You need 660 mL of a 80% alcohol solution. On hand, you have a 40% alcohol mixture. How much of the 40% alcohol mixture and pure alcohol will you need to obtain the desired solution?
You will need
mL of the 40% solution
and
mL of pure alcohol.
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Let a be the amount of 40% solution and b be the amount of pure alcohol that you will use.

You want to end up with a volume of 660 mL, so

a + b = 660

For each mL of solution used, the 40% solution would contribute 0.4 mL of alcohol, and the pure alcohol will contribute 1 mL of alcohol. In the desired solution, you want a concentration of 80% alcohol, so that it contains 0.8 • 660 mL = 528 mL of alcohol, and

0.4a + b = 528

Solve the system of equations above. Subtracting the second from the first eliminates b and lets us solve for a :

(a + b) - (0.4a + b) = 660 - 528

0.6a = 132

a = 220

Then

220 + b = 660

b = 440

So you will need 220 mL of the 40% solution and 440 mL of pure alcohol.