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      Question

      'A' and 'B' together can do a piece of work in 12 days

      while 'B' and 'C' together can do the same work in 16 days. 'A' started the work alone and worked on it for 1 day, then 'B' alone worked on it for 9 days and 'C' alone completed the remaining work in 18 days. Find the time taken by 'B' alone to complete the whole work.
      A 24 days Correct Answer Incorrect Answer
      B 48 days Correct Answer Incorrect Answer
      C 36 days Correct Answer Incorrect Answer
      D 60 days Correct Answer Incorrect Answer
      E 40 days Correct Answer Incorrect Answer

      Solution

      Let the total work be 48 units. {LCM of 12 and 16} Combined efficiency of 'A' and 'B' = 48 ÷ 12 = 4 units/day Combined efficiency of 'B' and 'C' = 48 ÷ 16 = 3 units/day Amount of work done by 'A' in 1 day = 1 × a units
      Amount of work done by 'B' in 9 days = 9 × b units
      Amount of work done by 'C' in 18 days = 18 × c units We have,
      a + b = 4
      b + c = 3 So, a = 4 − b and c = 3 − b Total work = 48
      (4 − b) + 9b + 18(3 − b) = 48
      4 − b + 9b + 54 − 18b = 48
      58 − 10b = 48
      10b = 10
      b = 1 unit/day Time taken by 'B' alone to complete the work = 48 ÷ 1 = 48 days

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