1.2 Homework Question 7, 1.2.63 , How many ounces of a \( 15 \% \) alcohol solution must be mixed with 13 ounces of a \( 20 \% \) alcohol solution to make a \( 16 \% \) alcohol solution? 0 of \( 1 \leftarrow \) Po.
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To solve this problem, we can use the formula for mixing solutions. Let's denote the amount of the \( 15\% \) solution as \( x \) ounces. The \( 15\% \) solution contains \( 0.15x \) ounces of alcohol, and the \( 20\% \) solution contains \( 0.20(13) = 2.6 \) ounces of alcohol. We want the final mixed solution to be \( 16\% \). The total volume of the solution will be \( x + 13 \) ounces, and the total amount of alcohol in the mixture will be \( 0.15x + 2.6 \) ounces. So, we set up the equation: \[ \frac{0.15x + 2.6}{x + 13} = 0.16 \] Cross-multiplying gives us: \[ 0.15x + 2.6 = 0.16(x + 13) \] Expanding and solving for \( x \): \[ 0.15x + 2.6 = 0.16x + 2.08 \] Rearranging terms leads to: \[ 2.6 - 2.08 = 0.16x - 0.15x \] \[ 0.52 = 0.01x \] Thus, \( x = \frac{0.52}{0.01} = 52 \) ounces. So, you need to mix **52 ounces** of the \( 15\% \) alcohol solution with 13 ounces of the \( 20\% \) alcohol solution to achieve a \( 16\% \) alcohol solution!