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If after 200 grams of water were added to the 24%-solution

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If after 200 grams of water were added to the 24%-solution [#permalink] New post 03 Aug 2009, 22:09
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A
B
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D
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Question Stats:

65% (02:54) correct 35% (02:24) wrong based on 50 sessions
If after 200 grams of water were added to the 24%-solution of alcohol, the strength of the solution decreased by one-third, how much of the 24%-solution was used?

A. 180 grams
B. 220 grams
C. 250 grams
D. 350 grams
E. 400 grams

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-after-200-grams-of-water-were-added-to-the-24-solution-99103.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 28 Oct 2013, 22:57, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: mixture problem [#permalink] New post 05 Aug 2009, 21:20
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i think wording of this question is little confusing ....answer choice cannot be in %
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File comment: Hence 400 gms of solution was taken to mix with 200 gms of water
mixture.GIF
mixture.GIF [ 2.5 KiB | Viewed 2232 times ]


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Re: mixture problem [#permalink] New post 13 Aug 2009, 12:52
There definitely is something wrong with this question....if the question is asking how much of the original solution was used...then it must be 400 gms...
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Re: mixture problem [#permalink] New post 26 Aug 2009, 17:55
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I too think 400 grams should be the answer.

My reasoning is

The h20 content in 200 gms of water is 1
h20 content in 24% solution is 0.76

Resulting new solution's concentration decreases by one third means. Now it becomes 16% solution which means h20 content is 0.84

Let amount of solution be X

Then the equation can be set up

200*1 + 0.76X = 0.84 (X+200)

0.08X = 32
8X = 3200
X = 400

Plain simple. I usually use the table and come up with this equation

Let me know if I am wrong.
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Re: mixture problem [#permalink] New post 26 Aug 2009, 19:11
I approach mixtures as weighted average problems.
I define y to be the amount (in grams) of 24% before water was added.
After adding water the solution is reduced by 1/3 and we have 2/3X24% strength of the new solutiion.
The equation will be the following"

\frac{24%\times Y+0%\times 200}{200+Y}=\frac{2}{3}\times 24%

\frac{24%\times Y}{200+Y}=\frac{2}{3}\times 24%

24% cancel out.

\frac{Y}{200+Y}=\frac{2}{3}

I think they ask about \frac{Y}{200+Y} which is 66.6%...I think there is a a mistake in the answer choices...

this corresponds to y=400 gram...So 400 gram out of new total 400+200=600 is 66.66%

If someone got something else I would like to know the way...
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Re: mixture problem [#permalink] New post 26 Aug 2009, 21:38
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Yeah, the answer should be in grams.

Another 10sec approach: decrease in concentration by 1/3 means increase in volume by 1/3. So, added water is 1/3 of new total volume and 24-% alcohol solution is 2/3 (twice as much as added water). Therefore, answer is 400g
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Re: mixture problem [#permalink] New post 27 Aug 2009, 05:58
walker wrote:
Yeah, the answer should be in grams.

Another 10sec approach: decrease in concentration by 1/3 means increase in volume by 1/3. So, added water is 1/3 of new total volume and 24-% alcohol solution is 2/3 (twice as much as added water). Therefore, answer is 400g


HI Walker,
decrease in concentration by 1/3 means increase in volume by 1/3 => this is very clear
May question may sound stupid but how do you come to he fact that water added is 1/3 of new total??
Thx for your help
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Re: mixture problem [#permalink] New post 27 Aug 2009, 06:15
walker wrote:
Yeah, the answer should be in grams.

Another 10sec approach: decrease in concentration by 1/3 means increase in volume by 1/3. So, added water is 1/3 of new total volume and 24-% alcohol solution is 2/3 (twice as much as added water). Therefore, answer is 400g


Why do you all assume that the answer should be in grams? I also got 400 gram...but they ask about percentage...so they want to know how much 400 gram represent out of total 600 gram (after water is added)...which is 66.6%...there is a typo in choice B (it should be 66.6% instead of 36.6%)....
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Re: mixture problem [#permalink] New post 27 Aug 2009, 06:40
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defoue wrote:
May question may sound stupid but how do you come to he fact that water added is 1/3 of new total??


Ok.
Concentration decreases by 1/3. In other words, new volume is larger than old volume by 1/3 of new volume.
Visually: XX - old volume, XXX - new volume.

Some example:

Concentration decrease by 4/5 (in 5 times). new volume is larger than old volume by 4/5 of new volume.
Visually: X - old volume; XXXXX - new volume.
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Re: mixture problem [#permalink] New post 27 Aug 2009, 06:49
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LenaA wrote:
...so they want to know how much 400 gram represent out of total 600 gram ....

They ask: "... how much of the 24%-solution was used?" It would be clear if we had 600g of solution and used 400g to prepare 1/3 reduced solution.
Honesty, I don't think it is a real GMAT question.
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Re: mixture problem [#permalink] New post 27 Aug 2009, 07:16
walker wrote:
LenaA wrote:
...so they want to know how much 400 gram represent out of total 600 gram ....

They ask: "... how much of the 24%-solution was used?" It would be clear if we had 600g of solution and used 400g to prepare 1/3 reduced solution.
Honesty, I don't think it is a real GMAT question.


The new reduced solution consists of 2/3 of 24% solution (400g) and 1/3 of water (200g)...I thought they were asking about the composition of the new solution...that 24% solution represents 2/3 of the new reduced one...My interpretation was wrong (English is not my native language), especially given the fact that 66.6% is not among the choices...Thanks for clarifying it to me.
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Re: mixture problem [#permalink] New post 27 Aug 2009, 08:12
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LenaA wrote:
My interpretation was wrong (English is not my native language)...

I'm not a native English speaker too :) What I know GMAT is trying to use clear unambiguous language. I guess it is a poorly written GMAT-like question, although it has a good idea.
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Re: mixture problem [#permalink] New post 27 Aug 2009, 08:58
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walker wrote:
LenaA wrote:
My interpretation was wrong (English is not my native language)...

I'm not a native English speaker too :) What I know GMAT is trying to use clear unambiguous language. I guess it is a poorly written GMAT-like question, although it has a good idea.

As a non native speaker to a non native speaker, what % they ment? I am confused. I got 400 gram but the answers are %! Percent of what?!
I really would like to know the answer if anyone has it...
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Re: mixture problem [#permalink] New post 27 Aug 2009, 13:53
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LenaA wrote:
As a non native speaker to a non native speaker, what % they ment? I am confused. I got 400 gram but the answers are %! Percent of what?!
I really would like to know the answer if anyone has it...


Don't worry, it is not a GMAT question. The base of a GMAT question of 50-51 level is a TRICK not misunderstanding.
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Re: mixture problem [#permalink] New post 28 Oct 2013, 19:47
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Re: If after 200 grams of water were added to the 24%-solution [#permalink] New post 28 Oct 2013, 22:58
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arvs212 wrote:
If after 200 grams of water were added to the 24%-solution of alcohol, the strength of the solution decreased by one-third, how much of the 24%-solution was used?

A. 180 grams
B. 220 grams
C. 250 grams
D. 350 grams
E. 400 grams


Let the weight of 24% solution used be x grams, weight of alcohol in it would be 0.24x. As in final solution strength decreased be 1/3 thus it became 24*2/3=16%.

Set the equation: 0.24x=0.16(x+200), the weight of 16% alcohol in x+200 grams of new solution comes only from (equals to) 24% alcohol in x grams of strong (initial) solution, as there is 0 grams of alcohol in water (0% alcohol solution) --> 0.08x=32 --> x=400.

Answer: E.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-after-200-grams-of-water-were-added-to-the-24-solution-99103.html
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Re: If after 200 grams of water were added to the 24%-solution   [#permalink] 28 Oct 2013, 22:58
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