Three taps $A, B$ and $C$ together can fill an empty cistern in $10$ $min.$ The tap $A$ alone can fill it in $30$ $min$ and the tap $B$ alone can fill it in $40$ $min.$ How long will the tap $C$ alone take to fill it? (in $minutes$)

$A$ 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:

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

Two taps $A$ and $B$ can fill a tank in $20 \text{ min}$ and $30 \text{ min}$,respectively. An outlet pipe $C$ can empty the full tank in $15 \text{ min}$. If $A, B$,and $C$ are opened alternatively,each for $1 \text{ min}$,how long (in $\text{min}$) will the tank take to be filled?

$A$ 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?

$A, B, C$ are pipes attached to a cistern. $A$ and $B$ can fill it in $20$ and $30$ minutes respectively,while $C$ can empty it in $15$ minutes. If $A, B, C$ are kept open successively for $1$ minute each,how soon will the cistern be filled? (in minutes)