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.

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:

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.