'B' and 'N' can collectively finish a task in 88 days, and 'N' and 'S' together can complete the same work in 66 days. If 'B' and 'S' work together, they can finish the job in 72 days. Determine the ratio of the efficiencies of 'B', 'N', and 'S', respectively.

Solution

ATQ, Let, the total work be 792 units (LCM of 88, 66 and 72) Amount of work done by 'B' and 'N' together in one day = (792/88) = 9 units Amount of work done by 'N' and 'S' together in one day = (792/66) = 12 units Amount of work done by 'B' and 'S' together in one day = (792/72) = 11 units Amount of work done by 'B', 'N' and 'S' together in one day = {(9 + 12 + 11) /2} = 16 units Amount of work done by 'B' alone in one day = 16 - 12 = 4 units Amount of work done by 'N' alone in one day = 16 - 11 = 5 units Amount of work done by 'S' alone in one day = 16 - 9 = 7 units So, the desired ratio = 4:5:7