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      Question

      Three workers P1, P2 and P3 got work from the

      contractor. In 30 days, ‘P1’ completed 30% of work. ‘P2’ is 20% less efficient than ‘P1’ but takes 5 days less than ‘P3’ to complete the work alone. Find the time taken by all three together to complete 33.4% of work.
      A 17 Correct Answer Incorrect Answer
      B 16 Correct Answer Incorrect Answer
      C 13 Correct Answer Incorrect Answer
      D 12.4 Correct Answer Incorrect Answer
      E 10 Correct Answer Incorrect Answer

      Solution

      Time taken by ‘P1’ to complete the work = 30/0.3 = 100 days
      Time taken by ‘P2’ to complete the work = 100/0.8 = 125 days
      Time taken by ‘P3’ to complete the work = 125 + 5 = 130 days
      Let the total work = 6500 units (LCM of 100, 125 and 130)
      Efficiency of ‘P1’ = 6500/100 = 65 units/day
      Efficiency of ‘P2’ = 6500/125 = 52 units/day
      Efficiency of ‘P3’ = 6500/130 = 50 units/day
      Required time taken = (0.334 × 6500)/167 = 13 days

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