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Single Variable Algebraic Mixture Problem

```Date: 05/03/2005 at 21:31:21
From: william

Brittany needs a mixture of 76% oats and 24% corn.  She currently has
25 pounds of a mixture that is 28% oats.  How many pounds of oats
should be added to the mixture?

How do I identify the x from the y?  In most mixture problems you have
a % or \$ or some coefficent.  I think you would need to have 2
equations with 2 variables, like this:

x + y = ?
Oats x + Corn y = 100%

I am sure there is something simple I am missing.  I am a 7th grade
teacher and asked the students to bring in a problem from the internet
to solve.  I explained to them I could not solve this but I told them
I would try to get help and share with them.  I know the answer is 50

- William

```

```
Date: 05/03/2005 at 22:52:26
From: Doctor Peterson

Hi, William.

First I'll visualize the story and make sure I understand the details.
We've got a bag containing 25 pounds of a mixture that is 28% oats,
and presumably 72% corn.  I could work out just how much of each grain
we have; I'm assuming that "percent" here means "by weight", so that
28% of the 25 pounds is oats.

We have another bag containing pure oats, and we want to add that in
to the mixture to increase the percentage of oats to 76% (with the
remaining 24% being corn).

Now, the next step is to choose a variable or two.  What don't we
know?  We need to figure out how much (by weight) of the oats to add;
I don't see any other obvious unknown, so that one variable may be all
we need:

Let x = amount of oats to be added, in pounds

(I like to define variables precisely, to make sure I always use them
the same way, and don't drift into using a different unit, for example.)

Now, if I knew how much oats I was adding, how would I figure out the
percentage of oats in the final mixture?  Well, I'd find the total
amount of oats and the total amount of the mixture (including the
corn), and then divide.  So let's write expressions for each step of
that process:

Amount of original mixture: 25 lb
Amount of oats in original mixture: 0.28*25 = 7 lb
Amount of oats in final mixture: 7+x lb
(since x lb of oats are added)
Total amount of final mixture: 25+x lb
(since x lbs of stuff are added in all)

(I like to say that the basic idea of algebra is to give a name to
anything you don't know, pretend you do know it, and then go through
whatever calculations the problem requires, using the variables and
writing expressions rather than getting actual numbers.  When you're
done, you'll have an equation or two.)

So the percentage of oats in the new mixture will be

(7+x)/(25+x)

as a fraction. The goal is for that fraction to be 76%, or 0.76. So we
can write an equation

(7+x)
------ = 0.76
(25+x)

or

7+x = 0.76(25+x)

Now I just have to solve this.  I'll simplify the right side, and then
isolate the variable:

7 + x = 19 + 0.76x

7 + x - 0.76x = 19

7 + 0.24x = 19

0.24x = 19 - 7

0.24x = 12

x = 12/0.24 = 50

And we're done, except for checking to see that the answer actually
works in the original problem.  If I add 50 lb of oats, I'll have a 75
lb bag containing 57 lb of oats, and the percentage is in fact 0.76.

If you have any further questions, feel free to write back.

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

```

```
Date: 05/03/2005 at 23:01:07
From: william

Doctor Peterson,

Thank you so very much.  The children will be thrilled that you

- William
```
Associated Topics:
Middle School Ratio and Proportion
Middle School Word Problems

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