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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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