Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Mixing Milk and Butterfat


Date: 09/04/2001 at 11:27:17
From: Nicky
Subject: Word problem 

Milk that has 5% butterfat is mixed with milk that has 2% butterfat. 
How much of each is needed to obtain 60 gallons of milk that has 3% 
butterfat?

This is what I did.

X = gallons 5% butterfat
Y =  "      2% Butterfat

X+Y = 60

5%X + 2%Y = 3%(60)

I really don't know.


Date: 09/04/2001 at 12:45:38
From: Doctor Peterson
Subject: Re: Word problem 

Hi, Nicky.

You have made the right first steps. Now you can rewrite the 
percentages as decimals:

    0.05x + 0.02y = 0.03*60 = 1.8

You might like to avoid decimals by multiplying the whole equation by 
100:

    5x + 2y = 180

Now you have a system of simultaneous equations:

     x +  y =  60
    5x + 2y = 180

You have probably learned some method of solving this. If not, the 
easiest way is to solve the first equation for y:

    y = 60 - x

and then replace y in the second equation with (60-x). This gives you 
a single equation to solve to find x.

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


Date: 09/04/2001 at 13:30:20
From: Doctor Greenie
Subject: Re: Word problem 

Hi, Nicky -

You have made a perfectly good start on this problem....

.. you have clearly defined your variables X and Y

.. you have correctly written the equation, which says the total 
   amount of milk is 60 gallons:  "X gallons with 5% butterfat, plus Y 
   gallons with 2% butterfat, results in a total of 60 gallons"

.. you have correctly written the equation describing the amount of
   butterfat in the mixture: "X gallons with 5% butterfat, plus Y
   gallons with 2% butterfat, results in 60 gallons with 3% butterfat"

For the traditional approach to the mixture problem, you now just need 
to solve these two equations simultaneously to find the values of X 
and Y. The immediate difficulty is with that second equation - it is 
very awkward to try to work with equations containing "%" signs.

So let's first change those percents to decimals in your second 
equation:

    .05X + .02Y = .03(60) = 1.80 = 1.8

Then your two equations are

       X +    Y = 60
    .05X + .02Y = 1.8

For me, at this point, the "ugly" part about this problem is the 
decimals; so I'm going to multiply the second equation by 100 to get 
rid of those decimals:

     X +  Y =  60
    5X + 2Y = 180

Now you can use your favorite method to solve these two equations; for 
me, with these two equations, a linear combination method looks easy 
to use...

    5X + 2Y = 180
    2X + 2Y = 120
   ---------------
    3X      =  60

So I have

    X = 20

from which it follows that

    Y = 40

So the mixture needs to be 20 gallons of milk with 5% butterfat and 40 
gallons of milk with 2% butterfat.


The preceding is the traditional approach to solving mixture problems; 
it is the method I have always seen taught. Following below is the 
solution to this same problem using an alternative approach that I 
find is usually far easier to use.

This alternative approach uses the idea that the relative closeness of 
the desired percentage for the mixture to the percentages of the two 
given batches determines the fraction of the mixture that is to be 
made up of each of the two batches. For example, if a "mixture" is 
made using only batch A, the "mixture" will have the same percentage 
as batch A. If a mixture is made using equal parts of batch A and 
batch B, the mixture will have a percentage halfway between the 
percentages of batches A and B. And if a mixture is made using 3 parts 
of batch A and 1 part of batch B, then the percentage of that mixture 
will be 3 times as close to the percentage of batch A as it is to the 
percentage of batch B.

So......

Here is what you are given:

The two batches of milk being mixed contain 2% butterfat and 5% 
butterfat; the desired mixture is to contain 3% butterfat.

Here is the reasoning you use to find what fraction of the mixture 
should be from each of the two given batches:

1) Think of 2%, 3%, and 5% on a number line. The "distance" from 2% 
   to 3% is 1%; the "distance" from 3% to 5% is 2%.

2) The desired percentage of butterfat is twice as close to 2% as it 
   is to 5%.

3) Therefore, the mixture should contain two parts of the 2% batch for 
   every one part of the 5% batch.

4) This means the ratio of the amount of the 2% batch to the amount 
   of the 5% batch should be 2:1.

5) This in turn means that 2/3 of the mixture should be the 2% batch 
   and 1/3 of the mixture should be the 5% batch.

6) Finally, with the mixture being 60 gallons, this means that the 
   mixture should contain (2/3)x60 = 40 gallons of the batch with 2% 
   butterfat and (1/3)x60 = 20 gallons of the batch with 5% butterfat.

I hope this helps.

Write back if you have any further questions about this type of 
problem.

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

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/