Question

A chemist has two acid solutions in stock: one is a 50% solution, and the other is a 80% solution. How much of each solution should be mixed to obtain 100 milliliters of a 68% solution?

Answer 1

Easy way, visual

50---68---80%
Difference 18 12
Invert order to give ratio of 50% and 80%
Ratio 12 18 = 2:3

So by simple proportion:
volume of 50% needed = 100*2/(2+3)=40 mL
volume of 80% needed = 100*3/(2+3)=60 mL

Algebra way
Let x=volume of 50% needed.
Then
0.50*x+0.80*(100-x) = 0.68(100)
Expand and solve for x
0.5x+80-0.8x = 68
0.8x-0.5x = 80-68
0.3x = 12
x=12/0.3 = 40 mL (volume of 50%)
100-x=100-40=60 mL (volume of 80%)