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Ratio of Anti-Freeze to Water

Date: 02/05/2003 at 10:53:37
From: Shelly
Subject: Ratio word problem

A barrel is filled with 45 gallons of anti-freeze. 9 gallons are 
removed and the barrel is refilled with water. 9 gallons of the 
mixture are removed and again, the barrel is refilled with water.  
What is the final ratio of anti-freeze to water in the barrel?


Date: 02/06/2003 at 16:24:49
From: Doctor Ian
Subject: Re: Ratio word problem

Hi Shelly,

The problem with using numbers is that it's tempting to simplify 
things too soon. Also, it's easy to get frustrated when the numbers
don't work out nicely.  

So let's avoid those problems and say that the original amount of 
antifreeze is A, and at each step we remove some fraction (1/p) of the 
mixture, to be replaced by water. 

In this particular case, the numbers are A=45 and p=5 (because 9 
gallons is 1/5 of 45 gallons), but let's forget about that for the
moment. 

It will be easier to understand what follows if you have a solid grip
on the fact that when you remove 1/p of something, the amount that
remains is ((p-1)/p).  For example, if you remove 1/10 of something,
you're left with 9/10 of it.  Okay?  Then let's go.

You start out with 

   Antifreeze    Water
   ----------    -----
            A        0   (Start)

We take out 1/p of the 'mixture', 

   Antifreeze    Water
   ----------    -----
            A        0             
   ((p-1)/p)A        0   (Remove 1/p of mixture)

and replace it with water:

   Antifreeze            Water
   ----------    -------------
           A                 0
  ((p-1)/p)A                 0
  ((p-1)/p)A            (1/p)A    (Replace with water)

Again, we remove 1/p of the mixture:

          Antifreeze              Water
   -----------------    ---------------
                   A                  0
          ((p-1)/p)A                  0
          ((p-1)/p)A             (1/p)A
 ((p-1)/p)((p-1)/p)A    ((p-1)/p)(1/p)A    (Remove)

And replace it with water:


          Antifreeze                       Water
   -----------------    ------------------------
                   A                  0
          ((p-1)/p)A                  0
          ((p-1)/p)A             (1/p)A
 ((p-1)/p)((p-1)/p)A    ((p-1)/p)(1/p)A    
 ((p-1)/p)((p-1)/p)A    ((p-1)/p)(1/p)A + (1/p)A    (Replace)


At this point, you're probably thinking: "Hey, I thought he said it
would be simpler!"

But now we start reaping the benefits of our patience. The ratio of
antifreeze to water is 

  ((p-1)/p)((p-1)/p)A 
  ------------------------
  ((p-1)/p)(1/p)A + (1/p)A    

The first thing we can do with this is combine the terms in the
denominator, using the distributive property,

  ac + bc = (a+b)c

to get

  ((p-1)/p)((p-1)/p)      * A 
  ---------------------------
  [(p-1)/p)(1/p) + (1/p)] * A    

which means that we can just get rid of the A's altogether. In other
words, the answer doesn't depend on how much antifreeze we started
with. So now we have

    ((p-1)/p)((p-1)/p)      
    --------------------- 
    (p-1)/p)(1/p) + (1/p) 


    (p-1)^2 / p^2      
  = -------------------------- 
    (p-1)  / p^2   +   p / p^2     

 
Using the distributive property again, we have

    (p-1)^2     / p^2      
  = ----------------- 
    (p - 1 + p) / p^2     


    (p-1)^2          
  = ----------- 
    (p - 1 + p)


    (p-1)^2          
  = ------- 
    (2p - 1)

We said that p was 5, so this comes out to 

    (5-1)^2          
  = ---------- 
    (2(5) - 1)
  

  = 16/9

Which is kind of cool, that everything cancels out that way, isn't it?  

Now, I have to admit, if you just had to solve this particular problem
one time, and you knew you'd never see one like it again, you'd be
better off doing the same steps, but using 45 and 1/5 and 4/5 instead
of A and 1/p and ((p-1)/p).  

But the nice thing about solving it this way is that if I want to
change the numbers around - say, I want to start with 36 gallons, and
replace 3 gallons each time, or start with 42 gallons, and replace 7
gallons each time - I can get the answer in about two seconds.  

And now I'm kind of curious to try a couple of generalizations: 

  (1) What if the fraction is q/p, instead of 1/p? 

  (2) What if I continue for k steps, instead of just two? 

Will it continue to work out nicely?  Who knows?  That's what makes it
interesting.   

Anyway, I hope this helps.  Write back if you'd like to talk more
about this, or anything else. 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
Middle School Algebra
Middle School Word Problems

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