A can do a piece of work in 15 days, B is 25% more efficient than A, and C is 40% more efficient than B. A and C work together for 3 days and then C leaves. A and B together will complete the remaining work in:

Solution

Let efficiency of A be 100 ⇒ Efficiency of B = 100 + (100 × 25%) ⇒ Efficiency of B = 100 + 25 = 125 ⇒ Efficiency of C = 125 + (125 × 40%) ⇒ Efficiency of C = 125 + 50 = 175 ⇒ Ratio of efficiency of A ∶ B ∶ C = 100 ∶ 125 ∶ 175 ⇒ Ratio of efficiency of A ∶ B ∶ C = 4 ∶ 5 ∶ 7 A work for 15 days with efficiency 4unit ⇒ Total work done by A = 15 × 4 = 60 unit Now, (A + C) together work for 3 days. ⇒ Total efficiency of A and C together = 4 + 7 = 11 unit/day ⇒ Total work done by A and C together = 11 × 3 = 33 unit ⇒ Remaining work = Total work - Work done by A and C together ⇒ Remaining work = 60 - 33 ⇒ Remaining work = 27 unit ⇒ Total efficiency of A and B together = 4 + 5 = 9 unit/day ⇒ Required number of days = (27/9) = 3 days ⇒ Time required to complete remaining work = 3 days. ∴ Time required to complete remaining work is 3 days.