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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mixture problem
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GmatMathPro
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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.
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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GMATGuruNY
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We can plug in the answers, which represent the amount of the original solution in the mixture.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.
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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CollegeKart
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x - is the quantity of solution
.76x+200=0.84x
solving we get x = 400
.76x+200=0.84x
solving we get x = 400
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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: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
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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