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?
9000
Solving Cistern Leak and Tap Capacity Problems
This problem involves understanding rates of work, specifically how a leak empties a cistern and a tap fills it, and how these rates combine.
We can think of the capacity of the cistern as the total 'work' to be done (filling it) or undone (emptying it). We'll use fractions representing the portion of the cistern filled or emptied per hour.
Understanding the Rates
- The leak empties the cistern in 6 hours. This means the leak empties \(\frac{1}{6}\) of the cistern per hour. Since it's emptying, we can consider this a negative rate: \(-\frac{1}{6}\) of the cistern per hour.
- The tap admits 10 litres of water per minute. To compare this with the leak's rate (per hour), we convert minutes to hours. There are 60 minutes in an hour. So, the tap admits \(10 \text{ litres/minute} \times 60 \text{ minutes/hour} = 600 \text{ litres/hour}\). This is a positive rate, filling the cistern.
- When both the leak and the tap are operating, the cistern empties in 10 hours. This means the combined rate is emptying \(\frac{1}{10}\) of the cistern per hour. This combined rate is also negative: \(-\frac{1}{10}\) of the cistern per hour.
Setting up the Equation
Let the capacity of the cistern be \(V\) litres.
The rate of the leak is \(\frac{V}{6}\) litres/hour (emptying). So its rate in terms of filling is \(-\frac{V}{6}\) litres/hour.
The rate of the tap is 600 litres/hour (filling).
The combined rate when both are working is \(\frac{V}{10}\) litres/hour (emptying). So the combined rate in terms of filling is \(-\frac{V}{10}\) litres/hour.
The combined rate is the sum of the individual rates:
Rate of tap + Rate of leak = Combined rate
\(600 \text{ litres/hour} + \left(-\frac{V}{6}\right) \text{ litres/hour} = -\frac{V}{10} \text{ litres/hour}\)
So, the equation is:
\(600 - \frac{V}{6} = -\frac{V}{10}\)
Solving for the Cistern Capacity (V)
Now, we solve this equation for \(V\):
- Rewrite the equation: \(600 - \frac{V}{6} = -\frac{V}{10}\)
- Move the terms involving \(V\) to one side: \(600 = \frac{V}{6} - \frac{V}{10}\)
- Find a common denominator for the fractions on the right side. The least common multiple of 6 and 10 is 30.
- Rewrite the fractions with the common denominator: \(\frac{V}{6} = \frac{5V}{30}\) and \(\frac{V}{10} = \frac{3V}{30}\)
- Substitute these back into the equation: \(600 = \frac{5V}{30} - \frac{3V}{30}\)
- Combine the fractions: \(600 = \frac{5V - 3V}{30}\)
- Simplify: \(600 = \frac{2V}{30}\)
- Simplify further: \(600 = \frac{V}{15}\)
- Solve for \(V\) by multiplying both sides by 15: \(V = 600 \times 15\)
- Calculate the final value: \(V = 9000\)
The capacity of the cistern is 9000 litres.
Summary of Calculation Steps
| Description | Rate (Cistern per hour) | Rate (Litres per hour) |
|---|---|---|
| Leak emptying | \(-\frac{1}{6}\) | \(-\frac{V}{6}\) |
| Tap filling | \(+\frac{600}{V}\) | \(+600\) |
| Combined (Tap + Leak) emptying | \(-\frac{1}{10}\) | \(-\frac{V}{10}\) |
Equation from combined rates: \(600 - \frac{V}{6} = -\frac{V}{10}\)
Solving: \(\frac{V}{6} - \frac{V}{10} = 600\)
\(\frac{5V - 3V}{30} = 600\)
\(\frac{2V}{30} = 600\)
\(\frac{V}{15} = 600\)
\(V = 600 \times 15 = 9000\)
Revision Table: Cistern Problems Key Concepts
| Concept | Explanation | Formula Idea |
|---|---|---|
| Work Rate | The amount of work (filling/emptying) done per unit of time. Often expressed as 1 / (Time taken). | If A takes T hours, Rate of A = \(1/T\) per hour. |
| Filling Rate | A positive rate, adding liquid to the container. | Represented with a positive sign (+). |
| Emptying Rate (Leak) | A negative rate, removing liquid from the container. | Represented with a negative sign (-). |
| Combined Rate | The sum of individual rates when multiple sources/leaks are active. | Combined Rate = Rate 1 + Rate 2 + ... |
| Time and Work | Total work = Rate \(\times\) Time. In these problems, total work is often the capacity of the cistern (represented as 1 unit or V litres). | \(1 = \text{Rate}_{\text{combined}} \times \text{Time}_{\text{combined}}\) or \(V = \text{Rate}_{\text{combined}} (\text{in litres/hr}) \times \text{Time}_{\text{combined}}\). |
Additional Information: Solving Similar Time and Work Problems
Problems involving pipes, cisterns, and leaks are common examples of 'Time and Work' questions in mathematics. The core idea is to determine the portion of the work (filling or emptying) done by each component in a unit of time.
- Assign Rates: Assign a rate to each tap or leak. A filling tap has a positive rate, an emptying leak has a negative rate.
- Convert Units: Ensure all rates are in the same units (e.g., per hour, per minute).
- Combine Rates: If multiple taps/leaks work simultaneously, add their rates to find the combined rate.
- Relate Rate to Time: The time taken to complete the work is inversely proportional to the rate. If the rate is \(R\) (fraction of work per unit time), the time taken is \(1/R\) units of time. If the rate is in litres/hour and the total volume is \(V\) litres, the time is \(V / \text{Rate}\).
- Formulate Equation: Set up an equation based on the information given about combined operation or specific time frames.
- Solve: Solve the equation to find the unknown quantity, which is often the time taken, the rate of a specific component, or the total capacity.
This method of using rates as fractions of the total work (or total volume) per unit time is a powerful approach for solving many types of time and work problems.
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 tank has two inlets, A and B, which can fill it in 15 hours and 20 hours, respectively. An outlet C can empty the full tank in 12 hours. If A, B and C are opened together when the tank is empty, then in how much time will the tank be filled?
