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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 20 days. 'A' started the work alone and worked on it for 2 days, then 'B' alone worked on it for 16 days and 'C' alone completed the remaining work in 18 days. Find the time taken by 'B' alone to complete the whole work.

      A 30 days Correct Answer Incorrect Answer
      B 20 days Correct Answer Incorrect Answer
      C 36 days Correct Answer Incorrect Answer
      D 40 days Correct Answer Incorrect Answer
      E 60 days Correct Answer Incorrect Answer

      Solution

      Let the total work be 60 units. {LCM of 12 and 20} Combined efficiency of 'A' and 'B' = 60 ÷ 12 = 5 units/day Combined efficiency of 'B' and 'C' = 60 ÷ 20 = 3 units/day Work done by A in 2 days = 2a units
      Work done by B in 16 days = 16b units
      Work done by C in 18 days = 18c units a + b = 5
      b + c = 3 So, a = 5 − b and c = 3 − b 2(5 − b) + 16b + 18(3 − b) = 60
      10 − 2b + 16b + 54 − 18b = 60
      64 − 4b = 60
      4b = 4
      b = 1 unit/day Time taken by 'B' alone = 60 ÷ 1 = 60 days

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