A chemist has one solution that is 80% acid and a second solution that is 30% acid. How many liters of each solution will the chemist need in order to make 50L of a solution that is 62% acid?

Answer 1

let amount of stronger solution to be used be x L
amount of weaker solution is 50-x L

.8x + .3(50-x) = .62(50)
.8x + 15 - .3x =31
.5x = 16
x = 16/.5 = 32

needs 32 L of the stronger and 36 of the weaker solution.

Answer 2

Ratio of 80% and 30% acids is
62-30:80-62
=32:18
=16:9
To make 50 L of 62%
Volume of 80% acid = 50*(16/25)=32 L
Volume of 30% acid = 50*(9/25)=18 L

Answer 3

looks like I can't do a simple subtraction

of course 50-32 = 18
as MathMate had