Question

s 16​% protein​; cornmeal is 8​% protein. How many pounds of each should be mixed together in order to get 320​-lb mixture that is 12​% protein​?
Question content area bottom
Part 1
How many pounds of cornmeal should be in the​ mixture?

Answer 1

To solve the problem of mixing a 16% protein source with an 8% protein source to create a 320-pound mixture that is 12% protein, we can set up the following equations.

Let:
- \( x \) = pounds of the 16% protein source
- \( y \) = pounds of the 8% protein source

From the problem, we have two key pieces of information:

1. The total weight of the mixture:
\[
x + y = 320
\]

2. The total protein content in the mixture must be 12% of 320 pounds:
\[
0.16x + 0.08y = 0.12 \times 320
\]

Calculating 12% of 320 pounds:
\[
0.12 \times 320 = 38.4
\]

Now we have our system of equations:

1. \( x + y = 320 \)
2. \( 0.16x + 0.08y = 38.4 \)

We can solve this system step by step.

**Step 1: Solve for \( y \) in terms of \( x \)** from the first equation:
\[
y = 320 - x
\]

**Step 2: Substitute \( y \) in the second equation:**
\[
0.16x + 0.08(320 - x) = 38.4
\]
\[
0.16x + 25.6 - 0.08x = 38.4
\]
\[
0.08x + 25.6 = 38.4
\]
\[
0.08x = 38.4 - 25.6
\]
\[
0.08x = 12.8
\]
\[
x = \frac{12.8}{0.08} = 160
\]

**Step 3: Calculate \( y \):**
Now we can substitute \( x = 160 \) back into the equation for \( y \):
\[
y = 320 - 160 = 160
\]

Thus, the amounts of each component in the mixture are:
- \( x = 160 \) pounds of the 16% protein source (not asked)
- \( y = 160 \) pounds of the 8% protein source (this is the cornmeal)

Therefore, to answer the question, **the amount of cornmeal needed in the mixture is**:
\[
\boxed{160} \text{ pounds}
\]