Two taps A and B can fill a tank in 4 and 6 hours respectively. The p…

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.