Let's walk through the problem:

We are given:

• Solution A: 40% alcohol, 1500 milliliters used.
• Solution B: 12% alcohol, we need to find the number of milliliters used.
• The resulting mixture: 32% alcohol.

We will use the concentration mixture formula for this:

$\text{Alcohol from A} + \text{Alcohol from B} = \text{Alcohol in the final mixture}$

Variables:

• Let $x$ = milliliters of Solution B used.
• Alcohol in Solution A: $1500 \times 0.40$
• Alcohol in Solution B: $x \times 0.12$
• Total alcohol in the final mixture: $(1500 + x) \times 0.32$

Setting up the equation:

$1500 \times 0.40 + x \times 0.12 = (1500 + x) \times 0.32$

Now, solve for $x$.

1. Calculate the terms: $600 + 0.12x = 480 + 0.32x$
2. Rearrange: $600 - 480 = 0.32x - 0.12x$ $120 = 0.20x$
3. Solve for $x$: $x = \frac{120}{0.20} = 600 \, \text{milliliters}$

Answer:

Maya needs to use 600 milliliters of Solution B.

Would you like me to explain any steps further or provide more examples? Here are some related questions for practice:

1. What if the target alcohol concentration was 35%? How would that change the answer?
2. How does the equation change if Solution B had a different concentration?
3. What if Solution A was only 1000 mL? How many milliliters of Solution B would be required then?
4. What happens if you mix two solutions with identical concentrations?
5. Can you use this method for non-alcoholic solutions, such as saltwater? How?

Tip: Always double-check the units when working with concentration problems to avoid simple mistakes.