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.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at
[email protected].
Student Review #1
Student Review #2
Student Review #3
