SOLUTIONS TO THE EXERCISES.] 



MATHEMATICS. ALGEBRA. 



501 



. . . (i) > x = 100 28 y = 72 52 = 20. 



Hence there must be 20 bushels of rye, and 52 bushels 

 of wheat. 



Otherwise. Suppose there were x bushels of rye ; then 

 there must have been 100 28 x bushels of wheat. 

 The worth of the x bushels of rye is 36* pence, and the 

 worth of the 72 x bushels of wheat is 48(72 x) 

 pence ; also the worth of the 28 bushels of barley is 

 28 x 28 pence. Hence, because the 100 bushels is 

 worth 4000 pence, by the question, we have 



36* + 48(72 x) + 28 X 28 = 4000 



Dividin" by 4, 9* + 12(72 x) + 28 X 7 = 1000 

 that is, Zx + 864 + 190 = 1000 

 .-. ar= 1000 1000 = 60, .-. x=20, bushels 

 of rye 



. . 100 28 x = 72 20 = 52, bushels of wheat. 



This latter solution is perhaps somewhat easier than 

 the former : you may vary both by employing shillings 

 throughout instead of pence ; representing the 2s. 4d. 

 and the 3s. 4d. by 2Js., and 31 3. respectively. 



10. Two persons, A and B, are engaged on a work 

 which they can finish in 16 days ; but after working 

 together 4 days, A is called off ; and B alone finishes it 

 iii3G days more In how many days could each dp it 

 separately ? Suppose A can do it in x days, and B in y 



days : then in one day they can do the and the - part 

 . * x y 



respectively, and consequently in 16 days they can do 



- + , which by the question, is the whole, 

 x y 



1 + 16 =1. 



x ^ y 



Also in 4 days they do the - + - part, that is, tlio J 



part ; so that after that time the J part remains to be 

 doue by B. Now, in 1 day B does the J- part of this ; 

 that is, the T ^j - fa of the whole ; but he does the 



- of the whole in 1 day ; hence 



y " 



Substituting this value of y in the former equation, we 

 have 



3+3- 



W1 o 

 i - . -* . . 



Hence A can finish the work in 24 days, and B iu 

 48 days. 



Otlterwiae, without Algebra. Since A and B have 

 worked together 4 days, they can complete what is left 

 iu 12 days ; and, therefore, -fa of it in 1 day, while B 

 alone can do only the fa in 1 d"ay ; so tliat the part of it 

 done by A in 1 day is 



1 _ 1 1 

 12 30 18 ; 



hence A is twice as good a workman as B, so that A 

 and B together are a* good as 3 B's ; and, therefore, 

 since 3 B's would do tho whole in 16 days, one B would 

 t:\ke 3 times as long, or 48 days ; and, consequently, A 

 would tako 24 days half the time. 



A'l'tin, with one unkwrwn quantity. By the question 

 A and B li.ivc done J of the work before they separate, 

 henco :1 of it remains to be done. Suppose A can do the 



1 

 whole in x days, then in 1 day he can do - of it ; but A 



and B together do -? 6 in a day ; hence B's part in a day 



i, ' - ; but B's part is also ^ ff of J ; hence 



J'j z 



!_! 1 



16 x = 48 

 1111 



x 16 48 24' 



24, A's days. 



And since B does the -fa part in 1 day, he can do the 

 whole in 48 days. 



11. A composition of copper and tin containing 100 



cubic inches, weighed 505 ounces ; how many ounces of 



j each metal did it contain, supposing a cubic inch of 



I copper to weigh 5J oz., and a cubic inch of tin to weigh 



4^oz.? 



Suppose there were x ounces of copper and y ounces 

 of tin : then x -\- y = 505. Also by the question, 



Mult. 1st equa. by 17, 



100,.'. + = 25 



17* + 21 iy = 8925 

 17* + 17y = 8585 



Subtracting, 4i/ = 340, . . y = 85 

 . . x = 505 y = 505 85 = 420. 

 Hence there were 420 oz. of copper and 85 oz. of tin. 



This is solved with only one unknown quantity, by 

 putting x for the number of ounces of copper, and 

 505 x for the number of ounces of tin. 



12. A cask is supplied by three spowta, which can till 

 it in a minutes, b minutes, and c minutes, respectively ; 

 in what time will it be tilled if all flow together 1 



Let x be the number of minutes, then the part of the 

 whole supplied in 1 minute is 



i . = I + 1 -f- - , by the question, 

 f a b c 



Clearins, ale = (at + ac + icl* 

 ale 



Questions of this kind may be treated rather differently, 

 thus : since the first cpout can fill the cask in a minutes, 

 it can fill 6c such casks in abc minutes. In like manner 

 the second spout can fill ac such casks in ale minutes, 

 and the third ab such casks in the sarno time ; so that 

 when all flow together, they can fill ab + ac + Ic casks in 

 abc minutes, and, consequently, one cask in 

 ate 



aA + ac + lie 



minutes. 



QUADRATIC EQUATIONS. EXAMPLES FOR EXERCISE. 

 PAGE 474. 



I.* 1 4j=45. Completing the square, 



* 4x + 4 = 4'J. 

 Extracting the root, jr 2 := + 7, 



..jr = 2+7 = 9 or 5. 



2. j: 2 + 8* = 33. Completing the square, 



Extracting the root, x + 4 = +7, 



,-.x = 4 + 7=3 or 11. 



3. x' 8jr = 9. Completing the square, 



a :_gj.+ 16 = 25. 

 Extracting the root, x 4 = + 5, 



,-.,r = 4 + 5 = 9 or 1. 



, 2 1C1 



4. 3j 7 + 2jp=:lGl. Dividing by 3, jr'+ ,.-.r = -^-' 



, , 2 1 161 1 481 



Completing the square, *- + - , 4. - = -^ 4. (j - = y . 



1 22 



Extracting the root, x + SBX ""' 



lj.22 

 - 





Or, by Rule !!., page 474, 



3o '. + 2* = 161, .'. 6^ + 2=^(1932 + 4)= -/193G 

 = + 44, 



2 + 44 _ . , 23 

 .'. * = -=/or 

 G o 



OA <14 



5. 7*- 20x = 32. Dividing by 7, x~ ~ f ~ ? . 



20 100 _ 32 100 324 

 Completing the square, t"- y * + ^g 7 + 4fl ^y. 



10 324 _ , 18 

 Extracting the root, x y = / ^.j IT 7 ' 



