How many gallons of 20% alcohol solution and 45% alcohol solution must be mixed to get 20 gallons of 25% alcohol solution?

1 Answer
Apr 18, 2018

20 gallons of 25% alcohol can be constructed with 16 gallons of 20% alcohol mixed with 4 gallons of 45% alcohol.

Explanation:

Note that the amount of alcohol in a solution is

amount of alcohol = number of gallons #xx# the fraction alcohol

For example, in the final mixture, we want 20 gallons of 25% alcohol. The amount of alcohol in this amount of solution will be

final amount of alcohol = #20xx0.25=5# gallons alcohol.

Let

#x# = the number of gallons of 20% alcohol solution required to make 20 gallons of the final mixture.

This means that the number of gallons of 45% solution is #20-x#.

Because this is a mixture problem and there is no chemical reaction taking place (no alcohol is created or destroyed in the mixture process) we must start with 5 gallons alcohol. This means that the amount of alcohol in the 20% solution plus the amount of alcohol in the 45% solution must add up to 5 gallons. We can express this mathematically as

#0.2x+0.45(20-x)=5#.

I prefer to work with fractions so I will rewrite this equation as

#x/5+9/20(20-x)=5#

Apply the distributive property.

#x/5+9-(9x)/20=5#.

Subtract 9 from both sides of this equation.

#x/5-(9x)/20=-4#.

Multiply both sides of this equation by #-1# just to make it a little less confusing.

#(9x)/20-x/5=4#

Multiply both sides of this equation by 20.

#9x-4x=80#

Combine like terms.

#5x=80#.

Divide both sides by 5.

#x=16# gallons of 20% solution

This means that we need

#20-x=20-16=4# gallons of 45% solution.