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    Question

    Tap 'A' can fill a tank 'X' in 18 minutes while it takes 27

    minutes when there is a leakage tank 'Y'. If tap 'B' is 33(1/3)% less efficient than tap 'A', then find the time taken by tap 'B' to fill the leakage tank 'Y'. (Capacity of tanks 'X' and 'Y' are equal)
    A 72 minutes Correct Answer Incorrect Answer
    B 45 minutes Correct Answer Incorrect Answer
    C 36 minutes Correct Answer Incorrect Answer
    D 54 minutes Correct Answer Incorrect Answer
    E 60 minutes Correct Answer Incorrect Answer

    Solution

    Let the total capacity of the tank be 54 units [L.C.M. of 18 and 27]
    So, the efficiency of tap 'A' = (54/18) = 3 units/minute
    Efficiency of tap 'A' with leakage = [54 ÷ 27] = 2 units/minute
    So, the efficiency of leakage = 3 - 2 = 1 unit/minute
    Efficiency of tap 'B' = 3 × (2/3) = 2 units/minute
    So, effective efficiency of tap 'B' along with leakage = 2 - 1 = 1 unit/minute
    Therefore, required time = (54/1) = 54 minutes

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