Discussion Forum

Two pipes can fill a cistern in $14$ hours and $16$ hours respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom it took $32$ minutes more to fill the cistern. When the cistern is full, in what time will the leak empty it ?

Answer:

Explanation:

Work done by the two pipes in $1$ hour $=\left(\frac{1}{14}+\frac{1}{16} \right)$ $=\frac{15}{112}$

Time taken by these pipes to fill the tank $=\frac{112}{15}$ hours = $7$ hours $28$ minutes

Due to leakage, time taken = $7$ hrs $28$ min + $32$ min = $8$ hrs

Work done by (two pipes + leak) in $1$ hour $=\frac{1}{8}$

Work done by the leak in $1$ hour $=\frac{15}{112}-\frac{1}{8}$ $=\frac{1}{112}$

Leak will empty the full cistern in 112 hours.