One of the two inlet pipes works twice as efficiently as the other. The two, working alongside a drain pipe that can empty a cistern all by itself in 8 hours, can fill the empty cistern in 8 hours. How many hours will the less efficient inlet pipe take to fill the empty cistern by itself?
Understanding the Cistern Filling Problem
This question involves understanding rates of work for pipes filling and emptying a cistern. We have two inlet pipes and one drain pipe working together. One inlet pipe is less efficient, and the other is twice as efficient. The drain pipe empties the cistern at a known rate. We need to find the time taken by the less efficient inlet pipe alone to fill the cistern.
Setting up the Rates of Work
Let the capacity of the cistern be \(V\). For simplicity in calculations, we can assume \(V = 1\) unit (representing the whole cistern).
- Let \(R_1\) be the rate at which the less efficient inlet pipe fills the cistern (fraction of cistern filled per hour).
- Let \(R_2\) be the rate at which the more efficient inlet pipe fills the cistern (fraction of cistern filled per hour).
- Let \(R_{drain}\) be the rate at which the drain pipe empties the cistern (fraction of cistern emptied per hour). Since it empties, its rate is negative.
Based on the problem description, we have the following information about the rates:
- The more efficient pipe works twice as efficiently as the other: \(R_2 = 2 R_1\).
- The drain pipe can empty the cistern by itself in 8 hours. Its rate is \(-\frac{V}{8}\). Assuming \(V=1\), \(R_{drain} = -\frac{1}{8}\) cistern per hour.
- The two inlet pipes and the drain pipe working together can fill the empty cistern in 8 hours. Their combined rate is \(\frac{V}{8}\). Assuming \(V=1\), the combined rate is \(\frac{1}{8}\) cistern per hour.
Combined Rate Equation
The combined rate when all three pipes work together is the sum of their individual rates:
Combined Rate = \(R_1 + R_2 + R_{drain}\)
We know the combined rate is \(1/8\) and \(R_{drain} = -1/8\). We also know \(R_2 = 2 R_1\). Substituting these into the equation:
\(R_1 + (2 R_1) + (-\frac{1}{8}) = \frac{1}{8}\)
Solving for the Less Efficient Pipe's Rate (\(R_1\))
Now, we solve the equation for \(R_1\):
\(R_1 + 2 R_1 - \frac{1}{8} = \frac{1}{8}\)
\(3 R_1 - \frac{1}{8} = \frac{1}{8}\)
Add \(\frac{1}{8}\) to both sides of the equation:
\(3 R_1 = \frac{1}{8} + \frac{1}{8}\)
\(3 R_1 = \frac{2}{8}\)
\(3 R_1 = \frac{1}{4}\)
Divide by 3 to find \(R_1\):
\(R_1 = \frac{1}{4 \times 3}\)
\(R_1 = \frac{1}{12}\)
So, the less efficient inlet pipe fills \(\frac{1}{12}\) of the cistern per hour.
Calculating the Time for the Less Efficient Pipe Alone
The time taken for a pipe to fill the cistern alone is the capacity of the cistern divided by its rate. Since we assumed the capacity \(V=1\), the time taken by the less efficient pipe is:
Time = \(\frac{\text{Capacity}}{R_1} = \frac{1}{R_1}\)
Substitute the value of \(R_1\) we found:
Time = \(\frac{1}{\frac{1}{12}} = 1 \times \frac{12}{1} = 12\)
Therefore, the less efficient inlet pipe will take 12 hours to fill the empty cistern by itself.
| Pipe | Rate (Cistern per hour) | Time (Hours) |
|---|---|---|
| Less Efficient Inlet (\(R_1\)) | \(\frac{1}{12}\) | 12 |
| More Efficient Inlet (\(R_2\)) | \(\frac{1}{6}\) (\(2 \times \frac{1}{12}\)) | 6 (\(1 \div \frac{1}{6}\)) |
| Drain (\(R_{drain}\)) | \(-\frac{1}{8}\) | 8 (to empty) |
| Combined (Inlets + Drain) | \(\frac{1}{8}\) (\(\frac{1}{12} + \frac{1}{6} - \frac{1}{8} = \frac{2}{24} + \frac{4}{24} - \frac{3}{24} = \frac{6-3}{24} = \frac{3}{24} = \frac{1}{8}\)) | 8 |
Conclusion
The less efficient inlet pipe takes 12 hours to fill the empty cistern by itself.
Revision Table: Cistern Pipe Problem
Here's a quick summary of the key elements in solving this type of problem:
| Concept | Explanation | Formula/Approach |
|---|---|---|
| Rate of Work | The amount of work done per unit of time (e.g., fraction of cistern filled per hour). | Rate = Work / Time |
| Work Done | The total task (e.g., filling the entire cistern, usually considered as 1 unit of work). | Work = Rate × Time |
| Time Taken | The duration required to complete the work. | Time = Work / Rate |
| Multiple Workers (Pipes) | When multiple entities work together, their rates are added (or subtracted for draining) to find the combined rate. | Combined Rate = Sum of Individual Rates |
| Efficiency | Related to the rate of work. Higher efficiency means a higher rate. If pipe A is twice as efficient as pipe B, RateA = 2 × RateB. | Efficiency \(\propto\) Rate |
Additional Information: Solving Work and Time Problems
Work and time problems often involve calculating the rate at which a task is completed. The fundamental principle is that Rate \(\times\) Time = Work. When dealing with pipes filling or emptying a tank, the "work" is usually filling (positive rate) or emptying (negative rate) the entire tank (which can be treated as 1 unit of work).
- If a pipe fills a tank in \(T\) hours, its filling rate is \(1/T\) of the tank per hour.
- If a pipe empties a tank in \(T\) hours, its emptying rate is \(-1/T\) of the tank per hour.
- When multiple pipes work together, their rates are combined. If pipes A and B fill at rates \(R_A\) and \(R_B\), and pipe C empties at rate \(R_C\), the net combined rate when all work together is \(R_A + R_B + R_C\).
- If the combined rate is \(R_{combined}\) and they complete the task (work = 1) in \(T_{combined}\) hours, then \(R_{combined} \times T_{combined} = 1\), or \(R_{combined} = 1 / T_{combined}\).
These principles allow you to set up equations based on the given information and solve for the unknown rate or time.
Similar Questions
Two pipes A and B can fill an empty cistern in 32 and 48 hours, respectively. Pipe C can drain the entire cistern in 64 hours when no other pipe is in operation. Initially, when the cistern was empty Pipe A and Pipe C were turned on. After a few hours, Pipe A was turned off and Pipe B was turned on instantly. In all it took 112 hours to fill the cistern. For how many hours was Pipe B turned on?
Pipes A, B and C are attached to an empty cistern. While the first two can fill the cistern in 4 and 10 hours, respectively, the third can drain the cistern, when filled, in 6 hours. If all the three pipes are opened simultaneously when the cistern is three-fifth full, how many hours will be needed to fill the cistern?
One pipe can fill an empty cistern in 4 hours while another can drain the cistern when full in 10 hours. Both the pipes were turned on when the cistern was half-empty. How long will it take the cistern to be full?
A pipe, working at full speed, can fill an empty cistern in 1 hour. However, during the first hour it worked at one-twelfth of its capacity, during the second hour at one-ninth of its capacity, during the third hour at one-sixth of its usual capacity, during the fourth hour at one- fourth of its usual capacity and during the fifth hour it was only one-third as efficient as it was supposed to be. A second pipe also displayed similar performance, but if it worked at full speed would have filled the empty cistern in 2 hours. Together with a drain pipe that drained water out of the tank at a constant rate, the empty cistern could be filled in 5 hours, all the three pipes working concurrently. How many hours will it take the drain pipe to empty the filled cistern if no other pipe was functioning during the time?
Two pipes, when working one at a time can fill a cistern in 2 hours and 3 hours, respectively while a third pipe can drain the cistern empty in 6 hours. All the three pipes were opened together when the cistern was 1/6 full. How long will it take for the cistern to be completely full?
A tanker can fill a cistern in 12 hours. After half the cistern is filled, 2 more similar tankers are opened. What is the time taken to fill the remaining half of the tank?
P, Q and R are channels which discharge solutions A, B, C respectively in a tank. When the tank is empty and all the three channels are opened what is the proportion of the solution C in the tank after 3 minutes, if the channels P, Q and R can fill the tank from empty to full in 30 minutes and 20 minutes and 10 minutes respectively when they are opened one at a time.
A sump is filled by three tankers with uniform flow. The first two tankers operating simultaneously fill the sump in the same time during which the sump is filled by the third tanker alone. The second tanker fills the sump 5 hours faster than the first tanker and 4 hours slower than the third tanker. The time required by the first tanker is:
Pipe A can fill a tank in 6 hours. Pipe B can empty it in 15 hours. If both the pipes are opened together, then the tank will be filled in how many hours?
- Two pipes fill a tank when working individually in 25 and 40 hours, respectively while a third pipe can drain the filled tank in 16 hours. If all the three pipes are turned on at the same time when the tank is empty, how long will it take to fill the tank completely?
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