A contractor assigned a job to three persons A, B and C. ‘A’ which is 25% less efficient than ‘B’ can complete 20% of a work in 8 days. ‘C’ takes 10 days more than ‘B’ to complete the same work. ‘A’ and ‘B’ started working together and after 4 days C joined them. Due to some personal emergency ‘A’ left after 5 more days and rest of the work is completed by 'B' and 'C', together. Find the total time taken to complete the whole work.

Solution

Time taken by ‘A’ to complete the work alone = 8/0.2 = 40 days Time taken by ‘B’ to complete the work alone = 0.75 × 40 = 30 days Time taken by ‘C’ to complete the work alone = 30 + 10 = 40 days Let, the total work = 120 units Efficiency of ‘A’ = 120/40 = 3 units/day Efficiency of ‘B’ = 120/30 = 4 units/day Efficiency of ‘C’ = 120/40 = 3 units/day Work completed in 4 days = 4 × (3 + 4) = 28 units Work completed in next 5 days = 5 × (3 + 4 + 3) = 50 units Time taken by ‘B’ and ‘C’ to complete remaining work together = (120 – 78)/7 = 6 days Therefore, total time taken = 4 + 5 + 6 = 15 days