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TAGS: GRE Instructor Joined: 10 Apr 2015
Posts: 1543
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Solution X is 30% alcohol by volume, solution Y is [#permalink]
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Expert's post 00:00

Question Stats: 84% (01:26) correct 15% (01:06) wrong based on 13 sessions
Solution X is 30% alcohol by volume, solution Y is 10% alcohol by volume, and solution Z is 40% alcohol by volume. If 41 gallons of solution X, 20 gallons of solution Y, and 20 gallons of solution Z are combined, which of the following best approximates the percentage of alcohol in the resulting mixture?

A) 20.13
B) 24.91
C) 25.07
D) 27.46
E) 27.53
[Reveal] Spoiler: OA

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Brent Hanneson – Creator of greenlighttestprep.com Sign up for our free GRE Question of the Day emails

Manager Joined: 26 Jan 2018
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Re: Solution X is 30% alcohol by volume, solution Y is [#permalink]
The change in approximation due to 41 gallon ads 0.3 to the numerator and 1 to the denominator. I had to calculate the complete problem to reach to the correct answer. Is there a better way and simpler way t select between last two options?? GRE Instructor Joined: 10 Apr 2015
Posts: 1543
Followers: 56

Kudos [?]: 1468  , given: 8

Re: Solution X is 30% alcohol by volume, solution Y is [#permalink]
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Expert's post
GreenlightTestPrep wrote:
Solution X is 30% alcohol by volume, solution Y is 10% alcohol by volume, and solution Z is 40% alcohol by volume. If 41 gallons of solution X, 20 gallons of solution Y, and 20 gallons of solution Z are combined, which of the following best approximates the percentage of alcohol in the resulting mixture?

A) 20.13
B) 24.91
C) 25.07
D) 27.46
E) 27.53

We can avoid tedious calculations by creating this mixture in parts

First combine 20 gallons of solution Y (10% alcohol) with 20 gallons of solution Z (40% alcohol)
Since we have EQUAL volumes of each solution, the percentage of alcohol in the resulting mixture will be the AVERAGE of the two concentrations.
So, concentration of alcohol in new mixture = (10% + 40%)/2 = 25%
So, now have 40 gallons of a solution that's 25% alcohol.

Next, add 41 gallons of solution X (30% alcohol) to this 40-gallon mixture (25% alcohol)
KEY POINT: If we were to add 40 gallons of solution X (30% alcohol) to the 40-gallon mixture (25% alcohol), then we'd have EQUAL volumes of each solution.
In this case the percentage of alcohol in the resulting mixture WOULD equal the AVERAGE of the two concentrations.
So, the concentration of alcohol in new mixture WOULD equal (30% + 25%)/2 = 27.5%

In reality we have 41 gallons of solution X (30%).
Since the volume of solution X (30%) is greater than the volume of the existing solution (25%), the concentration will be closer to 30% than it is to 25%
So, the concentration of the final solution will be a bit more than 27.5%

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Brent Hanneson – Creator of greenlighttestprep.com Sign up for our free GRE Question of the Day emails Re: Solution X is 30% alcohol by volume, solution Y is   [#permalink] 29 May 2018, 05:58
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