Question

How many gallons of 30% alcohol solution and how many of 60% alcohol solution must be mixed to produce 18 gallons of 50% solution?

Answer 1

x = 6 gallons (of 30% alcohol)

y = 12 gallons (of 60% alcohol)

Explanation:

Let

x = liters of 30% alcohol

y = liters of 60% alcohol

There are two unknowns, we need two equations

x + y = 18. (1)

0.30x + 0.60y = 0.50(x+y) (2)

From (1)

x + y = 18

y = 18-x

Substitute the value of y into (2) and solve for x:

0.30x + 0.60y = 0.50(x+y)

0.30x + 0.60(18-x) = 0.50(x+18-x)

0.30x + 10.8 - 0.60x = 0.50(18)

10.8 - 0.30x = 9

-0.30x = -1.8

Divide both sides by -0.30

x = 6 gallons (of 30% alcohol)

Substitute x=6 into (1) and solve for y:

x + y = 18

6 + y = 18

y = 12 gallons (of 60% alcohol)