A jar full of whiskey contains 40% alcohol. A part of this whiskey is replaced by another type containing 25% alcohol, and now the percentage of alcohol is found to be 32.5%. The proportion of whiskey replaced is:

Let’s assume that the jar initially contains 1 liter of whiskey. 

- The initial concentration of alcohol is 40%, so the amount of alcohol is \( 0.40 \times 1 = 0.4 \) liters.

Now, let \( x \) be the fraction of the original whiskey that is replaced by the whiskey containing 25% alcohol. After replacing \( x \) liters of the original whiskey with 25% alcohol whiskey:

- The amount of alcohol left from the original whiskey is \( (1 - x) \times 0.4 \).
- The amount of alcohol added from the new whiskey is \( x \times 0.25 \).

The total amount of alcohol in the mixture after the replacement is:
\[
(1 - x) \times 0.4 + x \times 0.25
\]

This new mixture contains 32.5% alcohol, so the total alcohol is \( 0.325 \times 1 = 0.325 \) liters. Therefore, the equation is:
\[
(1 - x) \times 0.4 + x \times 0.25 = 0.325
\]

Simplifying the equation:
\[
0.4 - 0.4x + 0.25x = 0.325
\]

Combine like terms:
\[
0.4 - 0.15x = 0.325
\]

Subtract 0.4 from both sides:
\[
-0.15x = 0.325 - 0.4
\]
\[
-0.15x = -0.075
\]

Solving for \( x \):
\[
x = \frac{-0.075}{-0.15} = 0.5
\]

Thus, the proportion of whiskey replaced is **0.5**, or 50%.

Would you like more details or have any questions? Here are 5 related questions:

1. What happens if a higher percentage of alcohol is used instead of 25%?
2. How does the percentage of alcohol change if no whiskey is replaced?
3. How would the calculation change if the total volume were different?
4. What would happen if the initial whiskey had 50% alcohol instead of 40%?
5. Can you create a formula for any percentages?

**Tip:** When dealing with mixtures, always define variables to represent unknown quantities and set up an equation based on the concentrations.

Learn how to solve a mixture problem involving whiskey with 40% alcohol being partially replaced by whiskey with 25% alcohol, resulting in a final alcohol concentration of 32.5%. This problem uses algebraic concepts and linear equations to determine the proportion of whiskey replaced. Suitable for students in Grades 8-10.

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