gmatpup wrote:The ratio, by volume, of soap to alcohol to water in a certain solution is 2:50:100. The solution will be altered so that the ratio of soap to alcohol is doubled while the ratio of soap to water is halved. If the altered solution will contain 100 cubic cm of alcohol, how many cubic cm of water will it contain?
A. 50
B. 200
C. 400
D. 625
E. 800
It says the answer is
E but I am completely lost. I thought I could plug in numbers but it is not working for me. Someone please explain
thank you!
Let S = soap, A = alcohol, and W = water.
Original S:A = 2:50.
With the amount of soap doubled, new S:A = 4:50 = 2:25.
Original S:W = 2:100.
With the amount of soap halved, new S:W = 1:100 = 2:200.
Combining the ratios, new S:A:W = 2:25:200.
Since the actual amount of alcohol = 100, and S:A:W = 2:25:200 = 8
800, the actual amount of water = 800.
The correct answer is
E.
Last edited by
GMATGuruNY on Tue Nov 08, 2011 6:57 pm, edited 1 time in total.
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800, the actual amount of water = 800.
