Q) If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:

a) 4 days              b) 5 days               c) 6 days                     d) 7 days

Ans:

Let m be the work done by a man in 1 day and b be the work done by a boy in 1 day

Hence, by given 1st condition: 6 m + 8 b = 1/10 ………… (i)

and by given 2nd condition: 26 m + 48 b = 1/2 ……………..(ii)

To solve, let’s multiple 1st equation (i) by 8 and subtract (ii) from it, we get:

6 (6 m + 8 b = 1/10) – (26 m + 48 b = 1/2)

∴ (36 m + 48 b) – (26 m + 48 b) = 6/10 – 1/2

∴ 36 m + 48 b – 26 m – 48 b) = 6/10 – 5/10

∴ 10 m = 1/10

∴ m = 1/100

From equation (i): 8 b = 1/10 – 6m

∴ 8 b = 1/10 – 6/100 =10/100 – 6/100 = 4 /100

∴ 2b = 1/100

∴ b = 1/200

Now work done by 15 men and 20 boys in 1 day: 15 m + 20 b

= 15/100 + 20/200 = 30/100 + 20/200 = 50/200 = 1/4

Since time taken by 15 men and 20 boys together to finish 1/4 work = 1 day

∴ time taken by 15 men and 20 boys together can finish full work = 1/(1/4) = 4 days

Hence, Answer is [A].

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