Solution for pipes and cistern question given below?

Question

Q.) Two pipes $P$ and $Q$ can fill a tank in $20$ hours and $25$ hours respectively while a third pipe $R$ can empty the tank in $30$ hours. 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.

Answer 1

We can look at the rate at which they fill each tank as shown below: $$W_P=\frac{1}{20},\,W_Q=\frac{1}{25},\,W_R=\frac{1}{30}$$ Now firstly we know that all three are open for $10$ hours together, meaning it will be filled by: $$L=10\left(\frac1{20}+\frac1{25}+\frac1{30}\right)$$ And for the rest of the time, which we will call $t_1$, only $W_P$ and $W_Q$ can be applied. This means the remaining amount needing to be filled takes: $$1-10\left(\frac1{20}+\frac1{25}+\frac1{30}\right)=t_1\left(\frac1{20}+\frac1{25}\right)$$ This can be rearranged to give: $$t_1=\frac{1-10\left(\frac1{20}+\frac1{25}+\frac1{30}\right)}{\left(\frac1{20}+\frac1{25}\right)}$$ But in the total time we must allow for the 10 hours that have already passed, and so: $$t=\frac{1-10\left(\frac1{20}+\frac1{25}+\frac1{30}\right)}{\left(\frac1{20}+\frac1{25}\right)}+10$$