Two pipes X and Y can fill a cistern in 21 hours and 24 hours, respectively. The pipes are opened simultaneously, and it is found that due to a leakage in the bottom it takes 48 minutes more to fill the cistern. When the cistern is full, in how much time will the leak empty it if no pipe is open during that time?

Correct option is D

Given:

Time taken by pipe A to fill the tank = 21 hours
Time taken by pipe B to fill the tank = 24 hours
Here, extra 48 minutes represents the extra time taken by the pipes due to the leak.

Solution:
Total units of water to be filled = LCM of (21, 24) = 168 units.
Pipe A can fill 8 units of water in 1 hr.
Pipe B can fill 7 units of water in 1 hr.
Normal Time is taken to fill the tank = 168/15 hrs = 11 hrs = 11 hrs 12 minutes.
With 48 min extra, the pipes would take 12 hrs to fill the tank.
Let the leakage can empty N units of water in 1 hr.
(7 + 8 - N) × 12 = 168,
15 - N = 14
N = 15 - 14 = 1
Leakage can empty 1 unit of water in 1 hr.
Leakage of 168 units of water takes 168 hrs.
Hence, the time taken to empty the tank is '168' hrs.