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# M03-04

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Joined: 12 Sep 2015
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04 Feb 2019, 06:29
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Mo2men wrote:
Dear GMATPrepNow Brent

When I see you explaination, I always want to say: Thanks to your effort and people like you who work hard to make it easier for students.

Indeed True tutor

Thanks for the kind words, Mo2men!

Cheers,
Brent
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Joined: 17 May 2018
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09 Mar 2019, 10:08
My solution:

Notice: "half alcohol/half water" means solution with 50% concentration.
Let x be the amount of the 25% solution and y the amount of the 50% solution.

Using aligation:

X Y
I----------------I-----------------I
0.25-------------0.3-------------0.5

x/y = 0.2/0.05 = 20/5 =4 so x = 4y.

Amount of the solution = y + x = y + 4y = 5y.
Percentage of the original alcohol replaced = part replaced/Amount of the solution = 4y/5y = 0.8 = 80%
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Location: Australia
Concentration: Technology, Strategy
GMAT 1: 560 Q41 V26
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WE: Management Consulting (Consulting)

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29 May 2019, 03:26
ScottTargetTestPrep would love your take on this using the table method?

I understand a lot of experts have already solved this.
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Joined: 28 Apr 2018
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04 Jun 2019, 12:59
First of all, l love question like this. Thanks Bunuel!
let x be the total volume of the original mixture of half alcohol and half water
let y be the portion that was removed from the mixture x
We must have in mind that the volume of the portion(y) removed is equal to the volume of solution that replaced it
Hence, % of alcohol in final solution =(0.5x - 0.5y + 0.25y)/(x - y + y) =30%
(0.5x - 0.25y)/(x) =0.3
0.5x - 0.25y =0.3x
0.2x =0.25y
y = 0.8x
or 80% of original mixture
VP
Joined: 14 Feb 2017
Posts: 1321
Location: Australia
Concentration: Technology, Strategy
GMAT 1: 560 Q41 V26
GMAT 2: 550 Q43 V23
GMAT 3: 650 Q47 V33
GMAT 4: 650 Q44 V36
GMAT 5: 650 Q48 V31
GMAT 6: 600 Q38 V35
GPA: 3
WE: Management Consulting (Consulting)

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02 Aug 2019, 21:46
Upon review, the easiest way to solve this is by picking numbers and thinking about it conceptually.

Any solution say a 20ml mix of water and alcohol (10ml water, 10 ml alchy)

(10-0.5x (removing the alcohol) + 0.25x (Adding the lower concentrate alcohol))/20 = 0.3
(10-0.25x)/20 =0.3
10 - 0.25x = 6
4=x/4
x= 16

16/20 = 4/5 = 0.8 0r 80% of original solution
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Joined: 19 Jan 2018
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11 Aug 2019, 09:45
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 B

If 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%

>> !!!

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Intern
Joined: 27 May 2019
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14 Aug 2019, 06:01
I think this is a high-quality question and I agree with explanation. Question is confusing it should be quoted 'original alcohol solution' and not 'original alcohol' for the solution to be as explained here.
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Joined: 03 May 2019
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05 Dec 2019, 03:55
I think this the explanation isn't clear enough, please elaborate.
Re M03-04   [#permalink] 05 Dec 2019, 03:55

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# M03-04

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