A is 40% more efficient than B and both together can complete a work in 9 3/8 days. If A works for the first five days alone and the remaining work is completed by B, then find in how many days the total work will be completed?

Explanation

Given: A is 40% more efficient than B. Total time taken by A and B to complete the work together = 9 3/8 days. A works for the first five days alone and the remaining work is completed by B. Concept used: Efficiency = Total work / Time. Calculations: Let the efficiency of B be 100. So, efficiency of A = 100 x 140/100 = 140. Ratio of A and B's efficiency = 140 : 100 = 7 : 5. Total efficiency of A and B = 7 + 5 = 12. Work done by A and B together in 9 3/8 days = 12 x 75/8 = 225/2. Work done by A in the first five days = 7 x 5 = 35. Remaining work = 225/2 - 35 = 155/2. Time taken by B to complete the remaining work = (155/2) / 5 = 31/2. Total time taken by A and B to finish the work = 5 + 31/2 = 41/2 = 20 1/2 days. Therefore, the total time taken is 20 1/2 days.