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by jzebra10 » Sun Dec 18, 2011 6:40 am
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?


* 180 grams
* 220 grams
* 250 grams
* 350 grams
* 400 grams

I don't understand the explanation give. Please help.
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by GmatMathPro » Sun Dec 18, 2011 7:20 am
If the 24% solution's strength decreased by 1/3, it is now a 24-24(1/3)=16 percent solution.

One approach:

Let x=the number of grams of 24% solution.

Therefore, we have .24x grams of alcohol in this solution.

When we add 200 grams, our total weight becomes 200+x grams, but the total amount of alcohol is still .24x

Now, if it is a 16% solution, that means that the amount of alcohol divided by the total weight is equal to 16% or:

.24x/(200+x)=0.16

Solve:

.24x=32+.16x

.08x=32

x=400.






jzebra10 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?


* 180 grams
* 220 grams
* 250 grams
* 350 grams
* 400 grams

I don't understand the explanation give. Please help.
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by GMATGuruNY » Sun Dec 18, 2011 8:31 am
jzebra10 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?


* 180 grams
* 220 grams
* 250 grams
* 350 grams
* 400 grams

I don't understand the explanation give. Please help.
We can plug in the answers, which represent the amount of the original solution in the mixture.

The percentage of alcohol in the mixture is to be reduced by 1/3:
24 - (1/3)24 = 16.
If equal amounts of the original solution (which is 24% alcohol) and water (which is 0% alcohol) are combined, the percentage of alcohol in the resulting mixture will be HALFWAY between 0 and 24:
(0+24)/2 = 12.
Thus, in order for the percentage pf alcohol in the resulting mixture to be considerably greater than 12%, the amount of original solution in the mixture must be considerably greater than the amount of water (200).
Eliminate A, B and C.

Answer choice E: original solution = 400 grams
Amount of alcohol = .24(400) = 96.
Percentage of alcohol in the mixture = 96/(400+200)* 100 = 16.
Success!

The correct answer is E.
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by CollegeKart » Sun Dec 18, 2011 4:10 pm
x - is the quantity of solution
.76x+200=0.84x

solving we get x = 400
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by Scott@TargetTestPrep » Fri Dec 15, 2017 10:04 am
jzebra10 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?


* 180 grams
* 220 grams
* 250 grams
* 350 grams
* 400 grams
Since 1/3 of 24 is 8, we know that after 200 grams of water were added to the 24%-solution of alcohol, the strength of the solution decreased to 16%. We can let the original amount of solution = x, and thus the original amount of alcohol = 0.24x. We can create the following equation and solve for x:

0.24x/(x + 200) = 0.16

0.24x = 0.16(x + 200)

0.24x = 0.16x + 32

0.08x = 32

x = 32/0.08 = 3200/8 = 400

Answer: E

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