Pipes A and B can fill a tank in 12 hours and 16 hours respectively and pipe C can empty the full tank in 24 hours. All three pipes are opened together, but after 4 hours pipe B is closed. In how many hours, the empty tank will be completely filled?
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Solving the Pipes and Cisterns Problem
This problem involves calculating the time taken to fill a tank when multiple pipes, including one that empties, are working together for a duration and then some pipes are closed. We will use the concept of work rate, where the total work (filling the tank) is represented by the tank's capacity, and the rate is the amount filled or emptied per unit of time (in this case, per hour).
Calculating Individual Pipe Work Rates
First, let's determine the work rate of each pipe. The capacity of the tank can be assumed as a common multiple of the times taken by the pipes individually. The times are 12 hours (A), 16 hours (B), and 24 hours (C). The least common multiple (LCM) of 12, 16, and 24 is 48. Let's assume the tank capacity is 48 units.
- Pipe A fills the tank in 12 hours. Its filling rate is $\frac{48}{12} = 4$ units per hour.
- Pipe B fills the tank in 16 hours. Its filling rate is $\frac{48}{16} = 3$ units per hour.
- Pipe C empties the tank in 24 hours. Its emptying rate is $\frac{48}{24} = 2$ units per hour. Since it empties, we consider its rate as negative when working with filling pipes.
Work Done in the First 4 Hours
Initially, all three pipes (A, B, and C) are opened together. Their combined work rate is the sum of their individual rates, considering filling as positive and emptying as negative.
Combined rate of A, B, and C = Rate of A + Rate of B - Rate of C
Combined rate = $4 + 3 - 2 = 5$ units per hour.
These three pipes work together for 4 hours. The amount of tank filled in these 4 hours is:
Work done in 4 hours = Combined rate × Time
Work done = $5 \text{ units/hour} \times 4 \text{ hours} = 20$ units.
Remaining Work After Pipe B is Closed
The total capacity of the tank is 48 units. After 4 hours, 20 units of the tank are filled. The remaining capacity to be filled is:
Remaining work = Total capacity - Work done in first 4 hours
Remaining work = $48 \text{ units} - 20 \text{ units} = 28$ units.
At this point, pipe B is closed. Only pipes A and C remain open.
Time Taken to Complete the Remaining Work
With pipe B closed, only pipes A and C are working. Pipe A is a filling pipe, and pipe C is an emptying pipe. Their combined work rate is:
Combined rate of A and C = Rate of A - Rate of C
Combined rate = $4 - 2 = 2$ units per hour.
This is the rate at which the tank is being filled by pipes A and C together. The remaining work is to fill 28 units.
Time taken for remaining work = Remaining work / Combined rate of A and C
Time taken = $\frac{28 \text{ units}}{2 \text{ units/hour}} = 14$ hours.
Total Time to Fill the Tank
The total time taken to fill the empty tank completely is the sum of the time spent in the first phase (with A, B, and C) and the time spent in the second phase (with A and C).
Total time = Time in first phase + Time in second phase
Total time = $4 \text{ hours} + 14 \text{ hours} = 18$ hours.
Summary of Steps
| Step | Action | Calculation | Result |
| 1 | Assume Tank Capacity (LCM of 12, 16, 24) | LCM(12, 16, 24) | 48 units |
| 2 | Calculate Rate of A | 48 / 12 | 4 units/hr |
| 3 | Calculate Rate of B | 48 / 16 | 3 units/hr |
| 4 | Calculate Rate of C (Emptying) | 48 / 24 | 2 units/hr |
| 5 | Combined Rate (A+B-C) for first 4 hours | 4 + 3 - 2 | 5 units/hr |
| 6 | Work done in first 4 hours | 5 × 4 | 20 units |
| 7 | Remaining Work | 48 - 20 | 28 units |
| 8 | Combined Rate (A-C) after B is closed | 4 - 2 | 2 units/hr |
| 9 | Time for remaining work | 28 / 2 | 14 hours |
| 10 | Total Time | 4 + 14 | 18 hours |
Therefore, the empty tank will be completely filled in a total of 18 hours.
Revision Table: Pipes and Cisterns Concepts
| Concept | Explanation | Formula/Relationship |
| Work Rate | Amount of work done per unit of time. For pipes, it's the fraction or amount of tank filled/emptied per hour/minute. | Rate = Total Work / Time |
| Total Work | The capacity of the tank to be filled. Often assumed as the LCM of individual times. | Usually represented as 1 (for the whole tank) or LCM of times. |
| Filling Pipe Rate | Positive contribution to filling the tank. | Rate = +Capacity / Time |
| Emptying Pipe Rate | Negative contribution (removing liquid) from the tank. | Rate = -Capacity / Time |
| Combined Rate | Sum of individual rates, with emptying rates subtracted. | Combined Rate = Sum of Filling Rates - Sum of Emptying Rates |
Additional Information: Solving Time and Work Problems
Pipes and cisterns problems are a common application of the 'Time and Work' concept. Here are some key points related to solving such problems:
- Unitary Method: You can calculate the fraction of the tank filled or emptied by each pipe in one hour. For example, if a pipe fills a tank in 12 hours, it fills $\frac{1}{12}$ of the tank in one hour.
- Efficiency Concept: Rate of work is often referred to as efficiency. Higher rate means higher efficiency and less time taken for the same work.
- LCM Method: Assuming the total work (tank capacity) as the LCM of the given times simplifies calculations, especially when dealing with multiple pipes. This avoids fractions until the final steps.
- Variable Rates: Some problems might involve pipes with changing rates or pipes opening/closing at different times, requiring calculations for specific time intervals.
- Leakage: A leak in the tank acts like an emptying pipe, reducing the effective filling rate.
Understanding how to calculate individual rates and then combine them based on whether pipes are filling or emptying is crucial for solving pipes and cisterns questions effectively in competitive exams.
Important Questions from Pipe and Cistern
Pipes A, B and C can fill a tank in 20, 30 and 60 hours, respectively. Pipes A, B and C are opened at 7 a.m., 8 a.m., and 9 a.m., respectively, on the same day. When will the tank be full?
There are two water taps in a tank which can fill the empty tank in 12 hours and 18 hours respectively. It is seen that there is a leakage point at the bottom of the tank which can empty the completely filled tank in 36 hours. If both the water taps are opened at the same time to fill the empty tank and leakage point was repaired after 1 hour, then in how much time the empty tank will be completely filled?
Two pipes A and B can fill a tank in 12 minutes and 24 minutes, respectively, while a third pipe C can empty the full tank in 32 minutes. All the three pipes are opened simultaneously. However, pipe C is closed 2 minutes before the tank is filled. In how much time (in minutes) will the tank be full?
Pipes A and B can fill a tank in 43.2 minutes and 108 minutes, respectively. Pipe C can empty it at 3 litres/minute. When all the three pipes are opened together, they fill the tank in 54 minutes. The capacity (in litres) of the tank is:
Pipes A and B can fill a tank in 12 minutes and 15 minutes, respectively. The tank when full can be emptied by pipe C in x minutes. When all the three pipes are opened simultaneously, the tank is full in 10 minutes. The value of x is:
