‘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?
4 minutes
Understanding Pipe Emptying Time and Flow Rate
This question involves calculating the time required to empty a tank using two pipes with different flow rates. The key factor here is how the diameter of a pipe affects its flow rate. A larger diameter means a larger cross-sectional area, which allows more liquid to flow through per unit of time.
Relating Diameter to Flow Rate
The flow rate of liquid through a pipe is directly proportional to the cross-sectional area of the pipe. The cross-sectional area of a circular pipe is given by the formula for the area of a circle, \(A = \pi r^2\), where \(r\) is the radius. Since the diameter \(d = 2r\), the area can also be expressed as \(A = \pi (\frac{d}{2})^2 = \frac{\pi d^2}{4}\). This shows that the area, and thus the flow rate, is proportional to the square of the diameter.
In this problem, pipe B has a diameter twice that of pipe A. Let \(d_A\) be the diameter of pipe A and \(d_B\) be the diameter of pipe B.
Given: \(d_B = 2d_A\).
The ratio of the areas is:
\( \frac{A_B}{A_A} = \frac{\frac{\pi d_B^2}{4}}{\frac{\pi d_A^2}{4}} = \frac{d_B^2}{d_A^2} \)
Substituting \(d_B = 2d_A\):
\( \frac{A_B}{A_A} = \frac{(2d_A)^2}{d_A^2} = \frac{4d_A^2}{d_A^2} = 4 \)
This means the cross-sectional area of pipe B is 4 times the area of pipe A. Therefore, the flow rate of pipe B is 4 times the flow rate of pipe A.
Calculating Individual Emptying Rates
Pipe A can empty the tank in 20 minutes. This means that in one minute, pipe A empties \(\frac{1}{20}\) of the tank.
Rate of pipe A (\(R_A\)) = \(\frac{1}{20}\) tank per minute.
Since the flow rate of pipe B is 4 times that of pipe A:
Rate of pipe B (\(R_B\)) = \(4 \times R_A = 4 \times \frac{1}{20} = \frac{4}{20} = \frac{1}{5}\) tank per minute.
This means pipe B can empty \(\frac{1}{5}\) of the tank in one minute.
Calculating Combined Emptying Rate
When both pipes A and B are attached to the tank, their emptying rates add up to find the combined rate.
Combined Rate (\(R_{A+B}\)) = \(R_A + R_B\)
\( R_{A+B} = \frac{1}{20} + \frac{1}{5} \)
To add these fractions, we find a common denominator, which is 20.
\( \frac{1}{5} = \frac{1 \times 4}{5 \times 4} = \frac{4}{20} \)
So, \( R_{A+B} = \frac{1}{20} + \frac{4}{20} = \frac{1+4}{20} = \frac{5}{20} = \frac{1}{4} \) tank per minute.
The combined rate of both pipes is \(\frac{1}{4}\) tank per minute.
Calculating Time to Empty Together
If the combined rate is \(\frac{1}{4}\) tank per minute, it means that in one minute, \(\frac{1}{4}\) of the tank is emptied. To empty the whole tank (which is 1 whole), the time required is the reciprocal of the combined rate.
Time taken (\(T_{A+B}\)) = \(\frac{1}{R_{A+B}} = \frac{1}{\frac{1}{4}} = 4\) minutes.
Therefore, if both pipes A and B are attached to the tank, it will take 4 minutes to empty the tank.
Summary of Calculation Steps
| Step | Description | Calculation |
|---|---|---|
| 1 | Find rate of pipe A | \(R_A = \frac{1}{20}\) tank/min |
| 2 | Find rate of pipe B (proportional to diameter squared) | \(R_B = (2)^2 \times R_A = 4 \times \frac{1}{20} = \frac{1}{5}\) tank/min |
| 3 | Find combined rate | \(R_{A+B} = R_A + R_B = \frac{1}{20} + \frac{1}{5} = \frac{1}{20} + \frac{4}{20} = \frac{5}{20} = \frac{1}{4}\) tank/min |
| 4 | Calculate time taken together | \(T_{A+B} = \frac{1}{R_{A+B}} = \frac{1}{\frac{1}{4}} = 4\) minutes |
Revision Table: Pipe and Tank Problems
Understanding the relationship between work rate, time, and quantity (like tank capacity) is crucial for solving pipe and tank problems.
- Work Rate: The amount of work done per unit of time (e.g., fraction of tank filled/emptied per minute). If a pipe fills/empties a tank in \(T\) time, its rate is \(1/T\).
- Combined Rate: If multiple pipes work together, their rates add up (for filling pipes) or subtract (if some are filling and some are emptying). If pipes 1 and 2 have rates \(R_1\) and \(R_2\), the combined rate is \(R_1 + R_2\) (if both filling/emptying).
- Time Taken: The time taken to complete a task is \(1 / \text{Rate}\).
- Diameter and Flow Rate: Flow rate is proportional to the square of the diameter of the pipe (\(R \propto d^2\)).
Additional Information: Work and Time Concepts
Problems involving pipes filling or emptying tanks are similar to work and time problems. The tank represents the total work to be done (filling or emptying), and the pipes are the workers. The rate of a pipe is the amount of work it does (fraction of the tank) in a unit of time.
- If a pipe fills a tank in \(t\) hours, it fills \(1/t\) of the tank in 1 hour.
- If a pipe empties a tank in \(T\) hours, it empties \(1/T\) of the tank in 1 hour.
- If pipes with rates \(r_1, r_2, \dots, r_n\) work together, the combined rate is \(R_{total} = r_1 + r_2 + \dots + r_n\). The total time taken is \(1/R_{total}\).
- In cases with filling and emptying pipes, the net rate is the sum of filling rates minus the sum of emptying rates.
Understanding these basic principles helps solve a variety of problems involving rates, time, and capacity.
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:
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?
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?
