# Suppose that you have a supply of a 20% solution of alcohol and a 80% solution of alcohol. How...

## Question:

Suppose that you have a supply of a 20% solution of alcohol and a 80% solution of alcohol. How many quarts of each should be mixed to produce 60 quarts that is 65% alcohol?

## Problems of Mixtures:

To solve the mixture problems of two mixtures, we assume two variables for the quantities of unknown mixtures, set up a system of two equations and solve them using an appropriate method.

The number of quarts of {eq}20\% {/eq} and {eq}80\% {/eq} alcohol solution is {eq}x {/eq} and {eq}y {/eq} respectively.

Since the total amount is {eq}60 {/eq} quarts, we get the equation:

$$x+y= 60\,\,\,\,\,\,\,\rightarrow (1)$$

Using the given information:

$$20\% \text{ of }x+ 80\% \text{ of }y = 65\% \text{ of }60\\[0.4cm] 0.20x+ 0.80y = 0.65(60)\\[0.4cm] 0.20x+ 0.80y=39\,\,\,\,\,\,\,\rightarrow (2)$$

Multiply both sides of (1) by {eq}-0.20 {/eq}, then we get:

$$-0.20x-0.20y= -12\,\,\,\,\,\,\,\rightarrow (3)$$

$$0.60y = 27\\[0.4cm] \text{Dividing both sides by 0.60}, \\[0.4cm] y=45$$

Substitute this in (1):

$$x+45=60\\[0.4cm] \text{Subtracting 45 from both sides}, \\[0.4cm] x=15$$

Therefore, the number of quarts of {eq}20\% {/eq} alcohol = {eq}\boxed{\mathbf{ 15 \text{ quarts}}} {/eq}

and the number of quarts of {eq}80\% {/eq} alcohol = {eq}\boxed{\mathbf{45 \text{ quarts}}} {/eq}.