# Mr. Fairwrench's radiator contains seven quarts of a mixture that is 35% antifreeze and 65% water. He needs a mixture that is 50% antifreeze and 50% water. How many quarts should he drain from his radiator and replace with pure antifreeze?

Oct 13, 2015

He should drain $1.62$ quarts of the initial mixture.

#### Explanation:

You know that Mr. Fairwrench's radiator contains $7$ quarts of a mixture that is 35% antifreeze and 65% water.

Let's say that this mixture contains $x$ quarts of antifreeze and $7 - x$ quarts of water.

The important thing to realize here is that the number of quarts of mixture does not change because you replace the amount of mixture you drain.

Now, you can use the known percent composition of the mixture to find the initial number of quarts of antifreeze and of water

7color(red)(cancel(color(black)("quarts mixture"))) * "35 quarts antifreeze"/(100color(red)(cancel(color(black)("quarts mixture")))) = "2.45 quarts"

This means that the mixture initially contained

$\left\{\begin{matrix}x = \text{2.45 quarts antifreeze" \\ y = 7 - 2.45 = "4.55 quarts water}\end{matrix}\right.$

You need the final mixture to contain 50% water and 50% antifreeze. Remeber that the mixture is still 7 quarts!

7color(red)(cancel(color(black)("quarts mixture"))) * "50 quarts antifreeze"/(100color(red)(cancel(color(black)("quarts mixture")))) = "3.50 quarts"

This time, you need

$\left\{\begin{matrix}x = \text{3.50 quarts antifreeze" \\ y = 7 - 3.50 = "3.50 quarts water}\end{matrix}\right.$

Determine how much mixture needs to be drained in order to remove $4.55 - 3.50 = 1.05$ quarts of water by using the initial mixture's percent composition of water

1.05color(red)(cancel(color(black)("quarts water"))) * "100 quarts mixture"/(65color(red)(cancel(color(black)("quarts water")))) = "1.62 quarts"

Keep in mind that draining this much of the initial mixture will also reduce the amount of antifreeze you have

1.62color(red)(cancel(color(black)("quarts mixture"))) * "35 quarts antifreeze"/(100color(red)(cancel(color(black)("quarts mixture")))) = "0.57 quarts"

The amount of antifreeze you're left with is

$2.45 - 0.57 = \text{1.88 quarts}$

Notice what happens when you add $1.62$ quarts of pure antifreeze

${V}_{\text{water" = "3.50 quarts}}$

${V}_{\text{anbtifreeze" = 1.88 + 1.62 = "3.50 quarts}}$

Now the mixture is 50% water and 50% antifreeze.