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

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 $\times$ 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 = $20 \times 0.25 = 5$ 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

$0.2 x + 0.45 \left(20 - x\right) = 5$.

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

$\frac{x}{5} + \frac{9}{20} \left(20 - x\right) = 5$

Apply the distributive property.

$\frac{x}{5} + 9 - \frac{9 x}{20} = 5$.

Subtract 9 from both sides of this equation.

$\frac{x}{5} - \frac{9 x}{20} = - 4$.

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

$\frac{9 x}{20} - \frac{x}{5} = 4$

Multiply both sides of this equation by 20.

$9 x - 4 x = 80$

Combine like terms.

$5 x = 80$.

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.