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by taneja.niks » Mon May 30, 2011 12:03 pm
If there are 10 liters of a 20%-solution of alcohol, how much water should be added to reduce the concentration of alcohol in the solution by 75% ?
25 liters
27 liters
30 liters
32 liters
35 liters
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by SoCan » Mon May 30, 2011 12:18 pm
The current solution has 2 liters of alcohol. If we want to reduce the concentration by 75%, we want to reduce the concentration from 20% to 5%. So:
2 liters of alcohol/(10 liters of original solution + x liters of water) = 0.05
2=0.5+0.05x
1.5=0.05x
30=x
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by cans » Mon May 30, 2011 12:22 pm
20% of 10 Litre solution => 2l alcohol.
reduce alcohol by 75% is same as keep alcohol by 25% = 20* 25/100 = 5%
Thus 5% of soln = 2 litre
100% = 40 litre
10 litre is already there and thus add 30 litre water more
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by taneja.niks » Mon May 30, 2011 12:23 pm
why 0.05?? shouldn it be 0.5???
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by SoCan » Mon May 30, 2011 12:38 pm
taneja.niks wrote:why 0.05?? shouldn it be 0.5???
Why is the right side of the equation 0.05? Because that's the decimal equivalent of 5%. If you solve for 0.5, you're going to have a negative number - how much water you need to take out of the solution so that alcohol comprises 50%.
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by galaxian » Mon May 30, 2011 12:39 pm
5% solution = 5/100 = .05 & not .5.
or you can say, 0.5 = 5/10 = 50 %
While, .05 = 5/100 = 5%.
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by GMATGuruNY » Mon May 30, 2011 7:53 pm
taneja.niks wrote:If there are 10 liters of a 20%-solution of alcohol, how much water should be added to reduce the concentration of alcohol in the solution by 75% ?
25 liters
27 liters
30 liters
32 liters
35 liters
Amount of alcohol in the original solution = .2*10 = 2 liters.
A 20% concentration reduced by 75% = 20 - .75*20 = 5%.
Thus, the 2 liters of alcohol must be 5% of the final mixure:
2 = .05x
x = 40 liters.

Since 40 liters are needed and the original mixture is 10 liters, the amount of water added = 40-10 = 30 liters.

The correct answer is C.
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