Solving the Pipe and Cistern Problem

This problem involves the concept of work rate, applied to pipes filling a cistern. The rate at which a pipe fills a cistern is the reciprocal of the time it takes to fill the cistern alone.

Let's define the variables based on the question:

• Let the time taken by Pipe A alone to fill the cistern be \(t_A\) minutes.
• Let the time taken by Pipe B alone to fill the cistern be \(t_B\) minutes.

Based on these times, their filling rates are:

• Rate of Pipe A = \( \frac{1}{t_A} \) of the cistern per minute.
• Rate of Pipe B = \( \frac{1}{t_B} \) of the cistern per minute.

Setting Up Equations for Pipe Filling Times

We are given two key pieces of information:

1. Pipe B takes 5 minutes more than Pipe A to fill the cistern.
2. Pipe A and Pipe B together can fill the cistern in 6 minutes.

From the first statement, we can write the equation:

\( t_B = t_A + 5 \)

From the second statement, their combined rate is the sum of their individual rates, and this combined rate completes the work (fills the cistern) in 6 minutes. So, the combined rate is \( \frac{1}{6} \) of the cistern per minute.

This gives us the equation:

\( \frac{1}{t_A} + \frac{1}{t_B} = \frac{1}{6} \)

Solving for the Individual Pipe Times

Now we have a system of two equations. We can substitute the first equation (\( t_B = t_A + 5 \)) into the second equation:

\( \frac{1}{t_A} + \frac{1}{t_A + 5} = \frac{1}{6} \)

To solve for \(t_A\), we find a common denominator on the left side, which is \(t_A(t_A + 5)\):

\( \frac{(t_A + 5) + t_A}{t_A(t_A + 5)} = \frac{1}{6} \)

\( \frac{2t_A + 5}{t_A^2 + 5t_A} = \frac{1}{6} \)

Now, we can cross-multiply:

\( 6(2t_A + 5) = 1(t_A^2 + 5t_A) \)

\( 12t_A + 30 = t_A^2 + 5t_A \)

Rearrange the terms to form a standard quadratic equation (\(at^2 + bt + c = 0\)):

\( t_A^2 + 5t_A - 12t_A - 30 = 0 \)

\( t_A^2 - 7t_A - 30 = 0 \)

We can solve this quadratic equation by factoring. We look for two numbers that multiply to -30 and add up to -7. These numbers are -10 and +3.

\( (t_A - 10)(t_A + 3) = 0 \)

This equation gives two possible solutions for \(t_A\):

• \( t_A - 10 = 0 \implies t_A = 10 \)
• \( t_A + 3 = 0 \implies t_A = -3 \)

Since time cannot be negative, the solution \( t_A = -3 \) is not valid in this context. Therefore, the time taken by Pipe A alone is \( t_A = 10 \) minutes.

Now we can find the time taken by Pipe B using the relationship \( t_B = t_A + 5 \):

\( t_B = 10 + 5 = 15 \)

So, the time taken by Pipe B alone is \( t_B = 15 \) minutes.

Verifying the Solution for Pipe Times

Let's check if these times work when the pipes run together. If Pipe A takes 10 minutes and Pipe B takes 15 minutes:

• Rate of A = \( \frac{1}{10} \) per minute.
• Rate of B = \( \frac{1}{15} \) per minute.

Combined rate = \( \frac{1}{10} + \frac{1}{15} \)

The least common multiple of 10 and 15 is 30.

Combined rate = \( \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6} \) per minute.

If the combined rate is \( \frac{1}{6} \) of the cistern per minute, the time taken to fill the entire cistern (1 whole) is \( 1 \div \frac{1}{6} = 1 \times 6 = 6 \) minutes.

This matches the information given in the question that the pipes together fill the cistern in 6 minutes.

Conclusion

The time in which Pipe A and Pipe B will fill the cistern separately are 10 minutes and 15 minutes, respectively.

Revision Table: Key Concepts

Additional Information: Pipe and Cistern Problems

Pipe and cistern problems are a common application of time and work concepts. They often involve calculating the time taken by pipes (or taps) to fill or empty a tank. Key things to remember:

• Filling pipes have a positive work rate.
• Emptying pipes (leaks) have a negative work rate.
• If a pipe fills a tank in \(T\) hours, it fills \(1/T\) of the tank in 1 hour.
• If a pipe empties a tank in \(T\) hours, it empties \(1/T\) of the tank in 1 hour (rate is \(-1/T\)).
• When pipes work together, their rates are added (or subtracted if there's an emptying pipe) to find the combined rate.
• The total work is usually considered as 1 unit (representing the full cistern).

Solving these problems often leads to linear or quadratic equations, as seen in this example. Always check if the solutions make sense in the context of the problem (e.g., time must be positive).