A reservoir has two pipes, A and B. A can fill the reservoir 5 hours faster than B. If both together fill the reservoir in 6 hours, the reservoir will be filled by A alone in-

<h1>Understanding the Solution to the Problem</h1><h2>Problem Statement</h2>

A reservoir has two pipes, A and B. Pipe A can fill the reservoir 5 hours faster than pipe B. If both pipes together fill the reservoir in 6 hours, how long will pipe A alone take to fill the reservoir?

<h2>Answer Choices</h2>

• 8 hours

• 10 hours

• 11 hours

• 12 hours

<h2>Correct Answer: 10 hours</h2><h2>Reasoning and Detailed Explanation</h2>

Let's denote the time taken by pipe A to fill the reservoir alone as $ t_A $ hours, and the time taken by pipe B to fill the reservoir alone as $ t_B $ hours.

According to the problem, pipe A can fill the reservoir 5 hours faster than pipe B. This gives us the equation:

$ t_A = t_B - 5 $

It is also given that both pipes together can fill the reservoir in 6 hours. The combined rate of both pipes filling the reservoir is the sum of their individual rates. We can express the rates as:

The rate of pipe A is $ \frac{1}{t_A} $ reservoirs per hour, and the rate of pipe B is $ \frac{1}{t_B} $ reservoirs per hour.

When both pipes work together, their combined rate is:

$ \frac{1}{t_A} + \frac{1}{t_B} = \frac{1}{6} $

Substituting $ t_A = t_B - 5 $ into the equation, we get:

$ \frac{1}{t_B - 5} + \frac{1}{t_B} = \frac{1}{6} $

To solve for $ t_B $, we need a common denominator on the left-hand side:

$ \frac{t_B + (t_B - 5)}{t_B(t_B - 5)} = \frac{1}{6} $

Simplify the numerator:

$ \frac{2t_B - 5}{t_B^2 - 5t_B} = \frac{1}{6} $

Cross multiply to eliminate the fractions:

$ 6(2t_B - 5) = t_B^2 - 5t_B $

This simplifies to:

$ 12t_B - 30 = t_B^2 - 5t_B $

Rearrange to form a quadratic equation:

$ t_B^2 - 17t_B + 30 = 0 $

To solve this quadratic equation, we can use the quadratic formula:

$ t_B = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $

Where $ a = 1 $, $ b = -17 $, and $ c = 30 $.

Substituting these values, we get:

$ t_B = \frac{17 \pm \sqrt{289 - 120}}{2} = \frac{17 \pm \sqrt{169}}{2} = \frac{17 \pm 13}{2} $

This gives us two solutions:

$ t_B = \frac{17 + 13}{2} = 15 $ and $ t_B = \frac{17 - 13}{2} = 2 $

Since it is not practical for pipe B to fill the reservoir in 2 hours (as pipe A would then take -3 hours, which is impossible), we discard the solution $ t_B = 2 $.

Thus, $ t_B = 15 $. Using the equation $ t_A = t_B - 5 $, we find:

$ t_A = 15 - 5 = 10 $

Therefore, pipe A alone takes 10 hours to fill the reservoir. Hence, the correct answer is 10 hours.