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# OG PS - 196

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Senior Manager
Joined: 25 Nov 2006
Posts: 332
Schools: St Gallen, Cambridge, HEC Montreal

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19 Jun 2007, 11:29
2
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Hello,

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 centimeters of alcohol, how many cubic centimeters of water will it contain

A - 50
B - 200
C - 400
D - 625
E - 800

Does anybody know the method to solve the problem?
VP
Joined: 08 Jun 2005
Posts: 1145

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19 Jun 2007, 12:11
2
KUDOS
S2:A50:W100

S4:A50:W200 (the ratio S:A got bigger - S:W unchanged)

S4:A50:W400 (the ratio S:W got smaller - S:A unchanged)

100 cc of A

S8:A100:W800

Manager
Joined: 21 May 2007
Posts: 120

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19 Jun 2007, 12:14
Initial composition = 2:50:100
step 1: Double soap:alcohol ==> 4:50:100
step 2 :Halve soap:water ==> 4:50:400 (dont disturb soap:alcohol here. Just change water)

This solution has 100 cc of alcohol, and thus will have 800 cc of water.

NOTE:
The 2 steps may be done by initially halving alcohol and then changing water to get the required ratio.

best,
parsifal
19 Jun 2007, 12:14
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