A contract is to be completed in 85 days and 209 men are to work 15 hours per day. After 65 days, \(\frac{3}{7}\) 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 19 hours per day?

506

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

\(\frac{3}{7}\) of the work means 3 work out of 7 done by 1st group in 65 days, remaining 4 work to be done by 2nd group in remaining 20 days. 

Therefore,

⇒ \(\frac{209 × 65 × 15 × 4}{M_2 × 20 × 19 × 3}\) 

M₂ = 715

Additional men required = 715 - 209 = 506