Question

How many gallons each of 30% alcohol and 10% alcohol should be mixed to obtain 20 gal of 28% alcohol?

Gallons of solution Percent Gallons of Pure Alcohol
x 30%=0.3
y 10%=0.1
20 28%=

How many gallons of 30% alcohol should be in the mixture? ___ gal

Answer 1

.3x + .1(20 - x) = .28(20)

.3x + 2 - .1x = 5.6

.2x + 2 = 5.6

.2x = 3.6

x = 18 gallons of 30% alcohol

20 - 18 = 2 gallons of 10% alcohol