Question

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

Answer 1

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.