MANUAL 
 
 OF 
 
 SHORT METHODS IN ARITHMETIC 
 
 HUNTER
 
 THE LIBRARY 
 
 OF 
 
 THE UNIVERSITY 
 OF CALIFORNIA 
 
 LOS ANGELES
 
 This book is DUE on the last date stamped below 
 
 Si AIHG?M P .1AL SCHOOL, 
 
 liOS AflGEIiEIS, CPJU.
 
 SHOKT METHODS IN AEITHMETIC
 
 LON'DOS : PIUXTED BY 
 
 POTTISWOOnF A1TD CO., NKW-STREET SQUARE 
 ASD PARHAMKXT STBEKT
 
 V 
 
 A MANUAL 
 
 OF 
 
 SHORT METHODS IN ARITHMETIC 
 
 DESIGNED FOR THE USE OF SCHOOLS 
 
 AND TO FACILITATE THE ARITHMETICAL CALCULATIONS 
 OF BUSINESS AND SCIENCE 
 
 EEV. JOHN HUNTEE, M.A. 
 
 I Z4-0 
 
 LONDON 
 LONGMANS, GREEN, AND CO. 
 
 1884 
 
 OCT 1? 1900
 
 QA 
 
 PBEFACE. 
 
 A VERY great amount of useful progress has been made, 
 during the last thirty years, in the development of the 
 power of common arithmetic to deal with many problems 
 for which Algebra was wont, less intellectually, to be 
 employed. Algebra has a boundless field for reasoning 
 and operation, in which Arithmetic is only a very humble 
 though a requisite auxiliary; but the employment of the 
 more elementary processes of Algebra to solve arith- 
 metical questions used to be too often resorted to, as a 
 mechanical relief from the more useful work of purely 
 arithmetical investigation. 
 
 Arithmetic as now taught in the best Treatises ex- 
 hibits, perhaps, its highest degree of power to deal with 
 the principles involved in the various Rules and in mis- 
 cellaneous problems; but its operations still admit of 
 considerable improvement in elegance and ease ; and to 
 promote such improvement is the aim of this Manual of 
 Short Methods. And here let it be observed that, in 
 many instances, the Author's shortness of method consists 
 in the easiness of the workings rather than in the space 
 they occupy.
 
 Vi PREFACE. 
 
 The book is intended for the more advanced Arith- 
 metic classes in schools, as well as for private students, 
 and it may accompany the use of any general treatise of 
 Arithmetic. Let general principles, by all means, be 
 well understood before short methods of calculation are 
 sought after ; but at the same time let it be observed 
 that there is cultivation of intellect, as well as lessening 
 of labour, in the proper use of the ciphering artifices 
 that are exemplified and recommended in the present 
 publication. 
 
 It is hoped that to many of those warehouse clerks 
 that are much occupied in extending invoices, some 
 portion of the chapter on the Calculation of Prices will 
 be of special service. 
 
 The aim and construction of this Manual are so 
 peculiar that, in order to secure thoroughly the advantage 
 to be derived from the study of it, the publication of a 
 Key to its numerous unworked examples has been re- 
 commended. The Author has accordingly prepared one ; 
 and he hopes that the student, having first worked with- 
 out its aid, will, by comparison of his own work with the 
 method presented in the Key, have either the satisfaction 
 of finding that his own skill is adequate, or the benefit 
 of learning how his skill may be improved.
 
 CONTENTS. 
 
 CHAPTER 
 
 I. THE GREATEST COMMON MEASURE . . . . 1 
 
 II. THE CALCULATION OF PRICES 4 
 
 III. SHORT METHODS IN SIMPLE MULTIPLICATION . . 15 
 
 IV. SHORT METHODS IN DIVISION . . . .22 
 V. THE CALCULATION OF SIMPLE INTEREST FOR DAYS 26 
 
 VI. THE CALCULATION OF COMPOUND INTEREST . 28 
 
 VII. SHORT METHODS IN THE WORK OF A CIVIL SER- 
 VICE EXAMINATION PAPER 33 
 
 VIII. MISCELLANEOUS QUESTIONS . . .40 
 
 IX. APPENDIX OF ADDITIONAL PROBLEMS IN HIGHER 
 
 ARITHMETIC, WITH SOLUTIONS . . . . 47 
 
 ANSWERS TO THE EXERCISES . . . 53
 
 MANUAL 
 
 OP 
 
 SHORT METHODS IN ARITHMETIC. 
 
 CHAPTER I. 
 
 THE GREATEST COMMON MEASURE. 
 
 1. THE Rule for finding the Greatest Common Measure 
 is often helpful for the reduction of a fraction's numerator 
 and denominator to lowest terms ; but it is resorted to 
 oftener than is necessary. Fractions, seeming at first 
 sight to require the usual process of determining the 
 G.C.M., may in very many instances be easily reduced to 
 lowest terms by trial, as we shall now exemplify. 
 
 Let the following be proposed for simplification : 
 
 301 637 234 957 1 O 1 
 
 U2T TTTSfT' T5TFT> TT3T> TT2TS' 
 
 In the first of these we can discover no small divisor of 
 the numerator ; but if we take down the denominator by 
 itself and break it up into factors, we shall find that it 
 contains 3 and 209, and that 11 goes in 209 and gives 
 19 ; so we have easily ascertained that the denominator 
 is divisible by 3, or 11, or 19. Now, as neither 3 nor 
 11 will go exactly in 361, we should try 19, which will 
 be found to go 19 times ; and thus the fraction is found 
 reducible to ^. 
 
 In the second fraction, the numerator shows itself 
 
 B
 
 2 SHORT METHODS IN ARITHMETIC. 
 
 obviously divisible by 7, giving 91 ; and 91 is easily seen 
 to be also divisible by 7, giving 13. The upper number, 
 then, will divide by 7 or 13, but the lower number will 
 not divide by 7; therefore, it must divide by 13, if we 
 are to have any cancelling at all. And 13 is found to 
 measure both terms of the fraction, reducing it to the 
 simpler form |^. 
 
 When we examine the third fraction, its numerator 
 seems easier than its denominator for breaking up ; we 
 therefore take it apart and divide it successively by 2 
 and 3 and 3, thus obtaining 13 as a trial divisor for 
 1001 ; for neither 2 nor 3 will suit that number. Divi- 
 ding the two terms by 13, we get ^f. 
 
 The fourth fraction obviously allows 3 to be struck 
 out of both terms, reducing it to f \ -f ; and now 1 1 may 
 be seen to go in the numerator 29 times ; and 29 divides 
 both terms down to \^. 
 
 In the last of the proposed fractions the numerator 
 seems to resist a first trial ; but the denominator plainly 
 shows itself equal to 2 x 5621 = 2 x 7 x 803 = 2 x 7 
 X 11 x 73; and though neither 2, nor 7, nor 11 can be 
 struck out of the numerator, 73 being tried is found to 
 be contained exactly 137 times; so that the resulting 
 
 137 
 
 = = - 
 
 fraction is 
 
 2. The usual mode of procedure in seeking the 
 Greatest Common Measure of two numbers might in 
 many instances be much shortened. Compare the two 
 subjoined ways of working Ex. 1. 
 
 Ex. 1. Find the G.C.M. of 5063 and 55029. 
 
 On the left we present the ordinary operation, but 
 securing some economy of space by writing the succes- 
 sive remainders vertically under the first divisor, as they 
 arise. 
 
 On the right, however, is shown a much better 
 method. When we obtain the second remainder, 664, 
 and have to find the G.C.M. of 664 and 4399, we can 
 readily judge that the factor 8 may be struck out of 664, 
 as 2 is not even once contained in 4399, and consequently
 
 THE GREATEST COMMON MEASURE. 
 
 8 cannot be a factor in the required common measure. 
 This cancelling shortens veiy much the rest of the work. 
 
 OtJicnvige. 
 
 5063 
 4399 
 664 
 415 
 249 
 166 
 G.C.M. = 83 
 
 55029 
 50630 
 
 10 
 
 1 
 6 
 1 
 1 
 
 1 
 2 
 
 5063 55029 
 50630 
 
 5063 
 4399 
 
 4399 5063 
 
 8)664 4399 
 G.C.M. = 83 415 
 
 4399 
 3984 
 
 249 
 249 
 
 664 
 415 
 415 
 249 
 249 
 166 
 
 
 
 166 
 166 
 
 10 
 
 Ex. 2. Find the G.C.M. of 7371 and 12441 ; also of 2449 and 
 5609. 
 
 7371 
 
 10)5070 
 
 507 
 
 273 
 
 2)234 
 
 117 
 
 G.C.M. = 39 
 
 (i.) 
 
 12441 
 
 1 
 
 7371 
 7098 
 
 14 
 
 1 
 
 507 
 
 273 
 234 
 
 2- 
 3 
 
 117 
 
 117 
 
 (ii.) 
 
 2449 
 711 
 
 4)316 
 G.C.M. = 79 
 
 5609 ! 2 
 
 4898 
 
 2449 
 2133 
 
 711 
 
 711 
 
 'Ex. 3. Find the G.C.M. of 47759 and 38957; also of 290413 
 and 906019. 
 
 (ii.) 
 
 167 
 
 * When the quotient is 1, the divisor need not be inserted 
 under the dividend to get the remainder. 
 
 Bl 
 
 38957 
 6)8802 
 1467 
 
 5)815 
 G.C.M. = 163 
 
 4775!i 
 
 1 290413 
 26 
 20)34780 
 G.C.M. = 1739 
 
 9 
 
 906019 
 871239 
 
 38957 
 2984 
 
 8802 
 
 290413 
 173!) 
 
 11651 
 10434 
 
 1467 
 1467 
 
 1217;; 
 
 12173 

 
 SHORT METHODS IN ARITHMETIC. 
 
 Exercises 1. 
 
 Find the G.C.M. of 
 
 1. 4371 and 17391. 
 
 2. 2668 and 8188. 
 
 3. 73033 and 118007. 
 
 4. 39330 and 29735. 
 
 5. 54385 and 57670. 
 
 6. 137897 and 78329. 
 
 7. 56498, 67691, and 30094. 
 
 8. 22661, 266815, and 42527. 
 
 9. 425383, 348467, and 228269. 
 
 Exercises 2. 
 
 Reduce to simplest terms the following fractions : the 
 first eight by trial, the last four by first calculating the 
 G.C.M. of numerator and denominator : 
 
 1 JL?JL 9 852 O 7o3 A 559 K 2765 a 2.118 
 -> 1639' J> 105H' "' 925' *' 3311' "' 6241' " 2773' 
 
 7 0503 Q J-L604_ O _323_ 1 n 41.71 1 1 32B7 1 O 2461 
 ' 4740' 0> 14043' " 1349' 1U< 6499' 11< 6031' 1 ^ > 13T89' 
 
 CHAPTER II. 
 
 THE CALCULATION OF PRICES. 
 
 1. THE Rule of Practice proposes to substitute shorter 
 work for the ordinary process of compound multiplica- 
 tion of money, when the given multiplier is a large 
 number. And it is very serviceable when the given 
 multiplicand is an easy aliquot part, as 3s. d., 4s., 5s., 
 &c., or consists of two easy aliquots, as 8s. 4d. t 2s. Qd., 
 135. 4d, &c. 
 
 Thus, to find the amount of 269 articles at 6s. 3d. 
 
 each, the method of Practice, 
 
 i ii i / iibit at fas. od. 
 
 as shown iri the annexed form 
 
 5s. Od. = i 
 
 67 5s. Od. 
 16 16 3 
 
 of operation, cannot, perhaps, 1*3=1 
 be surpassed in convenience by ~~$i Ts~3d 
 
 any other method of calcula- 
 tion ; but if the given price of each article were 15s. S^d., 
 Practice supplies no simpler mode of operation than the 
 tedious and clumsy 'one which we subjoin, and which,
 
 THE CALCULATION OF PRICES. 
 
 though not quite so bulky, is not so easy as the multiply- 
 ing process which we place beside it for comparison : 
 
 15*. 8$d. x 5 
 
 12 
 
 983 
 
 103 10 9~ 
 
 2_ 
 
 207~i 6~ 
 3 18 51 
 210 19*. 
 
 
 269 at 15*. 8d. 
 
 10*. Od. = A 
 
 l34 10 
 
 50=1 
 
 67 5 
 
 6 =i 
 
 6 14 6 
 
 2 = ^ 
 
 2 14 10 
 
 OJ =| 
 
 5 7 
 
 210 19*. 
 
 We shall presently direct attention to a much better 
 method than either of these, when we have first exempli- 
 fied a little the mode of dealing with small given multi- 
 pliers. 
 
 2. When the given number of articles of which the 
 cost is required is 13, or under 12, and the given price 
 is not an exact aliquot, it will generally be best to use 
 multiplication in the ordinary way. 
 
 Ex. 1. Find the cost of 7 yards of cloth at 14*. 6d. a yard, 
 that of 11 yards of linen at 2*. 5d., and that of 13 yards of silk 
 at 4*. 8<f. 
 
 14*. 
 
 5 1*. 6d. Ans. 
 
 2*. 5\d. 
 11 
 
 4*. 
 
 13 
 
 26*. 9f<2. Ans. 3 1 2\d. Ans. 
 
 Note. It is to be wished that the Multiplication 
 Table included products of 13. Let the pupil con these 
 products, if not already familiar to him. They are as 
 follows : 
 
 13 times 2 are 26 
 3 39 
 
 52 
 65 
 
 78 
 91 
 
 13 times 8 are 104 
 
 9 
 
 10 
 11 
 12 
 13 
 
 117 
 130 
 143 
 156 
 169 
 
 3. When the given number of things is 12, the 
 amount is just as many shillings as there are pence in the 
 price.
 
 SHORT METHODS IN ARITHMETIC. 
 
 Ex. 2. Calculate the cost of a dozen copies of a book at 4s. 
 . per copy, and that of a pound weight of gold at 3 17s. 
 per ounce. 
 
 4s. 9fd. 77s. IQirf. 
 
 12 
 
 57s. = 2 17s. 9<2. Ans. 934^. = 46 14s. Gd. Ans. 
 
 4. When the given number of articles is greater than 
 13, but less than 24, multiply by 12 + 2, 12 + 3, 
 12 + 4, &c. ... 12 + 11. 
 
 Ex. 3. Find the amount of 17 things at 2 17s. 9^. each, 
 and that of 23 things at 12s. 10f^. each. 
 
 2 17s. 9^. x5 12s. lOfd. x 11 
 
 12 12 
 
 34 13 6 
 14" 8 11 
 
 49 2s. 5\d. Ans. 
 
 9 
 10* 
 
 14 16s. 7kZ. Ans. 
 
 Here the advantage of first multiplying by 12 is that 
 we say at once 12 times 9-J>rf. is 9^ shillings, or 9s. 6d., 
 set down 6d. and carry 9s. ; and again, 12 times 10|d. is 
 10s., or 10. 9d., set down $d. and carry 10s. 
 
 Note. Special instances must frequently occur, for 
 which ways of calculating, simpler than that of a general 
 rule, are available. If, for example, the cost of a ton at 
 '2 13s. 4fZ. per cwt. is required: Expressing the price 
 in the single denomination of shillings, we have 20 things 
 at 53^ shillings each. Now, 20 at 53^- shillings each is 
 the same as 53-^ at 20 shillings each ; that is, 53^, or 
 53 6s. Sd. is the cost required. 
 
 Exercises 3. 
 
 s. d. 
 
 s. d. 
 
 1. 18 things at 7 3f 
 
 12. 21 things at 2 11 9| 
 
 2. 19 
 
 3 18 Hi 
 
 13. 20 
 
 1 10 6 
 
 3. 15 
 
 5 19 10| 
 
 14. 12 
 
 
 9 9 10 
 
 4. 17 
 
 17 8^ 
 
 15. 13 
 
 
 6 13 HJ 
 
 5. 20 
 
 065 
 
 16. 20 
 
 
 14 9 
 
 6. 12 
 
 13 6^ 
 
 17. 15 
 
 
 6 11 
 
 7- 13 
 
 845 18. 19 
 
 
 14 51 
 
 8. 21 
 
 1'i 10$ 19. 18 
 
 
 3 7 3f 
 
 9. 12 
 
 1 10 8i 20. 22 
 
 
 12 11 1 
 
 10. 13 
 
 17 9| 
 
 21. 23 
 
 
 1 16 l| 
 
 11. 12 
 
 15 7| 
 
 22. 21 
 
 
 4 17 6
 
 THE CALCULATION OF PRICES. 7 
 
 5. For those cases in which the given number of things 
 exceeds 24, and the price is not an exact aliquot, or does 
 not merely consist of two, or at most three, convenient 
 aliquots, we have now an easy and concise general method 
 to recommend ; and the method will be best taught by 
 beginning with examples in which the^iven multipliers 
 are large. 
 
 Let us, then, repropose for calculation an example 
 previously worked both by Practice and by ordinary 
 multiplication in 1. 
 
 Ex. 4. Calculate the cost of 269 articles at 15*. 8yd. 
 
 The multiplier 269 is equal to 240 + 24+5, hence, 
 if we should add together 240 times, 24 times, and 5 
 times the given price, we should obtain the amount re- 
 quired. 
 
 Now, in the annexed form of calculation we use the 
 above aggregate parts of the given 
 multiplier in the following advan- 
 tageous manner. We first obtain '?, 
 the product by 5 in the ordinary I8g 5 Q* 
 way; then, reducing the given price 24 = -'J 18 16 G 
 to pence, viz. 188j pence, we multiply 21(TT9 ll 1 
 that quantity by 240, by simply call- 
 ing it pounds, and write 188 under the 3 pounds, and 
 5 shillings, for the of a pound, under the 18 shillings. 
 We have now only to add 24 times the given price, or 
 one-tenth of the 240 times, already obtained. 
 
 Ex. 5. If a yard of linen be worth 3s. 7^d., what is the value 
 of 529 yards 1 
 
 Here 529 may be observed to be made up of 240, 240, 
 48, and 1 . Reducing, therefore, the price to pence, and 
 calling them pounds, we can calculate as follows : 
 
 .. 43 5*. 0</. = price x 240 
 
 43 5 = x 240 
 
 48 = I = 8 13 = x 48 
 
 3 7i = x 1 
 
 Ans. 95 6 7 = x 52'.
 
 8 
 
 SHORT METHODS IN ARITHMETIC. 
 
 Ex. 6. What would lls. 8|<Z. per day amount to in a year ? 
 
 Here the multiplier may be divided in either of the 
 two following ways, the latter of which will be thought 
 by some calculators the more convenient : 
 
 365 = 240 + 120 + 5 ; 
 or 365 = 240 + 80 + 30 + 15. 
 
 Accordingly, the calculation may be presented in the 
 two following forms : 
 
 365 at Us. 8fd. 
 5 
 
 365 
 
 at 
 
 5 
 
 
 2 
 
 18 
 
 7* 
 
 
 240 
 
 
 140 
 
 15 
 
 
 
 
 120 = 
 
 2 
 
 70 
 
 7 
 
 6 
 
 
 365 
 
 214 
 
 1 
 
 If 
 
 Ans. 
 
 240 
 
 
 140 
 
 15 
 
 
 
 80 
 30 
 15 
 365 
 
 _ i 
 
 3 
 
 8 
 
 46 
 
 17 
 8 
 
 18 
 11 
 15 
 
 4 
 10* 
 1H 
 
 214 
 
 1 
 
 1 a 
 it 
 
 Ans. 
 
 It should be noticed, however, that while the multipli- 
 cand here is a very awkward one for the application of 
 the method of Practice, the multiplier 365 can be very 
 conveniently applied in the ordinary way of compound 
 multiplication, as being equal to 5x12x6 + 5, and thus 
 enabling us to add in the line of 5 times while multiply- 
 ing by 6. We subjoin both forms of calculation. 
 
 365 at 11s. 8M 
 
 = 
 
 06=1 
 2 = i 
 OOf = 1 
 
 182 10 
 
 18 5 
 
 926 
 
 3 10 
 
 1 2 9| 
 
 11 8| 
 
 5_ 
 
 2 18 7f = 
 12 
 
 35 3 9 = 
 6 
 
 5 times 
 60 ; 
 
 214 1 l%Ans. 5214 1 If = (360 + 5) 
 
 *** We do not see that the given price can be dis- 
 tributed into more convenient parts for Practice than 
 those used above. Observe that in obtaining the last line 
 by the second method the calculation proceeds thus : 
 6 times 9 are 54 and 7f make 61 1 pence = 5 shillings 
 and l|c?., set down If and carry 5, &c. 
 
 Ex. 7. Find the product of 6 11*. 9d. by 327. 
 Here 327 may be distributed into 240, 80, and 7, or
 
 THE CALCULATION OF PRICES. 9 
 
 into 240, 60, 15, and 12; and hence the following choice 
 of ways to find the required product : 
 
 327 at611*. 9rf. 327 at 6 11 9 
 
 11 n 11 
 
 7 240 141 10 
 
 7 4 2 6 60 = = 35 7 6 
 
 240 141 10 15 = i = 8 16 10 
 
 80 = |= 47 34 12 = |= 716 
 
 6 x 327 = 1962 327 x 6 = 1962 
 
 2154 15 10| Ans. 2154 15 10 
 
 6. When the multiplier is under 240, we seek aliquots 
 in the multiplier, always in the first instance regarding 
 the number of pence in the price as pounds that express 
 240 times the price. 
 
 Ex. 8. What is the value of 217 cwt. of sugar at 35*. dd. per 
 cwt.? 
 
 Here we have 217 = 1 + 120 + 80 + 16 ; but the price, 
 which may at first appear to be inconvenient for Practice, 
 has accidentally the advantage that it consists of 14s. and 
 Is. 9cL, the latter of which is one-eighth of the former, 
 while the value at 14s., or ^ths of a pound, is easily got 
 by multiplying 217 by 7, and doubling the first figure 
 that arises for shillings. 
 
 We will, therefore, show the two methods, allowing 
 that this is a question for which the method of Practice 
 is rather better than the other way. 
 
 217 at 35*. 9rf. 217 at 1 15*. 9d. 
 
 240 429 217 
 
 1 
 
 120 = i = 
 
 80 = l = 
 16 = |- 
 
 1 
 
 214 
 
 143 
 
 28 
 
 15 
 10 
 
 
 12 
 
 9 
 
 
 
 
 
 14*. 
 1 
 
 Od. 
 9 
 
 = ^=151 
 
 = =18 
 
 18 
 19 
 
 
 
 9 
 
 387 
 
 17*. 
 
 9d 
 
 217 387 17*. 9d. Ans. 
 
 7. When the multiplier is seen to want only 1 of 
 being a number convenient for distribution, we may
 
 10 
 
 SHORT METHODS IN ARITHMETIC. 
 
 assume the convenient number and from its resulting 
 amount deduct the given price. 
 
 Ex. 9. What sum will buy 159 articles at the average price 
 of Is. H|<#. each ? 
 
 Here we have 159 = 120 + 40 1 ; but it is also 
 equal to 120 + 24 + 15. We will exhibit each of the 
 processes thus suggested. 
 
 159 at 7*. ll\d. 
 
 159 at 7s. 
 
 240 
 
 95 5 
 
 
 
 120 = 
 40 = 
 
 |=47 12 
 1 = 15 17 
 
 6 
 6 
 
 160 
 
 63 10 
 
 
 
 159 
 
 63 2*. 
 
 OM. Ans. 
 
 240 
 
 95 5 
 
 120= 1= 47 12 6 
 24 = I- = 9 10 6 
 
 16 -| 
 
 159 
 
 5 19 
 
 63 2s. Wd. 
 
 Ex. 10. A person bought 100 articles at 13*. l\d. each, and 
 sold them at 15*. i~d. each ; what did he give for 87 of them, 
 and what did he get for 74 of them ? 
 
 For this question we have 87 = 80 + 8 1, and 
 74 = 60 + 15 1 ; and hereby the following modes of 
 working are suggested. 
 
 87 at 13s. 1U. 
 
 240 
 
 163 10 
 
 = l = 54 10 
 .= Tn = 5 9 
 
 87 
 
 59 19 
 59 5*. ted. Ans. 
 
 240 
 
 184 10 
 
 
 
 60 = 
 15 = 
 
 75 
 
 i = 46 2 
 1=11 10 
 
 6 
 
 57 13 
 
 1* 
 
 56 17s. 
 
 Ofl. Ans. 
 
 Exercises 4. 
 
 1. 260 a 
 
 t 15 2 
 
 2. 264 
 
 9 11 
 
 3. 320 
 
 4 ll| 
 
 4. 247 
 
 13 7 
 
 5. 344 
 
 16 31 
 
 6. 340 
 
 1 8 5 
 
 7. 136 
 
 1 9 8 
 
 
 s. d. 
 
 8. 368 
 
 A 10 8| 
 
 9. 197 
 
 2 14 10 
 
 10. 294 
 
 17 8 
 
 11. 271 
 
 5 33 
 
 12. 119 
 
 13 11 
 
 13. 287 
 
 18 10 
 
 14. 93 
 
 3 17 101
 
 THE CALCULATION OF PRICES. 
 
 11 
 
 
 
 
 
 s. 
 
 d. 
 
 15. 
 
 291 a 
 
 fc 1 
 
 9 
 
 71 
 
 'a 
 
 16. 
 
 366 
 
 5 
 
 13 
 
 
 
 17. 
 
 184 
 
 
 
 19 
 
 5 
 
 18. 
 
 79 
 
 
 
 4 
 
 9| 
 
 19. 
 
 89 , 
 
 
 
 5 
 
 12 
 
 20. 
 
 375 
 
 
 
 12 
 
 7| 
 
 21. 
 
 372 
 
 
 
 7 
 
 9 
 
 22. 
 
 127 
 
 3 
 
 13 
 
 2 
 
 23! 
 
 348 
 
 
 
 6 
 
 9f 
 
 24. 
 
 411 
 
 3 
 
 6 
 
 5 
 
 25. 
 
 265 
 
 
 
 11 
 
 2- 
 
 26. 
 
 113 
 
 1 
 
 12 
 
 7* 
 
 27. 
 
 452 
 
 
 
 19 
 
 3 
 
 
 s. d. 
 
 28. 365 a 
 
 t 4 9i 
 
 29. 855 
 
 4 6 4| 
 
 30. 299 
 
 11 92 
 
 31. 547 
 
 12 llf 
 
 32. 351 
 
 4 17 7 
 
 33. 390 
 
 1 16 10* 
 
 34. 558 
 
 358 
 
 35. 503 
 
 1 1 8i 
 
 36. 615 
 
 2 13 6 
 
 37. 365 
 
 15 8| 
 
 38. 139 
 
 1 16 2i 
 
 39. 857 
 
 2 11 4 
 
 40. 169 
 
 1 14 3 
 
 8. We have now to direct attention to some ex- 
 pedient methods of dealing with certain small prices of 
 frequent occurrence in business. 
 
 In many commercial houses there is daily demand for 
 the extension of invoices requiring such calculation as 
 that of 
 
 135 (yds., Ibs., &c ) at 22|<2., 
 or 86| 2*. 5lrf., 
 or 216| 9^. 
 
 In general, we should expect the best method of doing 
 such work to be suggested by the form of the particular 
 price or quantity ; and, therefore, we shall here supply a 
 variety of worked examples, sufficient to beget an aptitude 
 for easy and expert calculation with such quantities and 
 prices generally. 
 
 Twelfths of a penny sometimes occur among the prices 
 customary in many commercial houses, but not so often 
 as eighths. Let the learner, therefore, acquire readiness 
 in expressing eighths of a pound as amounting to so many 
 half-crowns, and twelfths of a pound as so many times 
 Is. Sd. 
 
 Ex. 11. Calculate, to the nearest farthing, 135 yards, at 22|j<2. 
 per yard. 
 
 Here there can be no better way of working than 
 calling the pence pounds, and dividing 135 into 120 
 and 15.
 
 12 SHORT METHODS IN ARITHMETIC. 
 
 We set down the work as an extending clerk might 
 do it on his rough-work paper, dividing by 2 for the 
 120, and by 8 for the 15, as taught 
 in former examples. The pence 22 12s. 6/Z. 
 arising in the division by 8 is 3f eZ., 120 ~Tl 6 IT" 
 and the nearest farthing is, in favour 15 1 8 3| 
 
 of the buyer, f = id. ; in favour of Ant. 12 14s. \d. 
 the seller it would be f = %d. 
 
 Ex. 12. Find, to the nearest farthing, the cost of 86 Ib. 6 oz. 
 of tea, at 2s. 5\d. per Ib. 
 
 | of 86 7s. 6d. =10 15s. ll\d. 
 03 ld. 
 
 Am. 10 12s. id. 
 
 Here, as the price wants only a halfpenny of half-a- 
 crown, we first assume 2s. Qd. for the price, making the 
 amount equal to an eighth of 86-f ; then we have to 
 deduct for 86f Ib. at a halfpenny, which comes to 43 T ^ 
 pence, and the nearest farthing to -f$d. is ^ = \d. ; so 
 we subtract 
 
 Ex. 13. Calculate, to the nearest farthing, 216| yards, at $^ 
 per yard. 
 
 2163 at IQd. = 2167f& = 180s. l\d. 
 i of 216M. = 27d. =23 
 
 4ns. 8 18s. \d. 178 4| 
 
 Here again the calculation is accidentally made easy, 
 as 9 1 is only | short of 10. The eighth of 216|d. is 
 27^2^-j where the value of the fraction is much below 
 one farthing, and therefore is neglected. 
 
 Ex. 14. What will 377 Ib. of cotton come to at 10|rf. per Ib. ? 
 3770<Z. 
 
 471 
 
 3817^. = 318s. Id. = 15 18s. Id. Ans. 
 
 Here 377 is not made up of a convenient set of 
 aggregate parts : calculation by means of 240 + 48 + 48 
 + 40 + 1, though it would be easy, would take up too 
 much time. We have simply multiplied 377c. by
 
 THE CALCULATION OF PRICES. 13 
 
 (10 + ^), and reduced to shillings and pounds succes- 
 sively in a continuous line, not showing the divisors 
 12 and 20. 
 
 Ex. 15. Calculate 38 yards at 17 T |d. per yard. 
 
 17 8*. id. (Otherwise.) 
 
 1 14 10 3s. 2\d. x (12 + 5) 
 
 17 5 1 18 3 
 
 2 101 12)0 15 111 
 
 4i 1 3| 
 
 "2 15i.~ 6d. Ans- 2 log. Qd. Am. 
 
 24 
 
 12 
 2 
 
 By the first of these methods we begin with 
 as the value of 240 yards ; then we take 24 yards = T L, 
 12 yards = ^, 2 yards = %, and \ yard = . 
 
 By the second method we regard the question as re- 
 quiring the cost of 17-j^ yards at 38^. or 3s. '2\d. per 
 yard, and first taking 12" times the price = 38 shillings 
 = 1 18s. 3d., we add 5 times 3s. 2%d. as the cost of 5 
 yards, and T T 7 of that as the cost of ^ of a yard. 
 
 There is little, if any, superiority in either of the 
 methods over the other. 
 
 Ex. 16. Find the cost of 185^ yards at 2Q) 2 d. per dozen yards, 
 and also at 27%d. per dozen yards. 
 
 2 5s. 5d. 
 120 1 2 81 
 
 60 11 4|- 
 6 1 l| 
 
 1 15 2-$d. 
 
 = 2Qs. 5d. Ans. 1 Ws. l^d. Ans. 
 
 For the first part of the above question we have 
 multiplied 185'75rf. by (10 + 10 + i), and then divided 
 by 12 for dozens. This reserving of the 12 till the last 
 should be a general practice when neither the price nor the 
 number of articles is convenient for division by 12 at the 
 outset. 
 
 But for the second part of the question, which we 
 have worked by the method of calling the pence pounds, 
 we can conveniently begin with a twelfth part of 27 5s.,
 
 14 SHORT METHODS IN ARITHMETIC. 
 
 viz. '2 5s. 5d. Then assuming 186 for the number of 
 yards = 120 + 60 + 6, or ^, ^ and ^, we have only to 
 deduct the value of of T V of 27^., which for the near- 
 est approximation is ^ of 2 id. \d. 
 
 *%* As examples like those with which we are now 
 dealing cannot be referred to any general Rule, we must 
 trust to the exhibition of a variety of specimen workings 
 appropriate to particular questions as the best means of 
 promoting general skill in the learner. We will there- 
 fore extend preparatory illustration a little farther, before 
 prescribing the next set of Exercises. 
 
 Ex. 17. What will 35 J yards of net come to at 16^^. per 
 yard? 
 
 The simplest way of calculation here is that which 
 regards 35f yards at IQ^d. as equivalent to 16 T 8 7 yards 
 at 35|cZ. We will work out this together with another 
 method, which, though not so convenient in the present 
 instance, would in many cases be more suitable ; it takes 
 4 times the yards at a fourth of the price, and calls the 
 pence pounds. 
 
 2s. l\\d. 143 at \d. 
 
 1 3 10 120 
 
 12)1 3 10 20 
 
 1 lift 
 
 218 
 
 o 6 I 
 
 1 Qi 
 
 2 9s. Sd. An*. 2 9s. 8d. Ans. 
 
 Ex. 18. Find the value of 25 J yards at 3*. lO^d. per yard. 
 
 There can be no better way for this 
 
 question than to multiply the given Q ^ s iO- 5 -d. 
 price by (1 + 12 + 12 + ^) which are 26 5 12 
 aggregate parts of 25^. The conveni- 265 
 ence of the multiplier 12, converting, 
 
 as it does, the pence into 10^ shillings, i 18s - 7 2^- Ans - 
 is obvious. 
 
 Ex. 19. What is the value of a mixture of 13| gallons of 
 spirits at 17s. 9rf. and 8{j gallons at 19s. Id. ? and what is the 
 mixture worth per gallon ?
 
 THE CALCULATION OF PK1CES. 15 
 
 17*. 9d. x (10 + 3 + 1) 19*. Id. x (3 + 5 + f ) 
 
 2 18 9 
 8)4 17 11 
 I* 21 
 
 8 17 6 
 8)2 13 3 
 
 6 7 
 11 17*. 4fd. 8 Ss. 
 
 Hence, value of 22 gallons = 20 6*. 3$d. Ans. 
 
 11 
 1 
 
 = 10 3 
 
 18*. 5 Ans. 
 
 Exercises 5. 
 Calculate to the nearest farthing 
 
 Yards 
 
 ft 
 
 Yards 
 
 1. 210 at 
 
 Hi 
 
 12. 28 a1 
 
 2. 1101 
 
 9^ 
 
 13. 52 
 
 3. 317 
 
 6 I 
 
 14. 19| 
 
 4. 133i 
 
 | 
 
 15. 432 
 
 5. 131 f 
 
 '10 
 
 16. 37A 
 
 6. 358 
 
 4 
 
 17. 292 
 
 7. 280 
 
 7s 
 
 18. 37| 
 
 8. 171 
 
 13f 
 
 19. 19J 
 
 9. 276|- 
 
 ^1 
 
 20. 90f 
 
 10. 1565 
 
 2| 
 
 21. 8l| 
 
 11. 113| 
 
 l^ra 
 
 22. 47A 
 
 23 
 
 21J- 
 
 23 
 
 23. Find the value of 190| yards of lace at Is. 
 per dozen yards. 
 
 24. What is the cast of 89 yards at 2s. 
 dozen yards 1 
 
 per 
 
 CHAPTER III. 
 
 SHORT METHODS IN SIMPLE MULTIPLICATION. 
 
 1. To multiply a number by 25. This multiplier is 
 one-fourth of 100; therefore the amount of 25 times a 
 number is one-fourth of 100 times the number. 
 
 Hence, if you conceive two ciphers annexed to the 
 number, and divide by 4, you will obtain 25 times the 
 number.
 
 16 SHORT METHODS IN ARITHMETIC. 
 
 Ex. 1. Multiply 298, 437, and 147, severally, by 25. 
 4)298 . . 4)437 . . 4)147 . . 
 
 Answers, 7450, 10925, 3675. 
 
 Note. Similarly, to multiply by 125, divide 1000 
 times by 8. 
 
 2. To multiply a number by 75. This multiplier is 
 three-fourths of 100, or one-fourth less than 100; hence 
 the amount of 75 times a number must be one-fourth less 
 than 100 'times the number. 
 
 Conceive, therefore, two ciphers annexed to the 
 number, divide by 4, and subtract, and you will obtain 
 75 times the number. 
 
 Ex. 2. Multiply 365, 5278, and 1527, severally, by 75. 
 
 4)365.. 4)5278.. 4)1527.. 
 
 9125 131950 38175 
 
 Answers, 27375, 395850, 114525. 
 
 3. When a multiplier begins or ends with 25, the 
 short process described in 1 may be used. 
 
 Ex. 3. Multiply 1737, severally, by 825 and 258. 
 
 1737 1737 
 
 825 258 
 
 43425 43425 
 
 13896 13896 
 
 1433025 An*. 448146 A/is. 
 
 4. To multiply a number by 99, 999, &c. This is to 
 take once less than 100, 1000, &c. times the number. 
 
 Hence, by annexing two, three, &c. ciphers to the 
 given multiplicand, and then subtracting the given 
 multiplicand, you will obtain the required product. 
 
 Ex. 6. Multiply 467, severally, by 99 and 999. 
 
 46700 467000 
 
 467 467 
 
 46233 Am. 466533 Ans. 
 
 Note. By reference to this kind of abridgment a
 
 y. -5 M 
 
 ,-, 
 
 SIMPLE MULTIPLICATION. 17 
 
 common rule given in treatises on Arithmetic, about re- 
 curring decimals, may be illustrated. 
 
 Suppose it is required to convert '2768292 to a vulgar 
 fraction. 
 
 The usual Rule for such conversion is as follows : 
 
 To find the numerator : From the given decimal con- 
 sidered as an integer subtract the non-repeating part 
 considered as a separate integer. Then to form the de- 
 nominator : Write as many nines as the circulating 
 period contains figures, and annex as many ciphers as the 
 non-repeating part contains figures. 
 
 Following this Rule we have 
 
 2768292 -27 _ 307588 - 3 = 61517 . 
 9999900 " 1111100 222220 ' 
 
 then dividing the terms by their greatest common mea- 
 sure, 271, we obtain ff^ as the required fraction in its 
 simplest form. 
 
 The reason of the Rule will be seen from the following 
 demonstrative illustration : 
 
 2768295 = of 27 68292 
 100 99999 
 
 of [27 x (100000-1) + 68292] 
 
 9999900 
 
 of (2700000 + 68292-27). 
 
 9999900 
 
 And we shall generally do better than what the Rule 
 dictates, if we adopt the first line of the above demon- 
 strative process, and try to simplify the vulgar fraction 
 that first arises ; that fraction in the present instance is 
 
 68292 _ 7588 = 28. 
 99999 lllll~4l' 
 hehce rf, of 27!f = lig = ?g, as before. 
 
 5. It very often happens, when a multiplier is a large 
 number, that one of the partial products enables us to 
 obtain easily in one line the sum of two other partial 
 products. In the examples which we are about to give 
 of this source of abridgment, we of course have to frame 
 
 C
 
 18 SHORT METHODS IN ARITHMETIC. 
 
 multipliers to show the facilities ; but experience proves 
 accident to be, in at least seven cases out of ten, favour- 
 able to the application of one or another of the shortening 
 devices which we are about to exemplify ; so that these 
 devices are well worth noting. 
 
 (a) When in the multiplier there occur two con- 
 tiguous figures that could not easily be used at once, but 
 that express an exact multiple of another figure, higher 
 than unity, already used, the product by the figure al- 
 ready used, multiplied by the other factor of the multiple, 
 will obtain the work of the two figures in one line. 
 
 Ex. 6. Multiply 39478 by 486, and 37409 by 763. 
 
 39478 37409 
 
 486 763 
 
 236868 x 8 261863 x 9 
 
 1894944 2356767 
 
 19186308 Ans. 28543067 Am. 
 
 Here the figures 48 in the first multiplier express 8 
 times the 6 ; having therefore worked the line of 6 times, 
 we take 8 times that line for 48 times the given multi- 
 plicand, observing to place the first figure, 4, under the 
 units of its understood multiplier, 48. 
 
 In the second multiplier, the figures 63 express 9 
 times the 7 ; having therefore first worked the line of 7 
 times, we take 9 times that line for 63 times the given 
 multiplicand, observing to place the first figure, 7, under 
 the units of its understood multiplier, 63. 
 
 Ex. 7. Find the product of 9865 by 728. 
 
 9865 9865 
 
 728 728 
 
 78920 x 9 C9055 x 4 
 
 710280 276220 
 
 7181 720 Ans. 7181720 Ans. 
 
 Here we have the choice of beginning to multiply 
 from the right of the multiplier or from the left ; for 72 
 is 9 times 8, and 28 is 4 times 7.
 
 SIMPLE MULTIPLICATION. 19 
 
 Ex. 8. Multiply 597867 by 8472. 
 
 597867 597867 
 
 8472 8472 
 
 1195734 4782936 x 9 
 
 4185069 x 12 2391468 
 
 50220828 43046424 
 
 5065129224 Ans. 5065129224 Ans. 
 
 Here, again, we have the choice of beginning to 
 multiply from either end, since 12 times 7 is 84 and 9 
 times 8 is 72. 
 
 (6) Sometimes it is expedient to make the proposed 
 multiplicand and multiplier exchange places. 
 
 Ex. 9. Multiply 19036 by 8795, and 35756 by 37908. 
 
 8795 37908 
 
 19036 35756 
 
 8795 265356 
 
 79155 x 4 2122848 = 7 times x 8 
 
 316620 1326780 =7 x 5 
 
 167421620 Am. 1355438448 Ans. 
 
 Here the proposed multiplier does not present any 
 advantage in either case ; but by transposing the factors 
 we can shorten the work in both cases. 
 
 (c) Sometimes conceiving a figure of the multiplier to 
 be lessened, and the portion taken away to be annexed to 
 the next figure in the multiplier secures advantage. 
 
 Ex. 10. Multiply 8367 by 492, and 87058 by 7854. 
 
 Here the first multiplier = 480 4-12, and the second 
 = 6000 + 1800 + 54 ; accordingly, in the former case 
 we may take 12 times and 12 times x 4, and in the 
 latter case we may take 6 times, 6 times x 3, and 18 
 times x 3. 
 
 8367 87058 
 
 492 7854 
 
 100404 = 12 times 522348 = 6 times 
 
 401616 = 48 1567044 = 18 
 
 4116564 Ans. 4701132 = 54 
 
 683753532 Ans.
 
 20 SHORT METHODS IN ARITHMETIC. 
 
 Ex. 11. Multiply 175096 by 9724. 
 Here we have a choice of two equally good artifices. 
 
 175096 175096 
 
 9724 9724 
 
 1575864 = 9 times 700384 = 4 times 
 
 12606912 = 72 2101152 = 12 
 
 700384 = 4 16809216 = 96 
 
 1702633504 Ans. 1702633504 Am. 
 
 Ex. 12. Multiply 36981 by 4697. 
 
 Here, again, we have a choice of methods ; we may 
 resolve 4697 into 4200 + 490 + 7, or we may transpose 
 the factors. 
 
 36981 4697 
 
 4697 36981 
 
 258867 = 7 times 42273 = 9 times 
 
 1812069 = 49 380457 = 81 
 
 1553202 = 42 169092 = 36 
 
 173699757 Ans. 173699757 Ans. 
 
 Ex. 13. Find the product of 72996 and 6579. 
 
 Here we may make either of the numbers multiplier 
 with advantage ; for we have 72996=70000 + 2800 + 196, 
 the last of these heing 7 times 28 ; and we have 6579 
 = 6300 + 270 + 9. 
 
 72996 6579 
 
 6579 72996 
 
 656964 = 9 times 46053 = 7 times 
 
 1970892 = 27 184212 =28 
 
 4598748 = 63 1289484 = 196 
 
 480240684 Ans. 480240684 Ans. 
 
 Ex. 14. Multiply together the treble of 96012, the double of 
 52869, and the half of 365. 
 
 This, in effect, requires to multiply the continual pro- 
 duct of the three numbers by 3 x 2 x ^, that is, by 3. 
 We shall take 365 as equal to 10 + 5 + 350.
 
 SIMPLE MULTIPLICATION. 
 
 52869 
 96012 
 634428 
 5075424 
 
 5076058428 
 
 x 8 
 
 2)50760584280 
 
 25380292140 x 7 
 177662044980 
 1852761326220 x 3 
 
 5558283978660 Ans. 
 
 (d) Various other devices might be exemplified, in 
 the contrivance of which practice and experience pro- 
 mote expertness and save labour. We will work only 
 one example more. 
 
 Ex. 15. Multiply 4 times 5869 by 227825. 
 
 It is often useful to remember that numbers ending 
 with 25 or 75, when multiplied by 4, give a product 
 ending with two ciphers. As, therefore, the product re- 
 quired in the present question is equal to 227825 x 4 x 
 5869 = 911300 x 5869, the work is thus found to be 
 more simple than by following the direct order of the 
 question. Moreover, we have here an accidental ad- 
 vantage for those who can readily multiply at once by 
 13, for 91 is 7 times 13. The operation will then be as 
 follows : 
 
 5869 
 227825 x 4 = 911300 
 
 7629700 x 7 
 53407900 
 
 5348419700 Am. 
 
 Exercises 6. 
 
 Multiply 
 
 1. 62057 
 
 2. do. 
 
 3. 807439 
 
 4. dor. 
 
 5. 148703 
 
 6. do. 
 
 7. 27738 
 
 8. do. 
 
 9. 2507386 
 
 10. 7363891 
 
 11. 891257 
 
 12. 3926 x 8723 
 
 13. 9819x7465 
 
 
 
 Multiply 
 
 
 248 
 
 14. 
 
 8297 x 365 h 
 
 7 321 
 
 549 
 
 15. 
 
 611736 
 
 , 36120 
 
 420. 
 
 16. 
 
 993865 , 
 
 , 51236 
 
 819 
 
 17. 
 
 83754 
 
 2643 
 
 726 
 
 18. 
 
 9738 x 8379 
 
 25 
 
 497 
 
 19. 
 
 9586 x 6954 
 
 99 
 
 742 
 
 20. 
 
 56749 
 
 8759 
 
 856 
 
 21. 
 
 12786 x 2786 
 
 125 
 
 654 
 
 22. 
 
 630728 
 
 , 187648 
 
 648 
 
 23. 
 
 42308 x 7854 
 
 , 5296 
 
 3204 
 
 24. 
 
 13296 
 
 , 54387 
 
 25 
 
 25. 
 
 4962 x 9356 
 
 365 
 
 99 
 
 26. 
 
 170625 x 9587 
 
 8
 
 22 SHORT METHODS IN ARITHMETIC. 
 
 CHAPTER IV. 
 
 SHOBT METHODS IN DIVISION. 
 
 1. IT is familiarly known that when a dividend and its 
 divisor contain a common factor, cancelling or striking 
 out that factor from each, before dividing one by the 
 other, will cause no alteration of value in the result. 
 
 But, except in cases where cancelling can be used, the 
 application of short methods to division is very limited. 
 In the abridgments that we shall teach in the present 
 chapter we shall deal only with dividends that contain 
 remainders ; though of course the methods of abridgment 
 will be applicable generally. We shall divide remainders 
 decimally. 
 
 2. To divide a number or a simple quantity by 25, or 
 125, or 75. Multiply by the reciprocal of the divisor, 
 viz. by -04, '008, or -01. 
 
 Ex. 1. Divide 51067, severally, by 25, 125, and 75. 
 
 51067 51067 51067 x -01 
 
 01 008 17022i 
 
 2042-68 Ans. 408-536 Ans. 680'89i Ans. 
 
 3. To divide a number by 99. Add the quotients ob- 
 tained by dividing by 100 as often, in succession, as will 
 ensure the accuracy of the second decimal figure (if any) 
 in the result. 
 
 Note. For the division of a, simple quantity contain- 
 ing a decimal fraction, the process must be carried so far 
 as to ensure the accuracy of the second decimal place 
 beyond the number of places in the given decimal. 
 
 Ex. 2. Divide 532701623 and 762-218, severally, by 99. 
 
 5327016-23 7-62218 
 53270-162 -076222 
 
 532-702 -000762 
 
 5-327 -000007 
 
 _ -P"-'* 7 7 69917 
 
 5380824-47
 
 DIVISION. 
 
 The Answers therefore are 5380824-47 and 7-69917, 
 for 47 and 17 would be found to be the remainders, if we 
 divided by 99 in the ordinary way, not carrying the 
 division beyond the figures given. 
 
 4. The number 365, being the number of days in a 
 year, occurs frequently, in simple interest and other 
 calculations, as a divisor; and the following easy 
 Rule has been investigated to lessen the labour of the 
 division. 
 
 To divide a number or a simple quantity by 365. 
 Multiply by -002, and to the product add ^ + -^ + ^^j- 
 of itself; the sum diminished by its 10000th part will 
 give an extremely near approximation to perfect ac- 
 curacy. 
 
 Ex. 3. Divide 27493 by 365. 
 
 Ex. 4. If a person's income is 248 16*. 6d. a year, how much 
 is it per day ? 
 
 (i.) (ii.) 
 
 27493 4976-5*. 
 
 C02 -002 
 
 3 
 10 
 10 
 
 10000 
 
 54-986 
 18-32866$ 
 l-83286f 
 18328* 
 
 75-330820 
 007533 
 
 75-323287 Am. 
 
 3 
 10 
 10 
 
 10000 
 
 9-9630 
 
 3-31766f 
 33176f 
 03317f 
 
 13-63561 
 00136 
 
 13-63425*. 
 13*. 7-61d. Ans. 
 
 Note. It should be noticed that the above operations, 
 though they make a bulky appearance, are extremely 
 easy, being very little more than a succession of mere 
 copyings; and for practical purposes the decimals need 
 not be extended to the degree of accuracy to which we 
 have carried them. For such questions as the 4th, which 
 requires to divide a given sum of pounds, shillings, and 
 pence by 365, an answer true to the nearest farthing will 
 always represent the daily income as accurately as our 
 money denominations will permit, and hence the given 
 pence may be neglected.
 
 24 
 
 SHORT METHODS IN ARITHMETIC. 
 
 976*. x -002 
 
 3 
 
 10 
 10 
 
 9-952 
 
 3-317 
 
 332 
 
 033 
 
 In working the 4th example, 
 then, the sum to be divided by 365 
 is only 4976 shillings, and the opera- 
 tion should be conducted in the ap- 
 proximative way here shown, dis- 
 pensing with the deduction of the 
 lOOOOtf/i, and finding the answer to be simply 13s. l^d. 
 per day. 
 
 As this method of finding daily from yearly income 
 will often be found very useful in practice, we will give 
 two more worked examples. 
 
 Ex. 5. A man's yearly income is 1359 15*. ; how much is 
 that per day ? 
 
 JEte. 6. Find the average daily rate of an income amounting 
 to 939 9*. Wd. in a year. 
 
 3 
 10 
 10 
 
 (i-) 
 27195*. x 
 
 002 
 
 54-390 
 .18-130 
 1-813 
 181 
 
 74-514*. 
 12 
 
 6-068^. 
 
 (ii.) 
 
 18789*. x 
 
 002 
 
 3 
 
 10 
 10 
 
 37-578 
 
 12-526 
 
 1-253 
 
 125 
 
 51-482*. 
 12 
 
 or, 3 14*. 6d. Ans. 
 
 or, 2 11*. 5|<Z. Ans. 
 
 In the investigation of the Rule we'used the recipro- 
 cal of 365, for, dividing by any number will produce the 
 same result as multiplying by the reciprocal of the 
 number. 
 
 Now, the reciprocal of 365, viz., -g-^y, is found to be 
 equal to -00274 - ^ of T7ro i nnr = -00274 - Tir i w of 
 ^ = -00274 - -rvfan of -00274 very nearly. 
 
 Again, -00274 is equal to -002 + -00066f + -00006f 
 + -OOOOOf , that is, equal to -002 (1 + + '& + -jfor). 
 
 Hence the Rule. 
 
 5. It is often required to convert a large number of 
 cubic inches to gallons, the divisor for which is 277'274, 
 the number of cubic inches in a gallon ; and we have in- 
 vestigated a Rule similar to the preeeding, which makes.
 
 DIVISION. 25 
 
 division by 277'274 very nearly as easy. The Rule is as 
 follows : 
 
 To divide a number or a simple quantity by 277*274. 
 Multiply by -003, and to the product add i + -g^ + 
 
 Tnnnr - arrow of itself 
 
 The reciprocal of 277'274, as worked out by ordinary 
 
 division, is '00360654, and the Rule now given will be 
 found to be verified by working out the reciprocal accord- 
 ing to its dictates, as required in the first of the two 
 following questions. 
 
 Ex. 6. Find by the above Rule the reciprocal of 277-274. 
 Ex. 7. In 4873956 cubic inches of water how many gallons ? 
 
 (i.) (ii.) 
 
 5 -00300000 4873956 c. in. 
 
 100 -00060000 -003 
 
 10 -00000600 5 
 
 - 10 -00000060 100 
 
 00360660 10 
 
 - -00000006 - 10 
 
 14621-868 
 2924-373 
 29-244 
 2-924 
 
 00360054 Ans. 17578'409 
 
 - -292 
 Ans. 17578-117 gall. 
 
 Exercises 7. 
 
 1. What weight of salmon at Is. Id. per Ib. is worth 
 7 16*. 6d.1 
 
 2. On what quantity of brandy would the duty 
 amount to 6 guineas, at the rate of 10s. 5d. a gallon? 
 
 3. If 69 3f cubic feet of clay can be dug out in 25 
 hours, how much can be dug out in 24 hours 1 
 
 4. The produce of a field of 75 acres is 25 6f quarters 
 of wheat ; how many bushels is the produce per acre 1 
 
 5. Calculate the 75th part of 25 '9 34, and the 99th 
 part of 8550828. 
 
 6. Calculate the 365th part of 25908, and the 99th 
 part of 123456789. 
 
 7. What daily rate of expenditure amounts to 
 120 6s. 8d. in a year 1 
 
 8. Iu 76880'9 cubic inches of oil how many gallons?
 
 26 SHORT METHODS IN ARITHMETIC. 
 
 9. Divide 117-918 ounces by 8^ ounces, and by 62^-, 
 severally. 
 
 10. If a man's yearly income amount to 704 10s. Qd., 
 what is his daily rate of income 1 ? 
 
 11. What decimal quantity multiplied by 9|- will pro- 
 duce -01234? 
 
 12. Divide 1234 by 365, and the result by -75. 
 
 13. How much per day is a yearly income of 350 
 11s.? 
 
 14. A cask contains 9999 cubic inches of liquid : how 
 many gallons does it contain 1 
 
 15. Calculate the 99th part of 6477778. 
 
 16. Divide 111111 farthings by 7*. 7|d. 
 
 17. A cistern holds 5000 cubic inches of water : 
 reduce its contents to gallons. 
 
 18. A clerk's salary is 35 10s. a quarter : how much 
 is it per day 1 
 
 19. What number multiplied by '0365 will produce 
 30391 1 
 
 20. Divide 2681217 by 99, and the result by 365. 
 
 21. A rectangular tank, 19'58 feet long and 14-07 
 feet wide, contains 5f feet depth of water. Find the 
 volume of water in gallons. 
 
 22. Find the 99th part of the square of 3'65. 
 
 CHAPTER V. 
 
 SHORT METHODS IN THE CALCULATION OF SIMPLE 
 INTEREST FOR DAYS. 
 
 1. IN the calculation of simple interest for days, the 
 number 365 occurs as a divisor, and we may therefore in 
 such calculation avail ourselves of the short method of 
 dividing by 365 exemplified and explained in the pre- 
 ceding chapter. Also, as a given number of days occurs 
 as a multiplier, its employment can often be simplified 
 by the use of easy divisors as taught in the first chapter.
 
 SIMPLE INTEREST FOR DAYS. 
 
 27 
 
 And it should be noticed that in using such divisors we 
 can limit approximation with very few figures, because 
 we work, not from right to left, under the necessity of 
 writing several useless figures, but from left to right, 
 with assurance of the accuracy of each figure to wherever 
 we extend the approximation. 
 
 Ex. 1. 
 days, at 4 : 
 
 Calculate the simple i 
 per cent, per annum. 
 
 3875s. Id. 
 
 nterest on . 
 
 3 
 
 10 
 10 
 
 10000 
 
 or 7 
 
 re have 
 by inse 
 
 est on 74 
 urn. 
 
 3 
 10 
 10 
 
 10000 
 or 4 
 
 193 15s. Id., for 313 
 2577-747 x -002 
 
 465-07 x 4 
 
 5-15549 
 1-71850 
 17185 
 01718 
 
 1860-28 
 116-2675 
 
 240 
 
 60 
 12 
 
 1 
 
 1976-5475 
 494-1369 
 98-8274 
 8-2356 
 
 7-06302 
 00070 
 
 7-06232 
 Is. 2-957</. Am. 
 
 called the pence 
 rting the decimal 
 
 L 4s. lOd. for 1 year 
 15788 4475 x -002 
 
 313 2577-7474 -=-365 
 
 Note. At the outset v 
 pounds, and divided by 10 
 point. 
 
 Ex. 2. Find the simple inter 
 203 days, at 3f per cent, per ann 
 
 14824s. Wd. 
 1778-98 x 3| 
 
 1334-235' 
 
 31-576895 
 10-52563 
 1-05256 
 10526 
 
 240 
 240 
 80 
 8 
 
 6671-175 
 6671-175 
 2223-725 
 222-3725 
 
 43-26034 
 00432 
 
 43-25602 
 3 os. l'44rf. ^l/;s. 
 
 568 
 
 1578S-4475-5-365 
 
 %* The above question worked in the best style of 
 the common method is here subjoined for comparison, 
 all usual means of cancelling and simplification being 
 taken advantage of. 
 
 741 f s x 568 x 7^ _ 88949 x 568 x 15 
 
 73000 
 
 _88-949x71. 
 7372 
 
 73000x120x2 
 44-4745x71 
 73
 
 28 SHORT METHODS IN ARITHMETIC. 
 
 44-4745 x 71 
 3113215 
 
 73)3157-6895(43-256 
 292 20 
 
 237 s. 5-120 
 
 219 12 
 
 186 d. 1-44 
 
 146 Ans. 43 5s. 
 
 ~408 
 365 
 
 439 
 438 
 
 "15 
 
 Exercises 8. 
 
 Find the simple interest on 
 
 & s. d, days p.c. 
 
 1. 114 15 for 83 at 5 
 
 2. 462 18 6 113 4 
 
 3. 129 11 8 206 3 
 
 s. d, days p.c. 
 
 4. 84 11 9 for 278 at 3 
 
 5. 491 16 9 223 2| 
 
 6. 1108 13 9 191 5i 
 
 7. Calculate the simple interest on 1329 14s. lid!, 
 from February 10 to December 3, 1883, at 4^ per cent. 
 per annum. 
 
 8. What is the mercantile discount on a bill for 
 94 17s. 4cZ., payable on March 1, 1884, discounted on 
 October 10, 1883, at 3^ per cent, per annum ? 
 
 CHAPTER VI. 
 
 SHORT METHODS IN THE CALCULATION OF 
 COMPOUND INTEREST. 
 
 1. WHEN questions in Compound Interest are included 
 in Arithmetic Examination Papers, the number of years 
 proposed is always small, rarely exceeding 5, and usually 
 limited to 3 or 4. This is on account of the heavy 
 amount of calculation requisite for longer periods,
 
 COMPOUND INTEREST. 
 
 29 
 
 For the requirements of business Compound Interest 
 Tables are published. It is, however, necessary to the 
 attainment of a thorough knowledge of Arithmetic that 
 the principle involved in Compound Interest, and the 
 process of computing it, should be intelligently known. 
 And for this purpose the given period need not exceed 
 four or five years. 
 
 But there is no Rule of Arithmetic of which the 
 examples are more commonly worked in a tedious and 
 confused manner than Compound Interest. We will 
 here, therefore, exemplify some good methods of econo- 
 mising time and work and seeming distinct arrangement 
 in the working of this Rule. 
 
 Ex. 1. What would be the amount of 147 11s. 3d. in 5 years 
 at 2 per cent, per annum, compound interest ? 
 
 The given principal multiplied by 1'02 3 will produce 
 the amount required. Now, to raise TO 2 to the 5th 
 power is to find the amount of 1-02 x 1'02 x 1'02 x 1'02 x 
 T02, which may be done as in the left-hand portion of 
 the subjoined work, either dividing successively by 50, 
 or else multiplying successively by -02, for the interest 
 to be added, and observing in the multiplication by '02 
 to keep the first figure of the product two places to the 
 right of the multiplicand. 
 
 Then we have to obtain the product of 147 11s. 3d. 
 by 1-1040808, for the amount at the end of 5 years. 
 
 1-1040808 
 147 
 
 1-02 
 
 x -02 
 
 . = s l = -0204 
 
 - 
 
 1-0404 
 
 x -02 
 
 020808 
 
 
 1-061208 
 
 x -02 
 
 0212242 
 
 
 1-0824322 
 
 x -02 
 
 0216486 
 
 
 1-02 5 * 1-1040808 
 
 77285656 
 154571312 
 
 162-29988 
 
 10*. Od. = > = -55204 
 1 3 = i = -06900 
 
 162-92092 
 or 162 18*. 5-Q2d. Ans.
 
 SHORT METHODS IN ARITHMETIC. 
 
 Othertvise and more conveniently : 
 
 Let the given principal have 
 its value expressed in the de- 
 nomination of pounds ; then add 
 successively ^L, or multiply succes- 
 sively by -02, as in the form an- 
 nexed, to obtain the successive 
 amounts from year to year. 
 
 12 
 20 
 
 147-5625 
 
 2-95125 
 150-51375 
 
 3-01027 
 153-52402 
 
 3-07048 
 156-59450 
 
 3-13189 
 159-72631) 
 
 3-19453 
 162-92092 
 
 x -02 
 
 Aug. 
 
 Given principal = 147*5625 
 
 2. When it is required to find the principal that will 
 amount to a given sum in a given number of years at a 
 given rate by compound interest, or. in other words, to 
 find what would be the present worth of a given sum 
 payable so many years hence, were compound interest 
 allowed at a certain rate, we must employ the amount of 
 1 as found in the first of the preceding methods of opera- 
 tion. Thus, to find the principal that amounted to 
 162 18s. 5'02c. by compound interest in 5 years at 2 
 per cent, per annum, we have to divide I62'92092 by 
 1-02 5 , that is, by 1-104081. 
 
 The principal could be traced back by successive 
 divisions by 1'02 ; but the process would be too laborious. 
 
 3. Now, to involve the amount of 1 for a year to the 
 power denoted by the number of years, it is a general 
 practice with young students to proceed as in ordinary 
 multiplication. But much greater ease and expedition 
 are secured by the method we have exemplified in the 
 involution of 1'02; and similarly, the involution for the 
 rates 5, 4, 3, 2^, 3^, 4^-, should be begun as follows : 
 
 5 p.c. = 
 
 2 p.c 
 1 p.c. 
 
 1-05 
 5 1 ,, = -0525 ' 
 
 1-04 
 2 p.c. = i = -0208 
 0208 
 
 {Otherwise.") 
 
 1-04 x -04 
 0416 
 
 1-1025 
 
 1-03 
 = J - = -0206 
 
 = m = ' 103 
 
 1-0816 
 
 ( Otherivise.} \ 
 1:03 x -03 i 21 
 0309 
 
 1-0816 
 1-025 
 
 1-050625
 
 COMPOUND INTEREST. 31 
 
 1-0325 x -03 
 030975 
 
 1-045 x -04 
 04180 
 
 \ P- c. = ^ = -0025812 i p.c. = s = -005225 
 
 1-0660562 1-01)2025 
 
 Otherwise, 3^ per cent, may be calculated by adding 
 to T0325, the aliquot parts 2 p.c. =3^, 1 "p.c. = 
 and ^ of I p.c. ; also 4^ p.c. may be divided into 2^ = 
 and 2 = -$. This way is preferable when the index of 
 the power is high, because division ensures accuracy as 
 far as we choose to carry the approximation. 
 
 The following worked examples may now be ex- 
 amined. 
 
 Ex. 2. Of what sum payable 3 years hence would 
 888 19. ll^d. be the present worth, if compound interest were 
 allowed at 4 per cent, per annum ? 
 
 The question is easily seen to be 12 I H-125^. 
 
 only an indirect way of asking the 20 i 19-927083s. 
 
 amount of the given sum in 3 years at 888-996354 x -04 
 4 per cent, per annum. We might 
 
 have divided the rate into two equal 924-5o621 x -04 
 
 parts, each 2 p.c. = r ^. 
 
 961-53846 x -04 
 
 38-46154 
 
 1000-00000 Ans. 
 
 Ex. 3. What principal would in 4 years amount to 600 by 
 compound interest at 7 per cent, per annum ? 
 
 1 075* = 1-335469)600-000000 
 5 p.c. = i = -05375 449-28036 
 
 21 p.c. = 
 
 026875 
 
 6581240 
 
 1-155625 
 0577812 
 0288906 
 
 12393640 
 
 374419 
 
 107325 
 
 1-2422968 
 0621148 
 0310574 
 
 487 
 87 
 449-28036 
 20 
 
 1-3354690 
 Am. 
 
 5-6072 
 12 
 449 os. 7'2864rf.
 
 32 SHORT METHODS IN ARITHMETIC. 
 
 Ex. 4. On what principal would 53 lls. 9d. be the compound 
 interest for 5 years, at 4^ per cent, per annum ? 
 
 1-0433333 5 = 1-236276 
 i = -0347777 1-0 
 
 = iuo = '0104333. -236276 = int. on 1 ; 
 
 1-0885444 
 0362848 20)11-755. 
 
 0108854 236276)53-5875 
 
 1-135715 226-8 
 
 037857 47255 
 
 011357 6332 
 
 1-184929 1606 
 
 039498 189 
 
 011819 
 
 1-236276 2268 = =226 16s. Ans. 
 
 Exercises 9. 
 
 1. Find the amount by compound interest of 186 5s. 
 in 4 years, at 3 per cent, per annum. 
 
 2. To what sum will 654 3s. 3d. amount by com- 
 pound interest in 4 years, at 3| per cent, per annum ? 
 
 3. Find the compound interest on 1179 14s. Id. for 
 3 years, at 3^ per cent, per annum. 
 
 4. What principal would amount to 400 by com- 
 pound interest in 5 years, at 2^- per cent, per annum 1 
 
 5. What principal would amount to 1178 4s. Id. 
 in 3 years, by compound interest, at 5^ per cent, per 
 annum 1 
 
 6. On what principal would 6 years' compound 
 interest, at 4^ per cent, per annum be 17 5s. Id.
 
 CIVIL SERVICE EXAMINATION PAPER. 33 
 
 CHAPTER VII. 
 
 SHORT METHODS IN THE WORK OF AN EXAMINATION 
 PAPER. 
 
 [Set, under direction of the Civil Service Commissioners, at a 
 competition for Men Clerkships, Lower Division, in May 
 1883.] 
 
 QUESTIONS. 
 
 1. Express as a decimal the value of ^^ of the pro- 
 duct of 6-307692 and 1-428571. 
 
 2. How many plots, each containing 7 sq. ft. 38 ins., can 
 be formed out of a field of 8 poles 19 sq. yds. 4 ft. 72 ins. ? 
 
 3. If 9 gallons of brandy, together with l^j- cwt. of 
 sugar, cost as much as 12 tons of coals, while ^ gallon of 
 brandy costs the same as ^ cwt. of sugar, how much sugar 
 is equal in value to 10 tons of coals ? 
 
 4. Supposing a building in the form of a cube to contain 
 1259712 cubic feet, what would be the cost of carpeting 
 its floor with carpet | yard wide, worth 3.9. 4.^. per yard ? 
 
 5. Determine by duodecimals the area of a rectangle 
 whose adjacent sides respectively measure 10 ft. 9^ ins. 
 and 5 ft. 2^ ins. How many such rectangles are there 
 in one which contains 12125 sq. ft. ? 
 
 6. If f of the difference between a certain fraction 
 and lie between 1^- and 1^, find between what limits 
 the fraction itself must lie. 
 
 7. If it be known that the decimal -2421 or some por- 
 tion of it repeats, calculate as a vulgar fraction the differ- 
 ence between the greatest and least values which it can 
 have. 
 
 8. Extract the square root of 25400544 in the scale of 
 6 ; transform 3065 "263 from the scale of 8 to that of 10 ; 
 and reduce T ^ from the scale of 10 to a duodecimal fraction. 
 
 9. Find what is the first year in which a sum of 
 money will become more than doubled in amount, if put 
 out at compound interest at the rate of 10 per cent, per 
 annum.
 
 34 SHORT METHODS IN ARITHMETIC. 
 
 10. Two clocks point to 5 P.M. at the same instant. 
 One loses 7| seconds and the other gains 8^ seconds in 24 
 hours. Find the interval which will elapse before one 
 will be precisely \ an hour before the other, and the times 
 which each will then indicate. 
 
 11. If to some wheat bought at 39s. per qr. some 
 more is added costing 6s. per bushel, in what proportion 
 must the better quality be mixed with the other, in order 
 that, if the whole be sold at 57*. 6d. per quarter, the rate 
 of profit may be 25 per cent. 1 
 
 12. A. invests one-third of his money in the 2^ per 
 cents, at 90, the other two-thirds he lends to B. at 4 per- 
 cent, simple interest. B. pays the interest duly for 4 
 years ; at the end of another 4 years he becomes bankrupt 
 and pays a dividend of 13s. 4c. in the pound to A. on the 
 principal and interest owing. At the same time A. sells 
 out his stock at 80, and then finds that the whole sum 
 received during the 8 years as principal and interest is 
 10 less than that with which he began. What was his 
 original amount ? 
 
 13. Divide 2259 10s. Id. amongst four persons, 
 giving to the second half as much again as to the first, to 
 the third one- third less than to the first and second to- 
 gether, and to the fourth one-third less than to the 
 second and third together. 
 
 14. If by delaying an investment in 2^ per cent, stock 
 while the price paid falls from 90 to 89 1, the annual in- 
 come to be received is increased by 6 5s., what is the 
 sum invested ? 
 
 15. A. sells gold lace costing 375 per piece to B. at 
 518 15s. per piece, and gives him 9 months credit. 
 At the same time B. sells silk to A. which has cost him 
 125 per piece, and gives him 6 months credit. 
 
 If the two invoices are made out for the same amount, 
 at what price per piece must B. invoice the silk, in order 
 that, at the moment of sale, the total amount of A.'s 
 profit may equal that of B. ; the present values of the 
 invoices being determined upon the supposition that the 
 interest of money is 5 per cent, per annum 1
 
 CIVIL SERVICE EXAMINATION PAPER. 35 
 
 SOLUTIONS. 
 
 1 307G92 V 1*28571 3 4188 y 1 47fi 1 9 
 * ~9"5~ 9"~9 9 ~9~ "o~ff & o" ~jr ^s" ^ l 1 l 1 I" T" 111111" 
 
 = 65*3 x If ; hence ^ of f| x y = ^ 
 
 = -142857$ -T- 13 = -142857 -t- 13 
 = -610985. Ans. 
 
 Note. A candidate would be required to show, on 
 some space apart from the direct solution, the work of 
 finding greatest common measures for the reduction of 
 the two fractions on the right of the first line. We sub- 
 join the work : 
 
 Strike out 4)34188 ; I Strike out 7)111111 
 
 G. C. M. 8547)111111(13 G. C. M. 15873)47619(3 
 
 2. 8 sq. po. 19 1 yds. -7- 7jf sq. ft. 
 
 9 
 
 i. yds. -T- sq. yds. = 523 x 
 
 72 x 9 
 = 36 x 9 = 324 plots. Ans. 
 
 3. Since 12 tons of C. = 9 gall, of B. + f cwt of S., 
 dividing throughout by 27 shows that 
 
 ton of C. = i gall, of B. + ^ cwt, of S., 
 
 I + 5o> or i cwt - f S., 
 
 .*. 10 tons, or | x f x 10 tons, of C. = i x | x 10 cwt. of S. 
 = 45 cwt. -j- 8 = ojj cwt. of S. Ans. 
 
 4. ^/ 125971 2 = 108, the number of feet for each side 
 of the floor, =36 yards ; 
 
 hence, length of carpet required = 36 yds. x 36 -i- f 
 = 144 yds. x 12 at 40d. = 144 at iO^s. 
 = 4 1*. x 72 = 291 12*. Ans. 
 
 Note. We annex the work 
 
 of extracting the cube root. It io 2 x 300 11259712(108 
 is easy to see that the first two = 30000 1000 
 
 figures of the root will not ex- 308 x 8 = 2464 259712 
 ceed 10, which we therefore 32464 259712 
 
 write at once. 
 
 D2
 
 SHORT METHODS IN ARITHMETIC. 
 
 5. 
 
 10 ft. 9 in. 
 5 2 
 
 4' 
 6 
 
 8 
 8 
 
 12125 -r 56| T 
 12125 x 12 
 
 i 
 x 2 
 
 53 10 
 1 9 
 6 
 
 8 
 6 
 4 
 
 12125 x 12 
 
 56 1 
 12 
 iQ\ sq. in. = 
 
 T 1 
 
 >a. Ans. 
 
 an 
 
 = 12x2x9 = 
 
 216 
 
 6. I of the difference referred to is >^f and <ff ; 
 
 .*. the difference is > \- ~- 4, that is, > 1^, 
 /. the fraction itself is >(1^ + ), or >2 ; 
 but the difference is <ff -f- |, that is <lf, 
 .*. the fraction itself is <(lf + f), or <2J ; 
 Hence, the fraction lies between 2 and 2^. Ans. 
 
 7. The recurring form of highest value is '2421, and 
 that of lowest value is '242 1. 
 
 242124212421 + &c. 
 242111HU11 + &c. 
 difference, -000013101310 + &c. = -0000131, 
 
 8. (i.) V 25400544 senary = 4112 sen. Ans. 
 
 24 
 
 121 140 
 121 
 
 1221 jloOS 
 il221 
 
 12222 24444 
 124444 
 
 (ii.) The given quantity = 3065^^ octonary; and 
 now -we shall convert each of the three octonary numbers 
 to the denary or common scale ; 
 
 3065, 
 8 
 
 263, 
 
 8 
 
 21 
 8 
 
 8 
 1589 
 
 179 
 
 1000; 
 
 8 
 
 8 
 8 
 
 64 
 
 8 
 
 512
 
 CIVIL SERVICE EXAMINATION PAPER. 
 
 37 
 
 the given quantity, therefore, is equal to 1589 1-{ | denary ; 
 and the fraction here may be reduced to the decimal 
 form by dividing by 8x8x8, successively, as under ; 
 then, prefixing the integral part to the result gives the 
 answer. 
 
 8)179 
 
 8)22-37fi 
 
 (iii.) 
 
 1589-319609375. Ans. 
 =-fQ duodenary. Ans. 
 
 263 octonary 
 10 
 
 3-376 
 4-754 
 9-470 
 6-060 
 0-740, X-c. 
 
 Xote. The octonary radix fraction 
 might be reduced to the required form of 
 tenths, hundred ths, thousandths, &c. by 
 multiplying the fractional part succes- 
 sively by 10, observing to divide the pro- 
 duct of each figure by 8 as we go along. 
 We annex only a few steps of this process, 
 inserting the multiplier only once. The figures pre- 
 ceding the point are those of the required decimal. 1 
 
 9. Here we have to find what power of 1*1 first 
 exceeds in amount the number 2 ; and this is here meant 
 to be done by trial, the successive amounts of 1 at the 
 end of each year being calculated as follows : 
 
 1st vear, 1-1 
 ik- -11 
 
 2ml 
 
 1-610 
 i= -161 
 
 121 
 
 3rd l^SSl 
 133 
 
 4th 1-464 
 146 
 
 6th year, 1-771 
 177 
 
 7th 1-948 
 195 
 
 8th 2-143 
 Ans. In the 8th year. 
 
 5th 1-610 
 
 Xote. The general form of solution for this problem 
 is that which determines the value of x so that l'l c may 
 
 1 Sef the author's ' Supplementary Arithmetic,' chap. vi. 5.
 
 38 SHORT METHODS IN ARITHMETIC. 
 
 be equal to a given quantity. The solution is worked by 
 means of logarithms. 
 
 Here we have 1 -\ x = 2 + , to find x. 
 
 XT log 24- log 2+ 30103 + 
 
 Now x = = ~s = . 
 
 log 1-1 log 11-1 4189 
 = more than 7. Hence, in the 8th year. Ans. 
 
 10. The differences of time indicated will be 16 
 seconds the 1st day, 32 seconds the 2nd day, 48 seconds 
 the 3rd day, and so on ; and this series will grow to 30 
 minutes, or 1800 seconds, in 1800^16 = 112^ days of 24 
 hours : which is the first Answer. 
 
 Again : At the end of 225 half-days the slower clock 
 will indicate 3J seconds x 225 before 5 A.M. =(675 + 
 168|) sec.=843| sec. = 14 min. 3| sec. to 5 A.M.; and, 
 therefore, the faster clock will indicate 30 min. 14 min. 
 3| sec., or 15 min. 56^ sec. past 5 A.M. : which ate the 
 other Answers. 
 
 11. Prime cost of the mixture=|<$ of 57^s. = 230s. 
 -4-5= 46s. 
 
 The two qualities cost 39s. and 48s. per quarter, 
 respectively ; and these rates are, respectively, 7s. below 
 and 2s. above the average rate ; therefore the proportions 
 mixed must have been 7 of the better quality to 2 of the 
 inferior. Ans. 
 
 12. For every 90 that he invests in stock he lends 
 180; together=270. 
 
 Now, for the 90 invested he gets 2^x8., = 20 
 of dividends, and 80 by selling out, which are together 
 =100. 
 
 For the loan of 180 he gets 
 
 J 5 8 o of 180 + 1 of l T \f 5 of 180, 
 = 28-8t-l^of 120, 
 = 28-8 + 139-2 = 168; 
 
 his total return?, therefore, are 100 + 168=268 for an 
 investment of 270 ; and the question now is, if on 270 
 he loses 2, on what sum does he lose 10 ? 
 
 Ans.
 
 CIVIL SERVICE EXAMINATION PAPER. 39 
 
 13. The shares are as 1, H, If, and 2-^, or as 18, 27, 
 30, and 38. Hence the first person will have ^\\ of 
 2259 10s. Id , the second -ffa of that sum, the third 
 T 3 T %, and the fourth 
 
 113)2259 10s. 7<2.( 19 19s. lid. 
 
 113 _ 3 
 
 1129 59 19 9 = 
 
 1017 359 18 6 = 3 
 
 112 539 17 9= 
 
 "2250(19 59917 6j 
 
 113 1499 13 9 
 
 1120" 2259 10 7 
 
 1017 759 16 10 
 
 1243(11 
 1243 
 
 Hence the shares are 359 18*. &d.~\ 
 
 539 17 9 I A 
 599 17 6 ( 4<1 ' 
 759 16 10 J 
 
 14. The rates of income per pound of investment are, 
 
 j>2-L 2 
 
 respectively, ^- and | or ^ and -1^5', hence 
 
 no - ^ po un <3s invested^6^ or, multi- 
 plying throughout by 36, we have 
 
 (i"_l) of the sum invested = 225 ; 
 
 175. Ans. 
 
 15. Let us deal with one piece as given by A., and 
 find the corresponding quantity given by B. 
 
 The amount of A.'s invoice is 51 8|; that of B.'s is 
 the same ; but at the moment of sale the present worth 
 of A.'s invoice is that of 518| for 9 months at 5 p*-r 
 
 cent, per annum, viz. , or |^, of 
 
 = -fgof 
 and the present worth of B.'s invoice at the same mo-
 
 40 SHORT METHODS IN ARITHMETIC. 
 
 ment is that of 518J for 6 months, viz. j-, or AJ of 
 
 2 -V- 5 
 
 = Jf of 2075 = 2 -j 7 I ! L ; 
 
 A.'s present gain is 500 minus 375 =125; B.'s 
 present gain is also 125, which must be equal to 
 2 -^"- minus whatever the proportionate quantity of 
 silk, at 125 per piece, cost him; hence 20 ? y J1 125 
 is the prime cost of B.'s no. of pieces at 125 per piece ; 
 so that B.'s proportionate quantity is equal to 
 
 ( 2 --Y-P- 125) -r 125 = W ~ l = '-a pieces ; 
 and therefore he invoiced the silk at 518f -r 1 ^ 5 
 
 5] 8 * x 8 x 41 = 4-15 x 41 = 170 3g. Am. 
 1UOU 
 
 CHAPTER VIII. 
 
 MISCELLANEOUS QUESTIONS. 
 
 1. Find the highest common factor in 4389 and 4218. 
 
 2. What would be the cost of 19 tons at 1 Is. 10>rJ. 
 per cwt. 1 
 
 3. Find the cost of 23 cwt. at 6 11s. 8rf. per ton. 
 4 Calculate 75 times 239, and the 75th part of 
 
 1426054. 
 
 5. Multiply 6834 by 5397, and 1368 by 99. 
 
 6. Calculate 460 things at 1 14s. 9rf., and 365 at 
 2 15s. 4W- 
 
 7. Multiply -01236 by 6-321, and 4567 by 125. 
 
 8. Find the greatest common measure of 39780 and 
 370188. 
 
 9. Calculate 289 cwt at 16s. 10^., and 193^ cwt. at 
 12s. 3%d. 
 
 10. Multiply 732-84 by 4789, and 578 by 75. 
 
 11. A person's yearly income is 225 13s. : how 
 much is that per day 'i 
 
 12. Calculate 365 days at 9s. ll^rf. per day.
 
 MISCELLANEOUS QUESTIONS. 41 
 
 13. Find the G.C.M. of 15987 and 13505. 
 
 14. Find the amount of 25 times 9999 X 99, and 
 that of 999 times 674 x 75. 
 
 15. Calculate 310 Ibs. at 8*. 5d., and 477 at 6s. I$d. 
 
 16. Divide the product of 8526 and 3794 by 125. 
 
 17. What would an allowance of 4s. 5^d. a day 
 amount to in the year 1884 ? 
 
 18. Divide the squire of 109'26 by 365. 
 
 19. Reduce 27365 gallons to cubic inches. 
 
 20. Calculate 47 yards at 20-Jd., and 53 at 19& 
 
 21. Find the continued product of 2968, 2809, and 
 7926, and the quotient of 7792 divided by 75. 
 
 22. Calculate -^ of ^ of 23791. 
 
 23. Find the G.C.M. of 21669 and 27028. 
 
 24. The product of two numbers is 294426, and the 
 square of one of them is 9801 : Find the other. 
 
 25. Calculate 87 things at 13*. 2$d. 
 
 26. Divide 23706J, by the 6th part of 365. 
 
 27. Find the simple interest on 217 14s. lOdf. for 
 230 days, at 6f per cent, per annum. 
 
 28. What is the amount of 93 times 3 6s. 9Jd., 
 and that of 511 times 37s. 8d. ? 
 
 29. Multiply 4 times 123456789 by ^th of 625. 
 
 30. Work out 147321 x 123741, and 119| X 25. 
 
 31. When 15 16s. 9rf. is the interest on 396 for 
 a year, how much is a year's interest on 428 1 
 
 32. What is the highest common factor in 9729, 
 13113, and 113646] 
 
 33. Find the simple interest on 1385 15s. llfZ. for 
 118 days at 2i per cent, per annum. 
 
 34. Calculate 3 years' compound interest on 
 364 15*. 6rf. at 4 per cent, per annum. 
 
 35. The volume of a sphere is -5236 of the cube of 
 its diameter : Find the volume of a sphere of which the 
 diameter is 7'32 feet. 
 
 36. Find the amount of 215 yards at 2s. 5|d., and 
 that of 1 1 7f stone at Is. Gd. 
 
 37. A rectangular cistern is 1 1 feet long and 7 feet 
 wide, and its capacity is 3165 gallons : What is its 
 depth 1
 
 42 SHORT METHODS IN ARITHMETIC. 
 
 38. Find the simple interest on 74 3s. !(%/. for 1 
 year 179 days at 3 per cent, per annum. 
 
 39. What is the value of 117^ cwt. of cheese, at the 
 rate of 391 for 99 cwt.? 
 
 40. Calculate 504 things at 7s. S^d., and 154 at 
 13s. 6*d. 
 
 41. Find the greatest common measure of 92463, 
 156604, and 131495. 
 
 42. Calculate 1821 yards at ll|d.,and 76 at Is. lOjfd 
 
 43. Find the amount of ,1236 18s. by compound 
 interest for 4 years, at 5 per cent, per annum. 
 
 44. If 163 ounces of tea be worth as much as 365 
 ounces of coffe% find, in ounces and the decimal of an 
 ounce, what quantity of tea is worth as much as 217 
 ounces of coffee 1 
 
 45. Reduce to a decimal, accurate to five places 
 
 _L + _ * h _ _ + &c.i 
 
 2.5 2.5.8 2.5.8.11 
 
 46. Divide the product of 365 and 563 by 99. 
 
 47. Find the simple interest on 415 17s. 3</. for 
 343 days at 2^ per cent, per annum. 
 
 48. Calculate 365 Ibs. at 7ft?., and 149^ yards 
 at Q^d. 
 
 49. Convert 89102 cubic inches to gallons. 
 
 50. A reservoir holds 5617 gallons of water : What 
 is the weight of the water in tons, at the rate of 1000 
 ounces per cubic foot ? 
 
 51. What will 159 4s. Id. amount to, by compound 
 interest at 6 per cent, per annum, in 3 years ? 
 
 52. The produce of a field of 11 '27 acres being 365 
 bushels of grain, what extent of it yields 42 quarters? 
 
 53. Find the G.C.M. of 2002, 3198, and 4745. 
 
 54. Simplify ||| of 100123 of 4768406. 
 
 55. How many gallons of water will weigh a ton, if 
 a cubic foot of water weigh 1000 ounces ? 
 
 56. From 365 times 56493 subtract ^th of 394650. 
 
 1 Points placed like a full stop between figures denote multi- 
 plication.
 
 MISCELLANEOUS QUESTIONS. 43 
 
 57. Calculate 63| yai-ds at \$\d. per dozen yards, 
 and 35 1 yards at 17|^7. per dozen yards. 
 
 58. Find the amount of 93 Is. 5d. by compound 
 interest in 4 years at 3^ per cent, per annum. 
 
 59. Find the simple interest on 251 5s. from 
 June 15, 1883 to August 27, 1884 at 3^ per cent, per 
 annum. 
 
 60. A rectangular field is 896 links long and 479 
 broad : How many acres (each 10000 square links) does 
 it contain 1 
 
 61. Find a 4th proportional to 75, 381-52, and 7'385. 
 
 62. What is the value of 4 J of 48. 6s. 1 
 
 63. When 36^ yards of si'lk are worth 9 14s. 
 what is the value of 276 yards ? 
 
 64. Calculate 137f ounces of gold at 3 17s. 
 and 52^ ounces of silver at 4s. lO^cZ. 
 
 65. What is the compound interest on 1056 lls. Id., 
 for 3 years, at 3^ per cent, per annum ? 
 
 66. Calculate 21 4| yards at 3s. 4 T 7 ^. per dozen 
 yards, and 147 yards at 5^d. per dozen yards. 
 
 67. A cubical cistern can hold 4653 gallons : Find 
 its depth in feet. 
 
 68. The interest on a certain sum of money in 225 
 days was 30 16s. 5^d. : How much interest did it bear 
 in 193 days? 
 
 69. What principal would make 13 10s. more by 
 compound than by simple interest in 2 years at 5 per 
 cent, per annum ? 
 
 70. Multiply the double of 163853 by 49|f. 
 
 71. Calculate 26 yards at ll\id., and 7lf at Is. 5fd. 
 
 72. The diameter of a cylindrical tank is 27 feet 
 7 inches, and its depth 15 feet 11 inches : How many 
 gallons of water can* it contain? [Area of a circle 
 = '7854 of that of its circumscribing square.] 
 
 73. A bookseller buys 68 copies of a book published 
 at 6s. Qd. He is allowed 25 per cent, discount 011 the 
 publishing price, and is charged for the 68 copies the 
 net cost of 63. At what rate per cent, of discount on 
 the publishing price can he retail the book, and make a 
 total profit of 1 12s.
 
 44 SHORT METHODS IN ARITHMETIC. 
 
 74. Calculate to five decimal places the value of 
 
 2 + I + 3J, + i + 3-5X9 *> 
 
 75. When the base of an isosceles right-angled tri- 
 angle is 99, each of the equal sides is almost exactly 70 : 
 Hence find each of the equal sides in a similar triangle 
 whose base is 23'456. 
 
 76. Suppose a piece of metal 5'12 inches long, 3'75 
 inches wide and 2 '43. inches thick, to be melted, without 
 loss, into a cube : What would be the length of the edge 
 of the cube 1 and what difference would be made in the 
 amount of surface ? 
 
 77. When 1595 yards of calico cost as much as 427 
 yards of linen, and 1308 yards of flannel as much as 
 1025 yards of linen, find, to two places of decimals, the 
 quantity of calico that costs as much as 365 yards of 
 flannel. 
 
 78. A rectangular tank is 32 feet long and 18 feet 
 8 inches wide : What will be the depth of water in the 
 tank when the weight of the water, at 62^ Ib. per cubic 
 foot, is 56 tons 5 cwt. ? 
 
 79. Suppose the rate of a clock to be '05 per cent, 
 too fast, and to continue at that rate from July 1 to 
 December 31 ; how much time would it gain in the 
 half-year 1 
 
 80. A gas-meter, after being used for 96 days, regis- 
 ters 11280 cubic feet as the consumption. The meter is 
 then tested, and found to register 17^ per cent, in excess 
 of the gas actually consumed. Find the amount that 
 ought to be charged, at 3s. Qd. per thousand cubic feet. 
 
 81. How many cubic inches of copper are equal in 
 weight to 963 cubic inches of iron, if 4214 cubic inches of 
 copper weigh as much as 3257 of lead, and 1460 of iron 
 as much as 986 of lead ? 
 
 82. When the side of an equilateral triangle is 40 
 inches, its area is 693 square inches; hence find the side 
 of an equilateral triangle of which the area is 1171 - 24 
 square inches ; the areas of similar triangles being as the 
 squares of their corresponding sides.
 
 MISCELLANEOUS QUESTIONS. 45 
 
 83. The floor of a rectangular room, 16 feet high, is 
 twice as long as it is wide, and requires 60^ square yards 
 of carpet to cover it : What will the painting of its walls 
 come to at 3s. 5^d. per square yard, covering allowance 
 for windows and fireplace ? 
 
 84. If I invest my money in Dock shares paying 7 
 per share, when the 100 share is at 122*, I find that I 
 get ;t34 10*. 6c7. a year more than if I invest in 5^ India 
 Bonds at 105 : What is my capital ? 
 
 85. At what price per gallon must I invoice my sales 
 of brandy that cost me 1 7s. id. a gallon that I may have 
 12i per cent, profit after allowing 2^ per cent, discount? 
 
 86. The population of a country increases 3 per cent, 
 annually, but is lessened annually by emigration to the 
 extent of J per cent, on the whole : What will be the 
 increase per cent, at the end of three years 1 
 
 87. When each of the equal sides of an isosceles 
 right-angled triangle is 99, what is its altitude ? 
 
 88. Reduce the nonary fraction -j^VrVTr * denary 
 terms, then to lowest denary terms, and transform the 
 latter result to express lowest nonary terms. 
 
 89. The three sides of a triangle are AB 25, BC 29, 
 and AC 36 : Find its area, and the length of BD, a per- 
 pendicular on AC. 
 
 90. The product of three numbers is 3057^|; the 
 greatest of them is 1 7f ; and of the others the greater is 
 not more than 3 T J of the other, nor less than 2^ of it : 
 Find the limits within which the least must lie. 
 
 91. I buy cloth at 13*. Qd. a yard, and am allowed 
 3 months' credit; and I sell it immediately at 14s. 8d., 
 allowing a certain term of credit, and find that, rating the 
 interest of money at 5 per cent, per annum, the value of 
 my gain at the moment of sale is 9^ per cent. How 
 long credit do I give ? 
 
 92. With the earth excavated to form a uniform 
 trench, 53^ feet long and 12^ feet wide, an embankment 
 is made of 5136 cubic feet. Supposing the earth to have 
 been increased 6^ per cent, in bulk by the removal, what 
 is the depth of the trench ] 
 
 93. A mixture of 25 gallons of spirits costs me
 
 46 SHORT METHODS IN ARITHMETIC. 
 
 26 11s, 3cZ., and is a blend of two qualities, for one of 
 which I paid 25s. 3d. a gallon, and for the other 19s. a 
 gallon : How many gallons were there of the better 
 quality ? 
 
 94. Two unequal circles touch each other, the area of 
 the larger being thrice that of the smaller : If the" distance 
 between their centres be 34'15 inches, what is the diameter 
 of each ? 
 
 95. Extract the square root of 2523'4323 in the scale 
 of seven, working throughout according to septenary 
 notation, and represent the fractional part of the answer 
 as a septenary radix fraction. 
 
 96. If 27132 be the area of a rectangle of which the 
 diagonal is 365, what is the area of a similar rectangle of 
 which the diagonal is 132 1 
 
 97. An alloy of silver is mixed with an alloy of gold 
 in the proportion of 11 '4 to 2 - 6. The percentage of 
 dross in the silver alloy is 13 5, and in the gold 17'35 : 
 Find the percentage of dross in the mixture. 
 
 98. A cistern 5 feet long and 3 feet 6 inches wide, 
 filled with water, loses '0005 of its depth of water by 
 evaporation, the loss being found to amount to a quart of 
 water : What is the depth of the cistern 1 
 
 99. A. and B. together can perform a piece of work 
 in 77 hours : In what time could each do it alone if A. 
 could do B.'s hourly amount of work in "9604 of the 
 time in which B. could do A.'s hourly amount of work ? 
 
 100. A. lends B. a certain sum ; at the same time he 
 insures B.'s life for 737 12s. 6cZ., paying annual pre- 
 miums of 20. At the end of three years, and just 
 before the fourth premium is to be paid, B. dies, having 
 never repaid anything : What must A. have lent B. in 
 order that he may just have enough to recoup himself, 
 together with 5 per cent, compound interest on the sum 
 lent and 011 the premiums ?
 
 APPENDIX. 47 
 
 CHAPTER IX. 
 
 APPENDIX OF ADDITIONAL PROBLEMS IN HIGHER 
 ARITHMETIC, WITH SOLUTIONS. 
 
 PROBLEMS. 
 
 1. The square root of a vulgar fraction is equal to 
 7231, very nearly, its denominator being 153 : Find its 
 numerator. 
 
 2. Find three fractions of which the sum is |, the 
 first being equal to 72 per cent, of the second, and the 
 second equal to ^ of the third. 
 
 3. A. bought property for ^1041 13s. 4d., and sold 
 it to B. at a certain rate of profit ; B. sold it to C. at the 
 same rate of profit, and C. sold it to D. for 1105 Ss. Qd., 
 gaining still at the same rate : What was that rate ? 
 
 4. The sum of the circumferences of two circles touch- 
 ing each other is 99 inches : What is the distance between 
 their centres'? [The diameter of a circle is '31831 of the 
 circumference.] 
 
 5. Reduce to an undenary improper fraction, in 
 lowest terms, the undenary expression 13'864l, working 
 throughout according to undenary notation, [t is used 
 as a numeral for ten.] 
 
 6. Three trees, A, B, C, stand on a horizontal plane, 
 C being exactly east from A, and south from B ; more- 
 over, C's distance from B is 37 '86 yards, and A is 3'65 
 yards farther from B than from C : Find the distance 
 from A to C. 
 
 7. If the areas of the squares described on the three 
 sides of a triangle be 5, 9, and 20 square inches, what 
 must be the exact area of the triangle I 
 
 8. Three bricklayers, A., B., C., build a wall. The 
 whole would be built by A. alone in 14 days, or by B. in 
 18 days, or by C. in 21 days. They all begin together, 
 and A. and B. continue till the wall is finished ; but C.
 
 48 APPENDIX. 
 
 leaves off a day before the completion of the work : How 
 many days does the work last 1 
 
 9. There are two cubical water tanks, the larger of 
 which can hold 3192 gallons. The supply pipe of the 
 smaller delivers 27|- gallons of water per minute, that of 
 the larger 70 gallons per minute. If the tanks are empty, 
 and the supply pipes are set open together, both tanks 
 will be filled at the same time : What must the depth of 
 the smaller tank be if 11 gallons of water occupy 3050 
 cubic inches ? 
 
 1 0. A certain amount of digging is done by A . and 
 B. together in 4 days : What time would each take to do 
 the whole work by himself, if the time that A. would 
 require, in order to do as much as B. does in a day, is to 
 the time that B. would require, in order to do as much 
 as A. does in a day, as 95'663 is to 99 1 
 
 11. A piece of metal weighing 12 cwt. 66 Ibs. has 
 been formed by compounding three metals in quantities 
 which, by measure, are as 5 : 3 : 2 ; but the weights of 
 equal volumes of them would be as 7 : 11 : 13. What 
 weight of each of the component metals has been used ? 
 
 12. A merchant, having bought 400 tons of coal, 
 reckons that by selling them at 17s. Q^d. per ton he will 
 make 5^ per cent, on his outlay ; after selling 300 tons 
 at that rate, he disposes of the remainder at a price which 
 reduces his profit on the whole to 5 per cent. Find (1) 
 what he gave for the 400 tons of coal, and (2) at what 
 price per ton the second lot was sold. 
 
 1 3. A vessel making for a harbour fires a signal gun ; 
 the Bash is seen from the harbour, and the sound follows 
 in 22 J seconds; a tug puts off immediately and steams in 
 a straight coarse towards the vessel at the rate of 12 
 miles an hour ; and from the tug, five minutes afterwards, 
 the flash of a second gun is seen, the sound of which 
 follows in 15 seconds. If sound travels 13 miles per 
 minute, at what rate is the vessel approaching the har- 
 bour, and how soon after starting will the tug meet her ?
 
 ADDITIONAL PROBLEMS AND SOLUTIONS. 49 
 
 SOLUTIONS. 
 
 7231 2 = the fraction = numerator -f- 153 ; 
 
 7231 -52287 
 
 3 153 
 
 156861 x 5 
 784305 
 
 52287361 79'99911 = 80. Ant. 
 
 2. The 1st is to the 2nd as 72 is to 100 ; 
 
 the 2nd is to tjie 3rd as 100 is to 100 -=- ; 
 .-. the fractions are as T2, 18&, 2334, 
 or as 216, 300, 700, or as 54, 75, 175 ; 
 and the sum of these proportional parts is 304 ; 
 
 now -g^f of |-p = ^27 ; which severally multiplied 
 
 by 54, 75, 175, gives |, ^, *|. Arts. 
 
 3. 1041| multiplied by R 3 = 1105-425 ; 
 
 1105-425 x 12 _ 132651 = 1061208 . 
 
 1041f x 12 = 1250UO 100UUUO ' 
 
 ^1-061208 = 1-02 = R ; hence 2 per cent. Ans. 
 
 4. The sum of the diameters is '31831 of the sum of 
 the circumferences ; .*. sum of the radii, or distance be- 
 tween centres 
 
 = of 99 x -3183100 
 31831 
 
 = $ of 31-51269 in. = 15-756 in. Ans. 
 
 5. The expression is =13 undenary and 
 
 99*99 
 
 now we can find the G.C.M. of numerator and denomin- 
 ator to be 2469 undenary, as worked out below : 
 
 8641* 
 
 ttttt 
 H541 
 
 1 G.C.M. = 2469 
 
 8641* 
 7295 
 1357* 
 as not being a factor in 864 If. f 
 
 24690, from which strike out 10, 
 
 36 
 
 The G.C.M. thus found reduces the expression to 
 undenary = Vc : ' undenary. Ans. 
 

 
 50 APPENDIX. 
 
 6. The distances between the trees form a right- 
 angled triangle, in which AB is the hypotenuse. 
 
 BC 2 = the difference of the squares on AB and AC = 
 the product of the sum and difference of AB and AC = 
 (AB + AC) x 3-65 ; /. BC 2 -j- 3'65 = the sum of AB 
 and AC. 
 
 37-86 143338 x -002 
 
 37-86 3 
 
 22716 x 3 10 
 
 68148 x 2 10 
 
 136296 
 
 1433-3796 10000 
 
 286-676 
 95-558 
 9-556 
 956 
 
 392-746 
 040 
 
 or 1433-38 = BC 2 AB + AC = 392-706 
 
 1433-38 -H 3-65 AB - AC = 3-650 
 
 = 143338 -J- 365 2)389-056 = 2 AC 
 
 Ann. AC = 194-528 yds. 
 
 7. The three sides are equal to \/5, 3, and 2^/5; and 
 the half-sum of the sides is equal to li\/5 + 1| ; but to 
 avoid fractions, let us first find the area of a triangle four 
 times as large as the one given; then, the sides are to be 
 taken twice as great as those given, viz. 2\/5, 6, and 
 4>/5; the half-sum is now 3\/5 + 3; and subtracting 
 from this the three sides severally, and proceeding 
 according to the well-known Rule, we have the quadruple 
 area 
 
 - A/{(3+ A/5)(3--v/5)(3-/5-3X3y5 + 3J- ; 
 
 and here recognising, as in the preceding solution, that 
 the sum of two quantities multiplied by their difference 
 produces the difference of the squares of the quantities, 
 the expression for the quadruple area becomes 
 
 A/(9-5)(45-9) = A/4T36 = 12 ; 
 
 hence the area of the given triangle is exactly 3. Ans. 
 
 8. Suppose the whole work to consist of 126 equal 
 measures, that number being an exact common multiple 
 of 14, 18, arid 21 ; A. did 9 of these measures per day, 
 B. 7, C. 6. Now, if C. had not left off a day before the 
 completion of the 126 measures, he would have added six 
 extra measui'es, making 132 measures to have been done
 
 ADDITIONAL PROBLEMS AND SOLUTIONS. 51 
 
 in the required time at the rate of 9 + 7 + 6 or 22 measures 
 per day ; hence 132 -r- 22 = 6 days. Ans. 
 
 9. The smaller tank holds fL or ft of 3192 gallons 
 
 3192 
 = ^ of 11 gallons = 114 times 3050 cub. inches 
 
 = 347700 cub. in , the cube root of which is 70-318 in. 
 = 5-86 feet. Ans. 
 
 10. 95-663 -f- 99, or -966293 of the time in which 
 B. can do A.'s daily amount 
 
 is equal to the time in 
 
 + -0000956 , ? A , -r> , j -i 
 
 + -0000010 which A. can do B. s daily 
 
 amount. 
 
 Now, if A.'s daily amount of work be called a measures, 
 and B.'s be called b measures, B. could do A.'s daily 
 
 amount in - of a day, and A. could do B.'s daily amount 
 
 in _ of a day : hence - = '966293 of - 
 a a b 
 
 966293 
 
 = ( * V ; /. - = ^-966293 = -983; 
 V a / a 
 
 accordingly, A.'s daily amount of work is to B.'s as 1000 : 
 983 ; and as they jointly do ^ of the whole per day, 
 
 1000 f 983 
 
 A. does of J and B. of per day ; 
 
 /. A. would do the whole in 198 x = 7-932 days, 1 . 
 
 1000 I AM 
 
 B L- =8-069 days, 
 
 1 1 . There were 5 units of volume each 7 units of weight, 
 Q 1 1 
 
 J5 ** > 
 
 and 2 13 
 
 = 35 + 33 + 26 = 94 units of weight ; 
 hence, ^| of 1410 Ibs. = 15 Ibs. x 35 = 525 Ibs. 
 $$ =15 Ibs. x 33 = 495 Ibs. 
 
 |f =15 Ibs. x 26 = 390 Ibs. ; 
 
 or the weights used were 4 cwt. 77 Ibs., 4 cwt. 47 Ibs. 
 and 3 cwt. 54 Ibs. Ans. 
 
 E2
 
 52 APPENDIX. 
 
 12. (i.) Gave for the whole $%$ k of 17s. 6Jd. X 400 
 =-|-Sf of % 2 J x 20 = $ of 1000 = 333^. Ans. 
 
 (ii.) Bought 400 tons for 333 6s. Sd. 
 Gain on the whole 2 V == 16 13 4 
 Sold 400 tony for 350 
 
 300 at 17s. 6^. =263 2 6 
 .'. 100 sold for 86 17 6 
 
 which is 17s. 4^o?. per ton. Ans. 
 
 13. Sound being supposed to travel -^ of a mile per 
 second, the tug is at first ^f miles X *-/ = \ 9 miles, or 
 4| miles from the vessel. 
 
 The tug sails 1 mile in 5 minutes, and is then 
 ;'{- miles x 15 = 3i miles from the vessel; therefore 
 the vessel sailed 4|- 3i or If miles in 5 minutes, which 
 i* at the rate of 19 miles an hour. Ans. 
 
 Secondly, at starting there were 4|- miles to be sailed 
 over between the vessel and the tug, at the respective 
 rates of 19 1 and 12 miles an hour, 31^ miles an hour 
 together ; therefore the tug would meet the vessel in 
 * J r j r ~sT f an hour, = 65 min. -=- 7 = 9f inin. Ans.
 
 53 
 
 ANSWERS TO THE EXERCISES. 
 
 Exercises 1. 
 
 1. 
 
 93. 
 
 
 2. 92. 
 
 3. 199. 
 
 4. 95. 
 
 
 5 
 
 365. 
 
 6. 
 
 73. 
 
 
 7. 41. 
 
 8. 43. 
 
 9. 67. 
 
 
 
 
 Exercises 
 
 2. 
 
 
 
 
 1. 
 
 en 
 
 
 n 12 
 
 3. 
 
 l'- 
 
 4 13. 
 
 
 
 5. !. 
 
 
 149' 
 
 
 0. ,, 3 . 
 
 
 J.V 
 
 * TT* 
 
 
 
 
 6. 
 
 54 
 59* 
 
 
 7 Si 
 ' 42' 
 
 8. 
 
 II 
 
 K.l 
 
 Q IT 
 J. 7 j. 
 
 10. |f. 
 
 11. 
 
 115 
 349' 
 
 
 12. 23 . 
 
 
 
 
 
 
 
 Exercises 
 
 3. 
 
 
 
 
 
 
 
 s. 
 
 d. 
 
 
 
 
 S. 
 
 d. 
 
 
 
 
 s. 
 
 d. 
 
 l. 
 
 6 
 
 11 
 
 7 
 
 9. 
 
 18 
 
 8 
 
 3 
 
 17. 
 
 5 
 
 3 
 
 9 
 
 2. 
 
 74 
 
 19 
 
 9J 
 
 10. 
 
 11 
 
 11 
 
 3 
 
 18. 
 
 13 
 
 1 t 
 
 3| 
 
 3. 
 
 89 
 
 18 
 
 11 
 
 11. 
 
 9 
 
 7 
 
 9 
 
 19. 
 
 60 
 
 11 
 
 7 
 
 4. 
 
 15 
 
 
 
 84 
 
 12. 
 
 54 
 
 8 
 
 Of 
 
 20. 
 
 14 
 
 5 
 
 1 
 
 5. 
 
 6 
 
 8 
 
 4 
 
 13. 
 
 30 
 
 10 
 
 
 
 21. 
 
 41 
 
 10 
 
 4^ 
 
 6. 
 
 8 
 
 2 
 
 6 
 
 14. 
 
 113 
 
 18 
 
 
 
 22. 
 
 102 
 
 7 
 
 6 
 
 7. 
 
 106 
 
 17 
 
 5 
 
 15. 
 
 87 
 
 1 
 
 si 
 
 
 
 
 
 8. 
 
 17 
 
 14 
 
 4} 16. 
 
 14 
 
 15 
 
 
 
 
 
 
 
 Exercises 
 
 4. 
 
 
 
 
 
 
 
 s. 
 
 d. \ 
 
 
 
 *. 
 
 d. 
 
 
 
 
 s. 
 
 d. 
 
 1. 
 
 197 
 
 3 
 
 4 
 
 15. 
 
 431 
 
 
 
 10^ 
 
 29. 
 
 3692 
 
 10 
 
 7 
 
 2. 
 
 120 
 
 9 
 
 
 
 16. 
 
 2067 
 
 18 
 
 o~ 
 
 30. 
 
 176 
 
 11 
 
 Uf 
 
 3. 
 
 79 
 
 
 
 
 
 17. 
 
 178 
 
 12 
 
 8 
 
 31. 
 
 354 
 
 lit 
 
 71 
 
 4. 
 
 167 
 
 15 
 
 1 
 
 18. 
 
 19 
 
 
 
 2J 
 
 32. 
 
 1712 
 
 11 
 
 9 
 
 5. 
 
 279 
 
 17 
 
 2 
 
 19. 
 
 22 
 
 17 
 
 
 33. 
 
 718 
 
 13 
 
 H 
 
 6. 
 
 483 
 
 1 
 
 8 
 
 20. 
 
 237 
 
 2 
 
 2j 
 
 34. 
 
 1832 
 
 g 
 
 
 
 7. 
 
 201 
 
 14 
 
 8 21. 
 
 144 
 
 3 
 
 
 
 35. 
 
 545 
 
 19 
 
 3i 
 
 8. 
 
 197 
 
 
 
 8 
 
 22. 
 
 464 
 
 12 
 
 2 
 
 36. 
 
 1645 
 
 2 
 
 6" 
 
 9. 
 
 540 
 
 2 
 
 2 23. 
 
 118 
 
 10 
 
 9 
 
 37. 
 
 286 
 
 5 
 
 Hi 
 
 10. 
 
 269 
 
 14 
 
 j 24. 
 
 1364 
 
 17 
 
 3 
 
 38. 
 
 251 
 
 1.! 
 
 llj 
 
 11. 
 
 71 
 
 8 
 
 4f 
 
 25. 
 
 148 
 
 4 
 
 84 
 
 39. 
 
 2199 
 
 12 
 
 8 
 
 12. 
 
 82 
 
 16 
 
 1 
 
 26. 
 
 184 
 
 1 
 
 11 
 
 40. 
 
 289 
 
 - 
 
 3 
 
 13. 
 
 270 
 
 5 
 
 2 
 
 27. 
 
 435 
 
 19 
 
 10 
 
 
 
 
 
 14. 
 
 362 
 
 2 
 
 4i 
 
 28. 
 
 87 
 
 8 
 
 11A 
 
 
 

 
 54 
 
 ANSWERS TO THE EXERCISES. 
 
 Exercises 5. 
 
 s. d. 
 
 s. d. 
 
 s. d. 
 
 s, d. 
 
 1. 13 3f 
 
 7. 8 6 3 
 
 13. 4 15 
 
 19. 14 7 
 
 2. 4 12 2f 
 
 8. 9 15 11J 
 
 14. 1 3 5| 
 
 20. 3 6 3 
 
 3. 8 8 5 
 
 9. 6 9 6 
 
 15. 4 7 
 
 21. 3 2 
 
 4. 4 8 
 
 10. 1 15 lOf 
 
 16. 4 18 Si 
 
 22. 5 7 2i 
 
 5. 4 19 4 
 
 11. 7 15 7,t 
 
 17. 2 12 8| 
 
 23. 1 11 If 
 
 6. 8 11 6 
 
 12. 19 6 
 
 18. 2 1 11 
 
 24. 18 0^ 
 
 Exercises 6. 
 
 1. 15390136 
 
 
 10. 4771801368 
 
 19. 
 
 6599443356 
 
 2. 34069293 
 
 
 11. 2855587428 
 
 20. 
 
 497064491 
 
 3. 343969014 
 
 
 12. 856162450 
 
 21. 
 
 4452724500 
 
 4. 661292541 
 
 
 13. 7256584665 
 
 22. 
 
 118291774944 
 
 5. 107958378 
 
 
 14. 972118005 
 
 23. 
 
 1759792121472 
 
 6. 73905391 
 
 
 15. 18483904320 
 
 24. 
 
 723129552 
 
 7. 20581596 
 
 
 16. 50921667140 
 
 25. 
 
 16944932280 
 
 8. 23743728 
 
 
 17. 221361822 
 
 26. 
 
 13086255000 
 
 9. 1639830444 
 
 18. 2039867550 
 
 
 
 Exercises 7. 
 
 1. 75-12 lb. 
 
 
 8. 277-274 gall. 
 
 15. 65432- 16 
 
 2. 12-096 gall. 
 
 
 9. 14-15016 times; 
 
 16. 304-4137- 
 
 3. 665-664 c. ft. 
 
 
 1-886688 oz. 
 
 17. 18-0327 
 
 4. 27-373 bu. 
 
 
 10. 38s. 7-249d. 
 
 18. 
 
 7s. 9-37^. 
 
 5. -34578|; 86372 
 
 11. -0013163- 
 
 19. 832630 
 
 6. 70-9808; 
 
 
 12. 4-50776 
 
 20. 27083; 74-2 
 
 1247038 T 3 T 
 7. 6s. 7-123d. 
 
 
 13. 19s. 2$d.- 
 14. 36-0612 gall. 
 
 21. 9657-48 gall. 
 - 22. -134570 
 
 Exercises 8. 
 
 *. rf. 
 
 s. d. 
 
 
 s. d. 
 
 1. 1 6 1-13 
 
 
 4. 2 5 1-17 
 
 7. 
 
 48 10 6-3 
 
 2. 5 14 7-83 
 
 
 5. 7 10 2-955 
 
 8. 
 
 1 4 1-9 
 
 3. 2 3 10-56 
 
 
 6. 30 9 2-047 
 
 
 
 Exercises 9. 
 
 s. d. 
 
 s. rf. 
 
 
 *. d. 
 
 1. 209 12 6-24 
 
 3. 118 16 0-56 
 
 5. 
 
 1010 10 10 
 
 2. 750 13 3-96 
 
 4. 353 10 10-03 
 
 6. 
 
 57 3 3-97
 
 ANSWERS TO THE EXERCISES. 
 
 55 
 
 MISCELLANEOUS QUESTIONS. 
 
 1. 57 ' 
 
 31. 17 2*. l\d. 
 
 66. 3 0*. 5\d.; 
 
 2. 529 12*. 6d. 
 
 32. 141 
 
 5 5| 
 
 3. 7 11 5$ + 
 
 33. 11 4*. 
 
 67. 9-07 ft. 
 
 4. 17925; 
 
 34. 45 10*. lid. 
 
 68. 26 8*. 9rf. 
 
 19014-051 
 
 35. 205-368 cubic 
 
 69. 5400 
 
 5. 36883098; 
 
 feet 
 
 70. 16219791-91 
 
 135432 
 
 36. 26 15*. 3d.; 
 
 71. 1 6*. 2^.; 
 
 6. 799 5*.; 
 
 44 2 2| 
 
 5 6 li 
 
 1010 11s. 10%d. 
 
 37. 6 ft. 7-146 in. 
 
 72. 59275 gallons 
 
 1. -07812756; 
 
 38. 3 6*. 4d. 
 
 nearly 
 
 570875 
 
 39. 464 Is. 3d. + 
 
 73. 23^ p.c. disc. 
 
 8. 468 
 
 40. 194 5*. 0^. ; 
 
 74. 2-82137 
 
 9. 243 16*. W\d. ; 
 
 104 8 7| 
 
 75. 16-5850 
 
 118 19 4fd.- 
 
 41. 119 
 
 76. 3-6 in.; 3'7482 
 
 10. 3509570-76; 
 
 42. 8 10*. lO^d. ; 
 
 sq. in. 
 
 43350 
 
 7 1 8| 
 
 77. 1068-42 yds. 
 
 11. 12*. 4-37d. 
 
 43. 1503 9*. *\d. 
 
 78. 3|ft. 
 
 12. 181 14*. 9^. 
 
 44. 96-9068 oz. 
 
 79. 2 hrs. 13min. 
 
 13. 73 
 
 45. -11372 
 
 80. 36*. 
 
 14. 24747525; 
 
 46. 2075-70 
 
 81. 841-45 c. in. 
 
 50499450 
 
 47. 8 15*. 101^. 
 
 82. 52 inches 
 
 15. 130 15*. l\d. ; 
 
 48. 11 11*. lid.; 
 
 83. 30 8*. Sd. 
 
 146 11 $%d. 
 
 5 17 3| 
 
 84. 4833 10*. 
 
 16. 258781-152 
 
 49. 321-35 gall. 
 
 85. 20*. 
 
 17. 81 11*. 9d. 
 
 50. 25-148 tons = 
 
 86. 7-6418 p.c. 
 
 18. 32-70616- 
 
 51. 189 12*. 10d. 
 
 87. 70. 
 
 19. 7587603 c. in.+ 
 
 52. 10-3746 ac.- 
 
 00 -202 
 881 3310- 
 
 20. 4 1*. 3d.; 
 
 53. 13 
 
 89. 360; 20. 
 
 44 54 + 
 
 54. 4932 5 \ 
 
 90. 7 and 8| 
 
 21. 67491411792; 
 
 55. 223-36 gall. 
 
 91. 2 months. 
 
 103-89i 
 
 56. 20615958 , 7 T 
 
 92. 7-23 ft. 
 
 22. -869077 
 
 57. 8*. 6Jrf. 
 
 93. 9 gall. 
 
 23. 233 
 
 4 5i 
 
 94. 25 in. ; 43'3 in. 
 
 24. 2974 
 
 58. 107 2*. 10-84^. 
 
 95. 42-5412. 
 
 25. 57 7*. 3f<2. 
 
 59. 10 1*. 5d. 
 
 96. 3548-5. 
 
 26. 389-69 + 
 
 60. 4-292 ac. 
 
 97. 14| 
 
 27. 9 5*. 2|d. 
 
 61. 37-567 
 
 98. 4 ft. 7-015 in. 
 
 28. 310 13*. 6^.; 
 
 62. 48 3*. 4d. 
 
 99. A. 152-46 hrs. 
 
 962 7 8 
 
 63. 73 8*. l\d. 
 
 B. 155) hrs. 
 
 29. 34293552500 
 
 64. 536 7*. 3{d. 
 
 100. 580 
 
 30. 18229647661; 
 
 12 14 10 
 
 
 2981i 
 
 65. 106 7*. llirf. 
 
 
 
 Spotlisicoode & C Printert, Acir-rfrat Square, London.
 
 UNIVERSITY OF CALIFORNIA LIBRARY 
 
 Los Angeles 
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