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?

45

Formula to be used here,

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

where, M = No. of Men, D = No. days, W = Work,

⇒ Putting in the value,

⇒ \(\frac{105\;×\;25}{M_2\;×\;35}\) = \(\frac{1/3}{2/3}\),

⇒ \( { M}_{2 } \) = 150,

⇒ Therefore, 150 - 105 = 45 men more are needed to complete the work in time.