Question

How many gallons of a 60% alcohol solution and a 30% alcohol solution must be mixed to get 12 gallons of a 50% alcohol solution?

Answer 1

If there are a gallons of the 60% solution and b of the 30% solution, a+b=12.

The amount of alcohol in the mixture is 50% of 12=6 gallons.

So 0.6a+0.3b=6. Since b=12-a, we can substitute:

0.6a+0.3(12-a)=6, 0.6a+3.6-0.3a=6, 0.3a=6-3.6=2.4.

a=2.4/0.3=24/3=8 gallons, making b=4 gallons.

So we have 8 gallons of 60% solution and 4 gallons of 30% solution.