professor baird wants to mix a solution containing 10% acid with one containing 15% acid to obtain a 20-ounce solution containing 12% acid. How many ounces of the 10% solution should the professor use?

Answer 1

To solve this problem, let's denote:
- \( x \) as the number of ounces of the 10% acid solution.
- \( y \) as the number of ounces of the 15% acid solution.

We know the following:
1. The total volume of the final solution is 20 ounces:
\[
x + y = 20
\]

2. The final solution has 12% acid,
which means the total acid in the final solution is \( 0.12 \times 20 = 2.4 \) ounces.

3. The amount of acid in the 10% solution is \( 0.10x \) ounces.
4. The amount of acid in the 15% solution is \( 0.15y \) ounces.

Since the total acid in the final solution is the sum of the acid from both solutions, we have:
\[
0.10x + 0.15y = 2.4
\]

We now have a system of two equations:
1. \( x + y = 20 \)
2. \( 0.10x + 0.15y = 2.4 \)

First, let's solve the first equation for \( y \):
\[
y = 20 - x
\]

Substitute \( y \) in the second equation:
\[
0.10x + 0.15(20 - x) = 2.4
\]

Distribute the 0.15:
\[
0.10x + 3 - 0.15x = 2.4
\]

Combine like terms:
\[
-0.05x + 3 = 2.4
\]

Subtract 3 from both sides:
\[
-0.05x = -0.6
\]

Divide both sides by -0.05:
\[
x = \frac{-0.6}{-0.05} = 12
\]

So, the professor should use 12 ounces of the 10% acid solution.

To verify, let's find \( y \):
\[
y = 20 - x = 20 - 12 = 8
\]

Now, check the total amount of acid:
\[
0.10 \times 12 + 0.15 \times 8 = 1.2 + 1.2 = 2.4 \, \text{ounces}
\]

Everything checks out, so the solution is correct. The professor should use 12 ounces of the 10% acid solution.