Let's define the problem:

• You want to mix two solutions to make a final mixture of 1200 mL that is 40% alcohol.
• One solution is 20% alcohol, and the other is 65% alcohol.

We need to determine how much of each solution is required to reach the final 40% alcohol solution.

Step 1: Set up the variables

Let:

• $x$ = the volume (in mL) of the 20% alcohol solution.
• $1200 - x$ = the volume (in mL) of the 65% alcohol solution.

Step 2: Set up the equation for the alcohol content

The total amount of alcohol in the final solution is given by the sum of the alcohol in each of the two solutions:

• Alcohol from the 20% solution = $0.20 \times x$
• Alcohol from the 65% solution = $0.65 \times (1200 - x)$

The total alcohol in the final 1200 mL solution should be 40%, which is $0.40 \times 1200 = 480$ mL of alcohol.

Step 3: Set up the equation

The equation for the total amount of alcohol is:

$0.20 \times x + 0.65 \times (1200 - x) = 480$

Step 4: Solve the equation

Let's expand and simplify the equation:

$0.20x + 0.65(1200 - x) = 480$ $0.20x + 780 - 0.65x = 480$ $-0.45x + 780 = 480$ $-0.45x = 480 - 780$ $-0.45x = -300$ $x = \frac{-300}{-0.45}$ $x = 666.67 \, \text{mL}$

So, $x \approx 667 \, \text{mL}$ of the 20% alcohol solution is required.

Step 5: Calculate the amount of the 65% solution

Since the total volume is 1200 mL, the volume of the 65% alcohol solution is:

$1200 - 667 = 533 \, \text{mL}$

Final Answer:

• 667 mL of the 20% alcohol solution
• 533 mL of the 65% alcohol solution

Would you like more details or have any further questions? Here's a tip:

Tip: You can use this same method for mixing any two solutions to create a desired concentration, just by adjusting the alcohol percentages and the target volume.