Question

How many liters of a 20% alcohol solution must be mixed with 60 liters of a 90% solution to get a 80% solution?

Answer 1

x = liters of 20% solution

Mixing the 20% solution with the 60 L of 90% solution gives a new mixture with volume (x + 60) L. Each L of the 20% solution contributes 0.2 L of alcohol and thus a total of 0.2x L of alcohol, while the 60 L of 90% solution contributes 54 L of alcohol.

We want the concentration of the new mixture to be 80%, so we require

$\dfrac{0.2x+54}{x+60}=0.8$

Solve for x:

$0.2x+54=0.8x+42$

$12=0.4x$

$\implies\boxed{x=30}$

So you will need 30 L of the 20% solution.