This problem is about mixing two solutions of different alcohol concentrations to form a new solution with a desired concentration.

Let’s break it down:

• You are asked to mix a 12% alcohol solution and a 20% alcohol solution to form a 15% alcohol solution.
• The amount of the 20% alcohol solution is given as 3 ounces.
• Let the amount of the 12% alcohol solution be x ounces.

The key idea is that the total amount of alcohol from both solutions must equal the amount of alcohol in the final mixture.

Set up the equation:

1. The amount of alcohol in the 12% solution:
   $0.12x$ (since 12% means 0.12 times the amount of the solution).

2. The amount of alcohol in the 20% solution:
   $0.20 \times 3 = 0.60$ ounces (since you have 3 ounces of a 20% solution).

3. The total amount of alcohol in the final 15% solution:
   The total solution is $x + 3$ ounces, and 15% of that is alcohol:
   $0.15(x + 3)$.

Now, set up the equation:

$0.12x + 0.60 = 0.15(x + 3)$

Solve for $x$:

1. Distribute on the right-hand side: $0.12x + 0.60 = 0.15x + 0.45$

2. Subtract $0.12x$ from both sides: $0.60 = 0.03x + 0.45$

3. Subtract 0.45 from both sides: $0.15 = 0.03x$

4. Divide both sides by 0.03: $x = 5$

Final Answer:

You need to mix 5 ounces of the 12% alcohol solution.

Would you like further details or clarification? Here are five related questions for practice:

1. How do you solve a similar problem if the desired alcohol percentage is 18% instead of 15%?
2. What happens if you mix unequal amounts of two solutions with the same percentage?
3. How can this type of problem be modeled using a system of equations?
4. What would the equation look like if both solutions had the same concentration?
5. How do you adjust the equation if there is a different type of solution, like a sugar solution instead of alcohol?

Tip: Always keep track of your units (ounces in this case) and percentage (converted into decimals) in mixture problems to avoid errors.