A cistern has a hole in the bottom through which the water is leaking. A tap can fill the cistern in 6 hours and the hole in the bottom can empty the fully filled cistern in 15 hours. If both the tap and the hole are open, then what will be the time taken to completely fill the empty cistern?
10 hours
Cistern Filling Problem Explained
This problem involves calculating the time taken to fill a cistern when there is both an inflow (from a tap) and an outflow (due to a leak). We need to determine the net rate at which the cistern is filled.
Calculating Filling and Emptying Rates
First, let's determine the rate at which the tap fills the cistern and the rate at which the hole empties it.
- Tap's Filling Rate: The tap can fill the cistern in 6 hours. So, in 1 hour, the tap fills $\frac{1}{6}$ of the cistern.
- Hole's Emptying Rate: The hole can empty the fully filled cistern in 15 hours. So, in 1 hour, the hole empties $\frac{1}{15}$ of the cistern.
Combined Effect Calculation
When both the tap and the hole are open, the water is entering and leaving the cistern simultaneously. To find the net rate of filling, we subtract the emptying rate from the filling rate.
Let the rate of filling by the tap be $R_{fill} = \frac{1}{6}$ cistern/hour.
Let the rate of emptying by the hole be $R_{empty} = \frac{1}{15}$ cistern/hour.
The net rate of filling when both are open is:
$$ R_{net} = R_{fill} - R_{empty} $$
Substituting the values:
$$ R_{net} = \frac{1}{6} - \frac{1}{15} $$
To subtract these fractions, we find a common denominator, which is 30:
$$ R_{net} = \frac{1 \times 5}{6 \times 5} - \frac{1 \times 2}{15 \times 2} $$
$$ R_{net} = \frac{5}{30} - \frac{2}{30} $$
$$ R_{net} = \frac{5 - 2}{30} $$
$$ R_{net} = \frac{3}{30} $$
Simplifying the fraction:
$$ R_{net} = \frac{1}{10} \text{ cistern/hour} $$
This means that when both the tap and the hole are open, $\frac{1}{10}$ of the cistern is filled every hour.
Final Time Calculation
To find the total time taken to fill the empty cistern, we use the formula:
$$ \text{Time} = \frac{\text{Total Work}}{\text{Net Rate}} $$
In this case, the total work is filling 1 cistern. So:
$$ \text{Time} = \frac{1 \text{ cistern}}{\frac{1}{10} \text{ cistern/hour}} $$
$$ \text{Time} = 1 \times \frac{10}{1} \text{ hours} $$
$$ \text{Time} = 10 \text{ hours} $$
Therefore, if both the tap and the hole are open, it will take 10 hours to completely fill the empty cistern.
Important Questions from Pipe and Cistern
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?
