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m03 #4

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Manager
Joined: 18 Aug 2010
Posts: 88

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14 Jan 2011, 03:36
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...

Kudos [?]: 20 [0], given: 22

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Joined: 02 Sep 2009
Posts: 42322

Kudos [?]: 133093 [0], given: 12408

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14 Jan 2011, 04:03
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...

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%.

Other solutions: mixture-problem-can-someone-explain-this-100271.html
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Kudos [?]: 133093 [0], given: 12408

Re: m03 #4   [#permalink] 14 Jan 2011, 04:03
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m03 #4

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