Question

Two pipes can fill 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 leak empty it.

Answer 1

Works done by two pipes in $1$ hours $=(\frac{1}{14}+\frac{1}{16})=\frac{8+7}{112}=\frac{15}{112}$

Time taken by the pipes to fill the tank $=\frac{112}{15}$ hrs $=7$ hrs $28$ min

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

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

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

$\therefore$ leak will empty the full as term in $112$ hours.