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?

$$10\times\ \ \ $$From the formula,

Total work = $$\frac{mdh}{w1}$$

here, m = man/woman

d =days

h = hour

w = work

Total work = $$75 \times 187 \times 15$$ = 210375

Work done in 65 days = 210375 $$\times \frac{3}{5}$$ = 126255

Remaining work = 210375 - 126255 = 84150

$$(187 + m) \times 10 \times 17$$ = 84150

(187 + m) = 495

m = 495 -187 = 308

$$\therefore$$ 308 additional men may be employed, so that the work may be completed in time.