There are two water taps in a tank which can fill the empty tank in 12 hours and 18 hours respectively. It is seen that there is a leakage point at the bottom of the tank which can empty the completely filled tank in 36 hours. If both the water taps are opened at the same time to fill the empty tank and the leakage point was repaired after 1 hour, then in how much time the empty tank will be completely filled?

Question

YESPYQ · Accepted Answer

Correct answer: c) 7 hours 24 minutes. Let the capacity of the tank be 36a litres. Then, efficiency of tap 1 = 36a/12 = 3a litres/hour, Efficiency of tap 2 = 36a/18 = 2a litres/hour And efficiency of leakage point = -36a/36 = -a litres/hour Combined efficiency of all taps = 3a + 2a = 5a litres/hours Capacity filled in first hour = 5a – a = 4a litres Remaining capacity = 36a – 4a = 32a litres Time taken to fill the remaining capacity = 32a/5a = 6.4 hours Total time taken = 1 + 6.4 = 7.4 hours = 7 hours 24 minutes Hence, Option C is the correct answer.

There are two water taps in a tank which can fill the empty tank in 12 hours and 18 hours respectively. It is seen that …

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