Question

If 15 men, 24 women and 36 boys can do a piece of work in 12 days, working 8h per day, how many men must be associated with 12 women and 6 boys to do another piece of work $2\frac{1}{4}$ times as large in 30 days working 6 h per day?

• A. 6
• B. 4
• C. 8
• D. 10

Answer 1

Answer: (C)

8

Answer 2

If M1 men can do W1 work in D1 days working H1 hours per day and M2 men can do W2 work in D2 days working H2 hours per day (where all men work at the same rate), then

$\frac{M1D1H1}{W1}=\frac{M2D2H2}{W2}$………………(1)

So,given 15M=24W=36B...................(2)

where M,W,B are work done by one man, one woman and one boy respectively in one day.

Also given $W2=\frac{9}{4W1}$….....(3)

for W2 let x Men(M) are required.
So total Human-power(for W2) = xM + 12W + 6B

now using relations from (2) Human-power(for W2) = xM + $\frac{15}{2M}$ + $\frac{15}{6M}$ = (10+x)M

now using relations from (2) Human-power(for W2) = xM + $\frac{15}{2M}$ + $\frac{15}{6M}$ = (10+x)M

now using relations from (2) Human-power(for W2) = xM + $\frac{15}{2M}$ + $\frac{15}{6M}$ = (10+x)M

=> $\frac{15M\times12days\times8Hrs}{W1}=\frac{(10+x)M\times30days\times6Hrs}{(W2)}$

=> $\frac{15M12days8Hrs}{W1}=\frac{(10+x)M\times30days\times6Hrs}{(\frac{9}{4W1})}$

⇒ x=8 Men
So the correct option is (C)