$A$ big tanker can be filled by two pipes $A$ and $B$ in $60$ and $40$ $min$ respectively. What time will it take to fill an empty tanker if tap $B$ is used for half of the total time and both taps $A$ and $B$ together are used for the remaining half of the time?

Pipes $A$ and $B$ can fill a tank in $5$ $hours$ and $6$ $hours$ respectively. Pipe $C$ can empty it in $12$ $hours.$ If all the three pipes are opened together,then the tank will be filled in

$A$ tap can fill a tank in $6$ $h$. After half the tank is filled,three more similar taps are opened. What is the total time taken to fill the tank completely?

$A$ 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 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?

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: