You need a 45% alcohol solution. On hand, you have a 105 mL of a 40% alcohol mixture. You also have 80% alcohol mixture. How much of the 80% mixture will you need to add to obtain the desired solution?
You will need
mL of the 80% solution
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You have 105 mL of a 40% alcohol solution, which contains 0.4 • 105 mL = 42 mL of alcohol.

If you mix it with x mL of 80% solution, you end up with a (105 + x) mL solution that contains (42 + 0.8x) mL of alcohol.

In this solution, you want to have a concentration of 45%, so

((42 + 0.8x) mL) / ((105 + x) mL) = 0.45

Solve for x :

(42 + 0.8x) / (105 + x) = 0.45

(42 + 0.8x) / 0.45 = 105 + x

280/3 + 16/9 x = 105 + x

7/9 x = 35/3

x = 15

So you need 15 mL of the 80% solution.

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