SOLUTIONS TO THE EXERCISED] 



MATHEMATICS. ALGEBRA. 



489 



hence the horseman has ridden 8 hours, so that each 

 person must have travelled 9 X 8 = 72 miles. 



6. A person has 264 coins sovereigns and florins 

 he has 4j times as many florins as sovereigns : how many 

 of each coin has he 1 



Suppose he had x sovereigns ; then, by the question, 

 he must have had 4]x florins ; so that 



x + 41* = 264, that is, 5\x = 264, 

 I or Hi = 528, .-. x = 48. 



Consequently he had 48 sovereigns, and . . 48 X 4} = 

 216 florins. 



7. A person spends Jth of his yearly income in board 

 and lodging, }th in clothes and other expenses, and he 

 lays by 85 a-year : what is his income 1 



To avoid fractions, suppose 28x to be the number of 

 pounds he receives yearly, 28 being chosen because it is 

 divisible by both 4 and 7 : then, by the question, he 

 spends in board and lodging 7z pounds, and in clothes, 

 itc., 4z pounds. Consequently, since he lays by 85, 

 we have the equation 



1x + 4* + 85 = 28x, that is, llx + 85 = 28*. 

 Transposing, 85 = 17x, . '. x = 5, .-. 28* = 140. 

 Hence his yearly income was 140. 



8. What number is that whose third part exceeds its 

 fifth part by 72 ? 



To avoid fractions, let \5x represent the number ; 

 then, by the question, 



5x Sx = 72, .: 2x - 72, .% x = 36, 

 /. the number required is 36 x 15 = 540. 



9. I have a certain number in my thoughts. I mul- 

 tiply it by 7, add 3 to the product, and divide the sum 

 by 2. I then find that if I subtract 4 from the quotient 

 I get 15 : what number am I thinking of ? 



Let x represent the number ; then, by the question, 



7x + 3 



~5~" 



or 



7x 



= 19. 



Multiplying by 2, 7z + 3 = 38, .-. 7x - 36, .'. x = 5, 

 the number thought of. 



10. A man, 40 years old, has a son 9 years old : the 

 father is therefore more than four times as old as his 

 son. In how many years will the father be only. twice 

 as old as his son ? 



Suppose in x years : the father will then be 40 + x 

 yean old, and the son 9 + x years : by the question, 

 the former number is to be double of the latter : hence 

 the equation 



40 + x = 18 + 2z, . \ 40 18 = * = 22. 



Therefore the father will be twice as old as his son in 

 22 years ; in which time the father will be 62, and the 

 son 31. 



11. Two persons, A and B, 120 miles apart, set out 

 at the same time to meet each other. A goes 3 miles an 

 hour, and B 5 miles : what distance will each have tra- 

 velled when they meet 1 



Suppose that A has travelled x hours, then B also 

 must have travelled x hours. A must therefore have 

 gone 3x miles, and B 5x miles ; and since together they 

 must have travelled 120 miles, we have the equation 

 3x -f 5x - 120, that in, 8x = 120, .'. x = 15. 



The time occupied by each is therefore 15 hours ; so 

 that A must have travelled 15 X 3 = 45 miles, and B 

 15 X 5 = 75 miles. 



Otherwise. Suppose A goes x miles, then B goes 

 120 z miles : the time occupied by A, at 3 miles an 



hour, is therefore a hours ; and the time occupied by B, 



at 5 miles an hour, is g hours ; but the times are 



equal. Hence the equation 



x 120 x x 



5 6 =24 -6' 



Multiplying by 3, x - 72 . 



VOL. i. 



Multiplying by 5, 5x = 300 3x, .'. 8x, = 360, 



. . x = 45, A's distance. 

 . . 120 45 = 75, B's distance. 



12. Divide 250 among A, B, and C, so that B may 

 have 23 more than A, and C 105 more than B. 



Let A's share be x pounds, then B's is x -j- 23 ; and C's 

 x + 23 + 105 ; and the sum of the shares is 250 : hence 

 the equation 



Collecting, and transposing, 3x = 99, . . x = 33 : hence 

 the shares are as follow : 

 A's share 33 ; B's 56 ; C's 161 ; and their sum is 250. 



13. A can execute a piece of work in 3 days, which 

 takes B 7 days to perform : in how many days can it be 

 done if A and B work together ? 



Suppose they can do it in x days ; then since A can do 



one- third of it in 1 day, he can do ^ in x days : in like 



3 



manner B can do * of the whole in x days : hence, when 



nf ff 



working together, they do-^-\-^inx days ; but thesepetrf* 

 make up the whole work since they do it all in x days ; 



1 whole 

 3x 



* , z 



s+r 



Multiply by 3, 



by7, 7x + 3x = 21, .'. 10a; = 21,.-. x = 2J . 



Hence they complete the work in two days and one- 

 tentli. 



14. A cistern can be filled by three pipes ; by the first 

 in 2 hours, by the second in 3, and by the third in 4 ; in 

 what time can it be filled by all the pipes running 

 together 1 



Suppose it can be filled in x hours ; then since the first 



can supply one- half in 1 hour, it can supply s in x hours; 

 in like manner the second can supply -3 in x hours, and 



the third ^ ; and by the question the sum of these parts 

 is the whole. 



'" 2+3 + 4 =L 



Multiplying by 4, in order that the first and third 

 fractions may be removed at the same time, we have 



2x + * + x = 4, that is, 

 3 



= 4. 



12 



Multiplying by 3, 9^+4ar = 12, .. 13* = 12, .-. x = 



hence, when they all run together, the pipes will fill the 



12 

 cistern in ^h. = 55min. 23^-iSec. 



15. Solve the preceding question, when the first pipe 

 fills the cistern in 1 hour 20 minutes ; the second in 3 

 hours 20 minutes, and the third in 5 hours. 



Imitating the foregoing solution, x being the required 

 number of hours as before, the part of the whole sup- 



x 

 plied by the first pipe is JT, the part supplied by the 



second is STI an d the part supplied by the third is -=-, in 



**1T 



x hours ; hence, since these parts make up the whole, we 

 mve 



ll + l+f ~ L 



Or, multiplying the terms of the first pair of fractions 



3x 



1. 



SB 



