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:
20
Understanding Pipe and Cistern Problems
Pipe and cistern problems are a type of time and work problem. The key concept is to determine the amount of work (filling or emptying a tank) done per unit of time by each pipe. If a pipe can fill or empty a tank in 't' units of time, then in one unit of time, it can fill or empty \( \frac{1}{t} \) of the tank.
Setting Up the Rates for Each Pipe
Let's break down the given information:
- Pipe A fills the tank in 12 minutes. Its rate of filling is \( \frac{1}{12} \) of the tank per minute.
- Pipe B fills the tank in 15 minutes. Its rate of filling is \( \frac{1}{15} \) of the tank per minute.
- Pipe C empties the tank in x minutes. Since it's an emptying pipe, its rate is negative: \( -\frac{1}{x} \) of the tank per minute.
Combined Rate of All Three Pipes
When all three pipes (A, B, and C) are opened simultaneously, the tank is full in 10 minutes. This means their combined rate of filling (or emptying) is \( \frac{1}{10} \) of the tank per minute.
The combined rate is the sum of the individual rates, keeping in mind that emptying rates are subtracted from filling rates.
Combined Rate = Rate of A + Rate of B + Rate of C
\( \frac{1}{10} = \frac{1}{12} + \frac{1}{15} + \left(-\frac{1}{x}\right) \)
\( \frac{1}{10} = \frac{1}{12} + \frac{1}{15} - \frac{1}{x} \)
Solving for x
Now we need to solve the equation for \( x \). Let's isolate the term with \( x \):
\( \frac{1}{x} = \frac{1}{12} + \frac{1}{15} - \frac{1}{10} \)
To add and subtract these fractions, find a common denominator for 12, 15, and 10. The least common multiple (LCM) of 12, 15, and 10 is 60.
- Convert \( \frac{1}{12} \) to a fraction with denominator 60: \( \frac{1 \times 5}{12 \times 5} = \frac{5}{60} \)
- Convert \( \frac{1}{15} \) to a fraction with denominator 60: \( \frac{1 \times 4}{15 \times 4} = \frac{4}{60} \)
- Convert \( \frac{1}{10} \) to a fraction with denominator 60: \( \frac{1 \times 6}{10 \times 6} = \frac{6}{60} \)
Substitute these equivalent fractions back into the equation:
\( \frac{1}{x} = \frac{5}{60} + \frac{4}{60} - \frac{6}{60} \)
Now combine the numerators:
\( \frac{1}{x} = \frac{5 + 4 - 6}{60} \)
\( \frac{1}{x} = \frac{9 - 6}{60} \)
\( \frac{1}{x} = \frac{3}{60} \)
Simplify the fraction on the right side:
\( \frac{3}{60} = \frac{1}{20} \)
So, the equation becomes:
\( \frac{1}{x} = \frac{1}{20} \)
Since the numerators are equal (both are 1), the denominators must also be equal.
\( x = 20 \)
Thus, pipe C can empty the full tank in 20 minutes.
Final Answer
The value of \( x \) is 20.
| Pipe | Time to Fill/Empty (mins) | Rate (Tank per minute) |
|---|---|---|
| A (Filling) | 12 | \( \frac{1}{12} \) |
| B (Filling) | 15 | \( \frac{1}{15} \) |
| C (Emptying) | x | \( -\frac{1}{x} \) |
| A + B + C (Combined) | 10 | \( \frac{1}{10} \) |
Revision Table: Key Concepts in Pipe and Cistern Problems
| Concept | Explanation | Formula/Relation |
|---|---|---|
| Work Rate | The amount of work done per unit of time. For pipes, it's the fraction of the tank filled or emptied per minute/hour. | Rate = \( \frac{1}{\text{Time taken}} \) |
| Filling Pipe Rate | Positive rate, adds water to the tank. | \( +\frac{1}{\text{Time to fill}} \) |
| Emptying Pipe Rate | Negative rate, removes water from the tank. | \( -\frac{1}{\text{Time to empty}} \) |
| Combined Rate | Sum of individual rates (filling rates are positive, emptying rates are negative). | Combined Rate = Sum of individual rates |
| Total Time (Combined) | If combined rate is R, time taken to fill/empty the tank is \( \frac{1}{R} \). | Time = \( \frac{1}{\text{Combined Rate}} \) |
Additional Information: Tips for Solving Pipe Problems
Here are some useful tips when tackling pipe and cistern questions:
- Always determine the rate of work done in one unit of time (usually per minute or per hour).
- Represent filling pipes with positive rates and emptying pipes with negative rates.
- When multiple pipes work together, their rates are added to find the combined rate. Be careful with the signs (+ for filling, - for emptying).
- If the question asks for the time taken for a certain fraction of the tank to be filled or emptied, calculate the time to fill/empty the whole tank first, and then multiply by the required fraction. For example, to fill half the tank, it takes half the time needed to fill the whole tank.
- If a pipe works for only a part of the time, calculate the fraction of work done by that pipe during that specific duration.
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 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?
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:
