Question 511931
How many liters of a mixture containing 70% alcohol should be added to a mixture containing 20% alcohol to obtain 16 liters of a mixture containing 50% alcohol?
.
Let x = amount (liters) of 70% alcohol
then
16-x = amount (liters) of 20% alcohol
.
.70x + .20(16-x) = .50(16)
.70x + 3.2-.20x = 8
.50x + 3.2 = 8
.50x = 4.8
x = 9.6 liters (70% alcohol)
.
20% alcohol:
16-x = 16-9.6 = 6.4 liters