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 IN MEMORIAM 
 FLOR1AN CAJORI 
 
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 TREATISE 
 
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 (1 
 
 ARITHMETIC, 
 
 lit 
 THEORY AND PRACTICE. 
 
 WITH 
 
 AN APPENDIX, 
 
 CONTAINING AN INTRODUCTION TO MENSURATION. 
 
 BY JAMES THOMSON, LL.D. 
 
 PROFESSOR OF MATHEMATICS IN GLASGOW COLLEGBf 
 
 AUTHOR Ol' ELEMENTS OF PLANE AND SPHERICAL TRIGONOMETRY; 
 INTRODUCTION TO MODEKN GEOGRAPHY; ETC. 
 
 Adapted to the Present Syttem of Weights, Measures, and Currency. 
 
 BELFAST: 
 
 SIMMS .AND M'INTYRE, DONEGALL SI REET. 
 1837. 
 
ADVERTISEMENT TO THE SECOND EDITION. 
 
 THE object of the following Treatise, is to present a full and regular course of 
 whatever is useful in Arithmetic. With this view, rules are given for performing 
 all the requisite operations, and are illustrated by many examples; and numerous 
 exercises are prescribed, to afford the pupil that practice which alone can produce 
 expertness and accuracy in the management of numbers. 
 
 The Definitions and Rules, which it may perhaps be proper for the learner to 
 commit to memory, are in the largest type employed in the work; the Examples 
 and Exercises, and the principal illustrations, are in a character somewhat smaller; 
 the less important illustrations and remarks are in a type still smaller, and may 
 perhaps be omitted by the younger pupil ; and the notes contain miscellaneous 
 information, which may often be interesting to the reader. 
 
 The reasons of the rules and operations are explained, not in strict, formal 
 demonstrations, but generally by simple and easy illustrations of particular^ cases 
 and examples; and it is hoped, that the subject will thus be rendered as intelligible 
 and attractive as possible. This part of Arithmetic is too generally neglected, both 
 in treatises on the subject, and in teaching: and thus one of the principal divisions 
 of Mathematical Science is converted into a mere practical art; and what is cal- 
 culated to call forth and improve the reasoning powers of the pupil, is degraded into 
 a dry exercise of memory. 
 
 Of the Examples and Exercises, some are proposed in purely abstract terras, 
 being intended merely to afford practice to the learner in the rules and modes of 
 calculation. To these are subjoined, in those parts of the work in which it conid 
 be conveniently done, other questions, which will not only afford the pupil farther 
 exercise on the rules which precede them, but will also furnish him with many 
 important facts in Commerce, Geography, Astronomy, Chronology, Chemistry, and 
 other departments of knowledge. As the information contained in these questions 
 has been all derived from authentic sources, its correctness may be depended on; 
 and it is hoped, that what is thus presented may excite in the young reader a desire 
 to enrich his mind, by the acquisition of farther information of a similar nature. 
 
 Some subjects are omitted, which in several works on Arithmetic are treated at 
 considerable length. Such is Barter, an application of the Rule of Proportion, 
 which is of scarcely any use as a part of Mercantile Arithmetic. Neither has the 
 method of calculating Annuities at Simple Interest been introduced; as it is unjust 
 in principle, and, in real transactions, all calculations respecting annuities are made 
 according to the principles of Compound Interest. Many applications of the rules to 
 the circumstances of bodies in motion, and to other subjects in Natural Philosophy, 
 which have been given in works on this science, have also been omitted, as they 
 require a degree of knowledge to render them intelligible and useful, which is to be 
 acquired only by the study of a considerable course of mathematical and physical 
 science. 
 
 Instead of these, however, other subjects have been introduced, which are little 
 known, but which cannot fail to be interesting to the more advanced pupil. Of 
 this kind are Continued Fractions, and the Theory of Arithmetical Scales. The 
 principal and most important parts of the work, also, have been explained and 
 exemplified more largely than the rest; and in these parts a greater number of 
 exercises have been left unwrought, for the improvement of the pupil. Thus, the 
 Simple and Compound Rules, Proportion, Practice, Interest, and Exchange, have 
 
been treated at considerable length ; and much care has been taken to render them 
 as simple and easy as possible. Specimens of the most common and useful kinds 
 of merchants' accounts, such as Invoices, Sales, Accounts Current, &c, according 
 to modern and approved forms, are likewise given, and will be found to form useful 
 exercises for pupils who may be intended for the counting-house. 
 
 In preparing the present edition for the press, considerable changes have been 
 made in the manner of expressing and illustrating the rules, and in various other 
 respects. These have partly resulted from the Author's own experience, and have 
 partly been suggested by several teachers of the first respectability in this country, 
 who have kindly favoured him with hints for the improvement of the work. Some 
 additional tables have likewise been introduced, and a tract given on Mensuration, 
 which cannot fail to be useful to the majority of learners. At the desire also of 
 many teachers, the answers are, in this edition, annexed to all the questions, except 
 a few of the simplest, which are of such a kind, that the idle pupil might copy the 
 answers, without working the exercises : the answers to these are given at the end 
 of the book. With respect to the entire work, no exertion has been spared to attain 
 neatness of appearance ; and, above all, to secure that accuracy which is so desira- 
 ble both to the teacher and the pupil, in every work which contains calculations. 
 
 As so much care and labour have been employed in improving this edition, it 
 has been STEREOTYPED; that is, plates have been cast for the entire work, which 
 will in future be employed, instead of moveable types, in printing it. This mode of 
 printing prevents the possibility of errors creeping into future impressions ; and as 
 any mistakes which may escape observation in a first impression, can be corrected 
 in the plates, it affords the means of ultimately rendering the typography of a book 
 completely perfect. 
 
 College, Belfast, Jan. 1, 1825. 
 
 CONTENTS. 
 
 
 Page 
 
 Notation and Numeration 
 
 1 
 
 Simple Addition 
 Subtraction 
 
 6 
 12 
 
 Multiplication 
 
 15 
 
 23 
 
 Abbreviations in the Fundamental 
 
 
 Rules 
 
 30 
 
 Tables of Money, &c 
 
 37 
 
 Reduction 
 
 43 
 
 Compound Addition 
 
 53 
 58 
 
 Multiplication 
 
 59 
 65 
 
 Vulgar Fractions 
 
 70 
 8-2 
 
 85 
 
 Decimal Fractions 
 
 106 
 
 Practice 
 
 123 
 
 Tare and Tret 
 
 145 
 
 
 147 
 
 Discount 
 
 159 
 
 Commission, &c 
 
 163 
 
 Equation of Payments 
 
 168 
 
 Profit and Loss 
 
 170 
 
 Exchange 174 
 
 Proportional Parts 189 
 
 Simple Fellowship and Bankruptcy 192 
 
 Compound Fellowship 195 
 
 Alligation 196 
 
 Involution 200 
 
 Square Root 203 
 
 Cube Root 207 
 
 Roots in General 209 
 
 Equidifferent Series 211 
 
 Continual Proportionals 214 
 
 Harmonical Proportion 219 
 
 Single Position 220 
 
 Double Position 2:21 
 
 Compound Interest 224 
 
 Annuities Certain 229 
 
 Life Annuities 235 
 
 Continued Fractions 241 
 
 On Arithmetical Scales 246 
 
 Questions, with their Solutions 251 
 
 Miscellaneous Questions 255 
 
 Mensuration 263 
 
 Tables 276 
 
 Answers to Exercises 2^1 
 
TREATISE ON ARITHMETIC, 
 
 IN 
 
 THEORY AND PRACTICE. 
 
 NOTATION AND NUMERATION. 
 
 ARIT 
 
 THMETIC is a science which explains the proper- 
 ties, and shows the uses of numbers. 
 
 Unity, or a unit, is the number one. Every other num- 
 ber is an assemblage of units.* 
 
 For facilitating the management of numbers in Arithmetic, 
 they are generally expressed by signs or characters. 
 
 The method of expressing numbers by characters, is 
 termed Notation. 
 
 The method of discovering the values of numbers already 
 expressed by characters, is called Numeration. 
 
 In modern arithmetic, all numbers are expressed by 
 means of ten characters : the cipher, or zero, 0, which has 
 no value; and the nine significant figures, or digits, 1, 2, 
 3, 4, 5, 6, 7, 8, 9, which denote respectively the numbers 
 one, two, three, four, five, six, seven, eight, nine. 
 
 When any of these figures stands by itself, or when it is 
 followed by no other figure, it expresses merely its simple 
 value ; but when it is followed by one figure, it signifies 
 ten times its simple value ; when by two, one hundred 
 times ; when by three, one thousand times ; and so on, by 
 a tenfold increase for each figure that follows it. The in- 
 creased value thus denoted by a figure in consequence of 
 its position, is called its local value. 
 
 * This definition, though it might appear otherwise, is applicable to fractions as well a; 
 to whole numbers. Thus, the fraction ^'o i 3 an assemblage of seven units, each of which 
 is one-tenth of the integer. The use of the term unity to denote the number one, ap- 
 pears to be improper ; if it be, however, it has the sanction of long custom. Some late 
 writers, probably to avoid this use of the word, employ in the same sense the term unit, 
 but without the indefinite article ; perhaps because it is not used l>efore two, three, ten, 
 &c. But, besides the sanction of custom, may not the article be used before it with the 
 same propriety as before the words hundred, thousand, mill/on, &c. ? It seems to be as 
 awkward, and as incorrect, to talk of increasing a number by unit, as of increasing it by 
 hundred or thousand* 
 
2 NOTATION AND NUMERATION. 
 
 Thus the figure 5, when it stands by itself, or is followed by no other 
 figure, denotes simply five; but when placed to the left of one figure, 
 it expresses ten times five, or fifty ; when to the left of two figures, it 
 expresses ten times fifty, or five hundred ; when to the left of three, ten 
 times five hundred, or five thousand, &c. ; as in the number 5555. In 
 like manner, in the number 7854, the 4 signifies simply four units, or 
 four ; the 5, five tens, or fifty ; the 8, eight hundred ; and the 7, seven 
 thousand. The names of the local values of figures will be known 
 from the following table : 
 
 NUMERATION TABLE. 
 
 From this table it appears, that if a line of figures be divided into 
 periods of three figures each, commencing at the right hand, the first 
 period will contain units, the second thousands, the third millions, &c. ; 
 and it is usual and convenient thus to divide the figures by which large 
 numbers are expressed, for the purpose of facilitating their numeration. 
 The periods succeeding those contained in the table, are quintillions, 
 trxlUlions, septittions, octiUwns, and nonittions ; and analogical names might 
 be formed for the still higher periods. Those already given, however, 
 are more than sufficient to express any number which it is ever necessary 
 to designate in language. The local value of any figure used in expres- 
 sing a number, is at once discovered from this table. Thus, 6 in the 
 eighth place from the right hand, expresses six tens of millions, or sixty 
 millions; and. conversely, sixty millions will be expressed by the figure 
 6 in the eighth place.* 
 
 * This method of dividing lines of figures into periods, and of naming those periods, 
 ia that which is used by the French and Italians. It is strongly recommended by its 
 simplicity and elegance ; and has been adopted in the treatise on Arithmetic by Mr. 
 Anderson, in the Edinburgh Encyclopedia ; and in Professor Leslie's " Philosophy of 
 Arithmetic." In other English works, the periods are made to consist of six figures 
 each, and have the same names as those in the table given above, except thousands, for 
 which there is not a distinct period. The two methods agree as far as hundreds of mil- 
 lions, and it is rarely necessary to name larger numbers. For the use of those who pre- 
 
NOTATION AND NUMERATION. 3 
 
 The cipher, or zero, having no value, is used in combinations of figures, 
 to fill places where no value is to be expressed, and thus to make the 
 other figures occupy those places in which they will express the intended 
 values. Thus the figures 365, combined in this order, denote three 
 hundred and sixty-five; but the expression 306050, which contains the 
 same significant figures, means three hundred thousand, no tens or 
 thousands, six thousand, no hundreds, five tens, and no units ; or three 
 hundred and six thousand, and fifty. From these principles we have 
 the following rule : 
 
 To express in Words the Numbers denoted by Lines of Figure** 
 
 RULE (1.) Commencing at the right hand side, divide the 
 given figures into periods of three figures each, till not more 
 than three remain. (2.) Then the first period towards the 
 right hand contains units or ones ; the second, thousands ; 
 the third millions, &c. as in the Numeration Table : and 
 therefore commencing at the left side, annex to the value 
 expressed by the figures of each period, except that of the 
 units, the name of the period. 
 
 Thus, the expression 37053907, becomes by division into period?, 
 37,053,907, and is read thirty seven millions, Jifty three thousand, nine 
 
 fer the English method, the following table is subjoined ; and the answers of the Exer. 
 cises are given according to both methods at the end of the work. 1 1 is scarcely necessary 
 to say, that the rules and directions given in the text will be applicable in this method, 
 if the periods be made to consist of six figures each instead of three, and if the second 
 period be called millions, the third billions, &c. as in the following table : 
 
 COMMON NUMERATION TABLE 
 
 * 9 1 I 
 
 f t I ! f I 
 
 I 
 
 Jill ISff *sll 
 
 if 51 A 
 
 1 8- || |l 
 
4 NOTATION AND NUMERATION. 
 
 hundred and seven, the term units or ones at the last being omitted. By 
 practice the pupil will soon find it unnecessary to divide into periods 
 any lines of figures except those of considerable magnitude. 
 
 Exercises in Numeration. 
 
 Write down in words, or name, the numbers signified by the 
 following expressions : 
 
 Ex. 1. 24, 
 
 2. 144 
 
 3. 365 
 
 Ex. 4. 1000 
 
 5. 1728 
 
 6. 2240 
 
 Ex. 7. 9790 
 
 8. 37048 
 
 9. 30009 
 
 Ex. 10. 4055070 
 
 11. 300405 
 
 12. 79503046 
 
 Ex. 13. 8005600480 
 
 14. 557290000 
 
 15. 680000042 
 
 16. 93090093 
 
 17. 113355 
 
 18. 785398 
 
 19. 7030462 
 
 20. 24902490 
 
 21. 9003008005 
 
 Ex.22. 40657200 
 
 23. 100000001000 
 24. 60606060700707 
 25.. 102030405060708 
 
 26. 901001101201301 
 
 27. 200030040538 
 
 28. 73820760005192645 
 
 29. 10001000010000100 
 
 30. 40050060700809000 
 
 Ex. 31. 7946289006400300047264791 
 
 To express Numbers ly Figures. 
 
 RULE (1.) Make a sufficient number of ciphers or dots, and divide 
 them into periods ; (2.) Then, commencing at the left, place in 
 their proper positions beneath the dots or ciphers, the significant 
 figures necessary for expressing the proposed number. (3.) If any 
 places remain unoccupied, let them be filled with ciphers. 
 
 Thus, the method of expressing the number, two hundred and 
 five millons, twenty thousand, seven hundred and nine, will be 
 found in the following manner : 
 
 000,000,000 ...,...,..., 
 
 2 5 2 7 9, or 2 5 2 79, and thence by filling 
 the unoccupied places, 205,020,709. By practice the learner will 
 soon be able, in most cases, to dispense with the dots or ciphers. 
 
 Exercises in Notation. 
 Express the following numbers in figures : 
 
 Ex. 1. Fifty-two. 
 
 2. Three hundred and thirteen. 
 
 3. Five hundred and four. 
 
 4. One thousand and twenty-four. 
 
 5. Two thousand and forty-eight. 
 
 6. One thousand, eight hundred and fifteen. 
 
 7. Seven thousand, eight hundred, and fifty-four. 
 
 8. Three thousand and eight. 
 
NOTATION AND NUMERATION. 5 
 
 Ex. 9. Five thousand and seventy. 
 
 10. Four thousand, five hundred, and four. ' 
 
 11. Twenty thousand and eighty-four. 
 
 12. Six hundred and fifty thousand, and ninety. 
 
 13. Seven millions, seven thousand, and ten. 
 
 14. Sixty-four millions, three hundred. 
 
 15. Eleven millions, two thousand. 
 
 16. One hundred and ten millions, and twenty thousand. 
 
 1 7. One billion, ten millions, two hundred thousand. 
 
 18. One trillion, one hundred billions, two hundred millions, 
 
 19. One million, and fifty thousand. 
 
 20. One billion, two hundred thousand. 
 
 21. One trillion, seven hundred thousand. 
 
 22. One quadrillion, and nineteen millions. 
 
 23. Seventy billions, ten thousand, and eighty-eight. 
 
 24-. Nine hundred billions, sixty-eight millions, and twenty. 
 
 25. The following numbers express the distances of the prin- 
 cipal primary planets from the sun, in miles; express 
 them in figures : Mercury, thirty-seven millions; Venus, 
 sixty-nine millions; the Earth, ninety-five millions; Mars, 
 one hundred and forty-five millions; Jupiter, four nun- 
 dred and ninety-four millions; Saturn, nine hundred and 
 seven millions; and the Georgium Sidus, or Herschel, 
 one thousand eight hundred and ten millions.* 
 
 Such is the facility with which large numbers are expressed, both by 
 figures and in language, that we have generally a very limited and in- 
 adequate conception of their real magnitudes. The following consider- 
 ations may, perhaps, assist in enlarging the ideas of the pupil, on this 
 subject: To couui a million, at one per second, would require between 
 twenty-three and twenty-four days of twelve hours each. The seconds 
 in six thousand years, are less than one-fifth of a trillion.f A qua- 
 drillion^ of leaves of paper, each the two-hundredth part of an inch in 
 thickness, would form a pile, the height of which would be three hundred 
 and thirty times the moon's distance from the earth. Let it also be re- 
 membered, that a million is equal to a thousand repeated a thousand 
 times ; and a billion equal to a million repeated a thousand times. 
 
 In the ancient Roman notation, I signified one, V five, X ten, L fifty, 
 and C one hundred. To these characters were added, at a later period, 
 D, signifying five hundred, and M, one thousand. When any charac- 
 
 * In sersral of the subsequent rules, examples are introduced which are calculated 
 to exercise the pupil in notation. To familiarize him with numeration, the teacher 
 should frequently require him to read off in words, the answers of the exercises which 
 he performs. 
 
 f A billion, in the common notation. 
 
 J A thousand billions, in the common notation. 
 
 ^ In the other notation a billion is equal to a million repeated a million of times. 
 
6 SIMPLE ADDITION. 
 
 ter was followed by another of equal, or of less value, the compound 
 value was equal to the simple values of both taken together ; but when a 
 character preceded one of greater value, both together expressed a value 
 equal to the difference of their simple values. Thus, II expressed two ; 
 XI, eleven, and IX, nine; CX, one hundred and ten, and XC, ninety. 
 We find also 13 put for 500 ; and by every such 3 annexed, the value 
 is made ten times as great. Thus, 133 signifies 5000 ; 1333, 50,000, 
 &c. CI3 was also used to express 1000, and the prefixing of C and the 
 annexing of 3, increased its value ten times. Thus, CCI33 signified 
 10,000; CCCIp33, 100,000, &c. A line drawn over the top of a 
 letter, made it signify as many thousands as the letter itself expressed 
 units. Thus, V expressed 5000; C 100,000, &c. The following 
 table, together with the preceding observations, wiM give an adequate 
 idea of the Roman notation : 
 
 I .... 
 
 1 
 
 IX . 
 
 ... 9 
 
 LXXX . . 80 
 
 II .... 
 
 2 
 
 X . 
 
 ... 10 
 
 XC ... 90 
 
 in. 
 
 3 
 
 XX . 
 
 ... 20 
 
 C .... 100 
 
 IV, or 1III . 
 
 4 
 
 XXX 
 
 ... 30 
 
 D, or l> . 500 
 
 V ... 
 
 5 
 
 XL . 
 
 ... 40 
 
 M, or do . 1000 
 ^^ 
 
 VI . . . . 
 
 fi 
 
 L . 
 
 ... 50 
 
 Tvnvr rr 1 T 9nnn 
 
 VII ... 
 VIII . . . 
 
 7 
 8 
 
 LX . 
 LXX 
 
 ... GO 
 ... 70 
 
 iviivi, or 11 . .\}\j\j 
 133, or V^ . 5000 
 
 MDCCCXXIV, or Cl3l O CCCXXIV, 1824. 
 
 SIMPLE ADDITION. 
 
 ' THE object of ADDITION is to find the number which is 
 equivalent to two or more given numbers taken together. 
 
 The number which is equivalent to two or more numbers 
 taken together, is called their SUM. 
 
 When the given numbers are all of the same denomina- 
 tion, as all yards, or all gallons, the operation is termed 
 
 SIMPLE ADDITION. 
 
 When the numbers to be added express quantities of the 
 same kind, but of different denominations, the operation is 
 termed COMPOUND ADDITION. 
 
 To add Quantities of the same Denomination. 
 
 RULE (1.) Place the numbers so that units may stand 
 under units, tens under tens, &c. (2.) Find the sum of the 
 column of units, set down the last figure of it below that 
 column, and carry to the next the number expressed 
 by the remaining figure, or figures, if there be any. (3.) 
 Proceed as before with The remaining columns, and at the 
 last column set down the entire sum. 
 
SIMPLE ADDITION. 7 
 
 Thus, to add together 9468, 2956, and 79, let 9468 
 them be set as in the margin, and proceed thus : 2956 
 9 and 6 are 15, and 8 are 23; set down 3, and car- 79 
 
 ry 2 to the column of tens. Then 2 and 7 are 9, 
 
 and 5 are 14, and 6 are 20; set down a cipher, and 12503, sum 
 carry 2. 2 and 9 are 11, and 4 are 15; set down 5, 
 and carry 1. 1 and 2 are 3, and 9 are 12; set down 12, and the 
 sum, or answer required, is twelve thousand, five hundred, and three. 
 
 The sign -{-, commonly called plus, is employed in arithmetic 
 and other parts of mathematics, to signify that the quantities be- 
 tween which it is placed, are to be added together; and the sign 
 = , called the sign of equality, is used to denote, that the quanti- 
 ties between which it stands, are equal to one another. Thus, 
 the expression, 12-}-9=21, means, that 12 and 9 added together, 
 are equal to 21 ; or that the sum of 12 and 9 is 21. 
 
 Reason of the Ride. 
 
 The rule for performing addition depends on the nature of no- 
 tation, and on the obvious principle, that tJie whole is equal to the 
 sum of all its parts. By placing units under units, tens under tens, 
 &c. we are enabled the more easily to add together the figures of 
 the same local value; and one is carried for every ten, because, by 
 the nature of notation, ten in any column is equivalent only to one 
 in the column immediately to the left of it. We commence with 
 the units merely for the convenience of carrying to the next co- 
 lumns. Thus, in the preceding example, the sum of the column 
 of units is 23; and therefore, after setting down 3, we have 20 
 remaining. But, by the nature of notation, 2 in the next column 
 is equal to 20 in this; and therefore we carry only 2. 
 
 Some teachers may, perhaps, consider it pioper to make pupils com- 
 mit the following table to memory : 
 
 ADDITION TABLE. 
 
 2 and 
 
 2 and 
 
 3 and 
 
 4 and 
 
 5 and 
 
 6 and 
 
 7 and 
 
 2= 4 
 
 9 =11 
 
 7 = 10 
 
 6= 10 
 
 6=11 
 
 7 = 13 
 
 9 = 16 
 
 3=5 
 4=6 
 
 3 and 
 
 8=11 
 9 = 12 
 
 7= 11 
 8= 12 
 
 7 = 12 
 8 = 13 
 
 8 = 14 
 9 = 15 
 
 8 and 
 
 
 
 
 
 
 
 816 
 
 6=8 
 
 4=7 
 
 4 and 
 
 
 
 7 and 
 
 9 = 17 
 
 7=9 
 
 5 = 8 
 
 4=8 
 
 5 and 
 
 6 and 
 
 7 = 14 
 
 9 and 
 
 8=10 
 
 6=9 
 
 5=9 
 
 5= 10 
 
 6 = 12 
 
 8 = 15 
 
 9 = 18 
 
 To enable the learner to acquire accuracy and despatch in addition, it 
 is proper to train him to add in the following manner, till he can do it 
 with facility: Since 6 and 6 are 12, 26 and 6 are 32 : (here it should 
 be pointed out to him, that 12 and 32 end in the same figure :) since 9 
 and 7 are 16, 39 and 7 are 46 : since 8 and 6 are !4, 88 and 6 are 94 j 
 since 6 and 9 are 15, 16 and 9 are 25, &c. 
 
8 SIMPLE ADDITION. 
 
 Methods of Proof. 
 
 1. Add the several columns, according to the rule, comment 
 ing at the top and proceeding downward, and if the result be the 
 same as was obtained by adding them upward, it may be presumed 
 that the work is right. 
 
 2. Separate the given numbers into two or more divisions. 
 Find thf sums of these divisions, severally, and add these partial 
 sums together. If the last result be equal to that found by the 
 common method, the work is right. 
 
 This will appear obvious from the following example : 
 
 37928 
 
 93640 
 
 23574 
 
 75849 
 
 Entire sum, 230991 
 
 Sum of the 1st division, 131568 
 
 2d, 99423 
 
 Entire sum, 230991, roof. 
 
 This method may also be employed with advantage in finding 
 the sums of large columns, instead of adding the whole at a single 
 operation. The first method is very convenient and useful, 
 when the columns are not very large. 
 
 3. Commencing at the left hand, add the several columns with- 
 out carrying, and set down the full sum of each column with the 
 units in their proper place, and the tens below the figure immedi- 
 ately to the l^ft. Add together the two lines thus resulting, and 
 if the last result agree with that obtained by the common method, 
 it may be concluded, that both are right. 
 
 Thus, in the annexed example, the sum of the 
 left hand column is 25, which is set down in full : 
 the sum in the next column is 30; the cipher is set 
 in its proper place, and 3 under the 5; and so with 
 the rest. The sum of the two lines thus obtained, is 
 equal to the sum found by the ordinary method.* 28249, sum. 
 
 This method of addition might be used instead of 
 
 the common method; and as it requires nothing to 25029 
 be carried, it might be employed with advantage 322 
 when the calculator is liable to interruptions. 
 
 28249, proof 
 
 * Addition may also be proved by casting the nines out of each of the given numbers, 
 as will be explained in multiplication ; and then by casting the nines out of the sum of 
 the sxcesses, and out of the sum of the numbers. If these latter excesses be equal, the 
 work is generally right ; otherwise, it must be wrong. This method however, \s of lit. 
 l)c use in proving addition. 
 
SIMPLE ADDITION. 
 Exercises in Simple Addition. 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 3789 
 
 92864 
 
 486759 
 
 17896 
 
 258111 
 
 4236 
 
 79784 
 
 537192 
 
 570937 
 
 4174456 
 
 7483 
 
 4759 
 
 468013 
 
 784947 
 
 6880921 
 
 6047 
 
 28936 
 
 16975 
 
 9678 
 
 9911604 
 
 6. 94753 + 2847 -f 793688 + 9386 -f- 258 + 3456. 
 
 7. 8289364 -j-275224 + 6875144+ 12897 + 7650368 + 94986347 
 
 + 42682 + 3749286 + 7676. 
 
 8. 294796 + 489276+ 16759284+4938 + 5713245 + 3348675 + 
 
 798426 + 9482 + 39867. 
 
 9. 275 15436 + 8937549 + 37246375 + 48795 + 378 + 2863487 + 
 
 864937 + 3894+7863927 + 826957. 
 
 1 0. 986759 + 4976346 + 29483 + 898647 + 3984753 + 6489778 + 
 
 57893 + 2468 144 + 576989 + 498653. 
 
 1 1 . 4683795 + 24867593 1 + 94986473 + 2849758 + 533S8336 + 
 
 7788995 + 2137485+ 6758927 + 4926431 + 27729512 + 
 7842634+949867 + 343/216 + 78934. 
 
 12. Add together 9466495,375573735,754547,3789284,29886799 
 
 992984,293675,2684487,3592873,8847599, 738873, 7849376 
 334486,123845,672849,73554,8674. 
 
 1 3. Required the sum of 978+749+4764+8967 + 70889294 + 
 
 7759286 + 939723 + 864937 + 99375847 + 29-1-886 + 94623 
 + 924086 + 794867 + 935279423+9738413208 + 2468975 
 + 945237 + 3834975. 
 
 14. 28674 + 39257 + 3834 + 92751+92503+86759 + 394875 + 
 
 34938 + 375396 + 759394 + 267934 + 6846 + 94657835 + 
 1926 + 484673 1 9 + 2488 + 9357. 
 
 15. Add together seven thousand and ninety-four; two thou- 
 sand, one hundred, and nine ; eight thousand, nine hundred, and 
 sixty; eighty-seven thousand and sixty-two; three hundred and 
 seventy-five; nine thousand and thirty; thirty thousand and forty- 
 six; fifty-four thousand, seven hundred, and seventy-five; seven 
 thousand, eight hundred, and fifty-four. 
 
 16. Sir Isaac Newton was born in the year 1642, and died in 
 his eighty-fifth year. In what year did he die ? 
 
 17. William the Conqueror began his reign in England in the 
 year 1066, and reigned 21 years; William II. reigned 13 years; 
 Henry I. 15 years; Stephen, 39 years; Henry II. 35 years; Richard 
 I. 10 years; John, 17 years; Henry III. 56 years; Edward I. 35 
 years; Edward II. 20 years; Edward III. 50 years; Richard II. 
 22 years: in what year was this last prince dethroned? 
 
 18. In 1821, the population of the following towns in England, 
 Scotland, Ireland, and France, (the three largest in each) were as 
 follows: 
 
1O ' SIMPLE ADDITION. 
 
 London, 1,274,800 
 
 Manchester, 133,788 
 
 Liverpool, 118,972 
 
 Dublin, 186,276 
 
 Cork, 100,535 
 
 Limerick, 66,012 
 
 Glasgow, 147,043 
 
 Edinburgh, 138,235 
 
 Paisley, 47,003 
 
 Paris, 720,000 
 
 Lyons, 115,000 
 
 Marseilles, 102,000 
 
 Required the number of inhabitants contained in the three largest 
 towns, in each country. 
 
 19. In the year 1810, the weight, in pounds of the cotton 
 wool imported into England from the rest of Europe, was 
 16,725,708; from the United States, 55,194,616; from the British 
 West Indies and conquered Colonies, 17,889,184; from foreign 
 American Colonies, 22,137,397; from the East Indies, 23,144,907; 
 from other sources, 1,478,291. Required the entire quantity. 
 
 20. The Pyramids of Egypt are thought to have been built 
 337 years before the founding of Carthage, Carthage to have 
 been founded 49 years before the destruction of Troy, and Troy 
 to have been destroyed 431 years before Rome was founded; 
 Carthage was destroyed 607 year^ after the founding of Rome, 
 and 146 years before the commencement of the Christian era; the 
 Western empire of Rome ended in the year 476 of the Christian 
 era, and 590 years before the Norman conquest; Constantinople 
 was taken by the Turks 387 years after the Norman conquest, and 
 348 years before the union of Great Britain and Ireland in 1801. 
 How many years elapsed between the first and last of these events? 
 
 21. Captain Cook, in his first voyage round the world, sailed 
 from Portsmouth to the Madeiras, a distance of 1451 British 
 miles; thence to the Canaries, 339 miles; from these to the Cape 
 Verd islands, 985 miles ; and thence to Rio Janeiro, 3058 miles; from 
 that to Cape Horn, 2659 miles, and thence to Otaheite, 4919 
 miles; from Otaheite to the most southern point of the voyage, 
 
 1619 miles; and thence to Cook's strait in New Zealand, 1988 
 miles; from Cook's strait to Green Cape in New Holland, 1368 
 miles; and thence along the eastern coast of New Holland to the 
 most northern point of that island, 2176 miles; thence to the 
 straits of Sunda, 2487 miles; and thence to the cape of Good 
 Hope, 5818 miles; from that cape to St. Helena, 1884 miles; 
 and thence to Ascension island, 822 miles; from Ascension to 
 Corvo in the Azores, 3462 miles; and thence to Portsmouth, 
 1598 miles. How far did he sail in all, exclusive of numerous 
 deviations from these courses ? 
 
 22. The following is the number of barrels of porter brewed in 
 London by the twelve principal houses, between 5th July, 1814, 
 and 5th July, 1815: what is the entire quantity? 337,621; 182,104; 
 172,162; 161,618; 123,100; 119,333; 105,081; 72,080; 56,922; 
 51,294: 38,107; 32,256. 
 
 23. Required the entire population of England, from the fol- 
 
SIMPLE ADDITION. 
 
 H 
 
 lowing statement of the population of its several counties, accord- 
 ing to the Census of 1821 : 
 
 Bedford, 83,716 
 vjerks 131,97 r < 
 
 Hertford, 129,714 
 Huntingdon, 48,771 
 Kent 426016 
 
 Shropshire,.. 206,153 
 Somerset,.... 355,314 
 Southampton, 283,298 
 Stafford, 341,040 
 Suffolk, 270,542 
 Surrey, 398,658 
 Sussex, 233,019 
 Warwick,.... 274,392 
 Westmoreland, 51,359 
 Wilts 222 157 
 
 Bucks, 134,068 
 Cambridge,... 121,909 
 Cheshire, 270,098 
 Cornwall, 257,447 
 Cumberland,. 156,124 
 Derby, 213,333 
 Devo'n .. 439 040 
 
 Lancashire, 1,052,859 
 Leicester,.... 174,571 
 Lincoln, 283,058 
 Middlesex, 1,144,531 
 Monmouth,.. 71,833 
 Norfolk, 344,38b 
 Northampton, 162,483 
 Northumberland, 198,905 
 Nottingham,. 186,873 
 Oxford 136971 
 
 Dorset, 144,499 
 Durham, 207,673 
 Essex, 289,424 
 Gloucester,.. 335,843 
 Hereford,.... 103,243 
 
 Worcester,... 184,42-i 
 York, E. Riding, 190,449 
 , N. Riding 183,381 
 
 , w. Riding', 799,357 
 
 Rutland, 18,487 
 
 24. The following are the Irish acres, and the inhabitants of the 
 several counties of Ireland.* Required the number of acres, and 
 the population of each province, and of the whole kingdom; Lough 
 Neagh, in Ulster, containing 58,200 acres : 
 
 PROVINCE OF ULSTER. 
 
 PROVINCE OF LETNSTER. 
 
 
 Acres. 
 
 Population. 
 
 
 Acres. 
 
 Population. 
 
 Antrim, 
 
 387,200 
 
 269,856 
 
 Carlow, 
 
 137,050 
 
 81,287 
 
 Armagh, 
 
 181,450 
 
 196,577 
 
 Dublin, 
 
 142,050 
 
 346,550 
 
 Cavan, 
 
 301,000 
 
 194,330 
 
 Kildare, 
 
 236,750 
 
 101,715 
 
 Ronegall,.... 
 
 679,559 
 
 249,483 
 
 Kilkenny,... 
 
 300,350 
 
 180,326 
 
 Down, 
 
 348,550 
 
 329,348 
 
 King's co. ... 
 
 282,200 
 
 132,319 
 
 Fermanagh, . 
 
 283,450 
 
 130,399 
 
 Longford,... 
 
 134,150 
 
 107,702 
 
 L. Deny,.... 
 
 318,500 
 
 194,099 
 
 Louth, 
 
 110,750 
 
 119,188 
 
 Monaghan, ,. 
 
 179,600 
 
 178,183 
 
 Meath, 
 
 327,900 
 
 174,716 
 
 Tyrone, 
 
 463,700 
 
 259,691 
 
 Queen' sco. . 
 
 235,300 
 
 129,3^1 
 
 
 
 
 Westmeath,. 
 
 231,550 
 
 128,042 
 
 
 
 
 Wexford,.... 
 
 342,900 
 
 169,304 
 
 
 
 
 \Vicklow,.... 
 
 311,600 
 
 115,162 
 
 PROVINCE OF MUNSTER. 
 
 Acres. Population. 
 
 Clare, 476,200 209,595 
 
 Cork, 1,048,800 802,535 
 
 Kefry, 647,650 205,037 
 
 Limerick,.... 386,750 280,328 
 
 Tipperary,... 554,950 353,402 
 
 Waterford,... 262,800 154,466 
 
 PROVINCE OF CONNACKHT. 
 
 Acres. Population . 
 
 Gahvay, 969,950 314,748 
 
 Leitrim, 255,950 105,976 
 
 Mayo, 790,600 297,538 
 
 Roscommon, 346,650 207,777 
 
 Sligo, 247,150 127,8,51 
 
 * Th? acres a: e given from Beaufort, and the population from the Census of 1821. 
 A* that Census was, in some instances, in a slight degree defective, it is thought, tint ti.e 
 P"l>iila inn e*c.viled seven millions. The acres given alme are Irish, T<!I of wbuii aic 
 *quiv<ikMt :o 19-i uf tiie stanUur I acrc= u>w iu ue. 
 
12 SIMPLE SUBTRACTION. 
 
 25. The number of inhabitants contained in the several coun- 
 ties of Scotland, in 1821, was as follows. Required the popula- 
 tion of the whole kingdom: 
 
 Aberdeen,... 155,387 
 Argyle, 97,316 
 Ayr, 127,299 
 
 Elcpn, 31,162 
 
 Orkney & Shetland. 53, 1 24 
 Peebles, 10,046 
 Perth, 139,050 
 Renfrew, 112,175 
 
 ^To^, \ *> 
 Roxburgh, 40,892 
 Selkirk, 6,637 
 Stirling 65376 
 
 Fife, 114,556 
 
 Forfar, 113,430 
 
 Banff, 43,561 
 
 Haddington, 35,127 
 Inverness,... 90,157 
 Kincardine,.. 29,118 
 Kinross, 7,762 
 Kirkcudbright, 38,903 
 Lanark, 244,387 
 Linlithgow, .. 22,685 
 Nairn, 9,006 
 
 Berwick, 33,385 
 Bute 13 797 
 
 Caithness, ... 30,238 
 Clackmannan, 13,263 
 Dumbarton,. 27,317 
 Dumfries,... 70,878 
 Edinburgh,.. 191,514 
 
 Sutherland,... 23,840 
 ;Wigton, 33,210 
 
 SIMPLE SUBTRACTION. 
 
 THE object of SUBTRACTION is to find the difference be- 
 tween two numbers. 
 
 The number found in Subtraction is called the REMAIN- 
 DER, the DIFFERENCE, or the EXCESS.* 
 % When the given numbers are of the same denomination, 
 the process is termed SIMPLE SUBTRACTION. 
 
 When the given numbers express quantities of the same 
 kind, but of different denominations, the process is termed 
 
 COMPOUND SUBTRACTION. 
 
 Rule for Simple Subtraction. 
 
 (1.) Place the less number below the greater, with units 
 under units, tens under tens, &c. as in Addition. (2.) Be- 
 ginning with the units, take, if possible, each figure in 
 the lower line, from the figure above it, and set down the 
 remainder. (3.) But if any figure in the lower line be 
 greater than the figure above it, add ten to the upper; then 
 subtract as before, and carry one to the next figure in the 
 lower line. 
 
 The sign , generally called minus, when set between two num- 
 bers, denotes, that the latter is to be taken from the former 
 Thus, 16 9=7 denotes, that if 9 be taken from 16 the remainder 
 is 7. 
 
 * The number to be subtracted it sometimes called the subtrahend; and that from 
 which it is to be taken, the minuend. 
 
SIMPLE SUBTRACTION. AS 
 
 Methods of Proof. 
 
 1. Add the remainder and the less of the given numbers toge- 
 gether: if the sum be equal to the greater, the work is correct. 
 
 2. Subtract the number found from the greater of the given num- 
 bers ; if the remainder be equal to the less, the work is correct. 
 
 Examples in Simple Subtraction. 
 
 Ex. 1. From 7854 take 4513. 7854 
 
 Set the numbers as in the margin, and pro- 4513 
 ceed thus: 3 from 4, and 1 remains; 1 from 
 
 5, and 4 remain; 5 from 8, and 3 remain; 4 Remainder, 3341 
 
 from 7, and 3 remain : the remainder therefore 
 
 is 3341. Proof, 7854 
 
 To prove the work, to the less of the given 7854 
 
 numbers add the remainder, and the sum will 4513 
 
 be 7854, the greater; or, as in the second me- 
 thod, subtract the remainder from the greater Remainder, 3341 
 number, and the result will be 4513, the less. 
 
 Proof, 4513 
 
 Ex. 2. Required the difference of 3712 and 1831. 
 In this example proceed thus: 1 from 2, From 3712 
 
 and 1 remains; 3 from 11, and 8 remain; car- take 1831 
 
 ry 1 to 8, and then 9 from 1 7, and 8 remain ; 
 
 carry 1, and then 2 from 3, and 1 remains. Remainder, 1881 
 
 The difference, therefore, is 1881, and the 
 
 operation would be proved in the same manner as before. 
 
 Reason of the Rule. 
 
 The rule for Subtraction depends on the principle, that the dif- 
 ferences of the several parts of two numbers are, when taken to- 
 gether, equal to the difference of the numbers themselves. The 
 reason of placing units under units, tens under tens, &c. w, that 
 figures may be subtracted from others of the same local value with 
 more facility. By carrying one to the lower figure, we increase 
 the lower line as much as we increased the upper, and thus the 
 difference will be the same as if neither had been increased. Thus, 
 in the second of the above examples, when in the tens* place we 
 subtract 3 from 11, we thus add 10 to the 1 in the upper line; 
 then the lower line is increased by the same quantity by adding 1 
 to the 8; because, by the nature of notation, 1 in the third coluir.r 
 is equivalent to 10 in the second. Thus, therefore, both the given 
 numbers are equally increased, and consequently the difference 
 must be the same as if they had received no increase. 
 
14 SIMPLE SUBTRACTION. 
 
 As a farther illustration of Subtraction, let it be 
 required to find the difference of 83 and 57. Here, as 83 
 7 cannot be taken from 3, we may consider 83 as equal 57 
 to 70 and 13; and subtracting 7* from 13, and 5 from 
 7, we find the difference to be 26. In this simple and 26 
 natural method, the values of the given numbers under- 
 go no change; and, with only one exception, it might be employed 
 with as much facility as the common method, the next figure in 
 the upper line being always diminished by a unit, when one 
 would be carried to the figure below it, in the common method. 
 The exception is the case in which the next figure in the uppct 
 line is a cipher: in this case the common method is considera- 
 bly preferable ; and, as in practice, that method is in no case in- 
 ferior, it is universally preferred. 
 
 Exercises in Simple Subtraction. 
 Ex. 1. 4507932048 Ex.8. 915161718151617189 
 
 2. 3345617748 
 
 3. 65934 48566 
 
 4. 9040158270 
 
 5. 62341732686 
 
 6. 8463192177825 
 7 44444441234567 
 
 9. 202122223192021222 
 
 10. 357912468 24680135 
 
 11. 750304657134992884 
 
 12. 376995145 19490718 
 
 13. 15342517853845248 
 
 14. 100000000 1000 100 1 
 
 15. Take four thousand and four from four millions. 
 
 16. Required the difference between three millions and three 
 thousand. 
 
 17. Subtract nineteen thousand and nineteen, from twenty 
 thousand and ten. 
 
 18. Required the difference between three, and three hundred 
 thousand. 
 
 19. Subtract one million, nine thousand, and six, from two 
 millions, twenty thousand, nine hundred, and thirty. 
 
 20. Required the excess of nine hundred and twelve thousand, 
 and ten, above fifty thousand and eighty-two. 
 
 21. La Place, the celebrated French Mathematician and Phi- 
 losopher, was born in 1749: required his age in 1824. 
 
 22. The height of Mont Blanc, the highest mountain in Eu- 
 rope, is 15680 feet; and the height of Chimborazo, the highest 
 mountain in America, is 21427 feet: how much is the latter higher 
 than the former ? 
 
 23. The height of the Peak of Tibet, thought to be the highest 
 mountain in the world is 24,235 feet; and the height of the Peak 
 of Teneriffe is 12,340 feet: required the difference of their heights 
 
 24. In the years 1707, 1708, 1709, and 1710, the neat revenue 
 arising from the British Post-Office, was, at an average, 58,052; 
 in 1722, it was 98,010; in 1783, 159,858; in 1792, 368,784; 
 and in 1801, 755,299. Required the increase between the first of 
 these periods and the second, between the second and t 
 
SIMPLE MULTIPLICATION. 15 
 
 25. The population of London in 1821, was 1,274,800; of 
 Glasgow, 147,043; Edinburgh, 138/235; Manchester, 133,788; 
 Liverpool, 118,972; Birmingham, 106,722; Leeds, 83,796. Re- 
 quired the excess of the population of the first of these cities above 
 that of the second, of that of the second above that of the third, &c. 
 
 26. In 1821, 1822, and 1823, the imports of Ireland from Great 
 Britain and elsewhere, were 5,1 97, 1 93,6,407,428,and6,607,488, 
 respectively: required the excess of the third above the second, 
 and of the second above the first. 
 
 27. The exports of Great Britain and Ireland, in 1821, 1822, 
 and 1823, were 46,980,565, 47,289,320, and 46,196,554, respec- 
 tively: required the excesses of the second above the first and third, 
 
 28. In 1795, the quantity of tea sold by the East India Com- 
 pany, for home consumption, was 18,498,569 pounds; and in 1800, 
 20,780,724 pounds. How much more was sold in the latter year 
 than in the former ? 
 
 29. The following are the years of the Christian era in which 
 the undermentioned events happened : required the number of years 
 from each till the year 1824. Commencement of the Hegira, or 
 era of the flight of Mahomet, 622; The Arabic, or modern nota- 
 tion in Arithmetic, introduced from Arabia into Europe by the 
 Saracens, 991; First Crusade, 1096; Magna Charta signed by 
 King John, 1215; Linen first made in England, 1253; Termina- 
 tion of the Crusades, 1291; Spectacles invented by a monk ot 
 Pisa, 1299; Gunpowder first used in Europe, 1330; University 
 of St. Andrews founded, 1411; Algebra introduced into Europe 
 from Arabia, 1412; Printing invented, 1440; Constantinople taken 
 by the Turks, 1453; America discovered by Columbus, 1492; 
 Vasquez de Gama's discovery of the route to the East Indies by 
 the Cape of Good Hope, 1497; Commencement of the Refor- 
 mation, 1517; Spinning Wheel invented, 1530; Copernicus died, 
 1543; Spanish Armada destroyed, 1588; Telescopes invented, 
 1590; University of Dublin founded, 1591; English East India 
 Company established, 1600; Decimal Fractions invented, 1602; 
 Thermometers invented, and Satellites of Jupiter discovered, 
 1610; Logarithms published by Napier, 1614; Circulation of the 
 blood discovered by Harvey, 1619; Barometer invented, 1643; 
 Air Pump invented, 1654; Newtonian Philosophy published, 
 1686; Georgium Sidus discovered, 1781; Union of Great Britain 
 and Ireland, 1801; Battle of Trafalgar, 1805; Battle of Water, 
 ioo, 1815. 
 
 SIMPLE MULTIPLICATION. 
 
 THE object of MULTIPLICATION is to find the amount ot 
 a number repeated a certain number of times. It is, there- 
 fore, only an abridged method of performing Addition, 
 when the numbers to be added are equal to one another. 
 
16 SIMPLE MULTIPLICATION. 
 
 The number to be repeated is called the MULTIPLICAND; 
 the number which shows how often the multiplicand is to 
 be repeated, is called the MULTIPLIER; and the number 
 found is called the PRODUCT. Both the multiplicand and 
 multiplier are sometimes called FACTORS, from their making 
 or producing the product. 
 
 When the multiplicand expresses a quantity of only one 
 denomination, the process is termed SIMPLE MULTIPLICATION. 
 
 When the multiplicand expresses a quantity of the same 
 kind, but of more denominations than one, the process is 
 termed COMPOUND MULTIPLICATION. 
 
 MULTIPLICATION TABLE.* 
 
 Twice 
 
 3 times 
 
 4 times 
 
 5 times 
 
 6 times 
 
 7 times 
 
 1= 2 
 
 1= 3 
 
 1= 4 
 
 1= 5 
 
 1= 6 
 
 1= 7 
 
 ' 2= 4 
 
 2= 6 
 
 2= 8 
 
 2= 10 
 
 2= 12 
 
 2=14 
 
 3= 6 
 
 3= 9 
 
 3= 12 
 
 3= 15 
 
 3= 18 
 
 3 = 21 
 
 4 = 8 
 
 4=12 
 
 4=16 
 
 4=20 
 
 4 = 24 
 
 4 = 28 
 
 5 10 
 
 5=15 
 
 5 = 20 
 
 5=25 
 
 5 = 30 
 
 5= 35 
 
 b = 12 
 
 0= 18 
 
 6=24 
 
 6=30 
 
 6=36 
 
 6=42 
 
 7 = 14 
 
 7=21 
 
 7 = 28 
 
 7 = 35 
 
 7 = 42 
 
 7=49 
 
 8=16 
 
 8=24 
 
 8 = 32 
 
 8 = 40 
 
 8=48 
 
 8= 56 
 
 9= 18 
 
 9=27 
 
 9 = 36 
 
 9=45 
 
 9=54 
 
 9=63 
 
 10 = 20 
 
 10=30 
 
 10 = 40 
 
 10 = 50 
 
 10=60 
 
 10=70 
 
 1 1 = 22 
 
 11 =33 
 
 11=44 
 
 11 = 55 
 
 11 = 66 
 
 11 = 77 
 
 12 = 24 
 
 12 = 36 
 
 12 = 48 
 
 12=60 
 
 12=72 
 
 12=84 
 
 8 times 
 
 9 times 
 
 10 times 
 
 11 ti;res 
 
 12 times 
 
 1 = 8 
 
 J = 9 
 
 1= 10 
 
 1= 11 
 
 1= 12 1 
 
 i 2=16 
 
 2= 18 
 
 2= 20 
 
 2= 22 
 
 2= 24 
 
 3=24- 
 
 3= 27 
 
 3= 30 
 
 3= 33 
 
 3= 36 
 
 4=32 
 
 4= 36 
 
 4= 40 
 
 4= 44 
 
 4= 48 
 
 5 = 40 
 
 5= 45 
 
 5= 50 
 
 i= 55 
 
 5= 60 
 
 6 = 48 
 
 6= 54 
 
 6= 60 
 
 6= 66 
 
 (i= 72 
 
 7 = 56 
 
 7= 63 
 
 7= 70 
 
 7= 77 
 
 7= 84 
 
 8 = 64 
 
 8= 72 
 
 8= 80 
 
 8= 88 
 
 8= 96 
 
 9=72 
 
 9= 81 
 
 9= 90 
 
 9= 99 
 
 9= 11)8 
 
 10=80 
 
 10= 90 
 
 10= 100 
 
 10= 110 
 
 10= 120 
 
 I 11 = 88 
 
 11= 99 
 
 11= 110 
 
 11= 121 
 
 11= 132 
 
 12=96 
 
 12= 108 
 
 12= 120 
 
 12= 132 
 
 12= 144 
 
 * Though the part of the Multiplication Table given in the text, is quite enouh for 
 the pupil to commit to memory at first ; yet after he has made some proficieucy in 
 
SIMPLE MULTIPLICATION. 
 
 17 
 
 It will perhaps enable the pupil to commit the preceding Table to 
 memory with more ease, if he be caused to construct it by Addition. 
 Thus, to find the products by 7, let him set in one column the figure 1 
 seven times, in another the figure 2 seven times, in another the figure 3 
 as often, &c. Then, the sums of these columns will be the products 
 by 7. He will thus be taught the construction and nature of the Table, 
 and will perhaps find it easier and more interesting to commit the pro- 
 ducts which he has formed himself, than those which he finds without 
 explanation in the Table. It will also assist the learner, if his attention 
 be turned to the relations subsisting between some of the successive pro- 
 ducts in the Table. Thus he will see that the products by 10 are form- 
 ed simply by the addition of a cipher ; that the first nine products by 
 11 are formed by repeating the figure; that the products by 5 terminate 
 in 5 and 0, alternately ; that in the products by 9, the first figure gene- 
 rally increases, and the second decreases, by a unit, &c, 
 
 Rule for Simple Multiplication. 
 
 (1.) Place the multiplier below the multiplicand, with 
 units under units, tens under tens, &c. (2.) If the multi- 
 plier do not exceed \ C 2, multiply, by means of the Multi- 
 plication Table, each figure of the multiplicand by it, be- 
 ginning with the units, and setting down and carrying as 
 in Addition. The result will be the product required. (3.) 
 But if the multiplier be greater than 12, find the products 
 of the multiplicand by the several figures of the multiplier, 
 successively ; setting the right hand figure of each product 
 under that figure of the multiplier which produces it. The 
 sum of all these products will be the total product required. 
 
 If either the multiplicand, or the multiplier, or both, end 
 in ciphers, the significant figures may be arranged and mul- 
 tiplied according to the rule, and as many ciphers annexed 
 to the product, as are found at the end of both factors. 
 Ciphers in any other part of the multiplier, are to be ne- 
 glected. 
 
 Arithmetic, it may he found of advantage to require him to commit what follows, as it 
 Till enable him, in many cases, to shorten his work in a considerable degree. The la- 
 bour of committing a more extended table, would be scarcely compensated liy the id 
 vantage resulting. 
 
 13 times 
 
 14 times 
 
 lj timea 
 
 16 times 
 
 17 times 
 
 18 times 
 
 19 time* 
 
 = 26 ' 
 
 <z= 28 
 
 2n- 30 
 
 2 = 'J 
 
 2= 34 
 
 2= 36 
 
 2= 3s 
 
 3= 39 
 
 3= 4* 
 
 3= 45 
 
 S= 48 
 
 3= 51 
 
 3= .54 
 
 3= 57 
 
 4= 52 
 
 4= 56 
 
 4= 60 
 
 4= 64 
 
 4 = 68 
 
 4= 72 
 
 4= 76 
 
 5= 65 
 
 5= 70 
 
 5= 75 
 
 5= 80 
 
 5= 85 
 
 5= 90 
 
 5= 9^ 
 
 6= 78 
 
 6 = 84 
 
 6= BO 
 
 6= 96 
 
 6=102 
 
 6 = 108 
 
 6=1H 
 
 7= 91 
 
 1= 98 
 
 7 = 105 
 
 7=112 
 
 7 = 113 
 
 r = 126 
 
 7 = 133 
 
 8 = 104 
 
 8=112 
 
 8 = 120 
 
 8=1123 
 
 8 = 156 
 
 8 = 1*1 
 
 8 = 152 
 
 9=117 
 
 9=126 
 
 9 = 135 
 
 9 = H4 
 
 9 = 133 
 
 y = i<:2 
 
 3 = 171 
 
ls SIMPLE MULTIPLICATION. 
 
 The sign X , called the sign of Multiplication, is frequentl) 
 used to denote the multiplication of the numbers between which 
 it is placed. Thus, 20X12=240, denotes, that the product of 
 20 and 12 is 240. 
 
 Examples in Simple Multiplication. 
 Ex. 1. Multiply 324 by 8. 
 Set -the numbers as in the margin, and Multiplicand. ..324 
 
 proceed thus : 8 times 4 are 32; set down Multiplier 8 
 
 2, and carry 3: 8 times 2 are 16, and 3 are 
 
 19; set down 9, and carry 1; 8 times 3 are Product. ..2592 
 
 24, and 1 are 25 ; set down 25 ; and the pro- 
 duct is two thousand, five hundred, and ninety-two, the same 
 that would have been obtained by adding together eight lines, each 
 324: and thus it would be well to cause the pupil to work this 
 and a few similar questions, to make him sensible of the identity 
 of Multiplication and Addition. 
 
 Ex. 2. What is the product of 29 and 365 ? 
 Set the numbers as in the margin, multi- Multiplicand.. .365 
 
 ply by 9, and set 5, the first figure of the Multiplier 29 
 
 product, under the 9 ; then multiply by 2, 
 
 and set 0, the first figure, under the 2. 3285 
 
 The sum of these partial products will give 730 
 
 ten thousand, five hundred, and eighty-five, 
 
 the entire product required. Product... 10585 
 
 Ex. 3. Multiply 36407 by 40206. 
 Here the first figure of the" first partial Multiplicand. ..36407 
 
 product, is set below the figure 6 in the Multiplier 40206 
 
 multiplier; the first figure of the second 218442 
 
 partial product, below 2; and the first of 72814 
 
 the third, below 4; the ciphers in the 145628 
 
 multiplier being neglected. Product... 146377984 2 
 
 Ex. 4. Multiply 320 by 2400. 
 
 The numbers' being arranged as in the Multiplicand... 320 
 
 margin, multiply as in the last example 32 Multiplier 2400 
 
 by 24: to 768 the product, add three ci- f^g 
 
 phers, and the entire product is seven 64 
 
 hundred and sixty-eight thousand. Product... 768000 
 
 Reason of the Rule. 
 
 The principle on which this rule depends, is, that the products 
 of the several parts of the multiplicand by the multiplier, or its 
 parts, are, when taken together, equal to the product of the whole 
 multiplicand by the whole multiplier. The reason for the mode 
 of arranging the partial products, is, that figures of the same lo- 
 cal value may stand in the same column. This will be understood 
 from the following illustration. 
 
SIMPLE MULTIPLICATION. 19 
 
 In the example in the margin, we have first 7 times 497 
 7=49=4 tens and 9; the 9 is set down and the 4 7 
 
 tens reserved. Then 7 times 9=63, and 4- are 67, 
 
 or 6 tens and 7. The 4- added to 63, is, by the nature 3i79 
 of notation, equivalent to the 4 tens reserved, since 
 4 in the second place is equivalent to 40 in the first; 7 is then set 
 down, and 6 tens, or properly 6 hundreds, reserved, and 6 added 
 for them in the hundreds' place. 
 
 The annexing of one cipher to a number expressed by the modern 
 notation, -makes its value 10 times what it ivas before; two cipfier* 
 100 times; three ciphers, 1000 times, &c. This is evident from the 
 principles of notation, since the annexing of a cipher removes 
 each figure one place to the left. Thus, the figure 7, with one 
 cipher annexed, expresses 70, or 10 times 7; with two ciphers, 
 700, or 100 times 7, &c. 
 
 From this the operation in the margin 6758, Multiplicand 
 
 will be understood. The multiplier is 346, Multiplier. 
 
 equivalent to 300, 40, and 6, taken to- 
 
 gether; and, therefore, the multiplicand 40548= 6758 X 6 
 
 is multiplied by 6, 40, and 300, succes- 270320=0758 X 40 
 
 sively, and the sum of the products taken. 2027400=6758 X 300 
 
 In multiplying by 40, the product by 4 
 
 is taken, and a cipher annexed; and in 2338268=6758X346 
 multiplying by 300, two ciphers are an- 
 nexed to the product by 3; in both in- 6758, Multiplicand 
 stances according to the principle just 346, Multiplier 
 
 delivered. In the second, or common 
 
 form of the operation, these additional 40548 
 
 ciphers are omitted, and the omission 27032 
 
 is compensated by making the figures 20274 
 
 occupy the same places which they do 
 
 when the ciphers are annexed ; and 2338268, Product, 
 hence the reason of setting the figures 
 
 in the manner prescribed in the rule, is evident. When the pupL 
 is learning the reason of the rule for Multiplication, it may be 
 well to cause him for some time to annex the ciphers, in the mode 
 above explained. 
 
 Methods of Proof. 
 
 1. Let the multiplier be taken as multiplicand, and the multi- 
 plicand as multiplier; and if the product thus obtained agree with 
 the product found before, the process is right. 
 
 64 45 
 
 Thus, when 64 is multiplied by 45, 45 64 
 
 the product is found to be 2880; and "3^5 1~80 
 
 when 45 is multiplied by 64, the same 256 270 
 
 result is obtained. 2h80 2880 
 
20 SIMPLE MULTIPLICATION. 
 
 2. Add together all the digits of the multiplicand, except 9, 
 which is to be neglected when it occurs; reject also 9 from the 
 successive sums of the digits, M'hen those sums exceed 9, and re- 
 serve the final excess. Proceed in like manner with the multiplier 
 and the product. Then multiply together the excesses found in 
 the multiplicand and the multiplier, and, in the same manner as 
 before, find the excess in this product. If this be the same as the 
 excess in the product of the given factors, the work is, in most 
 instances, correct: if it differ, the work must 
 
 * From the great brevity and facility of this mode of proof, it is very convenient in 
 practice. In some cases, indeed, operations which are incorrect may appear by this 
 method of proof, to be correct ; an occurrence however which is extremely rare, unless 
 it be either purposely effected, or arise from the misplacing of figures ; and hence, for 
 the purposes of the experienced arithmetician, it is perhaps preferable to any other me- 
 thod of proof. It is scarcely necessary to say, that the using of a cross, (x) in this me- 
 thod of proof, is both awkward and unnecessary. 
 
 The principle on which this process depends, is, that {f any number and the sum of 
 tts digits be both divided by 9, the remainders in both cases, are the same. Thus, 1000= 
 999+1, where the remainder must be 1, since 999 is evidently divisible by 9, without 
 remainder. Hence, since 3000=1000x3=999x3+3, the remainder here will evidently 
 be 3, since 999x3 is divisible by 9, without remainder. In like manner it might be 
 shown, that in dividing by 9, the remainder in 5000 would be 5 ; in 400, 4 ; in 20000, 2, 
 &c. ; being in all cases, the same as the significant figure contained in the number. 
 Hence if the number 487 be proposed, it is equivalent to 400+80+7 ; which parts, if 
 divided successively by 9, would leave, by the above principle, the remainders 4, 8, 7 ; and 
 therefore the remainder of the entire number will evidently be the same as the sum of these, 
 diminished by the rejection of 9 as often as possible. Now, suppose it were required to 
 multiply 112 by 48 ; 112=12x9+4, and 48^5x9+3 : the product therefore will be equal 
 to 12x9+4 multiplied by 5x9+3; which, by multiplying the several terms, becomes 
 12x9x5x9+5x5x4+12x9x3+4x3. In this product the three first terms are evidently 
 divisible by 9, Without remainder : consequently, the remainder obtained by dividing 
 the whole product by 9, will be the same as the remainder obtained by dividing 4x3, the 
 product of the excesses of the factors, by 9 ; and thus the reason of this method of proof 
 is evident. The same property belongs to the digit 3 ; but it is more convenient in prac- 
 tice to use 9. 
 
 Another method of proof, which is nearly as easy, and is perhaps more certain than 
 the preceding, depend-; on a property of the number 11 in the decimal notation. This 
 property might, be proved in a manner similar to that given above. It would be too te- 
 dious however to be introduced here; and indeed both are better and more easily given 
 by Algebra. The following is the method referred to : Commencing at the units of the 
 multiplicand, add together the digits in the odd places, rejecting 11 as often as possible, 
 end reserve the result ; proceed in the same manner with the digits in the even placet, 
 and from the former result, increased if necessary by 11, take this result and place the 
 excess opposite to the multiplicand. In the same manner find Ike excesses of the multi- 
 plier and the product ; then multiply the excesses of the two factors together, and find 
 in the same manner, the excess of their product : if this excess, and the excess of the 
 product of the factors be the same, the work is in most cases correct , if they differ, it 
 must be wrong. 
 
 In the annexed example, (the work of which, forbrcv- Multiplicand. ..84963 10 
 
 ity, is omitted,) 3 and 9 are 12, which exceeds 11 by 1 ; 1 Multiplier...44085 8 
 
 and 8 are 9 ; then 6 and 4 are 10, which being taken from 
 
 20 (=9+11,) the excess is 10. Then, in the multiplier. Product, 3745593855 3 
 
 the sum of 5, 0, and 4, is 9, from which 1, the excess of 
 
SIMPLE MULTIPLfCATION. 21 
 
 In the annexed example, 6 and 8 are 14, 68597 8 
 
 which exceeds 9, by 5; 5 and 5 are 10, which 897 6 
 
 exceeds 9, by 1 ; 1 and 7 are 8, which excess 
 we set opposite to the multiplicand. In the 480179 
 
 multiplier, in like manner, 8 and 7 are 15; 617373 
 
 which exceeds 9, by 6, which excess is set op- 548776 
 
 posite to the multiplier. In the product, 6 
 
 and 1 are 7, and 5 are 12, which exceeds 9 61531509 3 
 
 by 3; 3 and 3 are 6, and 1 are 7, and 5 are 
 
 12, which gives an excess of 3, to be set opposite to the product. 
 
 Then the product of the two excesses, 6 and 8, is 48, the sum of 
 
 the digits in which is 1 2, which gives an excess of 3, the same as 
 
 the excess of the product; and hence we judge the work to be 
 
 correct. 
 
 Exercises in Simple Multiplication. 
 
 Ex. 1. 144X3 
 
 2. 795X4 
 
 3. 729X5 
 
 4. 936X6 
 
 5. 1599X7 
 
 Ex. 6. 125X8 
 
 7. 3046X9 
 
 8. 7308X11 
 
 9. 1729X12 
 10. 1816X10 
 
 Ex. 11.40376X40 
 12.40376X4000 
 13.819200X700 
 14.7854X900 
 15.4096X12000 
 
 Answers. 
 Ex. 16. 148X53 =7844 
 
 17. 958X34 =32572 
 
 18. 7198X216.... =1554768 
 
 19. 31416X 175 =5497800 
 
 20. 8862X189 =1674918 
 
 21. 7071 X 556 =3931476 
 
 22. 93186 X 4455 =415143630 
 
 23. 40930X779 =31884470 
 
 24. 12345 X 686 =8468670 
 
 25. 46481 X 936 =43506216 
 
 26. 16734X708 = 11847672 
 
 27. 7575X7575 =57380625 
 
 28. 8320900 X 1328 =11050155200 
 
 29. 17500X732 =12810000 
 
 30. 15607X3094 =48288058 
 
 31. 7422153X468 =3473567604 
 
 32. 9264397X9584 =88789980848 
 
 33. 4687319X1987 =9313702853 
 
 S-f-4 above 11, being taken, the excess is 8. In the product, 5 and 8 are 13 ; 2, the ex- 
 cess, and 9 are 11, which is rejected; 5 and 7 are 12, the excess of which is 1 ; then 5 
 and 3 are 8, and 5 are 13 ; 2 and 4 are 6, and 3 are 9 ; 9 from 12, (=1+11,) and the ex- 
 .cess is 3. The product of the first two excesses is 80, the excess of which (found by 
 taking 8 from 11,) is 3, the same as the excess of the product of the factors. 
 
 By applying both these modes of proof to the same operation, an almost absolute ce- 
 tainty of iU correctness will be obtained. 
 
22 SIMPLE MULTIPLICATION. 
 
 Answers. 
 
 Ex. 34. 204053 X 1617000 =32995370100(7" 
 
 35. 9507340X7071 =67-226401140 
 
 36. 39948123 X 6007 =239968374861 
 
 37. 73885246X6079 =4491484104,34 
 
 38. 57902468 X 5008 =289975559744 
 
 39. 57902468X5080 =294144537410 
 
 40. 57902468 X 5800 =335834314400 
 
 41. 12481632 X 1509 = 18834782688 
 
 42. 79068025 X 1386 = 109588282650 
 
 43. 92948789 X 7043 =654638320927 
 
 44. 58763718X6754 =396890151372 
 
 45. 73084163X7584 =554270292192 
 
 46. 144X 144X 144 =2985984 
 
 47. 3851X3851X3851 =57111104051 
 
 48. 79094451 X 764095 =60435674536845 
 
 49. 79548050X97280 =7738434304000 
 
 50. Multiply fifty-six millions, seven thousand eight hundred, 
 and fifty-four, -by eighty millions, six hundred thousand, nine 
 hundred and seventy-six. Answ. 451427696065504. 
 
 51. Multiply five hundred and seven millions, five hundred and 
 forty-six thousand, two hundred and fifty-six, by seven hundred 
 and eight millions, nine hundred thousand, and fifty-eight. Answ. 
 359799570316082848. 
 
 52. Multiply eighty millions, seven thousand, six hundred, by 
 eight millions, seven hundred, and sixty. Answ. 640121605776000. 
 
 53. Required the amount of seven hundred and nine millions, 
 four hundred and sixty-five thousand, nine hundred and eight, re- 
 peated eight hundred thousand, three hundred and sixtv-five times. 
 Answ. 567831681456420. 
 
 54. Multiply eight hundred, and seventy-seven millions, five 
 hundred and ten thousand, eight hundred and sixty-four, by five 
 hundred and forty-five thousand, three hundred and fiftv-seven. 
 
 Answ. 478556692258448. 
 
 55. Multiply one hundred and twenty-three millions, four hun- 
 dred and fifty-six thousand, seven hundred and eighty-nine, by 
 nine hundred and eighty-seven millions, six hundred and fifty-four 
 thousand, three hundred and twenty-one. Ans. 1 2 1 93263 1 1 1 2635269. 
 
 56. How many yards of linen are in 759 pieces, each containing 
 25 yards? Ausw" 18975. 
 
 57. Sound is known to move about 1130 feet per second; how 
 many feet will it move in 69 seconds ? A.isw. 77970. 
 
 58. It is found by microscopic observations, that in each square 
 inch of the human 'skin there are about 1000 pores; and the sur- 
 face of the body of a middle sized man contains about 2304 inches, 
 (or 16 square feet.) Required the number of pores in the surface 
 of such a bod}', 999 being supposed to be contained in each inch. 
 Answ. 2301696. 
 
 59. An elm will live 100 years; and it has been computer that 
 
SIMPLE DIVISION 23 
 
 each year, at an average, it produces three hundred and twenty- 
 nine thousand grains or seeds, each of which, if properly treated, 
 would produce a tree. How many trees might be produced from 
 one elm in 67 years, and how many in 100 years? Answ. 22043000, 
 and 32900000. 
 
 60. In London, of morning newspapers which are printed every 
 day, except Sunday, there are sold about fifteen thousand each 
 day ; and of evening papers published the same days, about thir- 
 teen thousand; of papers published three times a-week, about ten 
 thousand of each impression are sold. There are likewise sold cf 
 Sunday papers about twenty-six thousand, six hundred, weekly; 
 and of other kinds about twenty thousand, weekly. How many 
 single papers are sold in London weekly, and how many yearly, 
 the year containing 52 weeks? Answ. 244600, and 12719200. 
 
 61. During the 50 years between 1700 and 1750, the quantity 
 of linen exported from Ireland, each year at an average, was 4 
 millions of yards; during the next six years, 11,796,361 yards in 
 each; during the next seven years, 14,511,973 yards annually ; dur- 
 ing the next seven, 17,776,862 yards per year; and during the suc- 
 ceeding seven years ending 1777, the average quantity was 
 20,252,239 yards annually. Required the whole quantity ex- 
 ported from 1700 till 1777." Answ. 638565684. 
 
 62. Rees's " Cyclopaedia," consists of 39 volumes, each con- 
 taining at an average, 774 pages, of two columns. In each co- 
 lumn there are 67 lines, each containing at an average, 10 words, 
 and in those 10 words, there are, at an average, 47 letters. Re- 
 quired the number of pages, lines, words, and letters, contained 
 in the entire work. Answ. 3Q18Q pages, 4044924 lines, 40449240 
 words, 1 90 1 1 1 4280 letters. 
 
 SIMPLE DIVISION. 
 
 DIVISION is the method of finding how often one giver 
 number, called the DIVISOR, is contained in another, called 
 the DIVIDEND.* 
 
 The number which shows how often the divisor is con- 
 
 * Or, Division is the method of finding a number which will be contained a given 
 number qf times, in another given number; or, Division it the method of finding cr* 
 qf the factors of a given product, when the other factor is given. 
 
 These two definitions, and that given in the text, result from the different points of 
 view in which Division is considered, and are easily reduced to one another, The last 
 is applicable to numbers of every kind, whether whole or fractional, and the one before 
 it is alike applicable to Simple and Compound Division. The one in the text, however, 
 is the most simple for the learner, and has the advantage of showing very easily the 
 identity of Division and Subtraction. 
 
24 SIMPLE DIVISION, 
 
 tained in the dividend, is called the QUOTIENT ; and if any thing 
 be left after the operation is finished, it is termed the RE- 
 MAINDER.* 
 
 When the dividend expresses a quantity of one denomina- 
 tion, the process is termed SIMPLE DIVISION. 
 
 When the dividend expresses a quantity of the same kind, 
 but of different denominations, the process is termed COM- 
 POUND DIVISION. 
 
 Rule for Simple Division. 
 
 (1.) Place the divisor to the left of the dividend, with a 
 line between them, and leave a space to the right of the 
 dividend for containing the quotient. (2.) Find by the Mul- 
 tiplication Table how often the first figure of the divisor is 
 contained in the first figure, or the two first figures of the 
 dividend, and set the figure denoting the number of times 
 in the quotient. (3.) Multiply the divisor by the figure thus 
 found, set the product below the leading figures of the di- 
 vidend, and subtract it from them. (4.) To the remainder 
 annex the next figuref of the dividend, divide the result as 
 before, and thus proceed till the operation is finished. 
 
 If any product be greater than the number which stands 
 above it, the last figure in the quotient must be changed 
 for one of smaller value ; but if any remainder be greater 
 than the divisor, or equal to it, the last figure of the quo- 
 tient must be changed for a greater. If any of the succes- 
 sive dividends be less than the divisor, a cipher must be 
 put in the quotient, and another figure, if any remain, brought 
 down from the given dividend.^ 
 
 * When one number contains another a certain number of times exactly, or without 
 remainder, the greater is called a multiple of the less, or it is said to be divisible by the 
 less ; and the less is said to measure the greater, or to be a measure, or a divisor, or an 
 aliquot part of the greater. Thus, 21 is a multiple of 7, and 7 is a measure, a divisor, or 
 an aliquot part of 1. 
 
 When a number is a multiple of two or more numbers, it is called a common multiple 
 of these numbers; and a number which is a measure of two or more numbers, is called 
 a common measure of them. Thus, 12 is a common multiple of 2, 3, 4, and 6 ; and 3 is 
 a common measure of 6, 21, and 33. 
 
 When a whole number is a multiple of any other whole number, greater than unity, 
 it is called a composite number. Thus, 4, 6, 8, 9, &c. are composite numbers. But 
 when a whole number is not divisible by any other whole number, greater than unity, it 
 is called a prime number. Thus, 2, 3, 5, 7, 11, &c. are prime numbers. 
 
 t It is proper, for preventing mistakes, to put a dot beneath each figure of the given 
 dividend, when it is brought down. 
 
 t It is carefully to be observed, that for each figure of the dividend which is brought 
 down, cither a significant figure or a cipher mutt be put in the quotient, besidet the first 
 figure that MS put in it. 
 
SIMPLE DIVISION. 25 
 
 When the divisor is below 13, the several multiplications 
 aiid subtractions may be performed mentally, and the quo- 
 tient set under the dividend.* 
 
 The sign -=-, called the sign of Division, plnced between t\vo 
 numbers, denotes that the former is divided by the latter. The di- 
 vision of one number by another, is also frequently denoted by 
 writing the dividend above the divisor, with a line drawn between 
 them. Thus, 144 -=-9= 16, or 1 ^ 4 rrl6, denotes that 144 is di- 
 vided by 9, and that the quotient is 16. When there is a remain- 
 der, after the operation is finished, the division of it by the divisor 
 is generally expressed in the latter mode: thuts, if 76 be divided 
 by 9, the quotient is written 8$. 
 
 Methods of Proof. 
 
 \. Find the product of the divisor and quotient, and add to it 
 the remainder; if the sum be equal to the dividend, the work is 
 correct. 
 
 2. Subtract the remainder from the dividend, and divide the re- 
 sult by the quotient; if the quotient thus found be the same as 
 the original divisor, the work is right. 
 
 3. Cast the nines out of the divisor, dividend, quotient, and re- 
 mainder; then, to the product of the excesses of the divisor and 
 quotient, acid the excess of the remainder, and cast the nines out 
 of the sum; if this excess be equal to the excess of the dividend, 
 the work is generally correct. 
 
 Examples in Simple Division. 
 
 Exam. Divide 136 by 8. 
 
 Here we say, 8 in 13 once, and 5 remain; 8) 136 
 8 in 56, 7 times,. and nothing remains. The ~l?7Quotient. 
 
 quotient therefore is 1 7 ; and by multiplying Q 
 
 it by the divisor 8, we obtain 1*36, the divi- T 
 
 iii-i i * 1O 
 
 dend, which proves the operation. 
 
 Division may be regarded as a contracted mode of performing 
 successive subtractions of the same number. Thus the preceding 
 example might be wrought by subtracting 8 from 136, 8 from the 
 remainder, 8 from that remainder, &c. till, after 17 subtractions, 
 the dividend would be exhausted, there being no remainder. The 
 Dumber of subtractions shows how often the divisor is contained 
 in the dividend, and agrees with the result already found. The 
 pupil might, with advantage, be required to work, in this manner, 
 a few examples in which the quotients would be small numbers. 
 
 * When the oper.u'wi is performed in this manner, it is usually termed Short Divisivn, 
 wliile the other method is called Long Division. 
 
 c 
 
26 SIMPLE DIVISION. 
 
 Ex. 2. Divide 15967 by 57. 
 
 Let the divisor, 57, be set before the 57) 15967 ( 280^ 
 dividend, 15967, as in the margin, and 114" 57 
 
 proceed thus: How often is 5 contain- 
 
 ed in 15? twice:* place 2 in the quotient, 456 1967 
 
 multiply the divisor by it, and set the 456 1400 
 
 product below 159, the leading part of 
 
 the dividend. This being subtracted 7 15967,proof 
 
 from 159, the remainder is 45, to which 
 
 6, the next figure of the dividend, is annexed. Again, how often 
 5 in 45? 8 times: place 8 in the quotient, proceed as before, and 
 there is no remainder. Then 7, the remaining figure of the di- 
 ridend, containing 57 no times, a cipher is placed in the quotient, 
 and the remainder is written in the quotient over the divisor 57. 
 The quotient, therefore, is 280 Z V 
 
 If the divisor end in ciphers, cut them off for a contrac- 
 tion, and cut off the same number of figures from the right 
 of the dividend ; then proceed with the remaining figures, 
 according to the rule, and to the last remainder annex the 
 figures cut off in the dividend, to find the true remainder : 
 thus 
 
 Ex. 3. Divide 782967 by 3700. 
 
 In this exercise two ciphers 37|00) 7829(67 ( 21 Iff 
 are cut off from 3700, and two 74 
 
 figures from the dividend; then 4? 
 
 the division by 37 proceeds in 37 
 
 the common way, and to the "~^g 
 
 last remainder, 22, the figures 37 782967, Proof, 
 
 cut off are brought down, and - ^, T, . , 
 
 the true remaincler is 2267. 2 ~ 67 > Remainder. 
 
 Hence, when the divisor is unity, with one or more ciphers an- 
 nexed the quotient will be found by cutting off from the dividend 
 as many figures for remainder as there are ciphers in the divisor : thus, 
 If it be required to divide 53826 by 100, the quotient is simply 538 T V<j. 
 
 * It would seem here at first sight, that 3, and not 2, should be put in the quotient 
 Were the divisor multiplied by 3, however, the product, 171, would be greater than 159 
 The learner often finds difficulty, in this way, in discovering what figure heshouid place in 
 the quotient In this example, 5 is contained in 15 twice, with the remainder 5, which, 
 with 9, the next figure in the dividend, makes 59. In this, 7, the second figure of the 
 divisor, is contained 8 times, which being more than 2, the number of times 5 was 
 found in 15, we are certain that the divisor is contained twice. Had 5 been taken 3 
 times in 15, however, we should have had no remainder to prefix to 9 ; and 7, the second 
 figure of the divisor, is not contained 3 times in 3, and consequently, the whole divisor 
 is not contained 3 times in the leading figures of the dividend. This method may per- 
 haps be employed with some advantage when the divisor is large ; practice, however, 
 will soon render it unnecessary. When the second figure of the divisor is above 5, in 
 
SIMPLE DIVISION 27 
 
 When the divisor can be resolved into two factors not 
 exceeding 12, the dividend may be divided, by one of them 
 by Short Division, and the result by the other. In pro- 
 ceeding thus, if there be remainders, the true remainder 
 on the whole division, is found by multiplying the last re- 
 mainder by the first divisor, and adding to the product the 
 first remainder.* 
 
 Ex. 4. Divide 947351 by 56. 
 
 Here the first remainder is 6 and the second 7) 947351 
 7, and by the rule we obtain 55 for the remain- 
 der, in dividing by 56. The reason of this is 8) 135335$ 
 
 best shown by means of fractions. It is cvi- 
 
 dent, however, since the first line is 7 times 16916f 
 
 the second, that whatever remains in dividing, 
 the second line is only a seventh part of what it would have been 
 in the first, and hence the reason will be manifest. 
 
 When the pupil has had some practice in the methods already 
 explained, he may be taught to omit writkig the products, which 
 will at least save much room in his operations. This method will 
 be understood from the following excimple: 
 Ex. 5. Divide 436951 by 173. 
 
 Here the first figure put in the 1731 436951 (2525-1? J 
 
 quotient is 2; then we say, twice 3 * 
 
 are 6; 6 from 6, and nothing re- 909 
 
 mains; twice 7 are 14; 14 from 23, 
 
 and 9 remain; twice 1 are 2, and 2 445 
 
 (carried,) are 4; 4 from 4, and 
 
 nothing remains. We then bring 991 
 
 down 9, and place 5 in the quotient; 
 
 then 5 times three are 15; 15 from 126 
 
 19, and 4 remain; 5 times 7 are 
 
 trying for the figure to be placed in the quotient, the first may be increased by unity. 
 Thus, in the example before us, we might have said, how often 6 in 15 ? &c. 
 
 The French place the divisor to the right of the dividend, and 59122 f 82 
 
 the quotient below it, as in the example in the margin. This 574' '1.721 
 
 mode gives the work a more compact and neat appearance, and 
 
 possesses the advantage of having the figures of the quotient near 172 
 
 the divisor, by which means the practical difficulty of multiplying 164 
 
 the divisor by a figure placed at a distance from it, is removed. 
 This difficulty every one must have felt, particularly in long ope- 82 
 
 rations; and hence this method might, with much propriety, be 82 
 
 adopted in preference to that which is employed in this country. 
 
 * It may be proper to observe, that this method is not applicable when the divisor * 
 a prime number, as the factors must be such as exactly to produce the divisor. It may 
 be farther remarked, that when there are more than two factors, the true remainder 
 will be found by applying the rule given in the text to the two last remainders, then to 
 the result and the remainder before them, &c. 
 
 C 2 
 
23 
 
 SIMPLE DIVISION. 
 
 35, and 1 are 36; 36 from 40, and 4 remain; 5 times 1 are 5, and 
 4 (carried,) are 9; 9 from 9, and nothing remains, &c. 
 
 Reason of the Rule. 
 
 The principle on which the operations in Division depend, is, 
 that apart of the quotient is found, and the product of it and the 
 divisor deducted from the dividend: then another part of the quo- 
 tient is found, and its product by the divisor deducted from the 
 remainder found before; and thus the operation proceeds till 
 nothing remains, or till the remainder is less than the divisor. This 
 will be understood from the following example : 
 Ex. 6. Divide 31272 by 73. 
 
 Here, the first part of the quotient is 
 400, the product of which by 73, is 29200. 
 This taken from the dividend, leaves 
 2072 to be divided by 73. The next part 
 of the quotient is 20; the product of 
 which by 73, is 1460, which still leaves 
 a remainder of 612 to be divided by 73. 
 This gives 8, with the remainder 28. 
 Hence, it appears that 73 is contained 
 in the dividend 400 -f- 20 -j- 8 times, or 
 428 times, with the remainder 28. By 
 comparing this and the common process 
 subjoined, it will be found that the latter 
 is merely an abbreviation of this, the ci- 
 phers being omitted in the one and re- 
 tained in the other. 
 
 73) 31272 (400 
 
 29200 
 73)2072 (20 
 
 1460 
 73) 612 (8 
 
 584 
 
 428 
 
 28 
 
 73) 31272 (428?S 
 292- 
 
 207 
 
 146 
 612 
 584 
 
 28 
 
 Exercises in Simple Division.* 
 
 Ex. 1. 470850^-3 
 
 2. 1829765-^4 
 
 3. 4265983-^-5 
 
 Ex. 4. 3782047 -f-6 
 
 5. 7 1 65537 -f- 7 
 
 6. 27459332 -f-8 
 
 Ex, 7. 74593822-4-0 
 
 8. 53248675 -f- 11 
 
 9. 49275 189 -j- 12 
 Answers. 
 
 Ex. 10. 51846734-f-102 =508301^ 
 
 11. 727346489408 = 178271 If g| 
 
 12. 980263711809 = 1211698^ 
 
 13. 536819237907 =591862f* 
 
 * The pupil should be required to prove the work of the several exercises in Division, 
 ty Multiplication. Every operation in Long Division, thus proved, affords him in re- 
 aiity an exercise in each of the four, fundamental rules; as, in finding the quotient he 
 tiuploys Division and Subtraction, and in the proof, Multiplication and Addition. 
 Hence, perhaps tne exercises here given will not seem, on due consideration, too nu- 
 merous, though at nrst sight they might appear to be so. It is of the first importance 
 that the pupil should have acquired both accuracy and despatch in performing the opera- 
 tions in the fundamental rules, before he proceeds to apply them in the more advanced 
 parts of Arithmetic. 
 
 
SIMPLE DIVISION. 
 
 Answer*, 
 
 Ex. 14. 1457924651-4-1204 = 12 10900 t*$i 
 
 15. 28101418481 -T-l 107 = 2538520 1-fitfr 
 
 16. 513513513513917 =559992926^ 
 
 17. 643751624367731 = 8806451 7 G^\ 
 
 18. 4G5465465465 644 =722772461 Hi 
 
 19. 347382600435727 =477830201$!! 
 
 20. 1700649160000-1-759 =r 224064447 9fH 
 
 21. 571824753344839. = 681555129Bi 
 
 22. 245379633477 1263 = 194283161}y ! 
 
 23. 536847555555 1138 =471746533^84 
 
 24. 234516447519759 =308980826ff| 
 
 25. 1111111111111-4-854 =1301066874fii 
 
 26. 7890123456767384 =106854326^|| 
 
 27. 75843639426 -v- 8593 =882621 l||fg 
 
 28. 65358547823 -r-2789 =23434402^% 
 
 29. 33333333333 -=-5299 =6290495 ? WV 
 
 30. 321987653321 -r-7766 =41461196HM 
 
 31. 542713060315^-4444 =122I22650|H* 
 
 32. 2652104208416-4-7539 =351 7846 14ffH 
 
 33. 7314617334753-4-6784 =1078215998|fU 
 
 34. 3146173847837-4-9387 =335162868ff-|i 
 
 35. 555777999444777891 =623768798478f ^ 
 
 36. 765809034537648793 =96571 1266756^2 
 
 37. 58239017 1945J 10 693 =840389858506|^ 
 
 38. 582390171945110593 =982108215759^ 
 
 39. 582390171945110493 =1181318S07190|f2 
 
 40. 582390171945110393 = 1481 908834465 gf-j 
 
 41. 582390171945110293 =1987679767730f |-g 
 
 42. 582390171945110193 =3017565657746H3 
 
 43. 3333333333333333 -r-483 =690 13 11 249 137^1? 
 
 44. 3333333333333333 -r-484 =688705234 1 597$|f 
 
 45. 1 000000000000000 4-81 =1 23456790 12345|f 
 
 46. 1 000000000000000 -r- 729 =1371 742 112482^3 
 
 47. 1 000000000000000 -i-1 11 =9009009009009 r | 1 
 
 48. 1 000000000000000 -f-1111 =900090009000|^'? 
 
 49. 1000000000000000 --11111 = 90000900009 IT ^ M 
 
 50. 555555555555 123456 = 4500028 jW^Ve 
 
 51. 555555555555654321 =849056||HH 
 
 52. 102030405060123456 = 82645 1 T W^ 
 
 53. 908070605040654321 =1387805gt!tf 
 
 54. 3784926474826 -4-384305 = 98472 19|fff$ 
 
 55. 4678179384732100 =222770446ff J t } 
 
 56. 36781793842936500 =10077203f|fg 
 
 57. 2678179384738760 =30572824*^ 
 
 58. 167817938176957000 =175358ff ?--& 
 
 59. 16781793847387700 =1913545|H^g 
 
 60. 2678179381151360000 =196924tMA 
 
 61. 367817938473-1-87000 =4227792f4,i 
 
SO ABBREVIATIONS. 
 
 62. Devis, the highest mountain in the neighbourhood of Bel- 
 fast, is 1542 feet high; and Mont Blanc 15,680 feet: how many 
 mountains, each as high as the former, must be piled one above 
 another to equal the height of the latter? Answ. 10 T Y*V 
 
 63. In 1821, the population of Belfast and the suburbs, was 
 about 40,000, and that of London was 1,274,800: how many 
 towns, each containing as many inhabitants as Belfast, must be 
 combined, to form one containing the same population as London? 
 Amu. Sl^n- 
 
 64. The linen exported from Ireland in 1809, was 43,904,382 
 yards: of how many pieces, each containing 25 yards, did this 
 quantity consist? Answ. 1,756,175 ^. 
 
 65. The entire quantity of tea sold by the East India Company 
 in 1799, was 24,853,503 pounds: how many chests, each contain- 
 ing 87 pounds, would this quantity fill? Answ. 285,672$f. 
 
 66. If it be supposed, as in common circumstances is found to 
 be nearly true, that as many persons die in 33 years as are equal 
 to the entire population, it is required to find how many persons 
 die each year, at an average, in the British dominions in Europe, 
 the population (in 1821,) being 21,481,152. Answ. 650,944. 
 
 67. How many lessons of ninety-five lines each, are contained 
 in Virgil's ^Enaid, the number of lines contained in that poem 
 being nine thousand, eight hundred, and ninety-two? Answ. 104f. 
 
 68. The following is the quantity, in hundreds weight, of the 
 butter exported from Ireland, from 1800 till 1811 inclusive: re- 
 quired the quantity exported each year, at an average : in 1800, 
 263,288; in 1801, 178,196; in 1802, 304,666; in 1803, 396,353; 
 in 1804, 334,251; in 1805, 320,155; in 1806, 294,415; in 1807, 
 338,508; in 1808, 333,998; in 1809, 346,856; in 1810, 385,953; 
 in 1811, 390,833. Answ. 323,981. 
 
 69. The following is the quantity, in pounds, (avoirdupois,) of 
 cotton-wool imported into England, from 1805 till 1810 inclusive: 
 required the quantity imported each year, at an average: in 1805, 
 59,862,406; in 1806, 58,176,283; in 1807, 74,925,306; in 1808, 
 43,605,982; in 1809, 92,812,282; in 1810, 136,570,103. Answ. 
 77,658,727. 
 
 ABBREVIATIONS IN THE FUNDAMENTAL RULES. 
 
 THE rules already given for performing the fundamental opera- 
 tions in Arithmetic, are of a nature completely general, and are 
 fully adequate to the performance of all operations that can occur. 
 Hence the pupil should be made substantially acquainted with 
 them, before he proceeds to any thing else. As the operations, 
 however, are often of a tedious and laborious nature, it is desira- 
 ble to be able to employ easy and expeditious methods of perform- 
 ing them* when such can be obtained Some of the most useful 
 
IN MULTIPLICATION. 31 
 
 of these will be found in what follows. Others might have been 
 added, but not so easy or useful, or so frequently applicable. With 
 respect to those which are given, it will be well to proportion to 
 the capacity and proficiency of the pupil, the number of them 
 which he will be required to learn. 
 
 The principal abbreviation admissible in Addition, is the adding 
 of two or more figures at once. This is peculiarly convenient, when 
 the sum of two or more figures is exactly ten. Thus, 
 in the annexed exainple, we may say 10 and 14? 41257 
 are 24, and 10 are 34. Then, carrying 3, we may 63783 
 proceed, 10 and 10 are 20, and 10 (5 + 5) are 30, 56558 
 and 8 are 38. 10 (3-f-G-f 1) and 16(4-f 5-f 7) are 38416 
 26, and 2 are 28. 14 (2 -j- 5 + 7) and 14 (8 -f 6) are 77198 
 28, and 4 are 32. 14 (3 + 4 -f 7) and 14(3+54-6) 45672 
 
 are 28, and 4 are 32. When, by attention and 
 
 practice, facility in this mode of proceeding is ac- 322884 
 quired, it will be found to be of more value than 
 might at first be imagined. 
 
 To multiply by 5 : Add a cipher, or rather conceive it to be added, 
 to the multiplicand, and lake half the result. 
 This is evidently the same as multiplying by 384729 X 15 
 10, and taking half the product. To multi- 1923645 
 
 ply by 15 : Conceive a cipher to be added to the 
 
 multiplicand, and to the result add half of it- 5770935 
 self. To multiply by 25 : Conceive two ciphers 
 
 to be added to the multiplicand, and take one-fourth of the result. This 
 is the same as multiplying by 100, and taking one-fourth of the 
 product, which is done because 25 is one-fourth of 100. In like 
 manner, because 125 is one-eighth of 1000; to multiply by 125: 
 Conceive three ciphers to be added to the multiplicand, and take one- 
 eighth of the result. To multiply by 75 : Con- 
 ceive two ciphers to be added to the midtipli- 587 X 75 
 cand, and from the result take one-fourth of 14675 
 
 itself. To multiply by 35 : Conceive two '- 
 
 phcrs to be added, take one-fourth, and set it so 44025 
 
 that it may be added to the multiplicand, with 
 
 one cipher annexed. In a similar manner, the 463 X 35 
 
 product by 225 may be found from the pro- 11575 
 
 duct by 125. To multiply by 175: Divide 
 
 700 times the multiplicand by 4. To multiply 16205 
 
 by 275 : Divide \ 100 times the multiplicand by*. 
 
 To multiply by 9 : Conceive a cipher to be added to the multipli- 
 cand, and from the result subtract the multiplicand; or, simply, con- 
 ceive a cipher to be added, and from each figure of the result take the 
 one before it. Thus, 5 from 10 and 5 remain: 
 
 and 7 aie 8; 8 from 15, and 7 remain: 1 6475X9 
 and 4 are 5; 5 from 7, and 2 remain: 6 
 from 14, and 8 remain: carry 1; 1 from 6, 58275 
 
32 ABBREVIATIONS IN MULTIPLICATION. 
 
 and 5 remain. To multiply by 1 1 : Conceive a cipher to be annexed to 
 the multiplicand, and to the result add the multiplicand i or, simply, con- 
 ceive a cipher to be annexed to the multiplicand, 
 and to each figure of the result add the one 6483 X 1 1 
 
 which immediately precedes it. Thus, 3 and 
 are 3: 8 and 3 are 11: carry 1; 1 and 4 are 71313 
 5, and 8 are 13: 1 and 6 are 7, and 4 are 
 11:1 and 6 are 7. To multiply by 90: Con- 1438 X 99 
 
 ceive two ciphers to be annexed to the multipli- 
 cand) and from the result subtract the multipli- 142362 
 caiid; or, simply, conceive two ciphers to be 
 
 annexed to the multiplicand, and from each figure in the result sub 
 tract the figure nearest it on the left side, except one. Thus, 8 from 
 10, and 2 remain: 1 and 3 are 4; 4 from 10, and 6 remain: 1 and 
 4 are 5 ; 5 from 8, and 3 remain : I from 3, and 2 remain : from 
 4, and 4 remain: from 1, and 1 remains. To multiply by 101 : 
 Conceive two ciphers to be annexed to the multiplicand, and to the rc- 
 sidt add the multiplicand; or, simply, conceive 
 two ciphers to be annexed to the multiplicand, 5938 X 1 1 
 
 and to each figure of the result add the figure 
 
 nearest it on the left side, except one. Thus, 8 599738 
 and are 8; 3 and are 3; 9 and 8 are 17; 
 1 and 5 are 6, and 3 are 9; and 59 are 59. In general, to mul- 
 tiply by 9, 99, 999, &c. : To the multiplicand annex as many ciphers 
 as there are 9*5, and from the result subtract the multiplicand: and 
 to multiply by 11, 101, 1001, &c: To the multiplicand annex as 
 many ciphers as there are digits in the multiplier, wanting one, and to 
 the result add the multiplicand. 
 
 The principle on which these last rules depend, may be gene- 
 ralized in its application in the following manner: If the multiplier 
 be a little less than 100, 1000, or any other number expressed by a 
 unit with ciphers annexed, which number may be called the approxi- 
 mate multiplier; or if it be a little greater than one of these numbers, 
 take the difference between it and whichever of them is nearest to it; 
 in the former case, call this difference the complement, in the latter, 
 the excess; then, to the multiplicand annex as many ciphers as there 
 are in the approximate multiplier; and in the one case subtract from 
 the result the product of the midtiplicand and the complement; in the 
 other, add to the result the product of the midtiplicand and the excess.* 
 
 * It is obviou* that this rule will be principally useful, when the complement or ex- 
 cess does not exceed 12. It may indeed be employed with considerable facility, when the 
 complement or excess is between 12 and 20, if the pupil have committed to memory the 
 additional part of the Multiplication Table given in the note in page 17, or by means of 
 the method of multiplying by 13, M, &c. illustrated in the following 
 example : Here, 7 times 9 are 63; carry 6; 7 times 2 are 14, and G 7329x17 
 Are 20, and 9 ;the following figure,) arc 29 ; carry 2 : 7 times 3 are 21, 17 
 
 and 2 are 23, and 2, (the following figure,) are 25; carry 2 : 7 times 
 
 7 are 4-9, ana 2 are 51, and 3 are 54 ; carry 5 : 7 times is 0, but 5 
 juvu 7 are 12. It is obvious that we may multiply by any number i 
 
ABBREVIATIONS, S3 
 
 Thus, in the first of the annexed exam- 
 ples, the complement is 11, and, there- 74953 X 989 
 fore, we multiply 74953 by 11, and set 884483 
 
 the product, and subtract in such a man- 
 
 ner as if the upper line were 74953000. 74128517 
 
 The remainder, 74128517, is the product 
 
 required. In the second example, the 7936 X 112 
 
 excess is 12; therefore, the multiplicand 95232 
 
 is multiplied by 12, and the product set 
 
 in such a manner as to be added to 888832 
 
 793GOO. The sum is the product required. 
 
 Another abbreviation, which may perhaps be more frequently 
 useful than any of the above, consists in deriving a product from 
 others already found ; and in doing this, and indeed in all cases, it 
 should be considered, that it is a matter of indifference by which 
 figure of the multiplier we multiply first, by which secondly, fyc. pro- 
 'vided we assign to the partial products their proper places. In the 
 annexed example, where the multiplier is 
 1 68428, the product by 4 is first found ; this 673849 
 
 product is then multiplied by 7, to find 168428 
 
 the product bv 28, because 264x7. 
 The product tnus found for 28 is then 2695396 
 
 multiplied by 6, to find the product by 18867772 
 
 168, because 168=28X6. The sum of 1 13206632 
 
 these partial products, properly arranged, 
 
 is the product required. Even when no 1 1 3495039372 
 farther abbreviation can be employed, it 
 
 -ametimes renders the work less laborious to derive the product 
 single digit from one found already: thus, the product by 6 
 will be found by doubling the product by 3; the product by 8 is 4 
 times the product by 2, &c. It is scarcely necessary to say, that 
 when the same digit occurs twice, or oftener, in the multiplier, it 
 is sufficient to multiply once by it, as the product thus found may 
 be copied, when that digit recurs in the multiplier. 
 
 To divide by 5 : Double the dividend, and cut off the last figure of 
 the result, the half of which figure will be the remainder. This is ob- 
 viously nothing else than doubling the dividend, and dividing by 10. 
 To divide by 15, 3.3, 45, or 5o: Double the dividend, divide the re- 
 sult by 30, 70, 90, or 110, respectively, and for the true remainder take 
 half the remainder thus found. To divide by 25: Multiply by&, cut off 
 two figures from the remit, and take one-fourth of the number cx- 
 pretced by them for the remainder. To divide by 125: Multiply the 
 dividend by 8, cut off three figures from the result, and take one-eighth 
 
 tween 20 and SO, by multiplying by the units' figure, and adding each time, along with 
 what is carried, twice the following figure of the multiplicand ; and the principle ma-' 
 be readily extended to numbers above 50, but the operation becomes more t 
 and difficult. G 3 
 
34 ABBREVIATIONS IN DIVISION. 
 
 of the number expressed by them for the remainder. To divide by 
 75 : Multiply the dividend by 4, divide the product by 300, and for 
 the true remainder take one-fourth of the remainder thus found. To 
 divide by 175, 225, and 275: Multiply the dividend by 4, divide 
 the result in the first case by 700, in tfie second by 900, and in the 
 third by 1 100, atid taJce one-fourth of the remainder, in each case, for 
 the true remainder. 
 
 To divide by a number expressed by any number of 9's: Divide 
 the dividend by the number expressed by a unit, with as many ciphers 
 annexed as there are figures in the divisor, divide the quotient thus 
 found by the same number, and thus proceed as long as possible; then 
 add together all the remainders and all the quotients, carrying from 
 the sum of the remainders to the sum of the quotients, and adding also 
 to tJie sum of the remainders what is carried to the quotients ; the sums 
 thus found will be the quotient and the remainder required. Should 
 the remainder thus found be equal to the divisor, it is evident that 
 there is really no remainder; but that the quotient must be in- 
 creased by unity. The operation will be facilitated by arranging 
 the lines in the manner adopted in the 
 
 annexed example. Here a vertical 5739284651-7-999 
 line is drawn, cutting three figures 5739 284 
 
 from the dividend; then 4, the last 5739 
 
 figure of the quotient, thus obtained, 
 
 is set below the unit figure of the re- 5745029 679 
 mainder, and the other figures are 
 
 680 
 
 set in the corresponding places. The 
 vertical line being continued, we have 
 5739 for the second quotient, the unit figure of which is set in 
 the units' place of the remainder, and the other figures as before; 
 the third quotient, 5, being set in the same manner, the lines are 
 added together: the sum of the column of hundreds in the remain- 
 der is 16, from which 1 is carried to the units' column of the quo- 
 tients, and also added to the remainder, 679: the true quotient 
 and remainder, therefore, are 5745029, and 680. 
 The second example, in which 3467581 
 
 581-;-989 
 
 137 
 
 418 
 
 136 
 11 
 
 is divided by 989, shows the mode of ge- 3467 
 
 neralizing this principle. Here the com- 38 
 
 piement, (see page 32 J is (1000989,) 
 11, and the first quotient is 3467, which, 
 instead of being simply set beneath the 3500 
 
 given dividend, is multiplied by the com- 
 plement, and the product is set as in the 
 preceding example; we have then the quo- 147 
 
 tient 38, which is multiplied, in like man- 
 ner, by the complement, and the product (418,) set as before. 
 In the addition, the sum of the hundreds' column of the remain- 
 ders is 1 1, from which one is carried to the quotient, and is also 
 multiplied by the complement; the product thus found is also ad- 
 ded to the remainder; hence, the true quotient and remainder are 
 3506 and 147. 
 
ABBREVIATIONS IN DIVISION. 3.5 
 
 The reason of these operations will be 
 understood from the following illustration. 8053|93.-i-97 
 The quotient by 100, in the annexed ex- 241 59 
 
 ample, is 8053, and the remainder 93. 
 
 - -r 1 L A 1~ j. i-l_" 1 .*. * T 
 
 Now, it is evident, that were this multipli- 
 ed by 97, instead of 100, the product 
 
 would be less than 805300 by 3 times 8053, 8302 96 
 and therefore, 3 times 8053, or 24159, still 1 
 
 remains to be divided by 97, and the quo- 
 
 tient to be added to 8503,the part of the quo- 8303 99 
 tient already found; this being divided by 97 
 
 100, the quotient is 241, and the remainder 
 59; then, as before, were 241 multiplied by 
 97 (100 3,) it is obvious that the product 
 would be less than 24100 by 3 times 241; whence, the reason of 
 multiplying 241 by 3 is evident: and thus the reason of all the 
 similar multiplications is shown. The sum of the partial remain- 
 ders is 196, from which 1 is carried for 100, because 97 is con- 
 tained once in 100, with the remainder 3; this remainder being 
 annexed to 96, the other part of the remainder, the sum is 99, 
 which exceeding 97 by 2, one is added to the quotient. Hence, 
 the true quotient and remainder are 8303, and 2. It is evident 
 that for each hundred in the remainder, in the present example, 3, 
 which is equal to the complement, must be added to the number 
 expressed by the units and tens of the remainder; and hence the 
 reason of completing the remainder by the product of the com- 
 plement and the number carried from the remainder to the quotient. 
 
 In long operations in Division, or 
 when the same divisor is to be fre- 73)341949853(4684244 
 
 quently employed, it is of fen of advan- 292' 
 
 tage to form a table containing the seve- 499 
 
 ral products' of the divisor and the nine 433 1... 73 
 
 digits* Thus, to divide 341949853 -^ 2". 146 
 
 by 73, let a table of the products of KCJ, 3. ..218 
 
 73 be formed, as in the margin; then, 4... 292 
 
 by looking in the table we find that 5".*365 
 
 the nearest product below 341 is 292, 2J '"" , 6^438 
 
 the product by 4; we place 4, there- 7. ..5 1 1 
 
 fore, in the quotient, and subtract 292 146^ s!..584 
 
 from 341; the remainder is 49, to 325 9. ..657 
 
 which 9 is brought down : we then 292 
 
 see in the table that 438, the product "~333~ 
 
 for 6, is the nearest product below 292 
 
 499 ; 6 is, therefore, put in the quo- 
 tient, and 438 subtracted from 499, &c. 
 
 * In Multiplication also, the multiplicand may be tabulated in like manner, when it 
 is to be frequently used. The advantage, however, is not so great as in Division. Such 
 a table may be readily formed by successive additions of the multiplicand in the one ca:>c. 
 and of the divisor in the other. 
 
SS ABBREVIATIONS IN DIVISION. 
 
 In the article on Division, the method of dividing by the factors 
 of the divisor, has been given. In most cases in which this me- 
 thod is useful, the factors are readily discovered by any one who 
 is well acquainted with the Multiplication Table. In some cases, 
 however, which may be of use, they are not so easily discovered ; 
 and, besides, the discovering of the factors of composite numbers 
 is a problem of some interest in the theory of numbers, in frac- 
 tions, &c. Hence, the following observations on this subject may 
 be found useful. 
 
 2 is a factor of every number which ends in 0, or in any of the 
 even* digits, 2, 4, 6, 8: thus, 2 measures 226; for, since it mea- 
 sures 10, and consequently the multiples of 10, it must measure 
 220; and measuring 220 and 6, it must measure their sum. It 
 may be proved in the same manner, that 5 is a measure of every 
 number that ends in or 5. 3 is a measure of every number, the 
 sum of whose digits is divisible by 3; and 9 is a measure of every 
 number, the sum of whose digits is measured by 9. (See p. 20. ) 
 
 A composite number, suppose 600, is decomposed into all its 
 prime, or simple factors in the following manner: Divide by 2, 
 (the least prime number;) divide the quotient, 300, 
 by 2, and the next quotient, 150, by -2, also; let 2600 
 the quotient, 75, which is not divisible by 2, be di- 2 300 
 vided by 3, (the next prime number ;) then the quo- 2 
 
 150 
 75 
 25 
 5 
 
 tient, 25, which is not measured by 3, being divided 3 
 by 5, (the prime number next in order,) and the 5 
 result by 5, we have the quotient 1, and the de- 5 
 composition is completed; COO being equal to the 
 continual product of 2, 2, 2, 3, 5, 5. If none of 
 the prime numbers, 2, 3, 5, 7, 11, &c. measure the given number, 
 it is prime. In this case it is unnecessary to carry the divison any 
 farther, when the quotients begin to be less than the divisors which 
 produce them. The decomposition will be often facilitated by the 
 remarks in the preceding paragraph : thus, 245 has evidently the fac- 
 tor 5; and dividing by it, we have 49, the factors of which are 7 and 
 7; and therefore the simple, or prime factors of 245 are 5, 7, and 7. 
 Having found the simple divisors of a number, we may thence 
 find all the divisors which it admits. 
 
 Thus, to resume the same number, 600, L_?__?__J 5 
 
 let all its simple factors, together with 1 i 4 8 3 5 25 
 unity, (which is a divisor of every whole 6 10 50 
 
 number,) be set as in the margin; then 12 20 100 
 
 set unity as the first divisor, and, for the 24 40 200 
 
 second, multiply it by 2, the second sim- 15 75 
 
 pie divisor,and the result is 2 for another 30 150 
 
 divisor. Then, the next simple factor CO 300 
 
 being 2, v/c multiply I and 2 (the divi- 120 600 
 
 * F-.'cn numbers arr those which are divisible by 2, without remainder; all other* 
 
 arc caticd odd number*. 
 
TABLES. 57 
 
 SOTS already found,) by it; the products are 2 and 4, the former 
 of which is rejected, as it is the same as one already found. We 
 next multiply the divisors, 1, 2, and 4, by the next simple factor 
 2, and reject the products 2 and 4, as they are the same as two 
 divisors already found. The next divisor is 3, and by this the di- 
 visors 1, 2,4, 8, are multiplied: the products, 3, 6, 12, 24, are 
 other divisors. Proceeding thus with the two remaining factors, 
 5 and 5, we find the divisors of 600 to be, J, 2, 3, 4, 5, 6, 8, 10, 
 12, 15, 20, &c. being twenty-four in number; and these are the 
 only divisors which it admits. 
 
 TABLES OF MONEY, WEIGHTS, MEASURES, &c.* 
 
 TABLE OF MONEY. 
 
 2 farthings =1 halfpenny 
 
 2 halfpence, or 4 farthings =1 penny rf.f 
 
 12 pence = 1 shilling ...s.or/ 
 
 20 shillings =1 pound 
 
 21 shillings =1 guinea.^ 
 
 Hence, a pound contains 240 pence, 480 halfpence, or 960 far- 
 things; and a guinea, 252 pence. A half-guinea is 10/6. 
 
 * Before the pupil proceeds to Redviction, the Compound Rules* &c. he should be 
 acquainted with the divisions of ihe money, weights, and measures which are most ge- 
 neniily used. The teacher will exercise his own judgment in determining which of the 
 Tables here given should be committed to memory: perhaps the parts that are printed 
 in the largest type may be sufficient. The Tables fixed on for this purpose should be se- 
 verally committed immediately before the pupil commences those part* of Reduction in 
 which they are respectively employed , and when he commences the corresponding parts 
 of the Coinpoi^d Rules, it will have a good effect to require him to revise them. 
 
 f This character, and others similarly placed, in this and some of the following Tables, 
 are u^l as abbreviations for the names that precede them. The marks of abbreviation 
 are generally omitted, when they are sufficiently understood without explanation. 
 
 Farthings were formerly denoted by q ; but now , annexed to pence, denotes a farth- 
 ing 4, a halfpenny, or 2 farthings; and f, three farthings. . s. d. and q. are the 
 initial letters of the Latin words, libra, solidus, denarius, and qttadrans; which denote 
 pound, shilling, penny, at d farthing, or quarter, respectively. The character / is a 
 corruption cf the long/, arising from rapidity in making it. 
 
 t Other coins (some of them mentioned chiefly in old authors,) are the groat, value 
 4d. ; the crown, 5s. ; the nob'e, 6s. Sd. ; the angel, 10*.; the mark, or tnerk, 15*. 4rf. ; 
 the pistole, about 16. lO.i. ; the moidore, 27s. 
 
 The guinea (so called because the gold of which the first guinea* were mwie, was 
 prorured from Guinea, in Africa) orgina'ly bore the figure of an elephant, as a > em. 
 blem of the place from which the gold was brought It passed at first for 2(k. British ; 
 but, in consequence of a scatcity of go'd, it afterwards rose to 21s 6d.: at a subsequent 
 poriod it fell to it- present value. 44 guineas weigh a pound troy, and hence the 
 wpSght of each is a Tittle more than 5 penny-weights, 9 grains. 'Ihe pound is so called, 
 because anciently the silver for it weighed a pound troy. 
 
 la ibv year 1825, the currency of Ireland was assimilated to that of Great !5i itain Tlfl 
 
38 TABLES. 
 
 TABLE OF TROY WEIGHT.* 
 
 24- grains, ~l penny-weight, dwt. 
 
 20 penny-weights = ] ounce, oz. 
 
 12 ounces rrl pound It). 
 
 Hence, an ounce contains 480 grains j and a pound, 240 penny 
 weights, or 5760 grains. 
 
 This weight was formerly used for weighing articles of every kind: 
 it is now used in weighing gold, silver, jewels, and liquors; and in 
 philosophical experiments. 
 
 It is also employed by apothecaries in mixing their medicines, though 
 they buy and sell them by avoirdupois weight. When troy weight is 
 thus used, it is called APOTHI-C ARIES' WEIGHT; but in this case the 
 ounce (?) is divided into 8 drams, ( 5 ) the dram into 3 scruples, ( '^ ) and 
 the scruple into 20 grains. 
 
 TABLE OF AVOIRDUPOIS WEIGHT. 
 
 
 ounce, oz. 
 
 pound, Jb. 
 
 quarter of a hundred, q. or qr. 
 
 hundred weight, cwt. or c. 
 
 ton 
 
 stone 
 cwt. 
 
 16 drams 
 
 16 ounces = 
 
 28 pounds = 
 
 4- quarters, or 112 tbs = 
 
 20 hundreds = 
 
 Also, 14 pounds*. = 
 
 8 stone 
 
 The hundred weight here mentioned is sometimes called the great 
 hundred, or the standard hundred, to distinguish it from hundreds of 
 different magnitudes, which are used in particular places. One of the 
 most general of these is the long hundred, or the hundred in long weight, 
 which contains 120 pounds: hence, 
 In Long Weight, 30 pounds avoirdupois ... rrl quarter; 
 
 4- quarters, or liO pounds rrl hundred weight. 
 The use of this weight, however, is now illegal. 
 
 that time, though accounts were kept in both countries in the same denominations, the 
 Irish pounds, shillings, pence, and farthings, were oflessvaiue than those of Britain, 
 13 of each denomination in the former country being equivalent only to 12 of the 
 corresponding denomination in the latter. Thus, the I ritish shilling was equivalent to 
 IS pence Irish ; and, while the guinea was 110 British, it was equivalent to l 2 9 
 Irish. 
 
 * Troy weight was introduced into Europe from Cairo in Egypt, about the time of 
 the Crusades, and was first adopted in Tro) es, a city in France, where great fairs were 
 held, and whence it has its name. 
 
 The weights and measures used in the British Empire, as well as in almost all other 
 places, were derived originally from very imperfect standards. As the origin of weights, 
 a grain of wheat was taken from the middle of the ear, and being well dried, was used 
 as a weight, and called by its original name, a grain. A weight equal to 32 grains wa 
 called a penny- weight, from its being the weight of the silver penny then in currency. A 
 weight equal to i!0 penny- weights, was called an ounce, a woid of the same origin (the 
 Latin word, uncia,} and import as the word inch, each signifying a tiuetfth-ptirt, the one 
 appropriated to denote the twelfth part of a pound, and the other the twelfth part of a 
 foot. At a later period the penny-weight came to be divided, not into 32, but into 24 
 
TABLES. 99 
 
 The stone, in the greater number of places, is 14 pounds, which alone 
 is the legal one; but in different parts of England it is of various mag- 
 nitudes, from 8 to 16 pounds. In Ireland, also, in the sale of some 
 articles, the stone of 16 pounds, or one-seventh of the standard hundred, 
 is used.* A ton of stones is 21 hundreds, Long weight. 
 
 TABLE OF LONG MEASURE. 
 12 lines =1 inchj- J 40 perches z=l furlong 
 
 8 furlongs n=l mile 
 
 3 miles =1 league. 
 
 12 inches zzl foot 
 
 3 feet = 1 yard 
 
 5^ yards =1 perch J 
 
 A fathom is 2 yards, or 6 feet; a hand, (used in measuring 
 horses,) is 4? inches; a span, 9 inches. 
 
 From this Table, by an easy reduction, it will appear that a mile con- 
 tains 320 perches, 1760 yards, or 5280 feet. 
 
 Till the year 1826, the perch in Ireland contained 7 yards instead of 
 5^, so that 11 Irish miles were equivalent to 14 British ones, and the 
 Irish mile contained 2240 yards, or 6720 feet. 
 
 TABLE OF CLOTH MEASURE. 
 
 4 nails =1 quarter j 4 quarters =1 yard. 
 
 A Flemish ell is 3 quarters of a yard; an English ell, 5 quarters, or a 
 yard and a quarter; and a French ell, 6 quarters, or a yard and a half. 
 
 Cloth measure is a species of long measure, and the yard is the same 
 in both. Hence, a quarter of a yard is 9 inches; and a nail, 2 inches 
 and a quarter. 
 
 TABLE OF SQUARE MEASURE, OR OF THE MEASURES 
 OF SURFACES. 
 
 144 square || inches = 1 square foot 
 
 9 square feet rzl square yard 
 
 30 square yards rrl square perch. 
 
 equal parts; each of which, however, was called a grain, though really a third part 
 greater than the original grain. 
 
 * In England, 14 pounds of wool=rl stone, 2 stone=l tod, 6 tods and a half=l wejr, 
 2 weys=l sack, 12 sacks=l last : ami a pack of wocl=210 pounds. 
 
 t 3 barleycorns make an inch. The barleycorn, however, is never employed now as 
 a measure. Instead, also, of being divided into lines, the inch U now generally divided 
 into tenths. 
 
 J The perch is also sometimes called a pole or rod. Each of the names given to this 
 measure is expressive of the instrument by which it was measured a rod, a pole, or 
 perche, a French word of the same import. In some counties of England the perch is 
 6 yards, in some 7 yards, and in others 8 yards. In Cunningham measure it is t>^ yards ; 
 in forest measure, 8 yards ; and in woodland, or Burlcigh measure, 6 yarrti. Nor>e of 
 these, however, is now legal. The yard is said to have been taken from the length of 
 the arm of Henry I. of England, and seems to be the origin of long measure. 
 
 I! A Square is a figure which has four equal sides, each perpendicular to the adjacent 
 ones A square inch i* a square each of whose sides is an inch in length ; a square yard, 
 a square, each of whose side:, is a yard in length, &c. The table of square measure if 
 
40 TABLES. 
 
 TABLE OF LAND MEASURE. 
 40 square perches =rl rood | 4 roods r=l acre.* 
 
 This is obviously a continuation of the Table of square measure. 
 
 TABLE OF CUBICf OR SOLID MEASURE, OR OF THE 
 MEASURES OF CAPACITY. 
 
 1728 cubic inches. =:! cubic foot | 27 cubic feet r=l cubic yard. 
 
 LIQUID MEASURE. 
 
 4 naggins, or gills 1 pint j 2 pints =1 quart 
 
 4 quarts, or 8 pints =1 gallon.^ 
 
 This is evidently a species of cubic measure. 
 
 Formed from the table of long measure, by multiplying each lineal dimension by itself; 
 thus, a square foot is=12x!2=144 square inches, &c. 
 
 * In measuring land, surveyors use a chain, which is 4 perches in length, and is d.- 
 vided into 100 equal parts, called links. They also compute by chains and links, but ex- 
 flibit the result in acres, roods, and perches. 10 square chain*, or 100,000 square links, 
 are an acre. It may be observed, also, that 640 acres are a square mile j and that a hide 
 of land, mentioned by old wiiters, is 100 acres. 
 
 In Irish measure, or, as it is often called, Irish plantation measure, 64 acres are equi- 
 valent to 49 acres in Forest measure ; 625 to 784 in Cunningham measure ; 56 to 41) in 
 Woodland, or Burle igh measure ; 121 to 196 in English Statute measure ; and 1369 to 
 1764 Scotch acres. These numbers are found by multiplying each of the numbers ex. 
 pressing the length of the perches, in the different kinds of measure, by itself. Thux, 
 the English perch being 11 half yards, and the Irish 14 half yards, we have 121=11x11, 
 and 196=14x14. Hence, to pi event a common mistake, it maybe proper to remark, 
 that the difference in the comparative magnitudes of the acres is much greater than that 
 of the perches or miles. The chain in Scotland, prior to 1826, was fixed at 74 feet and 
 hence, the chain in Ireland being 84 fert, we find the equivalent numbers for Irish and 
 Scotch acres, by multiplying the halves of 74 and 84, respectively, by themselves. 
 
 From these principles the following Table is constructed, which is useful in reducing 
 Irish measure to any other. The first line shows the quantities equivalent to 1 Irish acre, 
 the second those equivalent to 10, &c. ; and by means of Compound Multiplication, and 
 Compound Addition, the values of other numbers of acres may be readily found. Thus, 
 for 729 Irish acres, add together 7 times the numbers in the third line, twice those in the 
 second, and 9 times those in the first. It may be observed that, now, the only legal 
 measure is the English Statute measure 
 
 Irish. 
 
 1'orut. 
 
 Cunningham. 
 
 Scotch. 
 
 n'ooJlantt. 
 
 Englit/i. 
 
 a. 
 
 a. r. p.lOtte- 
 
 a. r.p.lOthi. 
 
 a.r. p.lVthi. 
 
 a. r. p.lOtto. 
 
 a. r. p. Kit ** 
 
 I 
 
 0325 
 
 1107 
 
 1162 
 
 1 1 17 8 
 
 1 2 19 2 
 
 10 
 
 7 2 25 
 
 12 2 7 
 
 12 3 21 6 
 
 13 2 17 8 
 
 16 31 7 
 
 100 
 
 76 2 10 
 
 125 1 30 -i 
 
 128 3 16 5 
 
 136 17 8 
 
 161 3 37 4 
 
 1000 
 
 7oS 2 20 
 
 1254 1 24 
 
 1288 2 5 1 
 
 1361 17 8 
 
 1619 5 13 o 
 
 f A cube is a figure contained by six equal squares. (Dice afford a familiar instance 
 of this figure.) A cubic inch is a cube whose sides are each a square inch ; a cubic foot, 
 a cube whose sides are each a square foot, &o. It may be remarked, that 1728 is equal 
 to 12x12x12, and 27=3x3x3. 
 
 t Tht: English hogshead, in wine measixre, contains 65 gallons ; the pipe, 2 h 
 
 jogbhcads, 
 
TABLES. 
 
 TABLE OF DRY MEASURE. 
 
 2 pints =1 quart 
 
 2 quarts =1 pottle 
 
 2 pottles, or 
 
 4 quarts 
 
 I =1 galh 
 
 -2 gallons =1 peck 
 
 4 P^ or i = 1 bushel 
 
 8 gallons $ 
 
 8 bushels z=l quarter 
 
 5 quarters =1 wey 
 
 2 weys, or 10 qrs..z=l last. 
 
 This measure, which is another species of cubic measure, is used in 
 measuring grain, seeds, and various kinds of dry articles. In many 
 places, however, these are bought and sold by weight. 
 
 TABLE OF TIME. 
 
 60 seconds.. rrl minute 
 
 60 minutes = 1 hour 
 
 24- hours rrl day 
 
 7 days rrl week 
 
 52 weeks and 1 day, or 365 days =1 common year 
 
 52 weeks and 2 days, or 366 days z=l leap year. 
 
 The year is divided into 12 portions, called calendar months, the names 
 of which are January, February, March, April, May, June, July, 
 August, September, October, November, December. Of these, April, 
 June, September, and November, have 30 days each; and the rest, ex- 
 cept February, have 31 days each. In leap years* February has 29 
 days; in common years, 28 days; } so that a leap year contains 366 days, 
 and any other 365. The precise length of the year is found to be 365 
 day:*,, 5 hours, 43 minutes, 48 seconds: it is, therefore, 365 days, 6 
 hours, nearly. 
 
 or 12fi gallons ; and the tun, 2 pipes, or 4 hogsheads, or 252 gallons. In the measure of 
 foreign wines, however, ihere are great varieties. A tierce (the third part of a pi|>e) 
 is 42 gallons, and a puncheon is 2 tierces, or 84 gallons. 
 
 In beer measure, 9gallons=l firkin; 2 firkins, or 18 gallons^l kilderkin ; 2 kilder- 
 kins, or 36 gailonsrrl barrel 54 gallotis=il hogshead . 2 hogsheads, or 108 gallons 1 butt 
 
 The hothead of ale, in London, contains 48 gallons; the barrel, 32; the kilderkin, 
 16; and the firkin, 8. 
 
 In England, in the country, both in ale and beer measure, the hogshead contains 51 
 gallons; the barrel, 34 ; the kilderkin, 17 the firkin, 4 8}. 
 
 * Leap years occur at intervals of 4 years, and may be known by dividing by 4 the 
 number expressed by the last two figures in the number of the year, according to the 
 Christian era : if there be no remainder, it is leap year ; otherwise, the remainder .shows 
 how rmny ye.irs it is after le.ip year. To this there is one exception, as the exact cen 
 tnrie^ are not leap years, except when the number of centuries is divisible by 4, without 
 remainder Thus, the year 1820 was leap year, because 20 is divisible by 4 but 1833 was 
 the third year after leap year, because 3 remain when 23 is divided by 4. Aiso, the year 
 000 will be a leap year, but 1900 not, as 20 is divisible by 4, but 19 not. 
 
 f Learners may easily remember the number of days contained in each month, hy 
 recollecting that the months are long and short alternately t with the exception of August, 
 which u long, while the months after it follow the rule. 
 
42 TABLES. 
 
 TABLE OF THE DIVISION OF THE CIRCLE. 
 
 The circumference of every circle is supposed to be divided into 
 360 equal parts, called degrees;* each degree is subdivided into 
 60 equal parts, called minutes; and each minute into 60 equal 
 parts, called seconds. 
 
 MISCELLANEOUS TABLE. 
 
 12 articles =rl dozen 
 
 12 dozen =1 gross 
 
 20 articles rzl score 
 
 
 5 score =1 hundred 
 
 6 score or 120.. ^ 
 
 { long hundrei 
 24 sheets of paper ml quire 
 
 20 quires ........... =1 ream. 
 
 An Act of Parliament, already referred to, " for Ascertaining and 
 Establishing Uniformity of Weights and Measures," came into opera- 
 tion in January, 1826. By this act, "the standard yard of 1760," in 
 custody of the Clerk of the House of Commons, is to be the standard 
 yard, when the temperature is at 62 of Fahrenheit's thermometer; and 
 this is to be the origin of all other measures of length. The thirty-sixth 
 part of this yard is an inch; and the length of a pendulum vibrating 
 seconds in the latitude of London is found to be 39*1393 such inches; 
 that is, according to the notation of decimal fractions, 39 inches and 
 1393 ten thousandths of another. This affords the means of recovering 
 the standard yard, should it be lost. 
 
 It is enacted, in like manner, that "the Troy pound of 1 758," in custody 
 of the same officer, shall continue to be the standard unit of weight; 
 and, tbis pound containing 5760 grains, the avoirdupois shall contain 
 7000 grains. Hence, 144 pounds avoirdupois are equivalent to 175 
 pounds troy; and 1 Ib. avoirdupois is equal to 1 Ib. 2 oz. 11 dwts. 
 16 grs. troy; but the ounce troy exceeds the ounce avoirdupois by 
 42^ grains. The weight of a cubic inch of distilled water is 252*458 
 grains troy, the barometer being at 30 inches, and the thermometer at 
 62 ; and thus there is a means of recovering the standard pound, if it 
 should be lost. 
 
 It is farther enacted, that the standard measure of capacity for all 
 liquids, and for dry goods not measured by heaping, shall be a gallon 
 containing 10 pounds avoirdupois of distilled water in the same state 
 
 * Degrees, minutes, and seconds are marked thus : , ', ". Hence, the expression, 
 4i 24* 54", is read, 41 degrees, 24 minutes, 54 seconds. The reason of these marks being 
 employed will appear evident from the consideration, that minutes and seconds ate 
 only abbreviated expressions for first minutes, or minutes of the./?/ order, and second 
 minutes, or minutes of the second order; mimttes, in each instance, signifying small 
 parts. It may be proper to remark, that the circumference of a circle is the line which 
 contains it ; that all straight lines drawn from the centre to the circumference are equal ; 
 that any of these lines 'is called a radius; and that a line drawn through the centre, and 
 terminated both ways by the circumference, is called a diameter. It may be farther re- 
 marked, that seconds, both in time and in the circle, were formerly divided each into 00 
 thirds, but that they are now divided into tenths or hundreds. 
 
REDUCTION. 43 
 
 of the barometer and thermometer. This gallon, therefore, contains 
 277'274 cubic inches; and the bushel weighs 80 pounds, and contains 
 2218*192 cubic inches. It is a curious coincidence, that the cube of 
 one-sixth of the length of the seconds' pendulum is so nearly equal to 
 the gallon, as to exceed it by only about three-tenths of an inch. 
 
 In Heaped Measure, the diameter is to be at least double the depth, 
 and the height of the cone or heap is to be three-fourths of the depth. 
 The measures are to be made cylindrical, with a plain and even bottom; 
 and the outside of the measure is to be the extremity of the base of the 
 cone. The internal capacity is to be the same as that of the correspond- 
 ing liquid measure. The bushel is to be the standard of this measure, 
 and its diameter is to be 19^ inches from outside to outside.* 
 
 REDUCTION. 
 
 WHEN a given quantity is expressed in any denomina- 
 tion, REDUCTION shows the method of expressing it in an- 
 other. 
 
 Reduction consists of two branches, or problems ; one in which 
 quantities are reduced to a lower denomination ; the other in which they 
 are reduced to a higher. The former is often called REDUCTION DE- 
 SCENDING, and the latter REDUCTION ASCENDING. In the former Mul- 
 tiplication is employed ; in the latter Division, according to the following 
 rules : 
 
 Problem 1. To reduce a quantity to a lower denomination. 
 RULE Multiply the number which expresses the quan- 
 tity by the number which shows how many of the lower 
 denomination make one of the higher; and if any part of 
 the given quantity be already of the lower denomination, 
 add to it the product. 
 
 Problem 2. To reduce a quantity to a higher denomina- 
 tion. 
 
 RULE. Divide the number expressing the quantity, by 
 the number which shows how many of the denomination 
 in which it is, make one of the higher denomination : the 
 quotient will be of the higher denomination, and if there 
 be any remainder, it will be of the lower. 
 
 When there are intermediate denominations between that 
 in which the quantity is given, and that to which it is to 
 
 * It may be useful for the more advanced pupil to know, that the o'd French foot is 
 12'78933 English inches, or 12 inches, 9| lines, nearly ; the toise, or fathom, or 6 French 
 feet, 6 feet, 4 736 inches, English, or 6 feet, 4 inches, 9 lines, nearly; and the metre* 
 59-3702, or 1 yard, 3 inches, and 3 eighths, nearly. 
 
44 REDUCTION. 
 
 be reduced, it is generally better to reduce it by successive 
 steps ; first, to one or more of the intermediate denomina- 
 tions, and then to the required denomination. 
 
 Methods of Proof. 
 
 1. To the answer found by either of the preceding rules, apply 
 the other rule, and if the result be the same as the given number, 
 die work is conrect. 
 
 2. The operation may be proved by the methods of proving 
 Multiplication or Division. 
 
 The reasons of the rules will he explained in the illustrations of the 
 examples. 
 
 REDUCTION OF MONEY. 
 Exam. 1. Reduce 59 to farthings. 
 
 In this example the pounds are multi- 
 plied by 20, to reduce them to shillings, 
 because there are 20 shillings in each 
 pound. The shillings, in like manner, 
 are multiplied by 12, to reduce them to 
 pence, and the pence by 4, to reduce 
 them to farthings, because there are 12 
 pence in each shilling, and 4, farthings in 
 each penny. Hence, it appears, that in 
 59 there 1180 shillings, 14160 pence, or 
 56640 farthings. The operation is proved 
 by dividing successively by 4, 12, and 20, 
 the former multipliers, in a reversed order: 
 rect, since the final quotient, 59, is the same as 
 given in the question. 
 
 Exam. 2. Reduce 94 12s. 8d. to farthings. 
 
 In this example in the 
 multiplication by 20, 12 
 shillings are added to the 
 product; in the multipli- 
 cation by 12, 8 pence are 
 added, and in the multipli- 
 cation by 4, 1 farrffing is 
 taken in. Hence, the an- 
 swer is 90849 farthings. 
 
 
 59 
 20 
 1180 shillings. 
 
 12 
 14160 pence. 
 
 4)56640 farthings 
 12)14160 pence. 
 2|0) 11810 shillings. 
 59, proof. 
 
 and the work is cor 
 the quantity 
 
 d. 
 
 s. 
 
 94 12 8 
 
 20 4)90849 
 
 1 892 Shillings 1 2)22712 j 
 
 12 2lO)'Tfe9]2 8^ 
 
 22712 pence 94 12 8^, proof 
 
 4 
 
 90849 farthings. 
 
 Exam. 3. Reduce 83918 farthings to pounds, &c. 
 In this example, the farthings are di- 
 vided by 4, because in the same sum there 
 are 4 times as many farthings as there are 
 pence: for a similar reason, the pence 
 are divided by 12, to reduce them to shil- 
 lings, and the shillings thus found, by 20, 
 to reduce them to pounds. Hence, it ap- 
 pears that 83918 farthings are equivalent 
 
 4) 83918 farthin ; 
 12) 20979 
 2|0) 174|8 3. 
 67 8 3 
 
 In* 
 
REDUCTION. 45 
 
 to 20979 pence, with two farthings, or a halfpenny; to 174-8 shil- 
 lings, and 3 pence halfpenny; or finally, to 87 8 3. The ope- 
 ration would be proved by reducing the answer to farthings, in the 
 manner exhibited in the foregoing exam-pie. By this means we 
 should obtain 83918, which being the same as the given number 
 of farthings, it follows that the answer is correct. 
 
 Exercises. Answers. 
 
 1. Reduce 341 4 to pence 81844- 
 
 2. 97 17 3 to halfpence 46975 
 
 3. 783 2 3 to farthings 751789 
 
 4. 481 to pence 115440 
 
 5. 33333 pence to pounds 138 17 9 
 
 6. 1023 16 to shillings 20476 
 
 7. 113 15 3 to pence 27303 
 
 8. 1 2 9 to pence 273 
 
 9. 2300 pence to pounds 9118 
 
 10. 1 9 3 to pence 351 
 
 11. 463 19 7 to farthings 4454-22 
 
 12. 1 14- U to farthings 1638 
 
 13. 95283 ^aWpence to pounds 198 10 1 
 
 14. 38 14 to pence 9288 
 
 15. 133 6 8 to farthings 128000 
 
 16. 7853 19 llf to farthings 7539839 
 
 17. 47589 pence to pounds 198 5 9 
 
 18. 1234567 farthings to pounds... 1286 1| 
 
 19. 53 14 to farthings 51552 
 
 fcO. ~-~~~ 75396 shillings to pounds ....*.. 3769 16 
 
 21. 13/4 to pence 160 
 
 V 22. 47474747 farthings to pounds... 49452 17 2| 
 
 23. .. . 348 guineas to pence 87696 
 
 24. 967 guineas to pounds* 1015 7 
 
 25. 497 12 8 to farthings 477728 
 
 26. 69173 pence to guineas 274 gs. and 10/5 
 
 27. 1000 to guineas 952 gs. and 8/0 
 
 28. 371 14 10 to halfpence 178436 
 
 29. .~~_ 823903 farthings to guineas 817 gs.and 7/7f 
 
 * To reduce guineas to pound*, multiply them by 21t the product is shillings which 
 reduce to pounds. To reduce pounds to guineas, reduce them to shillings, and ... do 
 those .shilling.-, by 21. 
 
4* REDUCTION. 
 
 REDUCTION OF TROY WEIGHT. 
 
 Exam. 4. Reduce 1 Ib. 5 oz. 12 dwts. 13 grains, to grains. 
 
 lt> oz. dwt. grs. 
 
 In this example the multi- 1 5 12 13 
 pliers are 12, 20, and 24, be- 12 
 
 cause in each pound there 24)8461 
 
 are 12 ounces, in each ounce 17 
 
 20 penny-weights, in each 20 2|0)35|2 13 
 
 penny-weight 24 grains; and 
 
 the 5 ounces, 12 penny- 352 12)17 12 13 
 
 weights, and 13 grains, are 24 
 
 taken in, as hi the former Ib i 5 12 13, 
 
 examples. 1421 Proof. 
 
 704 
 
 8461' grains, Answ. 
 
 Exam. 5. Reduce 111111 grains to pounds. 
 
 In this example the grains are grains 
 
 divided by 24, (or by 6 X 4,) and 6)1 1 1 1 1 1 
 the result is 4629 penny-weights, 
 
 15 grains; these penny- weights 4) 18518 3 
 
 are again divided by 20, and the 
 
 result is 231 oz. 9 dwts. 15 grs.; 2|0) 462|9 15* 
 
 and these ounces being divided by 
 
 12, the final result is 19lbs. 3 oz. 12)231 9 15 
 
 9 dwts. 15 grs. the answer requir- . 
 
 ed. The operation would be prov- Ibs. 19 3 9 15, 
 
 ed in the manner in which the last Answ. 
 example was wrought. 
 
 ' 
 Exercises. Answers. 
 
 30. Reduce lloz. 12dwts. 12grs.tograins.5580 
 31. ~_~~~ 3 Ibs. 7 oz. to penny-weights.860 
 
 32. ,....... 1785 dwts. to pounds 7 Ibs. 5 oz. 5 dwts. 
 
 33. ,~.~-~ 1785 grains to ounces 3 oz. 14 dwts. 9 grs. 
 
 34 785398 grains to pounds 136lbs.4oz.4dwts.22grs. 
 
 35. ....- 29 pounds to grains ...^ 167040 
 
 , \ 
 
 For the irotle of fijidhig the true remainder here, see page 577. 
 
REDUCTION. 
 
 47 
 
 REDUCTION OF AVOIRDUPOIS WEIGHT. 
 
 Exam. 6. Reduce 27 cwt. 2 qrs. 22 Ibs. to pounds. 
 
 cwt. qrs 
 27 2 
 
 Ibs. 
 22 
 
 110 
 28 
 
 902 
 220 
 3102 pounds, Answ. 
 
 4)3102 
 7J "775 2 
 4) _HO 22 
 cwt. 27 - 
 proof. 
 
 In this example, the hun- 
 dreds being multiplied by 4, 
 the product is quarters, be- 
 cause there are 4 quarters 
 in each hundred; and the 
 quarters being multiplied by 
 28, the product is pounds, 
 because in each quarter there 
 are 28 pounds. The odd 
 quarters and pounds are 
 taken in, as the operation proceeds. 
 
 Hundreds may be very easily reduced to pounds by the following rule, 
 derived from the principles explained in pages 32 and 33: 
 
 Multiply the hundreds by 12, without writing the multi- 
 plier, and, as the work proceeds, take in the odd pounds, 
 and 28 pounds if there be 1 quarter, 56 pounds if there be 
 2, or 84 pounds if there be 3 : set the product below the 
 hundreds, two places to the right hand, and add the two 
 lines together. 
 
 Thus, in working the preceding example 
 in this manner, 27 is multiplied by 12, and 
 the product increased by 22 and 56, and the 
 result, 402, being added to 2700, (which 
 is evidently done in consequence of the 
 
 cwt. qrs. Ibs. 
 27 2 22 
 402 56 
 3 1 02 pound s, Answ. 
 
 way in which the numbers are set,) the sum, 3102 Ibs. is the an- 
 swer, the same as before. 
 
 Exam. 7. Reduce 591241 pounds to tons. 
 
 tn this example, as in 
 the proof of the last, the 
 pounds are divided by 28, 
 (or, which is equivalent, 
 by 4, and the quotient by 
 7,) to reduce them to quar- 
 ters ; the quarters by 4 to 
 reduce them to hundreds; 
 and the hundreds by 20 
 to reduce them to tons. 
 From the operation, it 
 appears, that 591241 
 pounds are equivalent to 
 21115 qrs. 2 1 Ibs.; or to 
 
 Ibs. 
 4)591241 
 
 7)147810 1 
 
 263 18 3 
 20 
 
 5278 
 
 21 
 
 4) 21115 21 
 
 TJ 
 
 
 
 
 21115 
 
 2|0) 527|8 3 21 
 
 28 
 
 Tbw*263 18 3 21, 
 
 168941 
 
 A*W. 
 
 42230 
 
 
 Ibs. 591241, proof. 
 5278 cw't. 3 qrs. 21 Ibs.; or to 263 tons, 18 cwt. 3 qrs. 21 Ibs. 
 
REDUCTION. 
 
 36. 
 
 37 
 
 38 
 
 39 
 
 40 
 
 41 
 
 42 
 
 43 
 
 44 
 
 45 
 
 46 
 
 47 
 
 48 
 
 49 
 
 50 
 
 51. 
 
 52. 
 
 Exercises. 
 Reduce 137 tons to hundreds ............... 2740 
 
 47 cwt. 3qrs. 241bs. to pounds .... 5372 
 
 135 cwt. 3qrs. lllbs. to pounds... 15215 
 484cwjt. Iqr. 20 Ibs. to pounds ... 54256 
 _ 13 Ibs. 4oz. 5drams, to drams .... 3397 
 
 _35cwtt 2qrs. 191bs. to pounds ... 3995 
 313cwt. Iqr. 25lbs. to pounds ... 35109 
 1 ton to ounces . .................... 35840 
 
 214cwt. Sqrs. to poun&s .......... 24052 
 
 94cwt. Iqr. lllbs. to pounds .... 10567 
 
 59 tons, 11 cwt. 8 Ibs. to pounds... 133400 
 
 57386 pounds to hundreds ........ 512cwt. 1 q. 141bs. 
 
 57386 quarters to tons ............ 717t. 6 cwt. 2qrs. 
 
 57386 drams to hundreds ......... 2 c. 2 oz. 10 drams. 
 
 1000000 pounds to tons ............ 446 1. 8c. 2q. 8lbs. 
 
 1000000 ounces to hundreds ...... 558cwt. Oqr. 4 Ibs. 
 
 38489 Ibs. to cwt. long weight ...... 320 cwt. 2q. 2M !U. 
 
 40865 pounds to hundreds ......... 364-cwt. 3q. i 
 
 92950 pounds to tons ............... 41 1. 9c. 3q. IS ibs. 
 
 238 c. 2q. 10 Ibs. long wt. to Ibs. . 28630 
 
 783914 pounds to tons ............. 349 1. 19c. 20 Ibs. 
 
 Exam. 8. Reduce 53 miles, 3 furlongs, 12 perches, 4 yards, 
 to yards. 
 
 In this example, because the 
 English perch contains 5 yards, 
 the perches are multiplied by 5, 
 and half of them taken, and the 
 4 yards are added in with the 
 two results. It may be observ- 
 ed, that if the last figure be re- 
 jected from the product by 5, 
 there will remain half the num- 
 ber of perches; and if the 
 figure rejected be 5, it will be 
 half a yard, or a foot and a 
 naif. This example would be 
 proved m the itr-t^er in which 
 
 M. F. P. Y. 
 
 53 3 12 4 
 8 
 
 427 furlongs. 
 40 
 
 17092 perches. 
 
 85460 
 8546 
 
 54010 vards, Anew. 
 
REDUCTION. 49 
 
 Exam. 9. Reduce 231278 yards to mile*. 
 In thi^ example, the yards 
 are multiplied by 2, to re- Yards, 
 
 duce them to half yards, and 231278 
 
 the result divided by 11, the 2 
 
 half yards in a perch, to re- 
 
 duce'it to perches. The re- 11)462556 half yards. 
 
 mainder is 6 half yards, or 3 
 
 yards; and the answer is found 4[0)4205|0 6 half yards, or 3yds. 
 
 to be 131 miles, 3 furlongs, 
 
 10 perches, and 3 yards. 8)1051 10 perches. 
 
 The proof would be perform- 
 
 ed in the same manner in Miles 131 3 10 3, Ansu>. 
 which the last example was 
 
 wrought. The pupil should be required to perform it. Had the 
 remainder been 7 half yards, instead of 6, it would have been 
 equivalent to 3 yards, and 1 foot, 6 inches. Hence, instead of 3 
 yards at the end of the answer, we should have had 3 yards, 1 foot, 
 and 6 inches. 
 
 Exercises. Answers. 
 
 57. Reduce 51 feet 4 inches to inches ... 61 C 
 
 os 3 miles 3 furlongs to yards... 5940 
 
 59. 94 miles, 1 fur. 6 per. to per. 30 ! 26 
 
 60 751823 feet to leagues 471. 1m. 3f. 4 p. 5y. 2f. 
 
 61 758246 inches to miles llm. 7f. 29p. 2y. 2f. 8i 
 
 62. 571 leagues 2 miles to miles 1715 
 
 G3 57391 1 yards to miles 326m. Of. 27p.2y. If. Gi. 
 
 64 23456 feet to miles 4m. 3f. 2)p. 2y. 3f. 6i. 
 
 65 25 miles and 34 per. to feet... 13256 1 
 
 C6 1000000 inches to miles 15m. 6f. 10p.2y.2f. 4i. 
 
 67 100 miles to inches 6336000 
 
 REDUCTION OF CLOTH MEASURE. 
 
 Excrcixes. Answers. 
 
 68. Reduce 28 yards, 3 qrs. 2 nails to nails*...462 
 
 69. 5247 nails to English ells 2G2 e. 1 q. 3 n. 
 
 70 58 yards to nails 928 
 
 71. , 285 nails to yards 17 y. 3 q. 1 n. 
 
 72 1241 English ells, 2 qrs. 3 nails, ? ^g^ 
 
 to nails ) ' 
 
 73 '4'iB yards to English ells 374 a. 2 q. 
 
 71. 15GO" Flemish ells to vurds..: 1170 
 
 \ 
 
 * MuKip'v tli? ' irds by 4, and take in the quarters ; and \r.\ 
 
 k the ncxrexrr<-:-e, divide the naiis l>*i; the ., 
 rfwer. 
 
50 REDUCTION. 
 
 Exercises. Answers. 
 
 75. Reduce 1123 English ells to French ells 935e. 5q. 
 
 76 985321 inches to yards 27370 y. lin. 
 
 REDUCTION OF SQUARE MEASURE. 
 
 Exercises. Answer*. 
 
 77. Reduce 245 square perches to square inches .. 9604980 
 78 ^^^ 1325419 square inches to sq. yds 1022 y. 6f. 43 i. 
 
 REDUCTION OF LAND MEASURE. 
 
 A. R. P. 
 
 37 3 12 
 4 
 
 151 roods. 
 Exam. 10. Reduce 37 acres, 40 
 
 3 roods, 12 perches, to square 
 
 yards. 6052 perches. 
 
 30^ 
 
 181560 
 1513 
 
 183073 yards, Anstff. 
 Exam. II. Reduce 111111 square yards to acres. 
 
 yards, 
 llllll 
 
 4 
 
 C 1 1)444444 qrs. of yards. 
 
 131 J 
 
 ( 11)40404 
 
 4lO)367|3p. 11 qrs. or 2} yards. 
 4)91 r. 33 p. 2f yards. 
 
 Answ. 22 a. 3r. 33 p. 2f yards. 
 
 In this example, to find the acres, the yards are reduced to 
 quarters, and divided by 121, the quarters in a perch, or 30| yards. 
 The remainder is 11 quarters of a yard, or 2 yards and 3 quarters. 
 
 Exercises. Ansivers. 
 
 79. Reduce 234 acres, 1 rood, 13 per. to per 37493 
 
 80 93827 perches to acres 586 a. Ir. 27 p. 
 
REDUCTION. 51 
 
 Exercises. Answers. 
 
 I. Reduce fi^'^ns, 8_hm|j yd,. j ^ ^ ^ ^ 
 
 82. ~~^^~ 256 acres, 15 perches, to yds. . 1 239493| 
 
 REDUCTION OF TIME.* 
 Reduce Exercises. Answer*. 
 
 83. 17 days to minutes 24480 
 
 84. 1 day, 4 hours, 12 seconds, to seconds 100812 
 
 85. 12345678 seconds to days ; 142d.21h.21m.18s. 
 
 8G. 34652 hours to weeks 206 w. 1 d. 20 h. 
 
 87. Eleven millions one hundred seconds to days... 127 d. 7 h. 35 m. 
 
 31 days in May 
 
 _8 
 
 Exam. 12. How many days 23 May 
 
 are there from the 8th of May 30 June 
 
 till the 23d of July ?f 23 July 
 
 76 days, Answ. 
 
 88. How many days are there from the 12th of August, 1817, 
 till the 24th of April, 1818? Answ. 255. 
 
 89. How many days are there from the Sth of January, 1816, 
 till the 12th of December in the same year? Amw. 339. 
 
 90. How many days are there between the 17th of March, and 
 the 25th of December ? Answ. 283. 
 
 91. In the mint in London there are eight coining presses, 
 which, with a child to supply each, strike 19000 coins in an hour. 
 Now, if these were employed 12 hours each day, for 313 days, in 
 coining halfpence, what would be the number and the value of 
 all the pieces coined during that tune? Answ. Number, 71364000; 
 value, 148675. 
 
 * Reduction of cubic measure, of liquid measure, and of dry measure, Ls so cr. -; - , 
 af;cr what has gone before, that it is unnecessary to give particular sets of exercises 1 jr 
 each Und. 
 
 f This, and the simi'ar questions which follow, though not strictly of tne same kind 
 as the others in Reduction, are inserted on account of their utility. 
 
 J Questions of the following kind are frequently useful : 
 
 1. If the Sth of August be on Saturday, on what day of the week will the 1st of N - 
 vembcr be ? 
 
 The number of days between these dates is found to be 85, which being divided by ',', 
 becomes 12 weeks and 1 day ; counting one day, therefore, after Saturday, we find th;.; 
 the first of November must be on Sunday. 
 
 2. If a common year begin on Friday, on what day will the 18th of June, the anniver- 
 sary of the battle of Waterloo, happen ? AKSVU. Friday. 
 
 3. The 9th of July, 1824, was on Friday; on what day did the year commence. Ansia. 
 On Thursday. 
 
 4. If a leap year commence on Wednesday, what day of September will be the fir-i 
 Monday of that month ? APSIU. 7lh. 
 
 Here, let ihe day be found on which September will commence, and the rest b easy. 
 
52 REDUCTION. 
 
 92. The quantity of linen exported from Ireland to the United 
 States in 1806, was two millions, six hundred and seventy-five 
 thousand, six hundred and nineteen yards. How many miles in 
 length was the whole? Answ. 1520m. If. 36 p. ly. 
 
 93. Required the amount of a collection of one halfpenny each, 
 from the inhabitants of the counties of Down and Antrim. (See 
 exercise 24, page 11.) Answ. .1248 6 10. 
 
 94-. "In the year 1815, Catharine Woods, of Dunmore, near 
 Ballynahinch, then about thirteen years of age, spun a hank of 
 linen yarn, of 12 cuts, each cut 120 threads, each thread two yards 
 and a half, which weighed ten grains," being at the rate of 700 
 hanks to the pound avoirdupois.* It is required to determine the 
 ength into which, at this rate, a pound of flax would have been 
 extended. (See Stuart's "Historical Memoirs of the City of Armagh" 
 page 424.) Answ. 1431 m. 6 f. 21 p. 4 y. 
 
 95. In the salmon fishery on the Bann, near Coleraine, 320 tons 
 of salmon were taken in 1760. How many stones of 14 pounds, 
 are in this quantity? Answ. 51200. 
 
 96. The quantity of cotton wool imported into Ireland, in 1811, 
 was 53,133 cwts. of which 16,148 cwts. were exported. Required 
 the number of pounds in each of these quantities, and in the quan- 
 tity consumed in the country. Answ. 5950896 Ibs.; 1808576 Ibs.; 
 and 41 42320 Ibs. 
 
 97. Suppose one person to lie in bed nine hours each day at an 
 average, and another only six hours and a half; and suppose the 
 latter to employ the time thus gained in reading and study, for 
 forty years; to how many years* study, of 12 hours each day, 
 would the entire time gained be equivalent? Answ. 8years, 121 day*, 
 8 hours. 
 
 98. The whole surface of the terraqueous globe contains 
 196,649,494 square miles; and of these, Europe is supposed to 
 contain 4,456,065. Required the number of acres contained in 
 each. Answ. 125855676160, and 2851881600. 
 
 99. In what time would a body move from the earth to the 
 moon, at the rate of thirty-one miles per day; the mean distance 
 being 238,545 miles? Answ. 21 years, 30 days. 
 
 100. In what time would a body, moving with the velocity of 
 sound, pass from the earth to the sun, the distance being 95 mil- 
 lions of miles? (See Ex. 57, page 22.) Answ. 14y. 27 d. 15 h. 
 50m. 
 
 * For this extraordinary, and perhaps unequalled performance, a premium of fifteen 
 guineas was awarded by the Linen Board of Ireland. 17J Ibs. of such yarn would cdto- 
 tain a thread more than equal to the circumference of the earth. Stuart's Memoirs. 
 
 The same year, another premium, of the same amount, was awarded to Elizal>f>th 
 Hl'Cance of the same neighbourhood, for having spun a hank which weighed 12 grains, 
 and which was consequently at the rate of (7000 grains-rl2) 583 hanks and 4 cuts in 
 pound. 
 
 
COMPOUND ADDITION. 53 
 
 101. In how long time would a cannon ball, with the velocity 
 of 1960 feet per second, move from the sun to the Georgium 
 Sidus? (See Ex. 25, page 5.) Answ. 154 years, 224 days, 5 hours, 
 46min, 7 T V 6 9 o sec - 
 
 102. By the latest measurements, the earth's mean diameter is 
 fouad to be 7911 miles and three quarters, nearly. How many 
 yards, feet, and inches, are contained in it? Answ. 13924680 yds.; 
 4 1 774040 feet ; 50 1 288480 in. 
 
 103. In England there are 50,535 square miles; and in Wales, 
 7425 such miles. Required the number of acres contained in both. 
 Answ. 37,094,400. 
 
 104. In the city of Pekin there are said to be seven bells, each 
 weighing 120,000 Iba ; and in Moscow there is one bell which 
 weighs 127,836 Ibs.; another which weighs 288,000 Ibs.; and a third 
 which weighs 432,000 Ibs. Required the weight of each in tons. 
 Answ. 53 t. lie. 1 q. 20 Ibs.; 57 t. 1 c. 1 q. J6 Ibs.; 128 t. 11 c. 
 1 q. 20 Ibs.; and 192t. 17c. 16 Ibs. 
 
 COMPOUND ADDITION.* 
 
 RULE. (1.) Arrange the given quantities so that those 
 in each column may be of the same denomination. (2.) 
 Add the numbers of the lowest denomination together; re- 
 duce their sum to the next higher denomination ; set the 
 remainder below the column added, and carry the quotient 
 to the next. (3.) Proceed thus with all the other denomi- 
 nations, except the highest, which is to be added in the 
 same manner as numbers in Simple Addition. 
 
 Either of the two first methods of proof given in Simple Ad- 
 dition, may be employed in Compound Addition. 
 
 Exam. 1. 
 
 In this example, the sum of the farthings . *. d. 
 
 is 14, which being divided by 4, the farthings 39 18 7 
 
 in a penny, the quotient is 3 pence; and 51 12 4i 
 
 the remainder, 2 farthings, or a halfpenny, 79 19 n 
 
 is set down. The quotient, 3, is then ad- 87] 
 
 ded with the pence; the sum, 54, being di- 43 13 9j 
 
 vided by 12, the pence in a shilling, the 375 16 10 
 
 quotient is 4 shillings, with a remainder of 
 6 pence, which is set down. The quotient, Sum, 599 9 6Jf 
 
 * For the definitions of the several Compound Rules, see the corresponding Simple 
 Rules, pages 6, 12. 16, and 24. 
 
 f The proof is Irft to exercise the learner ; and it will be proper to require him to 
 perform it. not only in this examples but in all the exercises in this rule. 
 
o-i COMPOUND ADDITION. 
 
 4, is then added with the units of the shillings: the sum 1 
 which the latter figure is set down, and the tens being carried to 
 the tens of the shillings, the sum is 8, which being divided by 2, 
 because 2 tens, or 20 shillings, make a pound, the quotient is 4, 
 which is added with the pounds, as in Simple Addition.* 
 
 . s. d. 
 
 In this example the halfpence amount Exam. 2. 15 3 8 
 to 5, or 2 pence halfpenny. In such ex- 16 7* 
 
 amples, where all the fractional parts are 
 halfpence, it is easier to call each a half- 
 penny, than 2 farthings. ^_9* 
 
 Sum, 234 18 9 
 
 When the columns are very long, the work becomes heavy and la- 
 borious; and therefore, in such a case, the given quantities may be sepa- 
 rated into two or more divisions, as is suggested in page 8. 
 
 Exercises in Compound Addition 
 
 
 1 
 
 
 
 2 
 
 
 
 3 
 
 
 
 4 
 
 
 
 s. 
 
 d. 
 
 
 
 s. 
 
 d. 
 
 
 
 s. 
 
 d, 
 
 
 
 s. 'd. 
 
 485 
 
 12 
 
 7f 
 
 3 
 
 14 
 
 8i 
 
 413 
 
 13 
 
 10* 
 
 4 
 
 13 6* 
 
 49 
 
 16 
 
 si 
 
 
 
 19 
 
 ^ 
 
 1245 
 
 10 
 
 9 
 
 3 
 
 15 7| 
 
 186 
 
 13 
 
 llj 
 
 
 5 
 
 17 
 
 
 7085 
 
 15 
 
 1 IA 
 
 7 
 
 10 11 
 
 787 
 
 10 
 
 8; 
 
 
 2 
 
 12 
 
 6 
 
 8519 
 
 6 
 
 4* 
 
 I 
 
 12 9| 
 
 239 
 
 9 
 
 9- 
 
 
 2 
 
 16 
 
 10* 
 
 3456 
 
 14 
 
 10 
 
 2 
 
 8 7 
 
 843 
 
 11 
 
 4 
 
 
 1 
 
 5 
 
 8| 
 
 90 
 
 12 
 
 5 
 
 4 
 
 3" 
 
 374 
 
 16 
 
 7 
 
 3 
 
 8 
 
 3 
 
 69 
 
 15 
 
 2 4 
 
 ' 6 
 
 16 8J 
 
 285 
 
 4 
 
 9J 
 
 5 
 
 2 
 
 4 A. 
 
 179, 
 
 18 
 
 11* 
 
 7 
 
 6 2 
 
 599 
 
 19 
 
 8 
 
 10 
 
 14 
 
 *4 
 
 788 
 
 9 
 
 9 
 
 9 
 
 15 10* 
 
 5. Add together 59 12 7f, 05 14 2*, 345 5 9$, 88 16 
 8J, 186 17 4*, 347 7 6, 3 2 9*, 7 14 7J, 52 8 6f, 
 59 3 4, 42 18 10*, 187 10 10, 954 16 5f 
 
 6. Add together 324 14 10*, 518 5 9*, 39 15 6, 54 
 11 11*, 49 1 8i, 9 7 1H, 1000, 86 6 3*, 324, 7?) 
 11 6*, 5 13 9, 611 4 2*, "186 13 1*, 476 "8 5. 
 
 *PENCE TABLE, 
 
 d. s. d. 
 
 d. s. d. 
 
 d. s. d. 
 
 d. *. d. 
 
 d. s. d. 
 
 rf. 5. <i. 
 
 12=1 
 
 40=3 4 
 
 72=6 
 
 100= 8 4 
 
 132=11 
 
 160=13 4' 
 
 20=1 fc 
 
 48=4 
 
 80=6 8 
 
 108= 9 
 
 140=1 1 S 
 
 168=14 
 
 24=t2 
 
 50=4 2 
 
 84=7 
 
 110= 9 fi 
 
 144=iy 
 
 170=14 2 
 
 30=2 6 
 
 60=6 
 
 90=7 6 
 
 ls!0=10 
 
 i.oO=ia 6 
 
 180=15 
 
 36=3 U 
 
 70=5 10 
 
 96=8 
 
 130=10 10 
 
 156=13 
 
 200=16 8 
 
 This Table has been inserted, lest some teachers should consider the want of it an im. 
 perfection. Jt seems better, however, not to impose upon the learner the labour of 
 committing '.t to memory, except perhaps a small part at the beginning. If, instead of 
 using it, he divide the amounts of the pence in his operations by 12, he will soon 
 acquire a readiness In this species of Addition, which will be highly valuable ; and by 
 practice he will soon innke himself acquainted TvUh the Table, without, formally com- 
 mitting tt especially if his attention be occasionally directed to the subject by the teacher* 
 
COMPOUND ADDITION. 55 
 
 7. Required the sum of ,21 1(H, 73 18 9, 22 3 7$, 
 64 16 9, 19 18 1$, 78 9 9, 10 9 10^, 250 9 6, 17 
 12 7*, 54 7$, 797 12 9, 1 14 1$, 60 5 9, 113 15, 
 170 12 6, 98" 19 3. 
 
 8. Add together one thousand and six pounds, fifteen shillings, 
 and three farthings; three hundred pounds, seventeen shillings, 
 and a halfpenny; four thousand and ninety-six pounds, eight, shil- 
 lings, and three farthings; seven thousand pounds, eighteen shil- 
 lings, and eleven pence; two pounds, and three halfpence; five 
 hundred and eleven pounds, and sixpence halfpenny; eighty-six 
 pounds; five hundred and eight pounds, seven shillings, and a 
 farthing; five thousand pounds; two hundred and ninety pounds, 
 ten shillings; two hundred and nine pounds, and ten pence. 
 
 ' 9. Required the sum of 15 13 8$, 53 16 7$, 199 6 3|, 
 548 8, 458 8, 1 2 9, 2 5 6, 3 8 3, 1 14 1$, 2 
 16 10$, 3 19 7$, 180 18 8, 31 13 4f. 
 
 10. Add together 15/8$, 13/10$, 16/6}, 19/3, 16/7$, 5/8, 13/9, 
 12/7$, 15/1 J, 11/4$, 9/4, 10/9$, 17/4$, 18/4, 4/5$, 3/10}. 
 
 11. 513 15 10^-f 34 2 10$+183 17 3$+548 13 8$ 
 + 8 15 9$+88 15 9$+176 15 6f + 79 13 11 + 5 
 6$ + 189 13 8$+195 8 10 + 179 16 6 + 348 17 9f + 
 59. 
 
 12. Add together 18 14 8$, 12 13 9 2r 21 12 10, 32 
 9 10$, 63 13 9$, 16 4 8$, 35 14 9$, 17 16 7$, 23 15 
 9$, 35 17 2$, 8 19 8, 12 10 0$, 13 8 8$, 17 18 4. 
 
 13. Find the sum of one hundred and two pounds, and ten pence; 
 fifty-eight pounds, and three farthings j forty pounds, seventeen 
 shillings, and five pence halfpenny, nine pounds, fifteen shillings, 
 and eleven pence; seventy-three pounds, twelve shillings; thirty- 
 nine pounds, fourteen shillings, and a farthing fifty-nine pounds; 
 eighty-two pounds, and nine pence halfpenny ; fifty pounds, and 
 a penny. 
 
 1'4. 918 12 7i+51 16 8$+519 14 4$+83 16 10$+ 
 55 12 8f + 183 7 9 + 188~ 18 2$+375 12 3$ +491 11 
 4$ + 318 15 9|+780 10 4+10 19 9f+508 16 9 + 
 374 19 5$+348 6 8$. 
 
 15. 459 16 11 +583 11 11$+15 14 8$+191 14 8} 
 + 99 13 6$ + 195 12 6$ + 17 18 4}+ 473 18 0$ + 18 
 16 7$ + 5 8 9$ + 31 12 2$ + 63 3 ll + 5 11 8+l 2 
 9 + 1005 11 9. 
 
 16. Add together 3 4 7}, 41 12 4, 186 17 9$, 3 8 
 10$, 67 18 9$, 112 16 6}, 73 15 5$, 139 7 4$, 510 
 6 ?$, 706 7 7$, 581 4 4}, 608 19 6$, 519 12 & 588 
 6 11, 137 1 2$. 
 
 17. Required the sum of 13 6 8, 88 14 7$, 175 16 4$, 
 1245 13 5|, 512 16 6, 91 12, 78 14 5, 475 19 9, 
 839 16 6}, 512 11 1H, 447 16 0}, 43, 74 7 11, ^/i 
 14 7$, 9 4 7}, 907 I*" 6, 10 3 0}. 
 
56 COMPOUND ADDITION. 
 
 18. Add together 118 16 lOf, 514 4 11, 880 17 
 644 15 4, 359 16 6, 346 3 10|, 435 10 3^, 634 17 
 6, 364 16 8$, 463 14 8, 643 16 2, 5 13 9, 3018 15 
 6, 998 4 10, 580 6, 14/9, 533 16 7, 85 18 9, 
 375 15 8, 90 14 8f, 187 16 6, 57 3 11, 348 6 10, 
 4048 14 9|, 798 16 5, 8 15 6, 89 19 8J, 175 18 8, 
 311 17 10i, 73 14 9f, 864 13 7^, 364 14 10, 819 14 
 L 59 17 8, 1044 5 
 
 
 *' 
 
 Exam. 3. 
 
 
 
 tons. cwt. q. 
 
 Ibs 
 
 In this example the sums of the pounds, 
 
 35 16 
 
 20 
 
 quarters, and hundreds, are respectively di- 
 
 42 14 2 
 
 18 
 
 vided by 28, 4, and 20, the remainders set 
 
 18 9 1 
 
 16 
 
 below their respective columns, and the 
 
 17 18 3 
 
 7 
 
 quotients carried to the next columns, res- 
 
 31 5 3 
 
 19 
 
 pectively. The hundreds may be added as 
 
 45 12 2 
 
 5 
 
 shillings in Addition of Money, the divisor 
 
 
 
 
 
 
 being the same in both cases. Sum, 191 17 2 1 
 
 Ex. 19. 
 
 20. 
 
 21. 
 
 
 22. 
 
 
 cwt. qrs. Ibs. 
 
 Ibs. oz. drs. 
 
 t. cwt. 
 
 qrs. 
 
 Ibs. oz. 
 
 
 53 2 12 
 
 16 12 13 
 
 75 13 
 
 1 
 
 1 11 
 
 
 17 1 15 
 
 5 15 3 
 
 83 17 
 
 2 
 
 53 14 
 
 
 16 3 19 
 
 12 12 5 
 
 17 8 
 
 
 
 47 10 
 
 
 19 18 
 
 3 11 9 
 
 16 16 
 
 1 
 
 86 9 
 
 
 25 3 18 
 
 19 1 11 
 
 61 15 
 
 2 
 
 94 7 
 
 
 48 3 6 
 
 14 4 8 
 
 39 9 
 
 3 
 
 29 3 
 
 
 42 2 7 
 
 24 7 14 
 
 88 7 
 
 3 
 
 42 10 
 
 
 23. Add together 55 c. 3 q. 18 Ibs.; 34 c. 2 q. 22 Ibs.; 63 c. 1 q. 
 23 Ibs.; 71 c. q. 19 Ibb.; 16 c. 3 q. 20 Ibs.; 3 c. 3 q. 26 Ibs.; 27 c. 
 
 2 q. 23 Ibs.; 41 c. 3 q. 9 Ibs.; 35 c. 1 q. 18 Ibs.; 43 c. 2 q. 24 Ibs.; 
 95 c. q. 10 Ibs.; 29 c. 2 q. 17 Ibs.; 32 c. 2 q.; 59 c. 3 q. 24 Ibs.; 
 43 c. 1 q. 21 Ibs.; 37 c. 3 q. 1 Ib. 
 
 24. Add together 17 c. 2 q. 17 Ibs.; 22 c. 1 q. 27 Ibs.; 55 c. 3 q. 
 19 Ibs.; 13 c. 1 q. 15 Ibs.; 73 c. 2 q. 13 Ibs.; 88 c. 2 q.; 48 c. q 
 23 Ibs.; 32 c. 3 q. 19 Ibs.; 52 c. 2 q. 22 Ibs.; 43 c. 3 q. 20 Ibs.; 
 53 c. 1 q. 18 Ibs.; 31 c. 2 q. 26 Ibs.; 45 c. 1 q. 14 Ibs.; 36 c. 3 q. 
 
 25. Required the sum of 2 Ibs. 3 oz.; 10 Ibs. 5 oz.; 1 Ib. 14 oz.; 
 
 3 Ibs. 13 oz.; 5 Ibs.; 7 Ibs. 7 oz.; 8 Ibs. 12 oz.; 7 Ibs. 15 oz.; 3 Ibs. 
 6 oz.; 1 Ib. 11 oz.; 12 Ibs. 10 oz. 
 
 26. Required the sum of 58 tons, 12 c. 3 q. 21 Ibs.; 32 t. 11 c. 
 
 2 q. 20 Ibs.; 19 t. 15 c. 1 q. 12 Ibs.j 17 t. 17 c. q. 17 Ibs.; 5 t. 
 
 3 c. 1 q. 25 Ibs.; 73 t. 15 c. 1 q. 12 Ibs.; 98 t. 16 c. 2 q. 22 Ibs.; 
 35 t. 16 c. 3 q. 19 Ibs.; 16 1. 14 c. 2 q. 16 Ibs.; 70 t. 13 c. 1 q. 24 Ibs. 
 
 27. Add together the following quantities, in long weight : 57 c. 
 3 q. 11 Ibs.; 39 c. 3 q. 26 Ibs.; 51 c. 3 q. 18 Ibs.; 26 c. 2 q. 24 Ibs.; 
 67 c, 1 q. 18 Ibs.; 18 c. 2 q. 21 Ibs ; 132 c. 2 q. 13 Ibs.; 73 c. 3 q. 
 
COMPOUND ADDITION 37 
 
 28. Add together 13 acres, 3 roods, 27 perches; 45 a. 1 r. 2? 
 p.; 63 a. 2 r. 17 p.; 26 a. 2 r. 26 p.; 16 a. 3 r. 34 p.; 21 a. r. 
 8 p.; 55 a. 2 r. 31 p.; 37 a. 2 r. 18 p.; 44 a. 2 r. 20 p.; 57 a. r. 
 19 p.; 61 a. 3 r. 18 p.; 39 a. 2 r.; 5 a. 1 r. 30 p.; 27 a. 2 r. 14 p.* 
 
 29. 19 a. 2 r. 28 p.; 10 a. r. 20 p.; 11 a. 3 r. 16 p.; 19 a. 2 r. 
 18 p.; 14 a. 2 r. 25 p.; 17 a. 3 r. 22 p.; 15 a. 2 r. 19 p.; 4 a. 3 r. 
 27 p.; 24 a. 2 r. 24 p.; 149 a. 1 r. 14 p.; 719 a. 2 r. 16 p.; 107 a, 
 
 1 r. 23 p.; 149 a. 2 r. 17 p.; 6 a. 3 r. 15 p.f 
 
 30. The following are the quantities and values of the linen ex- 
 ported from Ireland to America in the undermentioned years: 
 required the value of the whole, and the number of miles con- 
 tained in the entire number of yards: 
 
 Yards. s. d. 
 
 In 1798, ... 2,361,483 183,343 4 
 
 1799, ... 1,376,382 106,818 16 1 
 
 1800, ... 1,156,467 89,948 11 11 
 
 1801, ... 2,519,575 248,762 2 6 
 
 1802, ... 1,089,223 106,671 18 6 
 
 1803, ... 1,873,423 184,180 18 6 
 
 1804, ... 2,258,176 223,034 5 9 
 
 1805, ... 2,221,606 209,404 6 2 
 
 1806, ... 2,675,619 309,525 9 9 
 
 1807, ... 1,657,446 191,413 3 6 
 
 Ansu>. Value, 1,853, 10 * 16 8 j length, 10903 ro. Of. 21 p. 4y. 
 
 31. The amount of the duties paid on goods imported into Li- 
 merick in 1764, was 19,809 15 9; in 1765, 21,332 4 8; in 
 1766, 16,729 8 2; in 1767, 16,316 10j in 1768, 16,571 12 
 8; in 1769, 20,237 12 7; in 1770, 22,138 4; in 1771, 
 20,213 12 6; in 1772, 22,003 2 0; in 1773, 20,606 15 7; 
 in 1774, 17,317 9; in 1775, 16,979 10 6. What was the 
 entire amount during these twelve years ? Ansiv. 230,315 5 6. 
 
 32. The following are the several quantities of land, in English 
 measure, contained in the Royal Forests in England: New Forest, 
 66,942 a. 3 r. 26 p.; Dean Forest, 23,015 a. 3 r. 29 p.; Aliceholt 
 and Woolmer, 8,694 a. 1 r. 31 p. ; Whittlevvood, 4,850 a. 3 r. 32 p.; 
 Which wood, 3,709 a. 3 r. 5 p.; Waltham, 3,278 a. 3 r. 2 p.; Salcey, 
 1,847 a. r. 23 p.; Sherwood, 1,466 a. 3 r. 10 p.; Bere, 926 a. 
 
 2 r. 13 p.; Rockingham, 860 a. 3 r. 23 p. Required the sum. 
 Answ. } 15594 a. r. 34 p. 
 
 * The units of perches may be added as in Simple Addition, and the sum of the tens 
 divided by 4, the tens in 40. In like manner the units of minutes, seconds, thirds, &c. 
 may be added as in Simple Addition, and the sum of the tens divided by 6, the tens in 60. 
 
 f Several parts of Compound Addition usually delivered in books on Arithmetic, are 
 omitted, as they are in general of no use ; and if they should at any time be require*!, 
 they can be performed without difficulty by the application of the same principles on 
 which the others depend. 
 
 D2 
 
58 
 
 COMPOUND SUBTRACTION. 
 
 RULE. (1.) Place the less number below the greater,* so 
 that the numbers in each column may be of the same de- 
 nomination. (2.) Then, beginning with the lowest deno- 
 mination, subtract, if possible, each number in the lower 
 line from that which stands above it. (3.) But when this 
 cannot be done, subtract the number in the lower line from 
 a unit of the next higher denomination; to the remainder 
 add the upper number, for the true remainder, and carry 
 one to the next higher number in the lower" line. (4.) Pro- 
 ceed thus with all the denominations, except the highest, 
 in which the work is to be performed as in Simple Subtrac- 
 tion. 
 
 s. d. 
 
 Exam. 1. From 33 17 10 J 33 17 lOf 
 
 take 18 8 '4. 18 8 4^ 
 
 15 9 6i, Amw. 
 
 Exam. 2. Required the difference between 159 9 4 and 
 86 17 8^. 
 
 In this example, as a halfpenny is greater *. d. 
 
 than a farthing, it is taken from a penny, and 159 9 4^ 
 the remainder being added to the farthing, 86 17 8 
 
 the sum, three farthings, is set down; a pen- 
 
 ny is then carried to 8 pence, and the sum 72 1 1 1i%,Ans. 
 being taken from one shilling, the remainder 
 3 is added to 4- pence, and the amount set down. We then 
 proceed thus: 1 and 7 are 8; 8 from 9 and 1 remains; 1 from 2, 
 (the tens' figure in the shillings in a pound,) and 1 remains; 1 and 
 6 are 7; 7 from 9 and 2 remain, &c. 
 
 Exercises in Compound Subtraction. 
 
 t. d. s, d. 
 
 . 1. From 
 2. 
 
 19 3 10 
 575 15 H 
 
 take 8 15 3| 
 124 13 4 
 
 3. 
 4. 
 
 192 11 4| 
 511 3 2 
 
 88 16 9 
 247 10 64 
 
 5. 
 
 12 4 9 
 
 524} 
 
 * It is more usual, and generally more convenient, to set the less number below the 
 greater, in Subtraction : it is by no means essential, however; and the pupil should lie 
 accustomed, both in Simple and Compound Subtraction, to subtract downward as v/ull 
 as upward, that he may be able to do so, when it may happen in complex operations, 
 that the numbers are so arranged. 
 
 The methods of proof, and the principles on which the operations depend, are iho 
 sninr : in Siir.ple Subtraction. 
 
COMPOUND MULTIPLICATION. 
 
 8. d. 
 
 
 
 *. d. 
 
 Ex.6. 
 
 From 
 
 100 
 
 
 
 
 
 take 
 
 1 
 
 2 9 
 
 
 7. 
 
 JW+--M 
 
 513 
 
 5 
 
 8| 188 17 4J 
 
 
 8. 
 
 
 
 1516 
 
 19 
 
 si 
 
 847 12 9* 
 
 
 
 
 cwi. 
 
 q. 
 
 #>s. 
 
 
 cwt. 
 
 q. tbs. 
 
 Ex 
 
 . 9. 
 
 From 
 
 13 
 
 3 
 
 20 
 
 take 
 
 9 
 
 2 12 
 
 
 10. 
 
 jj.-j--jj.-j 
 
 23 
 
 1 
 
 5 
 
 ^-r-nrjjxjjj 
 
 17 
 
 3 22 
 
 
 11. 
 
 ~v~~~. 
 
 105 
 
 
 
 
 
 
 
 79 
 
 1 13 
 
 tons. cwt. q. Ibs. 
 
 tons. cwt. 
 
 q. Ibs. 
 
 Ex 
 
 .12. 
 
 From 
 
 15 7 
 
 
 
 24 
 
 take 
 
 5 12 
 
 1 10 
 
 days. h. min. 
 
 5<?C. 
 
 
 <%s. . 
 
 min. sec. 
 
 Ex. 
 
 13. 
 
 From 
 
 5 10 
 
 27 
 
 15 
 
 take 
 
 2 4 
 
 13 29 
 
 
 14. 
 
 ^^ r 
 
 83 17 
 
 3 
 
 
 
 rw^w* 
 
 59 7 
 
 12 30 
 
 15. The latitude of Rome, (St. Peter's,) is 41 53' 54" North; 
 of Paris, (Observatory of the Military School,) 48 51' 6" 
 
 Required 
 and second, second and third, &c. of these capitals. 
 
 16. The latitude of Gibraltar is 36 & 30" N. and that of the 
 North Cape in Lapland, 71 10 7 N. Required the difference. 
 
 17. The latitude of the Cape of Good Hope is 33 55 / 15" S. 
 and that of Cape Horn, 55 58> 30" S. Required their difference, 
 
 18. The latitudes of Belfas* and Glasgow, are 54 36' N. and 
 55 51' 32" N. respectively. Required their difference. 
 
 19. The following are the times in which the principal planets 
 perform their revolutions about the sun; required the differences 
 of the first and second, of the second and third, &c. : 
 
 Days. h. m. 
 
 Mercury, 87 23 16 
 
 Venus, 224 16 49 
 
 Earth, 365 6 9 
 
 Mars,.. .. 686 23 31 
 
 Days. k. m. 
 
 Jupiter,.. 4332 14 27 
 
 Saturn, 10759 ) 51 
 
 Georgium 
 
 Sidi 
 
 us, 
 
 ... 30737 18 
 
 COMPOUND MULTIPLICATION. 
 
 Problem 1. To multiply a number of more denominations 
 than one, by a number not exceeding 12.* 
 
 RULE. Commencing with the lowest denomination, mul- 
 tiply successively the several numbers in the multiplicand 
 by the multiplier, dividing, setting down, and carrying us 
 in Compound Addition. 
 
 * Or not exceeding 19, if the learner have committed to memory the supplement to the 
 Multiplication Table, page 17. The same circumstance will also modify the succeeding 
 P ..'.cms, in a similar manner. 
 
&> COMPOUND MULTIPLICATION. 
 
 Exam. 1. Multiply . I H 7 by 9. 
 In this example the farthings, pence* and s. d. 
 shillings are multiplied successively by 9; the 1 14- 7| 
 
 products, as they ure found, are respectively 9 
 
 divided by 4, 13*30, (or the tens of the shil- 
 lings by 2;) the several remainders are writ- 15 11 9f 
 ten down, and the quotients carried. The 
 pounds ure multiplied as in Simple Multiplication; and the pro- 
 duct is found to be .5 11 9|. 
 
 Exercises in Compound Multiplication. 
 
 s. d. 
 
 Ex.l. 3 14 9^X2 
 
 2. 1 17 8|X3 
 
 3. 18 11^X4 
 
 4. I 10 4X5 
 
 s. d. 
 
 5. 3 15 10! X 6 
 
 6. 1 2 9 X7 
 
 7. 2 18 9X8 
 
 8. 6 13 4X9 
 
 s. d. 
 
 9. 15 10 X 10 
 
 10. 9 7 8X11 
 
 11. 3 19 7x 12 
 
 12. 1 16 7 X12 
 
 Problem 2. Jo multiply by a number which exceeds 12, but 
 is the product of two or more Jactors, each less than 13.* 
 
 RULE. (1.) By the preceding problem, multiply the given 
 multiplicand by one cf the factors. (2.) Multiply the result 
 by another. (3.) Multiply this last result by another, if there 
 be so many ; and thus proceed, whatever is their number. 
 Exam. 2. Multiply 18/3J by 42. 
 
 In this example the multiplicand is s. d. 
 
 multiplied by 6, and the product is 5 18 3^ 
 
 9 7|. This again is multiplied by 7 6 
 
 and the product is 38 7 4J. Tht rea- 
 son of tne operation is sufficiently obvi- 597^ 
 ous, since 42 is the product of 6 and 7. 7 
 The work might be proved by multi- 
 plying the multiplicand by 7, and the 38 7 4 
 result by 6. 
 
 When the multiplicand contains one or more farthings, if one of the 
 factors be even, it is better to use it first, as the farthings may thus dis- 
 appear, and the rest of the work be easier. But if the multiplicand 
 end in even pence, without farthings, and one of the factors be 3, 6, 
 or 9, it is better to use that factor first, as the pence may thus disap- 
 pear : and in all cases in multiplication of money, 
 when 12 is one of the factors, it should be used l 13 7^ 
 first, as part of the operation may be performed by 12 
 
 inspection, by setting down 3 pence for each farth- 
 ing, and carrying to the shillings 1 for every penny JL'20 3 3 
 in the multiplicand. Thus, in multiplying l 13 
 7^, by 12, set down 3 pence, and carry 7 to the shillings, saying, 12 
 times 3 are 36, and 7 are 43, Ac. 
 
 * For the method of finding the factors in thf more difficult cases, see page 
 u^i. iy .vcvi,'*', Mill s-joii learn to and there by iu-pecuon, in ail useful cases. 
 
COMPOUND MULTIPLICATION. 61 
 
 Exercises. Answer.-,. 
 
 
 
 
 s. 
 
 d. 
 
 
 
 s. 
 
 d. 
 
 Ex. 13. 
 
 1 
 
 17 
 
 9i X 14 
 
 26 
 
 9 
 
 1 
 
 14. 
 
 
 
 13 
 
 4.iX 15 
 
 , 10 
 
 
 
 7 
 
 15. 
 
 
 
 2 
 
 8^X 16 
 
 2 
 
 3 
 
 4 
 
 16. 
 
 3 
 
 14 
 
 OjX 18 
 
 66 
 
 12 
 
 H 
 
 17. 
 
 1 
 
 5 
 
 3 X 21 
 
 , 26 
 
 10 
 
 3 
 
 18. 
 
 
 
 14 
 
 1 X 22 
 
 , 15 
 
 9 
 
 10 
 
 19. 
 
 
 
 13 
 
 2 X 24 
 
 15 
 
 16 
 
 
 
 20. 
 
 
 
 6 
 
 HfX 25 
 
 , 8 
 
 14 
 
 5f 
 
 21. 
 
 
 
 12 
 
 7^X 27 
 
 17 
 
 
 
 3j 
 
 22. 
 
 2 
 
 1 
 
 X 30 
 
 61 
 
 11 
 
 3 
 
 23. 
 
 2 
 
 15 
 
 5 X 32 
 
 88 
 
 13 
 
 4 
 
 24. 
 
 1 
 
 16 
 
 6^X 33 , 
 
 60 
 
 5 
 
 2 
 
 25. 
 
 1 
 
 19 
 
 4 X 36 
 
 70 
 
 16 
 
 
 
 26. 
 
 
 
 8 
 
 6|X 40 
 
 17 
 
 1 
 
 8 
 
 a?. 
 
 1 
 
 11 
 
 3 X 42 
 
 65 
 
 12 
 
 6 
 
 28. 
 
 2 
 
 2 
 
 6X 44 
 
 93 
 
 11 
 
 10 
 
 29. 
 
 
 
 8 
 
 4X 45 , 
 
 18 
 
 16 
 
 10$ 
 
 30. 
 
 2 
 
 12 
 
 
 126 
 
 18 
 
 
 
 31. 
 
 
 
 16 
 
 4|X 49 
 
 40 
 
 3 
 
 *i 
 
 32. 
 
 3 
 
 1 
 
 3|X 53 
 
 168 
 
 12 
 
 
 33. 
 
 
 
 3 
 
 lO^X 56 
 
 10 
 
 15 
 
 10 
 
 34. 
 
 
 
 6 
 
 HtX 60 
 
 20 
 
 16 
 
 3 
 
 35. 
 
 2 
 
 10 
 
 5 X 63 
 
 158 
 
 16 
 
 3 
 
 36. 
 
 1 
 
 14 
 
 4 X 'i 
 
 113 
 
 6 
 
 
 
 37. 
 
 
 
 4 
 
 5^X 72 
 
 15 
 
 19 
 
 6 
 
 38. 
 
 1 
 
 6 
 
 9fX N> 
 
 107 
 
 5 
 
 
 
 39. 
 
 
 
 5 
 
 31 X i^l . ... 
 
 21 
 
 6 
 
 Hi 
 
 40. 
 
 1 
 
 6 
 
 7 X *4 
 
 Ill 
 
 13 
 
 
 
 41. 
 
 4 
 
 
 
 94 X W?i 
 
 355 
 
 * 7 
 
 10 
 
 42. 
 
 1 
 
 10 
 
 8 x :;o 
 
 lo% 
 
 
 
 
 43. 
 
 1 
 
 15 
 
 llf X 9o 
 
 iW 
 
 ^4 
 
 
 
 44. 
 
 
 
 8 
 
 7JX 99 
 
 42 
 
 15 
 
 I] 4 
 
 45. 
 
 1 
 
 17 
 
 2 X100 
 
 185 
 
 16 
 
 8 
 
 46. 
 
 
 
 16 
 
 1HX108 
 
 91 
 
 11 
 
 6 
 
 47. 
 
 
 
 18 
 
 9 XI 10 
 
 103 
 
 2 
 
 6 
 
 4M. 
 
 
 
 4 
 
 8^X 1^0 , 
 
 28 
 
 2 
 
 6 
 
 49. 
 
 2 
 
 19 
 
 6 X121 
 
 359 
 
 19 
 
 6 
 
 50. 
 
 1 
 
 16 
 
 7^X132 
 
 , 241 
 
 14 
 
 6 
 
 51. 
 
 2 
 
 19 
 
 2^X144 
 
 426 
 
 3 
 
 
 
 52. 
 
 1 
 
 12 
 
 5 X*75 
 
 , 121 
 
 11 
 
 jj 
 
 53. 
 
 I 
 
 18 
 
 8fXll2 
 
 216 
 
 17 
 
 8 
 
 54. 
 
 
 
 14 
 
 2xl28 
 
 91 
 
 1 
 
 4 
 
 * In this and each of the six following exercises, the multiplier is the product of 
 three factors; 75 being=3x5x5 ; 112=8x2x7, or 4x4x7; 128=8x8x2, or 4x*x8; 
 fcc. In like manner, 140=2x7x10; 19ii=12x4x4j 240=12x'-'xlO; 17!=li'xIXi2, 
 *tc. &c. 
 
62 COMPOUND MULTIPLICATION. 
 
 Exercises. Answer*. 
 
 s. d. *. d. 
 
 Ex.55. 1 2 7X147 166 2 9f 
 
 56. 11 4X168 95 11 
 
 57. 1 16 8 X175 320 16 8 
 
 58. 215 2X196 54016 9 
 
 Problem 3. To multiply by a number which exceeds 12 but 
 is not produced by factors below 13. 
 
 RULE. (1.) Use those factors whose product is nearly 
 equal to the multiplier. (2.) Increase or diminish the re- 
 sult, as the case may require, by the product of the multi- 
 plicand and the difference between the multiplier and the 
 product of the factors employed. 
 
 Exam. 3. Multiply 14/10 by 38. g. d. 
 
 In this example, 38 not being the pro- 14 10 
 duct of any two factors not exceeding ii?, 12 
 
 we multiply by 36, as before, and to the ITIs 6^ prod, bv 12. 
 product we add twice the multiplicand, 
 to find the product by 38. The answer oT"!" 7 " 
 would have been obtained with nearly the ** o 9 
 
 same facility, had we multiplied by 40, _ . ? ' 
 
 (4X10,) and subtracted twice the mul- ~& ** 3, ... 38 
 tiplicand; and thus the operation might 
 be proved. 
 
 Exam. 4. Multiply 1 4 10 by 61. 
 
 The operation in this, example is con- 
 tracted by taking in the multiplicand, in 
 the multiplication by 5. This contmc- 
 tion may always be employed when 'K 
 product of the factors is one less t an the ^75 
 multiplier. A similar contraction might 
 be employed, though not with the same 
 facility, when the excess is 2, 3, &c. 
 
 Exam. 5. Multiply 2 7 8^ by 79. 
 
 In this example it is better to find the s. d. 
 
 product by 80, and subtract the multipli- 2 7 8^ 
 
 cand from it, than to multiply by 77, and 8 
 
 add twice the multiplicand to the product, 
 
 as in the one way there is one multiplica- 19 1 8, prod, by 8. 
 
 tion fewer than in the other. A like con- 1 
 
 traction may be applied with advantage 
 
 in every case in which factors can be I DO 16 8, 
 
 found, whose product is ow more than ^ 7 8^-,tobesubtr. 
 
 the given multiplier, while factors can- 
 
 not be found whose product is only OP.L 188 8 il, prod. by 79 
 less than it. 
 
COMPOUND MULTIPLICATION. 
 
 Exercises. 
 
 Aiisivfrs. 
 
 
 
 
 5. 
 
 (I. 
 
 
 
 
 .9. 
 
 d. 
 
 Ex. 59. 
 
 5 
 
 3 
 
 X X 
 
 13 
 
 .... 67 
 
 1 
 
 2 
 
 60. 
 
 2 
 
 8 
 
 7|X 
 
 23 
 
 
 IS 
 
 10J 
 
 61. 
 
 
 10 
 
 1 X 
 
 29 
 
 .... 43 
 
 12 
 
 5 
 
 63. 
 
 
 
 17 
 
 3iX 
 
 31 
 
 .... 26 
 
 1G 
 
 (H 
 
 (#. 
 
 
 
 11 
 
 
 39 
 
 .... 22 
 
 1 
 
 ^i 
 
 64. 
 
 
 
 18 
 
 8 X 
 
 40 
 
 .... 42 
 
 IS 
 
 h 
 
 65. 
 
 1 
 
 11 
 
 o^x 
 
 47 
 
 .... 72 
 
 17 
 
 1 I? 
 
 60. 
 
 4 
 
 6 
 
 5 X 
 
 52 
 
 .... 120 
 
 13 
 
 8 
 
 v 67. 
 
 
 
 13 
 
 0X 
 
 53 
 
 .... 34 
 
 10 
 
 \^ 
 
 v 68. 
 
 2 
 
 5 
 
 4 X 
 
 58 
 
 .... 131 
 
 9 
 
 4 
 
 , 69. 
 
 
 12 
 
 MX 
 
 65 
 
 .... 104 
 
 
 
 9| 
 
 70. 
 
 
 
 19 
 
 
 68 
 
 .... 67 
 
 7 
 
 3 
 
 71. 
 
 
 
 12 
 
 8*X 
 
 09 
 
 .... 43 
 
 10 
 
 1<H 
 
 72. 
 
 ' 
 
 7 
 
 11 X 
 
 70 
 
 .... 258 
 
 1 
 
 8 
 
 73. 
 
 
 
 
 9^ X 
 
 78 
 
 .... 81 
 
 1 
 
 9 
 
 74. 
 
 
 
 14 
 
 5 X 
 
 82 
 
 .... 59 
 
 2 
 
 2 
 
 75. 
 
 
 13 
 
 6 X 
 
 89 
 
 .... 149 
 
 1 
 
 6 
 
 76. 
 
 
 3 
 
 22 X 
 
 91 
 
 .... 105 
 
 13 
 
 10* 
 
 77. 
 
 
 16 
 
 4 X 
 
 94 
 
 .... 170 
 
 15 
 
 4 
 
 78. 
 
 2 
 
 2 
 
 11{X 
 
 95 
 
 .... 203 
 
 19 
 
 o* 
 
 79. 
 
 3 
 
 9 
 
 7 X 
 
 98 
 
 .... 340 
 
 19 
 
 2 
 
 60. 
 
 
 
 15 
 
 6jX 
 
 103 
 
 .... 80 
 
 o 
 
 14 
 
 81. 
 
 
 
 11 
 
 4iX 
 
 101- 
 
 .... 59 
 
 3 
 
 
 
 82. 
 
 1 
 
 10 
 
 OjX 
 
 105 
 
 .... 100 
 
 4 
 
 8* 
 
 83. 
 
 2 
 
 5 
 
 10 X 
 
 Ill 
 
 .... 254 
 
 "t" 
 
 6 
 
 84. 
 
 3 
 
 15 
 
 3^X 
 
 117 
 
 .... 440 
 
 11 
 
 <>* 
 
 85. 
 
 3 
 
 17 
 
 8.f X 
 
 122 
 
 .... 471- 
 
 2 
 
 m 
 
 86. 
 
 2 
 
 T3 
 
 MX 
 
 125 
 
 .... 331 
 
 IS 
 
 
 
 87. 
 
 
 10 
 
 8 X 
 
 130 
 
 .... l ( W 
 
 6 
 
 8 
 
 88. 
 
 
 
 6 
 
 94 X 
 
 139 
 
 .... 47 
 
 4 
 
 Of 
 
 89. 
 
 
 
 17 
 
 4X 
 
 145 
 
 .... 120 
 
 2 
 
 4 
 
 90. 
 
 
 
 15 
 
 7 X 
 
 150 
 
 
 17 
 
 6 
 
 91. 
 
 2 
 
 4 
 
 8|x 
 
 155 
 
 .... 340 
 
 9 
 
 9* 
 
 92. 
 
 2 
 
 11 
 
 5iX 
 
 150f.. 
 
 .,, . 4U1 
 
 4 
 
 3 
 
 * In multiplving by 13, if the pupil have committed the Multiplication Table no {.. 
 Iher than 1-' times 12, he may multiply by 12, and take in the 
 multiplicand as he proceeds, as in the annexed example. In this 
 way the multiplication by any multiple of 13, not exceeding 13 
 times 13, may be performed in two lines, as also by any number 
 
 which is greater by unity than any of these multiples. Hence, - 
 
 only two lines are necessary in multiplying by 6, 3f>, 52, 53, 65, d ly i^, _-^s?u. 
 78> 79, 91, !, 104, 105, 117, 118, 130, 131, 143, 156. 157, 16", 170. 
 
 J- The subjoined example will show how Compoilnd Multiplication maybe pcVI'onncd, 
 however great the multiplier is. This method, however, will scarcely be employed by 
 any one who has learned the modes of computation usually taught ill the Rule of P> at- 
 tics. It is therefore given in a note, and not in the text. 
 
64 
 
 COMPOUND MULTIPLICATION. 
 
 Exam. 6. Multiply 33 c. 3 q. 22 Ibs. by 7.* 
 
 In this example the pounds, when mul- 
 tiplied, are divided by 28, and the quar- 
 ters by 4. In long weight the divisors are 
 30 and 4. 
 
 cwt. q. Ibs. 
 33 3 22 
 7 
 
 237 2 U,Answ. 
 
 Exercises. Answers, 
 
 cwt. q. Ibs. cwt. q. Ibs. 
 
 Ex.93. 1 2 17X27 44 2 11 
 
 94. 3 22X86 81 1 16 
 
 95. 2 2 19 (long weight J X 59 156 3 11 
 
 96. 1 2 27 (long weight J X 96 165 2 12 
 
 97. Required the cost of a chest of tea, containing 97 lt>s. at 
 6/JO ^ Ib. Answ. 33 6 10 J. 
 
 98. Required the amount of a box of linen cloth, containing 
 as under: 
 
 ' piec. yd. per yd. 
 3 75 at 3s. Id. 
 3 75 39 
 3 68 3 11 
 3 75 42 
 Answ. 140 2 0. 
 
 99. Required the amount of a hogshead of rum containing G6 
 gallons, at 4/10 ty gallon, and a puncheon containing 116 gallons, 
 at 4/9 ^ gallon. Answ. 15 19 0, and 27 11 0. 
 
 100. What cost a hundred weight of indigo, at H/4 V pound? 
 Answ. 63 14 0. 
 
 piec. 
 
 yd. 
 
 per yd. 
 
 piec. 
 
 yd. 
 
 per yd. 
 
 2 
 
 49 
 
 at 2s. Id. 
 
 3 
 
 75 
 
 at 2s. 
 
 lid. 
 
 2 
 
 49 
 
 2 3 
 
 3 
 
 75 
 
 3 
 
 I 
 
 2 
 
 48 
 
 2 5 
 
 3 
 
 75 
 
 3 
 
 3 
 
 3 
 
 73 
 
 2 7 
 
 3 
 
 75 
 
 3 
 
 5 
 
 3 
 
 75 
 
 2 9 
 
 
 
 
 
 Exam. What cost 2485 yards of broadcloth, 
 at 15/7$ per yard. 
 
 t. d. 
 
 15 ~l~ price of 1 yard. 
 
 10 
 
 "7~~16 3 = ... 10 yards. 
 10 
 
 100 .. 
 
 10 = 
 
 312 10 = 
 
 62 10 = 
 
 3 18 1|= 
 
 ... 1000 
 
 2000 
 400 
 
 In this example we find successively the pricei 
 ef 10, 100, 1000. We then multiply the price of 
 1000 by 2; of 100 by4; of 10 by 8; and of 1 by 5. 
 We have thus the prices of 2000, of MX), of 80, 
 and of 5, the sum of which is 1941 8 1}, the 
 answer. 
 
 The following exercises may be wrought in a si- 
 ncilar manner : 
 
 Exercises. 
 s. d. 
 3 6 Six 3178 
 
 1 11 31x159.34 24913 9 5 
 
 2 6 9|x 938 2194 10 7 
 
 17 1|X63491 54430 6 1 
 
 1941 8 l|=z .. 
 
 5^ 
 
 2485 
 
 Answers. 
 
 s. d. 
 
 10556 18 4J 
 
 Ex.1. 
 
 2. 
 3. 
 4. 
 
 * Compound Multiplication is seldom employed, except in relation to money ; but if 
 it should be necessary to use it hi cases not illustrated here, no difficulty can arise, as 
 the method is similar in '^ l *-'*.'* 
 
COMPOUND DIVISION. $ 
 
 101. If a carpenter receive 18/4$ ^ week, what is his salary in 
 52 weeks? Answ. 47 15 6. 
 
 102. How many pounds are there in 91 guineas? Answ. 95 
 11 0. 
 
 103. What is the amount of the duty on 100 gallons of brandy, 
 at 13/7 V gallon? Answ. 67 18 4. 
 
 104. What is the duty on 63 gallons of rum, at 10/10 ^ gallon ? 
 Answ. .34 2 6. 
 
 105. What is the duty on 58 cwt. of raw sugar, at 1 126^ 
 cwt.? Answ. 94 5 0. 
 
 106. What is the amount of the duty on 149 Ibs. of West India 
 coffee, at 7d. V ft>.? Answ. 4 16 2f. 
 
 107. From 1783 to 1793, both inclusive, the money paid for 
 slaves, imported into the West Indies in Liverpool vessels, was, 
 at an average, 1,380,622 16 4 each year. What was the entire 
 amount? Answ. 15,186,851 1. 
 
 It may perhaps be proper to caution learners against the absurdity of 
 attempting *<> multijfy mcney by money. This caution will not appear 
 unnecessary, if it be considered that whole pages have been filled with 
 instructions how to perform this problem ; and it has been attempted to 
 be shown, even with the semblance of geometrical demonstration, that if 
 2/6 be multiplied by 2/6, the product may be either Sfd. or 6/3 ! 
 Let it be considered, however, that in Multiplication a quantity is sim- 
 ply repeated a given number of times : thus, if 2/6 be repeated 4 
 times, the amount is 10/; if 5 times, 12/6 &c. To talk, therefore, 
 of multiplying 2/6 by 2/6, or, which is precisely the same, of re- 
 pealing 2/6 2/6 times, is absolute nonsense. In the Rule of Pro- 
 portion, indeed, we sometimes appear to multiply such quantities. Thus, 
 in finding the interest of a sum at a given rate, for a year, we multiply 
 by the rate, and divide by 100. In this case, however, both 100 and 
 the rate are divested of their characters as expressions for money, and 
 are merely to be regarded as abstract numbers, used as the terms of a 
 ratio. By multiplying by the rate, suppose 5, we merely re]ieat the 
 principal 5 times, or find a principal 5 times as great; and then, as 
 there must be one pound ot" interest for each hundred pounds in this 
 increased principal, we try by Division how often it contains 100 ; and 
 we thus find the pounds of the interest. We see from the nature of 
 Division, that there is no absurdity in dividing money by money ; that 
 is, in finding how often one sum is contained in another. 
 
 COMPOUND DIVISION. 
 
 Problem 1. To divide a number of more denominations than 
 one, by a number not exceeding 12. 
 
 RULE. (I.) Divide the highest denomination by the given 
 divisor, by Short Division (2.) Reduce the remainder, if 
 
66 COMPOUND DIVISION, 
 
 there be any, to the denomination next lower, and add to 
 the result what was given of that denomination. (3.) Di- 
 vide the sum by the divisor ; and thus proceed to the lowest 
 denomination, or till nothing remains. 
 
 Exam. 1. Divide 14 16 7 by 10. 
 In this example, after dividing 14 
 by 10, we have remaining 4, or 80 shil- 
 lings; which, increased by 16, becomes 
 96 shillings. Hence, we find the next 
 part of the quotient to be 9 shillings, and 
 the remainder is 6 shillings, or 72 pence, 
 which, increased by 7, becomes 79. This 
 being divided by 10, we have the remain- 
 der 9 pence, or 36 farthings, to which the odd farthing is annexed; 
 and continuing the division, we find the entire quotient to be 1 
 9 7f, and the remainder 7 farthings, or Ifd. The proof is per- 
 formed as in Simple Division. 
 
 Exercises in Compound Division. 
 
 14> 
 
 s. d. 
 
 s : d. 
 
 s. d. 
 
 Ex.1. 10 11 8- 
 
 -2 
 
 6. 1 19 5- 
 
 - 7 
 
 11. 20 1 3- 
 
 -12 
 
 2. 2 16 7 - - 
 
 -3- 
 
 7. 3 15 7~- 
 
 - 8 
 
 12. 1 0- 
 
 - 6 
 
 3. 4 5 9f- 
 
 -4 
 
 8. 5 17 11- 
 
 - 9 
 
 13. 1 0- 
 
 - 7 
 
 4. 9 9 6- 
 
 5 
 
 9. 7 13 6 - 
 
 -10 
 
 14. 1 10 0- 
 
 - 8 
 
 5. 8 13 4 - 
 
 -6 
 
 10. 5 12 9 - 
 
 -11 
 
 15. 1 9 3- 
 
 - 9 
 
 Problem 2. To divide by a number which is greater than 
 12, but is the product of two or more factors, each less than 13. 
 
 RULE. (1.) Divide the given number, by Short Division, 
 by one of the factors. (2.) Divide the quotient by another 
 factor. (3.) Divide the result thus obtained by another fac- 
 tor, if there be so many: and thus proceed, whatever may 
 be their number. 
 
 Exam. 2. Divide 59 13 3^ by 66. 
 
 In this example the factors are 6 and 
 11. In the division by 6 the quotient is 
 9 18 10^, and the remainder 2 far- 
 things; and in the division of this quo- 
 tient by 11, the quotient resulting is 
 18/Of, and the remainder 9. This re- 
 mainder being multiplied by 6, the first 
 divisor, and the product increased by the 
 former remainder 2, (see page 27,) the 
 true remainder is found to be 56 farthings, 
 or , 1/2. In the proof by Multiplication, 
 this remainder must be added to the final product. 
 
 
 
 6)59_ 
 119 
 
 5 8 
 
 59 12 
 
 59~3 
 
 d. 
 
 Jt 
 
 10^.. .2 
 
 Of...5Gfar 
 6 [or 1/2. 
 
 proof 
 
COMPOUND DIVISION. 57 
 
 In the use of this Rule, if there be no pence in the dividend, or if 
 it tjnd|p 3, 6, or 9 pence, and if one of the factors be 3, 6, or 12, it 
 is better to use that factor first, as in the division by it there will be no 
 remainder. If ne factor be 2, 4, or 8, and there be no farthings in 
 the dividend, it is generally better to begin with that factor. If the 
 shillings and pence of the dividend be any multiple of 10 pence, (as 
 1/8, 2/6, 3/4, 15/10, &c.) and one of the factors be 5 or 10, it is 
 best to begin with that factor. The same may be observed in rela- 
 tion to multiples of 5d. and 2d. (and, when 5 is a factor, in relatio" 
 to multiples of l^d. ;) but these are not so easily discovered. 
 
 Exercises. 
 
 Answers. 
 
 150 ,, 18 
 
 6 180 12 
 
 8d. 
 5|d. 
 
 2 id. 
 6d. . 
 
 74 
 
 V2* 
 1/0* 
 9d. 
 
 6d. 
 
 4d. 
 
 1/0 
 
 2/5* 
 
 1/6 
 
 2/6 
 1/3 
 
 Problem 3. To divide by a number which is greater than 
 12, and is not produced by factors below 13. 
 
 RULE. The process is to be conducted as in problem 1, 
 except that Long Division is to be employed instead of 
 Short. 
 
COMPOUND DIVISION. 
 
 Exam. 3. Divide 2074 6 9^ by 597. 
 Here, the first part of 
 
 the quotient is 3, and 
 the remainder 283 6, 
 or 5666 shillings, from 
 which, as in Simple Di- 
 vision, we obtain for quo- 
 tient 9 shillings, and a re- 
 mainder of 293 shillings 
 and 9 pence, or 3525 
 pence. Dividing this by 
 the divisor, we obtain 
 for quotient 5 pence, 
 and a remainder of 540^ 
 pence, or 2161 farthings, 
 which gives a quotient of 
 3 farthings, and a remain- 
 der of 370 farthings, or 
 7/8. The work is proved 
 by multiplying the quoti- 
 ent by 600(6X10X10,) 
 and subtracting three time 
 597, and finally, by addir 
 
 Exerci 
 s. d. 
 Ex. 46. 61 - 
 
 
 597)2074 
 1791 
 
 ~283 
 20 
 
 s. d. s. d. 
 6 9 (3 9 b'i,Amw. 
 6 
 
 20 16 iutj 
 10 
 
 5666 
 5373 
 
 208 8 9 
 
 293 
 12 
 
 2084 7 6 
 10 8 5| 
 
 3525 
 
 2985 
 
 2073 19 0} 
 7 8 
 
 540 2074 6 94, 
 4 Proof. 
 2161 
 1791 
 370 farthings, or 7/8 . 
 
 ;s the quotient, to obtain the product by 
 g the remainder, 7/8, to the result. 
 
 ses. Answers. 
 s. d. Rem. 
 
 _ 13 4, 13 10 9(\ 
 
 47. 937- 
 48. 50 4 2 - 
 49. 99 - 
 50. 45 10 Oi- 
 51. 115 12 6 - 
 52 93 9 4^ 
 
 - 17 . 
 
 10 9^ lAd 
 
 - 19 
 
 .. 2 12 10 4d 
 
 _ 26 
 
 .. 3 16 1} 2d 
 
 _ 29 
 
 ..111 4}.. .. l|d 
 
 _ 37 
 
 ..326 
 
 _ 4,3 
 
 ..23 5^... . 8d 
 
 53. 82 6 11 - 
 54 67 3 10} 
 
 _ 51 
 
 .. 1 12 3| d. 
 
 _ 57 
 
 .. 1 3 6| lOd 
 
 55. 54 13 2 - 
 56. 108 18 1|- 
 57. 329 4 - 
 58. 167 16 - 
 59. 53 12 - 
 60. 288 5 - 
 61. 344 14 
 
 _ 65 
 
 .. 16 9^ 4|d 
 
 _ 71 
 
 .. 1 10 8 . 9Ad 
 
 _ 78 
 
 ..44 4| i/fj 
 
 _ 85 
 
 .. 1 19 5 a . 3^d 
 
 _ 91 
 
 ..Oil 9^ 10 jd 
 
 _ 97 
 
 .. 2 19 5 1/7 
 
 _ 103 
 
 ..3 611 1/7 
 
 62. 179 6 - 
 63. 599 24- 
 64. 342 11 4|- 
 65. 400 - 
 66. 9846 13 4 - 
 67. 2045 16 5}- 
 68. 3982 2 8i 
 
 _ 133 
 
 .. 1 6 11 . 6|d 
 
 _ 201 
 
 .. 2 19 7^.. . 1/10 ;; 
 
 _ 313 
 
 . 1 1 10 1 . 4/6 
 
 _ 365 
 
 .. 1 in 5d. 
 
 5375 
 
 1 17 4 
 
 
 .. 10 0^ 1/4| 
 
 1347 .. 
 
 ..219 I 1 . ]/* 
 
COMPOUND DIVISION. 69 
 
 70. If the duty on a pipe of port wine, containing 138 gallons, 
 be .52 6 6, how much is the duty on each gallon? Answ. 7/7. 
 
 71. If a chest of tea, containing 96 tbs. cost 33, what cost 
 1 ft.? Answ. 6/10i. 
 
 72. If a hundred weight of sugar cost 4: 8 8, what are the 
 costs of a pound and of a stone? Answ. 9^d. and 11/1. 
 
 73. If a contribution of 354 12 'is to be made up in equal 
 shares by 26 persons, how much must each contribute ? Answ. 
 13 12 9, rem. 6d. 
 
 74. If prize-money to the amount of 495 5, is to be divided 
 equally between 75 seamen, how much will each receive ? Answ. 
 6 12 Of, rem. 3fd. 
 
 75. If a person spend 200 a-year, how much does he spend, 
 at an average, each day, and how much each week ? Answ. 10/1 L}, 
 rem. 2d.; and 3 1611, rem. 4d. 
 
 76. The following are the amounts of the duties paid at the 
 custom-house of Belfast, during the undermentioned years: re- 
 quired the annual sum, at an average. 
 
 In 1800, 
 
 62,068 
 
 In 1806, 
 
 1801, 
 
 182,314 
 
 1807, 
 
 1802, 
 
 270,434 
 
 1808, 
 
 1803, 
 
 201,180 
 
 1809, 
 
 1804, 
 
 207,402 
 
 1810, 
 
 1805, 
 
 228,645 
 
 1811, 
 
 In 1812, 395,254 
 1813, 450,498 
 1814, 
 1820, 
 
 373,721 
 306,263 
 386,709 
 
 207,382 
 320,981 
 318,121 
 425,174 
 
 321,325 1821, 
 344,449 
 
 Answ. 294,265 17 71 f. 
 
 77. In 1821, the population of Great Britain and Ireland was 
 2 1 ,238,580, and the net amount of the public revenue of the United 
 Kingdom was 58,108,855 2 2|. What was the quota of this 
 amount paid by each individual, at an average ? Answ. 2 14 8i, 
 rem. 12489 8 Of. 
 
 78. In the year 1821, there were coined in the British mint 
 203,761 pounds of gold, value 9,520,732 14 6. Required the 
 value of each pound. Ansiv. 46 14 6. 
 
 79. How much land is there, at an average, for each individual 
 in England? (See page 1 1, Ex. 23, and page 53, Ex. \Q3.J Answ. 
 *a.3r. lOfiWrWrP- 
 
 80. How much land is there, at an average, for each individual 
 in Ireland? (Set- page 11, Ex. 24.^ Answ. 1 a. 2 r. 39 B V&V:& P- 
 late Irish measure. 
 
 It may not be improper here to shrw the method of multiplving and 
 dividing, when the multiplier or divi or contains a fraction, though 
 these oj erations really belong to the di. urine of fi action*. 
 
70 SIMPLE PROPORTION. 
 
 In the annexed example, to multiply 
 
 by 22, we first multiply by 22, in the 1 16 6 X 22 J 
 
 way already explained. Then, for f , we 
 multiply in a separate place 1 16 6 by 
 3, and divide the product by 4: the re- 
 sult, is 1 7 4, which being added to 
 the product by 22, the sum is 41 10 
 4, the product required. The work 40 3 
 
 for the fractional part is left to the 1 7 4 
 
 learner to perform. If the multiplier 
 had been 22 or 22|, we should have 41 10 4 
 found the product by 22, as before; and 
 
 in the one case, have added to it one-fourth, in the other one-half 
 of the multiplicand. 
 
 As an example in this 22f)30 2 10$ 
 
 kind of division, let it be _4_ *_ s - d - 
 
 required to divide 30 2 91)12011 6(1 6 6 
 
 104 by 22|. Here we 91 
 
 multiply 22 by 4, the low. 39 
 
 er term of the fraction, 20 
 
 and to the product 88, we ' VQI 
 
 add 3, the upper term. 
 
 We multiply the given di- _ 
 
 vidend also by 4. Then, 4*5 
 
 dividing this product by 
 91, we find 166 for 546 
 
 the required quotient. 546 
 
 It may be observed, that 
 these methods are equal- 
 ly applicable in Simple and Compound Multiplication and Division. 
 
 SIMPLE PROPORTION, 
 
 IN comparing the magnitudes of two quantities of the 
 same kind, the flsotieni, which is obtained by dividing the 
 first by the second, is called the RATIO of the first to the 
 second. The two magnitudes are called the TERMS of the 
 ratio, the first the ANTECEDENT, the second the CONSE- 
 QUENT. 
 
 Thus, the ratio of 42 to 14 is 3, or is that of 3 to 1 ; and the ratio 
 of 7 to 4 is If, or is that of if to 1. In like manner, if the second 
 term be divided by the first, the quotient will be the ratio of the second 
 to the first. 
 
 Equality of ratios constitutes PROPORTION or ANALOGY; 
 and the terms of equal ratios are called PROPORTIONALS : 
 the first and last the EXTREMES, the rest 'the MEANS. 
 
SIMPLE PROPORTION. 71 
 
 Thus, since 8 is one-half of 16, and 6 one-half of 12, the fofer 
 numbers, 8, 16, 6, and 12, are proportionals, and constitute a propor- 
 tion or analogy. When considered in this relation, they are usually 
 written thus, 8 : 16 : : 6 : 12, and are read, S is to 16, as 6 is to 12 ; or, 
 they are written, as 8 : 16 : : 6 : 12, and read, as 8 is to 16, so is 6 to 12. 
 
 SIMPLE PROPORTION is the equality of the ratio of two 
 quantities, to that of two other quantities. 
 
 COMPOUND PROPORTION is the equality of the ratio of 
 two quantities to another ratio, the antecedent and conse- 
 quent of which are respectively the products of the ante- 
 cedents and consequents of two or more ratios. 
 
 The object of that part of Simple Proportion which is 
 usually taught in courses of Arithmetic, is to find the num- 
 ber which has the same ratio to one of three given numbers, 
 that there is between the other two ; or, to find a fourth 
 proportional to three given numbers. 
 
 Three numbers being given, to find a fourth proportional.* 
 
 RULE. (1.) Arrange the three* given terms in the same 
 line, in succession, placing the one which is of the same 
 kind with the required term, the third in order ; and if, by 
 the nature of the question, the required term is to be 
 .greater than the third term, put the greater of the other 
 two terms in the second place ; otherwise, put the less in 
 that place. (2.) Then, if the two first terms be not of the 
 same simple denomination, reduce them to the same deno- 
 mination, usually the lowest mentioned in either. (3.) Find 
 the product of the second and third terms, and divide it by 
 the first ; the quotient is the fourth proportional in the same 
 denomination as the third term, and may be reduced to a 
 higher denomination, if necessary. 
 
 Reason of the Rule. 
 
 By the definitions of ratio and proportion, if the first of fotir 
 proportionals be divided by the second, and the third by the fourth 
 the quotients are equal: and if the first of these quotients be mul- 
 tiplied by the second and fourth terms, and the second of them 
 by the fourth and second terms, the first result will be the pro- 
 
 * The resolution of this problem is unquestionably the most important result of the 
 doctrine of proportion. On account of its great utility and very general application, it 
 is often called in the older works on Arithmetic, the Golden Rule. From the circum- 
 stance of three terms being given to find a fourth, it .Mas al<n been very generally called 
 the Rule of Three, a name which should be disused, oi it conveys no idea of thon.it ur 
 or Direct of the problem. 
 
11 SIMPLE PROPORTION. 
 
 duct of the extremes, and the second that of the means. But 
 these products must be equal, since it is evident, that if equals be 
 multiplied by equals, the products are equal. Hence, it follows 
 that the product of the extremes is equtd to ike product of the means: 
 and therefore if the product of the means (the second and third 
 terms,) be divided by one of the extremes, the quotient must be 
 the other.* 
 
 Thus, in the proportionals 9 : 3 : : 6 : 2, if the quotient of 9 by 3, 
 be multiplied by 3, the product is 9, which must be equal to three times 
 the quotient of 6 by 2, or to the quotient of 3 times 6 by 2 : and mul- 
 tiplying these equals by 2, we have '2X9=3X6; that is, the products 
 of the extremes and means are equal. This will be better understood 
 if the pupil has learned fractions; since, if the two equal fractions 
 and , be multiplied successively by 3 and 2, the products are 2X9 an d 
 3X6, which must evidently be equal. 
 
 # It is of the utmost importance, that the pupil should, as early as possible, acquire 
 an accurate idea of proportion. The definitions of ratio and proportion given in tl^e 
 text, are the fame in substance as those adopted by Lacroix and others, and will per. 
 haps convey to the mind of the learner, a clearer idea of proportion than any other, 
 particularly after he has acquired some knowledge of fractions. It will be found to be 
 of advantage in enabling him to understand the subject, to require him, after having 
 weight an exercise, to arrange iu succession the three given terms, and the fourth, 
 when found, and to examine them according to the definitions. In a short time he will 
 thus acquire a correct idea of this interesting subject, and be able to apply it with ease 
 and success, not on'y in Arithmetic, but in any other part of the Mathematics with the 
 principles of which he may be acquainted. 
 
 It, is scarcely necessary to state to the arithmetical reader, that while all writers are 
 perfectly agreed with respect to the nature, operations, and uses of proportion, they 
 differ very considerably in cheir modes of defining it, and establishing its principles. 
 Those who wish for farther information on this subject considered as a branch of 
 mathematical science, may have recourse to the Elements of Euclid, and the com- 
 ments of the later editors of that work, particularly Simson, Playfair, and Elrington. 
 The last of the.-e writers substitute*, instead of Euclid's definition, one wliieh is per- 
 haps more convenient than any other, as a basis for establishing the doctrine of propor- 
 tion on strict principles. 
 
 If four magnitudes a, 6, c, d, be proportional, according to the definition given iu 
 
 a c 
 
 the text, we have -= ; and hence, multiplying by b and d, we find ad=bc t which af- 
 
 b d 
 
 fords a proof to the reader, who is acquainted with Algebra, that the product of the 
 extremes is equal to that of the uu-aiis. 
 
 In the method of arranging the terms which is delivered in almost all the books on 
 Arithmetic, and which is very generally employed in practice, the term which is of the 
 sain,; kind with the answer, is put in the second place. This method is entirely subver- 
 sive of the principles of proportion, and is calculated to prevent the learner 
 from acquiring just views of this subject, as in it a ratio is, in many cases, instituted 
 between quantities entirely heterogeneous. The method above explained, which has 
 been adopted by Mr. Walker, and some other late writers, is founded on principles 
 etrictly Mathematical ; and besides being very simple and easy in practice, it possesses 
 the advantage of training the mind to accurate thinking, and of preparing it for tht 
 subsequent study of Mathematics. It also precl-ides the necessity of what is generally 
 called tlic Rule of Three Inverse, as r.H the questions usually solved under that head, 
 may lie folved by the rule above delivered 
 
SIMPLE PROPORTION. 73 
 
 When the first and second terms are in the same denomination, 
 they are evidently in the same ratio as the numbers which express 
 them; and therefore they are to be reduced to the same denomi- 
 nation, if they be not so already. 
 
 Exam. 1. If 12 yards of broadcloth cost 15, what would 
 8 yards cost at the same rate ? 
 
 The answer to this question must evidently be in money, and 
 therefore 15, the money given in the ques- 
 tion, must occupy the third place: and, as yds. yds. . 
 8 yards will cost less than 12 yards, 8 yards, As 12 : 8 : : 15 : 
 the less of these two terms, must be put in 8 
 
 the second place. Hence, the terms will be 
 
 arranged, and the operation performed, as in 12)120 
 
 the margin; and it appears, that at the rate 
 specified in the question, 8 yards would Aiisw. 10, 
 
 cost 10. 
 
 In this example, it is evident that as often as 8 yards are contained 
 in 12 yards, so often would the answer be contained in 15. Hence 
 the quantities are proportionals, and the reason of the process is evi- 
 dent from what goes before. 
 
 The reason of the operation will also be plain if it be considered, 
 th it the price of one yard would be found by dividing .15 by 12, and 
 tnat the price of 8 yards would be found by multiplying the quotient 
 by 8 ; and the result will evidently be the same that would be obtained 
 by multiplying 15 by 8 and dividing the product by 12* The method 
 According to the rule is in general preferable however, as it has the ad- 
 v mtage of freeing the operation, as much as possible, from fractional 
 quantities. 
 
 The operation may also be illustrated very simply in the following 
 manner: since 12 yards cost '15, by multiplying by 8 we obtain '120, 
 which must be the price of 96 (8X12) yards; and therefore dividing 
 this by 12, we get 1O, the price of 8 yards. 
 
 Ex. 1. If 57 cwt. of sugar cost 216, what would 95 cwt. cost 
 at the same rate ? Answ. 360. 
 
 2. If 385 yards of linen cost 63, how much might be bought 
 for .18 at the same rate? AHSW. 110 yards. 
 
 3. How much wine may be bought for 396, if 90 gallons cost 
 72 ? Answ. 495 gallons. 
 
 4. If the yearly rent of a farm of 182 acres be 273, what is 
 the rent of a part of it containing 42 acres ? Answ. 63. 
 
 5. If 275 reams of paper cost 330, what would 990 reams 
 cost? Answ. 1188. 
 
 6. If 96 men reap 40 acres of grain in a week, how many would 
 reap 65 acres in the same time? Answ. 156. 
 
 7. How many men would perform, in 168 days, a piece of work 
 which 108 men can perform in 266 days?* Answ. 171. 
 
 * This question belongs to v, hat is commonly called the Rule of Three Inverse as the 
 
SIMPLE PROPORTION. 
 
 8. If 84 sheep can be grazed in a field for 12 days, how long 
 might 112 sheep have been grazed equally well in the same field 
 Answ. 9 days. 
 
 Exam. 2. If 12 cwt. of wheat cost 6 10 8, how much 
 might be bought for 11 8 8 ? 
 
 In this example, according 
 to the second part of the rule, 
 the first and second terms are 
 reduced to the same denomina- 
 tion, (pence,) and the rest of 
 
 the work proceeds as before. 1568)32928(21 cwt. 
 
 3136 Answ. 
 1568 
 1568 
 
 Ex. 9. If 148 gallons of mm cost 119 10, how much maybe 
 bought for 89 12 6? Answ. Ill gallons. 
 
 10. If 114 be paid for 52 cwt. 1 q. 4 Ifes. of flour, what would 
 122 cwt. cost at the same rate? Answ. 266. 
 
 11. What is the rent of 21 acres, 3 roods, 20 perches of ground, 
 if the rent of 36 acres, 3 roods be 42 ? Answ. 25. 
 
 12. If a person walk 17 miles in 5 hours, 12 minutes, 31 se- 
 conds, how far would he walk at the same rate in 3 hours. 40 
 minutes, 36 seconds ? Answ. 12 miles. 
 
 13. If 51 pay for 10 cwt. 2 qrs. 1 4 Ibs. of sugar, wjiat would 
 4 cwt. 1 qr. 14 Ibs. cost at the same rate ? Answ. 21. 
 
 f4. If 19 cwt. 3 qrs. 21 Ibs. of Beef (long iveightj cost 36, 
 what would 48 cwt. 1 qr. 29 Ibs. cost at the same rate ? Answ. 84%, 
 
 15. If the fare of a coach for 88 miles be 1 9 4, for what 
 distance would 2 3 4 pay at the same rate? Answ. 130 miles. 
 
 16. If. the earth move 69000 miles in its orbit in an hour, 
 through what space does it move at the same rate in 16 minutes, 
 48 seconds? Answ. 19320 miles. 
 
 If there be a remainder after the division by the first 
 term, it is of the same denomination as the third term ; and 
 if it admit, it may be reduced lower, and the operation 
 continued as in Compound Division. 
 
 greater the number of the men is, the less must be the number of days. In this case, 205 
 is said to have to 168 the reciprocal or inverse ratio of 108 to 171, and the numbers are 
 aid to be reciprocally proportional. The direct analogy is 168 days : 266 days : : 108 
 men : 171 men. It is obvious, that the terms must be thus arranged : since 266x10? 
 would be the number of days in which the work would be performed by one man ; arxi 
 if this be divided by 168, the quotient must be the number of days required : conse- 
 quently, 266 and 108 must be made the second and third terms, that their product may 
 be taken in the operation. The next question is of. the same kind. 
 
 n is of the 
 
S1IMPLE PROPORTION. 
 
 Exam. 3. If a web of linen con- 
 taining 26 yards cost 3, what would 
 732 yards cost at the same rate ? 
 
 Here, after we have found 84, the 
 first part of the answer, there is a 
 remainder of 12, which being reduced 
 to shillings, the rest of the operation 
 proceeds by Compound Division, and 
 there is a final remainder of 2 far- 
 things. 
 
 In the next exercise, the first and se- 
 cond terms are first reduced to the 
 same denomination, and then the work 
 proceeds as in this example. When 
 the third term is a unit, the second is 
 simply to be divided by the first. 
 
 yds. 
 As 26 : 
 
 d. 
 
 2* 
 
 Answ. 
 
 
 32 : : 3 : 
 
 _3 *. 
 26)2196(84 9 
 208- 
 116 
 
 ~12 
 20 
 
 240 shillings. 
 234 
 
 12 
 
 20 
 
 80 farthings. 
 78 
 
 Ex. 17. If the rent of 5 acres be 4 13 4, how much land 
 could be rented at the same rate for 70 10 6 ? Answ. 75 a. 2 r. lOp. 
 
 18. How much wheat, at 18/3 ^ cwt. may be purchased for 
 149 16 4? Answ. 164 c. q. 20?f Ibs. 
 
 19. How much oats, at 8/8 ^ cwt. may be bought for 64 
 3 2? Answ. 148 c. q. 7 Ibs. 
 
 20. If 6 c. 3 q. 12 Ibs. of flour cost 9, what would 4 c. Sqrs. 
 cost at the same rate? Answ. 5 18 1. 
 
 21. If 39 c. 1 q. 10 Ibs. (long weight J of pork cost 59 0, 
 what cost 13 cwt.? Answ. 19 10. 
 
 When the third term is of more denominations than one, 
 it is generally necessary, or at least proper, to reduce it to 
 the Jowest denomination mentioned in it. When the first 
 an^ second terms are not large, however, it is often bet- 
 ter not to perform this reduction, but to employ Compound 
 Multiplication and Division, especially if the third term be 
 money. 
 
 The work is often much abbreviated by dividing the first 
 and second, or the first and third terms, (but never the se- 
 cond and third,) by any number which exactly measures 
 them, and employing the quotients instead of the numbers 
 themselves. 
 
A. 
 
 As 1 
 160 
 
 R. 
 
 3 
 
 A. 
 
 47 
 4 
 
 191 
 
 40 
 
 7661 
 294 
 
 30644 
 68949 
 15322 
 
 4)22523314 
 4) 56308... 14 
 12) 14077... 14 
 2lO)117|3 1 
 58 13 ! 
 
 
 
 : 1 
 20 
 
 7"> SIMPLE PROPORTION, 
 
 Exam. 4. What is the yearly rent of 47 a. 3 r. 21 p. at 1 4 6 
 <t? acre? 
 
 In this example, the first 
 and second terms are reduced 
 to perches, the lowest deno- 
 mination mentioned in either, 
 
 and the third to pence, the 191 24 
 
 lowest mentioned in it. 40 12 
 
 Then, by the usual process, 
 the answer is found to be 
 14077 pence, with 14 re- 
 maining ; and this, by proper 
 reduction, becomes 58 13 
 1 rW I n tne division by 
 160, a cipher is cut from the 
 divisor, and a figure from the 
 dividend; and then the work 
 is contracted, by using the 
 factors of 16, instead of it- 
 self. With respect to the 
 reason of the operation, it is 
 evident, that were the third 
 term, 294d. divided by the 
 first term, 160 perches, the 
 
 quotient would be the rent ,58 13 Ij 1 ^, Answ. 
 
 of one perch ; and were this 
 
 multiplied by 7661, the perches in the second term, the product 
 would be the rent of 47 a. 3r. 21 p. But as in other cases, the 
 same result is found, and more easily, by dividing the product of 
 the second and third terms by the first. 
 
 yds. yds. 
 As 15 : 55 
 
 Exam. 5. If 15 yards of linen 
 cost 2 1 10, what will 55 yards 
 cost at the same rate? 
 
 s. d, 
 1 lOi 
 
 
 
 2 
 
 20 
 
 41 shillings. 
 12 
 
 In this example, after the terms 
 are arranged according to the rule, 
 the last is brought to farthings, the 
 lowest denomination mentioned in 
 it; these farthings are then multi- 
 plied by 55, the second term, and 
 the product divided by 15, the first 
 term; the quotient is 7370, which 
 is farthings, because the third term 
 was reduced to farthings ; and these 
 farthings, by reduction, become 7 
 13 6, the price of 55 yards, as 
 was required. 
 
 502 pence. 
 
 4 
 
 20 10 farthings. 
 
 55 
 
 10050 
 10050 
 5)110550 
 3)22110 
 
 4) 7370 farthings. 
 12)1842$ . 
 2lO)15|3 6 
 7 13 6, Ans. 
 
SIMPLE PROPORTION. 77 
 
 This operation may be much con- 
 tracted in the following manner: 
 let the numbers be arranged as be- 
 fore, and let the first and second 
 . terms be divided by 5, which is a 
 common measure; the quotients 
 are 3 and 1 1, which are to be used 
 instead of 15 and 55; and the ope- 
 ration is still farther abbreviated 
 by employing Compound Multipli- 
 cation and Division. + 
 
 The reason of the contraction employed in this example is, that 
 if the divisor and dividend be multiplied or divided by the same 
 number, the quotient is not changed. Thus, if 12 be divided by 
 4, the quotient is the same as if 24 be divided by 8, or 3 by 1 : 
 and hence the ratio of two numbers is the same as the ratio of 
 two numbers which are obtained by dividing or multiplying them 
 by the same number- This principle, and indeed the whole doc- 
 trine of Proportion, will be better understood by the pupil who is 
 acquainted with fractions. 
 
 Ex. 22. If 63 gallons of wine cost 41 10 6, what cost 10 
 gallons?* Answ. 6 11 9f, rem. 1/2%. 
 
 23. What cost 4 tons of logwood, if 36 tons cost 316 73? 
 Answ. 35 3 0^, rem. % f. 
 
 24. If a lot of sugar, containing 56 cwt. cost 214 18 4, what 
 does it cost V ton? Answ. 76 15 \\, rem. f f. 
 
 25. If the yearly rent of 18 acres be 24 18 6, what is the 
 rent of 29 acres at the same rate? Answ. 40 3 1 rem. f. 
 
 In working this exercise, and the nine following, the third term in the analogy in 
 each may be reduced to the lowest denomination mentioned in it, and the product of 
 rie result and the second term divided by the first In all of them, however, and in 
 several others that follow, the reduction may be omitted, and the work performed by 
 Qunpourid Multiplication and Division. In exercise 23, the first and second terms may 
 lie divided by 4 ; and then as the second term will be 1, the answer will be found by di- 
 Aiding the third term by 9. In exercise 24, divide the first and second terms, in hun- 
 dreds weight, by 4; and in exercise gf;, divide them by 2. In hke manner, in exercise 
 1, the first and second terms may be divided by 19 ; in exercise 2, by 9 ; in exercise 3, 
 the first and second, or the first and third terms by 18, &c. 
 
 In this and the following exercises, the remainders, when there are any, are annexed 
 to the answers, and are expressed in their simplest form by dividing each remainder, 
 and the divisor which produces it, by any number that is contained in them with. 
 out remainder. (See the preceding paragraph.) It is generally sufficient, however, 
 to require the learner to find the answer, if it be money, true to the nearest farthing. 
 To require all answers to be brought out true to the smallest fractional parts, would oc- 
 cupy a great deal of time which maybe more advantageously sjwnt ; and such a degree 
 IK accuracy is scarcely ever aimed at, and indeed can scarcely ever be attained, hi any 
 practical application of Arithmetic. In viewing he matter in this light, when farthings 
 aro wrought for, if the remainder be more than half the divisor, it may be rejected and 
 one added to the farthings. Thus, in exercise 22, the answer might be taken 611 10 ; 
 it, exercise 24. 76 15 1J ; in exercise 25, 40 3 1| in exercise 27, 96 18 1J, &c. 
 
78 SIMPLE PROPORTION. 
 
 26. If 10 cwt. of oatmeal cost 9 15, what will 252 cwt. cost 
 at the same rate? Answ. 246 14 0. 
 
 27. What is the value of 1 10 gallons of Spanish brandy, at the 
 rate of 5 3 4 for 7 gallons ? Answ. 96 181, rem. }d. 
 
 28. What 'cost 126 cwt. of oatmeal, if 5 cwt. cost 4 17 6? 
 Answ. 122 17 0. 
 
 29. L 19 gallons of whiskey cost 1 169, what will 37 gallons 
 tost at die same rate? Answ. 15 5 3. 
 
 x 30. What cost 79 cwt. of flour, if 19 cwt. cost 24 12 6? 
 
 Mtuw. 102 7 9, rem. &d. 
 4^31. If 23 yards of broadcloth cost 25 18 6, what cost 98 
 
 Yards? ^n^110 9 3, rem. ^ y d. 
 
 32. How much cambric may be bought for 186 2 4, if 13 
 yards cost 8 6 3J? Answ. 291 yards. 
 
 33. At 1/8 V pound, how much coffee may be bought for 
 17 12 7$? Answ. 1 c. 3 q. 15$$ Ibs. 
 
 34. How much wine, at the rate of 31 15 4 for 46 gallons, 
 may be bought for 117 16 8? Answ. 170f-g gallons. 
 
 35. How much linen may be bought for 41 12 6, if 405 yards 
 cost 69 76? Answ. 243 yards. 
 
 36. How much cotton-wool, at 2/10 V pound, may be bought 
 for ,39 14 8? Answ. 276ff ibs. 
 
 37. How much barley, at 13/9 V cwt. maybe bought for 
 62 7 9? Answ. 90 c. 3 q. 
 
 38. If 6 yards of muslin cost 1 6 6, what will 79 yard? 
 cost at the same rate? Answ. 16 2 7. 
 
 39. What cost 78 yards, 3 quarters, and 2 Hails of broadcloth 
 at 17/4 F-yard? Answ. 68 7 2. 
 
 40. If 17 c. 3 q. 19 Ibs. of barley cost 8 18 9, how much 
 may be bought for 5 12 6, at the same rate? Answ. '11 c. 1 q. 
 3-^ Ibs. 
 
 41. How much tea, at 7/6 W lb. may be bought for 37 26? 
 Answ. 99 Ibs. 
 
 42. If a farm containing 396 a. 3 r. 10 p. be let at the yearly 
 rent of 740 14 4, what is the rent of each acre ? Answ. 1 17 4. 
 
 43. What cost 4 cwt. 2 qrs. 20 Ibs. (long weight J of pork, at 
 3 14 6 tf* cwt.? Answ. 17 7 8. 
 
 44^1? 172 cwt. 2 qrs. 18 Ibs. of rye cost 87 6 3, how much 
 may be bought at the same rate for 3 1 9 4 ? Answ. 7 c. 3 q. 11 Ibs. 
 
 45T What cost 124 cwt. 2 qrs. 14 ibs. of ricei at 2 1 6 V 
 rt.? Answ. 258 11 11. 
 
 46. If the prime cost of 247 cwt. 3 qrs. 14 Ibs. of flour, togtv 
 ther with the charges, amount to 423 9 Of, what does it stand 
 3?" cwt.? Answ. 1 14 2. 
 
 47. ^HoV much wheat, at 15/2 V cwt. may be bought for 113 
 15? Answ. 150 cwt. 
 
 43. If 2 cwt. 3 qra. 21 Ibs. of sugar cost 12 3 4, what is the 
 value of 17 cwt. 2 qrs. 14 ibs. at the same rate? Answ. 73, 
 
SIMPLE PROPORTION. 79 
 
 49. At 9 2 ^ cwt. what cost 3 Ibs. of raisins ? Answ. 4/10$. 
 
 50. If C cwt. qrs. 29 Ibs. (long weight J of beef, cost 13 7 6, 
 how much may be bought for 17 12 6 ? Answ. 8 c. a. 27 Ibs. 
 
 51. What cost 39 yards, 3 quarters, 3 nails of broadcloth, if 
 3 yards, 2 quarters, 1 nail cost 3 2 10 ? Answ. 35 4 4f , nearly. 
 
 52. If the rent of 175 a. 2 r. 30 p. be 278 3^, how much 
 land can be rented at the same rate for 1 11 8 ? &nsw. 1 acre. 
 
 53. What cost 365 bottles of wine at 2 13 ^ dozen? Answ. 
 81 7 3$. 
 
 54. What cost 311 sheep, at 37 12 6 V scpre? Answ. 585 
 1 4. 
 
 55. How much sugar maybe bought for 113 13 4, if 7 c. 3q. 
 14 Ibs. cost 31 68? Answ. 28 c. 2 q. 7|f Ibs. 
 
 56. If the rent of 12 a. 2 r. 30 p. be 28 8 9, what is the rent 
 of 59 a. 3 r. 20 p. at the same rate? Answ. 134 4 0, rem. ^ f. 
 
 57. How much muslin, at 2/8 ^ yard, is equal in value to 
 169 yards of cambric, at 7/8 fc yard? Answ. 481 yards. 
 
 58. If the assessment on a townland, containing 445 acres, be 
 14 14 9f, how much is it on each score of acres? Answ. 13/3. 
 
 59. If a clerk have a salary of 89 12 6 V year, commencing 
 at the first of May, how much has he to receive on leaving hj^ 
 situation on the 18th of December? Answ. 56 14 5f. 
 
 60. If a person save 8/1 V week, in what time will he save 
 112 10? Answ. 5 years, 16Jf weeks. 
 
 61. If the shilling loaf weigh 3 Ibs. 6 oz. when flour sells aV 
 1 13 6 If cwt. how much should it \veighwhenfloursellsat 
 176^ cwt. ? Answ. 4 Ibs. lf$ oz. 
 
 62. If the digging of a British mile qf a canal cost 1347 7 6, 
 what cost the digging of an Irish mile at the same rate ? Answ, 
 1714 16 9|, rem. & f. 
 
 63. When wheat sells at 54/7 <p (quarter (of 456 Its.), how 
 much is it f cwt.? Answ. 13/5. 
 
 64. What is the value of a ton of gold, the guinea weighing 
 5 dwts. 9 grs.? Answ. 127,627 18 I ; rem. |f. 
 
 65. At 7/6 ^ ounce, what is the value of a silver bowl weigh- 
 ing 9 oz. 13 dwts. 8 grs. ? Answ. 3 12 6 
 
 66. If a person travelling 14 hours *& day, finish the first half 
 of a journey in 9 days, in what time will he finish the remaining 
 half, travelling 10 hours ty day at the same rate each hour ? Answ. 
 12| days. 
 
 67. If a person walk from Belfast to Newry, a distance of 30 
 miles, in 10 hours, 42 minutes; in what time will he walk at the 
 same rate from Belfast to Dublin, the distance being 80 miles ? 
 Answ. 28 h. 32 m. 
 
 68. How many British miles are equivalent to 128 Irish, 11 
 Irish miles being equivalent to 14 British? Answ. 162 m. 7 fur. 
 10 p. 5 vds. 
 
 69. 'fhe mean length of a -.iej^ee on the earth's surface, 
 
80 SIMPLE PROPORTION. 
 
 British miles and 80 yards, nearly: how much is it in Irish miles? 
 
 Answ. 54 in. 2 fur. 
 
 70. It is found, that the diameter of every circle is to its cir- 
 cumference very nearly in the ratio of 113 to 355: what then is 
 the earth's circumference, its diameter being 791 1| miles, nearly? 
 Amw. 24855^ miles, nearly. 
 
 71. If the girt of a round tree be 6 feet 10 inches, what is its 
 diameter? Answ: 2 feet, 2// r inches. 
 
 72. The earth revolves round the sun in an orbit, the mean 
 diameter of which is one hundred and ninety millions of miles. 
 Were this orbit an exact circle, as it nearly is, what would be its 
 circumference? Answ. 596,902,654 r 9 T V miles. 
 
 73. If a farmer lend his neighbour, for 34 days, a cart horse 
 which draws 12 c. 3 q. how long should he have in return a horse 
 that draws 10 c. 3 q. ? Answ 40^f days. 
 
 74. After A has travelled 51 miles on a journey, B sets out to 
 overtake him, and for 16 miles travelled by A, B constantly travels 
 19 miles. How far will each have travelled before A will be 
 overtaken ? Answ. 323 miles. 
 
 75. In a copy of Milton's Paradise Lost, containing 304 pages, 
 the combat of Michael and Satan commences at the end of 
 the 139th page; at what page may it be expected to commence in 
 a copy containing 328 pages ? Answ. The fourth proportional is 
 149|, and hence the passage maybe expected to commence near 
 the bottom of the 150th page. 
 
 76. A pound, Troy weight, of silver is coined into 66 shillings: 
 what is the value of a pound, avoirdupois weight ? (See page 4&.J 
 Answ. 4 2. 
 
 77. How much should^a gold coin, worth one pound, weigh, 
 the weight cf the guinea being 5 dwts. 9 grs.? Answ. 5 dwts. 
 2* grs. 
 
 78. The length of a wall, being tried by a measuring hnc, ap- 
 pears to be 1287 feet, 4 inches; but on examination, the line is 
 found to be 50 feet, 10 inches in length, instead of 50 feet, its 
 supposed length. Required the true length of the wall. Answ. 
 1309 feet, 10fc$ inches. 
 
 79. If a shopkeeper use a false weight of 14| oz. for a pound, 
 how many pounds will 112 tbs. of just weight appear to be, when 
 weighed by his weight? Amw. 12 Iff- ibs. 
 
 80. If a dealer in spirits use, instead of a gallon, a measure 
 which is deficient by half a pint, what will be the true measure of 
 100 of these false gallons ? .Answ. 93- gallons. 
 
 81. If the port wine contained in a vessel, weigh 15 c. 3 qrs. 
 24 Ibs. how much weight of pure water, and how much of cow's 
 milk, would the same vessel contain, the weight of equal bulks of 
 port wine, of water, and of inilk, being as the numbers 997, 
 1000, and 1032? Answ. 16 c. q. HH Ibs- and 16 c. 2 q. 2Jf Ibs. 
 
/ SIMPLE PROPORTION. 81 
 
 82. If a stone column weigh 69 cwt. 3 qrs. 10 Ibs. what will a 
 pillar of oak of the same dimensions weigh, the weight of equal 
 bulks of the stone and oak being as the numbers 2484 and 1170? 
 Answ. 32 c. 3 q. 16J& J bs. 
 
 83. A cubic foot of fresh water weighs 1000 ounces avoirdu- 
 pois nearly : what is the weight of a vessel of such water, con- 
 taining 217f cubic inches ? Answ. 1 Ibs. 13\$% oz. 
 
 84. The quantity of rum imported into England from the British 
 .West India Islands, in 1798, was four millions, one hundred and 
 ninety-six thousand, one hundred and ninety-three gallons; the 
 duty on which amounted to 95,996 5 5. Of this quantity, 
 333,093 gallons were exported. What was the amount of the 
 duty on the quantity consumed in England ? Answ. 88,376 2 1 \, 
 nearly. 
 
 8.5. In 1798, the quantity of sugar imported into England from 
 the West Indies, was 2,361,715 cwt. qr. 8 Ibs.; the duty paiti 
 3n which was 2,070,377 2 7. What was the duty^" cwt.? Answ. 
 
 86. In what time would wind move from the pole to the equator, 
 at the rate cf 2f miles ^ hour, the distance being 62 14 miles ? 
 Answ. 94 days, 3 hours, 38 minutes, lOj? seconds. 
 
 87. The earth describes its orbit round the sun in 365 days, 5 
 hours, 48 minutes, and 48 seconds : through what space does it 
 move each hour, at an average, the circumference of the orbit be- 
 ing 596,902,655 miles ? Answ. 68094 BHH- 
 
 88. The ecliptic contains 360 degrees, and is apparently de- 
 scribed by the sun in 365 days, 5 hours, 48 min. and 48 sec. 
 through what part of it does he appear to move at an average each 
 day, and each hour? Answ. 59 8" 19"', and 2' 27" 51'", nearly. 
 
 89. The value of British manufactures and produce exported 
 from Britain in 1808, was 40,881,871; and that of Irish manu- 
 factures and produce exported from Ireland the same year was 
 12,577,517 10 11. It is hence required to determine how much 
 the latter sum affords for each 100 contained in the former. Answ. 
 30 15 3|, nearly. 
 
 The pupil having been thus presented with illustrations and exercises 
 itl the practical application of the doctrine of Proportion, what follows 
 respecting the changes in the order and magnitude of proportionals. 
 may be found useful, especially to such pupils as may afterwards study 
 Mathematics. 
 
 It has already been shown, that if four numbers be proportional, the 
 product of the extremes, is equal to the product of the means. From 
 the converse of this principle, it follows that whatever changes are made 
 in the order or magnitude of proportionals, the resiiks will be propor- 
 tionals, if the product of the extremes be equal tt^hat of the means. 
 If four magnitudes be proportional, other proportionals will be ob- 
 tained, if either the antecedents, or consequents, or both, be respec- 
 tively increased or diminished by like parts of themselves ; or if the 
 
 E3 
 
52 COMPOUND PROPORTION. 
 
 two terms of either, or of both ratios be respectively increased or di- 
 minished by quantities in the same ratio with themselves. This will he 
 understood from the following view of the most common and useful 
 changes which proportionals admit; and from this view also, the tech- 
 nical terms used to denote these changes will be understood. 
 
 Let the four numbers 8, 6, 20, and 15, be proposed, which arc 
 evidently proportionals; that is, 
 
 8 : 6 : : 20 : 15. Then, 
 
 Alternately, 8 : 20 : : 6 : 15; 
 
 By Inversion, 6 : 8 : : 15 : 20; 
 
 By Composition, 8+6 : 6 
 
 By Division, 8 6 : 6 
 
 By Conversion, 8 : 8 6 
 
 : 20+15 : 15, or 14 : 6 : : 35 : 15 ; 
 : 2015: 15, or 2:6 : : 5: 15; 
 : 20 : 20 15, or 8 : 2 : : 20 : 5 ; 
 
 To these changes, the following, out of many to which no specific 
 names have been applied, may be added : 
 
 Let8 : 6: : 20: 15; then 
 
 8 : 8-1-6 : : 20 : 20+15, or 8 : 14 : : 2O : 35 ; 
 
 8+6 : 86 : : 20+15 : 2015, or 14 : 2 : : 85 : 5 ; 
 
 8+20 : 8 : : 6+15 : 6, or 28 : 8 : : 21 : 6 ; 
 
 20+8 : 208 ; : 15+6 : 156, or 28 : 12 : : 21 : 9. 
 
 Also, if 8 : 6 : : 20 : 15 : : 12 : 9 : : 4 : 3 ; then 8 + 20+12+i*: 
 6+1,5+9+3 : : 8 : 6 : : 20 : 15, &c. ; that is, the sum of ail the an- 
 tecedents of any number of equal ratios, is to the sum of all the consequents* 
 as any one of the antecedents is to its consequent. 
 
 Let, again, 8 : 6 : : 20 : 1 5, 
 and 14: 4 :: 21 : 6; 
 
 then 8X14 : 6X* : : 20X21 : 15X6, or 112 : 24 : : 420 : 9O. 
 It may be observed, in conclusion, that all these changes and modi 
 fications hold, without exception, in respect to abstract numbers ; but 
 when the given numbers express real quantities, the terms of every ra- 
 tio, and also the numbers whose sum or difference is taken, must always 
 be of the same kind. Thus, in the analogy, 8 yards : 6 yards : : 2C 
 shillings : 15 shillings, we are not at liberty to take the terms alternately; 
 for if we did, we should thus fall into the absurdity of instituting a ra- 
 tio betwee'rt 8 yards and 20 shillings, and between 6 yards and 15 shillings. 
 
 COMPOUND PROPORTION. 
 
 RULE I. (1.) By the rule for Simple Proportion, find a 
 fourth proportional to two given terms of the same kind 
 with one another, and to the term which is of the same 
 kind as the answer. (2.) To two other terms of the same 
 kind, and to the last obtained find a fourth proportional ; 
 and thus proceed if there be more terms : the final result 
 will be the answer. 
 
 RULE II. (1.) Place the term which is of the same kind 
 
COMPOUND PROPORTION. 33 
 
 as the required term, in the last place. ( C 2.) Comparing 
 the other given terms by pairs, place each as antecedent or 
 consequent, according to the rule for Simple Proportion. 
 (3.) Divide the continual product of all the consequents 
 and the last term, by the continual product of all the ante- 
 cedents ; the quotient will be the answer. 
 
 As a contraction in the use of this latter rule, divide an 
 antecedent and any consequent by any number which will 
 measure them, and employ the results instead of those 
 terms : or, if an antecedent and any consequent, or an 
 antecedent and the last term be the same, reject them. 
 
 Exam. 1. If 40 gallons of ale serve 17 persons 5 days, how 
 many gallons will serve 9 persons for a year, at the same rate ? 
 
 L As 5 days : 365 days : : 40 gallons : 2920 gallons, 
 
 As 17 persons : 9 persons : : 2920 gallons : 1545}^ gallons. 
 
 In this operation, we first proceed as if the number of persons 
 in both cases were 17, and on this supposition we find that 2920 
 gallons would be used by those persons in a year. But the num- 
 ber of persons being 9, instead of 17, we find by the second an- 
 alogy, that if 17 persons would use 2920 gallons in a year, 9 per- 
 sons would use 1545-jf gallons in the same time. 
 
 II. As 5 days : 365 days ) An 
 
 17 persons : 9 persons \ : : 40 ^ lons : 1545 " S allons ' 
 
 In the second method that term is placed last which is of the 
 same kind as the required term. Then, were the number of per- 
 sons the sanie, it is evident that the answer would be greater than 
 40 gallons ; and therefore we put 365 days in the second, and 5 
 days in the first place: but were the number of days the same, it 
 is obvious that the required term would be less than 40 gallons ; 
 and sve therefore put 9 persons in the second place, and 17 in the 
 first. We then multiply the product of the consequents by the 
 common third term, 40, and divide the result by the product of 
 the antecedents, and the same answer is found as before. This 
 operation is in effect the same as that in the first method; for 
 the result of the first single analogy, without the actual mul- 
 
 365 y 40 
 tiplication and division, is > and, consequently, the 
 
 second analogy becomes 17:9: : ^^l*^i? ; whence 
 
 it appears that in both methods the same multiplications and di- 
 visions are in reality performed, and consequently the one is only 
 a modification of the other. In this method 5 and 365, or 5 and 
 40, might be divided by 5 as a contraction. 
 
 The principles on which the operations in Compound and us 
 Simple Proportion depend are the same. 
 
*4 COMPOUND PROPORTION. 
 
 It may be remarked, that one very considerable advantage which tne 
 second rule possesses, is, that by it the operation is kept entirely free 
 from fractions till the conclusion ; while in the other mode, fractions of- 
 ten arise from the first analogy, and render the remaining part of the 
 work more intricate and difficult. 
 
 Exam. 2. If the carriage of 66 standard hundreds, for 4 2 
 
 British miles, be .6 5, what is the carriage of 36 hundreds, long 
 
 tveighf., for 90 Irish miles, at the same rate ? 
 
 As 66 stand, cwt. 36 ions; cwt. 
 
 112 lt)s. 
 42 miles, Brit. 
 11 
 
 120 tbs. 
 
 90 mile's, Irish, f' : *6 5 : 9 
 
 In this example, since the answer is to be in money, 6 5 is 
 placed as the common third term. Then it is evident, that were 
 all things alike, except the number of hundreds, the answer 
 would be less than Q 5; and therefore, 36 is put as consequent, 
 and 66 as antecedent. But had the number of hundreds been 
 equal, the answer must evidently have been greater than 6 5, 
 on account of the difference in their magnitudes; arid therefore, 
 112 is made antecedent, and 120 consequent. The arrangement 
 of the remaining terms proceeds on similar principles, and the an- 
 swer is found by dividing the continual product of .6 5 and the 
 consequents, by the continual- product of the antecedents. The 
 operation is much abbreviat- 
 ed, as in the margin, by di- 11:6 ~) 
 viding the terms of the first (14): 15 I .. fi , . o q . 
 and third ratios by 6, and 7 : 15 f" 
 those of the second by 8, and 11 :(14) J 
 neglecting in the work 14, 
 
 which occurs as an antecedent in one ratio, and a consequent iu 
 another. 
 
 This question might also have been wrought by four analogies in 
 Simple Proportion. Everyone of these, however, would have given 
 origin to fractions, the neglecting of which would have prevented the 
 precise result from being obtained. The work might also have been 
 nodified by reducing the hundreds and the miles, respectively, to the 
 same kind. The using of 112, indeed, as an antecedent, and 120 as a 
 consequent, serves this purpose, as the hundreds are thus reduced to 
 pounds : and the multiplication by 11 and 14 serves a similar purpose, 
 in relation to the miles. 
 
 Ex. 1. If the carriage of 59 cwt. 19 miles, cost 2 16, how far 
 may -1-3 cwt. be carried at the same rate for 2 4? Ansio. 20m. 
 
 2. If the carnage of 13 cwt. 65 miles, cost ? 5, how many 
 hundreds may be carried 40 miles at the same rate for 3 15 ? 
 Answ. 35 j 5 ? cwt. 
 
 3. if 12 horses in 5 days plough 11 acres, how many horses 
 would plough 33 acres in 18 days? Answ. 10. 
 
 4. If a person walking 12 hours each day, perform a journey of 
 
FRACTIONS. 85 
 
 250 miles in 9 days, in how many days, walking 10 hours each at 
 the same rate, would he complete a journey of 400 miles ? Answ. 
 17^V days. 
 
 5. If the expenses of a family of 8 persons amount to .42, in 
 16 weeks, how long will 100 support a family of 6 persons, at 
 the same rate? Answ. 50|$ weeks. 
 
 6. If 29 men, in 5 days of 12 hours each, reap 32 acres, in how 
 many days of 13 hours each, will 20 men, working equally, reap 
 40 acres ? Anxw. 8^| days. 
 
 7. If 15 12 pay 16 labourers for 18 days, how many labour- 
 ers, at the same rate, will 35 2 pay for 24 days ? Answ. 27. 
 
 8. If 36 yards of cloth, 7 quarters wide, cost 25 4, what 
 cost IzQ yards of the same quality, but only 5 quarters wide? 
 A)isw. 60. 
 
 9. If a tradesman earn 16 guineas in 108 days, how many so- 
 vereigns would he earn at the same rate, in 270 days; 20 guineas 
 being equivalent to 21 sovereigns? Answ. 42. 
 
 10. If the rent of a farm containing 26 a. 2 r. 23 p. be 50 8 9, 
 what would be the rent of another farm containing 17 a. 3 r. 2 p. 
 if 6 acres of the latter be worth 7 of the former ? Answ. 39 4 7. 
 
 11. If a puncheon of rum containing 85 gallons, cost 58 8 9, 
 what would be the value of a hogshead containing 63 gallons, and 
 composed of four parts of the same rum, and one part of water ? 
 Answ. 34- 13 0. 
 
 12. In what time would 23 men reap a field which 40 women 
 would reap in 6 days, if 7 men can reap as much as 9 women ? 
 Answ. 8^-5 days. 
 
 13. If a person walking 13 hours each day, travel 191 miles in 
 7 days; in how many days of 10 hours, will he complete the re- 
 mainder of a journev of 500 miles, at the same rate each hour ? 
 Answ. 14Hi- 
 
 14. If 63 Its. of tea cost 20 10 6, what cost 70 tbs. of a 
 different quality, 9 Jbs. of the former being equal in value to 10 
 I os. of the latter ? Answ. 20 10 6.* 
 
 FRACTIONS.f 
 
 IF a quantity, considered as a whole, be divided into any 
 number of equal parts, a FRACTION in respect to that quan- 
 
 * Many of the questions usually proposed under the head of Compound Proportion 
 arc quite misplaced, as some of them are merely anticipations of what is afterwards de- 
 livered in Interest, and others belong to Mensuration, or other subjects, with the prin- 
 ciples of which the pupil is supposed to be unacquainted. Hence, the number 21 d 
 variety of exercises here given, are purposely less than in several other works on Antii- 
 nietic. 
 
 t With respect to the position which the doctrine of Fractions should occupy in a 
 
86 FRACTIONS. 
 
 tity, signifies one or more such parts : and the quantity di- 
 vided is called, in respect to the fraction, the INTEGER, or 
 UNIT. 
 
 A fraction is expressed by two numbers, or terms, called 
 the numerator and the denominator. 
 
 The DENOMINATOR is written below the numerator, and 
 expresses the number of equal parts into which the integer 
 is divided ; and the NUMERATOR expresses the number of 
 such parts denoted by the fraction. 
 
 Thus, A, which is read four-jifths, is a fraction, and signifies that 
 a unit is divided into five equal parts, and that four of these parts are 
 taken. 
 
 If the numerator of a fraction be taken as an integer, 
 and divided into as many equal parts as there are units in 
 the denominator, the fraction may also be regarded as ex- 
 pressing one of these parts. 
 
 Thus, if 4 be divided into 5 equal parts, the fraction f expresses one 
 of these parts, and is therefore the same as one-jiflh of 4. 
 
 A PROPER FRACTION is that whose numerator is less than 
 its denominator. 
 
 An IMPROPER FRACTION is that whose numerator is not 
 less than its denominator. 
 
 Thus, and are proper fractions; and |, f, and 1 ^ are improper 
 fractions, 
 
 A number consisting of a whole number, with a fraction 
 annexed, as 5f, is termed a MIXED NUMBER. 
 
 course of Arithmetic, writers seem to be much divided in their opinions ; some intro- 
 ducing it immediately after Simple Division, some after the doctrine of Proportion, and 
 others at a more advanced part of the course. The first of these positions is certainly 
 the natural one, both because Fractions for the most part arise from the remainders in 
 Division, and because on this account they are of the same nature with the quantities 
 treated of in the Compound Rules. The principal objection to this arrangement is, 
 that the study of Fractions is too difficult and intricate to be prosecuted with success by 
 the majority of pupils, when they have learned only the fundamental rules, without 
 having acquired that ease and readiness in managing numbers, which are derived from 
 more varied and extemled practice. The following article is so composed, that tne pupii 
 may study it in the place where it stands ; or if the teacher should think it better, after 
 Simple Division, after the mercantile rules, or indeed at almost any other part of the 
 course before the extraction of roots. It may be observed, in general, that the several 
 parts of Arithmetic tend to illustrate each other. Thus, if the pupil have learned the 
 doctrine of Proportion, he will more easily and more successfully study Fractions ; and 
 if he have studied Fractions before Proportion, he will be able to acquire a more e~- 
 Umcted and adequate knowledge of the latter. For this reason his attention should be 
 frequently recalled to the rules he has passed ; and in this case he should bero.-.de to ;ivai* 
 himself of tuch illustrations, as the subsequent rules which li* lias learned may afford 
 
FRACTIONS. 87 
 
 A SIMPLE FRACTION expresses one or more of the equal 
 
 parts into which a unit is divided, without reference to any 
 
 other fraction. 
 
 A COMPOUND FRACTION expresses one or more of the 
 
 equal parts into which another fraction, or a mixed number, 
 
 is divided. 
 
 Thus, | is a simple fraction ; but f of \%, and % of f of 1$, are 
 compound fractions. Compound fractions have the word of interposed 
 between the simple fractions of which they are composed. 
 
 A COMPLEX FRACTION is that which has a fraction either 
 in its numerator or denominator, or in each of them. 
 
 Thus, , , and , are complex fractions. 
 9 9$ 6 y 
 
 For the definitions of several other terms that will be employed in 
 what follows, see the first Note in page 24. 
 
 It is of the greatest consequence that the pupil should acquire a cor- 
 rect idta of the nature of fractions, on commencing the study of this 
 important part of Arithmetic The following illustrations may perhaps 
 be found useful in assisting him in this respect. 
 
 Fractions generally have their origin from the division of a number 
 by another which does not measure it. Thus, if 13 be divided by 5, 
 the quotient is 2, with a remainder of 3 to be divided by 5. To 
 have an accurate idea of the nature and value of the latter part, we are 
 to conceive 3 divided into 5 equal parts; and, by the nature of divi- 
 sion, any of these parts will be its correct value. Each of these parts 
 is evidently less than l, the unit under consideration, and is therefore 
 called a fraction, from its being, as it were, a broken part of the unit 
 or integer. Arithmeticians have agreed to denote it, as has been 
 already remarked in Division, by the expression , which may be 
 read, one-Jtfih of 3. In like manner, the quotient of 4 cwt. by 7, is 
 expressed C wt. and may be read one-seventh of 4 cwt. 
 
 Resuming the former fraction, if we suppose each pound to be di- 
 vided into 5 equal parts., 5 being the number of units in the divisor, 
 each part will be one-fifth of \, and in S there will be 5X3, or 15 
 such parts. Dividing these by 5, we obtain 3 of these parts, or three- 
 fifths of l, for the value of the fraction : whence it appears, that one- 
 Ji/iJi of three pounds is equivalent to three-fifths of one pound ; and in a 
 similar manner it would be shown, that whatever may be the integer, 
 one-fifth of three is equivalent to three-fifths of one, or three times the 
 fifth part of one ; one-seventh of four to four-sevenths of one, &c. 
 When fractions are considered in the abstract, without reference to any 
 particular integer, we say briefly four-sevenths, instead of four-sevenths 
 of one; the former expression being understood to be equivalent to the 
 latter. Hence we see, that the two definitions of a fraction given in 
 the text, though tliey suggest to the mind ideas which aie apparently 
 different, are in effect the same. Agreeably to either, the lower num- 
 ber may properly be called the denominator, as it gives name (such as 
 seventh, &c.) to the parts into which the integer is divided ; and, 
 
88 FRACTIONS. 
 
 agreeably to the first definition in the text, the upper number is called 
 
 the numerator, as it numbers the parts expressed by the fraction. 
 
 According to these views, every fraction would have its numerator 
 less than its denominator, and wouhi consequently be less than its unit, 
 as its name imports. Such a view is too limited however to answer the 
 general purposes of calculation ; and it is often necessary to consider 
 as fractions, quantities which equal or exceed the integer ; or which 
 indicate the quotients of numbers not only by greater, but also by equal 
 or by smaller numbers, without the division being actually performed. 
 Quantities of the latter kind are therefore called improper fractions ; and, 
 for the sake of distinction, others, in respect to them, are called projyer 
 fractions. It is evident, from either view of the nature of a fraction, 
 that it is less or greater than a unit accordingly as its numerator is less 
 or greater than its denominator, and that it is equal to a unit, if its nu- 
 merator and denominator be equal. 
 
 In addition to the preceding illustrations, it may be proper to state, 
 thut fractions are of the same nature as the subordinate quotients in 
 Compound Division. Thus, if 27 be divided by 20, the quotient is 
 l 7, or 1-^3, each expressing precisely the same quantity, and each 
 of the 7 remaining pounds being reduced into 20 equal parts in each 
 case. 
 
 From these principles we may readily derive some of the chief pro- 
 perties of fractions. For this purpose., let us compare the fractions $ 
 and jfcj, the denominator of the second of winch is treble of the deno- 
 minator of the first, while their numerators are the same. Now it is 
 easy to see that the former is three times as great as the latter : for in 
 the former, the integer 2 is divided into 5 equal parts, while in the other 
 it is divided into three times as many equal parts ; and consequently, 
 each part in the latter case, is only one-third of each part in the fox- 
 mer ; that is, the fraction ^ is one-third of the fraction |. Hence, it 
 appears, that by multiplying the denominator of the fraction by S, 
 we obtain a fraction equal to one-third of -| ; and that by dividing the 
 denominator of the fraction T \ by 3, we find a fraction equal to three 
 times ^ 5 . Now, since, by the preceding illustrations, the quotient of 
 
 2 by 15 is the same as two-fifteenths, the product of -fa by .'3 is six- 
 fifteenths, in the -same manner as the product of 2 shillings by 3, is 6 
 shillings; and since ^ ^ s one-third of J, and also one-third of T 6 ^, it 
 follows, that T \ must be equal to f. Hence, by multiplying each term 
 of the fraction ^ by 3, we obtain an equal fraction T ^ ; and by dividing 
 each term of the fraction T 6 5 by 3, we get an equal fraction -|. We see 
 also, that by multiplying the numerator T 2 5 by 3, we obtain a fraction 
 
 3 times as great as T 2 -^; and by dividing the numerator of the fraction 
 jSg. by 3, we find a fraction which is equal to one-third of T 6 ? . It is 
 evident, that similar properties might be proved in a similar manner to 
 hold respecting any other fractions; and hence we have the following 
 important properties which belong to all fractions: 
 
 1. If the terms of any fraction be both multiplied, or both divided by tlie 
 ssifnc number, the value of the fraction is not changed. 
 
 2. A fraction is multiplied by any number either by multiplying its nume- 
 rator, or dividing its denominator, by that number. 
 
 3. A fraction is divided by any number either by divitlifig its numerator, 
 or multiplying its denominator, by that number. 
 
REDUCTION OF FRACTIONS. 89 
 
 4. It appears also, from the principles here explained, and from the 
 principles of proportion, that if two fractions be equal, the numerator qf 
 the one is to its denominator, as the mimerator of the other is to its denomir- 
 nator ; or, by inversion, that tlie denominator of the one is to its numera- 
 tor, as the denominator of the other is to its numerator. 
 
 The same operations can be performed on fractional as on integral 
 quantities. Hence, the doctrine of Fractions comprises the Addition, 
 Multiplication, Subtraction, and Division, of Fractions. Before entering 
 on these operations, it is proper to show how such quantities may be 
 modified, without changing their value, so as to fit them for the several 
 operations to be performed on them, and for the different uses to which 
 they are to be applied. This will constitute Reduction of Fractions.* 
 
 REDUCTION OF FRACTIONS.f 
 
 Problem I. To jlnd the greatest common measure of t\oo 
 given numbers. 
 
 RULE. (1.) Divide the greater number by the less. (2.) 
 If there be a remainder, divide the less by it ; and thus 
 proceed, always dividing the last divisor by the last remain- 
 der, till nothing remains. The divisor which leaves no re- 
 mainder is the common measure required. 
 
 If in the operation any divisor be a prime number, and 
 leave a remainder, it is unnecessary to proceed farther, as 
 there is no common measure greater than unity. 
 
 Exam. 1. Required the greatest common measure of 247 
 and 323. 
 
 In this example, the first remainder is 247)323(1 
 76 ; and the less number, 247, being di- 247 
 
 vided by this, the remainder is 19. Di- 76)247(3 
 
 viding 76, the last divisor, by this, we 228 
 
 find that there is no remainder. Hence, F<h76f4 
 
 19 is the common measure required, and '^ 
 
 would be found by division to be contain- _ 
 
 ed 13 times in 247, and 17 times in 323, 
 both without remainder. 
 
 * One great class of fractional quantities are called decimal fraction* from the nature 
 of their denominators. Hence, the doctrine of fractions is generally divided into 
 two branches ; one treating of fractions in genera), and the other of the peculiar ma- 
 nagement and properties of decimal fractions. Fractions which are not decimals, are 
 usually called vulgar, or comftton fractions. 
 
 f Reduction of Fractions consists of two branches : one which treats of the changes 
 that may be made in the forms of abstract fractional quantities without changing their 
 values ; and the other which treats of fractions as referred to particular unite, such ns 
 the denominations of money, weights, and measures. To the former branch belong 
 problems II. IV V. and VI.; and to the other, problem VII. &c. 
 
90 REDUCTION OF FRACTIONS. 
 
 With respect to the reason of the ride, since 19 is a measure of 
 the last dividend, 76, it must also be a measure of the preceding 
 dividend, 247, because 247zr3X 76-f-19; and 247 is one of the 
 given numbers. Then, since 19 measures 76 and 247, it must 
 measure their sum, 323, which is the other given number: and a 
 similar illustration is applicable in all cases. 19 is also the greatest 
 common measure. For suppose, if possible, a greater number, 
 as 20, to measure 247 and 323; then it is evident, it must mea- 
 sure their difference, 76, and also 3 times 76, or 228; and since 
 it is supposed also to measure 247, it must measure 19, the dif- 
 ference of 228 and 247: that is, if 20 measure. 247 and 323, it 
 must measure 19, a number less than itself, which is evidently 
 impossible : therefore 20 does not measure both the given num- 
 bers; and in the same manner it may be shown, that no other 
 number greater than 19 can measure them; therefore 19 is the 
 greatest common measure. The remark which follows the rule 
 would be proved in nearly a similar manner. 
 
 This problem and problem III. are given in this place, because, 
 though they respect whole numbers, they are chiefly useful in the se- 
 cond and fourth problems in this article. It is easy to see, that the 
 greatest common measure of three or more numbers, would be found 
 by applying the preceding rule to two of the numbers; then to the re- 
 sult and another of the numbers, &c. 
 
 Exercises. Find the greatest common measures of the following 
 numbers : 
 
 1. 285 and 465 .... Answ. 15 
 
 2. 532 1274 .... Answ. 14 
 
 3. 888 2775 .... Azsw. Ill 
 
 4. 2145 and 3471 .... Anna. 39 
 
 5. 1879 _ 2425 .... Answ. 1 
 
 6. 693 1815 .... Answ. 33 
 
 A fraction is said to be in its LOWEST TERMS, or in its 
 SIMPLEST FORM, when it is expressed in the least whole 
 numbers possible. 
 
 Problem II. To reduce a fraction to its lowest terms. 
 
 RULE I. (1.) Divide the terms of the given fraction by 
 any number that will measure both : the quotients will be 
 the numerator and denominator of an equivalent fraction 
 in lower terms. (2.) Let this fraction, if possible, be re- 
 duced in like manner; and proceed thus, till a fraction is 
 obtained for whose terms no common measure can be 
 found. 
 
 The application of this rule will often be facilitated by the following di- 
 rections and remarks : 1. If the terms of the fraction end in ciphers, 
 cut an equal number from both. 2. If they end each in 5, or one in 
 5, and the other in a cipher, divide them both by 5 ; or double them, 
 and reject a cipher from each of the results. 3. If 2 measure the last 
 figure of each term, it will measure the terms themselves. In like 
 manner, if 4 measure the number expressed by the two last figures, or 
 8 that expressed by the three last, 4 in the one case, and 8 in the other, 
 
REDUCTION OF FRACTIONS. 91 
 
 will measure both terms. 4. If 3 or 9 measure the sum of the digita 
 of each term, 3 in the one case, and 9 in the other, will measure l>oth 
 terms. See page 36. 
 
 Exam. 2. Reduce V?l to i ts lowest terms. 
 
 Here, we divide the given terms by 2 ; 9) 3) 
 
 those of the result by 9 ; and those of that 2)^|f=|^=f=:|. 
 result by 3; and we see that ^ is equal 
 
 to f $ , f, or . The last of these has evidently no common mea- 
 sure greater than unity ; it is therefore the expression required, 
 being the simplest form of the given fraction. The same result 
 would have been obtained rather more quickly by dividing by 6 
 and 9. 
 
 Exam. 3. Reduce Hi 8 to ' lis simplest form. 
 
 Here, two ciphers are cut from the end of 7) 
 
 each term, which is equivalent to the divi- 6)HS!S=S!=I- 
 sion of each by 100. The quotients are 
 then divided by 6, and the results by 7, and the fraction is re- 
 duced to the form , which is its simplest form. 
 
 The reason of this rule is evident from the first of the princi- 
 ples established in page 88. 
 
 Exercises. Reduce the following fractions to their simplest forms : 
 
 Answ. I Ex. Answ. 
 
 7. 3k I 
 
 8. M T S 
 
 9. m f 
 
 10. tfg i 
 
 11. if! I 
 
 12. H 
 
 Ex. Answ. 
 
 is. 
 
 15. 
 
 This method of reducing fractions to their lowest terms is very con- 
 venient and easy in practice, when the terms of the proposed fractions 
 are not very large, or when the divisors are readily discovered. It la- 
 bours under the disadvantage, however, of not determining, in many 
 cases, whether the fraction is in its lowest terms or not ; and besidts 
 its being a process which depends on trial, and which therefore is not 
 strictly mathematical, the measures are generally discovered with diffi- 
 culty, unless they are some of the smaller numbers. Thus, if the 
 fraction ^-|^ were proposed, we should readily discover that it may be 
 reduced, by division by 3, to the form ^g|, which we would perhaps 
 conceive to be its simplest form, not knowing that it is still farther re- 
 ducible by 19, and that the simplest form is -fa. The following me- 
 thod, though often tedious, is perfect in principle, a'wtys reducing the 
 fraction to its simplest form, and not depending on trial. 
 
 RULE. II. (1.) Find, by problem I, the greatest com- 
 mon measure of the numerator and denominator. (2.) Di- 
 vide them both by this, and the quotients will be the nu- 
 merator and denominator of the required fraction. 
 
 It is often of advantage to carry the reduction as far, by 
 the first rule, as can readily be done, and then to apply 
 the second rule to the result. 
 
92 REDUCTION OF FRACTIONS. 
 
 Exam. 4. Reduce the fraction -^V^ to i ts lowest terms. 
 In this example, by dividing the denominator by the numerator, 
 the numerator by the remainder, &c. we find the greatest com- 
 mon measure to be 97: and dividing both terms of the given frac- 
 tion by this, we obtain r ' ff , which is the equivalent fraction in its 
 lowest terms. 
 
 Exam. 5. Reduce ^V^V to its lowest terms. 
 Here, because the terms end in 5 and 0, 5 is a measure, and 
 reduces the fraction to fff. This again is reducible by 9, be- 
 cause 9 measures the sum of the digits of each term ; and the re- 
 sult is f|, which, by the second rule, is reduced to T V 
 
 Exam. 6. Reduce T Vs% * i ts most simple expression. 
 
 We see here immediately, that 6 is a common measure ; for 3 
 measures the sum of the digits of each term, and 2 must be a 
 measure, since the unit figures are even. After division by 6, 
 therefore, the fraction becomes ? 7 r 8 T . In applying the second rule 
 to this, 55 at the third division is to be divided by 23, which is a 
 prime number, and is not a measure of 55. ^ 7 -/ T , therefore, we 
 conclude to be the simplest form of the fraction; a;id thus we are 
 saved the trouble of three or four divisions. 
 
 It may be proper for the pupil to recollect, that 
 
 At the conclusion of evert/ operation, if there be a fraction^ 
 it should be reduced to its simplest form. 
 
 Reduce the following fractions to their lowest terms: 
 
 Ex. Answ. 
 
 17. 
 
 if 
 
 18. f H* .......... If* 
 
 Ex. Answ. 
 
 19. mi if* 
 
 20. 
 
 21. 
 
 Ex. 
 
 22. 
 23. 
 24- JJIf | 
 
 Problem III. To find the least common multiple of t<ux> or 
 more given numbers. 
 
 RULE I. (I.) Arrange the given numbers in succession, 
 ami find by inspection a number which will measure as 
 many of them as possible. (2.) By this number divide all 
 the given numbers which it measures, and write the quo- 
 tients and the undivided numbers in a new line. (3.) Pro- 
 ceed, if possible, in the same manner with the numbers in 
 this line ; and thus continue the process till no number 
 greater than a unit will measure any two or more of the 
 numbers last found. (4-.) Then multiply all the numbers 
 in the last line, and all the divisors employed in the opera- 
 tion continually together, and the result will be the com- 
 mon multiple required. 
 
 The work is often much shortened by rejecting, in any 
 
REDUCTION OF FRACTIONS. 93 
 
 line, any number that is contained without remainder in 
 any other number in the same line. 
 
 If no two of the given numbers have any common mea- 
 sure greater than unity, their continual product will be the 
 least common multiple. 
 
 RULE II. (1.) Find by problem I, the greatest common mea- 
 sure of two of the given numbers. (2.) By this, divide one of 
 those two numbers, and multiply the quotient by the other. (3.) 
 Perform a similar operation on the product and another of the 
 given numbers. (4-.) Continue the process thus with each succes- 
 sive quotient, till all the numbers have been used, and the final 
 result will be the least common multiple required. 
 
 Exam. 7. Required the least common multiple of 24, 10, 9, 
 32, 6, 45, and 25. 
 
 Here, the given numbers be- By Rule I. 
 
 ing placed in the same line, as 2) 24 10 (9) 32 (6) 45 25 
 we see by inspection, that 9 is 3) \% (5) \Q 45 25 
 
 contained exactly in 45, and 6 ^ -TO ^Q 15 25 
 
 in 24, these are neglected. Then ' ' 
 
 using 2 as a divisor, we obtain 
 
 the quotients 12, 5, and 16, 2X3X5X 16X3X5=7200,^n. 
 and we place in the line with 
 
 them the undivided numbers, 45 and 25 : and since the quotient 
 5 is contained exactly in either of these, it is rejected. We then 
 divide by 3, and obtain the quotients 4 and 15, which with the 
 undivided numbers, 16 and 25, are placed in a new line. 4 is 
 then rejected as it is a measure of 16: and dividing by 5, we 
 place in the next line the quotients 3 and 5, and the undivided 
 number 16. Then, as no two of these have any common measure 
 greater than unity, the three divisors and the three numbers in 
 the last line are multiplied continually together, and the product, 
 7200, is the common multiple required. 
 
 By Rule II. Rejecting 9 and 6 as before, we divide 24 by 2, the 
 greatest common measure of it and 10, and we multiply the quo- 
 tient 12 by 10: the product, 120, is the least common multiple of 
 24 and 10. Then, the greatest common measure of 120 and 32 
 being 8, we have 32-^-8=4, and 120X4=480, the least common 
 multiple of 24, 10, and 32. In the next place, the greatest com- 
 mon measure of 480 and 45 is 15; and, since 45-r-15=3, we have 
 480X3=1440, the least common multiple of 24, 10, 32, and 45. 
 Lastly, since the greatest common measure of 1440 and 25 is 5, 
 and since 25-f-5=5, we have 1440X5=7200, the least common 
 multiple of 24, 10, 32, 45, and 25, and consequently of 9 and 
 6, of which 45 and 24- are multiples. 
 
 With respect to the reason of these rules, it is difficult to give a 
 strict, and, at the same time, an easily comprehended proof; and 
 
94 REDUCTION OF FRACTIONS. 
 
 for most learners the following illustration may perhaps be pre- 
 ferable : 
 
 In the operation by the first rule, 2 X 12 X 5 X 16 X 45 X 25, the 
 product of the first divisor and the numbers in the second line, is 
 evidently a multiple of each of the given numbers, 24 being con- 
 tained in it 5 X 16 X 45 X 25 times; 10, the second number, 12 x 
 16 X 45 X 25 times, &c. Again, 2 X 3 X 4 X 1 6 X 15 X 25, the 
 product of the two first divisors and the numbers in the third 
 line, is also a multiple of each of the given numbers, 24 beinir 
 contained in it 16X15X25 times; 10, the second number, 3>^ 
 4 X 16 X 3 X 25 times, (since 2 X 15=30, and 30= 10 X 3;) 32, the 
 fourth number, 3X4X15X25 times, &c.: and the illustration 
 mav be extended in a similar manner to the rest of the operation. 
 That the given number 6 may be rejected in the operation, will 
 readily appear from considering, that 6 is contained exactly 4 
 times in 24, and will therefore be contained without remainder in 
 any multiple of 24, and 4 times as often as 24. In like manner 
 it would appear, that 9 may be rejected, as 45 is a multiple of it. 
 That 5 and 4 may be rejected in the succeeding lines will be evi- 
 dent from this, that they would disappear were the lines that con- 
 tain them divided respectively by 5 and 4. 
 
 The proof of the second rule depends on the principle, that if 
 the product of any two numbers be divided by any factor which is 
 common to both, the quotient will be a common multiple of the two num- 
 bers. Thus, if 48, the product of 6 and 8, be divided by 2, a 
 factor of both, the quotient 24, will be a multiple of both, since 
 it may be regarded either as 8 multiplied by the quotient of 6 by 
 the factor 2, or as 6 multiplied by the quotient of 8 by the same- 
 factor. Now this being so, it is obvious that the greater the com- 
 mon measure is, the less will be the multiple; and consequently, 
 the greatest common measure will produce the least common mul- 
 tiple. When the common multiple of the two first numbers is 
 found, it is evident, that any number which is a common multiple 
 of it and of the third number, will be a multiple of the first, se- 
 cond, and third numbers : and thus the reason of every part of the 
 second rule is manifest. 
 
 It may be remarked with respect to the practical application of these 
 rules, that the second always gives the least common multiple. The 
 multiple found by the first is not always the least possible; but it will 
 be such, if care be always taken to use such a divisor as will measure 
 more of the given numbers, than any other divisor would. This rule 
 therefore, being very easy in practice, is generally preferred to the 
 other. It may be farther remarked, that by practice the pupil will 
 gradually become able to discover, in many instances, the common 
 multiples by inspection, especially when the given numbers are not 
 large. 
 
REDUCTION OF FRACTIONS. 95 
 
 Exercises. Required the least common multiples of the follow- 
 ing numbers: 
 
 Exercises. Answ. 
 
 25. 6 10 15 18 90 
 
 26 7 11 13 3 3003 
 
 Exercises. Answ. 
 
 28. 63 12 84 7 252 
 
 29. 54 81 63 14 ... .. 1134 
 
 27. 8 12 20 24 25 .... 600 30. 2 3 4 5 6 7 8 9... 2520 
 
 Problem IV. Any number of fractions having different 
 denominators being given ; to Jind equivalent fractions hav- 
 ing a common denominator* 
 
 RULE I. (1.) Find, by problem III. a common multiple of 
 all the denominators : this will be the common denomina- 
 tor. (2.) Then, divide the common multiple by the first 
 of the given denominators, and multiply the quotient by 
 the first of the given numerators : the product will be the 
 first of the required numerators. The other numerators are 
 found in a similar manner. 
 
 RULE II. (1.) Multiply each numerator by all the denominators, 
 except its own, and the product will be the numerator of the equi- 
 valent fraction. (2.) Multiply all the denominators continually 
 together for the common denominator. 
 
 Exam. 7. Reduce $> T \, H, and ^ 9 D , to fractions having a 
 common denominator. 
 
 By Hide I. Here, by 8)360 12)360 18)360 20)360 
 
 problem in. the least com- 
 
 mon multiple is found to 45 30 20 J 8 
 
 be 360 ; and the rest of 5 7 11 9 
 
 the work will stand as in 
 
 the margin. The cor- 225 210 220 162 
 
 rectness of the results ^ 5 . 
 
 would be proved by re- *** *** * io > Anno ' 
 
 ducing them to their 
 
 simplest forms, as the given fractions would thus be obtained. 
 
 Thus, H& would be reduced to f ; f to T \, &c. 
 
 By Rule II. 5 X 12 X 18 X 20=21600, the first numerator; 7 X 
 8XL8X20=: 20 160, the second numerator; 11X8X12X20= 
 21120, the third numerator; 9X8X12X18=15552, the fourth 
 numerator; and 8X12X18X20=34560, the common denomi- 
 nator. Hence, the equivalent fractions found by this rule, are 
 Sim, 3HIS, fiWfc and ifi!, which being reduced to their 
 lowest terms, would in like manner, be shown to be equivalent to 
 the given fractions. 
 
 In this example the results found by the second rule are expressed in 
 numbers inconveniently large; and the same is very generally the case. 
 Hence, the other rule should always be preferred, except when tto 
 
95 REDUCTION OF FRACTIONS. 
 
 given denominators are prime to each other; that is, when they have no 
 common measure greater than a unit. 
 
 With respect to the reason of these rules, it is evident, that the 
 operation by the second is nothing else than the multiplication of 
 both the terms of each fraction by all the denominators, except 
 its own; and in this way the rule might have been expressed. 
 Thus, the first of the given fractions is |, which by the operation 
 
 , , . 5X12X18X20 21600 
 is transformed into ^ ^ ^ 2Q ,or ^ ; and this by the 
 
 first of the properties proved in page 88, is equal to the proposed 
 fraction |. 
 
 In the first rule, the division of the common multiple by th 
 given denominator is evidently nothing else than the finding of the 
 number by which if both the numerator and the denominator of 
 the given fraction be multiplied, the denominator of the result 
 will be the same as the common multiple; and, by the principle 
 before referred to, the result will be equal to the given fraction. 
 Thus, in the preceding example, when 360 is divided by 8, the 
 quotient is 45; by which if both terms of the fraction - be mul- 
 tiplied, there results ff , for the equivalent fraction required. 
 
 Exercises. Reduce the following sets of fractions to other frac- 
 tions having common denominators: - 
 
 Exercises. Answers. 
 
 31. &, A, H, H, t m HI, HI, m, HI 
 
 32- H, H, I, T 7 *, t, i 
 
 33. ij, H, H, 4*, |, A- 
 
 34. H, H. H, TtW, if 
 
 35. i,|, i, rH 
 
 36. H, ii, n, H, TV 
 
 37. T^T, -riirr* TwVw 
 
 38. H> U, H, i*7 t 
 
 39. 4, A, H, A, *V> A VWJr, iV*V **, TV/Ty, fV 4 s %, ifgff 
 
 40. i, t, T v, /T, &, TVV-. w, m w, T%, T 4 ^, TV? 
 
 Problem V. To reduce an improper fraction to a whole or 
 mixed number. 
 
 RULE. Divide the numerator by the denominator; the 
 quotient will be the whole number required : and if there 
 be any remainder, write it over the given denominator for 
 the fractional part of the required result. 
 
 The reason of this rule and of that for the next problem, is evi 
 dent from the nature of fractions. 
 
 Exam. 8. Reduce V and ^ 
 to whole or mixed numbers. ' ?. ^ 20 
 
 8* dnsiu. 22$, Answ. 
 
REDUCTION OF FRACTIONS. 
 
 Exercises. Reduce the following fractions to whole or mixed 
 numbers: 
 
 Ex. Answ. \ Ex. Answ. Ex. 
 
 41. H 1 43. VW 20^ 45. SV 
 
 42. 6_4^ 21HI I 44. 1 i??.... W^ft 46. VV 30 
 
 Problem VI. Jo reduce a mixed number to an improper 
 
 fraction. 
 
 RULE. Multiply the whole number by the denominator 
 of the fractional part, and to the product add the numera- 
 tor; the sum will be the required numerator, below which 
 write the given denominator. 
 
 A whole number may be expressed in a fractional form 
 by writing a unit below it ; or by multiplying it by any 
 whole number, and writing that number below the product 
 as denominator. Thus, 9=i= 3 4% &c. 
 
 6H 
 Exam. 9. Reduce 5}- to an improper 12 
 
 fraction. ff, Answ. 
 
 Exercises. Reduce the following numbers to improper fractions : 
 Ex. Ansiv. Ex. Answ. 
 
 Ex. Answ. 
 
 47. 15* V 
 
 48. 3 W 
 
 49.46| 
 50. 
 
 51. i^ftfe \m 
 
 52. 19f ^ 8 
 
 Problem VII. To express any given quantity as a fraction 
 
 of another given quantity of the same kind) considered as an 
 
 integer. 
 
 RULE. Make the integer the denominator of the required 
 
 fraction, and the other given quantity its numerator, both 
 
 being reduced to the same denomination, if they be not ot 
 
 the same already. 
 
 Exam. 10. Reduce 13/9 to the fraction of 1. 
 In 13/9 there are 165 pence, and in 1, 240 pence; the fraction 
 therefore is $ which, by reduction to its lowest terms, becomes 
 \%. The reason of this is evident, since 1 penny is 7 ^ of a 
 pound, and in 13/9 there are 165 pence, or of "a pound. 
 
 Exam. 11. Express a pound troy weight as a fraction of a 
 pound avoirdupois. 
 
 A pound avoirdupois weight contains 7000 grains, and a pound 
 troy weight 5760 grains. (See page 39. ) Hence, the fraction is 
 ?o> r in its lowest terms, {$$. 
 
 This rule enables us to find the ratio of two quantities. Thus, the 
 ratio of 13/9 to l is that of f to 1, or 11 to 16 ; and a pound avoir- 
 dupois is.to a pound troy, as 1 to \%> or as 175 to 144. (Seepage 89, 
 
98 REDUCTION OF FRACTIONS. 
 
 Exercises. Answers. 
 
 53, Reduce 12/6 to the fraction of 1 | 
 
 54 32 10 100 tf 
 
 55 2/8 1 2 9 & 
 
 56 Iqr. 201bs . ~~ 1 cwt. long weight ..,. fc 
 
 57 2 hours 23 h. 56m. 4s J^ 
 
 58 2/1 6/8 ft 
 
 59. The height of Ben Nevis, the highest mountain in Britain, 
 is 4350 feet, and that of mount Ararat 9500 feet: express the 
 former as a fraction of the latter in its simplest form. Answ. T W 
 
 60. Express 96 pages as a fraction of a book containing 432 
 pages. Answ. f. 
 
 61. If one farm contain 37 a. 2r. 14 p. and another 168 a. Or. 
 28 p. what is the simplest form of the fraction expressing their 
 comparative magnitudes? Answ. ^V 
 
 Problem VIII. To find the value of a fraction in the de- 
 nominations contained in the integer. 
 
 RULE. Consider the numerator as expressing the integer 
 taken as often as there are units in the numerator, and di- 
 vide it by the denominator. 
 
 This problem is, in strictness, a particular case of the next. On ac- 
 count of its frequent use, however, it is perhaps better in a separate 
 form. 
 
 Exam. 12. Required the value of . 
 
 Here the integer is 1; and the numerator, s. d. 
 
 being regarded as 4 times that integer, or 7)4 
 
 4, is divided by the denominator: the quo- 
 tient ll/5 is the value of the fraction. The 11 5^ 
 reason of the rule is manifest from the nature 
 of fractions. 
 
 Ex. 13. Required the value off of 5 13 9. 
 
 s. d. 
 
 Here the integer is 5 13 9; then, the nu- 5 13 9 
 
 merator expressing three times this quantity, 3 
 
 we multiply by 3, and divide by the denomi- 
 
 nator 8; the quotient, 2 2 7|, is the required 8)17 1 3 
 
 value. 
 
 2 2 71 
 
 Exam. 14. The height of M'Gillicuddy's Reeks, the highest 
 mountain in Ireland, is 3610 feet: if a person have ascended 
 through f of this space, to what height has he ascended? 
 
 Here, the height of the mountain is the in- 
 teger; and the numerator expressing three 
 times this quantity, we multiply by 3 and di- Q\7nQn 
 vide by 8, and we find for the answer 1353f ^ 
 
 1353* feet. 
 
REDUCTION OF FRACTIONS. 99 
 
 Exercises. Aiiswers. 
 
 62. f ......... 
 
 63. T \ ........ =3/9 
 
 64. T VVo- ..... =6/9ft 
 
 shilling, = lO^d. 
 
 Exercises. Answers. 
 
 60. W fot = 5i's inches. 
 
 67. ^ Irish mile. = lf.26p.-iy.2f. 
 
 68. | of \ 2 9.. =13/7$ 
 
 G9. A of -148 5. =80 17 3^ 
 
 of 175 tons ........... = 145 tons, 16 cwt. 2 qrs. 18J Its. 
 
 71. The earth revolves once on its axis in 23 hours, 56 minutes, 
 4 seconds; in what time does it perfofm f of a rotation ? Anm\ 10 
 hours, 38 minutes, 15 seconds. 
 
 72. By the articles of the Union of Great Britain and Ireland, 
 which took place in 1801, during the first twenty years Britain 
 was to contribute jf, and Ireland -j^V, of the amount of the pub- 
 lic expenditure. How much did each country contribute in making 
 a million sterling? Answ. 882,352 18 9if-, and 117,647 1 2 T V 
 
 Problem IX. To reduce a given fraction to another of a 
 lower denomination. 
 
 RULE. As in common Reduction, reduce the given nu- 
 merator, considered as an integer, to the denomination to 
 which the fraction is to be reduced, and write below it the 
 given denominator. 
 
 Exam. 15. Reduce T e 3 oo to tne fraction of a penny. 
 Here, by reducing 3 to pence, and writing 1600 below it, we 
 have for the required fraction tV^d. which by reduction to its 
 lowest terms, becomes /od. The reason of this is manifest, since 
 by the nature of fractions, the given fraction expresses a sixteen- 
 hundredth part of 3, and it must therefore express a sixteen- 
 hundredth part of the pence in 3. 
 
 Exercises. Answers. 
 
 73. Reduce -fc to the fraction of a shilling ............. f shil. 
 
 74-. - -3 cwt. to the denomination of pounds .... 16^ fbs. 
 
 75* What part of a second k-nny&jzro- dav? .............. AV 
 
 The converse of this problem, or the reduction of a fraction to a 
 higher denomination, is seldom of use. It is performed by operating 
 on the denominator in the way mentioned in the rule. Thus, if it be 
 required to reduce ^d. to the fraction of a pound, let the denominator 
 be multiplied by 20 and 12, and the numerator be retained : the result, 
 4a 9 00> or, in its simplest form, ^QQ, is the result required.* 
 
 * The method of reducing compound to simple fractions will be given in Multiplica- 
 tion of Fractions, and that of reducing complex fractions to simple ones, in Division of 
 Fractions. These are the natural and proper positions of these problems ; but should 
 any teacher conceive their introduction necessary here, with a view to prepare fractional 
 quantities for Addition, Subtraction, &c. it will be easy for the learner to turn over far 
 them to Multiplication and Division. 
 
 F -2 
 
100 
 
 ADDITION OF FRACTIONS. 
 
 RULE. (1.) Reduce the given fractions to others, having 
 a common denominator, if they be not such already. (2.) 
 When they are in this state, add all the numerators toge- 
 ther, and below their sum write the common denominator : 
 the result will be the sum of the fractions ; which, if it be 
 an improper fraction, must be reduced to a whole or mixed 
 number. (3.) If some of the given quantities be mixed 
 numbers, the fractions are to be added by the preceding 
 part of the rule; and the whole numbers, with any integral 
 part that may be obtained by the adding of the fractions, 
 are to be added by Simple Addition. 
 Exam. r. Add together if, \l, and T V 
 
 Here, the sum is f f, which becomes, by 14 ) 
 
 reduction to a mixed number, 2 T 3 ^, or, by 11 > 15 
 
 the reduction of the fractional part to its 8 ) 
 
 lowest terms, 2, the sum required. The 
 reason of the process is so obvious as 33 
 
 scarcely to require illustration. The first 
 
 of the given fractions expresses fourteen- 1 5)33(2 &, or 2$, 
 
 fifteenths of the integer; the next, eleven of Answ. 
 
 these parts: consequently, the two taken 
 together must express twenty-five of these parts. In like manner 
 the addition of any others may be illustrated. 
 
 Exam. 2. Add together 3&, ff , and 6|f. 
 In this exercise, by reduc- 3|& .... 1395 ) 
 
 ing the given fractions to .... 1450 > 1800 
 
 equivalent ones having a com- 6|f .... 1656 ) 
 
 mon denominator, we find 
 the first to be tffcfc, the se- llffik 4501 
 
 cond H&, & c - Then, by Answ. - 
 
 adding the numerators, we 1800)4501 (2 flfo 
 
 find the sum of the fractions - 
 
 to be ff g, or 2-^, by re- 901 
 
 duction to a mixed number. 
 
 We then write the fractional part beneath the given fractions; and 
 carrying 2 to the whole numbers, we find the required sum to be 
 
 Exercises. Answers. 
 
 1- 
 
 2. 
 
 3. 
 
 * 
 
 5. *+.H*i**t+ tt-Mt-HH 
 
6. 
 7. 
 8. 
 
 9- 
 10. 
 11. 
 
 12. 
 
 15. 
 16. 
 17. 
 
 SUBTRACTION OF FRACTIONS. 101 
 
 Exercises. Answers. 
 
 3H-12i+13|i ...... .......................... 29H* 
 
 f+H 
 
 II* 
 
 SUBTRACTION OF FRACTIONS. 
 
 RULE. (1.) Set the less quantity below the greater, and 
 prepare them both as in Addition of Fractions. (2.) Then, 
 if possible, subtract the lower numerator from the upper ; 
 below the remainder write the common denominator, and if 
 there be whole numbers, find their difference as in Simple 
 Subtraction. (3.) But if the lower numerator exceed the 
 upper, subtract it from the common denominator ; to the 
 remainder add the upper numerator ; write the common 
 denominator beneath the sum ; and carry one to the whole 
 number in the lower line. 
 
 The reason of this rule is evident from the explanations already 
 given of Simple Subtraction, and of Addition of Fractions. 
 
 Exam. 1. 
 From 87 A 
 
 take 42 A 
 Rem. 45 A 
 
 Exercises. 
 1- ^-^f 3 - 
 
 3*. ISfldfi" 
 
 4. \ A 
 
 5. 12^ 
 6. 
 
 Exam. 2. 
 
 73| 25 ) 
 
 17A.....J 5 
 56H 17 
 Answ. 
 
 Exam. 3. 
 
 10A ...... 14 > 
 
 3ff. ..... UO 
 
 6 T V* 79 
 
 175 
 * 
 
 Exercises. Answ. 
 
 7. A A 
 
 8. TV- A tft 
 
 9. H<WA* -B^ 
 
 10. H^/3 T*T 
 
 11. 5| f 2f 3,V 7 
 
 12. llf + SJ 9H .... 10W 
 
 * This sign, v-^- , is used to denote the difference of two quantities, without indicat- 
 ing which of them is the greater. In the exercises in which it occurs, therefore, the 
 fractions are to be reduced to others having a common denominator, and the less taken 
 from the greater. 
 
102 
 
 MULTIPLICATION OF FRACTIONS. 
 
 RULE. (1.) If any of the quantities be whole or mixed 
 numbers, reduce them to improper fractions. (2.) Find 
 the product of the numerators for the numerator of the re- 
 quired result, and the product of the denominators for its 
 denominator. 
 
 As a contraction, if a numerator and any denominator 
 be the same, reject them ; or divide a numerator and any 
 denominator by any number which will measure them, and 
 instead of them employ the results. 
 
 Compound fractions are reduced to simple ones in the 
 same manner. 
 
 Exam. 1. Multiply by }f. | X tt=&V=3k Amu,. 
 
 Here, by taking the products of the numerators, and of the 
 denominators, we find for the required result &%, or, in its sim- 
 plest form, H. 
 
 Exam. 2. Multiply by ft. 4 X T\=? f = iV 
 
 In this example, as one of the numerators is 7, and also one 
 of the denominators the same, by 
 
 neglecting this numerator and de- 4> (7)__ 4 
 
 nominator, we have T \ for the CO^ ^f IT 
 
 required result, the same as before. 
 
 The reason of this rule will appear evident from 'the principles 
 established in page 88. Thus, since -}f is the same as one-nine- 
 teenth of 12, it is evident, that the product, of -J by if i TIT f 
 its product by 12. Hence it appears, that to find the required 
 product, we are to multiply $ by 1 2 and to divide the result by 
 19; and by the principles referred to, this will be effected by mul- 
 tiplying the numerator by 12, and the denominator by 19: that 
 is, the numerator of the final result is the product of the given 
 numerators, and its denominator the product of the given deno- 
 minators. The contraction evidently produces the same result 
 that would be obtained if the terms of the product found by the 
 rule, were divided by the number rejected, or by the common 
 measure employed in the contraction. 
 
 These principles will assist in removing the difficulty which learners 
 often feel in accounting for the product of a quantity by a fraction b< - 
 ing less than the quantity itself. From what we have seen above, it ap- 
 pears, that in every case, in what is called Multiplication of Fractions, 
 there is not only a multiplication but also a dimsion. Hence, in the first 
 example, siure we multiply -^ by 12, and divide the product by 19, the 
 result must obviously be less than the multiplicand |~ From this also 
 it is evident, that when one factor is less than a unit, the product is less 
 than the other ; and consequently, when both factors are less than units, 
 the product is less than either. 
 
8 
 
 DIVISION OF FRACTIONS. 10S 
 
 Exam. 3. Multiply 17f by 12|. 
 
 These quantities, by reduction to improper fractions, become 
 g 7 and V- Then, 1 f 7 X V = 5 tl 7z:: l V 9 > by reduction to its 
 lowest terms; and this being reduced to a mixed number, becomes 
 227|, the product required. 
 
 The product may also be found, as 17| 
 
 in the margin, without reducing the 12f 
 
 factors to improper fractions. In this 
 mode, 12 times f= 6 6j or 10, which 214 
 
 is carried to the product of 17 and 12.... 6 
 
 12. We have then to find the pro- 
 ducts of 17 and f, and of f and f. 
 The former is 12|, and the latter f. 227f V, or If. 
 
 The sum of all the three partial pro- 
 ducts is 227f, the same as the answer found already.* 
 
 Exam. 4. Reduce the compound fraction f of f of 3^ to a 
 simple fraction. 
 
 Here the mixed number 3 is reduced to y. Then, by taking 
 the product of the numerators, 3, 8, and 10; and also that of 
 the denominators 4, 9, and 3, we obtain f-gf, or in its lowest 
 terms 2 -, the simple fraction required, which might be reduced 
 to the mixed number 2f. In this example, the work might be 
 contracted by rejecting the numerator and the denominator 3, and 
 by dividing the denominator 4 and the numerator 8 by 4, and the 
 same result would be obtained. 
 
 Exercises. Answ. 
 
 AXtt .............. rifr 
 
 H 
 .............. It 
 
 ............ rift 
 
 5. 
 6. 
 
 7. 
 
 8. 19 T VXlT\ 
 
 9. 34HXW 
 
 10. 21fX2HX21| 
 
 11- 
 12. 
 
 HI 
 
 164UI 
 10252}! 
 
 iWo 3 o 
 rift: 
 
 13. 
 
 14. 
 
 Exercises. 
 Hi 
 
 Answ. 
 
 24f| 
 
 S T V, 
 
 16.HXI* ............... W 
 
 17. 12^-XlVr 
 
 18. | of f 
 
 19. A of ^ ............. ft 
 
 20. ^of 9| ............... 6^ 
 
 21. f off of | off ..... $ 
 
 22. 1 of |- of 7 ........... 5 
 
 23.fXTV-ffXrV .... It 
 
 24. XiXH ........ 
 
 DIVISION OF FRACTIONS. 
 
 RULE (1.) Reduce mixed or whole numbers to improper 
 fractions, and compound fractions to simple ones, if any 
 
 * When the pupil has learned the method of aliquot parts, he will see, that it may 
 often be employed with ease and advantage in the multiplication of fractional quantities. 
 
104 DIVISION OF FRACTIONS. 
 
 such be given. (2.) Then invert the divisor,* or rather con- 
 ceive it to-be inverted, and proceed as in Multiplication. 
 
 A complex fraction is reduced to a simple one by divid- 
 ing the numerator by the denominator, according to the 
 preceding rule. 
 
 Exam. 1. Divide fo by fo &-KV= /&=? Answ. 
 
 Here, the divisor being conceived to be inverted, and the frac- 
 tions being multiplied, we have for the answer T \X V=i^% ii- 
 
 The rtason of this Deration will be understood from consider- 
 ing, that j 7 7 is only one-twelfth of 7; and consequently, the quo- 
 tient obtained by dividing by T \ must be twelve times the quotient 
 obtained by dividing by 7. We must therefore divide by 7, and 
 multiply the result by 12, which is the same that is done by the 
 inversion of the divisor. 
 
 From this it appears, that in Division of Fractions, there is in reality, 
 both a division and a multiplication. It will also readily appear, that if 
 the divisor be a proper fraction, the quotient will be greater than the 
 dividend. 
 
 Exam. 2. Divide 5^ by 2|. 
 
 Here, 5&=}f, and 2r= V 3 : we have therefore for the required 
 result, f f X *= V X A= W=2f 
 
 By the application of the principles established in page 88, a 
 fractional number may be more easily divided by a whole number, 
 than it is by the general rule. Thus, 
 
 Exam. 3. Divide 75^ by 9. 
 
 Here, 9 is contained 8 tunes in 75, with the 9)75 fa 
 
 remainder 3. Then 3 1 7 , by reduction to an ~~8f| 
 
 improper fraction, becomes f$, which by divi- 
 sion by 9, becomes f$, the remaining part of the quotient.f 
 
 Exam. 4. Required the value of the complex fraction \ 
 Here, 5|=V> and 6f=Yi an ^ tne simple fraction required is 
 
 v-M?<*m- 
 
 It may be useful on some occasions to know, that if the given frac- 
 tions be reduced to equivalent ones having a common denominator, the 
 quotient of the resulting numerators will be the required quotient. 
 Thus, the fractions in the first example are equivalent to g and f i : 
 
 * That is, take its numerator as denominator, and its denominator as numerator. 
 It may be remarked, that the fraction resulting from the inversion of another fraction, 
 is called its reciprocals since any two numbert whose product is a unit, are called !hr 
 reciprocals of each other. Thus, ^ and j 7 5 , | and f or 8, are the reciprocals of each 
 other, as y x T ' a =1. and 1x8=1. 
 
 t This mode of dividing furnishes an explanation of the rule in page 27, for finding 
 the true remainder in dividmtt t>y the factors of a given divisor. 
 
THE RULE OF PROPORTION IN FRACTIONS. 105 
 
 Dividing therefore the numerator 10, by the numerator 21, we have 12, 
 the same quotient as before ; and thus may all the exercises in this rule 
 be performed. 
 
 7. 
 
 8. 
 
 9. 
 10. 
 11. 
 
 Exercises. 
 
 Answ. 
 
 =4320 
 
 Exercises. 
 
 13. f-5-A 
 
 14- ! 3 T-f 
 15. ef 
 
 20. 
 21. 
 22. 
 
 Answ. 
 = T ^ 
 = 2* 
 
 =A 
 =Uf 
 
 =1^ 
 =6? 
 
 THE RULE OF PROPORTION IN FRACTIONS. 
 
 IN fractions, in the rule of Proportion, whether Simple 
 or Compound, the terms are arranged in the same manner 
 as if they were whole numbers; but they are reduced, mul- 
 tiplied, and divided by the rules for the reduction, multi- 
 plication, and division of fractions. 
 
 When only three terms are given, it is evident from the 
 rules for the multiplication and division of fractions, that if 
 the first term, after the necessary reduction, be inverted, 
 the continual product of the result and the other two terms, 
 will be the fourth proportional required. 
 
 The work may also be performed by reducing the first 
 and second terms to fractions having a common denomina- 
 tor. The common denominator may then be rejected and 
 only the numerators employed. 
 
 Exam. 1. If 3f cwt. of sugar cost 17^*3, how much may be 
 bought for 4i ? 
 
 Here, as 11 P T \ : 1\ : : 3f cwt.; or, by reduction to improper 
 
 15X29X25 10875 
 
 tractions, as W : *? : y cwt.; g59x?x7 ^^l^ 
 
 cwt. the fourth proportional required, the value of which by pro- 
 blem VIII. page 98, is found to be 3 qrs. IHtfi ^* s - 
 
 Exam. 2. If % purchase 1| Ifts. of indigo, how much may 
 be bourht for |f- shilling? 
 
 As | : \%s. : : 1|- Ibs.; or, by reducing the first tenri to the 
 denomination of shillings, and the third to an improper fraction, 
 
106 DECIMAL FACTIONS. 
 
 9X16XH * 3X II 
 
 as >rs. : ifs. : : V ft. 160xl9><6 
 ^y Ibs.; or, by multiplying by 16, = lff oz. 
 
 Exam. 3. If f lb. of tea cost 6/8, what cost f lb. ? 
 As lb. : I lb. : : 6/8 : X $ X 6/8, or H X 6/8, or 6/2$. The 
 work may also be as follows, since 6/8 i^y, or $: as It). : 
 
 fi v 7 
 fr lb. : : * : * =^U; or by problem VIII. page 98, =r 
 
 > X " X <j 
 6/2f, 
 
 Since ^f, and |- }f, the answer might also be obtained 
 very easily by the following analogy: as 15 : 14 :: 6/8 : 6/2$. 
 
 Exercises. 1. If -^ cwt. cost . T 4 T , what cost 1 cwt. ? Answ. 
 \ 2 Hf. 
 
 2. How much wine may be bought for .17 8 4,* if 5 r \ gal- 
 lons cost ,-776? Answ' 12|f gallons. 
 
 3. How much sugar may be bought for .81 14 6, if If cwt. 
 cost 6f x ? Answ. 19 cwt. 3 qrs. 16$ff Ibs. 
 
 4.. If lb. cost -BS. what cost 2H cwt.? 4/z0. 17 6 8. 
 
 5. If the yearly rent of 45 a. 2 r. 20 p. be 44^, what rs the 
 rent of If acre? ,4i/. 1 11 7ff. 
 
 6. If 3f shillings pay for 1 T V yard, how much may be bought 
 for 6\i? Answ. 42 yds. 1 qr. 1 T V nail. 
 
 7. What cost 49 T 3 T yards of broadcloth, if 7| yards cost 7 
 18 4? Answ. 51 3 
 
 DECIMAL FRACTIONS.f 
 
 A DECIMAL FRACTION, or a DECIMAL, is a fraction whose 
 denominator is 10, or some number produced by the con- 
 tinued multiplication of 10 as factor, such as 100, 1000, &c.| 
 
 Hence all the rules for the management of fractions in 
 general, are applicable to decimal fractions. 
 
 In the notation of decimals, the denominator is usually 
 omitted ; and, to indicate its value, a point is placed to the 
 left of as many figures of the numerator, as there are ci- 
 
 * This, by problem VII. page 97,becomes 17 T \ : and 7 7 fi is equivalent to 7|. 
 
 f- The following article on decimal fractions is merely an adaptation of the general 
 principles of fractional quantities, already delivered, to a particular and important class 
 of fractions. It therefore contains no principles that can be properly called new ; but 
 rather short and easy modes of applying those already given. Every operation on deci- 
 mal fractions, indeed, may be performed by reducing them to common fractions, as will 
 enpear hereafter, and then applying to the results, the rules in the preceding articles. 
 
 I It will be seen hereafter, that the number which results from the continued multi- 
 plication qf any number as factor, is called a power of that factor. Thus, 100 is the 
 second power of 10; 1000, its third power, txc. 
 
DECIMAL FRACTIONS. 107 
 
 phers in the denominator. Should there not be a sufficient 
 number of figures in the numerator, as many ciphers are 
 prefixed as supply the deficiency. 
 
 Thus, T V, yfcy, TO 06 00' TO> are decimals ; and -& is written -7 ; 
 Y&y '9; TiH&OT' '00003; $g, 4'75, &c. Hence, conversely, 
 
 The denominator of a decimal, thus expressed, is the 
 number denoted by a unit with as many ciphers annexed, 
 as there are figures in the given number. 
 
 Thus, -37i3 T 3&j '004, ^; -00083, ^MtRy &c. 
 
 From this notation it is evident, that the figure immedi- 
 ately following the decimal point, denotes tenths; the next 
 figure, hundredths; the third, thousandths, &c. 
 
 Thus, since 476 is equivalent to 400+70-J-6, the fraction '476, or 
 -^j is equivalent to T$8&+ T 3&y+ 1 iftnr or to T%+T&H~n&ff> b X 
 dividing the terms of the tirst of these fractions by 1OO, and those of 
 the second by 10. In a similar manner it would appear, that "709 i,s 
 equivalent to seven tenths, no hundredths, and nine thousandths. 
 
 Hence, the values of figures in decimals, as well as in 
 whole numbers, are increased in a tenfold ratio by remov^Jjjy 
 ing them one place towards the left hand, and diminished 
 in the same ratio by removing them one place to the right. 
 
 Thus, in the decimal '004, by removing the point one place to the ' 
 right, and consequently the figures of the decimal one place to the left, 
 u e have O4, which denotes T ^, or ?$$$, and is ten times^the given 
 fraction, -j^g- ; but by introducing a cipher after the point, which rt> 
 . moves the figures a place to the right, we have -OO04 or T ^^Q, whicn 
 is evidently only a tenth part of the given fraction j^W, or i^,, 
 From these principles it will readily appear, that 
 
 A decimal is multiplied by 10, if the separating point be 
 removed one place towards the right hand; by 100, if two 
 places; by 1000, if three places, &c.: and, conversely, a 
 decimal is divided by 10, if the separating point be removed 
 one place towards the left hand; by 100. if two places; by 
 1000, if three places, &c. ; vacant places, when such oc- 
 cur, being supplied in both cases by ciphers. 
 
 Thus, -7248X10=7-248, or 7 T <&& ; 6 '347X100=634 -7 ; 6'3X 
 1000=6300. Also, 78-33-i- 10=7 -833; '736-:- 100= -00736 ; 7'3~ 
 100='073, &c. It appears also from the same principles, that 
 
 The value of a decimal is not changed by annexing a 
 cipher to the end of it, nor by taking one away, as in each 
 case the significant figures retain the same positions in re- 
 lation to the separating point. 
 
 Thus, '50=-5=-500, each being equivalent to one half. 
 
 From this view of the nature of decimal fractions, it appears, that 
 there is in every respect the closest resemblance between them and whott 
 numbers j and hence all operations on decimals are penorm^d in x- 
 
108 DEDUCTION OF DECIMAL FRACTIONS. 
 
 actly the same manner as those on whole numbers due attention being 
 paid to the position of the separating point. This last circumstance, 
 indeed, demands the utmost care, as the point is the characteristic of the 
 decimal ; and from what precedes, it is evident how much depends 011 
 its proper position. 
 
 Exercises in Notation and Numeration of Decimal Fractions. 
 
 Write the following fractions according to the notation of deci- 
 mal fractions: 
 
 1. Three hundred and six ten-thousandths. 2. Three hundred- 
 thousandths. 3. One ten-millionth. 4. One ten-thousandth. 5. 
 One tenth. 6. Fifty-seven hundred-thousandths. 7. Two hun- 
 dred ten-millionths. 8. Five hundred and nine hundredth^. 
 
 Express the following decimal fractions in words: 
 
 1. -58 I 3. -007 I 5. 31-73 | 7. -00004 I 9. -034567 
 
 2. -106 4. -0007 6. 3-173 8. -00041 10. -00000011 
 
 REDUCTION OF DECIMAL FRACTIONS. 
 
 Problem I. To reduce a common fraction to a decimal. 
 
 RULE. (1.) Let the numerator, with a cipher annexed, 
 be divided by the denominator, and to the significant figure 
 or cipher placed in the quotient, prefix a point. (2.) Then, 
 if there be a remainder, annex to it more ciphers, and con- 
 tinue the division till nothing remain, or till the result con- 
 sist of as many figures as may be thought necessary. 
 
 A given quantity may be reduced to the decimal of an- 
 other given quantity of the same kind, considered as an in- 
 teger, by first reducing it to a common fraction by problem 
 VII. page 97, and then reducing the fraction thus found to 
 a decimal. 
 
 Exam. 1. Reduce ? f 3- to a decimal. 5 1 2)3000(-005859375 
 Here, 3 by the addition of one cipher, 
 becomes 30, in which 512 is not contain- 4400 
 
 ed; and therefore a cipher is put in the 
 quotient, and a point prefixed. ~By the 3040 
 
 annexing of another cipher, 30 becomes 
 300, in which the divisor not being con- 4800 
 
 tained, another cipher is put in the quo- 
 tient. After this the division proceeds 1920 
 in the usual way, a cipher being added 
 each time, and the decimal is found to be 3840 
 005859375, or T ^Wo 7 <5W> which by 
 reduction to its lowest terms, would be- 2560 
 come -S^TT, the given fraction; and thus 
 the work is proved to be correct. 
 
REDUCTION OF DECIMAL FRACTIONS. 109 
 
 With respect to the reason of tJie rule, the given fraction, by an- 
 nexing to each of its terms nine ciphers, (the number that was 
 annexed in the preceding operation,) becomes 5itr8-(ji and 
 this, by dividing each term by 512, becomes TuV^Wo s jn5j which 
 in the notation of decimals, is '005859375, the same as before. 
 Hence it is obvious, that the preceding reduction is in reality the 
 multiplication of each term of the given fraction by 1000000000, 
 and the division of each of the results by 512; and it is evident 
 from the principles established in page 68, that the result must be 
 equivalent to the given fraction. 
 
 This operation might also be illustrated in other modes. Thus, 
 according to the fourth principle in page 89, we should have 512 : 
 3 : : 1000000000 : 5859375, the numerator of the required frac- 
 tion; which in relation to the denominator 1000000000 would be 
 expressed by '005859375, according to the notation of decimals. 
 Another illustration might also be derived from the latter part of 
 the rule for problem IV. page 95. 
 
 Exam. 2. Reduce ^ to a decimal. 
 In this example, each remainder alter 12)500000 
 
 the two first is 8, and hence the same 
 figure must be repeated perpetually. *41666, &c. 
 
 Exam. 3. Reduce T 6 T to a decimal. 
 Here, after two figures have been ob- 11)60000 
 
 tained in the quotient, the same series 
 
 of remainders, and consequently the same '5454, &c. 
 
 series of figures in the quotient, must 
 
 recur; and therefore, were the work pursued, the first two figures 
 of the decimal would recur without end. 
 
 Exam. 4. Reduce 13/8 to the decimal of a pound sterling. 
 Here, 13/8=164 pencezri|= 
 f; and <, reduced to a decimal, 12)80000 
 
 in the manner already shown, becomes 
 
 68333, &c. The same result may 20)13-6666, &c. 
 
 be obtained by writing, as in the mar- . 
 
 gin, 13 shillings below 8 pence; then '6833, &c. 
 
 dividing by 12 and by 20, we find 13/8 
 
 equivalent to 13-6666, &c. shillings, or to '68333, &c. as before. 
 
 The first three figures of the decimal of a pound may be found very 
 readily, thus : Take half the number of shillings for the first figure ; 
 and for the other figures increase the farthings in the given pence and 
 farthings by 1, if they amount to 24 or more, and by 50, if there were 
 a remainder of 1 shilling. Thus, if 13/8 be given, the half of 13 is 6, 
 and 1 remains ; and in 8 pence there are 32 farthings, which exceeding 
 i'4, is made 33, and this, because 1 shilling remained, is increased by 
 50, and becomes 83 : hence the three first figures are '683. 
 
 A decimal which cannot be exactly expressed, but which 
 may be continued to an unlimited number of figures, is 
 
110 REDUCTION OF DECIMAL FRACTIONS. 
 
 called an INTERMINATE DECIMAL, to distinguish it from 
 
 others, which in respect to it, are called TERMINATE, 
 
 When a decimal is expressed either by the continual re- 
 petition of the same figure, or of the number expressed by 
 two or more figures, it is called a PERIODICAL or a CIRCU- 
 LATING DECIMAL ; and the figure, or number, so repeated 
 is called the PERIOD. 
 
 Thus the decimals in the second and third examples are periodical, 
 the period in the one consisting of one figure, and that in the other of 
 two. When only one figure is repeated, the application of the terms 
 yerwd and periodical, though convenient, and perhaps best on the whole, 
 is scarcely correct, as each of the terms suggests the idea of more 
 figures than one. On this account, some writers term such decimals 
 repeaters, or recreating decimals. 
 
 A periodical decimal is said to be MIXED, if it consist of 
 one or more figures prefixed to a periodical part ; others 
 are called PURE. 
 
 For the sake of brevity, in writing decimals of this kind, 
 it will often be sufficient to write the period but once, and 
 to denote its continuation by putting a trait, or stroke, over 
 the first figure of the period, and another over the last, or 
 one over the repeating figure, if there be but one figure in 
 the period. 
 
 Thus, -47333, &c. may be expressed by '473', and '5637637, c. by 
 56'37'. 
 
 The following considerations will tend to explain still farther the na- 
 ture of these fractions. By the preceding rule we find the following 
 results : ='1111, &c. ; ^-=-010101, &c. ; ^3= -001001, &c. ; ^y 
 = 00010001, &c. These decimals are all periodical; and if any of 
 them be multiplied by a whole number, die result will also be period- 
 ical. Thus, if we multiply the third by 128, we find =! 28 128, 
 &c. ; if by 998, we get ff|= -998998, c. Here we see, that the 
 value of the periodical fraction is the common fraction, whose denomi- 
 nator is the number expressed by as many 9's as there are figures in the 
 period, and whose numerator is the period itself: and the same may be 
 shown in every case. Hence, we have the following rule : To "find 
 the value of a pure periodical decimal, take the period for numerator, and at 
 many 9*s as there are figures in the period, for denominator. Thus, -8181, 
 c.=fi= T T ; 2'97'=fH==H; '3'==f=h- '^9, &c.=|=l, &c. 
 
 If the decimal be mixed, its value may be easily found by the same 
 principle. Thus, if it were required to find the common fraction 
 which would produce the decimal -12436436, &c. by multiplying by 
 100 we should find 12'4'36'V or 12-fig. Dividing this by 100, we ob- 
 tain, for the value of the proposed fraction, 1 ^ ? -f"at^o> or D y reduc- 
 
 12X999 . 4H6 12424 3106 
 t,on to a common denominator, ___+__=-_=__. 
 
 From this vre may readily derive a rule which will be easier iu yrae- 
 
REDUCTION OF DECIMAL FRACTIONS. Ill 
 
 ttce. For in the last result, since 12 is multiplied into 999, or into 
 1000 1, the product is 12000 12; to which if the other numerator be 
 added, we have for the required numerator 12436 12, and the deno- 
 minator remains the same as before. From a due consideration of this 
 result, we may derive the following rule : To find the value of a mixed 
 periodical deci?nal, from tlie number expressed by ike finite jmrt with the 
 period annexed, subtract the finite part for the numerator ; and, for the de- 
 nominator, to as many 9*5 as there are figures in the period, annex as many 
 ciptiers as there are figures in the finite part. Thus, to ftnd the value of 
 83', from 83 take 8, and there will remain 75, and the required frac- 
 tion is %%, or f. Again, if -5416' be given, we have 5416-7-541, or 
 4875 for numerator, and the fraction is ^, or 1-|. In like manner, 
 to find the value of '263'Gl', we have for numerator 26301 26=26275, 
 and.for denominator 99SOO. Hence the required fraction is ggggg, 01 
 iggs- The value of such fractions may also be found by the summa- 
 tion of series, as will appear hereafter. 
 
 By the method thus shown, interminate decimals may be reduced to 
 common fractions, and subjected to the rules for managing such quan- 
 tities. Unless when complete precision is required however, this is not 
 necessary: and indeed in all useful cases their values, instead of being 
 found with entire accuracy, are to be approximated, by carrying the de- 
 cimals out to as many figures as may be necessary in any particular case. 
 In thus approximating the value of a decimal, if it be carried to two 
 places, the error irill be less than a hundredth part of the integer; if to three 
 places, less titan a thousandth; if to four places, less than a ten-thousandllt, 
 c. Thus, if the decimal -2 f V, be carried only to "27, the error will 
 be less than yi^, since the true value is more than T -^j, and less than 
 ^$j ; if to the three figures '272, the error will be less than -J^JQ, since 
 the true value is greater than T %\j^- and less than T Vo%> & c. 
 
 It may be remarked here, that when a vulgar fraction is in its lowest 
 terms, and its denominator contains no simple factors except 2 or 5, 
 (the two prime factors of 10,) the equivalent decimal is finite, but in 
 every other case it is interminate. The cause of this is, that neither JO, 
 100, nor any similar number, is divisible by any of the nine digits ex- 
 cept 2 and 5. It may also be observed, that the number of circulating 
 figures must always be less than the units in the denominator. This is 
 obvious from the consideration, that the number of remainders different 
 from each other, which can arise in any operation in division, nust be 
 less than the units in the divisor. Thus, in dividing by 7, it is evi- 
 dent, that the only possible remainders are 1, 2, 3, 4, 5, and 6 ; and 
 since, in reducing to decimals, a cipher is annexed to each remainder, 
 there cannot be more than six dividends, and consequently, six figures 
 in the quotient, all different. 
 
 It may be added, on this part of the subject, that any prime number, 
 except 2 and 5, is contained, without remainder, in the number expressed 
 in the common notation, by as many 9's as there are units, wanting one, in 
 tiie prime number itself.* Thus, 3 is a measure of 09 ; 7 of 999,999; 
 and 13 of 999,999,999,999. It is easy to show from this property, that 
 every prime number, except 2, 3, and 5, isa'measure of the number 
 
 * For a demonstration of the general proposition of which this is a particular cac**, 
 ice Legendie's Theorie des Nombres, No. 129 
 
112 REDUCTION OF DECIMAL FRACTIONS. 
 
 expressed in the common notation, by as many 1's as there are units, 
 wanting one, in the prime number itself. Thus, 7 -is a measure of 
 111,111; and 13 of 111,111,111,111, Sec. It may also be remarked, 
 that while the prime number is always a measure of the number expres- 
 sed by the number of 9's or 1's above specified, it is sometimes con- 
 tained exactly in the number expressed by one-half, one-third, or some 
 other submultiple, or exact part, of the number of those figures. Thus, 
 while 11 is a measure of 9,999,999,999, it is also contained exactly in 
 99, the number expressed by two (one-fifth of 10,) 9's ; and 13 is a mea- 
 sure of 999,999, or 111,111. 
 
 Exercises. Reduce the following fractional quantities to decimals : 
 
 Exercises. Answers. 
 9. & -(X-l/ 
 
 10. $ -4/683544303797 
 
 11. vfo 0'044 / 
 
 12. ^ B 0'00444' 
 
 Exercises. Answers. 
 
 1. T V -4375 
 
 2. ^r -09375 
 
 3. nfr* -0048828125 
 
 4. H '275 
 
 5. TV / 76923 y 13. ^ -O^S^ 
 
 6. T \ '583' 14. !' 
 
 7. yfa -01'04895' 15. -fa 0/12345679' 
 
 8. TTy fo T -0'0059994' 16. ^ 022 / 7' 
 
 Exercises. Answers. 
 
 17. 10/9 to the decimal of 1 -5375 
 
 18. 0/10 1 -04375 
 
 19. 17/7 __ 1 -87916' 
 
 20. 3 r. 11 p an acre -81875 
 
 21. 2 qrs. 8 Ibs cwt 5'71428 / 
 
 22. 37 perches, mile -115625 
 
 23. 3 hours, 30 minutes, day -14583' 
 
 24. 15 minutes, 30 seconds, hour , -2583' 
 
 25. 16/8 1 2 9 l f 326W 
 
 26. 3 c. 1 q. 7 Ibs. ton -165625 
 
 27. 6d. ,^vv~~~~~~~.,~~~~~~. shilling -5416' 
 
 Ex. 28. If the diameter of a circle be 1, the circumference is 
 3}- nearly, or 3 T a T V more nearly. Express each of these decimally.* 
 Answ. 3-l'42857', and 3*1415929, &c. 
 
 29. If the circumference of a circle be 1, the diameter is 5 \ 
 nearly, or $$$. more nearly. Express each of these decimally. 
 Answ. -31'8', and -318309859, &c. 
 
 30. An English acre is if of an Irish acre : required the equi- 
 valent decimal. Answ. -61734693877551, &c. 
 
 31. Reduce a quarter of wheat containing 456 Ibs. to the decimal 
 of a hundred weight. Answ. 4-07'14285'. 
 
 32. The old Irish liquid gallon contained 217-6 cubic inches, 
 and the old English wine gallon 231. Reduce the former to the 
 decimal of the latter. Answ. -94/19913'. 
 
 * The circumference true to twenty decimal places, is S-HljD'.^SS^STSSSSfjitv 
 
REDUCTION OF DECIMAL FRACTIONS. 113 
 
 Problem II. To Jind the value of a given decimal in the 
 parts contained in the integer. 
 
 RULE. Multiply the given decimal by the numbers which, 
 were it an integer, would reduce it to the lower denomina- 
 tions contained in it, and, after each multiplication, point 
 off for decimals as many figures towards the right hand as 
 there were ^figures in the given decimal. The figures re- 
 maining on the left will express the required value. 
 
 Exam. 5. Required the value of '3945 of a day in hours, &c. 
 
 Here, the given decimal is multiplied '3945 day 
 
 by 24, the number of hours in a day, and 24; 
 
 four figures being cut off towards the -^ 
 
 right, it appears that -3945 day is equal 1^780 
 
 to 9 T V$& or 9 T V<& hours. The deci- 7890 
 
 mal -4-68 (the cipher being omitted, see 
 
 page 107,) is then multiplied by 60, and 9-4080 hours 
 
 three figures being cut off, there results 60 
 
 28-080 or 28-08 minutes. By continuing 
 
 the process, the value of the given deci- 28-080 minutes 
 
 mal is found to be 9 h. 28 min. 4 sec. 48 60 
 
 thirds^ With respect to the reason of 
 
 this process, it is only necessary to ob- 4*80 seconds 
 
 serve, that it is exactly the same as find- 60 
 
 ing the value of V$&& day by problem 
 
 VIII. page 98, the pointing off of the 48'0 thirds. 
 decimals serving the purpose of dividing 
 by die denominator. 
 
 Exam. 6. Required the value of -5937. 
 
 Here, by multiplying by 20, we find, -5937 
 
 that the given decimal is equivalent to 20 
 
 llTVok shillings. Then, by multiply- 11-8740 shillings 
 
 ing by 12, we find *874s. to be equiva- 12 
 
 lent to 10-488 pence. In like manner, ~7n~Tft 
 
 488d.is shown to be equivalent to Ifjfo ^ P< 
 
 farthing, or a halfpenny, very nearly; 
 
 and consequently, the required value is 1-952 farthings. 
 
 11/10$, nearly. Answ. 11/10, nearly. 
 
 The following rule is easy and useful in practice : To Jind tlie value 
 of the decimal of a pound sterling to the nearest farthing ; (l.J Take a fifth 
 of the number expressed by the two first figures of the decimal, for the tfril- 
 lins,s of the result : (2.J Diminish the number expressed b>/ the remainder 
 with the third figure of the decimal annexed, by a twenty-fifth of itself, and 
 what remains will be the farthings in the rest of the required value. Thus, 
 in finding the value of -5937, the fifth part of 59 is 11, the sniffings of the 
 answer ; the remainder is 4, which being prefixed to 4, (the third figure in- 
 creased by a unit, because the next figure 7, is greater than five,) we have 
 
114 REDUCTION OF DECIMAL FRACTIONS. 
 
 44, the twenty-fifth part of which is more nearly 2 than 1; we therefore re- 
 icct 2, and have remaining 42 farthings, or IQi; and hence, the answer is 
 11/10^, nearly, the same as before. With respect to the reason of this 
 process, it is evident that the given decimal is more nearly equal to 
 594O than -5930; we use therefore the former, or its equal 594. Now 
 this is equivalent to '55-|-'044, the value of the former part of which 
 would be found by the general rule for this problem, by multiplying by 
 20 and dividing by 100, which is the same as dividing l>y 5, since 20 
 is one-fifth of 100. Then, for the value of -044 or T ofo>> by di- 
 minishing the denominator by one twenty-fifth part of itself, we have 
 960 ; and by diminishing the numerator by a similar part of itself, we 
 have 42 nearly : hence 1^0^^ nearly equal to -^g%, or 42 farthings, 
 since gJ^=l farthing. Practice will soon enable the learner to esti- 
 mate with sufficient correctness, the effect of the fourth figure of the 
 decimal, in respect to the quantity to be rejected. In this example, in- 
 stead of using 44 farthings, we might use 43'7, and it would be easy 
 to see, that the twenty-fifth of this would be about 1'7 ; and rejecting 
 this, we should have 42, as before. 
 
 Exam. 7. Required the value of *805''of a yard in long mea- 
 sure. 
 
 This, and similar exercises, may be 8055' of a yard, 
 
 wrought either by converting the pro- 3 
 
 posed decimal into a common fraction, 
 
 m the way shown in page 111, or more 2*4166 feet, 
 easily by employing an approximative 12 
 
 process, as in the margin. Here, we . 
 
 carry 1 to the product of 3 and 5, be- 4*9999 inches. 
 
 cause, had the decimal been continued 
 
 farther, it is evident 1 must have been Answ. 2 feet, 5 inches. 
 
 carried from the preceding product. 
 
 For a similar reason, 7 is carried to the product of 12 and 6. The 
 
 result is found to be 2 feet, 4*9 / inches, which is equivalent to 2 
 
 feet, 5 inches. ( See page IIO.J Another 5 was added to the 
 
 given decimal, that the result might be more distinct and certain. 
 
 Exercises. Required the values of the following decimals : 
 Exercises. Answers. 
 
 33. -0675 of a cwt ?1|- Ibs. 
 
 34. '4625 of a ton 9 cwt. 1 qr. 
 
 35. -0484 
 
 36. *8845 of an acre 3 r. 
 
 37. '00213 of a day 3 minutes, 4?^% s sec. 
 
 38. '!' 13/<H> nearly. 
 
 39. '&857l4f of a cwt 1 qr. 4 tbs. 
 
 40. IIS'O' of a mile 36 perches, 2 yards. 
 
 4J. '615 of a shilling 7gd. 
 
 4?. '483'5 / 9/8 nearly. 
 
 43. -2385 of a degree 14> 18" 36"' 
 
 44. -GS'ltf of a cwt. Iwig weight 2qrs. 15$ tbs. nearly 
 
ADDITION OF DECIMAL FRACTIONS. 115 
 
 Exercises. Answers. 
 
 45. '06' 1/4 
 
 46. -47916' fos, troy 5 oz. 15 dwts. 
 
 47. -2'& 4 8/6, nearly. 
 
 48. -4375 of a shilling 5^d. 
 
 49. -09375 of an acre , 15 perches. 
 
 50. '4/ foot, long measure 5 inches. 
 
 ADDITION OF DECIMAL FRACTIONS. 
 
 RULE. (1.) Arrange the given numbers so that the sepa- 
 rating points may all be in the same column. (2.) Find 
 the sum as in Simple Addition. (3.) Point off as many 
 places of decimals, as there are in the given number which 
 contains most. 
 
 If any of the given numbers contain periodical decimals, 
 let these be carried out to as many places as there are in 
 the longest of the finite decimals; or, if greater accuracy 
 be required, let them be carried as far as may be judged 
 necessary. 
 
 In the application of decimals to practical purposes, it is generally 
 known, from the nature of the case under consideration, to how many 
 places it is necessary, that the result should be true. When a result is 
 thus required to be true to an assigned number of places of decimals, it is //- 
 per to carry the decimals which consist of more places, to at le;tst one place 
 beyond the 9>sig>ied number, and to reject the last , figure. In this case, it 
 is proper to observe, that when a decimal is not carried out to its full length, 
 the last figure of the part retained, should be increased by a. unit, if the suc- 
 ceeding figure be 5, or greater than 5. 
 
 Exam. 1. Add together 81-4632, 9'75, and 47-388. 
 
 Here, the numbers are arranged as in 81 '4632 
 
 the margin, and added as in common 9-75 
 
 Addition. The reason of the arrange- 47-388 
 
 ment and operation is manifest, those 
 
 figures being added together which are of 138*601 2, sum. 
 
 tht same local value. 
 
 ^xam. 2. Add together 3*7'3', '873, 51-7', 108-2, and 73-4-63 128, 
 so that the result may have four places of decimals true. 
 
 In this example, the first, third, and 3*73737 
 
 fifth numbers are carried to five places, -873 
 
 each, and the last figure of the third is 51-77778 
 
 made 8, because the next figure would 108*2 
 be 7. In like manner the fifth figure of 73-46313 
 
 the last line is madesJ, because the sue- 
 
 ceeding figure is 8. : The reason of this 238*0513, sum. 
 
116 SUBTRACTION OF DECIMAL FRACTIONS. 
 
 is evident, since 30 is nearer 28 than 20 is, and 30, by the re- 
 jection of the last figure, becomes 3. In the addition, 'the sum 
 of the last column is 18, from which 2 is carried, because 18 is 
 nearer 20 than 10. The correct sum, found by carrying the deci- 
 mals farther, is 238-0512795'!', which by retaining only four figures 
 of the decimal, and increasing the last of them by a unitjji because 
 it is followed by 79, &c. becomes 238-0513, the same as before. 
 
 Exercises. 1. Add together }-83, 5'674, -3125, 18'3, 100, 38'62, 
 4-3957, and -5. Answ. 169-6322. 
 
 2. Required the sum of 93-617843, 7-836, 12-25, -71375, 4-391, 
 7-839, 3-7674285, and -8693. Answ. 131-2843215. 
 
 3. Required the sum, true to six places, of 51*25, 3'4', -63'7 , 
 7-885', 7-875, 7*8'75', and 11-1'. Answ. 90'0793'60724'. 
 
 4. Required the sum of -7354, -7354', '735'4', 7 / 354 / , -07354, and 
 0735'4', Answ. 3-088857'991'. 
 
 5. Add together -3, -3', -45, -4'5', -3'51', -6468, 6468', -646'8', and 
 64W. Answ. 4*47663'456I8'. 
 
 6. Add together -8', *8'7', and -&7&. Answ. 2*6'44553'. 
 
 7. 8, and 9. Required the sum, true to five places, of the num- 
 bers given in exercises 5th, 6th, and 13th of Addition of Frac- 
 tions, the several fractions being previously reduced to decimals. 
 Answ, 6-0078125, 3'907'14285', and 1*56101 1'90476'. 
 
 SUBTRACTION OF DECIMAL FRACTIONS. 
 
 RULE, (1.) Set the less number so that each figure in it 
 may stand below a figure of the same local value in the 
 greater. (2.) Then find the .difference as in Simple Subtrac- 
 tion, and place the separating point as in Addition of De- 
 cimals. 
 
 Exam. 1. From S'5'4 7 take 1-34265. 
 
 3-546 
 
 Here, the greater number is ex- . 1-34265 
 
 tended, and the remainder is found 
 to be 2-202804'5'. 2-202804 / 5 / , diff. 
 
 Exam. 2. Required the difference of 8'6 and 2-7'. 
 
 Here, the, less number is carried to From 8-6 
 
 four places, that the true answer may Take 2*7777 
 
 be discovered with greatgr certainty. 
 In the subtraction, ciph&s are con- Rem. 5-8222 
 
 ceived to be annexed to* the greater 
 
 number, and 1 is carried to the repeating figure first used, be- 
 cause this must have been done, had the less number been carried 
 one place farther. The answer is found to be 5'S^. 
 
MULTIPLICATION OF DECIMAL FRACTIONS. 117 
 
 Exercises. Answers. 
 
 1. 3-4681-2591 = 2*2089 
 
 2. 6-45 l-WSr = 5-104'5' 
 
 3. 56-4297 29-68534= 26--7443G 
 
 4. 34-52810-63475. =23-89325 
 
 5. -682 -09647 ='58553 
 
 6. 13-6' 4-345 =9'3216 / 
 
 Exercises. Answers* 
 
 7. 5-83 4-l'7'.... =l-658 / 2' 
 
 8. 17-4' -4/8' .. .. = 16-9/5' 
 
 9. 3-3'42' l-7'5 y . = 1'5'84766/ 
 
 10. 17| 7f* =9-94642857 
 
 11. 7V* ff 4e 7 -5 =2-9617521 
 
 12. 15 T V 13 ? V.. =1-8188235 
 
 MULTIPLICATION OF DECIMAL FRACTIONS*: 
 
 RULE I. Multiply the factors as in Simple Multiplication, 
 
 and point off in the product as many places of decimals as 
 
 there are in both factors, supplying the deficiency, when 
 any occurs, by' prefixing ciphers. 
 Exam. 1. Multiply -582 by 66-3. 
 
 Here, because there are three places '582 
 
 of decimals in the one factor, and one 66*3 
 in the other, there must be four places 
 
 of decimals in the product. 1746 
 
 The reason of the rule will be under- 3492 
 
 stood from considering, that when the 3492 
 
 denominators are supplied, the first fac- 
 
 tor becomes $&&> an d the second 66 T 3 ff , 38*5866, product, 
 or 0% which, by the rule for the mul- 
 
 tQQ vj* fV'Q 
 
 tiplication of frdHbns, page 102, give for product 
 
 10000 
 
 whence it appears, that the product of 582 and 663 must be di- 
 vided by 10000, 'which is effected by cutting off four figures. It 
 is evident, that the divisor must contain as many ciphers as there 
 are in both denominators; that is, as there are decimal figures in 
 both factors. 
 
 Exam. 2. Multiply '13, by -7. -13 
 
 Here a cipher must be prefixed to the pro- *7 
 
 duct 91, ad there are two places of decimals 
 in the one factor, and one in the other. 
 
 Exercises. Answers. 
 
 1. -78 X '42 =-3276 
 
 2. 7-8X4-2 =32-76 
 
 3. 7-49X63-1 =472-619 
 
 Exercises. 
 4. -144X*144 
 
 091 
 
 Answers. 
 .. =-02073 
 
 5. 1-05 X 1-05 X 1'05= 1-15762 
 
 6. 36-48 X '475 =17-328 
 
 Exercises. Answers. 
 
 7. -IX'IX'1 X'lX'l =-00001 
 
 8. 13-825X5-128 =70*8946 
 
 9. -08 X "036 =-00288 
 
 * In this exercise and the two next, the given tractions are to be reduced to decimals, 
 and the difference taken according to the rule. 
 
118 MULTIPLICATION OF DECIMAL FRACTIONS. 
 
 Exercises. Answers. 
 
 10. -31X*32 = '0992 
 
 11. 3-18X41-7 =132-606 
 
 12. 62-38X7 =436-66 
 
 When the number of decimal figures is great, or the factors numer- 
 ous, the decimal figures resulting from the application of the preceding 
 rule, are, in many cases, unnecessarily and inconveniently numerous. 
 The following approximative rule will be found extremely useful in 
 such cases. 
 
 RULE II. (1.) Count off, after the separating point in the 
 multiplicand, (annexing ciphers, if requisite,) as many figures 
 of decimals as it is necessary to have in the product. (2.) 
 Below the last of these, write the unit figure of the multi- 
 plier, and write its other figures in reversed order. (3.) 
 Then multiply by each figure of the multiplier thus inverted, 
 neglecting all the figures of the multiplicand to the right of 
 that figure, except to find what is to be carried ; and let all 
 the partial products be so arranged, that their right hand 
 figures may stand in the same column. (4-.) Lastly, from 
 the sum of these partial products, cut the assigned num- 
 ber of decimal places. 
 
 In carrying from the rejected figures, we should always 
 take what is nearest the truth, whether it be too great or 
 too small. 
 
 Exam. 3. Multiply 7-24651 by 81-4632, so that there maybe 
 only three places of decimals in the product. '% 
 
 Here 1, the unit figure of the multiplier, is written below 6, 
 the third decimal figure of the multiplicand; 8, the figure which 
 precedes 1, is written after itj 4, the figure which follows it, is 
 set before it, &c. We then sav, 8 times 
 5=40, and 1, carried for 8 times 1,=41: 7-24651 
 
 1 is then set down, and 4 carried, and the 2364' 18 
 
 rest of the multiplication by 8 proceeds in 
 the usual way. Then, in multiplying 7-246 
 by 1, we add 1 to the product for 51, be- 
 cause 51 is nearer 100 than 0, and there- 
 fore it is nearer the truth to carry 1 than 
 0. In multiplying 7-24 by 4, three is car- 
 ried for the product that would have' re- 
 sulted for the rejected figures : for going 
 two places back, we have 4 times 5rz20; 590-325, product. 
 
 4 times 6=24, and 2z=26, which being 
 
 nearer 30 than 20, we carry 3. For a similar reason, in multiply- 
 ing 7*2 by 6, we carry 3 from the rejected figures; and thus we 
 proceed in similar cases. In finding what is to be carried for the 
 rejected figures, it is generally sufficient to go one figure back : 
 but in .doubtful eases it is better to go farther. 
 
MULTIPLICATION OF DECIMAL FRACTIONS. 119 
 
 The reason of the preceding operation will 7-246,3 I 
 
 be seen from the adjoined work at full length, 81-4(iti2 
 
 in which a vertical line is drawn, cutting off 
 
 ihe part rejected in the abbreviated process. 11449302 
 
 The result in this way is 590-323, or rather 21 173953 
 
 590-324, on account of the following figures, 434J7906 
 
 and is less by a thousandth part of a unit, 2898 604 
 
 than the result before obtained. The rea- 724651 
 
 son of this difference is, that all the partial 579720 8 
 products in the contracted mode, except the 
 
 last, happen to be rather too great. If, as 590-323 893432 
 
 in the preceding example, the results which 
 are too great be marked by the sign , and those that are too 
 small by -f- > i fc mav enable us in some degree to judge of the ac- 
 curacy of the result, as we may suppose it to be nearly correct, 
 if the -number of the signs of each kind be nearly the same, since 
 the excesses and the defects will then probably balance each other, 
 
 Exam. 4. Multiply -681472 by 
 
 01286, so that the decimal in the '681472 
 
 -product may contain five figures. 68210-0 
 
 In this example, since the multipli- 
 er contains no integer, a cipher is 
 placed below the fifth figure of the 
 multiplicand; and then, the multiplier 
 being written in reversed order, the 
 work proceeds as in the last example. *00876, product. 
 
 Exam. 5. Multiply 7-94 / by 3*69, 7-9444' 
 
 so that there may be four places of 96-3 
 
 decimals in the product. 
 
 Here the multiplicand is carried out 238333 
 
 to four places, and by a process simi- 47666 
 
 lar to those which precede, the answer 7150 
 
 is found to be 29-315, which is quite 
 
 correct. 29-3150, product. 
 
 Exercises. Answers. 
 
 13. 1-123674 X M23674 to 6 places =1-262643 
 
 14. 7-2 / 85714 / X 36-74405 to 6 places =267-706650 
 
 15. 24-6'3' X '2347 to 6 places =5'78215 / 4' 
 
 16. -863541 X -10983 to 5 places =-09484 
 
 17. -1347866 X '288793 to 7 places =-0389254 
 
 18. -26736 X '28758 to 4 places ='0769 
 
 19. 2-65641 9 X l'7 / 23 / to 6 places =4*578932 
 
 20. l-e'o'X 1'^S 7 to 5 places =2-45975 
 
 21. -053437 X '047 126 to 6 places =-002521 
 
 22, 23, 24, 25. Required the product, true to six places of de- 
 cimals, of the numbers fciven in exercises 4, 6, 14, and 17 in 
 
120 DIVISION OF DECIMAL FRACTIONS. 
 
 Multiplication of Fractions; the several fractions being previously 
 reduced to decimals. Answ. -113445, -467480, 2-393162, and 
 150-289438. J 
 
 DIVISION OF DECIMAL FRACTIONS. 
 
 RULE. (1.) If the divisor and dividend do not contain 
 the same number of places of decimals, supply the defici- 
 ency by annexing ciphers, or in a periodical decimal, the 
 next figures of the period. (2.) Then rejecting the sepa- 
 rating points, divide as in whole numbers, and the quotient 
 will be a whole number, p.) If there be a remainder after 
 all the figures of the dividend have been employed, ciphers 
 or periodical figures may be annexed, till there be no re- 
 mainder, or till as many figures be found as maybe judged 
 necessary. The part of the quotient thus obtained will be 
 a decimal. 
 
 If after the rejection ofSthe separating points, the divi- 
 sor be greater than the dividend, the quotient will contain 
 no whole number, and the work will proceed according to 
 the rule for problem I. in Reduction of Decimals. 
 
 When the divisor is large, the work will be shortened, if, 
 instead of annexing a cipher or periodical figure to each re- 
 mainder, a figure be cut off from the divisor. In this case, 
 each product is to be increased by carrying from the pro- 
 duct of the figure last cut off, and of the figure last placed 
 in the quotient. 
 
 It may facilitate the use of this contraction, if after the rejection of 
 the separating points, so many figures be annexed to the divisor and di- 
 vidend, or taken from them, that the divisor may contain one, or for 
 greater accuracy, two figures more than are required to be in the quo- 
 tient. Other directions might be given ; but the following examples 
 and illustrations will perhaps be preferable. 
 
 Exam. 1. Divide 1346-5 by 43-68. 
 
 Here, by annexing a cipher to the dividend, and rejecting the 
 points, we have for divisor 4368, and for dividend 1 34650. Hence, 
 dividing in the common way, we find 30 for the integral part, and 
 annexing ciphers to the remainders, and continuing the operation, 
 we get -826465, c.; and the entire answer is 30-826465, &c. The 
 work is left for the learner to perform. 
 
 With respect to the reason of the operation, the value of 1346*5 
 is not changed by the annexing of a cipher; and the removal of 
 the points merely multiplies each of the given numbers by 100. 
 
DIVISION OF DECIMAL FRACTIONS. 121 
 
 (See page 107. J It is evident therefore, that the value of the 
 quotient will not be affected, sinee, while the dividend is multi- 
 plied by 100, the divisor is increased in the .same ratio. We might 
 also consider the dividend as the numerator, and the divisor as the 
 denominator of a fraction ; and then the reason of the process 
 would depend on the first of the principles established in page 88. 
 The reason of removing the points, is to make the dividend and 
 divisor whole numbers, and thus to render the operation as much 
 as possible the same as in Simple Division. 
 
 Exam. 2. Divide -1342 by 67-1. 
 
 Here, by annexing three ciphers to the divisor and rejecting 
 the points, we get for divisor 671000, and for dividend 1342. 
 Then the dividend being the greater, the quotient will contain no 
 integral part; and the annexing of a cipher to the dividend, gives 
 one cipher for the quotient: the annexing of a second gives ano- 
 ther cipher; but the addition of a third gives 2. Hence the quo- 
 tient is -002. 
 
 When the number of places of decimals In the divisor is not greater 
 than in the dividend, the number of figures of decimals in the quotient 
 is to be equal to the difference between the number of places in the 
 divisor and dividend, as is evident from Multiplication of Decimals; 
 and in this way the number of decimal figures in the quotient is often 
 very easily determined. 
 
 Exam, 3. Divide 2-3748 by 1-473G, so that the quotient may 
 contain three places of decimals. 
 
 In this example, the 14736)23748(1-61 1 1 4736)23743(1-0 1 1 
 numbers being prepar- " 14736 14736 
 
 ed according to the 
 
 rule, and $xi first fi- 9012 90I2|0 
 
 gure of the quotient 8842 8841 6 
 
 being found, instead of 
 
 adding a cipher to the 170 170f40 
 
 remainder 9012, we 147 14736 
 
 omit the last figure of ! 
 
 the divisor, to denote 23 23|04Q 
 
 which, a point is plac- 15 141736 
 
 ed below it. Then 6 
 
 being put in the quo- ' 8 b 304 
 
 tient, we multiply 6, 
 
 the figure cut off, by it, and without setting any thing down, we 
 carry 4, because the product 3G is nearer 40 than 30. After that, 
 3 is cut off in like manner, and then 7. The quotient is found to 
 be 1-611, or more nearly 1-612, because the remainder 8 is rather 
 more than the half of 14. The annexed operation at full 
 will explain the reason of the contracted process, the vert: 
 cutting off the rejected part. 
 
 G 
 
122 
 
 DIVISION OF DECIMAL FRACTIONS, 
 
 Exam. 4. Divide 73*64 by '43^, so that the quotient may 
 have four places of decimals. 
 
 43232323)7364000000(1 70-33oo 
 43232323 
 
 Here it is easy to see, that 
 the quotient will contain three 
 places of whole numbers. 
 (This would be seen by divid- 
 ing 736 by 4.) Hence the 
 quotient must contain seven 
 figures. Extending therefore 
 the divisor to eight places, and 
 the dividend to the same num- 
 ber of places of decimals, the 
 process will stand as in the 
 margin. In the work, instead 
 of bringing down the two last 
 ciphers of the dividend, two 
 figures may be cut in succes- 
 sion from the divisor, and the 
 rest of the operation will pro- 
 ceed as before. 
 
 30407677 
 30262626 
 
 145051 
 129697 
 
 15354 
 12970 
 
 2384 
 2162 
 
 222 
 216 
 
 6 
 
 The pupil should work the following exercises by both methods, par- 
 ticularly by the abbreviated process : 
 
 Exercises. Answers. 
 
 1. 47-58 -i-26- 175 =1-8 1776504 
 
 2. -341 2 -j-8-4736= -040266239 
 
 3. 468-7' -r-3-365 ..= 139'309889 
 
 4. -o'8'-;-77-482... = -00756 122 
 
 5. 75-347 -K3829 = 1 96-779838 
 
 6. 6-5' -=-7-06249..:= -928222 
 
 Exercises. Answers. 
 
 7. 1-f- 10-473654..= -09547766 
 
 8. 7-5-J-37-38 =-20004-205 
 
 9. 5-09'-j-6'2' = -8T 
 
 10. 
 
 11. 
 
 12. 2-r62'-f-3-125..= -69 18' 
 
 Exercises. Answers. 
 
 13. -079085-^-83497 =-094716 
 
 14. -6'l'-f-I3-543516 =-04549495 
 
 15. 23-6-f--037538 =628-69625-:. 3 
 
 16. 7-1 2649 1H--531 , =13-420887005 
 
 17. -879454-T--897 =-98043924 
 
 18. 52-7'3'-r-52-734567 =1-000053224 
 
 19. 2-3 / 70 / -i-4-9 / 23076 / =-4 / 81 / 
 
 After the full illustration of the multiplication and division of deci- 
 mals, which has been given in the preceding pages, it appears unneces- 
 sary to give their application in the Rule of Proportion ; as in thus ap- 
 plying them the pupil can encounter no difficulty, the terms being arrang- 
 ed in the manner already explained, and the product of the second aiul 
 third terms, in like manner, divided by the first. Should it be thought 
 necessary however, the exercises in page 106 may be wrought in this 
 manner, by reducing the common fractions which they contain, to de- 
 c-imal fraciioiis. 
 
123 
 
 PRACTICE. 
 
 An ALIQUOT PART of a number is such a part as, when 
 taken a certain number of times, will exactly make that 
 number. Thus, 5 is an aliquot part of 20, S of 12, &c. 
 
 What is generally called PRACTICE is only an abridged 
 method of performing operations in the Rule of Proportion 
 by the use of aliquot parts; and is generally employed in 
 calculating the prices of commodities. 
 
 TABLES OF ALIQUOT PARTS. 
 
 10/0 
 
 1 
 
 2/0 . .. JTS 
 
 4d .... J of 1/0 
 
 6/8 .. 
 
 ___? 
 
 1/8 .... J* 
 
 3d .... \ of 1/0 
 
 5/0 
 
 
 
 
 2d .... = i of 1/0 
 
 4/0 
 
 -4- 
 
 1/0 I 
 
 l|d . i of 1/0 
 
 3/4 
 
 2/6 .., 
 
 
 1/0 .... = & 
 6d -^- of 1/0 
 
 Id .... = T V of 1/0 
 
 Jd ...=!* of 1/0 
 
 2 qrs 
 1 qr 
 16 IDS 
 
 = \ cwt. 
 
 \ CWt. 
 
 =| cwt. 
 
 COMMON WEIGHT. 
 
 14 Ibs = i cwt. 
 
 8 Ibs = T ycwt. 
 
 7 Ibs z^iV cwt. 
 
 4 Ibs.. r:.jV cwt. 
 14 Ibs.. = of aqr. 
 7 Ibs.. = of aqr. 
 
 LONG WEIGHT. 
 
 2 qrs =^ of a cwt. 
 
 1 qr rz^ of a cwt. 
 
 20 IDS =1 of a cwt. 
 
 15 tbs rr | of a cwt 
 
 12 Ibs =-^ s of a cwt. 
 
 10 Ibs =^5 of a cwt. 
 
 2 roods 
 
 Irood 
 
 = | acre 
 rr^acre 
 
 LAND MEASURE. 
 
 20 perches | acre 
 16 perches^ acre 
 
 10 perches ! rood 
 8 perches^: rood 
 
 These tables may be constructed by problem VII. page 97 ; or rather 
 by dividing l, 1 acre, &c. by 2, 3, 4, &c. and selecting such of the 
 quotients as are free from fractions. The following continuation of 
 them may often be found useful. By its means the pupil may frequently 
 be assisted in discoveting what aliquot parts may be most advantage- 
 ously employed in many of the operations which he may be required to 
 perform. The more obvious and less useful parts are omitted. 
 
 2/6 : 
 
 1/8 : 
 
 MONEY. 
 
 ( $ of 10/0 
 tJf 5,0 
 
 Uof JO/0 
 of 5/0 
 
 1/4...= 
 1/3...= 
 
 i of 4/0 
 C | of 10/0 
 ] i of 5/0 
 C i of 2/6 
 
 
 (1 .> 
 
 of 100 
 
 10d....= 
 
124 
 10d....= 
 
 8d.. .= 
 
 PRACTICE. 
 
 % of 2/6 
 iof 1/8 
 of 4/0 
 i of 3/4 
 J of 2/0 
 
 i of 5 / 
 i of 2/6 
 
 . , 
 5d "" 
 
 i T Vof 5/0 
 i of 3/4 
 i of 2/6 
 i of 1/8 
 T V of 4/0 
 .A of 3/4 
 . i of 2/0 
 iV of 2/6 
 i of 2/0 
 
 COMMON WEIGHT. 
 
 ld....= 
 
 TV Of 2/6 
 
 J of loci 
 ' T Vof 1/3 
 
 i of i/ 
 % of 6d. 
 I- of 6d. 
 1 of 3d. 
 
 14 
 
 7 lbs = 
 
 of 2 qrs. 
 i of 2 qrs. 
 4^ of 16 lbs.* 
 I of 2 qrs. 
 i of 14 lbs. 
 
 o lu 
 
 2lbs ..... = 
 
 of 1 qr. 
 of 16 U*. 
 
 of 16 lbs - 
 of Bibs. 
 
 In the calculation of prices, the quantity of the commodity may be 
 of one denomination, or of more than one : and accordingly, the subject 
 divides itself into two branches, with several varieties, as will uppear 
 from the following rules and illustrations. . 
 
 RULE I. In finding the price of a commodity, when the 
 price of each article^ as well as the quantity, is of one deno- 
 mination, the product of the given price and of the number 
 of articles, will be the price required. 
 
 Exam. 1. Required the price of 289 .cwt. of beef at 2 ^ cwt. 
 
 Here the price of 
 cwt. at \ ty 
 c\vt. is evidently 
 289; and the price 
 of the same at 2 
 & xr.vt. must obvi- 
 ously be twice that 
 amount. 
 
 Exercises. 
 
 289 cwt. at 2 W cwt. 
 
 289...=price of 289 cwt. at \ 
 2 
 
 cwt. 
 
 578. ..=~ 
 
 .2 
 
 Answ. 
 
 1. 311 at 3 933 
 
 2. 1286 at 5 ... 6*30 
 
 Exercises. Answ. 
 
 3. 197 at 4- 788 
 
 4. 309 at 1 309 
 
 RULE II. When the price is an aliquot part of a higher 
 denomination, take a like part of the number of articles, 
 and the result will be the price in the higher denomination. 
 
 Exam. 2. What cost 532 lbs. of tea at 6/8 F tb. ? 
 532 tbs. at 6/8 W tb. 
 
 532 rrpriceiof 532 tbs. at 1 each 
 
 6/8=: J of XI... 177 6 8=^L 6/8 
 
PRACTICE. 
 
 IP this example, since 532 articles, at 1 each, would cost 
 532, it is evident, that at 6/8, the same number of articles would 
 cost one-third of that amount, 6/8 being one-third of 1. We 
 therefore divide 532 by 3, and the quotient, 177 6 8, is the 
 required price. 
 
 Exam. 3. Required the price of 537 yards of ribbon at 3d. 
 ^ yard. 
 
 537 yards at 3d. W yard. 
 
 537 s =price of 537 at 1/0 # yard. 
 
 3d.= of 1/0 134/3 =__ at 3d. 
 
 6 14 3, Answ. 
 
 Here, the price of 537 yards at 1 shilling would be 537 shil- 
 lings; and the price of the same at 3d. would evidently be one 
 fourth of that, or 134/3, which by reduction becomes 6 14 3. 
 
 Exerdses. Amiv. 
 
 7. 739 at 3/4 123 3 4 
 
 8. 645 at 2/6 ....... 80 12 6 
 
 Exercises. Answ. 
 
 5. 389 at 10/0 194 10 
 
 6. 538 at 5/ 134 10 
 
 RULE III. When the price of each articled not an aliquot 
 of ajtigher denomination, it is to be divided into such parts, 
 that the price of the whole quantity at each of these prices, 
 may be found by the first or second rule ; and the sum of 
 prices thus obtained will be the whole price required. 
 
 Exam. 4. Required the price of 479 cwt. of sugar at 4 9 6 
 cwt. 
 
 479 cwt. at 4 9 6 V cwt. 
 
 479 =price of 479 cwt. at? 1 ^ cwt. 
 
 4 
 
 5/0= i 
 4/0=i 
 
 of 1 
 of 1 
 5 of 5/0 
 
 Answer. 
 
 1916. 
 
 95 
 11 
 
 15 
 
 16 
 19 
 
 
 
 A 
 
 
 
 
 
 
 
 6=ZZZZ^ 
 
 x^-^^-u-a 
 
 
 
 2143 
 
 10 
 
 6, 
 
 4 9 
 
 In this example, at 1 ^ cwt. 479 cwt. would cost 479, and 
 hence, to find the price at 4 ^ cwt. we multiply by 4. Again, 
 since 5/=r one-fourth of a pound, and 4/zr one-fifth of a pound, 
 the prices at these rates will be found by rule II. by taking the 
 fourth and the fifth of 479. It still remains to find the price at 
 6d.: now, since 6d. is one-tenth of 5/, the price at Cd. will evi- 
 dently be one-tenth of the price at 5/, and therefore we take on-3- 
 
izo PRACTICE. 
 
 tenth of 119 15 the price at 5/, and the result is 1! 19 0, thf 
 price of 479 cwt. at 6d. ty cwt. Finally, the price of the whole 
 quantity at 4 V cwt. being 1916; at 5/, 119 15; at 4/, 95 
 16; and at 6d. 11 19 6; the price at 4 9 6 will be 2143 10 
 6, the sum of these. 
 
 Exam. 5. Required the price of 647 yards of linen at 3/9 V yd 
 647 yards at 3/9 V yard. 
 
 647 rrpriceof 647 yards at 1 V yard. 
 
 2/6= of 1 I 80 17 6= 2/6 
 
 1/3= I of 2/6 I 40 8 9= 1/3 
 
 Answer, 121 6 3rr 3/9 
 
 In this example, the price at 1 IT yard is obviously 647, and 
 as 2/6 is one-eighth of a pound, the price at 2/6 will be one-eighth 
 of 647, or 80 17 6. Now, 2/6 with i/3 make up the given 
 price 3/9; and 1/3 being one-half of 2/6, the price at 1/3 will be 
 one-half of the price at 2/6, and hence we take the half of 8C 
 17 6. Then the price at 2/6 being 80 17 6, and that at 1/3 
 being 40 8 9, the price at 2/6 and 1/3, or at 3/9, will be 121 6 
 3, the sum of these. This exercise might be wrought with nearly 
 equal facility by finding the prices at 3/4 and 5d. and taking their sum. 
 
 Exam. 6. Required the price of 247 ewt. of flour at 1 5 ^ c. 
 247 cwt. at 1 5 ty cwt. 
 
 247 =price at 1 V cwt. 
 
 5/0=|ofl... 61 15= 5/ 
 
 Answer, 308 15 = 1 5 
 
 This and the preceding examples have been wrought at full 
 length for showing the reason of the operations, and thus the 
 pupil should be accustomed 
 
 to work similar questions, 247 cwt. at 1 5 V cwt. 
 
 till he thoroughly understand 5/=^... 61 15 
 
 the reason of the process. 
 
 Afterwards, however, the 308 15, Answer. 
 
 work may be properly con- 
 tracted as in the margin, which is the mode commonly employed. 
 Exam. 7. Required the price of 195 Ibs. of raisins at 1/3 V lb. 
 195lbs. at 1/3 V lb. 
 
 ) 4)195 =price at 1 lb. 
 
 1/3=^ of !...$ 
 
 ) 4)48 15 
 
 Answer, 12 3 9= 1/3, 
 
PRACTICE. 127 
 
 Or thus, 195fos. at 1/3 perJb. 
 
 195s.. ..= price at 1/0 perlb. 
 3d.= ^ of 1/O..J48 9 = 0/3 
 
 2lO)24|3 9 = 1/3 
 
 ,12 3 9, Answer, as before. 
 
 In the first of the preceding operations, since 1/3 is one-six 
 teenth of a pound, we divide 195, the price at a pound sterling 
 per pound, by 4, and 
 
 the result again by 4. 195lbs. at 1/3 perlb. 
 
 In the second method, 3d.=: of 1/0. ..48 9 
 
 the price at one shil- 
 
 ling being 195 shillings, 2|0)24|3 9 
 
 and the priceat 3d. one- 
 fourth of that amount, 12 3 9, Answer. 
 or 48/9, the sum of 
 
 both is 243/9, the price required, which by reduction becomes 
 12 3 9, the same as before. This operation may also stand as 
 in the margin. 
 
 Example 8. What cost 1257 yards of ribbon at Gfd. per 
 yard? 
 
 1257 yards at 6|d. per yard. 
 
 6d.= of 1/0 I 628 6 =r price at 6d. per yard. 
 fd.= of6d. I 78 6f= fd 
 
 2|0)70|7 Of- 6|d 
 
 35 7 Of, Answer. 
 
 Exam. 9. Required the price of 347 cwt. of coffee at 7 11 6$ 
 per cwt. 
 
 347 cwt. at 7 11 6| per cwt. 
 
 7 
 
 10/=i 
 J/3=i 
 3/3=i 
 Jd.=i 
 
 ofl 
 of 10/ 
 of 1/3 
 of 0/3 
 
 f^ffX&9 
 
 173 
 21 
 4 
 
 1 
 
 10 
 13 
 6 
 
 
 9 
 9 
 
 8 
 
 = 
 
 
 
 
 
 
 10 
 
 
 
 
 r 
 
 o 
 
 3 
 3 
 0| 
 
 2629 12 2= ^7 11 Cf 
 
PRACTICE. 
 Or thus, 347 cwt. at 1 11 6f per cwt. 
 
 10/= of 1 
 
 i/ = ^of io/ 
 
 0/6= of 1/0 
 d.= of 6d. 
 
 = price at 7 per cwt. 
 
 173 10 = 10 
 
 17 7 ^.-^.y...... 010, 
 
 8 13 6 = 006, 
 
 1 1 8|= Of, 
 
 Answ. 262912 2= 711 6| 
 
 Exam. 10. If a tradesman have 3/9 per day, how much is 
 his yearly salary, the days of labour in the year being 313? 
 313 days at 3/9 per day. 
 
 3/4= i ofl 
 3d. = of 3/4. 
 
 52 3 4 = amount at 3/4 per da} 7 . 
 6 10 5 = 0/5 , 
 
 13 0= ^, 0/0| 
 
 Answ. 59 6 9*= 3/9^ " 
 
 This exercise might have been wrought by distributing 3/9^ into the 
 parts, 2/6, 1/3, and *d. ; but those employed above are much piefer- 
 able, as, in the other case, *d. being one-thirtieth of 1/3, we must have 
 divided by a number inconveniently large. Unless, indeed, in particu- 
 lar 'circumstances, we should, "if possible, avoid taking parts that would 
 L require us to divide by any number greater than 12. The seventh ex- 
 ample affords an instance in which no inconvenience or difficulty arises 
 from employing a larger divisor; and every person's experience will 
 point out others. 
 
 In taking aliquot parts, it sometimes shortens the work to take the 
 Kime part twice, as the result of the first operation may be. 1 copied with- 
 out working for it again. Thus, 18/6 may be distributed' into 10/, 4/, 
 4/, and 6d. Sometimes also the price at a small rate may be found, and 
 from it the price of a greater may be obtained by multiplication. Thus, 
 16/4 may be divided into 2/, 14/, and 4d. the price at H/ being 7 times 
 the price at 2/. In like manner, for, 17/1 we may take 1/8, 15/, and 5d. 
 Other remarks on this subject will be found at the end of this article, 
 
 Exercises. Answers. 
 
 
 
 
 
 J. 
 
 d. 
 
 
 
 s. 
 
 d. 
 
 9. 
 
 1625 
 
 at 
 
 2 
 
 8^ , 
 
 220 
 
 I 
 
 0. 
 
 10. 
 
 1429 
 
 I 
 
 2 
 
 9 
 
 1625 
 
 9 
 
 9 
 
 11. 
 
 1973 
 
 
 
 6 
 
 10 
 
 674 
 
 2 
 
 2 
 
 12. 
 
 749 
 
 
 
 5 
 
 8 
 
 212 
 
 4 
 
 4 
 
 13. 
 
 1689 
 
 
 
 4 
 
 10* 
 
 411 
 
 13 
 
 10 
 
 14. 
 
 2476 
 
 
 
 18 
 
 6 
 
 , 
 
 6 
 
 
 
 15. 
 
 5926 
 
 
 
 11 
 
 8 
 
 34of> 
 
 16 
 
 8 
 
 16. 
 
 313 
 
 
 
 a 
 
 8 
 
 135 
 
 12 
 
 B 
 
 17. 
 
 5934. 
 
 1 
 
 5 
 
 10 
 
 7664 
 
 15 
 
 
 
 18, 
 
 3576 
 
 
 
 11 
 
 4^ 
 
 2033 
 
 17 
 
 
 
 19. 
 
 958 
 
 1 
 
 18 
 
 8 , 
 
 18.52 
 
 2 
 
 8 
 
% 
 m 
 
 PRACTICE. 129 
 
 Exercises. Answers. 
 
 i. d. t. d. 
 
 1898 at 6 10$ ........................... 632 8 9 
 
 21. 1594 13 6 ........................... 1075 19 
 
 22. 695 14- 10 ........................... 515 9 2 
 
 23. 2386 1 6 6 ........... . ............... 3161 9 
 
 2k 725 178 ........................... 1002 18 4 
 
 25. - 589 1 11 6 ........................... 927 13 G 
 
 26. 286 12 1 ........................... 172 15 10 
 
 27. 7649 5f ........................... 183 5 If 
 
 28. 5728 1\ ............. ~ ............ 173 8 
 
 29. 6491 lOf ......... . ................. 290 14 10| 
 
 30. 991 1 3j ........................... 64 
 
 31. 436 ' 4 17 8 ........................... 2129 2 s" / 
 
 32. 3725 5 8 ...... . .................... 1059 5 \\\ // 
 
 33. 1677 5 1 ........................... 426 4 9 
 
 34. 7913 2 16 10 ........................... 22502 11 I0j 
 
 35. 4265 1 14 \\ ..... ...................... 7277 3 Ij 
 
 36. 249 5 13 9 ...... . .................... 1416 3 9 - 
 
 37. 576 18 11| .......... . ................ 546 U 
 
 38. 6485 2 10 2| ..... . ..................... 16280 1 Ok 
 
 RULE IV. The price of any number of articles at 2 shU~ 
 tings each, is found by doubling the last figure of the num- 
 ber for shillings, and taking the number expressed by the 
 preceding figures as pounds. 
 
 Exam. 11. What cost 647 yards of muslin, at 2 shillings F 
 yard? 
 
 The reason of this rule is, that 2 shillings 647 yards at 2/ 
 are one-tenth of a pound, and the work is - 
 no more than a contracted division by ten. 64 14 0, Answ. 
 Thus, in the annexed example, by dividing 
 
 by 10, we should have .64, with the fraction fa or y doub- 
 ling both the terms, %$, or 14 shillings, 
 
 RuLBy. If the rate be an even number of shillings, mul- 
 tiply by half that number, and in multiplying, double the 
 last figure of the product for shilling^ ; the rest will be pound?. 
 
 Exam. 12. Required the price of 273 fts. of indigo. 
 shillings ^ Ib. 
 
 This rule and operation are evi- 273 Jbs. at 8/ v 
 
 dently nothing more than an ex- 4 
 
 tension of the last. In the annex- - 
 ed example, the price is ^ or T 4 ^ <10& 4 
 of a pound; and hence \*;e multi- * 
 
 ply by 4, and the doubling of the last figure 
 shillings, is equivalent to a division by 10. 
 
 ^ 3 
 
787 Ifes. at 17/ ^ ib. 
 8 
 
 J30 PRACTICE. 
 
 RULE VI. When the number of shillings is odd, we may 
 find, by the last rule, the amount at one shilling less than 
 the given rate, and for that shilling take aV of the price at 
 one pound. 
 
 Exam. 13. What cost 787 Ibs. of nutmegs, at 17 shillings 
 
 pound? 
 
 In this example, the price 
 is first found at 16/ W pound, 
 by multiplying by 8, and 
 doubling the last figure of the 
 product for shillings; then, 
 for the remaining shilling, a 
 twentieth part of 787, the 668 
 
 price at 1 per lb. is taken; 
 and the sum of both results is 668 19, the price at 17 shillings. 
 
 Exam. 14. Required the amount of 233 cwt. of pearl ashes, 
 at 3 9 per cwt. 
 
 233 cwt. at 3 9 per cwt. 
 3 4 
 
 699 
 93 
 
 =price at 3 per lb. 
 4 _ 8/ ...-.....-. 
 3= - I/ - 
 
 Answer, 803 17=. 
 
 ,3 9 
 
 In this example, we 
 first find the price at 3 
 by multiplying by 3; 
 then the price at 8/ by 
 multiplying by 4, and 
 doubling the last figure 
 of the product ..for shil- 
 lings ; and -lastly, the 
 price a? I/, by taking a 
 twentieth part of 233, the price at 1 per cwt. It. is evident, 
 that we might have taken one eighth of the price at eight shillings, 
 for the price at one shilling. 
 
 Answers. 
 
 s. d. 
 
 13/ 438 15 
 
 14/ 358 8 
 
 1G/ .255 4 
 
 17/ 737 16 
 
 18/ 423 
 
 3 4...S145 12 
 
 5 13..4090 12 
 
 RULE VII. The price may often be determined very 
 easily, byjinding the amount at a rate higher than the given 
 rate, a?td deducting from the amount the price at the difference 
 beiiueen the given and the assumed rates. 
 
 'This method is generally of little use, unless the difference between 
 the given and the higher rate be an aliquot part of the higher. This 
 ditkrence may be called the COMPLEMENT of the given rate. 
 
 Exercises. 
 39 397 at 2/ 
 
 Answer 
 s. 
 39 14 
 62 14 
 228 18 
 132 6 
 336 8 
 421 13 
 569 8 
 
 s. 
 d. 
 .0 
 
 
 
 
 
 
 
 Exercises. 
 
 46. 675 at 
 47. 512 
 48. 319 
 49. 868 
 50. 470 
 51. 983 
 52. 724 
 
 40 418 3/ ....... 
 
 41. 763 6/ 
 
 4.0 378 7/ . 
 
 43 841 8/ 
 
 44. 937 9/ 
 
 45. 949 12/ ... 
 
PRACTICE. 131 
 
 Exam. 15. What cost 189 tons of coals at 17/6 per ton? 
 
 189 tons at 17/6 per ton. 
 
 Or simply thus: 
 
 189 =price at \ per ton. 189 at 17,0 
 
 2/6=23 12 6= 2/6 2/6=.. .23 12 6 
 
 Aruw. 165 7 6= 17/6 166 76 
 
 In this example, since 17/6 is less than a pound by 2/6, which 
 is an aliquot part of a pound, the required price is found by :^k- 
 ing from 189, the price at 1 per ton, 23 12 6, the price at 
 2/6 per ton. 
 
 Exam. 16. What cost 257 feet of plank, at lOd. per foot ? 
 
 Here, from 257 shil- 257 feet at lOd. W foot, 
 
 lings, the price at I/ per 2d.= of I/... 42 10 
 
 foot, we take 42/10, the 
 
 price at 2d. per foot: 20)214 2 
 
 the remainder is 214/2, 
 
 or by reduction 10 14 10 14 2, Answer. 
 
 2, the price at lOd. per 
 
 foot. 
 
 Exam. 17. What cost 514 gallons of seal oil at 3/1 1 ty gallon? 
 514 gallons, at 3/11 V gallon. 
 
 2056/ =price at 4/ V gallon. 
 
 ld.= &of I/... 42 10...= Id 
 
 210)201 13 2...= 3/11 
 
 100 13 2, Answer. 
 
 Exam. 18. Required the price of 193 cwt. of Turkey fiee. at 
 3 18 r cwt. 
 
 lu this example, 193 cwt. at 3 18 V cwt. 
 
 the price at / V 4 
 
 cwt. (found by 
 
 doubling the last 772 =price at 4 V cwt. 
 
 figure, &c.) is taken 19 6...= 2 
 
 from the priffe at . 
 
 4. 752 14...= 3 18 
 
 Exercises. Answers. 
 
 s. d. f . d. 
 
 53. 358 at 13 4 238 13 4 
 
 54. 599 10$ 26 4 11 
 
 55. 967 008 32 4 8 
 
 56. 275 2 15 750 5 
 
132 PRACTICE. 
 
 Exercises. Answers. 
 
 s. d. s. d. 
 
 57. 361 at 3 12 **.... 1299 12 
 
 53. 889 009 33 6 9 
 
 5'0. 483 5 16 8 2817 10 
 
 oO. 189 047 43 6 3 
 
 til. 997 11 45 13 11 
 
 62. 753 019 65 17 9 
 
 63. 649 1 5. 47 6 5 
 
 64. 721 19 634 19 0" 
 
 We now proceed to problems of the second class ; and in resolving 
 them, we may employ either of the two following general rules. 
 
 RULE VIII. When the quantity is not expressed by a whole 
 number of one denomination, find the price of the integral 
 quantity according to the method already illustrated, and 
 then find the price of the fractional parts, or lower deno- 
 minations, from the given rate, by means of aliquoj parts, 
 or otherwise : the sum of all will be the whole price re- 
 quired. Or, 
 
 RULE IX. When the quantity is not expressed by a whole 
 number of one denomination, find the price of the entire 
 given quantity at \-\ for each unit of the integral part, 
 valuing the subordinate parts at the same rate ; and then 
 the work will proceed in the manner already explained, with- 
 out the necessity of farther work for the subordinate parts. 
 
 Exam. 19. Required the price of 79| yards of broadcloth, at 
 1 2 11 Vyard. 
 
 By Ride VIII. 
 79| yards, atl 2 11. 
 
 5 
 
 ^6rrJ 
 
 
 9 
 
 17 
 
 6 = price at 2/6 
 
 5< 
 
 1 = 1 
 
 of 2/6 
 
 1 
 
 12 
 
 11 = 5d. 
 
 * 
 
 of 1 2 11 
 
 
 
 11 
 
 5i^^ J.-^JJ-KJ of Jt vard 
 
 i 
 
 of 1 
 
 i AH 
 
 
 
 5 
 
 8|= ^1 
 
 
 
 91 
 
 7 
 
 7^, Answer. 
 
 * .> 12 is the difference between 4 and 8 shillings, and 8/=one-tenth of 4. --Ex. 
 ercie t52 will be wrought by finding the price at two shillings, and then proceeding in 
 the usu.il manner ; and the answers of exercises 60 and 63 may be derived from the 
 prices ;it 5/0 and 1/8. 
 
 f Thus rule is restricted to pounds, which is almost the only useful or necessary appli 
 cation of the principle. The principle however, is universal in its nature, and might be 
 extcutiod to any other rate in a similar manner. 
 
 W,th respect to the eighth and ninth rules, it may be observed, that the ninth (of 
 which rules X. XL XII. and XIII. are particular applications,) is very elegant, on ac 
 
PRACTICE. 133 
 
 In this mode of resolving this exercise, the price of 79 yards is 
 first found, (or rather the parts of which it is made up are found ;) 
 and then for yard tJife half of 1 2 1 1 is taken, and for ^ yard 
 the half of that is taken : then the sum of all these parts is 91 
 7 74, the result required. 
 
 By Ride IX. 
 79f yards at 1 2 11. 
 
 79 15 
 
 2/6= | I 9 19 
 
 5d. = of2/6 I 1 13 
 
 = price at \ 
 44 _ 2/6 
 
 gl ; JJJWJJWJX . 5 
 
 yard. 
 
 91 1 7=. 
 
 2 11. 
 
 In this method we first find the price of 79f yards at \ ty 
 yard. This is obviously 79 15; for the price of 79 yards is 79, 
 and the price of a quarter of a yard being evidently 5/0, the price 
 of | of a yard is 15/. Then, the price at 1 ^ yard being 79 
 15, the price at 2/6 ^ yard will be one-eighth of this, or 9 19 
 4 ; and the price at 5d". one-sixth of the price at 2/6, or 1 13 2. 
 The sum of these prices is 91 7 7^, the whole price, as before. 
 
 Exam. 20. What cost 69 yards of cambric, at 13/10 & yard ? 
 By Rule VIII. 
 
 69$ at 13/10 
 
 3/4= i of 10/ 
 
 .= yd. 
 = iof f yd. 
 
 34 
 
 11 
 
 1 
 
 
 
 
 
 
 
 6 
 
 11 
 
 8f 
 
 Here, after finding, in the 
 way already illustrated, the 
 parts whose sum will be the 
 price of 69 yards at 13/10 
 W yard, we take for f of a 
 yard the half of 13/10, the 
 price, of one yard, and for the 
 remaining eighth we take ^ 
 of 6/11, and the result is 1/8|, 
 the price of ^ of a yard. The 
 sum of nil these partial prices is 48 3 If, the whole price re- 
 quire .i. 
 
 By Ride IX. 
 69f yards at 13/10 V yard. 
 
 Answer, 48 3 If. 
 
 69 12 6 =price at 1 ^ yard. 
 
 10/= 
 3/4=4- of 10 
 
 48 3 lf=. 
 
 13/10 
 
 count of the principle on u hich it depends ; and in many cases it affosoa tne neatest ajKl 
 most concise mode of resolving problems. In some cases, however, it gives ori^ji) to 
 
1&4 PRACTICE. 
 
 In this method, since, at \ $*" yard, one-eighth of a yard would 
 cost 2/6, five-eighths would cost 5 times 2/6, or 12/6; and con- 
 sequently, at that rate, 69|- yards would cost .69 12 6. The rest 
 of the work proceeds in the usual manner. With respect to both 
 modes, it may be observed, that ou account of the large divisor, 
 20, the work would have perhaps been as short, had we taken 10/, 
 3/4, 5d. and Id. though it would appear to be a line longer. 
 
 Exercises. Answers. 
 
 65 
 
 328f at 
 
 
 
 
 6 
 
 <f. - 
 6 
 
 
 106 
 
 s. 
 16 
 
 d. 
 UU 
 
 66 
 
 675A 
 
 
 14 
 
 4 
 
 , 1835 
 
 9 
 
 A "2 
 
 67 
 
 7538 
 
 
 
 9 
 
 4 , 
 
 , 879 
 
 10 
 
 ^ 
 
 68 
 
 176c. 3q. 
 
 
 
 1H 
 
 8 
 
 164 
 
 19 
 
 4 
 
 
 164 
 
 g 
 
 5 
 
 6 
 
 375 
 
 1 
 
 
 70 
 
 239a. 3r. 
 
 
 
 10 
 
 10 
 
 129 
 
 17 
 
 rU 
 
 71 
 
 172 T 7 o 
 
 3 
 
 15 
 
 10 
 
 , 654 
 
 16 
 
 5 
 
 72. 
 
 25744 
 
 2 
 
 9 
 
 4.. 
 
 .. 636 
 
 3 
 
 10* 
 
 RULE X. In calculating the prices of hundreds, quarters, 
 and pounds, common weight, multiply the pounds by 2}-, and 
 consider the product as pence ; and multiply the quarters 
 by 5, and consider the product as shillings : the rest of the 
 operation will be performed by rule IX. 
 
 The reason of this rule is evident, since at 1 W cwt. each quar- 
 ter would evidently cost 5/, and each pound a twenty-eighth of 
 this, or 2} pence. 
 
 Exam. 21. Required the price of 198 c. 2qrs. 21 fbs. of cast 
 steel, at 4 8 8 V cwt. 
 
 c. q. Ids. / 
 
 198 2 21 at 4, 8 8 per cwt 
 
 792 =r price of 198 cwt. at 4 V cwt. 
 
 = 8/ 
 
 r-.^.^.-..-.-..-...-.-......-.....-.-.........,. 8d. 
 
 4 = price of 2 qrs. at 4 8 8 ^ cwt. 
 
 1 = i4ft>s 
 
 = a Ibs.. 
 
 880 16 11 = price r/quired. 
 
 fractions, which, when perfect accuracy is required, render theJfperation tedious in a 
 considerable degree. In such cases the eighth rule/is sometimra not so objectionable. 
 For mercantile p\trposes, however, as it is unnecessary to work for the precise fractions 
 the ninth rule, when managed as will be shown in the succeeding rules abovementioned, 
 will perhaps be found superior in all the principal and more difficult calculations. Let 
 the teacher and calculator judge, however, and employ whichever they may consider 
 preferable. 
 
c. q. Ibs. 
 
 198 2 21 at 4 8 8 
 5 2* 
 
 PRACTICE. 
 By Ride X. 
 
 V cwt. 
 
 135 
 
 198 13 9 =price of 198 c. 2 q. 21 Ibs. at \ V cwt. 
 
 794 15 = 
 
 79 9 6 =. 
 
 V 6 12 5|= 
 
 880 16 1H= 
 
 400 
 
 080 
 a 
 
 8 8 
 
 The first mode of working this exercise is sufficiently explained 
 in the operation itself. In the second mode, we say, one-seventh 
 of 21 is 3, and twice 21=42, and 3=45 pence, or 3/9, which is 
 the value of the pounds at 1 ^ cwt. We then set down 9d. 
 and carry 3 to 5 times 2, or 10 shillings, the price of 2 quarters; 
 and we thus find the price of 2 q. 21 Jtbs. at 1 ^ cwt. to be 13/9, 
 to which 198, the price of 198 cwt. at the same rate, is pre- 
 fixed. The remaining part of the operation proceeds in the usual 
 manner. 
 
 Exercises. 
 
 73. 
 
 74. 
 75. 
 
 76. 
 77. 
 78. 
 79. 
 80. 
 
 cwt. 
 
 75 
 285 
 117 
 
 84 
 134 
 836 
 812 
 176 
 
 q. Ibs. s. 
 3 21 at 4 14 
 
 3 
 
 
 
 12 
 
 
 2 
 
 2 16 
 
 Answers. 
 s. 
 358 16 
 18 
 
 .1109 
 . 101 
 .1068 
 . 123 
 .2021 
 .5391 
 . 501 
 
 13 5 
 
 2 
 
 4 8* 
 
 19 10} 
 
 13 \} 
 
 3 m 
 
 Exam. 22, What cost 319 c. 3 q. 16 Jbs. of glue, at 2 12 6 
 cwt. ? 
 
 "By Ride VIII. 
 c. q. Ibs. 
 
 319 3 16 at 2 12 6 per cwt. 
 2 
 
 638 = price of 319 cwt. at 2 per cwt, 
 T;Q T n n in/ 
 
 2/6= ^ of 10/ 
 2 q.= of 1 c. 
 1 q.-~| of 2q. 
 
 39 
 1 
 
 
 
 17 
 6 
 13 
 
 7 
 
 6 
 
 
 
 ,9/fi ^ , 
 
 3 = price of 2 qrs. at 
 
 U -1 nr 
 
 2 
 
 12 6percvrt, 
 
 16lb.=} of 1 c. 
 
 Si 
 
 6 
 
 16 Ibs. 
 
 
 
 
 
 
 
 Answer, 839 14 4, the price required. 
 
136 PRACTICE. 
 
 By Ride IX. 
 c. q. Ibs. 
 
 319 3 16 at 2 12 6 per cwt. 
 5 2} 
 
 319 17 10$ = price at 1 per cwt. 
 
 2 
 
 639 
 
 10/= I 159 
 
 2/u=iof 10/ I 39 
 
 839 14 4^ ~ __ 2 12 6 _ 
 
 The latter mode of performing this example gives origin to frac- 
 tions, which render the operation more tedious. This may be ob- 
 viated by conducting the operation on the same principle, but con- 
 verting the fractions into decimals, which it will be found to be 
 sufficient to carry out to two places each. Thus, the work mrv 
 be as follows : 
 
 c. q. Ibs. 
 
 319 3 16 at 2 12 6. 
 5 24 
 
 319 17 10-29 
 
 2 
 
 639 15 8-58 
 
 10/= I 159 18 ll-14i 
 
 2/6=r| of 10/ I 39 19 8-78 
 
 Answer, 839 14- 4'50, or 839 14- 4, as before. 
 
 Here, in taking ^ of 16, we have 2 to carry, and 2 remaining; 
 then, conceiving a cipher annexed to the remainder, and dividing 
 20 by 7, we set down the quotient 2, and conceiving a cipher an- 
 nexed to the remainder 6, we have 7 contained most nearjy '.* 
 times in 60. We then proceed in the same manner as before, and 
 find for the price at 1 per cwt. 319 17 10-29 nearly. After 
 this the work proceeds exactly as before, except that in each line 
 the [>ence and the decimal are multiplied and divided as if they 
 were a single whole number, the point alone being preserved. 
 Thus, in finding the price at 2/6, after having found 39 19, we 
 have 2/11, or 35d. remaining; we then divide 35-14 by 4, as if it 
 were all one number, and find for the quotient 878, or, the point 
 being placed before the two last figures, 8-78. The final result is 
 839 14 4'50, or 839 14 4, the same as by the other processes. 
 
 In working by this method, it should be recollected, that '25d. is a 
 ; '50d. a Italfpenny ; and '15(1. three farthings : and in valuing th 
 
PRACTICE. 
 
 137 
 
 decimal found in the ansiver, the pupil should consider to which of these it ii 
 nearest, and value it accordingly, 
 
 It may be observed, that in consequence of the method of managing 
 the decimals in all operations of this kind being uniformly the same, 
 any pupU may practise this method, whether he have studied decimal fractkms 
 or not. 
 
 The following exercise is wrought several ways for the purpose of 
 showing their comparative advantages and disadvantages, and of afford- 
 ing to the pupil an additional example of the mode ct' performing cal- 
 culations of this kind. 
 
 Exam. 23. What cost 212 c. 3q. 19 Ibs. of barilla, at 1 
 13 2F cwt.? 
 
 ByRult 
 
 ? VIII. 
 c. <l?>s. 
 212 3M9 at c< 
 106 
 21 4 
 10 12 
 1 15 4 
 16 7 
 8 3 
 4 8f 
 7^ 
 3f| 
 
 On 
 
 fee. 
 10/ 
 
 hus: ^ / 
 c. q. Ibs, 
 212 3 19 at &c. 
 106 
 21 4 
 10 12 
 1 15 4 
 16 7 
 8 3-50 x> 
 4 8-86 
 7-11 
 3-55 
 
 i/=iof2/*.;; 
 
 2d.=ofl/... 
 2q.=4oflc... 
 Iq.=of2q... 
 
 2lb'.=|ofl6lb 
 ltb.= ^of 2lb 
 
 i/=frf2/::: 
 
 2d.=iofl/... 
 Iq.= of2q.... 
 2ib'.= ofl6rb! 
 
 Hb.=of 2lb. 
 
 353 1 
 
 Answer, 353 1 10'02, or 
 353 1 10, nearly, 
 
 
 By Rule X. 
 
 
 Or thus: 
 
 
 
 c. q. Ibs. 
 
 
 c. q. 
 
 Ibs. 
 
 
 212 3 19at&c. 
 
 212 3 
 
 19 at i 
 
 
 5 2* 
 
 
 5 
 
 2$ 
 
 212 18 44 
 
 212 18 
 
 4-7' 
 
 10/=.... 
 
 106 9 2 T \ 
 
 
 10/=.... 106 9 
 
 2-35 
 
 2/=&... 
 
 21 5 10 T \ 
 
 
 2/= &... 21 5 
 
 10-07 
 
 l/=iof2/ 
 
 10 12 ll^a 
 
 
 l/=of2/ 10 12 
 
 11-03 
 
 2J.=ofl/ 
 
 1 15 5| 
 
 
 2d.=iofl/ 1 15 
 
 5-84 
 
 Answer, 353 1 10^ 
 
 
 Answer, 353 1 
 
 10-00 
 
 
 Exercises. 
 
 Answers. 
 
 c. 
 
 q. Ibs. s. 
 
 d. 
 
 s. d. 
 
 ci 75 
 
 1 16 at 1 9 
 
 9 H2 
 
 V? | 1 _1_ 
 
 1 * *J 
 
 82. 538 
 
 2 17 10 
 
 4 278 
 
 6 of 
 
 83. 346 
 
 14 1 12 
 
 7 531 
 
 j v 4. 
 
 3 If 
 
 84. 786 
 
 28 18 
 
 9 
 
 8 2+ 
 
 R.T 647 
 
 2 11 6 10 
 
 8 4,930 
 
 > *T2 
 
 19 6 
 
 oO. \rv 
 
 86. 238 
 
 03 3 19 
 
 7 947 
 
 12 10 
 
 8?! 181 
 
 3 13 2 13 
 
 4 .. .. 484 
 
 19 6i 
 
j 
 
 ., V 
 
 1S 
 
 Exercises. 
 
 
 c. 
 
 ? 
 
 Ibs. 
 
 
 
 s. 
 
 88. 
 
 251 
 
 2 
 
 1 
 
 at 1 
 
 17 
 
 89. 
 
 103 
 
 
 
 27 
 
 5 
 
 14 
 
 90. 
 
 418 
 
 2 
 
 17 
 
 2 
 
 
 
 91. 
 
 179 
 
 3 
 
 25 
 
 3 
 
 10 
 
 92. 
 
 246 
 
 3 
 
 24 
 
 3 
 
 5 
 
 93. 
 
 319 
 
 1 
 
 9 
 
 
 
 18 
 
 94. 
 
 90 
 
 2 
 
 10 
 
 5 
 
 2 
 
 PRACTICE. 
 
 Answers, 
 
 d. s. d. 
 
 ] 466 6 9^ 
 
 14 10 592 15 6^ 
 
 8 851 5 2 
 
 3 .. 632 3 U 
 
 4 806 15 
 
 7 297 7 
 
 4A 463 14 
 
 
 
 RULE XI. In computing prices in long weight, multiply 
 the pounds ^by 2 for pence, and the quarters by 5 for shil- 
 lings ; and perform the rest of the work by rule IX. 
 
 The reason of iJiis rule is evident, since at 1 ^ cwt. long 
 weight, each quarter would cost 5/, and each pound one-thirtieth 
 of this, or 2d. 
 
 Exam. 24. Required the price of 218 c. 2 q. 17 Ibs. (long 
 weight ) of beef, at 1 17 3 ^ cwt. 
 
 By Rule VIII. 
 
 c. q. Ibs. 
 518 2 17 at &c. 
 
 10/=. 
 
 .= | cwt.... 
 .=iof 2q.. 
 
 09 
 
 54 10 
 
 21 16 
 
 2 14 
 
 18 
 
 3 
 
 1 
 
 -i - 
 6| 
 
 Answer, 407 4 
 
 By Rule XL 
 
 c. q. Ibs. 
 218 2 17at&c. 
 5 2 
 
 218 12 10 
 
 ..|109 6 5 
 
 54 13 2*50 
 
 21 17 3-40 
 
 2 14 7-92 
 
 407 4 4-82,or 
 
 407 4 4f,nearty 
 
 Exercises in long weight. 
 c. q. Ibs. s. d, 
 281 3 3 at 113 
 
 Answers. 
 *. 
 
 95. 281 3 3 at 113 7 47314 
 
 96. 86 3 17 1 14 l| 148 5 
 
 97. 598 3 13 2 12 6 1572 
 
 98. 238 2 19 1 16 6 435 11 
 
 99. 216 2 1 13 10 365 8 
 
 100. 411 1 28 12 3 252 17 
 
 101. 384 17 108 396 18 
 
 102. 277 2 23 383 947 12 
 
 RULE XII. In computing the values of acres, roods, and 
 nerches, multiply the perches by 1 i for pence, and the roods 
 by 5 for shillings, and the rest of the work will proceed ac- 
 cording to rule IX. 
 
 \ 
 
PRACTICE. 139 
 
 The reason of this ride is also evident, since at 1 V acre, each 
 rood would be worth 5/, and each perch worth one-fortieth of 
 this, or l^d. 
 
 Exam. 25. What is the yearly rent of 136 a, 3r. 29 p. of 
 land, at 1 4 3 V acre? 
 
 The work by rule VIII. A. R. p. 
 
 136 3 29 at 1 4 3. 
 5 H 
 
 136 
 
 I 27 
 
 18 7-50 
 
 7 8-70 
 
 14 2-79 
 
 Answer, 166 
 166 
 
 <L 
 
 in this exercise and some 
 others, is left for the pupil 
 to perform, if it be thought 
 necessary. The exact an- 
 swer to this question is 
 166 6}|, as would 2 
 be found by employing 
 common fractions instead 
 of decimals. 
 
 Exercises. 
 a. r. p. 9. 
 
 103. 561 2 20 at 1 17 
 
 104. 45 2 35 16 
 
 105. 77 1 30 12 
 
 106. 586 1 31 17 
 
 107. 674 2 29 11 
 
 108. 311 2 26 12 
 
 109. 1268 3 17 16 
 
 110. 139 3 39 2 7 
 
 RULE XIII. In calculating the price of tons, hundreds, and 
 quarters, at a given rate per ton, take the tons and hundreds 
 as pounds and shillings, and multiply the quarters by 3 for 
 pence : and in calculations in troy weight, at a given rate 
 ver ounce, take the ounces as pounds, the penny weights 
 as shillings, and half the grains as pence. 
 
 The reason of this rule depends on the principle, that at 1 ^ 
 ton, every hundred would cost i shilling, and every quarter 3d.; 
 and that at 1 V ounce, every penny weight would cost a shil- 
 ling, and every grain a halfpenny. 
 
 Exam. 26. Required the price of 98 tons, 14 c. 3 q. of bark, 
 at 9 14 V ton. 
 
 t. c. q. 
 
 98 14 3 at 9 14 V ton. 
 3 
 
 d. 
 6 
 6 
 
 
 3* 
 *$ 
 
 9 
 
 6 
 
 10 
 
 6-90, or 
 7, nearly. 
 
 Answers. 
 
 s. 
 
 ...1053 
 
 ... 37 1* 
 
 ... 46 9 
 
 ... 800 5 
 
 ... 383 14 
 
 ... 354 10 
 
 ...1681 4 
 
 ... 334 16 
 
 98 14 9 =price at 1 
 _ 9 
 12 9 
 7 4 
 
 14 ill, nearly. 
 
 15 0|, nearly, Answer. 
 
 ton. 
 
4 
 
 > - - - 
 
 HO PRACTICE. 
 
 Exam. 27. Required the price of a set of silver article* 
 weighing 216 oz. 14 dwts. 18 grains, at 9/2 ^ oz. 
 02. dwt. grs. 
 216 14 18 at 9/2 $" oz. 
 
 216 14 9 price at 1 f oz. 
 
 72 4 11 
 2/6= 27 1 10 
 
 99 6 9, Answer. 
 Exercises. Answers. 
 
 111. 175 t. 18 c. 1 q. at 38 13 6799 
 
 112. 219 t. 16 c. 3q. 11 7 6 250013 
 
 113. 93 oz. 7dwt. 15 gr. 10 4 48 4 ll| 
 
 114. 263 oz. IGdwt. 9 gr. 11 3 148 7 tl| 
 
 The preceding are the most useful applications of the principle ou 
 which the ninth and the succeeding rules are founded. Various others 
 might be given, which however are omitted, as they are of less impor- 
 tance, and as the pupil who is well acquainted with those already ex- 
 plained, will find it easy to apply the same principles in other cases. 
 
 RULE XIV. In many calculations, instead of multiply- 
 ing the quantity by the price, it is better to multiply the 
 price by the quantity. This is often the case, when Com- 
 pound Multiplication can conveniently be employed. 
 
 Exam. 28. Required the price of 12c. 3q. 8 Ibs. of ho: 
 23 IS 6 f cwt. 
 
 c. q. Ibs. . s. d. 
 12 3 8, at 23 18 6 #" cwt. 
 12 
 
 115. 
 116. 
 117. 
 118. 
 
 119. 
 
 1*0. 
 
 
 
 2 qrs.= c 
 1 qr.=|of2q. 
 8 tbs.=* of 2 q. 
 
 Answer, 
 Exercises. 
 
 8 c. 2 qr. 12 Ibs. at 
 8 a. 3 r. 19 p. 
 9 tons, >3 cwt. 
 12 cwt. 1 q. 10 Ibs. 
 (long weight}.... at 
 1 1 a. 1 r. 23 p. 
 1 yd. 3 n. 2 nails 
 
 287 2 =rprice of 12 cw 
 11 19 3 = 2 
 5 19 7 = 1 
 1 14 21 
 
 t. 
 qrs. 
 qr. 
 8 Ibs. 
 
 
 
 ;306 15 0^=r_ 
 
 *. d. 
 1 15 9 
 
 12C.3 
 
 Am 
 
 15 
 
 q. 8 Ibs. 
 
 *wers. 
 *. cL 
 
 7 8$ 
 
 i oj 
 
 14 9 
 
 18 2 
 9 2j 
 H 10* 
 
 18 10 
 
 8 
 
 5 19 8 
 
 57 
 
 2 10 H .. 
 
 30 
 
 1 3 7 
 
 13 
 
 193. 
 
 , 2 
 
PRACTICE. i'4j 
 
 121. Tlie quantity of rum imported into Great Britain from Ja- 
 maica in isil, was 4,604,771 gallons. Required the amount oi 
 the priaie cost at 3/9 j, and oi' the importation duty at it; 
 gallon. Required also the drawback, (or sum refund*! on ex- 
 portation,) at 7/7 V gallon, oh one-third of the same quantity. 
 
 ^72987 1(5 8^; 2494250 19 2; and 581991 17 9." 
 
 122. The quantity of wool exported from Ireland in 1806, was 
 4337 c. 3 q. 8 Ibs.: what was the value at 3 19 7 & cwt.? 
 Answ. 17*69 19 0|. 
 
 123. What is the importation duty on 359 gallons of port wine, 
 
 -iiion? Answ. <136 2 5. 
 
 What is the duty on 710 gallons of Madeira wine, at 7/8 J 
 F gallon? Answ. 273 12 11. 
 
 125. Required the amount of the duty on 124 gallons of French 
 : L 1/4J W gallon. 'Answ. 70 10 6. 
 What is the duty on a cwt. of opium, at 8/9 I 5 " lb. ? Ansiv. 
 . 
 
 127. Required the cmty on 47 c. q. 19 Ibs. of mother-o '- 
 :i8lls, at 4 1341^ cwt.; and on 197 fbs. of tortoise shell, 
 llf f lb. Ansiv. 220 2 6, and 3 ;9. 9|. 
 . 128. What is the duty on 517 Ibs. of j*st fndia coffee, at 5 
 28^ cwt.? Answ. 23 13 11. 
 
 129. Required the duty on 179 c. 2. q. 12 Ibs. of Muscatel rai- 
 at 2" 3 6 & cwt' Answ. 391 4f. 
 
 130. In 1SOO, the quantity of foreign calicoes ^and muslins 
 printed in England and Wales, was 1,577,536 yards; of British 
 calicoes and muslins, 28,692,790 yards; and of linens and stufis, 
 3,232,073 yards. Required the amount of the duty paid on each 
 quantity, the first at 7d. and the second and third at 3^ ty yard. 
 Answ. 46011 9 4; 418436 10 5; and 47134 7 11. 
 
 J3I. The quantity of tea imported by the East India Company, 
 in 1791, was 16,299,854 Ibs. the average prime cost of which in 
 India was 1/6| per pound. Required the entire prime 
 Aii&w. 1273426 1 10i, 
 
 132. The quantity o"f wheat consumed in England and Wales, 
 :), is computed to have been 7,876,100 quarters. Required 
 rh-j value at 4 15 7 per quarter, the average price of wheat 
 during that year: required also the value, at the same rate, of 
 46,598 quarters, the part of that quantity which was imported 
 from Ireland. Answ. 3 7,64 1,1 94 11 8, and 222,699 12 2. 
 
 133. Belfast, 1 3th G 
 
 Jenry Wilson, 
 
 Bought of S.i-.. . 
 
 yards fine white linen, at 4/! *..; 
 
 >7 yards cambric, at 12/10 
 
 ^17 yards muslin, at 3/9 ,. ,, , 
 
 S8f * J 
 
14* PRACTICE. 
 
 134* 
 Joseph O'Reilly, Esq. 
 
 1823. To David White & Co. Br 
 
 Sept. 2th, To fine scale sugar, 4c. 3q. 22lfes. 
 
 at 4 17 4^cwt 
 
 Dec. I st, To tea, for 1 chest, containing 83 Jbs. 
 
 at 7/4 ^lb 
 
 54 10 1 
 
 135. Dublin, 8th December, 1824. Mr. William Joyce buys 
 from Patrick M'Neale, 138 gallons port wine, at 16/7^ ^"gallon; 
 130 gallons sherry, at 16/4; 110 gallons Madeira, at 26/9; and 
 120 gallons Teneriffe, at 14/10. Required the whole amount. 
 Ansu<.457 Q \. 
 
 136. Mr. Edw. Stone buys of, Hugh Sinclaire of Cork, Jan. 3d, 
 1824, 156 tierces prime betf, at 5 9 8; 313 barrels ditto, at 
 3 5 8; Feb. 8th, 93 barrels prime pork, at 3 8 3; Feb. 26th, 
 f>4 barrels inferior ditto, at 3 3 6. Required the entire amount. 
 Answ. 2403 12 11. 
 
 137. Belfast, Jan. 2d, 1830. Mr. Alexander Jefferson buys from 
 William Fitzpatrick, 218 yards linen at 3/2 per yard; 173 yards 
 muslin, at l/4 per yard; 2 pieces printed calico, containing 
 56 yards, at 1/2 per yard; and one piece ditto, containing 27| 
 yards, at lOd, per yard: and he pays in part 32 12 6. How 
 much remains due? Answ. 18 4 9. 
 
 138. Mr. Robert Bellingham buys from Patrick Cunningham of 
 Limerick, Feb. 5th, 1830, 3 cwt. of fine scale sugar, at 8/10 per 
 stone; 2 cwt. coarse ditto, at 7/4 per stone; a chest of tea con- 
 taining 86lbs. at 4/6 per tt>; 38 gallons whiskey, at 6/10 per gal- 
 lon; and 10 gallons rum, at 13/6 per gallon. Required the 
 amount. Answ. 55 11 0. 
 
 The method of aliquot parts in its application in -finding prices, hav- 
 ing been fully developed and illustrated in the preceding pages, it may 
 be proper to conclude this article with some miscellaneous matter that 
 could not with propriety have been intermixed with the general princi- 
 ples already exemplified. 
 
 * In this exercise and the preceding, which are called Bills of Parcels, the price* of 
 the several articles are to be found and set in the blank spaces towards the right hand .- 
 the sum of these partial amounts is the entire amount required. The form of the first 
 of thc-Ae exercise* is that which is usually employed, when all the articles are bought at 
 the same time: but when the times are diflerent, the form is generally that of the next 
 exercise. The remaining exercises of this kind are left for the pupil to write oui, in 
 ;orm, for his own improvement. 
 
PRACTICE. 143 
 
 Operations in the Rule 
 of Proportion may often 
 be abbreviated by the me- 
 thod of aliquot parts, whe- 
 ther the first term is a 
 unit or not. Thus, if it 
 were proposed to find how 
 muchflour might be bought 
 for 6 5 8, if 7c. 2q. 
 IGtbs. cost .11 ; the terms Answer, 4 1 13, 
 
 being arranged in the usual nearly, 
 
 way, we may multiply the 
 
 third term by 6, and take parts of it for 5/8. By this means, 
 the product of the second and third terms is found to be 48 c. Oq. 
 2 T 8 3 tbs ; ami this being divided by 11, by Compound Division, 
 the quotient is 4c. 1 q. 13lbs. nearly, the quantity required. 
 
 Thus, also, if it be re- b. b. *. d. 
 
 quired to find the price As 12 : 365 :: 2 13 6 
 
 of 365 bottles of wine, at 2 
 
 2 13 6 per dozen (ques- 
 
 tion 53, page 79), the work 730 
 
 will stand as in the mar- 10/=r^ 182 10 
 
 gin. Experience will ena- 2/6= ^ of 10/ 45 12 6 
 
 b!e the student to judge !/0= r VoflO/ 18 5 
 
 when this method may be 
 
 employed with advantage, 12)976 7 6 
 
 and when the common 
 
 method is preferable. 81 7 3, Answ. 
 
 As an application of this method to quantities of another kind, 
 
 let it be required to find the sun's mean apparent motion in 10 
 days, 7 hours, 20 minutes, 
 
 the mean space which he 5^ 8" - 3 
 
 apparently describes each 10 
 
 day in the ecliptic, being 
 
 8"'3. Here the daily h. m. 9 51 23 
 
 14 47-1 
 2 27-8 
 49-3 
 
 space being multiplied by 6 0= day 
 10 by Compound Multi- 1 Q= of 6h. 
 plication, there results 9 20=4 of * n - 
 
 51' 23", for the space de- 
 
 scribed in 10 days. Then 10 9 27 2 
 
 for 6 hours, a fourth of the 
 
 daily space is taken; for one hour, a sixth of the result; and for yo 
 minutes, a third of this last result. The sum of all these partial 
 results is 10 9 f 27"'2 nearly, the mean space required. 
 
 The following is another example of the application' of t!-.! 
 principle. When the radius of a circle, that is, half ITS di 
 is I, the half of the circumference is 3-14159265; hence, let it ' 
 
3-14159265 
 
 30 0' 0"=i of 180 
 6 = i of 30 
 2 =r $ of 6 
 40 = of 2 
 4 0=^3 of 40' 
 30 = of 4' 
 3 =r,kof 30" 
 Answ 
 
 52359677 
 10471975 
 03490658 
 01163553 
 00116355 
 0001454-i 
 00001454 
 67618416 
 
 144 PRACTICE. 
 
 required to find the length of a part of the circumference con- 
 taining 38 Ml 33". 
 
 Here, half the circumfe- 
 ference being 180, the rea- 
 son of the process is obvious; 
 and the answer is true, ex- 
 cept the last figure, which 
 should be 3, the difference 
 being occasioned by the re- 
 jection of remainders in the 
 divisions. 
 
 Various abbreviated modes of finding prices in particular cases, 
 are given in works on Mercantile Arithmetic. These may be use- 
 ful ibr the pupil who has had considerable experience in arithme- 
 tical calculations, and who is well acquainted with the more com- 
 mon and general rules for such purposes; but for the less experi- 
 enced pupil they are quite unfit, as from their variety and want 01 
 connexion they must tend only to perplex and puzzle him, and to 
 make him conceive the subject to be more difficult than it really 
 is. It has been thought better, therefore, that no abbreviations 
 except such as are of the most general and obvious nature, and 
 such as are likely to be most frequently useful, should be intro- 
 duced in the preceding part of this article, but that such others 
 as might appear to be worthy of notice, should be inserted at the 
 end of the article, that the attention of the pupil might be di- 
 rected to them or not, as the teacher should reckon best. 
 
 To find the price of any number of articles at 2d. each; since 
 2d.= T ^, we divide by 
 
 120: then since 2d.= s. 4532 at 2<L 
 
 we take one-sixth of the 
 
 remainder-, regarded as 2J. T TJ=: of Is. ... .37 15 4 
 shillings, for the remain- 
 ing part of the answer. This can all be done in a single line, the 
 division by 120 being performed by conceiving the last figure to 
 be cut off, and then dividing by 1 2. In like manner the price at 
 3d. 4-d. or 6d. may be found. 
 
 To find the price at 17/4; find the 3729 at 17/4- 
 
 price at 1G/ by the method given in 8 
 
 page 129, and for the price at 1/4 take 
 
 a twelfth of the result: -and to find 2983 4 for 16/ 
 
 the price at 14/8, from the price at 248 12 1/4 
 
 16/ take a twelfth of itself. In like 
 
 manner, to find the price at 15/2, to 3231 16 
 the price at 14/.add a twelfth of it- 
 self; and to find the price at 12/10, from the price at 14/ take 
 twelfth of itself: and lastly, to find the price at 19/6, to the price 
 add a twelfth of itself, and to find the price at 16/6, from 
 ice at 18/ take>a twelfth of itself. 
 
TARE AND TRET. 145 
 
 The price at 5/7 may be found by adding to the price at 5/ an 
 eighth of itself; the price at 3/1 by adding to the price at 2/6 a 
 fourth of itself; aud the price at 2/9 J by adding to the price at 
 2/6 an eighth of itself. The prices at 4/4, 1/10|, and 2/2, may 
 be found by the same rules, except that we are to subtract, instead 
 of adding. 
 
 To find the price at Q 15, to the price at 6, add an eighth 
 of itself; and to find the price at 6/9, to the price at 6/, add one- 
 eighth of itself. In like manner, to find the price at .3 7 6, to 
 the price at 3, add an eighth of itself; and to find the price at 
 2 12 6, take an eighth of itself from the price at 3. 
 
 The price of 24 articles may often be found very readily by tak- 
 ing each penny in the price as 2/j the price of 48, by taking each 
 halfpenny as 2/; and the price of 96, by taking each farthing as 
 2/. Hence, the price of 25 yards at 3/6, is readily found to be 
 <-i 7 6; for in 3/6 there are 42 pence; and by doubling the last 
 figure of this, we have <4 4 for the price of 24 yards; to which 
 3/6, the price of one yard, being added, the ,sum ,4 7 6 is the 
 required price. 
 
 Because 112 farthings 2/4, to find the price of 112 articles, 
 reduce the price of one to farthings, and doubling the last figure 
 for shillings, take the rest as pounds, and to the result add a sixth 
 of itself. Thus, since in 9fd. there are 39 farthings, to find the 
 price of 1 12 at 9|d. each, by doubling the last figure of 39, we 
 have 3 18, to which a sixth of itself being added, we have .4 
 11, the price required. 
 
 To find the price of 120, reckon every penny in the given rate 
 10/. Thus, 120 at 4d. each, amount to 40/, or 2; and at 5<l, 
 to 55/j or .2 15. The same results would be obtained by taking 
 the farthings in the given rate as pounds, and dividing by 8. 
 
 Almost numberless other abbreviations might be added. Should any 
 person choose, however, to make himself master of the preceding, he 
 will find it easy to discover many others. 
 
 TARE AND TRET. / 
 
 THE whole weight of any commodity, together with the 
 weight of the box, barrel, c. which contains it, is called 
 
 itS GROSS WEIGHT. 
 
 The weight of the bo^x, barrel, &c. which contains any 
 commodity, is called its TARE. 
 
 The SUTTLE WEIGHT is what remains after the tare is 
 deducted. 
 
 TRET is an allowance of 4 Ibs. for each 104 ibs. of the 
 suttle, in goods that are liable to waste. 
 
 H 
 
146 TARE AND TRET. 
 
 CLOFF is an allowance of 2 Ibs. in every 3 cwt. of what 
 remains after the other deductions are made. This allow- 
 ance is for the turning of the beam in weighing the goods. 
 
 The NEAT or NET WEIGHT is what remains after all de- 
 ductions have been made. 
 
 The deductions made in different places vary ; the following examples, 
 however, will illustrate the mode in which they are to be made, what- 
 ever they may be. > 
 
 Exam. 1. Required 
 the neat weight of 39 c. 
 3 q. 21 Ibs. tare 16 Ibs. ^ 
 cwt. tret and cloff as usual. 
 In the several parts of 
 this and the following ope- 
 rations, for the purpose of 
 avoiding fractions, for the 
 last figure of eac^ quoti- 
 ent, that which u 
 the truth has been tifcen. 
 If greater accuracy 
 quired, one or two plj 
 of decimals may be wrought 
 for. The reason of the 
 process is sufficiently ma T 
 mfest. 
 
 / 
 
 c. 
 
 q. 
 
 los. 
 
 
 39 
 
 3 
 
 21 
 
 16 lbs.=i- of 1 c. 
 
 5 
 
 2 
 
 23 
 
 2 lbs.=4 of 16 Ibs. 
 
 
 
 2 
 
 24 
 
 Tare, 
 
 6 
 
 1 
 
 19 
 
 Suttle, 
 4lbs.=A of 104 Ibs... 
 
 33 
 
 , 1 
 
 2 
 1 
 
 2 
 4 
 
 
 32 
 
 
 
 26 
 
 2 lbs.= T ^ of 3 c... 
 
 
 
 
 
 22 
 
 Neat weight, 32 4 
 
 Exam. 2. Required 
 the neat weight of 59 8 &< 
 casks of butter, weigh- 4 ^ 
 i ng gross, 40 o. q. 4 IDs. 
 tare 12 Ibs. V cask. 
 
 Exam. 3. Required 
 the neat weight of 24 c. 
 1 q. 16 Ibs. tare 2 $f 
 cent 
 Here, 2 being a for- 
 tieth of 100, the tare is 
 a fortieth of the gross weight, 
 
 59 casks. 
 s.= T Voflc. 4 24 
 3.= of 8 Ibs. 2 12 
 
 Tare, t> 
 Gross, 40 
 
 1 8) 
 4 ] 
 
 subt. 
 
 Ibs. 
 16 
 12 
 
 Neat weight, 33 
 
 2=r, l a of 100. 
 Neat weight, 
 
 2 24 
 
 c. q. 
 24 1 
 . 02 
 
 23 3 
 
 4 
 
 Exercises. Find the neat weights of the following quantities: 
 
 1. 166 c. q. 8 Ibs. tare 4 V cent. Annv. 159 c. 1 q. 20 Ibs. 
 
 2. 164- c. 1 q. 12 Ibs. tare 5 ^ cent, tret as usual. Amw. 150 c. 
 q. 15 Ibs. 
 
 3. 104 cwt. tare 12 Ibs. ^cwt. tret as usual. Anstu 89 c. 1 q. 4 Jbs. 
 
 4. 125 c. 2 q. 17 Jbs. tare 15 Ibs. #" cwt. Answ. 108 c. 3 q. 8 tbs, 
 
 5. 75 c. 1 q. 26 tbs. tare 13 Ibs- ^ cwt. tret as usual. Answ. 64 c. 
 q, |?i Ibs. nearly. 
 
 2 
 
6. 9i c. 3 q. 4 Ibs. tare 12 Ibs. ty cwt. tret and cloff as usual. 
 Answ. 80 c. 3q. 16 Ibs. 
 
 7. 133 c. q. 10 Ibs. tare 17 Ibs. ^ cwt. tret as usual 4nsip t - lOSc. 
 9. q..5 Ibs. 
 
 8. 91 c. 1 q. tare 5 V cent. Answ. 86 c. 2 q. 21 Iba. 
 
 9. 88 c. 2 q. tare 7 f cent. Answ. 81 c. 3 q. .13 Ibs. 
 
 10. 24- c. 2 q. 7 Ibs. tare 6 #* cent. JJWH;. 23 c. q.'lO Jbs. 
 
 11. 19 c. 1 q. tare -i ^ cent. Answ. 18 c. 1 q. 26 Ibs. / 
 
 12. 22 bags weighing gross 45 c. I q. 19 Ibs. tare 5 Ibs. & bag, 
 tf.it as usual. Answ. 42 c. 2 q. 26 Ibs. 
 
 13. 39 barrels, weighing gross 83 c. 2q. tare 18 Ibs. W barrel, 
 tret as usual. Aasw. 74 c. 1 q. 1 Ib. 
 
 14. 96 tierces, weighing gross 213 c. tare 23 Ibs. W tierce, cloff 
 -; usual. Answ. 192 c. Oq. 15 Ibs. / 
 
 15. 141 c. Oq. 25 Ibs. tare 6| F cent, tret and cloiF as usual. 
 Answ. 126 c. q. 23 Its. nearly". / 
 
 INTEREST. 
 
 THE sum to be paid by a person for the use of money 
 which he owes, is called the INTEREST of that money. 
 The money due is called the PRINCIPAL. 
 The sum of the principal and interest is called the 
 
 AMOUNT. 
 
 The RATE is the money allowed for die use of one hun-- 
 dred pounds for any given time, but usually for a year. 
 
 When interest is charged on the original principal only, 
 it is termed SIM; LE INTEREST. 
 
 When interest is charged, not only on the original prin- 
 cipal, but also on the interest a-s it becomes due, it is called 
 
 COMPOUND INTEREST. 
 
 It is scarcely necessary to remark, tint per cenL means per hundred, 
 and per annum, per year- 
 
 SIMPLE'INTEREST.* '/^ f 
 
 RULE K& To find the interest of a given si, >-. , - : at a give* 
 
 rate per cent, per annum ; multiply the principal Lv'iae rate, and 
 divide the product -by loo. Or, 
 
 As 100 are to the rate F cent. & annum, so is the pri 
 to its interest for one yeai. 
 
 * In Interest five quantities are concerned, tV ;,r;>u. 
 f crest, and the amount; and any throe of these, except tliv 
 
 .at, being given, the rest can befm,.;l. Hence, caic', 
 
 H2 
 
148 INTEREST. 
 
 Exam. 1. Required the interest of .576 5 8 for 1 year, at 
 6 W cent. ty annum.* 
 
 Or thiLs: (See note, last page.) 
 
 576 5 8 576 5 8 
 
 6 6 
 
 34|57 14 3 100)3457 14 3 
 
 20 
 
 Answer^ 34 11 6 
 
 Here, by the first, or exact me- 
 thod, the answer is 34 11 6^, with 
 T&O or ITS of a farthing. 
 2,04 
 
 The reason of the operation is quite evident, as it is nothing 
 more than this: as the principal 100 is to its interest 6, so is 
 
 mil of ^several problems. The most useful however, and consequently that which claims 
 the greatest degree of attention, is that in which the principal, the time, and the rate, 
 are given, to find the interest or amount. This problem may be resolved in all cases by 
 means of the first or second rule : the third and fourth, present modified, and in many 
 ases, shorter methods of effecting the same. The following rule may also be found 
 useful : 
 
 To divide money by 100, for the pounds of the quotient, take the pounds of the divi- 
 dend, except the two last figures, which are to be divided by 5 for shillings : from the 
 remainder with half the shillings of the dividend annexed, reject a twenty-fifth part, 
 and regard what remains as farthings. If there be pence, or an odd shilling, their e- 
 feet in modifying the quotient, may be estimated as nearly as 
 possible. Thus, let it be required to divide 8i)47 13 8 by 100. t. cL 
 
 Here, by cutting off' two figures, we have 89; and one- fifth 100)8947 13 8 
 of 47 is 9, the shillings required, and the remainder is 2. This 
 remainder with 7, the half of 14 shillings, annexed, becomes 89 9 6} 
 
 27, from which 1, nearly its twenty-fifth part, being rejected, 
 
 we have 26 farthings, or 6|d. "We use 14 shillings in this example, because 13/8 is 
 nearly 14/. 
 
 The reason of ihe rule will be understood from the paragraph commencing near the 
 bottom of page 113. 
 
 As another example, let it be required to divide 2658 16 10 2658 16 10 
 
 by 100. Here, after cutting off two figures, we have 26, and 
 
 the fifth of the remainder is 11 for shillings, and the remainder 26 11 9J 
 
 3. This remainder being prefixed to 8|, the half of I"/, to 
 
 which 16/10 is nearly equal, we have 38, the twenty-fifth part of which is obviously 
 about 1 J. Then 38j| being diminished by this quantity, the remainder is 37 farthings, or 
 <><d. which completes the quotient 
 
 * The rale of interest has varied much at different periods, and in different countries, 
 tut it lias uc'cn generally oLscrvud to diminish as commerce extends. In Italy, about 
 thy beginning of the the thirteenth century, it varied between 20 and 30 per cent. ; and 
 <-!:m<!s it wa- fixed by Charles V. in 1560, at 12 per cent. By an act of the 
 ilt-my VIII. interest in England, was not to exceed 10 per cent. By 21st 
 iucjLHl to 8 per cent. Soon after the Restoration, it was reduced still 
 T cent. ; and in the 12lh of Anne, to 5 per cent, the present rate. Tb 
 interest in Ireland is at present 6 i>ur cent, per annum. 
 
INTEREST. 1*3 
 
 the principal, .576 5 8, to its interest; and it is evident, that as 
 often as the one principal contains its interest, so often will the 
 other contain its interest: that is, by the nature of proportion, 
 the interest will be proportional to the principal. See the com- 
 mencement of Simple Proportion. 
 
 Exam. 2. Required the interest and the amount of 619 9 6 
 for 1 year, at 5 W cent. ^ annum. 
 619 9 6 at 5 V cent. 
 
 5 The division at full length 
 
 is as follows : 
 
 3097 7 6 for 6 V cent. 34(07 2 3 
 
 309 14 9 for .__ 20 
 
 100)3407 2 3 for 5 i 1|42 
 
 12 
 
 . , , ( 34 1 5=interest. 
 
 <619 9 6=princrpal. 5|07 
 
 653 10 11:= amount. 
 
 In the work of this example the principal is multiplied by 5 for 5 
 per cent, and, for ^ per cent, half the principal is added to the product. 
 At the conclusion of the contracted division, it gives the result more 
 nearly true to reject one than nothing from 21, though less than 25, and 
 more especially as there are three pence in the dividend. In the exam- 
 ples that follow, the division at full length will be omitted ; it may be 
 proper however for the pupil occasionally to work exercises both ways. 
 
 Exam. 3. What is the interest of 1374 1 9 for 1 year, at 
 5 1 per cent, per annum ? 
 
 1374 1 9 at 5| tf* cent. 
 
 Since 5=,& of 100, 
 
 this example might be 6870 8 9 
 
 wrought by taking a twen- = of 5. ..858 16 1 
 tieth of the principal, and - 
 
 increasing the result by 100)7729 4 10 
 
 an eighth of itself. - 
 
 77 5 10, Answer. 
 
 Exercises. Find the interests of the following sums for 1 year, at 
 the given rates per cent, per annum. 
 
 Exercises. Answers. 
 
 s. d. s. d. 
 
 1. 774 11 3 at 5 .............................. 38 14 6| 
 
 2. 53912 6 5 ............................. 2913 7 
 
 3. 288 16 6| ...................... ....... 18 15 5 
 
 4. 468 16 8 3j ............................. 16 8 2 
 
 5. 254 14 8 5 ............................. 13 . 
 
 6. 87612 6 51 ............................ 50 8 i} 
 
150 
 
 INTEREST. 
 
 Exercises. 
 
 
 
 
 s. 
 
 d. 
 
 7. 
 
 589 
 
 13 
 
 4 
 
 8. 
 
 376 
 
 12 
 
 8 
 
 9. 
 
 175 
 
 8 
 
 2 
 
 10. 
 
 286 
 
 4 
 
 7 
 
 11. 
 
 637 
 
 11 
 
 4-- 
 
 12. 
 
 411 
 
 1 
 
 10' 
 
 13. 
 
 2617 
 
 7 
 
 3 
 
 14. 
 
 899 
 
 10 
 
 7i 
 
 15. 
 
 937 
 
 5 
 
 II 5 
 
 16. 
 
 534 
 
 3 
 
 
 17. 
 
 39 
 
 16 
 
 
 18. 
 
 671 
 
 19 
 
 5J 
 
 Answers. 
 
 s. 
 
 d. 
 
 at 3| 22 2 
 
 3 
 
 4 15 1 
 
 3f 
 
 9 15 15 
 
 8f 
 
 71 22 10 
 
 9f 
 
 5| 37 3 
 
 10 
 
 11 *. 45 4 
 
 4J 
 
 3i 91 12 
 
 
 4| 42 14 
 
 64 
 
 4^ 42 3 
 
 (ji 
 
 7 ? 38 14 
 
 7 
 
 5 .., 2 
 
 9i 
 
 4 4 28 11 
 
 2 
 
 RULE II. To Jind the interest of a given principal for any 
 a year; (1.) Find the interest for a year by 
 rule I. : (2.) as one year is to the given time, so is the inte- 
 rest for one year to the interest required. 
 
 The work may often be abbreviated by finding the inte- 
 rest for months or fractional parts of a year, by the method 
 of aliquot parts. In using this method, the answer will 
 often be found with more ease, or with a greater degree of 
 correctness by multiplying by the rate ; then multiplying 
 or taking aliquot parts for the time ; and last of all divid- 
 ing by 100. 
 
 Exam. 4. Required the interest of 99 2 4 for 2| years at 
 4 per cent, per annum. 
 
 In this example the in- 99 2 4 
 
 terest for 1 year is first 4 
 
 found, which is 3 19 3 
 nearly. This is then mul- 
 
 100)396 9 4 
 
 3 19^ 3 4, int. for 1 year. 
 
 tiplied by 2|, the number 
 of years. It might have 
 been done by multiplying 
 by 3, and subtracting a 
 fourth of 3 19 3. 
 The formal analogy would 
 have been : as 1 y. : 2f y. 
 ::3 19 3:iO 18 Of. 
 
 10 18 Of, Answer. 
 
 Exam. 5. Required the interest of 179 12 11 for 1 year 
 and 7 months, at 5 per cent, per annum. 
 
 The interest for 1 year is found, by the method already ex- 
 plained, to be 8 19 7J. The rest of the work by aliquot parts, 
 
 is us follows: 
 
 '1 ! 
 
6 months= year 
 1 month := of 6 mo. 
 
 In this method the divi- 
 sion by 100 As delayed till 
 the end of the operation : 
 everything else is as before. 
 By this means the error 
 that often arises from ne- 
 glecting the remainders in 
 the division by 100, is done 
 away, and the answer thus 
 found is more nearly true, 
 than that which in many 
 cases- would be obtained by 
 the other method.* 
 
 INTEREST. . 
 
 8 19 7f=interest for 1 year 
 
 151 
 
 9 
 14 
 
 6 months 
 1 month 
 
 4> S^^r interest required. 
 
 Or thus: 
 179 1241 
 5 
 
 6 
 
 4i month = 
 
 898 
 44-9 
 
 4 7 
 2 3i 
 
 74 17 
 
 100)1422 3 
 
 !* 4 
 
 Exercises. Find the interests of the following principals for thee 
 given times, and at the given rates per cent, per annum. 
 
 Exercises. Answers. 
 
 s. d. s. d. 
 
 812 10 10 for2y. 5 in. at 4| 93 5 5 
 
 ~8|m. at5 :.... 26 15 
 
 y. 10m. at 4f 33 12 
 
 y. 8m. at 6 49 
 
 Om. at 6 41 
 
 y. 5 m. af 5| 34 16 
 
 y. 10m. at 6 94 9 
 
 y. 9 m. at 5 21 19 
 
 7 m. at 44 17 I 
 
 44_ 1 y. 10 rn"at 5 78 13 
 
 2y. at 5| 20 16 11 
 
 1 y. 9m. at 3 31 19 9^ 
 
 5 m.at 6 49 2 2\ 
 
 2y. 4m. at 5| 48 15 10| 
 
 19. 
 
 812 
 
 10 
 
 10 
 
 JO. 
 
 719 
 
 18 
 
 4 
 
 21. 
 
 419 
 
 7 
 
 9 
 
 22. 
 
 493 
 
 16 
 
 8 
 
 23. 
 
 824 
 
 18 
 
 9 
 
 24. 
 
 427 
 
 8 
 
 8 
 
 25. 
 
 792 
 
 12 
 
 3 
 
 26. 
 
 250 
 
 18 
 
 4 
 
 27. 
 
 651 
 
 
 
 
 
 28. 
 
 780 
 
 1 
 
 4 ; 
 
 29. 
 
 193 
 
 18 
 
 2; 
 
 30. 
 
 584 
 
 18 
 
 8' 
 
 31. 
 
 1964 
 
 7 
 
 6 
 
 32. 
 
 365 
 
 4 
 
 10: 
 
 * The interest of a sum for any number of months at 6 per cent, per annum, may 
 be ocry easily found by multiplying the sum by half the number of /no.-Ms, and dividing 
 the result by 100 : and hence the interest at other rates may be derived by means rf nit- 
 quit partt. Tims, to find the interest of 250 for 10 months, at 4 per cent, pci ai.ru;:,i, 
 we multiply 250 by 5 ; and dividing the product by 100, we find the inteicst at 6 per 
 cent, to be 12 10 0. We then take from this a third of itself, and there rcii.a:; - 
 8 6 8, the interest required. This method is neat and short, b>it is seU: > 
 real business. Other contractions iu calculating interest for months, are iiai. 
 same objection. 
 
 *>. 
 
152 INTEREST. 
 
 Exam. 6. Required the interest of 342 11 8 for 86 dayv 
 at 4 per cent, per annum. 
 
 In this exercise the interest for a year is found to be 13 14 J 
 nearly; and as 365 days : 86 days : : 13 14- Of : 3 4 7 nearly, 
 the interest required. 
 
 Exam. 7. Required the interest of 29 17 4, from June 
 29, 1818. till Feb. 12, 1819, at 5 per cent, per annum. 
 
 The number of days from the 29th of June, till the 12th of 
 February followmg, is found, by the method shown in page 51, to 
 be 228; and the interest of 429 17 4 for a year, is found, by 
 the method already explained, to be 21 9 10^. Then, as 365 
 days : 228 days : : 21 9 10 : 13 8 6, the interest required. 
 
 It will readily be seen, that this and all similar questions may 
 be wrought by Compound Proportion. The terms will be arranged 
 thus : 
 
 AS ?6 5 day S fit days } : : * 29 * : *I3 8 <fc A, Un ,er. 
 
 Find the interests of the following sums for the proposed times, 
 and at the assigned rates per cent, per annum. 
 
 Exercises. Answers. 
 
 . s. d. days. $. ft. 
 
 118 9^ 
 8 710| 
 3 3f 
 
 8 10 *$ 
 2710 1\ 
 16 10 9 
 
 12 1 2 
 
 9 19 llf 
 11 1 
 23 9 4 
 
 7 15 8 
 
 RULE n. To Jind the interest of a given sum for any 
 number of days, multiply the principal by twice the rate, and 
 the product by the days, and divide the result by 73,000. 
 
 The division by 73,000 rn ay be performed by the following rule: 
 Below theMiidend write one-third of itself, one-tenth #f that third, and one- 
 tenth of that tenth, rejecting shillings and remainders,': then add the four lines 
 ^ogether, divide the sum by 100,000 (or cut off jite figures), and reject a 
 farthing for each lO in the result.* 
 
 . * The exact correction is a farthing for 10 8 4$. The reason of this rule will appear 
 fi-om performing the operation indicated in it on 73,000, the result being 100,010. Now, 
 IP, die exces^pf this above 100,000, is contained in it 1U,OC1 tiaies, and 10,001 farthing* 
 arc f. I.. 
 
 33 
 
 456 10 
 
 for 31, at 5 .... 
 
 
 
 34 
 
 729 16 
 
 3 73, 5| .... 
 
 
 
 S5 
 
 1000 
 
 ~, . I, 6 . 
 
 
 
 36 
 
 698 14 
 
 8 89 5 . 
 
 
 
 37. 
 
 788 14 
 
 10 196, 6| .... 
 
 
 
 3ft 
 
 381 1 
 
 8" 264, 6 .... 
 
 
 
 39. 
 40. 
 4i. 
 42. 
 43. 
 
 447 12 
 
 487 18 
 474 15 
 
 9-13 i 
 
 6 from July 8, till 
 9 Ap. 21, _ 
 3 June 20, 
 8 May 1, _ 
 8 Aug. 24, _ 
 
 Dec. 26, 
 Sept. 4, 
 Nov. 8, 
 Oct. 21, 
 Jan. 1, 
 
 at 5f 
 
 5 * 
 6 
 
 6 
 
f 
 
 INTEREST. 953 
 
 The division by 73,000 may also be facilitated by cutting off the ci- 
 phers, and using the table given in page 35. 
 
 Exam. 8. What is the interest of 372 10 10, from Feb- 
 ruary 12, till December 17, 1824, at 4 per cent, per annum? 
 
 1824 being a leap year, the num- .372 10 10 
 
 ber of days is found to be 309; and 9 
 
 the product of this, of the given 
 
 principal, and of 9, (twice the given 3352 17 6, or 
 
 rate,) being found, as in the margin, 3353 0, nearly* 
 
 is 1036077, nearly; which being 309 
 
 divided by 73,01)0, the quotient is - 
 
 .143 lO^the interest required. 30177 
 
 The division by 73,000 by the second 10059 
 
 mode wil; *tand as in the margin, - 
 below the ^.iswer as found by the 73000)1036077 
 
 first method. After cutting off five 
 
 figures, we have 14 remaining to 14 3 10, Answ* , 
 
 the left. Then, according to the 
 
 method shown at the bottom of Or thus: 
 
 page 1 13, we divide the number ex- 1036077 
 
 pressed by the next two figures by J 345359 
 
 5, and obtain for quotient 3 shil- 34536 
 
 lings with the remainder 4. Pre- 3453 
 
 fixing this to the next figure, we 
 
 have 44; and rejecting 2, for a 14,19425 
 
 twenty -fifth of 44, \ve have remain- 
 
 ing 42 farthi..gs or lO^d. We then 14 3 10 
 
 correct the result by rejecting 1 Correc 0^ 
 
 farthing, and we find the same re- 
 
 suit as before. 14 3 10, Answ. 
 
 TJie reason of the rule will be evident from the operation by Com- 
 pound Proportion, if, instead of 100 and the rate IT cent, their 
 doubles be employed. Thus, we should have, in this exercise, 
 As 200 :9 > . ? . 
 
 365 days : 309 days }' 1U . J4 d J0 , 
 
 and in working this, we should, by the rule for Compound Pro- 
 nortion, multiply together the principal, the days, and twice the 
 rate, and divide'the product by 365 X 200, or 73000, 
 
 It is easy to see, that when the rate is 5 per cent, since the double 
 if 5 is 10, it will be sufficient to divide the product of the principal 
 and days by 7300. 
 
 Exercises. Answers. 
 
 s. d. s. d. 
 
 44. 648 15 6 from June 2, till Nov. 25, at 5 15 1 
 
 io. 1120 10 Mar. 23, Nov. 2, 6 4 1 
 
 i6. 688 18 4 Mar. 10, Aug. 25, _ 6 19 ' t 
 
 H3 
 
154 1 7 INTEREST. 
 
 Exercises. lA Ansivcrs. 
 
 s. d. s. d. 
 
 47. 884 8 8 from Mar. 3, till Oct. 28, at 5 ..... 28 19 1| 
 
 48.486815 June 8, Nov. 1, 6| .....126 11 9 
 
 49. 597 10 8 May 6, Aug. 21, 5| 9 12 ,8^ 
 
 The following rule will be found very easy and practicable in Interest 
 and Discount : 
 
 RULE IV.* To Jind the interest of a given principal for 
 any number of days, at 4- per cent, per annum ; (1.) multiply 
 the principal by the days : (2.) to the product add one-tenth* 
 of itself: (3.) from the sum take four times the same pro- 
 duct, wanting the three last figures : (4.) divide what re- 
 mains by 10,000 (or cut off four figures ;) the quotient will 
 be the answer nearly. 
 
 When the interest is large, reject a farthing for each 10 
 contained in it. 
 
 For other rates than 4 per cent, increase or diminish the 
 product of the principal and days, by the method of ali- 
 quot parts, and then proceed by the rule. 
 
 Exam. 9. Required the interest of .8985 14 for 12 days, 
 at 4 ty cent. ^ annum. 
 
 Here the product of the principal and 8985 14 
 
 days is 107828 nearly, and the tenth of 12 
 this (found by setting each figure one 
 
 place nearer the right hand side, and in- 107828 8 
 
 creasing the unit figure by 1, because 28 10783 
 
 is nearly 30,) being added to it, the sum W- 
 
 is 118611. After this, we multiply 107 118611 /' 
 
 by 4, and increase the product by 3, (car- 431 
 ried for 4 times 8, the first of the figures 
 
 cut off:) the result, 431, is then subtract- 1 1,8180 
 ed, and the remainder, 118180, divided 
 
 by 10,000, in the way pointed out in the 11 16 
 last example: the quotient KS 11 16 4^, 
 from which, because it is nearly 10, a 
 
 farthing is subtracted, and the remainder, 11 16 4, Answ. 
 J1 16 4, is the interest required. 
 
 * The following method of finding interest for days at 6 per cent, per annum, may :>:? 
 fout'd useful, when the interest to be found is not very large. 
 
 Divide the product of the principal and days by 100 j take one third of the quc'.jpi.t 
 Kir shillings, and one-sixth of the remainder for farthings ; and from the sum thus ol>- 
 t.Tiued, reject a penny for each six shillings contained in it ; the remainder will be the 
 .merest required, very neatly. 
 
 .or correction may sometimes be made, by adding to the result obtained by the 
 
 ' part of the rule, a penny for each six shillings contained in the first correc- 
 
 tioi , or a penny for each 20 in the entire interest, will give nearly the same correction. 
 
INTEREST. 155 
 
 Had the rate been 5 V cent, we must have increased 107828 8, ' 
 by one-fourth of itself, and then have added to the result oi^e- 
 tenth of itself, &c. 
 
 With respect to the reason of this easy and expeditious rule, 
 the reader who has studied decimal fractions, will find by the last 
 rule, that the interest of ], for 1 day, at 4- W cent. ^? annum, is 
 873000, or -0001 096, nearly, or -0001 1 -0000004, nearly : 
 and it will appear on a little consideration, that the operation by 
 the rule is nothing else than multiplying by '00011 and -0000004, 
 and takingthe difference of the results. The correction is neces- 
 sary, because the decimal '0001096 is not the exact interest of 
 one pound for a day. 
 
 Exercises. Answers, 
 
 s. d. *. d. 
 
 50. 773 6 8 from June 9, till Dec. 6, at 4 15 5 1 
 
 51. 59312 6 May 12, Oct. 29, 4 11 1 2 
 
 52. 374 5 April J, Dec.29, 4 11 3 U 
 
 53. 246 18 10 Mar. 14, June 8, 6 3 9 <jj 
 
 54. 511 16 8 May 13, Sept. 29,^ 6 11 13 I'pf 
 
 55. 176 11 4 ^Mar.17, Aug.25 r .5J 4 5 b 
 
 The calculation of interest on accounts current, affords a useful 
 application of the preceding principles. An ACCOUNT ccRHtxr 
 contains a statement of the mercantile transactions of one person 
 with another, when immediate payments are not made. It "is usu- 
 ally written on two pages, in the manner of a ledger account; the 
 page towards the left hand containing every payment made by the 
 merchant who furnishes the account, and the other page every 
 thing that is paid to him.- At the foot of pages 156 and T57, is a 
 specimen of this kind of account, and the following is the method 
 of calculating the interest on it, at 4 ^ cent. W annum. 
 
 s. d. days. Dr. Cr. 
 
 Feb. 11, To 186 7 6X303. ..56472 
 
 26, To 214 10 OX288...61776 / 
 
 June 20, To 415 8 4X 173...71867 ^~~/ 
 
 March 24, By 166 13 4X261 43500 / 
 
 April 6, By 347 18 0X248, 86279 // 
 
 Sept. 26, By 200 OX 75..,^^ 15000 / / 
 
 190115 144779 
 144779 
 
156 
 
 INTEREST. 
 
 For rightly understanding this calculation, it is necessary to con- 
 sider, that the account is made up till the I Oth of December, and 
 the interest calculated on it till that date. We place in a column, 
 as above, all the sums on the debit side, and then all those on 
 the credit side, prefixing to both their dates, and to the former 
 the word to, and to the latter by, for the sake of distinction. We 
 next find, successively, the number of days that elapse between 
 Feb. 1 1 and Dec. 10, between Feb. 26 and Dec. 10, &c. and place 
 them in the next column. A debit column and a credit one are 
 then formed, and all the sums on the debit side are multiplied by 
 the corresponding numbers of days, and the products are placed 
 in the debit column. In like manner, the products of the sums 
 on the credit side by the days which follow them, are placed in 
 the credit column. The sums of the two columns are then taken, 
 and the debit side is found by subtraction to exceed the other side 
 by 4-5336; which by means of rule IV. (or of rule 111.1 gives 
 4: 19 44, the interest due on the entire account. This is placed 
 on the debit side of the account; and then the sum of all on the 
 credit side is taken from the sum of all on the debit side, and the 
 remainder .106 13 10|, is placed on the credit>side, and is the 
 sum due by the person to whom the account is furnished. It is 
 scarcely necessary to say, that the two last lines, in Italics, form 
 the answer of the account, being found by the calculator. 
 
 It may be proper to require the pupil not only to perform the calcu- 
 tion of the interest on the following accounts current, but also to write 
 the accounts out in proper form on a sheet of paper, after the manner 
 of the specimen given at the foot of this page and the succeeding. The 
 answers are in the lines which are printed in Italics. 
 
 Exercises 57, 58, 59. Required the principal and interest due on 
 each of the following accounts current, till the date at the end of 
 each, the first at 6, the second at 5, and the third at 6 ^ cent. ^ an. 
 
 Dr. Mr. J. Fox, in Account Current, with S. BELL. Cr. 
 
 1824. 
 
 s. d. 
 
 1824. 
 
 s. d. 
 
 May 19, 
 
 To goods ... 512 12 6 
 
 June 13, 
 
 By cash .... 400 18 
 
 Aug. 23, 
 
 To tea 273 8 
 
 Nov. 8, 
 
 By wheat... 680 
 
 Oct. 4, 
 
 To goods ... 186 10 
 
 Dec. 1, 
 
 By bill 73 5 8 
 
 ]Sov. 18, 
 
 To sugar ... 272 5 
 
 1825.. 
 
 
 1825. 
 Jan. 18, 
 
 To balance* J . n , . , 
 of interest] 10 4 ll 
 
 Jan. 18, 
 
 By balance ) m , 6 g 
 tonewac. } 
 
 ,1255 5 
 
 1255 5 
 
 Kx. 5G. Dr 
 
 Mr. JOHN JARDINE, Newry, in Account 
 
 J riiM . 
 Feb. 11, 
 
 
 
 
 s. 
 
 7 
 
 d. 
 6 
 
 
 
 914 
 
 10 
 
 
 
 June 20 
 
 To amount of rum and sugar . 
 
 415 
 
 8 
 
 4 
 
 btv. 10, 
 
 fo interest, due on 1 his aecount 
 
 4 
 
 
 
 
 
 "821 
 
 r~ 
 
 24 
 
 
V 
 
 INTEREST. 
 
 Dr. 
 
 Mr. C. JOHNS, Dublin, in Account Current, with 
 THOMAS LINN & Co. Belfast. 
 
 Cr. 
 
 1824. 
 
 
 
 s. 
 
 d. 
 
 1824. 
 
 
 
 
 *. 
 
 a. 
 
 Sept. 3, 
 
 To balance 1280 
 
 3 
 
 11 
 
 Sept. 19, 
 
 By sugar 
 
 1510 
 
 10 
 
 
 
 Dec. 21, 
 
 ToJinen... 793 
 
 18 
 
 
 
 1825. 
 
 
 
 
 
 18*25. 
 
 
 
 Mar. 20, 
 
 By goods 
 
 1248 
 
 8 
 
 
 
 Jan. 27, 
 
 To goods...l040 
 
 5 
 
 
 
 May 25, By "bill... 
 
 ..912 
 
 16 
 
 8 
 
 Mar. 26, 
 
 To linen... 838 
 
 14 
 
 2 
 
 June 2Q\By balance 
 
 ?306 
 
 16 
 
 Ql 
 
 June 26, 
 
 To balance ) 9t , 
 of interest J ~ 
 
 9 
 
 1* 
 
 \ tonewac " 
 
 * 
 
 
 ^2 
 
 
 
 
 3978 
 
 10 
 
 in 
 
 3978 
 
 10 
 
 iii 
 
 Dr. Mr. T. HART, in Account Current with H. ORR. 
 
 /Cr. 
 
 1824. 
 
 s. d\ 
 
 1824. 
 
 s. d. 
 
 Feb. 9, 
 
 To linen... 768 8 9 
 
 April 1, 
 
 By sales of > 
 
 
 May 7, 
 
 To lawn. ...436 17 6 
 
 
 flaxseed J 
 
 4 8 
 
 July 8, 
 
 To goods.. 948 5 10 
 
 June 27, 
 
 By sugar 500 
 
 
 
 Aug. 18, 
 
 To linen... 673 11' 
 
 Sept. 11, 
 
 By bill on ) 
 
 
 Oct. 29, 
 
 To balance ? , A ]0 
 of interest. ] 3 * 
 
 
 Ash & Co. C 1533 
 London ) 
 
 12 6 
 
 / 
 
 Oct. 29, 
 
 By balance ? ^ nfl 
 
 16 9 
 
 / 
 
 to new ac. J 
 
 
 2861 13 11 
 
 2861 13 11 
 
 Ex. 60. On the fifth of January, 1822, the public funded debt 
 of Great Britain and Ireland, consisted of 
 
 534,355,086 6 1 1 at 3 F cent. V annum ; 
 29,547,003 19 3_3i __ 
 75,947,763 19 44 ___ 
 and, 153,056,763 7 9_5 _ 
 Required the annual interest of the whole. Answ. 27,755,546 9 1. 
 
 The following rules serve for the resolution of the remaining cases of 
 Interest. As these cases are of minor importance, the rules are given 
 without proof, and without illustration by examples. They are easily 
 proved however, by the principles of proportion ; and the pupil will 
 iind it easy to apply them iu the resolution of the subjoined exercises. 
 
 Current, with CHARLES CAULFIELD, Belfast. 
 
 Cr. 
 
 1824. 
 Mar 24 
 
 
 
 166 
 
 s. 
 
 d. 
 4 
 
 April 6 
 
 
 347 
 
 18 
 
 
 
 Sept 26 
 
 By bill on Cavan & Co. Dublin 
 
 200 
 
 
 
 
 
 Dec. 10, 
 
 By balance to your debit in a new account 
 
 106 
 
 13 
 
 10 
 
 
 
 82' 
 
 5 
 
 * 
 
158 INTEREST. 
 
 RULE V. To find what principal, in a given time, would produce 
 a given interest) at a given rate per cent, per annum; 
 
 As rate : 100 > . principal. 
 
 Given time : 1 year $ 
 
 Ex. 61. How much money must be lent on the 2d of April, at 
 6 W cent. W annum, to bring in for interest 24-, on the 18th of 
 November following ? Answ. 634 15 7J. 
 
 62. What principal, at 5 V cent. W annum, will bring a yearly 
 income of 341 5 ? Answ. 6825. 
 
 03. What principal lent from the 24th of March, 1824, till the 
 i7th of November, 1825, at 4^ IP" cent, "t?" annum, will gain 1200 '/ 
 Ar.sw. 1614-1 10 3. 
 
 64). What principal at 5 IT cent. ^ annum, would be equivalent 
 to the funded national debt of Great Britain and Ireland ? (Ses 
 ike answer to question 60 of this rule.) Answ. 555,1 10,929 2 6. 
 
 RULE VI. To find what principal, in a giiien time, would increase 
 to a given amount, at a given rate per cent, per annum; (1.) To the 
 product of the time and rate, add the product of 100 and 1 year 
 in the same name as the given time: (2.) Thui, as the sum is to 
 the abovementioned product of 100 and * ^ear, so is the amount 
 to the principal. 
 
 65. What principal lent on the 1st of January, 1824, at 5 ^ 
 cent. ^ annum, would amount to 1000, on the 29th of Septem- 
 ber in the same year ? Answ.96Q 12 6i. 
 
 66. What sum must be lent at simple interest, at 4- IT cent. ^ 
 annum, that the amotmt, at the end of 2 years, 10 months, may 
 be 627 186? Answ. 564 1. 
 
 RULE VII. To find the time in winch, at a given rate, per cent, pet 
 annum, a given principal would produce a given interest: 
 
 As principal : 100 ) 
 
 ik [ : interest J '' V^ar : time required. 
 
 67. In what time will 460 amount to 500, at 4 V cent. $* 
 annum ? Answ. 1 year, 340 days. 
 
 68. How long must 2000 be lent at simple interest, at 8 
 cent, per annum, to amount to 2280 ? Answ. 4 years. 
 
 69. How long must 887 5 be lent at 5 4 per cent, 
 simple interest, to gain 120? Answ. 2 years, 210 days. 
 
 RULE VIII. To find at what rate a given principal would gain '. 
 given inter (st in a given time: 
 
 70. At what rate per cent, per annum, simple interest, 
 1500 amount to 1850, in 4 years ? Answ. 5 3 8$. 
 
DISCOUNT. 159 
 
 71. If a merchant, with a capital of 5000, gain 2000 in 2| 
 years, at what rate per cent, per annum, simple interest, has he 
 gained? Answ. 14 10 11, nearly. 
 
 72. If 1 amount to 1 2 9 in 3 years, at simple interest, at 
 what rate per cent, per annum must it have been lent? Answ. 
 4 4, 7 
 
 DISCOUNT. 
 
 DISCOUNT is an abatement made for advancing money 
 before it becomes due. 
 
 The money which is received as the full payment of any 
 debt or bill due some time after, is called its PRESENT 
 WORTH. 
 
 RULE. To find ike present "worth of a If ill or debt, (1.) 
 find the interest of the debt at the given rate, and for the 
 given time : (2.) consider this interest as discount, and sub- 
 tract it from the debt to find the present worth. 
 
 Exam. 1. Required the present worth of a bill of 170, due 
 at the end of 3 months, at 5 per cent, per annum. 
 
 Here, by the method already explained in Interest, the discount 
 is readily found to be 2 2 6; and this being taken from 170, 
 the remainder, 167 17 6, is the present worth. 
 
 Exam. 2. What is th^resent worth of a bill of 39 5, due 
 on the 1st of September, but paid on the 3d of July preceding, 
 discount being allowed at 5 per cent, per annum ? 
 
 The time here is 60 days, for which the interest of 39 5 is 
 found to be 6/5^; and by subtracting this from 39 5, we have 
 remaining 38 18 6^, the present worth. 
 
 In Great Britain and Ireland three days, called DAYS OF GRACE, 
 are always allowed after the time a bill is nominally due y before it is 
 legally due. Thus, suppose a bill were drawn on the 8th of April, 
 at 4 months, it would be due, not on the 8th, but on the llth of 
 August. . 
 
 It may be remarked, that if, withdyt the days of grace, a bill should 
 appear to be due on the thirty-first o a month which contains only thirty 
 days, the last day of that month is to be taken, and not the first of thy 
 next ; and consequently, the third of the next month will be the day on 
 which, by the addition of the days of grace, the bill will be really due. 
 Thus, a bill d-rawn on the thirty-first of August, at three months, 
 would be due on the third of December. In like manner, a bill, 
 which, without tlie addition of the days of gr,ace, would be due on the 
 twenty-ninth, ^hirtieth, or thirty-first of February, if that month con- 
 tained so many days, would be really due on the third of March. It 
 may be farther remarked, that bills which fall due on Sunday* are paid 
 in Britain on Saturday, but in Ireland on Monday. 
 
160 DISCOUNT. 
 
 Exam. 3. Required the present worth of a bill of .-677, drawn 
 8th March, at Q months, and discounted 3d June, at. 5 per cent, 
 per annum. 
 
 By counting forward 6 months and 3 days from the 8th of March, 
 we find this hill to be due on the 1 1th of September. The num- 
 ber of days from the 3d of June till this date is 100, and the in- 
 terest of .77 for 100 days, at cTper cent, per annum, is found by 
 any of the methods formerly explained, to be 1 1 !, and con- 
 sequently, the present worth is 75 18 lOf . 
 
 Exercises. Required the present worths of thd following bills, at 
 the given rates & cent. ^ annum: I'M 
 
 Exercises. Zj_ ' Answers. 
 
 s. d. Drawn, Discounted ', s. d, 
 
 1. 416 3 4, Mar. J, at 7 mos. June 9, at 4 .... 410 16 7| 
 
 2. 533 6 8, Sept. 5, 5 Nov. 12, 4|... 527 10 11^ 
 
 3. 218 11 8, Au?. 14, 4 Oct. 3, 4 ....216 15 8 
 
 4. 607 34, May 22, 5 July 10, 5.... 597 7 6 
 
 5. 895 12 0, Jan. 5,~~11 May 9, 5 .... 869 9 
 
 6. 284 8 8, Sep. 25, 5 Nov. 30, 6.... 280-1 
 
 7. 588 12 8, Mar. (5, _ 6 June 11, 6 ....579 18 6 
 
 8. 486 18 8, Mar. 25, 10 June 19, 5 .... 472 1 2 
 
 9. 875 5 8, Feb.25, 7 ~ June 4, 5 .... 861 7 6 
 
 JO. 388 2 6, Dec. fe,_ 6 Mar. 25, 6 .... 383 ' 2 11| 
 
 11.1000 0,Fcb.JG, 11 Sep. 12, 5$.... 98011 2| 
 
 12. 568 12 9, Apr. 27, 7 June 3, 5".... 554 12 4 
 
 13. 447 12 6, June 23, 6 July 8, 5f.... 43511 4- 
 
 14. 511 3 4, Man 31, 7 .May 8, 6.... 494 17 5^ 
 
 15. 649 13 4, Nov. 9, 9 April 19, 5|.... 638 8 2 
 
 Ex. 16. Required the discount of 284 13 OJ, for one year, 
 at 5 per cent, per annum. Amw. 15 13 H. 
 
 17. What is the discount of 549 9 5^, "for 32 days, at 5 per 
 cent, per annum? Amw. 2 8 2. / 
 
 18. What is the* resent worth of 970 18 4, due at the end of 
 19 months, at 4 per cent, per annum ? Amw. 897 17 11. 
 
 The rule which has been givei. above for the calculation of discount, 
 is that which is always empK y-.d by merchants. It is founded however 
 on a principle radically fal^e ; and always nines the discount too large, and 
 consequently the present irm tit too small, Ini the interest of the true discount. 
 This will appear manifest, it' we consider, that the true present worth of 
 any debt, is such a sum as twiild, if tf.nl at interest at the assigned rate. 
 Amount to that debt at the time ut winch it u'ould liace been due : and con- 
 fequently the discount, or the difference hetween the present worth aiul 
 'he debt, should be, not the inteu^t of the debt, but the interest of the 
 present worth; and therefore the interest of the debt will exceed the 
 'rue discount, that is, the interest of the present worth, by the interest 
 t discount. 
 
 lie present worth will be found by the following analogy: At 
 the amount y \QQ fo r t / ie g i v( , n t i me aHf i al t / ie j^oposed rate, is to . 10O, 
 
DISCOUNT. 161 
 
 to is the debt to its true present w^rth; and the present worth being sub- 
 tracted from the debt, the remainder is the discount. 
 
 The reason of, this rule v>ill Le evident from the consideration, that 
 100 is the present worth of its amount regarded as a debt : and conse- 
 quently the analogy given above will become .simply this: as the amount 
 of 100 considered as a debt, is to 100, the present worth of that 
 debt, so is any other debt to its present worth. It is obvious al.-^o, that 
 for the two first terms of the analogy, we might use the amount of any 
 turn whatever, and that sum itself j but it is generally more simple and 
 easy to employ 100 and its amount. 
 
 To exemplify this rule, let it be required to find the true present 
 worth of 200, due at the end of a year, at 5 per cent, per annum. 
 In this case, the amount of 100 being 105, we have, by the rule, 
 this analogy: as j105 : 100 : : ='200 : lSO 9 6^, the present 
 worth required. By the common rule the result would have been lSO: 
 consequently the error is 9/6^. 
 
 Again, let it be required to find the true present worth of 463, for 
 7 months, at 5 per cent, per annum. Here the amount of aIGO is 
 102 18 4 ; and therefore, as ,102 18 4 : =100:: -i63: =449 17 6f 
 nearly. By the common method the result would be 449 9 1 1, and 
 the error 7/7J. 
 
 This question, and all similar ones, may be very easily wrought by 
 the following rule : Multiply the months by the rate, and add the product 
 to 1200 ; then, as the sum is to 1200, so is the debt to its true present worth. 
 Thus, in the preceding example we should have this analogy: as 1235 t 
 1200: : 463 : 449 17 6f. The reason of this rule may be thus shown : 
 as 12 months > 7 months : : 5 : |, the interest of 100 for 7 months. 
 Then, as =100f : .100, or by reduction of both to twelfths, as 1235 
 : 1200: : &c. 
 
 As another example, let it be required to find the correct present 
 worth^^f 512, due on the 19th of September, but paid- op the 8th of 
 May* preceding. 
 
 Here, the number, of days being 134, we have this analogy: as 365, 
 (Jays : 134 "days : : 5 : l 16 8^, the interest of 100 for 134 days: 
 and then/\as\tlOi' 16 8 : =100 : : ,512 : 502 15 5^, answer. 
 
 This .queWion, and all others in which it is required to find the correct 
 present worth of a sum for a given number of days, may be v rougbl 
 more easily and more accurately by the following rule :* Multiply tlte 
 days by the rate, and add the product to 36500 (=365X100:) then, as 
 this su?n is to 36500, so is the debt to its true present worth. Thus, in the 
 present example 134X5=670, and 36500-}- 670=3.71 70 : then, as 
 37 170 : 36500 : : .512 : =502 15 5, the present worth. 
 
 This rule depends on the same principle as the last ; and the reason of 
 it may be thus illustrated. As 365 days : 134 days : : &S : %$%> tl)e 
 interest of 100 for 134' days, at 5 per cent per annum. Then, as 
 100|-2 : ;100; or by reduction to three-hundred-and-sixty-fifths, as 
 37170 : 36500 : : the dabt to its present worth. Both these rules are iu 
 reality the same as rule VI. in Interest.* 
 
 * Unless the time be great, tlie true present worth may be readily derived by approxi- 
 mation from that found by the common method, by finding the interest cf the interest 
 fiirt foiled, and adding it to the present worth, found by the common method ; then by 
 
162 DISCOUNT. 
 
 The following examples and considerations will 'perhaps be useful in 
 showing the falsity of the common method of discount. 
 
 If a person have a bill for .100 payable at the end of a year, at 5 
 per cent, he will receive, according to the common method of discount, 
 only 95 for it ; and were he to lend this sum for a year at the same 
 rate, instead of 100, to which it obviously should amount, he would 
 receive only 99 15. The true present worth is 95 4 9], and con- 
 sequently the error 4/9-f. Again, had the same bill been payable at the 
 end of two years, the present worth, by the common method, would 
 have been 90, while it should be ,90 18 2 T 2 T . The error is conse- 
 quently 18/2^, and '90, instead of amounting to ,100 at the end of 
 two years, would amount to no more than 99. Had the time been 
 four years, the present worths would have been 80, and 83 6 8, 
 and the error 3 6 8. The amount also of the present worth, '80, 
 would be 96) and consequently '4 less than it should be. If the time 
 had been ten years, the present worths would have been 50 and 
 66 13 4, where the error is 16 134; and the amount of the pre- 
 sent worth 50 would be 75 instead of .100. Finally, were the time 
 20 years, the present worth, according to the common method, would 
 be nothing, while it should be =50: and were the time greater than 20 
 years, the present worth would be unassignable, as it would appear to 
 be less than nothing ; or if any meaning could be attached to the result 
 of the operation, it would be, that the person who held the bill, instead 
 of receiving any thing for it, would be required to pay something to get 
 it off' his hands. 
 
 From these examples, it will appear how very erroneous the com- 
 mon method of computing discount is, when the time is long.* In 
 every case, indeed, the discounter of the bill has a greater rate of in- 
 terest for his money, than the nominal rate; and the longer the time, 
 
 finding the interest of this last interest, and subtracting it from the approximate present 
 worth ; and so on by adding and subtracting alternately, the interest of the last, interest 
 till the correction becomes so small, that it would be unnecessary to carry the operation 
 farther. When the time i very short, the true result will often be obtained as easily iu 
 this way, as by the principles above explained. 
 
 As an example, let it be required to find the present worth of a bill of 140, due at 
 the end of 6 months, at 4 per cent, per annum. 
 
 Here the interest of 140 is 2 16 ; that of 2 16 is !/!, and that of 1/1| is a farthing. 
 Then, by subtracting 2 16 from 140; by adding 1/1-J to ths remainder, and lastly, by 
 subtracting a farthing from that result, we find the present worth to be l'Ji 5 l\, which 
 is correct. 
 
 * It might be shown algebraically, that the error in the common method of calcu- 
 lating the discount or present worth of a given debt, at a given rate, is nearly propor- 
 tional to the square of the time, when the time is small, or more properly when the dis- 
 c< unt is small compared with the debt. Thus, at 5 per cent, per annum, the error on 
 a bili of 1000, for g'months, is nearly 1/4-J, while for 4 months it is nearly 5/5$, or very 
 nearly 4 times 1/4J. 
 
 It might be shown in nearly the same manner, that the error in the sums to which 
 the present worth of a given sum found by the common method, would amount at the 
 given rate, would be exactly proportional to the squares of the times. Thus, in the ex- 
 an pies in the text in this page, it appeared, that in ca e of a bill of 100, payable in a 
 year, at 5 per cent, per annum, the amount would be 99 15, while, if it were payable 
 in two years, the amount would be only 99 the error being in the one case 5/, and in 
 the other, 1. or 4 times 5J 
 
COMMISSION, INSURANCE, &c. 163 
 
 the greater is this rate. Thus, to recur to the last series of examples, 
 since by paying 95 at present, the discounter will be entitled to lOO 
 at the end of a year, he obviously gains 5 on 95 ; and therefore, as 
 95 : 5 : : 100 : 5 5 3 T ^ his gain per cent, In like manner, if 
 the time were two years, the gain per cent, would be found to be 
 5 11 1^-; if four years, 6 5 ; if 10 years, 10 per cent. ; and if 19 
 years, cent, per cent. 
 
 It is true indeed, that when the time is short, as it generally is in 
 roal business, the difference between the results found by the two me- 
 thods, is inconsiderable; and therefore the common method, the calcu- 
 lation for which is so easy, may be employed without much error. Still 
 however, the principle is false, as it gives profits to the discounter, which 
 are not proportional to the times. It may be said, indeed, that those 
 who keep money for the purpose of discounting, are entitled to more 
 than the simple common rate. This may be true ; yet at the same time, 
 it does not prove the correctness of the principle on which this mode of 
 computation depends ; since, if the discounter is to have a gj-eater rate, 
 it should be some fixed rate, and not a variable one, depending on the 
 ^ime the hill has to run. 
 
 Should it be thought advisable for the learner to work Discount in the 
 correct method, the exercises at the beginning of this article will serve 
 his purpose as well as any others ; and the following are their answers 
 by tii at method : 
 
 True Answers. 
 s. d. 
 Ex. 1. 410 17 11 Ex. 7. 580 
 
 12 2 J 8. 472 
 
 16 0| 9. 801 
 
 10 7f 10. 383 
 11. 980 
 
 2. 527 
 
 3. 216 
 
 4. 597 
 
 5. 870 
 
 6. 280 
 
 2 3| 
 
 12. 554 
 
 s. d. 
 
 9 11| 
 11 10 
 
 18 7| 
 19 1 
 
 s. d. 
 Ex. 13. 435 17 8 
 
 14. 495 7 6 
 15. 638 12 
 16. 14 16 9i 
 17. 2 7 ll| 
 18. 903 Of 
 
 COMMISSION, INSURANCE, &c. 
 
 COMMISSION is the sum which a merchant charges for 
 buying or selling goods for another. 
 
 BROKERAGE is a smaller allowance of the same nature, 
 paid usually for negotiating bills, or transacting other mo- 
 ney concerns. 
 
 INSURANCE or ASSURANCE is a contract by which one 
 party, on being paid a certain sum or premium by another, 
 on account of property that is exposed to- risk, engages in 
 case of loss, to pay to the owner of the property the sum 
 insured on it. 
 
 RULE I. To compute the commission, brokerage, insurance, 
 or any other allowance on a given sum, at a given rale per 
 
164 COMMISSION, INSURANCE, &c. 
 
 cent.; multiply the sum by the rate per cent, and divide the 
 product by 100: or, as 100 are to the rate per cent, so is 
 the given sum to the allowance required. 
 
 The work may often be abbreviated by the method of 
 aliquot parts. 
 
 Exam. 1. Find the commission on .791 11 8, at 2 V cent. 
 
 Here, by multiplying by 2, and dividing by 100, we find for 
 the answer .19 15 9^. The same result would be obtained by 
 dividing the given sum by 40, since 2 is a fortieth of 100. 
 
 Exam. 2. Find the brokerage on 829 14- 6, at 3/6 V cent. 
 
 Here, aliquot parts are taken s. d. 
 
 for 3/6, and the result being di- 829 14- 6 
 
 vided by 100, the quotient is 
 
 1 9 o|, the brokerage required. 2/6=% j 103 14 3| 
 
 In like manner, had the rate been 1/0= ^ \ 41 9 8f 
 
 , f, or any other fractional 
 
 rate V" cent, aliquot parts must 100)145 4 
 
 have been taken for the rate, and 
 
 the result divided by 100. 1 9 
 
 Exam. 3. Required the premium of insurance on 512 9 4, 
 "at 6166 per cent. 
 
 Here, by multiplying by g. cL 
 
 6, taking parts for 16/6, and 512 9 4 
 
 dividingby 100, the insurance 6 
 
 is found to be 34 19 6. 
 The result, in this particular 307416 
 
 exercise, might be found 12/= T \yof6 
 
 more easily by multiplying 4/= ^ of 12/ 
 
 by 6 and dividing by 100, by 6d.= | of 4/ 
 
 307 9 7$ 
 
 102 9 10: 
 
 12 16 
 
 which means, the insurance 
 
 at 6 ^cent. would be found; 100)3497 11 
 
 and thence the insurance at 
 
 6 16 6 Wcent. would be 34 19 
 
 deduced by adding to the 
 result an eighth of itself, and a tenth of that eighth; because 
 6 16 6 is six times 129, the best parts for which are 2/6 ={ 
 of 1, and 3d.= T \y of 2/6. ijj^ 
 
 RULE II. Tojindhotu much must be insured on propertifquvrth 
 a given sum, so that, in case of loss, both the value of the pro- 
 perty and the premium of insurance may be repaid: (1.) sub- 
 tract the rate from 100; (2.) as the remainder is to 100, 
 so is the value of the property to the sum to be insured. 
 
 Exam. 4. How much must be insured, at 8^ ^Pcent. on an 
 adventure of 600, that, in case of loss, not only the value of the 
 adventure, but also the premium of insurance, may be repaid? 
 
 Here, as 100 8^ : 100 : : 600 : 655 14 9. The truth 
 
COMMISSION, INSURANCE, c. 165 
 
 of this operation is proved by finding the premium on 655 14 9, 
 at 8 ^ cent. This is found to be 55 14 9, Hence, in 
 case of the adventure being lost, the owner will receive not only 
 600, the value of the adventure, but also 55 14 9, the pre- 
 mium; and thus he will sustain no loss whatever. The reason will 
 appear manifest from considering, that in receiving 100 which 
 had been insured at 8 ty cent, the owner would receive but 91 
 in lieu of the property, 8 having been paid for the insurance. 
 
 Ex. 1. What is the commission on 942 16 3, at 4 <^cent.? 
 Answ. 42 8 6. 
 
 2. Required the brokerage on 946 18 10, at 5/6 ^cent. 
 Answ. 2 12 1. 
 
 3. Kequired the commission on 569 14 9, at 7^ ^fcent. 
 Answ. 42 14 1\. 
 
 4. What is the premium of insurance on 675 11 8, at 5 13 9 
 tf'cent.? Answ. 38 8 5J. 
 
 5. What is the expense of insuring a vessel and cargo, worth 
 3649 8 0, at 3^ cent.? Answ. 118 12 !. 
 
 6. Required the premium of insurance on 1486 13 9, at 
 2 16 lOi^cent. Answ. 42 5 6|. 
 
 7. What must be the sum insured, at 5f W cent, on goods 
 worth 1938 12 6, that, in case of loss, the owner may be repaid 
 both the value of the goods and the premium of insurance? 
 Answ. 2056 17 11. 
 
 8. What is the brokerage on 681 4 10, at 5/3 ^ cent. Answ. 
 1 15 9. 
 
 9. At 2 56^ cent, what will be the cost of insuring goods 
 worth 1560, that, in case of loss, the owner may be entitled to 
 the value of the goods and the premium? Answ. 36 6 3f. 
 
 10. Required the commission on 863 12 6, at 2 W cent. 
 Answ. 21 11 9f. 
 
 11. Add to 579 16 10 the commission on itself at 7| ^ cent, 
 and find the insurance of the sum at 4| #" cent. Answ. 27 5 5. 
 
 12. At 4f guineas ^cent. what will be the premium of insuring 
 property worth 592 6 8, so that, in case of loss, the owner 
 may be entitled to the premium and the value of the* property? 
 Answ. 31 1 10. 
 
 13. At 5 guineas $f cent, how much must be insured on goods 
 worth 813 ll 3, so that, in cae of loss, the owner may receive 
 the value of the goods and the pre"h*nin? Answ. 863 8 6. 
 
 The following accounts of mercantile transactions will serve as ex- 
 ercises in calculating insurance, &c. ; and may be found useful, particu- 
 larly for pupils who are intended for business in the counting-house. 
 The learner should be required to write out the fifteenth, seventeenth, 
 and eighteenth, in proper form, and to find the neat proceeds. 
 
 An ACCOUNT SALES is an account of goods sold by a factor. It is 
 furnished by him to his employer, and contains the quantity and price 
 of the goods sold, the charges attending the sales, and the neat proceeds, 
 
166 COMMISSION 7 , INSURANCE, &c. 
 
 or the sum which the owner of the goods is to receive, after all charges 
 are deducted. 
 
 An o lNVOlCE is an account containing the quantity, prime cost, and 
 charges of goods sent from one person to another, usually by sea. 
 
 The following account contains the names of the purchasers, the 
 quantities sold, and the rates. The amounts of the several quantities 
 are left for the pupil to find and set in the money columns. The sum of 
 the several amounts will be found to bei'1146 16 10. The several 
 charges, and the commission on l 146 16 10, are then added together, 
 and the sum, 116 8 9, taken from the gross amount, "1146 16 10. 
 The remainder, =1030 8 1, is the neat proceeds. 
 
 Ex. 14?. Account sales butttr received per the Dolphin, from 
 Cork, sold per order and on account of John Evans, Esq. Belfast. 
 
 ' 
 
 fir- 
 tins 
 
 tt>s. 
 
 
 1824. 
 
 Jan. 
 
 Feb. 
 Mar. 
 
 16 
 
 20 
 25 
 3 
 
 5 
 
 6 
 13 
 
 15 
 16 
 
 15 
 
 Jos Hyman 
 
 5 
 5 
 2 
 
 4 
 19 
 20 
 10 
 10 
 10 
 10 
 22 
 20 
 10 
 25 
 3 
 15 
 10 
 
 200 
 
 312 
 
 318 
 128 
 251 
 1200 
 1274 
 646 
 634 
 636 
 642 
 1402 
 1266 
 628 
 1603 
 ;187 
 ,'920 
 \ 647 
 
 12694 
 
 at 2/1 
 
 C2 2/2 
 2/0 
 1/9 
 
 
 16 
 
 8 
 
 10 
 
 9 
 1 
 
 Elizabeth Naylor .. 
 John Hughes ....... 
 
 VV'ii Roberts .. 
 
 M. A. Hvnds 
 
 Aaron Moreno 
 Eml. Emanuel 
 
 Mathew Shannon .. 
 S.Touzalin &Co.... 
 Soloman Marks .... 
 Thos Hill 
 
 22 4d. 
 
 Eml. Emanuel 
 J. C. Clarke 
 
 jrl. A. Hyams 
 Jos. Isaacs & Co.... 
 B. Daly & Co 
 
 1/9 
 22|d. 
 
 
 
 11-16 
 116 
 
 Fieigh 
 wha 
 Adverl 
 negi 
 r Comm 
 
 ^ f 7}p 
 
 Neat 
 Eva 
 
 t and primage, 22 1, 
 *fage and landing, 5 5, 
 isipg, 17/6, drayage and 
 
 28 6 
 ... 2 2 6 
 
 ission on 1146 16 10, at 
 er cent 
 
 >roceeds, to credit of J. 
 
 1 
 
 1030 
 
 8 
 
 Errors excepted. 
 
 Kingston, Jamaica, 5-pt. 14, 1824. 
 
 MORROW, KEVIN, and EVANS 
 
 \ 
 
COMMISSION, INSURANCE, &c. 167 
 
 Ex. v ]5. Account sales of soap received per the Waterloo from 
 Belfast, sold by order, and on account of David Sanford, Esq, 
 1S27, March 2. To John Panillo, 2055 Ibs. at 5 per 100 Ibs.; 
 to A. Maxwell, 43? Ibs. at -* March 23, to A. Maxwell, 1976lbs. 
 at 3 3 4; to' William Moliere, 2431 ibs. at 3 2 6; April 1, 
 to Marcusa Hyams, 1198 ibs. at 3 10; May23,-to D. S. Davis, 
 1963 Ibs. at 3; to A. Grant & Co. 463 Ibs. "at 3; June 18, to 
 John Angus, 3392 Ibs. at 2 13 4; June 26, to Carvallo and 
 Rivers, 4091 ibs. at 2 13 4; June 23, to J. Antonio, 2372 Ibs. 
 at 2 13 4; Aug. 22, to John Berry, 279 Ibs. at 3', to R. 
 M'Clela-nd, 1997 Ibs. at 3 10; Sept. 4* to Leah .Levy, 3988 ibs. 
 at 3 6 8; Sept. 10, to Sarah Myers, 3194 its. at 3 6 8. 
 CHARGES Advertising for sale, Drayage, and Negro hire, 2 10; 
 Commission on , at 1\ pta cent. Kingston^ Jamaica, 
 
 Sept. 12, 1824. (Signed) Mason "and Newsam. Amw. 887 6. 
 
 16. INVOICE of 3 whole and 40 half barrels beef, and 20 half 
 barrels pork, shipped by John Marsden & Co. on board the Ro- 
 sanna, Robert Bishop, master, for Glasgow, on account and risk 
 of Mr. Andrew Blakely, merchant, Greenock. 
 
 A. B. 
 
 3 barrels Mess Beef at 4 1 1 
 
 40 half do. do. do. 25 
 
 20 do. do. Pork 212 
 
 CHARGES. 
 
 Paid carriage and shipping ) _ 
 
 7/2, quayage 1/6.. \ 
 
 Stamp 3/4, Bills of Lading I/ 
 
 Commission on 
 
 at 2 per cent 
 
 Errors excepted. 
 
 Belfast, Dec. 4, 1824. 
 
 In the preceding invoice, the lettersTM^Mj| i are the 
 
 letters with which the barrels were ma.i<H| K are left to be 
 
 filled up by the learner. The follow! rs^WPwOT^ire to be written out 
 in proper form by the pupil, and the amount of the cost and charges 
 found. 
 
 Ex. 17. INVOICE of 100 firkins butter, shipped by Pollock and 
 Blackwell in the Henry of Belfast, James iVL'Millan,, master, for 
 Jamaica, to address of Messrs. Mackay and Dunn, Kingston, by 
 order and for account of John Maxwell & Co. Belfast. r 100 fir- 
 kins, (marked I. M.) neat 55 c. 1 q. 17 Ibs. at 5 17 per cwt. 
 CHARGES. Duty 4 per cent, (on the prime cost:) Cooperage, Iron 
 Hoops, Nails, Cartage, Shipping, Inspection, &c. .16 5: Com- 
 mission at 2^ per cent, on 353 6 3. (Signed) Pollock and 
 Blackwell. Cork, June 28, 1824. Answ. 362 2 11. 
 
 18. INVOICE of 10 bags sea island cotton, shipped on board the 
 
 * Per 100 Ibs ; and so in all the rest. 
 
168 EQUATION OF PAYMENTS. 
 
 Dovvnshire, William Garnett, master, for Belfast, by order, and 
 for account and risk of James Morton, merchant there. (The 
 bags marked P. B. and their number and weight as follows:) No. 
 132, Icwt. 3q. 26 IDS.; No. 290, Ic. 3q. 2 libs.; No. 382, lc. 3q. 
 20 Ibs.; No. 12, 1 c. 3q. 26 IDS.; No. 125, lc. 3q. 271bs.; No. 
 250, lc. 3q. 23lbs.; No. 286, to. 3q. 221bs.; No. 149, lc. 3q. 
 211bs.; No. 225, 2c. Oq. 2lbs.; No. 385, 2 c. q. lib.; draft lib. 
 per bag, tare 2q. 22 Ibs.: price 2/3 per Ib. neat. CHARGES En- 
 try, Town duty, &c. 10/4; Cartage and Porterage, 10/7; Bills of 
 Lading, 3/6; Insurance on .256, at 1 guinea per cent.; Policy on 
 300, at 3/ per cent.; Commission on .243 3 10, at 2 per cent. 
 (Signed) Croker & Co. Liverpool, May 1, 1824. Answ. 249 5 5. 
 
 The draft mentioned in the preceding invoice, is a deduction on cot- 
 ton wool, of the same nature as tare, tret, &c. ; and the policy is an ex- 
 pense attending the effecting of an insurance, and charged on even 
 hundreds of pounds. 
 
 EQUATION OF PAYMENTS. 
 
 WHEN one person owes another several debts payable at 
 different times, the rule which determines the just time for 
 a single payment of the whole, is called EQUATION or PAY- 
 MENTS. 
 
 RULE I. (1.) Multiply each debt by the time that muse 
 elapse before it will become due. (2.) then divide the sum 
 of the products thus obtained by the sum of the debts, and 
 the quotient will be the time required. 
 
 When the products are taken, all the times, and likewise all the debts, 
 must evidently be in the same denomination. 
 
 Exam. I. If one person owe to another' ,300 payable in 4 
 months, 500 payable in 6 months, and 400 payable in 10| 
 months; in what time may the whole be paid without loss to ei- 
 ther party ? 
 
 Here, 300X4+500X6+ 400 X 1<H= 8400, and 300 + 500 + 
 400=1200: then 8400-^-1200:= 7. Hence, the time required, ac- 
 cording to the preceding rule, is 7 months. 
 
 The work may be proved, according to the principle on which the 
 preceding rule is founded, by finding the interest of the first debt at 
 any assumed rate fov> 8 months, the interest of the second at the same 
 rate for 1 month, and the interest of the third at the same rate still 
 for 3|- months : the last of these interests will be found to be exactly 
 equal to the sum of the other two. The times used in the proof, are 
 the differences between the equated time, and the times at which the se- 
 veral debts are due. 
 
EQUATION OF PAYMENTS. 169 
 
 Exam. 2. If a person owe .100 payable at present, and 000 
 payable in 7 months; at what time may both be justly paid at a 
 single payment? 
 
 Here, 100 XO-f- 600 X 7=4200, and 100 + 600=700: ilien 
 4200 -=-700=6, the months required. 
 
 Ex. 1. Required the equated time for the payment of two debts, 
 one of,350 due at the end of 8 months, and another of 600 
 due at 13 months. Answ. llfV months. 
 
 2. If a debt of 45 be payable a't 6 months, another of 70 at 
 
 1 1 months, and a third of 75 at 13 months; what is the equated 
 time for the payment of the whole? Answ. lOff months. 
 
 3. If a debt of 1200 be payable, one-half at 18 months, one- 
 fourth at 15 months, one-sixth at 10 month.s, and the remainder 
 at 3 months, what is the equated time for the payment of the 
 whole ? Answ. 14rj months. 
 
 -k What i.s the equated time for the payment of four debts, the 
 first for 120 due at 1 month, the second for 135 due at 7 
 months, the third for 160 due at 10 months, and the fourth for 
 iOO due at 1 year? Answ. 7 T 5 ^j- months. 
 
 5. What is the just time for the payment of two debts, the one 
 of 139 6 due at the end of 7 months, and the other of 170 
 
 12 6 due at the end of 2J years? Answ. 21 T W 6 T months. 
 
 6. If a debt be payable, one-third at present, one-fourth at 6 
 months, one-fifth at 12 months, and the remainder at 18 months; 
 what is the ecjuated time for the payment of the whole ? Answ. 
 7* months. 
 
 The rule above given, is that which is generally preferred in work- 
 ing questions in Equation of Payments. No rule in Arithmetic, how- 
 ever, has given origin to so warm disputes ; some writers arguing 
 strongly in support of its accuracy, and others entirely condemning the 
 principles from which it is deduced. The following rule is used by those 
 who consider the preceding one erroneous: 
 
 RULE II. (1.) Find the present worths of the several debts at the 
 given or common rate of interest: (2.) then find, by rule VII. page 
 158, in what time, at the same rate, the sum of the present worths 
 thus found, would amount to the sjam of the debts. The result thus 
 obtained will be the time required. 
 
 To renew the dispute on this unimportant subject, and to echo fie 
 arguments that have been advanced on both sides of the question, is in- 
 consistent with the plan of this work. It may suffice to say, that the 
 principle on which the first rule depends, is, that the interest of the 
 money, the payment of which is delayed beyond the time at which it is 
 due, is to be equal to the interest of that which is to be paid, before it 
 becomes due, and that this principle i-> little to the same objections as 
 the common rule for Discount. Jn the second rule, on the contrary, 
 the result depends on tlie present values of the debt; and it appears to 
 be an obvious principle, that all such transactions and agreements should 
 be regulated in conformity to the present value of the money, and the 
 improvement of which it is susceptible. This latter principle, which is 
 
170 PROFIT AND LOSS. 
 
 entirely neglected in the use of the first rule, is acted on in the second 
 in using the rate of interest. When the times are short, however, the 
 difference of the results by the two rules is small, and the first being 
 easier in its application, may be employed with as much propriety as 
 the common rule for Discount. 
 
 The following are the answers of the exercises according to the se- 
 cond rule ; the first at 5, and the rest at 6 per cent, per annum : 
 
 1. 11 mos. 4 days I 3. 14 mos. 17 days I 5. 20 mos. 16^ days 
 
 2. 10 mos. 17 days | 4. 7 mos. 13 days j 6. 7 mos. 17 days. 
 
 PROFIT AND LOSS. 
 
 THAT branch of Arithmetic which treats of the gains or 
 losses on mercantile transactions, is called PROFIT AND 
 Loss. 
 
 The method of performing the first example and the first six exercises, 
 is so obvious as not to require a formal rule. 
 
 Exam. 1. If a hun- 112 Ibs. at 6d. 
 
 dred weight of rice be 
 
 bought for 2 8, and 6d.= \ of 1/0 j 56 
 sold at 6d. per pound, $d,= fa of 6d. | 48 
 what is the gain ? 
 
 Here, the hundred 60/8, or 
 
 weight being bought for Selling price 3 8 
 2 8, and sold for 3 First cost.... 280 
 8, the gain is evident* 
 
 ly 12/8, the difference Gain 12 8, Answer. 
 
 between them. 
 
 Ex 1. If a merchant purchase 93 c. 3 q. 12 Ibs. of rozin, at 9/4 
 per cwt. and pay for charges 3 4 0; what does he gain by sell- 
 ing 27 c. 2 q. at 12/4 per cwt ; 29 c. 1 q. 20 Ibs. at 12/8 per cwt, j 
 and the reminder at 12/9 per cwt.? Answ. 12 2 3|. 
 
 Tin's exerciseJBfeiy be wrought MBfnding the price of 93 c. 3 q. 12 Ibs. 
 at 9/4 per c>t<flaBfc increasing theresult by 3 4 6 ; then by finding 
 the price ?*2w9B^ at 12/4; of 29 c. 1 q. 20 Ibs. at 12/8; and of the 
 remainder, 36-cJ^B^^p Ibs. at 12/9 : and finally, by subtracting the re- 
 sult formerly ohM^Htrom the sum of these prices. It may be wrought 
 more easily howew^^fcr finding the amount of 27 c. 2 q. at 3/0, (the 
 difference 'hetweeviroHaud 9/4 ;) of 29 c. 1 q. 20 Ibs. at 3/4, (the dif- 
 ference between l^/8 ; i^d 9/4 ;) and of 36 c. 3 q. 20 Ibs. at S/5, (the dif- 
 ference between 12/9 and 9/4 :) and finally, by taking 3 4 6 from the 
 sum of tbv'st;. 
 
 2. If a piece of linen containing 25} yards, cost j3 8 3, what 
 is gained by selling it at 3/9} per yard"? Amw. \ 8 5^. 
 
 3. If a pipo of wine containing 138 gallons, cost 113 15, what 
 is gained by selling it at 18/6 per gallon ? Answ. 13 18. 
 
PROFIT AND LOSS. 171 
 
 4. If a chest of tea containing 113 Ibs. be sold at 7/10 per lb.: 
 what is the gain, the first cost and charges being 37 10? Ansio. 
 6 15 2. 
 
 5. If 246 c. 1 q. 11 Ibs. '(long weight,) of pork be bought at 
 2 48^ cwt. and the charges amount to 41; what is the gain 
 or loss, the neat proceeds being 504 ? Answ. Loss, 87 3 3. 
 
 6. If a roll of tobacco containing 25 Ibs. be bought at 3/6 V lb. ; 
 what is gained by selling it at 3d. W ounce, a pound and a half 
 being lost by weighing it out in small quantities, and by drying ? 
 Answ. 
 
 HOLE I. From the prime cost and the selling price, to find 
 the gain or loss per cent. ; as the prime cost is to the gain or 
 loss on that cost, so are 100 to the gain or loss per cent. 
 
 RULE II. To jlnd how a commodity must be sold to vain or 
 lose a certain rate per cent.; as 100 are to the gain or loss 
 per cent, so is the prime cost to the gain or loss on that cost ; 
 and from -this and the prime cost, the selling price will be 
 found by addition or subtraction. 
 
 It should be particularly remarked, that by the gain or loss per cent, is 
 to be understood the sum tliat would be gained or lost at the given rales, not 
 on a hundred pounds' worth sold, but on a hundred pounds laid out in 
 prime cos/, and in charges, if there be any. 
 
 Exam. 2. If tea be bought at 6/6 V lb. and sold at 7/4 V lb. 
 what is the. gain ^^4t. ? 
 
 Here, ^/S^jjjfK'd. the gain on 6/6; then, as 6/6 (the first 
 cost of 1 lb.) : lW(the gain on 6/6) : : 100 (regarded as first 
 cost) : 12Jf, or 12 16 4||, the gain V cent, or the gain on 
 100. 
 
 Exam. 3. How must nutmegs, which cost 17/6 W lb. be sold 
 to gain 16.^"; cent.? 
 
 As 100 (regarded as first cost) : 16 (the gain on 100) : : 
 17/6 (the 'first cost of 1 lb.) : 2/9f, the gain V lb. which being 
 added to 17/6, the amount is 1 3f, the required rate. 
 
 Exam. 4. If cotton wool be bought at 3/1 ^ lb. and sold at 
 2/1 1 V lb. what is the loss ^ cent. ? 
 
 In this exercise there is a loss of 2d. upon the first cost, 3/1 : 
 therefore, as 3/1 : 2d. : : 100 : otf or 5 8 1^, the loss V 
 cent. 
 
 Exam. 5. If sheet lead, which cost 226^ cwt. be sold 
 at a loss of 7 ^ cent.; at what rate ^ cwt. is it sold ? 
 
 As 100 : 7 (the loss on 100) : : 2 2 6 : 2/1 Ifo the loss 
 ^cwt, which being taen from 2 2 6, the remainder, 1 19 6/g, 
 is the selling price required. 
 
 This question may also be wrought thus: as 100 : 93 : : 
 2 2 6 : 1 19 6^ tne selling price as before. The reason of 
 
172 PROFIT AND LOSS. 
 
 this latter mode of operation is, that since there is a loss of 7 V 
 cent, what cost 100, is sold for 100 7, or 93. 
 
 Ex. 7. If whiskey which cost 6/8 V gallon, be sold for 7/JO" 
 gallon, what is gained ^ cent.? ^TWW. 17. 
 
 8. How must wine which cost 15/ V gallon, be sold to gain 
 24 f cent.? Answ. 18/2 1. 
 
 9. If paper which cost 19/2 V ream, be sold for 17/11 Vream, 
 how much is lost ^ cent.? Answ. 6 10 5^ 3 . 
 
 10. How must linen which cost 3/l| per yard, be Sld to gain 
 
 16 per cent.? Answ. 3/7^. 
 
 11. If rum cost 14/6 per gallon, and be sold for 18/ per gal- 
 lon, what is gained per cent. ? Answ. 24 2 9^. 
 
 12. If broadcloth cost I 3 6 per yard, how much is gained 
 per cent, by selling one parcel at 1 8 per yard; and how much 
 by selling another at 1 9 per yard? Answ. 19 2 11|4, and 
 23 8 1*V 
 
 13. If hemp cost 48 7 6 per ton, and be sold for 45 1 per 
 ton, how much per cent, is lost, and how much is lost in the sale 
 of 39 tons, 17 c.? Answ. 5 18 10^* and 114 11 4$. 
 
 14. Isinglass which cost 8/6 per Ib. is sold at a loss of 1 1 per 
 cent.: at what rate is it sold, and how much is lost on the sale of 
 
 17 c. 1 q. at the same rate ? Answ. 7/6$$, arid 90 6 5^. 
 
 15. How must bees* wax which cost 14 5 per cwt. be sold to 
 gain 21 per cent, and how much must be sold at that rate to gain 
 100? Answ. 17 4 10, and 33 c. 1 q. 18lbs. 
 
 16. If paper which cost 129 per ream, be sold at 1 5 6 
 per ream, how much is gained per cent.? Answ. 12 1 9^. 
 
 17. If soap cost 3 5 per cwt. how must it be sold to gain 10 
 per cent, and how to gain 20 and 30 per cent. ? Answ. 3 11 6, 
 3 18, and 4 4 6. 
 
 18. How must butter which cost 5 14 10 per cwt. be sold to 
 gain 19 per cent.? Answ. 6 16 7&. 
 
 19. Bought 2048 yards of linen at 3/2 1 per yard, and sold the 
 whole for 360 19 9: required the whole gain, and the gain per 
 cent. Answ. 32 9 1, and 9 17 6|. 
 
 20. Bought 125 c. 3q. 12lbs. of flour at 1 7 10 per cwt and 
 sold it at 1 10 8 per cwt.: required the whole gain and the gain 
 per cent. Answ. 17 16 7, and 10^f. 
 
 21. Bought 2688 yards of cambric at 8/8 per yard; and sold 
 one-fourth at 10/2, one-third at 10/11 , and the remainder at 
 1 1/4| per yard. Required the whole gain, and the gain per cent. 
 Answ. 304 14 8, and 26 3 2|. 
 
 The following rule, the reason of which will^ appear evident from the 
 exercises, will solve many questions that are usually proposed in Profit 
 and Loss. 
 
 RULE III. From the gain per cent, and the selling price > to 
 find the Jirst cost ; as aSlOO, together with the gain per cent. 
 
PROFIT AND LOSS. m 
 
 or diminished by the loss per cent, are to jglOO, so is the 
 selling price to the prime cost. 
 
 Exam. 6. What was the first cost of flaxseed, which being 
 sold at 3 10 6 per hogshead, the seller clears 13 per cent.? 
 
 As 113 : 100 : : 3 10 6 : 3 2 4 T \V, the first cost re- 
 quired. In this example it is evident, that what cost 100, is 
 sold for } 13; and the analogy used above is no more than this- 
 as the selling price, 113, is to its first cost, 100, so is the sell 
 ling price, 3 10 6, to the corresponding first cost, 3 2 4^. 
 
 Exam. 7. If a merchant, by^ selling pearl ashes at 4- 14. 6 
 per cwt. lose 18 per cent.: what was the prime cost? 
 
 As .100 18 : 100 : : 4 14 6 : 5 15 2ff, the first cost 
 required. In this example it is evident, that what cost 100, 
 must have been sold for 82 ; and therefore the preceding analoev 
 is simply this: as the selling price, 82, is to the corresponding 
 first cost, 100, so is the selling price, 4 14 G, to the corres- 
 ponding first cost, 5 15 2|f. 
 
 Ex. 22. If 11 per cent, be lost by selling 128 yards of broad- 
 cloth for 98 188, what was the prime cost per yard ? Ans. 17/4$. 
 
 23. Suppose a book to be sold for 4 19 9, and 17 per cent. 
 to be gained; what was the first cost? Answ. 4 5 3 T V- 
 
 24. If 13 per cent, be gained by selling tea at 7/4 per lt>. what 
 was the first cost, and what is gained by the sole of 34-9 Ibs. at the 
 same rate? Answ. 6/5 tW, and 14 14 5 
 
 25. What was the first cos of tar which being sol d at l</4 
 per barrel, the merchant loses 9 per cent.; and how much is lost 
 on the sale of 63 barrels ? Answ. 19/0^, and 5 8. 
 
 26. If 13 13 be lost by the sale of 112lbs. of raw silk at 
 2 8 per lb. what was the prime cost, and what is the loss per 
 cent. ? Answ. 2 10 5, and 4|f |. 
 
 Exam. 8. If 7 per cent, be gained by selling linen at 2/9 per 
 yard, how must it be sold to gain 25 per cent. ? 
 
 This exercise might be wrought by finding the first cost, and 
 thence the selling price, as in the preceding examples. It will be 
 wrought more easily however by this analogy: as 107 : 125 : : 
 2/9 : 3/2^y 9 Tj the price required. The reason of this will be ma- 
 nifest, if it be considered, that what cost 100, is in the one case 
 sold for 107, and in the other for 125; and therefore the above 
 analogy is no more than this : as the selling price of one hundred 
 pounds' worth in the one case, is to its selling price in the other, 
 so is the selling price of a yard in the first case, to its selling price 
 in the other. The jpode of working the following exercise de- 
 pends on the same principle. 
 
 Exam. 9. If 15 percent, be gained by selling molasses at 
 2 10 per cwt. how much is gained per cent, by selling it at 
 256 per cwt. ? 
 
174 EXCHANGE WITH 
 
 As 2 10 : 2 5 6 : : 115 : 104 13; and 104 13 100 
 = 4 13, answ. 
 
 Ex. 27. If 6 per cent, be gained by selling broadcloth at 19/4| 
 per yard, how much per cent, would be gained by seJling the same 
 at 1 2 9 per yard ? Answ. 24 9 3$$. 
 
 28. If by selling muslin at 3/1 $ per yard, 2 per cent, be lost, 
 how must it be sold to gain 25 per cent. ? Answ. 3/11 iff. 
 
 29. If 17 per cent, be gained by selling flour at 1 58 per cwt. 
 how much per cent, is gained or lost by selling it at 1 9 per t 
 cwt.? Answ. 5 8 2^, lost. 
 
 30. Bought 21,300 yards of linen at 2/9 per yard, and paid 
 for various charges 88 15 1; and sold one-third at 3/0, one- 
 third at 3/4, and one-third at 3/2 per yard : required the whole 
 gain, and the gain per cent. Answ. 310 12 5, and 10 2 lOf. 
 
 EXCHANGE. 
 
 THE object of EXCHANGE is to find how much of the 
 money of one country is equivalent to a given sum of the 
 money of another. 
 
 By the PAR OF EXCHANGE between two countries, is 
 meant the intrinsic value of the money of the one compared 
 with that of the other, and estimated by the weight and 
 fineness of the corns.* 
 
 The COURSE OF EXCHANGE at any particular time, is the 
 sum of the money of one country, which at that time is 
 given for a fixed sum of the money of another country. 
 This is seldom at par, but is continually varying according 
 to the circumstances of trade.f 
 
 * The par continues the same, so long as the coins in each country are of the same 
 weight, fineness, and nominal value. There is an exception to this, when the standard 
 coins are of different metals. IH this case the par varies as the comparative values of 
 the metals vary. Thus, gold is the standard coin in this country, but in most of the 
 Continent silver is the standard. Now, from the gold mines in America having of late 
 years become less productive, and the silver ones more productive than formerly, the 
 comparative value of bilver has fallen in a small degree ; so that tTie par between Hol- 
 land and England is now about 11| florins for the pound sterling, though it was formerly 
 only 11 ; and a similar reduction has taken place in other countries under similar cir- 
 cumstances. 
 
 f The variation in the course of exchange depends on several circumstances, which, 
 as well as severa^ther questions connected with exchanges, it does not accord with the 
 plan of this Plication to explain, such discussions belqpging properly to Political 
 Economy. Suffice it to say, that whenever, in consequence of a deficient crop in any 
 country, such quantities of corn and other provisions are to be imported, as will exceed 
 the value of the exports to the places from which these articles were brought ; or when, 
 in consequence of war, subsidies are to be paid to foreign powers, or remittances are to 
 be made abroad for the payment of forces en foreign service ; or in a word, when any 
 
AMSTERDAM. 175 
 
 In foreign exchanges, one place always gives another a 
 
 Jixed sum or piece of money for a variable one. The former is 
 
 called the CERTAIN RATE; the latter, the VARIABLE RATE. 
 
 Thus, in exchanges with Portugal, London gives for one milree some- 
 times 65 pence, sometimes 68 pence, &c. In this case, 1 milree is the 
 Jixed price, and 65 pence, 68 pence, &c. the variable price. 
 
 All the calculations in Exchange may be performed by 
 the Rule of Proportion: and the operation may often be 
 abbreviated by the method of aliquot parts. 
 
 RULE I. or GENERAL RULE. Place as the second term 
 in the analogy, that sum whose value is to be found in the 
 money of another country: make that term of the rate 
 which is of the same kind with it, the first term of the ana- 
 logy, and the remaining term of the rate the third term; 
 then work the analogy in the usual way. 
 
 The tables that precede most of the subdivisions of the remaining part 
 of this article, must be employed in the work of almost every exercise. 
 It may be a matter for consideration, whether they should be committed 
 to memory, or only referred to in the book when necessary. If they be 
 committed, the learner will find considerably more facility in performing 
 the calculations; but most pupils will probably soon forget them. 
 
 EXCHANGE WITH AMSTERDAM. 
 
 In Amsterdam accounts are kept in florins, stivers, and pen- 
 nings; and also in pounds, skillings, and grotes, or pence, Flemish. 
 
 TABLE. 8 pennings=l grote or penny Flemish; 2 grotes or 
 pence, or 16 penningszzl stiver; 20 stivers, or 40 grotes=l florin 
 or guilder; 6 florinsrrl pound. 
 
 Also, 12 grotes or pencerzl skilling; 20 skillingszr 1 pound. 
 
 Also, 2 guilders, or 50 stiversrz 1 rix dollar. 
 
 In exchanges between London and Amsterdam, London gives the 
 
 cause makes the imports from a country greater than the exports to it : in all these 
 cases, the bills of the foreign state must increase in value, as none of them will be re- 
 quired to be sent abroad to pay the balance; and thus the money in that country will 
 increase in value, and will continue above par, till coin or bullion is remitted to make 
 up the deficiency ; or till, by a greater exportation or a smaller importation, or both, on 
 the part of the debtor country, She equilibrium of trade is restored. It is of consequence 
 to observe, that the course of exchange can never differ very much from the par, as 
 coin or bullion will be remitted instead of bills, whenever the course of exchange Is 
 uch, that the value of coin or bullion, and the expense of remittin^nd insuring it, 
 would be less than the cost of bills. Besides, when the course of excfiWge is against a 
 country, it affords an inducement to merchants to export to the other country, as the 
 bills which they will get in return will be more valuable, hi consequence of the money 
 of the fore gn country being above par; and thus they can procure a better and sutei 
 market for their commodities, as they wiU be enabled to sell them at a lower price: 
 
176 EXCHANGE WITH 
 
 certain, and Amsterdam the variable sum. Thus, London for 1 pound 
 receives sometimes more and sometimes less Flemish money. 
 
 Exam. 1. Reduce 386 17 4 British to Dutch money, ex- 
 
 change at 37/2 Flemish W pound sterling. 
 
 As .1 : 386 174:: 37/2, or as 240d. : 92848d. : : 3568 pen- 
 nings : 1380340 pennings. By division by 16 and 20, this will 
 become 4313 flor. 11 st. 4 pennings, the answer in florins, &c. 
 but if it be divided by 8, 12, and 20, it will become 718 pounds, 
 18 sk. 6 grotes, 4 pennings, or, as it maybe written, 718 18 6> 
 the answer in pounds, &c. 
 
 This exercise has been wrought without any contraction ; and it may 
 perhaps be proper for the learner to proceed thus for some time, as in 
 this way he will better understand what he does. Abbreviations will 
 soon however readily suggest themselves, particularly in the mode ot 
 reducing the terms. These are left unexplained, because in far the 
 greater number of instances, the use of aliquot parts is much to be pre- 
 ferred, especially in conjunction with the following rule : 
 
 RULE II. To reduce pounds, shillings, and grotes, Flemish, 
 to jlorins and stivers ; Multiply the pounds and skillings by 
 6, adding to the skillings half the grotes or pence before 
 found. 
 
 Thus, resuming the last 386 17 4 at 1 17 2 
 
 example, and taking ali- 193 8 8 for 10 skillings 
 
 quot parts, we find 718 96 14 4 5 _ 
 
 18 6|, the same as before, 38 13 8f 2 _ 
 
 and by multiplying the 3 4 5~~. 2 grotes 
 
 pounds and skillings by 6, il8~18 6%, 
 
 and adding to the skillings g 
 31, the half of 6i: the same 
 re's'ultisfouodL before. 
 
 , pennings 
 
 Exam. 2. Reduce 2475 florins, 10 stivers, 9 pennings to 
 British money, exchange at 36/10 Flemish V pound sterling 
 As 36s. lOg. : 2475fl. lOst. 9p. : : 1 : 224 7. 
 12 20 
 
 442 grotes 49510 stivers 
 8 16 
 
 3536 pen. 792169 pennings 
 
 After the preceding reductions, which are performed according to the 
 table, the answer is found by simply dividing the second term by tlie 
 first, since the third is a unit. 
 
 There are two kinds of money in Amsterdam, called banco, or 
 bank money, and currency, or current money. In the former of 
 these, all bills of exchange are valued and paid. It is of purer 
 
AMSTERDAM. 177 
 
 metal than the currency; and hence it bears a premium of 3 or 4, 
 and sometimes 5 ty cent.: that is, ,100 of bank money is valued 
 at 103, 104, &c. of currency. This premium is called the Agio. 
 
 The method of reducing bank to current money, and current to bank 
 money, will appear from the two following examples : 
 
 Exam. 3. How much currency is equivalent to 798 florins, 17 
 stivers banco, agio at 3 ^ cent. ? 
 
 Here the last term is As 100 : 103 : : 798 f. 17 st, 
 multiplied by 20, to re- 20 
 
 duce it to stivers. Then, 
 
 the operation being con- Stivers 15977 > , . , 
 
 ducted in the usual way, 103 5 m 
 
 the result is 16536 stivers, 
 
 with the remainder 19, 100)1653619! 
 
 which is multiplied bv 
 
 16, and divided by 100', 20)16536 3 
 
 to find pennings, because 
 
 I6pennings=i stiver. Florins 826 16 3y Answer* 
 
 Then, the stivers being 
 
 divided by 20 to reduce them to florins, the answer is found to be 
 826 florins, 16 stivers, 3 pennings. The operation might also 
 have been performed by multiplying the given sum by 3, dividing 
 the product by 100, and adding the result to the given sum. 
 
 Exam. 4. Reduce 439 florins, 14 stivers, currency, to banco* 
 agio at 3 1 IT cent. 
 
 Here, after the necessa- As 103| : 100 : : F.439 14 
 ry reductions and contrac- 44 10 
 
 tionsof the first and second 
 
 terms, the third is multi- 5)415 5)400 
 
 plied by 80 by Compound 
 
 Multiplication ; and by di- 83 80 
 
 viding the product by 83, 83)35 176(423ff.r6s.2p. 
 
 the answer is found to be Answer. 
 
 423 florins, 16 stivers, and 2 pennings^ 
 
 Exam. 5. tleduce 125 14 6 British to Dutch currency r , ex- 
 change at 36/3 Flemish ^ pound sterling, and agio at 4 V cent* 
 
 125 14 6 at 1 16 3 
 In this example, by the 62 1? 3 for 10/ 
 
 method of aliquot parts, 31 g 7 , ^ 
 
 the value in bank money, is 7 17 * j/3 
 
 found to be 227 17 6|. ,.-- .^ .."". ~ 
 
 The agio on this (found by *** l ' | f . 
 
 multipljing by 4 and di- 9 2 "* tor ^ 1O 
 
 viding by 100) is 9 2 3^, ^236 19 10, Answer. 
 
 which being added to 6 
 
 227 17 6, the sum, Flor. 1421 19 stivers, Amwer, 
 
178 EXCHANGE WITH 
 
 236 19 10, is the result in pounds, skillings, and grotes, currency, 
 which by reduction, according to rule IV. becomes 1421 florins, 
 19 stivers. 
 
 Ex. 6. Reduce 2084 florins, 16 stivers, currency, to British 
 money, agio at 4| ^f cent, and exchange 37/8 Flemish W pound 
 sterling. 
 
 By mtiltiplication by 20 and 2, the given sum is reduced to 
 83392 grotes or pence: then, as 104| :"lOO : : S3392d. : 79992d. 
 the value of the given sum in bank money; and, as 37/8 : 79992d. 
 : : 1 : 176 19 5, which is the answer. The reason of the ope- 
 ration is obvious. 
 
 Exercises. Reduce the follwing sums at the given rates Flemish 
 ty pound British. 
 
 Exercises. Answers. 
 
 1. 3486 fl. 17stiv. 1 gr. to British, at 35/3 .... 329 14 6| 
 
 2. 345 1 8 to Flemish, at 33/11 351 1 fl. 4st.7pen. 
 
 3. 4GOOfl. 7stiv. to British, at 35/10 ...&,. 427 8 10| 
 
 4. 159 to Flemish, at 35/3 ..;... 1681fl. 8st. 1 gr. 
 
 5. 235 to Flemisk at 35/6 2502fl. lost. 
 
 6. 722 18 4 tojfeih, at 34/6 ,.... 7482 fl. 3st. 
 
 7. 2406 fl. 17^ st v fc. .British, at 35/4 227 1 3| 
 
 . 5409 fl. 5 st. 5m*. to British, at 34/7 521 7 6 
 
 9. 6270fl. 11 st. ffi-pen. to British, at 34/11... 598 12 6 
 
 ]0. 2617fl. 17stjJkjD British, at 35/4 246 19 4 
 
 Ex. 11. What Mrlie value, in Flemish money, of goods sold in 
 England for gHJifa 6, exchange at 34/6 Flemish V pound 
 British? Answ. Hfisfl. 17st. 4 pen. 
 
 12. Required the value, in British money, of a bill for 7767 fl. 
 17|st. exchange between England and Holland at 35/7 Flemish 
 f pound British. Answ. 726 16 4. 
 
 13. Reduce 1863 rix dollars,* 48| stivers, to British money, 
 exchange between England and Holland at 35/l Flemish V pound 
 sterling. Answ. 442 4 5. 
 
 14. Reduce 279 9 5 British to florins, &c. currency, exchange 
 at 37/10 Flemish W pound British, and agio at 3f ^cent. Answ. 
 3279 fl. 1st. nearly. 
 
 15. Reduce 5179 florins, 16 stivers, currency, to British money, 
 exchange at 36/11 Flemish ^ pound sterling, and agio at 4 
 ^cent, Answ. 449 3 5 *-. 
 
 * R\x dollars being reduced to stivers by multiplying them by 50, and conversely, sti- 
 vers to rix dollars by dividing them by 50, we can as readily reduce rix dollars, o find 
 the value of rix dollars, as we can manage florins. 
 
HAMBURGH. 179 
 
 EXCHANGE WITH HAMBURGH. 
 
 Accounts are kept in Hamburgh in marks, schillings, and pfen- 
 nings ; and also in pounds, skillings, and pence Flemish. 
 
 TABLE. 6 pfennings^ I grote, or penny Flemish; 12 grotes or 
 pencerzl skilling; 20 skillings 1 pound. 
 
 Also, 12 pfennings, or 2 grotes or pence Flemish 1 schilling, 
 Hamburgh money; 16 schillings 1 mark.* 
 
 Hence, 1 skilling Flemish 6 schillings Hambro' money, and 
 the schilling is equal to the stiver. 
 
 In Hamburgh, as in Amsterdam, there are two kinds of money, 
 banco, or bank money, in which exchanges are reckoned, and cur- 
 rency. The agio on the former is high, varying from 18 to 25 $T 
 cent. 
 
 Exam. 7. Reduce .318 7 9 British to Hambro' money, 
 exchange at 35/6 Flemish ^f pound British. 
 
 As 1 : 318 79:: 35/6, or, by reduction, as 240d. : 76413d. 
 : : 4-26 grotes : 135633 grotes, which being divided, according to 
 the table, by 2 and by 16, is reduced to 4-238 marks, 8 schillings, 
 or 4238 m. 8 sch. 6 pf. 
 
 This may also be wrought in the following manner: In 35/6, 
 the rate Ffemish, multiply the skillings by 6, and add to the pro- 
 duct half the grotes; the sum 213 is the schillings, Hambro' mo- 
 ney, which are equivalent to 35/6 Flemish, as is evident from the 
 table. Then by multiplying this by 318 for the pounds of the 
 given sum, and taking aliquot parts for 7/9, we find for the an- 
 swer 67816 schillings, 6| pfennings, or, by reduction, 4238 m. 8 sch. 
 6 pf. the 3fene as before. 
 
 Exam; >. Reduce 5127 marks, 5 schillings, Hambro', to 
 British money, exchange at 36/2 Flemish ^ pound sterling. 
 
 As 36/2 : 5127 m. 5 sch. or, by reduction, as 217 schillings : 
 82037 schillings : : 1 : 378 1, Answer. 
 
 Exam. 9. Reduce 86 17 1 British to Hambro' currency ', 
 exchange at 35/8 Flemish per pound sterling, and agio at 22 per 
 cent. 
 
 This being wrgught as the 7th example, there will result 18586 
 schillings, 9 pfennings, the value in bank money. Then for cur- 
 rency, multiply by 22 and divide by 100, and there will result for 
 the agio 4089 sch. 1 pf. which being added to the bank money al- 
 ready found, the sum, 22675 sch. 10 pf. or 1417 m. 3 sch. 10^ pf. 
 is the sum required in currency. 
 
 Exam. 10. Reduce 7854 marks, 7 schillings, Hambro' cur- 
 
 * Also 2 marks 1 dollar of exchange, 3 marksrrl rix dollar. Hamburgh money was 
 formerly distinguished by the word tubs, from Lubec, the place where it wa< comet!. 
 Thus, in most bunks on Arithmetic, .we find mark lubs instead of mark, frc. This how- 
 ever appears now to be laid aside among men of business, the word Hamitro' bcin_: jjc^-- 
 rally used injreferec"e. By the examples it will be seen, tltat Londou gives the cerium, 
 and HamUufehMf u/tCfrfeJtfl price 
 
180 EXCHANGE WITH 
 
 rency, to British money, exchange at 34/1 1 Flemish per pound 
 sterling, and agio at 21 per cent. 
 
 The given sum is equivalent to 125671 schillings; then, as 12! : 
 100 :: 125671 sch. : 103860 schillings nearly, banco; and, again, 
 as 34/11, or its equivalent, 209 schillings : 103S60 schillings : : 
 1 : 495 15 Of British, Answer. 
 
 Exercises. Reduce the following sums at the given rates, Flem- 
 ish per pound British: 
 
 Exercises. Answers. 
 
 16. 7563 m. 8 sch. 8 pf. to British, at 34/6... 584 12 5 
 
 17. 275 12 6 to marks, &c. at 35/3 3643m. 6 sch. If g. 
 
 18. 346 18 10 to marks, &c. at 36/1 .... 4700 m. 
 
 19. 2468 m. 5 sch. to British, at 33/8 195 10 2 
 
 20. 496 12 8 to Hambro' at 35/6 6611 m. 6 sch. If g. 
 
 21. 6487 m. 13 sch. 4 pf. to British at 35/H.. 492 11 
 
 22. 543 16 10 to Hambro' at 35/8 "... 7273m. 14 sch. 
 
 23. 8236 m. 10 sch. to British, at 35/11 611 10 8 
 
 Ex. 24.. Reduce 428 14 9 British to Hambro' currency, ex- 
 change at 36/4 Flemish per pound British, and agio at 22 per 
 cent. Answ. 7155 m. 14 sch. g. 4 pf. 
 
 25. Reduce 5364 m. 13 sch. Hambro' currency to British money, 
 exchange at 35/9 Flemish per pound British, and agio at 23 per 
 cent. Answ. 325 6 10. 
 
 26 Required the value, in Hambro' bank money, of a bill for 
 721 13 10 British, exchange between England and Hamburgh 
 at 35/3 Flemish per pound British. Answ. 9539m. 13 sch. HJ-g. 
 
 27. Reduce 6748m. 1 1 sch. Hambro' currency, to British money, 
 exchange at 36/10 Flemish per pound sterling, and agio at 23 per 
 cent. Answ. 397 4 7$, 
 
 EXCHANGE WITH FRANCE. 
 
 Accounts are kept in France in francs and centimes. They were 
 formerly kept in livres, sous, and denier s. 
 
 The livres, sous, and deniers were formerly used exclusively in accounts 
 in France, but now the francs and centimes are employed, except in 
 small transactions. The livre also had formerly the name of franc ; l>ut 
 in a coinage during the time of the Republic, by some mistake in the 
 mint, the 5 livre pieces were made too heavy, each being worth 101^ sous 
 instead of 100, and the error made in the one coin was extended to the 
 rest. The coin corresponding to the former livre was called the Jrai^ 
 and the name livre appropriated exclusively to the old coin ; and to fa- 
 cilitate calculations, the old division, (similar to the divisions of our 
 coins) was laid aside, and the franc was divided decimally, as will appear 
 from the first of the following tables. 
 
 TABLES I. 10 centimesirl decime; 100 centimes, or 10 deci- 
 
 flics=l tranc. 
 
FRANCE. 181 
 
 II. 12 cleniers = 1 sou; 20 sous =r 1 livre j 3 livres = 1 ecu, or 
 crown Tournois; 81 livres = 80 francs. 
 
 In exchange with France, England gives the certain for the variable 
 price. The great advantage of the decimal division of money, will i* 
 felt in the work of the examples and exercises in exchange with France 
 and Portugal. 
 
 Exam. 11. Reduce 274 5 9 British to francs, &c. exchange 
 at 23 francs, 57 centimes & pound sterling. 
 
 As 1 : 274, 5 9 : : 23 f. 57 c. or, by reduction, as 240d. : 
 65829 d. : : 2357 centimes : 646496 centimes, or 6464 francs, 96 
 centimes. 
 
 2357 
 274 5 9 
 
 This and other exer- 
 
 cises of the same kind, 9428 
 
 may be wrought by the 16499 
 
 method shown in the 4714 
 
 margin, the division by 589^ for 5/ 
 
 100 being performed 59 0/6 
 
 merely by cutting off the 29 0/3 
 
 two figures. 
 
 646495f cents, or 
 6464 f. 95| c. 
 
 Exam. 12. Reduce 2867 fr. 48 centimes to British money, 
 exchange at 24 f. 36 c. W pound sterling. 
 
 As 2436 centimes : 286748 centimes : : 1 : 117 14 3, Answ. 
 
 Exercises. Reduce the following sums at the given rates, French, 
 ty pound British : 
 
 Exercises. Answers. 
 
 28. 184 18 4 to French, at 23 f. 49 e. 11390 f. 69| c. 
 
 29. 9564 f. 33 c. to British^ at 23 f. 57 c 405 15 8 
 
 30. 675 18 3 to French, at 23 f. 15 c 15647 f. 37 c. 
 
 31. 4260 f. 75 c.to British, at 23 f. 50 c. 181 6 2 
 
 32. 6499 f. 67 c. to British, at 23 f. 40 c 277 15 3 
 
 33. 868 17 6 to French, at 23 f. 60 c 20505 f. 45 c. 
 
 34. 21 167 f. 83 c. to British, at 23 f. 34 c. . 906 IS 8 
 
 35. 488 19 lO^to French, at 24 f. 1 c. ... 11260 f. 54 c. 
 
 Ex. 36. Required the value, in French money, of a bill for 
 483 16 11, exchange at 24 francs 37 centimes ^ pound British. 
 Answ. 11791f. 34c. 
 
 37. Required the value, in British money, of goods sold for 
 75647 francs, 18 centimes, exchange at 24f, 51 c. per pound sf.er- 
 lin. Answ. 3086 7 7. 
 
182 EXCHANGP; WITH 
 
 EXCHANGE WITH PORTUGAL. 
 
 In Portugal, accounts are kept in milrees and rees. 
 
 TABLE. 1000 rees zz 1 milree. 
 
 Also 400 rees 1 crusado, and 4800 rees zz 1 moidore : con- 
 sequently, 2^ crusados zz 1 milree. 
 
 Exam. "13. Reduce 64 10 10 British to milrees, &c., ex- 
 change at 5/3 sterling per milree. 
 
 As 5/3^ : 64 10 10 : : 1 milree : 243 milrees, 937 rees, Answ. 
 Exam. 14. Reduce 5236 milrees, 266 rees, to British money 
 exchange at 5/4 per milree. 
 
 As 1000 rees : 5236266 rees : : 64d. : 1396 6 9, Answer. 
 
 rp, . , . ., 5236-266 
 
 This and similar ques- 
 
 1047-2532 
 349-0844 
 1396-3376, or 
 
 tions may be very easily tw 1 f 
 
 wrought as in the margin, V^ 1 3 * 
 
 ;by the method of aliquot 
 
 parts, and by decimals. 1396 6 9, Answ. 
 
 Exercises. Reduce the following sums at the given rates per 
 milree : 
 
 Exercises. Answers. 
 
 38. 479 17 6 to Portuguese, at 64^d 1785 m. 581 r. 
 
 39. 2486 milrees, 560 rees, to British, at 63|d 655 6 2f 
 
 40. 594 ra. 480 r, to British, at 64d 158 16 9 
 
 41. 512 8 9 to Portuguese, at 62|d 1971 m. 703 r. 
 
 Ex. 42. If goods worth 236 14 8 British be consigned to 
 
 Portugal, what is their value in Portuguese money, at 5ld. per 
 milree? Answ. 962m. 990 r. 
 
 43. If port wine be bought for 3245 milrees, 435 rees, \\hat is 
 the cost in British money, at 6 Id. per milree, 6 per cent, being 
 paid for insurance? Answ. 874 7 5|. 
 
 EXCHANGE WITH SPAIN. 
 
 In Spain there are two kinds of money called plate and vellon ; 
 and accounts are kept in both in piastres, reals, and maravedies. 
 The piastre is also called the pezza, the dollar of exchange ; and 
 the piece of eight, or of f . 
 
 Plate money is more valuable than vellon in the ratio of 3-2 to 
 17. Thus, 17 reals plate are equivalent to 32 reals vellon. Plate 
 only is used in exchanges with England. 
 
 Hence, as 17 is to 32, so is any sum, plate, to the sum, vellon, 
 equal to it; and as 32 is to 17, so is any sum, vellon, to the equiva- 
 lent sum, plate. 
 
 TABLE. 34 maravedies zz 1 real ; 8 reals zz piastre. 
 
 Also 4 piastres zz 1 pistole of exchange ; 375 maravedies 1 
 ducat. 
 
 Exam. 15. Reduce 406 3 9 British to Spanish money, ex- 
 change at 39^-d. per piastre. 
 
AMERICA AND THE WEST INDIES. 183 
 
 As 39 : .06 3 9, or by reduction, as 79 halfpence : 194970 
 halfpence : : 1 piastre : 2467 piast. 7 reals, 27 mar. Here England 
 gives the uncertain price. 
 
 Exam. 16. Reduce 7511 piast. 5 reals to British money, ex- 
 change at 40d. per piastre. 
 
 As 1 p. : 7511 p. 5 r. : : 40d. : rfJ267 11 8J. 
 
 The second mode would 7511 n ^Vlt q/4,1 
 
 have been wrought rather /o11 ?' r * at 3 /4i 
 
 more easily by first finding ^25116 8 for 3/4 
 
 the value at I per piastre, , ~ , , , . ' 
 
 (See RULE IX. page 132.) lo ! f ^f - f; 
 
 This value would have been 4 * r aJS * 
 
 7511 12 6; 5 reals being __ _ 
 
 five-eighths of a piastre. J26? n 
 
 Exercises. Reduce the following sums at the given rates per 
 piastre : 
 
 Exercises. Answers. 
 
 44. 936 p. 4 r. 15 m. to British at 38d ......... 150 4 9 
 
 45. 167 15 4 to Spanish, at 39d ............... 1025 p. 6 r. 23 m. 
 
 46. 809 9 8 to Spanish, at 40^d .............. 4767 p. 4 r. 2 m. 
 
 47. 5318 p. 5 r. 24 m. to British at 38|d ........ 858 15 
 
 Ex. 48. Reduce 263 18 9 British to Spanish money, vellon, 
 exchange at 41fd, per piastre, plate. Answ. 2855 p. 7r. 31 m. 
 
 49. Reduce 8429 piastres, 7 reals, vellon, to British money, ex- 
 change at 40d, per piastre, plate. Amw. 760 7 9. 
 
 50. Required the value, in Spanish money, of a bill for 
 363 18 10| British, exchange at 38|d. per piastre. Answ. 
 2254 p. Or. 28m. 
 
 51. Reduce 1927 piastres, 3 reals, vellon, to British money, 
 exchange at 39fd. per piastre, plate. Answ. 169 11 Sf. 
 
 EXCHANGE WITH AMERICA AND THE WEST INDIES 
 
 In the United States, accounts are pretty generally kept" in dol- 
 lars, which are subdivided each into 100 cents, or into 10 dimes, each 
 containing 10 cents. In several parts of the United States, how- 
 ever, and in the British possessions in America, accounts are kept, 
 as in Britain and Ireland, in pounds, shillings, and pence. 
 
 American pounds, shillings, c. are generally distinguished by the 
 term currency annexed. These pounds, in consequence of the scarcity 
 of coins, and the use of paper money instead of them, are of much 
 smaller value than pounds British. 
 
 When exchanges with America are made in pounds, the calculations 
 are conducted in the same way as in exchanges between England and 
 Ireland : but when they are made in dollars, the calculations uroceed in. 
 the same manner as in exchanges with France or Portugal. 
 
1R4 EXCHANGE WITH 
 
 Exam. 17. Reduce 946 17 10 British, to American cur- 
 rency, at 77 ty cent. 
 
 As 100: 177:: 946 17 10: 1675 19 11^, currency. The 
 answer might also be found by taking aliquot parts ; or by multi- 
 plying by 77, dividing the product by 100, and adding the quotient 
 to the given sum. 
 
 Exam. 18. Reduce 796 13 6 American currency, to Brit- 
 ish money, exchange at 69 V cent. 
 
 As 169 : 100 : t 796 13 6 ; 471 8 l, British, Answer. 
 
 Exam. 19. Reduce 2746 dollars, 30 cents, to British money, 
 exchange at 4/3 British, V dollar. 
 
 As 100 cents : 274630 cents : : 4/3| : 589 6 2|, British, Answ. 
 The answer might also be found very easily by the method of 
 aliquot parts. 
 
 Exam. 20. Reduce492 3 4 British, to dollars, &c. exchange 
 at 4/7 British <$" dollar. 
 As 4/7 : 492 3 4 : : 1 dollar : 2147 dollars, 64 cents, Answer. 
 
 Exercises. Answers. 
 
 52. Reduce 382 8 Brit, to Amer. cur. at 71, &c. 653 18 1 
 53 5611 dol. 42 cents, to Brit, at 4/5|F d. 1250 17 7 
 
 54. 846 10 9 currency, to Brit, at 66, &c. 509 19 3 
 
 55. 936 5 4 Brit, to dollars at 4/6 ^ d. 4123 d. 1 c. 
 
 Ex. 56. What sum, American currency, is a bill for 367 4 
 British, exchange between England and America at 52 ^cent.? 
 Answ. 558 2 10. 
 
 57. Reduce 7593 dollars, 70 cents, to British money, exchange 
 between England and America at 4/7 ^ dollar. Answ. 1740 4 5. 
 
 If the pupil be made fully acquainted with what precedes respecting 
 exchanges, he will find little difficulty, with the assistance of tables, in 
 vplying the same principles to similar cases, which it would exceed the 
 limits of the present Publication to illustrate individually. To facilitate 
 this, the following tables are annexed, which will be found to contain 
 what is most useful and necessary on the subject* 
 
 LEGHORN. 12 denari* di pezza =r 1 soldo di pezza : 20 soldi di 
 pezza = 1 pezza of 8 reals. 
 
 Also 12 denari di lira = 1 soldo di lira; 20 soldi di lira = 1 
 lira ; 5f lire, moneta buona = 1 pezza of 8 reals. 
 
 GENOA. The same table as for Leghorn serves for Genoa. Be- 
 sides this, 4 lire and 12 soldi = 1 scudio di cambio, or crown of 
 exchange; 10 Ike and 14 soldi = 1 scudio d'oro marche, or gold 
 crown. 
 
 NAPLES. 10 grains == 1 carlin ; 10 carlins, or 100 grains rr I 
 ducat regno. 
 
 * Denari is the plural ofdenaro, soldi of soldo, lire oflira t and pezxe of pezaa 
 
ITALY, &c. 185 
 
 VENICE. 12 denari = 1 soldo; 20 soldi = 1 lira; 6 lire and 4 
 soldi :=: 1 ducat current, or of account; 8 lire = 1 ducat effective. 
 
 PETEESBURGH. 100 copecs == 1 ruble. 
 
 VIENNA. 4 pfenings =: I creutzer ; 60 creutzers == 1 florin ; 90 
 creutzers, or 1 florin = 1 rix dollar of account. 
 
 STOCKHOLM. 12 fenings, or oers = 1 skilling; 48 skillings =r 1 
 rix dollar. 
 
 COPENHAGEN. 12 pfenings = 1 skilling; 16 skillings = 1 mark; 
 6 marks Danish, (or 3 marks Hambro') = 1 rix dollar. 
 
 The following exercises, which the pupil will find to be easily resolved, 
 will serve to illustrate these tables. 
 
 Ex. 5a Reduce 2467 pezze, 12 soldi, 6 denari, of Leghorn, to 
 British money, exchange at 50 pence sterling ^ p. Arts. 51 4 1 9^. 
 
 59. Reduce 467 14 8 British to pezze, &c. of Leghorn, ex- 
 change at 49 pence V pezza. Answ. 2290 p. 18 s. 9d. 
 
 60. Reduce 947 8 6 British to pezze, &c. of Genoa, exchange 
 at 47 pence sterling V pezza. Answ. 4837 p. 18 s. 3 d. 
 
 61. Reduce 4263 pezze, 16 soldi, 8 denari of Genoa to British 
 money, exchange at 44d. sterling W pezza. Answ. 790 11 8i. 
 
 62. Reduce 786 4 8 British to Naples money, exchange at 
 41 d. sterling, V ducat regno. Answ. 4546 d. 8 c. 9 gr. 
 
 63. Reduce 3794 ducats, 4 carlins, 5 grains, Naples currency, 
 to British money, exchange at 40f d. fy ducat. Answ. 644 5 3|. 
 
 64. Reduce 170 10 British to lire, &c. of Venice, exchange at 
 47 pence sterling V lira. Answ. 870 1. 12 s. 9 d. 
 
 65. Reduce 793 lire, 12 soldi, and 6 denari of Venice, to British 
 money, exchange at 45 pence sterling ty lira. Answ. 148 16 1. 
 
 66. Reduce 190 19 3 British to rubles, &c. exchange at 3/1 
 sterling fr ruble. Answ. 1222 r. 16 c. 
 
 67. Reduce 31416 rubles, 20 copecs, to British money, ex- 
 change at 3/4 sterling V ruble. Answ. 5301 9 8. 
 
 68. Reduce 255 118 British to rix dollars, &c. of Vienna, ex- 
 change at 4/4 sterling ty rix dollar. Answ. 1179 rix d. 55 cr. 1 pf. 
 
 69. Reduce 867 rix dollars, 25 creutzers, of Vienna, to British 
 money, exchange at 4/7 ^ rix dollar. Answ. 198 15 0^. 
 
 70. Reduce 1045 6 8 British to Swedish money, exchange at 
 4/3 ^ rix dollar. Answ. 4919 rix d. 10 sk. 4 f. 
 
 71. Reduce 1318 rix dollars, 20 skillings, 6 fenings, Swedish, to 
 British money, exchange at 4/6 sterling V rix d. Answ. 296 12 1 U 
 
 72. Reduce 313 9 7 British to Danish money, exchange at 
 per rix dollar. Answ. 1433 rix d. 4 sk. 7 pf. 
 
 73. Reduce 576 rix dollars, 4 Danish marks, 4 skillings, to 
 Brit, money, exchange at 4/4 sterling per rix cl. Ans. 124 19 Of. 
 
 74. In 1795, the values of the imports and exports of Peters- 
 burgh were 31,767,952 and 23,019,175 rubles respectively. Re- 
 quired the values in British money, exchange between Petersburgh 
 and London at 2/1 1 per ruble. Answ. 4,699,009 11 4, and 
 3,404,919 12 8. 
 
186 ARBITRATION, &c. 
 
 75. At the end of the reign of Louis XIV. the values of the 
 imports and exports of France, were 71,000,000, and 105,000,000 
 livres respectively ; and at the period of the French Revolution, they 
 had respectively increased to 380,000,000, and 424,000,000 livres. 
 Reduce these several values to British money, exchange at 25 livres, 
 9 sous, 9 deniers, per pound sterling. Answ. 2,785.679 5 1 ; 
 4,119,666 10 Of; 14,909,269 4 llfj and 16,635,605 13 9i, 
 
 76. In 1784, the value of the exports from Spain to America was 
 computed at 4,348,078 British ; and the value of the imports was, 
 in money and jewels, 9,291,237, and in merchandise 3,343,936. 
 Required the value of each of these sums in Spanish money, 
 exchange at 38^ d. British per dollar. Answ. 27,104,90 I 
 57,919,399f3 del.; and 20,845,3 1 
 
 ARBITRATION OF EXCHANGES. 
 
 WHEN the courses of exchange between the first and the 
 second, the second and the third, the third and the fourth, 
 &c. of any number of places, are given; the method ot 
 finding the course of exchange between the first place and 
 the last, corresponding to these courses, or of valuing any 
 sum of the money of the first place in that of the last 
 through the medium of the others, is called ARBITRATION 
 
 OF EXCHANGES. 
 
 As the actual course of exchange between the first place and the last 
 is almost always, from various circumstances, different from the arbi- 
 trated course, this method is of use in enabling a merchant in one place 
 to discover whether he should draw and remit directly between his own 
 place and another, or circuitously through other places. 
 
 All the operations may be performed by one or more 
 analogies in the Rule of Proportion. 
 
 The method by the Rule of Proportion is easy and intelligible, when 
 there are only three places concerned, or in what has been termed SIMPLE 
 ARBITRATION. But when more than three places are concerned, or in what 
 has been called COMPOUND ARBITRATION, the following rule, usually 
 called the CHAIN RULE, is generally preferable. 
 
 RULE (1.) Let all the quantities of the same kind be re- 
 iuced to the same denomination, ir they be not so already. 
 (2.) Let a blank* be left for the required quantity, and to 
 
 The interrogative mark (?) may properly be placed in the blank. 
 
 This rule will be found considerably more easy in its application, than those usually 
 given for the same purpose in books on Arithmetic. It is also expressed in general terms, 
 o as to serve not only for the purposes of exchange, but for the comparison of weigbu 
 and measures, and for any other uses to which it can be applied. 
 
ARBITRATION, &c. J87 
 
 the right of it place as consequent the term to which it is to 
 be equivalent: then, below the blank, place as antecedent 
 the other given term which is of the same kind as the last 
 consequent, and to the right of it place as consequent the 
 term which is equivalent to it. (3.) Proceed thus, till 
 all the terms are arranged in two vertical columns: then, 
 divide the continual product of the consequents by the 
 continual product of the antecedents, and the quotient will 
 be the result required. 
 
 The operation may often be much abridged by striking 
 out any antecedent and consequent that are equal, and by 
 reducing to their lowest terms such as admit a common 
 divisor.* 
 
 Exam. 21. When the exchange between England and America 
 is at 80 per cent, and between England and Amsterdam at 36/4 
 Flemish per pound British; what is the arbitrated course of ex- 
 change between America and Amsterdam ? 
 
 Here the American Flemish ? =.1 American, 
 
 money is in pounds, American j180=100 English, 
 
 and also the English. English .1 =436d. Flemish. 
 
 Then the Flemish be- 100X436_20X I09_ , , __ 2f)A2 . 
 
 ing reduced to pence, jgO 9 ' ' 
 the answer is found in 
 
 pence Flemish, and becomes, by reduction, 20/2$ Flemish per 
 pound American. 
 
 This question might have been solved by a single analogy. Thus, 
 as 180 : 100 t 436d. : 242fd. In this mode, we multiply and 
 divide by the same quantities as in working by the chain rule; and the 
 same is the case in all applications of this rule. This not only shows 
 the correctness of the rule, but also that it is nothing else than Simple 
 or Compound Proportion exhibited in a different, and, in many cases, 
 in a more convenient form. The learner will perceive, that, in the ar- 
 rangement of the terms by the chain rule, the first antecedent and the 
 last consequent are always of the same kind. 
 
 Ex. 77. What is the value of a franc in American money, ex- 
 change between England and America at 50 per cent, and between 
 England and France at 24 francs, 87 centimes, per pound sterling? 
 Answ. Hfffd. 
 
 Exam. 22. Suppose a merchant in England owes a merchant 
 in Portugal 572, whether is it better for the Portuguese mer- 
 
 * The application of logarithms greatly facilitates, in many cases, the operations by the 
 chain rule. In using them in this way, the contraction above mentioned is generally of 
 little use, unless when terms can be rejected. 
 
188 PAR OF EXCHANGE. 
 
 ciiant to have a direct remittance from London to Lisbon at 68d. 
 per milree, or a circular remittance through Amsterdam and Paris, 
 exchange between London and Amsterdam being at 37/3 Flemish 
 per pound sterling ; between Amsterdam and Paris at 56 pence 
 Flemish for 3 francs; and between Paris and Lisbon at 460 
 rees for 3 francs, an expense of 1 per cent, being incurred in the 
 circular course ? 
 
 Portuguese ? = 572 English. 
 English 1 = 37/3 Flemish =r 447d. 
 Flemish 56d. = 3 Francs. 
 Francs 3 = 460 rees, Portuguese. 
 
 100 98 
 
 572 X 447 X 3 X 460 X 98_ 143 X 447 X 46 X 197 
 
 56 X 3 X 100 28 X 10 
 
 3 
 
 the sum Portuguese by the circular course. Again, as 68d. : 
 572 : : 1 milree : 2018 milrees, 824 rees, the sum Portuguese by 
 direct remittance. Hence the circular exchange will be more ad- 
 vantageous to the Portuguese merchant, as by it he will receive 49 
 milrees, 933 rees more than by the direct. Below the antecedents 
 and consequents we place 100 and 98, and multiply by them, to 
 modify the result according to the loss W cent. ; 98 being equal 
 to 100 1^. 
 
 Ex. 78. Suppose, that a merchant in Russia owes a merchant 
 in London 12000 rubles, and that the course of exchange between 
 London and Petersburgh is 36d. V ruble : it is required to find 
 how much more or less profitable it is for the London merchant to 
 draw directly on Petersburgh, or to draw through Paris, Amsterdam, 
 Hamburgh, and Vienna, the course of exchange between London 
 and Paris being at 24 f. 55 c. $P pound sterling ; between Paris 
 and Amsterdam at 55 grotes Flemish for 3 francs ; between 
 Amsterdam and Hamburgh at 33 stivers for 2 marks Hambro' ; 
 between Hamburgh and Vienna at 100 rix dollars Hambro' for 
 299 rix dollars of Vienna ; and between Vienna and Petersburgh 
 at 185 creutzers $* ruble. Answ. The direct remittance is better 
 by 10 7 11. 
 
 The following useful table, with the remarks which are subjoined to 
 it, will form a proper conclusion to this article. It will be easily under- 
 stood by a reference to the tables already given. 
 
 " TABLE of the intrinsic par of exchange between London and 
 the principal places in Europe, 'gold against gold, and silver 
 against silver,' calculated from assays lately made both in London 
 and Paris :" 
 
PROPORTIONAL PARTS. 
 
 189 
 
 Amsterdam, banco 
 Rotterdam, cur. ... 
 Hamburgh, banco. .. 
 Paris, old coins ... 
 , new coins ... 
 
 GOLD. 
 
 SILVER. 
 
 35/10-8 
 
 37/1-7 
 11 flor. 14 st. 
 35/1 Flemish 
 25 liv. 9s. 9 den. 
 24 f. 87 c. 
 45*82 pence, ster. 
 4fi-/7 
 
 11 flor. 4 st 
 
 34/2-4. Flemish 
 25 liv. 9s. 9 den... 
 25 francs, 26 c 
 45-52 pence, ster.... 
 49-0 .._.... _ . 
 
 
 
 42 
 
 41 1 
 
 
 fifi-5 ____ 
 
 fiS'4, 
 
 Madrid and Cadiz 
 Venice 
 
 3fi-n* 
 
 39 ., 
 
 46-38 
 
 48-9 
 
 . In addition to the information contained in the preceding table, 
 it may be remarked, that the American dollar, and the rix dollars 
 of Vienna, Copenhagen, and Stockholm, are each equivalent to 
 about 4/6 British at par. 
 
 The rubles in use in Russia prior to 1762, were of different 
 values, from 3/7 to 3/9 British, valued according to their weignts ; 
 but those coined since that time have been made lighter, and their 
 values have been about 3/2 or 3/3 British. 
 
 The East India coin which is most frequently mentioned, is the 
 rupee. Of this there are two kinds the current rupee, and the 
 sicca rupee. The value of the former, according to the mint price 
 of silver in England, is 21d., or more accurately, 21-177d. ; and 
 that of the latter 24d., or more accurately, 24-566d. The cur- 
 rent is to the sicca rupee, as 100 to 116. In India, the market 
 price of rupees is generally much higher than their intrinsic value, 
 the current rupee being worth about 2/, and the sicca rupee nearly 
 2/6. A lack of rupees is 100,000 rupees, or about 10,000 British, 
 and a crore is 100 lacks. 
 
 DIVISION INTO PROPORTIONAL PARTS. 
 
 RULE. To divide a given quantity into parts which shall 
 have to each other given ratios : as the sum of the numbers 
 expressing the ratios, is to any one, of these numbers, so 
 is the entire quantity to be divided, to the part correspond- 
 ing to the number used as the second term of the analogy. 
 
 The operation is proved by adding the several results together : 
 if the sum be equal to the quantity to be divided, the work is right. 
 
 When all the parts, except one, have been determined, that one may 
 be found by adding the rest together, and taking the sum from the 
 number to be divided. In this case the operation will be proved by 
 finding that part by the general rule : the agreement of the two results 
 will prove the accuracy of the work. 
 
190 DIVISION INTO 
 
 The reason of this rule, which is of frequent use In Arithmetic, 
 and other branches of the Mathematics, is obvious from the princi- 
 ples delivered in page 82. 
 
 Exam. A farm of 97 a. 3 r. 5 p. is to be divided into two parts, 
 such that the one may be three-fourths of the other. What are 
 the parts ? 
 
 Here, the parts are evidently in the ratio of 4 and 3, the sum of 
 which is 7. Therefore, as 7 : 4 : : 97 a. 3 r. 5 p. : 55 a. 3 r. 20 p., 
 the greater part j and as 7 : 3 : : 97 a. 3 r. 5 p. : 41 a. 3 r. 25 p., 
 the less part. The sum of these parts is 97 a. 3 r. 5 p., which 
 proves the operation. 
 
 Ex. 1. Suppose a traveller to proceed from Belfast for KJllarney 
 at the rate of 6 miles ^ hour, and another at the same time from 
 Killarney for Belfast at the rate of 5 miles ^ hour : where 
 will they meet, the distance between the two places being 224 
 miles? Answ. 122 T \ miles from Belfast. 
 
 2. Divide 398 into three parts which will be to one another as 
 the numbers 5, 7, and 11. Answ. 86|f, 121 5 3 g-, and 190 5 8 5 -. 
 
 3. Divide 80 miles into four parts in the ratio of 10, 9, 8, 
 and 7. Answ. 23 T 9 T , 21 T 3 T , \8tf-, and 16W- 
 
 4. Divide 5000 among three persons in such a manner, that 
 the share of the second may be one-half greater than that of the 
 first, and the share of the third, one half greater than that of the 
 second. Answ. 1052 12 7^, 1578 18 H T V,and 2368 8 5 T V 
 
 In this exercise it is easj T to see, that the parts will be as the numbers 
 1, 1, and 2^, or as 4, 6, and 9. 
 
 5. Pure water is composed of two gases, or kinds of air, called 
 oxygen and hydrogen, in such proportions, that the weight of the 
 former is to that of the latter as 15 to 2 Required the weight of 
 each contained in a cubic foot, or 1 000 ounces, avoirdupois weight, 
 of water. AIISUJ. 882 T V oz., and 117J4 oz. 
 
 6. How much copper and how much tin will be required to make 
 a cannon weighing 16 c. 1 q. 20 Ibs., gun-metal being composed 
 of 100 parts of copper, and 11 of tin? Answ. 14 c. 3 q. 5 T 7 T 3 i tt>s., 
 and 1 c. 2 q. 14 T \\ tbs. 
 
 7. The British standard gold for coinage consists of 11 parts of 
 pure gold, and 1 part of alloy ; (usually a mixture of silver and 
 copper:) how much pure gold and how much alloy are contained in a 
 guinea? (See exercise 77, page 80.) Antw. 4 dvvts. 22^ gr. & 10$ grs. 
 
 8. The British silver coin consists of 37 parts of silver and 3 
 of copper; how much of each does the half crown (2/6) con- 
 tain, each pound, troy weight, being coined into 66 shillings ? 
 Answ. 8 dwts. 9^ T grs., and 16 T 4 T grs. 
 
 9. How much tin and copper are contained in a bell weighing 
 )50 Ibs, bell-metal being composed of three parts of copper and I 
 of tin ? Answ. 112 Ibs., and 37 Jbs. 
 
 
PROPORTIONAL PARTS. 191 
 
 10. Pewter is composed of 112 parts of tin, 15 of lead, and 6 of 
 brass ; how much of each ingredient is requisite to make a ton of - 
 pewter ? Answ. 16 c. 3 q. 10 T % tbs.,2c. 1 q. 0\% ft), and 3 q. 17^ tbs. 
 
 11. Proof spirits are composed of 48 parts of alcohol, or pure 
 spirit, and 52 parts of water. How much of each of these is con- 
 tained in 84 gallons of proof spirits? Answ. 40^ gal. and 43f g. 
 
 12. 76 parts of nitre, 14 of charcoal, and 10 of sulphur, compose 
 gun-powder : how much of these ingredients will be requisite to 
 form a hundred weight of powder? Answ. 3 q. I fa ftx 15 tbs., 
 and 1 1 tbs. 
 
 The following questions are illustrative of some applications of this 
 rule which are often very useful to the Mathematical student. They 
 are not useful however, nor even intelligible, to the mere Arithmetical 
 pupil. By him, therefore, they should be omitted. 
 
 Exam. 2. Required the natural sine of 54 35' 43", those of 
 54 '6af and 54 36' being -8149593 and -8151278. 
 
 Here the difference of the two sines is '0001685; then, as 1' or 60", 
 (the difference of the arcs whose sines are given) :'43" (the excess of the 
 intermediate arc above the less,) : : -0001685 : -0001208, the part of the 
 difference corresponding to 43"; and the sum of this and the less sine is 
 8150301, the sine required. The ciphers might have been rejected, 
 and 1685 used as a whole number, and the result 1208 added to the 
 smaller sine, as if it had been a whole number. Had the second of the 
 given numbers been less than the first (as is the case in the cosines,) the 
 result of the analogy must evidently have been subtracted. 
 
 Ex. 13. The logarithm of 3-1415 being -4971371, and that of 
 3- 141G, -497 1509, required the logarithm of 3 r 141593.^;u;. -4971499. 
 
 l'-k The logarithmic tangent of 23 27' is 9*6372646, and that of 
 23 28' is 9-6376106: required that of 23 ' 27' 54". Answ. 9-6375760. 
 
 15. Required the sun's declination at Greenwich on the 10th of 
 July 1818, at 35 minutes past 7 o'clock in the evening, his decli- 
 nation on the 10th at noon being 22 18' 47", and on the llth at 
 noon 22 11' 10" Answ. 22 16' 23". 
 
 16. The moon passed the meridian of Greenwich, on the 10th 
 of October 1818, at 36 minutes past 9 o*clock in the evening, and 
 on the llth at 23 minutes past 10 ; at what time did she pass the 
 meridian of Mexico, in longitude 99 5' 15" W.? Answ. 49 minutes 
 
 . past 9 o'clock, on the 10th. 
 
 17. On the 19th of August, 1818, the declination of Venus, at 
 noon, at Greenwich, was 2 0' S.,and on the 25th at noon 5 5'S. 
 Required the declination on the 23d at noon, in the island of Owy- 
 l.ee, in longitude 155 58' 45" \V. Ansm. 4 - \& 41" S. 
 
192 
 FELLOWSHIP. 
 
 FELLOWSHIP is the method of determining the respective 
 gains or losses of the partners in a mercantile company. 
 
 Fellowship is usually distinguished into two kinds, Simple 
 and Compound, or Single and Double. 
 
 In SIMPLE or SINGLE FELLOWSHIP the stocks or sums 
 contributed by the several partners, all continue in trade 
 for the same time. 
 
 In COMPOUND or DOUBLE FELLOWSHIP the stocks continue 
 in trade for different periods. 
 
 SIMPLE FELLOWSHIP AND BANKRUPTCY. 
 
 RULE I. As the whole stock is to the whole gain or 
 loss, so is the stock of any partner to his gain or loss. 
 
 In the same way, the estate of a bankrupt may be di- 
 vided among his creditors by this analogy : As the sum of 
 all the claims on the estate is to its value, so is the claim of 
 any creditor, to his dividend or share of the estate. 
 
 This rule is merely a particular application of that contained in the 
 last article, and therefore requires no separate illustration. The method 
 of proof is also the same. 
 
 Exam. 1. Three merchants, A, B, and C, form a joint capi- 
 tal, of which A contributes 700, B 1000, and C 1600. What 
 is the share of each in a gain of 880 ? 
 
 Here, the sum of the stocks is 3300, the whole capital. Then, 
 as 3300 : 880, or by contraction, as 15 : 4 : : 700 : 186 13 4, 
 A's share; and as 15 : 4 : : 1000 : 266 13 4, B's share; and 
 lastly, as 15 : 4 : :'1600 : 426 13 4, C's share. The sum of 
 these shares is exactly 880, which proves the operation to be 
 correct. 
 
 Exam. 2. A bankrupt owes to A 900; to B 860; to C 
 640; to D 150; to E 70; and to F 30; but his whole 
 estate amounts only to 1250. Required the share of each creditor. 
 
 Here the sum of the debts is 2650. Then, as 2650 : 1250, 
 or by contraction, as 53 : 25 : : 900 : 424 10 6f, A's dividend; 
 and as 53 : 25 : : 860 : 405 13 2, B's dividend. In the same 
 manner we find C's share to be 301 17 8| ; D's 70 15 1 ; 
 E's 33 4|; and F's 14 3 0|. The sum of all these is 
 1249 19 11 1, a farthing being lost by neglecting the remainders. 
 
 In the division of a bankrupt's estate, it is usual first to find how 
 much in the pound he can pay, that is, how much the creditors will 
 receive for each pound of their respective claims. Thus, resuming 
 
 
AND BANKRUPTCY. 193 
 
 the same example, we have this analogy : as .2650 : .1250, or 
 as 53 : 25 : : 1 : 9/5 nearly, the sum that each creditor is to re- 
 ceive in the pound. Then, 
 
 by using the method of ali- 300 at 9/5 V pound, 
 
 quot parts as in the margin, 
 
 A's share will be found to 5/ $ 225 
 
 be 4,23 15 0, agreeing 4/ = i 180 
 
 nearly with the result al- 5d = T V of 5/. ..18 15 
 
 ready found, the difference 
 
 arising from taking 9/5 in- 423 15 
 
 stead of 9/5|. In the same way the rest of the dividends would 
 
 be found. 
 
 The following rule, which may be readily employed by a person who 
 has not learned a regular course of decimal fractions, will be found to 
 be preferable, in a considerable degree, to that given above, in the more 
 complicated questions in Fellowship ; and it is evident, that it is equally 
 applicable in every case of dividing into parts in a given ratio. 
 
 RULE II. (1.) Reduce the whole stock and gain to the 
 same denomination, if they be not so already : (2.) To the 
 latter annex six ciphers, or more if the stocks be large, and 
 divide by the former ; the quotient will be a decimal : 
 (3.) Multiply this decimal by the several stocks succes- 
 sively, taking aliquot parts for shillings and pence; and 
 from each product cut off as many figures as there were ci- 
 phers annexed : (4.) Value the figures cut off by the method 
 shown in Reduction of decimals,. problem II. 
 
 Exam. 3. If a bankrupt, whose property amounts to 2100, 
 owe to A S6 12, to B 1263 9 6, to C 724 15 10, to 
 D 1000, and to E 242 16 4; how much can he pay in the 
 pound, and what is the dividend of each creditor ? 
 
 In this example, the sum of the debts is 4057 13 8; and the 
 pence in this and in 2100 are 973844 and 504000. Annexing 
 ciphers to the latter, and dividing by the former, we obtain 
 5175367, the value of which is 10/4^ nearly, the sum which Ike 
 can pay in the pound. We then, as in the margin, multiply 
 0175367 by the first debt -5175367 
 
 826 12, and obtain for 826 12 
 
 A's dividend 427 15 11. 
 
 By proceeding in a similar 31052202 
 
 manner, we should find for 10350734 
 
 B's dividend 653 17 lOf, 41402936 
 
 for C's 375 2 1 1 ; for D's 10/= .... 2587683 
 
 .1517 10 8f; and for E's 2/ = 1 1 3 517537 
 
 125 13 4. The sum of 
 
 a 1 these is 2100, which 427-7958362, or 
 
 proves the correctness of 427 15 11. 
 
 the operation. The learner who has studied decimal fractious, 
 
 K 
 
194 SIMPLE FELLOWSHIP, &c. 
 
 will see that in finding the decimal, besides other contractions, figures 
 may often be cut from the divisor, and a smaller number of ciphers 
 annexed than are prescribed in the rule. 
 
 Several of the following questions are expressed briefly, and it may be 
 a useful exercise for the pupil to write them out at large. 
 
 Ex. 1. A's and B's stocks are 375 and 425 respectively : re- 
 quired the share of each in a gain of 240. Answ. A's .112 10, 
 B's 127 10. . . 
 
 2. Three merchants, A, B, and C, enter into partnership : A 
 puts into the joint stock 329, B 289 10, and C 317. Re- 
 quired the share of each in a gain of 583 12 6. Aiisw. A's 205 5 0|, 
 B's 180 12 2, C's 197 15 3|. 
 
 3. A's stock 1750, B's 1250 : whole gain 565 12. 
 Answ. A's 329 18 8, B's 235 13 4. 
 
 4. A's stock 349 16 7, B's 520 : whole gain 346 18 9. 
 Answ. A's 139 10 7f, B's 207 8 \\. 
 
 5. A's stock 384 18 10, B's 186 17 4, C's 811 17 6* : 
 whole gain 396 13 7. Answ. A's 110 7 1^, B's 53 11 5^, 
 C's 232 15 0. 
 
 0. A's stock 348 16 6, B's 804 11 4, C's 621 12 2: 
 whole gain 795 18 8. Answ. A's 156 8 4^, B's 360 15 6$ 
 C's *78 14 9. 
 
 7. B's stock 595 12 8, C's 701 116: whole gain 588 I 9. 
 Answ. B's 270 7, C's 318 1 2. 
 
 8. X's stock 448 19 3, Y's 582 13 4, Z's 261 1 4: 
 whole gain 718 18. Answ. X's 249 13 7, Y's 324 h| 
 Z's 145 3 81. 
 
 9. M's stock 475 15 8, N's 346 12 4, O's 396 17 6 : 
 v.'holegain 279 4 10. Answ. M's 108 19 3, N's 79 7 7 a, 
 O's 90 17 10. 
 
 10. A's stock 178 18 8, B's 236 15 8, C's 493 18 B, 
 D's 213 17 6: whole gain 583 10 9. Answ. A's 92 18 8-j, 
 B's 122 19 7|, C's 256 10 9, D's 1111 7f. 
 
 i 1. A debtor, the value of whose effects is only 1075 12 6, 
 owes to A 586 13 7, to B 348 10, to C 674 5, and to D 
 1000. What is the dividend of each, and how much is paid in 
 the pound ? Answ. A's 241 16 8, B's 143 13 1, C's 277 18 7. 
 D's 412 4 H; and 8/2 in the pound. 
 
 12. A, B, C, and D enter into partnership, and A puts in 987 
 - gallons of whiskey at 9/'10| IT gallon; B, 789 gallons of rum at 
 I6/H: C, 238 gallons of brandy at 1 4 9, and 183 gallons of 
 geneva at 1 111; .md 1) 497 gallons of wine at 18/4: and 
 they contribute besides 120 in money, in equal shares. Required 
 the share of each in a gain of 318 10 8. Answ. A's 75 1 4, 
 B's .96 13 2f, C's 76 3 9, D's 70 12 3. 
 
 13. A, B, and C, enter into partnership with a joint stock of 
 7500, of which 3(>00 belong to A, 3000 to B, and the re- 
 mainder to C. At the end uf a year, the gain is found to be 
 1079 4. Required the s>hare of this gain which each is to re- 
 
COMPOUND FELLOWSHIP. 195 
 
 ceive, a clear salary of 511 17 6 ^annnm hei allowed to C 
 as acting partner. Answ. A's 560 6 3|, B's 466 18 71, C's 
 651 19 1. 
 
 14. If two persons purchase a house jointly for j2000, and after- 
 wards let it for the yearly rent of 183 6 8 ; what share of the 
 yearly profit rent is each to receive, the one having contributed 
 350, and the other 1 150 of the purchase money, and the ground 
 rent being 44 8 V year ^ Aiis.-v. 59 11^, and 79 17 8f. 
 
 COMPOUND FELLOWSHIP. 
 
 HULE (1.) Let all the times be of the same denomination, 
 and multiply each stock by the time of its continuance in 
 trade : (2.) Then using the products as stocks, proceed ac- 
 cording to either of the rules for Simple Fellowship. 
 
 Exam. A and B enter into partnership : A contributes 
 600 for 13 months, and B 800 for 10 months. Required the 
 share of each in a gain of 650. 
 
 Here the products are 7800 and 600 X 13 =. 7800 
 8000, the sum of which is 15800. 800 X 10 = 8000 
 
 Then, as 1 5800 : 650 : 7800 : : or, 
 
 by contracting the first and second 15800 
 
 terms, as 316 : 13 : : 7800 : 3:30 17 8|, A's share : and as 
 316 : 13 : : 8000 : 329 2 3$, B's share. The sum of these 
 is 650, which proves the correctness of the operation. 
 
 In proceeding by RULE II. of Simple Fellowship, we divide 650 
 with ciphers annexed by 15800, or 13 with ciphers annexed by 
 316, and we find for quotient '04-1 13924- ; and multiplying this 
 successively by 7800 and 8000, and cutting eight figures from the 
 products, we obtain 320-886072 and 329-1 1392, or 320 17 8f 
 and 329 2 3, the same as before. 
 
 The reason of this rule will be evident from the consideration, 
 that a stock of 600 for 13 months, would be the same as 13 
 times 600 for 1 month ; and one of 800 for 10 months, the 
 same as 10 times 800 for 1 month. Hence, if these increased 
 stocks be employed, it is evident, that since the times are then to 
 be regarded as equal, the operation will proceed in the same 
 manner as those in Simple Fellowship. 
 
 Ex. 1. A's stock 280 for 5 months, B's 266 13 4 for 6 
 months: whole gain 331 12 6. Answ. A's gain 154 15 2, 
 B's 176 17 4. 
 
 2. A's stock 170 for 8 months, B's 80 for 6 months : 
 whole gain 250. Answ. A's 111 16 10$, B's 138 3 If. 
 
 3. A's stock 248 12 6 for 10 months, B's670 for 3 months, C's 
 512 7 6 far 6 months: whole gain 439 18 8. Ans. A's 144 9 7, 
 B's 116 161, C's 178 12 11J. 
 
 K 2 
 
19$ ALLIGATION. 
 
 4. C's stock .178 6 8 for 18 months, D's 237 17 6 for 12 
 months, E's 536 5 for 10 months : whole gain 370. AJUW. C's 
 103 18 9 : i, D's 92 8 6, E's 173 12 8|. 
 
 5. A's stock 485 IS 4 'for 1 year, B's 279 10 for 9 months, 
 C's 675 11 8 for 8 months: whole gain 386 15. Answ. A's 
 163 19 11, B's 70 14 11^, C's 152 1-j. 
 
 6. A's stock 576 15 for 11 months, B's 365 4 10 for 15 
 months, C's 582 6 8 for 9 months: whole gain 568 15. 
 Answ. A's 211 9 If, B's 182 12 If, C's 174 13 8. 
 
 7. M's stock 1038 13 9 for 6 months, N's 692 9 2 for 9 
 months, O's 1384 18 4 for 6 months: whole gain 686 1 2 
 Answ. M's 180 10 10, N's 216 13, O's 288 17 4. 
 
 8 Three merchants A, B, and C, entered into partnership, and 
 or, the 1st of March each contributed 1000. On the 3d of May 
 k took out 300; on the 8th of June B put in 360, and on the 
 20th of August C withdrew 280. On the 1st of September A 
 put in 450; and on the 16th of October each took out 180. 
 On the 8th of January of the following year, on making up accounts, 
 it is found, that they have gained 1250. How is this gain to 
 be divided among them ? Answ. A's share 384 1 1 1 f , B's 
 512 12 U, C's 353 5 103. 
 
 ALLIGATION. 
 
 ALLIGATION is a rule which is chiefly employed in cal- 
 culations respecting the compounding or combining of 
 articles of different kinds. 
 
 This rule has its name from a Latin word, which signifies to bind, be- 
 cause in the practical application of the rule, the quantities are usually 
 linked or connected together by lines. It is a rule which is of little 
 practical utility ; being principally used in the solution of questions, 
 which are of rare occurrence in real transactions. Besides, every thing 
 that can be effected by this rule can be done in general in a better and 
 easier way by Algebra. Hence this article will be more circumscribed 
 in its limits than it might otherwise have been. The following is the 
 principal problem in this rule, and indeed the only one that belongs to 
 it exclusively. 
 
 Tojind in ivhat proportions, quantities of given values must 
 be taken, to form a compound of a given value. 
 
 RULE. (1.) Let the rates of the ingredients, all in the 
 same denomination, be written in a line ; and let the mean 
 rate in the same denomination . be written above them. 
 (2.) Take two of the rates, one of which is greater, and 
 the other less, than the mean rate, and write the difference 
 between each of them and the mean rate, below the other. 
 (3.) Proceed thus with the rates two by two, if there 
 
 ence 
 Lher. 
 - be 
 
ALLIGATION. 197 
 
 more than two, till one or more differences stand below 
 each. (4.) Then, if only one difference stand below any 
 rate, it will be the quantity required at that rate ; but if 
 there be more than one, their sum will be the required 
 quantity. 
 
 The connecting or linking of the rates with crooked or curved lines, 
 in the use of this rule, is attended with little advantage. Should that 
 method be preferred, however, it can present no difficulty, as each rate 
 less than the mean rate is to be connected with one greater, and each 
 greater with one less, and the differences are to be set below the rate to 
 which the line directs. 
 
 Exam. 1. In what ratios must two kinds of flour, worth 2^d. and 
 d. V lb. respectively, be taken, to make a mixture worth 
 
 ib.? 
 
 Here the mean rate, 13 farthings, is 
 set above the other rates, 10 farthings 
 and 15 farthings. Then, the difference 
 between 10 and the mean is set below 
 15, and the difference between 15 and 
 
 the mean below 10. Hence we find, that the quantities must be 
 in the ratio of 2 to 3; that is, for every 2 Ibs., or 2 cwts. at 2|d. 
 ^ Ib., 3 Ibs., or 3 cwts. at 3|d. V Ib. must be taken, to form a 
 compound worth 3d. V Ib. 
 
 The correctness of this operation, and of the principle on which it 
 depends, will appear manifest from the consideration, that in selling 
 3 Ibs. at 3d. per Ib., instead of 3fd., there is a loss of 11<L ; but in selling 
 2 Ibs. at 3d. per Ib., which cost, only 2id. per Ib., there is a gain of 
 lid. : and therefore, the gain on the one quantity balancing the loss on 
 the other, the value of the compound must be exactly the mean rate. 
 
 Exam. 2. In what proportions must wines worth 10/, 14/, 17/, 
 and 18/, V gallon respectively, be mixed, so that the compound 
 maybe worth 16/ W gallon ? 
 
 Here by setting tha difference be- 
 tween 10 and 16 below 17, and the 16 
 difference between 17 and 16 below 10, 14, 17, 18 
 
 10 ; and likewise by setting the dif- 
 
 ference between 14 and 16 below 18, 1, 2, 6, 2 ; or 
 
 and the difference between 18 and 16 2. I, 2, 6 
 
 below 14, we find, that for 1 gallon 
 
 at 10/, we must take 2 at 14/, 6 at 17/, and 2 at 18/. By using 
 10/ with 18/, and 14/ with 17/, a second answer is obtained, from 
 which it appears, that if 2 gallons at 10/, 1 at 14/, 2 at 17/, and 6 
 at 18/, be mixed together, the compound will also be worth lf>/ 
 ^ gallon. It is scarcely necessary to say, that any quantities in the 
 same ratios will serve the same purpose. 
 
 With respect to the reason of the operation, it is obvious from what 
 was said respecting the preceding exercise, that 1 gallon at 10/, and 6 
 
198 ALLIGATION. 
 
 at 17/ each, would make a mixture worth 16/ per gallon, and likewise, 
 that a mixture of 2 gallons at 14 /, and 2 at IS/, would be worth the same 
 per gallon ; and It is evident, that both mixtures taken together must 
 make a mixture of the same value per gallon also ; and in the same way 
 every operation in this rule may be explained. 
 
 From these principles it is also manifest, that if we should mul- 
 tiply or divide I and 6 by any number, and 2 and 2 by any number, 
 we should still have results that would satisfy the conditions of 
 the question. Thus, multiplying the former by 3, and dividing the 
 latter by 2, we find for answer 3 gallons at 10/, 1 at 14/, 18 at 17/, 
 and 1 at 18/. Different answers 
 may also be found by connecting 16 
 
 the rates differently. Thus, by 10, 14, 17, 18 
 
 using 10 and 17 we get 1 and 6, 
 
 and by connecting 10 and 18, we 1 1 ' 6 6 
 
 have 2 and 6 ; and then by using 2 2 
 
 14 and 17 we get 1 and 2. Alter 
 
 this; by the requisite addition, we 3, 1, 8, 6 
 
 find 3, 1, 8, 6 for the required 
 
 quantities ; and it is obvious, that by a still farther application of 
 these principles, different answers might be found without limit. 
 
 The correctness of these results is proved by adding together the prices 
 of 3 gallons at 10/, of 1 at 14/, of 8 at J7/, and of 6 at 18/. This 
 will be found to be 288/, which divided by 3-f- 1 -f <8 ~h 6 g a " on s, 
 gives exactly 16/ for the mean rate ; and thus the proof may be conducted 
 in every case. This method of proof has been generally made a separate 
 case of this rule, and called with no great propriety Alligation Medial. 
 The operation according to the preceding rule, is usually called Attign- 
 tion Alternate. 
 
 Exam. 3. How much linen at 2/ and at 2/5 ^ yard, must be taken 
 with 216 yards at 3/4, that the whole may be worth 2/6 & yard 
 at an average ? 
 
 Here by taking the differences, &c. 30 
 
 as in the margin, we find that the quan- 24, 29, 40 
 
 tides may be in the ratio of 10, 10, and 
 
 7. Then as 7 : 10 : . 216 : 308f It 10 10 6 
 
 appears therefore, that 308f yards at 1 
 
 2y, the same quantity at 2/5, and 216 
 
 yards at 3/4, will compose a parcel 10, 10, 7 
 
 worth 2/6 W yard, at an average : and 
 
 various other answers might be found. This question belongs to 
 
 what is usually called Alligation Partial. 
 
 Exam. 4. What quantities of tea, worth 8/, 7/6, and 6/6 V lb. re- 
 spectively, must be mixed together, to form a parcel containing 
 I12fts. worth 7/4 rib ? 
 
ALLIGATION. 199 
 
 Here, in finding the ratios, to make 88 
 
 the first two terms different, the differ- 96, 90, 78 
 
 cnces between 88 and 90, and 88 and 
 
 78, are set down twice. In this way it 10 10 8 
 
 is found, that the quantities may be as 102 
 
 10, 20, and 12, or, by halving each term, 2 
 
 as 5, 10, and 6. Hence, by the method 
 
 of dividing into parts in a given ratio 10, 20,12 
 
 (see page 191,) as 21, the sum of these, 
 
 : 5 : : 112 Ibs,; or by contracting, as 3 : 5 : : 16 Ibs.: 26| Ibs.; 
 and as 3 : 10 : ? 16 tbs. : .53* Ibs. ; and lastly, as 3 : 6 : : 16 ibs. : 
 32 Ibs. It appears therefore, that 26 Ibs. at 8/, 53^ Ibs. at 7/6, 
 and 32 Ibs. at 6/6, will form a compound of 112 Ibs., worth 7/4 
 V lb. This question belongs to what has generally been called 
 Alligation Total. This name for this particular case of Alligation, 
 as well as those already mentioned for the other cases, is properly 
 falling into disuse. 
 
 As many questions in this rule admit of several answers, the pupil 
 should prove his results in working the following exercises, particularly 
 when his answers may differ from those here given. 
 
 Ex. 1. In what proportions must sugars, worth 13d., lld.,and 
 9d. IT lb. respectively, be compounded, that the mixture may be 
 \vorth lO^d r lb. ? Answ. 3, 3, 7 ; 1, 2, 3 ; &c. 
 
 2. How much water must be added to a cask of spirits con- 
 taining 84 gallons, worth 13/6 ^ gallon, to reduce the value to 
 1 1/4^ ty gallon ? Answ. 1 5W gallons. 
 
 3. What quantities of three different kinds of raisins worth lid., 
 15d., and 22d. ^ lb. respectively, must be mixed together, to fill 
 a cask containing 200 Ibs., and to be worth 16d. V lb. Answ. 6 1 Ibs. ; 
 61 ^s.; and 77$ Ibs. , 
 
 4. How much land worth 1 7/6 ^ acre, must be added to a farm 
 containing 51 a. 2 r. 20 p., worth \ 14 6 V acre, to reduce the 
 average value of both together to 1 2 9? Answ. 115 a. 2 r. 6fp. 
 
 5. A box of linen, containing 1200 yards, worth at an average 
 3/ V yard, consists of two kinds, one worth 2/8J, and the other 
 worth 3/9 IT yard. How much of each kind does it contain ? 
 Answ. 876 H yards, and 323 T V yards. 
 
 6. How much spirits, at 14/ W gallon, must be added to a mix- 
 ture consisting of 41 gallons at 9/6, and 59 gallons at 10/8, to 
 make the compound worth 11/6 ^ gallon? Answ. 52 ^3- gallons. 
 
 7. How much first flour worth .1 11 6 per cwt. second flour 
 wort 1 9, third flour worth 1 7 6, and fourth flour worth 17/6, 
 must be taken, to form a ton worth 25 168? Ans. 2 c. 1 q. 20 \ Ibs.; 
 7 c. 1 q. 5i Ibs.; 4- c. 3 q. 13 Ibs. ; and 5 c. I q. 17 Ibs. 
 
 8. A gentleman's labourers xronsist of men at 1/4, and women 
 at lid. per day; and the amount of the wages of the whole is the 
 same as if each of them hail \f'i\. Required the number of the 
 men, the number of the women being 21. Answ. 49. 
 
200 
 INVOLUTION. 
 
 A POWER of any number is the product obtained by the 
 continual multiplication of that number, repeated a certain 
 number of times as factor. 
 
 A number, in relation to any power of it, is called the 
 ROOT of that power. 
 
 When the proposed number is used twice as factor, the 
 product is called the SECOND POWER, or the SQUARE,* of 
 that-number ; when three times, the THIRD POWER or CUBE; 
 when four times, the FOURTH POWER ; when Jive times, the 
 
 FIFTH POWER, &C. 
 
 Powers are often denoted by writing after the proposed 
 number, and a little higher, the number which shows how 
 often the proposed number is repeated as factor. This 
 number is called the INDEX, or the EXPONENT, of the power. 
 
 Thus, 5X-5 = 25, is the second power, or the square of 5, and may 
 be written 5 2 , where 2 is the index ; 7X"X?X7 = 2401, is the fourth 
 power of 7, and may be written 7 4 , where 4 is the indox, &c. 
 
 The method of finding any assigned power of a given 
 number, or, as it is also expressed, the method of raising a 
 number to any proposed power, is called INVOLUTION. 
 
 From the preceding definition of a power we have the following rule 
 for Involution : 
 
 RULE. To find any assigned power of a given number ; or to 
 raise a given number to any proposed power : find the con- 
 tinual product of the given number repeated as factor, as 
 often as there are units in the index of the proposed power. 
 
 The process may often be abbreviated by multiplying to- 
 gether powers already found. In this case, the index of the 
 power thus found is equal to the sum of 'the indices of the 
 powers multiplied together. 
 
 * The second power of a number is very improperly called the square of the number. 
 The square of a. line, or the square described on aline, is a geometrical figure havu;<; 
 four sides each equal to the proposed line, and having its adjacent sides perpendicular to 
 each other : but this is evidently by no means applicable to an abstract number. The, 
 mistake has arisen from the circumstance, that the area of a square is found numerically 
 by multiplying the number expressing the length of the side, by itself, which is the same 
 as the process by which the second power of that number \s determined. It might be 
 Sliown in a similar manner, that the third power of a number is with equal impropriety 
 called the cube of that number, the cube being, not a power of a number, but a solid 
 body. The terms square and cube, however, as well as square root and cube root, have 
 been so long and so generally used in this improper sense, that it would perhaps be vain 
 to attempt to correct the error any farther than by frequently using the legitimate CM* 
 pressions, se cond powe r, second root, &c. instead of thsm. 
 
INVOLUTION 201 
 
 When the given number is either wholly or partly a deci- 
 mal, the operation may often be much abbreviated, by the 
 rule for contracting the multiplication of decimals given in 
 page 118. 
 
 Exam. 1. Required the fifth power of 23. 
 
 Here, by multiplying 23 Multiply ...... $ 23= 1st power. 
 
 by itself, we find 529 for the 
 
 second power of 23. By Multip i y ....... 
 
 multiplying tins by 23, we _ 
 
 find 12*167 for the third M . . . ( 12167 = 3d _ 
 
 power. By proceeding in Miply.... ^ 23 
 
 like manner, we find the c 2?9841=4th WJWWJJM 
 
 fourth power to be 27984*1, Multiply.. .< 23 
 
 and the fifth to be 6136343. ^ ^jjjjH&fl _ 
 
 The same result might also have been found by multiplying the 
 second power 529 by itself, and the product 279841, which is the 
 fourth power, by 23. The same result would also be obtained by 
 multiplying the third power by the second. 
 
 In like manner, if it were required to find the twelfth power of a 
 number; multiply the number by itself to find the second power ; the 
 second power by itself to find the fourth power ; the fourth by itself to 
 find the eighth ; and, lastly, the eighth by the fourth to find the twelfth. 
 
 Exam. 2. Required the fifth power of f . 
 
 The fifth power of 3 is 243, and the fifth power of 8, 327G8 : 
 the answer therefore is Tiff?' '-^ ie reason of this is evident front. 
 Multiplication of Fractions. 
 
 Exam. 3. What is the third potter of !.? 
 This, by reduction fro an improper fraction, becomes ; and b? 
 involving the numerator and denominator each to the third power, 
 we find for answer VV> r Ifi- Each of the last examples might 
 have been wrought by reducing the fractions to decimals, and then 
 working by the general rule. 
 
 Exam. 4. Required the sixth power 1-12, true to five places 
 of decimals. 
 
 1-404928 = 3d power. 
 8 29404-1 
 
 1404928 
 
 By raising this to the third power, in 561971 
 the way already shown, we find 1-404928, 5620 
 
 and this being multiplied by itself, as in 1264 
 
 the margin, we find for the sixth power 28 
 
 1-973822. U 
 
 1-973822= 6th 
 
 Kxercises. Involve the following numbers to the powers denotec 
 by their respective indices : 
 
 K3 
 
202 
 
 EVOLUTION. 
 
 Ex. Answ. 
 
 1. 678 s . . ............... 4:59684? 
 
 2. 119 3 ............... 1685159 
 
 3. 75 4 .............. 31640625 
 
 86*... ........ 4704270176 
 
 9 9 ............ 387420489 
 
 7 12 ........ 13841287201 
 
 II 10 ........ 25937424601 
 
 (f) 6 * 
 
 4. 
 5. 
 
 6. 
 
 7. 
 
 8. 
 
 9. 
 
 10. 
 
 3) 4 .............. H6UM 
 
 Ex. 
 
 Answ. 
 
 11. (t) 7 
 
 12. 4-S67 4 ......... 363-691179 
 
 13. 1'03 17 ............ 1-652848 
 
 14. 1-035 18 ........... 1-857489 
 
 15. l'04 16 f .......... 1-800943 
 
 16. 1-0475 24 ......... 3-045767 
 
 17. 1-05 31 ............ 4-538039 
 
 18. 1-055 26 ......... 4-0231290 
 
 19. 1-06 34 ........... 7-251025 
 
 20. 1-07 32 ..... 8-7152708 
 
 EVOLUTION, OR THE EXTRACTION OF ROOTS. 
 
 EVOLUTION is the method of finding an assigned root of a 
 given number. J 
 
 The I^DEX of a root is a fraction whose denominator de- 
 notes the order of the root, and whose numerator is a unit. 
 The root of a number is also expressed by prefixing to the 
 number the sign,^ with the number above it which- de- 
 notes the order of the root. In case of the square or second 
 root, however, the number 2 is omitted. 
 
 Thus, the fourth root of 10 is denoted by 10*> or ^/ 10, and means a 
 number whose fourth power is 10; and the second, or square root of 7 
 is written 7a" or y^7, and means a number, such, that if it be multiplied 
 by itself, the product will be 7. 
 
 To facilitate the extraction of the square and cube roots, it may be 
 proper for the pupil to be familiar with the following tables : 
 
 TABLE I. The square of 1 = 1 ; of 2 = 4; of 3 = 9 ; of 4 = 16; 
 of a = 25 ; of 6 == 36; of 7=49 ; of 8 = 64 ; of 9 = 81. 
 
 TABLE II. The cube of 1 = 1 ; of 2 = 8 ; of 3 = 27; of 4 = 64; 
 of 5= 125; of 6 =216; of 7= 343; of 8 = 512; of 9 = 729 
 
 * The brackets enclosing tbig fraction, which in this use of them constitute the a)ge- 
 liraic v'inculum in one of its forms, denote, that the entire fraction, and not its numerator 
 alone, is to be involved to the sixth power. The object in the next exercise, in like 
 manner, is to find the fifth power of 2|. 
 
 f Perhaps the easiest mode of working this exercise will be to find the sixteenth power, 
 and divide it by 1'04. In like manner, in the 17th exercise, the thirty-second power may 
 be divided by 105 
 
 It is unnecessary to inform the Mathematical student, that involution, when great ac- 
 curacy is not required, is immensely facilitated by the use of logarithms. 
 
 t Evolution may also be defined to be the method of finding a number, the continual 
 product of which repeated a given number of times as factor, will amount to a given 
 number. 
 
 \ This sign is the letter r, the initial of the Latin word radix, a root, changed in form 
 by rapidity in making it, and by its appropnatiou to a particular use. 
 
203 
 EXTRACTION OF THE SECOND, OR SQUARE ROOT. 
 
 RULE I. To extract the second, or square root of a given 
 number: (1.) Commencing at the unit figure, cut off periods of 
 two figures each, till all the figures of the given number are 
 exhausted.* (2.) The first figure of the required root will 
 be the square root of the first period, or of the greatest 
 square contained in it, if it be not a square itself. (3.) Sub- 
 tract the square of this figure from the first period ; to the 
 remainder annex the next period for a dividend ; and, for 
 part of a divisor, double the part of the root already ob- 
 tained. (4.) Try how often this part of the divisor is con- 
 tained in the dividend wanting the last figure, and annex 
 the figure thus found to the parts of the root and of the 
 divisor already determined. '5.) Then multiply and sub- 
 tract, as in Division; to the remainder bring down the next 
 period ; and, adding to the divisor the figure of the root 
 last found, proceed as before. (6.) Continue the process 
 till all the figures in the given number have been used ; and 
 if any thing remain, proceed in the same manner to find 
 decimals, adding, to find each figure, two ciphers, or if the 
 given number end in an interminate decimal, the two 
 figures that would next arise from its continuation. 
 
 Exam. 1. Required the square root of 365. 
 Here, by placing a separating mark 3 / 65(19*1049 
 
 between 3 and 6, the given number is 1 
 
 divided into the two periods, 3 and 
 65. 1, the root of the greatest intcg- 29) 265 
 
 ral square contained in 3, is then put 9 261 
 
 in the quotient, and its square taken 
 from the first period. To the remain- 381 ) 400 
 
 der the next period is brought down, 1 381 
 
 which makes for dividend 265. The 
 first part of the divisor is found by 38204) 190000 
 doubling 1, the first part of the quo- 4 155816 
 
 tient. In finding, in the next place, 
 
 what figure must be annexed to the 382089)3718400 
 part of the root already found, though 9 343880 1 
 
 2 would be contained 13 timss in 26, 
 
 yet we try 9, as we know the next 382098 ) 279599 
 
 figure cannot be greater than 9. We 
 annex 9, therefore, to the parts of the root and of the divisor al- 
 
 * In dividing a decimal, or a number consisting of a whole number with a decimal, 
 Into periods, the division must also commence at the unit figure or the decimal peuu, 
 and must be continued both ways, if there be a whole number, and if there he an od.i 
 iiiivire at the end of the decimal, a cipher, or if it be a periodical decimal, the figure that 
 would next arise from its continuation, must be annexed. Thus, 4T7'24~> will be divided 
 thus, V17/-2V50 ; -H'SWBS. &c. thus. *!/' W'66; and -567 thus, -S&lQi && 
 
204 EVOLUTION. 
 
 ready found, and multiplying 29 by 9, we find for product 261, and 
 for remainder 4-. Hence, we have for root 19, and for remainder 
 4. Now, to find decimal figures, a point is put in the quotient, 
 two ciphers annexed to the remainder, and 9, the figure last found, 
 added to 29, the former divisor. We have then for dividend 4'00, 
 and for part of a divisor 38. This part of the divisor is contained 
 once in 40 ; and therefore the first figure of the decimal is 1, which 
 is also annexed to 38. In working for the next figure, we have the 
 divisor 382, which not being contained in 190, a cipher is annexed 
 to the parts of the root and divisor already found, and two ciphers 
 annexed to the dividend to find another figure. The rest of the 
 work proceeds in the same manner, and the root, true to four 
 places of decimals, is found to be 19-1019. The truth of this re- 
 sult is proved by multiplying 19*1049 by itself, and adding the re- 
 mainder to the product ; as the result will be exactly 365. 
 
 In this, ay well as in every other case in extracting roots, in which there is a 
 remainder, after all the significant figures have been used, the fractional part of 
 the root would be an inlerminate decimal, differing from the interminaty 
 decimals which we have seen already, in its not repeating or circulating, 
 and thus presenting no law by which it can be continued.* In this case, 
 in any practical application, the extraction, is to be carried on, till as 
 many decimal figures are obtained as the degree of accuracy necessary 
 in the result may require. 
 
 The principle on which the preceding rule depends, is, that the 
 square of the sum of two numbers is equal to the squares of the numbers 
 with twice their product. Thus, the square of 34 is equal to ttit- 
 squares of 30 and of 4 with twice the product of 30 and 4 ; that is, 
 to 900-f 2X30x4-f-16n 1156. Hence, in extracting the second 
 root of 1156, we separate it into two parts, 1100 and 56. Then 
 1100 contains 900, the square of 30, with the remainder 200 ; the 
 first part of the root therefore is 30, and the remainder 200 -j- 56 
 or 256. Now, according to the principle abovementioned, this re- 
 mainder must be twice the product of 30 and the part of the root 
 still to be found, together with the square of that part. Now, di- 
 viding 256 by 60, the douWe of 30, we find for quotient 4. Then, 
 this part being added to 60 the sum is 64, which being multiplied 
 by 4, the product, 256, is evidently twice the product of 30 and 4, 
 together with the square of 4. In the same manner the operation 
 may be illustrated in every case. The rule is best demonstrated, 
 however, by Algebra. 
 
 RULE II. When the root is to be extracted to manijjlgure.s^ 
 the process may be much contracted by the following ride : Find, 
 by rule I., half, or one more than half, the number of figures 
 required : then to the remainder annex one figure instead of 
 
 * Continued fractions, as will appear hereafter, show the law of continuation of square 
 roots : in respect to other roots, however, there is no method known which shows the 
 lav. of their i ontinuation. 
 
SQUARE ROOT. 205 
 
 two, and having found the divisor in the usual way, proceed 
 according to the contracted method of dividing decimals. 
 
 Thus, suppose it had been 382098)2795990(1 9-1 049 7-3174 
 required, in the preceding ex- 2674691 
 
 ample to find the answer true 
 
 to nine places of decimals; 121299 
 
 these and the two places of 114-6:30 
 
 whole numbers are eleven 
 
 figures in all ; and, therefore, 6669 
 
 before commencing the con- 3821 
 
 traction, it is necessary to 
 
 find six figures. These have 2848 
 
 been already found to be 2675 
 
 19-1049. Taking therefore 
 
 382098 and 279599, the divi- 173 
 
 sor and remainder already 153 
 
 found, and also the quotient 
 
 19-1049, the continuation of 20 
 
 the preceding work will stand as in the margin, a cipher being 
 
 added to the dividend, according to the rule. The first figure thtrf 
 
 results from the division is 7, which, in working the operation at 
 
 full length, must have been annexed to the divisor : we therefore 
 
 carry 5, for 7 times 7, to 56, the product of 7 and 8. After this, 
 
 tile work proceeds exactly as in the contracted mode of dividing 
 
 decimals. By this means the root is found to be 19-104973174. 
 
 RITLE III. To extract the root of a vulgarfraction, reduce 
 it to its simplest form, if it be not so already, and extract 
 the roots of both terms, if they be complete powers : other- 
 wise divide (he root of their product by the denomina- 
 tor. 
 
 The root may also be found by reducing the fraction to 
 a decimal, if it be not such already, and taking the root of 
 the decimal. 
 
 Thus, the second root of is L. This result may be obtained either 
 by taking the roots of both terms, or by reducing the given fraction to 
 the decimal -25, the second root of which is *5, or T %, or , the same as 
 before. 
 
 In like manner, the root of 2^, or f , is , or 1 . This result might 
 also be obtained by extracting the root of ~2'25. This would be found 
 to be 1-5, or !, as before. 
 
 Again, if it be required to find the second root of ^; let the square 
 root of 35 ( = 5X7,) which will be found to be 5 '91 60798, be divided 
 by the denominator 7, and there will result -84515425, the root required. 
 The same result would be obtained by extracting the root of 71'4i?85 / 71, 
 the decimal equivalent to the given fraction. 
 
 The more advanced pupil may sometimes find the following contrac- 
 tions useful : 1. {f the denominator be an exact square, and tlie numerator 
 
206 EVOLUTION. 
 
 not, divide the square root of the numerator by the square root of the 
 denominator. 2. If Ike numerator be an exact square and the denominator 
 not, divide the product of the square roots of the numerator and denomi- 
 nator by the denominator. 
 
 The following rule, which is only a particular application of the gene- 
 ral rule to be given hereafter for finding the roots of powers in general, 
 may be found useful in carrying a root out to a great number of figures, 
 after it has been carried to a considerable number by the common rule: 
 
 (1.) Find the square of the part of the root already found. (2.) Then 
 make the sum of the given number and three times this square the Jtrst term 
 of an analogy ; the sum of this, square and three times the given number, the 
 second term ; and the port of liie root already found, the third term: the 
 fourth term will be the required root true to nearly three times tlie number of 
 figures in the part before found. 
 
 Thus, the square root of 2 is found by the common method to be 
 1-414, the square of which is 1*999396. Then, as 3X 1 '999396+2 : 
 3X2+1-999396: : 1-414 : i '4 1421356237, which is true, except the 
 last figure; and if we should repeat the operation, using this number 
 and its square, the root would be found true to about thirty places. 
 
 We might employ this rule from the beginning of the operation, by 
 estimating the root as nearly as possible. It will in general be easier, 
 however, to find the leading figures of the root by the common method. 
 That method also, contracted in the way already shown, will for the 
 most part be preferable to this, when the root is not required to be carried 
 beyond ten or twelve figures. 
 
 Exercises. Required the square roots of the following numbers : 
 
 Exercises. Answers. I Exercises. Answers. 
 
 1. 5 2-236068 | 16. 33 5-74456-26465 
 
 2. -5 -7071068 I 17. 333 18-24828759 
 
 3. '7 .. .. -8366600 ' 18. 666 25-806975801 
 
 4. -07 -2645751 
 
 3. -06 -2449490 
 
 6. -006 -0774597 
 
 7. 785 28O1785 
 
 8. 78-5 8-8600226 
 
 9. 562 23-70653918 
 
 10. | -6123724357 
 
 1 1. 13 ... 3-633180425 
 
 12. 1728 41-56921938 
 
 13. & -19364916731 
 
 14. 1| 1-172603940 
 
 15. IT 1-018577439 
 
 19. $ -74535599250 
 
 20. T 4 T -60302268915 
 
 21. -81649658093 
 
 22. H| 1-16', or H 
 
 23. lli 3-3', or 3$ 
 
 24. ii -82915619759 
 
 25. 6} 2-4784787961 
 
 26. 794^ 28-181554251 
 
 27. 3333333 1825-7417671 
 
 28. 105VS 10-269106361 
 
 29. 25381| 159-31624734 
 
 30. 57000^ 238-74727068442 
 
 31. 123456789 11111-1110605555 
 
 32. 987654321 31426-968052932 
 
 33. 207JI 14-411607975672 
 
 34. 822650 907-00055126775 
 
 35. 34967W 186-99513848809 
 
 36. 29524'+ 244* 29525*008247247 
 
207 
 EXTRACTION OF THE THIRD, OR CUBE ROOT. 
 
 RULE I. To extract the third, or cube root of a given 
 number : (1.) Find by trial a number nearly equal to the re- 
 quired root, and call it the assumed root : (2.) Find the cube 
 of this root : (3.) Then as twice this cube added to the 
 given number, is to twice the given number added to the 
 same cube, so is the assumed root to the true root, nearly: 
 (4-.) By employing the approximate root thus found, and 
 repeating the process, a number still nearer the true root 
 will be obtained ; and thus the process may be repeated as 
 often as may be thought necessary.* 
 
 The operation is proved by involving the root, when found, to 
 the third power: and the more nearly the result agrees with the 
 given number, the more nearly correct is the root. 
 
 In the use of this rule, each result will generally be true to between 
 two and three times the number of figures to which the assumed root 
 was true. It may be observed also, that each remit is too nearly eyuai to 
 the assumed root; or, in other words, the correction is too small, as if 
 the assumed root be too great, the result is also too great; otherwise, it 
 is too small. 
 
 Exam. 1. Required the third or cube root of 34567. 
 
 Here, by cutting off three figures towards the right hand, we 
 find for the first period 3-t, the root of which is above 3, since the 
 third power of 3 is only 27 ; and hence, since the root must have 
 
 * This rule, as well as the approximative rule already given for the extraction of ih 
 square root, is only a particular application of the general rule to be given hereafter, for 
 the extraction of roots. The rule which is commonly employed in extracting the cube 
 root is here subjoined for the use of any who may be disposed to prefer it. The rule giver, 
 above, however, abridges the labour very greatly, as will appear by the application of 
 both rules to any particular exercise. The labour by the common i vle, indeed, of finding 
 the cube root of a number true even to seven or eight figures, is so very great as to ren- 
 der the operation formidable in a high degree, and to make any shorter method very 
 desirable. The following Is the 
 
 Common Rule. (1.) Commencing at the units, divide the given number into periods 
 of three figures each. (2.) The first figure of the required root will be the cube root of 
 the first period, or of the greatest cube contained in it, if it be not a cute itself. (3.) Sub- 
 tract the third power of this figure from the first period, and to the remainder annex the 
 next period for a dividend, and for part of the divisor take 300 times the square of the 
 part of the root already obtained. (,-i.) Try how often this part of the divisor is contained 
 in the dividend, and annex the figure thus found to the part of the root already deter, 
 mined. (5.) Then, to find the complete divisor, add to the part already found 30 times 
 the last figure of the root multiplied by the part of the root before it, and add also the 
 square of the last figure. (6.) Then multiply and subtract as in Division ; to the remain- 
 der bring down die next period, and proceed as before. (7.) Repeat the operation, till all 
 the figures in the given number have been used ; and, if any thing remain, continue the 
 operation in the same manner to find decimals, adding, to find each figure, three ciphew : 
 r, if there be an interminate decimal, the three figures that would next arise from iu 
 continuation. 
 
208 EVOLUTION. 
 
 two figures, (one for each period,) it must be greater than 30.* 
 We may suppose it therefore to be about 32 or 33. The cube of 
 the former we find to be 32768, and that of the latter 35937. Hence, 
 the given number differing from each by pretty nearly the same 
 quantity, we may suppose its root to be about 32'5. The cube of 
 this, is found to be 34328-125. The first part of the remaining 
 work may conveniently 34328-125 34567 
 
 stand as in the margin. 2 2 
 
 The work is then to 
 
 proceed as in the Rule 68656-250 69134 
 
 of Proportion and the 3456? 34328-125 
 
 result is found to be 
 
 32-5752101, which is the As 1032 9 3 .250 : 103462-125 : : 32-5: 
 
 required result, true ex- 32-5752101, Answ. 
 
 cept the last figure, the 
 
 true result being 32-57521043, &c., as would be found by repeat- 
 ing the operation with 32'575, as the assumed root. 
 
 Exam. 2. Required the cube root of 100. 
 
 The root here is evidently between 4 and 5, but nearer the lat- 
 ter, as 100 differs less from 125, the third power of 5, than from 
 64, the third power g? . 336 1QO 
 
 of 4. Let then 4-6, 
 
 the cube of which '_ 
 
 is 97-336, be assum- . Q4 . rt79 200 
 ed After this the 6 ' 2 . 33(j 
 work may stand as 
 
 &>S?5tfg As a* 678 ' 297-336:: 4-6:4-0^8 
 4-64158, the required root nearly. By repeating the operation with 
 4-64, as the assumed root, there will result 4-641588833 for the 
 required root still more correctly. 
 
 Exam. 3. Required the cube root of 782140. 
 Here the root proper 778688 7 82 140 
 
 to be assumed is readily 2 2 
 
 found to be 92, the cube 
 
 of which is 778688. The 1557376 1564280 
 
 work will then stand as 782140 778688 
 
 in the margin, and there 
 
 will result 92-13575 for As 2339516 : 2342968 : : 92 : 92-13575 
 the required root nearly. 
 
 By repeating the operation. with 92-136, there will be found, for 
 the required root, very nearly true, 92-135747933. 
 
 * It is scarcely necessary to remind the Mathematical pupil, that the root prom r to 
 be assumed, both in this article and the next, will be found wilh great ease by moans of 
 logarithms. The finding of it will also be facilitated, as in the preceding exanmi?, in- 
 dividing the given number .into periods as in the Square Root, but each consisting of 
 tiireo figures instead of two, and by considering that there must be a figure in the root 
 for each period. 
 
ROOTS IN GENERAL. 209 
 
 RULE II. To extract the cube root of a vulgar fraction, re- 
 dace it to a. decimal, and then extract the root ; or multiply 
 the numerator by the square of the denominator, find the 
 cube root of the product, and divide it by the denominator. 
 The cube root of a mixed number is generally best found 
 by reducing the fractional part to a decimal, if it be not so 
 already, and then extracting. It may also be found by re- 
 ducing the given number to an improper fraction, and then 
 working according to the preceding directions. 
 
 Exercises. Find the cube roots of the following numbers : 
 
 Exercises. Answers. 
 
 6. 1234567 107-276572 
 
 7. 44-6 3-546323 
 
 8. T \ -64-365958974. 
 
 9. f -9614997135 
 
 10. 376 .. ,.. 7-217652 
 
 Exercises. Answers. 
 
 1. 123 , 4-973190 
 
 2. 517 8-025957 
 
 3. 900 9-654894 
 
 4. 1 23456789 . 4-97-9338592 
 
 5. 12345678 .... 231-120418 
 
 EXTRACTION OF ROOTS IN GENERAL. 
 
 The following article is intended only for the Mathematical or 
 the more advanced Arithmetical pupil. For others, besides it be- 
 ing in a considerable degree difficult, it is seldom useful. 
 
 RULE. To extract any root whatever: (1.) Call the index of the 
 given power n ; and find by trial a number nearly equal to the re- 
 quired root, and call it the assumed root. (2.) Raise the assumed 
 root to the power whose index is n. (3.) Then, as n-\-l times 
 this power added to n-^-l times the given number, is to n 1 times 
 this power added to -j- I times the given number, so is the as- 
 sumed root to the true root nearly. (4.) The number thus found 
 may bs employed as a new assumed root, and the operation repeated 
 to find a result still nearer the true root. 
 
 For the mode of investigating this rule, which is perhaps the best and 
 most convenient that has been discovered, see the teruth of Dr. Hutlon's 
 Tracts, where it was first given. 
 
 Exam. 1. Required the 365th root of 1-06. 
 
 Here we may take 1 for the assumed root, the 365th power of 
 which is 1 ; and n being 365, we have n-\- 1 =: 366, and n I 364. 
 The work will then proceed in the following manner, and the an- 
 swer is found to be 1-0001596. 
 
 1 X 366 == 366 I X 364 = 364 
 
 1-06 X 364 = 385-84 1-06 X 366 = 387-96 
 
 As 751-84 : 751-96 : : 1 : 1-0001596. 
 
 In extracting the fourth root, we may either use the preceding rule, 
 or we may extract the second root of the given number, and the second 
 root of the result. In extracting the sixth root also, we may eithet use 
 
210 SERIES. 
 
 the preceding rule, or we may extract the third root of the giren number, 
 and the second root of the result: and in this way we may proceed in 
 every case in which the index of the root to be extracted is a composite 
 number. When the index is a prime number, however, the root must 
 be found by the general rule. 
 
 Exam. 2. Find the value of 111 
 
 This expression means the third root of the second power of 1 i, 
 and therefore we may extract the cube root of 121 for the required 
 result. The answer might also be obtained by the general rule by 
 faking 5 for the assumed root, and I, the reciprocal of I, for n. 
 The pupil will probably find the former method the more simple. 
 The answer will be found to be 4-946088. 
 
 Exercises. Find the roots of the following numbers signified by 
 their several indices: 
 
 Exercises. Answers. I Exercises. Answers. 
 
 1 987654321* 19-27274 3. I05 T ^ 1-004074 
 
 .. 1-047128 4. 9^... .. 52-19591 
 
 SERIES. 
 
 A SERIES is a succession of quantities, or terms, that de- 
 pend on one another according to a certain law. 
 
 In every series, the first and last terms are called the 
 EXTREMES, and the rest the MEANS. 
 
 Writers on Arithmetic usually treat of only two kinds of series, 
 equidifferent series and continual proportionals. These are of more frequent 
 and general use, than other kinds of series, and on this account claim 
 more particular attention. Quantities in equidifferent series are also 
 said to be in arithmetical progression ; and continual proportionals are said 
 to be in geometrical progression. * 
 
 * These names for equidifferent quantities and continual proportionals, are highly im- 
 proper. Series of both kinds belong equally to Arithmetic and Geometry. The appel- 
 lations, arithmetical progression and geometrical progression, should, thi-refore, be entirely 
 disused, as tending to impress false ideas on the mind respecting the nature of the quanti- 
 ties. The term proportion is applied, if possible, still more improperly to equidifferent 
 quantities, as this term is alv/ays expressive not of equality of differences, but of equality 
 qf ratios. The latest and best continental writers have accordingly rejected these terms, 
 and substituted more appropriate ones, calling them by the names above given, or others 
 of similar import, such as progressions by difference*, and progressions by quotients. 
 Bor.nycastle, in his larger work on Algebra, though he still retains the terms, acknowledges 
 their impropriety. W.th respect to the name, continual proportionals, here applied to 
 the second kind of quantities, it may be observed, that, besides its being perfectly ex. 
 pressive of the nature of such quantities, it has long been thus applied in works oa 
 Geometry, and it is equally applicable in Arithmetic. 
 
211 
 
 EQUIDIFFERENT SERIES, OR ARITHMETICAL 
 PROGRESSION. 
 
 An EQUIDIFFERENT SERIES is that in which the terms 
 either all increase, or all decrease, by the same quantity.* 
 This quantity is called the COMMON DIFFERENCE. 
 
 Thus, 5, 7, 9, 11, 13, is an equidifferent series, in which the succes 
 sive terms increase by the addition of the common difference 2 : and 20, 
 17, 14, 11, 8, is another, in which the successive terms decrease by the 
 common difference 3. 
 
 The following are the most useful rules for the management of quan- 
 tities of this kind. 
 
 RULE I. The Jirst term and the common difference being 
 given, to Jind any other assigned term: (I.) Multiply the 
 common difference by the number which is equal to the 
 number of the terms preceding the required term : (2.) Then, 
 if it be an increasing series, add the product to the first 
 term ; otherwise, subtract it. 
 
 Exam. 1. Required the thirty-fifth term of the increasing 
 equiditferent series, whose first term is 7, and common difference 3. 
 
 Here, 34- times precede the required term; wherefore, 34 X 3 -f- 
 7 = 109, is the term required. 
 
 The reason of this operation will be manifest from the consider- 
 ation, that were the series to be continued to the thirty-fifth term, 
 the first term must be increased by 34? additions of the common 
 difference. 
 
 Ex. 1. Required the fifty-fourth term of the decreasing equidif- 
 ferent series whose first term is 100, and common difference 
 }\. Answ. 33|. 
 
 2. First term of an increasing series 36, common difference 3f ; 
 required the hundredth term. Answ. 392f . 
 
 3. First term of a decreasing series 329, common difference $ ; 
 required the ninty-ninth term. Answ. 24>3|. 
 
 RULE II. The extremes and the number of terms being given, 
 to Jind the sum of the series . Multiply the sum of the ex- 
 tremes by the number of terms, and take half the product. 
 
 Exam. 2. The first term of an equidifferent series is I, its 
 last term 312, and the number of terms 193. What is its sum ? 
 
 Here, 1+312 := 313, and 313 X 193 = 60409, tne half of which 
 is 30204^, the sum of the series. 
 
 The definition of eqi:idifferent quantities would stand more correctly, though n >t 
 so simply, thus : An equtdiffereni scries is that in which the terms either ail increase, or 
 nil decrease, and the difference of any <Uw adjacent terms is the same as that of nay other 
 tio adjacent terms ; and the cciimtton difference may defined to be the difference 'iftutfeit 
 any (wo .idjaceft tera. 
 
212 EQUID1 FFK RENT SERIES. 
 
 The reason of this ride will be understood from the following 
 property of equidifferent quantities : 
 
 In any equidifferent series, the sum of the extremes is equal to the 
 sum of any two terms that are equally distant from them, or to double 
 the middle term, if the number of terms be odd. 
 
 Thus, in the series 5, 8, 1 1, 14, 17, 20, 23, the sum of 5 and 23 
 is equal to the sum of 8 and 20, and of 11 and 17, and is double 
 of the raiddle term 14. The reason of this will appear manifest 
 if it be considered that 8 and 11 respectively exceed the less ex- 
 treme by the same quantities by which 20 and 17 are respectively 
 less than the other extreme. 
 
 Hence, with respect to the sum of this latter series, it is evi- 
 dent that though each term were made 14, half the sum of the 
 extremes, still the sum of the whole would be the same ; and con- 
 sequently the sum of the series will be obtained by multiplying 
 half the sum of the extremes by the number of terms, or which is 
 the same by multiplying the sum of the extremes by the number 
 of terms, and taking half the product. 
 
 Ex. 4-. Given the greater extreme r= 1000, the common differ- 
 ence 2, and the number of terms 367 ; required the sum of the 
 series. Answ. 221464. v 
 
 In working this exercise, the less extreme will be found by rule I. 
 and the sum by rule II. 
 
 Ex. 5. Given the greater extreme =r 1, the common difference 
 == rtji7> an d the number of terms zr 51 ; required the sum of the 
 scries. Answ. 38. 
 
 RCJLE III. The extremes and the common difference being 
 given, to find the number of terms : Divide the difference of 
 the extremes by the common difference, and add a unit to 
 the quotient. 
 
 The reasons of this rule and of the next will be obvious from com- 
 paring them with rule I. 
 
 Ex. 6. Given the greater extreme = 500, the less rr 70, and the 
 common difference 10 ; required the number of terms, Answ. 44. 
 
 7. Given the less extreme rr 3, the greater, 579, and the 
 common difference =r 9; what is the sum of the series? Ansiv. 18915. 
 
 Here, let the number of terms be found by this rule, and the sum of 
 Oie series by rule II. 
 
 Ex. 8. With, the common difference 12, how many equidif- 
 ferent means can be inserted between the extremes 8 and 1700? 
 Ansiv. 140. 
 
 RULE jtV. The extremes and the number of ferns being 
 tojind the common difference ; Take a unit from the 
 
EQUIDIFFERENT SERIES. 213 
 
 number of terms, and by the remainder divide the difference 
 of the extremes. 
 
 Ex. 9. Given the extremes = 3 and 300, respectively, and the 
 number of terms =: 10; required the common difference. Answ. 33. 
 
 10. What is the common difference of a series consisting of 1001 
 terms, the extremes being 1 and 100001 ? Answ. 100. 
 
 RULE V. The extremes being given, to find any assigned 
 number of equidifferent means : Find the common difference 
 by rule IV, and add it continually to the less extreme, or 
 subtract it from the greater ; the several results will be the 
 required terms. 
 
 One mean may be found by taking half the sum of the 
 extremes. 
 
 Exam. 3. Given the first term 1, the last =: 99, and the 
 number of terms 8 ; required the complete series. 
 
 By rule IV. the common difference is found to be 14, by the 
 continual addition of which to 1, the entire series is found to be 
 1, 15, 29, 43, 57, 71, 85, 99. 
 
 Ex. 11. Insert 5 equidifferent means between 20 and 30. 
 An*iv. 2 If, 23i, 25, 26f, 28^ 
 
 12. Required the several terms of a series the extremes of which 
 are 4 and 49, and the number of terms 6. Answ. 4,13,22,31,40,49. 
 
 RULE VI. The sum of the series, one extreme, and the num- 
 ber of terms beinv given j to find the other extreme : Divide 
 twice the sum of the series by the number of terms, and 
 from the quotient take the given extreme. 
 
 The reason of this rule is evident from rule II.* 
 
 Ex. 13. Given the first term of an equidifferent series consisting 
 of 24- terms, =r 1 ; required the last term, the sum of the series 
 being 576. Ansiu. 47. 
 
 * If the greater extreme be denoted by g, the less by /, the common difference by J, the 
 number of terms by n, and the sum of the series by*: then g = l+(n l}d, and * = 
 n(g-+4) ; from which twoequation.sanythreeoftlie.se quantities being given, the rest ran 
 De fuu'.id. The full resolution of these equaMons will, among other results, give the 
 rules contained in the text. 
 
 The consideration of such equidifferent quantities as are usually, but very improperly 
 said to be in arithmetical proportion, has been omitted from its comparative inutility. 
 These are quantities of such a nature, that the differences of the first and second, of the 
 third and fourth, of the fifth and sixth, &c., are equal. Of thi kind are the following 
 quantities : 2, 5 ; 10, 13 ; 21, 4, &c. The principal property of four such quantities is, 
 that, as in the continued equidifferent quantities already explained, the sura of the ex- 
 tremes is equal to the sum of the means ; whence it follows, that if there be four such 
 quantities, and from the sum of the means either extreme be taken, the remainder Will 
 be Uie other extreme. 
 
214 SERIES. 
 
 14-. Giver, the number cf terras = 50, and the strm of the 
 series = 1275; required the greater extreme, the less being 3. 
 Answ. 47. 
 
 15. Required the sum of the first ten thousand numbers in the 
 natural series, 1, 2, 3, 4, &c. Answ. 50,005,000. 
 
 16. Required the sum of the first ten thousand odd numbera, 
 l,3,5,7,&c. Answ. 100,000,000. 
 
 17. Required the sum of the first ten thousand even numbers, 
 2,4,6,8,&c. Answ. 100,010,000. 
 
 18. Required the sum of the first ten thousand numbers that 
 are divisible by 3, (3, 6, 9, 1 2, &c.) Aw>w. 1 50,0 1 5,000. 
 
 19. If a person on a journey travel the first day 30 miles, and 
 each succeeding day a quarter of a mile less than he did the day 
 before, how far will he travel in 30 days? Answ. 791 5 miles. 
 
 20. How many strokes does a common clock strike in a year ? 
 Answ. 56940. 
 
 21. If 120 stones be laid in a straight line, each at the distance 
 of a yard and a quarter from the one next it ; how far must a per- 
 son travel who picks them up singly and places them in a heap at 
 the distance of 6 yards from the end of the line, and in its continu- 
 ation ? Answ. 8 Irish miles, 4 furl. 35 per. 5 yds. 
 
 22. A body falling by its own weight, if it were not resisted by 
 the air, would descend in the first second of time through a space 
 of 16 feet and 1 inch; in the next second, through three times 
 that space ; in the third, through five times that space ; in the 
 fourth, through seven times, &c. Through what space would it 
 fall at the same rate of increase in a minute ? Answ. 57900 feet. 
 
 CONTINUAL PROPORTIONALS, OR GEOMETRICAL 
 
 PROGRESSION. 
 A SERIES OF CONTINUAL PROPORTIONALS IS that in which 
 
 the successive terms all increase by a common multiplier, 
 or all decrease by a common divisor.* 
 
 The common multiplier, or common divisor, is called 
 
 THE RATIO OF THE SERIES. Or THE COMMON RATIO. 
 
 Thus, 3, 6, 12, 24, 48, are continual proportionals, in which the suc- 
 cessive terms increase by the ratio 2 ; and 192, 48, 12, 3, f, &c. are con- 
 tinual proportionals decreasing by the ratio 4. 
 
 It may be observed, that we might regard every series of this kind, 
 whether increasing or decreasing, as being produced by multiplication, 
 the ratio in a decreasing series being a proper fraction. Thus, in the 
 
 * This definition would stand more correctly, though not so simply, thus : A seriet of 
 continual proportionals is that in which the terms all increase, or all decrease ; and 
 the quotient obtained by dividing the greater of any two adjacent terms by the less, is 
 equal to that obtained by dividing the g> eater nfimy other two adjacent terms by the less. 
 
CONTINUAL PROPORTIONALS. 21o 
 
 series last given, the ratio or common multiplier might be considered to 
 he ^. In what follows, however, the ratio will be taken always greater 
 than a unit, according to the definition already given.* 
 
 The following are the most useful rules for the management of quan- 
 tities of this kind. 
 
 RULE I. The first term and the ratio being given, to find 
 any other proposed term : (1.) Raise the ratio to a power 
 whose index is equal to the number of the terms which pre- 
 cede the required term. (2.) Then, if it be an increasing 
 series, multiply the first term by the result before found ; 
 otherwise divide it by that result. 
 
 Exam. 1. Required the 8th term of the series of continual 
 proportionals, whose first term is 6, and ratio 2. 
 
 Here, the 7th power of 2 is found to be 128; which being mul- 
 tiplied by the first term 6, the product is 768, the 8th term. 
 
 The reason of this operation will be manifest if it be considered, 
 that in finding the successive terms up to the 8th, the first term 
 must be multiplied by 2, the product by 2, that product by 2, and 
 so on, till the 8th term would be found after seven such multipli- 
 cations : and it is evident, that the same result will be found by a 
 single multiplication by the 7th power of 2. A similar illustration 
 serves in case of a decreasing series. 
 
 Exam. 2. Required the 20th term of the series, whose first 
 term and ratio are each T06. 
 
 Here, we are to multiply the 19th power of 1 06 by 1*06, or, 
 which is the same, we are to involve 108 to the 20th power. 
 This is found by involution, to be 3-207135. 
 
 Exam. 3, Required the 6th term of the "decreasing series 
 whose first term is 100, and ratio 1. 
 
 The 5th power of 1, or 1-5, is 7-59375, and 100 being divided 
 by this, the quotient is 13-16872428, the term required. 
 
 Ex. 1. Given the first term of an increasing series 12, and its 
 ratio 3; to find the 18th term. Answ. 15-19681956. 
 
 2. The first term of a decreasing series is 1, and the ratio 1*07; 
 required the 14th term. AKSW. '4 149644. 
 
 3. The first term of an increasing series is 194-3, and the 
 ratio 1-05; required the 3 1st term. Answ. 839-75333. 
 
 4. Given 'the first term of a decreasing serjes r= 500, and the 
 ratio = 1-04; to find the 14th term. Answ. 300-287. 
 
 * It is proper for the learner to know, that in a series of continual proportionals, the 
 proiluct of the extremes is equal to the product of any two terms equally distant frwn 
 than ; or to the second power of the middle term, if the number of terms be odd. The rea- 
 son of this is evident, since the greater extreme exceeds the term next it in the same 
 ratiu in which the other extreme is less than the term next it. 
 
:216 SERIES. 
 
 5. Given the first term of an increasing serfes 1, and the 
 ratio 2 ; required the 3oth term. Ansiv. ^34359738368. 
 
 6. Given the first term of an increasing series 1, and the 
 ratio = 3; required the 36th term. Answ. 50031545098999707. 
 
 RULE II. To find the sum of a series of continual propor- 
 tionals ; multiply the greater extreme by the ratio, and di- 
 vide the difference between the product and the less ex- 
 treme, by the difference between the ratio and a unit. 
 
 When the series is a decreasing one, and the number oj 
 terms infinite, divide the product of the ratio and the great- 
 est term by the difference between the ratio and a unit. Or, 
 
 Divide the ratio by the difference between it and a unit, 
 and multiply the quotient by the first term.* 
 
 Exam. 4. Given the first term of an increasing series rr 4, the 
 ratio = 3, and the number of terms = 6; to find the sum of the 
 series. 
 
 Here, by rule I. we find the last term to be 972. Multiplying 
 this by the ratio, we obtain 2916 : and dividing 2912, the dif- 
 ference between this and the first term, by 2, the difference, be- 
 tween the ratio and a unit, we obtain 1456, the required sum. 
 
 The reason of this operation is best shown by Algebra; it may be 
 illustrated in the following manner however : let the terms of the 
 
 ed ri a S s b fn P 't a h C ; *+ 12 + 36 + '"8+324+972 = sum 
 mar^n ' then 12 + 36 + 108+324+972 + 2916 = sum X 3 
 
 let each term be multiplied by the ratio, and let the products be re- 
 moved each one place towards the right hand. If the upper line 
 be then subtracted from the lower, there will remain 2912 rz sum 
 X2; and consequently the sum is equal to 2912-f-2zr 1456. 
 Now 2916 is evidently the product of the ratio and the greater 
 extreme, and 2912 is the difference between this and the less ex- 
 treme ; also the divisor 2 is the difference between the ratio and 
 a unit : and a similar illustration may be given in any other case. 
 In^a decreasing infinite series, the last term is to be regarded as 
 nothing; and hence the reason of the rule for its summation is 
 manifest. 
 
 * The following rules, which may be illustrated in the same manner as the rule given 
 io the text, may also on some occasions be employed with advantage : 
 
 I. To sum an increasing series : (1.) Raise the ratio to a power whose index is equal to 
 the number of terms : (2.) Divide the difference between die result and a unit, by the 
 difference between the ratio and a unit: (3.) Multiply the quotient by the first term. 
 
 II. To sum a decreasing series : (1.) Raise the ratio to a power whose index is equal 
 to the number of terms : (-2.) From the power thus obtained, take a unit, and divide the 
 remainder by the same power : (3.) Divide the quotient by the difference between the 
 ratio and a unit, and multiply the result by the firt term. 
 
CONTINUAL PROPORTIONALS. 217 
 
 Exam. 5. Required the sum of the series whose least term is 
 i5, and ratio 1-04, the number of terms being 31. 
 
 This exercise may be wrought in the same manner as the last ; 
 or perhaps more easily by the first rule in the note. Thus, by in- 
 volution, we find the thirty-first power of 1-04 to be 3-37313 
 Then 3-37313 1 = 2-37313, and 1 -04- 1 r= -04 ; also 2-37313 
 -r--04 = 59-32825, the product of which by 45, the first teim is 
 2669-77125, the required sum. 
 
 Exam. 6. Required the value of the interminate decimal -1'8'. 
 
 This is the same as ^+ToVoo + T<joVo<yo-f- &c., continued 
 without limit, where the ratio is evidently 100. Multiplying the 
 first term therefore by 100, and dividing the result by 100 1, 
 we obtain for the sum of the series, or the value of the decimal, 
 , or in its lowest terms, T 2 T . 
 
 Ex. 7. Required the sum of the series whose first and last 
 terms are 100, and 13-16872428 respectively, and its ratio H. 
 Answ. 273-66255144. 
 
 8. Given the extremes = 1 and 18*42015, and the ratio = 1-06; 
 required the sum of the series. Answ. 308-755983. 
 
 9. Find the sum of the infinite series whose greatest term is 
 100, and ratio 1'04. Answ. 2600. 
 
 10. Given the first term of an increasing series = 6, the ratio 
 = 4, and the number of terms = 8; to find the sum of the series. 
 Answ. 131070. 
 
 11. Given the first term of a decreasing series = 1, the ratio 
 rr 3, and the number of terms zr 12 ; to find the sum of the series. 
 Answ. IT $ TI\V> or 1^ nearly. 
 
 12. Given the least term of a series = 1, the ratio = 1|, and 
 the number of terms := 16: required the sum of the series. 
 Answ. 1311-68167114. 
 
 13. Given the least term := , the ratio = 4, and the number of 
 terms = 14 : required the sum of the series. Answ. 11184S10|. 
 
 14. Given the greatest term =-12, the ratio := 1 J-, and the number 
 of terms = 12; to find the sum of the scries. Answ. 53-2706. 
 
 RULE III. The extremes and the number of terms being 
 given, to jind the ratio : Divide the greater extreme by the 
 less, and extract that root of the quotient whose index is 
 one less than the number of terms. 
 
 Exam. 7. Given the extremes of a series r= 3 and 192, and 
 the number of terms = 7 : required the ratio. 
 
 Here, 192-f-S = 64, the 6th root of which is 2, the ratio of the 
 series. In the use of this rule we must generally employ either 
 logarithms, or the rule given in page 211. 
 
 Ex. 15. Given the extremes == 1 and 10, and the number of 
 terms = 9 : required the ratio. Answ. 1-333521. 
 
 L 
 
218 SERIES. 
 
 RULE IV. Tojind any proposed number of mean propor- 
 tionals between two given numbers : (1.) Take the two given 
 numbers as extremes ; take also the number of terms in the 
 series two greater than the required number of means, and 
 find the ratio by rule III : (2.) Then the product of the ratio 
 and the less extreme will be one of the means; the product oi 
 this mean and the ratio will be another ; and thus all the 
 means may be found, whatever is their number. 
 
 When only one mean is required, it is most easily found by 
 
 extracting the second root of the product of the extremes. 
 
 Exam. 8. Find three mean proportionals between 5 and 1280. 
 
 Here, the series would consist of 5 terms, and the extremes are 
 5 and 1280; and hence the ratio is found, by the last rule, to be 
 4; and by the repeated multiplication of this and of the first term, 
 the means are found to be 20, 80, and 320. 
 
 Exam. 9. Find a mean proportional between 5 and 10. 
 
 Here, 5 X 10 = 50, the square root of which is 7-0710678, &c. 
 the mean required. 
 
 Ex. 16. Find two mean proportionals between 1 and 2. 
 Answ. 1-259921, and 1-587401. 
 
 17. Find a mean proportional between -^ and 100. Answ. 
 3-16227766. 
 
 Ex. 18. If a thrasher agree to work 18 days for a farmer, on 
 condition of receiving two grains of wheat for the first day's work, 
 6 grains for the second, 18 for the third, &c. : what would be the 
 value of all he would be entitled to receive, supposing 7680 grains 
 to fill a pint, and the wheat to be worth 7 shillings a bushel ? 
 Answ. 275 17 5J. 
 
 19. Suppose a house, having 20 windows, to be sold at 
 the rate of 4/0 for the first of these windows, 6/0 for the second, 
 9/0 for the third, and so on ; the value of each being increased by 
 one half of itself, to find the value of the next ; for how much 
 would it be sold ? Answ. 1329 14 0$. 
 
 20. If a father give as a portion to his daughter, aged 19, a far- 
 thing for the first year of her life, three farthings for the second, 
 9 farthings for the third, &c. ; . to how much does her portion 
 amount? Answ. .605,344 10 3^. 
 
 21. It is said that an Indian discovered the game of chess, and 
 showed it to his sovereign, who was so much pleased with it, that 
 he desired the inventor to ask any reward he chose. The inves- 
 tor then asked one grain of wheat for the first square of the chess 
 table, two for the second, four for the third, and so on ; doubling 
 continually to 64, the number of squares. Now, suppose it had 
 been possible for the prince to pay this reward, what would have 
 
HARMONICAL PROPORTION. 219 
 
 been the value of the whole at 12/6 W cwt., 10,000 grains being 
 supposed to weigh a pound avoirdupois ? 
 Answ. 10,293,942,005,418 5 6^. 
 
 22. Find the value of the interminate decimal "IG'S'. Answ. ^^. 
 
 23. Required the value of the infinite decimal '51'85'. Answ. \$. 
 
 24. Find the sum of the infinite series i, , -, T V &c. Answ. 1. 
 
 25. Find the sum of the infinite series 3-, ^, ^ , B* r , &c. Answ. . 
 
 HARMONICAL PROPORTION. 
 
 It may be proper to subjoin to what lias been said respecting equidif- 
 ferent quantities and continual proportionals, a few observations on har- 
 monical proportion, a subject which, though of minor importance, should 
 not be entirely overlooked. 
 
 Three or four numbers are said to be in HARMONICAL PROPORTION, 
 when the first is to the last, as the difference of the first and second is to 
 the difference of the last and the last but one. 
 
 Thus, 2, 3, and 6, are three numbers in harmonical proportion, 2 being 
 to 6 as 3 2 to 6 3: and 15, 12, 6, and 5, are four numbers in har- 
 monical proportion, since 15 : 5 : : 15 12 : 6 5. 
 
 The reciprocals of three equidifferent numbers are in harmonical pro- 
 portion. Thus, A, , and , the reciprocals of 2, 5, and 8, are in har- 
 monical proportion : and if these fractions be reduced to equivalent ones 
 having a common denominator, the numerators 20, 8, and 5, will be of 
 the same nature. 
 
 To find^/bur numbers in harmonical proportion, find three such num- 
 bers in the manner just shown ; and then let the mean ancione extreme, 
 and the mean and the other extreme, be multiplied or divided by any 
 numbers whatever, and the four numbers thus found will be the terms 
 required. Thus, taking 20, 8, and 5, the numbers last found, we have, by 
 halving the first term and the mean, and trebling the mean and the third 
 term, 10, 4, 24, and 15, which are in harmonic proportion, since 10 j 
 15 :: 104 : 2415 ; and in this manner we may find as man,y such 
 numbers as we please.* 
 
 * It is worthy of remark, that the reciprocals of continual proportionals are also con- 
 tinual proportionals, while the reciprocals of three equidifferent numbers are ia hanuoni- 
 cal proportion, and, consequently, the reciprocals of three numbers in harmonical pro- 
 portion are equidifferent numbers. Farther also, if any two numbers be taken as ex- 
 tremes, the harmonical mean, themean proportional, and the equidifferent mean between 
 them are three continual proportionals Tims, the three means between 20 and 5 ; .ire 
 8, 10, and 12 : the second of which is a mean proportional between the other two. 
 
 I o 
 
POSITION.* 
 
 POSITION is a rule by which, from the assumption of one 
 or more false answers to a problem, the true one is obtained. 
 It admits of two varieties, Single Position^ and Double 
 Position. 
 
 In SINGLE POSITION the answer is obtained by one as- 
 sumption : in DOUBLE POSITION it is obtained by two- 
 Single Position may be employed in resolving problems, in which the 
 required number is any how increased or diminished in any given ratio ; 
 such as when it is increased or diminished by any part of itself, or when 
 it is multiplied or divided by any number. 
 
 Double Position is used, when the result obtained by increasing or 
 diminishing the required number in a given ratio, is increased or 
 diminished by some number which is no known part or multiple of the 
 required number ; or when any root or power of the required num- 
 ber, is either directly or indirectly contained in the result given in the 
 question. 
 
 SINGLE POSITION.f 
 
 RULE. (1.) Assume any number, and perform on it the 
 operations mentioned in the question as being performed on 
 the required number : (2.) Then, as the result thus obtain- 
 ed, is to the assumed number, so is the result given in the 
 question, to the number required. 
 
 Exam. 1. Required a number to which if one half, one third, 
 one fourth, and one fifth of itself be added, the sum may be 1644. 
 
 Suppose the number to be 60 : then, if to 60 one half, one third, 
 one fourth, and one fifth of itself be added, the sum is 137. Hence, 
 according to the rule, as 137 : 1644 : : 60 : 720, the. number re- 
 quired. The truth of the result is proved by adding to 720, one 
 half, one third, &c. of itself, and the sum is found to be 1644. 
 The number 60 was here assumed, not as being near the truth, but 
 as being a multiple of 2, 3, 4, and 5 ; and in this way the operation 
 was kept free from fractions. By the assumption of any other 
 number, however, the answer would have been found correctly, 
 but often not so easily. The reason of the operation is obvious 
 from the principles of proportion. 
 
 Ex. 1. Divide 2000 between A, B, and C, giving A as much 
 as B and a fifth part more, and C as much as both together. 
 Answ. A's part 545 9 1, B's 454 10 11, C's 1000. 
 
 * This rule is sometimes called the Rule of False, or the Rule of False Position, or the 
 Rule of Trial and Error. It might properly be called the Rule of Supposition. 
 
 f Every question that can be resolved by this rule, may also be resolved by the rule for 
 Double Position, or without Position, by some of the preceding rules; and hence this 
 rule is of little importance. 
 
DOUBLE POSITION. 221 
 
 2. One third ~of a ship belongs to A, and one fifth to B, and 
 A's part is worth 1000 more than B's : required the value or 
 the ship. Answ. 7500. 
 
 3. It is required to divide 252 into three parts, such that one 
 third of the first, one fourth of the second, and one fifth of the 
 third, shall all be equal to one another. Answ. 63, 84, and 105. 
 
 4. To find a number such that if it be multiplied by 10, and the 
 product be divided by 13, the quotient, increased by the number 
 itself, and by 80, will amount to 1000. Answ. 520. 
 
 5. Required a number to which if one half of itself, one third of 
 that half, and one fourth of that third, be added, the sum will be 
 500. Answ. 292Jf. 
 
 6. A father bequeaths to his three sons 7000 in such a man- 
 ner, that if the share of the eldest be multiplied by 5, that of the 
 second by 6, and that of the third by 7, the products are all equal. 
 What are their shares ? Answ. 2747 13 ^, 2289 14 42, and 
 1962 12 4. 
 
 7. The number of a gentleman's horses is two fifths of the num- 
 ber of his black cattle, and for every four of the latter he has eleven 
 sheep. Required the number of each, the number of the sheep 
 exceeding that of the horses by 141. Answ. 24 horses, 60 black 
 cattle, 1 65 sheep. 
 
 DOUBLE POSITION. 
 
 RULE I. (I.) Assume two different numbers, and perform 
 on them separately the operations indicated in the question: 
 (2.) Then, as the difference of the results thus obtained, is 
 to the difference of the assumed numbers, so is the difference 
 between the true result and either of the others, to the cor- 
 rection to be applied, by addition or subtraction, as the case 
 may require, to the assumed number which gave this result. 
 
 This rule, which was first published in substance by Mr. Bonnycastle 
 in his larger work on Arithmetic, in 1810, is the simplest and easiest that 
 has yet appeared for the resolution of questions in which the given result 
 is a known number, independent on tJie required number : and these ques>- 
 tions are generally the most useful. Mr. Bonnycastle appears, however, 
 not to have been aware, that this rule fails in relation to the whole class 
 of questions, in which the result of the operations to be performed, ac- 
 cording to the question, on the required number, is not a known, determi- 
 nate number, but tlte required number, or one depending on it, such as some 
 multiple or part of it. In that case the following rule will be necessary. 
 This rule has also the advantage of being applicable in every case whatever. 
 
 RULE II. (1.) Having assumed two different numbers, per- 
 form on them separately the operations indicated in the 
 question, and find the errors of the results. (2.) Then, as the 
 difference of the errors, if both results be too great or both 
 too little, or as the sum of the errors, if one result be too 
 
222 DOUBLE POSITION. 
 
 great and the other too small, is to the difference of the as- 
 sumed numbers, so is either error to the correction to be 
 applied to the number that produced that error. 
 
 Exam. 1. Required a number, from which if 2 be subtracted* 
 one third of the remainder will be 5 less than half the required 
 number. 
 
 Here, suppose the required number to be 8, from which take 2, 
 and one third of the remainder is 2. This being taken from one 
 half of 8, the remainder is 2, the first result. Suppose again, the 
 number to be 32, and from it take 2 : one third of the remainder 
 is 10, which being taken from the half of 32, the remainder is 6, 
 the s&cond result. Then, the difference of the results being 4, the 
 difference of the assumed numbers 24, and the difference between 
 5, the true result, and 6 the result nearer it, being 1 ; as 4 : 24 : : 
 1 : 6, the correction to be subtracted from 32, since the result 6 
 waa too great. Hence, the required number is 26. 
 
 Exam. 2. If one person's age be now only four times as great 
 as another person's, though 7 years ago it was six times as great : 
 what is the age of each ? 
 
 Here, suppose the age of the younger to be 12 years; then would 
 the age of the older be 48. Take 7 from each of these, and there will 
 remain 5 and 41, then* ages 7 years ago. Now, 6 times 5 is 30, 
 which taken from 41, leaves an error of 11 years. By supposing 
 the age of the younger to be 15, and proceeding in a similar man- 
 ner, the error is found to be 5 years. Hence, as 6, the difference 
 of the errors, (both results being too small,) is to 3, the difference 
 of the assumed numbers, so is 5, the less error, to 2, the correc- 
 tion ; which being added to 15, the sum, 17, is the age of the 
 younger, and consequently that of the older must be 70. 
 
 Both the rules above given for Double Position depend on the prin- 
 ciple, that the differences between the true and the assumed numbers, are 
 proportional to the differences between the result given in the question 
 and the results arising from the assumed numbers. This principle is 
 quite correct in relation to all questions which in Algebra would be re- 
 solved by simple equations, but not in relation to any others ; and hence, 
 when applied to others, it gives only approximations to the true results. 
 la this case the assumed numbers should be taken as near the true an- 
 swer as possible. Then, to approximate the required number still more 
 nearly, assume for a second operation the number found by the first, and 
 that one of the two first assumptions which was nearer the true answer, 
 or any other number that may appear to be nearer it still. In this way, 
 by repeating the operation as often as may be necessary, the true result 
 may be approximated to any assigned degree of accuracy. When ap- 
 plied in this way, Double Position is of considerable use in Algebra, 
 affording in many cases a very convenient mode of approximating the roots 
 of equations, and finding the values of unknown quantities in very com- 
 plicated expressions, without the usual reductions. 
 
 Exam. 3. Required a number to which if twice its square be 
 added, the sum will be 100. 
 
DOUBLE POSITION. 223 
 
 It is easy to see that this number must be between 6 and 7. 
 These numbers being assumed, therefore, the sum of 6 and twice 
 
 twice its square be added, and the result is 99-8448. Then, as 
 105 99-8448:7 6-82 : : 105 100: -1746 ; which being taken 
 from 7, the remainder is 6-8254-, the required number still more 
 nearly : and if the operation were repeated with this and the for- 
 mer approximate answer, the required number would, be found 
 true for seven or eight figures. 
 
 Ex. 1. A merchant increased his capital each year by a fourth 
 of itself, except an expenditure of .300 ^ annum, and at the end 
 of four years found himself worth 5000. What was his original 
 capital? Answ. 2756 9 7$. 
 
 2. Suppose every thing to be as in the last exercise, except that 
 at the end of four years the merchant found himself possessed of 
 twice his original stock: how much had he to begin with? 
 Answ. 3918 11 8. 
 
 3. Required as in the two preceding exercises, supposing every 
 thing as before, except that at the end of the time the merchant 
 finds himself possessed of only half his original capital. Ansiu. 
 890 18 11. 
 
 4. Required a number from which if 81- be taken, three times 
 the remainder will exceed the required number by a fourth of 
 itself. Answ. 144. 
 
 5. Required a number such that if it be multiplied by 1 1, and 
 320 be taken from the product, the tenth part of the remainder will 
 be 20 less than the number itself. Answ. 120. 
 
 6. Required a number whose half is as much less than 1000, as 
 its double is greater than 999. Answ. 799$. 
 
 7. A farmer engaged a labourer on condition of paying him 1/4 
 a day for every day he should work, and of charging him 9d. for 
 his I)oar4ing every day he should be idle. Now, at the end of a 
 
 -*^ar (313 days) the man was entitled to 12 : how many days then 
 did he work ? Answ. 227%$ days. 
 
 8. How many guineas of 1 2 9, and moidores of I 9 3 each, 
 will pay a bill of 130, the number of pieces of both kinds being 
 100? Answ. 50 of each. 
 
 9. If 1 be added to a number, and 100 divided by the sum, the 
 quotient is 3 less than if 1 had been subtracted from the number, 
 and 100 divided by the remainder : required the number. Answ. 
 8-2259751. 
 
 10. Required a number which exceeds 3 times its square root 
 by 11. Answ. 26-4201648. 
 
 11. Required a numbei to which if twice its square and 3 times 
 its cube be added, the sum will be 2000. Answ. 8-506744. 
 
 13. Given the sum of two numbers == 20, and the sum of their 
 squares =z 324 : required the numbers. Answ. 17*8740079, &<-' 
 
224 
 COMPOUND INTEREST. 
 
 The method that naturally presents itself for finding the amount 
 of a sum at compound interest, is to find its amount at simple 
 interest at the end of the first year; then to take this amount as 
 a new principal, and find its amount in like manner, which would 
 be the amount at compound interest at the end of the second 
 year, and the principal for the third year, the amount of which 
 must be found in like manner. Continuing the process, we should 
 thus find the amount at the end of the proposed time. This will 
 be illustrated in the following example. 
 
 Exam. 1. Required the amount of 2500 at the end of 4 
 years, at 6 W cent. ^ annum, compound interest. 
 
 Here, the amount for 1 year is 2650 ; the amount of which 
 for 1 year also is 2809, the amount at compound interest for two 
 years. The amount of this again for 1 year, or the amount of the 
 given sum at the end of the third year, is 2977 10 9 : the 
 amount of which for the same time is 3156 3 10, the amount of 
 2500 for four years. The amount at simple interest would have 
 been 3100, which is less than the amount at compound interest 
 by 56 3 10. 
 
 When the time is short, this method may be practised without 
 much trouble; but when the time is long, the labour would be- 
 come very great. In this case, the methods that follow should be 
 employed. 
 
 RULE I. Tojind the amount of one pound sterling for any 
 number of years, at compound interest: (1.) Divide the 
 amount of 100 for 1 year by 100, and the quotient will be 
 the amount of one pound for 1 year: (2.) This amount in- 
 volved to a power denoted by the number of years, will be 
 the amount of 1 pound for that number of years. 
 
 The contracted mode of multiplication of decimals is peculiarly use- 
 ful in this rule, and in computations in compound interest and annuities 
 in general. So also is the contracted method of dividing decimals. 
 118, 121, and 203.) 
 
 Exam 2. Required the amount of one pound sterling for 20 
 years at 4 ty cent. ty annum, compound interest. 
 
 Here, the amount of 100 for 1 year is 104*5, the hundredth 
 part of which is 1*045, the amount of 1 for a year. The second 
 power of this is 1*092025, the amount of 1 at the end of the 
 second year. The product of this by itself, by the contracted 
 method of multiplication, is 1-192518, the amount at the end of 
 the fourth year. The square of this again is 1*4-22099, the 
 amount for eight years ; the square of which is 2*022366, the 
 amount at the end of the sixteenth year. Finally, the product of 
 
COMPOUND INTEREST. 2X, 
 
 this by ^ei-192518 (the amount for four years) is .2-41 1708, or 
 2 8 2f nearly, the amount of ,1 for 20 years. At simple inter- 
 est, the amount would have been only 1 13. 
 
 Had the products herebpen found at full length, the labour would have 
 been immense. In the last multiplication, one of the factors would have 
 contained 49 figures, the other 1C-, and the pioduct 61. It should be 
 carefully remarked, however, that the decimal part of the amount found 
 as above, will rarely be true in all its places. A trifling error in reject- 
 ing or over-estimating a figure at the end of s decimal may accumulate, 
 and render the accuracy of the last figure, or the last two figures doubt- 
 ful. Thus, in the preceding result, the two last figures should have 
 been 14 instead of OS, which would occasion an error of rather more 
 than a penny in the amount of .1000. When great accuracy is required, 
 therefore, the amounts should be brought out to a greater number of 
 places, and the last figure or two of the final amounts rejected, or not de- 
 pended on. The larger the sum also whose interest is required, and 
 the longer the time, this is the more necessary, as the effect of the error is 
 the more perceptible. 
 
 Exam. 3. What is the amount, true to six places of decimals, 
 f '1 for 6 years, payable half-yearly, at 5 ^ cent. & annum 
 compound interest ? 
 
 Here, the payments being half-yearly, the amount of 100 for 
 half a year is ,102 10, or 102-5 ; and, consequently, that of < 1 
 for the same time is 1-025: the square of this is "1-050625, or 
 1*0506250, the amount for two half years, or one year c Multiply- 
 ing this by 1-050625, by the contracted method, we obtain 
 1-1038129, the fourth power of 1-025, or the amount of \ for 
 two years ; the third power of which is 1-3448888, the twelfth 
 power of 1-025, or the amount of \ for six years or if only six 
 figures of decimals be retained, 1-344889. 
 
 Exercises. Find the amounts of t in the following exercises, 
 at the given rates W cent. W annum : 
 
 Exercises. Answers. 
 
 1. For 10 years, at 10$* cent., &c 2-593742 
 
 2. 17 6 2-802799 
 
 3. 100 3 ~~- 19-218632 
 
 4. 100 4 50-504948 
 
 5. ~~. 100 *~. 5 131-501258 
 
 6. 100 6 339-302083 
 
 RULE II. To find the amount, or the interest, of any sum, 
 it compound interest, for a given time, and at a given rate : 
 Find the amount of \ for the given time by rule I., and mul- 
 tiply it hy the given sum ; the product will be the amount 
 required. 
 
 If the principal be subtracted from the amount, the re- 
 mainder will be the interest. 
 
 L3 
 
226 COMPOUND INTEREST. 
 
 Exam. 4. Required the amount of 760 14. 4 for 12 years, 
 at 5 If cent. ^ annum, compound interest. 
 
 By rule I., the amount of 1 is found to be 1-795856. 
 This being multiplied by 760, ] '795856 
 
 and aliquot parts taken for 14/4, 144 
 
 as in the margin, there results 
 for the amount .1366 2 9; and 107751360 
 
 the principal being subtracted I2570QQ2 
 
 from this, there remains for the QQ'-OOQ f in/ 
 
 interest, 005 8 5. The siin- 359171 _ 4/ 
 
 >le interest would have been you^i ' * J.A 
 
 456 8 7. The same result 
 would have been obtained bv 
 
 would in that case have been found by multiplication alone. 
 
 Exercises. Find the amounts of the following sums at the given 
 rates ty cent. IP" annum : 
 
 Exercises. Answers. 
 
 s. d. 3. d. 
 
 7. 251 16 6 for 9 years, at 5 V cent., c 390 13 3 
 
 8. 212 0_l5 4 _ 381 16 
 
 9. 213 13 4 14 5 452 2 9 
 
 10. 463 10 10 12 6 932 14 8 
 
 11. 295 12 6 17 4^ 624 15 4 
 
 12. 495 7 6 13 3^ ___ 774 14 lOf 
 
 13. 649 13 6 16 5 1418 3 2 
 
 14. 582 7 6 5 5 761 2 9| 
 
 15. If a boy 12 years old, have a legacy of 1396 16 8 left to 
 him, how much will he have to receive at the age of 21, the legacy 
 being improved by compound interest at 5 ^ cent. W annum. ? 
 Answ. 2166 18*11. 
 
 16. Required the amount of 589 10 5 from the third till the 
 twenty-first year of a boy's life, at 5 ^cent. V annum, compound 
 interest. Avsw. 1418 15 0. 
 
 17. 648 from the 6th till the 21st year of a boy's life, at 4f, &c. 
 Answ. 1299 16 6. 
 
 18. If a merchant commence trade with a capital of 1200, and 
 eaoh year, after paying all expenses, increase the capital of the 
 former year by a fifth part of itself; how much will he be worth at 
 the end of 30 years ? Answ. 284,851 11 <j. 
 
 RULE III. Tojind the principal, which at a given rate, and 
 in a given time, will amount to a given sum : Or, to Jind the pre- 
 sent worth of a sum at compound interest for a given time, and 
 fit a given rate : Divide the given sum by the amount of 1, 
 found by rule I., and the quotient will be the principal, or 
 present worth required 
 
COMPOUND INTEREST. 227 
 
 The present worth of 1 may be found by dividing it 
 by its amount for the given time. 
 
 Exam. 5. What sum must be lent at compound interest, at 
 5 $F cent. & annum, at the birth of a child, so that the amount 
 may be 3000 6 8 at the end of 21 years? 
 
 Here, the amount of 1 for 21 years being 2-785962, we have 
 for answer 3000'3'-:-2-785962 = 1076-9469 = 1076 18 14. 
 
 Exercises. Required the present worths of the folio wi ig sums, 
 or the principals that would produce them, at compound interest, 
 at the given rates W cent. V annum : 
 
 Exercises. Anstvert. 
 
 s. d. s. d. 
 
 19. 324 18 6 for 9 years, at 5 V cent, &c 209 81 If 
 
 20. 264- U8 12 4 15112 4 
 
 21. 554- 184 27 4. 192 9 1 
 
 22. 1000 ~~. 22 8 ...-^ ..-.-+- 183 18 9 
 
 23. A brother is to pay his sister a portion of 4500 at the end 
 of 11 years: how much will discharge the debt at the end of 4 
 years, compound interest being allowed at 4 W cent ^ annum 
 on the sum he pays? Answ. 3306 14 6f. 
 
 24. With what capital must a merchant commence trade, to be 
 worth 15000 at the end of 12 years, if he may be expected to 
 clear annually an eighth of his capital? Answ. 3649 14 7f. 
 
 25. Whether is it better to sell a farm for 1000 payable at 
 present, 1000 payable at the end of 5 years, and 1000 payable 
 at the end of 10 years ; or to sell it for 3000 payable at the end 
 of 5 years, compound interest being allowed at 4 V cent. ^ an- 
 num ? Answ. Better at three payments, by 31 14< 2. 
 
 The reason of the first rule will appear from the following con- 
 siderations. The amount of 1 for a year will evidently be a hun- 
 dredth part of the amount of 100 : and as 1 is to its amount 
 for a year, so is any other principal to its amount for the same 
 time. Hence, to take a particular instance, the amount of 1 for 
 a year at 5 per cent., will be 1-05 ; and by the nature of compound 
 interest, this will be the principal for the second year. Then, as 
 the principal 1 : 1'05, its amount : : the principal 1-05 : 
 1'05% which will be the amount at the end of the second year, and 
 the principal for the third year. Again, as 1 : 1-05, its amount : : 
 the principal 1'05 2 : 1'05 3 , the amount at the end of the third 
 year, and the principal for the fourth year. . In this manner it will 
 appear, that the amount of one pound for any number of years 
 will be equal to 1'05 raised to the power denoted by the number 
 of years. The amount of 1 being thus determined, it is evident, 
 that the amount of any other principal will be had by multiplying 
 the amount of 1 by that principal, since the amount will evidently 
 be proportional to the principal ; which proves the second ru<e. 
 
228 COMPOUND INTEREST. 
 
 The third rule is evidently the converse of the second, and hence 
 
 its correctness is evident.* 
 
 It may be proper here to observe, that, if interest be payable yearly t 
 the amount of l at the end of 6 months will be the square root of its 
 mount for a year ; its amount for 4 months, or one-third of a year, the 
 cube root of the same ; for three months the fouith root; for a day the 
 365th root, &c. Farther, also, the amount of =1 for 2 days will be the 
 quare, for 3 days the cube, &c. of its amount for one day ; its amount 
 Or 8 months will be the square of its amount for 4 months ; its amount 
 for 9 months the cube of its amount for 3 months : and, finally, its 
 amount for a year and a quarter will be the product of its amounts for 
 a year, and for 3 months ; and its amount for 6 years and a half will be 
 the product of its amounts for 6 years, and for half a year; and so in 
 similar cases. All this will appear obvious from a due* consideration 
 of the nature of compound interest. 
 
 If the interest were payable half-yearly, however, at a given rate, per 
 cent, per annum, as in the third example, the amount at the end of a 
 year would be more than if it were payable yearly. Thus at 4 per cent, 
 per annum, payable half-yearly, the amount at the end of half a year 
 would be 1 -02 ; at the end of a year, or two half years, 1 -02 2 , or 1 -0404 ; 
 at the end of a year and a half, or 3 half years, 1 '02 3 ; at the end of 2 
 years, or 4 half years, 1-02 4 , &c. Had it been payable quarterly, the 
 
 * If r be put the amount of \ for a year, p = the principal, t = the time, and a =. 
 
 the amount ; the second rule expressed algebraically will become a =pr* (]) , or, by 
 
 taking the logarithms of both sides, log a = \ogp-)-t log r (2.) From equation 1 
 
 a 
 we have p = f, which is the algebraic expression for rule III. : also from equation 2. 
 
 ve have log p =log a t log r, which is a logarithmic formula for the same purpose. 
 
 _L 
 
 From equation 1, we have alsor. ( p ) ; and from equation 2, log r= ^~ 
 
 Each of these serves for the resolution of the problem in which it is required to find the 
 rate at which a given principal will amount to a given sum in a given time. The first 
 of them expressed in words gives this rule : Divide the amount by the principal, and that 
 root of the quotient which isdenoted by the number of years, will be the amount oflfor 
 1 vear, whence the rate will be known. Thus, if it be required to find at what rate 1000 
 would amount to 1360 17 S in seven years, let the latter, with its shillings and pence 
 reduced to a decimal, be divided by the former : the quotient is 1-3608625, the seventh 
 root of which is T045 nearly. Hence, the rate must be 4|- per cent, per annum. The 
 difficulty of the extraction of high roots, renders this mode of calculation inferior almost 
 beyond comparison to that by the logarithmic formula above exhibited, unless, which is 
 rarely the case, a degree of accuracy may be required which cannot be attained by logar- 
 ithmic tables. In general, indeed, the great facility afforded by logarithms in calulation, 
 will be as much felt in Compound Interest as in almost any other rart of Mathematics. 
 
 We have also, from equation 2, t ='? a r- >0 8 f .- a formula which will serve to find 
 
 i gr 
 
 the time in which a given principal at a given rate will amount to a given svm. This 
 may also be approximated without logarithms by Position : for if the given amount be 
 divided by the principal, the quotient will be the amount of 1 for the required time : then 
 this and the amount of 1 for a year, being known, the t-me will be found, either exactly 
 or nearly so, without much difficulty. 
 
ANNUITIES CERTAIN. 229 
 
 s mounts at the end of 1, 2, 3, &c. quarters would have been 1*01, 1 % 01 S , 
 1 Ol 3 , &c. In such cases, to find the amount of L at the end of the pro- 
 l*)ted time, raise its amount at the end of the first payment, to the power de- 
 noted by the number of jtayments. 
 
 As calculations in Compound Interest are much facilitated by the use 
 of interest tables, a table constructed by rule I., is given at the end of 
 the book showing the amount of l at compound interest for any number 
 of years not exceeding 50, at the most useful rates. The pupil, after hav- 
 ing wrought the preceding exercises by the rules already given, should 
 be taught, instead of finding the amount of ,1 by rule I, to take it, 
 when he can, from the table. This table will also be useful in finding 
 by inspection, in many instances, the time, from the principal, rate, and 
 amount ; and the rate, from the principal, time, and amount. 
 
 ANNUITIES. 
 
 An ANNUITY is a fixed sum of money payable at the ends 
 of equal periods of time, such as years, half-years, or quar- 
 ters. 
 
 Annuities are of two kinds, Certain and Contingent. 
 
 ANNUITIES CERTAIN are those which commence at a 
 fixed time, and continue for a determinate number of years. 
 
 ANNUITIES CONTINGENT are those whose commencement, 
 or continuance, or both, depend on some contingent event, 
 usually the life or death of one or more individuals. 
 
 The PRESENT* VALUE of an an annuity at compound in- 
 terest, is such a sum as would, if lent at compound interest 
 for the given time, amount to the same sum to which the 
 annuity itself would amount, if forborn during the same time. 
 
 When an annuity does not come into possession till a 
 given time has elapsed, or some particular event has taken 
 place, it is said to be an ANNUITY JN REVERSION. 
 
 ANNUITIES CERTAIN 
 
 RULE I. To find the amount of an annuity, payable y early > 
 tJie payments oj which are forborn for a given time, compound 
 interest being charged on them as they become due : (1.) Sub- 
 tract a unit from the amount of \ for a year, and from its 
 amount for the given time at compound interest : (2.) Di- 
 
 * An annuity is commonly said to be worth as many years' purchase as there are wmnd* 
 in the present value of an annuity of l. Tnus in the case of an annuity for 20 years at 
 5 per cent, per annum, because the present valcie of an annuity of l is l2-*t>2, &c. 
 the annuity is said to be worth about 12$ yeans' purchase. 
 
230 ANNUITIES CERTAIN, 
 
 vide the latter remainder by the former, and the quotient 
 will be the amount of an annuity of \ forborn for the pro- 
 posed time: (3.) Multiply this amount by the given an- 
 nuity, and the product will be the amount required. 
 
 When the payments are not yearly, instead of the amount 
 of 1 for a year, use its amount for the interval between the 
 payments ; and instead of the number of years, use the 
 number of payments that would have been made during the 
 time they were remitted, and then proceed as before. 
 
 Exam. 1. If a person save 120 ^ annum, and improve it 
 at 5 ^ cent. ^ annum, compound interest, how much will he be 
 worth at the end of 20 years ? 
 
 The amounts of 1 for 1 year and for 20 years, at 5 ^ cent. 
 per annum, are (as found by rule I. Compound Interest) 1-05 and 
 2-6532977; from each of which if a unit be subtracted, there remain 
 05, and 1-6532977. Let the latter of these be divided by the for- 
 mer, and the quotient, 33-065954, is the amount of an annuity of 
 \ for 20 years; then let this be multiplied by 120, and the pro- 
 duct, 3967-91448, or 3967 18 3, is the amount required. In 
 this case, the gain by interest is 1567 18 3, since the person's 
 savings without interest, would have been 120X20, or 2400. 
 
 Exam 2. Let every thing be as in the last example, except 
 ttot the annuit , is payable half-yearly, instead of yearly. 
 
 Here, since the payments are half-yearly, there would have been 
 40 payments ; and the amount of 1 at the end of half a year, in 
 these circumstances, is 1-025, the 40th power of which is 
 2*6850723, the amount of 1 at compound interest at the end of 
 20 years. Then, 1-6850723 -f- -025 = 67-402892, is the amount, at 
 the end of 20 years, of an annuity of 1 payable at the end of each 
 period of 6 months. Multiply this by 60, the sum payable each 
 half year, and the product, 4044-17352, or 4044 3 5, is the 
 amount required, which is 76 5 2 more than the answer of the 
 last question. It is evident, that the more frequent the payments 
 are, the greater is the amount : for the several gains by interest 
 are thus put sooner to gain more interest. (See exercise \\.J 
 
 The theory of the preceding rule is much more easily and satisfactorily 
 explained by an algebraic investigation, than it can by common Arith- 
 metic. For the use of those, however, who are unacquainted with 
 Algebra, the following illustration of a particular case is annexed. Let 
 it be required to find the amount of an annuity of l for 8 years at 5 
 per cent, per annum, compound interest. At the end of the time, 
 the eighth payment woula be simply l : the value of the seventh 
 would be l'Q5, as it would remain at interest for 1 year : that of the 
 sixth l'05 2 , as it would remain at interest 2 years : that of the fifth 
 jl-05 3 ; of the fourth al-05+; of the third 1-05* ; of the second 
 l -05 s j and of the first l -05 7 . Hence, the entire amount to be re- 
 
ANNUITIES CERTAIN. 231 
 
 ceived at the end of the time would be the sum of the scries of continual 
 proportionals 1, 1 '05, 1'05, 1'05 3 . TO5 4 , 1'05 3 , 1 05 , and 1"05 7 . 
 But, by rule II., page 216, the sum of this series is (1 '05 6 Ij -- *05, 
 which agrees with the rule here given for finding the amount of an an- 
 nuity of 1. The rest is obvious. 
 
 It' may serve to illustrate the nature of annuities, to show another 
 method of resolving the first exercise, which method might also be em- 
 ployed in solving all questions of a similar nature. As 5 : 100: : 
 aJ20 : : '2400, the principal which would gain 120 per annum. Then, 
 at compound interest, the amount of 2400 for 20 years is s6367 183^; 
 from which 2400 being subtracted, we have remaining 3967 18 3, 
 for the interest, or improvement of this imaginary principal, which is 
 also the amount of the annuity. 
 
 When the pupil shall have learned to perform the exercises on this 
 rule and the next, he may be taught to use Tables II. and III. at the 
 end of the book as often as they are applicable. By this means the labour 
 will often be greatly abridged, in the same manner as operations in com- 
 pound interest are often-much shortened by the use of Table I. 
 
 Exercises. Find the amounts, at compound interest, of the fol- 
 lowing annuities, payable yearly, and forborn during the given 
 times, and at the given rates per cent, per annum : 
 
 ^ Exercises. Answers. 
 
 1. Annuity 
 
 2. 
 
 3. 
 
 7 
 
 8. If a person rent a farm at 57 14 3 ^ year, payable yearly, 
 and forbear paying rent for 16 years: how much will he owe 
 to the proprietor at the end of that time, allowing him compound 
 interest at 5 IT cent. W annum? Answ. 1365. 6 8. 
 
 9. Suppose a person who has a salary of 75 a year, payable 
 half-yearly, to allow it to remain unpaid 17 years : how much will 
 he be entitled to receive at the end of that time, compound interest 
 being allowed at 6 IT cent, ty annum? Answ. 2164 17 7$. 
 
 10. Suppose a salary of 1 1 7 6 a year, payable at periods of 
 two years each, to be forborn for 12 years at 5 v cent. IT annum, 
 compound interest : what will be its amount? (See exercise 4. ) 
 Answ. 175 10 7f 
 
 11. Suppose a salary of 120 V annum, payable quarterly, to 
 be forborn at compound interest for 20 years : to what sum will 
 it amount at 5 V cent. V annum ? Answ. 4083 11 3. 
 
 RULE II. To Jlnd the present value of an annuity at com- 
 pound interest : (1.) Find by the last rule the amount of an 
 
 s. d. 
 100 
 13 15 9 
 
 forborn 10 years, 
 14 
 
 at 4 
 
 6 . . . 
 
 s. d. 
 1200 12 2i 
 9g9 14 103 
 
 56 17 6 
 
 9 
 
 6 
 
 653 11 4^ 
 
 11 7 6 
 
 12 
 
 
 181 1 is 
 
 34 2 6 
 
 8 
 
 5* . 
 
 331 14 11| 
 
 14 15 9 
 
 15 
 
 
 371 11 10| 
 
 51 2 8* 
 
 14 
 
 ~ 4A... 
 
 968 2 OJ 
 
ANNUITIES CERTAIN. 
 
 annuity of 1 forborn for the given time, and at the given 
 rate : (2.) Divide this by the amount of \ at compound 
 interest for the given time, and the quotient will be the 
 present value of an annuity of 1 for the given time : (3.) 
 Multiply this by the given annuity to find the present value 
 required.* 
 
 In case of an annuity to continue for ever, or, as it is 
 called, a perpetuity, subtract a unit from the amount of 1 
 for a year, or for the interval between the payments, and di- 
 vide a unit by the remainder : the quotient is the present 
 value of a perpetuity of 1 ; which multiply by the given 
 perpetuity. 
 
 Or, as the given rate : .100 : : the perpetuity : its pre- 
 sent value. 
 
 Exam. 3. Required the present value of a house held on a 
 lease of which 22 years are unexpired, and bringing a profit rent 
 of 5 10 per annum, payable yearly, compound interest being 
 allowed at 6 per cent, per annum. 
 
 Here, the amount of 1 for 22 years is 3-603537. Then divid- 
 ing 2-603537 by -OG, we get 43-3923; the quotient of which by 
 3-603537, is 12-041583, the present value of an annuity of 1 for 
 22 years at 6 per cent, per annum. Let this be multiplied by 
 .45 10, and the product is 547-892026, or 547 17 10, the re- 
 quired value. 
 
 Exam* 4. Let every thing be the same as in the last example, 
 except that the annuity is payable Jialf-yearly instead of yearly. 
 
 In this case the amount of 1, at the end of 6 months is 
 1-029563, the square root of 1*06; the 44th power of which (44 
 being the number of payments,) or its equal, the 22d power of 1-06, 
 is 3-603537. From this take 1, and divide the remainder, 2-603537, 
 Iry -029563, and the result by3"-603537: the quotient, 24-43916, is 
 the present value of 1 of each payment, which being multiplied 
 by 22 15, the half-yearly payment, the product is 555-99089, 
 or 555 1 9 9|, the amount required. 
 
 Exam. 5. Required the value of a perpetuity of 80 a year, 
 at 6 per cent, per annum. 
 
 * Perhaps some might prefer the following very simple rule for this problem : As the 
 given rate is to 100, so is the annuity to the imaginary principal which would annually 
 produce the annuity. Then from this principal subtract its present worth found by rule 
 III. for Compound Interest : the remainder will be the present worth of the annuity. 
 
 It may be remarked, that when the payments are not yearly, different writers have viewed 
 the subject in different lights, and given different rules for computing the present value. 
 The method employed in example 4, is perhaps preferable; but the reader who is well 
 acquainted with the principles of compound interest will feel it easy to form other rules 
 fbunded on diflerent suppositions. The work of example 2, proceeds on a different 
 principle. 
 
ANNUITIES CERTAIN. 283 
 
 As 6 : 100 : : 80 : 1333 6 8 : or 80-^-06 = 1333$, 
 the value required. 
 
 Exam. 6. Suppose the same perpetuity as in the last question, 
 payable half-yearly : what is its present value ? 
 
 At 6 per cent., the amount of 1 at the end of half a year, is 
 1 '029563 j and in this question each payment is 40 : therefore 
 the value of the perpetuity is 40-7-'029563 =r 1353-0426, or 
 1353 10, exceeding that found in the last question, in conse- 
 quence of the frequency of the payments, by 19 14 2^. 
 
 The reason of the first part of this rule is evident from the defi- 
 nition of the present value of an annuity (page 229,) and from rule 
 I. in this article, and rule III. in Compound Interest. It might 
 also be easily shown from rule III. of Compound Interest, that at 
 any particular rate, as 5 per cent, the present value of an annuity 
 of 1 is the sum of the decreasing series of continual proportion- 
 als, whose terms are the present worths of 1 at compound inter- 
 est for I year, 2 years, 3 years, and so on; the first term being 
 1 _j_ 105, the ratio 1 '05, and the number of terms equal to the num- 
 ber of years. Now the summation of this series, according to rule 
 II. in the note, page 218, agrees exactly with the first part of the 
 rule given above: and the summation of the infinite series of pre- 
 sent worths, according to the second part of the rule in the text 
 of the same page, agrees with the rule here given for a perpetuity. 
 
 Exercises. Find the present values of the following annuities, 
 for the given times, and at the given rates per cent, per annum, 
 compound interest : 
 
 Exercises. 
 
 Answers. 
 
 
 
 
 s. d. 
 
 
 
 s. 
 
 d. 
 
 12. 
 
 Annuity 84 
 
 7 
 
 9 
 
 to continue 9 years, at 
 
 5 
 
 599 
 
 16 
 
 2f 
 
 13. 
 
 JJ-JJUU-XJJUU-JJM 4S 
 
 12 
 
 9 
 
 19 
 
 6 
 
 391 
 
 
 
 Oi 
 
 14. 
 
 75 
 
 o 
 
 o 
 
 ^^ 6 
 
 2i 
 
 L 413 
 
 2 
 
 4 
 91 
 
 15. 
 
 58 
 
 10 
 
 
 
 5 
 
 3: 
 
 I 264 
 
 2 
 
 7! 
 
 16. 
 17. 
 
 ~~~~~~. 113 
 95 
 
 15 
 5 
 
 
 G 
 
 JJWJWJJJ 1 
 
 4' 
 10 
 
 922 
 676 
 
 12 
 15 
 
 4 
 
 si 
 
 18. How much must a person pay to have a salary of 224 per 
 annum for 19 years, being allowed compound interest at 5 per cent, 
 per annum ? Answ. 2707 2 2f . 
 
 19. Suppose a widow to be entitled to an annuity of 40 a year, 
 payable half-yearly, from a fund, for 8 years : what is it worth at 
 6 per cent, per annum, compound interest ? Answ. 252 1 3^. 
 
 20. If a farmer have a lease of 65 acres for 36 years, at 1 12 
 per acre, what fine must he pay to reduce the rent to 15/ per acre, 
 compound interest being allowed at 6 per cent, per annum ? 
 Answ. 807 16 2. 
 
 21. Required the present values of a perpetuity of 1, and of 
 
234 ANNUITIES CERTAIN. 
 
 another of 68 5, payable annually, at 4f per cent, per annum 
 
 Answ. 21 I OJI, and 1436 16 10tV. 
 
 22. A perpetuity of 126 6 8 at 4 per cent., &c. Answ. 
 2972 10 IHf. 
 
 23. A perpetuity of 96 7 6, and another of 1, at 3 per 
 cent., &c. Answ. 2753 11 5^, and 28 11 5}-. 
 
 24. Required the present values, at 5 per cent, per annum, of a 
 perpetuity of 1, payable yearly; of the same payable half- 
 \early; and of the same payable quarterly. Answ. 20; 
 20 4 11; and 20 7 5. 
 
 RULE III. Tojind the present value of an annuity in rever- 
 sion : (1.) Find by rule II. the present value of the annuity 
 from the present time till the end of the period of its con- 
 tinuance : (2.) Find also its value for the time before it 
 comes into possession : (3.) The difference of these two re- 
 sults will be the present value required* 
 
 The reason of this rule is so obvious as to require no explanation. 
 The following rule is also founded on obvious principles, and may per- 
 haps be preferred by some. 
 
 RULE IV. (1 .) Find by rule II. the present value of the an- 
 nuity during the time it is to be possessed : (2.) Then the 
 present value of this result, found by rule III. of Compound 
 Interest, will be the present value of the reversion. 
 
 The next rule which is in principle the same as the last, will perhaps 
 be found preferable for the purposes of calculation to either of the 
 preceding. 
 
 RULE V. Subtract a unit from the amount of 1 at com- 
 pound interest during the period in which the annuity is to 
 be possessed: divide the remainder by the amount of \ 
 at compound interest during the period from the present 
 time till the termination of the annuity ; and the quotient by 
 the amount of \ for a year, diminished by a unit ; the 
 result will be the present value of a reversion of \. 
 
 Exam. 7. A father leaves to his eldest child for 8 years a 
 profit rent of 280 per annum, payable yearly, and the reversion 
 of if for the 12 years succeeding to his second child. What is the 
 present value of the legacy of the second at 4 per cent, per annum, 
 compound interest? 
 
 By rule III. Here, by rule II., the value of an annuity of 1 
 for 20 years at 4 per cent, is 13-590325, and for 8 years 6'73274o; 
 the difference of which is 6-85758, the present value of a reversion 
 of 1 in the proposed circumstances. This being multiplied by 
 280, the product, 1920-1224, or 1920 2 5, is the value re- 
 quired. 
 
LIFE ANNUITIES. 235 
 
 By rule IV. The present value of an annuity of 1 for 12 years 
 is found by rule II. to be 9-385073 ; and by rule III. of Com- 
 pound Interest, the present value of this for 8 years is found to be 
 6-85757l>, us before. 
 
 By rule V. The amount of 1 at compound interest for 12 
 years, diminished by a unit, is -601032; and the amount of \ for 
 20 years is 2-191123. Divide the former by the latter, and the 
 result by -04, and there will finally result 6-857579, the same 
 present value as before. 
 
 Exercises. Required the present values of the following an- 
 nuities in reversion, at the given rates per cent, per annum : 
 
 Exercises. Ansrvers. 
 
 *. d. 3. cL 
 
 25. Ann. 135 10 9, after6yrs. & for 8 yrs. at5L. 622 13 
 
 26 . 79 12 6, 4 ^ 5 ,.332 9 
 
 27 58 9 10, 3 '. 7 44.. 302 
 
 28._^_5412 3, 4 8 7 .. 248 15 8$ 
 
 29. A Perpetuity of 84, 7 6, after 8 years, at 7 .. 701 10 7 
 
 30 136 17 9, 8 5 ..1853 4$ 
 
 31. What fine must be paid, to change into a perpetuity, a 
 lease for 16 years, which brings a profit rent of .71 13 3 per an- 
 num, payable yearly, compound interest being allowed at -ij per 
 cent, per annum ? Answ. 866 6 8. 
 
 32. What fine must be paid to add 25 years to a lease which 
 brings a profit rent of .1 12 10, and of which 14 years are unex- 
 pired, compound interest being allowed at 5 V cent. & annum ? 
 Answ. 800 16 4f. 
 
 33. What is the present value of the reversion of a perpetuity of 
 .60 W annum, payable yearly, but not to come into possession 
 till the expiration of 100 years, compound interest being allowed 
 at 6 ^ cent. ^ annum?* Answ. 2 18 11. 
 
 ANNUITIES CONTINGENT, OR LIFE ANNUITIES. 
 
 LIFE ANNUITIES are those whose commencement or ter- 
 mination, or both, depend on the extinction of one or more 
 lives. 
 
 When life annuities are in possession, they are often called 
 simply ANNUITIES ON LIVES ; but when they are in rever- 
 sion, they are generally called ANNUITIES ON SURVIVOR- 
 SHIPS. 
 
 * This question may tend to correct a mistake that pretty generally prevails respecting 
 the comparative values of long leases and perpetuities, the latter being supposed to ex- 
 ceed the former in value in a far greater degree than they really do. In the case on 
 which this exercise is founded, the difference of present values U no more than 2 13 (1. 
 
236 LIFE ANNUITIES. 
 
 The VALUE OF A LIFE is the present value of an annuity 
 of \ to continue during that life. 
 
 The EXPECTATION OF A LIFE of a given age, is the mean 
 period during which persons of that age live. 
 
 The COMPLEMENT OF A LIFE is double the expectation 
 of the same life. 
 
 The calculation of life annuities depends on the joint application of 
 the rules of compound interest, and of the doctrine of chances, to 
 tables deduced from observations on the duration of human life. In 
 what follows on this subject, a selection of the rules most generally use- 
 ful will be given./ ^oV the theory of these rules, which is of a nature 
 too complicated to be given in a work like the present, the reader who 
 wishes to become thoroughly acquainted with this interesting and diffi- 
 cult subject, may have recourse to the writings of Simpson, De Moivre, 
 and more particularly of Dr. Price and Mr. Morgan, where the subject 
 will be fo-und treated at great length, both in theory and practice. 
 
 The duration of life being different in different countries, calculations 
 have been founded on the registers of births and deaths kept in London, 
 Breslaw, Northampton, and various other places. Tables IV. and V 
 at the end of the book, which are employed in what follows, are founded 
 o the Northampton register, which is thought to serve, for the gene- 
 rality of places, better than any other. 
 
 RULE I. Tojind the present value of an annuity to continue 
 during the life of a person whose age is given : Take from 
 table IV. the value of \ for the given age and rate, and 
 multiply it by the given annuity. 
 
 Exam. 1. What should be given at 6 ^ cent. $T annum, for a 
 farm worth 36 ty annum, held on a lease of one life aged 58 years? 
 
 Here, by the table, the value of an annuity of \ on the life of 
 a person aged 58 years, is, at 6 ^ cent., 8' 173, the product of 
 which by .36 is 294-228, or 294 4 6f, the value required. 
 
 Ex. 1. If a person aged 38 years, have a salary of 138 10 ^ 
 year for life, what is its present value at 5 ^ cent. W annum ? 
 Ansu>. ,1678 1 3f. 
 
 2. What should a person aged 32 years, pay, at 4 W cent, # 
 annum, to have for life a yearly salary of 180? Answ. 2609 2. 
 
 3. If a farm of 28 acres be held at 1 14 6 ^ acre, on a life 
 aged 41 years, what fine must be paid, at 5 per cent, per annum, 
 to reduce the rent to 10 shillings per acre? Answ. 401 2 9^. 
 
 RULE II. To Jind the present value of an annuity which 
 is to continue during the joint lives of two persons, and to 
 cense when either of them dies: In table V. find the age of 
 the younger, or of either if they be equal, in the first column ; 
 and in the same division of the table, in the second column, 
 find the age of the other; opposite to the latter is the value 
 of an annuity of l, which multiply by the given annuity. 
 
LIFE ANNUITIES. 237 
 
 Exam. 2. What is the value at 4 per cent., of an annuity of 
 .90 per annum, to continue during the joint lives of two persons, 
 whose ages are 15 and 50 years respectively? 
 
 Here, by table V., the value of an annuity of 1 is 9-872, the 
 product of which by .90 is 888-480, or 888 9 7|, the value 
 required. 
 
 Exam. 3. Required the present value, at 6 per cent., of an 
 annuity of 120 per annum, which is to cease, when either of two 
 persons, aged 14 and 57 years respectively, shall die. 
 
 Neither of these ages being in the table, recourse must be 
 had to the method of proportional parts. (See page 191.) 
 Thus, the table gives, for 10 and 55, 7*951, and for 15 and 55, 
 7-812. The difference of these is -139, four fifths of which being 
 subtracted from 7-951, the remainder, 7-840, is the value of 1 
 for the ages 14 and 55. In like manner, the table gives for 10 and 
 60, 7-250, and for 15 and 60, 7-135; four fifths of the difference 
 of which being taken from 7-250, the remainder, 7-158, is the value 
 of 1 for the ages 14 and 60. Hence we have, for 14 and 55, 
 7-840, and for 14 and 60, 7-158; two fifths of the difference of 
 which being subtracted from 7-840, the remainder, 7-567, is very 
 nearly the present value of 1 for the ages 14 and 57 : and this 
 being multiplied by 120, the product, 908-04, or 908 9, 
 is the value required. 
 
 Ex; 4. Required the present value, at 6 per cent., of an annuity 
 of 39 10, on the joint continuance of two lives of 25 and 73 
 years. Answ. 182 8 10^. 
 
 5. Required, at 5 per cent., the present value of an annuity of 
 43 126 on the joint lives of two persons, whose ages are 44 
 and 51 years respectively. Answ. 342 9 5|. 
 
 RULE III. To find the present value of an annuity to con- 
 tinue during the longer of two lives : From the sum of the 
 values of the single lives (found in table IV.) subtract the 
 value of the joint lives (found by rule II. :) the remainder 
 is the present value of an annuity of 1 on the longer of 
 the two lives. 
 
 Exam. 4. For how much should a house be sold which brings 
 a profit rent of 41 5 per annum, and is held by a lease on the 
 longer of two lives aged 20 and 35 years, compound interest being 
 allowed at 6 per cent, per annum ? 
 
 By table IV. the values of the lives are 12-398 and 11-236, the 
 sum of which is 23-634. Also, by table V., the value of them 
 iointly is 9-45 1, which taken from the preceding sum, leaves 14-183, 
 the value of an annuity of 1 on the life of the longest liver j 
 which being multiplied by 41 5, the product, 585-049, or 
 5S5 llf. is the present value required. 
 
 Ex. 6. What should be paid, at 5 per cent., for the purchase cf 
 
238 LIFE ANNUITIES. 
 
 a profit rent of .68 5 per annum, to continue during the longer 
 
 of two lives, aged 38 and 42 years ? Answ. .1001 18 4. 
 
 7. What should a man, aged 44, pay, at 5 per cent., to secure, 
 during his own life, and that of his wife,* aged 39, an annuity of 
 ,200 a year ? Answ. 2892 9 3^. 
 
 RULE IV. Tojind the value of an annuity during the joint 
 continuance of three lives ; or which is to terminate on the ex- 
 tinction of any one oj them : Find by rule II., the value of 
 the two eldest jointly ; and, by table IV., find what single 
 life would have this same value. Then find, by rule II., the 
 value of the joint continuance of the single life thus found, 
 and of the youngest, and this will be the value of the three 
 proposed lives jointly. 
 
 Exam. 5. Required the present value at 4 per cent., of an an- 
 nuity of 140 on the joint continuance of three lives aged 15, 30, 
 and 35 years. 
 
 Here, at 4 per cent., the value of the joint continuance of two 
 lives aged 30 and 35, is 10-948; which in table IV. is found to he 
 the value of a single age of 51 years nearly. Then, by rule II., 
 the value of the joint continuance of two lives, of 15 and 51, is 
 9-6335, the value of an annuity of \ on the joint continuance of 
 the three given lives. The product of this by 140 is 1348-69, 
 or 1348 13 9, the value of the given annuity. 
 
 RULE V. Tojind the value of an annuity on the longest of 
 three lives ; or which is to continue, till they are all extinct : 
 From the sum of all the values of the single lives, found in 
 table IV. and of the value of the three jointly, found by rule 
 IV. ; subtract the sum of the joint values of the lives com- 
 bined two and two, by rule II. ; and multiply the remainder 
 by the given annuity. 
 
 Exam. 6. Required the present value, at 4 $T cent., of a house 
 and farm which yield a profit rent of 32 10 ty annum, held by 
 a lease of three lives, aged 35, 30, and 15 years. 
 
 Here, by the last example, the value of the three lives jointly is 
 0*6335, and their values singly are 14-039, 14*781, and 16*791. Also, 
 by rule II., the value of the first and second jointly is 10-948; of 
 the first and third 11*787; and of the second and third 12*246. 
 From the sum of the first four of these numbers take the sum of 
 the others, and the remainder, 20*2635, is the value of an annuity 
 
 * The duration of the lives of females is found to be somewhat greater than that of 
 males. It is inccnsk-tent with the natureof this work, however, to take into calculation 
 this cirouuuKaittCt which in genera) produces but a ;1 ght difference in the result. 
 
LIFE ANNUITIES. 239 
 
 of 1 on the life of the longest liver of the three. This multiplied 
 by 32 10, gives for the value required, 658-56375, or 658 1 1 3, 
 
 Ex. 8. How much should a man pay at 5 W cent. ty annum, 
 to purchase an annuity of 320 a year to continue till himself 
 aged 55, his wife aged 49, and his son aged 20, shall all be extinct? 
 Answ. 5103 11 0. 
 
 RULE VI. To find the value of the reversion of an annuity 
 ajter the death of the present possessor : From the present 
 value of an annuity of 1 for the entire continuance of the 
 annuity, subtract the value of an annuity of 1 during the 
 life of the possessor: the remainder will be the value of a 
 reversion of \ in the given circumstances, which multiply 
 by the given annuity. 
 
 If there be two or three lives, the same rule will serve, 
 their values being used instead of the value of the single life. 
 
 Exam. 7. If a person aged 47 years, possess a perpetuity of 
 500 per annum j what is its present value at 5 ^ cent, to his 
 son, who is to possess it at his death ? 
 
 Here, the value of 1 in perpetuity is 20, and, by table IV, 
 the value of an annuity of I on a life of 47 years is 10*784. Then 
 20 10-784 =9-216; which multiplied by 500 produces 4608, 
 the value required. 
 
 Exam. 8. A man leaves an annuity of 165 $" annum, of 
 which 30 years are yet to come, to his nephew, aged 38 years, 
 during life, or till its termination ; and the reversion of it to a 
 charitable institution, in case the nephew die before the termina- 
 tion of the annuity. How much is the present value of the re- 
 version at 4 ^ cent. ? 
 
 In this case, the present value of an annuity of 1 for 30 years, is 
 by rule II. of Annuities Certain, 17-292 ; and, by table IV. the pre- 
 sent value of 1 on a life of 38 years, is 13-548. Then, 17-292 
 13-548 = 3-744 ; and 3-744 X 165 = 617-76, or 617 15 2, 
 the value required. 
 
 Ex. 9. What is the present value at 5 V cent., of the reversion 
 of a perpetuity of 400 ty annum, not to come into possession till 
 the death of the present incumbent aged 70, and of his intended 
 successor aged 30? Answ. 2538 16 0. 
 
 RULE VII. To find the present value of a proposed sum 
 payable at the decease of a person 'whose age is given : From 
 the value of \ in perpetuity, take the value of an annuity 
 of \ on the life ; divide the remainder by the value of the 
 perpetuity increased by a unit ; and multiply the result by 
 the given sum. 
 
240 LIFE ANNUITIES. 
 
 If the present value thus found be divided by the value 
 of the life, the quotient will be the annuity payable yearly* 
 
 By using the value of two or three lives found by the 
 preceding rules, we may apply the same rule. 
 
 Exam. 9. How much must a person aged 32 years, pay, at 5 
 ty cent., to secure to his family at his death the sum of .1000? 
 
 At 5 W cent., the perpetuity of 1 is .20, and, by table IV. 
 the value of a life of 32 years, is 12*854. The difference of these 
 is 7-146; and this being divided by 21, the quotient is '340286, the 
 present value of 1 payable at the decease of the person. Then, 
 340286 X 1000= 340 5 8, the value required: and 
 340-286 -;- 12-854= 26'473 = 26 9 5, the annuity. Hence 
 it appears, that at 5 ^Tcent. ty annum, if a person either at one 
 payment deposit 340 5 8, or pay each year during life the sum 
 of 26 9 5, his heirs at his death will be entitled to receive 1000. 
 
 Ex. 10. At 6 V cent. W annum, how much ty year, must a 
 person aged 66 years, pay during life, to entitle his heirs at his 
 death to receive 1500? Answ. 128 13 10. 
 
 11. If a husband and his wife be each aged 45 years; how 
 much V annum must be paid during the longer of the two lives, 
 to entitle the family after the decease of both, to receive 4000, 
 compound interest being allowed at 6 & cent. ^ annum ? 
 Answ. 75 8 5|. 
 
 Exam 10. How much must a man, aged 39, pay per year, 
 at 6 & cent., during his marriage, or during life, to entitle his 
 wife aged 32 years, in case she survive him, to an annuity of 50 
 a year during the remainder of her life ? 
 
 By rule II., the value of the two lives jointly is 8-774, and the 
 value of a life of 32 years is, by table IV., 1 1 -512. The difference 
 of these, 2-738, is the present value of the reversion of an annuity 
 of 1 to be paid to the wife after her husband's death, in case she 
 survive him. The product of this by 50 is 136-9, or 136 18, 
 the sum to be paid at a single payment. Let this be divided by 
 S'774, the value of their lives jointly, the quotient 16*603, or 
 15 12 Of, is the annual payment. Hence, if there be paid 
 annually, during the joint continuance of both lives, the sum of 
 15 12 Of, or at present, in a single payment, 136 18, the 
 wife, if she survive the husband, will be entitled to an annuity of 
 50 per annum during life. 
 
 This solution proceeded on the supposition in regard to the 
 annual payments, that the first of these was not made till the end 
 of the first year. If it be made at present, however, we must di- 
 vide by the value of the joint continuance increased by a unit j 
 in the present case by 9-774. In this case the annual payment 
 would be found to be J4 i. 
 
CONTINUED FRACTIONS. 241 
 
 Exam. 11. If a number of persons form themselves into an 
 association for providing annuities of 4:0 & annum for their 
 surviving widows, and each person, at the age of 30, contribute 
 15 to the fund, and equal annual contributions during the rest of 
 his life: how much must each of these annual contributions be, 
 money being improvable at 6 ^ cent. ^ annum, and the wife of 
 each being supposed to be at an average three years younger than 
 himself? 
 
 The value of a life of 27 years is 11*917, and the value of two 
 lives of 27 and 30 jointly is 9-481. The difference of these, 3-436, 
 multiplied by 40, gives for the entire sum to be paid at present, 
 at a single payment, 97-44. But as only 75 are to be paid at 
 present, there will still remain 22*44 to be paid by annual contri- 
 butions, by dividing which by 8-481, we obtain 2 7 3% for each 
 of these contributions. 
 
 Exam. 12. Suppose that, as in the last question, a man aged 
 30 years, pays 50 into a fund ; what annual contribution must 
 he pay during life to entitle his family to an annuity of 40 per 
 annum for ^ryears after his death, interest being at G & cent ? 
 
 The present value of an annuity of 40 for 8 years at 6 $T cent. 
 is 248-3918, (by rule II. of Annuities Certain.) Then, by rule 
 VII. of this article, the present value of this sum is found to be 
 70-084 from which 50 being subtracted, the remainder, -20*084, 
 is the sum still remaining to be paid at present ; and this being 
 divided by 11*682, the value of a life of 30, the quotient, 
 1 14 4, is the annual contribution required. 
 
 CONTINUED FRACTIONS.* 
 
 IT often happens that fractions, even when reduced to their 
 simplest forms, are expressed in numbers inconveniently large; 
 and hence it is often desirable to approximate their values in 
 .smaller numbers. Thus, if the fraction %%%$, which is in its 
 lowest terms, be proposed, we may wish, even for the purpose 
 of forming a more correct idea of its magnitude, to find other 
 fractions, in smaller terms, which will be nearly of the same 
 value. The method that most naturally presents itself for this 
 purpose is to divide both terms by the smaller of them, as the 
 smaller will by this means become 1, and we shall thus be 
 enabled to compare the other with that number with which the 
 
 * For full information respecting the nature and applications of continued fractions, 
 see Bonnycastle's larger Treatise on Algebra, Lacroix's Complement des Rumens 
 t? Atgebre, Legendre'a Essai sur la Theorie dot Nombres, Birlow's Theory of Numbers, 
 l-agrwige's Additions to Euler's Algebra'and several other works. What is here given 
 is one of the most intere.-ting applications-, and one which cannot fai! to be usolui to the 
 more advanced arithmetician. 
 
 M 
 
2452 CONTINUED FRACTIONS. 
 
 mind can most easily compare it in respect of magnitude. By 
 
 this means the fraction becomes g , the denominator of which, 
 
 being between 2 and 3, we conclude that the value of the give n 
 fraction is between and ^; and therefore is a first approxima- 
 tion to its value, being too great. Divide again both terms of the 
 the fraction in the denominator by 217, and it will become 
 
 r-rr- , which is between and . By taking instead of 
 
 we shall have, for a second approximate value of the prop osed 
 fraction, O r ft, which is too small, as instead of ft*fr, we use d in 
 
 the denominator , which is greater than T YrV To continue the ap- 
 proximation still farther, we divide botb terms of the fraction ^, 
 
 by 30 : the result is , which is less than \ and greater than 
 
 ' 3~o~ I 
 
 Hence, if instead of using in the preceding fraction, we use -rr 
 
 or its equal 3 \, we shall have for the next approximation to the va- 
 
 lue of the given fraction - or *f which is too great, because 
 I 2 3 7 ^ 
 
 T-J-, or its equal 3 V> is too small, in consequence of $ being too 
 J T \ 
 
 great; and therefore ( - must be too great, since the denomina- 
 
 2 & 
 
 tor is too small. By continuing the process in a similar manner, 
 we find |ff, T V<nr> and H, f r tne succeeding fractions, the 
 last of which is the given fraction,* and thefirst an approximate frac- 
 tion smaller, and the second another greater, than the given frac- 
 tion. These successive fractions have the remarkable property that 
 each of them approaches more nearly than the one which precedes 
 it, to the value of the given fraction. Thus, >$, is nearly equal 
 to |; more nearly equal to ft; more nearly equal to 4f-J still 
 more nearly equal to iff; and, finally, more nearly still to /ijVff* 
 That there is this continual approach to the true value of the frac- 
 tion, will appear evident, if it be considered that in each of the re- 
 sults in the preceding operations, a correction is made on the re- 
 sult which goes before it 
 
 The several converging fractions above obtained, if the last or 
 supplementary simple fraction be rejected from each, except when 
 ?ts numerator, like that of the rest, is a unit, may be written at 
 full length, as follows : 
 
 
 * If the given fraction be finite, the last of the converging fraction! wffl ahnqm be 
 
 equal to it, as in this example. 
 
 
CONTINUED FRACTIONS, 243 
 
 1 
 
 In this form they are called continued fractions. It appears, there- 
 fore, that a CONTINUED FRACTION is that which has for Us denominator 
 a whole number with a fraction annexed, which fraction hat also for 
 its denominator a whole number with a fraction annexed, the denomi- 
 nator of which latter fraction is also a whole number with a fraction ; 
 and so on, however far the fraction may be continued ; and each nu- 
 merator is a unit.* 
 
 It will be seen by a review of the preceding processes, that the 
 denominators of the continued fraction are the quotients which 
 would be found in using the rule given in page 89, for finding the 
 greatest common measure. Hence, we have the following rule : 
 
 RULE I. To convert a given simple fraction into a continued frac- 
 tion: Divide the greater term by the less, the less by the remain- 
 der, &c. as in finding the greatest common measure : the quotients 
 will be the denominators of the several fractions in the continued 
 fraction, and the numerator of each will be a unit. 
 
 The successive fractions which approach continually to the va- 
 lue of a given fraction expressed in large numbers, may be found 
 bv reducing it to a continued fraction, and perating in the way 
 already employed in approximating the value of if : the following 
 method^ however, will be found much preferable. 
 
 RULE 1 1. " A fraction expressed by a great number of figures be- 
 ing given, to find all the fractions in less terms, which approach si 
 near the truth, that it is impossible to approach nearer without employ- 
 ing greater terms :" Reduce the proposed fraction to a continued 
 one, or at least find the quotients, by the preceding problem. Then 
 write the several quotients in a line ; and if the given fraction be 
 greater than a unit, take the first quotient as the numerator, and 
 a unit as the denominator, of the first fraction, which set below 
 the second quotient ; but if the given fraction be less than a unit, 
 make the first quotient the denominator, and a unit the numerator 
 f the first fraction. Then, for the second fraction, multiply both 
 the terms of the first by the quotient which stands above it, and 
 add a unit to the product of that term which was the first quo- 
 tient : the result is the second fraction, which is to be set below 
 the third quotient. To find the succeeding fractions, multiply the 
 terms .of each fraction, when found by the quotient, which stands 
 above it, and to the products add separately the terms of the pre- 
 ceding fraction. 
 
 Exam. 1. The height of Slieve Donard, in county Down, is 2654 
 
 * It is not essential to the nature of continued fractions that each numerator be a 
 unit ; but fraction* of this kind only are used as instruments of calculation. 
 
 M2 
 
244 CONTINUED FRACTIONS. 
 
 feet, and that of Ben Lomond 3262 feet : it is required to find a 
 series of converging fractions, expressing as nearly as possible the 
 ratio of the heights of these mountains. 
 
 Here the fraction expressing the 1327)1631(1 
 
 height of Slieve Donard in relation to 
 
 that of Ben Lomond, is f||f, or, in its 304)1327(4- 
 
 lowest terms, IfH, and the quotients 1216 
 
 found in the manner shown in the 
 
 margin, are I, 4, 2, 1, 2, 1, 4, 1, 4: 111)304(2 
 
 consequently the continued fraction is &c. 
 
 , 
 +T.L-L 
 
 The successive converging tractions will be found thus : 
 14212141 4 
 
 *. t, A, H, u, ii, m. m, wi, 
 
 Here, the quotients, or denominators, being arranged in succey- 
 sion, we take for the denominator of the first fraction, 1, the first 
 quotient, and we make its numerator 1 also. Then, by multiply- 
 ing both terms by 4, the quotient which stands above them, and 
 adding 1 to the product resulting from the denominator, because 
 it was the first quotient, we obtain f, for the second fraction. We 
 then multiply 4. by 2, the figure above.it, and add to the product 
 the preceding numerator ; and we multiply 5 by 2, and add to the 
 product the preceding denominator ; and thus we obtain T 9 T for the 
 third fraction. We next multiply the terms of this fraction by 1, 
 and add to the results 4 and 5 severally : we thus find the next 
 fraction to be . In this way, rejecting the first fraction, -f, 
 which is evidently far from the truth, we find that the height of 
 Slieve Donard is nearly | that of Ben Lomond ; more nearly -fa 
 
 it: still more nearly, -^l; more nearly still, -, &c. ; f- being 
 too small, -ft too great, \% too small, ff too great, &c.* 
 
 E cam. 2. The circumference of the circle whose diameter is 1, 
 is found to be greater than 3-1415926, but less than 3-1415927: 
 
 * Thus, 4 of S262=2fi09|-; T 9 T of 3262=2668lf > -ff of 3262=26 50f ; f$ of 3262= 
 2655^- ; if of 362=265S*f ' wl ch are evidently approaching 654, the true value, be- 
 ing alternateV greater and less. It may be observed, that in all cases, the error of ea*h 
 fraction is a less part of the integer than a unit divided by the product of its own and tli 
 succeeding denominator ; but a greater part than a unit divided by the product of its 
 own denominator, and the sum of that and the succeeding denominator. Thus, in the 
 
 preceding example, the error of f is less than-~ 5 n X3262, but greater than 
 
CONTINUED FRACTIONS. 245 
 
 required the series of fractions converging to the ratio of the cir- 
 cumference and the diameter. 
 
 In questions like this, in which one term of the fraction or ratio 
 is not precisely given, but is contained between given limits, as 
 when one of the terms is an infinite decimal, it is proper to work 
 for the quotients by both limits, and to use those only which result 
 from both. Thus, in the proposed exercise, by dividing 3* 14 15926 
 by 1 '0000000, this divisor by the remainder, &c. and by proceed- 
 ing in like manner with 3*1415927, we find in both cases, the quo- 
 tients 3, 7, 15, 1, after which the quotients would be different, and 
 are therefore not to be used. Hence 
 the converging fractions are found as in 3 7 15 1 
 the margin. f, y, ?M iff, 
 
 It appears, therefore, that the diame- 
 ter of a circle is to its circumference 
 
 nearly as 1 to 3 ; more nearly, as 7 to 22; more nearly again, as 
 108 to 333; and still more nearly, as 113 to 355. The degree of 
 approximation of each of these to the given ratio, will be discover- 
 ed by dividing the first term of each by the second. In this way 
 the last gives 3' 14 159292, which exceeds the truth by less than a 
 ten-millionth part of the circumference. Had the circumference 
 been taken to a greater number of places, (see note, page i 12,) 
 the succeeding fractions would have been found to be 
 ^ 
 
 Exam. 3. Let it be required to approximate the ratio of 
 1-41421* to 1. 
 
 Here the quotients are found to , 1 . 
 be 1, 2, 2, 2, 2, &c. and consequently ^~2~^JL J^ 
 
 the continued fraction, the same as in 2 -f- -j-j- , _ 
 
 the margin. The converging fractions 2 ~T 
 
 also are thus found ; & c - 
 
 12222 2 
 
 }> f, i, H, f *, ft, &c. 
 
 Ex. 1. The height of mount Etna is 10963 feet, and that of 
 mount Vesuvius 3900 feet; required the approximate ratios of their 
 heights. Answ. fc J, T \, H, rV*, &, W- 
 
 2 The height of mount Hecla is 4900 feet, and that of mount 
 Perdu, the highest of the Pyrenees, 11,283 feet : required the ap- 
 proximate ratios of their heights. Ansiv. , f , f, f, -fife, & c - 
 
 3 Find the approximate ratios of 1 and 3'60555 13, (that is, >f 1 
 to the square root of 13.) Answ. fr. *, f, T a r , &, VW, yW> j&, ; -' c 
 
 4. Required the series of ratios approaching the ratio of Engl^h 
 and Irish acres. Answ. \, , f, |, i, &, H, I?, \U- 
 
 5. The weights of equal bulks of pure water and fluid mercury 
 
 * This is the square root of 2, which therefore is expressed by a unit, with a continued 
 fraction, each of whese denoimiiatori is 2 Hence ttic fraction may be continue,! with- 
 
ON ARITHMETICAL SCALES. 
 
 are as 1 to 13-568 ; required the series of fractions converging to 
 this ratio. Answ. i,^, tS> A* TroV> *V\ TrWV- 
 
 6 It has been computed, that between the years 1696 and 1800, 
 the value of money decreased so much, that in the former year 
 1 would have procured as much of the necessaries of life as 
 2 7 1 1 in the latter. Required the series of fractions approach- 
 ine to the latio of the values of money at these periods. Answ. , 
 
 ON ARITHMETICAL SCALES. 
 
 THE system of notation for which we are indebted to the Arabi- 
 ans, and which has been employed and exemplified in the preceding 
 pages, proceeds according to the combinations of the number ten, 
 and is such as to correspond to the names given to numerals in 
 almost all languages. The language of every civilized nation, an- 
 cient or modern, furnishes names for ten times ten, and for ten 
 *iines that product : but none of them furnishes distinct names, of 
 general use, for the powers of seven, eight, twelve, &c. ; these, as 
 well as other numbers which are not powers of ten, being denomi- 
 nated from a combination of the names of the powers of ten, with 
 the names of units,* when necessary. Thus, for the second power 
 of eleven we have no distinct name, but we call it one hundred and 
 twenty one, that is, the second power O/*TEN with twice TEN, and one 
 unit : and the third power of fifteen we call three thousand, three 
 hundred, and seventy jive ; that is, three times the third power of TEN, 
 three times the second power of TEN, seven TENS, and Jive units. In 
 like manner, in the natural succession of numbers, the first after 
 one hundred, is called one hundred and one, the next one hundred 
 and two, &c., no new names being given, but merely combinations 
 of those previously formed. 
 
 This remarkable agreement in the numerical language of almost 
 ail nations, seems evidently to arise from the use that is made of 
 the fingers in arithmetical computations, in the ruder periods of 
 society, and very generally by those who have not been instructed 
 in better modes of calculation. Such persons, in counting a num- 
 ber of objects, would naturally distribute them into parcels, each 
 consisting of ten, from the number of the fingers employed in reck- 
 
 out limit, the law of continuation being manifest; r 1=34. * i 
 
 which is not the case when the root is expressed 7 ' "S~_j_~ - J i 
 
 decimally. The same holds respecting the square 3 4-~~ 
 
 root of every number which is not a square. 
 
 Thus, the square roots of 11 and 35 are expressed 
 
 by the continued fractions in the margin, the ^ 
 
 law of continuation in each of which is mani- ' 1 H~ 1 4- &e. 
 
 test. 
 
 * The mathematical reader will know, that the number oru may be regarded as that 
 power of ten ^or indeed of any number) whose index is zero. 
 
ON ARITHMETICAL SCALES. 247 
 
 onmg them : and, if the number of these parcels should be great, 
 it would be natural, in ascertaining their number, to form them into 
 larger parcels, each containing ten of the smaller ; and it is easy 
 to see how this principle would be extended to the numeration 
 of any number of objects however great. It is also evident, that 
 when names would be invented, by some process of formation now 
 unknown, for all the numbers so far as ten, and also for the powers 
 of ten, the names of all other numbers would be obtained by a pro- 
 per combination of these. Thus, the number eleven might be 
 called one and ten ; twelve, two and ten ; thirteen, three and ten ; 
 twenty, two tens; thirty, three tens, &c. : and we find this method 
 of denominating numbers strictly followed, except in some of the 
 smaller numbers, such as eleven, twelve, twenty, sixty, &c. which, 
 from their frequent use, have been more liable to have their names 
 corrupted and altered, but which, when their derivations can be 
 discovered, are always found to be formed on correct anological 
 principles, according to the foregoing explanation. 
 
 Of the advantages and excellence of this system of notation we 
 can scarcely be duly sensible. Instructed in its use from the 
 earliest lessons we receive in Arithmetic; never comparing it, 
 or comparing it but slightly, with other modes of expressing num- 
 bers by characters ; and finding no deficiency, no need of im- 
 provement, nothing to call our thoughts to the subject, we use it 
 without feeling its superiority, and with a very inadequate idea of 
 its power. We do not reflect, that merely by means of the differ- 
 ent positions and combinations of no more than ten simple cha- 
 racters, we can adequately and correctly express any number, how- 
 ever great. With the Roman, or even with the Greek notation, 
 on the contrary, we find it impossible to express numbers that ex- 
 ceed a certain magnitude : and, even were additional characters 
 formed to supply this defect, we would find, that calculations, 
 which are performed with great facility and despatch by the deci- 
 mal notation, would, by either the Greek or Roman system, be ex- 
 cessively tedious and intricate, while the performance of many 
 others would be almost impracticable. 
 
 But, though the decimal system of notation has so far the ad- 
 vantage over these and all other systems not depending on the 
 same principle, we must not conclude, that no other system of 
 equal excellence could be invented. There may be an indefinite 
 number of systems of notation founded on the same principle, and 
 possessing various degrees of excellence. In the decimal nota- 
 tion, we distribute numbers into classed or parcels of ten each ; 
 these classes again into higher classes, each containing ten of the 
 lower ; these into still higher classes, each containing ten classes 
 of the second order, and so on, till the numbers are exhausted. 
 But if we proceed still in the same manner, only making the 
 classes consist of two, instead of ten, each, we have the binary 
 scale of notation, in which only the two characters, 1 and 0, are 
 requisite for expressing all numbers : and if the classes be made 
 
248 ON ARITHMETICAL SCALES. 
 
 to consist of three each, we have the ternary scale, in which only 
 three characters, 1, 2, 0, are requisite. In the t^arse manner, it 
 is obvious we may have a quaternary, quinary, duodecimal, trigest- 
 mal, sexagesimal, centesimal, or any other scale, by merely taking 
 4, 5, 12, 30, 60, 100, or any other assigned number, as the number 
 contained in each class.- This number may be called the RADIX, 
 ROOT, or BASE, of the system; and it is obvious, that in each system 
 there will be as many distinct characters required, as there are 
 units in the radix. Thus, in the decimal scale ten characters are 
 necessary, but. in the duodecimal twelve would be required, which 
 number would be made up by adding to the characters at present 
 in use, two others to denote ten and eleven. 
 
 In what follows, D will be used to denote ten, and H to denote 
 eleven ;* and, in the duodecimal scale, twelve will of course be 
 written 10. 
 
 Hence, to express a given number in any atxigned scale : JJiiide 
 the given number by the radix of the scale ; divide Lite reside also by the 
 radix, and the result arising from this again by the radic. Continue 
 the division in this manner as long a,-; possible, and to the final quotient 
 annex the several remainders in 
 a retrograde order, placing d- i^vovq^'- 
 
 phers where there is no remainder. 
 Thus, the expression of 592835 
 in the decimal scale, will be 
 2470DH in the duodecimal scale, 1 ; >V 1 Hi 
 
 as will appear from the annexed 
 operation. ~ 
 
 Here it is evident, that by 
 dividing the given number by IO^OQ 
 
 12, it is distributed into 49402 
 Classes each containing 12, with 
 :he remainder 11. By the se- 
 cond division by 12, these 
 
 classes are distributed into 41 1G classes, each containing 12 times 
 12, or the second power of 12, with a remainder of 10 of the for- 
 mer classes, each containing 12. By the third operation, the 
 classes last found are distributed into 343 classes, each containing 
 12 of the latter, which were each the second power of 12, and 
 therefore these are each the third power of 12 ; and the remainder 
 is 0. In like manner, the next quotient expresses 28 times the 
 fourth power of 12, and the remainder, 7 times the third power of 
 12; and the final quotient expresses twice the fifth power of 12. 
 with a remainder of 4 times the fourth power of 12. Hence the 
 
 * These characters, which may serve the intended purpose vis well as any others in the 
 few instances in which they will be employed in what follow*, may be easily recollected 
 by conceiving the fiist to be formed by i mining together 1 ai;d 0, the characters winch 
 express 10 in the common nutation aiw 1 by joining wuh a I-L< 1 anu 1, v.Lkh express 
 ektw. 
 
\ 
 
 ON ARITHMETICAL SCALES. 219 
 
 iriven number is analysed into 2 X 12 3 4-4x 12 4 4-7 X 12 a -f- 
 X 12- 4~ 10 X 12-{-ll,or 2470DH, according to the notation above 
 adopted. 
 
 It will be easily found by proceeding in the same manner, that 
 for seven thousand, eight hundred, and fifty four, the expression 
 iu the binary scale will be 1111010101110; in the- ternary, 
 10120*2220; in the quaternary, 1322232; in t\\Q quinary, 2224O4j 
 in the senary, 1 00210; in the septenary, 31620; in the ociary, 
 172-31); in the nonary, 11686; in the denary or decimal, 7854; in 
 the undcnary or undecimal, 59oO J in the duodenary or duodecimal, 
 4060; in the vigesimal, or that whose radix is twenty, (19) (12) (14); 
 in the trigesimal (radix thirty), 8 (21) (24);. in the quinquagesimul 
 (radix fifty), 374; in the sexagesimal (radix sixty), 2D(54); ami in 
 the centesimal (radix one hundred), (78) (54): where each pair of 
 the figures enclosed in brackets would be represented by a single 
 character, were there a sufficient number of distinct characters 
 for each scale. 
 
 The converse of this problem, or the reduction of a number to 
 the decimal scale from any other, will be performed by finding Hie 
 indues of the several digits, and collecting those values into one sum: 
 or, more cattily, by multiplying the left hand digit by the radix, and 
 adding to the product the next 
 
 digit; then by multiplying this sum 4,-p' : 
 
 by iJie radix, and adding to the pro- 
 duct the next digit, and so on till all 
 the digits shall have been employed.* ^a 
 
 Thus, 4503142 in the senary scale 
 is equivalent to 4 X 6 8 -f- 5 X 6 5 + 
 0X6*4-3 X 6" -f-1 X 6*-|-4X 
 64-2, or 226214; which result 
 will be obtained more easily by 
 the operation in the margin. 
 
 Thus it appears that the deci- 
 mal scale is only one out of 
 many on the same principle; and, _ 
 
 upon due consideration, it will be 
 found not to be the best scale 
 that might be adopted. It does . ~ 
 
 not suit the plan or the limits of 
 
 the present work. to add much to 
 
 what has been said on this curi- 99fi9i4, 
 
 ous and interesting subject. l;or 
 farther information, recourse may 
 
 be had to various works, especially to Professor Leslie's Philo- 
 sophy of Arithmetic, to Mr. Anderson's Treatise on Arithmetic 
 in the Edinburgh Encyclopaedia, and Barlow's Theory of Kuiu- 
 
 * It is scarcely necessary to remark, that both this rule and the pieceding are \:<K 
 Kime in principle. a the rule* for the reduction of quant, ties of different denomination*. 
 
 M3 
 
250 ON ARITHMETICAL SCALES. 
 
 bers. it may be sufficient to observe, that the binary is chiefly 
 important from its unfolding some curious properties of numbers ; 
 that from its employing no character of higher value than a unit, 
 operations would be performed by it, though tediously, yet with 
 great facility, and with little exertion of memory : but an insu- 
 perable obstacle to the general use of this scale, is that in ex- 
 pressing numbers, the characters 1 and 0, must generally be re- 
 peated a great number of times. The same advantages, and the 
 same disadvantages belong, but in a less degree, to the tenary 
 scale, and in a still less o gree to the quaternary. The quinary, 
 septenary, and undenary, aii labour under the disadvantage of hav- 
 ing their bases prime numbers, and of thus giving origin to a great 
 number of interminate fractions.* The octary and nonary scales 
 are not so objectionable on this account, but they have no ad- 
 vantage of a different kind to recommend them. The decimal 
 scale does not give origin to so many interminate fractions as 
 either of the latter, and has besides the advantage of expressing 
 numbers more concisely. The senary and duodenary scales, having 
 each so many integral aliquot parts in proportion to their magni- 
 tudes, and those of so convenient a kind, give origin to so much 
 fewer interminate fractions, than any of the above mentioned. 
 The radix of the senary scale is so small however, that numbers 
 are not expressed by it with sufficient conciseness, an objection 
 which does not apply to the duodenary. The latter therefore, for 
 this and other .reasons, is much preferable to any ether scale that 
 could be adopted. To introduce this scale now, however, when 
 men are accustomed to the decimal scale ; when the languages of 
 all civilized nations are suited to it ; and when so many valuable 
 works, particularly tables, in which it is adopted, would be ren- 
 dered comparatively useless, would be unadviseable, and perhaps 
 impracticable : but we must regret that the decimal scale was 
 adopted at a time, when any other scale might have been intro- 
 duced with equal facility. 
 
 The following exercises are annexed for the use of those who 
 may wish to familiarize themselves with this subject : 
 
 Ex. 1. Reduce 1000000 in the decimal scale, to the ternary scale, 
 and also to the nonary. Ansv. 1,2 12,2 1 0,202,001, and 1,783,661. 
 
 2. Reduce 1000000 in the quaternary scale, to the denary and 
 binary scales. Answ. 4096 ; and 1 ,000,000,000,000. 
 
 3. Reduce 123454321 in the senary scale, to the duodenary 
 scale. Answ. 997,3Dl. 
 
 4. How will 476897 in the decimal scale, be expressed in the 
 duodecimal scale? Answ. Imi,ii95. 
 
 5. How will 6666 in the decimal scale, be expressed in the bi- 
 nary and quinary scales? Amw. 1,101,000,001,010; and 203,131. 
 
 That is, interminate fractions having for their denominators the radix of the 
 or i* powers, such as interminate decimals in ihe common scale. 
 
QUESTIONS WITH THEIR SOLUTIONS. '.51 
 
 . Reduce 13579 in the duodecimal scale, to the undecimai 
 
 scale. Answ. 190o3. 
 
 7. Express nine hundred and three pounds, eighteen shillings, 
 
 and eleven pence, in the quinary scale. Answ. 12,103 33 21. 
 
 QUESTIONS WITH THEIR SOLUTIONS.* 
 
 1. WHAT number taken from the square of 48, will leave 16 
 tieies 54? 
 
 Solution. t8 2 =2304, and 54X16=864; then 2304864= 
 l 440, the answer. 
 
 2. Divide 1000 between A, B, and C, and give A 120 more 
 than C, and C 95 more than B. 
 
 Sol. A's share is evidently to be 215 (= 120 + 95) more than B's: 
 therefore 1000 215 95=690; and 690---3=230=B's share. 
 Hence, 230+95= 325, C's share; and 325 +120= 445, A's 
 share. 
 
 3. A father left to his eldest son f f- of his property, to his se- 
 cond %% of the remainder, and to his third son what was left. 
 What was the share of each, the shares of the first and second 
 differing by 500? 
 
 Sol. $$ of H IHi> tne P art f tne whole property belonging 
 to the second son. Then, Hf HH=*i, and 1 !=!?, 
 the part of the youngest. Also J| i_HI=HH> tne difference 
 of the parts of the first and second. Then, as *$} : $f, or as 
 2251 : 3476 : : 500 : 772^^4-, the share of the eldest : as 2251 : 
 1225 : : 500 : 272^ 2 5 r , the share of the second : and as 2251 : 
 1540 :: 500 : 342 Y y i s 8 T, the share of the third. 
 
 4. A gets 4 of a legacy for 3 that B gets, and C 5 for 6 
 that B gets, and A's share is 5000. What is the whole legacy ? 
 
 Sol. As 3 : 6 : : 4 : 8, A's part, when B gets 6, and 
 C 5. Then as 8 : 8-f-6-f5 : : 5000 . 11875, the whole legacy. 
 
 a. A person possessed of | of a ship, sold of his share for 
 1260. What is the value of the whole ship at the same rate? 
 
 Sot of f=|; and as : 1 : : 1260 : 5040, the answer. 
 
 6. A person being asked the hour of the day, said, that the time 
 past noon was f of the time till midnight. What was the hour? 
 
 Sol. As l-J-3 : J, or as 9 : 4 : : 12 hours : 5 hours, 20 minutes; 
 and therefore the time was 20 minutes past 5 o'clock in the after- 
 noon. 
 
 7. A, B, and C purchase a ship ; A pays $, B i, and C 2000, 
 the cost. What are the sums paid by A and B ? 
 
 * These questions are selected from various authors. Their solutions are annexed for 
 the purpose of showing the more advanced student, how they and similar arithmetH a> 
 questions may be resolved. They will be found useful in preparing the pupil for v.-rk.j y 
 the series of questions that follow them, and for resolving any of the more 
 questions that may be proposed for resolution by common Arithmetic . 
 
252 QUESTIONS WITH THEIR SOLUTIONS. 
 
 Sol. $4-f=i, and 1 |f |i, C's part. Then, as J : ?, or 
 as 31 : 14 : : 2000 : 903 U 7 T ; the part paid by A; and as : -?, 
 or 31 : 18 : : 2000 : 1161 ? 9 T , the part paid by B. 
 
 8. Required the stocks of A and B, A's gain being 160, and 
 B's 130, and A's stock 175 more than B's. 
 
 Sol. 160 130=30. Then, as 30 : 175 : : 160 : 
 933, A's stock; and as 30 : 175 : : 130 : 758^, B's 
 stock. 
 
 9. A's stock 240, B's 210; whole gain 120, and C's part 
 of it 30. Required A's and B's gains, and C's stock. 
 
 Sol. 120 3Q=9Q, the sum of A's and B's gains; also 
 240 -{-210:= 450, the sum of their stocks. Then, as 450 : 
 90, or 5 : 1 : : 240 : 48, A's gain ; as 5 : I : : 210 : 42, B's 
 gain; and as 1 : 5 : : 30 : 150, C's stock. 
 
 10. A gains 12 in 6 months, B 15 in 5 months, and C 21 
 in 9 months. What is the whole stock, C's part of it being 40? 
 
 Sol. In compound fellowship the gains are proportional to the 
 products of the stocks and times ; and, conversely, the stocks are 
 proportional to the quotients obtained by dividing the gains by the 
 times. Hence, as V : + + *0 : 125$, the whole 
 stock. A's stock would be found thus : as V : y, or as 2^- : 2, 
 or as 7 : 6 : : 40 : 34f . 
 
 11. A and B gain 13 10, B and C 12 12, and A and C 
 11 16 6. What is -the gain of each ? 
 
 Sol. From 18 19 3, half the sum of the given monies, which 
 'is evidently equal to the gains of all the three, take 12 12, and 
 Aere will remain 6 7 3, A's part. In like manner, 18 193 
 il 166=7 2 9, B's share; and 18 193 13 10=r5 9 3, 
 C's share. 
 
 12. If A can do a piece of work in 10 days, and B in 13; in 
 what time will both do it, at the same rate ? 
 
 Sol. In one day A does fe and B T V f ^ ne work ; therefore 
 both together do in one day iV4-YV T 2 3 3 o f it- Hence, as T Vo 
 of the work : 1, the whole work ; or as 23 : 130 : : 1 day : 5-J-| 
 days, the time required. It appears from this work, that the 
 answer will be found by dividing the product of 10 and 13. by 
 their sum. 
 
 13. A, B, and C, can trench a field in 12 days ; B, C, and I), 
 in 14- days; C, D, and A, in 15 days; and D, A v and B, in 18 
 uays. In what time will it be done by all of them together, and 
 by each of them singly ? 
 
 Sol. In one day, A, B, and C, will do ^ of the whole work 
 B, C, and D, -& ; C, D, and A, T V ; and D, A, and B, T V Lee 
 these fractions be added together, and the sum, iVe%> * s tne P art 
 done by all working 3 d-ays, since in each of the three parts, ^ 
 y*5, and T^, of the whole work, there is one day's work of A ; 
 in^each of the three parts, ^ ^ f and T l , one day's work of B, &c. 
 Dividing the sum by 3, therefore, we have / 7 Yo the part done bv 
 all four in one day. Hence, as yVW of the work: 1, the whole 
 
QUESTIONS WITH THEIR SOLUTICKNS. 253 
 
 work ; or as 349 : 3780 : : 1 day : 103$ clays, the time in which 
 it would be performed by all of them working together. Now, from 
 3 : W<y tne P art done by A, B> C, arid D, take -fa, the part done by 
 B, C, and D, and the remainder g^fcu is tne part done in one day 
 by A. Then, as T f w : 1, or as 7i> : 3780 : : 1 day : 47*$ days, 
 the time in which the work would be finished by A alone. By 
 proceeding in the same manner, we should find, that B would 
 perform it 38$4 days ; C in 27 T Vg- days ; and D in 1 i 1 T \ days. 
 
 14. A an.l B,at the opposite extremities of a wood 135 fathoms 
 in compass, begin to go round it the same way at the same time; 
 A at the rate of 11 fathoms jn 2 minutes, and B of 17 fathoms in 
 3 minutes. How many rounds will each make, before the one will 
 overtake the other ? 
 
 Sol. As 2 minutes : 3 minutes : : 11 fathoms : 16 fathoms, 
 the space gone by A in three minutes. Hence it appears, that B, 
 in going 17 fathoms, gains | fathom on A; and the object of the 
 question being to find how many rounds he will make in gaining 
 half a round, we have this analogy; as fathom : 17 fathoms : : 3 
 round : 17 rounds, the space to be gone over by B: consequently 
 A will make 16^ rounds. 
 
 15. Suppose A, B, and C, to start from the same point, and to 
 travel in the same direction, about an island 73 miles in compass, 
 A at the rate of 6, B of 10, and C of 16 miles per day: in what 
 time will they be next together ? 
 
 Sol. Since B gains 4- miles each day on A, as 4 miles : 73 
 miles : : 1 day : 18^ days, the time in which B would gain a round 
 on A, or in which these two would first be together again. Abo, 
 ns 6 miles, the space gained each day by C on B : 73 miles : : 1 
 day . lii^ days, the time in which B and C will first be together, 
 Now the least common multiple of 18 and 12^, which are both 
 divisible by G^, is readily found to be 3G|, the number of days 
 required. 
 
 16. A person remarked that when he counted over his basket of 
 nuts two by two, three by three, four by four, five by five, or six 
 by six, there was one remaining ; but when he counted them by 
 sevens there was no remainder. How many had he ? 
 
 Sol. The least common multiple of 2, 3, 4, 5, and 6, being 60, 
 it is evident, that if 01 were divisible by 7, it would answer the 
 condition? of the question. This not being the case, however, let 
 60 X 2-f 1, 60 X 3-f- 1, 60 X 4-f 1, &c. be tried successively, and 
 it will be found that 301 = 60X54-1, is divisible by 7 ; and con- 
 sequently this number answers the conditions of the question. If 
 to this we add 420, the least common multiple of 2, 3, 4, 5, 6, and 
 7,. the sum, 721, will be another answer; and by adding perpetu- 
 ally 420, we may find as many answers as we please. 
 
 I 7. At what time, between twelve and one o'clock, do the hour 
 and minute hands of a common clock or watch, point in directions 
 exactly opposite ? 
 
254 QUESTIONS WITH THEIR SOLUTIONS, 
 
 So/. This is the same, as to find in what time after twelve, at 
 which time the hands are together, the minute hand will have 
 gained half a round on the hour hand. Now it is evident, that in 
 12 hours the minute hand gains eleven rounds ; and consequently 
 one round is gained in the eleventh part of 12 hours, and half a 
 round in half that time, or the eleventh part of 6 hours ; that is, in 
 32?\ minutes. The time required, therefore, is 32^ minutes after 
 twelve o'clock. 
 
 18. If one ship, containing 150 hogsheads of wine, pay for toll 
 at the Sound, the value of 2 hogsheads wanting .6 ; and another, 
 containing 240 hogsheads, pay at the same rate, the value of 2 
 hogsheads, and .18 besides; what is the value of the wine $T 
 hogshead ? 
 
 Sol. The tolls must evidently be in the ratio of 150 to 240, or 
 of 5 to 8. Hence the value of 2 hogsheads less by 6, must be 
 to the value of 2 hogsheads together with .18, or, the half of 
 each being taken, the value of 1 hogshead wanting 3, must be to 
 the value of 1 hogshead together with 9, as 5 to 8. of the va- 
 lue of 1 hogshead with | of 9, therefore, must be equal to the 
 value of 1 hogshead wanting 3; that is, f of the value of 1 hogs- 
 head, with 5f , must be equal to the value of 1 hogshead want- 
 ing 3. Hence, 5% must be equal to f of the value of 1 hogs- 
 head wanting 3 : and consequently | of the value of a hogshead 
 must be equal to 8 3 . Hence, as |- hhd. : 1 hhd. . : S% : 23, 
 the value of 1 hogshead. 
 
 19. If 3 men, or 4 women, can do a piece of work in 56 days, 
 in what time will one man and one woman together, perform it ? 
 
 Sol. In 56 days, one man will do J, and one woman of the 
 work ; and consequently, in the same time, one man -and one wo- 
 man will do i -|-:j, or T \, of the work. Hence, as T V of the work : 
 1, the whole work : : 56 days : 96 days, the time required. 
 
 20. If 7 gallons of brandy cost as much as 9 gallons of rum, and 
 9 gallons of mm as much as 12 gallons of geneva, and the price of 
 3 gallons of these, taking 1 of each kind, was 2 26; what was 
 the value of each V gallon ? 
 
 Sol. It appears from the question, that the prices of 1 gallon 
 of t>randy, 1 of rum, and 1 of geneva are as ^, , and -fa; whence, by 
 reducing these fractions to equivalent ones having a common de- 
 nominator, and using the numerators, we find, for the ratios ot 
 the prices, 36, 28, and, 21. We have, therefore, by the method of 
 dividing into proportional parts, the following analogies : as 36 -f- 
 28-4-21, that is, as 85 : 36 : : 2 2 6 : 18/, the value of the brandy 
 V gallon ; as 85 : 28 : : 2 2 6 : 14/, that of the rum ; and as 
 85 T 21 : : 2 2 6 : 10/6, that of the geneva. 
 
255 
 MISCELLANEOUS QUESTIONS.* 
 
 ' 1. IF a person gain 83- per cent, by selling apples at the rate of 
 8 for Gd., how much does he gain per cent, by selling them at the 
 rate of 3 for 2^d ? Answer, 1 1 ^. 
 
 2. If eggs be bought at the rate of 5 for a penny, how must they 
 be sold to gain 40 per cent.? Answ. At the rate of 25 for 7d. 
 
 3. If 150 apples cost 9/4, how many of them must be sold at 
 the rate of 8 for 6d., and how many at the rate of 3 for 2^d., that 
 the gain on the whole may be 10 per cent. ? Answ. 90 at 3 for 
 2-$d., and 60 at 8 for 6d. 
 
 4. A merchant engages a clerk at the rate of 20 for the first 
 year, 25 for the second, 30 for the third, &c., thus augmenting his 
 salary by 5 each year. How long must the clerk retain his situation, 
 so as to receive on the whole as much as he would have received, 
 had his salary been fixed at 52 10 ty annum? Ansuu. 14 years. 
 
 5. Three gentlemen contribute 164 5 towards the building of 
 a church at the distance of 2 miles from the first, 2-J- miles from 
 the second, and 3^ miles from the third ; and they agree that their 
 shares shall be reciprocally proportional to their distances from 
 the church. How much must they severally contribute? Ansu> 
 72 9, 50 8, and 4-1 8. 
 
 6. If a person purchase pins, when there are 18 in the row, 
 which sells for a farthing, and sell them when there are only 1 1 
 in the row, how much is his gain fT cent.? Answ. 63 fY 
 
 7. A hosier sells 90 pair of stockings and gloves for 12 10, the 
 stockings at 3/, and the gloves at 2/6 IT pair. Required the num- 
 ber of each. Answ. 50 pair of stockings, and 40 pair of gloves. 
 
 8. A son having asked his father's age, the father replied : 
 " Your age is twelve years; to which if five-eighths of both our 
 ages be added, the sum will be equal to mine." What was the 
 father's age? Answ. 52 years. 
 
 * The qupstions contained in this article are intended to exercise the advanced stu- 
 dent in the use of the several rules and modes of operation, exhibited both in the text 
 and in the notes of the preceding part of this work. They are not adapted for the ica- 
 jority of arithmetical pupils ; as for their purpose they are too difficult, acd possess too 
 little practical utility. It i* hoped, however, that, besides affording much practice in cal- 
 culation, and in the application of the rules already delivered, they will form useful exer- 
 cises for the reasoning powers of tnost who have taste or ability for such speculations. 
 The great and principal object with every teacher of Arithmetic, should be, to make hit 
 pupils acquire an extensive and substantial practical knowledge of this science, without 
 occupying their time and attention with many puzzling or difficult questions. At the 
 same time, however, when he meets with pupils of capacity, and of considerable profici- 
 ency, if may be very proper to direct their attention to such questions as are contained m 
 this article. By this means he will have a farther proof of their capacity, and he may 
 lay the foundation of future proficiency in other departments of mathematical science. 
 
 Some of the following questions may be solved perhaps most easily by Position. They 
 tnciy ail, however, be wrought without Position : and the student should endeavour to do 
 them without this ruie, as they will thus, form a much better exercise for his thinking 
 powers, and fit him in a greater degree, for solving questions of difficulty, either ui 
 rithmetic, or m other departments of Mathematics. 
 
266 MISCELLANEOUS QUESTIONS. 
 
 9. The population of Great Britain was 10,820,100 in 1801, and 
 12,596,803 in 1811. Hence, it is required to find its amounts in 
 1810 and 1821, the yearly increase being supposed to be propor- 
 tional to the population. Answ. 12,406,733 in 1810, and 14,665,248 
 in 1821. 
 
 10. Three merchantsJb&dng formed a joint stock of ] 064, A's 
 stock continues in tra|w months, B's 8 months, C's 12 months; 
 and A's share of the gph is 114, B's 133 4, and C's 165. 
 What was the stock of each? Answ. A's 456, B's 333, and 
 C's 275. 
 
 1 1. The stocks of three partners, X, Y, and Z, continue in trade 
 8, 10, and 7 months respectively; and their respective gains are 
 115 10, 204 15, and 183 15. Hence, it is required to find 
 their several stocks, the difference between those of Y and Z be- 
 ing 220. Answ. X's stock 550, Y's 780, and Z's 1000. 
 
 12. The joint sum of two series* of continual proportionals, con 
 sisting of five terms each, and having a common mean, is 80^, ami 
 their ratios are 1 and 2-. Required the series'. Answ. 2%, 3^, 5, 
 7$, 11; and f, 2, 5, 12, and 31^. 
 
 13. If A, B, and C could pave a street in 18 days; B, C, and 
 D in 20 days ; C, D, and A in 24 days ; and D, A, .and B in 27 
 days ; in what times would it be done by all of them together, and 
 by each of them singly? Answ. By all in 16-^ days; by A in 
 87f f days ; by B in 50|- days ; by C in 41 T V days ; and by D in 
 170jg days. 
 
 14. If A can reap a field in 13 days, and B in 16 days, in what 
 time would both together reap it? Answ. In 7 -./y days. 
 
 15. If A and B, with C working half time, could build a wall in 
 21 days ; B and C, with D working half time, in 24 days ; C and 
 D, with A working half time, in 28 days ; and D and A, with B 
 working half time, in 32 days ; in what times would it be built by 
 all of them together, and by each of them singly ? Answ. A would 
 finish it in 52 i days ; B in 57|| days ; C in 44 T 4 7 days ; D in 280 
 days ; and all in 16 days. 
 
 16. X, Y, and Z can build -^ of a wall in 10 days ; Y, Z, and V, 
 ! 4 , of it, in 6 days ; Z, V, and X, T 4 ff of it in 7 days ; and V, X, 
 and Y, the remainder of it in 9 days. Hence, it is required to find 
 in what times it would by done by all of them together, and by 
 each of them separately. Answ. By all in 23 ^0^ days ; and by 
 X, Y, Z, and V respectively, in 292^*%, 79|HSS 93HIII, 
 and 624flB days. 
 
 1 7. The stocks of three partners, A, B, and C, are 350, 220, 
 and 250, and their gains 1 12, 88, and 120, respectively ; and 
 B's stock continued in trade 2 months longer than A's. Re- 
 quired the time the money of each continued in trade. Answ. S, 
 10, and 12 months respectively. 
 
 18. In what arithmetical scale would five hundred and fifty four 
 be expressed by 95 ? Answ. In the scale whose radix is 61. 
 
MISCELLANEOUS QUESTIONS. 267 
 
 19. The population of Great Britain uas 0,523,000 in 1700; 
 7,8(30,000 in 1 750 ; 10,820,100 in 1801; and 12,59(3,803 in 181 J. 
 Hence it is required to find the annual raies of' increase on each 
 million of the population between the first and second, the second 
 and third, and the third and fourth, of the above-mentioned years, 
 the rate of increase at any time during each period, being sup- 
 posed to be proportional to the population at that time. Answ. 
 During the first period, 3736 ; during th&second, 6287 ; and dur- 
 ing the third, 15320. 
 
 20. If a gallon of water were resolved into the oxygen and hy- 
 drogen of which it is composed, it is required to detennine the 
 bulk into which it would thus be expanded, water beins 74-1 times 
 heavier than an equal bulk of oxygen, and 9699 times heavier than 
 an equal bulk of hydrogen. (See excr. 5, p. 192 J Answ. The 
 two gases would fill 1794? {% gallons. 
 
 21. The Neperian logarithm* of any number, is to th-2 common 
 logarithm of the same number as 1 is to -4-3429448, nearly. Re- 
 quired the series of ratios converging to this ratio. Anxiv. 2 to 1 ; 
 7 to 3 ; 23 to 10 ; 76 to 33; 99 to 43 ; 175,to 76 ; (or 700 to 304 ;) 
 624 to 271 ; 3919 to 1702; 12331 to 5377; 16300 to 7079, &c. 
 
 22. Reduce the fraction whose numerator is the square root of 
 5, and denominator the square root of 11, to a continued fraction, 
 and find the first seven of the series of fractions converging to its 
 value. Answ. i.'f, ff, f$, &, fH, and H$\. 
 
 23. A person, in discounting a bill, at 6 per cent, per annum, 
 according to the common or false method, finds that he has 6^ per 
 cent, per annum for his money. How long must the bill have 
 been discounted before it was due ? Answ. 1 year and 103 days, 
 nearly. 
 
 24. A person pays 54 for the insurance of goods at 3f per 
 cent. ; and he finds, that in case of the goods being lost, he will by 
 this means be entitled to the value of the goods, the premium of 
 insurance, and 5 besides. What is the value of the goods ? 
 Answ. .1381. 
 
 25. Reduce five ninths to a fraction in the septenary scale of 
 notation, whose denominator in that scale may be expressed by a 
 unit with as many ciphers annexed as there are figures in the nu- 
 merator. -Answ. The numerator will be -381361361, &c. or -S'ei'. 
 
 26. If a merchant each year increase his capital by a fifth part 
 of itself, except an expenditure of 400 per annum, and at the 
 end of 15 years be worth 12000 ; find, without Position, his ori- 
 ginal capital. Answ. 2649 1 1. 
 
 27. A servant draws off one gallon each day for 20 days, from 
 cask containing 10 gallons of rum, each time supplying the de- 
 ficiency by the addition of a gallon of water ; and then, to escape 
 
 * Neperian logarithms are also frequently, but improperly, called hyperbolic loga- 
 rithms. The number -43i'29319, &c- is called the tnjdutus uf the system of cuimuaa 
 logarithms. 
 
 
258 MISCELLANEOUS QUESTIONS. 
 
 detection, he again draws off 20 gallons, supplying the deficiency 
 each time by a gallon of rum. It is required to determine how 
 much water still remains in the cask. Answ. 1*0679577 gallon, or 
 rather more than a gallon and half a pint. 
 
 28. A sells a quantity of tea, which cost him 246 12, to B ; 
 and B sells it to C, who' disposes of it for 391 1 1 10. Required 
 the prices at which A and B sold it, each of the three merchants 
 having gained at the same rate per cent. Answ. A geld it for 
 281 14, and B for 335 13. 
 
 29. A and B set out from the same place, and in the same di- 
 evtion. A travels uniformly 18 miles per day, and after 9 days 
 
 turns and goes back as far as B has travelled during those 9 days; 
 he then turns again, and pursuing his journey, overtakes B 22 
 days from the time they first set out. It is required to find the 
 rate at which B uniformly travelled. Answ. 10 miles per day. 
 
 30. A merchant every year gains 50 per cent, on his capital, of 
 which he spends 300 per annum in house and other expenses, 
 and at the end of 4 years he finds himself possessed of a capital 
 
 4 times as great as what he had at commencing business. Find 
 his original capital without using the rule of Position. Answ. 
 2294,V. 
 
 31. At what time does the sun set, when the length of the day 
 (from sunrise till sunset) is four times the length of the morning or 
 evening twilight, and the evening twilight two sevenths of the time 
 from its termination till day-break ? Answ. At 3 r V minutes past 
 
 5 o'clock. 
 
 32. How will 13579 in the trigesimal scale, be expressed in the 
 duodecimal scale ? Answ. 372433. 
 
 33. Find the first nine fractions approaching to the ratio of 1 
 to the cube root of 2. Answ. \, f, $, ff, H, M, Hl> IH. and 
 
 504. 
 
 34. Required, the approximate ratios of the English foot to the 
 French metre, and also to the toise. (See note, page 43.) Answ. 
 The foot to the metre, as 1 to 3, 3 to 10, 4 to 13, 7 to 23, 25 to 
 82, 32 to 105, 57 to 187, 89 to 292, 146 to 479, &c. ; and the foot 
 to the toie, as 1 to 6, 2 to 13, 3 to 19, 5 to 32, 33 to 21 1, &e. 
 
 35. How will that fraction be expressed in the decimal scale, 
 whose denominator in the octary scale is a unit with as many ci- 
 phers annexed as there are figures in its numerator, and its nu- 
 merator 644 repeated without end ? Answ. f j. 
 
 36. What is the difference between ^ in the quinary scale, and 
 5 7 T in the nonary scale ? Answ. Twenty-seven one-hundred-and 
 thirty -thirds. 
 
 37. What is the product of \ 4 in the duodecimal scale, and J, 
 in the octary scale ? Answ. One and five-elevenths. 
 
 38. It is required to find a sum of money, of which, in the space 
 4< years, the true discount, at simple interest, is 5 more at the 
 rate of 6 than of 4 per cent, per annum. Am>w. 89 18. 
 
 39. One third of a quantity of flour being sold to gain a certain 
 
MISCELLANEOUS QUESTIONS. 259 
 
 rate per cent., one-fourth to gain twice as much per cent., and 
 the remainder to gain three times as much per cent. : it is required 
 to determine the gain per cent, on each part, the gain upon the 
 whole being 20 per cent. Answ. The gains per cent, are 9|, 19|, 
 and 28$. 
 
 40. A man travels from his own house to Belfast in 4 days, 
 and home again in 5 days, travelling each day, during the whole 
 journey, one mile less than he did the preceding. How far does 
 he live from Belfast ? Answ. 90 miles. 
 
 41. What is the radix of the arithmetical scale> in which 
 9(20) (12) 609 in the trigesimal notation will be expressed by 
 5000004? Answ. 19. 
 
 42. The men employed by a gentleman work 12 hours, the wo- 
 men 9 hours, and the boys 8 hours, each day : for labouring the 
 same number of hours, each man receives a half more than each 
 woman, and each woman a third more than each boy : the entire 
 sum paid to all the women each day is double of the sum paid to all 
 the boys ; and for every five shillings earned by all the women 
 each day, twelve shillings are earned by all the men. Hence it is 
 required to find the number of each class employed, the entire 
 number being 59. Answ. 24 men, 20 women, and 15 boys. 
 
 43. A man leaves to his eldest child one-fourth of his property ; 
 to his second, one-fourth of the remainder, and 350 besides ; to 
 his third one-fourth of the remainder and 975 ; to his youngest 
 one-fourth of the remainder and .1400; and what still remains 
 he bequeaths to his wife, whose share is found to be one-fifth of 
 the whole. Hence it is required to find the value of the whole 
 property. Answ. 20,000. 
 
 44. Ttie less of two bales of cloth is bought at the rate of 
 twice as many pence per yard as it contains yards, and costs 
 .31 2 more than the greater, which contains 4 yards for every 
 3 in the less, and is bought at the rate of as many pence per yard 
 as it contains yards. How many yards are contained in each ? 
 Answ. 244 yards in the greater, and 183 in the less. 
 
 45. It is required to find a sum of money such that its true dis- 
 count for one year at 5 per cent, will be 1 more than the sum 
 of the true discounts of one-half of it at 4 per cent, and the rest 
 at 6 per cent. Answ. 11575 4. 
 
 46. A property of 10,000 is left to four children whose ages 
 are 6, 8, 10, and 12 years respectively; and it is divided among 
 them in such a manner, that their several shares being improved 
 at 4 per cent, per annum, compound interest, they shall all have 
 equal properties at the age of 21. It is required to determine the 
 sum left to each. Answ. 2180 3 4, 2380 15 11, 2599 17 9, 
 and 2839 2 lOf. 
 
 47. A property is left to four children, one aged 6 years, two 
 aged 9 years each, and one aged 1 1 years, in such a manner that 
 all their properties are to be equal on their coming to age, com- 
 pound interest being allowed at 4 per cent, per annum. Now, one 
 
260 MISCELLANEOUS QUESTIONS, 
 
 of the twins dying before the period at which the eldest would be 
 of age, his share is divided among the rest in such a manner that 
 their properties may be still equal, when they come to age ; arid, 
 in consequence, each is found at that period to have 1000 more 
 than he would otherwise have had. What was the value of the 
 entire property bequeathed ? Answ. 7367 2 11^. 
 
 48. A man owes a debt, to be paid in four equal instalments at 
 the end of 4, 9, 12, and 20 months respectively; and he finds that, 
 discount being allowed, according to the true method, at 5 " 
 cent. $T annum, 750 paid at present will discharge the whole 
 
 lebt. How much did he owe? Answ. 784 1 V/ T 1 ^. 
 
 49. If a merchant commence trade with a capital of 5000, and 
 uin so much, that, after paying all expenses, his capital, each year, 
 is increased by a tenth part of itself wanting 100, how much will 
 he be worth at the end of 20 years ? Answ. 27910. 
 
 50. A man borrows 500, and agrees to pay simple interest at 5 $*" 
 cent. IP annum. At the end of 1 1 months he pays one part of the 
 principal with its interest; 8 months after he pays another part with 
 its interest from the time it was borrowed; and 11 months after that 
 he pays the remainder of the principal with its interest in like man- 
 ner from the time it was borrowed. What was the amount of 
 each payment, each of the two last being double of the first ? 
 Answ. The first payment 108ff^ J, and each of the others 
 
 51. Mercury revolves round the sun in 87 days, 23 hours, 15 
 minutes, 44 seconds, and the earth in 365 days, 6 hours, 9 minutes, 
 12 seconds. Required the seven first approximate ratios of these 
 periods. Answ. , *\, &, H, TT, iVi ^d fi- 
 
 52. Venus revolves round the sun in 224 days, 1C hours, 49 
 minutes, 1 1 seconds, and the earth in the time stated in the pre- 
 ceding question. Required the first eight fractions approaching 
 the ratios of these periods. Answ. \> , f, f, T %, $f |, f^f, and 
 
 Htff-- 
 
 53. 35 Ibs. of tea being mixed with 20 Ibs. of a better quality, 
 the mixture is found to be worth 7/4 per Ib. Required the value 
 of each kind, the difference of their values being 1/10 ty Ib. 
 Ansiv: 8/6 and 6/8 V Ib. 
 
 54. Express three hundred and fifty seven in the scale of nota- 
 tion whose radix is 3. Answ. 2110K*. 
 
 55. If a person lend 7000 at 6 ^ cent. IP annum, compound 
 interest, and allow the interest to accumulate in the hands of the 
 creditor, except 240 ^ annum, which he lifts for family expen- 
 diture ; how much will the creditor owe him at the end of 1G 
 years? Answ. 11621 1 1. 
 
 56. If a boy read each day 2 lines more of Virgil than he did 
 the day before, and find, that having read a certain quantity in 
 18 days, he will read, at this rate of increase, the same quantity 
 in the next 14 days; how much will he read in the whole time? 
 Answ, 4032 lines. 
 
MISCELLANEOUS QUESTIONS. 261 
 
 f 57. Two men, A and B, are on a straight road, on the opno- 
 
 :e sides of a gate, and distant from it 308 yards and 277 yards 
 effectively, and travel each towards the original station of the 
 other. How long must they walk till their distances from the 
 gate will be equal, B travelling 2 yards, and A 2J- yards, per second? 
 Answ. 1 minute, 33 seconds, or 2 minutes, 15 seconds, 
 _ 58. Every thing being supposed to be as in the preceding ques- 
 tion, at what time will each be at the same distance from the ori- 
 ginal station of the other, as the other is from his ? Answ. In 44 
 minutes after starting. 
 
 59. Suppose a person to mix 1 1 Ibs. of tea with 5 tfcs. of an in- 
 ferior quality, and to gain 16 per cent, by selling the mixture at 7/3 
 per Ib. : it is required to determine the prime costs, when a pound 
 < f the one cost a shilling more than a pound of the other 
 Answ. 6/6$ and 5/6f per 15. 
 
 00. If a person borrow 1100 at 6 per cent, per annum, coin- 
 pound interest, and agree to pay both principal and interest in 
 eleven equal annual payments, how much must each payment be, 
 the first being made at the end of the first year? Answ. 1 39 9 5. 
 
 64. If a farm of 84- acres be held at 1 76 per acre, on a lease 
 of which 48 years are unexpired, what fine must be paid at present 
 to reduce the rent to 10/ per acre during the last 30 years of the 
 lease, compound interest being allowed at 6 per cent, per annum? 
 Answ. 354- 8 11|. 
 
 62. A man aged 45 has a pension of 300 a year during his 
 own life ; but he wishes to exchange it for another to continue 
 not only during his own life, but also during that of his wife, 
 aged 40. What will this pension be, money being supposed to be 
 improvable at 6 per cent, per annum, compound, interest ? Answ. 
 236 14 7. 
 
 63. Required the converging fractions approaching to the ratio 
 of 5 hours, 48 minutes, 48 seconds, and 24 hours. Answ. ^, 2 7 9 , 
 
 A, iV^and ,Vr- 
 
 64. If the acting partner in a mercantile concern contribute 
 1000 to the original joint stock of the company, and annually 
 increase this sum by 150 saved from his salary; to how much 
 will his share of the joint stock amount, at the end of 1 1 years, 
 on the supposition, that, after all expenses are paid, there is a 
 clear gain of 10 per cent, per annum on the entire capital ? Answ. 
 5632 15 10. 
 
 65. Suppose 17 gallons of spirits, at 10/6 per gallon, to be 
 mixed with 7 gallons at a different price. What was the price of 
 the latter per gallon, when 20 per cent, is gained by selling the mix- 
 ture at 13/ per gallon? Answ. 11/7^-. 
 
 66. If a grocer mix together 13 parts of better and 7 parts of 
 worse sugar, the price of the latter per cwt. being only f of that of 
 the former ; v/hat were their prime costs, if by selling 17cwt. 3qrs. 
 20lbs. at 4 6 6 per cwt., he gain 12 11*? Answ. The better 
 3 16 1 IVB, and the worse 3 4 il?f per cwt. respectively. 
 
262 MISCELLANEOUS QUESTIONS. 
 
 67. A merchant bequeaths 1000 among six clerks in pro- 
 portion to their salaries, and the periods they have held their 
 situations. Now, one of them has held his situation five years, 
 and his salary is 120; two of them four years, with salaries of 
 75 each ; and the rest two years, with salaries of 60 each. 
 Required their several shares. Answ. The share of the first 
 384 12 3^; of the two next 192 6 1H each; and of the 
 rest 76 18 5 T 7 3 each. 
 
 68. If a shopkeeper each year double his capital, except an 
 expenditure of 240 per annum, and, at the end of 4 years, 
 be worth only three-fourths of his original capital ; find without 
 Position, what he had at commencing trade, 
 
 Answ. 236 I 3H 
 
 69. What sum will amount to 1 more at simple than at 
 compound interest, in four months, at 5 per cent, per annum ? 
 Answ. 3699 9 2. 
 
 70. Required the sum of the infinite series, f r 4 5~f"^5 i^5~i~ 
 <fa &c. Answ. ^. 
 
 71. Given the extremes of three numbers in harmonical propor- 
 tions 10 and 18 : required the mean. Answ. 12f. 
 
 72. Given the first and second of three numbers in harmonical 
 proportions 70 and 126 respectively: required the third term. 
 Ansiv. 630. 
 
 73. Given the first, second, and third of four numbers in har- 
 monical proportions 6, 7, and 8 respectively : required the fourth. 
 Answ. 9|. 
 
 74. Given the first, third, and fourth of four numbers in har- 
 monical proportions 5, 9, and 10 respectively: required the se- 
 cond term Answ. 5%. 
 
 75. If a bank borrow 10,000 at 4, and employ it in discount- 
 ing bills at 6 ^ cent. ty annum, and afterwards borrow money a* 
 2, and discount bills with it at 4 V cent. V annum ; how much 
 must be borrowed, and so employed, in the latter case, that the 
 gain may be the same as in the former, the bills being discounted 
 in both cases six months before they would be due ? Answ, 
 
APPENDIX, 
 
 CONTAIHINC 
 
 AN INTRODUCTION TO MENSURATION. 
 
 DEFINITIONS. 
 
 I. AN ANGLE is the inclination of two straight lines 
 which meet one another in a point, which is called the 
 VERTEX of the angle : or it is the degree of their opening or 
 divergence. 
 
 II. When one straight line 
 
 standing on another, makes ^ 
 
 with it two angles which are % 
 
 equal, each of these angles is 
 called a RIGHT ANGLE ; and _ 
 
 the straight line which stands 
 on the other is said to be PER- 
 PENDICULAR, Or AT RIGHT ANGLES, tO it, or tO be a PER- 
 PENDICULAR to it. 
 
 III. An OBTUSE ANGLE is great- o 
 
 er than a right angle : an ACUTE % 
 
 ANGLE is less than a right angle. ^ 
 
 An angle is usually named by three 
 letters, the middle one being placed at c J5 
 
 the vertex, and the other two some- 
 where on the lines which contain the 
 
 angle. Thus, the obtuse angle in the last diagram, may be called the 
 angle ACB, or BCA and the acute one the angle A CD, or DCA. 
 When there is only one angle at the same point, it may be named by 
 a single letter placed at that point. 
 
 IV. PARALLEL STRAIGHT 
 
 LINES are those which have 
 every where equal perpendicu- . 
 
 lar distances from each other. 
 
 V. A SURFACE, or SUPERFICIES, has length and breadth, 
 without thickness. 
 
 VI. A BODY, or as it is often called, a SOLID, has 
 length, breadth, and thickness, or depth. 
 
 VII. A FIGURE is a portion of space inclosed by one or 
 more boundaries. 
 
 The small introductory tract on Mensuration which ia here given, is intended for 
 the use of such pupils as may not have time or opportunity for studying a more extended 
 course. ^ For this reason, the easiest and most useful parts are selected, and the defi- 
 nitions <nd illustrations are delivered in plain and familiar terms, rather than with a 
 view to Mathematical precision. The rules are also given without demonstrations, aj 
 ' the pupils for whom this abstract is intended, are not supposed to have read a preparatory 
 course of Mathematics. Those who may wish to prosecute the subject more extensively 
 may have recourse to the treatises of Bonnycastle, Button, and other* on the subject. 
 Button's larger treuise, in particular, ii a very extensive and valuable work. 
 
264 APPENDIX. 
 
 VIII. A FIGURE in said to be EQUILATERAL, if if have 
 equal sides; and EQUIANGULAR, if it have equal angles. 
 
 IX. If a figure be contained by three lines, it is culled a 
 TRIANGLE; if bv four, a QUADRILATERAL ; if by more than 
 four, a POLYGON. 
 
 X. An equilateral and equiangul-ar polygon is often called 
 
 a REGULAR POLYGON. 
 
 XL Polygons of five, six, seven, eight, nine, ten, eleven., 
 and twelve sides, are often called respectively, PENTAGONS, 
 
 HEXAGONS, HEPTAGONS, OCTAGONS, NONAGONS, DECAGONS. 
 HENDECAGONS, and DODECAGONS. 
 
 XII. A quadrilateral which has its opposite sides parallel, 
 is called a PARALLELOGRAM : and a parallelogram which 
 has its angles right angles, is called a RECTANGLE. 
 
 XIII. A quadrilateral which has its sides equal, and its 
 angles right angles, is termed a SQUARE ; and a quadrilateral 
 which has its sides equal, but its angles not right angles, is 
 called" a RHOMBUS. 
 
 XIV. A quadrilateral which has two sides parallel, and 
 the other two not, is called a TRAPEZOID. 
 
 XV. Any quadrilateral, except a parallelogram or trape- 
 zoid, is called a TRAPEZIUM. 
 
 XVI. A DIAGONAL of a figure is a line passing through 
 two of its remote angles. 
 
 XVII. A CIRCLE is a figure contained on a flat surface, 
 by one line which is called the CIRCUMFERENCE; and is 
 such that all straight lines drawn to the circumference 
 from a certain point within the figure, called the CENTRE, 
 are equal to each other. Any of those equal lines is called 
 a RADIUS : and a line drawn through the centre, and ter- 
 minated both ways by the circumference, is called a DIA- 
 METER. Hence, a diameter is evidently double of a ra- 
 dius. 
 
 XVIII. If the ends of a thread, 
 FPf, be fastened to two pins, 
 F,f, fixed at a less distance asun- 
 der than the length of the 
 thread, and if the point of a pen 
 or pencil, P, be carried round 
 in such a manner as to keep the 
 thread constantly stretched, the 
 figure inclosed b}' the curve l>ne 
 tlius described, is called an EL- 
 
 I.IPSE; the points F and f, wher. the pins are fixed, are 
 
APPENDIX. 265 
 
 called the FOCI Cand each of them a FOCUS) ; the line AB, 
 drawn through the foci, and terminated both ways by the 
 curve, is called the GREATER AXIS; and the line DE, 
 drawn perpendicular to this axis through its middle point, 
 and terminated by the curve,*fs called the LESS AXIS. 
 
 XIX. MENSURATION is the method of determining by 
 computation the comparative magnitudes of figures ; and 
 is divided into two branches, the mensuration of surfaces, 
 and the mensuration of bodies or solids. 
 
 XX. The AREA of a figure is the space which it contains. 
 In Mensuration, the magnitude of this space is ascertained 
 by the number of times that a given space, called the mea- 
 suring unity is contained in it. 
 
 XXI. The MEASURING UNIT which is adopted for sur- 
 faces, is a square whose side is some of the common mea- 
 sures of length, such as a square inch, a square foot, a square 
 yard, &c. (See the Table of Square Measure, page 39.) 
 
 MENSURATION OF SURFACE& 
 
 RULE I. Tojindthe area of a parallelogram : Multiply its 
 length by its perpendicular breadth. 
 
 Hence, to find the area of a square, we are only to mul- 
 tiply a side by itself. 
 
 Exam. 1. Required the area of A B 
 
 the parallelogram ABDC, the length 
 AB, or CD, being 30 feet, and the 
 perpendicular breadth, AE, 15 feet. j 
 
 C E D 
 
 Here the product of 36 and 15 is 540, the area in square feet. 
 
 Exam. 2. If the length of the floor of a rectangular room be 
 22 feet 5 inches, and its breadth 16 feet 11 inches, how many 
 square feet does it contain ? 
 
 This and similar examples, in which the dimensions are given in 
 feet and inches, may be wrought in different ways. 
 
 1. By reducing 5 inches and 11 inches each to the decimal of a 
 foot, we obtain for the dimensions, 22-4-16', and 16-9 16 7 ; the pro- 
 duct of which is 379-21527' square feet, or 379 square feet nearly, 
 the area required. 
 
 2. As a second method, by reducing the dimensions to inches. 
 we obtain 269 and 203 ; the product of which is 54607 quare 
 inches, the area required : and by dividing this bv 144, we get 379 
 square feet, 31 square inches, a result which agrees with that found 
 by the former method. 
 
 N 
 
266 APPENDIX. 
 
 3. Another mode, which is perhaps preferable to either of the 
 preceding, is the method of duodecimals, or, as it is often improperly 
 called, the method of cress multipli- Feet In. 
 
 cation. This will be understood from 22 5 
 
 the operation in the margin, in which 16 11 
 
 we multiply successively by 16 and 
 
 11. The product of 16 and 5 is 358 8 
 
 divided by 12, and the remainder is 20 6 7 
 
 set down, and the quotient carried. 
 In multiplying by 1 1, both the pro- 379 2 
 
 ducts are divided by 12, and the remainders set one place towards 
 the right hand. The sum of the two partial products is then taken, 
 and the result is found to be 379 square feet, 2 twelfths of a square 
 foot, or 24 square inches, and 7 square inches ; or 379 square feet, 
 31 square inches, as before. The answer might also be obtained 
 by multiplying by 16, and working for 11 inches by means of ali- 
 quot parts. 
 
 The results obtained by , this latter method are often called, very im- 
 properly,^^, inches, and parts, instead of square feet, twelfths of such 
 feet, and square inches. According to this mode, the last answer would 
 be read 379 feet, 2 inches, and 7 parts. It may be written, 379 f. 2* 7". 
 The multiplication of feet and inches might also be performed 
 very simply and easily by reducing the feet to the duodecimal scale, 
 and then carrying for 12 in the multiplication. Thus, by reduction 
 to the duodecimal scale, (see page 
 248) the dimensions in the foregoing 
 example become lD-5 and 14-H, 
 and the work will stand as in the 
 margin, the answer in the duodeci- 
 mal scale being 277-27, or by re- 
 duction to the decimal scale, 379 
 feet, 2 twelfths of a foot, and 7 in- 277-27, or 
 
 ches. The learner, after a little 379f. 2> 7", Answer, 
 
 practice, will perhaps prefer this method to any other. 
 
 Exam. 3. Required the area of a square field, each of whose 
 sides is 6 chains 30 Jinks. 
 
 Here, (note 1, page 40) each side is 630 
 links ; and multiplying this by itself we ob- 
 tain for the area 396900 square links ; or 
 3-969 acres, by division by 100,000; whence 
 by multiplying by 4 and 40, we get for the 3-96900 
 
 required Content' 3a. 3r. 35 p. with a small 4> 
 
 remainder. 
 
 35-040 
 
APPENDIX. 267 
 
 jx. 1 . Required the content of a field in form of a parallelo- 
 gram, the length and breadth of which are 12 chains 76 links, and 
 Q chains 43 links. Answ. 12 acres, roods, 5 per. 
 
 2. Given the length of a street = 937 feet 6 inches, and its 
 breadth=66 feet 8 inches ; required the cost of paving it at 8d. ^ 
 square yard. Answ. .245 18 llf. ,. 
 
 3. At 9f d. ^ yard, required the cost of painting a room, the 
 sum of the lengths of whose sides is 70 feet 10 inches, and its 
 height 10 feet I inch. Answ. 3 4 5f. 
 
 4-. Required the content of a rectangular garden whose length is 
 98 yards, and its breadth 81 yards. Answ. I acre, 2 perches, Irish 
 measure; or 1 acre, 2 roods, 22 perches, 12 yards, English mea- 
 sure. 
 
 5. What is the content of a deal-board, 9 feet 8 inches long, and 
 8 inches broad? Answ. 6f. 1CX 2". 
 
 6. If each side of a square table be 3 feet, 10 inches, what is its 
 content? Answ* 14-f. & V. 
 
 RULE II. To jind the area of a triangle : Multiply the base 
 by the perpendicular, and take half the product : or multi- 
 ply one of these lines by half the other. 
 
 Exam. 4. Required the area of the triangle ACD (see the dia- 
 gram page 265) whose base CD is 15 feet 4 inches, and its perpen- 
 dicular AE, 8 feet 7 inches. 
 
 Feet In. 
 
 Here, by the method of duo- 15 4 
 
 decimals already explained, and by 8 7 
 
 halving, we find for the area, 65 "j22 8 
 
 feet, 9 twelfths of a foot, and 8 8114 
 
 inches, or 65f feet, nearly. g - 7 4 
 
 65 98 
 
 Ex.7. Base= 13 chains 24 links, perpendiculatf = 8 chains 59 
 links. Answ. Areaz: 5 a. 2r. 30 p. 
 
 8. Base = 21 feet 7 inches, perpendicular = 17 feet 10 inches. 
 Answ. Area= 192f. 5' 5". 
 
 RULE III. To find the area when the three sides are given 
 {I.} Add the three sides together, and take half the sum: 
 (2.) From the half sum take the three sides severally: (3.) 
 Find the continual product of the half sum and the three 
 remainders : (4.) Extract the square root of this product* 
 
 The area of an equilateral triangle may be found by mul- 
 tiplying die square of one of the sides by '4330127. 
 
 E vim. 5. Required the area of a triangle, whose sides are 3, 
 3, and 4 feet respectively. 
 
 Here, half the sura of the sides is 4*5 ; and subtracting frc-m this 
 
2 f,S APFFNDIX'. 
 
 the three sides successively, we find the three remainders 2*5,1*5, 
 and -5. Then 4-5 X 2-5 X 1'5X '5 = 8-4375 ; and ^ 8'4375== 
 2-9047375, the area required. 
 
 Ex. 9. Sides, 9 chains 62 links ; 6 chains 38 links ; and 7 chains 
 20 links. Answ. Area, 2 acres, 1 rood, 7 perches, nearly. 
 
 10. Sides, 13, 14, and*15 feet. Answ. Area, 84 feet, or 9 yards. 
 
 11. Sides, 3 feet 8 inches ; 4 feet 7 inches; and 6 feet 5 inches. 
 Answ. Area, 8'233 feet. 
 
 RULE IV. To Jind the area of a trapezoid : Multiply the 
 sum of the parallel sides by the perpendicular breadth of 
 '.he figure, and take half the product. 
 
 Ex. 12. Given the parallel sides of a trapezoid =: 33 and 28 in- 
 ches, and its breadth =11 inches : required the area. Answ. 
 335 J inches. 
 
 13. Parallel sides = 75 and 33 feet, breadth = 20. Answ. Area, 
 1080 feet, or 120 yards, 
 
 RULE V. Tojindthe area of a trapezium : Multiply either 
 of the diagonals by the sum of the perpendiculars drawn to 
 it from the opposite angles, and take half the product. Or, 
 
 Find, by rule II. or III., the areas of the triangles that 
 compose the trapezium, and add them together. 
 
 Exam. 6 Given BD, the diagonal 
 of the trapezium ABCD, rr 15 perches 
 6 yards ; and the perpendiculars drawn 
 from A and C = 6 perches 5 yards, and 
 5 perches 4 yards ; to find the area. 
 
 Here, the sum of the perpendiculars is 12 perches 2 yards, or 
 86 yards; and the diagonal being 15 perches 6 yards, or 1 1 1 yards, 
 we have 8H X 111 = 9546, the naif of which is 4773, the area in 
 square yards, which, by division by 49 and 40, is reduced to ? 
 roods, 17 perches, and 20 yards. 
 
 Ex. 14. Perpendiculars 8 chains 82 links, and 7 chains 73 links ; 
 diagonal 17ch. 561, Answ. Area = 14 acres, 2 roods, 5 perches. 
 
 15. Given BD = 21 perches, BA=18 perches, AD=9 
 perches, BC= 12 perches, and CDrr 15 perches. Answ. Area 
 = 168-68008. 
 
 RULE VI. To find the area of a polygon : (].) Divide it 
 by diagonals into triangles and trapeziums : ('2.) Find the 
 areas of these by some of the foregoing rules : (3.) Add all 
 these areas together. 
 
APPENDIX. 269 
 
 Ex. 16. Given EU= ad 
 perches, and the perpendicu- 
 lars to it from A and D rr: 27 
 and 18 perches respectively; 
 given also AC = 4-6 perches, 
 and the perpendicular to it 
 from B = 20 perches ; to 
 find the area. Ans. lla.0r.5p. 13 
 
 17. Given AB = 6 feet, BC= 3 feet, CD = 4 feet, DE =r 5 
 feet, AE = 4 feet, ACzr 7 feet- M EC = 8 feet. Required the 
 area. Answ- 31"! 23574 fpp- 
 
 I?"~ ~z. V>'^ iiie area of a regular polygon is most easily 
 fou.iu uj multiplying the square of one of the sides by 
 die number standing opposite to the name of the polygon 
 in the annexed table. 
 
 Pentagon 1-7204774- 
 
 Hexagon 2-5980762 
 
 Heptagon 3-6339126 
 
 Octagon 4-8364272 
 
 Nonagon 6-1818240 
 
 Decagon 7-69420S8 
 
 Hendecagon 9-3656411 
 
 Dodecagon 11-1961524 
 
 Ex. 1 8. Side of a Hexagon = 5 inches. Answ. Area = 
 54-951905. 
 
 19. Side of an Octagonal inclosure = 60 yards. Answ. Area 
 = 2 a. Or. 34 p. 36yds. Irish measure, or 3 a. 2r. 14 p. 19yds 
 English measure. 
 
 RULE VIII. The diameter of a circle being given to find the 
 circumference: Multiply the diameter by 3*1416, or more 
 accurately, 3* 14- 1593 : Or, 
 
 As 113 is to 355, so is the diameter to the circumference : 
 or, when much accuracy is not necessary, as 7 is to 22, so 
 is the diameter to the circumference. 
 
 The diameter would be found from the circumference, 
 by reversing any of the foregoing processes. 
 
 Ex.20. Given the diameter = 13 inches. Answ. Circumfer- 
 ence = 40-8407. 
 
 21. If the diameter of a circular cask be 2 feet 9 inches, what 
 will be the length of a hoop for it ? Answ. 8 feet, 7-67257 inches. 
 
 22. If the girt of a round tree be 12 feet 5 inches, what is its 
 diameter? Answ. 3 feet, 11-42817 inches. 
 
 RULE IX. Tojind the area of a circle : Multiply the di- 
 ameter by the circumference, and take one fourth of the 
 product : Or, 
 
 Multiply the square of the diameter by 7S54-, or, for 
 greater accuracy, by '785398 : Or, 
 
 Multiply the square of the radius by 3-141593 : Or, 
 
270 APPENDIX 
 
 Multiply the square of the circumference by *07958. 
 
 The area of an ellipse is found by multiplying the product 
 of the axes by -7854, or more accurately, '785398. 
 
 Ex. 23. Required the area of a circular pond whose diameter is 
 31 yards. Answ. 754-7694- square yards. 
 
 24. Required the area of the space on which a horse may graze, 
 when confined by a cord 7 perches in length, having one of its 
 ends fixed at a certain point. Answ. 1 a. r. 16'7 p. 
 
 25. Required the content of a circular grove, 56'5 perches in 
 circumference. Answ. la. 2 r. 14- p. 
 
 26. What is the area of a circular table whose diameter is 5 feet 
 8 inches ? Answ. 25 feet 31$ inches, nearly. 
 
 27. What is the area of an elliptic ceiling, the axes of which are 
 33 feet 5 inches, and 20 feet 3 inches ? Answ. 59 yards, feet, 
 67 inches. 
 
 MENSURATION OF BODIES, OR SOLIDS. 
 DEFINITIONS." 
 
 I. A figure whose ends, or bases, are parallel, and its sides 
 parallelograms x is called a PRISM. Such a figure is termed 
 a RIGHT PRISM, if each of its bases be perpendicular to its 
 other sides ; and it is said to be TRIANGULAR, HEXAGO-NAL, 
 &c. if its bases be triangles, hexagons, &c. 
 
 II. A prism whose bases, as well as its other sides, are 
 parallelograms, is called a PARALLELEPIPED. A common 
 chest, bar of iron, brick, &c. afford instances of right parallele- 
 pipeds. 
 
 III. A right parallelepiped whose sides are equal squares, 
 is called a CUBE. Such are dice. 
 
 IV. A PYRAMID is a body bounded by plane surfaces 
 meeting in a point called the VERTEX of the pyramid, and 
 by a rectilineal base terminated by those planes. 
 
 V. A CYLINDER is a round body which is of equal thick- 
 ness throughout, and wnich has circular bases, parallel to 
 each other, such as a rolling stone for walks, a round pillar, &c. 
 
 VI. A body which has a circular base, and which tapers 
 uniformly to a point named the VERTEX, is called a CONE. 
 A sugar-loaf, and the top of a round pillar are nearly of 
 this form. 
 
 VII. A FRUSTUM of anybody is what remains, when the 
 top is taken away. 
 
 VIII. A GLOBE, or SPHERE, is a body of such a figure 
 
APPENDIX. / / I 271 
 
 that all points of the surface are equally distant from a 
 point within it called the CENTRE. 
 
 IX. If one of the parts into which an ellipse is divided 
 by either of the axes, revolve about that axis, the figure 
 which it describes is called a SPHEROID : PROLATE, if the 
 revolution be performed round the greater axis ; OBLATE, 
 if round the less. An egg is nearly of the former figure ; 
 and a watch, or a flat turnip, nearly of the latter. 
 
 X. The CONTENT, or VOLUME, or as it is often improperly 
 called, the SOLIDITY of a body, is the space contained with- 
 in it. The magnitude of this space is expressed by the 
 number of times that it contains a given space called the 
 measuring unit, 
 
 XI. The MEASURING UNIT which is adopted for bodies, 
 is a cube whose base is the measuring unit for surfaces, 
 such as a cubic inch, a cubic foot, &c. (See the Table of 
 Cubic Measure, page 40.) 
 
 RULE I. To Jind the content of a prism or cylinder : (1.) 
 Find the area of the base : (2.) Multiply this by the per- 
 pendicular height, or distance between the ends. 
 
 Hence, to Jind the content of a right parallelepiped, take 
 the continual product of the length, breadth, and thickness, 
 or depth : and, to Jind the content of a cube, find the third 
 power of one of the sides of its base. 
 
 Exam. 1. Required the content of a box of cloth, the length, 
 breadth, and depth of which are 4 feet 10 inches ; 2 feet 1 1 inches; 
 and 2 feet 2 inches, respectively. 
 
 Here, by the method of duodecimals, we multiply 4 feet 10 
 
 inches by 2 feet 1 1 inches : the Feet In. 
 
 product, 14 feet, 1 twelfth, and 4 10 
 
 2 inches, is the area of the base. 2 1 1 
 
 We then multiply in a similar 
 
 manner by 2 feet 2 inches, the 9 8 
 
 depth, setting each product in 452 
 
 multiplying by the inches, one 
 
 place towards the right hand. 14 1 2 
 
 The final product or content of 22 
 the box, is 30 cubic feet, 6 
 
 twelfths of a foot, 6 one-lmn- 28 2 4 
 
 dred-and-forty-fourths of a foot, 242 
 
 and 4 inches, or 30 \ feet, nearly. 
 
 In measures of capacity, since 30 & 6" 4 W . 
 
 the cubic foot contains 1728 cubic inches, the twelfths of a foot 
 are each 144 cubic inches, and the hundred-and-forty-fourths are 
 each 12 inches. Hence the foregoing answer is easily reduced to 
 
272 APPENDIX. 
 
 30 feet and 940 inches. The same result would be obtained by 
 
 reducing the dimensions to inches, finding their continual product, 
 
 and dividing it by 1728; or by reducing the inches to decimals, 
 
 and proceeding in a similar manner. The results in such cases as 
 
 tiie present, are often improperly called feet, inches, parts, and 
 
 seconds. 
 
 Ex. 1. Required the number of gallons of water, of 217*6 cubic 
 inches each, contained in a rectangular cistern, the length, breadth, 
 and depth of which are 16 feet, 10 feet 6 inches, and 8 feet 4 
 inches. Answ. 11117*647. 
 
 2. If each side of the base of a triangular prism be 2 inches, and 
 its length 14 inches, what is its content ? Answ. 24-2487 inches. 
 
 3. Required the number of cubic feet contained in a room 
 whose length, breadth, and height are 24 feet, 18 feet > inches, 
 and 10 feet 7 inches. Answ. 4699. 
 
 4. Given the diameter of the base of a cylindric column = 3 
 feet 1 inch, and its height =: 18 feet 9 inches :* required its content. 
 Answ. 140-00 16 feet. 
 
 5. Required the content of a bale, the length, breadth, and 
 thickness of which are, 4 feet 8 inches, 3 feet 3 inches, and 2 feet 
 6 inches. Answ. 37 feet, 1 1 twelfths. 
 
 6. Given the length, breadth, and thickness of a uniform plank 
 equal to 22 feet 7 inches, 1 foot 5 inches, and 6 inches, respec- 
 tively. Required the content. Answ. 17-32957 feet. 
 
 RULE If. To find the content of a pyramid or cone : Mul- 
 tiply the area of the base by the perpendicular height, and 
 take one third of the product. 
 
 Ex.7. Given each side of the base of a square pyramid =r 10 
 inches, and the perpendicular height, or altitude == 9 feet 9 inches : 
 required the content. Answ. 2f. S 7 1". 
 
 8. Given the diameter of the base of a conical glass-house = 
 37 feet 8 inches, and the altitude =r 79 feet 9 inches. Required 
 the entire space enclosed. Answ. 29622 feet, nearly. 
 
 9. The height of the largest of the Egyptian pyramids is 477 
 feet, and each side of its base, which is a square, is 720 feet. Re- 
 quired the content. Answ. 82425600 cubic feet, or 3052800 
 cubic yards. 
 
 10. Given the altitude of a conical sugar-loaf r= 17 inches, and 
 the diameter of its base ;= 9 inches. Required die content. Answ. 
 360'4986 inches. 
 
 11. How often may a conical glass, 3 inches deep, and If inches 
 in diameter at the mouth, be filled out of an Irish liquid gallon ? 
 Answ. 90^ times nearly. 
 
 RULE III. To find the content of a frustum of a pyramid, 
 made by a plane parallel to its base: (I.) To the product of two 
 
APPENDIX. 273 
 
 corresponding sides of the greater and less ends, add one 
 third of the square of their difference ; the sum will be the 
 square of a side of the mean base : (2.) From this find the 
 area of the mean base by some of the rules already given 
 for the measurement of plane surfaces : (3.) Multiply the 
 result by the height. 
 
 To find the content of a like frustum of a cone : (1.) To 
 the product of the diameters of the two ends add one third 
 of the square of their difference ; the sum will be the square 
 of a mean diameter : (2.) Multiply this square by '1854, 
 and the product by the height. 
 
 Exam. 2. Given the sides of the greater and less bases of a 
 regular octagonal pyramid == 19 and 10 inches respectively, and 
 the length 5 feet 6 inches : required the content. 
 
 Here, 19 10 =r 9, and 19 X 19 + 9 2 -H3= 217, the square of a 
 side of the mean base, which being multiplied by 4*8284272, the 
 tabular number (see page 27 1 ) we find for the area of the mean 
 base 1047-7687024 : and the product of this by 5-5, the length, 
 is 5762-7278632 : and dividing this by 144, we obtain 40-01894349 
 feet, the content required. 
 
 Ex. 12. If the length of a frustum of a square pyramid be 18 
 feet 8 inches, the side of its greater base 27 inches, and that of its 
 less 16 inches, what is the content? Answ. 61*228395 cubic feet. 
 
 13. Required the content of an ale glass in form of the frustum 
 of a cone, the diameter at the mouth being 2$ inches, that of 
 the bottom 1 inch, and the depth 5 inches. Answ. 12-76275 
 cubic inches. 
 
 RULE IV. To find the content of a globe: Multiply the 
 cube of the diameter by -5236* 
 
 To find the content of a spheroid: Multiply the fixed axis 
 by the square of the revolving axis, and the product by 
 5236. 
 
 Ex. 14. Required the contents of three globes, whose diameters 
 are 12, 15, and 21 inches respectively. Answ. 904-7808, 1767-15, 
 and 4849-0596 cubic inches respectively. 
 
 15. Required the content of a balloon in form of a prolate 
 spheroid, having its longest diameter 48 feet, and its shortest 38 
 feet. Answ. 36291-7632 cubic feet. 
 
 RULE V. To find the area of the surface of a body bounded 
 by plane surfaces : Find the areas of those surfaces separately, 
 and add them together. 
 
 To find the area of the curve surface of a right cone : Mul- 
 tiply the circumference of the base by the slant height, and 
 take half the product. 
 
274 APPENDIX. 
 
 To fold the area of the surface of a globe ; Multiply th 
 square of the diameter by 3'1416. 
 
 Ex. 1C. Required the area of the surface of a square pyramid 
 each side of the base of which is 2 feet 8 inches, and its slai?* 
 height, measured from the vertex to the middle of any side of the 
 base, 3 feet 9 inches. Answ. 27 feet, 1 twelfth, and 4 inches. 
 
 17. Required the area of the entire surface of a right cone, the 
 slant height of which is 4 feet 7 inches, and the diameter of its 
 base 2 feet 1 1 inches. Answ. 27-679896 square feet. 
 
 18. If the earth were a sphere 7912 miles in diameter, what 
 would be its superficial content? Answ. 1 96,663,355-7504 square 
 miles. 
 
 19. Required the superficial contents of the three globes men- 
 tioned in exercise 14. Answ. 452-3904, 706-86, and 1385-4456 
 square inches respectively. 
 
 RULE VI. Tojlnd the content of round or squared timber : 
 (1.) Take the girt of the tree at the middle, and divide it 
 by 4 (which may be done by halving tbe line used in taking 
 the girt, and then halving the half thus obtained) : (2.) 
 Multiply the square of the quarter girt thus found, by the 
 length. 
 
 If the breadth and depth of squared timber differ consider' 
 ably, measure them, and multiply their product by the 
 length. 
 
 If the tree do not taper uniformly ^ measure the girts at the 
 middle and ends, or at other equal distances ; add the re- 
 sults together, and divide the sum by the number of girts 
 taken ; use the quotient as a mean girt, and proceed as before. 
 
 If great accuracy were required, timber which tapers uniformly, 
 should be measured as the frustum of a pyramid or cone ; but when it 
 tapers slowly, as is generally the case, any dimension taken at the 
 middle may be used as a mean without much error. The measurement 
 of round timber, however, by taking, according to the foregoing rule, 
 which is universally adopted in practice, the quarter girt as the side of 
 a square to a mean section of the tree, is very erroneous, giving the 
 content far too small. If it should be wished to correct the result thus 
 fyund, it mny be done by adding to it one fourth of itself, and to the sum one 
 fool for every 54 contained in it. Another method of approximating the 
 true content is, to multiply the square of one fifth of the girt by twice the 
 length ; or, which is perhaps easier, to multiply the square of the girt by 
 08, and the product by the length. By both these methods the result is 
 rather too great, and requires to be diminished by one foot in 190. 
 
 Ex. 20. Given the mean girt of a square piece of timber ~ 6 
 feet 8 inches, and its length r= 44 feet 4 inches : required the con- 
 tent. Answ. 123- F48 7 cubic feet. 
 
 21. Required the content of a 
 
APPENDIX. 275 
 
 is 32 feet 6 inches, and its mean girt 5 feet 10 inches. Answ. 69*1 19 
 cubic feet; or, more correctly, 87-9 feet. 
 
 22. Given the length of a piece of squared timber = 3 1 feet 4 
 inches, and its breadth and depth at the middle = 2 feet 7 inches, 
 and one 1 foot 9 inches respectively. Required its content. Answ. 
 14-1-65 cubic feet. 
 
 23. Given the girts of a round tree at the ends, and at two in- 
 termediate points equally distant from them and from each other, 
 equal to 10 feet 6 inches, 5 feet 6 inches, 8 feet 8 inches, and 7 
 feet respectively, and the length 21 feet 5 inches. Required the 
 content. Answ. 83-89 cubic feet ; or, more correctly, 106-8 feet. 
 
 RULE VII. Tojind the content of a common cask in gallons: 
 (1 ) To the square of the head diameter add double the 
 square of the bung diameter, and from the sum take four- 
 tenths of the square of the difference of those diameters : 
 (2.) Multiply the remainder by the length : (3.) Then mul- 
 tiply the result by '0012, to find Irish liquid gallons ; by 
 0011-5, to fincl English wine gallons; orb"y0009|, to find 
 English ale gallons. 
 
 The result thus found in Irish gallons may be corrected by 
 adding to it a four-hundredth of itself. 
 
 Ex. 2-1, If the head diameter of a cask be 25 inches, the bung 
 diameter 34 inches, and the length 43 inches, how many gallons 
 does it contain? Answ. 150 Irish, 141^ wine gallons, or 115 
 ale gallons. 
 
 Ex. 25. Given the length of a rectangular field = 15 perches : 
 required its breadth so that it may contain an acre. Answ. lOf perches. 
 
 26. If a horse be bound in the middle of a field by a cord, one 
 end of which is fixed at a certain point ; what must be the length 
 of the cord, that the horse may be allowed to graze on exactly an 
 acre? Answ. 7-1365 perches. 
 
 27. Given the height of a stone column in form of a frustum of 
 a cone = 28 feet 6 inches, and the diameters of its ends 3 feet, and 
 2 feet 3 inches, respectively. Required its weight, a cubic foot of 
 the stone weighing 2568 ounces, avoirdupois. Answ. 11 tons, 2 
 cwt. 2qra. 3tbs. 
 
 28. Required the diameter of a circle whose area is a square 
 foot. Answ. 13-54054 inches. 
 
 29. Required the diameter of a globe whose content is a cubic 
 foot. Answ. 14-8884 inches. 
 
 30. Given the diameter of the base of a cone =r 50 inches, and 
 its content = 50 cubic feet : required its height. Answ. 1 1 feet 
 
HE AMOUNT O? 1, AT COMPOUND 1NTEKJB.S1 
 
 
 
 4 per cent. 
 
 5 per cent. j 
 
 6 per cent. 
 
 Yrs. 
 
 3 per cent. 
 
 1*040,000 
 
 ro50 ooo 
 
 1*060 000 
 
 1 
 
 1-030,000 
 
 1-OS1/H30 
 
 1-102)500 
 
 X \J\J\J)\J\J\J 
 
 1-123,600 
 
 2 
 
 1*060,900 
 
 1-124,804 
 
 1-157,625 
 
 1-191,016 
 
 
 1-09-2,727 . 
 
 1-169,859 
 
 215,506 
 
 1-262,477 
 
 4 
 
 I*12i),o09 , 
 
 1-216,653 
 
 276,282 
 
 1-338,226 
 
 5 
 
 1-159,274 ( 
 
 11 f\ 1 f\ * Ch u 
 
 1-265,319 
 
 340,096 
 
 1-418,519 
 
 
 ,194,052 1 
 
 1-315,932 
 
 407,100 
 
 1-503,630 
 
 
 1 -229,874 
 
 1-368,569 
 
 477,455 
 
 1-593,848 
 
 
 1-266,770 
 
 1-423,312 
 
 551,328 
 
 1-689,479 
 
 10 
 
 11 
 
 1-343*916 
 
 IO O A 1 O 1 
 
 1-480,244 
 1-539,454 
 
 628,895 
 710,339 
 
 1-790,848 
 1-898,299 
 
 I 
 12 
 13 
 
 384,234 
 1.425,761 
 1.468,534 
 
 1. r i o e n/" 
 
 1-601,032 
 1-665,074 
 1-731,676 
 
 795,856 
 885,6 9 
 979,932 
 
 2-012,196 
 2-132,928 
 2-260,904 
 
 1 e 
 
 512,590 
 
 1-800,944 
 
 2-078,928 
 
 2-396,558 
 
 10 
 
 16 
 
 n 1 
 
 1-604^706 
 
 I./? til O/1 O 
 
 1-872,981 
 1 947,900 
 
 2-182,875 
 2-292,018 
 
 2-540,352 
 2-692,773 
 
 1 
 
 "652,848 
 
 1_r*fAO >f OO 
 
 2-025,817 
 
 2*406,619 
 
 2-854,339 
 
 io 
 19 
 
 O/ 
 
 702,433 
 1-753,506 
 
 Ion/? lit 
 
 2-106,849 
 2-191,123 
 
 2-526,950 
 2-653,298 
 
 3-025,600 
 3-207,135 
 
 P 
 2) 
 22 
 2 
 B 
 2, 
 
 97 
 
 .806,111 
 1-860,295 
 1-916,103 
 V973,587 
 2-032,794 
 '093,778 
 2.- 156,592 
 
 2-278,768 
 2-369,919 
 2464,716 
 2-563,304 
 2-665,836 
 2-772,470 
 2-883,369 
 
 2'785,96 
 2-925,261 
 3-071,52 
 3-225,101 
 3-386,3&; J 
 3-555,673 
 3-733,456 
 
 3-399,564 
 3-603,537 
 3-819,750 
 
 4-048,935 
 4-291,871 
 4-549,383 
 4-822,346 
 
 1 
 
 C _~l,4O.> 
 
 2-998,703 
 
 3-920,129 
 
 5- 11J, 687 
 
 28 
 29 
 30 
 31 
 32 
 S3 
 34 
 35 
 36 
 37 
 38 
 39 
 40 
 41 
 
 2-287,928 
 2-356,566 
 2-427,262 
 2-500,080 
 8-575,083 
 2-652,335 
 2-731,905 
 2-813,862 
 2.898,278 
 2-985,227 
 3-074,783 
 3-167,027 
 3-262,038 
 3-359,899 
 
 O. \ Gf\ f*C\C* 
 
 3-118,651 
 3-243,398 
 3-373,133 
 3-508,059 
 3-648,381 
 3-794,316 
 3-946,089 
 4-103,933 
 4-268,090 
 4-438,813 
 4-616,366 
 4-801,021 
 4-993,061 
 5-192,784 
 
 4-116,136 
 4-321,942 
 4-538,039 
 4-764,941 
 5-003,189 
 5-253,348 
 5-516,015 
 5-791,816 
 6-081,407 
 6-385,477 
 6-704,751 
 7-039,989 
 7-391,988 
 7-761,588 
 
 6-418,388 
 6-743,491 
 6-088,101 
 6-453,386 
 6-840,590 
 7-251,025 
 7-686,087 
 8-147,252 
 8-636,087 
 9-154,252 
 9-703,50? 
 10-285,718 
 10-902,861 
 11-557,0:33 
 
 44 
 
 46 
 
 LQ 
 
 o 4oU,oyo 
 3-564,517 
 S-67 1,452 
 3-781,596 
 3-895,044 
 4-011,895 
 
 5-400,495 
 5-616,515 
 5-841,176 
 6-074,823 
 6*317,816 
 6-570,528 
 
 8-149,667 
 8-557,150 
 8-985,008 
 9-434,258 
 9-905,971 
 10-401,270 
 
 12-250,455 
 12-935,482 
 13-764.611 
 14-590,487 
 15-465,917 
 16-393,872 
 
 r 
 
 49 
 50 
 
 4-256^219 
 4- ,83,906, 
 
 6-833,349 
 7-106,683 
 
 10-921,333 
 11-467.400 
 
 17-377,504 
 18-420,154 
 
TABLE 11. SHOWING THE AMOUNI OF AN ANNUITY OF L 
 
 3 per cent. 4 per cent. } 5 per cent. , 6 per cent 
 
 1-OUO,000 ' 1-000,000 
 
 1-000,000 
 
 1-000,000 
 
 2-030,000 t 2-040,000 
 
 2-050,000 
 
 2-060,000 
 
 3-090,900 
 
 3-121,600 
 
 3-152,500 
 
 3-183,600 
 
 4-183,627 
 
 4-246,464 
 
 4-310,125 
 
 4-374,616 
 
 5-309,135 
 
 5-416,322 
 
 5-525,631 
 
 5-637,092 
 
 6-468,409 
 
 6-632,975 i 6-801,912 
 
 6-975,318 
 
 7-66?,462 
 
 7-898,294 8-142,008 
 
 8-393,837 
 
 8-892,336 
 
 9-214,226 9-549,108 
 
 9-897,467 
 
 10-159,106 
 
 10-582,795 11-026,564 
 
 11-491,315 
 
 11-463,879 
 
 12-006,107 
 
 12-577,892 
 
 13-180,794 
 
 12-807,795 
 
 13-486,351 
 
 14-206,787 
 
 14-971,642 
 
 14-192,029 
 
 15-025,805 
 
 15-917,126 
 
 16-869,941 
 
 15-617,790 
 
 16-626,837 
 
 17-712,982 
 
 18-882,137 
 
 17-086,324 
 
 18-291,911 
 
 19-598,631 
 
 21-015,065 
 
 18-598,913 
 
 20-023,587 
 
 21-578,563 
 
 23-275,969 
 
 20-156,881 
 
 21-824,531 
 
 23-657,491 
 
 25-672,528 
 
 21-761,587 
 
 23-697,512 
 
 25-840,366 
 
 28-212,879 
 
 23-414,435 
 
 25-645,412 
 
 28-132,384 
 
 30-905,652 
 
 25-116,868 
 
 27-67.1,229 
 
 30-539,003 
 
 33-759,991 
 
 26-870,374 
 
 29-778,078 
 
 33-065,954 
 
 36-785,591 
 
 28-676,485 
 
 31-969,201 
 
 85-719,251 
 
 39-992,726 
 
 30-536,780 
 
 34-247,969 
 
 38-505,214 
 
 43-392,290 
 
 32-452,883 
 
 36-617,888 
 
 *1-430,475 
 
 46-995,827 
 
 34-426,470 
 
 39-082,604 
 
 44-501,998 
 
 50-815,577 
 
 36-459,264 
 
 41-645,908 
 
 47-727,098 
 
 54-864,512 
 
 38-553,042 
 
 44-311,744 
 
 51-113,453 
 
 59-156,382 
 
 40-709,633 
 
 47-084,214 
 
 54-669,126 
 
 63-705,765 
 
 42-930,922 
 
 49-967,582 
 
 58-402,582 
 
 68-528,111 
 
 45-218,850 
 
 52-966,286 
 
 62-322,711 
 
 73-639,798 
 
 47-575,415 
 
 56-084,937 
 
 66-438,847 
 
 79-058,186 
 
 50-002,678 
 
 59-328,335 
 
 70-760,789 
 
 84-801,677 
 
 52-502,758 
 
 62-701,468 
 
 75-298,829 
 
 90-889,778 
 
 55-077,841 
 
 66-209,527 
 
 80-063,770 
 
 97-343,164 
 
 57-730,176 69-857,908 
 
 85-066,959 
 
 104-183,754 
 
 60-462.081 
 
 73-652,224 
 
 90-320,307 
 
 111-434,779 
 
 63-2 7 5, 944 
 
 77-598,313 
 
 95-836,322 
 
 119-120,866 
 
 66-1 i 4 322 
 
 81'702,246 
 
 101-628,138 
 
 127-268,118 
 
 69-159,449 
 
 85-970,336 
 
 107-700,545 
 
 135-904,205 
 
 72-234,232 
 
 90-409,149 
 
 114-095,023 
 
 145^058,458 
 
 75-401,259 
 
 95-025,515 
 
 120-799,774 
 
 154-761,965 
 
 78-663,297 
 
 99-826,536 
 
 127-839,762 
 
 165-047,683 
 
 82-023,196 
 
 104-819,597 
 
 135-231,751 
 
 175-950,544 
 
 85-483,892 
 
 110-012,381 
 
 1 42-993,338 
 
 187-507,577 
 
 89-048,409 
 
 115-412,876 
 
 151- 143,005 
 
 199-758,031 
 
 92-719,861 
 
 121-029,392 
 
 159-700,155 i 8 1 2-743,5; 3 
 
 96-501,4-57 126-870,567 
 
 168-685,163 j 226-508,124. 
 
 100-396,500 132-945,390 
 
 1^-119,421 HI -098,61 2 
 
 104-408,395 139-263,206 
 
 188-025,392 I 866-564,528 
 
 108-540,647 145-833,734 
 
 198-426,662 / 72-958,400 
 
 112-796,867 152*667,083 
 
 209-347,935 tfSO-335,904 
 
TABLE III. SHOWING THE PRESENT V.ii.trE OF AN ANNUITY OF 1. 
 
 ifrs. 3 per cent. 
 
 4 per cent. 3 per cent < 6 per cent. 
 
 1 
 
 970874 
 
 961538 
 
 952381 
 
 943396 
 
 2 
 
 1-913470 
 
 1-886194 
 
 1-859410 
 
 1-833392 
 
 3 
 
 2-828612 
 
 2-775190 
 
 2-723248 
 
 2-673011 
 
 4 
 
 3-717099 
 
 3-629994 
 
 3-545950 
 
 3-465105 
 
 5 
 
 4-579708 
 
 4-451821 
 
 4-329476 
 
 4-212363 
 
 6 
 
 5-417192 
 
 5-242136 
 
 5-075691 
 
 i-9 17324 
 
 7 
 
 6-230284 
 
 6-002054 
 
 5-786372 
 
 5-582381 
 
 8 
 
 7-019693 
 
 6-732744 
 
 6-46321 1 
 
 6-209793 
 
 9 
 
 7-786110 
 
 7-435331 
 
 7-107820 
 
 6-801691 
 
 10 
 
 8-530204 
 
 8-110895 
 
 7-721733 
 
 7-360086 
 
 11 
 
 9-252625 
 
 8-760576 
 
 8-306412 
 
 7-886874 
 
 1-2 
 
 9-954005 
 
 9-385073 
 
 8-863249 
 
 8-383843 
 
 13 
 
 0-634956 
 
 9-985647 
 
 9-393570 
 
 8-852682 
 
 14 
 
 11-296074 
 
 10-563122 
 
 9-898638 
 
 9-294983 
 
 35 
 
 11-937936 
 
 11-118487 
 
 10-379655 
 
 9-712248 
 
 16 
 
 12-561103 
 
 11-652395 
 
 10-837767 
 
 10-105894 
 
 17 
 
 13-166119 
 
 12-165668 
 
 11-274064 
 
 10-477258 
 
 18 
 
 13-753514 
 
 12-659396 
 
 11-689585 
 
 10 827602 
 
 19 
 
 14-323800 
 
 13-133938 
 
 12-085319 
 
 11-158115 
 
 20 
 
 14-877476 
 
 13-590325 
 
 12-462208 
 
 1 1-469920 
 
 21 
 
 15-415025 
 
 14-029159 
 
 12-821150 
 
 11-764075 
 
 22 
 
 15-936918 
 
 14-451114 
 
 13-163000 
 
 12-041580 
 
 23 
 
 16-443610 
 
 14-856840 
 
 13-488571 
 
 12-303377 
 
 24, 
 
 16-935544 
 
 15-246961 
 
 13-798639 
 
 12-550356 
 
 25 
 
 17-413150 
 
 15-622078 
 
 14-093942 
 
 12-783355 
 
 26 
 
 17-876845 
 
 15-982767 
 
 14-375183 
 
 13-003165 
 
 27 
 
 18-327034 
 
 16-329584 
 
 14-643031 
 
 13-210533 
 
 28 
 
 18-764111 
 
 16-663061 
 
 14-898125 
 
 13-406163 
 
 29 
 
 19-188457 
 
 16-983712 
 
 15-141071 
 
 13-590720 
 
 30 
 
 19-600444 
 
 17-292031 
 
 15-372448 
 
 13-764830 
 
 31 
 
 20-000431 
 
 17-588491 
 
 1 5-592807 
 
 13-929085 
 
 32 
 
 20-388768 
 
 17-873540 
 
 15-802673 
 
 14-084042 
 
 33 
 
 20-765794 
 
 18-147643 
 
 16-002546 
 
 14-230228 
 
 34- 
 
 21-131839 
 
 18-411195 
 
 16-192901 
 
 14-368140 
 
 35 21-487222 
 
 16-664610 
 
 16-374191 
 
 14-498245 
 
 30 21-832254- 
 
 J 8-908279 
 
 16-546848 
 
 14-620986 
 
 37 
 
 22-167237 
 
 19-142576 
 
 16-711284 
 
 14-736779 
 
 38 
 
 22-492463 
 
 19-367861 
 
 16-867889 
 
 14-846018 
 
 39 | 22-808217 \ 19-584482 
 
 17-017037 
 
 14-949074 
 
 40 23-114774 19-79277 J 
 
 17-159083 
 
 15-046296 
 
 41 23-412402 19-993049 
 
 17-294365 
 
 15-138015 
 
 42 i 23-701361 20-185624 
 
 1 7-423205 
 
 15-224542 
 
 43 I 23-981904 20-370792 
 
 17-545909 
 
 15-306172 
 
 44 24-254276 i 20-548838 
 
 17-662770 
 
 15-383181 
 
 45 24-518715 20-720036 
 
 17-774067 
 
 15-455831 
 
 46 , 24-775452 20-884650 
 
 17-880064 
 
 15-524369 
 
 47 25*024711 21-042933 
 
 17-981013 
 
 15-589027 
 
 48 25-266710 21-195128 
 
 \ 18-077155 
 
 15-650025 
 
 25-501660 : 21-341469 18-168719 
 
 15-707571 
 
 50 | 25-7297 '67 21-482 IP? ' 18-255923 
 
 15-761859 
 
TABLE IV. SHOW ING THE VALUE OF AN ANNUITY ON A SINGLE LIFE. 
 
 Ages. 
 
 4 per cent. 
 
 5 per cent. 
 
 6 per cent 
 
 Ages. 4 per cent 
 
 5 per cent. 6 per cent 
 
 1 
 
 13-4(55 
 
 11-503 
 
 10-107 
 
 49 
 
 11-475 
 
 10-443 
 
 9-563 
 
 2 
 
 15-633 
 
 13-420 
 
 11-724 
 
 50 
 
 11-264 
 
 10-269 
 
 9-417 
 
 3 
 
 16-462 
 
 14-135 
 
 12-348 
 
 51 
 
 11-057 
 
 10-097 
 
 9-273 
 
 4 
 
 17*010 
 
 14-613 
 
 12-769 
 
 52 
 
 10-849 
 
 9-925 
 
 9-129 
 
 5 
 
 17-248 
 
 14-827 
 
 12-962 
 
 53 
 
 10-637 
 
 9-748 
 
 S-980 
 
 6 
 
 17-482 
 
 15-041 
 
 13-156 
 
 54- 
 
 J 0*421 
 
 9-567 
 
 8-827 
 
 7 
 
 17-611 
 
 15-166 
 
 J 3-275 
 
 55 
 
 10-201 
 
 9-382 
 
 8-670 
 
 8 
 
 17-662 
 
 15-226 
 
 13-337 
 
 56 
 
 9-977 
 
 9-193 
 
 8-509 
 
 9 
 
 17-625 
 
 15-210 
 
 13-335 
 
 57 
 
 9-749 
 
 8-999 
 
 8-343 
 
 10 
 
 17-523 
 
 15-139 
 
 13-285 
 
 58 
 
 9-516 
 
 8-801 
 
 8-173 
 
 11 
 
 17-393 
 
 15-043 
 
 13-212 
 
 59 
 
 9-280 
 
 8-599 
 
 7*999 
 
 12 
 
 17-251 
 
 14-937 
 
 13-130 
 
 60 
 
 9-039 
 
 8-392 
 
 7-820 
 
 13 
 
 17-103 
 
 14-826 
 
 13-044 
 
 61 
 
 8-795 
 
 8-181 
 
 7-637 
 
 14 
 
 1G-950 
 
 14-710 
 
 12-953 
 
 62 
 
 8-547 
 
 7-966 
 
 7-449 
 
 15 
 
 16-791 
 
 14-588 
 
 12-857 
 
 63 
 
 8-291 
 
 7-742 
 
 T-263 
 
 16 
 
 16-625 
 
 14-460 
 
 12-755 
 
 64 
 
 8-030 
 
 7-514 
 
 7-052 
 
 17 
 
 16-462 
 
 14-334 
 
 12-655 
 
 65 
 
 7-761 
 
 7-27G 
 
 6-841 
 
 18 
 
 16-309 
 
 14-217 
 
 12-562 
 
 66 
 
 7-488 
 
 7-034 
 
 6-625 
 
 19 
 
 16-167 
 
 14-108 
 
 12-477 
 
 67 
 
 7-211 
 
 6-787 
 
 6-405 
 
 20 
 
 16-033 
 
 14-007 
 
 12-398 
 
 68 
 
 6-930 
 
 6-536 
 
 6-179 
 
 21 
 
 15-912 
 
 13-917 
 
 12-329 
 
 69 
 
 6-647 
 
 6-281 
 
 5-949 
 
 22 
 
 15-797 
 
 13-833 
 
 12-265 
 
 70 
 
 6-3G1 
 
 6-023 
 
 5-716 
 
 23 
 
 15-680 
 
 13-746 
 
 1 2-200 
 
 71 
 
 6-075 
 
 5-764 
 
 5-479 
 
 24 
 
 15-560 
 
 13-658 
 
 12-132 
 
 72 
 
 5-790 
 
 5-504 
 
 5-241 
 
 25 
 
 15-438 
 
 13-567 
 
 12-063 
 
 73 
 
 5-507 
 
 5-245 
 
 5-004 
 
 26 
 
 15-312 
 
 13-473 
 
 11-992 
 
 74 
 
 5-230 
 
 4-990 
 
 4-769 
 
 27 
 
 15-184 
 
 13-377 
 
 11-917 
 
 75 
 
 4-962 
 
 4-744 
 
 4-542 
 
 28 
 
 15-053 
 
 13-278 
 
 11-841 
 
 76 
 
 4-710 
 
 4-511 
 
 4-328 
 
 29 
 
 14-918 
 
 13-177 
 
 11-763 
 
 77 
 
 4-457 
 
 4-277 
 
 4-109 
 
 30 
 
 14-781 
 
 13-072 
 
 1 1-682 
 
 78 
 
 4-197 
 
 4-035 
 
 3-SS4 
 
 31 
 
 14-639 
 
 12-965 
 
 11-598 
 
 79 
 
 3-921 
 
 3-776 
 
 3-641 
 
 32 
 
 14-495 
 
 12-854 
 
 11-512 
 
 80 
 
 3-643 
 
 3-515 
 
 3-394 
 
 33 
 
 14-347 
 
 12-740 
 
 1 1-423 
 
 81 
 
 3-377 
 
 3-263 
 
 3-156 
 
 34 
 
 14-195 
 
 12-623 
 
 11-331 
 
 82 
 
 3-122 
 
 3-020 
 
 2-926 
 
 35 
 
 14-039 
 
 12-502 
 
 11-236 
 
 83 
 
 2-887 
 
 2-797 
 
 2-713 
 
 36 
 
 13-880 
 
 12-377 
 
 11-137 
 
 84s 
 
 2-708 
 
 2-627 
 
 2-551 
 
 37 
 
 13-716 
 
 12-249 
 
 11-035 
 
 85 
 
 2-543 
 
 2-471 
 
 2-402 
 
 38 
 
 13-548 
 
 12-116 
 
 10-929 
 
 86 
 
 2-393 
 
 2-328 
 
 2-266 
 
 39 
 
 13-375 
 
 11-979 
 
 10-819 
 
 87 
 
 2-251 
 
 2-193 
 
 2-138 
 
 40 
 
 13-197 
 
 11-837 
 
 10-705 
 
 88 
 
 2-131 
 
 2-080 
 
 2-031 
 
 41 
 
 13-018 
 
 11-695 
 
 10-589 
 
 89 
 
 1-967 
 
 1-924 
 
 1-882 
 
 42 
 
 12-838 
 
 11-551 
 
 10-473 
 
 90 
 
 1-758 
 
 1-723 
 
 1-689 
 
 43 
 
 12-657 
 
 1 1-407 
 
 10-356 
 
 91 
 
 1-474 
 
 1-447 
 
 1-422 
 
 44 
 
 12-472 
 
 1 1-258 
 
 10-235 
 
 92 
 
 1-171 
 
 1-153 
 
 1-136 
 
 45 
 
 12-283 
 
 11-105 
 
 10-110 
 
 93 
 
 0-827 
 
 0-816 
 
 0-806 
 
 46 
 
 12-089 
 
 10-947 
 
 9-980 
 
 94 
 
 0-530 
 
 0-524 
 
 0-518 
 
 47 
 
 11-890 
 
 10-784 
 
 9-846 
 
 95 
 
 0-240 
 
 0-238 
 
 0-236 
 
 48 
 
 11-685 1 
 
 10-616 
 
 9-707 
 
 96 
 
 o-ooo 
 
 o-ooo 
 
 o-ooo 
 
TABLE V. SHOWING THE PRESENT VALUE OF AN ANNUITY OF 1 
 ON THE JOINT CONTINUANCE OF TWO LIVES. 
 
 Ages. 4 per en 
 
 5 per cnt. 6 per en 
 
 Ages. 
 
 4 per cnt 
 
 5 per cnt. 
 
 G per cnt. 
 
 
 5 
 
 13-591 
 
 11-984 10-691 
 
 30 
 
 11-313 
 
 10-255 
 
 9-260 
 
 
 10 
 
 13-9J3 
 
 12-31. 
 
 i 11-010 
 
 
 35 
 
 1O948 
 
 9-954 
 
 9'112 
 
 
 15 
 
 13479 
 
 11954 
 
 1 10-716 
 
 
 40 
 
 10-490 
 
 9-576 
 
 8-7L5 
 
 
 20 
 
 12993 
 
 11-561 1O391 
 
 
 45 
 
 9-959 
 
 9-135 
 
 8-424 
 
 
 25 
 
 12 633 
 
 il-S!81 10-170 
 
 
 50 
 
 9-321 
 
 8-596 
 
 7-966 
 
 
 30 
 
 12 220 
 
 10-95S 
 
 9-913 
 
 3C 
 
 55 
 
 8-619 
 
 7-999 
 
 7453 
 
 
 35 
 
 11732 
 
 10-572 
 
 9-602 
 
 
 60 
 
 7-802 
 
 7-292 
 
 6 &5T 
 
 5 
 
 40 
 
 11-1.50 
 
 10-102 j 9219 
 
 
 65 
 
 6-844 
 
 6-447 
 
 6-069 
 
 
 45 
 
 10.500 
 
 9-571 i 8778 
 
 
 70 
 
 5-729 
 
 5442 
 
 5-180 
 
 
 50 
 
 9-742 
 
 8-941 j 8-248 
 
 
 75 
 
 4-5.37 
 
 4-365 
 
 4- 188 
 
 
 55 
 
 8-931 
 
 8-256 7-665 
 
 
 80 
 
 3-406 
 
 3-290 
 
 3-J81 
 
 
 60 
 
 8011 
 
 7-466 6-982 
 
 _ 
 
 
 
 
 
 
 65 
 
 6-963 
 
 6-546 6-171 
 
 
 35 
 
 10-612 
 
 9(580 
 
 8'b83 
 
 
 70 
 
 5-768 
 
 5-472 5-209 
 
 
 40 
 
 10-196 
 
 9-331 
 
 "8-589 
 
 
 75 
 
 4' 57 
 
 4-362 ! 4-181 
 
 
 45 
 
 9-706 
 
 8-921 
 
 8-242 
 
 
 
 ^ 
 
 T - 
 
 
 50 
 
 9-110 
 
 8415 
 
 7-809 
 
 * m 
 
 10 
 
 14-277 
 
 12-665 11-345 
 
 35 
 
 55 
 
 8-448 
 
 7-849 
 
 7-3.2 
 
 
 15 
 
 13-841 
 
 12-302 11-048 
 
 
 60 
 
 7-669 
 
 7-174 
 
 6-732 
 
 
 20 
 
 13-355 
 
 11*906 10-719 
 
 
 65 
 
 6-747 
 
 6-360 
 
 6-010 
 
 
 25 
 
 12-998 
 
 11-627 10-497 
 
 
 70 
 
 5663 
 
 5'382 
 
 5-125 
 
 
 30 
 
 12-586 
 
 11-304 10-239 
 
 
 75 
 
 4-5 16 
 
 4.327 
 
 4 -152 
 
 
 36 
 
 12-098 
 
 10-916 9925 
 
 
 80 
 
 3-383 
 
 3 1268 
 
 3360 
 
 
 40 
 
 11-513 
 
 10-442 9-537 
 
 
 
 
 . 
 
 _____ 
 
 
 10 
 
 45 
 
 10-851 
 
 9-900 9088 
 
 
 40 
 
 y-820 
 
 9016 
 
 isw 
 
 
 50 
 
 10-085 
 
 9-260 ; 8-548 
 
 
 45 
 
 9-381 
 
 8643 
 
 8-003 
 
 
 55 
 
 9-256 
 
 8-560 7-951 
 
 
 50 
 
 8'834 
 
 8177 
 
 7-607 
 
 
 60 
 
 8-314 
 
 7-750 7-250 
 
 40 
 
 55 
 
 8221 
 
 7651 
 
 7-146 
 
 
 65 
 
 7-236 
 
 6-803 6-414 
 
 
 60 
 
 7490 
 
 7-015 
 
 6 ;>!0 
 
 
 - 70 
 
 6-008 
 
 5-700 5418 
 
 
 65 
 
 6614 
 
 6240 
 
 5iM 
 
 
 75 
 
 4-725 
 
 4 522 4-350 
 
 
 70 
 
 5-571 
 
 5-i'98 
 
 5047 
 
 
 80 
 
 3-517 
 
 3-395 ; 3-281 
 
 
 75 
 
 4-457 
 
 4272 
 
 4101 
 
 m ^ 
 
 
 i 
 
 
 80 
 
 3-349 
 
 3236 
 
 3 1^0 
 
 < 15 
 
 13-411 
 
 11-960 10-767 
 
 
 
 
 
 
 
 
 20 12-961 
 
 11-585 f 10-453 
 
 
 45 
 
 8-990 
 
 8-312 
 
 771h 
 
 25 
 
 12-630 
 
 11-324 10244 
 
 
 50 
 
 8-503 
 
 7-891 
 
 7-353 
 
 SO 
 
 12-246 
 
 11-021 10-001 
 
 45 
 
 55 
 
 7-948 
 
 7-411 
 
 6-935 
 
 
 35 
 
 11-787 
 
 10-655 i 9-703 
 
 
 60 
 
 7-274 
 
 6-822 
 
 6418 
 
 
 40 
 
 11-234 
 
 10-205 9-333 
 
 
 65 
 
 6-453 
 
 6-094 
 
 5-769 
 
 15 45 
 
 10-607 
 
 9-690 : 8-905 
 
 
 70 
 
 5-460 
 
 5195 
 
 4-953 
 
 
 50 
 
 9-872 
 
 9-076 i 8-386 
 
 
 75 
 
 4-386 
 
 4206 
 
 4'( '-! 
 
 
 55 
 
 9-077 
 
 8-403 J 7-812 
 
 
 SO 
 
 3-308 
 
 3-191 
 
 3 (&,-> 
 
 
 60 
 
 8-170 
 
 7-622 I 7-135 
 
 
 _ 
 
 
 
 
 ____ 
 
 
 65 
 
 7-127 
 
 6-705 6-325 
 
 
 50 
 
 8-081 
 
 7522 
 
 7050 
 
 
 8 
 
 5933 
 4-695 
 
 5-631 
 4-495 
 
 5-355 
 4-310 
 
 
 55 
 
 60 
 
 7-593 
 6989 
 
 7098 
 6568 
 
 6658 
 
 6- ,89 
 
 
 80 
 
 3-492 
 
 3-372 
 
 3-259 
 
 50 
 
 65 
 
 6--V36 
 
 5897 
 
 5*590 
 
 _ ^ 
 
 __, 
 
 n -_ | T . . 
 
 
 
 _ 
 
 
 70 
 
 5-306 
 
 5054 
 
 4-82. 
 
 
 20 
 
 12-535 
 
 11-232 
 
 10-156 
 
 
 75 
 
 VS35 
 
 4-112 
 
 3-951 
 
 
 25 
 
 12229 
 
 10-989 
 
 9-960 
 
 ___ 
 
 80 
 
 3-247 
 
 3140 
 
 3-039 
 
 
 30 
 
 11-873 
 
 10-707 
 
 9-732 
 
 
 _ 
 
 
 
 . 
 
 
 35 
 
 11-445 
 
 10368 
 
 9-451 
 
 
 55 
 
 ?179 
 
 6735 
 
 fi-33-5 
 
 
 40 
 
 10-924 
 
 9-937 
 
 9-100 
 
 
 60 
 
 6-659 
 
 6272 
 
 5 "924 
 
 
 45 
 
 10-330 
 
 9-448 
 
 8-692 
 
 55 
 
 05 
 
 5-986 
 
 5671 
 
 5 384 
 
 20 
 
 50 
 
 9630 
 
 8-861 
 
 8-195 
 
 
 70 
 
 5-132 
 
 4893 
 
 4-674 
 
 
 55 
 
 8-869 
 
 8216 
 
 7-643 
 
 
 75 
 
 4-171 
 
 4-006 
 
 3-852 
 
 
 60 
 
 7-995 
 
 7-463 
 
 6-990 
 
 
 
 80 
 
 3-180 
 
 8-076 
 
 2-978 
 
 
 65 
 
 6-986 
 
 6-576 6-205 
 
 
 60 
 
 
 5.000 
 
 5"-579 
 
 1 
 i 
 
 70 
 75 
 80 
 
 5-826 
 4-619 
 3-443 
 
 5532 
 44C4 
 3-325 
 
 5-262 
 4-242 
 3-214 
 
 60 
 
 65 
 
 70 
 75 
 
 5-658 
 4-900 
 
 O OOo 
 
 5-372 
 
 *'680 
 3866 
 
 5112 
 
 4478 
 3-721 
 
 _ i 
 
 25 
 30 
 
 11-944 
 11-618 
 
 10-764 
 10-499 
 
 9-776 
 9-561 
 
 - 
 
 80 
 65 
 
 3-092 
 201 
 
 T9GO 
 
 2-699 
 
 4-7,'JH 
 
 
 35 
 
 
 10-175 
 
 9*295 
 
 M 
 
 70 
 
 4-573 
 
 4-378 
 
 4199 
 
 
 40 
 
 10-725 
 
 9*771 
 
 8960 
 
 
 75 
 
 3-806 
 
 S-66.-3 
 
 3*5'>3 
 
 
 45 
 
 10' 160 
 
 9-304 
 
 
 
 80 
 
 2-965 
 
 2-873 
 
 2-786 
 
 
 50 
 
 9'48S 
 
 8-739 
 
 8-089 
 
 
 70 
 
 4-087 
 
 3-930 
 
 S-7'81 
 
 25 
 
 55 
 
 8-754 
 
 8-116 
 
 7-555 
 
 
 75 
 
 3471 
 
 3347 
 
 S'-twfi 
 
 
 GO 
 
 7-9C>6 
 
 7-383 
 
 6-919 
 
 70 
 
 SO 
 
 2-757 
 
 2-675 
 
 2-598 
 
 
 65 
 
 70 
 "5 
 
 6920 
 5-780 
 
 6-515 
 
 5-489 
 4-396 
 
 5-223 
 4-216 
 
 """ 
 
 75 
 
 80 
 
 3-015 
 2-4-IS 
 
 "917 
 2 381 
 
 2827 
 2-323 
 
 ' 80 3-4i'h ' 3-S<'8 ' S-WS 
 
 80 
 
 80 2-068 *01 ^S9 
 
ANSWERS TO EXERCISES. 
 NUMERATION. 
 
 Ex. 1. TWENTY FOUR. 2. One hundred and forty four. 3. Three 
 hundred and sixty five. 4. One- thousand. 5. One thousand, 
 seven hundred, and twenty eight. 6. Two thousand, two hundred, 
 and forty. ?. Nine thousand, seven hundred, and ninety, 8. 
 Thirty seven thousand, and forty eight. 9. Thirty thousand and 
 nine. 10. Four millions, fifty five thousand, and seventy, li. 
 Three hundred thousand, four hundred and five. 12. Seventy 
 nine millions, five hundred and three thousand, and forty six. 13.* 
 Eight billions, five millions, six hundred thousand, four hundred, 
 and eighty. 14. Five hundred arid fifty seven millions, two hun- 
 dred and ninety thousand. 15. Six hundred and eighty millions, 
 and forty two. 16. Ninety three millions, ninety thousand, and ninety 
 three. 17. One hundred and thirteen thousand, three hundred, 
 and fifty five. 18. Seven hundred and eighty five thousand, three 
 hundred, and ninety eight. 19. Seven millions, thirty thousand, 
 four hundred, and sixty two. 20. Twenty four millions, nine 
 hundred and two thousand, four hundred, and ninety. 21.* Nine 
 billions, three millions, eight thousand, and five. 22. Forty mil- 
 lions, six hundred and fifty seven thousand, two hundred. 23.* 
 One hundred billions, and one thousand. 24.* Sixty trillions, 
 six hundred and six billions, sixty millions, seven hundred 
 thousand, seven hundred, and seven. 25.* One hundred and 
 two trillions, thirty billions, four hundred and five millions, sixty 
 thousand, seven hundred and eight. 26*. Nine hundred and one 
 trillions, one billion, one hundred and one millions, two hundred 
 md one thousand, three hundred and one. 27.* Two hundred 
 billions, thirty million.-:, forty thousand, five hundred and t& rt y 
 eight. 28.* Seventy three quadrillions, eight hundred and t\venty 
 trillions, seven hundred and sixty billions, five millions, one hun- 
 dred and ninety two thousand, six hundred and forty five. 29.* 
 
 According to theccnnmon Notation : 
 
 Ex. 13. Eight thousand and five millions, six hundred thousand, four hundred and 
 eighty. 21. Nine thousand and three millions, eight thousand and five. 5. One him- 
 dred thousand millions, one thousand. 24. Sixty billion?, six hundred and six thousand, 
 and sixty millions, seven hundred thousand, seven bundled and seven. 25. One hun- 
 dred and two bi II ions, thirty thousand, four hundred an<i five millions, sixty thousand 
 eeven hundred and eight. 2$. Nine hundred and one billions, one thousand one hun- 
 dred and one millions, two hundred and one thousand, three hundred and one. 27. Two 
 hundred thousand and thirty millions, forty thousand, five hundred and thirty eight. 
 8. Seventy three thousand eight hundred and twenty billions, seven hundred and sixty 
 Chouand and five millions, one hundred and ninety two thousand six hundred and forty 
 five. 29. Ten thousand and one billions, ten millions, one hundred. SO. Forty thousand 
 and fifty billions, sixty thousand seven hundred millions, eight hundred and nine thou- 
 sand. 31 Seven quadrillions, nine hundred and forty six thousand two hundred and 
 eighty nine trillions, six thousand four hundred billions, three hundred thousand and 
 forty seven millions, two hundred and sixty four thousand, seven hundred and ninety one. 
 
282 ANSWERS TO EXERCISES. 
 
 Ten quadrillions, one trillion, ten millions, one hundred. 30.* 
 Forty quadrillions, fifty trillions, sixty billions, seven hundred rail- 
 lions, eight hundred and nine thousand. 31.* Seven septillions, 
 nine hundred and forty six sextillions, two hundred and eighty 
 nine quintillions, six quadrillions, four hundred trillions, three 
 hundred billions, forty seven millions, two hundred and sixty foui 
 thousand, seven hundred and ninety one. 
 
 NOTATION. 
 
 Ex.f. 52 Ex.9. 5070 Ex.17. 1010200000* 
 
 2. 313 10. 4504 18. 1100200000000* 
 
 3. 504 11. 20084 19. 1050000 
 
 4. 1024 12. 650090 20. 1000200000* 
 
 5. 2048 13. 7007010 21. 1000000700000* 
 
 6. 1815 , 14. 64000300 22. 1000000019000000* 
 . 7. 7854 15. 11002000 23. 70000010088* 
 
 8. 3008 16. 110020000 24. 900068000020* 
 
 25. 37000000, 69000000, 95000000, 145000000 
 <gtf 94000000, 907000000, 1810000000. 
 
 \ 
 ^ SIMPLE ADDITION. 
 
 Ex.I. 21555 Ex.13. 1087189482.: 
 
 2. 206343 14. 145322086 
 
 3. 1508939 15. 207305 
 
 4. 1383458 16. In 1727. 
 .21225092 17. In 1399. 
 
 ^388 18. 1527560, 33228 1 
 121888988 352853, 937000 
 
 27457989 19. 136,570,103 
 
 86171735 20. 3371 
 
 10.20967445 21. 36633 miles. 
 
 11. 463140294 22. 1,451,678 
 
 12. 445684135 23. 11,261,437 
 
 Acres Population 
 
 Ulster ............ 3,201,200 ................... 2,001,966 
 
 Leinster ......... 2,792,550 .................... 1,785,702 
 
 Munster ... _____ 3,377,150 .................... 2,005,363 
 
 Oonnaught... 2,630,300 .................... 1,053,918 
 
 Total ............. 12,001,200 .................. 6,846,949 
 
 Ex.25. 2,093,456 
 
 According to the common Notation ; 
 
 * Ex. 17. 1000010200000 
 
 18. lOOOlOOOOUSOOOOOOOO 
 
 20. 1000000200000 
 
 21. 1000000000000700000 
 
 22. ICXJOOOOOOOOCXXXJUOISOOOOOO 
 
 23. 70000000010088 
 
 24. 900000068000020 
 
 ^ 
 
 Ex. 
 
ANSWERS TO EXERCISES. 
 SIMPLE SUBTRACTION. 
 
 E>.1. 
 
 13031 
 
 Ex.16. 
 
 2997000 
 
 14816 29 
 
 . 1202 
 
 281 
 
 2. 
 
 15708 
 
 17. 
 
 991 
 
 12250 
 
 833 
 
 236 
 
 3. 
 
 17368 
 
 18. 
 
 299997 
 
 22926 
 
 728 
 
 234 
 
 4. 
 
 32131 
 
 19. 
 
 1011924 Ex.26. 
 
 200060 
 
 609 
 
 233 
 
 5. 
 
 590731 
 
 20. 
 
 861928 
 
 1210235 
 
 571 
 
 224 
 
 6. 
 
 8285307 
 
 21. 
 
 75 27. 
 
 308755 
 
 533 
 
 222 
 
 7. 
 
 3209877 
 
 22. 
 
 5747 
 
 1092766 
 
 525 
 
 214 
 
 8. 
 
 763544529 
 
 23. 
 
 11895 28. 
 
 2282155tbs.- 
 
 494 
 
 210 
 
 9. 
 
 10101001 
 
 24. 
 
 39958 
 
 
 413 
 
 205 
 
 10. 
 
 333232333 
 
 
 61848 
 
 
 412 
 
 1S1 
 
 11. 
 
 7468053687 
 
 
 208926 
 
 
 384 
 
 170 
 
 12. 
 
 327504427 
 
 
 38651*4 
 
 
 371 
 
 138 
 
 13. 
 
 99579930 
 
 85. 
 
 1127757 
 
 
 332 
 
 43 
 
 14. 
 
 89998999 
 
 
 8808 <^. 
 
 *^*\ 
 
 327 
 
 23 
 
 15, 
 
 3995996 
 
 
 4447 
 
 
 307 
 
 19 
 
 
 
 
 
 
 294 
 
 9 
 
 SIMPLE MULTIPLICATION. 
 
 Ex.1. 432 Ex.9. 2074$ V 
 
 2. 3180 10. 18160 / 
 
 3. 3645 11. 1615040 
 
 4. 5616 12. 161504000 
 
 5. 11193 13. 573440000 
 
 6. 1000 14. 7068600 
 
 7. ;7414 15. 49152000 
 
 8. 80388 
 
 SIMPLE DIVISION. 
 
 Ex. I. 156950 Ex.6. 3432416* 
 
 2. 45744 1 i 7. 8288202$ 
 
 3. 853196| 8. 4840788^ 
 
 4. 6303411 9. 4106265ft 
 
 5. 1023648} , , 
 
 COMPOUND ADDITION. 
 
 
 Ex.1. 3852 
 
 2. 36 
 
 3. 21849 
 
 48 
 2431 
 3765 
 1861 
 19011 
 1503 
 10 
 
 2612 
 331 
 515 
 52J 1 
 3169 
 
 4. 
 
 5. 
 
 6. 
 
 7. 
 
 8. 
 
 9. 
 10. 
 11. 
 12. 
 13. 
 14. 
 15. 
 
 s. d. 
 
 15 10 
 
 12 3 : 
 
 18 2- 
 
 6 
 
 9 2; 
 
 15 1,. 
 
 19 W, 
 
 17 ?;. 
 
 4 6; 
 
 4 6 
 
 8 2$ 
 
 10 10J 
 
 1 2 
 
 11 9 
 8 6 
 
 s. d. 
 
 Ex.16. 4281 9f 
 
 17. 6219 7 l| 
 
 18. 20351 16 &l 
 
 19. 224c. Iq. lltbs. 
 
 20. 97tbs. loz. loclrs. 
 
 21. 383 tons. 8 cwt, 
 
 22. 362lbs. 
 
 23. 693 cwt. Oq. 23Ibs. 
 
 24. 616 cwt. 2q. 91bs.. 
 
 25. 65lbs. 
 
 26. 429t. 17c. 2q. 20lb s . 
 
 27. 468c. 3q. lllbs. 
 
 28. 517a, Or. 9p. 
 
 29. 127 la. 3r. 4p. 
 
284 ANSWERS TO EXERCISES. 
 
 COMPOUND SUBTRACTION. 
 . d. d. h. min. sec. 
 
 Ex.1. 10 8 G Ex.14. 24 9 50 -30 
 
 2. 451 1 9 15> 6 o 57> 12 * 
 
 3. 103 14 7 . 2 39 43 
 
 4. 263 12 7| 1 50 11 
 
 5. 7 2 *i g 36 57 
 
 6. 98 17 3 3 58 2 6 
 
 7. 324 8 3 16. 35 3 30 
 
 8. 669 6 51 17< 22 3 15 
 
 18. 1 15 32 
 N c. q. Ibs. 
 
 9. 418 Days h. m. 
 
 10. 5 1 11 19. 136 17 33 
 
 11. 25 2 15 140 13 20 
 
 12. 9t.l4 3 14 321 17 '22 
 
 3645 14 56 
 d. h. min. sc. .6426 1 1 24 
 
 13. 3 6 13 46 19978 16 9 
 
 COMPOUND MULTIPLICATION. 
 
 s. d. s. d. 
 
 Ex.1. 797 
 2. 5 13 21 
 3. 3 15 9 
 4. 7 11 8 
 5. 22 15 1| 
 6 7 19 3 
 
 Ex.7. 23 10 
 8. 60 
 
 9. 7 18 6; 
 10. 103 4 4 
 11. 47 15 6 
 12. 21 19 
 
 COMPOUND DIVISION. 
 
 s. d. Rem. s. d. Ren, 
 
 Ex.1. 5 5 101 Ex.9. 15 4 ...2d. 
 
 2. 18 10i...fd. 10. 10 3 
 
 3. 1 1 5i...fd. 11. 1 13 51 
 
 4. 1 17 lo|...|d. 12. 034 
 
 5. 1 8 10... Id. 13. 2 10$...$d 
 
 6. 5 7$...ld. 14. 3 9 
 
 7. 9 5j...ld. 15. 033 
 
 8. 13 H...-W. 
 
VA 02437 
 
 -. 
 
 1; 
 
 M306030 
 
 34 
 
 THE UNIVERSITY OF CALIFORNIA LIBRARY