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M25-30

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Bunuel
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M25-30 [#permalink] New post   16 Sep 2014, 01:24
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77% (01:35) correct 23% (02:25) wrong based on 338 sessions
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Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as the result, a 40% solution of acid was obtained, what part of the original solution was replaced?

A. \(\frac{1}{5}\)


B. \(\frac{1}{4}\)


C. \(\frac{1}{2}\)


D. \(\frac{3}{4}\)


E. \(\frac{4}{5}\)
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C
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Bunuel
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Re M25-30 [#permalink] New post   16 Sep 2014, 01:24
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Official Solution:

Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as the result, a 40% solution of acid was obtained, what part of the original solution was replaced?

A. \(\frac{1}{5}\)


B. \(\frac{1}{4}\)


C. \(\frac{1}{2}\)


D. \(\frac{3}{4}\)


E. \(\frac{4}{5}\)


Since the concentration of acid in the resulting solution (40%) is the average of 50% and 30%, exactly half of the original solution was replaced.

Alternatively, you can do the following, say \(x\) part of the original solution was replaced, then: \((1-x)*0.5+x*0.3=1*0.4\), which gives \(x=\frac{1}{2}\).


Answer: C
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gupta87
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Re: M25-30 [#permalink] New post   24 Jan 2017, 11:04
why are we doing (1−x)∗0.5? why it shouldn't be x*0.5?
pls explain
thnx
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Bunuel
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Re: M25-30 [#permalink] New post   25 Jan 2017, 06:23
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gupta87 wrote:
why are we doing (1−x)∗0.5? why it shouldn't be x*0.5?
pls explain
thnx


If x part of the original solution was replaced then (1 - x) part of the original solution is left in which 50% is acid.
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VSJ
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Re: M25-30 [#permalink] New post   22 Nov 2017, 01:38
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I agree with explanation. other method can be alligation:

since 50% and 30% solutions giving 40%. therefore, new ratio is 1:1 (meaning both original solutions formed equal part). therefore, 50% solution's only half part was replaced (1/2. as other half formed by 30% solution)
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RefError
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Re: M25-30 [#permalink] New post   10 Jul 2018, 16:55
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I like to do the "see-saw" approach for this.
You started at 50, replaced with 30. So 50-30 = 20. 20 is the total of the potential change.
The result was 40. So 50-40 = 10. 10 is the change in the composition of the new mixture vs. the composition of the old mixture.
If we take the actual change over the potential change, or 10/20, we get 1/2, which is the answer.
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amandesai17
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Re: M25-30 [#permalink] New post   22 Jun 2019, 01:44
Bunuel Can we use weighted averages to find a solution for this?
-Aman
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Bunuel
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Re: M25-30 [#permalink] New post   22 Jun 2019, 02:14
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amandesai17 wrote:
Bunuel Can we use weighted averages to find a solution for this?
-Aman


Check here: https://gmatclub.com/forum/some-part-of ... 68805.html
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Bunuel
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Re: M25-30 [#permalink] New post   12 Aug 2023, 08:44
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M25-30 [#permalink]
12 Aug 2023, 08:44
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