Solving Pipe and Cistern Problems

Pipe and cistern problems are common in quantitative aptitude tests. They are similar to time and work problems, where instead of individuals doing work, pipes fill or empty a tank (cistern).

Understanding Work Rate

The key concept is the work rate, which is the amount of work done (or fraction of the cistern filled/emptied) per unit of time. If a pipe can fill a cistern in \(N\) minutes, its work rate is \(\frac{1}{N}\) of the cistern per minute.

Calculating Individual Pipe Rates

• Pipe A fills the cistern in 9 minutes.
• Work rate of Pipe A = \(\frac{1}{9}\) of the cistern per minute.
• Pipe B fills the cistern in 6 minutes.
• Work rate of Pipe B = \(\frac{1}{6}\) of the cistern per minute.

Analyzing the Problem Scenario

Both pipes A and B are turned on simultaneously. The cistern is filled in a total of 4 minutes. Pipe B is closed at some point before the 4 minutes are over. Pipe A works for the entire duration of 4 minutes.

Setting up the Equation

Let \(t\) be the time (in minutes) for which Pipe B is open. This is the time from the start until Pipe B is closed.

• Pipe A works for 4 minutes.
• Work done by Pipe A in 4 minutes = Work rate of A \(\times\) Time A worked
• Work done by Pipe A = \(\frac{1}{9} \times 4 = \frac{4}{9}\) of the cistern.

Pipe B works for \(t\) minutes.

• Work done by Pipe B in \(t\) minutes = Work rate of B \(\times\) Time B worked
• Work done by Pipe B = \(\frac{1}{6} \times t = \frac{t}{6}\) of the cistern.

The total work done by both pipes is filling the entire cistern, which is represented as 1 unit of work.

Total work = Work done by A + Work done by B

So, the equation is:

\(\frac{4}{9} + \frac{t}{6} = 1\)

Solving for \(t\)

Now, we need to solve this equation for \(t\), which represents the time when Pipe B was closed.

Start with the equation:

\(\frac{4}{9} + \frac{t}{6} = 1\)

Subtract \(\frac{4}{9}\) from both sides:

\(\frac{t}{6} = 1 - \frac{4}{9}\)

Find a common denominator for the right side (which is 9):

\(\frac{t}{6} = \frac{9}{9} - \frac{4}{9}\)

\(\frac{t}{6} = \frac{9-4}{9}\)

\(\frac{t}{6} = \frac{5}{9}\)

Multiply both sides by 6 to isolate \(t\):

\(t = \frac{5}{9} \times 6\)

\(t = \frac{30}{9}\)

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 3:

\(t = \frac{30 \div 3}{9 \div 3}\)

\(t = \frac{10}{3}\)

The value of \(t\) is \(\frac{10}{3}\) minutes.

This value \(t\) is the duration for which Pipe B was open, starting from when both pipes were turned on. Therefore, Pipe B should be closed after \(\frac{10}{3}\) minutes.

Conclusion

To fill the cistern in exactly 4 minutes, starting with both pipes A and B open, Pipe B must be closed after \(\frac{10}{3}\) minutes.

Revision Table: Key Values

Additional Information on Pipe and Cistern Problems

Pipe and cistern problems often involve calculating the combined work rate of multiple pipes. If pipes are filling, their rates are added. If a pipe is emptying, its rate is subtracted.

• Combined Rate (Filling): If pipe 1 fills in \(N_1\) and pipe 2 fills in \(N_2\), their combined rate is \(\frac{1}{N_1} + \frac{1}{N_2}\). The time to fill together is \(\frac{1}{\text{Combined Rate}}\).
• Combined Rate (Filling and Emptying): If pipe 1 fills in \(N_1\) and pipe 2 empties in \(N_2\), their combined rate is \(\frac{1}{N_1} - \frac{1}{N_2}\) (assuming \(N_1 < N_2\)).
• Problems can involve pipes working for different durations, as seen in this question, or pipes being turned on/off at different times. Always calculate the fraction of work done by each pipe during its operational time and sum them up to equal 1 (for a full cistern).