$A$ cistern has two pipes. One can fill it in $8 \, h$ and the other can empty it in $5 \, h$. In how many $hours$ will the cistern be emptied if both the pipes are opened together when $\frac{3}{4}$ of the cistern is already full of water?

$A$ tap can fill the cistern in $8$ hours and another can empty it in $16$ hours. If both the taps are opened simultaneously,the time (in hours) to fill the tank is?

$A$ water tank has two pipes. The empty tank is filled in $12$ $min$ by the $1$st pipe and the full tank is emptied by the $2$nd pipe in $20$ $min$. What is the time required to fill the $1/2$ full tank when both pipes are in action (in $min$)?

Two taps $A$ and $B$ can fill a tank in $10$ $hours$ and $15$ $hours$,respectively. If both the taps are opened together,the tank will be full in (in $hours$):

$A$ reservoir is fitted with two pipes $A$ and $B$. Pipe $A$ can fill the reservoir $5$ $hours$ faster than pipe $B$. If both the pipes together fill the reservoir in $6$ $hours$,the reservoir will be filled by $A$ alone in:

There are two taps to fill a tank while a third tap is used to empty it. When the third tap is closed,the first two taps can fill the tank in $10$ $minutes$ and $12$ $minutes$ respectively. If all three taps are opened,the tank is filled in $15$ $minutes$. If the first two taps are closed,in what time can the third tap empty the tank when it is full?