You need a 40% alcohol solution. On hand, you have a 350 mL of a 15% alcohol mixture. You also have 90% alcohol mixture. How much of the 90% mixture will you need to add to obtain the desired solution?
You will need
mL of the 90% solution
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Let x be the amount of 90% solution you will use.

You have 350 mL of a 15% alcohol solution, so it contains 0.15 • 350 mL = 52.5 mL.

For each mL of the 90% solution that you use, you would add 0.9 mL of alcohol. So in the new mixture, the volume would be (350 + x) mL, and its concentration needs to be 40% so that

((52.5 + 0.9x) mL) / ((350 + x) mL) = 0.4

Solve for x :

(52.5 + 0.9x) / (350 + x) = 0.4

(52.5 + 0.9x) / 0.4 = 350 + x

131.25 + 2.25x = 350 + x

1.25x = 218.75

x = 175

so you will need to use 175 mL of the 90% solution.

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