Two taps A and B can fill a tank in 4 and 6 hours respectively. The pipes are opened simultaneously and it takes 6 minutes more to fill the tank due to leakage. If the tank is full, then find the time taken by the leakage to empty the tank.

60 hrs

The correct answer is option (2) : 60 hrs

Part of the tank filled by both the taps A and B in 1 hr $=\frac{1}{4}+\frac{1}{6}=\frac{5}{12}$

So, time taken by both the taps to fill the tank $=\frac{12}{15}hr$

$= 2\,  hrs\, \, 24 min $

Now, time taken to fill the tank due to leakage

Now, time taken its fill the tank due to leakage $= 2 hr\, 24 min\, + 6\, min $

$= 2\, hrs 30\, min $

$= 2\frac{1}{2}\, hrs $

So part of the tank filled by both the taps and the leakage in 1 hr $=\frac{1}{\frac{5}{2}}=\frac{2}{5}$

∴ Part of the tank emptied by the leakage in 1 hr  $=\frac{5}{12}-\frac{2}{5}=\frac{1}{60}.$

Hence, time taken by the leakage to empty, the full tank is 60 hrs.