You need 385 mL of a \( 15 \% \) alcohol solution. On hand, you have \( 5 \% \) alcohol mixture. You also have \( 40 \% \) alcohol mixture. How much of each mixture will you need to add to obtain the desired solution? You will need Enter your answer \( \begin{array}{lll}\sqrt{x} & \mathrm{~mL} \text { of the } 5 \% \text { solution } \\ \text { Enter your answer } & \sqrt{x} & \mathrm{~mL} \text { of the } 40 \% \text { solution }\end{array} \)

To obtain 385 mL of a 15% alcohol solution using a 5% alcohol mixture and a 40% alcohol mixture, let's set up the equations. Let \( x \) be the amount of the 5% solution. Then, \( 385 - x \) would be the amount of the 40% solution. Now, the equation for the total alcohol content is: \[ 0.05x + 0.40(385 - x) = 0.15 \times 385 \] Solving this gives: \[ 0.05x + 154 - 0.40x = 57.75 \] \[ -0.35x + 154 = 57.75 \] \[ -0.35x = -96.25 \] \[ x = \frac{96.25}{0.35} \approx 275 \, \text{mL} \] So, you would need approximately: \( 275 \, \text{mL} \) of the 5% solution and \( 385 - 275 = 110 \, \text{mL} \) of the 40% solution. Now, to specify how much of each solution you will need: You will need \( \sqrt{x} = \sqrt{275} \approx 16.58 \, \text{mL} \) of the 5% solution, and \( \sqrt{x} = \sqrt{110} \approx 10.49 \, \text{mL} \) of the 40% solution.