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The problem states:

1. The diluted wine mixture contains 8 liters of wine and the rest is water.
2. A new mixture with a 30% concentration of wine needs to be formed by replacing some of the mixture with pure wine.
3. Initially, there were 32 liters of water in the mixture.

Let's break this down step-by-step:

1. **Find the total volume of the initial mixture:**

   Since there are 8 liters of wine and 32 liters of water, the total initial volume of the mixture is:
   \[
   8 + 32 = 40 \text{ liters}
   \]

2. **Define variables:**

   Let \( x \) be the number of liters of the mixture replaced by pure wine.

3. **Set up the mixture after replacing:**

   When \( x \) liters of the mixture are replaced, the amount of wine added is \( x \) liters because pure wine is 100% wine. The amount of wine left from the original mixture will be \( 8 - \frac{8}{40}x \) since the mixture contains 20% wine (\( \frac{8}{40} \)).

   So, the new amount of wine in the mixture is:
   \[
   (8 - \frac{8}{40}x) + x = 8 + \frac{32}{40}x
   \]

   The new total volume of the mixture remains 40 liters.

4. **Set up the equation for the concentration of wine:**

   To achieve 30% concentration of wine:
   \[
   \frac{8 + \frac{32}{40}x}{40} = 0.3
   \]

5. **Solve the equation:**

   Let’s solve for \( x \).The solution is \( x = 5 \). This means that 5 liters of the original mixture need to be replaced with pure wine to achieve the desired 30% concentration of wine.

Would you like more details on the steps, or do you have any questions?

Here are some related questions to further explore:

1. How does the concentration change if more or less of the mixture is replaced?
2. What happens to the concentration if a different volume of water is replaced instead?
3. How would the solution change if the initial mixture had a different concentration of wine?
4. What if the total volume of the initial mixture were different?
5. How would you solve a similar problem where multiple replacements are made?

**Tip:** Always keep track of the initial concentrations and volumes when dealing with mixture problems; setting up a clear equation helps in solving such problems systematically.

The diluted wine contains only 8 liters of wine and the rest is water. A new mixture whose concentration is 30% is to be formed by replacing wine. How many liters of the mixture shall be replaced with pure wine if there was initially 32 liters of water in the mixture?

Learn how to calculate the volume of a mixture that needs to be replaced with pure wine to achieve a 30% concentration. This problem involves understanding mixture problems, concentration calculations, and algebraic equations. Suitable for students in Grades 10-12.

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