Two pipes P and Q can fill a tank in $20$ hrs and $25$ hrs respectively while a third pipe R can empty the tank in $30$ hrs. If all the pipes are opened together for $10$ hrs and then pipe R is closed, then in what time the tank can be filled?

Work they all do in one hour:

$\frac{1}{P}=\frac{1}{20}$

$\frac{1}{Q}=\frac{1}{25}$

$\frac{1}{R}=-\frac{1}{30}$

When they are all opened for 10 hours:

$10[\frac{1}{20}+\frac{1}{25}-\frac{1}{30}]$

$10[\frac{15+12-10}{300}]=\frac{170}{300}$

Now pipe R is closed, we need to find in how much time P and Q will fill the tank now

Remaining capacity of the tank $=1-\frac{170}{300}=\frac{130}{300}$

Let remaining time be $x$, then:

$x[\frac{1}{20}+\frac{1}{25}]=\frac{130}{300}$

$x[\frac{9}{100}]=\frac{130}{300}$

$x=\frac{130\times100}{300\times9}=\frac{130}{27}$

So, total time $=10+\frac{130}{27}=\frac{270+130}{27}=\frac{400}{27}$ hours