Question
Tap 'A' can fill a tank 'X' in 36 minutes while it takes 54
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)Solution
Let the total capacity of the tank be 108 units [L.C.M. of 36 and 54]
So, the efficiency of tap 'A' = (108/36) = 3 units/minute
Efficiency of tap 'A' with leakage = [108 ÷ 54] = 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 = (108/1) = 108 minutes
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