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A contractor takes a contract to complete a road in 60 days and employed 105 labours. After 25 days, he found that one third work is completed. How many more labours he requires to complete the remaining work in time?
Let's assume the more labour required after 25 days to complete the work on time is 'y'.
$$\frac{M_1\times\ D_1}{W_1}=\frac{M_2\times\ D_2}{W_2}$$
$$\frac{25\times\ 105}{\frac{1}{3}} = \frac{35\times\ \left(105+y\right)}{\frac{2}{3}}$$
$$\frac{5\times\ 105}{1}=\frac{7\times\ \left(105+y\right)}{2}$$
$$1050 = 7\times\ \left(105+y\right)$$
150 = (105+y)
y = 150-105
= 45
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