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Re: M21-21 [#permalink]
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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]
I think this is a high-quality question and I agree with explanation.
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Re: M21-21 [#permalink]
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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: M21-21 [#permalink]
Bunuel wrote:
Official Solution:

After 200 grams of water were added to the 24%-solution of alcohol, the strength of the solution decreased by one-third. What was the weight of the original 24% solution ?

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


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

Since the amount of alcohol in the original and the final solutions is the same, we can equate them and set the equation: \(0.24x=0.16(x+200)\).

Solving gives: \(x=400\).


Answer: E

­how did you infer that the amount of alcohol in the original and the final solution is same? 
Bunuel
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Re: M21-21 [#permalink]
Expert Reply
adityakaregamba wrote:
Bunuel wrote:
Official Solution:

After 200 grams of water were added to the 24%-solution of alcohol, the strength of the solution decreased by one-third. What was the weight of the original 24% solution ?

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


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

Since the amount of alcohol in the original and the final solutions is the same, we can equate them and set the equation: \(0.24x=0.16(x+200)\).

Solving gives: \(x=400\).


Answer: E

­how did you infer that the amount of alcohol in the original and the final solution is same? 
Bunuel

­
Adding water, which has no alcohol, to an alcohol solution will not increase the amount of alcohol in the final solution.
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Re: M21-21 [#permalink]
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