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If after 200 grams of water were added to the 24%-solution [#permalink]
03 Aug 2009, 22:09

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Difficulty:

45% (medium)

Question Stats:

63% (02:52) correct
37% (02:03) wrong based on 55 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

Re: mixture problem [#permalink]
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...

Re: mixture problem [#permalink]
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"

Re: mixture problem [#permalink]
26 Aug 2009, 21:38

1

This post received KUDOS

Expert's post

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 _________________

Re: mixture problem [#permalink]
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

Re: mixture problem [#permalink]
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%)....

Re: mixture problem [#permalink]
27 Aug 2009, 06:40

Expert's post

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. _________________

Re: mixture problem [#permalink]
27 Aug 2009, 06:49

Expert's post

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. _________________

Re: mixture problem [#permalink]
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.

Re: mixture problem [#permalink]
27 Aug 2009, 08:12

Expert's post

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. _________________

Re: mixture problem [#permalink]
27 Aug 2009, 08:58

1

This post received KUDOS

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...

Re: mixture problem [#permalink]
27 Aug 2009, 13:53

Expert's post

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. _________________

Re: mixture problem [#permalink]
28 Oct 2013, 19:47

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: If after 200 grams of water were added to the 24%-solution [#permalink]
28 Oct 2013, 22:58

1

This post received KUDOS

Expert's post

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) --> \(0.08x=32\) --> \(x=400\).

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