There are two pipes connected to a swimming pool. The first pipe fills the pool with water, and the second pipe drains the water. Additionally, it takes 2 hours longer to fill the pool using the first pipe than it takes to drain the pool using the second pipe. At the beginning of the workday, the pool was 40% full. Later, both pipes were opened, and after 9 hours and 36 minutes, the pool was empty.

We are given that two pipes are connected to a swimming pool, with the first pipe filling the pool and the second pipe draining it. The first pipe takes 2 hours longer to fill the pool than the second pipe takes to drain it. Initially, the pool is 40% full, and after both pipes are opened, the pool is empty after 9 hours and 36 minutes.

### Let's define the variables:
- Let \( t \) be the time (in hours) it takes for the second pipe to drain the pool.
- The first pipe fills the pool in \( t + 2 \) hours.

### Rate of filling and draining:
- The rate of the first pipe filling the pool is \( \frac{1}{t+2} \) pools per hour.
- The rate of the second pipe draining the pool is \( \frac{1}{t} \) pools per hour.

### Initial condition:
The pool is initially 40% full, which means 60% of the pool needs to be drained.

### Combined effect of both pipes:
When both pipes are opened, the net rate at which the pool is drained is the difference between the draining rate and the filling rate:
\[
\text{Net rate} = \frac{1}{t} - \frac{1}{t+2}
\]
This is the rate at which the pool is being emptied.

### Time condition:
The pool becomes empty after 9 hours and 36 minutes, which is equivalent to 9.6 hours. The total volume of the pool that needs to be drained is 60% of the pool, or \( 0.6 \) pools.

### Equation setup:
The net rate of emptying the pool multiplied by the time taken should equal 0.6 (since we are draining 60% of the pool):
\[
\left( \frac{1}{t} - \frac{1}{t+2} \right) \times 9.6 = 0.6
\]

Let's solve this equation step by step.

### Simplification:
The net rate can be simplified as:
\[
\frac{1}{t} - \frac{1}{t+2} = \frac{(t+2) - t}{t(t+2)} = \frac{2}{t(t+2)}
\]

Substitute this into the time equation:
\[
\frac{2}{t(t+2)} \times 9.6 = 0.6
\]

Now, solve for \( t \):
\[
\frac{2 \times 9.6}{t(t+2)} = 0.6
\]
\[
\frac{19.2}{t(t+2)} = 0.6
\]
\[
19.2 = 0.6 t(t+2)
\]
\[
32 = t(t+2)
\]
\[
t^2 + 2t - 32 = 0
\]

### Solving the quadratic equation:
The quadratic equation \( t^2 + 2t - 32 = 0 \) can be solved using the quadratic formula:
\[
t = \frac{-2 \pm \sqrt{2^2 - 4(1)(-32)}}{2(1)}
\]
\[
t = \frac{-2 \pm \sqrt{4 + 128}}{2}
\]
\[
t = \frac{-2 \pm \sqrt{132}}{2}
\]
\[
t = \frac{-2 \pm 11.49}{2}
\]
Taking the positive root (since time must be positive):
\[
t = \frac{-2 + 11.49}{2} = \frac{9.49}{2} \approx 4.745
\]

Thus, it takes approximately **4.75 hours** for the second pipe to drain the pool.

### Time for the first pipe to fill the pool:
Since the first pipe takes 2 hours longer, it will take:
\[
t + 2 = 4.75 + 2 = 6.75 \text{ hours}
\]

### Summary:
- The second pipe drains the pool in approximately 4.75 hours.
- The first pipe fills the pool in approximately 6.75 hours.

Let me know if you'd like more details or have any questions!

#### 5 Related Questions:
1. How would the solution change if the pool was 50% full at the start?
2. What if the first pipe took only 1 hour longer than the

This problem explores the rates at which two pipes fill and drain a swimming pool, using algebra and quadratic equations to solve for time. The pool starts 40% full and becomes empty after 9 hours and 36 minutes, with one pipe taking 2 hours longer to fill than the other takes to drain. Suitable for Grades 8-10.

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