Weighted Averages

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.

Problem:  An advertisement for an orange drink claims that the drink 
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.

I hope this helps.  Please write back if you'd like to talk about this 
some more.

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/