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To solve this problem, we’ll use the concept of a mixture to find the volumes of the two acid solutions needed to get a 37% acid solution.

Let:
- \( V_{30} \) be the volume of the 30% acid solution.
- \( V_{50} \) be the volume of the 50% acid solution.
  
The total volume of the mixture is 500 mL, so we have:
\[
V_{30} + V_{50} = 500
\]

We also know that the mixture should be a 37% acid solution. This gives us a second equation based on the concentration:
\[
0.30V_{30} + 0.50V_{50} = 0.37 \times 500
\]

### Step 1: Simplify the Concentration Equation
Calculate the right side of the concentration equation:
\[
0.37 \times 500 = 185
\]
So, we have:
\[
0.30V_{30} + 0.50V_{50} = 185
\]

### Step 2: Solve the System of Equations
We now have two equations:
1. \( V_{30} + V_{50} = 500 \)
2. \( 0.30V_{30} + 0.50V_{50} = 185 \)

From the first equation, we can express \( V_{50} \) in terms of \( V_{30} \):
\[
V_{50} = 500 - V_{30}
\]

Substitute \( V_{50} = 500 - V_{30} \) into the second equation:
\[
0.30V_{30} + 0.50(500 - V_{30}) = 185
\]

Expanding and solving for \( V_{30} \):
\[
0.30V_{30} + 250 - 0.50V_{30} = 185
\]
\[
-0.20V_{30} = -65
\]
\[
V_{30} = 325
\]

Now, substitute \( V_{30} = 325 \) back into the first equation to find \( V_{50} \):
\[
V_{50} = 500 - 325 = 175
\]

### Solution
The volumes required are:
- \( V_{30} = 325 \) mL
- \( V_{50} = 175 \) mL

### Answer
The correct answer is:
- \( V_{30} = 325 \) mL, \( V_{50} = 175 \) mL

Gabriel had 50% and 30% acid solutions. How many mL of each will he need to mix 500 mL of a 37% solution? Round to the nearest mL.

Learn how to determine the volumes needed for a mixture problem involving 50% and 30% acid solutions to achieve a 37% target concentration in a 500 mL solution. This solution includes algebraic methods, system of equations, and mixture concepts. Suitable for high school students in Grades 9-12.

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