Two taps $A$ and $B$ can fill a tank in $10$ $hours$ and $12$ $hours$ respectively. If both taps are opened at $10 \text{ am}$,at what time (in $\text{pm}$) should tap $A$ be closed so that the tank is completely filled at exactly $4 \text{ pm}$?

Three taps $P, Q$ and $R$ separately can fill a tank completely in $4, 8$ and $12$ hours respectively. Another tap $S$ can empty the filled tank in $10$ hours. Which of the following combinations will fill the empty tank in the least amount of time?

$A$ pipe can fill a tank in $10 \, h$, while another pipe can empty it in $6 \, h$. Find the time (in $hours$) taken to empty the tank when both the pipes are opened simultaneously.

Three pipes $A, B$ and $C$ can fill a tank in $30 \ min$,$20 \ min$ and $10 \ min$,respectively. When the tank is empty,all the three pipes are opened. If $A, B$ and $C$ discharge chemical solutions $P, Q$ and $R$ respectively,then the part of solution $R$ in the liquid in the tank after $3 \ min$ is

Three pipes $A, B$ and $C$ can fill a tank in $6$ hours. After working together for $2$ hours,$C$ is closed and $A$ and $B$ fill the remaining part in $7$ hours. The number of hours taken by $C$ alone to fill the tank is:

$A$ cistern can be filled by a tap in $4$ $hours$,while it can be emptied by another tap in $9$ $hours$. If both the taps are opened simultaneously,then after how much time (in $hours$) will the cistern get filled?