First, determine the rate at which each pipe empties the tank individually.
When both pipes work together, their rates add up.
Combined rate = Rate of Pipe 1 + Rate of Pipe 2
Combined rate = $\frac{1}{6} + \frac{1}{18}$
To add these fractions, find a common denominator, which is 18.
Combined rate = $\frac{1 \times 3}{6 \times 3} + \frac{1}{18} = \frac{3}{18} + \frac{1}{18}$
Combined rate = $\frac{3+1}{18} = \frac{4}{18}$
Simplify the combined rate:
Combined rate = $\frac{2}{9}$ of the tank per hour.
The time taken to empty the tank when working together is the reciprocal of the combined rate.
Time = $\frac{1}{\text{Combined Rate}}$
Time = $\frac{1}{\frac{2}{9}}$
Time = $\frac{9}{2}$ hours
Convert this improper fraction to a decimal or mixed number:
Time = $4.5$ hours
Therefore, if both pipes work together, they will empty the full tank in 4.5 hours.
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?