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      Question

      ‘A’ can complete 2/5 of a work in 6 days. ‘B’

      takes 9 days more than ‘A’ to complete the work and B is 25% more efficient than ‘C’. ‘A’ started the work alone and on 2nd day, he is joined by ‘B’ only, on 3rd day, he is joined by ‘C’ only and on 4th day, he works alone again and so on. They worked this way for 9 days. The sum of the time taken to complete the remaining work by ‘A’ alone and by ‘B’ alone is how much percent more/less than that by ‘C’ alone?
      A 20% Correct Answer Incorrect Answer
      B 30% Correct Answer Incorrect Answer
      C 50% Correct Answer Incorrect Answer
      D 80% Correct Answer Incorrect Answer
      E None of these Correct Answer Incorrect Answer

      Solution

      ATQ, A completes 2/5 of work in 6 days. Time for whole work = 6 ÷ (2/5) = 6 × 5/2 = 15 days. B’s time = 15 + 9 = 24 days. B 25% more efficient than C ⇒ C’s time = 24 × 5/4 = 30 days. Let total work = 600 units. A’s efficiency = 600/15 = 40 units/day. B’s efficiency = 600/24 = 25 units/day. C’s efficiency = 600/30 = 20 units/day. Work in 3-day cycle: 40 + 65 + 60 = 165 units. In 9 days: 165 × 3 = 495 units. Remaining = 600 − 495 = 105 units. Time by A = 105/40 = 21/8 days. Time by B = 105/25 = 21/5 days. Time by C = 105/20 = 21/4 days. Sum(A+B) = 273/40 days. Excess over C = 273/40 − 21/4 = 63/40 days. Required percentage = (63/40 ÷ 21/4) × 100 = 30% more.

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