Bunuel wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
A. \(3\%\)
B. \(20\%\)
C. \(66\%\)
D. \(75\%\)
E. \(80\%\)
Refer to Graph attached:
If the mixture was replaced with an equal amount of the 25% alcohol solution, it should result in a 37.5% alcohol solution instead of 30% alcohol solution.
--This means that more than 50% of the original solution was replaced with the 25% solution.
Eliminate A and BIf you subtract the differences, you get a ratio of 5x:20x => 1x:4x = 5x
--That means, for every 5 liters (1x+4x), you have 1 part of 50% alcohol solution and 4 parts of the 25% solution
4:5 = 80%
Answer E
>> !!!
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