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An electric pump can fill a tank in 6 hours. Due to a leakage in the tank, it takes 715\frac 1551​​hours to fill the tank. How much time will this lea
Question

An electric pump can fill a tank in 6 hours. Due to a leakage in the tank, it takes 715\frac 15​hours to fill the tank. How much time will this leak take to empty the full tank if water does not get in or out of the tank through any other point during this period?

A.

36 hrs

B.

38 hrs

C.

17 hrs

D.

18 hrs

Correct option is C

Given:
Time taken by the pump alone to fill the tank = 6 hours
Time taken to fill the tank with a leak = 7(1/5) hours = 36/5 hours
Solution:
Work done by the pump in 1 hour = 16\frac 16
Work done when both pump and leak are working together in 1 hour = 1365\frac {1}{\frac{36}{5}} = 536\frac{5}{36}​​
Work done by the leak in 1 hour = Pump’s work – (Pump + Leak)’s work = 16536=6536=136\frac{1}{6} - \frac{5}{36} = \frac{6 - 5}{36} = \frac{1}{36}​​
So, the leak can empty 136 \frac{1}{36}​ of the tank in 1 hour.
The leak will empty the full tank in 36 hours.
Alternate Solution:  
Let total work = LCM = 36 units
Efficiency of pump = 366=6 units/hour\frac{36}{6} = 6 \text{ units/hour} \\
Net efficiency with leak = 36365=5 units/hour \frac{36}{\frac{36}{5}} = 5 \text{ units/hour} \\
Leak's efficiency = 6 - 5 = 1 unit/hour (emptying)
Time to empty full tank  = 36 ÷ 1 = 36 hours
∴ Leak will empty the full tank in 36 hours


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