2 men and 7 boys can do a piecee of work in 4 days. The same work is done in 3 days by 4 men and 4 boys. How long would it take one man and one boy to do it alone?
Answers
HII!!
Time taken by one man = x days , and that by 1 boy = y days
work done in 1 day by one man = 1/x
work done in 1 day by one boy= 1/y
work done by 2 men in one day = 2/x
work done by 7 boys in one day = 7/y
so ,
2/x + 7/y = 1/4 -----> [ 1]
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Similarly ,
work done by 4 men in one day = 4/x
work done by 4 boys in one day = 4/y
4/x + 4/y = 1/3 ----> [2]
Multiplying equation 1 by , 2 ,
2(2/x + 7/y = 1/4 )
= 4/x + 14/y = 1/2 ---> [3]
solving equations 2 and 3
4/x + 4/y = 1/3
- 4/x + 14/y = 1/2
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-10/y = 2-3/6
-10/y = -1/6
-1/6y = -10
y = -10 × 6/-1
= 60
2/x + 7/y = 1/4
2/x + 7/60 = 1/4
7/60 - 1/4 = -2/x
-8/60 = -2/x
-120 = -8x
x = -120/-8
x = 15
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Time taken by 1 man to do the whole job = 15 days
time taken by 1 boy to do the whole job = 60 days
time taken by one boy and one man to do one day's work = 1/60 + 1/15
= 1/60 + 4/60
= 5/60 = 1/12
∴ Together they can do it in 12 days !