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M03-04

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CEO
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Re: M03-04  [#permalink]

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New post 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 :thumbup: :angel:


Thanks for the kind words, Mo2men!

Cheers,
Brent
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Re: M03-04  [#permalink]

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New post 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%
Director
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Re: M03-04  [#permalink]

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New post 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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M03-04  [#permalink]

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New post 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
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M03-04   [#permalink] 04 Jun 2019, 12:59

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