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m03 #4

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m03 #4 [#permalink] New post 14 Jan 2011, 02:36
can anyone explain ???
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
thanks...
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Re: m03 #4 [#permalink] New post 14 Jan 2011, 03:03
Expert's post
tinki wrote:
can anyone explain ???
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
thanks...


Please post answer choices with PS questions.

If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
A. 3%
B. 20%
C. 66%
D. 75%
E. 80%

Question can be solved algebraically or using allegation method.

Algebraic approach:

Initial solution is "half water/half alcohol mix" means it's 50% (0.5) alcohol solution.

Let the portion replaced be x and the volume of initial solution be 1 unit.

Then the amount of alcohol after removal of a portion will be 0.5(1-x) and the amount of alcohol added will be 0.25x, so total amount of alcohol will be 0.5(1-x)+0.25x. On the other hand as in the end 30% alcohol solution was obtained then the amount of alcohol in the end was 0.3*1.

So 0.5(1-x)+0.25x=0.3 --> x=0.8, or 80%.

Answer: E.

Other solutions: mixture-problem-can-someone-explain-this-100271.html
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Re: m03 #4   [#permalink] 14 Jan 2011, 03:03
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