If operated alone, an inlet pipe to a water tank takes 320 hours to fill. Starting from the second hour, two extra inlet pipes are added every hour, but one new leakage devel that each leakage is as efficient as an inlet pipe, during which hour will the tank be filled?

Let's work through the problem step-by-step:

1. **Understand the Inlet Pipe Efficiency**: 
   The main inlet pipe alone would fill the tank in 320 hours. So, each hour it fills:
   \[
   \frac{1}{320} \text{ of the tank.}
   \]

2. **Additional Pipes and Leakages**:
   - Starting from the second hour, two extra pipes are added every hour. This means that each additional inlet pipe also fills \(\frac{1}{320}\) of the tank per hour.
   - Simultaneously, a leakage develops each hour starting from the second hour. Each leakage has the same capacity as an inlet pipe, meaning it also "empties" the tank at a rate of \(\frac{1}{320}\) per hour.

3. **Filling Rate per Hour**:
   Let's calculate the net filling rate for each hour.

   - **Hour 1**: Only 1 inlet pipe is open, so the filling rate is:
     \[
     1 \times \frac{1}{320} = \frac{1}{320}
     \]

   - **Hour 2**: There are now 3 inlet pipes (1 original + 2 new), but also 1 leakage. The net filling rate is:
     \[
     (3 - 1) \times \frac{1}{320} = \frac{2}{320} = \frac{1}{160}
     \]

   - **Hour 3**: There are now 5 inlet pipes (1 original + 4 new), and 2 leakages. The net filling rate is:
     \[
     (5 - 2) \times \frac{1}{320} = \frac{3}{320}
     \]

   - **Hour 4**: There are 7 inlet pipes, and 3 leakages. The net filling rate is:
     \[
     (7 - 3) \times \frac{1}{320} = \frac{4}{320} = \frac{1}{80}
     \]

   This pattern continues each hour.

4. **General Formula for Hour \( n \)**:
   For any given hour \( n \), the number of inlet pipes is \( 1 + 2(n-1) = 2n - 1 \), and the number of leakages is \( n - 1 \).

   Therefore, the net number of pipes (inlet pipes minus leakages) in hour \( n \) is:
   \[
   (2n - 1) - (n - 1) = n
   \]

   So, the net filling rate in hour \( n \) is:
   \[
   n \times \frac{1}{320} = \frac{n}{320}
   \]

5. **Calculate the Cumulative Filling**:
   To find when the tank is completely filled, we need the cumulative filling rate to equal 1. Thus, we solve the following inequality:
   \[
   \sum_{n=1}^{k} \frac{n}{320} \geq 1
   \]

   The sum of the first \( k \) natural numbers is given by \( \frac{k(k+1)}{2} \). So we have:
   \[
   \frac{1}{320} \cdot \frac{k(k+1)}{2} \geq 1
   \]

   Simplifying, we get:
   \[
   \frac{k(k+1)}{640} \geq 1
   \]
   \[
   k(k+1) \geq 640
   \]

6. **Solve the Inequality**:
   Solving \( k(k+1) \geq 640 \):
   - For \( k = 25 \):
     \[
     25 \times 26 = 650 \quad (\text{which satisfies the inequality})
     \]
   - For \( k = 24 \):
     \[
     24 \times 25 = 600 \quad (\text{which does not satisfy the inequality})
     \]

   Therefore, the tank will be filled during the **25th hour**. 

### Answer:
The tank will be filled during the **25th hour**.

Would you like more details or have any questions?

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Here are 5 related questions you could explore:
1. How would the solution change if one more leakage occurred every two hours?
2. What if the inlet pipes were each 20% more efficient?
3. How long would it take if only one extra inlet pipe and one leakage were added every hour?
4. How does the solution differ if leakages reduce efficiency by half, rather than being fully equivalent to inlet pipes?
5. Can we express this solution as a general formula for any number of initial hours?

**Tip**: Summing sequences is often useful in problems involving cumulative rates over time.

If operated alone, an inlet pipe to a water tank takes 320 hours to fill. Starting from the second hour, two extra inlet pipes are added every hour, but one new leakage develops each hour such that each leakage is as efficient as an inlet pipe. During which hour will the tank be filled?

Explore a step-by-step solution to a complex water tank filling problem involving multiple inlet pipes and leakages. Learn how to calculate cumulative rates over time using sequences and series, and determine when the tank will be full. Suitable for Grades 10-12 students.

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