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M21-21

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M21-21  [#permalink]

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New post 16 Sep 2014, 01:11
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  75% (hard)

Question Stats:

58% (02:06) correct 42% (02:17) wrong based on 135 sessions

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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

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Re M21-21  [#permalink]

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New post 16 Sep 2014, 01:11
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1
Official Solution:

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, then the weight of alcohol in it would be \(0.24x\). Next, as in the final solution strength decreased by \(\frac{1}{3}\) thus the strength of the final mixture became \(24*\frac{2}{3}=16 \%\).

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). Solving gives: \(x=400\).


Answer: E
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Re: M21-21  [#permalink]

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New post 06 Dec 2014, 16:56
I solved it a little bit different, and I don't know whether I was lucky to get the right answer or not.
we had 0.24x = alcohol and 0.76x = water.
We have added 200 grams of water, and therefore have x+200 - total volume
we know that the concentration of alcohol has decreased by 1/3 => which means we have 16% alcohol in new solution
0.16(x+200) + 0.76x+200 = x+200
we then have:
0.92x +32 = x
32 = 0.08x
x = (32 * 100) / 8
x = 400

but I guess my method is more lengthy.
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Re M21-21  [#permalink]

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New post 26 Feb 2015, 22:53
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Hi,

I am unable to understand, what the question is asking. Please help.
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Re: M21-21  [#permalink]

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New post 12 Aug 2016, 10:44
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Bunuel wrote:
Official Solution:

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, then the weight of alcohol in it would be \(0.24x\). Next, as in the final solution strength decreased by \(\frac{1}{3}\) thus the strength of the final mixture became \(24*\frac{2}{3}=16 \%\).

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). Solving gives: \(x=400\).


Answer: E



Hi I use a different method and solve it really quickly :

After the water has been added the concentration decrease by \(\frac{1}{3}\).

It means that \(\frac{1}{3}\) of the final mixture contains no alcohol and therefore is water.

Hence, 200g of water = \(\frac{1}{3}\) of final mixture

Final Mixture = 600g wich 200g are water and 400 are the initial solution
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Re: M21-21  [#permalink]

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New post 07 Dec 2016, 12:58
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Just want to add to Richa07's question, what is meant by "used"? Used for what? All that I can tell is happening in the question is that the solution has been watered down, there does not seem to be any other actions indicated...is the question asking how much total solution there is after? before?
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Re: M21-21  [#permalink]

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New post 17 Jan 2017, 05:40
My approach was:

0,24x + 200*0 = 2/3 (0,24(x+200))
x=400

I guess the easiest way here is to recognize that we can create an equation based on two statements about the solution.
The question is telling us that 200 grams of water were added to a 24%-alcohol solution. The amount of of the solution is unknow, thus we can build 0,24x + 200*0 and x is the amount of the solution. If x would be given we could calculate the amount of pure alcohol.
Further, after adding 200 grams of water, the strength of the solution decreased by one-third. Thus we can add 2/3(0,24(x+200)) to our equation.

This kind of questions are very simple if you recognize that you can work with *amount-of-acid*=*amount-of-acid*
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Re: M21-21  [#permalink]

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New post 01 Dec 2017, 08:41
Question is confusing. I didn’t get what was asked in the question.

Posted from my mobile device
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Re: M21-21  [#permalink]

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New post 12 Oct 2018, 13:16
such kind of qn can be solved using quantity balance equation.
24% of X(THE ACID SOLUTION)=2/3(24%(X+200))

THATS ALL THE EQUATION OF X WILL BE 400.
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Re: M21-21 &nbs [#permalink] 12 Oct 2018, 13:16
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M21-21

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