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Percentage of Alcohol in a Solution


Date: 06/02/99 at 00:28:59
From: Nycole
Subject: Algebra

Suppose 30 liters of a solution with an unknown percentage of alcohol 
is mixed with 5 liters of a 90% alcohol solution. If the resulting 
mixture is a 62% alcohol solution, what is the percentage of alcohol 
in the first solution?

First, I tried to figure out an equation  -->  (30 + x) + (5 + .90) = 
.62 (or 62%).

I do not know if you should add the percentage of alcohol with the 
liters, then add that with 5 liters of a 90% alcohol solution. Is 
that right?

Please help!


Date: 06/02/99 at 12:05:23
From: Doctor Rob
Subject: Re: Algebra

Think of the amount of alcohol in each solution. In the first one, it 
is 30*(x/100) liters. In the second one, it is 5*(90/100). In the last 
one, it is 35*(62/100) (because the total amount of the mixture is 
35 = 30 + 5 liters). When the first two solutions are mixed, the total 
amount of alcohol is the sum of the amounts in the two ingredients, so 
you get an equation:

   30*(x/100) + 5*(90/100) = 35*(62/100).

This equation is pretty easy to solve for x, the percentage you are 
seeking.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Physics/Chemistry

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