Question 235297
    A chemist has two solutions: one containing 40% alcohol and another containing 70% alcohol. How much of each should be used to obtain 80 liters of 49% solution.
.
Let x= liters of 40% alcohol
then
80-x = liters of 70% alcohol
.
.40x + .70(80-x) = .49(80)
.40x + 56 - .70x = 39.2
56 - .30x = 39.2
-.30x = -16.8
x = 56 liters of 40% alcohol
.
70% alcohol:
80-x = 80-56 = 24 liters of 70% alcohol