This problem involves calculating the time taken to fill a tank when one pipe fills it and another empties it simultaneously.
The first pipe fills the tank in 5 hours. Its filling rate is the amount of the tank it fills per hour.
The second pipe empties the tank in 10 hours. Its emptying rate is the amount of the tank it empties per hour.
When both pipes work together, the net rate at which the tank fills is the filling rate minus the emptying rate.
This means that when both pipes are open, \(\frac{1}{10}\) of the tank is filled every hour.
To find the total time required to fill the tank, divide the total work (filling 1 tank) by the net filling rate.
Therefore, it will take 10 hours for both pipes to fill the tank when working together.
A pipe can fill a cistern in 20 minutes where as the cistern when full can be emptied by a leak in 28 minutes. When both are opened, The time taken to fill the cistern is:
‘A’ pipe can empty a tank in 20 minutes. The second pipe ‘B’ has a diameter twice as that of ‘A’. If both A & B pipe are attached to the tank how much time will be required to empty the tank?
Pipes A and B can empty a full tank in 16 hours and 24 hours, respectively. Pipe C alone can fill the empty tank in 4 hours. If A, B and C are opened together, the tank will be 35% full after :
Pipes A and B can fill a tank in 36 minutes and 45 minutes, respectively. Both these pipes were opened simultaneously. After 20 minutes, a leak at the bottom of the tank was spotted which was immediately sealed. The tank was full in another 15 minutes. The leak alone can empty the full tank in:
A cistern has a leak which would empty it in 6 hours. A tap is turned on which admits 10 litres of water per minute into the cistern. When it is full it is now emptied in 10 hours. What is the capacity (in litres) of the cistern?