Finding a Formula for Mixtures

Date: 11/10/2003 at 21:41:56
From: Gaye 
Subject: algebra story problem

Alfredo needs to make 250 ml of a 27% alcohol solution, using a 15% 
solution and a 40% solution.  How much of each should he use?

Date: 11/11/2003 at 12:57:35
From: Doctor Ian
Subject: Re: alegebra story problem

Hi Gaye,

As a guess, let's suppose that he starts with 125 ml of each kind of
solution.  The first solution will contain x parts alcohol, where

   15 parts alcohol      x parts alcohol
  ------------------ = ------------------
  100 parts solution   125 parts solution

        (0.15 * 125) = x

The second will contain y parts alcohol, where 

   40 parts alcohol      y parts alcohol
  ------------------ = ------------------
  100 parts solution   125 parts solution

        (0.40 * 125) = y

And the concentration of the resulting solution will be 

      (x + y) parts alcohol     27 parts alcohol
  -------------------------- = ------------------
  (125 + 125) parts solution   100 parts solution

Does this make sense? 

We can substitute the values we just found for x and y:

  (0.15 * 125 + 0.40 * 125) parts alcohol     27 parts alcohol
  ---------------------------------------- = ------------------
                (125 + 125) parts solution   100 parts solution

Now, this probably isn't going to work out to be true!  In fact, if we
evaluate each side, we get 

  68.75    27
  ----- = ---
   250    100

Cross-multiplying we get 

  100 * 68.75 = 250 * 27

         6875 = 3712.5

which is false.  So our guess was wrong. 

But what if we don't split the two solutions evenly?  For example,
suppose we use 100 parts of the 15% solution to 150 parts of the 40%
solution?  Then we get

  (0.15 * 100 + 0.40 * 150) parts alcohol     27 parts alcohol
  ---------------------------------------- = ------------------
                (100 + 150) parts solution   100 parts solution


This _still_ isn't right, probably.  (Work it out if you'd like.) 
However, we can start to see some patterns.  

For one thing, the two amounts have to add up to 250 parts, no matter
how we split them up:

  (0.15 * 100 + 0.40 * 150) parts alcohol     27 parts alcohol
  ---------------------------------------- = ------------------
                        250 parts solution   100 parts solution

And as soon as we choose the amount of 15% solution, the amount of 40%
solution is fixed.  That is, if we use P parts of the 15% solution, we get

  (0.15*P + 0.40*(250-P)) parts alcohol     27 parts alcohol
  ------------------------------------- = ------------------
                     250 parts solution   100 parts solution


We can generalize it a little further, by letting T stand for some
arbitrary total amount of solution:

  (0.15*P + 0.40*(T-P)) parts alcohol     27 parts alcohol
  ----------------------------------- = ------------------
                     T parts solution   100 parts solution

Does this make sense?  Can you take it from here? 

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