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### Alcohol Solutions

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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/
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Associated Topics:
High School Physics/Chemistry

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