A contract is to be completed in 75 days and 187 men are to work 15 hours per day. After 65 days, $\frac{3}{5}$ of the work is completed. How many additional men may be employed, so that the work may be completed in time, each man now working 17 hours per day?

528

Formula to be used here, 

\(\frac{M_1\;×\;D_1\;×\;H_1}{M_2\;×\;D_2\;×\;H_2}\)=\(\frac{W_1}{W_2}\)

where, M = No. of men, D = No. of days, H = No. of hours, W = Work

⇒ Putting values, 

⇒ \(\frac{187\;×\;65\;×\;15}{M_2\;×\;10\;×\;17}\) = \(\frac{3/5}{2/5}\)

⇒ \( { M}_{2 } \) = 715,

⇒ Extra Men needed = 715 - 187 = 528 men.