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.)

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To determine how much of the $$ 60\% $$ methyl alcohol solution and water Paul needs to mix to make a gallon (128 ounces) of a $$ 9\% $$ methyl alcohol solution, we can set up the following equation:

Let $$ x $$ be the number of ounces of the $$ 60\% $$ methyl alcohol solution.

The amount of pure methyl alcohol in the solution is:
$$
0.60x
$$

The total amount of pure methyl alcohol desired in the final solution is:
$$
0.09 \times 128 = 11.52 \text{ ounces}
$$

Setting these equal gives:
$$
0.60x = 11.52
$$

Solving for $$ x $$:
$$
x = \frac{11.52}{0.60} = 19.2 \text{ ounces}
$$

So, Paul needs to mix **19.2 ounces** of the $$ 60\% $$ methyl alcohol solution with water to make a gallon of $$ 9\% $$ methyl alcohol solution.

**Answer:**  
19.2
Paul needs to mix 19.2 ounces of the 60% methyl alcohol solution with water to make a gallon of 9% methyl alcohol solution.

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