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Let's solve this problem step by step.

### Problem Breakdown:
- Initially, the cask contains only water.
- **9 liters** of water is removed, and the same quantity of **milk** is added. This process is repeated **twice**.
- After these two operations, the ratio of water to milk in the cask is **16:9**.

We need to find the **volume of the cask**.

### Solution:
Let the total volume of the cask be **V liters**.

1. **First operation**:
   - 9 liters of water is taken out, so the remaining water in the cask is \( V - 9 \).
   - The fraction of water remaining after the first operation is:
     \[
     \frac{V-9}{V}
     \]
   - The rest of the volume (9 liters) is replaced by milk.

2. **Second operation**:
   - Again, 9 liters of the mixture (water + milk) is removed. The fraction of water remaining after the second operation (after removing the water-milk mixture) is:
     \[
     \frac{V-9}{V} \times \frac{V-9}{V}
     \]
     This gives the fraction of water left in the cask after both operations.

3. After two operations, the ratio of water to milk is given as **16:9**. The fraction of water in the cask is:
   \[
   \frac{16}{16+9} = \frac{16}{25}
   \]
   Therefore, after two operations, the fraction of water remaining in the cask is also \( \frac{16}{25} \).

4. **Equating the two fractions**:
   \[
   \left( \frac{V-9}{V} \right)^2 = \frac{16}{25}
   \]

5. **Solving for V**:
   Taking the square root of both sides:
   \[
   \frac{V-9}{V} = \frac{4}{5}
   \]
   Multiply both sides by \( V \):
   \[
   V - 9 = \frac{4}{5}V
   \]
   Now solve for \( V \):
   \[
   V - \frac{4}{5}V = 9
   \]
   \[
   \frac{1}{5}V = 9
   \]
   \[
   V = 9 \times 5 = 45
   \]

Thus, the **volume of the cask is 45 liters**.

Would you like more details or have any questions? Here are some related questions:

1. What would the water to milk ratio be after just one operation?
2. How would the solution change if the ratio was different, say 2:1 instead of 16:9?
3. Can you calculate how much milk would be in the cask after the two operations?
4. How does the fraction of water change if more than 9 liters is removed each time?
5. What would happen if the process is repeated more than twice?

### Tip:
In mixture problems like this, it's often helpful to think in terms of fractions remaining after each operation.

From a cask containing water, 9 L is taken out. It is replaced with an equal quantity of pure milk. This process is done twice. The ratio of water to milk in the cask now is 16:9. What is the volume of the cask?

Solve the classic mixture problem where water is replaced by milk in a cask. Given that 9 liters of water is replaced by milk twice, and the final ratio of water to milk is 16:9, find the volume of the cask. Learn the algebraic steps to reach the solution.

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