Correct Answer - Option 2 : 3
Given:
Five men finish work in 28 days working 3 hours a day.
The same work has to be completed by 10 boys such that the efficiency of a boy is double than the efficiency of a man.
Boys work for 7 hours a day.
Formula used:
1.) MDH/W is constant.
2.) M1 × D1 × H1/W1 = M2 × D2 × H2/W2
Where,
M → The number of men
H → The number of hours
D → The number of days
W → The amount of work
Calculation:
The efficiency of a boy is double than the efficiency of a man.
10 boys = (10 × 2) men = 20 men
Work done by 5 men = 5 × 28 × 3
Work by 10 boys = work by 20 men
Comparing the work done by 5 men and 10 boys,
Let them work for d days.
⇒ 5 × 28 × 3 = 20 × 7 × d
⇒ d = 3 days
∴ 10 boys can finish the work in 3 days.
