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Mixture Problem

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sfbay
 Post subject: Mixture Problem
 Post Posted: Tue Mar 22, 2011 6:55 pm 
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Posts: 26
Location: San Francisco
Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?

1/5
1/4
1/2
3/4
4/5


Question is from GMAT club but it is from their MGMAT challenge set.

(C) 2008 GMAT Club
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sfbay
 Post subject: Re: Mixture Problem
 Post Posted: Tue Mar 22, 2011 7:27 pm 
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Location: San Francisco
yes, i should have posted this in the other forum category. i don't see a 'delete' button or i would .....
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david.khoy
 Post subject: Re: Mixture Problem
 Post Posted: Tue Mar 22, 2011 7:32 pm 
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Students


Posts: 30
I think the anwer is 1/2.

It's a very intuitive answer. You can check it by weighing the concentrations:

1/2 * 50/100 + 1/2 * 30/100 = 25/100 + 15/100 = 40/100

To get a 40% solution, you have to make a solution that contains half the original 50% solution and half the 30% solution.

Now if you prefer a more classical way than the weighing, you can use the following equation:

v = original volume of solution
x = volume of solution that gets replaced

0.5(v-x) + 0.3x = 0.4v
0.5v - 0.5x + 0.3x = 0.4v
0.1v = 0.2x
x=0.1v/0.2

x=v*1/2

The volume of solution that gets replaced (x) is half of the volume (v).
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sfbay
 Post subject: Re: Mixture Problem
 Post Posted: Tue Mar 22, 2011 8:32 pm 
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Posts: 26
Location: San Francisco
yes, the answer is 1/2 but i was looking for a 'cleaner way' of doing it....the way i did it, it was ugly.

coincidentally someone has recently posted a similar problem on the general math forum
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jnelson0612
 Post subject: Re: Mixture Problem
 Post Posted: Wed Mar 23, 2011 7:49 pm 
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ManhattanGMAT Staff


Posts: 1857
munny, if you will move the problem over to the General Math forum I will delete this thread and we can resume our discussion over there.

_________________
Jamie Nelson
ManhattanGMAT Instructor
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