A contract is to be complete in 92 days and 234 men were set to work, each working 16 h a days. After 66 days, 4/7 of the work is completed. How many addition men may be employed, so that the work may be completed in time, each man now working 18 h a day?

Correct Answer: 162

Explanation:

Remaining work = 1 - (4/7) = 3/7
Remaining period = (92 - 66) = 26 days
Let the number of addition men = N
Given, M₁ = 234, D₁ = 66, T₁ = 16, W₁= 4/7,
M₂ = (234 + x), D₂ = 26, T₂ = 18, W₂ = 3/7
According to the question,
M₁W₂T₁D₁= M₂W₁T₂D₂
⇒ 234 x (3/7) x 16 x 66 = (234 x N) x (4/7) x 18 x 26
⇒ 234 + N = (3 x 66 x 16 x 234) / (4 x 26 x 18)
⇒ 234 + N = 36 x 11 = 396
∴ N = 396 - 234 = 162
Additional men to be employed = 162

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