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# M21-21

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Math Expert
Joined: 02 Sep 2009
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Math Expert
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Posts: 94412
Own Kudos [?]: 642247 [10]
Given Kudos: 86311
General Discussion
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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: $$x=400$$.

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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Joined: 01 Jun 2020
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I think this is a high-quality question and I agree with explanation.
Math Expert
Joined: 02 Sep 2009
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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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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: $$x=400$$.

­how did you infer that the amount of alcohol in the original and the final solution is same?
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 94412
Own Kudos [?]: 642247 [0]
Given Kudos: 86311
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: $$x=400$$.