Pipe $A$ can fill a tank in $4$ $hours$ and pipe $B$ can fill it in $6$ $hours$. If they are opened on alternate hours and if pipe $A$ is opened first,then in how many hours will the tank be full?

$A$ tap takes $36$ hours extra to fill a tank due to a leakage equivalent to half of its inflow. In how many hours can the inlet pipe alone fill the tank?

Two taps $A$ and $B$ can fill an oil tank in $15$ $min$ and $18$ $min$ respectively,while a third tap is used to empty the tank. The third tap is opened after $6$ $min$ of opening of taps $A$ and $B$. If the tank empties in $16 \frac{1}{2}$ $min$ after opening the third tap,then in how much time (in $min$) can the third tap empty the full tank?

$A$ pipe $P$ can fill a tank in $12$ $min$ and another pipe $R$ can fill it in $15$ $min.$ But,the $3$rd pipe $M$ can empty it in $6$ $min.$ The $1$st two pipes $P$ and $R$ are kept open for $5$ $min$ in the beginning and then the $3$rd pipe is also opened. In what time (in minutes) is the tank emptied?

$A$ tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank $5$ $hours$ faster than the first pipe and $4$ $hours$ slower than the third pipe. The time (in $hours$) required by the first pipe is:

Three pipes $A$,$B$,and $C$ can fill a cistern in $10$,$12$,and $15$ $hours$,respectively,while working alone. If all the three pipes are opened together,the time taken to fill the cistern will be: