Paul has an \( 60 \% \) methyl alcohol solution. He wishes to make a gallon of a solution by mixing his methyl alcohol solution with water. If 128 ounces, or a gallon, of solution should contain \( 9 \% \) methyl alcohol, how much of the \( 60 \% \) methyl alcohol solution and how much water must be mixed? The amount of \( 60 \% \) methyl alcohol solution is \( \square \) ounces. (Type an integer or a decimal.)

To solve this problem, let's denote the amount of the \( 60\% \) methyl alcohol solution that Paul will use as \( x \) ounces. Since he wants to make a total of \( 128 \) ounces of solution, the amount of water he will use can be expressed as \( 128 - x \) ounces. The total methyl alcohol content in the solution is given by the equation: \[ 0.60x \text{ (from the methyl alcohol solution)} + 0 \text{ (from the water)} = 0.09 \times 128 \text{ (total methyl alcohol required)} \] Calculating the total methyl alcohol required: \[ 0.09 \times 128 = 11.52 \text{ ounces of methyl alcohol} \] Now, we can set up the equation: \[ 0.60x = 11.52 \] To find \( x \), divide both sides by \( 0.60 \): \[ x = \frac{11.52}{0.60} = 19.2 \] So, Paul needs \( 19.2 \) ounces of the \( 60\% \) methyl alcohol solution. Now, to find out how much water to mix, we calculate: \[ 128 - x = 128 - 19.2 = 108.8 \] Thus, the amount of water needed is \( 108.8 \) ounces. To answer your question, the amount of \( 60 \% \) methyl alcohol solution is \( 19.2 \) ounces.