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by sukhman » Fri Oct 18, 2013 8:14 am
A vessel has 500 ml of alcohol . 50 ml of alcohol is removed and 50 ml of water is poured into the vessel ( bringing the volume of the mixture in the vessel back to 500 ml ). If this operation is recreated another 2 times , what is the percentage of alcohol in the vessel at the end ?
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by theCodeToGMAT » Fri Oct 18, 2013 9:16 am
Alcohol = 500ML

After Removal = 450ML
Water = 50ML

First Time Removal & Addition
Alcohol = 400
Water = 100

Second Time Removal & Addition
Alcohol = 350
Water = 150

Percentage of alcohol = 350/500*100 = 70%
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by sukhman » Fri Oct 18, 2013 9:30 am
its 72.9 % but I dont understand the logic behind this formula

solution is using a formula [(P-Q)/P ]^n=k
whwre P= volume of liquid initially and in each operation Q volume is taken out and replaced by Q volume of water then at the end of n such operations , concentration k of the liquid as a Proportion of total volume of the solution. In termes of percentage , it is equal to 100k
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by Brent@GMATPrepNow » Fri Oct 18, 2013 9:32 am
There's some ambiguity here around "this operation is recreated another 2 times"
Does this mean remove 50 ml of alcohol each time?
Or does it mean remove 50 ml of the resulting solution each time?
theCodeToGMAT has interpreted it as the first meaning.

If the answer is 72.9% then the question's author is expecting us to use the other interpretation.

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by sukhman » Fri Oct 18, 2013 9:37 am
Author has used the formula (500-50/500)^3 = (450/500)^3
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by Brent@GMATPrepNow » Fri Oct 18, 2013 9:43 am
sukhman wrote:A vessel has 500 ml of alcohol . 50 ml of alcohol is removed and 50 ml of water is poured into the vessel ( bringing the volume of the mixture in the vessel back to 500 ml ). If this operation is recreated another 2 times , what is the percentage of alcohol in the vessel at the end ?
If we're removing 50 ml of the resulting solution each time, then here's one approach:

Let's keep track of the alcohol with each move.

Start: 500 ml of alcohol

Remove 50ml from vessel: Left with 450 ml of alcohol
Add 50ml of water: Still have 450 ml of alcohol
NOTE: the mixture is now 90% alcohol [since 450/500 = 90%]

Remove 50ml from vessel: 90% of this is alcohol. So, of the 50ml of mixture, 45 mls are alcohol. So, we are removing 45 mls of alcohol
We're left with 405 ml of alcohol
Add 50ml of water: Still have 405 ml of alcohol
NOTE: the mixture is now 81% alcohol [since 405/500 = 81%]

Remove 50ml from vessel: 81% of this is alcohol. So, of the 50ml of mixture, 40.5 mls are alcohol. So, we are removing 40.5 mls of alcohol
We're left with 364.5 ml of alcohol
Add 50ml of water: Still have 364.5 ml of alcohol

364.5/500 = 729/1000 = 72.9/100 = [spoiler]72.9%[/spoiler]

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by theCodeToGMAT » Fri Oct 18, 2013 10:06 am
Another approach

Assuming question means 10% removal

= 500 ( 1 - 10/100) ( 1 - 10/100) ( 1 - 10/100)
= 500 (0.9)(0.9)(0.9)
= 500 * .729

So, 72.9%
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by GMATGuruNY » Fri Oct 18, 2013 11:49 am
I posted an approach to this sort of problem here:

https://www.beatthegmat.com/tough-question-t263754.html
https://www.beatthegmat.com/a-ps-from-gmat-t95960.html (3 posts)
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