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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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