2 men and 5 boys can finish a piece of work in 4 days,while 3 men and 6 boys can finish it in 3 days. find the time taken by one man alone to finish the work and that taken by one boy alone to finish the work.
let 1 men can finish the work in x days and 1 boy can finish the work in y days.
then 1 man's 1 day work = 1/x and 1 boy's one day work = 1/y
according to the question---
2/x + 5/y = 1/4 ------ (1)
3/x + 6/y = 1/3 --------(2)
put 1/x = u and 1/y = v in eq, (1) and (2)
2u + 5v = 1/4
= 8u + 20 v = 1 ----------- (3)
3u+ 6v =1/3
= 9u + 18 v = 1 ------------(4)
solving equation 3 and 4
multiply equation 3 with 9 and 4 with 8
9 (8u +20v =1)
8( 9u + 18v =1)
by multiplying we get
72u + 180v =9 ------ (5)
72u + 144v =8 -------(6)
now subtract eq. 5 from 6
72u + 180v =9
72u + 144v =8
- - -
_______________________
36V = 1
so v= 1/36
put v = 1/36 in eq. 3
8u + 20 X 1/36 = 1
8u =4/9
72u = 4
u = 1/18
u= 1/x = 1/18
x = 18 days
v = 1/y = 1/36
y = 36 days
so man can finish work in 18 days and boy can finsh work in 36 days