Question
Tap A can fill tank X in 20 minutes, but with a leakage
(tank Y), it takes 33(1/3) minutes to fill. Tap B operates at 20% less efficiency than Tap A. Determine the time required for Tap B to fill the tank Y with the leakage. (Capacity of tanks 'X' and 'Y' are equal)Solution
Let the total capacity of the tank be 100 units [L.C.M. of 20 and (100/3) ] So, the efficiency of tap 'A' = (100/20) = 5 units/minute Efficiency of tap 'A' with leakage = [100 ÷ (100/3) ] = 3 units/minute So, the efficiency of leakage = 5 - 3 = 2 units/minute Efficiency of tap 'B' = 5 X 0.8 = 4 units/minute So, effective efficiency of tap 'B' along with leakage = 4 - 2 = 2 units/minute Thererfore, required time = (100/2) = 50 minutes
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