Question 423843
One solution contains 50% alcohol and another solution contains 80% alcohol. How many liters of each solution should be mixed to produce 10.5 liters of a 70% alcohol solution?
.
Let x = amount (liters) of 50% solution
then
10.5-x = amount of 80% solution
.
.50x + .80(10.5-x) = .70(10.5)
.50x + 8.4 - .80x = 7.35
.50x - .80x = -1.05
-.30x = -1.05
x = 3.5 liters (50% alcohol)
.
80% alcohol:
10.5-x = 10.5-3.5 = 7 liters