Two pipes can fill a cistern in 14 and 16 hours respectively. Both the pipes are opened simultaneously and it is found that due to leakage in the bottom \(\frac{8}{15}\) (or 32 min) hours extra had been taken to fill the tank. If the cistern is full, in what time would the leak empty the tank?

112 hrs

Time taken to fill the tank by A and B = \(\frac{112}{15}\)

Extra time taken due to leakage = \(\frac{8}{15}\)

Total time taken by A, B and Leakage = \(\frac{112}{15}\) + \(\frac{8}{15}\) = \(\frac{120}{15}\) = 8 hrs.

So,

Efficiency of A + B - Leakage = \(\frac{112}{8}\) = 14

Efficiency of Leakage = 14 - 8 - 7 = -1

Hence, leakage can empty the full tank in = \(\frac{112}{1}\) = 112 hrs.