Question

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 1

See the attachment for detail solution
Hope it will help you

Answer 2

Answer: 112/ 13 hours

Step-by-step explanation:

first pipe unit work = 1/14 hours

second pipe unit work = 1/16 hours

so if they work simultaneously it took time =

$\frac{1}{14}+\frac{1}{16}\\\\=\frac{7+8}{112}\\\\=\frac{15}{112}$

it means both pipe will take 112/15 hours

due to leakage it takes 32 minutes more

so total time =

$=\frac{112}{15}+\frac{32}{60}\\\\=\frac{240}{60}\\\\=4$

let time t will be taken if leakage alone empty the tank.

$\frac{1}{14}+\frac{1}{16}-\frac{1}{t}=\frac{1}{4}\\\\\frac{1}{t}=\frac{1}{4}-\frac{1}{14}-\frac{1}{16}\\\\\frac{1}{t}=\frac{28-8-7}{112}\\\\\frac{1}{t}=\frac{13}{112}$

so it will empty the tank in 112/ 13 hours