A contractor under takes to repair loads with 120 men in 124 days. After 64 days, as $$\frac{2}{3}$$ of the work has been over already. He withdraws some men as surplus. If the work is completed in time, how many men were removed ?

Part of the work completed in 64 days = 2/3

Part of the work completed in 32 days = 1/3

Remainging work = 1/3

1/3 of the work can be completed in 32 days by 120 men

How many men can complete 1/3 of the work in 60 days?

32:60 = x:120

x = (32 x 120) / 60

x = 64

64 men can complete 1/3 of the work in 60 days.

120 - 64 = 56

Therefore, 56 men have to be removed to complete the remaining work in 60 days.

A is correct choice.