'A' and 'B' together can do a piece of work in 12 days while 'B' and 'C' together can do the same work in 16 days. 'A' started the work alone and worked on it for 1 day, then 'B' alone worked on it for 9 days and 'C' alone completed the remaining work in 18 days. Find the time taken by 'B' alone to complete the whole work.

Solution

Let the total work be 48 units. {LCM of 12 and 16} Combined efficiency of 'A' and 'B' = 48 ÷ 12 = 4 units/day Combined efficiency of 'B' and 'C' = 48 ÷ 16 = 3 units/day Amount of work done by 'A' in 1 day = 1 × a units
Amount of work done by 'B' in 9 days = 9 × b units
Amount of work done by 'C' in 18 days = 18 × c units We have,
a + b = 4
b + c = 3 So, a = 4 − b and c = 3 − b Total work = 48
(4 − b) + 9b + 18(3 − b) = 48
4 − b + 9b + 54 − 18b = 48
58 − 10b = 48
10b = 10
b = 1 unit/day Time taken by 'B' alone to complete the work = 48 ÷ 1 = 48 days