How much of each must be used to make 15 liters of a solution that is 50% alcohol?

Question

A Chemist has one solution that is 40% alcohol and another that is 55% alcohol.

Answer 1

Let the quantity of each solution be a and b.

Then, a+b=15 and 0.40a+0.55b=0.50×15=7.5.

We can write b=15-a then substitute for b: 0.40a+0.55(15-a).

0.40a+8.25-0.55a=7.5; 0.75=0.15a, so a=5L and b=15-5=10L.

So we need 5L of 40% solution and 10L of 55% solution.