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
_____________________________________________

Alcohol Solutions


Date: 04/22/97 at 15:38:02
From: John Samoylo
Subject: Algebra - solution/mixture problems

How many liters of a 14 percent alcohol solution must be mixed with 
20 liters of a 50 percent alcohol solution to get a 20 percent alcohol 
solution?

I don't know where to begin!  


Date: 05/22/97 at 22:58:09
From: Doctor Sydney
Subject: Re: Algebra - solution/mixture problems

Hi John,

I will help you set this up, and then when we have an equation with 
one variable, I'll let you solve for the variable. If you have 
trouble with that, feel free to write back to us and we will help.

Okay, so we want to know how many liters of a 14 percent alcohol 
solution must be mixed with 20 liters of a 50 percent alcohol solution 
to get a 20 percent alcohol solution, right?  Let's give this unknown 
quantity a letter name - let's call it x. So we want to figure out 
what x is when x is the number of liters of a 14 percent alcohol 
solution such that when mixed with 20 liters of a 50 percent alcohol 
solution give a 20 percent alcohol solution.  

Let's think about the amount of alcohol in our solution. If we mix x 
liters of a 14 percent alcohol solution with 20 liters of a 50 percent 
alcohol solution, how much alcohol do we have?  Well, how much alcohol 
is in x liters of 14 percent solution? We have 14 percent of x liters 
of alcohol in the solution, right? That means that we have .14x liters 
of alcohol in the solution.  

Similarly, we have (20)(.50) liters of alcohol in the 20 liters of 50% 
alcohol solution, so the total number of liters of alcohol in this 
mixture is: .14x + (20)(.50).  

Now we are halfway there. Let's go back and look at the problem.  
It says that if we mix x liters of 14 percent alcohol solution with 
20 liters of a 50 percent alcohol solution we get a 20 percent alcohol 
solution.  

We have already figured out that when we mix x liters of 14 percent 
alcohol solution with 20 liters of a 50 percent alcohol solution we 
get .14x + (20)(.50) liters of alcohol.  But the statement of the 
problem gives us more information.  It says that this mixture yields a 
20 percent alcohol solution.  

This additional piece of information allows us to calculate the 
amount of alcohol in the mixture in a different way.  If the mixture 
is 20 percent alcohol and is a total of 20 + x liters (remember we 
added 20 liters of solution to x liters of solution, so the total 
numer of liters of solution must be 20 + x), then the amount of 
alcohol in the solution must be (.20)(20 + x), right?  We just use the 
same reasoning we used above when we figured out that there were .14x 
liters of alcohol in x liters of 14 percent alcohol solution.  

Now we have that the number of liters of alcohol in the mixture is 
(.20)(20 + x).  But above we also said that the number of liters of 
alcohol in the mixture is .14x + (20)(.50).  So, these two numbers 
must be equal, and we can set up an equation:  

  (.20)(20 + x) = .14x + (20)(.50)

Now all you have to do to finish the problem is solve the equation 
for x.  Let me know if you need more help on this!  Good luck!

-Doctor Sydney,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Physics/Chemistry

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/