A contractor takes a contract to complete a road in 60 days and employed 70 labours. After 25 days, he found that one fourth of work is completed. How many more labours he requires to complete the remaining work in time?

$$\frac{\left(M_1\times\ D_1\right)}{W_1}\ =\ \frac{\left(M_2\times\ D_2\right)}{W_2}$$

A contractor takes a contract to complete a road in 60 days and employed 70 labours. After 25 days, he found that one fourth of work is completed.

Let's assume the more number of labour needed after 25 days to complete the work on time is 'y'.

$$\frac{\left(25\times\ 70\right)}{1}\ =\ \frac{\left(35\times\left(70+y\right)\right)}{3}$$

$$\frac{\left(25\times\ 2\right)}{1}\ =\ \frac{\left(1\times\left(70+y\right)\right)}{3}$$

$$50\times3 = (70+y)$$

(70+y) = 150

y = 150-70

the more number of labour needed after 25 days to complete the work on time = y = 80