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?
3 hours 20 minutes
Solving Cistern and Pipe Problems: Fill and Drain Rates
This question involves calculating the time it takes to fill a cistern when two pipes are working simultaneously: one filling and one draining. These types of problems are common in aptitude tests and require understanding the concept of work rates.
Understanding Individual Pipe Rates
First, let's determine the rate at which each pipe performs its task. The rate is usually expressed as the fraction of the cistern filled or drained per unit of time (in this case, per hour).
- The first pipe can fill the entire cistern in 4 hours. This means its filling rate is $\frac{1 \text{ cistern}}{4 \text{ hours}} = \frac{1}{4}$ of the cistern per hour.
- The second pipe can drain the entire cistern in 10 hours. This means its draining rate is $\frac{1 \text{ cistern}}{10 \text{ hours}} = \frac{1}{10}$ of the cistern per hour.
Calculating the Combined Rate
When both pipes are turned on, one is adding water (filling) and the other is removing water (draining). To find their combined effect, we subtract the draining rate from the filling rate, as the draining pipe works against the filling pipe.
Combined rate = Filling rate - Draining rate
Combined rate $= \frac{1}{4} - \frac{1}{10}$ per hour.
To subtract these fractions, we need a common denominator. The least common multiple of 4 and 10 is 20.
- $\frac{1}{4} = \frac{1 \times 5}{4 \times 5} = \frac{5}{20}$
- $\frac{1}{10} = \frac{1 \times 2}{10 \times 2} = \frac{2}{20}$
So, the combined rate is:
Combined rate $= \frac{5}{20} - \frac{2}{20} = \frac{5-2}{20} = \frac{3}{20}$ of the cistern per hour.
This means that with both pipes operating, the cistern fills up at a net rate of $\frac{3}{20}$ of its capacity every hour.
Determining the Volume to be Filled
The problem states that both pipes were turned on when the cistern was half-empty. This means the cistern was already half-full. The portion of the cistern that still needs to be filled is the remaining half.
Volume to be filled = Full capacity - Current volume
Volume to be filled $= 1 - \frac{1}{2} = \frac{1}{2}$ of the cistern.
Calculating the Time to Fill the Remaining Volume
Now that we know the combined rate and the volume that needs to be filled, we can find the time required. The relationship is:
Time = $\frac{\text{Volume to be filled}}{\text{Combined rate}}$
Time $= \frac{1/2}{3/20}$ hours.
To divide by a fraction, we multiply by its reciprocal:
Time $= \frac{1}{2} \times \frac{20}{3} = \frac{1 \times 20}{2 \times 3} = \frac{20}{6}$ hours.
This fraction can be simplified:
Time $= \frac{10}{3}$ hours.
Converting Time to Hours and Minutes
The time is $\frac{10}{3}$ hours. Let's convert this into hours and minutes. $\frac{10}{3}$ hours $= 3 \frac{1}{3}$ hours.
The 3 represents 3 full hours. The $\frac{1}{3}$ is a fraction of an hour, which we convert to minutes by multiplying by 60 (since there are 60 minutes in an hour).
Minutes $= \frac{1}{3} \times 60$ minutes $= 20$ minutes.
So, the total time required to fill the remaining half of the cistern is 3 hours and 20 minutes.
| Task | Rate per hour |
|---|---|
| Pipe 1 (Filling) | $\frac{1}{4}$ |
| Pipe 2 (Draining) | $\frac{1}{10}$ |
| Combined Rate (Filling) | $\frac{1}{4} - \frac{1}{10} = \frac{3}{20}$ |
| Volume to fill | $\frac{1}{2}$ |
| Time taken | $\frac{1/2}{3/20} = \frac{10}{3}$ hours |
| Time in Hours & Minutes | 3 hours 20 minutes |
Final Answer
It will take 3 hours and 20 minutes for the cistern to be full, starting from half-empty, with both pipes operating.
Revision Table: Cistern and Pipe Problems
| Concept | Explanation | Formula |
|---|---|---|
| Individual Rate | Amount of work (fraction of cistern) done by a pipe in one unit of time. | Rate = $\frac{1}{\text{Time taken to complete the job}}$ |
| Combined Rate (Filling) | Sum of individual filling rates when pipes work together to fill. | $R_{total} = R_1 + R_2 + ...$ |
| Combined Rate (Filling & Draining) | Difference between total filling rate and total draining rate. | $R_{total} = (R_{fill1} + ...) - (R_{drain1} + ...)$ |
| Time Taken | Total volume of work divided by the combined rate. | Time = $\frac{\text{Total Work}}{\text{Combined Rate}}$ |
Additional Information: Work and Time Concepts
Cistern and pipe problems are a specific type of 'Work and Time' problems. The core idea is that if a person or a pipe can complete a task (like filling a cistern) in 'T' hours, their rate of work is $1/T$ of the task per hour. When multiple entities work together, their rates are usually added (if they work towards the same goal) or subtracted (if they work against each other).
- If two pipes A and B fill a cistern in $T_A$ and $T_B$ hours respectively, their combined rate of filling is $\frac{1}{T_A} + \frac{1}{T_B}$ per hour.
- If a pipe A fills in $T_A$ hours and a pipe B drains in $T_B$ hours, their combined rate when working together is $\frac{1}{T_A} - \frac{1}{T_B}$ per hour (assuming $T_A < T_B$, otherwise the cistern won't fill).
- The total work is usually considered as '1 unit' (the full cistern). If only a fraction of the work needs to be done (like filling half a cistern), that fraction is used as the 'Total Work' value in the calculation for 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?
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?
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?
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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