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### Weighted Averages

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Date: 11/30/2001 at 22:32:26
From: Tyler
Subject: Weighted averages

I cannot figure out the correct formula to use. I think if I knew how
to set it up I could do it.

contains 10% orange juice. How much pure orange juice would have to
be added to 5 quarts of the drink to obtain a mixture containing 40%
orange juice?

Can you help me?
```

```
Date: 11/30/2001 at 23:50:22
From: Doctor Paul
Subject: Re: Weighted averages

You have 5 quarts of "drink" that is 10% orange juice. Thus the 5
quarts of drink contain 5*10% = 5*.1 = .5 quarts of orange juice, and
hence must contain 4.5 quarts of other stuff. Now we're going to add
100% orange juice and we want to know how much to add so that the new
mixture will be 40% orange juice.

Well, what if you add 1 quart of 100% orange juice? What would that do
to the concentration of orange juice in the mixture?

It would mean that we now have 1.5 quarts of pure orange juice mixed
in a carton that now contains 6 quarts of liquid. Thus the
concentration of orange juice in this 6-quart jug would be
1.5/6 = .25 = 25%

So we haven't added enough, but maybe you see what's going on.

In a more general situation:

If we add x quarts of pure orange juice to the original mixture that
contains .5 quarts of OJ and 4.5 quarts of something else, then what
we have is a mixture of 5+x quarts that contains .5+x quarts of OJ.

So after we've added x quarts of pure OJ, the concentration of OJ will
be (.5+x)/(5+x)

We want to know when this will be equal to .4

So solve for x:

(.5+x)/(5+x) = .4

.5+x = 2 + .4x

.6x = 1.5

x = 1.5/.6 = 2.5

Thus you should add 2.5 quarts of pure OJ to make the concentration of
the new mixture 40% OJ.

some more.

- Doctor Paul, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
Middle School Algebra
Middle School Ratio and Proportion
Middle School Word Problems

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