can be pur- 
 chased for $900 ? Ans. 200. 
 
PROPORTION. 119 
 
 PROPORTION. 
 
 (Art. 71.) From the comparison of numbers we de- 
 rive proportion, one of the most fruitful and valuable 
 principles in arithmetic, tlie mathematics, and natural 
 philosophy. Too much cannot be said of its importance 
 and utility. 
 
 Definitions. 
 
 1. When two numbers are directly compared, by di- 
 viding one by the other, as in Article 69, the quotient 
 expresses their relation. 
 
 2. If two other num'ners are compared in the same 
 way, and we find the same quotient the saine relation, 
 the four numbers may constitute a proportion. 
 
 3. Two numbers, thus compared, are called a couplet. 
 
 4. Tiie quotient, or relation, is technically called the 
 ratio. 
 
 For example : If we compare 2 to 4, by dividing 4 by 
 2 we find the intio, 2. Also, 3 compares with 6 by the 
 same ra'io. Therefore, 2 is to 4 as 3 is to 6. 
 
 'i'he words between the terms need not be written, as 
 points, thus, : : : : or, : = : express the same as the 
 written words, (Definition, Art. 5). Hence, 
 
 2 : 4 : : 3 : 6, 
 
 is a proportion. The ratio to this proportion is 2. Ob- 
 serve, that the product of the extreme terms is equal to 
 the product of the mean or middle terms ; that is 2X6 
 =4X3, and this is true in every proportion. If, in any 
 proposed case, this is not true, the proportion is false, 
 and is not, in fact, a proportion. 
 
 (Art. 72.) This property of extremes and means, 
 results from the equality of the ratio between the 
 couplets. 
 
 Thus, 2:4, i = ratio 2. 
 
 3 : 6, |=ratio 2. 
 Hence, {=|, because, things equal to the same thing 
 equal one another. 
 
120 ARITHMETIC. 
 
 6X2 
 Multiply both terms by 2, and 4= . Now mul- 
 
 o. 
 
 tiply by 3, and 3X4=6X2 ; that is, the product of the 
 extremes equals that of the means. 12 : 48 compare, as 
 1 to 4; that is, the ratio between the terms of this cou- 
 plet is 4.* 3 : 12 compares, as 1 to 4, same ratio ; there- 
 fore 12 : 48 :: 3 : 12 must be a perfect proportion ; hence, 
 12X12=48X3. 
 
 (Art. 73.) The terms of this proportion may be chang- 
 ed, and still constitute :i perfect proportion ; but will have 
 a ditl'erent ratio between the couplets. 
 
 1. Invert the means, and 12 : 3 :: 48 : 12. The ra- 
 tio is now \ ; but the product of the extremes and means 
 is the same as before. 
 
 2. If we treat both couplets exactly alike, no matter 
 what we do, the result will still be a proportion. Let 
 us subtract; the second from the first is to the 2d, as the 
 fourth from the third is to the 4th ; that is, 9 : 3 :: 36 : 12. 
 These couplets, we perceive, have the equal ratio, ~, and 
 the product of the extremes is equal to the product of the 
 means. Again : the second from the first is to the first, 
 as the fourth from the third is to the second. That is, 
 9 : 12 :: 36 : 48. V=|- ^atio ; Also, ff=J ratio. 
 Hence this proportion is true. 
 
 (Art. 74.) Two or more proportions may be multi- 
 plied together, term by term, and constitute a new pro- 
 portion. 
 
 Thus, as 2:4:: 5:10 
 
 Also, 6 : 1 :: 24 : 4 
 
 and 9:3:: 3:1 
 
 Then, 108 : 12 :: 360 : 40 
 
 * Some mathematicians, particularly the English, reduce the se- 
 cond term to unity, by dividing both terms by the second. In that 
 case, the ratio of 12 to 48, would be i, in place of 4. Either way 
 is correct, if we are uniform and consistent. We prefer the French 
 method, of reducing the first term to unity, conceiving it more sim- 
 ple than the other. 
 
PROrORTION. 121 
 
 Vt'e may divide the first couplet by 12, the second by 
 40, and we have 9 : 1 :: 9 : 1, a proportion manifestly 
 true. We may multiply or divide either antecedents or 
 consequents by the same number, and we still have a 
 proportion. 
 
 (Art. 75.) By retaining strong hold of the fact, that 
 the products of the extreynes and means arc equal, in 
 every geometrical proportion, we can find any term, or 
 factor of a term, that may be wanting or lost. Take 
 the proportion, 2 : 4 :: 5 : 10. If the fourth term, the 
 10, be wanting, we still have the product of the means. 
 We have also one factor of the extremes, which, used 
 as a divisor to divide the common product, will give the 
 other factor, or the fourth term : thus, 
 
 4X5 
 
 =10. 
 
 2 
 
 When the fourth term of any proportion is wanting, we 
 then have three terms, and the application of Proporion 
 then becomes the Rule of Three, or Golden Rule. The 
 first technicality arose from the fact of there being three 
 given terms to tind a fourth, the last from the great utilily 
 and beauty of its application. 
 
 Resume the proportion, 2 : 4 :: 5 : 10, nnd suppose 
 the second term wanting; we then have 2 : [] :: 5 : 10 
 only. The extremes now give us the common product, 
 20 ; which, divided by 5, gives 4, the term wanting. On 
 this principle, if only one term is wanting, no matter 
 which one, it can be found. When the fourth term is 
 wanting, it is the rule of three direct. When the fust 
 term is wanting, it is the rule of three inverse. 
 
 (Art. 76.) In a direct proportion, we can obtam the 
 fourth term, by multiplying the third term by the ratio. 
 
 Example. 5 : 15 :: 11 : , to the fourth term. AVhat 
 is'that term? 5 : 15 are as 1 to 3. Then 3 is the ratio. 
 Hence, 11X3=33, Mns. In an inverse proportion, the 
 first term is found by dividing the second term by the 
 ratio. 
 
 Example. [] : 50 :: 24 : 4. What part of 24 is 4? 
 Ans. i=the ratio. Hence, divide 50 by ^=300 .fins. 
 
122 ARITHMETIC. 
 
 Or, in consideration of the common product, we have, 
 
 50X24 
 
 for the first term, =300 as before. 
 
 4 
 
 (Art. 77.) In any proportion, suppose 300 : 50 :: 24 : 
 4, we may conceive any term as composed of two or 
 more factors, and one of them wanting. It can be re- 
 stored by means of the common product^ divided by the 
 known factors, in the deficient extremes or means, as the 
 case may be. Let us suppose the second term, 50, to 
 be separated into factors, 5X 10, and the 5 a term sought. 
 We then have 300 : []X 10 :: 24 : 4; the extremes be- 
 ing perfect, gives tlie common product. The means are 
 imperfect ; but if we divide the common product by the 
 factor we have in the means, the quotient will give the 
 lost or sought factor. These multiplications and divi- 
 sions should be performed as directed in Art. 24. 
 
 ^6 
 6 ^^ 
 
 ^^^ on 6 30 
 
 ^00 30 ^ 
 
 5 Ans. 
 
 If we have 300 : 50 :: 2X [] : 4 : to find the lost 
 factor to the third term, the process is the same as be- 
 fore. If we have 3X[] : 50 :: 24 : 4, to find the lost 
 or sought factor, we observe that the means are now per- 
 fect, and they give the common product, and the ex- 
 treme factors are divisors. Thus: 
 
 %. -^-^o^- 
 
 APPLICATION OF THE FOREGOING PRINCIPLES. 
 
 (Art. 78.) Questions which should fall under propor- 
 tion, or the rule of three, have two terms of comparison 
 of the same name ; one in the supposition, the other in 
 the demand. The third term is of the same name as 
 the ansiver, or the term required. To state a question, 
 is to arrange the terms in order, agreeably to the laws of 
 proportion. When the question is direct — that is, more 
 requiring more, or less requiring less — state the question 
 by the following 
 
PROPORTION. 123 
 
 Rule. Of the two terms of comparison, place that 
 of supposition first, and that of the demand second, 
 and that of the same name as the answer required, the 
 third. Then find the fourth term or answer, as di- 
 rected in .irt. 75. 
 
 Example 1. If 12 bushels of wheat are wprth IG dol- 
 lars, wliat is the value of 48 bushels ? 
 
 Here are two terms of the same name, (bushels,) 
 which we call terms of comparison, and one term (dol- 
 lars) of the same name as the answer required. Of these 
 terms of comparison, 12 is that of supposition. We know 
 it by the word if, or one of like import, that may stand 
 before it, the 48 is that of demand, as the language most 
 clearly implies. Hence, by the rule, 
 
 12 : 48 : : 16 : to the fourth term. 
 
 In former times, when mechanical labor was consider- 
 (h1 no task, we would multiply 48 by 16, and divide the 
 i product bv 12 ; we now do it by canceling, according to 
 I Art. 24. 
 
 I Or, we may say, 12 to 48 as 1 to 4 ; therefore 16X4 
 j =64 answer, which is canceling ivithout the form, and 
 j which is generally most convenient in the rule of three. 
 Example 2. If 7 pounds of sugar cost 75 cents, how- 
 many pounds can be bought for 9 dollars ? 
 
 Two of the given terms in this question are of the same' 
 kind (money); but not of the same denomination. These 
 are the terms of comparison, after one of them is reduced 
 to the denomination of the other (Art. 70). 
 
 It is obvious that 75 cents is the term of supposition ; 
 hut the if, (jpfssicn cf condition is not directly before it ; 
 however, a litde change in the phraseology will bring 
 it so. 
 
 If 75 cents M^ill purchase 7 pounds of sugar, how ma- 
 ny pounds can be bought for 9 dollars ? 
 
 Slate metit, 75 : 900 : : 7 : to the answer. 
 Divide the couplet or terms of comparison, by 25, 
 
 and 3 : 36 : : 7 : to the answer ; 
 
 or, 1 : 12 : : 7 : 84. the anwser. 
 
124 ARITHMETIC. 
 
 $ $ lb. 
 Same question again, ^ : 9 : : 7 : the answer. 
 
 Divide by 3 l- : 3 : : 7 : the answer. 
 
 Mult'p'y. by 4. ... 1 : 12 :: 7 : 84 the answer. 
 
 As a specimen of the manner of doing things in for- 
 mer days, Kwe cut the following problem and solution, 
 from a well-known book. 
 
 Example. If 48 yards of cloth cost $67,25, what will 
 144 yards cost at the same rate ? 
 Operation, 
 yd. yd. $ 
 
 48 : 144 : : 67,25 : Ans. 
 144 
 
 26900 
 26900 
 6725 
 
 48)9684,00($201,75 
 96 
 
 84 
 48 
 
 S60 
 336 
 
 240 
 240 
 
 Now let it be observed once for all, that these opera- 
 tions are those of comparison or proportion, and the ac- 
 tual numbers given need not be used, provided we use 
 their proportion. 
 
 Here 48 is to 144, obviously, as 4 to 12 ; or, as 1 to 
 3. Hence 67,25 X 3=$201,75, Ans. 
 
 (Art. 79.) When the terms are of various denomina- 
 tions of weights, measures, &c., as a general rule, reduce 
 them all to the lowest denomination mentioned after 
 
F'ROPor.Tiox. 125 
 
 statement ; but by making use of fractions, we may very 
 often abridge. 
 
 Example^. If 3 pounds 12 ounces of tea cost $3,50, 
 what will 11 pounds 4 ounces cost? 
 lb. oz. lb. oz. $ 
 Statement,. • 3 12 : 11 4 :: 3,50 
 In place of reducing the Aveights to ounces, we observe 
 12 0^. = ;; lbs. 
 
 lb. lb. money. 
 ^ : \\\ : : 3,50 : Ans. 
 Multiply the terms of comparison by 4, and 
 15 : 45 : : 3,50 
 
 or, 1 : 3 : : 3,50 
 
 He^ce, 3,50X3= S10,50 Ans. 
 
 The following problem is commonly wrought by a 
 very tedious process, and, indeed, must be, if rules rath- 
 er than principles are to be our guide. 
 
 If 2 hundredweight 3 quarters 21 pounds of sugar 
 cost £6 1 s. 8 d., what cost 35 hundred weight 1 quar- 
 ter ? Ans. £73. 
 
 By the general rules, (which, as general rules, are good,) 
 we must reduce the quantities to pounds, and tlie money 
 to pence, either after or before the statement. 
 
 cwt. qr. lb. cwt. qr. £. s. d. 
 
 We proceed thus : 2 3 21 : 35 1 :: . . . 6 1 8 
 
 Reduce to qrs. .. 11} : 141 ::...6 1 8 
 
 Multiply by 4, .. 47 141X4 ::...6 1 8 
 
 Divide by 47, ... 1 .... ' 3X4 6 1 8 
 
 Hence, 12 
 
 Ans. £73 
 For further illustration, we extract the following : 
 If 7 pounds of coffee cost 87^ cents, what must I pay 
 for 244 pounds? ^ Ans. $30,50. 
 
 This is worked out, in the book from which we ex- 
 tracted it, in the longest and most formal manner, thus 
 teaching bad habits at first, which are difficult to eradicate ; 
 
 and until such are eradicated, it is impossible to be skillful 
 — 
 
126 
 
 ARITHMETIC. 
 
 As another specimen of our ordinary school opera- 
 tions, we cut the following problem and solution from a 
 modern popular aritlimetic : 
 
 " If 3 hundred weight of sugar cost £9 2 ^. d., what 
 will 4 hundred weight 3 quarters 26 pounds cost, at the 
 same rate ? 
 
 cwt. dqr. 26/6. 
 
 £ 9 
 20 
 
 3 cwt. 4 
 
 4 4 
 
 2 s. Od. 
 
 12 
 
 19 
 
 4X7 = 
 
 ("7 7 
 
 I 84 133 
 
 182 s. 
 12 
 
 336/6.558/6. 
 
 " We first reduce 
 the first and second 
 terms to pounds, then 
 the 3d term to pence. 
 The answer comes 
 out in pence, and is 
 afterwards reduced to 
 pounds, shillings and 
 pence. 
 
 2184 (/, 
 558 
 
 2184 
 
 ^ns. 
 
 17472 
 10920 
 10920 
 
 336)1218672(3627 d. 
 1008 
 12)3627 
 
 2106 
 2016 
 
 20)30 s. 3d. 
 
 907 ^15 - 2s. 
 672 
 
 2352 
 2352 
 
 A7is.£l5 2 s. 3d.'* 
 
 One object of this work is to teach, if possible, that 
 kind of tact possessed by business men, when they use 
 thejr judgments, and quickly arrive at results. The 
 above operation is strictly legitimate, and agreeable io 
 rule; but nevertheless, it is such as a business man, or a 
 
PROPORTION. 
 
 127 
 
 skillful arithmetician, wouM never go through. The first 
 caiiiion is strictly to examine numhers. We perceive 
 that 4 hundred weight 3 quarters 26 pounds, is 2 pounds 
 less than 5 hundred weight. Now let us compute the 
 cost of 5 hundred weight, and deduct the cost of 2 pounds. 
 cwt. cwt. £. 
 Therefore, as 3 : 5 :: 9,1 : the 4th term. 
 As there is a tenth in the 3d term, multiply the first and 
 third by 10, and 30 
 
 Or as 6 
 
 For the 2 pounds. 
 
 lbs. Ib'i. shillings 
 As 3X112 : 2 :: 182 
 or, 168 : 1 :: 182 
 or, 84 : 1 :: 91 
 
 We give another specimen from another book. 
 
 " If 8 bushels 2 pecks cost $4,25, how many bushels 
 can I purchase with $38,25 ? Jins. 76 bu. 2 pk. 
 
 $.cts. $. cts. bu. pk. 
 
 4,25 : 38,25 :: 8 2 
 34 4 
 
 5 :: 91 
 
 : the 4th term. 
 
 
 £. s. d. 
 
 1 :: 91 
 
 : V=15 3 4 
 
 
 £. s. d. 
 
 'S. 
 
 From 15 3 4 
 
 
 take 1 1 
 
 9 X 1 1 
 
 Ans. 15 2 3 
 
 1.5300 
 11475 
 
 34 pk. 
 
 425)130050(306 pk. 
 1275 
 
 pk. 
 
 2550 4)306 
 
 2550 
 
 76 bu. 
 
 $. $. 
 We operate thus : 41 : 38^ 
 
 or, 1 : 38'- 
 
 As two denominations 
 occur in the third term, it 
 is reduced to the less ; 
 hence the result is pecks, 
 which must be reduced 
 to bushels. 
 
 pk." 
 bu. 
 :: 8^ 
 : 2 
 
 : Ans. 
 
 : 76^- bu. .9ns. 
 
 (Art. 80.) There are some problems, where more ap- 
 parendy requires less, and at first view demands another 
 rule (jf statement from that given in Art. 79. 
 
 Example. If 6 men build a wall in 20 days, how long 
 
128 ARITH3IETIC. 
 
 will it require 30 men to do the same? Here it is evi- 
 dent that more men will do the work in less time ; and 
 more requires less, apparently. By the same nature of 
 things, more money will require less time to produce the 
 same interest. But this is all apparent; there is no 
 such thing in nature as more in reality, more cause, more 
 force, producing less in effect 
 
 Lithe last example, the mere e^ris^ence of men can- 
 not be compared to wall, or to any thing they may do. 
 Days'* 7vork may be compared to wall, or to any thing 
 accomplished by labor. Money, of itself, cannot pro- 
 duce interest ; it must be compounded with time ; and 
 all such problems are really problems in compound pro- 
 portion, and only appear to be in simple proportion, by 
 one term in each branch of the question being the same. 
 (See Art. 84.) If we should state the question, thus : 
 
 men. men. days. 
 As .... 6 : 30 :: 20 : days required, as a care- 
 less or unphilosophical operator might, it would em- 
 body no sense, as it would imply that men compare with 
 men, as days compare with days, without any reference 
 to what may have been done. But if we multiply 6 men 
 by the number of days they worked, (20,) we shall have 
 120 days' work, which built the wall , but when this 120 
 days' work was performed by 30 men, it w^ould require 
 them 4 days. We can state all questions by direct pro- 
 portion^ provided ive use philosophical terms. The 
 question under consideration should be stated thus : 
 days^ ivork. days^ work. wall, ivall. 
 6X20 : 30X[] :: 1 : 1. 
 Here is a proportion wanting a factor in the second 
 term, as in Art. 77, and can be found in the same way as 
 there explained. 
 
 Example 2. If a man perform a journey in 6 days, 
 when the days are 8 hours long, how many d-iys will he 
 require to perform the same journey when the days are 
 12 hours long? Here the journey evidently refers, or is 
 measured, by the number of hours taken to perform it; 
 but the hours in both branches of the question, is the 
 compound of days and hours. The statement is thus : 
 
PROPORTION. 129 
 
 hoin's. hours. dis. (lis. 
 
 6X8 : []X12 :: 1 : 1 
 
 Solved by Art. 77. ^ns. 4. 
 
 If the distances had been diflerent, not exactly the 
 same journey, the compound nature of the question 
 would have been apparent at once. 
 
 (Art. 81.) In stating a question, the mind should rest 
 upon tilings and denominations, and not on their numeri- 
 cal values. But after the question is stated, we should then 
 look on the terms as abstract numbers, and nothing else; 
 and, as the first is a divisor, and (he other two, factors 
 in a dividend, we may divide the first by any number 
 that will divide it, and either one of the others, without a 
 remainder, (Art. 24.) 
 
 Example 1. What will 26 yards of cloth come to, if 
 •$6,90 are paid for 13 ells French ? To compare ells with 
 yards, we must reduce both to quarters; but in many 
 such reductions, the multiplications or divisions may be 
 only formal, not reed. Observe the following: 
 (jr. qr. cts. 
 
 6X13 : 26X4 :: 690 
 Divide 1st and 2d 
 by 13, and 
 
 or, 3 : 4 : : 690 
 
 or, 1 : 4 :: 230 
 
 As we have before observed, this is the same as can- 
 (^eling, (Art. 24 ;) but in proportion, this appears to be 
 most convenient. 
 
 Example 2. If 15^ bushels of clover cost $1561;, 
 what quantity can be bought for $95 '^^^ ? 
 
 .^ns. 9,575 bush. 
 Stateinent. 
 156,25 : 95,75 :: 15,625 bu. e^^^s. 
 or, 10 : 95,75 :: 1 : 9,575 bu. .^ns. 
 
 ^^'>- <^ MISCELLANEOUS EXAMPLES. 
 
 1. What will 20 cords of wood cost, if 48 cords are 
 worth 120 dollars? ./Ins. $50. 
 
 2. If 20 cords are worth 50 dollars, what will 48 cords 
 cost? .^ns. $120. 
 
 2X4 :: 690 
 
 Jins. 
 
 Ans. 
 
 Ans. 
 $9,20 A. 
 
130 ARITHMETIC. 
 
 3. If 120 dollars will purchase 48 cords of wood, how 
 iPiany cords can be bough. t for 50 dollars ? A)Vi. 20. 
 
 4. Thirty-six gallons of wine were purchased for 108 
 dollars, what should be given for 4 gallons ? Ans. $12. 
 
 5. If 7 men can do a certain piece of work in 12 days, 
 how long will it take 15 men to do one-half of it? (See 
 Art. 80.) Ans. 2i days. 
 
 6. If 8 yards of cloth cost 3^ dollars, how many yards 
 can be bought for 50 dollars ? Ans, 1 14| yd. 
 
 7. A garrison has provision for 8 months, at the rate 
 of 15 ounces per day ; what must be each man's allow- 
 ance, that the same provision may last them 12 months ? 
 
 Ans. 10 ounces. 
 By Art. 83, this problem is solved thus : 
 
 oz. oz. provis. pi'ovis. 
 
 8X15 : 12X[] :: 1 : 1. Hence, 12 
 
 , _ ^ =Ans. 
 
 Keep the same Art. in mind in solving the two follow- 
 ing: 
 
 8. When a quarter of wheat affords 60 ten-penny 
 loaves, how many eight-penny loaves may be made from 
 the same ? Ans. 75. 
 
 9. If 10 dollars' worth of provision serve 7 men 4 
 days, how many days will the same provision serve 9 
 men ? Ans. 3^ days. 
 
 IQ. If 15 degrees of longitude cause a difference in 
 time of 1 hour, what will be the hour at Washington, 
 when it is 12 o'clock at a place situated 36° 45' east of it? 
 
 Ans. 9 o'clock 33 min. 
 
 11. If a man's yearfy income b# $2000, and his aver- 
 age expenditures be S25,87^ per week, how much can 
 he lay up in 6 years, including one leap-year ? 
 
 Ans. ^4057,621. 
 
 12. If 17 men do a piece of work in 20 days, in what 
 time will 20 men do double the work ? Ans. 34 days. 
 
 13. If a staff 3 feet 8 inches long, cast a shadow 1 foot 
 6 iuciies, what is the height of a steeple that casts a shad- 
 ow 75 feet at the same time ? Ans. 183'^. 
 
 14.^4f I of an ell English cost \ of a pound, what will 
 12| yards cost? Ans. £5 lis. l\ d. 
 
PROPORTION. 131 
 
 15.il If 12!, hundred weight of iron cost $42] , what will 
 48;'5- liundred weight cost? ./Ins, $163,50. 
 
 16. What quantity of Hning, f yard wide, will it re- 
 quire to line 9|. yards of clotli, 1-^ yard wide ? 
 
 .r]ns. 15^ yards. 
 
 17. If 12 gallons cost $30, what will 63 gallons cost ? 
 
 JIns. $157,50. 
 
 18. How many yards of clolli, 3 quarters wide, are 
 equal to 30 yards 5^ quarters wide ? .^ns. 50 yards. 
 
 lO.'How'many yards of paper, | of a yard wide, will 
 be sufficient to "hang a room 20 yards square, and 3^ 
 yards high ? ./^ns. 400 yards. 
 
 20. If 28 dollars are paid for 15 gallons of wine, what 
 is the price per pint ? *^ns. 23} cents. 
 
 21. If 240 bushels of wheat are purchased at the rate 
 of $22V for 18 bushels, and sold at the rate of $33| for 
 22 V bushels, what is the profit of the whole ? 
 
 Ms. 60 dollars. 
 
 22. If $100 gain $7 interest in a year, what will $49,- 
 75 gain in the same time ? ^n-s. $3,48 'j. 
 
 23. If a hogshead of brandy cost ^61 6 16 shiUings, what 
 Mdll 12 gallons cost ? ^ns. £3 4 s. 
 
 24. If 27 yards of Irish linen cost £5 12 shillings 6 
 pence,. how many ells English can be bought for £100 ? 
 
 Ms. 384. 
 
 25. If 300 cwt. be carried 660 miles, for 4 dollars, how 
 many hundred can be carried 60 miles for the same mo- 
 ney? ^ns. 3300. 
 
 i^26. What quantity of water added to 31-^- gallons of 
 whisky, which cost $13,50, would enable the purchaser 
 to sell it for 40 cents per gallon, at a profit of 10 cents a 
 gallon on his purchase? Ms. 10^^ gallons. 
 
 27. If it require 90 yards of carpeting, J wide, to car- 
 pet a floor, how many yards, 11 wide, would be suffi- 
 cient ? *^^i^' 60 yards. 
 
 28. If 9 1 yards of linen cost 2 pounds 12 shillings, 
 what will 32 1' yards cost in dollars, at 4 shillings 6 pence, 
 each? .^775. $382. 
 
 29. At $2,75 for 14 pounds of sugar, what would be 
 the price of I hundred weight? Ms. $22. 
 
 / 30. If A can Aow ^, B i, and C j of an acre of grass 
 
 I ; — 
 
132 ARITHMETIC. 
 
 in an hour, in how many hours can ihey together mow 
 'S\ acres ? ^ns. 6 h. 30 m. 30;, sec. 
 
 J ^ 3 Ip* IX A^, and B together can do a piece of work in 7 
 "X. 1 9^ysv^t;id;^ alone in 12 days, in how many days can A 
 alone do f of it? | ^9ns. 1^ days. 
 
 32. If 5 hundred weight 3 quarters 14 ppunds of su- 
 gar cost 6 pounds 1 shilling 8 pence, how many dollars, 
 at 6 shillings each, will 35 hundred weight 28 pounds 
 cost? ^.; c^n*. $121f. 
 
 N. B. See example 2, Art.^TQ. 
 
 33. A merchant bought of a farmer 120 bushels of 
 wheat at 75 cents per bushel, and sold it at 20 per cent, 
 advanced price ; what did he receive for the whole, and 
 how much did he gain ? 
 
 Statement, 100 : 120 : : 75 : to the merchant's price, 
 90 cents ; whole gain, $18. .\, | 
 
 34. A merchant bought sugar at $4,75 per 100 pounds; 
 how much shall he sell it per pound to gain 25 'per cent.? 
 
 ^<^ns. 6 cents, very nearly. 
 
 35. ^Bought tea at 62^ cents per pound, but by a fall 
 -ra the market, I am willing to sell it at 10 per cent, loss ; 
 'what shall be the price of it ? Jins. 56^ cents. 
 
 N. B. The three preceding problems could properly 
 be put under the technical head of loss and gain ; but it 
 is as propp7' that they should be under proportion. 
 
 36. 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 ? ^^ns. 338 ft. 4 in. 
 
 37. A plane of a certain extent having supplied a body 
 of 3000 horses with forage for 18 days ; how long would 
 it have supplied 2000 horses? .^liis. 27 days. 
 
 38. If a carriage- wheel in turning twice round, advance 
 33 feet 10 inches ; how far would it go in turning round 
 63360 times ? .^ns. 203 miles. 
 
 39. Sound flies at the rate of 1142 feet in 1 second of 
 time: I heard the report of a gun 1 minute and 3 seconds 
 after I saw the flash ; how far distant was it ? 
 
PROPORTION. 133 
 
 40. If a carter haul 100 bushels of coal at. every 3 loads, 
 how many days will it require for him to load a boat with 
 3600 busiiels, suppose he hauls 9 loads a day ? 
 
 ^ns. 12 days. 
 
 41. If the moon move at the average, rate of 13° 10' 
 35" in one day, how lonjr will it require to perform a re- 
 volution ? ' Ans. 27 d. 7 h. 43 m. nearly. 
 
 42. The mean motion of the moon being-, as stated in 
 the last problem, 13° 10' 35" per day, the mean apparent 
 motion of the sun in the same time 59' 8"33, wliat time 
 will be required for the moon to gain a revolution on the 
 sun ? Ans. 29 d. 12 h. 44 m. 3 s. 
 
 Questions. 
 
 What is meant by comparison of numbers? Ans. An 
 examination of their relative magnitudes or values. 
 
 When such magnitude is expressed, what is the ex- 
 pression called ? Ans. Proportion, or ratio. It is called 
 ratio, when one number is expressed by unity. 
 
 Is the ratio of two numbers the same as the quotient 
 resulting from the division of one of them by the other ? 
 
 What name is given to the first, and what to the second 
 term of a couplet ? 
 
 When two pairs of numbers have the same ratio, what 
 do they constitute? 
 
 What name is given to the two outer terms, and what 
 to the two middle terms of a proportion ? 
 
 What relation is there between the product of the 
 means, and the product of the extremes of a proportion ? 
 
 Why is the product of the means equal to the product 
 of the extremes ? 
 
 AVhat rule is founded on this equality ? 
 
 Explain the common method of stating in the Single 
 Rule of Three. 
 
 How do v.e proceed after a question is stated ? 
 
 Why must the two like terms be brought to the same 
 denomination ? 
 
 When we have the 1st, 2d, and 4th terms of a propor- 
 tion, how do we find the third ? 
 
 When a factor of any one term is wanting, how is it 
 found ? 
 
 M 
 
134 ARITHMETIC. 
 
 COMPOUND PROPORTION: 
 
 OR, 
 
 THE DOUBLE RULE OF THREE. 
 
 (Art. 82.) The following rule is common to all modern 
 authors, though they may make slight variations in ex- 
 pressing it. We give it in the words of Mr. G. "Wilson, 
 v/ith an example and its accompanied explanation, as ex- 
 pressed in great clearness and brevity. 
 
 Compound Proportion is compounded of two or more 
 ranks of proportionals, five, seven, nine, &;c., terms be- 
 ing given to find a sixth, eighth, &c. 
 
 Rule. Work by two or more statements in Simple 
 Proportion ; or, set that term, which is like the term 
 sought, in the third place, and consider each pair of 
 similar terms and this third one, as the terms of a state- 
 ment in Simple Proportion, and set them severally, in 
 the first and second places, agreeably to the directions 
 under that rule. 
 
 TVhen the question is thus stated, reduce the similar 
 terms to the like denominations, and then multiply all 
 the terms in the second and third places together, arid 
 divide the product by the product of those in the first 
 place; the quotient will be the answer, or term sought. 
 
 The method of statement will best be illustrated by an 
 example. 
 
 1. If 8 men can build a wall 20 feet long, 6 feet high, 
 and 4 feet thick, in i2 days, in what time v/ill 24 men 
 build one 200 feet long, 8 feet high, and 6 feet thick ? 
 
 Here the answer sought men 24 : 8"^ , 
 
 is days. We, therefore, length, 20 : 200 ! ^^ 
 place 12 days for the third ^ • ' ■ 
 term, and comparing the 
 number of men, it is evi- 
 dent that 24 men will re- 
 quire less time to do the same piece of work than 8 men. 
 We therefore place the less of the two numbers last. 
 
 men 24 . 
 
 8^ 
 
 length, 20 : 
 
 200 ! 
 
 height, 6 : 
 
 thickness, 4 : 
 
 6J 
 
COMPOUND PROPORTION. 135 
 
 We next compare the length of the two walls, and see 
 that one 200 leet long will require more clays than one 
 20 feet long, and place the longest last. For the same 
 reason, we arrange the heights of the two walls in the 
 same order; and so also of the thickness. The con- 
 tinued product of all the middle terms and the third term, 
 divided by the product of all the first terms, gives the an- 
 swer 90 days. 
 
 But the labor of multiplication (as in simple propor- 
 tion) may be shortened by reducing any of the first terms, 
 and any of the other terms proportionally by division. 
 Or we may draw a perpendicular line, and cancel as in 
 Art. 24; the full multiplications and divisions should 
 never be accomplished, unless the numbers are prime to 
 each other. 
 
 The preceding method is very good, as a 7nechanicol 
 contrivance^ but, as a philosophical expression, it will not 
 bear criticism. We prefer the following, by reason of its 
 greater simplicity and more extensive application. 
 
 (Art. 83.) It is an axiom in philosophy, that equal 
 causes produce equal effects; and that effects are always 
 proportionate 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 an uni- 
 versal rule, applicable to all cases which can arise un- 
 der 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 required 
 effect. 
 
 The only difficulty the pupil can experience in this 
 system of proportion, is readily to determine what is 
 cause, and what is effect. But this difilculty is soon 
 overcome, when we consider that all action, of whatso- 
 ever nature, must be cause — and wdiatever is accomplish- 
 ed by that action, or follows such action, must be eifect. 
 
136 ARITHMETIC. 
 
 EXAMPLES. 
 
 1. If 16 horses, in 50 days, consume 128 bushels of 
 oats, how many bushels will 5 horses consume in 90 
 days ? Ans. TZ. 
 
 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 : 
 
 n 
 
 50 
 
 
 90 
 
 
 We write the factors, one under another, as above ; 
 their multiplication is understood, but rarely or never ac- 
 tually accomplished. The second effect is an unknown 
 term, or answer, required — a bracket, or blank, or point, 
 is left to represent it. When found, the four terms above 
 would be 2. perfect geometrical proportion, and the pro- 
 duct 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, namely — the 
 answer. 
 
 Therefore, agreeably to our mode of performing mul- 
 tiplication and division, we draw a line thus, and cancel 
 down : 
 
 16 
 50 
 
 128 
 5 
 90 
 
 2. If $480, in 30 months, produce $84 interest, what 
 capital, 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 
 
COMPOUND PROPORTION. 137 
 
 Here one element of the second co.use is wanting; that 
 is, tlie answer to the question. 
 
 In this case, the extremes ar& complete ; we will there- 
 fore, divide the product of the extremes by the factors in 
 the mean, and tlie 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 teet 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. Jins. 2| days. 
 
 4. If 6 men build a wall in 12 days, how long will it 
 require 20 men to build it? Ans. 3| days. 
 
 (Art. 84.) 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 de- 
 mand, it may be omitted. That term, in the present ex- 
 ample, is, one wall. If we make the number different 
 in the two brandies of the question, or connect any con- 
 ditions 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 grass, by 
 working 8{ hours a "day, how many acres will 15 men 
 mow in 3| days, by working 9 hours a day? 
 
 Ans. 40 yy acres. 
 
 ITT- 
 
138 ARITHMETIC. 
 
 Cause. Eject. Cause. Effect. 
 
 4 : 6f :: 15 : [] 
 
 ^ 3J 
 
 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 ounces of wool make H yards of cloth, |- of 
 a yard wide, how many yards, l\ wide, will 16 pounds 
 of wool make ? Ans. 22 1 yards. 
 
 With this example, some might hesitate as to arranging 
 it under cause and effect, as the actors, those who 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 : H :: 16> [] 
 
 i 16^ • H 
 
 7. If the transportation of 124 hundred weight, 206 
 miles, cost $25,75, how far, at the same rate, may 3 tons 
 and 3 quarters 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, 825,75, 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 
 which for eflect. The above example may be stated thus : 
 
 Cause. Effect. 
 : : 243 : 60| cwt. 
 [] 
 
 Cause. Eff'ect. 
 60J : 243 
 206' ^ [] 
 
 Cause. 
 
 Effect. 
 
 25^- 
 
 : 124 cwt. 
 
 
 206 miles 
 
 Or thus : 
 
 
 Cause. 
 
 Effect. 
 
 ' 12± 
 
 : 25| 
 
COMPOUND PROPORTION. 139 
 
 Or thus : 
 
 
 Cause. 
 
 12i 
 206 
 
 Cause. 
 : 60?- 
 
 [] 
 
 Or thus : 
 
 
 Effect. 
 243 
 
 Effect. 
 : 251 
 
 Effect. Effect. 
 251 : 243 
 
 Cause. Cause. 
 : 60| : 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 scnne 
 terms are multiplied together. 
 
 The most that can be said for the common mode of 
 statement, in the Double Rule of Three, is, that ihe pro- 
 ducts, when the terms are midtiplied out, are propor- 
 tional. 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, five or six special rules, 
 which encumber every arithmetic. We give a few ex- 
 amples, to apply this rule. 
 
 8. What is the interest of 240 dollars for 3-2 years, at 
 6 per cent.? 
 
 Cause. Effect. Cause. Effect. 
 
 100 : 6 :: 240 : [] 
 
 1 3t 
 
 To obtain the answer from this statement, we perceive 
 that we must multiply the means together — i. e. 240 X 6, 
 the rate, and that by the 3i years, the time — and divide 
 by 100; and this is the common ride. 
 
 9. The interest of a certain sum of money, at 6 per 
 cent., for 15 months, was 60 dollars ; what was the sum ? 
 
 ^ns. $800. 
 Cause. Effect, Cause. Effect. 
 
 100 : 6 :: [ ] : 60 
 
 12 15 
 
140 ARITHMETIC. 
 
 10. If 800 dollars, in 15 months, should gain 60 dol- 
 lars, 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 dollars 4 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 
 vnthdravm, and the residue be finished in 15 days more, 
 if the laborers are employed only 7 hours a day ? 
 
 Ans. 4 men. 
 
 13. The amount of a note, on interest for 2 years and 
 6 months, at 6 per cent., is 690 dollars ; required the 
 principal. Ans. S600. 
 
 14. What is the interest of 1248 dollars, for 16 days? 
 
 Ans. $3,28. 
 
 Cause. Effect. Cause. Effect. 
 
 100 : 6 :; 1248 : [] 
 
 12? , 16 
 
 30^ days. 
 
 This can be canceled down, and made very brief. 
 
 15. What is the interest of 1200 dollars, for 15 days, 
 at 6 per cent.? Ans. $3,00. 
 
 16. What is the interest for 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 
 
 17. The interest of a certain sum, for 36 days, is 
 $1,12 — the rate per cent, is 7 ; what is the sum ? 
 
 Ans. $160, 
 
 18. The interest of 160 dollars for 36 days, is $1,12 ; 
 what was the rate per cent.? Ans. 7. 
 
COMPOUND PROPORTION. 141 
 
 19. The interest of 160 dollars, at 7 per cent., was 
 $1,12; what was the tune? JIns. 36 days. 
 
 20. In what time will any sum double itself at 6 per 
 cent.? At any per cent.? 
 
 JIns. At 6 per cent., in 16| years; at any per cent., 
 divide 100 by the per cent. 
 
 21. If 2^ yards of cloth, 1| wide, cost 3 dollars 37^ 
 cents, how much will 36", yards cost, 1 i yards wide ? 
 
 ^Ans. $52,79. 
 
 22. If 1 men spend f of f of f of Vi of £30, in -fy 
 of ^- of 14 of y of 9 days, how many dollars, at 6 shil- 
 lings each, will 21 men spend in | of \^ of ^ of yS of 45 
 days? Ans. $630. 
 
 23. I lend a friend 200 dollars for 6 months ; how 
 long ought he to lend me 1000 dollars, to requite the fa- 
 vor — allowing 30 days to a month? Ans. 36 days. 
 
 24. 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?^ Ans.^S\ ounces. 
 
 25. If a foot-man, in 12^days, traveling ^lours a day, 
 perform a journey of 240'miles, in how many days will 
 he perform one of 720^Tiiles, if he travel 8 hours a day? 
 
 Ans. '21 days. 
 ^^ 26. If 20 men, in 12 days, working 5 hours a day, 
 can perform a piece of work, how many hours a day 
 must 15 men work, in order to perform 3^- times as much 
 work in 30 days 1 Ans. 8-^ hours. 
 
 27. In what time will $627,50, loaned at 7 per cent., 
 pro<luce as much interest as $2510, at 3^ per cent., will 
 produce in 1 year and 8 months? Ans. 3|- years. 
 
 28. If a block of marble 2 feet 6 inches long, 1 foot 9 
 inches broad, and 1 foot 3 inches thick, weigh 9 hundred 
 weight 2 quarters, what would it weigh, if each of its di- 
 mensions were doubled? Ans. 3 tons, 16cwt. 
 
 29. Eight workmen, laboring 7 hours a day for 15 
 days, were able to execute \ of a job on which they were 
 engaged ; in how many days can they complete the res- 
 
142 ARITHMETIC. 
 
 idiie. by working 9 hours a day, if 4 workmen are added 
 to their number? Jlns. 15| days. 
 
 30. How many men will reap 417,6 acres in 12 days, 
 if 5 men reap 52,2 acres in 6 days ? Ans. 20 men. 
 
 31. 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 days. 
 
 32. If 54 men, in 24^ days, working 12^ hours each 
 day, can build a fort, in how many days will 75 men do 
 the same, when they work but 10| hours each day 1 
 
 Ans. 21 days. 
 
 33. If 24 men, in 189 days, working 14 hours each 
 day, dig a trench 33 1 yards long, 3^- deep, and 5| wide, 
 how many hours per day must 217 men work, to dig a 
 trench 23i yards long, 21 deep, and 3| wide, in 5^ days? 
 
 Ans. 16 hours. 
 
 34. If a cistern, 16 feet long, 7 broad, and 15 deep, 
 cost 36 dollars 72 cents, how much, in proportion, would 
 another cistern cost, that is 17^ feet long, 10^ broad, and 
 16 deep? Ans. %M,2Q. 
 
 35. If a cistern 17^ feet long, 10^ broad, and 13 deep, 
 hold 546 barrels, how many barrels will that cistern hold, 
 that is 16 feet long, 7 broad, and 15 deep? 
 
 Ans. 384 bbls. 
 (Art. 85.) With the exception of square and cube 
 root, and a few obvious principles in mensuration, we 
 have now explained every arithmetical operation. The 
 four groimd rules, with reduction, and proportion, in- 
 clude all the principles that can be used in Practice, In- 
 terest, Barter, Loss and Gain, Exchange, &c. &c.; hence 
 when we come to these subjects, we have no new prin- 
 ciples to explain. It only remains to give clear defini- 
 tions under each head, and teach how to apply the knowl- 
 edge we are supposed already to possess ; and now all 
 the judgment and tact of the jpupil will be called into full 
 exercise. 
 
PRACTICE. 
 
 143 
 
 PRACTICE. 
 
 (Art. 86.) Practice is literally what the term implies: 
 short and expeditious methods, practiced bv merchants 
 and business men, to find tlie amount of bills, and the 
 aggregate cost of articles bougJal or sold. The principle 
 involved in all such calculations, is proportion ; but the 
 price of the unit is generally given, which renders a i'or- 
 mal statement unnecessary; and the computations ;ire 
 usually made by taking the number of times, and the 
 aliquot parts of the unit. 
 
 TABLE OF ALIQUOT (OR EVEN) PARTS. 
 
 Cents. 
 
 Parts 
 
 of$l. 
 
 
 
 6 
 
 Parts 
 of 1 year. 
 
 Parts of £1. 
 
 Parts of 1 
 
 shilling-. 
 
 Parts of 1 cwt. 
 
 50 
 
 V 
 
 i 
 
 lOs. =V 
 
 6d. =i 
 
 561b. = !, 
 
 33L 
 
 4 
 
 4 
 
 3" 
 
 6s.8d.=i 
 
 4d. =• 
 
 28lb. = L 
 
 25 
 
 4 
 
 3 
 
 4 
 
 5s. =L 
 
 3d. =• 
 
 161b.=l 
 
 20 
 
 T 
 
 2 
 
 6 
 
 4s. i=i^ 
 
 2d. =J 
 
 14lb. = i 
 
 12\ 
 
 ■r- 
 
 1 
 
 tV 
 
 3s.4d.=i. 
 
 lid.=i 
 
 7lb. = J, 
 
 6't 
 
 Te 
 
 
 or i of 
 
 2s.6d.=i 
 
 Id. =yV 
 
 
 5 
 
 I 
 
 20 
 
 
 3 mo. 
 
 ls.8d.=J^ 
 
 
 
 (Art. 87.) The principles of computation being sup- 
 posed to be well fixed in the mind, 2Lny formal rule that 
 would, in any degree, restrain the free exercise of judg- 
 ment, in the almost endless variety of cases, would prove 
 an nijury rather than an aid ; therefore we shall explain 
 only by examples. 
 
 Example 1. What will 64 bushels of oats cost, at 25 
 cents per bushel ? Ans, \ as many dollars as bushels, 
 16 dollars. 
 
 2. What will 47^ bushels of wheat come to, at 87^, 
 cents ? 
 
 At$l, $47,50 
 
 iless, 5,94 
 
 Jlns. $41,56 
 
144 ARITHMETIC. 
 
 3. What will 78 gallons of wine cost, at 1 dollar 62 J- 
 cents per gallon ? 
 
 At $1, ... • $78,00 
 
 At 50 cts 39,00 
 
 At 12^, 9,75 
 
 Ans, $126,75 
 4. AVhat will 1 bushel 3 pecks 6 quarts of beans cost, 
 at 1 dollar 12 1 cents per bushel ? This is 2 quarts less 
 than 2 bushels : hence, 2 bushels cost $2,25 
 
 of 1,12|, less, or 7 cents, nearly, 7 
 
 Ans. $2,18 
 
 5. What will 75 bushels cost, at 331 cents a bushel ? 
 
 3)75 
 
 25 dollars, Ans. 
 
 6. What will 224 barrels of flour come to, at 3 dollars 
 43^ cents per barrel? 
 
 224 at $3, 672 
 
 at 50 cts 112 
 
 784 
 -r'^ of 224, less, 14 
 
 Ans. $770 
 
 7. What will 462 yards of cloth cost, at $1,1 2 V per 
 yard? Ans. $519,75. 
 
 8. What will 185 yards of cloth cost, at $1,20 per 
 yard? Ans. $222. 
 
 9. What will 150 yards of cloth cost, at $1,3^ per 
 yard? Ans. $196,87^-. 
 
 10. What is the cost of 144 pounds of rice, at 3V pence 
 a pound ? This had better be done by canceling ; 3^ = |. 
 
 4^ 12 
 
 Thus, ^1 ' 
 
 20 Ans. £2 2 s. 
 
 11. What will 45^ barrels of flour come to, at 6 dol- 
 lars 25 cents a barrel ? ' Ans. $284,37|. 
 
PRACTICE. 145 
 
 12. What will 18^ cords of wood cost, at 4 dollars 75 
 cents a curd I " Jlns. $87,87 ^ 
 
 13. What will "3 J tons of coal cost, at 4 dollars 25 
 cents per ton ? Jins. $15,93^. 
 
 14. Wliat will 2 hundred weight 2 quarters 14 pounds 
 of sugar come to, at $8,50 per hundred weiglit? 
 
 Jim. $22,31 V. 
 
 15. What will 3 bushels 3 pecks 6 quarts of beans 
 cost, at $1,60 per bushel ? [Find what 4 bushels cost.] 
 
 Am. $G,30. 
 
 (Art. 88.) In some States, particularly in New Eng- 
 land and New York, the prices of articles arc still given 
 in sliillings and pence; yet they never render a bill in 
 })ound3, sliillings and pence, but always chaufj-e it to dol- 
 lars and cents. In New England, as there are 6 shillings 
 to a dollar, 6 yards or 6 pounds will cost as many dol- 
 lars as there are shillings and parts of a shilling to one 
 yard, pound, &.c. 
 
 Excnnple 1. At 5 shillings 6 pence per yard, what will 
 6 yards cost, in dollars and cents ? Jlns. $5,50, 
 
 N. B. The 5 shillings are called 5 dollars, and pence 
 are called 50 cents. 
 
 Example 2. At 7 shillings 3 pence. New Egland cur- 
 rency, per yard, what will 18 yards of cloth cost? 
 
 6 yards will cost $7,25 
 
 18 yards is 3 times as much 3 
 
 Am. $21,75 
 
 Example 3. What will 15 bushels of wheat cost, at 5 
 f hillings 9 pence (N. E.) per bushel ? 
 
 6 bushels cost $5,75 
 
 6 bushels more cost 5,75 
 
 3 bushels cost 2,87-^ 
 
 15 bushels cost $14,37-^- c^;-25. 
 
 In New York currency, 8 yards, &e., cost as many 
 dollars and parts of a dollar, as one yard, pound, &;c. 
 costs shillings and parts of a shilling. 
 
 N 
 
146 ARITHrilETIC. 
 
 Example 1. At 4 shillings 3 pence per gallon, N. Y. 
 currency, what will 12 gallons cost? 
 
 8 gallons will cost .......:. $4,25 
 
 4 gallons will cost 2,12^ 
 
 12 gallons will cost $6,37^ Ans. 
 
 Example 2. At 3 shillings 6 pence per bushel, N. Y. 
 currency, what will 31 i bushels cost? 
 
 8 bushels will cost $3,50 
 
 4X8=32 bushels will cost $14,00 
 
 \ a bushel is yV of 8 bushels ; hence yV of 3,50, which 
 
 s 10 ' lb'' 
 
 is 22 cents nearly, must be deducted. Ans. $13,78. 
 
 [Note. To notice very small fractions of a cent in 
 such problems as these, is being more nice than wise.'] 
 
 1. What sum must be paid for 19 firkins of butter, 
 each containing 42 pounds, at 16| cents per pound ? 
 
 Ans. $133. 
 
 2. At 12^2 cents per pound, what must be paid for 4 
 boxes of sugar, each containing 125 pounds ? 
 
 Ans. $62,50. 
 
 3. What must be paid for a cask of rice, containing 2 
 cwt. 1 qr. 17 lb. at 5 cents per pound? Ans, $13,45. 
 
 Run out the items of the following bills : the first is N. 
 England, the second N. York currency. 
 
 Mr. Charles Lardner, 
 
 Bought of James Fisher, 
 1840. 5. d. 
 
 Jan. 3, 9 yds. Bleached cotton, at • • 1-3 
 
 6 lbs. Souchong tea • • " • • 4 4 
 12 « Black do. . . " . . 3 9 
 18 " Coffee " • • 1 3 
 
 7 yds. Irish linen ...".. 4 8 
 24 " Calico " . . 1 8 
 
 North Hampton, Jan. 3, 1840. 
 Received payment, 
 
 $29,57 
 James Fisher. 
 
PRACTICE. 
 
 147 
 
 Mr. John Schoolcraft, 
 
 To C. P. Smith, Di 
 
 1844. 
 
 
 
 s. 
 
 d. 
 
 Jan. 8. 
 
 To 10 lbs. 
 
 Coffee, ... at . 
 
 I 
 
 9 
 
 a 
 
 " 12 " 
 
 Suirar. . . .at . 
 
 . 1 
 
 3 
 
 " 12. 
 
 " 18 " 
 
 Sabnoii ... at • 
 
 
 6 
 
 Feb. 1. 
 
 » 9 " 
 
 Black tea . • at • 
 
 5 
 
 3 
 
 (4 
 
 " 24 " 
 
 Buckwheat . at . 
 
 
 3 
 
 *« 6. 
 
 u 9 <. 
 
 Corn'd beef at • 
 
 
 4 
 
 New York, March 4, 1844. 
 
 Received payment, 
 
 $12,21 
 
 C. P. Smith. 
 
 Lexington, June 3, 1840. 
 Mr. William Davis, 
 
 Bought of Samuel Merchant, 
 * $ cts. 
 
 25 'j lbs Imperial tea, at 1 
 
 1 5 pair worsted hose, at 
 
 7* dozen women's gloves, at. . -5 
 
 16 dozen knives and forks, at ... 1 
 2', cwt. brown sugar, at 7 
 
 '25 lbs. Loaf sugar, at 
 
 75 
 
 87^ 
 
 .37^ 
 
 75" 
 
 50 
 
 30 
 
 Received payment in full. 
 
 $152,31 
 
 Samuel Merchant. 
 
 The following problems are a compound, or rather 
 combination of reduction and practice, yet all strictly 
 practical. They, moreover, serve to explain the modern 
 method of canceling, which is no less useful than beauti- 
 ful. 
 
 1. What is the value of a piece of gold 'weighing 1 lb. 
 3 pemiyweights, at 12^ cents per grain? „in6-. $729. 
 243 dwt. 
 M 3 
 
148 
 
 ARITHMETIC. 
 
 2. At 5 cents a pound, what will 6 hundredweight 1 
 quarter cost ? Ans. $35. 
 
 5 
 
 3. At $2,25 a quarter, what will 1 ton 2 hundredweight 
 cost? ^^ns. $198. 
 
 22 cwt. 
 
 4 
 
 9 
 
 cost? 
 
 
 
 
 7 lbs. 
 
 8 oz. 
 
 =^n 
 
 = V lbs. 
 
 
 
 i 
 
 ^0 ^00 
 
 4. At 5 cents per ounce, what will 7 pounds 8 ounces 
 
 Therefore, 
 
 1^ 3 
 
 /l^ 2 
 
 But, says an objector to this, mode of operation, it is 
 very fortunate that this happened to be 8 ounces, just 
 half a pound ; what would you have done had it been 9 
 or 7, or 10 oz. ? Compute it just as we have 
 at 8 ounces ; and if it v/ere 9 ounces, it would be 5 cents 
 more, as it is 5 cents per ounce; if 2 ounces more, 10 
 cents ; if 1 ounce less, 5 cents less, &c.: no difficulty 
 with a willing disposition. 
 
 5. What will 16 yards 2 quarters of cloth cost, at 12^ 
 cents a nail? ' .'t??i5. $33. 
 
 66 qrs. 
 
 V 
 
 8 
 
 . How many coat patterns, each containing 3 yards 
 2 quarters, can he cut out of a piece of cloth containhig 
 140 ells Flemish? ^^ns. 30. 
 
 14 ^^^ 
 ^^ 3 
 
 7. What will 1 hogshead of wine cost, at 6^- cents a 
 gill? Ans. $1^6. 
 
 Observe, that 6'^ •■=-/? ^"^' as 4 is a divisor to 25, it 
 must be put on the opposite side of the line. 
 
PRACTICE- 
 
 149 
 
 100 
 4 
 
 63 
 4 
 2 
 4 
 
 25 
 
 or, 
 
 16 
 
 63 
 4 
 2 
 4 
 
 8. Suppose a hogshead of inolasses, wliich cost 2'i 
 dollars, be retailed at 12^ cents a quart, wliat is the pro- 
 fit on it ? 
 
 63 gallons. Sale, 31,50 
 
 ^ ^ Cost, 23 
 
 2 ^ 
 
 Profit, 8,50 
 9. What will 5 barrels of flour cost, at 3.', cents per 
 pound? .6lni. ^'Si,SO. 
 
 10. How many times is | of a pint contained in | of a 
 gallon ? 
 
 or. 
 
 .'his. 6|. 
 
 20 
 
 6i; 
 
 We have already remarked, that denominators of frac- 
 tions must go over the line from, the term to ivhich they 
 belong. 
 
 1 1. How many times can a vessel, holding y^ of a quart, 
 be filled from ^ of a barrel, containing 3U gallons^ 
 
 3U 
 
 ^/i 7 
 10 
 
 Ans. 46|. 
 
 12. If 1 acre and 20 rods of ground produce 45 bushels 
 of wheat, at that rate, how much will 9 acres produce ? 
 
 Jlns. 360. 
 
 la. 20r. = 180. ^0/1^0 /l^C) 8 45X8=360. 
 45 
 
 N. B. We shall plan no more problems. The fol- 
 lowing require no real labor., save correct reasoning. 
 When the numbers are properly arranged, a few clips 
 with the pencil,* and perhaps a trijiing multiplication, 
 
 will suffice. 
 
 — 
 
150 ARITHMETIC. 
 
 13. At H cents a gill, how many gallons of cider can 
 be bought for $24 ? Ans. 50. 
 
 14. How many casks, each containing 12 gallons, can 
 be filled out of half a tun of wine? Ans. \^\. 
 
 15. A man retailed 6 barrels of ale, and received for it 
 $129,60 ; at what price did he sell it a pint? 
 
 An8. 1\ cents. 
 
 16. How much butter, at 9 cents per pound, will pay 
 for 12 yards of cloth, at 2 dollars 19 cents per yard? 
 
 .8ns. 2921b. 
 
 17. At 45^ dollars per acre, what will 32 rods of land 
 come to? " Am. $9,10. 
 
 18. How long must a laborer work, at 62 ^^ cents a day, 
 to earn $25 ? Jins. 40 days. 
 
 19. A merchant sold 275 pounds of iron, at 5| cents a 
 pound, and took his pay in oats at 50 sents a bushel ; 
 how many bushels did he receive ? Jins. 28|. 
 
 20. How many yards of cloth, at $4,66 a yard, must 
 be given for 18 barrels of flour, at $9,32 a barrel? 
 
 Ans. 36. 
 
 21. How long will it require one of the heavenly bo- 
 dies to move through a quadrant, at the rate of 43' 12" 
 per minute ? Ans, 2^^^ hours. 
 
 22. If a comet move through an arc of 7° 12' per day, 
 how long would it be in passing an arc of 90° ? 
 
 Ans. VZ\ days. 
 
 23. What is the cost of 8 hogsheads of wine, at 5 cts. 
 per pint? .^/?5. $201,60. 
 
 24. There are 30 J^ square yards in one perch of land ; 
 how many perches are there in 363 square yards ? 
 
 Am. 12. 
 
 25. What will 18| yards cost, at 75 cents per yard? 
 
 Am. $14,06' 
 
 26. If 10 persons receive 1032 dollars for 43 days' 
 work, how much does each man earn per day? 
 
 Ans. $1,50. 
 
 27. How many times will a wheel, which is 12 feet in 
 circumference, turn round in going 2 miles ? 
 
 Am. 880. 
 28 What number of revolutions will a measuring wheel 
 
SIMPLE INTEREST. 151 
 
 of 18 feet 6 inches in circiimferenoe, make, in running 
 150 miles? Mns. 42810, and 5 yd. left. 
 
 29. How many times will a wheel, which is 9 feet 2 
 inches in circumference, turn round in going 65 miles? 
 
 ^ns. 37440. 
 
 30. If a man can travel 39 miles 20 rods a day, how 
 long would it take him to walk round the globe, the dis- 
 tance being about 25000 miles ? ^ns. 640 days. 
 
 31. How many steps, of 30 inches each, must a person 
 take in walking 21 miles? ^0?is. 44352. 
 
 32. How many suits of clothes, each containing 5 yards 
 1 quarter, can be cut out of 168 yards ? Jins. 32. 
 
 33. How many coat patterns, each containing 2 yards 
 1 quarter 1 nail, can be cut out of 39 yards 1 quarter 1 
 nail? .^ns. 17. 
 
 34. How many times can a vessel, holding 4 gallons 2 
 quarts, be filled out of a hogshead of wine ? ^7is. 14. 
 
 35. A man distributed £3 3 s. 9 d. among a company 
 of boys, giving each 3 s. 9 d.; how many boys were 
 there in company? ^^us. 17. 
 
 N. B. The last four problems correspond to Note 
 under Art. 22. 
 
 SIMPLE INTEREST. 
 
 (Art. 89.) Simple Interest is a sum allowed for the 
 use of money, goods, or property. 
 
 The sum, for which interest is allowed, is called the 
 principal. 
 
 The rate of allowance for 100 dollars, for 1 year, is 
 called the rate per cent. — per cent meaning per 100. 
 
 The amount is the principal and interest, added to- 
 gether.* 
 
 * The rate of interest is, in almost all countries, established by 
 law ; and asking, or taking more than such rates, is termed usnry, 
 and is subject to legal penalties. In New England and Ohio, the 
 legal rate is 6 per cent,, in New York 7; but. in many countries of 
 Europe, money can be obtained at 4 and 5 per cent. 
 
152 ARITH3IETIC. 
 
 Every question of simple interest, is propcrlv an ex- 
 ample in proportion, or the Rule of Three. Three terms 
 are oiven to tind a fourth term. 'J'hus: let it be required 
 to tind the interest of any sum, say $240. for 1 year, at 
 6 per cent. The term 6 per cent, means that 100 dollars 
 yields 6 dollars in one year; and 240 dollars must yield 
 in that proportion. 
 
 Hence, as 100 : 240 : : 6 : to a fourth term, 
 6 
 
 14,40 
 This gives $14,40 for one year's interest; and if tlie 
 interest were required for two or three, or more years, 
 we should multipl}^ this sum by 2, 3, or more, as the 
 case may require; hence v/e have the following 
 
 General Rule. Multipli^ the principal by the rate 
 per ce7if., and that product by the time, (a year being 
 the unit,) and divide by 100. 
 
 Let the pupil remember that this is a general and uni- 
 versal rule, equally applicable to any per cent, or any re- 
 quired time, and all other rules must be reconcilable to it; 
 and, in fact, all other rules are but modifications of this. 
 
 (Art. 90.) When the interest is Q per cent., we 7nay 
 midliply the principal by half the ninnber of months, 
 and divide by 100. Let us reconcile this rule with the 
 former. What is the interest of $800 for 16 montlis, at 
 6 per cent.? As a year is the irnit of time, we must di- 
 vide the 16 by 12, to bring it into years and parts of a 
 year; or the result must be the same, if we multiply by 
 16 and divide by 12 (Art. 24). Hence, by the General 
 Rule, and under the form of Art. 24, we have 
 
 300= principal. 
 
 12 
 
 100 
 
 6=rate. 
 16= time. 
 
 Now cancel 6 in 12, and the quotient 2, into 16; it wil 
 then stand thus: 
 
 300 
 
 100 
 
 8 Ans. $24. 
 
SIMPLE INTEREST. 153 
 
 That is, midliply the princijml by half the number of 
 months, and divide by 100. 
 
 (Art. 91.) When the rate of interest is 4 per cent. 
 midliply the principal by ^ the number of months, and 
 divide by 100. 
 
 Let us investigate this rule. What is the interest of 
 $720, for 2 years and G months, at 4 per cent.? Take 
 the general rule : thus, 
 
 720 
 
 100 
 
 .4 
 ^^ months. 
 
 10 = 1 the months. 
 
 When thus far canceled, the remaining part of the op- 
 eration, put in words, is the rule above cited. In iJiis 
 way, we can show that 3 per cent, will require us to 
 multiply by \ of the number of months, and divide by 
 100 ; 2 per cent, corresponds with ^ of the number of 
 months; and 1 per cent, with y'^ of the number of 
 months, &lq. 
 
 (Art. 92.) When the per cent, is complex or mixed, 
 as, 4|,, 5|, 7|, &c., it may be most convenient to first 
 cast the interest at 1 per cent., by multiplying the prin- 
 cipal by -j'j of the number of months, and dividing by 
 100 ; and afterwards, multiplying by the per cent., what- 
 ever it may be ; or we may proceed by the general rule. 
 The first will be most brief, if the months will divide by 
 12, or if we find any large common factors, between 100 
 and the months. 
 
 N. B. When days are in question, they must be con- 
 sidered as parts of a month, 30 days to a month ; there- 
 fore, 3 days is ,1 of a month, 6 days ,2, 9 days ,3, &;c. ; 
 or, in vulgar fractions, 1 day is -3'^, 2 days y^, &c. 
 
 EXAMPLES. 
 
 1. What is the interest of $75, for 1 year 6 months 
 and 12 days, at 6 per cent. ? • 
 y. m. d. 
 1 6 12 = 18,4 months, half 9,2. 
 
154 ARITHMETIC. 
 
 By rule (Art. 00)— 
 
 ■ji 3 4)27,6 
 
 4 
 
 9,2 
 
 $6,90 ^ns. 
 
 2. Find the interest of $467,17 for 3 years 5 months, 
 at 7 per cent. 
 
 This had better be done by the general rule. 
 
 467,17 
 
 7 
 
 Interest for 1 year, 32,7019 
 
 3 
 
 Interest for 3 years, 98,1057 
 
 Interest for j of a year, or 4 months, . . . 10,9006 
 Interest for 1 month, or j^ of 4 months, • • 2,725 
 
 .^ns. $111,731 
 
 3. Required the interest of $64,50 for 3 years 5 mon. 
 and 10 days, at 7 per cent.? 
 
 The days here make the operation more brief, if we 
 take the canceling form, and cast the interest at 1 per 
 cent. (Art. 92.) 
 
 Months, 41i = ^r. 3 ^^ 
 
 100 
 
 ^^,li6 21,50 
 
 21,50 
 31 
 
 21,50 
 645 
 
 300)666,50 
 
 Interest at 1 per cent. .... 2,2225 
 
 7 
 
 Interest at 7 per cent. . . . 15,5575 or $15,56 r^ns. 
 
SIMPLE INTEREST. 155 
 
 4. What is the interest on $125, for 17 days, at 6 per 
 cent.? 
 
 First find the interest for 15 days, i- of one month, 
 (Art. 90.) 
 
 4 
 
 m 
 
 4 
 
 ^^A i5 
 
 That is y^g of a dollar, or ,31'^ cents is the interest for 15 
 days ; /^ of this in addition will give the interest for 17 
 days. Ans. 35 cents 4 mills. 
 
 5. AVhat is the interest of $204, for 40 days, at 6 per 
 cent.? 
 
 40 days = l|- month=|^, half =f . 
 
 ^ m 68 
 
 100 "^ 
 
 Jins $1,36. 
 
 6. Find the interest on $420, for 2 years 1 month and 
 27 days, at 6 per cent. 
 
 Time 25,9 months, p, 
 100 
 
 ^^0 21 
 25,9 
 25,9 
 21 
 
 259 
 518 
 
 10)5439 
 
 $54,39 Ans. 
 
 7. What is the amount of $560, for 9 months and 20 
 days, at 6 per cent.? .dns. $587,066+ 
 
 8. What is the interest of $50,11, for 1 year and 21 
 days, at 6 per cent.? Ans. $3,18 + . 
 
 9. What is the interest of $340,50, for 3 months, at 6 
 percent.? e^n.s. $5,1 1. 
 
 10. What is the interest of $180, for 1 year 2 months 
 and 6 days, at 6 per cent.? Ans. $12,78. 
 
 11. What is the interest of 25 dollars, for 3 years 6 
 months and 20 days, at 6 per cent.? A7is. $5,33^. 
 
156 ARITHMETIC. 
 
 12. What is the interest of 125 dollars, for 5 years 5 
 months 5 days, at 6 per cent.? ' M)i.s. 40,73. 
 
 13. Wliat is the amount of 56 dollars, for 9 years 8 
 months 3 days, at 6 per cent.? Jins. $88,51. 
 
 14. What is the amount of 84 dollars, for 5 years 5 
 months and 9 days, at 5 per cent.? ^^ns. $106,85. 
 
 15. What is the interest of $72,05, for 3 montlis, at 2 
 per cent.? e/??i5. 36 cents. 
 
 16. What will 625 dollars produce in 3 years 2 months 
 and 7 days, at 10 per cent.? .^?75.'$199,13-[-. 
 
 17. How much interest will $865,25 produce in 9 
 years 4 months and 15 days, at 6 per cent.? 
 
 Jns. $486,70+. 
 
 18. What is the interest on 9000 dollars, for 1 year 4 
 months and 8 days, at 5 per cent.? ^ns. $610. 
 
 19. What is the interest on 998 dollars, for 3 years 10 
 months and 18 days, at 6 per cent.? Compute for 4 
 years, and deduct the interest of the sum for 1 m. and 12 d. 
 
 Thus, ... 998 For the days, 998 
 
 ,24 ^ the number of months, ,7 
 
 1996 Deduct $2,986 
 
 3992 
 
 $239,52 
 
 (Art. 93.) When the principal is in pounds, shillings 
 and pence, the pound being the unit, reduce the shillings 
 and pence to the decimal of a pound, by Art. 69. 
 
 20. What is the interest of £75 12 s. 6d. for 2 years 
 4 months and 10 days, at 5 per cent.? 
 
 12 
 20 
 
 6 , 
 12,5 
 
 Form, ^ 12 
 
 3 
 
 Principal=75,625 
 Time 28J- months. 
 
 ^^,0^ 3,25 
 
 (4 100 
 
 1 '1 
 
 5 
 
 85 
 
SIMPLE INTEREST. 
 
 157 
 
 3,25 
 5 
 
 16,25 
 
 85 
 
 8125 
 13000 
 
 £ 
 
 144)1381,25(9,599 nearly. 
 1296 20 
 
 852 11,980 
 720 12 
 
 1325 11,76 
 1296 
 
 1290 
 
 £ s. d. 
 Ans. 9 11 IH. 
 
 2 1 . Find the amount of £320 8 s. for ten years, at 6 
 per cent. 
 
 O A. — 2 — To — »^ 
 
 £ S. cl. 
 Principal, 320 8 
 Interest, .192 4 91 
 
 100 
 
 Amount, . 512 12 9i 
 
 320,4 
 60 
 320,4 
 6 
 
 £192,24 
 20 
 
 4,80 
 12 
 
 Form, 100 
 
 9,60 
 
 22. Find the amount of £1230 4 s. for 8 years and 
 6 months at 4 per cent. ? 
 
 1230,2 
 
 34 .^ns. £1648 9 s. 4 d. 
 
 34 being \ of the number of months, corresponding to 4 
 per cent. (Art. 91). 
 
 23. What is the interest of £21 18 s. 4 d. for 3 years 
 and 4 months, at 7 per cent.? Ans. £5 2 s. 6^'d. 
 
 O 
 
158 ARITH3IETIC. 
 
 24. What is the amount of £156 9 shillings 3 pence, 
 for 1 veai* and 9 months, at 6 per cent.? 
 
 Ans. £172 17 s. 9d. 3 qr. + 
 
 25. What is the amount of £27 2 shillings 6 pence 3 
 quarters, for 1 year and 10 months, at 6 per cent.? 
 
 Am. £30 2s. 2d. 3 qr.-|- 
 
 26. What is the interest of £75 10 shillings, for 2 
 years and 3 months, at 6 per cent.? 
 
 .^715. £10 3s. lOd.-l- 
 
 27. What is the interest of 450 dollars, for 6 months 
 and 20 days, at ^\ per cent.? 
 
 Per cent. 5^=^. Time, 6| months=\?, years=§^ 
 
 2 
 By the General Rule, 2 ^d 
 
 Art. 89, . .• A00 
 
 m 50 
 11 
 
 \6 4)55 
 
 Ans. 13,75 
 
 (Art. 94.) When interest is required on any sum for 
 days only, it is a universal custom to consider 30 days a 
 month, and 12 months a year; and, as the unit of time 
 is a year, the interest of any sum for one day is ^^^, 
 what it would be for a year. For 2 days, ^^f^, &c.; hence, 
 if we multiply by the days, we must divide by 360. 
 
 EXAMPLES. 
 
 1. What is the interest of $327,30, for 25 days, at 6 
 per cent.? 
 
 100 
 
 327,30") Gives interest for 1 year, by General 
 6 5 Rule, Art. 89. 
 25 
 
 The 6 'will cancel in the 360: then the operation will 
 stand thus : 
 
 360 
 
 100 
 00 
 
 327,30 
 25 
 
 This gives the following special rule, when the inter- 
 est is six per cent., and the time, days. 
 
 Rule. Multiply the principal by the number of days; 
 divide the product by 60, and remove the decimal point 
 two places to the left. 
 
SIMPLE INTEREST. 159 
 
 To obtain the result in this particular example, we ob- 
 serve the numbers will still further cancel. 
 
 2. What is the interest of 854 dollars, for. 30 days, at 
 6 per cent, per ainuiin ? Ana. $4,27. 
 
 3. AVhat is the interest of 1100 dollars, for 48 days, at 
 6 per cent, per annum ? Jlns, $8,80. 
 
 4. What is the interest of 3459 dollars, for 75 days, 
 at 6 per cent, per annum ? .^n.?. $43,23 cts. 7 m.-4- 
 
 5. What is the interest of 1500 dollars, for 60 days, 
 at 5 per cent, per annum? Ans. $12,50. 
 
 6. WHiat is the interest of 1000 dollars, for 29 days ? 
 
 Ans. $4,833 -f- 
 
 7. What is the interest of 204 dollars, for 40 days ? 
 
 Ans. $1,36. 
 
 8. W^hat is the interest of 472 dollars, for 18 days ? 
 
 Ans. $1,416. 
 When no rate is mentioned, 6 per cent, is understood. 
 
 (Art. 95.) To compute interest on Notes, Bonds, 
 and Mortgages, on which partial payments have been 
 made, two or three rules are given. The first of the follow- 
 ing is called the common rule, and applies to cases where 
 the time is short, and payments made within a year of 
 each other. This rule is sanctioned by custom and com- 
 mon law ; it is true to the principles of simple interest, 
 and requires no special enactment. The other rules are 
 rules of law, made to suit such cases as require (either 
 expressed or implied) annual interest to be paid, and of 
 course apply to no business transactions closed within a 
 year. 
 
 Rule 1. Compute the interest of the principal sum 
 for the tvhole time to the day of settlement, and find the 
 amount. Compute the interest on the several payments 
 from the time each was paid to the day of settlement ; 
 add the several payments and the interest on each to- 
 gether, and ccdl the sum the amount of the payments. 
 Subtract the amount of the payments from the amount 
 of the principal, ivill leave the surn due. 
 
 The second rule which follows, is the rule of the Su- 
 preme Court of the United States, and with some trifling 
 
1 60 arith:\ietic. 
 
 variation, adopted by several of the States. We give it 
 in the language of Chancellor Kent, as expressed in the 
 New York Chancery Reports : 
 
 Rule 2. '' .Hpply the payment, in the first place, to 
 the discharge of the interest then due. If the paipntnt 
 exceeds the interest, the surplus goes toivnrds discharg- 
 ing the principal, and the subsequent interest is to be 
 computed on the balance of principal remaining due. 
 If the payment be less than the interest, the surplus of 
 interest must not be take?! to augment the principal ; 
 but interest continues on the former principal, until the 
 period when the payments taken together exceed the in- 
 terest due, and then the surplus is to be applied toward 
 discharging the principal ; and interest is to be compu- 
 ted on the balance, as aforesaid.''^ 
 
 Rule 3d, which follows, is one adopted by the Su- 
 preme Court of the State of Connecticut, is more expli- 
 cit and guarded against granting partial compound inter- 
 est, than the Chancellor's rule. The Connecticut rule 
 does not require interest to be paid until the principal has 
 run a year. It supposes that interest is not due under 
 that time, though payments may have been made every 
 few months, and the debtor is allowed interest on his pay- 
 ments up to the end of the year, agreeably to our first 
 rule. But let the rule speak for itself: 
 
 CONNECTICUT RULE. 
 
 Rule 3. ''Find the amount of the principal to the 
 time of the first payment, if it be a year or more after 
 interest commenced, and subtract this payment from it. 
 The remainder will be a neiv principal, on which cal- 
 culate the amount to the jiext payment, and from it sub- 
 tract the payment. Proceed in this way from payment 
 to payment, until they are all employ ed, provided a year 
 or rnore intervenes between each two payments. But, 
 if the intervening time be less than a year, find the 
 amount of the last principal for a year, and of the pay- 
 ment up to the same time, and subtract the latter fr on 
 the former. If however, a year overruns the time of 
 settlement, find the amounts up to that time, instead of 
 for a year. 
 
SIMPLE INTEREST. 161 
 
 If any payment is less than the interest then due^ 
 continue the former principal to the next payment, and 
 put both payments together. 
 
 EXAMPLES. 
 
 1. For value received, I promise to pay Smith Jones, 
 or order, one thousand dollars on demand, with interest 
 from date. William Smith. 
 
 Boston, January 1st, 1836. 
 
 On taking up this note, Deo. 1st, 1836, the following 
 endorsements were found on the back: 
 *1. 1836. March 1. Received, $250. 
 
 2. 1836. July 1. Received, 300. 
 
 3. 1836. Oct. 1. Received, 125. 
 
 What was due on the note? ,/ins. By Rule 1, $360. 
 
 By Rule 2, 360,90. 
 
 It will be observed, however, that this note should be 
 settled by Rule 1. To show the harmony and discrep- 
 ancy of these rules, let us suppose the following case : 
 
 2. A gave his note to B, for 10,000 dollars ; at the end 
 of 4 months, A paid 6000 dollars; and at the expiration 
 of another 4 months, he paid an additional sum of 3000 
 dollars ; how much did he owe B, at the close of the 
 year? 
 
 By the Common Rule, 
 
 Principal, $10,000 
 
 Interest for the whole time, 600 
 
 Amount, $10,600 
 
 1st payment,. • . $6000 
 Interest, 8 months, 240 
 2d payment, 3000 
 
 Interest, 4 months, 60 
 
 Amount, $9300 9,300 
 
 Due, $1300 
 
 It will be observed that the Connecticut rule must work 
 
 * We find the time elapsed between two payments, by com- 
 pound subtraction, (Art. 29.) 
 
 o2 
 
162 ARITHMETIC. 
 
 this question liie same way, and arrive at the eame re- 
 sult. Now let us take the"2d, or Chancellor Kent's rule. 
 
 Principal, $10,000 
 
 Interest, 4 months, 200 
 
 Amount, 10,200 
 
 Subtract payment, 6,000 
 
 New principal, 4,200 
 
 Interest, 4 months, 84 
 
 Amount, 4,284 
 
 Second payment, 3,000 
 
 New principal, 1,284 
 
 Interest, 4 months, 25,68 
 
 Remains due, $1,309,68 
 
 The dilTerence resulting from the different rules, as 
 applied to this case, is $9,68. To thoroughly under- 
 stand this, let us consider that $200 interest was paid to 
 B., out of the 6000 dollars, 8 months before it was pro- 
 perly due ; and B. had the opportunity to put that sum 
 out at interest again, and thereby receive partial com- 
 pound interest. This is the reason that the Connecticut 
 rule requires interest to run a year on the principal, and 
 interest on the payments up to the end of a year.* That 
 this is a correct view of the case, let us cast the interest 
 on the $200 for 8 months, and on the $84 for 4 months, 
 which was paid out of the $3000, 4 months before it 
 was due. Interest on 200 dollars, 8 months,=8. 
 
 " on 84 dollars, 4 months, =1,68, 
 
 $9,68. 
 
 *NoTE. Annual mieresi, actually paid when due, is in fact [or 
 in effect] compound interest, and these State rules seem to be a 
 sort of compromise between simple and compound interest ; they re- 
 cognize the principles of simple interest, when they sutler interest to 
 overrun a year, and give compound interest, liy not allowing interest 
 on the payments, or by requiring interest to be paid until the day of 
 I settlement. 
 
SIMPLE INTEREST. 163 
 
 3. A note for 2000 dollars was given January 4, 1827, 
 on which there were the lollowing payments : 
 
 February 19, 1828, $400. 
 
 June 29, 1829, 1000. 
 
 November 14, 1829, .520. 
 
 How much remained due Dee. 24, 1830, interest at 6 
 per cent.? 
 
 Principal, $2000 
 
 Interest from Jan. 4, '27, to Feb. 19, '28, 
 
 (13^ months,) 135 
 
 First amount, 2135 
 
 First payment, 400 
 
 Balance, forming a new principal, 1735 
 
 Interest from Feb. 19, '28, to June 29, '29, 
 
 (16^ months,) 141,69 
 
 Second amount, 1876,69 
 
 Second payment, 1000 
 
 Balance, forming a new principal, 876,69 
 
 Interest from Ju"ne 29, to Nov. 14, (4| mos.) 19,72 
 
 Third amount, 896,41 
 
 Third payment, 520 
 
 Balance, forming a new principal, 376,41 
 
 Interest from Nov. 14, to Dec. 24, (131 m.) 25,09 
 
 Balance due on taking up the note, $ 401,50 
 
 This is by the United States Court Rule, which is 
 sometimes called the New York and Massachusetts rule 
 (Rule 2). By the Connecticut rule, the balance due on 
 
 this note is $401,16. 
 
 The difference, as the reader will observe, arises by pay- 
 ing (according to the New York rule) $19,72 interest, 
 71 months before the Connecticut rule considers it due ; 
 and by the Connecticut rule, we cast interest on $870,69 
 for one year, and then on the payment of $520, for 7J- 
 
164 ARITIIMETIC>^^\yx\.^^ 
 
 months, up to the end of the year, aiicl subtract the 
 amount for the amount of the principal sum. Then we 
 cast interest on the remainder, to the day of setllement. 
 By Rule 1, no interest would be paid until the day of 
 setdement; hence, with heavy sums and long trjnsac- 
 tions, or in any case where annual interest is expected, 
 Rule 1 is not* proper; but where a note or bond does 
 not run two years, and several payments are made witli- 
 in a few months of each other, Rule 1 corresponds, both 
 in letter and spirit, to the principles of simple interest, 
 opposed to usury. A correct understanding of this mat- 
 ter among the people, may prevent considerable litigation 
 and some injustice/-' 
 
 4. $2500.' One year from date, I promise to pay James 
 Vance, or order, two thousand tive hundred dollars, with 
 interest, for value received. George Kidd. 
 
 Bosloih Jan'y 1, 1836. 
 
 As a personal favor to James Vance, George Kidd 
 paid on this note, 
 
 Mav 1, 1836, ^400. 
 
 Julv 1, 1836, 300. 
 
 No'vember 1, 1836, 600. 
 
 February 1, 1837, 800. 
 
 George Kidd not being able to pay more at this time, 
 and James Vance being pressed for money, disposed of 
 the note at under value, to a note-shaver. Kidd took up 
 the note Julv 1, 1837; what was then due? 
 
 Ans. Bv Rule 1, S535. 
 
 By Rule 2, 541,77. 
 
 Difference, 6,77. 
 
 A difference of feeling now arose, as to which rule 
 should decide the balance ; the note-shaver, like Shylock, 
 
 * On making inquiry of a lawyer, of very considerable practice, 
 how he made up balances on notes, mortgages, &c., cm which several 
 payments had bten made, he replied— When my cUent is plaintiff, 
 I go from payment to payment: when defendant, and is to pay the 
 balance, I reckon on the whole time, and then on the payments up 
 to the day of settlement ; and, if the opposing lawyer makes no ob- 
 jections, all is well ; if he does, I must conform to law. 
 
SIMPLE INTEREST. 165 
 
 contended for the law, the time being more than a 
 year. 'J'he other contended that he paid part of the note 
 before it was due, to accommodate tlie former holder, 
 and the present holder bonirht it under vahie ; and more- 
 over, according to the principles of equation of payments, 
 no partial compound interest could be exacted. Which 
 rule should decide the balance? The technicalities of the 
 law favor Rule 2d ; the true spirit of the law would es- 
 timate by Rule 1st. 
 
 5. For value received, I promise to pay Joseph Linn, or 
 order, two hundred and seventeen dollars and tifty cents, 
 in four months, with interest afterwards. G. Davis. 
 
 Hartford, July 11, 1831. 
 
 On tins note were the three following endorsements : 
 
 1831. Nov. 16, $93. 
 
 18.32. Feb. 12, 50. 
 
 1832. Aug. 2, 67,50. 
 
 1832, Oct. 4, the note was taken up; how much was 
 
 due upon it? ^n.5. $11,05. 
 
 This being in Connecticut, must be computed by the 
 Connecticut rule. 
 
 6. A note of hand, dated New York, April 4, 1838, 
 was given for the payment of six hundred dollars, on 
 which there were endorsements as follows: — July 10, 
 
 1838, $84,64; November 22, 1838, $10,00; April 30, 
 
 1839, $14; December 5, 1839, $309. What was the 
 balance due on taking up the note, April 5, 1840 ? 
 
 Ans. $251,03. 
 
 Neiv York, July 10, 1830. 
 
 7. Four years from date, we jointly and severally pro- 
 mise, for value received, to pay to the order of George 
 R. Guernsey, eight hundred and sixty-four dollars, with 
 interest. Robert C. Duncan. 
 
 David Johnston. 
 On this note are endorsements as follows: 
 
 April 6, 1831, $34. 
 
 June 21, 1832, 300. 
 
 February 26, 1833, 180. 
 
 January 1, 1834, 40. 
 
 What was the balance due at the maturity of the note ? 
 
166 ARITHMETIC. 
 
 COMPOUND INTEREST. 
 
 (Art. 96.) When interest is payable at definite and 
 stated periods, such as rents and special loans, for a deti- 
 niie time, the time comes, and neither principal nor in- 
 terest are paid — both due. A new obligation throws 
 them into one sum, forming a new principal, the interest 
 being on interest. Another period elapses; nothing is 
 paid, and a like obligation takes place, interest again be- 
 ing on interest, which is called compound interest. This 
 explanation must give us the following 
 
 Rule. Compute the interest up to the time that it be- 
 came payable, and add it to the principal; then cast 
 the interest on that amount for the next period, and 
 add it to its principal, and so on. The first principal 
 subtracted from the last amount, will give the compound 
 interest for the whole time. 
 
 Example. What is the compound interest of $250 for 
 3 years, at 7 per cent.? The compound interest of $250, 
 is 250 times greater than the compound interest for $1,00. 
 We therefore compute on 1 dollar. 
 
 Principal and interest, 1st year, 1,07 
 
 Multiply by 1+07, or, 1,07 
 
 7,49 
 1,07 
 
 Principal and interest, 2 years, 1,1449 
 
 Multiply by 1,07 
 
 80143 
 1,1449 
 
 Amount of $1, for 3 years, 1,225043 
 
 Drop $l,and we have ,225043 for the compound interest 
 of 1 dollar; which, multiplied by 250, gives $56,36, Ans. 
 The above operation might be expressed thus: 
 (1,07=^— 1)X 250 = 56,26. 
 
COMPOUND INTEREST. 167 
 
 2. Find the amount of 1 dollar every year, for 4 years, 
 at compound interest at 6 per cent. 
 
 Amount at the end of the 1st year, $1,0G 
 
 1,00 
 
 Interest, 2d year, ,00 3G 
 
 Principal, 1,00 
 
 Amount at the end of 2d year, 1,12 30 
 
 1,06 
 
 Interest, 3d year, ,0674 16 
 
 Principal, 1,12 36 
 
 Amount at the end of 4th year, 1,19 10 16 
 
 1,06 
 
 Interest, 4th year, ,07 14 60 96 
 
 Principal, 1,10 10 16 
 
 Amount at the end of 4th year, • • • 1,26 24 769 
 (Art. 97.) In this manner, we can compute the amount 
 of one dollar at any per cent., for any number of years, 
 and it is evident that two dollars will amount to twice as 
 much as one dollar; three dollars, three times as much ; 
 ten dollars, ten times as much, &c. Hence, if we have 
 a table with the amount of one dollar, at several rates, 
 and for a number of years, we can take such amount 
 from the table, and multiply it by the number of dollars 
 in question, and the product will be the amount of that 
 number of dollars, for the time required. 
 
 TABLE 1, 
 
 Showing flip amovjif of One Dollar for one month and 
 quarters, at the several rates mentioned. 
 
 1 month. 
 
 1 quarter. 
 
 2 quarters 
 
 3 quarters 
 
 4 qr. 1 yr. 
 
 4 per ci'iii.4-i- per cent.; 5 per cent, jo^ per cent. 6percen 
 
 1,003333 
 1,009853 
 1,019804 
 1,029852 
 1,04 
 
 1,003750 1,0041G7 1,004583 1,005 
 1,011065 1,012272 1,013475 1,014074 
 1,022252 1,024694 1,027132 1,029563 
 1,033563 1,037270 1,040973,1,044671 
 1,045 11,05 1,055 |l,06 
 
168 
 
 ARITHMETIC. 
 
 TABLE 2,*' 
 
 Showing the amount of One Dollar, from one to twen- 
 ty years. 
 
 r 4 percent. I 4^ per cenl. 5 per cent. SJ per cent. G per 
 
 1,0400000 
 1,0816000 
 1,1248640 
 1,1698585 
 1,2166529 
 1,2653190 
 1,3159317 
 1,3685690 
 1,4233118 
 1,4802442 
 1,5394540 
 1,6010322 
 1,6650735 
 1,7316764 
 1,8009435 
 1,8729812 
 1,9479005 
 2,0258161 
 2,1068491 
 2,1911231 
 
 1,0450000 
 1,0920250 
 1,1411661 
 1,1925186 
 1,2461819 
 1,3022601 
 1,3608618 
 1,4221006 
 1,4860951 
 1,5529694 
 1,6228530 
 1,6958814 
 1,7721961 
 1,8519449 
 1,9352824 
 2,0223701 
 2,1133768 
 2,2084787 
 2,3078603 
 2,4117140 
 
 1,0500000 
 1,1025000 
 1,1576250 
 1,2155062 
 1,2762815 
 1 1,3400956 
 1,4071004 
 1,4774554 
 1,5513282 
 ; 1,6238946 
 1,7103393 
 '1,7958563 
 i 1,8856491 
 :i,9799316 
 2,0789281 
 |2,1828745 
 1 2,2920 183 
 ,2,4066192 
 2,5269502 
 2,6532977 
 
 1,0550000 
 1,1130250 
 1,1742413 
 1,2388246 
 1,3069598 
 1,3788426 
 1,4546789 
 1,5346862 
 1,6190939 
 1,7081440 
 1,8020919 
 1,9012069 
 2,0057732 
 2,1160907 
 2,2324756 
 2,3552617 
 2,4848011 
 2,6214652 
 2,7656458 
 2,9177563 
 
 1,0600000 
 1,1123600 
 1,1910160 
 1,2624769 
 1 1,3382256 
 1,4185191 
 1,5036302 
 11,5938480 
 1,6894789 
 11,7908476 
 1 1,8982985 
 2,0121964 
 2,1329282 
 1 2,2609039 
 '2,3965581 
 2,5403517 
 12,6927727 
 '2,8543391 
 3,0255995 
 |3,2071355 
 
 * By inspection of Table 2, it will be seen that any sum doubles 
 itself at 4 per cent., compound interest, in less than 18 years; at 4^ 
 per cent., in less than 16; at 5 per cent., in less than 15 ; and at 6 
 per cent., in less than 12 years. The amount, at every rate, in- 
 creases faster and faster, as the time increases. In 19 years, at 6 per 
 cent., the amount is more than 3 times the original amount ; and in 
 40 years, more than ten times that sum. These facts show that it 
 would work much injustice and ruin, to allow compound interest. 
 Though in times of prosperity, and in schemes of speculation, mo- 
 ney may be worth 6, and even a greater, per cent.; yet, in the sober 
 realities of life, in unsettled estates in law, men in a state of bank- 
 ruptcy, or other contingency, which must leave debts in an unsettled 
 condition, ruin must follow, if compound interest were allowed by 
 law. Capital would swallow up competency, the widow and the 
 orphan would lose their inheritance, and the unfortunate could 
 never recover from misfortune. There is much small logic abroad, 
 concerning the abstract justice of compound interest ; but this sub- 
 ject has engaged the attention of the most gifted minds, in all ages, 
 and in every civilized country ; and the result has invariably been 
 a condemnation of eorapound interest 
 
COMPOUND INTEREST. 109 
 
 3. Find the compound interest for 475 dollars, for 5 
 years, at 5 per cent. From the table, opposite 5 years, 
 
 and under 5 per cent., we find 1,34009 
 
 For the amount of $1. We multiply by 475 
 
 6,70045 
 9,38003 
 5,30036 
 
 Amount, . . . 036,54275 
 Principal, • . 475 
 
 ^ns. 161,54 + 
 N. B. When the interest alone is required, as in the 
 preceding example, neglect one unit in the table, and ihe 
 product will be the interest at once. 
 
 4. Wliat will 786 dollars amount to, in 3 years and a 
 quarter, at 6 per cent.? ^^ns. $949,87 -{- 
 
 l,1910ft)=amount of 1 dol. for 3 years. 
 l,014674=am't of 1 dol. for 3 mos. or 1 qr. 
 
 4764064 
 8337112 
 7146096 
 4764064 
 1191016 
 1191016 
 
 1 ,208492968784=the amount of $1 for the given time. 
 78.^=: the given sum. 
 
 7250957812704 
 9607943750272 
 8459450781488 
 
 949,875-}- =the amount for the time required. 
 
 5. What is the compound interest of 629 dollars, for 
 7 years, at 6 per cent.? ^^ns. $316,78 -f 
 
 6. What will be the compound interest on a bond, for 
 I $750, for 4| years, at 5^ per cent.? Ans. $217,18 + 
 
 — 
 
170 ARITHMETIC. 
 
 7. What is the compound interest of 3200 dollars, for 
 8 years, at 5} per cent.? 
 
 8. What is the compound interest of 720 dollars, for 
 7 years, at 6 per cent.? Ans. $362,61 -\- 
 
 9. A broker loans money at the rate of 6 per cent, per 
 annum, but renews quarterly ; how much interest does 
 he receive per annum, on 100 dollars ? 
 
 Principal, 100 
 
 Interest, 1st quarter, 1,50 
 
 Principal, 2d quarter, 101,50 
 
 Rate per quarter, ,015 
 
 50750 
 1015 
 
 Interest, 2d quarter, 1,52250 
 
 Principal, 101,50 
 
 Principal, 3d quarter, 103,0225 
 
 ,015 
 
 5151125 
 1930225 
 
 Interest, 3d quarter, 1,5453375 
 
 103,0225 
 
 Principal, 3d quarter, 104,56783 
 
 ,015 
 
 52283915 
 10456783 
 
 Interest, 4th quarter, 1,56851745 
 
 Principal, 104,56783 
 
 106,13634 
 100 
 
 Ans. $6,136 -f 
 
DISCOUNT AND BANKING. 171 
 
 On 1000 dollars, the interest would be 61 dollars and 
 30 cents. It will be observed, that this problem does not 
 correspond, in number, to Table 2 ; nor should it, though 
 the rate per annum is tbe same; the imit of time, in that 
 table is one year, and the unit of time in this example, is 
 one quarter. 
 
 DISCOUNT AND BANKING. 
 
 (Art. 08.) Discount, — counting back, — a deduction 
 from an account, applicable to demands due at a future 
 time not drawing interest, and applicable to notes on 
 which the interest is paid in advance ; the drawer re- 
 ceiving a sum in hand, which, at the specified rate of in- 
 terest, will amount to the face of the note in the speci- 
 (lecl time. The problem, then, of discount, is, from the 
 raff, lime and amount, to find the principal. Interest 
 is proportion (Art. 89) ; hence discount is proportion also. 
 A double sum, to run the same time and at the same 
 rate, will require a double discount, &c. &c. Now, if 
 we assume a principal, and compute the amount on it 
 for the given rate and time, that amount will be to its 
 principal, as the given amount to its required principal. 
 If we take unity for the assumed principal, and compute 
 the amount corresponding to the given rate and time giv- 
 en in any problem, that amount will be the 1st term of 
 a proportion, one (or unity) the 2nd term, and the given 
 sum the ,3rd term. But, as the 2nd term is one, we 
 find the 4th term by dividing the 3rd term by the 1st. 
 Therefore, to obtain the present worth of a future pay- 
 ment, we have the following 
 
 Rule. Divide the given sum by the amount of one 
 dollar, or one pound, for the given rate and time. The 
 difference between the given sum and present worth, is 
 the discount. 
 
 EXAMPLES. 
 
 ]. What is the present luorth and discount on $150, 
 due 9 months hence, discounting at the rate of 6 percent.? 
 
172 
 
 ARITH3IETIC. 
 
 The amount of 1 dollar for 9 months, at 6 per cent., is 
 1,045; hence, as 1,045 : 1 : : 150 : present worth. 
 By rule : 
 
 1,045)150,000(143,53 -j- 
 1045 
 
 4550 
 4180 
 
 Given sum, . 
 Present value 
 
 Discount, . . . 
 
 
 3700 
 3135 
 
 . 143,53=Prin. 
 
 . $6,47 
 
 5650 
 5225 
 
 3250 
 3135 
 
 
 115 
 2. A holds a note of 375 dollars, due in 90 days, but 
 is much in want of ready money, which the debtor 
 agrees to pay, provided A will allow discount at 7 per 
 cent.; what would be the discount? Amount of $1, 3 
 months, at 7 per cent., is 1,0175. By rule, 
 
 1,0175)375,0000(368,55 + 
 30525 
 
 69750 
 61050 
 
 87000 
 81400 
 
 Given sum, • • • 375= Amount. 
 Present worth, 368,55 =Prin. 
 
 Discount, 
 
 . $6.45, Jns. 
 
 50000 
 
 50875 
 
 51250 
 
 50875 
 
 3750 
 
DISCOUNT AND EANKIXG. 173 
 
 Let the pupils notice, that the sums we call present 
 values, in the precedinof examples, are principals, which, 
 put at interest for their respective rates and times, would 
 give the amounts stated in the questions. 
 
 3. A person pays a debt of 394 dollars, 1 year and 8 
 months before it is due, discounting at per cent.; what 
 must he pay? Ans. $358,18 -h 
 
 4. What is the present worth of 775 dollars 50 cents, 
 due in 4 years, at 5 per cent, per annum? 
 
 Jln.'i. t|;564G,25. 
 
 5. What is the present worth of 580 dollars, due in 8 
 months, at 6 per cent, per annum? Jhi^. $557,69 -\- 
 
 6. What is the present worth of 954 dollars, due in 3 
 years, at 41, per cent, per annum .' .uUh. $840,528 + 
 
 7. What is the discount of 205 dollars, due in 15 
 months, at 7 per cent, per annum ? .^ns. $16,495 + 
 
 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. $746,987. 
 
 9. B owes A to the value of 1005 dollars, to pay as fol- 
 lows: viz. 475 dollars in 10 months, and the remainder 
 in 15 months; what is the present wortli, allovv^ing dis- 
 count at 6 per cent, per annum ? Ans. $945,404. 
 
 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 per 
 cent.? Ans. $156,462 + 
 
 (Art. 99.) Discount and Bank interest ought to be 
 the same, and are the same in the state of New York; 
 but elsewhere, even in England, as far as the author's in- 
 formation extends, bank interest is simple interest, on 
 the face of a note paid in advance ; not interest on the 
 present worth — not interest on what the drawer actually 
 receives — but interest on what he promises to pay at a 
 specified time. For example, A presents his note at a 
 bank, properly endorsed to run a year, the legal interest 
 of the state being 6 per cent.; the bank gives him $94, 
 reserving 6 dollars for discoxmt. Now, 94 dollars, put 
 out at interest for a year, at 6 per cent., will not amount 
 
 to 100 dollars; therefore, the rate per cent, that a bank 
 _ 
 
174 ARITHMETIC. 
 
 draws interest, is more than the nominal rate — in fact, 
 more than lawful interest.- Banks rarely discount paper 
 that has more than 4 months to run ; they always prefer 
 short periods, for two reasons. One is, that it is difficult 
 to look far into futurity, and decide on future circumstan- 
 ces ; and the other is!^ that short intervals, and advance 
 payments of interest, give them all the advantages of 
 compound interest. Bank notes — or rather, notes dis- 
 counted by a bank — run three days longer than tJie time 
 specilied, called three days of grace; but the interest is 
 computed for the three days, and the note need not be 
 taken up until the last day. 
 
 From the preceding explanations, it is evident that Bank 
 interest must be compu ^d as simple interest, on the face 
 of the note adding three days to llie time — except in New 
 York, where bank interest is discount, as computed by 
 the rule under Art. 98. 
 
 EXAMPLES. 
 
 1 . A note for 350 dollars, for 90 days, was discounted at 
 bank ; what was the discount ? 
 
 Operation. 
 Rule, under Art. 94, . . • 20 /t0(^ 1 3^^0 7 
 2 ^0 1 ^^ 31 
 
 4,0)217 
 
 .fins,' • .$5,42',. 
 
 2. What is the bank interest on a note of 135 dollars, 
 payable 90 days after date? Ans. $2,091 -f 
 
 3. What is"^the bank interest on a note of 1850 dol- 
 lars, payable 90 days after date? Ans. $28,671. 
 
 4. What is the bank interest on a note of 475 dollars, 
 payable 30 days after date ? Ans. 82,61. 
 
 5. AVhat will be the proceeds of a note of 1370 dol- 
 lars, payable 30 days after date ? Ans. 1362,46|. 
 
 * Yet, the supreme courts ha '3 decided, that this mode of cal- 
 culating discount is not usury Banks are, therefore, authorized 
 by judicial authority to discount at a higher rate than that establish- 
 ed by law. We know not on what principles this decision has been 
 made. 
 
DISCvOUNT AND BANKING. 175 
 
 6. What will be the proceeds of a note of 880 dollars, 
 payable 90 days after date ? ^2ns. $860,3G. 
 
 MISCELLANEOUS PROBLEMS IN INTEREST. 
 
 (Art. 100.) The following problems are best nnder- 
 stood by cause and effect, as we have already explained 
 (Art. 84). We only give a few more examples in this 
 place, as being in local order. 
 
 1. A loaned 1600 dollars at 6 per cent., until it amount- 
 ed to 2000 dollars ; what was the time ? 
 
 Ans. 4 years 2 montlis. 
 Cause. Effect. Cause. Effect. 
 100 : 6 : : 1600 : 400 
 Statement, • • • 1.2 fl 
 
 2. A sum of money was put out at interest, at 7 per 
 cent., and remained 3 years 5 months and 10 days, and 
 the interest was $15,57; what was the principal ? 
 
 Ans. $64,58. 
 
 Cause. Effect. Cause. Effect. 
 
 Q, , , 100 : 7 :: [] : 1^,57 
 
 Statement, ... ,^ k-[ 
 
 3. At what rate per cent, per annum, will 1200 dollars 
 amount to 1476 dollars, in 5 years and 9 months? 
 
 Ans. 4 per cent. 
 
 4. If 834 dollars, at interest 2 years and 6 mouths, 
 amount to 927 dollars 82^ cents, what was the rats per 
 cent, per annum ? Jlns. 4,1^ per cent. 
 
 5. Sixty-three dollars was at interest 1 year 3 mouths, 
 and yielded 4 dollars 72^ cents; what was the rate? 
 
 .Sns. C. 
 
 6. The interest on a certain sum, for 6 years and 3 
 months, at 6 per cent., was 28|- dollars ; what was the 
 sum? Ans. $75. 
 
 7. The interest on 730 dollars, at 6 per cent., was 253 
 dollars 31 cents; what was the time? 
 
 Ans. 5y. 9 m. 12 d. 
 
 8. Six hundred and thirty dollars was put at interest, 
 at 6 per cent., and the interest received was 25 dollars 
 20 cents; what was the time out? Ans. 8 months. 
 
 9. Seven hundred and fifty dollars was at interest 9 
 
170 ARITHMETIC. 
 
 months, and the amount received was $789,375 ; what 
 was the rate per cent? ^8ns. 7. 
 
 10. If 300 doUars gives 45 dollars interest in 5 years, 
 what was the rate per cent.? Ans. 3. 
 
 11. The amount of 750 dollars, at 6 per cent., was 
 1301 dollars 25 cents ; what time was it out? 
 
 Ans. 12 y. 3 m. 
 
 12. The interest on a certain svim at 4 per cent., for 4 
 months and 18 days, was 9 dollars 20 cents ; what was 
 the sum ? Ans. $600. 
 
 13. In what time will 837 dollars amount to 1029 dol- 
 lars 51 cents, at 5 J per cent.? .dns. 4 years. 
 
 14. What sum will amount to 571 dollars 20 cents in 4 
 years, at 5 per cent.? [This comes under discount.] 
 
 Ans. $476. 
 
 15. The interest on 576 dollars, at 6 per cent., is 
 $60,48; what was the time? Ans. 21 months. 
 
 PERCENTAGE. 
 
 (Art. 101.) This indxdes Commission, Broherage and 
 Insurance; all of which is paid at so much per hundred; 
 hence, it is called per-centage, and must fall back on pro- 
 portion ; or, it may be more naturally compared to in- 
 terest, wanting the element of time, or, as in the case of 
 insurance, considering no other than the iinit of time ; 
 but commission and brokerage have no reference to time 
 whatever; we only consider the proportion to the given 
 rate per 100 : hence the following 
 
 Rule. Jis 100 is to the given rate per cent., so is the 
 given sum to the required commission, brokerage, or 
 insurance.'- Or, proceed in the same manner as though 
 
 ■* Commission is compensation for selling or buying goods for eth- 
 ers. Brokerage is allowance made to dealers in money or stocks, 
 for the purchase of stock or the exchange of money, or commercial 
 paper. Insurance is money paid to a joint-stock company, for them 
 to assume the hazard and risk of damage by fire, or loss of any kind 
 by sea; and, in case of damage or loss, the company or insurers 
 are to make such loss good, as expressed in the contract. The written 
 
PER-CENTAGE. 177 
 
 you were required to find the interest of the given sum 
 for one year. 
 
 EXAMPLES. 
 
 1. An agent sold goods for his employer to the amount 
 of 1200 dollars ; what was his conmiission at 5 per cent.-? 
 Statement: as 100 : 5 :: 1200 : to answer. J]ns. $60. 
 
 2. Sold goods to the amonnt of 975 dollars 50 cents, 
 on a commission of 7^^ per cent.; what was niv compen- 
 sation? " /]ns. $73, J 6^. 
 
 3. A merchant exchanges 3250 dollars of unbankable 
 notes with a broker, allowing him 1^^^ per cent, for par 
 funds ; how much did the broker pay him ? 
 
 By practice ; thus, at 1 per cent., the broker's bonus 
 
 or commission would be, $32,50 
 
 At i per cent., 16,25 
 
 1 " " 8,125 
 
 Broker's full commission, $56,875 
 
 Therefore, the merchant must receive 3193 dollars 12^ 
 cents, the remaining part of the 3250 dollars. 
 
 4. An agent sells 750 bales of cotton, at 48 dollars per 
 bale, and received H per cent. coRimission ; how much 
 did he receive? 
 
 750 
 
 Q 1 ,. 100 
 
 Solution : 
 
 2 
 
 48 
 3 .^ns. $540. 
 
 5. What must be paid to insure a steamboat and cargo, 
 from Pittsburgh to New Orleans, at a premium of | of 1 
 per cent., the valuation of the whole being 47500 dollars? 
 
 /Jns. $356,25. 
 
 6. A gentleman in New York has a house valued at 
 15000 dollars ; he can obtain insurance lor 12000 dollars, 
 at a premium of 2| per cent.: what is his insurance tax ? 
 
 .^ns. $285. 
 
 7. The sales of certain goods amount to 1680 dollars; 
 what sum is to be received for them, allowing 2^ per 
 cent, for commission? ^^ns. $1633,80. 
 
 instrument expressing such contract, is called a policy The amount 
 of insurance paid, is called the premium. 
 
178 ARITHMETIC. 
 
 8. What is the insurance of 760 dollars, at 6!,- per 
 cent.? ^'liis. $49,40. 
 
 9. What is the insurance of 5630 dollars, at 7^ per 
 cent.1 ^ns. 436,325. 
 
 10. A merchant sent a ship and cargo to sea, in time of 
 war, valued at 17654 dollars; what would be the amount 
 of insurance, at 18} per cent.? .^ns. $3310,12',. 
 
 11. What is the brokerage on 21'50 dollars, at 2 per 
 cent.? ^^ns. $43. 
 
 12. A merchant shipped, on board a vessel bound 
 from Boston to St. Thomas, goods to the amount of 2464 
 dollars ; what must he pay to be insured against the dan- 
 gers of the sea, the premium being 2j per cent.? 
 
 ^)is. $55,44. 
 
 13. A merchant having a house, valued at 10650 dol- 
 lars, and goods in his store amounting to 6740 dollars ; 
 what must he pay to be insured against the danger of tire, 
 the premium being U per cent.? ^ns. $260,85. 
 
 14. A sends 4500 dollars to an agent, to purchase rail- 
 road stock, and allows the agent 2 per cent, for his trou- 
 ble in making the investment ; how much of the money 
 shall the agent retain? .^ns. $88,23 -f 
 
 (Art. 102.) In this case, it is evident that the agent 
 must not compute per-centage on that portion of the mo- 
 ney which he may retain. It is only 2 per cent, on the 
 portion spent for his employer. For every 100 dollars 
 the agent spends, he is to have 2 dollars ; then for every 
 102 dollars, 100 is to go to the principal, and the remain- 
 der to the agent. Therefore, as 102 : 100 : : 4500 : 
 a 4th term. This 4th term is $4411,76 4-? spent for the 
 employer, and the remaining $88,23 -{-, is retained by the 
 agent. Strictly speaking, this is diaeount. 
 
 (Art. 103.) Neither individuals nor insurance compa- 
 nies will ever insure to the full value of property, as this 
 might create negligence in those who have care of it; or 
 worse, tempt the unprincipled to destroy, or suffer its de- 
 struction, to turn it into cash. Yet, in tiie various turns 
 of business, it is sometimes necessary for individuals to 
 secure a policy for a given specified amount^ even at the 
 expense of paying a higher rate. Such a policy may be 
 
STOCKS. 179 
 
 wanted, to send before a vcsse' to a foreign port, lo sus- 
 tain the credit of a partner, agent, otr friend, until the in- 
 sured siiip or cargo sliall arrive. 
 
 To secure a given amount, in case of tlie destruction 
 of the property, we must draw for a higher sum, such a 
 sum as when the per-centage is taken off, will leave the 
 given amount. 
 
 Now, as all these operations are based on proportion, 
 it is evident that we must take the per-centage from 100: 
 then say, 
 
 ,fis the remainder is to 100, no is the desired amount 
 to the policy to be taken out. 
 
 Example 1. A merchant wishes to secure $1920 on 
 an adventure to London ; what sum must be mentioned 
 in the policy to cover that amount, the premium being 4 
 per cent.? 
 
 As 96 : 100 :: 1920 : fourth term ; 
 
 or, 1 : 100 :: 20 : 2000. Jins. $2000. 
 
 2. A gentleman has produce on hand to the amount of 
 $14000, which he intends to ship to Soutli America. 
 The marine insurance compaay are willing to insure to 
 the amount of $9000 for 2\ per cent. ; but he wishes 
 to secure $11000, and for this sum they demand a pre- 
 mium of 2| per cent.; what sum must be mentioned in 
 the policy ? Ans. $ 1 1 3 1 1 ,05. 
 
 3. A lady has property on which she wishes to secure 
 the sum of $15000; what must be the policy to cover 
 that sum, the premium being 6^ per cent.? 
 
 Ans, $16042,79. 
 
 STOCKS. 
 
 (Art. 104.) Stock is the term given to the aggregate 
 or any number of shares of the capital belonging to any 
 legal association or company, created for any special pur- 
 pose, such as banking, insurance, making public improve- 
 ments, or establishing manufactories. 
 
 To organize a bank, such as the people will have con- 
 
180 ARITHMETIC. 
 
 fidence in, it is necessary that it should have ready capi- 
 tal at all times, to redeem its notes ; and to secure this 
 end, a certain amount of capital is designated in the char- 
 ter or legal act of corporation : this capital is divided into 
 shares, H^enerally consisting of $100 each, for which indi- 
 viduals subscribe and pay. These individuals are called 
 stockholders, and receive the profits of the bank, in pro- 
 portion to the shares they hold. Certificates of stock are 
 bought and sold like individual notes or other property ; 
 and when the profits of the company are large, the stock 
 will command more than its nominal value in the mar- 
 ket, and is then said to be above par ; and when the pro- 
 fits are small, the stock sinks below par. 
 
 The rates above and below par are estimated at so 
 much on, or so much taken ofT of the 100. Hence, we 
 compute the value of stocks by per-centage, as in Art. 
 101, or Art. 103. 
 
 EXAMPLES. 
 
 1. A sells 25 shares of bank stock at 3 per cent, ad- 
 vance, the par value per share being $100 ; how much 
 money did he receive? ^^ns. $2575. 
 
 Statement : If 100 give 103, what will 2500 give ? 
 
 2. The stock in a certain rail-road company is 15 per 
 cent, below par; what are 12 shares worth, the original 
 value being $50 per share ? ^ns. $510. 
 
 Statement: 100 : 85 : : 12X50 : .^ns. 
 
 3. At a time of great fluctuation in stocks, a gentleman 
 bought 36 shares of the United States Bank stock, nom- 
 inal value $100, at 17|- per cent, advance, and sold the 
 same day at 181^ advance; what did he realize by the 
 transaction ? 
 
 ^7is. His gain was | of 1 per cent., or $27. 
 
 4. If the stock of an insurance company sells for 3i 
 per cent, below par, what are $1200 of the stock worth ? 
 
 Ans. $1159,50. 
 
 5. I directed a broker to purchase for me 75 shares in 
 a certain rail-road company, if he could obtain them at 
 
EQUATION OF PAYMENTS. 181 
 
 22 per cent below par, promising him f per cent, for his 
 trouble ; what did the whole cost me ? 
 
 ^ns. $5886,56 -}- 
 
 [Note. "When the original or nominal value of a share 
 is not mentioned, 100 is understood.] 
 
 6. Bought 56 shares in a New York state bank, at 7/, 
 per cent, advance, the original value being $75 per share ; 
 what will the whole cost me, allowing | per cent, to the 
 broker, who made the purchase ? Ans. $4530,75. 
 
 EQUATION OF PAYMENTS. 
 
 (Art. 105.) By this is meant the equitable time of 
 paying several debts due at ditferent times, so that neither 
 party siiall sustain any loss of interest. 
 
 To explain more fully, let us suppose that a person 
 owes his neighbor as follows : 
 
 2 dollars, to be paid in 2 months ; 
 2 dollars, to be paid in 3 months ; and 
 2 dollars, to be paid in 5 months. 
 In what time shall he pay the whole 6 dollars, so that 
 neither himself nor creditor shall lose interest? We an- 
 alyze it thus : 
 
 2 dollars in 2 months will gain as much interest as 
 
 4 dollars in 1 month ; 
 2 dollars in 3 months will gain as much interest as 
 
 6 dollars in 1 month ; 
 2 dollars in 3 months will gain as much interest as 
 10 dollars in 1 month. Then 
 
 $6 for the several times will g'ain as much interest as 
 $20 for 1 month. 
 But there are only 6 dollars to be paid, not 20 ; and as 
 interest is always in proportion to the compound of mo- 
 ney and time, we must find ivhat number of months, 
 
182 ARITHMETIC. 
 
 multiplied by G, will give the same product ari 20, multi- 
 plied by 1 ; and that evidently is -^' =3^- months. 
 From this we may derive the following general 
 
 Rule. Mulliply each payment by its thne, and di- 
 vide the sum of the several products by the sum of the 
 jjayments, and the quotient will be the equated time for 
 the payment of the whole, 
 
 EXAMPLES. 
 
 1. If 600 dollars are now due, 600 dollars in 4 months, 
 and 600 dollars in 8 months, 600 dollars in 12 months, 
 what is the equitable time for paying the whole ? 
 
 N. B. It is not necessary in this or any other prob- 
 lem, that we should use the sums of money actually giv- 
 en. It is sufficient that Ave use the same proportional 
 parts. In this case we will use one dollar in place of 600, 
 and the operation will stand thus : 
 
 IX 0= 
 IX 4= 4 
 IX 8= 8 
 1X12=12 
 
 4) =24(6 months, ,^ns. 
 
 2. If 750 dollars are to be paid, | of it in H years, 
 ~ of it in 2 years, and the residue in 2^ years, what is 
 the equated time of paying the whole at once ? • 
 
 ./???5. 23 J months. 
 In this example, it is not necessary to use 750 dollars; 
 we had better use 10. 
 
 3. A owes B 100 dollars, to be paid in 6 months ; 120 
 dollars, to be paid in 10 months, and 160, to be paid in 
 14 months; what is the equated time for paying the 
 whole? .^ns. 10|^r months. 
 
 4. A merchant haUi owing to him 300 dollars, to be 
 paid as follows: 50 dollars at 2 months, 160 dollars at 5 
 months, and the residue at 8 months; and it is agreed to 
 make one payment of the whole ; when must that time 
 be? Jlns. 6 months. 
 
 5. F owes H 2400 dollars, of which 480 dollars are 
 
EQUATION OF PAYMENTS. 183 
 
 lo be paid present, 960 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? 
 
 A.rins. 6 montlis. 
 
 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 make one payment of the whole ; what 
 is the mean or equated time I Ans. G months. 
 
 N. B. To solve the following, we had better fall back 
 on the general principle, and not attempt to apply the 
 rule. 
 
 7. A owes B 600 dollars, to be paid in 2 years from 
 the date of the note ; but, at the expiration of 6 months, 
 A agrees to pay 150 dollars, if B will wait enough long- 
 er for the balance, to compensate for the advance ; how 
 long ought B to wait? 
 
 Ans. 6 months after the 2 years. 
 
 Here 150 dollars was paid 18 months before the time ; 
 how long shall the remaining 450 dollars remain after the 
 expiration of the 2 years, to give the same interest. 
 
 Statement: 150X18 : 450X[] : : 1 : 1. 
 
 m 
 
 8. 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 forborne a longer time ; it is 
 agreed on; tlie time of forbearance, therefore, is required? 
 Ans. 7 months ; that is, 1 month in addition to the 6. 
 
 9. A young gentleman has a legacy of 1500 dollars, 
 to be paid him in 16 months ; but, being in want of ready 
 money, agrees to defer the payment of the balance the 
 proper time, if he can have 500 dollars in hand ; when 
 shall the balance be paid ? Ans. 2 years hence. 
 
 See Art. 80. ^ ^^0 
 
184 ARITHMETIC. 
 
 PROFIT AND LOSS PER CENT. 
 
 (Art. 106.) When articles are bought and sold, it is 
 often desirable to know the rate of gain or loss per 
 cent., corresponding to any definite or assumed price. 
 This of course can be done hy proportion, Practice, and 
 proportion. Interest and discount are all the arithmeti- 
 cal principles that can be brought into requisition, under 
 this head ; and practice, interest, and discount, are all re- 
 solvable into joro/)or/20?i; hence proportion alone is the 
 sole i^rinciple, 
 
 EXAMPI.ES. 
 
 1. A merchant bought cotton cloth at 1 shilling 6 
 pence per yard, and sold it at 1 shilling 10 pence; what 
 was his gain per cent.? 
 
 pence, pence, gain. 
 Statement: If 18 : 4 : : 100 : how much? 
 
 Ans. 22f . 
 
 2. A grocer bought tea at 60 cents per pound, and sold 
 the same at 75 cents; what was his profit per cent.? 
 
 Ans. 25. 
 
 As 60 : 15 :: 100 : Answer 
 Statement: ^.^ 4 . 1 : : 100 : 25 
 
 3. I bought Irish linen at 56 cents per yard ; what shall 
 I sell it for, to gain 20 per cent.? Ans. 67 fo- 
 
 Statement: As 100 : 120 : : 56 : Answer. 
 
 4. A merchant sold broadcloth at 4 dollars 75 cents 
 per yard, making a profit of 32 per cent.; what did the 
 cloth cost him ? Ans. $3,60, nearly. 
 
 Statement: 132, that he now receives, originally cost 
 him 100 ; in that proportion, what did 475 cost? or, as 
 132 : 100 :: 475 : Answer. 
 
 5. A merchant bought English broadcloth ; the first 
 cost, duties and transportation, amount to 14 shillings 8 
 pence per yard ; what must be his price, in dollars and 
 cents, to gain 20 per cent., the dollar being equal to 4 
 shillings 6 pence sterling? Ans. $3,91 -|- 
 
PROFIT AND LOSS PER CENT. 185 
 
 Statement: 100 : 120 :: 14| : price in shillings sterl- 
 ing. But to bring shillings, sterling money, into dollars, 
 we must divide by 4\, or multiply by 2 and divide by 9. 
 Tiie whole combined in one c peration (Art. 24), stands 
 thus : 
 
 100 
 9 
 
 44 
 
 ^0 4 
 2 
 
 6. If I purchase Irish linen at 2 shillings 4 pence per 
 yard, sterling money, what must I sell it at per yard, in 
 federal money, to gain 30 per cent.? ^^ns. 67^4. 
 
 7. A man bought 300 sheep, at 2 dollars 15 cents per 
 head, and his expense in making the purchase, was 45 
 dollars 50 cents ; he sold them at 3 dollars 20 cents per 
 head; what was his whole gain, and what was his gain 
 percent.? ^ns. Whole gain, $269,50; gain per cent. 
 39, nearly. 
 
 8. If I buy 12'2 hundred weight of sugar for 140 dol- 
 lars, how much must it sell at per pound, to make 25 per 
 cent.? ^ns. 12^- cents. 
 
 Solved as in Practice (Art. 88), 
 
 > cents. 
 
 
 
 140 
 100 
 
 
 100 
 
 125 
 
 CWf.' 
 
 . 12L 
 
 
 qr. . . 
 
 . 4 
 
 . 
 
 lb.. . 
 
 . 28 
 
 
 This cancels down to the answer. 
 
 9. Bought 126 gallons of wine for 150 dollars, and 
 retailed it at 20 cents per pint ; what was the whole gain, 
 and what the gain per cent.? 
 
 Ans. Whole gain, $51,60; gain per cent. 34|. 
 
 10. Bought a hogshead of molasses for 26 cents per 
 gallon, and suppose it lost 5 per cent, in waste and leak- 
 age ; how must I sell the remainder, per gallon, to gain 
 25 per cent.? Ans. 34yV. 
 
 11. If, by selling 1 pound of pepper for 10^ cents, 
 there are % cents lost, how much is the loss per cent.? 
 
 Ans, 16. 
 
 J2 
 
186 ARITHMETIC. 
 
 12. A merchant bought 180 casks of raisins at 2 dol- 
 lars 13 cents per cask, and sells them at 3 dollars 68 
 cents per hundred iveif(ht, and gains 25 per cent.; re- 
 quired the average weight in each cask. 
 
 jlns. 8 1/2 pounds. 
 cost. sale. 
 
 Solution: 180X213 : 368X[] :: 100 : 125 
 A proportion wanting a factor to one term, that factor is 
 the number of hundred weis^ht, Avhich must be reduced 
 to pounds, and divided by 180. By Art. 24 : thus, 
 368 ^^0 
 4 ^00 213 
 
 ;t^0 ^^^ ^ 
 112 
 
 13. A merchant receives raisins, which cost 2 dollars 
 50 cents per cask, and, by selling them at 5 cents a pound, 
 lie gains 20 per cent, on the first cost ; what was the av- 
 erage weight of each cask ?* ^^ns. 60 lbs. 
 
 14. What is the loss, and what is the loss per cent., 
 in laying out 70 dollars for hats, at 1 dollar 75 cents each, 
 and selling them at 25 cents a-piece less than cost? 
 
 Jns. Whole loss, $10, loss per cent. 14f . 
 
 15. A merchant bought 1200 pounds of coffee for 135 
 dollars ; what must he sell it at per pound, to gain 35 
 dollars on the whole? »^ns. 14^ cents. 
 
 16. Bought 2 hogsheads of wine, at 1 dollar 25 cents 
 per gallon, and sold the same at 1 dollar 60 cents ; what 
 was the whole gain, and the gain per cent.? 
 
 .^ns.- Whole gain, $44,10; gain per cent. 28. 
 
 17. If, by selling cloth at 4 dollars 50 cents a yard, I 
 lose 20 per cent., what was the prime cost? 
 
 .^715. $5,621. 
 
 18. Sold corn at 25 cents per bushel, and lost 3 cents: 
 what was the loss per cent.? ^ins. 10+. 
 
 (Art. 107.) When persons sell on credit, it is evident 
 that discount must be taken off, before we have the true 
 
 * This problem is extracted. Its author mentions 200 casks; we 
 omit the number, as it will cancel itself, whatever it be, as seen in the 
 preceding problem. 
 
PROFIT AND LOSS PER CENT. 187 
 
 profit; also, when goods have been on Imnd some time, 
 the interest of tirst cost shoukl be added on, before we 
 really have the prime cost. 
 
 19. Sold goods to the amount of 425 dollars, on 6 
 months credit, which was 25 dollars more than the goods 
 cost me ; what was the true profit, money being worth G 
 per cent.? JJns. $12,62. 
 
 20. A merchant sold goods for 667 dollars, to be paid 
 in 8 months : the same goods, one year ago, cost him 
 600 dollars, apparently gaining 67 dollars ; what was his 
 true gain, money commanding 6 per cent.? 
 
 An8. $5,346 -}- 
 [Note. That is, he gained 5 dollars and .34 cents, on 
 the supposition that he traded on borrowed capital, or 
 on the supposition that mterest is 7iot gain. But this is 
 not true ; interest is gain, and legal interest is as much 
 as mere capital ought to gain. In the case under consid- 
 eration, the merchant gained 5 dollars and 34 cents in 
 trading in goods, more than it is supposed he could have 
 done by trading in, or loaning money.] 
 
 21. A merchant has had a certain lot of goods on his 
 hands 6 months, and now has an opportunity to sell on 
 10 months credit, at an advance of 30 per cent, on the 
 original cost ; what are his profits per cent., when he is 
 paying 5 per cent, interest on capital ? 
 
 .%is. 21i 
 
 22. Sold goods to the amount of 723 dollars, ready 
 cash, at 15 per cent, above first cost ; what was my real 
 gain, having had the goods on hand 1 year and 6 months, 
 money being worth 6 per cent.: that is, how much more 
 do I gain by the purchase and sale of the goods than I 
 should have gaified by putting the original cost of the 
 goods at interest? Jhis. $37,73. 
 
 23. A grocer bought 3 barrels of sugar, the whole nett 
 weight being 690 pounds, at 8^^ cents per pound, and 
 sells it at 9'^ ; what is his whole gain, and gain per cent.? 
 
 .^ns I Wliolegain, $10,35. 
 ' 5 Gain per cent. IBfy. 
 
 24. A grocer bought 1'. begs of coff'ee, nelt weight of 
 the whole being 1350 pounds, a 12^ cents per pound; 
 
188 ARITHMETIC. 
 
 what must be his price per pound, to gain $50 on tiie 
 whole ? Ans. 10|- cents. 
 
 25. Bought a cask of molasses, containing 42 gallons; 
 6 gallons leaked out, and I sold the remainder lor 37 \ 
 cents per gallon, and gained 20 per cent. ; what was the 
 original cost per gallon I Ans. 26^^ cents. 
 
 REDUCTION OF CURRENCIES. 
 
 (Art. 108.) By this is meant the changing of any sum 
 of money from one sfundard to its equivalent value in 
 another, and involves no other principle than general re- 
 duction ; and, as in general reduction, the operator is 
 guided only by the given tabular values. 
 
 TABLE OF STATE CURRENCIES. 
 
 In New England and Virginia, Kentucky and 
 Tennessee. 
 
 3 
 3 
 18r/. = 0,25 J J 12^ cents=9 pence 
 
 In New York, Ohio and North Carolina. 
 2 
 8 
 24 d. ■■= ,25 (_ J 25 cents=2 shillings 
 
 In Pennsylvania, N. Jersey, Delaware and Maryland. 
 
 3 5.= ,40 -j or, [.$1=7 5. 6 fZ. =90(7. 
 9d.= ,10 ( \ ,50=3 s. 9 d.=45 d. 
 
 £=$10,00'] ■)Sl=,3i3 
 
 .s. = 0,50 lor, I $1=6 shillings. 
 
 d.= 0,25 J J 12^ cents=9 pe: 
 
 w York, Ohio and North Carolin 
 £=$5,00r "]$1=,4£ 
 5. =$1,00 4 or, [>$1=8 shillings. 
 d. ■■= ,25 (_ J 25 cents=2 shil 
 
 inia, N. Jersey, Delaware and Jh 
 
 £=$8,oor ~)$i=-2je 
 
 s.= ,40ior,[$l=7s.6d.={ 
 d.= ,10 t J ,50=3 5. 9c/.= 
 
 South Carolina and Georgia. 
 7^ =$30,00 ~) •)U = ^\£ 
 7 s. = 1,50 lor, I $1=4 5. 8 ^Z.=i 
 4 d.= ,25 J j ,25 = 1 5. 2 c?. 
 
 , . , =56ff. 
 
 14 
 
 [Note. In former days, before congress established 
 our national currency, all accounts were kept in pounds, 
 shillings and pence, 4 shillings 6 pence corresponding to 
 
REDUCTION OF CURRENCIES. 189 
 
 the Spanish dolhir. The several colonies, as these states 
 were then called, issued bills of credit, and passed them 
 as money ; but they soon depreciated, and moie in some 
 
 tates than in others, and at the general adoption of fede- 
 ral money, the values of the pounds, shilling, Sic. were 
 left as noted in the table. These currencies are legally 
 dead, and ought to be buried; but long habit becomes se- 
 cond nature to most people, and the detestable spirit of 
 avarice serves, in some places, to keep up the distinction. 
 For instance, in New York, where the penny and the 
 cent are very near in value to each other, the petty mer- 
 chants and shopkeepers purchase in cents, and sell in 
 pence, and thereby gain 4 per cent. 
 
 From the preceding table, to change pounds, shillings, 
 and pence into dollars and cents, we have the following 
 
 Rules. — 1. Reduce the shillings and pence lo the de- 
 cimal of a pound [Art. 69). Then state the probleyn in 
 proportion, by the relation of £ and $, taken from the 
 table. 
 
 Or, divide the given number of £^s, and its decimal, 
 by the decimal of a pound that makes a dollar. 
 
 Or, if it is a vulgar fraction, multiply by its denom- 
 inator, and divide by the numerator, taking care to can- 
 cel, if possible. 
 
 2. Reduce the given pounds, shillings and pence, to 
 shillings and decimals of a shilling, and divide by the 
 number of shillings in a dollar; or, if more convenient, 
 reduce all to pence, and divide by the number of pence 
 in a dollar, 
 
 examples. 
 
 1. Reduce 25 pounds 8 shillings 6 pence, New Eng- 
 land currency, to dollars. 
 
 Operation, 
 6 
 
 12 
 
 20 
 ,3 
 
 8,5 
 
 25,425 =pounds and decimals of a pound. 
 
 Ans, $84,75 
 
190 
 
 ARITHMETIC. 
 
 2. Reduce 42 pounds 10 shillings 3 pence, New York 
 currency, to dollars. 
 
 1st Operation. 
 
 12 
 20 
 ,4 
 
 3 
 
 10,25 
 
 42,5125 
 
 2nd Operation. 
 42 1 5. 3 c/. 
 20 
 
 8)850,25 
 
 Ans. $106,281 
 
 Ans. $106,281 
 3. Reduce 120 pounds 7 shillings 9 pence. New Jer- 
 sey currency, to dollars. 
 
 120 7 9 
 20 
 
 Is. 6rf.=7,5)2407,75($321,03-f 
 
 4. Reduce 38 pounds 5 shillings 6 pence, Georgia cur- 
 rency, to dollars. 
 12 6 
 
 20 
 
 5,5 
 
 £38,275. Di\dding by /„, gives $161,178+ 
 
 5. In 11 shillings 6 pence, Georgia currency, how ma- 
 ny dollars and cents ? Am. $2,464+ 
 
 6. In iElOO, New York currency, how many dollars ? 
 
 Ans. $250. 
 
 7. In £100, New England currency, how many dol- 
 lars 1 
 
 £ £ $ 
 
 Statement, . . 3 : 100 :: 10 : $333,33^- Ans. 
 
 8. In £100, New Jersey currency, how many dollars ? 
 
 Ans. $266,66|. 
 
 9. Reduce 54 pounds 6 shillings, New England cur- 
 rency, to dollars ? A?is. $181. 
 
 10. Reduce 304 pounds 9 pence, New England cur- 
 rency, to dollars ? Ans. $1013,458. 
 
 11. Reduce 364 pounds 2 shillings 6 pence, New York 
 currency, to dollars ? Ans. $910,31 + 
 
REDUCTION OF CURRENCIES. 191 
 
 (Art. 109.) The reverse operations will, of course, 
 bring dollars to pounds. In New England currency, 
 1 dollar equals f^^ of a pound; in New York currency, 
 1 dollar equals y\ of a pound ; in New Jersey currency, 
 1 dollar equals f of a pound ; hence, to reduce dollars to 
 pounds, we must take the following 
 
 Rule. Multiply the dollars ayid decimals of a dollar 
 by the value of a dollar^ expressed in parts of a pound. 
 
 EXAMPLES. 
 
 1. In 45 dollars 75 cents, how many pounds, shillings 
 and pence, New England currency ? 
 45,75 
 ,3 
 
 13,725 
 20 
 
 14,500 
 12 
 
 6,000 Ans. ^13 14 s. Q d. 
 2. Reduce 90 dollars to pounds, <fec. Georgia currency. 
 Operation. 
 90 
 
 90X/o' or 30 
 
 Ans. ^621. 
 
 3. Reduce 640 dollars to pounds. New Jersey cur- 
 rency. Thus, 
 
 8 
 
 640 
 3 Jins. £2i. 
 
 4. Reduce 500 dollars, to New England, New York, 
 New Jersey and Georgia currencies. 
 
 f$500=£l50 New England currency. 
 J $500=^2200 New York currency. 
 •^^^* ] $500=-£187 10 5. New Jersey currency. 
 L$500=^eil6 13 5. 4 f/. Georgia currency. 
 
 [Note. From these answers, we perceive that ^6150, 
 New England currency, is equal to ^6200, New York 
 
192 
 
 ARITHMETIC. 
 
 currency; and in the same manner, we may compare any 
 one with any other. Also, we observe, that to change 
 New England currency to its equivalent value, N. York 
 currency, we must add its i ; and to reduce New York 
 to its value expressed in New England currency, we 
 must subtract its \. But such problems never occur in 
 business.] 
 
 REDUCTION OF FOREIGN CURRENCIES AND 
 EXCHANGE. 
 
 In different countries, different modes of reckoning mo- 
 ney always have existed, and no doubt always will exist; 
 and it is important for commercial men to he ready and 
 familiar with the most common and practical operations ; 
 but there can be no difficulty with those who understand 
 the general principles of reduction ; and all the exam- 
 ples of money exchanges that can now be brought, are 
 but illustrations and applications of those principles al- 
 ready supposed to be learned: hence we shall only add 
 a few examples after the following 
 
 TABLE OF FOREIGN CURRENCIES. 
 
 English Money, 
 ^1=$4,44, before 1832; 
 since 1832, — 
 
 £1=$4,80, or,^5 = $24,00 
 1 English crown • =$ 1,10 
 Or, 10 crowns • . =-$11,00 
 1 English shilling . . =,222 
 Irish Currency. 
 
 £\ Irish, $4,10 
 
 1 shilling, ,205 
 
 1 penny, ,017 
 
 French Currency. 
 French crowns same as 
 English. 
 
 1 Five franc piece, • • $0,93 
 
 1 Franc, ,186 
 
 1 Decime, ,0180 
 
 Currencies of other nations. 
 The Roman crown, Span- 
 ish and Mexican dollar, and 
 the rix-dollar, of Sweden, 
 are the same in value as the 
 dollar, federal money. 
 1 Milree, of Portugal, $1,24 
 1 Russian Rix-dollar, ,600 
 1 Ducat of Naples, . . . ,80 
 1 Florin, of Trieste, • • ,48 
 1 Rix-dollar, of Trieste, ,96 
 1 Rupee, of Bengal, • • ,55^ 
 
REDUCTION OF FOREIGN CURRENCIES. 193 
 
 1 Rupee, of Bombay • • ,50 
 1 Tale, of Canton, • -1,48 
 1 Tale, of Sumatra, . • 4,1G 
 
 Canada Currency. 
 1 shilling, ,20 
 
 1 Florin, of Java, ... ,40 
 1 Mark Banco, of Ham- 
 burg, ,331- 
 
 1 Guilder, of Amster- 
 dam, ,40 
 
 * 
 
 EXAMPLES. 
 
 1. Reduce ^5 pounds 7 shillings 10 pence, sterling mo- 
 ney, to dollars and cents. Solution : as j6 1 = 480 cents, 1 
 shiiling=24 cents, and 1 penny=2 cents. Therefore, 
 45X480+7X24-1-10X2=21788 cents, or, $217,88. 
 
 2. Reduce 13 pounds 10 shillings 4 penc^e, sterling mo- 
 ney, to dollars, cents, &c. £ns. $64,88. 
 
 3. Reduce 13 pounds 13 shillings G pence, sterling, as 
 valued before 1832, to dollars and cents. 
 
 Solution: 13,675 X V =^60,77 +, Ans. 
 
 (Art. 111.) From these operations, v/e perceive that, 
 to reduce dollars and cents to pounds, shillings and pence, 
 sterling, at the old valuation, we must first inulliply the 
 dollars and decimal by ^''^ ; and secondly, reduce ike de- 
 cunal part to shillings and pence. For the modern val- 
 uation, we have the ibllowing 
 
 Rule. Divide the sum in cents by 2, and the quotient 
 is pence, ivhich reduce to shillings and pounds. In all 
 opercdiunSs, cancel ivhen practicable. 
 
 4. [\\ 480 dollars, how many pounds sterling, as valued 
 prior to 1832? 
 
 Solution: 480X /„ =12X 9=£108, Ans. 
 
 5. In $720,46, how many pounds, shillings and pence, 
 as valued since 1832? 
 
 Operation. 
 2)72046 
 
 12)36023=peiice. 
 
 20)3001 = sliillings, and 11 over, 
 
 i^lSO Is. lid. Ans. 
 - 
 
I 194 ARITil.-IETIC. 
 
 6. Reduce 1200 dollars to pounds, &c., sterling money, 
 as now valued. c'i/i.S". £250. 
 
 7 Reduce 600 pound 10 shillings sterling, to dollars 
 and cents. ^^ns. $2882,40. 
 
 8. In 47 pounds 2 shillings 6 pence, Irish currency, 
 how.many dollars and cents ? ^ns. S193,21 + 
 
 9. Reduce 720 dollars 40 cents to pounds, &c., Irish 
 currency. 
 
 410)72040(17512.^8; or, .6175 9 s. 3 d. -\~ ^ 
 
 10. In 3405 milrees of Portugal, how many dollars ? 
 
 .^ns. 84222,20. 
 
 11. Reduce 6246 milrees to dollars. 
 
 ,.^ns. $7745,04. 
 
 12. Reduce 248 dollars 50 cents to milrees of Portu- 
 gal, ^^ns. 200^. 
 
 13. In 330 Russian rix-dollars, how many Spanish. 
 Mexican, or Federal dollars? ^■9ns. 8220. 
 
 14. Reduce 960 dollars to Russian rix-dollars. 
 
 j3ns, 1440. 
 
 15. Reduce 5272 francs to federal dollars. 
 
 Jns. S980,59 + 
 
 16. Reduce 1200 dollars to francs. 
 
 .^ns. 6451,61 -f 
 
 17. In 2400 rupees of Bengal, liow many dollars ? 
 
 .^ns, $1335,33, 
 
 18. In 4282 ducats of Naples, how many dollars ? 
 
 Ans, $3425,60. 
 
 19. Reduce 5240 dollars to ducats of Naples. 
 
 .^ns. 6550. 
 
 20. Reduce 320 dollars to marks hanco of Hamburg. 
 
 .'2?7.s. 960. . 
 
 21. In 4560 tales of Canton, how many dollars? 
 
 .diis. 86748,80. 
 
 22. In 735 florins of Java, or guilders of Amsterdam, 
 how many dollars and cents ? »^??s. $294,00. 
 
 23. In 40 pounds 12 shillings 8 pence, Canada cur- 
 rency, how many dollars ? .■^n.s. $162,53;?-. 
 
 24. Reduce 740 dollars 42 cents, to pounds, shillings 
 and pence, Canada currency, .^ns. ^6185 2 5. lyV d. 
 
 25. In 428 tales of Sumatra, how many dollars ? 
 
 ^ns. $1780,48. 
 
EXCHANGE AND PER-CENTAGE. 195 
 
 EXCHANGE AND PERCENTAGE, 
 
 COMBINED. 
 
 (Art. 112.) By some writers, tliis is simply called ex- 
 change ; but we hope lo be more clear by being (in this 
 case) moie technical. The preceding problems come lit- 
 erally under the head of reduction of currencies, which 
 may be called exchange without premium. But as a ge- 
 neral thing, when it is necessary to reduce money from 
 one currency to another, it is done in consequence of some 
 commercial transaction which requires profit, loss or com- 
 mission, at the same time. Also, in the same operation, 
 we may make allowances for waste, shrinkage, reduc- 
 tions for weight of boxes, bales, &c., all by multiplica- 
 tion and division. As all reductions of every kind and 
 name must have some given proportion, and reductions 
 by proportion are effected by multiplications and divis- 
 ioi^.s ; therefore, we can combine all these difTerent al- 
 lowances under one general operation, as in Art. 24. The 
 allowances for weight of bales, boxes, &c., have been 
 called tare and tret ; but this technicahty is of no conse- 
 quence. 
 
 Under this head the problems and conditions can be so 
 varied, tliat definite rules would only be cumbersome : 
 the pupil must rely on general principles and good judg- 
 ment. "\Ye shall explain by the following 
 
 EXAMPLES. 
 
 1. Purchased, in London, 300 yards of broadcloth, 
 which cost me, including transportation and duties, 260 
 pounds. I must meet another charge of 2 per cent, for 
 commission ; and before my capital returns I expect the 
 average time must be 9 months, and money is worth 4 
 per cent. : taking these circumstances into consideration, 
 how must I sell the cloth per yard, in Federal Bluney, 
 to gain 1 5 per cent.? Ans. $5,286 -f 
 
 The first operation is to reduce the 260 pounds to dol- 
 lars. Draw a perpendicular line, (Art. 24,) thus ; 
 260 
 24 
 
196 ARiTHMrnc. 
 
 Tliis may now be considered as done ; and the next 
 step is to increase it 2 per cent., and that is done by mul- 
 tiplying by 102, and dividing by 100. Then we shall 
 have, as here represented. 
 
 260 
 
 5 
 100 
 
 24 
 102 
 
 "We must now give interest on this money, at 4 per 
 cent, for 9 months ; that is, multiply by 103, and divide 
 by 100. After that, we must increase it 15 per cent., 
 and we shall have the whole amount of dollars to be re- 
 ceived for the 300 yards of cloth. We then divide by 
 300, and we shall have the sum corresponding to one 
 yard, and the work is done, all but the canceling. When 
 completed, in form it stands thus : 
 
 260 
 
 5 
 
 100 
 100 
 100 
 300 
 
 24 
 
 102 
 103 
 115 
 
 This will cancel in part, and as far as it will cancel, it 
 is so much labor saved ; but the chief merit insisted up- 
 on, is the plan, going through the entire operation in form 
 without multiplying or dividing, going through the solu- 
 tion as a reasoner, and not as a mechanical ope'rator 
 only. 
 
 Let this be a jyattern or a ride for all other problems 
 of the like kind. 
 
 2. A merchant bought 280 yards of sheeting in Liver- 
 pool for 25 pounds, and duties and transportation cost 7 
 pounds 10 shillings more ; what shall be the price per 
 yard in cents, to gain 20 per cent, on first cost ? 
 
 .4'??.?. 66 cents 8 mills. 
 
 3. Received 600 yards of Irish linen, the whole cost, 
 including transportation, duties and commission, was 75 
 pounds 10 shillings, Irish currency ; what must be my 
 retail price per yard, in Federal Money, to gain 20 per 
 cent.? »^ns. 60,4 cents, nearly. 
 
EXCHANGE AND PER-CENTAGE. 197 
 
 4. A merchant bought sugar in New York at 6 pence 
 per pound, New York currency ; and while on his iiand 
 the wastage was estimated at 5 per cent., and interest on 
 tlie first cost to the time of sale, at 2 per cent, ; how ma- 
 ny cents shall he ask per pound to gain 25 per cent.? 
 
 Ans. 8^ cents. 
 
 5. A merchant bought 3 hogsheads of Lisbon wine at 
 I of a milree per gallon : on arrival at New York, it was 
 found that one hogshead had lost 21 gallons by leakage ; 
 the duties and transportation amounted to 18 per cent, on 
 the original cost ; what shall he sell it per gallon, to gain 
 20 per cent.? Jins. 79 cents. 
 
 6. Received from Malaga six hogsheads of wine, 
 which, including duties and transportation, cost me 175 
 Spanish dollars : it has suffered a leakage of about 4 per 
 cent. ; how must I sell it per gallon to gain 12 V per cent.? 
 
 A71S. 54} cents. 
 
 7. I bought 560 yards of Irish linen, which cost me, 
 including all expenses, 90 pounds Irish currency, and I 
 sold it at 75 cents per yard ; what was my gain per cent.? 
 
 Ans. 14 per cent, nearly. 
 cost. sale. 
 
 Solution: as 90X410 : 560X75 :: 100 : 4th term. 
 8. If 108 dollars were given for 9 hundred weight of 
 sugar, nett weight, tare 16 pounds per hundred weight, 
 what must be the price per pound, to gain 12 per cent.? 
 
 Ans. 14 cents. 
 
 9. Received from my agent, 300 barrels of flour, which 
 cost me 1400 dollars, and 75 dollars incidental expenses; 
 I now ship the same to Brazil, and pay $1,25 per barrel 
 for transportation; what must it bring in that market to 
 yield me 15 per cent., exclusive of duties at that port? 
 
 Ans. S7,09 -I- 
 
 10. A merchant bought, in the British West Indies, 3 
 casks of molasses, each containing 54 gallons, for 12 pounds 
 13 shillings 6 pence; transportation and duties, ^63 4 s. 
 6 d. in addition ; he then estimates a wastage of 4 per 
 cent.; what shall be his price per gallon, in federal money, 
 to gain 20 per cent.? Ans. ,58 + cts. 
 
 11. What is the cost of 15 chests of tea, each contain- 
 
 ^ __ 
 
198 ARITHMETIC. 
 
 ing 147 pounds gross, at 4 shillings 6 pence, New York 
 currency, per pound, tare 16 pounds on every 112 pounds? 
 
 Ans. $1063,12^. 
 
 12. Sold cloth at 12 shillings 6 pence, New York cur- 
 rency, and gained 15 per cent.; what would have been 
 the price per yard, in dollars and cents, had I gained 27 
 per cent.? " .^ns. $1,725 -f 
 
 12,5X100 
 
 Solution: ■ : fl :: 100 : 127 
 
 8X115 "--^ 
 
 Dividing by 8, reduces shillings to dollars, New York cur- 
 rency. 
 
 13. A merchant received from Amsterdam, 620 yards 
 of carpeting, whole cost 2480 guilders ; what must be 
 the retail price per yard, in federal money, to gain 15 per 
 cent.? Anb'. $1,84. 
 
 14. jYew York, March 30, 1842. This day received 
 from Liverpool, 900 yards of broadcloth, whole cost 1200 
 pounds sterling; how must I sell it per yard, in federal 
 money, to gain 20 per cent.? ' Aiis.- $7,68. 
 
 15. Boston, July 1, 1843. This day received from 
 France, 1200 bottles of Champagne, for which I paid 400 
 French guineas, each $4,60; how must I sell this wine 
 per botde, to gain 35 per cent.? Ans. $2,07. 
 
 16. " Received 300 ells of cloth from Hamburg, which 
 cost 1500 marks banco ; how must the same be sold per 
 yard, in federal money, to gain 12 1 per cent., the ell 
 Hamburg being 2\ qr." " Ans. $3,00. 
 
 17. A man shipped flour to Dublin: whole cost, in- 
 cluding purchase, transportation, commission, and duties, 
 was 4500 dollars ; he directs his agent to sell it not un- 
 der an advance of 10 per cent.; how many pounds, Irish 
 currency, could he draw for? Ans. £1207 6 s. 4 d.-{- 
 
 (Art. 113.) Connected with the higher order of ex- 
 changes, is a certain class of problems which trace a 
 result through several proportions: yet different from 
 compound proportion, as that which is generally consid- 
 ered compound proportion, is, in reality, in respect to 
 things, single proportion, and compound only, in respect 
 to nwnericcd factors ; but the proportions we are about 
 
EXCHANGE AND PER-CENTAGE. 199 
 
 to introduce, change from one thing to another, in fact, 
 and, after being condensed and properly linked together, 
 the single operation is called Conjoined Proportion : or, 
 The Chain Rule. We shall treat it, however, as simple 
 reduction ; but reduction is proportion, as we may see 
 by the following example : 
 
 Reduce 3 tons to pounds, by proportion ; thus, 
 
 First. As 
 
 ton. 
 
 tons. 
 
 cwt. 
 
 1 
 
 3 :: 
 
 20 
 
 20 
 
 60 
 
 Second 
 
 As 
 
 cwt. 
 
 1 
 
 cwt. 
 : 60 :: 
 
 4 
 
 240 qr. 
 
 qr. 
 4 : 
 
 Third. 
 
 As 
 
 qr. 
 1 : 
 
 qr. 
 240 : : 
 
 28 
 
 1920 
 
 48 
 
 6720 Ib.-- 
 
 lb. 
 
 28 
 
 ^Ans. 
 
 Thus we perceive that reduction is proportion ; but the 
 formal statements, one by one, step by step, render all 
 such simple problems complex. It is obvious that these 
 three proportions are linked together ; they are conjoined, 
 or enchained ; but, as a whole, are entirely unlike com- 
 pound proportion. We call particular attention to this, 
 as some authors have m?de the sad mistake of placing 
 problems in conjoined, under rules of compound, pro- 
 portion. 
 
 Let us return to our example again. We observe that 
 the first terms of the several proportions are units : hence 
 there will be no dividing; the whole operation will be 
 
200 ARITHMETIC. 
 
 actual multiplication, and, by Art. 24, the entire operation 
 will stand thus : 
 
 3 
 1 20 
 1 4 
 1 28 
 This may be read as follows: *What number of 
 pounds in 3 tons, if 1 ton =20 cwt., 1 cwt.=4 quar- 
 ters, and 1 quarter=28 pounds ? We may render this 
 same question much more complex, M-ithout changing its 
 nature. It will still be reduction, stated thus : 
 
 If 2 tons in London make 40 hundred weight in New 
 York, and 3 hundred weight in New York make 12 quar- 
 ters in Havana, and 4 quarters in Havana are equal to 112 
 pounds in Brazil, how many pounds in Brazil are equal 
 to 3 tons in London ? 
 
 This solution will stand thus : 
 
 3 tons, term of demand. 
 If tons, . 2= 40 hundred weight; 
 and cwt., 3= 12 quarters; 
 and qr., . 4=112 pounds. 
 Divide or cancel out the numbers on the left hand side of 
 the line, and it will be the same as before expressed. 
 
 This is rendered complex, by mentioning places, 
 which need not, yea, should not be mentioned in connec- 
 tion with this problem, but which must be mentioned in 
 connection with some others ; we mention them here, to 
 show how mere names of places, or different names to 
 things, will throw confusion into the mind, when in real- 
 ity the problem may be simple. 
 
 For a further illustration, let us take the following well 
 known problem. 
 
 * In view of these equalities, and the subsequent arrangements of 
 the terms each side of a perpendicular line, some have been led to 
 say that tliis perpendicular line, in all cases, represents equality — an 
 unphilosophical assertion. Indeed, it is to be apprehended that this 
 whole system of canceling may, for a time, suffer in the estimation 
 of the public, by the immodest pretensions of some of its pro- 
 fessors. 
 
EXCHANGE AND PER-CENTAGE. 201 
 
 If 50 pounds Englisli, make 45 pounds Plcmish, and 
 22 pounds Flemish, make 28 pounds Bologna, how ma- 
 ny pounds English, are equal to 5G pounds Bologna? 
 
 Ans. 48 1. 
 This may be done by two statements in proportion, 
 thus : 
 
 FL Fl. F)ig. Eng. 
 22 X 50 
 
 As 45 : 22 : : 50 : =28 lb. Bologna. 
 
 45 ^ 
 
 22X50 
 
 Again, as 28 : 56 : : : Answer. 
 
 45 
 
 56 term of demand, Bologna. 
 
 22 Flemish. 
 
 50 EnMish, odd term. 
 
 ByArt.24.^P^^^_^ 
 
 By paying attention to the order of arrangement of the 
 terms in the solutions of the preceding problems, the pu- 
 pil will understand the rationale of the following 
 
 Rule. Distinguish the terms or condition /rom the 
 one o/' DEMAND. Draw a perpendicidar line ; first, and 
 to the right of it place, the term of demand. Place the 
 conditional term of the same name in a line below and 
 to the left of the perpendicidar, and opposite to it on the 
 right put the term ecjucd to it. On the left agcmi, place 
 the conditional term of the same name, as the last one 
 on the right; and so on, until all the terms are nsed. 
 The last term on the right will be cm odd term, of the 
 same name as the answer. 
 
 Then cancel down, and the resxdt of the division 
 of the right hand factors by those on the left hand 
 of the line, will be the answer sought. 
 
 N. B. If we had the answer at first to put on the left 
 of the perpendicular line, the product of the factors on 
 both sides would then balance. 
 
 EXAMPLES. 
 
 1. If 12 pounds at Boston make 10 at Amsterdam, and 
 100 at Amsterdam, are equal to 120 pounds at Paris; 
 
202 ARITHMETIC. 
 
 Solution : 
 
 B. 40 
 H. 90 
 
 liow many pounds nt Boston are equal to 100 pounds at 
 Pai-is ? -'^'f'"' 1^0 pounds. 
 
 Solution— 100 pounds at Paris, term of demand ; 
 
 Paris,. . • 120 100 pounds, Amsterdam ; 
 Amsterdam, 10 12 pounds, Boston, odd term. 
 
 2. If 14 braces at Venice be equal to 15 braces at Leg- 
 horn, and 7 braces at Leghorn equal to 4 American yards, 
 how many braces of Venice are equal to 30 American 
 yards ? " ^ns. 49. 
 
 3. If 40 pounds at Boston make 36 in Holland, and 
 90 pounds in Holland make 116 in Poland, how many 
 pounds in Poland are equal to 250 at Boston? 
 
 £ns. 290 lbs. 
 250 demand, B. 
 36 H. 
 
 116 P. (odd term.) 
 The pupil will perceive tliat the odd term, in every case, 
 is balanced by the answer, when found. 
 
 4. A merchant in Russia owes 950 ducats in Berlin, 
 which he wishes to pay by an order for rubles, by the 
 way of Holland ; he has this information, that 2 rubles 
 make 95 stivers, and 20 stivers make 1 florin, and 5 flo- 
 rins make 2 rix-dollars of Holland; 100 rix-doilars of 
 Holland are equal to 142 rix-dollars of Prussia, and 3 
 rix-dollars of Prussia are equal to 1 ducat of Berlin ; how 
 many rubles will pay the debt? .^??s. 21124f rubles. 
 
 5. If A can do as much work in 3 days as B can do 
 in ih days, and B can do as much in 9 days as C in 12, 
 and C as much in 10 days as D in 8, how many days' 
 work of D are equal to 5 days of A ? .4^5. 8. 
 
 6. If 2 pounds of tea are worth 11 pounds of coflee, 
 and 3 pounds of coffee worth 5 pounds of sugar, and 18 
 pounds of sugar worth 21 pounds of rice, how many 
 pounds of rice can be purchased with 12 pounds of tea? 
 
 ./???5. 1281. 
 We found the following under the rule of compound 
 proportion; but we think it wandered out of its place: 
 
 7. " If 25 pears can be bought for 10 lemons, and 28 
 lemons for 8 pomegranates, and 1 pomegranate for 48 al- 
 monds, and 50 almonds for 70 chestnuts, and 108 chest- 
 
FKLLOWSHIP. 203 
 
 nuts for 2\ cents, how many pears can I buy for 135 
 cents?" " . 'i 22V. 759a. ^ 
 
 8. If 7 pounds of raisins are worth 10 pounds of cof- 
 fee, and 10 pounds of coffee worth 3 g-nllons of molasses, 
 and 2 gallons of molasses worth 62 .'^ cents, how many 
 pounds of raisins can be bought for 8 dollars? 
 
 ,/lns. 59\l lbs. nearly. 
 
 FELLOWSHIP. 
 
 (Art. 114.) Fellowships are of two kinds, single and 
 double; the former includes such cases as take no account 
 of time; the latter admits time as an element in the trans- 
 action. When men associate for any single act of spec- 
 ulation or adventure, or when they, in con>mon, are to 
 sustain a specific loss, or raise a specific tax, tlie opera- 
 tion of determining each man's portion of the gain or 
 loss, is called single fellowship. This is not applicable 
 to business partnerships, as generally conducfcd, in 
 which a fixed proportion of the profit or loss is usually 
 agreed upon between the parties, and the disparity in the 
 amount of capital is compensated by personal services, 
 or by an allowance of interest on the excess furnished by 
 either. In single transactions, such as fall under single 
 fellowship, it is plain that he who furnished most capital, 
 either iu money or labor, must have most of the profit; 
 and he who hazards most in any adventure, must sustain 
 most of the loss, when loss occurs ; and just in propor- 
 tion to that capital, or tliat hazard, compared to the 
 capital or hazard of others. Hence fellowship is but 
 another application of that pure 2>rinciple of pro- 
 portion, which, as applied to such cases, may be express- 
 ed thus : 
 
 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. 
 
204 
 
 ARITHMETIC. 
 
 exa:«ples. 
 ^l. Three workmen having undertaken to do a piece of 
 work for 275 dollars, agreed to divide their protits 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? 
 Solution: • • M = 50 days. 
 N=:65 " 
 0=85 " 
 
 Sum, 200 
 
 Divide by 25, and 8 
 
 Also, 8 
 
 And 8 
 
 : $275 
 
 : 50 : 
 
 : 11 
 
 : 50 : 
 
 : 11 
 
 : 65 : 
 
 : 11 
 
 : 85 : 
 
 $68,75 =M's 
 
 89,37?, = N's 
 
 116,87i = 0's 
 
 $275,00 Proof. 
 2. A man bequeathed to the eldest of his four children 
 2400 dollars ; to the second, 2000 dollars : to the third, 
 1800 dollars, and to the youngest, 1600 dollars, or in that 
 proportion of the value of his real estate; but, on the set- 
 tlement, only 4000 dollars were found to divide ; what 
 was the share of each ? 
 
 N. B. It is not necessary to use the identical numbers 
 given ; it is sufficient to use numbers that bear the same 
 relation. Hence we take, 
 1st. 12 
 2d. 10 
 3d. 9 
 4th. 8 
 
 39 : 4000 
 
 12 
 
 123J3- dollars = 1st. 
 
 * Here we compare days' work with dollars. It has been asserted 
 by many, in modern times, that we cannot compare unhke things, 
 and that such comparisons are absurd, &c.; but we think these views 
 over nice, at least. It is true we cannot measure unlike things by 
 each other, but we can compare their numerical vahies. In the pre- 
 sent instance, who can doubt the philosophy of comparing the days' 
 work to the dollars received, to see whether one is more or less innu- 
 vierical amount, than the other] Yet, as a general thing, we admit 
 it is more clear to compare quantities of like kind; and while we make 
 this admission, we do not concede that it is more philosophical. 
 
FELLOWSHIP. 205 
 
 3. A man is indebted to A $250,50, to B $500, to 
 C $349,50; but when he comes to nmke a settlement, it 
 is found he is worth but $960; how much will each one 
 receive, if it be in proportion to their respective claims? 
 
 fA $218,618-1- 
 
 Ans. \ B $436,363-f 
 
 (^C $305,018-1- 
 
 4. Four men hired a coach to convey them to their re- 
 spective homes, which were at distances from the place 
 of starting as follows: A's 16 miles, B's 24 miles, C's 
 28 miles, and D's 36 miles ; what ought each to pay, as 
 his part of the coach hire, which was 13 dollars? 
 
 Ans. A $2, B $3, C $3,50, and D $4,50. 
 
 5. Divide 360 dollars in the proportion of 2, 3 and 4. 
 
 Ans. 80, 120, 160. 
 
 6. In the year 1835, three men, P, Q, and R, made a 
 joint capital of 7500 dollars, to purchase city lots in Buf- 
 falo; P put in 1500 dollars, Q 2500, and R the remain- 
 der ; ihey soon gained paper obligations to the amount of 
 18000 dollars; what was each partner's share of the 
 gain? rP=$3600. 
 
 rp=i 
 
 LR=! 
 
 Ans. < Q=$8000. 
 
 :$8400. 
 
 (Art. 115.) Taxes. The taxable inhabitants of any 
 city, town, or county, are all in political fellowship, for 
 the payment of taxes and support of government. A 
 small tax in some states, and in most foreign governments, 
 is imposed on every male citizen above a certain age, 
 and being the same to all, independent of property, is 
 called per head, or per poll, or poll tax. In addition to 
 this, tliere is a tax on property, both fixed and movable : 
 i. e. real and personal : the amount of which property is 
 determined by official appraisers, called assessors, and 
 their list of persons, and property held by each, is called 
 the assessment-roll. All persons on this roll are, as we 
 before observed, in felloivship. When the amount of 
 tax is determined upon, (the poll-tax of the district, if 
 any) is taken out, and the remainder divided out among 
 the inhabitants, in proportion to their division of property. 
 
20G 
 
 ARITHMETIC. 
 
 EXAMPLES. 
 
 I. If a t:ix of 3000 dollars be assessed upon all the 
 taxable property of a toM'ii, the valuation of which, on 
 the assessment-roll, is 600,000 dollars, what will it be on 
 1, 2, 3, 4, &c., to 10 dollars ? what will it be on 100 and 
 1000 dollars? 
 
 Solution: As 600000 : 3000 :: 1 : 4th term. 
 
 Or, as 200 : 1 
 
 :: 1 : _i-.-=,005 = 5 mills. 
 
 
 r$l pays 
 
 ,005 
 
 $6 pays ,03 
 
 
 2 " 
 
 ,01 
 
 7 " ,035 
 
 Ans. < 
 
 3 " 
 
 ,015 
 
 8 " ,04 
 
 
 4 " 
 
 ,02 
 
 9 " ,045 
 
 
 L 5 '* 
 
 ,025 
 
 10 " ,05 
 
 100 " ,50 
 
 1000 " $5,00 
 
 The property of John Wells, in this town, is assessed 
 at $2780 ; what is his property tax ? 
 
 As $1000 pays $5, $2000 pays $10,00 
 
 700 pays 3,50 
 
 80 pays ,40 
 
 Therefore, 2780 pays $13,90, Ans, 
 
 In the same manner, estimate the tax of any other indi- 
 vidual inhabitant who possesses taxable property. 
 
 2. The corporation of a city vote to raise an extra tax 
 of$5400 for certain public improvements, and the assess- 
 ment-roll sums up to $2,700,000 ; what is the extra tax 
 on a dollar, and liow much extra tax must A pay, who is 
 taxed for $3757 ? 
 
 Ans. 2 mills on a dollar: A pays $7,515. 
 
 3. A tax of 3000 dollars is to be raised in a certain 
 town, where the taxable property amounts to 120000 dol- 
 lars, and the number of polls 320, at 50 cents each ; 
 what is paid on a dollar, and what is A's tax on a pro- 
 perty valuation of 2500 dollars ? 
 
 Ans. ,023| on $1 ; A's tax, $59,16|. 
 
COMPOUND FELLOWSHIP. 207 
 
 COMPOUND FELLOWSHIP. 
 
 (Art. 116.) There are certain cases of fellowship that 
 compel lis to take time into consideration. The produc- 
 tive vaUie of stock depends on the amount of capital, by 
 the time, as in interest, or the amount of labor depends 
 on the number of laborers, by the time they labor, &c. 
 All such cases must be considered under a compound, of 
 which time is an element; hence it is called compound 
 fellowship. The following example will fully illustrate 
 this principle : 
 
 Three men hired a pasture lot for 25 dollars ; A put 
 in 5 cows, 12 weeks; B 4 cows, 10 weeks; and C 3 
 cows, 15 weeks; what part of the 25 dollars shall ench 
 pay ? 
 
 Here it is evident, that the number of cows, each put 
 in, cannot decide the question, nor can the time alone de- 
 cide ; we must take tJie compound of both these elements, 
 thus : 
 
 5 cows 12 weeks =60 cows 1 week. 
 4 cows 10 weeks =40 cows 1 week. 
 3 cows 15 weeks =45 cows 1 week. 
 
 The whole pasturage is, then, the same as 145 cows 1 
 week ; and the time is thus reduced to unity, and then 
 solved by single fellowship; thus, 
 
 145 : 25 :: 60 : A's part. 
 
 or, . . 29 : 5 : : 60 : A's part. A=$10,344 -{- 
 
 29 : 5 :: 40 : B's part. B= 6,897 + 
 
 29 : 5 :: 45 : C's part. C= 7,758 4" 
 
 Froof, $24,999 + 
 lit the same way, we estimate labor, as well as con- 
 sumption, interest,' &c., and from this we draw the fol- 
 lowing 
 
 Rule. Multiply the active ager.ts by the time each 
 was in action; or, if the agent is money or stock, mul- 
 tiply each man's stock by its time, and add the several 
 products together. Then by proportion, 
 
208 ARITHMETIC. 
 
 As the sum of the products 
 
 Is to each particular product, 
 
 So is the wJiole 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 dollars; 
 what is each man's share ? 
 
 $ mo. 
 A's stock 120X 9 = 1080 
 B's stock 100X16=1600 
 C's stock 100X14 = 1400 
 
 Sum, 4080 
 Then, as 4080 : 1080 :: 100 
 or, as ... 102 : 27 : : 100 : $26,47 -f A's share, &c. 
 
 2. Two men commenced partnership for a year; one 
 put in 1000 at the commencement, the other put in his 
 capital 4 months afterwards ; at the close of the year 
 they divided equally ; what capital did the second put in ? 
 
 Ans. $15tl0. 
 N. B. Many problems must be solved, rather by the 
 general principle, than by the general rule. In the 
 above example, if they shared equally, the products of 
 their capital and time must be equa4. 
 
 3. A, B, and C, made a stock for 12 months; A put 
 in, at first, 873 dollars 60 cents, and 4 months after, he 
 put in 96 dollars more ; B put in, at first, 979 dollars 20 
 cents, and at the end of 7 months, he took out 206 dol- 
 lars 40 cents ; C put in, at first, 355 dollars 20 cents, and 
 3 months after, he put in 206 dollars 40 cents, and 5 
 months after that, he put in 240 dollars more. At the 
 end of 12 months, their gain is found to be 3446 dollars 
 40 cents ; what is each man's share of the gain ? 
 
 rA's share is $1362,88 
 .^ns. -i B's 1298,34-j- 
 
 [C's 785,13 
 
 4. Three men took a field of grain to harv'est and 
 thrash, for \ of the crop; A furnished 3 hands 4 days, 
 
COMPOUND FELLOWSHIP. 209 
 
 B 2 hands 5 days, and C 4 hands 6 days ; the whole 
 crop amounted to 320 bushels ; what was each partner's 
 share ? 
 
 Ans, 
 
 rA's=20|f. 
 J B's = 17i'3- 
 (.C's=4Ui. 
 
 5. A, B, and C, had a capital stock of 5762 dollars. 
 A's money was in trade 5 months, B's 7, and C's 9 ; 
 they gahied 780, which was divided in the proportion of 
 4, 5, and 3 ; that is, A 4, as often as B 5, and C 3 ; now 
 B received 2087 dollars, and absconded ; what did each 
 gain, and put in ; and what did A and C gain and lose by 
 B's misconduct? 
 
 f A's stock $2494,887 gain 260 
 ^ J B's do. 2227,577 do. 325 
 ^^^' 1 C's do. 1039,536 do. 195 
 
 LA and C gain $465,577 
 
 [Note. As money, in compound fellowship, is divided 
 in proportion to each man's capital, multiplied by its time 
 in trade; inversely, then, each man's capital must be in 
 proportion to his gain, divided by the time his money 
 was in trade. Therefore, as their shares of the gain are 
 as the numbers 4, 5, and 3, their capital stocks must be 
 in proportion to |-, 4 and i.] 
 
 6. In a certain factory were employed men, women and 
 boys ; the boys receive 3 cents per hour, the women 4, 
 and the men 6 ; the boys work 8 hours a day, the wo- 
 men 9, and the men 12 ; the boys receive 5 dollars as 
 often as the women 10 dollars, and for every ten dollars 
 paid to the women, 24 dollars were paid to the men ; how 
 many men, women and boys were there, the whole num- 
 ber being 59 ? Jlns. 24 men, 20 women, and 15 boys. 
 
 7. A, B, and C, united with 1930 dollars ; A's money 
 was in trade 3 months, B's 5, and C's 7 ; they gained 
 117 dollars, which was so divided that | of A's share 
 was equal to ^ of B's, and to \ of C's ; what did each 
 gain, and put in ? 
 
 ^ C A gained $26, B $39, and C $52. 
 *^"*' I A's capital was $700, B's $630, and C's $600. 
 
 8. A, B, and C, are employed to do a piece of work 
 for 26 dollars 45 cents ; A and B together are supposed 
 
 72 """" 
 
210 ARITHMETIC. 
 
 to do ^ of it ; A and C j\, and B and C i|, and are paid 
 proportionally to this supposition ; what is each man's 
 share of the money ? 
 
 ^7is. A $11,50, B $5,75, C $9,20. 
 
 BARTER. 
 
 (Art. 117.) Barter is striking the balance of trade, 
 or determining how much of one article is equal to a giv- 
 en quantity of another, or several others; and, of course, 
 involves no new principle of computation. The nature 
 of the case before us must determine the mode of opera- 
 tion. Rules. Practice. Rule of Three. 
 
 EXAMPLES. 
 
 1. A farmer bartered 3 barrels of flour, at 5 dollars 25 
 cents per barrel, for sugar and coflee, to receive an equal 
 quantity of each; how much of each must he receive, 
 admitting the sugar to be valued at 9 cents per pound, 
 and the coffee at 15 cents ? ^^n.s. 65|. 
 
 2. 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? ^ns. $125,75. 
 
 3. 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 ? 
 
 ^is. 28 bushels. 
 
 4. What is rice per pound, when 340 pounds are given 
 for 4 yards of cloth, at 4 dollars 25 cents per yard ? 
 
 ^ns. 5 cents. 
 
 5. Gave in barter 65 pounds of tea for 156 gallons of 
 rum, at 33^ cents per gallon ; what was the tea rated at ? 
 
 .^ns. 80 c. per lb. 
 
 6. How many pounds of butter, at 12^ cents per 
 pound, must be given in exchange for 9| pounds of In- 
 
 I digo, at 235 cents per pound ? ^?is. 183y\ lb. 
 
BARTER. 21 
 
 7. A merchMnt bought 80 yards of broadcloth, at 3-^^ 
 dollars per yard, and paid in coifee at 10 pence per pound, 
 New York currency, how many pounds did it require? 
 
 Arts. 2496 pounds, 
 
 8. Sold 24 hundred weight of hops at 4 pence per 
 pound, and took pay in sugar at 6 pence per pound ; how 
 many pounds did I require? Ayis. 16 cwt. 
 
 9. Sold 5 hundred weight 1 quarter of sugar at 8 dol- 
 lars 50 cents per hundred weight, and received for pay 
 21 yards of cloth; what was the cloth per yard? 
 
 Jlns. $1,12^. 
 All the examples thus far, except No. 2, can be solved 
 very briefly by Art. 24. No. 2 is also very brief, but 
 not in that form. 
 
 10. Q has coffee worth 16 cents per pound, but in bar- 
 ter, raised it to 18 cents; B has broadcloth worth 4 dol- 
 lars 64 cents per yard ; what must B raise his cloth to, 
 so as to make a fair barter with Q ? Ans. $5,22. 
 
 11. P and Q barter; P has Irish linen, worth 3 shil- 
 lings 7 pence, cash, but, in barter, he will have 3 shillings 
 10 pence ; Q delivers him broadcloth at 1 pound 16 shil- 
 lings 6 pence per yard, worth only 1 pound 13 shillings; 
 how much linen does P give Q for 138 yards of broad- 
 cloth, and which gains by the bargain ? and how much 
 in dollars and cents, the whole being in Irish currency ? 
 
 Jins. P gives 1314 yards ; Q gains $ 33-|- » because 
 his cloth is proportionally too high. 
 
 12. A and B barter; A has 150 gallons of brandy, at 
 7 shillings 6 pence per gallon, cash, but in barter he will 
 have 8 shillings ; B has linen at 3 shillings 6 pence per 
 yard, ready money; how much must it be per yard, to 
 meet A's bartering price, and how many yards are equal 
 to A's brandy ? 
 
 Ans. Barter price, 3 s. 8f d.; B must give A 321i 
 yards, nearly. 
 
212 
 
 ARITHMETIC. 
 
 SECTION V. 
 
 MENSURATION.* 
 
 (Art. 118.) Mensuration is finding, or defining, the 
 numerical contents of surfaces and solids, and its founda- 
 tion is the science of geometry; but its every-day appli- 
 cation to the mechanical business of life, demands a place 
 for its practical rules in every arithmetic. Some of these 
 rules are obvious, from inspection: otiiers must be taken 
 on trust, until the pupil studies geometry, from whence 
 they are drawn ; we shall, however, give some of the 
 common illustrations. All surfaces, of whatever figure, 
 are referred to squares for measurement, and the iinit 
 may be taken at pleasure ; it may be an inch, a foot, a 
 yard, a rod, a mile, &c., according as convenience and 
 common sense may dictate. The following figures will 
 render the truth of several rules apparent: 
 
 Fig. 1 . A square. 
 
 Fig. 2. A right angled 
 
 parallelogram. 
 12 3 4 5 6 7 
 
 The first, a square, the side of which is 3 units in 
 length ; and it is apparent to the eye, that the whole 
 square contains 9 square units ; and, considering the 3 
 
 * Most writers on arithmetic have placed mensuration after the 
 square and cube roots ; but we think it an injudicious arrangement, 
 as, in the operations of finding roots, mensuration is almost univer- 
 sally used. 
 
MENSURATION. 
 
 213 
 
 units in a side as feet, the whole square is one square 
 yard, containing 9 square feet, agreeably to the dictation 
 of the square, measure table. 
 
 By figure 2nd, we perceive that the number of litde 
 squares it contains, is equal to the iniits in lengtli, mul- 
 tiplied by the units in breadth, which must be true in 
 every case. 
 
 (Art. 119.) If the sides are not at right angles, as in 
 figure 3d, we must multiply its length by its jierpendic- 
 idar breadth, as it is proved in geometry, that the fiffure 
 ABCG = L!ie figure BCDE. 
 
 Fig. 3. A right and an oblique-angled parallelogram. 
 A E G D 
 
 A triangle is equal to half of a parallelogram, of the 
 same base and altitude, as is obvious, from inspecting fig- 
 ure 4, AC=the base, and BE=the altitude. 
 
 B D 
 
 Fig. 4. 
 
 A plain figure having 4 sides, and 2 of them parallel 
 and unequal^ is called a trapezoid. It is rigidly proved, 
 in geometry, that the area of such a figure is found by 
 multiplying its perpendicular altitude EF, by a medium 
 base between AD and BC. This truth is apparent, 
 from the inspection of Fig. 5. 
 
 BEG 
 
 D 
 
2U 
 
 ARITHMETIC. 
 
 Fig. 6. 
 
 
 / g\ 
 
 \ ^ 
 
 ; 
 
 (Art. J20.) When a circle is inscribed in a squaie, it 
 lias been ascertained by geometry, that the area or sur- 
 face of the circle occupies ,7854 (as near as can be ex- 
 pressed with 4 decimals,) of the whole square. Bu* the 
 square is the square of the diameter of the circle, '«s ap- 
 pears from figure 6. 
 
 Therefore, to find the area 
 of a circle, we may square 
 the diameter, and multiply it 
 by the decimal ,7854. 
 
 It is also shown in geome- 
 try, that the circumference 
 of a circle is 3,1416, nearly, 
 when the diameter is 1 ; and 
 when the diameter is 2, the 
 circumference will be double, 
 and so on in proportion : ten times in diameter giving 
 ten times for circumference, &c. &lc. 
 
 Again: in figure 6, from the centre C, we can draw 
 radii CD, CG, as near each other as we please; and 
 if we draw them extremely near each other, the small 
 part of the circumference GD, may be considered a 
 straight line, and the sector CDG, a triangle ; but the 
 area of this triangle is equal to CD, multiplied by \ of 
 GD. But the whole circle can be filled up with such 
 triangles : hence the area of the whole circle is equal to V 
 of its diameter, multiplied by \ of all the GD's, or ^ of 
 the whole circumference. Squares, circles, and all fig- 
 ures of a similar shape to one another, are, in area, as 
 the square, of their like dimensions. That is, a circle of 
 double diameter to another, is 4 times in surface, of .3 times 
 in diamefe^-r, 9 times in surface, &c. &:c. This may be 
 brought to the mind by inspecting figure 1. If we call 
 the whole square in fiffure 1 an unit, then the square of 
 1 is J^'; that is, to square a fraction, square its numera- 
 tor and denominator: geometrical figures and numericals 
 agree. 
 
.-VIE.NSl KATIOX. 215 
 
 RIXAPITULATION. 
 
 To find the surface or area of a square. 
 
 Rule 1. Multiply its equal length and breadth to- 
 gether. 
 
 To find the area of a parallelogram. 
 
 Rule 2. Multiply length and breadth together, (ta- 
 king care always to measure them at right angles.) 
 
 To find the area of a triangle. 
 
 Rule 3. TVlien the base and altitude are given, mul- 
 tiply one by half the other. 
 
 When not a right angle, and the altitude not known, 
 but the three sides given, the following is the rule which 
 the mere arithmetician must take on trust. 
 
 Rule 4. Add all the three sides together, and take 
 half that sum. Next, subtract each side severally from 
 the said half sum, obtaining three remainders. Then 
 multiply the said half sum and those three remainders 
 all together, and extract the square root of the last pro- 
 duct, for the area of the triangle. 
 
 To find the area of a trapezoid, or a figure of 4 sides, 
 and only two opposite sides parallel. 
 
 Rule 5. Add the two parcdlel sides together, and 
 take half the sum : this will be the mean or average 
 width. Multiply this mean vndth by the perpendicular 
 distance between the parallel sides> 
 
 To find the area of a circle. 
 
 Rule 6. Square the diameter, and multiply that 
 square by the decimal ,7854. 
 
 Or, when most convenient. Multiply half the diam- 
 eter by half the circumference. 
 
 [Note. Tlie circumference to the diameter of any cir- 
 cle, is found to be very nearly in the proportion of 22 to 
 7: a more accurate ratio is .3,1416 to 1.] 
 
 To find the surface of a globe or sphere. 
 
 Rule 7. Multiply the circumference and diameter 
 toor.ther : or, inultiply the square of the circumference 
 6y 3,1416. 
 
216 ARITHMETIC. 
 
 This rule must be taken on authority. For proof, see 
 Geometry. 
 
 Definitions. A right cylinder is a round body of 
 uniform diameter ; its two ends parallel and at right an- 
 gles with its length. Conceive the surface of such a cy- 
 linder cut lengthwise and rolled out, it will form a paral- 
 lelogram whose width is the circumference of the cylin- 
 der ; therefore, to find the surface of a cylinder, we have 
 the following 
 
 Rule 8. Multiply the length by the circumference. 
 
 A right cone may be covered by a multitude of equal 
 isosceles triangles, whose base is the base of the cone, 
 and whose altitude is the slant height of the cone ; there- 
 fore, 
 
 To find the surface of a right cone. 
 
 Rule 9. Multiply the circumference of the base by 
 half the slant height. 
 
 To find the area of an ellipse. 
 
 Rule. Multiply the two diameters together, and that 
 product by the decimal ,7854. 
 
 EXAMPLES UNDER THE FOREGOING RULES. 
 
 1. There is a square of 80 rods on a side, and a par- 
 allelogram also of 80 rods each side ; but from the base 
 to the opposite side is only 70 rods of perpendicular dis- 
 tance ; what is the difference in area of these two figures? 
 
 Jim. 800 rods. 
 
 2. How many yards in a triangle whose base is 148 
 feet, and perpendicular 45 feet ? .^ns. 370 yards. 
 
 3. The three sides of a triangle are severally 60, 50, 
 and 40 rods ; how many acres does it contain ? 
 
 ^ns. 61. 
 
 4. On a base of 120 rods, a surveyor wished to lay off 
 a rectangular lot of land containing 40 acres ; what dis- 
 tance in rods must he run from his base line ? 
 
 Jlns. 5S^ rods. 
 
 5. How many boards will it require to cover the body 
 of a barn 60 feet long, 40 feet wide and 20 feet high, to 
 
MENSURATION. 217 
 
 the eaves, the boards being on an average 15 inches wide 
 and 10 feet long? Ans. .S20 hoards. 
 
 6. How many feet in a board 22 inches wide at one 
 end, 8 inches wide at the other, and 14 feet long? 
 
 ^ns. 17^ feet. 
 
 7. A board is 9 indies wide ; how much in length 
 will it require to make 6| square feet? .'?».s-. 8| feet. 
 
 8. The bottom of a circular cistern is 5^ feet in diam- 
 eter ; how many square feet of surface ? 
 
 .^ns. 23,758+ feet. 
 
 9. What quantity of surface in a cylinder 30 feet long 
 and 3 feet in diameter, the two ends included ? 
 
 £ns. 296,88-}-feet. 
 
 10. A man bought a farm 198 rods long, and 150 rods 
 wide, and agreed to give $32 per acre ; what did the farm 
 come to ? ^ns. $5940. 
 
 N. B. Make no attempt to compute the number of 
 acres definitely. 
 
 11. If the forward wheels of a coach are 4 feet, and j 
 the hind ones 5 feet in diameter, how many more times 
 will the former revolve than the latter, in going a mile ? 
 
 .dns. 84. 
 N. B. In this problem use 7 to 22. 
 
 The solutions of the foregoing problems are all very 
 brief by canceling, except ihe 3d and 9th. 
 
 (Art, 121.) The foregaing rules apply to Plasterers^ 
 Pavers^ and Carpenters^ loork. These artificers sfener- 
 ally compute their work at so much a square yard, or^o 
 much for 100 square feet, which last is sometimes called 
 a carpenter's square. 
 
 To compute the number of square feet, yards, &;c., the 
 length, breadth and heighth of a room is taken in feet and 
 inclies, never measuring closer than inches. In books, 
 we often meet with feet, inches, thirds, fourths, <fec., de- 
 scending on a scale of 12, called duodecimals, the foot 
 being the unit ; hence inches are fractions, and two frac- 
 tions multiplied together, that is, taking part of a part, 
 makes a less part. Tiierefore, inches multiplied by indi- 
 es, must give less than inches, that is, give thirds, &c. 
 
218 ARITHMETIC. 
 
 As a study of numbers, a few examples in duodecimals, 
 might be of some advantage ; but with them we must re- 
 tain cumberous and clumsy modes of operations, which 
 are of no practical value, and thus cherish bad habits that 
 ought to be forgotten ; v:e shall ^ therefore, dispense with 
 duodecimals. AH practical men take no account of meas- 
 ures lower than inches. 
 
 EXAMPLES. 
 
 [Note. In the solution of problems, we always call 
 inches part of a foot; 6 inches is !,, 7 inches y^, 9 inches 
 \, &c.] 
 
 1. What will it cost to pave a foot-path 40 feet long, 
 and 7V feet wide, at 22 cents per square yard ? 
 
 Solution. 2 
 
 9 
 100 
 
 40 
 15 
 22 
 
 Ans. 7^ dollars. 
 
 2. What will it cost to paint a room which meas- 
 ures 82 feet 6 inches in compass and 12 feet 9 inches 
 high, at 16 cents a square yard? Ans. $18,70. 
 
 3. A floor is 36 feet 4 inches, by 14 feet 9 inclies ; 
 what will it cost to lay it at 3 dollars 2.5 cents per square ? 
 
 Ans. $17,42 nearly. 
 
 4. What will it cost to roof a building 42 feet long, 
 and the rafters on each side being 17 feet 6 inches long, 
 at 3 dollars and 25 cents per square of 100 feet ? 
 
 Ans. $47,771. 
 
 5. What will the plastering of a ceiling come to, at 15 
 cents per yard, it being 21 feet 7 inches long, and 12 feet 
 6 inches broad ? Ans. $4,47 -\- 
 
 6. What will the whitewashing of a room cost at 2 
 cents per yard, allowing it to be 30 feet 6 inches long, 24 
 feet 9 inches wide, and 10 feet 6 inches high, no deduc- 
 tions beino- made for vacuities ? Ans. $4,23 -}- 
 
 7. A room is 20 feet 6 inches long, and 13 feet 6 inch- 
 es in width, and 10 feet high ; what will it cost to paper 
 the walls, at 27 cents per square yard, deducting a fire- 
 place of 4 feet 4 inches, and 3 windows, each 6 feet by 
 3 feet 2 inches? Ans. $18,17. 
 
MENSURATION OF SOLIDS, ETC. 219 
 
 8. A room is 27 feet 6 inches by 22 feet 6 inches, and 
 10 feet 3 inches from the wash-board up. In said room 
 is a lire-place of 4 feet by 4 feet 6 inches ; 2 doors, each 
 8 feet by 4 feet 4 inches, and 2 windows, each 6 feet by 
 3 feet 4 inches. Wiiat will it cost to plaster said room, 
 at 18 cents per yard, sides and ceiling, deducting for va- 
 cuities '? Ans. $30,32. 
 
 9. What is the cost of smoothing and polishing a mar- 
 ble slab, whose length is 5 feet 7 inches, breadth 2 feet 3 
 inches, at 38 cents per square foot ? Ans. $4,52|. 
 
 MENSURATION OF SOLIDS AND CAPACITIES. 
 
 (Art. 122.) It will be recollected that surfaces are 
 measured by squares (Art. 118): so solids are measured 
 by unit-cubes, whatever size cube we take for the unit. 
 A nihe is a solid figure of six equal square surfaces, 
 and all its angles right angles. To measure a riglit 
 angular solid, such as a square stick of timber, for in- 
 stance, tlie surface of the end is measured by little squares ; 
 and if we saw off a length equal to a side of one of tliese 
 little squares, it is plain that we shall have as many cubes 
 as the surface of the end contains squares; twice this 
 length will give twice as many cubes ; three times the 
 length, three times as many cubes, &;c. &c. But to find 
 the number of squares in the end, we multiply the width 
 and depth of the end together ; then that product by th.e 
 length, and we have the number of unit-cubes in the 
 whole solid. All solid figures, of whatever shape, are 
 referred to regular right angular solids, directly or in- 
 directly. 
 
 To reduce one solid figure to its equivalent, in another 
 shape, is the office and object of solid geometry, not 
 arithmetic; hence our rules are drawn from geometry; 
 and, as arithmeticians, we must take them on authority. 
 We must advance to geometry to perfectly comprehend 
 them. 
 
220 ARITHMETIC. 
 
 BefinUion 1st. To find the cubical contents of a right 
 angular solid. 
 
 Rule 1. JMuUipJij length, breadth and depth together. 
 
 [Note. No school should be without a set of geomet- 
 rical solids. A bare inspection of them will give a bet- 
 ter idea of the regular solids, than the most exact defini- 
 tion, or the most minute description.] 
 
 Bef, 2. A prism is a solid, whose ends are any simi- 
 lar equal and parallel surfaces, and its sides rectangles ; 
 hence, to find the solidity of a prism, 
 
 Rule 2. Multiply the area of one end by the length. 
 
 Def. 3. A cylinder is a circular solid, equal at both 
 ends. A right cylinder has circular ends, at right angles 
 to the axis. To find the contents of such a cylinder. 
 
 RtTLE 8. Multiply the area of one of the circular ends 
 by the length. 
 
 To estimate the contents of a wedge formed solid. 
 
 Rule 4. Multiply the area of its end by half of its 
 perpmdicidar length. 
 
 Def. 4. A pyramid is a solid which decreases uniform- 
 ly from its base, until it comes to a point. The base of 
 a pyramid may be either a square, a triangle, or a circle. 
 Hence it is named a square pyramid, a triangular pyra- 
 mid, or a cone. The point in which the pyramid ends, is 
 called its vertex; a straight line from the vertex at right 
 angles to the base, is called the altitude. To find the 
 solidity of a pyramid or cone. 
 
 Rule 5. Multiply the area of the base by one-third 
 of the altitude. 
 
 Def. 5. When the smaller end of a pyramid or cone 
 is cut off parallel to its base, the residue is called the frus- 
 trum of a pyramid or cone. When the base is a square, 
 the following rule will give the solidity of a frustrum : 
 
 Rule 6. Square both upper and lower base : and, to 
 flu .nnn of these squares, add the product of the sides 
 
MENSURATION OF SOLIDS, ETC. 221 
 
 of upper and lower bane, and midtiphj this last sum by 
 one-third of the altifiide of the frutitrum. 
 
 When the base is a circle. 
 
 Rule 7. To the sum of the squares of the two diam- 
 eters, add their product ; multiply this sum by one-third 
 of the altitude, and this last product by the decimal 
 ,7854. 
 
 To find the solidity of a sphere or globe. 
 
 Rule 8. Find the surface by rule 7, Art. 120, then 
 multiply the surface by \ of the diameter ; or, which 
 is the same thing; multiply the square of the diameter 
 by the circumference, and divide the product by 6. Or, 
 we may cube the diameter, and multiply it by the deci- 
 mal ,5236. Or, if more convenient, cube the circum- 
 ference, and midtiply it by ,01688. 
 
 To find the solid contents of a spherical segment. 
 
 Rule 9. From three times the diameter of the sphere, 
 take double the height of the segment ; flien multiply 
 the remainder by the square of the height, and that 
 product by the decimal ,5236, and ive have the content. 
 
 Tlie foregoing rules have a variety of practical ap- 
 plications to business and the mechanic arts. Irregular 
 bodies may be measured approximately, by imagining 
 them cut and divided so that the parts in succession may 
 make some one, two, or more of the mathematical figures : 
 this is done by engineers in measuring excavations and 
 embankments. We can measure the cubical contents of 
 the most irregular stone in the field, by putting it into a 
 regular vessel of water, and observing the cubical contents 
 of the rise f or, by taking the specific gravity of a body, 
 we can estimate its contents from its weight; but this 
 cames us out of arithmetic into philosophy. 
 
 QUESTIONS UNDER THE FOREGOING RULES. 
 
 1. How many solid feet in a block 6 feet long, 4 feet 
 high, and 3 wide ? Solution : 6X 4X 3=72 feet, Ans. 
 
 * A boy once made a wager with one of his companions, that he 
 could measure the cubical contents of a certain pile of brush. He 
 did so by cutting it short, and sinking it m a regular tan-vat. 
 
 " ~^ f2 ~~~ 
 
222 
 
 ARITHMETIC. 
 
 2. How many cubic feet in a piece of timber 18 feet 
 long, 1 foot 6 inches high, and 1 foot 3 inches wide ? 
 
 .^ns. 33 ^ 
 
 Sohition : 
 
 8 
 
 
 
 18 
 
 U 
 
 or, 
 
 2 
 
 3 
 
 n 
 
 
 4 
 
 5 
 
 12 
 
 40 
 
 12 
 
 12 
 
 12 
 
 U 
 
 
 8 
 
 3. Tlie 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 1 Ans. 31^ feet. 
 
 Take this form for all solids, Art. 24. 
 
 inches. 
 
 N. B. To find how many cords in a given pile of wood, 
 to find how many bushels, gallons, &c., in any given ca- 
 pacity, we must, of course, divide the whole contents ex- 
 pressed in cubic feet, or inches, as the case may require, 
 by the feet, or inches, &c., which make one cord, or one 
 bushel, &;c., of the required dimensions. As an illustra- 
 tration, observe the solution of the following example. 
 The dimensions must be reduced to cubic inches. 
 
 3. There is a rectangular cistern, whose length is 14 
 feet, depth 16 feet, and breadth 9 feet; how many bar- 
 rels will it hold, allowing 231 cubic inches to a gallon, 
 
 and 32 gallons to a barrel ? 
 
 Solution : 
 
 Divisors, 
 
 { 
 
 231 
 
 32 
 
 Ans. 471yV 
 inches in length. 
 
 inches in depth. 
 
 inches in width. 
 
 This cancels down to 11, for a divisor, and 3 times 1728 
 for a dividend ; which quotient gives the answer. 
 
 4. A block of marble is 6 feet long, 3 feet wide, and 
 1 foot 4 inches thick ; how much will it cost, at 30 cents 
 per cubic foot? Ans. $7,20. 
 
 5. 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. 
 
MENSURATION OF SOLIDS, ETC. 223 
 
 6. The bin of a granary is 10 feet long, 5 feet wide, and 
 4 feet high ; allowing the cubical contents of a dry gal- 
 lon to be 2G8j cubic inches; how many bushels of grain 
 will it contain 1 Ans. 160,^. 
 
 7. If you wanted a bin to contain twice as much as 
 mentioned in the last problem, with a lengdi of 12 feet, 
 and a breadth of 6 feet, of what height must it be ? 
 
 Ans. ^l feet. 
 
 8. A canal contractor engaged to excavate 2 miles of 
 canal across a plane, at 8 cents 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. 
 
 9. What is the area of a circle of 5 feet in diameter? 
 
 Ans. 19,635 feet. 
 10. What is the area of a circle, whose diameter is 5 
 feet 6 inches ? Ans. 23,758-f 
 
 Remember, that similar figures are to each other as the 
 squares of their like dimensions (Art. 120). Apply this 
 principle to problem 10, and we have, as (5)^ : to (S't)^; 
 or as 10-: U^. That is, as 100 : 121 : 19,635 : 
 23,75835. 
 
 11. 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. 
 
 \ 246^- gallons increase. 
 
 12. A man wishes to make a cistern, of 8 feet in di- 
 ameter, to contain 60 barrels, at 32 gallons each., \\m\ 231 
 cubic inches to a gallon ; what shall be tlic deptli of the 
 cistern ? 
 
 60X32X231 gives the cubic inches the cistern i:; to 
 contain. This, divided by the circular end, expressed in 
 inches, will give the depth in inches. Ans. 61 J- inches. 
 
 13. A stick of timber is 8 inches wide, and 5 inches 
 thick ; how much in length will it require to make 3 cu- 
 bic feet ? 
 
 N. B. The principles contained in Rule 2 will solve 
 this problem. 
 
224 ARITHMETIC. 
 
 Solution, 
 
 2 ^ 
 5 
 
 3 
 
 i^ 3 
 
 12 
 
 ^^ Ans. 104 U%i. 
 
 14. How many bricks will it require to build a wall 
 44 feet long, 42 I'eet wide, (on the outbide,) and 25 feet 
 high, and the wall to be 1 foot thick, allowing 20 bricks to 
 the solid foot ? Ans. 84000. 
 
 15. What will it cost to build a brick wall 240 feet 
 long, 6 feet high, and 3 feet thick, at 3 dollars 25 cents 
 per 1000 bricks — each brick being 9 inches long, 4 inches 
 wide, and 2 inches thick ? Ans. $336,96. 
 
 16. A ship's hold is lb\ feet long, 18.V feet wide, and 
 1\ deep; how many bales of goods 3", feet long, 2^^ feet 
 deep, and 2^ wide, may be stowed therein, leaving a 
 gangway, the whole length, of 3^ feet wide? 
 
 Ans. 385,4 + 
 N. B. Do this by one operation ; after taking out the 
 gangway mentally ^ by subtraction. 
 
 17. 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. 221. 
 
 18. How many solid feet in a stick of timber 30 feet 
 long, 16 inches broad, and 9 inches thick? 
 
 Ans. 30 feet. 
 
 19. How many cubic feet in a piece of timber 19 feet 
 6 inches long, 1 foot 6 inches broad, and I foot 4 inches 
 thick ? Ans. 39 cubic feet. 
 
 20. 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 heigiit of 
 each part. 
 
 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. It is proved in geometry, that similar solids are 
 to one another as the cubes of their like dimensions. 
 This principle solves problem 20. 
 
MENSURATION OF SOLIDS, ETC. 225 
 
 21. The base of a cone is 8 feet in diameter, and 60 
 feet high ; how many cubic feet does it contain ? 
 
 .^iu\ 1005,6. 
 
 22. A stick of round timlier is 2 feet in diameter at 
 one end, and 10 inches at the other, and 20 feet long; 
 how many solid leel does it contain ? 
 
 Solution, Ihth 7. 
 
 (24--fl0=^+240)20X12X,7854 
 
 ^ ' =22,83+feet, Arts. 
 
 3.12.12.12 
 
 Rule 7th can be apph'ed to gai/gino;, or measuring of 
 casks. A common cask, of circular form, larger in di- 
 ameter in the middle than at the ends, may be considered 
 as composed of two equal fntshims of a cone, and its 
 contents found in inches accordingly; which contents, 
 divided by 231, for wine gallons, 282 for beer gallons. 
 For bushels, dividedby 2 150,4? or, for all practical purpo- 
 ses, by 2150 is sufficiently accurate. A bhshel of corn, 
 lime, &c., is 2688 inches. 
 
 EXAMrLES. 
 
 1. The bung-diameter of a cask is 38 inches, the head- 
 diameters inside the staves 28 inches, and the length 45 
 inches ; how many wine gallons will it contain ? 
 
 Jins. 107,89-1- 
 
 (38--|-28--|-38.28),7854 X 45 
 
 Solution, • . 
 
 3X231 
 
 Now observe, that the decimal ,7854, and the divisor 
 231, for wine gallons, will appear in any problem ; and 
 as the decimal is divisible by 231, and give the quotient 
 ,0034, we may give the following abridged ride for Jin d- 
 ing the wine gallons in a cask. 
 
 Rule. To the square of the head and hung-diame- 
 fers, add their product ; multiphj the sum by i of the 
 l&ngth, and that product by the decimal ,0034. 
 
 2. The head-diameter of a cask is 16 inches, the bung- 
 diameter 18 inches, and the length 27 inches; what is 
 the content of die cask in wine gallons ? Ans. 20,56. 
 
 3. In a circular vessel, whose greater diameter is 80 
 
226 ARITHMETIC. 
 
 inches, the less 71, and the depth 34, what is the con- 
 tents in liquid gallons, and also in bushels of grain ? 
 
 ^ 5 659,72+ gallons. 
 *^"'^^* I 70,86 bushels. 
 
 4. The greater diameter of a tub is 38 inches, the less 
 20,2, and the depth 21 ; M^hat is the content in gallons ? 
 
 ,.^ns. 62,34+ gallons. 
 
 5. The top diameter of a tube is 22 inches, the bottom 
 40, and the height 60 ; what is its contents in gallons, 
 also in bushels of grain ? /? S 201,55+ gallons. 
 
 '^^^^' I 21,64+ bushels. 
 
 6. How many barrels, of 32 gallons each, in a cistern 
 whose greater diameter is 8 feet 6 inches, less 8 feet, and 
 depth 7 feet 9 inches ? ^'^ns. 96 barrels 28+ gallons. 
 
 7. How many bushels of grain in a bin that is 8 feet 
 long, 4 feet wide, and 6 feet high ? 
 
 ^ .^ns. 123+ bushels. 
 
 8. How many bushels in a loaded boat 60 feet long, 
 16 feet wide, and 4\ feet deep ? j^ns. 2777+ bushels. 
 
 9. A man bought a grindstone, which was 48 inches 
 in diameter, and 5 inches in thickness, for 10 dollars. 
 When he had ground ofl" 3 inches from its circumference, 
 his neighbor proposed to purchase it from him, giving a 
 proportional part of the first cost for the residue ; deduct- 
 ing 4 inches every way from the center, allowed to be 
 the limit to which it could be used. What should he 
 have paid ■ ^ns. $7,58. 
 
 10. Find the solid contents of a sphere, whose diame- 
 ter is 12, by Rule 8 ? .^ns. 904,78 + 
 
 11. What is the sohd contents of a globe 48 inches in 
 diameter ? Ans. 64 X 904,78+ inches. 
 
 (Art. 123.) Connected with mensuration, as we have 
 already observed, are square and cube figures, the sides 
 of which represent square and cube roots. But to inves- 
 tigate these, we must commence with the general 
 
 INVOLUTION OF POWERS. 
 
 A POWER is the product of a number multiplied into 
 itself any number of times. 
 
 The number itself is called its 1st power. 
 
INVOLUTION" OF I'OWERS. 227 
 
 The number, multiplied into itself, is called the 2nd 
 power. 
 
 The 2nd power, multiplied again by tlie number, is 
 called the 3rd power. 
 
 The 3rd power, again multiplied by tlie number, is 
 called the 4th poM'er, &c. &c. 
 
 The number so multiplied into itself, is called a root ; 
 and the number of times a root is taken as a factor, de- 
 notes its power. Thus, 
 
 3= the root or first power. 
 3X3=3'=9 the square or 2nd power of 3. 
 3X3X3=3^=27 the cube or 3rd power of 3. 
 
 This is as far as numericals, with their roots and pow- 
 ers, can be made to correspond to geometrical figures. 
 The following are entirely numerical. 
 
 3X3X3X3=3^=81 the 4th power of 3. 
 3X3X3X3X3=3^=243 the 5th power of 3, &c. 
 
 The expression 3'' is simply an abbreviation of 3 X 3 X 
 3X3; the small figure 4 above and on the right of the 3, 
 is called an index or exponent, showing how many times 
 the root is taken as a factor. 
 
 If we take the powers indicated by any two exponents, 
 and multiply them into each other, their product will be 
 the power indicated by the sum of the two exponents. 
 Thus, 3^X3''=3^ the exponent of which is ^; or, if we 
 multiply any power into itself, the product will be the 
 power indicated by the addition of the exponent to itself. 
 Thus, 8^X 8=^=8^ Likewise, if a higher power be divi- 
 ded by a lower, the quotient will be the power indicated 
 by the difl^erence of the exponents. Thus, 16'^-^16''= 
 16^. Therefore, to multiply powers of the same root, we 
 simply add the indices, and to divide, we subtract the in- 
 dices or exponents. 
 
 JVeJind the power of any number by multiplication. 
 
 EXAMPLES. 
 
 1. What is the 3rd power of 2 ? Ans. 8. 
 
 2- What is the 5th power of 4 ? Ans. 1024. 
 
 3. What is the 6th power of 8 ? Ans. 262144. 
 
228 ARITHMETIC. 
 
 Problem third is performed thus : 8''=512 ; but S^X 
 83=8^ Hence 512X512=262144.^^15. 
 
 4. Find the square of 56. Jins. 3136. 
 
 5. Find the cube, or 3rd power, of 67. 
 
 Ans. 300763. 
 
 6. What is the cube of 186 ? Ans. 6434856. 
 
 7. What is the 3rd power of 9 ? Ans. 729. 
 
 8. What m the 9th power of 9 ? 
 
 Am. 729X729X729=387420489. 
 
 9. What is the 7th power of 9 ? 
 
 Ans. 729X729X9=4782969. 
 10. What is the 3rd power of 1? Ans. 1. The 4th 
 power of 1 ? Ans. 1. The 10th power of 1 ? Ans. 1. 
 Any power of 1 ? Ans. 1. 
 
 (Art. 124.) By the nature of multiplication and the 
 definition of powers, the pupil will perceive that any 
 power or root of 1, is neither more nor less than 1, for 
 the multiplication of one into itself any number of times, 
 cannot change it; but all numbers greater than one, gives | 
 higher numbers for powers, and the higher the power, \ 
 the higher the number above its root. 1 
 
 We now propose to examine the powers of a proper j 
 fraction. Suppose it required to find the square or se- \ 
 cond power of ~ ; that is, to multiply |- by l ; or, to take I 
 1, one-thiid of one time, which must be ^th the power i 
 less than its root. Now it must be very clear to all who ; 
 will take the trouble to call it to mind, that a fraction mul- 
 tiplied into itself, is taking a part of a part, ivhich must 
 be still less ; hence any power of a fraction must be less, 
 and of course, the reverse, any root of a fraction must 
 be greater than the power. We can further illustrate this 
 by fig. 1, Art. 118. If we take the whole line for unity, 
 the area of the figure is unity, and one section will re- 
 present 1, and its square is i- of the whole, as we may 
 see by inspection. 
 
 EXAMPLES. 
 
 1. Find the cube of \. Ans. \. 
 
 2. What is the 3rd power of ^ ? Ans. ■^\^. 
 
 3. Find the 4th power of ,3. Ans. ,0081. 
 
 4. Find the 4th power of ,05. Ans. ,00000625. 
 
EVOLUTION, ETC. 
 
 229 
 
 5. AVhat is the 2nd power of | ? 
 
 6. What is the 3d power of f ? 
 
 7. What is the 5th power of ^ ? 
 
 8. What is the cube of ,25 ? 
 
 9. What is the square of ,75 ? 
 10. What is the cube of li ? 
 
 Ans. i|A. 
 
 ^;i.s. ,015625. 
 Jlns. ,5625. 
 
 EVOLUTION, OR THE EXTRACTION OF 
 ROOTS. 
 
 (Art. 125.) Evolution is the reverse of Involution. 
 In a practical point of view, it is only important that we 
 should learn how to extract two roots, the square and the 
 cube roof. In the higher mathematics, we extract the 
 roots of higher powers by logarithms; and, in fact, math- 
 ematicians generally extract all roots by means of loga- 
 ritlims ; but it is important, on many accounts, that we 
 learn the common method of extraction. 
 
 As evolution is the reverse of involution, we must care- 
 fully notice how roots form powers by multiplication, that 
 we may trace back the operation from powers to their 
 roots. As a preliminary step, we give the following ta- 
 ble with the first 10 numbers, with their corresponding 
 squares and cubes, which the pupil should so familiarize 
 to his eye, as to recognize them as square or cube num- 
 bers, as the case may be, as soon as seen. 
 
 Numbers, 
 
 1 
 
 2 
 
 3 4 
 
 1 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 10 
 
 Squares, 
 
 1 
 
 4 
 
 9 
 
 16 
 
 25 
 
 36 
 
 49 
 
 64 
 
 81 
 
 100 i 
 
 ■Cubes. 
 
 1 
 
 8 27 64 
 
 1 
 
 125 
 
 216 
 
 343 
 
 512 
 
 729 
 
 1000 
 
 Any number will produce a perfect power, by involu- 
 tion ; yet there are many numbers that huve no perfect 
 roots. Imperfect roots are called surds, and such 
 
 can only be found approximately, but to a sufficient de- 
 gree of exactness for all priictical utility. For example, 
 36 and 49 are square numbers, and all the other whole, 
 and many mixed numbers between 36 and 49, have surd 
 
230 ARITH3IETIC. 
 
 roots. Very few numbers are perfect powers of more 
 than one root; thus, 8 is a cube number, but not a 
 sqvare number, &;c. 
 
 (Art. 126.) As a very important and useful fact, we 
 call the pupil to notice, that any root composed of one 
 Jigure, its square may be composed of two figures, and 
 no more than two; and its cube may be composed of 
 three figures, and no more than three, as will be seen by 
 inspecting the preceding table. 
 
 Again : every root composed of two figures, may have 
 more than 2, and not more than 4 figures, in its square ; 
 and may have more than 3, and not more than 6 figures 
 in its cube, as we may learn by involution ; thus. 
 The lowest root of 2 orders, is 10, the highest is 99 
 
 Squares, 100 9801 
 
 Cubes, 1000 970299 
 
 Lowest root of 3 places, is • -100, highest is . . 999 
 
 Squares. 10000 998001 
 
 Cubes, 1000000 .... 997002999 
 
 Hence, for every figure in a root, we may have two in its 
 square and three in its cube. This fact teaches us, in 
 preparing to extract the square root, to separate the pow- 
 er off inio periods oUwo figures each, and in preparing 
 to extract the cube root, to separate the power off into 
 periods of ^figures each, comynencing at the unit. The 
 number of periods, then, in both cases, will indicate the 
 number of figures in the root. 
 
 Now let us turn exclusive attention to 
 
 SQUARE ROOT. 
 
 (Art. 127.) There are two ways of explaining the 
 principle of the operation, in extracting square root: one 
 by retracing the numerical operation of squaring, the oth- 
 er by referring and comparing the power to a geometrical 
 square surface. Both of these should be understood. 
 To prepare the way for a clear understanding of this im- 
 portant branch of arithmetic, the pupil should square sev- 
 eral numbers, consisting of two figures, and scrutinize ev- 
 ery step of the operation : thus, 
 
SQUARE ROOT. 231 
 
 Let it be required to square 35 
 
 Write the number under itself, 35 
 
 Tlie units by the units, give 25 
 
 The units below by tens above, give • • • .150 
 The tens below by units above, give • • .150 
 The tens by the tens give liundreds, • . 900 
 
 Product or square, 1225 
 
 Now observe, that this square is made up of the square 
 of the tens, twice the product of the tens into the units, 
 and the square of the units. We must now suppose our- 
 selves ignorant of the real root, 35, and by the above 
 principle, find it. First, divide the power ofl" into pe- 
 riods of two figures each (Art. 126), and it will stand thus : 
 
 1215. 
 Being two periods, there will be two figures in the root, 
 a unit and a ten. Now, the square of the tens can form 
 no part of the Knit period, as the square of one 10 pro- 
 duces 100, and the square of tens generally produce 
 hundreds, and nothing less. Hence, the greatest square 
 contained in 12, the 2d period, is the square of the tens ; 
 but 9 is the greatest square number less than 12, and 3 
 its root; and so far, the operation may stand thus : 
 1225(3 
 9 
 
 325 
 
 But the tens and units are twice multiplied together, 
 and their product forms no part of the units in the pow- 
 er; hence, if we divide the 32 tens, by double the tens 
 in tlie root, we must have the other factor, or the units of 
 the loot. Now the operation stands thus : 
 
 12*25(35 
 9 
 
 Double of 3 is 6 65 325 
 6 in 32, 5 times, 325 
 5 times 65 is 325, 
 
 000 
 
232 
 
 ARITHMETIC. 
 
 120 
 
 9 
 
 H 
 
 X600 
 
 I 
 
 120 
 
 (Art. 128.) Let us now illustrate the mode of extrac- 
 tion, by means of a geometrical square. We cannot de- 
 mand the square root of any. number, without first sup- 
 posing it to represent the area of some square surface, 
 and tlie root is the length of a side of the supposed 
 square ; but to brina: it before the mind in a definite man- 
 ner, we demand the square root of 1849. This number, 
 then, must represent a square. 
 
 Suppose the figure A E B 
 
 ABCD contains 1849 
 square feet, and that the 
 number consists of two 
 periods ; then there must 
 be 2 figures in the 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 D 
 square of this is 16, (or 
 1600.) This taken from 
 the whole square ABCD 
 or 1849, leaves 249. 
 
 Now double GH, or 
 HI, which is 40, for a 
 divisor, omitting the ci^ 
 pher to leave place for 
 the next quotient figure, 
 to complete the divisor, 
 
 80 into 249 is con- 
 tained 3 times ; this 3 is 
 the widtli of the oblong 
 AEHG, orHFCI. But 
 the square is imperfect 
 without EBFH; then 
 annex the three to the 
 divisor. Now multiply 
 this perfect divisor by 
 the last fiffure of the root, 
 
 18,49(43 
 16 
 
 83)249 
 249 
 
 EHID =1600 
 AEHG = 120 
 HFCl ==■ 120 
 EBFH = 9 
 
 ABCD =1819 
 
SQUARE ROOT. 233 
 
 to get the quantity in the two oblontr (lornrcs, and the small 
 square whicli comprises the great square ABCD. 
 
 From either of the preceding modes of examining this 
 subject, we draw the following 
 
 Rule. \^ Separate the given number into periods^ of 
 two fici-ures each, beginning at the iiniCs place. 
 
 2. Find the greatest square contained in the left-hand 
 period, and set its root on the right of the given num- 
 ber; sulHract said square from the left-hand period, and 
 to the remainder bring down the next period for a di- 
 vidual. 
 
 3. Do2(ble the root for a divisor, and try how often 
 this divisor {with the figure used in the trial thereto an- 
 nexed) is contained in the dividual: set the number of 
 times in the root ; then, multiph/ and subtract as in di- 
 vision, and bring down the next period to the remain- 
 der 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. When all the periods are brought down, if 
 there be a remainder, annex ciphers to it for decimals, 
 and proceed till the root is obtained to a sufficient de- 
 gree of exactness. 
 
 Observe, that the decimal periods are to be pointed off 
 from the decimal point toward the right hand ; and that 
 there must be as many whole number figures in the root, 
 as there are periods of whole numbers, and as many de- 
 cimal figures as there are periods of decimals.] 
 
 EXAMPLES. 
 
 1. What is the square root of 729? Jlns. 27. 
 
 2. What is the square root of 106929 ? Ans. 327. 
 
 3. What is the square root of 429025 ? Ans. 655. 
 
 4. What is the square root of 444889 ? Ans. 667. 
 
 5. What is tlie square root of 776161 ? Ans. 881. 
 
 6. What is the square root of 994009 ? Ans. 997. 
 
 7. What is the square root of 29855296 ? 
 
 Ans. 5464. 
 
 8. What is the square root of 141225,64 ? 
 
 Ans. 375,8. 
 
 u"2 
 
234 ARITHMETIC. 
 
 9. What is the square root of 43046271 ? 
 
 .fins. 6561. 
 
 10. What is the square root of 904? 
 
 ^ns. 30,066 + 
 
 11. What is the square root of 670? 
 
 MS. 1^,884 -}- 
 
 12. What is the square root of 20? 
 
 ^ns. 4,472 + 
 
 13. What is the square root of 2 ? 
 
 ^ns. 1,4142 + 
 
 14. Find the square root of 3. .dns. 1.732 -f- 
 
 15. Find the square root of 5. ^ns. 2,236 -j- 
 
 16. Find the square root of 10. j9ns. 3,1822 + 
 N. B. When the power is a decimal, commence to 
 
 point off periods from the decimal point, and operate as 
 in whole numbers. 
 
 EXAMPLES. 
 
 1. What is the square root of ,00008836? 
 
 ^7is. ,0094. 
 
 2. Find the square root of ,00529. Jins. ,023. 
 
 3. Find the square root of ,002916. Jiyis. ,054. 
 Observe, the roots are greater than the powers, as they 
 
 should be in fractions. 
 
 (Art. 129.) Mixed numbers may be square, when re- 
 duced to the fractional form (improper fractions) ; but the 
 square form will commonly be indicated by the denomi- 
 nator of the fraction being a square number. In such 
 cases, reduce the mixed numbers to improper fractions, 
 and extract the roots of numerator and denominator. 
 
 EXAMPLES. 
 
 1. What is the square root of 37^f ? 
 
 2. What is the square root of 272 \- ? 
 
 3. Find the square root of 27yg. 
 
 4. Find the square root of 9i|. 
 
 5. Find the square root of 20'^. .^ns. 4i. 
 Fractions may be square in value, but not expressed 
 
 by square numbers; but can be reduced to such, and af- 
 
 Jlns. 
 
 61. 
 
 Mns. 
 
 16L. 
 
 Ans. 
 
 5|. 
 
 Arts. 
 
 31. 
 
SQUARE ROOT. 235 
 
 terwards the square root extracted. The following ex- 
 amples will illustrate : 
 
 6. What is the square root of -,~ ? ^dns. i. 
 
 7. What is the square root of f^^ ? Jinn, f . 
 
 8. Find the square root of =-|^ . Ans. ^. 
 
 9. What is the square root of ^^5.| ? ^ns. |. 
 
 10. What is the square root of f l|| ? Jins, |. 
 
 11. Find the square root of ||-|J. Ans. |. 
 
 When fractions or mixed numbers are surds, reduce 
 the fractions to decimals, and proceed as in whole num- 
 bers. 
 
 EXAMPLES. 
 
 1. What is the square root of 7/^ ? Ans. 2,7961 -f 
 
 2. Find the square root of 17|. Ans. 4,168-}- 
 
 3. Find the square root of ?. Ans. ,86602 -[- 
 
 4. Find the square root of ~\. Ans. ,9574-}- 
 
 5. Find the square root of 85{-i. Ans. 9,27-f- 
 
 APPLICATION. 
 
 1. A certain square pavement contains 20736 square 
 stones, all of the same size ; what number is contained in 
 one of its sides? Ans. 144. 
 
 2. If 484 trees be planted at an equal distance from 
 each other, so as to form a square orchard, how many- 
 will be in a row each way ? Ans. 22. 
 
 3. A certain number of men gave 30 shillings 1 pence 
 for a charitable purpose ; each man gave as many pence 
 as there were men : how many men were there ? 
 
 A7ts. 19. 
 
 (Art. 130.) To be scientific and skillful in the appli- 
 cation of square root, it is necessary to attend to the fol- 
 lowing properties of square numbers. 
 
 I. A square number, multiplied by a square number, 
 will produce a square number. 
 
 II. A square number, divided by a square number will 
 give a square. 
 
 III. If the square root of a number is a composite 
 
236 ARITHMETIC. 
 
 number, the square itself may be divided into integer 
 square factors ; but if the root is a prij7ie number, the 
 square cannot be separated into square factors, ivitkout 
 fractions. 
 
 IV. When two or more factors are to be multiplied to- 
 gether, we may often abbreviate the operation, by ob- 
 serving whether they are, or whether we can change them 
 into, square factors. If so, take the square roots of the 
 factors, and multiply them together for the result. 
 
 EXAMPLES. 
 
 1. 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? ^^ns. 320. 
 
 Very. many of our teachers 7,oouJd actually reduce the 
 acres to square rods, by multiplying by 160, and extract 
 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 mul- 
 tiplied by 1600, both square numbers, whose roots are 8 
 and 40— product 320, the answer. 
 
 2. A man has a farm in exactly a square form, con- 
 taining 50| acres ; how many rods on a side ? 
 
 Ans. 90 rods. 
 Solution, 50| = ^|-^ Then ^f^ x 160=405X20=81 
 XIOO. 
 
 3. A certain triangular field contains 10 acres of land ; 
 what is the length of the sides of a square containing the 
 same number of acres ? Ans. 40 rods. 
 
 4. 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 ? 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 3X 12=36, 
 the answer. 
 
 5. A has two fields, one containing 10 acres, and the 
 other 12 V ; what will be the length of the side of a field 
 containing as many acres as both of them? 
 
 Ans. 60 rods. 
 
SQUARE ROOT. 237 
 
 6. It is required to lay out 120 acres of land in the 
 form of a right-angled parallelogram, so that the length 
 shall be 3 times the breadth ; required the sides. 
 
 .^iis. 240 feet length, 80 feet i^readth. 
 
 7. Required the dimensions of a square, the area of 
 which shall be equal to the area of a parallelogram 18 
 feet in length, and 8 in breadth. Jlns, 12 feet square. 
 
 (Art. 131.) A mean proportional between two num- 
 bers, is a number requisite to form the means of a geo- 
 metrical proportion between those numbers. Thus, 
 
 3 : 6 :: 6 : 12 
 The two means in this proportion are expressed by the 
 some nwnber. That number is, therefore, called a mean 
 proportional, and its square is equal to the product of the 
 given extremes (Art. 72). 
 
 Hence, to find a mean proportional between two num- 
 bers, we may have this 
 
 Rule. Mulliply the numbers together, arid take the 
 square root of the product. 
 
 Observation. When the given numbers are square 
 numbers, take their roots at once, and multiply them to- 
 gether ; or, if convenient, change them into equivalent 
 square factors, &;c., as in Art. 130. 
 
 EXAMPLES. 
 
 1. Find the mean proportional between 4 and 25. 
 
 Ans. 10. 
 
 2. Find the mean proportional between 2 and 12,5. 
 
 Ans. 5. 
 
 3. Find the mean proportional between ^ and 4. 
 
 Jins. |. 
 
 4. Find the proportional between 24 and 96. 
 
 Jlns. 48. 
 As 96=4X24; the product of the two numbers is 
 equal to 4X24X24=4X 24^ root, is 2X24=48 Ans, 
 
 5. Find the mean proportional between 18 and 24. 
 
 Am. 1273. 
 
 6. Find the mean proportional between 18 and 32. 
 
 Arts, 24. 
 
238 ARITHMETIC. 
 
 7. Find the mean proportional between ,25 and 1. 
 
 Ans. ,5. 
 
 8. AVhat is the mean proportional between 3 and 6 1 
 
 Arts. SJ2. 
 
 9. What is the mean proportional between 4 and 40 ? 
 
 Ans. 4^10. 
 
 These surd answers may be taken more definitely by 
 extracting the root in decimals, or by taking the roots al- 
 ready extracted in Art. 128. 
 
 (Art. 132.) A right-angled triangle, is a triangle hav- 
 ing one right angle, and the other two angles acute. The 
 side opposite the right angle is called the hypotenuse. 
 
 C 
 
 The adjoining figure represents a right- 
 angled triangle — the right angle at B. It 
 is proved, in geometry, that the sum of 
 the squares of the two sides containing the 
 right angle, is equal to the square of the 
 side opposite the right angle. Therefore 
 the difference of the squares between the B A 
 
 hypotenuse and one of the sides, is equal to the square 
 of the other side. 
 
 It is also proved, in geometry, that triangles of the 
 same shape have their like sides proportional. 
 
 This last theorem enables us to abbreviate and reduce 
 large right-angled triangles to smaller ones, and the re- 
 verse. 
 
 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 ivhen the numbers have 
 a common divisor, or when the side sought is expressed 
 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. 
 
SQUARE ROOT. 239 
 
 The square root of 25 is 5, wliich, multiplied by 15, gives 
 75, the Miiswer. 
 
 2. Two brothers left their father's house and went, one 
 64 miles due west, the other 48 miles due north, and 
 purchased farms ; how far are they from each other ? 
 
 ^^ns. 80 miles. 
 
 Divide by ... 16)64 48 
 
 4 3 164-9=25, 5X16=80. 
 
 3. The hypotenuse of a riglu-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 being 
 broken off, struck the ground at the distance of 15 feet 
 from the foot, and measured 39 feet. Ans. 75 feet. 
 
 3. 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 distance, in 
 yards, did he shoot? Ans, 81,15+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? Ans. 24. 
 
 When the hypotenuse and one side is given, in place 
 of squaring them, and taking the difference, &c., we may 
 take the square root of the product of their sum and 
 difference, for the answer, as in the following example : 
 
 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. 14,142-+-yards. 
 
340 ARITHMETIC. 
 
 Sum, 50; diff. 4; ^50X4 = V200 = 14,142 + ^n.s. 
 
 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 witliout moving it at the foot, will reach a 
 window on the other side 21 feet high, what is the 
 breadth of the street ? ^iis. 56,64-4-feet. 
 
 9. A ladder 40 feet long stands close against the side 
 of a building; it is drawn out at the bottom sufliciendy 
 to lower the top 2 feet; how many feet is it drawn out 
 at bottom ? ^^^^ ^156= 12,49 feet, nearly. 
 
 10. A certain room is 28 feet long, 22 feet wide, and 
 12 feet high ; what is the diagonal distance from the low- 
 er to the opposite upper corner of the room ? 
 
 .^7is, 37,57+feet. 
 
 11. Wishing to know the distance from a station near 
 the side of a river to a tree below, but being prevented 
 by a curve in the stream from measuring it, I run out at 
 right angles to the line of direction 15 rods, and then 30 
 rods in a right line to the tree; what was the distance 
 required ? Jins. 26 rods, nearly. 
 
 12. Suppose the width of a barn to be 36 feet, of what 
 length must the rafters be, if the elevation of the ridge 
 above the level of their foot be 14 feet. 
 
 Ms. 22 feet 9| in. 
 
 Questions. 
 
 What is evolution ? 
 
 To extract the square root, why do you point off pe- 
 riods of 2 figures each? 
 
 In obtaining divisors, why do you double the root al- 
 ready found ? 
 
 Has every number an exact root ? 
 
 What is a right-angled triangle ? 
 
 What relation is there between the square of the hy- 
 potenuse, and the squares of the two sides ? 
 
 If the two sides are given, how do we find the hypo- 
 tenuse ? 
 
 If the hypotenuse and base are given, how is the per- 
 pendicular found ? 
 
CUBE ROOT. 241 
 
 If the hypotenuse and perpendicular are known, how 
 do we find the base ? 
 
 Can you ever reduce a right-angled triangle to smaller 
 dimensions ? 
 
 Can you reduce a right-angled triangle, v/lien the num- 
 bers expressing the sides are prime numbers? 
 
 How do you return to the original triangle ? 
 
 TO EXTRACT THE CUBE ROOT. 
 
 (Art. 133.) The cube of a number is the product of 
 that number, multiplied into its square. A cube is also 
 a geometrical solid, of six equal square surfaces. From 
 each of these definitions, or modes of considering a cube, 
 we can draw the same rule for extracting: the root; one, 
 by retracing the multiplication of the different figures of 
 the root, to form the cube: the other, by constructing and 
 increasing geometrical solids: we shall orive the analysis 
 of both. 
 
 In Art. 126, we have sufficiently explained that the 
 cube of a number may contain three times the number of 
 places of figures that are in the root ; and, therefore, we 
 must divide a cube off into periods of three figures each, 
 to fiiid the number of figures in the root; but t!iis we 
 deem of sufiicient importance, to take another view of 
 the subject. If we take any figure (say 8), and involve 
 it as units, tens, hundreds, and so on: that is, as 8, 80, 
 800, and so on; we shall readily discover the Inw v.'hich 
 governs tlie posiiion of the successive products, 8 X 8X 8 
 = 512. As a unit, its cube terminates with the unit's 
 place, and influences hut two other places, the tens and 
 liundreds. As a ten, its proper product is the same, but 
 is removed three places toward tlie left; thus, 80X80X 
 80 = 512000. As a hundred, its product, a cube, \\\\\ 
 fall three places higher, and so on — clearly indicating 
 three places in a cube to one in the root. 
 
 (Art. 134.) To trace back a cube, let us involve a 
 number, consisting of two figures, to its cube, keeping 
 the numbers separate by signs, and observing that a num- 
 ber multiplied by itself, gives its square, which may be 
 indicated by the exponent 2, the cube by the exponent 3. 
 
242 ARITHMETIC. 
 
 Cube 46. Express it thus, . . . • 40-r6 
 
 40+6 
 
 40X6-1-62 
 
 40^-1- 40X6 
 
 402-j- 2 X 40 X 6+63.=square. 
 
 40-1-6 
 
 40^X6+2X40X62-1-65 
 40^+2X40^X6+ 40X6^ 
 
 40^+3 X 40^ X 6+3 X 40 X 62+6-=cube. 
 The whole cube then consists of — 
 
 1. The cube of the tens 40^ =64000 
 
 2. Three times the square of the 
 
 tens into the units, 3 X 40"" X 6 =28800 
 
 3. Three times the tens into the 
 
 square of the units, 3X40 X 6'= 4320 
 
 4th, and last. The cube of the 
 
 units, 6^= 216 
 
 Entire cube, =97336 
 
 To retrace this operation — to extract the cube root of 
 97-336 — we must divide it off into periods, of three fig- 
 ures each, commencinfT at the units ; this gives 97 for the 
 superior period, in which the cube of the tens must ex- 
 ist. Take die greatest possible cube in 97, which is 64, 
 (whose root is 4,) and subtract it from the whole cube ; 
 thus, 
 
 97*336(4 
 
 64 
 
 33336= 1st dividual. 
 
 Let us call to mind the fact, that we know the root of 
 this cube already, and that its next superior constituent 
 part is 3.4016." Observe, that 402=4-X 100, and this, 
 multiplied by 3, is 4^X300; that is, the square of the 
 root already found, multiplied by 300. In retracing, the 
 6 is not yet discovered ; but 42X 300=4800, forms a par- 
 
CUBE ROOT. 2'13 
 
 lial divisor for tlie dividual. Tlie number of times this 
 will go in the dividual, is an index for the next figure. 
 Wlieii the divisor is small in relation to the dividual, the 
 next ligurc may be one or two uniis less than the quo- 
 tient, as the dividual must contain three terms. Now, 
 
 97336(46 
 
 64 
 
 Partial divisor, 4800 )33336=dividual. 
 
 4800X0 28800 = lstterm. 
 
 4 X 6- X 30 =3.40.62= 4320 =2d term. 
 
 6X6X6 = 216 = 3d and last term. 
 
 S3336=the dividual. 
 This operation, put in words, gives the following 
 
 RuLR. Separate the given number into periods of 
 three figures each, beginning at the iinii's place. Find 
 the greatest cube in the left-hand period, and set its 
 root in the quotient; subtract said cube from the pe- 
 riods, and to the remainder bring doivn tlie next pe- 
 riod for a dividual. Square the root cdready found, 
 and multiply it by 300, /or a partial or imperfect divi- 
 sor. See how often the partial divisor is contained in 
 the dividucd, and take the result for the next figure of 
 the roof. 
 
 [Note. As the divisor is partial, 7?o?/^//, this quotient, 
 ill some cases, will be too large for the next figure of th.e 
 root: there is no certain way of deciding except by trial.] 
 
 Multiply the particd divisor by the last foitnd figure 
 of the root ; also multiply the square of the last found 
 figure by the former part of the root, and that product 
 by 30 ; ab-o cube the last found figure. .,^dd these three 
 products together, and subtract their amount fr era tJie 
 dividual. 
 
 To the remainder, bring dcicn the next period, and 
 proceed as before, until the periods are all brought down. 
 
 TVhcn (I remainder occurs, annex periods of ciphers 
 to obtain decimals, which may be carried to any desired 
 number. 
 
244 
 
 ARITHMETIC. 
 
 N. B. Tlie cube root of a vulgar fraction is found by 
 reducing it to its lowest terms, and extracting the root of 
 tlie numerator for a numerator, and of the denominator 
 for a denominator. If it be a surd, extract the root of its 
 equivalent decimal to the desired degree of exactness. 
 
 (Art. 135.) We now illustrate by means of geometri- 
 cal solids. What is the cube root of 15625 ? 
 
 Operation, 
 
 2X2X2= 8 
 
 15,625 1 25 
 
 7625 
 
 In this number there 
 are 2 periods ; of course 
 there will be 2 figures 
 in the root. 
 
 ''The greatest cube 2X2X300 = 1200 | 7625 
 
 in the lefi-hand period 
 
 (15), is 8, the root of 6000 
 
 i^,'/u*cA fs 2 ;" therefore, 5X5X2X30= 1500 
 2 is the hrst figure of 5X5X5= 125 
 
 the root ; and, as we 
 
 shall have another fig- 
 ure in the root, the 2 
 stands for lens, 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, contains 20 
 X20X20=8000. 
 
 " Subtract the cube 
 thus found (8) from 
 said periods and to the 
 remainder bring doicn 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, Fig. 1, 20 feet long, 20 feet wide, and 20 feet 
 high. 
 
 The cube is to be enlarged by the addition of 7625 
 feet which remain. In doing this, the figure must be en- 
 larged on three sides, to make it longer, and ivider, and 
 higher, to maintain the complete cubic form. 
 
CUBE ROOT. 
 
 245 
 
 The next step is, to find a divisor; and tliis mnst be 
 the number of square feet contained in the three sides to 
 which the addition must be niadc. Hence we " muUlphi 
 the square of the, quotient /i<ri(re by 300." TJiat is, 
 2X2X300=1200: or, 20X26 X 3 = 1200 feet, which is 
 the superficial content of the tliree sides, «, b, and c. 
 
 This ''divisor (1200) is con- 
 tained inthc.dividiiaP' (7625) 
 5 times: tlien 5 is the second 
 quotient fitrure; that is, the ad- 
 dition to each of the three sides 
 is 5 feet thick; if 1200 feet co- 
 ver the three sides 1 foot thick, 
 5 feet thick will require 5 times 
 as many; that is, 1200X5 = 
 6000. 
 
 But when tlie additions are 
 made to three squares, there 
 will be a deficiency along the 
 whole length of the sides of 
 the squares between the addi- 
 tions, which must be supplied 
 before the cube will be com- 
 plete. These deficiencies will 
 be three, as may be seen at 
 NNN in Fig. 2; therefore it 
 is that we "■midtiply the square 
 of the last figure by tlie prece- 
 ding figure, andtnj 30," (ihat 
 is, 5 X 5 X 2 X 30,) or, 5 X 5 X 5 
 X20X3 = 1500, which is the 
 quantity required to supply the 
 three deficiencies. 
 
 Figure 3 represents the sol- 
 id, with these deficiencies sup- 
 plied, and discovers another de- 
 ficiency, where they approach 
 each other at 000. 
 
 Lastly, ''cube the last fg- 
 ure:^' this is done to fill the de- 
 
 Hi 
 
 x2 
 
246 
 
 ARITHMETIC. 
 
 Proof. 
 
 20X20X20= 8000 
 
 20X20X3X5= 6000 
 
 5X5X20x3= 1500 
 
 5X5X5= 125 
 
 25X25X25= 15625 
 quently, the cube, of 
 
 ficiency left at the corner, 
 in filling up the other defi- 
 ciencies. This corner is 
 limited by the three por- 
 tions applied to fill the for- 
 mer vacancies, which were 
 5 feet in breadth : conse- 
 
 5 will be the solid contents of the 
 corner. Fig. 4 represents this deficiency [eee) supplied, 
 and the cube is complete. 
 
 Observe, that 20 X 20 X 3 X 5 =2^ X 300 X 5 ; and thus, 
 by a Utile change of form, the numerical and geometrical 
 analysis agree, and will give one and the same rule. 
 
 EXAMPLES. 
 
 1. AVhat is the cube root of 99252847? 
 99'252'847(463 
 4X4X4=64 
 
 4X4X300=4800 
 
 Div. 4800X6= 
 
 6X6X4X30 = 
 
 6X6X6= 
 
 35252 
 
 28800 
 
 4320 
 
 216 
 
 .^ns. 463. 
 
 463 
 463 
 
 Subtrahend, 33336 
 
 46X46X300=634800 1 1916847 
 
 1389 
 
 2778 
 1852 
 
 214369 
 463 
 
 
 Div. 634800X3=1904400 613107 
 
 
 3X3X46X30= 12420 1286214 
 
 
 3X3X3= 27 857476 
 
 Subtrahend, 1916847 Proo/, 99252847 
 
 2. 
 
 What is the cube root of 84604.519? Ans. 439. 
 
 3. 
 
 What is the cube root of 259694072 ? 
 
 
 ^ ■ Ans. 638. 
 
 4. 
 
 What is the cube root of 32461759 ? 
 
 
 A71S. 319. 
 
 5. 
 
 What is the cube root of 27054036008 ? 
 
 i 
 
 Ans. 3002. 
 
CUBE ROOT. 247 
 
 6. What is the cube root of 22069810125? 
 
 Ans. 2805. 
 
 7. What is the cube root of 15926,9725 ? 
 
 Ans. 25,16+ 
 Wheii decimals occur, point the periods both ways, 
 beginning at the decimal point : and if the last period of 
 the decimal be not complete, arr.ex one or more ciphers. 
 
 A mixed number may be reduced to an improper frac- 
 tion, or a decimal, and then the root extracted. 
 
 8. What is the approximate cube root of ,000684134 ? 
 
 Ans. ,088-f 
 
 9. Find the approximate cube root of ,008649. 
 
 Ans. ,2052-1- 
 
 10. What is the cube root of /jVo? ^^»«- f 
 
 11. What is the cube root of |f^? Ans. |. 
 
 12. What is the cube root of W^l A71S. f. 
 
 13. What is the cube root of 12^^ 1 Ans. 2L. 
 
 14. What is the cube root of 31 3' 4V- ^^'^^' ^f- 
 
 APPLICATION. 
 
 1. There is a cistern, or vat, of a cubical lorm, which 
 contains 1331 cubical feet; what are the lenglh, breadth 
 and depth of it? Ans. Each 11 feet. 
 
 2. A certain stone of a cubical form contains 474552 
 solid inches ; what is the superticial content of one of its 
 sides, expressed in feet? Ans. 421 sq. feet. 
 
 3. The pedestal of a monument was a square block 
 of gi-anite, coniaining 373248 cubic inches; what was 
 the length of one of its sides ? Ans. G feet. 
 
 4. The statute bushel contains 2150,4 cubic incjies ; 
 what is the side of a cubic box that will contain 50 bush- 
 els ? Ans. 47,5 inches. 
 
 5. A man wishes to make a bin to contain 125 bushels, 
 of equal depth and width, and double in length ; what 
 shall be its dimensions ? 
 
 Ans. Width and depth, 5 1,22+ inches. 
 In lenglh, .... 102,44 inches. 
 
 (Art. 136.) It is proved in solid geometry, that all so- 
 lid bodies of similar shape, are to one another as the 
 cubes of their like dimensions. 
 
248 ARITHMETIC. 
 
 N. B. We need not take their real, if we can find 
 their more simple, relalive dimensions. 
 
 On this principle, solve the following problems : 
 
 1. Mercury is about 2000 miles in diameter, and the 
 earth about 8000 ; what is their relative magnitudes ? 
 
 .r2ns. As 1 to 64. 
 
 2. Mars is about 4000 miles in diameter, the earth 
 8000 ; what is their relative magnitudes ? 
 
 .^728. As 1 to 8. 
 
 3. The diameter of the earth, to that of the sun, is 
 nearly as 1 to 111!,; what is their relative magnitudes, 
 or bulks? " .^??5. As 1 to 13fGUi5, nearly. 
 
 4. If a ball, 6 inches in diameter, weigh 32 pounds, 
 what will be the weight of a ball, of the same metal, 
 whose diameter is 3 inches ? ^^tis. 4 pounds, 
 
 6 : 3 :: 2 : I 
 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. 
 
 ./^ns. 54 lbs. 4 oz. 
 
 6. If a cable, 12 inches about, require an anchor of 
 18 hundred weight, of what weight must an anchor be 
 for a 15 inch cable ? .^ns. 35 cut. 15 lbs. 
 
 7. Suppose the diameter of the earth to be 3952 miles, 
 and a portion of it is water sufficient to cover the whole 
 body to the average depth of 9\ miles ; what proportion of 
 the earth's bulk would then be water ? 1 
 
 »^ns. n. 
 
 69 
 
 8. If a ball of lead, 2 inches in diameter, Aveighs 2 
 pounds, what will be the weight of another, 10 incb^es in 
 diameter ? .^ns. 250 Ids. 
 
 9. In 2 cubic feet, how many cubes of 6 inches ? 
 
 »^ns. 16. 
 
SUPPLEMENT TO SqUARK AND CUBE ROOTS. 
 
 249 
 
 SUPPLEMENT TO SQUARE AND CUBE 
 ROOTS. 
 
 RECAPITULATION. 
 
 We again call to mind the following properties of num- 
 bers. Their importance cannot be exaggerated, if we wish 
 to insure skill, or even sound knowledge, on this subject. 
 
 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. 
 
 V. If any root is a composite number, its power (the 
 square or cube, as the case may be) may be separated 
 into square or cube factors; but, if the root is ^ prime 
 number, the power cannot be so separated. 
 
 VI. If the unit figure of a square number is 5, we may 
 multiply by the square number 4, and we shall have an- 
 otlier square, whose ?//n7 period will be ciphers. 
 
 VII. If the unit figure of a cube is 5, we may multiply 
 by the cube number 8, and produce another cube, whose 
 unit period will be ciphers. 
 
 N. B. If a supposed cube, whose unit figure is 5, be 
 multiplied by 8, and the product does not give three ci- 
 phers on the right, the number is not a cube. 
 
 (Art. 137.) V/e again present the following table, for 
 the pupil to compare the natural numbers with the unit 
 figure of their squares and cubes, that he may be able to 
 extract roots by inspection. 
 
 Numbers, 
 
 I 
 
 2 
 
 3 4 
 
 1 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 10 
 
 Squares. 
 
 1 
 
 4 
 
 9 
 
 IG 
 
 25 
 
 36 
 
 49 
 
 64 
 
 81 
 
 100 
 
 Cubes. 
 
 1 
 
 8 27 
 
 64 
 
 125 
 
 216 
 
 343 
 
 512 
 
 729 
 
 1000 
 
250 ARITHMETIC. 
 
 EXERCISES FOR PRACTICE. 
 
 1. What is the square root of G25? ^%is. 25. 
 If the root is an integer number^ we may know, by 
 
 the inspection of the 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 5561 ? Ans. 81. 
 
 As the unit Jigiire, in this example, is 1, and in the 
 line of squares in the 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 ? Ans. 327. 
 
 As the unit figure, in th^s example, is 9, if the number 
 is a square, it si divisible 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=lG9. 109, 
 multiplied by 3, the root of 9, gives 327 for the answer. 
 
 4. What is the mot of 451584 ? Ans. 672. 
 This number is obviously divisible by the square num- 
 ber 4: thus, 4)451584 
 
 Divide again by 4, 4)112896 
 
 And again, 4)28224 
 
 And again, 4^7056 
 
 And again, as long as the division is obvious, 4)1764 
 
 441 
 Now, the square rcot of the last quotient is 21, and 
 the square roots of the divisors are 2. 2, 2, 2, 2; the con- 
 tinued product of all these roots is 672, the ansv/er. 
 
ABBREVIATIONS IX CUBE ROOT. 251 
 
 5. Extract the square root of 21)25. Ans. Ad. 
 1st. Divide by the square number, 25, and we tind 
 
 tlie two factors, 25X81, as equivalent to the given num- 
 ber. Roots of these factors, 5X9 = 45, the answer. 
 
 2d. Divide by tlie square number, 100, and we have 
 20^X100 = ^X100,^X10 = 45, as before. 
 
 3d. Take . . . 2025 
 Multiply by . . . 4 
 
 8100 Root 90, divide by 2, because 
 we multiplied the power by 4, and we have 45 for the 
 result. 
 
 6. AVhat is the square root of 390625 ? d^u. 625. 
 
 390625 
 Multiply by 4, 4 
 
 1562500 
 Multiply by 4 again, 4 
 
 6250000 
 Multiply by 4 again, 4 
 
 25000000 
 The square root of this last product is obviously 
 5000; which, divide by 2, three times, or by 8, and 
 we have 625 for the answer, or square root of the origi- 
 nal sum. 
 
 7. What is the square root of 119025 ? Jins. 345. 
 
 8. What is the square root of 75625? JIns. 275. 
 
 (Art. 138.) ABBREVIATIONS IN CUBE ROOT. 
 
 1. What is the cube root of 91125 ? Ann. 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. 
 
252 ARITHMETIC. 
 
 2. The contents of a cubical cellar are 1953,125 cu- 
 bic feet ; what is the length of one of its sides ? 
 
 Ms. 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 Ave 
 multiplied by 8 twice, and we have 12,5, tlie answer. 
 
 When it is requisite to multiply several numbers to- 
 gether, and extract the cube root of their product, try to 
 change them into cube factors, and extract the root he- 
 fore multiplication. 
 
 EXAMPLES. 
 
 1. What is the side of a cubical mound equal to one 
 288 feet long, 216 feet broad, and 48 feet high ? 
 
 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 
 =2X 12X 12. and 48 = 4 X 12; therefore, the whole pro- 
 duct is 216X8X 12X 12X 12. Now, the cube root of 
 216 is 6, of 8 is 2, and of 12^ is 12, and the product of 
 6X 2 X 12 = 144, the answer. 
 
 2. Required the cube root of the product of 448X392, 
 in a brief manner. Jlns. 56. 
 
 3. What is the side of a cubical mound, equal to one 
 144 feet long, 108 feet broad, and 24 feet deep ? 
 
 .8ns. 72. 
 
 4. A pile of wood is 160 feet long, G feet wide, and 9 
 feet high ; what would be the side of a cubic pile con- 
 taining the same quantity? jj^^^ l2 3V5=20,5+feet. 
 
 5. What shall be the side of a cubical cistern, to con- 
 tain 04 hogsheads? Jins. 8,138 feet. 
 Solution: 64 X 63 X 231 =04 X 7X9X3X77=64.27.539 
 
 Cube root is 4X 3X='V539 = 12X8,1 38 in., or 8,138 ft. 
 
ABBREVIATIONS IN CUBE ROOT. 253 
 
 (Art. 139.) T'fe can extract the root of cube num- 
 bers, by inspection, when they do not contain more than 
 two periods. 
 
 EXAMPLES. 
 
 Find the cube root of 1951 12. This number consists 
 of two periods; compare the superior period with the 
 cubes in the table, and we find that 195 lies between 125 
 and 2,16. The cube root of tlie tens, then, must be 5. 
 The unit fio-ure of the given cube is 2, and no cube in tiie 
 table has 2 for its unit figure, except 512, whose root 
 is 8 ; therefore, 58 is the root required. 
 
 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 fio-ure that will ansv/er. 
 
 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 1 10592 ? Ans. 48. 
 
 5. What is the cube root of 357911 ? Ans. 71. 
 
 6. Wh.at is the cube root of 889017 ? Ans. 73. 
 
 7. What is the cube root of 571787? Ans. 83. 
 
 When a cube has more than two periods, it can gene- 
 rally be reduced to two by dividing by some one or more 
 of the cube numbers, unless the root is a prime number. 
 
 The number 4741632 is a cube; required its root. 
 Here we observe, that the unit figure is 2; the unit figure 
 of the root must, therefore, be the root of 512, as that is 
 the only cube of the 9 digits v/hose unit figure is 2. The 
 cube root of 512 is 8; therefore, 8 is the unit figure in 
 tlie root, and the root is an even number, and can be di- 
 vided by 2 — and, of course, the cube itself can be divi- 
 ded by 8, the cube of 2. 
 
 8)4741632 
 
 592704 
 
25i ARITHMETIC. 
 
 Now, ?,s the first number was a cube, and being divi- 
 ded by a cube, tlie 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 ? 
 
 ./??2.<?. 237. 
 
 As there are three periods, there must be three fiirures, 
 units, tens, and hundreds, in the root ; the hundreds must 
 be 2, the units must be 7. Let us then find the second 
 figure, or the tens, in the usual ivay, and w^e 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, w^hich, divided by 2, gives 265 for the root required. 
 
 TO FIND THE APPROXIMATE CUBE ROOT OF SURDS. 
 
 (x\rt 140.) The usual way of direct extraction, is too 
 tedious to 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 tliat event, 
 several eminent mathematicians bestowed much time and 
 labor to obtain short practical rules — and some of their 
 rides are too ingenious and useful to be lost, notw^ith- 
 standing the invention oflrgarithras has nearly superced- 
 ed their absolute value in practice. 
 
 The following method is from Dr. Halley's algebraic 
 formula, but more commodiously expressed ; and after 
 knowing the result of the analysis, -we can apparently 
 draw the same from mere observations on numbers, thus. 
 
ABBREVIATIOXS IN CUBE ROOT. 255 
 
 Let us take two cube numbers, say 125 and 216, whose 
 roots are 5 and G. There must be some law of compar- 
 ison between them, either simple or complex. We evi- 
 dently cannot make a proportional comparison between 
 them, as 125 is not to 216, as 5 to 6. But let us double 
 125 and add 216, which gives 466, and double 216 and add 
 125, which gives 557. Now 466 is to 557 as 5 to 6, 
 nearly. Here we have an approximale proportion. But 
 to have a very near approximation, our cubes nuist have 
 a nearer relative value ; 216 is nearly double of 125 : this 
 should never be the case in making practical use of ap- 
 proximate proportions. 
 
 To show that we can be more accurate, observe that 
 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 near- 
 ly 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, 
 
 668981 : 10981 : : 60 : 1 very nearly. 
 
 But 1 is the difference between the two roots, and if the 
 last root, or 61, were unknown, this proportion would give 
 it very nearly ^ by giving the difference between the two 
 roots. 
 
 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 
 
 64X2 = 128 2X66=132 
 63 64 
 
 194 : 196 : : 4 : to root of 66. 
 
 Or, 194 : 2 :: 4 : to a correction. 
 
256 ARITHMETIC. 
 
 194)8,0000(0,04124 
 7 76 
 
 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 
 
 2 238 Take 0,02681 
 
 3020 Ans. 4.97319 
 
 2984 
 
 860 
 
 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 cfnd add the assumed cube to it. Take the dif- 
 ference of these sums, then say, Jis double of the as- 
 
ABBREVIATIONS IN CUBE ROOT. 257 
 
 Sinned cube, added to the numher, is to this difference, 
 so is the assumed root to a correclion. 
 
 This correction, added to or subtracted from the assum- 
 ed root, us the case may require, will give the cube root 
 very nearly. 
 
 By repeating tlie operation with tlie root last found as 
 an assumed root, we may obtain results to any degree of 
 exactness ; one operation, however, is generally suffi- 
 cient. 
 
 3. What is the cube root of 28? £ns. 3,03658-f- 
 
 4. What is the cube root of 26 ? .fins. 2,96249-{- 
 
 5. What is the cube root of 214? Ans. ^,m\A2-{- 
 
 6. Wjiat is the cube root of 346 ? Ans. 7,02034-1- 
 
 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. 
 
 1. What is the cube root of 3214? Ans. 14,75758. 
 
 Suppose the root is 15 — its cube is 3375, which, be- 
 ing greater than 3214, shows that 15 is too great; the 
 correction will, therefore, be subtractive. 
 
 By the rule, 9964 : 161 : : 15 : 0,2423, the correc- 
 tion. 
 
 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 ? 
 
 Ans, 2453. 
 
 Suppose the root 2400, &c. Take the correction to 
 the nearest unit, and you will find it 53. 
 
 y2 
 
258 ARITIIMKTIC. 
 
 3. Wliat is the cube root of 980922617856 ? 
 
 .^7is. 9936. 
 Suppose the root to be 10000. 
 
 4. What is the cube root of 9 ? ^ns. 2,08008. 
 
 5. What is the cube root of 9^ ? ^/Ins. 2,092+ 
 
 6. What is the cube root of 41 ? ^ns. 3,4482-1- 
 
 7. AVliat is the cube root of 321 ? .^ns. 0,847-}- 
 
 8. What is the cube root of 32,1 ? ^ns. 3,178-j- 
 
 9. What is the cube root of 2,5 ? ./ins. 1,357-1- 
 10. What is the cube root of 7 ? Ans. 1,9129-}- 
 
 A GENERAL RULE FOR EXTRACTING THE 
 ROOTS OF ALL POAVERS. 
 
 (Art. 141.) Whenever we propose a root, we conceive 
 the given number to be a corresponding power. 
 
 If we propose or demand the cube root, we virtually 
 assert tliat the given number is a third power. If we de- 
 mand the fourth root, the given number is understood to 
 be a fourth power, &c. 
 
 The following rule is from mere inspection of a gene- 
 ral algebraic formula, first translated into words by Dr. 
 Hutton. 
 
 Rule. 1. Point off the given nxnnher into periods 
 consisting of as many places as correspond to the index 
 of the power, 
 
 2. Find the first figure of the root by the table of 
 powers, or by trial: subtract its power frcm the left 
 hand period, 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 mulliply it by the number denoting 
 the given power, for a divisor ; by which find a second 
 figure of the root. 
 
 ! 4. Involve the 2vhole ascertained root to the given 
 power, and sid)tract it from the first and second peri- 
 ods. Bring down the first figure of the next period to 
 the remainder, for a nexv dividend ; to which, fi^nd a 
 new divisor, as before ; and so proceed. 
 
ARITHMETICAL PROGRESSION. 259 
 
 N. B. Roots of component 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 9ih, take the cube root of the cube root. 
 For the 12th, take the cube root of the 4th root. 
 
 EXAMPLES. 
 
 1. What is the 5th root of 916132832 ? 
 
 9161'32832(62 ^ns, 
 
 7776 6X6X6X6X6=7776 
 
 6X6X6X6X5=6480 div. 
 
 6480)13853 
 
 916132832 62X62X62X62X62=916132832 
 916132832 
 
 2. What is the 7ih root of 194754273881 ? 
 
 ^ns. 41. 
 
 3. What is the 9th root of 2 ? ^ns. 1,080059. 
 
 ARITHxMETICAL PROGRESSION. 
 
 (Art. 142.) A series of numbers, increasing or de- 
 creasing by a common difference, is called Arithmetical 
 Progression. 
 
 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. 
 
 By inspecting the nature of a single series we may es- 
 tablish the following facts: 
 
 I. That the last term of any ascending series is made 
 up of successive additions of the common difference, to 
 the first term. 
 
260 ARITHMETIC. 
 
 II. In a descending series of successive subtractions, 
 one less times than the number of terms. 
 
 III. When tlie number of terms is odd, there will be 
 a middle or mean term ; when even, there will be two 
 middle terms, or the middle of the series will be between 
 two terms. 
 
 IV. x\s any series increases or decreases regularly, 
 the value of the middle term must be the average vahiu 
 of the whole, and the middle term must be equal to the 
 half sum of the extreme terms. 
 
 From these observations we draw the following rules. 
 
 [Note. The terms used in the rules show what must 
 be i{iven, and we deem the technicalities too obvious to 
 require explanation.] 
 
 Rule 1. To find the last term. Multiply the com- 
 mon difference' by the number of terms less one,"^ and 
 add the product to the first term, if an ascending series, 
 otherwise subtract it. 
 
 Rule 2. To find the sum of a series. Multiply half 
 the sum of the extreme terms, or mean term, by the 
 number of terms. 
 
 Rule 3. To find the common difference. Divide the 
 difference of tlie extremes by the number of terms less 
 one. 
 
 Rule 4. To find the number of terms. Divide the 
 difference of the extreme terms by the common differ- 
 ence, and add 1 to the quotient, 
 
 PROMISCUOUS EXAMPLES. 
 
 1. What is the 100th term of the series 2, 5, 8, &c.? 
 (Rule 1.) Solution,. . • . 99X3-1-2. »^n5. 299. 
 
 2. The first term is 5, the last 32, and the number of 
 terms 10; what is the svnn of the series? Ans. 185. 
 
 3. How many strokes does the hammer of a common 
 clock strike in 12 hours ? Anfi. 78. 
 
 4. What debt can be discharged in one year, by week- 
 
 * Observe, that the first term exists independenily of the common 
 difference. 
 
ARITHMETICAL TROGRESSION. 261 
 
 ly payments in arithmetical progression, the first being 
 $12, and the last, or fifty-second, payment $1230 ? 
 
 ^7is. 32418. 
 
 5. What debt can be discliarged in one year, by weekly 
 payments in arithmetical progression — the first being $12, 
 and the last $1230; what is the common difierence ? 
 
 Jins. $24. 
 
 6. A man traveling a journey, went 18 miles the first 
 day, and increased his distance each day by 2 miles ; and 
 die last day went 48 miles. How many days did he 
 travel, and what distance ? ^ C 10 days. 
 
 '^"*' i 528 miles. 
 
 7. A houee was leased for 7 years at $400 per annum, 
 and the rent unpaid until the end of the lease; how much 
 was then due, simple interest, at 6 per cent. ? 
 
 ^ns. $3304. 
 
 N. B. At the end of the 1st year, $400 was due; at 
 the end of the second year, $424 more ; at the end of the 
 3rd year, $448 more, &c. 
 
 8. What will be the amount of an annuity of $50, to 
 be paid annually, but forborne 20 years ; simple interest, 
 at 6 per cent.? ^7is. $1570. 
 
 9. 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 return 
 with each separately to the basket. 
 
 ^ns. 20200 yards. 
 
 (Art. 143.) An arithmetical series may be represent- 
 ed by the surface between two converging lines. Two 
 such surfaces put together, the widest end of one against 
 the narrowest end of the other, form a regular parallelo- 
 gram. 
 
 Thus, 3 5 7 9 11 
 
 Reversed,. . . 11 9 7 5 3 
 
 14 14 14 14 14 
 
 The sum of the two converging spaces make one equal 
 in width, &;c. This is another method of explaining 
 Rule 2. 
 
262 ARIT."i.MtTIC, 
 
 10. How many acres in a piece of land 80 rods wide 
 at oae end. and 60 ai the other, and 120 rods long ? 
 
 dfu. o'2\. 
 
 It may be observed, that the natural numbers, 1,2, 3, 
 4, 5, 6. 7, 6ic., 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 term the sum of cdl the precedinsr, and we shall have 
 I, 3, 6, 10, 15, 21,28, &c. 
 
 These are called triansrular rtvrribers, because ihey 
 may be represented by points, forming equilateral trian- 
 gles, thus : 
 
 Heace we perceive, iha: the sum of the natural numbers, 
 lo any degree, expresses the triangular number of ihe 
 same degree. 
 
 GEOMETRICAL PROGRESSION. 
 
 (Art. 144.) A series of numbers, increasing or de- 
 creasi-ng by a common ratio, is called a Geometrical 
 Prosrression. 
 
 Thus, 2, 4, 8, 16, 32, 64, 128, is an increasing se- 
 ries, whose common ratio is 2 ; 
 
 And 729, 243, 81, 27, 9, 3, is a decreasing series, 
 whose common ratio is ^. 
 
 In arithmetical progression, the terms vary by con- 
 stant atiditions or subtractions : here, they vary by con- 
 stant multiplicHtions or divisions. The first term exists 
 independendy cf the ratio. After we use the ratio once 
 we have two terms, using it twice, we have 3 terms, A:c. 
 
 AVhererer we slop is the last term : ience. to find the 
 last term we hive ;;..; 
 
GKOMF.TRICAL PROGRKSSIOX. 263 
 
 RuLc. Raise the ratio to a power one lesa than the 
 number of terms, and midtiphj it by the first term. 
 
 To investigate a rule to find the sum of any series, take 
 the following, 
 
 3, 12, 48, 192, 768, 3072, 12288. 
 Multiply by the ratio 4, and place the product one term 
 to tiie right, — 
 
 12, 48, 192, 768, 3072 12288, 49152. 
 Subtract the upper series from the lower, and we have 
 49152 — 3. 'J'his difference comprises 3 limes the orig- 
 inal series, because we multiplied it by 4, and then sub- 
 tracied the original, 4 — 1=3. Therefore, di. ide this re- 
 mainder by 3, and we have the sum of the series, 'i'he 
 remainder, 49152 — 3, consists of but two terms. One is 
 the last term of the original series, multiplied by the ra- 
 tio. The other term is the first term of the original se- 
 ries subtracted. We then divide the difference by the 
 ra'io less one, and we have the sum of the series. 
 
 Hence,.to find the sum of a series, we have the follow- 
 ing 
 
 Rule 2. Multiply the last term by the ratio; and 
 from, the product, subtract the first term: then divide 
 the remainder by the ratio less one. 
 
 When the extreme terms, and number of terms, are 
 given to find the ratio, the reverse of Rule 1 is used. 
 
 Rule 3. Divide one extreme by the otiier, and extract 
 such a root of the quotient as corresponds to the num- 
 ber of terms less one. 
 
 EXAMPLES. 
 
 1. The first term of a series is 4, the ratio 4, and the 
 number of terms 9 ; what is the last term ? 
 
 Ans. 262144. 
 
 2. The first term of a series is 2, the ratio 3, and the 
 number of terms 8 ; what is the last term, and what is 
 the sum of the terms • ^ 5 ^'^'"^^ term, .... 4374. 
 
 ' * I Sum of the term^, 6560. 
 
 3. What is the sum of ten terms of the series 1, f, ^, 
 &c. Ans. V^VV/. 
 
264 ARITHMETIC. 
 
 4. The first term of a series is 3, the last term 12288, 
 and the number of terms 8; what is the ratio, and what 
 are the other terms ? 
 
 Jlns. Ratio, 4 ; other terms, 12, 48, &c. 
 
 5. Sold 10 yards of velvet, at 4 mills for the first yard, 
 20 for the second, 100 for the third, &c.; what did the 
 piece cost? ^^ns. $^9765,624. 
 
 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, &lc..1 
 
 .ans. $42949672,95. 
 
 (Art. 145.) The sinn 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 ovfeiv, there is no variation in the rule. We may, 
 therefore, require the sum of the series, 6, 3, 1, |, f , &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 esthnated, the milhonth still less, and the infinite term 
 nothing; not, as some tell us, "extremely small," or, 
 "too little to be considered," &:c.; hwiab solid el ij nothing. 
 Remember, that it is the number of terms, not the sum 
 of them, that is infinite. 
 
 Let every decreasing series be inverted, and the first 
 term called the last; then the first term will be 0, and the 
 ratio greater, than unity. Then, by Rule 2, work the fol- 
 lowing 
 
 EXAMPLES* 
 
 1. What is the sum of the infinite series, H-j+j+ 
 y^^, (fee? Invert the series: \ is the last term, and 2 the 
 
 1X2—0 
 ratio ; hence, =1, the answer. 
 
 2. What is the sum of the infinite series y\, yf 7, yi?^^, 
 y\X10— 
 
 &c.? =!•» the answer. 
 
GKOMETRICAL PROGRESSION. 205 
 
 3. What is the value of i-, -^j, j},-., &c., to infinity ? 
 
 4. What is the value of I, i, i, f , &c., to infinity ? 
 
 JIns. '2. 
 
 5. What is the value of 1, |, y^., &c., to infinity? 
 
 ^ns. 4. 
 G. What is the value of f, /j, jt-^, &c., to infinity? 
 
 .^ns. f . 
 
 7. What is the value of ,777, Sic, to infinity? This 
 may he expressed by y\, yl^, y^Vo' <^c. ^'^yi^. ^. 
 
 8. What is the sum of ,6666, <fec., to infinity ? 
 
 ./?;2.5. f . 
 
 9. What is the value of ,232323,*&c., infinitely ex- 
 tended? This may be expressed by y^oj toVott* <S^c. 
 
 10. What is the value of ,71333, (fee, to infinity ? Ob- 
 serve, that the geometrical series does not commence, un- 
 til we pass fJ^. The first 3 is j^-~, tlie 2d is y-,f^^: 
 ratio 10. 
 
 Ans. Sum of the series, ^J-^: whole value, yJ--|-_i^ 
 
 (Art. 146.) Geometrical progression is itsedinfmd- 
 ino- (he amount of anmnfies, at compound inf eresf, when 
 remaining a number of years unpaid. The annuity is 
 the first term of a series. The next term is the first term, 
 with one year's interest. This amount now becomes 
 principal; and ihis again, with one year's interest, is the 
 next term, and so on. 
 
 Now to find the amount of any sum for one year, we 
 multiply it by the amount of 1 dollar, for one vear; then, 
 of course, this multiplier is the ratio to the serl^«, and the 
 sum is found by the combination of Rule 1 and Rule 2, 
 of Art. 144. But, to be less general, we give the fol- 
 lowing 
 
 Rule. Paise the ra'io to the power denoted J.y the 
 ninnler of years; multiply that power by the annidty ; 
 fiom Ike product .sidu'ract t/ie annuity, and divide the 
 rcmaiiidcr by the ratio less 1. 
 
 Z 
 
266 ARITHMETIC. 
 
 EXAMPLES. 
 
 1. If an annuity of 125 dollars be forborne 4 years, 
 what will be its amount, at compound interest? 
 
 Ans. $546,81. 
 (1,06)'X 125—125=32,8087 ,06)32,8087(546,81, J]ns. 
 
 2. Find the amount of an annuity of 260 dollars, in 
 arrears for 3 years, at 7 per cent, compound interest? 
 
 Ans. $835,87+ 
 [NoTE. Powers of the ratio can be found in Table 1, 
 Art. 97, as the numbers in that table are nothing but 
 powers of interest ratios ; but it is unnecessary to multi- 
 ply examples.] 
 
 ALLIGATIOx\. 
 
 Medial and alternate: a commingling, or throwing 
 together. Alligation medial, is finding the medium or 
 middle quality, or value, of certain amounts of given 
 things put togther. Alligation alternate is finding the 
 quality or value of several things, corresponding to a giv- 
 en m.edium; hence, in certain respects, one is the reverse 
 of the other. Neither of them are of much practical 
 value. But, as an improvement to the mind, and as a 
 study of numbers, they form a good lesson. 
 
 (Art. 147.) 1st. Alligation medial. From the defini- 
 tion of finding a medium, we recognize the following 
 
 Rule. Find the value of each ingredient^ and divide 
 their sum by the number of ingredients. 
 
 examples. 
 
 1. A grocer mixed 10 pounds of sugar, at 8 cents, with 
 12 pounds at 9 cents, and 16 pounds at 11 cents; what 
 was a pound of the mixture worth ? 
 
 Solution : Whole cost, 364 cents ; divided by 38 lbs., 
 gives 9~ cts. answer. 
 
 2. If 120 bushels of wheat be bought at 80 cents per 
 bushel, and 75 bushels at 86 cents a bushel, what is the 
 average cost per bushel of the whole purchase ? 
 
 Ans. 82-|-cents. 
 
ALLEGATION. 267 
 
 3. An innkeeper mixed 13 gallons of water with 52 
 gallons of brandy, which cost him $1,25 per gallon: 
 what is the value of 1 gallon of the mixture, and what 
 his profit on the sale of the whole at 6^ cents per gill ? 
 
 C$1 a gallon. 
 '^'^•^- ^$65 profit. 
 
 (Art. 148.) Alligation alternate. This requires more 
 explanation ; and, for the purpose of explaining, let us 
 take the following problem: 
 
 A grocer has sugars, at 9 cents and 16 cents per pound ; 
 he wishes to make a mixture worth 1 1 cents ; what por- 
 tions of each sort shall he take ? 
 
 If he takes 1 pound of each, and sells it at 11 cents, 
 it is evident he would lose 5 cents on the 16 cent sugar, 
 and gain only 2 cents on the 9 cent sugar. On that sup- 
 position, he would lose more than he would gain ; and to 
 j equalize the value, he must take more of the 9 cent su- 
 Igar; as much more as 9 is nearer ^o 11, flum 16 is 
 \ nearer to \l. That is, take the reciprocal ditference 
 I from the mean price. Ti.,,^. n S ^1 ^ 
 
 1 ^^''^' ^^ ^16J 2 
 
 [Note. If the pupil had studied Natural Philosophy, 
 I would compare the principle of this process to balanc- 
 ing weights on the long and short arm of a lever: the 
 smaller tlie difference the greater the quantity ; the short- 
 er the arm, the greater the weight, &c.] 
 
 The difference between 9 and 11 is 2, put opposite 16, 
 for the quantity; and the difference between 16 and 11 
 is 5, put opposite 9, for that quantity. The difference 
 between the values and the mean value, are alieniafed. 
 It may therefore be designated alligation alternate. To 
 prove this result correct, we take alligation medial: 5 
 pounds at 9 cents, cost 45 cents, 2 pounds, at 16 cents, 
 ^.ost 32 ; the whole, 7 pounds, cost 77 cents, or 11 cents 
 per pound, as required. 
 
 We may add one or two more qualities of sugar to die 
 same question, and have it read thus : 
 
 1. A grocer has sugars at 9, 10, 14, and 16 cents per 
 
268 ARITHMETIC. 
 
 pound, and wishes to make a mixture, worth 11 cents; 
 how much of each quality shall he take? 
 
 Take a balance between 9 and 16, f 9 — -. 5 
 
 as before; then between 10 and 14. ,. J 10"^ 3 
 The connectinglines show wliat terms "^ 14J I 
 
 are linked together. One term or in- 1^16 — -^ 2 
 
 gradient may be taken to balance se- 
 veral, when there is but one above or below the mean 
 price, and several on the other side. 
 
 To prove that the above result is correct, let us com- 
 pute the whole cost and find the average. 'J'hus, 5 
 pounds at 9 cents=45 cents, 3 at 10=30, 1 at 14 = 14, 
 2 at 16=32. Whole sum, 121 cents, divided by 11 
 pounds, gives 11 cents per pound. 
 
 2. A merchant would mix teas at 70, 85 and 100 cents 
 per pound, and sell the mixture at 75 cents ; what pro- 
 portion of each must he take ? 
 
 Make an alternation between 70 r70"| "^ 25-f 10 
 
 and 100 first; then between 70 75 < 85 J j 5 
 and 85. Thus, the answer is, 25 (_ 100--^ 5 
 
 pounds at 70, will correspond to 5 
 
 pounds at 100, and 10 pounds at 70, will correspound to 
 5 pounds at 85. Hence, we must take 35 pounds at 70, 
 to correspond to 5 of each of the others. Or, we must 
 take the quantities in this proportion, not necessarily 
 this quantity. For instance, 35 is to 5 as 7 to 1. Now 
 if we take 7 pounds at 70, and one pound each at 85 and 
 100, the whole 9 pounds will be worth 75 cents per 
 pound. Therefore, after making a required mixture, we 
 can proportion the different ingredients so that the mix- 
 ture shall contain any required amount of one of them. 
 From the foregoing we draw the following 
 
 Rule. Set the several ingredients in order, one under 
 the. other, and the mean on one side. Connect one less 
 than the mean ivith any one grea'er ; or, if the case re- 
 quires if, one less icith several greater, or vice versa. 
 
 Place the difference of each and the mean rate against 
 the ingredient ?oith wliich it is connected. 
 
 If only 07ie difference stand against any rate, it will 
 be the required quantity of that ingredient ; but if there 
 
POSITION-. 209 
 
 be more than one, iheir sum will be the. quantHy re- 
 quired of that ingredient. 
 
 EXAMPLES. 
 
 1. A merchant lias spioes at 32 cents, 10 cents and 64 
 cents per pound. He wisiies to mix 5 pounds of the rirsl 
 Willi tile others, so Uiat tlie compound may be worth 48 
 cen's. How mucii of each must he use? 
 
 ./Ins. 5 pounds of the second, and 7 pounds 8 ounces 
 of the third. 
 
 Malve a mixture first, without any reg-ard to the 5 lbs.; 
 then if against the first stands 5 pounds, the problem is 
 solved, if not, proportion the whole so as to make the 
 first 5 pounds. 
 
 2. A farmer wishes to mix 14 bii^hels of rye worth 50 
 cents per bushel, with corn at 40 cents, and oats at 30 
 cents per bushel, so that the mixture may be worth 37 
 cents per bushel ; what quantities must he take of each ? 
 
 ^(/ns. 14 bushels of corn and 32 of oats. 
 
 3. A goldsmith has gold of 15, 17, 20 and 22 carats 
 fine, and would melt togetlier of all tliese sorts, so much 
 as to make a mass of 40 ounces 18 carats fine. How 
 much of each sort is required? 
 
 .^ns. 16 ounces of 15, 8 of 17,4 of 20, and 12 of 22. 
 
 4. A man filled a wine hogshead containing 63 gallons 
 with a mixture of wine worth 120 cents per gallon, wi h 
 some worth 160 cents, and water worth nothing. The 
 mixture was worth 130 cents per gallon ; how much of 
 each did he take ? 
 
 JIns. 9/-; gallons of water, 9 /^ gallons of wine at 120 
 cents, 44 "n gallons of wine at 100 cents : or, he took 47 j 
 gallons of M'ine at 120 cents, and 15^ gallons at 160 cents, 
 and no water. 
 
 POSITION. 
 
 (Art. 149.) From time immemorial, formal rules have 
 been given in arithmetic, under Position, to solve a cer- 
 tain class of rather complex proportional questions, more 
 
270 AR1TH3IETI 
 
 properly belonging to algebra. Such rules were limited 
 in their application, and could cover no questions involv- 
 in<r powers and roots, as powers and roots are not pro- 
 portional to each other. 
 
 For example, 16 and 64 are square numbers, and their 
 roots are 4 and 8, or as 1 to 2; but the numbers them- 
 selves, 16 and 64, are to each other as 1 to 4, a different 
 relation. 
 
 1 . 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, make 220 dollars ; how much was in the 
 purse ? 
 
 It is evident, that this question must be excluded from 
 proportional operation ; for, unless Ave first suppose the 
 right number, the result of the supposition will not be to 
 the given result as the supposed number to the true num- 
 ber ; and when this proportion fails, supposition, that is, 
 position fails. 
 
 We prefer to leave the following questions without 
 rules other than general analysis and proportion. The 
 explanations following some one or two will be sufficient. 
 
 EXAMPLES. 
 
 1. One-half, one-third and one-fourth of a certain num- 
 ber, added together, make 130 ; what is the number ? 
 
 Jins. 120. 
 We take the position, that the required number is a 
 whole. Then, 
 
 L-{-|-|-i. = ||. Then, by proportion, VI : 130 :: i| : 
 Jlns. 
 
 2. A post is \ in the earth, ^ in the water, and 13 feet 
 above the water ; what is the length of the post ? 
 
 Ans. 35 feet. 
 Add ^ and ^ ; not \ and I of the number 1, but i and 
 ■| of the lohole post. These parts, added together, make 
 II ; the remaining ^1 m^^st be 13 feet. Then, by pro- 
 portion. 
 
 Or, 13 : 13 : : 35 : 35 the answer. 
 
POSITION. 271 
 
 3. A and B have the same income. A contracts an 
 annual debt amounting to i of it ; B lives upon i of it ; 
 at the end of ten 3'ears, B lends to A money enough to 
 pay off his debts, and has 160 dollars to spare; what is 
 the income of each ? Jlns. $280. 
 
 If B lives on j, he saves t ; out of this he pays A's 
 debts, i. Hence, from \ subtract i-, and there remains 
 /j. This, in 10 years, is worth 160 dollars; therefore, 
 in 1 year it is worth 16 dollars. Now, by proportion, 
 
 -"■J : 16 :: |i : the answer ; 
 Or, . . 2 : 16 : : 35 : the answer ; 
 Or, . . 1 : 8 : : 35 : 280, the answer. 
 
 4. Of the trees in an orchard, J are apple-trees, -^-^ 
 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. 
 
 5. A gendeman 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 ? 
 
 This problem is generally given under position ; but it 
 is really one in simple division. Divide the money into 
 shares : it will take one share to purchase the harness, 2 
 shares to purchase the horse, &c. 
 
 6. A person being asked what time it was, answered, 
 that the time past noon was \ of the time to midnight; 
 what was it ? Jlns. 2 o'clock. 
 
 Solution, 1^ : 12 : : \ : Ans. 
 
 7. There is a fish whose head weighs 9 pounds, the 
 tad weighs as much as his head and half of his body, and 
 the body weighs as much as his head and tail ; what is 
 the whole weight of the fish ? Ans. 72 pounds. 
 
 8. A man, after spending \ of his money and \ of the 
 remainder, had only 10 dollars left; how much had he 
 at first ? Ans. $30. 
 
272 ARITHMETIC. 
 
 PERMUTATION. 
 
 (Art. 150.) The only changes tliat can be made with 
 two things, as two lelters, «, /;, are two, ab and ba. If we 
 add anotlier thing or letter, as f, we can place it on the 
 right, on the left, and in the middle of ab, making 3 per- 
 mutations witli each of the above. Hence, with 2 things, 
 the numerical expression will be 1X2=2, for 3 things, 
 1X2X3 = 6. 
 
 If we take another letter, d, we can place it on the 
 right, on the left, and between each of the letters, abc, 
 making 4 new permutations, with each of the six ])erniu- 
 tations already made, or 24 in all. Hence, for four things, 
 the expression will be 1X2X3X4=24. For any num- 
 ber of things, then, we have the following 
 
 RuLK. Take, the, continued product of the natural 
 numbers, 1, 2, 3, 4, &c., 2(p to the given number of 
 things, for the number of pennicfatims [or changes) 
 that can be made ivilh that number. 
 
 EXAMPLES. 
 
 1. How many changes can be rung on 5 bells ? 
 
 Ans. 120. 
 
 2. Seven gentlemen agreed to remain together as long 
 as they could arrange lliemselves (hlTerently each day at 
 dmner ; how many days did they remain ? 
 
 Ans. 5040. 
 
 3. How many variations will ten letters of the alpha- 
 bet admit ? ' Ans. 3556800. 
 
 (Art. 151.) How many permulations can be made 
 with the five letters, a, b, c, d, e, taking two at a time ? 
 
 Take a, and place it by itself. To this, each of the 
 other 4 letters can be joined, making 4 permutations. 
 'J'iie same can be done with each of the 5 letters, making 
 5X4=20 permutations of 2 letters. 
 
 If we set apart each arrangement of 2 letters, found as 
 above, and connect each one with the 3 remaining letters. 
 
COiMniNATlON. 273 
 
 we must have 20 X .'^, or 5X4X3 = 60 permutations of 
 3 letters. Hence this 
 
 Ri'LK. The permuUi'ions ivhlch can he male of any 
 ymnil er cf ihim^s, taken a glve)i number ut a timc^ 
 are equal to the conlinued product of a decreasing na- 
 tural series, whose greatest term is the ivliole number 
 of tilings, and ivhone number of terms, the number to 
 be taken at a time. 
 
 4. How many numbers can be expressed by the nine 
 digits, taken four at a time ? Ans. 3024. 
 
 5. How many words of five letters each, may be made 
 from an alphabet of 20 letters, supposing that any num- 
 ber of consonants may make a word ? Ans. 7893600. 
 
 COMBINATION. 
 
 (Art. 152.) Combination — is finding how many dif- 
 ferent ways a less number of things can be taken out of 
 a greater number of the same kind. 
 
 How many combinations of two letters can be made 
 from tlie letters a, b, c. d? 
 
 Let the pupil observe that ab and ha, though different 
 permutations, are the same ccmbination, and the differ- 
 ent combinations are ab, ac, ad, be, cd, bd = Q. 
 
 No two are in any sense alike, and therefore they are 
 in fact different ccmbinadons. 
 
 Now tlie number of permutations of two, from 4 
 thinjjp, we have seen, (Art. 151,) is 4X3; and the num- 
 ber of permutations of tico is 1 X 2. The first divided by 
 the second, gives the number of comlinations ; and 
 thus, generally, to find the combinations of any given 
 number, taken out of a given number, we have the fol- 
 lowing 
 
 Rule. Find the number of permutations which can 
 be made from the proposed set, faking the given num- 
 ber of tilings at a time, and divide it by the number of 
 permufations which cctn be made on another set, con- 
 
274 ARITHMETIC. 
 
 sisting only of as many things as are to be taken at 
 a time. 
 
 1. How many combinations of two letters can be made 
 from 24 ? Jlns. 276. 
 
 2. How many combinations of 3 things can be select- 
 ed from 20? Ans. 1140. 
 
 3. How many syllables of 3 letters, can be found out 
 of 18 letters, no one of which occurs twice in the same 
 syllable ; and no two syllables containing all the same 
 letters ? Ans. 816. 
 
 MISCELLANEOUS EXAMPLES. 
 
 1. The divisor of a certain number is I, the quotient 
 is 120 ; what is the dividend ? 
 
 2. The dividend is ,42, the quotient 42 ; what is the 
 divisor 1 
 
 3. How many men must be employed to accomplish 
 a piece of work in 15 days, which would require 5 men 
 21 days? 
 
 4. My horse and saddle are worth $96 ; my horse is 
 worth 7 times as much as my saddle ; what is the value 
 of each (by simple division) ? 
 
 5. A can do a piece of work in 5 daj'S ; B can do the 
 same in 20 days : how long will they be about it if they 
 both work toijether ? .^ns. 4 days. 
 
 6. A and B can do a piece of work in 8 days ; A can 
 do it alone in 12 days ; in what time can B do it alone ? 
 
 Ans. 24 days. 
 
 7. A merchant bought 63 gallons of rum for $28,35 ; 
 how much Avater must be added to reduce the first cost 
 to 35 cents per gallon ? .^ns. 18 gallons. 
 
 8. $1600 was put at interest at 6 per cent, per an- 
 num, until it amounted to $2000 ; what was the time ? 
 
 Ans. 4 years 2 months. 
 
MISCELLANEOUS EXAMPLES. 275 
 
 9. 1 of a -certain number exceeds 4 of itself by 10 ; 
 what is the number ? ^ns. 560. 
 
 10. Divide a prize of $10200 among 60 men, 6 sub- 
 altern ofllcers, 3 lieutenants, and a commander ; giving to 
 each subaltern double the share of a man, each lieutenant 
 3 times as much as the subaltern, and to the commander 
 double that of a lieutenant ; how much is eacli one's 
 share? ^ C Captain $1200. 
 
 *^'^*- ^Each man $100. 
 
 The three following problems redeem the pledge given 
 in Art. 50. 
 
 11. Suppose 63 gallons jfill a hogshead, 42 a tierce, 
 and 36 a l3arre] ; what is the least quantity of liquid that 
 can be shipped in hogsheads, or tierces, or barrels, just 
 filling the vessels, without defect or redundancy ? 
 
 Ans. 252 gallons. 
 
 12. Three men start at the same time, for the same 
 point, to walk round a circle of 20 miles in circuit, tlie 
 first going 2 miles per hour, the second 4, and the third 
 6 ; how long before they come together again, and where, 
 from the starting point, will it be '] 
 
 Ans. They will come together in 10 hours, at the 
 starting point. 
 
 13. A ferryman has four boats; one will carry 8 bar- 
 els, another 9, another 15, and another 16 ; what is the 
 smallest number of barrels that will make full Ireight for 
 either one of the boats ? Ans. 720. 
 
 14. If a conic pyramid be 27 feet high, and its base 7 
 feet in diameter; how many cubic feet does it contain ? 
 
 Ans. 346 '3. 
 
 15. A farmer sold 17 bushels of barley and 13 bush- 
 els of wheat, for $31,55, the wheat at 35 cents a bushel 
 more than the barley ; what was the price of cacli per 
 bushel ? ■' /? S Barley $,90. 
 
 '^^'^•^ Wheat $1,25. 
 
 16. B's age is \\ the age of A, and C's is 2-^^ the age 
 of both, and the sum of their ages is 93 ; what is the age 
 of each ? [Fellowship.] Ans. A 12 years. 
 
 17. A merchant bought a pipe of wine for 80 dollars, 
 from which 16 gallons leaked out. What will he gain 
 
276 ARITHMETIC. 
 
 by selling the remainder at 12^ cents per pint? and how 
 much will he gain? ^Ins. J$30. 
 
 18. My horse and saddle are, together, worth 80 dol- 
 lars ; my horse is worth 7 times as mucii as my saddle : 
 what is tiie value of each? ,fins. Saddle, §<10. 
 
 The following question is said to be by Sir Isaac New- 
 ton. 
 
 19. If 12 oxen eat up 3^ acres of grass in 4 weeks, 
 and 21 oxen eat up 10 acres in 9 weeks, how many ox- 
 en will eat up 24 acres in 18 weeks, the grass being at 
 first equal on every acre, and growing equally ? 
 
 Jins. 36 oxen. 
 'I'his problem requires three statements in compound 
 proportion. Thus, 
 
 Cause. Effect. Cause. Effect. 
 Oxen, . . 12 : '3L : : 21 : [] 
 Weeks, • • 4 : 9 
 
 The result of this proportion gives 13^ acres for 21 
 oxen 9 weeks. Bui the question gives only 10 acres ; 
 therefore, ten acres become 13^ by having 5 weeks to 
 grow, that is, 5 weeks longer than the first quantity. 
 Cause. Effect. Cause. Effect. 
 Acres,. . • 10 : ^ : : 24 : [] 
 Weeks, . . 5 14 
 
 The result of this proportion gives 21 acres of growth, 
 to be added to the 24 acres. — Total, 45 acres. Lastly, 
 Cause. Effect. Cause. Effect, 
 Oxen,. . . 12 : 3^^ : : [] : 45 
 Weeks, • . 4 18 
 
 20. Suppose a meteor so high in the heavens, as to be 
 visible, at the same moment, at Boston, longitude 71° 3' 
 W.; at Washington, longitude 77° 43' W.; and at the 
 Sandwich Islands, longitude 155° W.: and suppose the 
 time of appearance at Washington to be 5 minutes past 
 10, P. M. What time is it by the clocks at the oilier 
 places? n SAt Boston, 10 h. 31 m. 40 s., P.M. 
 
 •^"*- I At S. Islands, 4 h. 55 m. 52 s., P. M. 
 
 21. What time is it, when the past interval from noon 
 is y'y of the time, onward to midnight? 
 
 Ans. 5h. 24m. 
 
MISCELLANEOUS EXAMPLES. 277 
 
 22. A gentleman bought several gallons of wine for 94 
 dollars ; and, after using 7 gallons himself, found that oue- 
 fourih of llie remainder was worth 21) dollars; iiow many 
 gallons were there at first ? ^^)is. 47. 
 
 [Solve questions 21 and 22, by proportion.] 
 
 23. Find three numbers in arithmetical progression, 
 such, that the least shall be to the greatest, as 5 to 9, -.md 
 ihe sum of the three, 63. .^ns. 15, 21, 27. 
 
 'Z4. A can produce a certain effect in 3 lioiirs ; B in 4 
 hours, and C in 5 hours ; in what time can the three to- 
 getiier produce the same effect? .^ns. ly^ hours. 
 
 25. A general, disposing his army into a square, found 
 he had 231 men over and above; but, increasing each 
 side with one soldier, he wanted 44 to till up the square; 
 how many men had he] Jlns. 19000 men. 
 
 N. B. Problem 25 is very simple, in fact: yet, from 
 some cause or other, it is very troublesome to students — 
 few of whom attempt an unaided solution. 2314-44 men 
 are required to place round one side, and one end, of a 
 square ; how many men does such a square contain ? is 
 the question. 
 
 26. A parallelogram is 6 rods longer than it is wide, 
 and its diagonal measures 30 rods ; required its area and 
 dimensions. 
 
 .dns. Area, 432 rods, length 24, breadth 18 rods. 
 
 27. Suppose a ball, 9 feet in diameter, cut down to a 
 cube ; what would be the length of a side ? 
 
 ^ns. 5.19+ 
 
 28. How many acres in a square field, the diagonal of 
 which is 20 rods longer than either side ? 
 
 ^ns. 14 acres 2 roods 11^- poles. 
 
 N. B. To solve this, assume any square, and find the 
 difference between its side and diagonal ; then make a pro- 
 portion. 
 
 29. Suppose the hind-wheels to a wagon to be 12 feet, 
 and the fore-wheels 9j feet in circumference; how far 
 must the wagon run, for the fore-wheels to gain 800 rev- 
 oluiicns over the hind-wheels? ^^ns. 6f| miles. 
 
 30. A hare starts 40 yards before a grayhound, and is 
 
 _ 
 
278 ARITHMETIC. 
 
 not perceived by him until she has been up 40 seconds ; 
 she scuds away at the rate of 10 miles an hour ; the dog 
 makes after her at the rate of 18 miles an hour; how 
 long will the hound be in overtaking the hare, and what 
 distance will he run ? 
 
 ^^ns. 60, f^ seconds, and Vv-ill run 530 yards. 
 
 31. A man owing the sum of 6480 dollars to several 
 creditors, orders his clerk to pay the first 40 dollars, and 
 the last 500, in arithmetical progression; required the 
 number of creditors, and the difference between each. 
 
 j9ns. 24 creditors, and difference $20. 
 
 32. Three men. A, B, and C, have a sum of money to 
 divide among themselves ; A is to have 9 dollars, B is to 
 have as much as A and \ of C's, and C is to have as 
 much as A and B both ; what sum is to be divided, and 
 how much is the share of each ? 
 
 ^ C Whole sum, 72 dollars. 
 •^^^*-^A$9, B$27, C$36. 
 
 33. Bodies fall by gravity, about 16 feet the first se- 
 cond of time, and the distance or extent of fall increases 
 as the square of the time in seconds ; how far, then, will 
 a body fall in 7 seconds ? Mns. 49 X 16=784 feet. 
 
 34. A body was observed to fall during 6 seconds ; 
 how far did it fall during the last second ? 
 
 Ans. 176 feet. 
 
 35. A body occupied 8^ seconds in falling through a 
 certain space ; how far did it fall during the last half se- 
 cond ? .^>25. 132 feet. 
 
 36. The ditch of a fortification is 1000 feet long, 9 feet 
 deep, 20 feet broad at bottom, and 22 at top ; how much 
 water will fill the ditch? ^ns. 1158127 gals, nearly. 
 
 37. Seven men bought a grind-stone, of 60 inches 
 diameter, each paying \ part of the expense ; what part 
 of the diameter must each grind down for his share ? 
 
 ■Ans. The 1st, 4,4508, 2d, 4,8400, 3d, 5,3535, 4th, 
 6,0765, 5th, 7,2079, 6th, 9,3935, 7th; 22,6778 inches. 
 
 38. How many rods less will it take to fence in an 
 acre, if it be laid out in the form of a circle, than in the 
 form of a square 1 ^^ns. 5|+rods. 
 
 39. How many pieces of cloth, at 20,8 dollars per 
 
PROMISCUOUS EXAMPLES. 279 
 
 piece, are equal in value to 240 pieces, at 12,6 dollars 
 per piece ? ^^ns. 145,38+pieccs. 
 
 40. If, when tlie price of M'heat is 74,6 cents per bush- 
 el, the penny roll weighs 5,2 ounces, what should it be 
 per bushel when the penny roll weighs 3,5 ounces ? 
 
 Ans. $1,108. 
 
 41. If a globe of stone, 9 inches in diameter, weifrli 
 36 pounds, liow much will another globe of the same kind 
 of stone weigh, whose diameter is 15 inches ? 
 
 Ans. 166,6-4-lbs. 
 
 42. Mix 6 bushels of oats, worth 20 cents per bushel, 
 with 8 bushels of oats, worth 25 cents per bushel, to rye, 
 worth 70 cents per bushel, and wheat worth 80 cents — 
 and sell the mixture at 75 cents per bushel ; how much 
 rye and wheat must there be in the mixture 1 
 
 Ans. Rye, 14 bush.; wheat 160 bush. 
 
 43. Seven men not agreeing with the owner of a board- 
 ing house, about the price of boarding, offer to give 100 
 dollars each, for as long time as they can seat themselves 
 every day differently at dinner; this offer being accepted, 
 how long may they stay ? 
 
 Ans. 5040 days, or 13 years 295 days. 
 
 44. What will be the expense of paving a rectangular 
 yard, whose length is 63 feet, and breadth 45 feet, in 
 M'hich there is laid a foot-path 5 feet 3 inches broad, run- 
 ning the whole length, with broad stones, at 36 cents a 
 yard ; the rest being paved with pebbles, at 30 cents a 
 yard ? Ans. $96,70^. 
 
 45. In exchanging 20^ yards of cloth, of 1^ yards 
 wide, for some of the same quality, of J of a yard wide, 
 what quantity of the latter makes an equal barter ? 
 
 Ans. 34^ yards. 
 
 46. Suppose a large wheel, in mill-work, to contain 72 
 cogs, and a smaller wheel, working in it, to contain 50 
 cogs ; in how many revolutions of the greater wheel, will 
 the lesser one gain iOO revolutions ? Ans. 227 j 
 
 1 1 
 
APPENDIX 
 
 MECHANICAL POWERS. 
 
 The mechanical powers properly belong to Natural 
 Philosophy, and not to Arithmetic. The problems per- 
 taining to them, in a numerical point of view, are usually 
 very trifling ; but the principles on which the computa- 
 tions are based, require exact thought, and should not be 
 passed over in a careless manner. We make these ob- 
 servations to guard the pupil against imbibing the error 
 of measuring the importance of a thing by the difficulty 
 or ease of numerical computations. 
 
 Properly speaking, there are but two fundameotal me- 
 chanical powers, viz.: the lever and inclined plane. 
 
 From the lever, we derive the pulley, and wheel and 
 axle. 
 
 From the inclined plane, we derive the screw and the 
 ivedo^e. 
 
 Theoretically, the lever is an imponderable and inflex- 
 ible bar, supported on a fixed point, called the fulcrum. 
 Let the line AB represent it, 
 resting on the point C, call- C 
 
 ed the fulcrum, and weights A j B 
 
 that balance each other at A 
 and B. 
 
 'I'he fiffure below represents the application of the 
 power of the lever as used in jjrying. 
 
 28Q 
 
ARITHMETIC. 281 
 
 In natural philosophy the term, momentum, means 
 quantity of force; and two bodies of unequal weight can 
 have the same momentum if their degrees of motion are 
 reciprocal to their weight. 
 
 A body in motion, will require a certain resistance to 
 stop or overcome its motion : if its motion were twice as 
 rapid, the resistance must be twice as great to produce 
 the same eflect. 
 
 If two bodies move with the same velocity, and one 
 of them double the weight of the other, it is evident that 
 it would require double the resistance to stop the heavier 
 body, that would be required to stop the other. This 
 amount of resistance is equal to the momentum, and evi- 
 dendy depends on the compound, or product of weight 
 multiplied by velocity. 
 
 When two bodies balance each other over the fulcrum 
 point of a lever, the momentums on each side of the ful- 
 crum must be equal. 
 
 Conceive the two bodies, A and B, (first figure on the 
 preceding page,) to balance on the point C, and give the 
 bar a slight motion round the fixed point C. If A is 
 twice as far from C as B is from C, it is evident that A 
 will move twice the distance that B will, in the same 
 time. If A is three times as far from C as B is from C, 
 then A will have three times the motion that B will 
 have, &LC. <fcc. That is, the relative motion depends on 
 the relative lengths of the arms of the lever. 
 
 But the motion, multiplied by the weight, gives the 
 momentum ; or, which is relatively the same, the weights, 
 multiplied by their distances from the fulcrum, give their 
 relative momenta, which must be equal to form an 
 equilibrium. 
 
 Therefore, when two bodies balance over a fulcrum. 
 The weight of one body, 7nidtiplied by its distance from 
 the fulcrum, is equal to the iveight of the other body 
 multiplied by its distance from the fulcrum. 
 
 Here are four things forming two equal products, 
 namely, the two bodies and the two arms of the lever, 
 and when one of them is sought, the other three being 
 given, we can find the sought term as a term in propor- 
 tion. (Art. 75.) 
 
 __ 
 
282 APPENDIX. 
 
 1. A lever 16 feet long-, the fulcrum 3 feet from one 
 end, is in a stale of equilibrium, liokliuff a weight of 390 
 pounds ; what is the power? Ana. 90 pounds. 
 
 2. 'J'he arms of a lever are, the one 20, the other 3 
 feet ; what weight will a power of 69 pounds balance ? 
 
 ^dns. 400 pounds. 
 
 3. One arm of a lever is IH feet, the other 6 inches, 
 what weight will balance a power of 23 pounds ? 
 
 Ann. 529 pounds. 
 
 4. Given a weight of 420 pounds, a power of 40, and 
 a lever 18 feet long; where shall we place the fulcrum, 
 that the weiglit and power may balance? 
 
 Ans. 1^1 feet from one end. 
 
 N. B. This last problem is the same as the follow- 
 ing : 
 
 5. Divide 18 into two such parts, that one may be to 
 the other, as 420 is to 40, whicii is a problem in fellow- 
 ship. Indeed, ihe weight caid power du form fellowship 
 to preserve an equilibrium. 
 
 6. A lever is 30 inches long, tlie weight at the two 
 ends are 12 and 20 respectively ; what are the lengths of 
 the arms when the lever is in equilibrium ? 
 
 An.s. 18^ and \l\ inches. 
 
 7. The difference in the lengths of the arms of a lever 
 IS 6 inches, the weight on one arm is 20 pounds, and the 
 weight on the other is 30; the lever is in equilibrium ; 
 what is its length ? Ans. 30 inches. 
 
 Weight may be applied at any point on the lever, and 
 the fulcrum at the end, as in the above figure ; this is 
 called a lever of the second kind. 
 
 'J'he weight, in this case, rests on botli ends of the le- 
 ver in reciprocal proportion to the distances on each side 
 of it. If twice as near one end as the other, double the 
 weight will rest on that end, &c. 
 
ARITHMETIC. 283 
 
 8. Example. A lever 12 feet loii'^, one end rcsliiitr 
 firmly on a point, a man lifiinjr with a force of 75 pounds 
 at tile other; what weight would he be able to raise 2 feet 
 from the stationary end ? ^ins. 450 pounds. 
 
 To analyze this we observe, 75 pounds must rest on 
 the man's hand, or on that end of the lever, and a certain 
 weight on the other end. But this certain weight must 
 bear the same relation to 75 pounds as 2 feet to 10 feet. 
 
 Or, as 2 : 10 : : 75 : 375. 
 
 That is, 75 pounds on the man's hand, and 375 pounds 
 on the fixed point ; therefore the whole weight is 450 
 pounds. 
 
 9. A man capable of lifting 375 pounds, takes hold at 
 the end of a lever 24 feet long ; what weight can he raise 
 6 inches from the other end as a fixed point ? 
 
 Ans. 18000 pounds. 
 
 Proposition. When several weights are suspended on 
 one side of a lever, at different distances from the ful- 
 crum. To find the force on that side, multiply each 
 weight by its distance from the fulcrum, and add the sev- 
 eral products together. We can then find what weight 
 will balance this on the other side of the fulcrum by di- 
 viding by that arm of the lever. 
 
 10. The two arms of a lever are 12 and 8 inches, re- 
 spectively; 2 inches from the fulcrum on the short arm is 
 suspended 20 pounds, at 6 inches is suspended 30 pounds, 
 and at 8 inches is suspended 40 pounds ; what weight on 
 the extremity of the other arm will balance these? 
 
 Ans. 45 pounds. 
 
 11. A man weighing 150 pounds rests on the short 
 arm of a lever H feet from the fulcrum, on the other side 
 20 pounds is suspended 5 feet from the fulcrum ; there is 
 another weight of 15 pounds; where must it be placed 
 on the lever to make an equilibrium? 
 
 Ans. S~ feet tVom the fulcrum. 
 
284 
 
 APPENDIX. 
 
 The wheel and axle is but a 
 continuation of the lever, or a se- 
 ries of levers, filiino- up the whole 
 circle round the fulcrum. The 
 point or line on which the wheel 
 turns is the fulcrum of the lever;, 
 the semidiameter of the wheel is 
 the long arm of the lever, the sem- 
 idiameter of the axle is the short 
 arm ; but circumferences are to 
 one another as their respective 
 semidiameters; therefore we can 
 compute for equilibrium between 
 weight and power, on the wheel and axle, the same as 
 we do on the long and short arms of a lever, considering 
 the circumference or diameter of the wheel the long arm, 
 and the circumference or diameter of the axle (according 
 as which may be given), the short arm of a lever. 
 
 Example 1. If the diameter of the axle be 6 inches, 
 and that of the wheel 6 feet, what weight, attached to the 
 axle, will balance 16 pounds attached to the M'heel ? 
 
 Ans. 192 pounds. 
 
 Solution: As i^ : 6 : : 16 : Answer. 
 
 2. A large stone, supposed to weigh about 2 tons, or 
 2000 pounds, is required to be raised by a wheel and 
 axle, the axle 2 feet 6 inches in circumference, and the 
 circumference of the wheel 18 feet; what power must be 
 applied to the wheel? Ans. 211} pounds. 
 
 N. B. The answers to this question is on the suppo- 
 sition that the machine is perfect, and experiences no fric- 
 tion ; but such can never be the case in practice : the 
 power must always bear a greater proportion to the weight, 
 some say ~ more. 
 
 3. The diameter of a wheel is 12 feet, to which GO 
 pounds are applied, and balances 3600 on the axle; what 
 is the diameter of the axle ? Ans. 2,4 inches. 
 
 4. If the diameter of an axle be 2 feet, and the di- 
 ameter of the wheel 20 feet, what power will balance a 
 weight of 4000 pounds. Ans. 400 pounds. 
 
ARITHMETIC. 
 
 285 
 
 I 
 
 The pulley is a single wheel, mova- 
 ble on its center, with a cord running 
 over it. A single pulley may be con- 
 sidered a perpetual lever, with the ful- 
 crum in the center, or the two arms 
 equal ; and, of course, the weight and 
 power must be equal, to produce an 
 equilibrio. Hen(;e, no power is gained 
 by a rope running over a single wheel. 
 
 Wlien a rope runs round a series of 
 wheels, as in the adjoining figure, call- 
 ed a system of pulleys, it is manifest 
 that the weight should be equally divi- 
 ded among the ropes. As the whole 
 machine is supposed to be free, the 
 wheels can turn, and the tension of the 
 rope can adjust itself; therefore, there 
 is no reason why one portion of the 
 perpendicular line of rope should bear 
 more of the weight than another. 
 
 Hence, to find the power, Divide the 
 weight by the number of lines of rope 
 sustaining it. Inversely, multiply the 
 power by the number of ropes to obtain 
 the weight, 
 
 N. B. Care must be taken not to count the line of rope 
 supporting the power P. 
 
 Example 1. A weight of 2240 pounds is sustained by 
 4 ropes, or which is the same thing, by one rope passing 
 round 4 pulleys ; what power is required to balance it ? 
 
 Ans. 560 pounds. 
 
 2. What weight will a power of 15 pounds balance by 
 a rope running round 6 pulleys ? Ans. 90 pounds. 
 
 3. By the aid of a system of 12 pulleys, how many 
 pounds will a man sustain who is capable of applying a 
 power of 150 pounds? Ans. 1800 lbs. 
 
 4. How many pounds would be required by the aid of 
 4 pulleys, to sustain 800 pounds ? Ans. 200 lbs. 
 
 5. By the aid of 10 pulleys, how many pounds would 
 be required to balance 2000 pounds ? Ans. 200 lbs. 
 
 W 
 
286 APPENDIX. 
 
 The Inclined Plane. 
 A complete explanation of the power or advantage of 
 an inclined plane would lead us too far; it is a })roblem 
 of the composition of forces in mechanics. AV^e can only 
 assert, that, if a j)lane, as in the ad- 
 joining figure, is twice as long as it 
 is high, a man can roll up on its 
 surface twice as much as he can 
 lift. If the plane be 3 times as long 
 us it is high, we can roll up, alontr its surface, three times 
 as much as we can lift. If 4 times, 4 times as much, 
 &.C.; hence, to find the power of a plane, 
 Divide its lenglh by its height, 
 
 [Note. This is on the supposition, however, that bo- 
 dies roll, or slide, without friction : which is far from be- 
 ing true. Nevertheless, tliese primary theoretical con- 
 siderations are essential and important.] 
 
 The amount of friction depends on a variety of cir- 
 cumstances, such as the smoothness and hardness of the 
 plane, the shape and smoothness of the moving body, 
 or the structure of the carriage, or machine on which the 
 weight is carried. Every elevation or depression along 
 the common highway, may be considered an inclined 
 plane ; and, in the construction of rail-roads, great atten- 
 tion is paid to the degree of elevation of the inclined 
 planes. 
 
 EXAMPLES. 
 
 1. An inclined plane is 40 feet long, and 5 feet high ; 
 Avhat is the advantage of the plane? .^ns. 8. 
 
 That is, any power acting along the surface of the 
 plane, will move 8 times its weight. Conceive a cord at- 
 tached to a body ascending a plane, and that cord to run 
 over a pulley at the vertex of the plane, and to it any 
 weight attached; this weight is called a power. Suppose 
 this power to be 20 pounds. Now, 20X8 = 160 pounds, 
 that may be drawn up the plane. 
 
 2. A plane is 45 feet long, and 3 feet high; w-hat is the 
 power of the plane ? ^8ns. 15. 
 
 That is, any power will draw 15 times its weVhtj or 
 
APHKNDIX. 
 
 'Zbl 
 
 j\, or Yj of the load and wagon would be a dead weight 
 against the draft. 
 
 3. At a certain place on a rail-road, the elevation is at 
 the rate of 30 feet to the mile; what pcu'iion of the trains 
 make dead weight in ascending the j.lane ? 
 
 c6'?iv. 5^i-o = TTTr» "early. 
 N. B. When the power acts horizontally, not along 
 the surface of the p!ane, but along its hase ; tlun the 
 power of the plane is found by dividing the base by the 
 height. 
 
 4. Suppose, on a rail-road, there is an inclined plane 
 300 rods in length, and rising lo the perpendicular height 
 of 50 leet; what jiower will be required lo sustain a 
 weight of 84000 pounds ? dns. 14000 lbs. 
 
 The 2ved2;e is formed by two equal in- \ 
 dined planes put together. Its power is 
 governed by the same principles as those 
 of the inclined jilane. The power applied 
 to the head or end, is to the power on the 
 side, as the wid'h of the wedge to its length. 
 But we cannot numerically measure the 
 force of a blow ; therefore, there is no ob- 
 ject in giving problems under this head. 
 
 The screw may he considered an inclined plane wound 
 round a shaft; one turn gives us the length of the plane, 
 and the perpendicular distance between the threads give 
 us the height of the plane. Therefore, to find the 
 power of a screw, divide the circumference by the dis- 
 tance between the threads. But, to move a screw, a le- 
 ver-power is applied ; and on the principles of momentinn, 
 the length of the turn of the screw need not be consid- 
 ered. By one turn of the lever, the screw is pushed 
 the distance between two consecutive threads, and the 
 power is moved through the circumference described by 
 the lever. But, on the principle of momentum, the 
 power, midliplied by the circimiference of its motion, is 
 equal to the weight multiplied by the distance between 
 the threads. 
 
 Eocample 1. If the threads of a screw be 2 inches 
 
288 APPENDIX. 
 
 apart, the lever 40 inches in length, and a power of 60 
 pounds be applied to the end of tlie lever, what force is 
 exerted by the screw? Ans. 7543 lbs., nearly. 
 
 2. Suppose the length of the lever to be 40 inches, the 
 distance of the threads 1 inch, and the weight to be rais- 
 ed, 8000 pounds ; required the power. Ans. 33^ lbs. 
 
 3. If the distance between the threads of a screw be 
 1 inch, and the lever 4 feet long, what weight could be 
 sustained by a power of 350 pounds, acting on the ex- 
 tremity of the lever? Ans. 105556 lbs., nearly. 
 
 Let the pupil remember that these results suppose no 
 friction ; but much of the power of the screw is destroy- 
 ed by friction, from | to j of the whole. 
 
 The mechanical powers may be variously combined, 
 and their combinations form our many curious and useful 
 machines. 
 
 The force applied to machines is directly, or indirectly, 
 referred to gravity, or pounds weight, drawn up in a per- 
 pendicular direction. The agents are various, such as 
 wind, water, animal and steam power. The power of a 
 man, to be for any considerable time pulling on a rope to 
 raise a weight, is only equal to about 14 pounds ; though 
 for a single exertion, to cease in a moment, it is much 
 greater. The powers of a horse, and to move at the rate 
 of 3 miles per hour, is only 125 of dead weight, much 
 less than generally supposed ; and this is the average 
 standard of horsepower. Steam power is computed, or 
 rather, compared, with horse power ; but many of these 
 forces are very variable and indefinite, being so modified 
 by circumstances, and the condition of the machinery. 
 
A 
 PRACTICAL SYSTEM 
 
 OF 
 
 BOOK-KEEPING, 
 
 FOR 
 
 MECHANICS AND RETAILERS, 
 
 Book-Keepihg is tlie method of recording a syste 
 matic account of business transactions. 
 
 It is of two kmds—Sinffle and Double Entry. Tli; 
 former, only, will be noticed in this work. On account 
 of the simplicity of Single Entry, it is, perhaps, the best 
 which can be recommended to farmers, mechanics, and 
 retailers. It consists of two principal books — the Day 
 Booh, or Waste Book, and the Leger, and one auxiliary 
 book, the Cask Book. 
 
 THE DAY BOOK. 
 
 This book is ruled Avith 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 the last two for dollars and cents. 
 
 This book exhibits a minute history of business trans- 
 actions in the order of time in which they occur, Avith 
 every circumstance necessary to render the transaction 
 plain and intelligible. 
 
Cincinnati, 1852. 
 
 Jan. 
 
 11 
 
 11 
 
 14 
 
 14 
 
 15 
 
 David Judkins, Dr. 
 
 To 10 lbs. coffee at 17 cts. $1 70 
 
 '* 25lbs. suo-ar,atlOcts. 2 50 
 
 Timothy ^Y. Coolidge, Dr. 
 To 1 bl. sugar, weighing 
 135lbs. neat, atSActs.^ll 47^- 
 *' 1 bag coffee, 98 lbs. at 
 15 cents, 14 70 
 
 George H. Eaton, Dr- 
 
 To 1 bl. flour, 83 87^ 
 
 " 1 lb. Y.H. Tea, 1 12^ 
 
 *' 1 keg lard, neat weight 
 
 60 lbs,, at 61 cts. 3 75 
 
 David Judkins, 
 By cash on account. 
 
 Cr. 
 
 James Wilson, Dr. 
 
 To 3 weeks boarding, at 2 dol- 
 lars per week. 
 
 Hiram Ames, Dr. 
 
 To 1 2 yards broadcloth at 6 dol- 
 lars "per yard, 87^ 00 
 *' 30 yds. muslin at 14 cts. 4 20 
 ^ Cr. 
 By an order on J. Jones for 
 
 Groceries, 851 00 
 
 *' Cash, 20 00 
 
 Timothy W. Coolidge, Cr. 
 By a bill of carpenter work. 
 
 James Wilson, 
 
 Dr. 
 
 83 25 
 1 keg lOd. nails, weight 
 1 1 1 lbs. at 8 cts. 8 88 
 
 To 1 bl. vinegar 
 
 cts. 
 
 20 
 
 26 
 
 76 
 
 71 
 
 25 
 
 12 
 
 75 
 
 00 
 
 00 
 
 20 
 
 00 
 
 00 
 
 13 
 
Cincinnati, 1G52. 
 
 Jan. 19 
 
 21 
 
 Feb. 
 
 25 
 
 29 
 
 30 
 
 31 
 
 George Hamilton, Cr. 
 
 By 1 set of Fancy chairs, 
 
 George H, Eaton, Cr. 
 
 By cash, to balance account, 
 
 Kobert Young, Dr. 
 
 To cash on account, $3 00 
 
 «' 10 lbs. N. 0. sugar, at 
 
 10 cents, 100 
 
 " 12 lbs. coffee at 16^ cts. 2 00 
 - J-ib.Y.H. Tea, 62^ 
 
 Jackson Moore, Dr- 
 
 To 21 lbs. Ham, at lOcts. 82 10 
 1 box soap, 30 lbs. at 
 5 cents, 1 50 
 
 Cr. 
 
 David Judkins, Di 
 
 To 12 bis. apples at 75 cts. 
 
 By 47 bu. corn, at 20 cents, 
 
 James Wilson, Cr. 
 
 By cash on account. 
 
 10 
 
 Thomas Hilton, Dr. 
 
 To cash, $5 00 
 
 ** 1 lb. Y. H. Tea, 1 06 
 
 *' 16 lbs. Rice, at 6^ cts. 100 
 
 Cr. 
 By 13 days labor, at 87| cts. 
 
 George Hamilton, Dr- 
 
 To 1 canister Imperal Tea, 
 14 lbs. at -^1 87iperlb. 
 
 •,? ds. 
 25 00 
 
 75 
 
 Hiram Ames, Cr. 
 
 1 3 ' By order on E. Disney for goods, 
 
 621 
 
 60 
 
 00 
 
 40 
 
 00 
 
 7 
 
 06 
 
 11 
 
 37i 
 
 26 
 
 25 
 
 10 
 
 00 
 
Cincinnati, 1852. 
 
 Feb. 
 
 1 
 15 
 
 21 
 
 27 
 1 6 
 
 6 
 
 8 
 8 
 
 16 
 
 Robert Young, Dr. 
 To 1 box sperm candles, 
 
 25 lbs, at 30 cts. perlb. ^7 50 
 '* 2 bu. dried fruit, at ^125 
 per bushel, 2 50 
 
 (< 
 
 James Vv^lson, Dr. 
 To 10 gals, molasses, at 40 
 
 cents, i4 00 
 '* 4 lbs. Old Hyson Tea, 
 
 at 93 cents, 3 72 
 
 a 
 
 Robert Young, Cr. 
 By 12 cords of wood at ^2 25 
 
 Marc! 
 
 Ames & Smith, Dr- 
 To 18 lbs. sole leather, at 
 
 25 cents, S4 50 
 '* 1 side upper leather, 2 75 
 ** 3calf skms, ^1 25, 3 75 
 
 tt 
 
 Hiram Ames, Dr. 
 To 500 feet white pine boards 
 
 at 812 50 per M. ^6 25 
 ** 25bu. potatoes, at50cts. 12 50 
 -1 ton of Hay, 10 00 
 
 it 
 
 David Judkins, Dr. 
 To 200 lbs. tlour, at 81 75 
 
 tt 
 
 Thomas Hilton, Dr. 
 To cash paid his order to. 
 Wilham Coohdge, 
 
 ti 
 
 George Hamilton, Dr. 
 To 1 copy Whelpley's Com- 
 
 pend, 81 25 
 *' 1 ream letter-paper, 4 50 
 '* 1 doz. Spelling books, 1 00 
 
 10 
 
 27 
 
 11 
 
 28 
 
Cincinnati, 1852. 
 
 Mar. 
 
 20 
 
 'i 
 
 (( 
 
 <c 
 
 23 
 
 Theodore P. Letton, Dr. 
 
 To sliarpenino- his plough,^! t)0 
 
 ** Shoeing his horse, 1 62^- 
 
 " Repairing chain, 25 
 
 Timothy W. Coolidge, Dr 
 To 2 qrs. tuition of hiniself at 
 evening school, at 83 per qr. 
 
 25 
 
 26 
 
 Thomas Hilton, Cr. 
 
 By the hire of his horse 10 ds. 
 at 621- cts. per day, 
 
 Ames & Smith, Cr. 
 
 Byl hhd. sugar, weight 13171b. 
 neat, at 7|- cents. 
 
 T. P. Letton, Dr. 
 
 To 1 ream wrapping paper, 
 
 81 C2^- 
 " 1 beaver hat, 5 00 
 
 " set silver tea-spoons, 6 00 
 
 30 
 
 31 
 
 31 
 
 Henry C. Sanxay, Cr. 
 
 By my order on him in favor of 
 Jno. Torrence for stationary. 
 
 Ames & Smith, Dr- 
 
 To 2000 ft. clear pine boards, at 
 
 820 per M. ^40 00 
 
 '* 500 common do. at $8, 4 00 
 
 *' 5000 shingles, at 82 25 11 25 
 
 " Cashtobalanceacc't, 32 52 
 
 Jackson Moore, 
 By painting my house. 
 
 Cr. 
 
 98 
 
 12 
 
 12 
 
 ds. 
 
 87J 
 
 00 
 
 25 
 77i 
 
 62i 
 
 871. 
 
 87 
 25 
 
 7/ 
 
 00 
 
 George Hamilton, Cr. 
 
 By an order on J. Hulse, 85 00 
 
 cash, 6 50 
 
 12 50 
 
Cincinnati, 1852, 
 
 March 31 
 
 31 
 
 31 
 
 31 
 
 Thomas Hilton, Dr. 
 
 To 4 bii. wheat, at ^1 25, $5 00 
 
 " 1 bl. mess pork, 9 00 
 
 *' 2 bu. salt at 50 cts.' 1 00 
 
 " 8 lbs. brown suo-ar, 1 1 cts. 88 
 
 George Hamilton. Dr. 
 
 To 12cedar posts, at 25c. ^3 00 
 
 '* 1 plough, 9 37| 
 
 '' 1 scythe, 1 62^ 
 
 Jackson Moore, Dr. 
 
 To repairing his wagon and 
 plough, 
 
 Thomas Hilton, Cr. 
 
 By 1 pair shoes, ^1 50 
 
 " 1 mahogany table, 12 50 
 
 15 
 
 14 
 
 14 
 
 cts. 
 
 00 
 
 00 
 
 00 
 
 THE LEGER. 
 
 This book is used to collect the scattered accounts of the Day 
 Book, and to arrange all that relates to each individual, into one 
 separate statement. The business 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 separate the debts from the credits. Each side has two col- 
 umns for dollars and cents, one for the page of the Day Book from 
 which the particular item is 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 whicli the ac- 
 count was posted, is written in the 2d column of the Leger. 
 
 In posting, begin with the first account in the Day Book, which 
 you will perceive is the name of David Judkins. Enter his name 
 in the first page of the Leger, in a large, fair hand, with Dr. on 
 the left and Cr. on tlie right.— As there are several articles charg- 
 ed to D. Judkins on the 4th of January, instead of specifying 
 each article in the Leger, we merely say. For Sundries, and en- 
 ter the amount in the proper columns — see Leger, page 1. 
 
 The Leger has an index or alphabet, in wliich the names 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 — 
 Ames & Smith . . 
 
 .2 
 .'.4 
 
 I & J 
 Judkins, David. 
 
 .1 
 
 R 
 
 B 
 Balance 
 
 .5 
 
 K 
 
 S 
 Sanxay, H. C. 
 
 ....5 
 
 C 
 Coolidge, T. "W.. 
 
 .2 
 
 Lctton 
 
 L 
 T. P.... 
 
 .5 
 
 T 
 
 
 D 
 
 Moore, 
 
 M 
 
 Jackson. 
 
 .4 
 
 U 
 
 E 
 
 Eaton, George H. 
 
 ..2 
 
 N 
 
 y 
 
 F 
 
 
 
 
 
 Wilson, Jas. , . 
 
 ....3 
 
 G 
 
 P 
 
 X 
 
 H 
 Hilton, Thomas,. 
 Hamilton, Geo.... 
 
 .3 
 .3 
 
 Q 
 
 Y 
 
 Young, Robt.. 
 
 ...4 
 
 Z 
 
 Dr. 
 
 David Judkins, 
 
 Cr. 
 
 1852 
 Jan. 4 
 *' 30 
 Mar. 7 
 
 Ap. 1 
 
 To sundries 
 '' Apples 
 *' Flour 
 
 To balance 
 of account 
 brot. down, 
 
 4;20 
 900 
 3,50 
 
 16 70 
 
 4 30 
 
 Jan. 8 
 
 " 30 
 
 By cash 
 '' corn 
 " Bal. 
 
 00 
 40 
 30 
 
 70 
 
 IfoTE. — 'The Dr. on the left hand side of the page signifies 
 debtor, and that the sums entered on that side of the page, are 
 those for articles sold to others, and for which theij owe you. The 
 Cr. on the right hand side signifies credit, or that tlie sums en- 
 tered on that side of the page are for articles received of the per- 
 son under vrhose account they stand, and forwhicli you otcehim. 
 
Dr. 
 
 Timothy W. Coolidge, 
 
 1852 
 Jan. 4 
 Ma.20 
 
 To sundries 
 '' Tuition 
 
 Apl. 1 To Balance 
 
 2,26 
 5 6 
 
 32 
 
 7 
 
 in 
 00 
 
 111 
 
 1 Ja. 14 
 
 Apl.4 
 
 By work 
 '* Balance 
 
 2 25 00 
 7 171 
 
 32 
 
 17] 
 
 Note. — The Dr. side of this account shows the amount of ar- 
 ticles Mr. Coolidge has received of me, and the Cr. side shows 
 what I have received of him. It appears that the total amount 
 of my account against him is $32 17)^, from which deduct the 
 $25 00 (which stands to his credit on the right of the account) 
 and there will remain a balance of $7 17}^ due me. 
 
 George H. Eaton. 
 
 Jan. 6 To sundries 2 8 75 Jan21 By cash 3 
 
 75 
 
 JS'oTE.— This account presents equal suins on both sides; hence 
 it is evident that I owe G. H. Eaton nothing, and that he owes 
 me nothing. The account is fully closed. 
 
 Hiram Ames. 
 
 Janll 
 Mar.6 
 
 Api. 1 
 
 To sund. 
 *' corn 
 
 To Bal. 
 
 2 
 
 76! 
 
 20 
 
 Jan. 11 
 
 4 
 
 28j 
 
 75 
 
 Feb. 13 
 
 
 1 
 104 
 
 95 
 
 Apl. 1 
 
 
 — - 
 
 = 
 
 
 
 23 
 
 1 
 
 1 
 
 95 
 
 
 By sund.|2( 71100 
 
 Order3 
 Bal. 
 
 lOjOO 
 2395 
 
 104195 
 
Dr. Thomas Hilton, Or. 3 
 
 1852 
 
 1 , 
 
 1 
 
 1 ' 1 
 
 
 
 Feb. 3 
 
 To simd. 3 7 
 
 06 i 
 
 Feb. 3 By labor, 
 
 3 
 
 11 
 
 37^ 
 
 Mar. 8 
 
 '* Cash. 
 
 4 18 
 
 75 
 
 Ma 23 
 
 "■ hire of 
 
 
 
 
 '* 31 
 
 *' do. 
 
 6 15 
 
 88 
 
 
 horse, 
 
 5 
 
 6 
 
 25 
 
 
 
 
 
 " 31 
 
 " Sund. 
 
 6 
 
 14 
 
 00 
 
 
 
 
 
 Apl.l 
 
 - Bal. 
 
 1 
 
 10 
 
 06 1- 
 
 
 
 41 
 
 69 
 
 
 
 
 41 
 
 69 
 
 Apl.l 
 
 To Bat. 
 
 |io 
 
 061 
 
 
 
 
 
 
 Jca7ies Wilson. 
 
 Jan 11 Toboard- 
 
 
 
 
 Jan31 
 
 By cash 
 
 3 
 
 |15 
 
 00 
 
 
 ing, 
 
 2 
 
 6 
 
 00 
 
 Apl.l 
 
 *' Bal. 
 
 
 'lO 
 
 85 
 
 " 15 
 
 *' simds. 
 
 2 
 
 12 
 
 13 
 
 
 
 
 
 
 Feb21 
 
 '' do. 
 
 4 
 
 7 
 25 
 
 72 
 85 
 
 
 
 
 S 
 
 85 
 
 Apl.l 
 
 To bal- 
 ance bro't 
 down, 
 
 
 10 
 
 85 
 
 
 
 
 
 
 ISToTE. — When au account is settled only, and wot fully paid, as 
 
 in the above, and several preceding accounts, the balance, ^vvlieth- 
 
 er it be in your favor or against you, is brought down and placed 
 
 distinctly by itself, and serves for the beginning of a new ac- 
 
 count, as you perceive has been done in the above example, the 
 
 balance being $10 8.5. 
 
 George Hamilton. 
 
 Feb 10 
 
 To Tea, 
 
 3 |26 
 
 25 
 
 Janl9 
 
 Bychairs, 
 
 3 
 
 25 
 
 00 
 
 Ma. 15 
 
 - Sund. 
 
 4 
 
 6 
 
 75 
 
 Ma.31 
 
 " Simds. 
 
 5 
 
 n 
 
 50 
 
 '' 31 
 
 - do. 
 
 G 
 
 14 
 
 47 
 
 00 
 00 
 
 Apl. 1 
 
 Balance, 
 
 
 K 
 47 
 
 50 
 00 
 
 Apl.l 
 
 To Bal. 
 
 
 10 
 
 50 
 
 
 
 
 
 
Br. 
 
 Robert Young, 
 
 Cr. 
 
 1852 
 
 Jan27 
 
 Feb 15 
 
 Apl. 1 
 
 To Sund- 
 ries, 
 " do. 
 '« Bol 
 
 62i 
 00 ' 
 371 
 
 00 
 
 Feb27 
 
 By wood. 
 
 Apl.l 
 
 By Bed. 
 
 00 
 
 00 
 
 37J 
 
 ISToTE. — 111 tlie above account the difference between the Dr. 
 and Or. side is $10 STi^, by whicli I perceive that the balance 
 against vie, in favor of Robert Young, is $10 37i,<. 
 
 Jackson Moore. 
 
 Jan29 
 
 Ma.31 
 
 Apl. 1 
 
 To Sun- 
 dries, 
 " Sunds. 
 - Bed. 
 
 1 
 3 
 6! 
 
 I 
 
 11 
 
 60 
 00 
 
 40 
 
 00 
 
 Ma.31 
 
 
 
 
 
 Apl. 1 
 
 By paint- 
 ino\ 
 
 By bal- 
 ance bro't 
 down, 
 
 2100 
 
 21 
 
 11 
 
 00 
 
 40 
 
 Ames & Smith. 
 
 Mar. 6 
 '' 30 
 
 To Sund. 
 '' do. 
 
 4 11 
 
 87 
 
 98 
 
 00 
 
 77 
 
 77 
 
 Ma. 25 1 By sugar. 
 
 98 77 
 
 9877 
 
 Note — This account, like the one on the second page, is fully 
 closed, the amount on the Dr. side being just equal to that on 
 the Or. 
 
T. F. Letton, 
 
 Cr 
 
 1852 
 
 Ma.20 
 *' 26 
 
 Apl. 1 
 
 I I 
 To sunds. 5 
 
 " do. 5 
 
 To Bed. 
 
 287i 
 
 12621 
 
 is! 50 
 15 50 
 
 Apl.l 
 
 By Bal 
 
 50 
 
 50 
 
 Note. — In the above account there is no sum on the Cr. side, 
 and tlie inference is, that T. P. Letton owes me $15 50 for sun- 
 dry articles expressed in detail, in the Day Book, page 5. 
 
 Henri/ C. Sanxay. 
 
 Apl.l 
 
 To Bal. 
 
 12871 
 
 Ma.28 
 
 By my 
 
 order, 
 
 5 
 
 1 
 
 12 071 
 
 
 
 12 871 
 
 
 
 
 12 
 
 871 
 
 
 
 1 ; 
 
 1 1 
 
 Apl.l 
 
 By Bal. 
 
 . 
 
 12 
 
 ^^ 
 
 Note. — In this account it will be perceived that there is no 
 amount charged to H. C. Sanx»y, on the Dr. side, I owe him 
 $12 87>^. 
 
 J)r. 
 
 Balance. 
 
 Cr. 
 
 Apl.l 
 
 To D. Judkins, 
 T. Yr. Coolidge, 
 H. Ames, 
 T.Hilton, 
 J. Wilson, 
 G. Hamilton, 
 T. P. Letton, 
 
 4 30 
 
 7,17 
 
 •2; 23:95 
 
 1006 
 1085 
 10.50 
 15 50 
 
 82.33 I 
 
 Apl.l 
 
 By R. Young, 
 " J. Moore, 
 " H.Sanxay, 
 
 34 .54 
 
 Note. — This account exhibits the exact state of your books. 
 It is made from the preceding accounts in the Leger. The Dr. 
 side is an exhibit of the amounts due to you by others, and the 
 Cr. side the amounts due hy you to others. It is not strictly ne- 
 cessary that this account should be introduced in the Leger in 
 single entry : it will be found convenient, however, to balance the 
 
.1 
 
 The Cash Boole. 
 
 book at stated intervals, and transfer tlie balances to the new 
 accounts below, as in the preceding Leg-er, 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 PRECEDIXG. 
 
 Mr. David Judkins, 
 1852 To Edward Thomson, Br. 
 
 To 10 lbs. coffee at 17 cts $1 70 
 
 25 lbs. suo-ar at 10 cts 2 50 
 
 Jan. 
 
 4 
 
 (C 
 
 30 
 
 Mar 
 
 8 
 
 Jan 
 
 8 
 
 
 30 
 
 12 bbls. apples at 75 cts. 
 200 lbs. flour at $175,.. 
 
 Cr. 
 By Cash, P 00 
 
 47 bushels corn at 20 cts 9 40 
 
 Errors excepted. Balance due,' 4.30 
 
 Rec'd Payment in full, EDWARD THOMSOJS". — '— 
 
 4 20 
 9 00 
 3 50 
 
 16 
 
 12 
 
 70 
 
 40 
 
 THE CASH BOOK. 
 The Cash Book is used to record the daily receipts and pay- 
 ments of money. It is ruled nearly the same as the Leger ; the 
 Dr. side exhibits the amonntof money received, and the Cr. side, 
 the amount paid mit. Subtraat the sum of the Cr. 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. 
 
 a-. 
 
 1852 
 
 Jan. 1 
 
 " 1 
 
 " 1 
 
 " 1 
 
 To cash on hand. 
 Cash of J. Young, 
 H. Sanxay, 
 ' D.Juclkins, 
 
 I ',1852 i 
 73 81 j 'Jan. 1 By house rent pd. 
 16 40 ' 
 
 25 00! 
 lOiOO 
 
 T. P. Lctton, 
 iPd. note R. Hand, 
 1 Family expenses, 
 1 By cash on hand, 
 
 ! 125 21 
 
 Jan.2iTo cash on hand, 1 
 " 2CashofTCoolidge 
 •' 2 Cash found. 
 
 Jan. 3 To cash on hand. 
 
 52 84 By cash pd. Ames 
 
 2316 Jan. 2 & Smith, 
 
 29 00 ! " 2 By cash on hand, 
 
 105,00; 
 
 "85 001 
 
 18 
 
 00 
 
 
 50 
 
 00 
 
 
 4 
 
 37 
 
 
 52 
 
 34 
 
 
 125 
 
 21 
 
 
 = 
 
 = 
 
 
 ! 20 
 
 00 
 
 
 ! S5 
 
 00 
 
 
 — 
 
 , — 
 
 
 105 
 
 00 
 
 
 ^^ 
 
 ^^^ 
 
 i 
 
p 
 
 VB 35835 
 
JACOB EENSJl 
 
 ir 
 
 iiei irrr;}» mi'eei, ViticimiuU, Ohio, 
 
 }3ubli£ii!cr of Hobinson s i^lathcuialkal Scriee. 
 
 i 
 
 do. I7aii;\;.i ; ;;'losO; 
 
 do. (^eon'eii_ 
 
 do. .vstiYx;. chooi FditJL 
 
 ISi""l:: • ■ of AlAiKU 
 
 W"Ai--;cu ;; r^nooi Dictioriu; 
 
 N'^" "^ n"=! Eieiner.t? ofScicr 
 
 1'he • ^ rraoiiiu. a collectl'^ri oi i ,;:<y o ^ ; . 
 
 School 'ros*.?.mc-uts, ll'ino. 
 
 ALL THU L£ii:. IT. 
 
 Blank Ecoir.CcpyBockr 
 position Bcl'.>-. Sera "^ 
 i Cap, Comr'i- 
 I Drawing i'oi . > .^ ^ 
 i Blanks f ai! i:^.e varlcn- 
 
 ■3 voices, 38 llj 
 
 ■ '' - TE 
 
 .u-iCoiu' :j 
 
 .JJ!