A cistern has 2 pipes. One can fill it with water in 10 hours and the other can empty it in 6 hours. In how many hours will the cistern be emptied if both the pipes are opened together when $\frac{3^{th}}{4}$ of the cistern is already full of water ?

Part of  the tank filled by pipe 1 hour $=\frac{1}{10}$

Part of  the tank emptied by pipe 2 in hour $=\frac{1}{6}$

Cistern emptied when both are together opened  $\Rightarrow \frac{1}{6}-\frac{1}{10}=\frac{10-6}{60}=\frac{4}{60}=\frac{2}{30}$

$\frac{2}{30}$ part is emptied in 1 hour

 $\Rightarrow \frac{3}{4}th$ of  the tank

$\therefore \frac{30}{2}\times \frac{3}{4}\Rightarrow \frac{90}{8}=11\frac{1}{4}=11hrs15\text{\hspace{0.17em}\hspace{0.17em}minutes}$