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Intern Joined: 16 Feb 2017
Posts: 17
Followers: 0

Kudos [?]: 2 , given: 4 00:00

Question Stats: 56% (00:48) correct 43% (00:41) wrong based on 16 sessions
x > y

 Quantity A Quantity B The average (arithmetic mean) of x, x, x, y, and y The average (arithmetic mean) of x, x, and y

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

Last edited by GreenlightTestPrep on 06 Apr 2017, 10:37, edited 3 times in total.
Edited by Carcass
Retired Moderator Joined: 07 Jun 2014
Posts: 4808
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
Followers: 145

Kudos [?]: 2282 , given: 393

Expert's post
Lets examine our claim put

Put x=10 and y= 5

Quantity A: 8

Quantity B: 8.33333

Quantity B is greater.

x = -5 and y= -10.

Quantity A: -7

Quantity B: -6.6666667

B is still greater.

So B is correct.
_________________

Sandy
If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test GRE Instructor Joined: 10 Apr 2015
Posts: 2569
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Kudos [?]: 2731  , given: 40

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Expert's post
gre99 wrote:
x > y

 Quantity A Quantity B The average (arithmetic mean) of x, x, x, y, and y The average (arithmetic mean) of x, x, and y

We can also solve this by using the strategy of matching operations

Quantity A: average of x, x, x, y, and y
Quantity B: average x, x, and y

Apply average formula to get:
Quantity A: (x + x + x + y + y)/5
Quantity B: (x + x + y)/3

Simplify:
Quantity A: (3x + 2y)/5
Quantity B: (2x + y)/3

To eliminate the fractions, multiply both quantities by 15 (the least common multiple of 3 and 5):
Quantity A: 15(3x + 2y)/5
Quantity B: 15(2x + y)/3

Simplify:
Quantity A: 3(3x + 2y)
Quantity B: 5(2x + y)

Expand:
Quantity A: 9x + 6y
Quantity B: 10x + 5y

Subtract 9x from both quantities:
Quantity A: 6y
Quantity B: x + 5y

Subtract 5y from both quantities:
Quantity A: y
Quantity B: x

Since we're TOLD that x > y, the correct answer is B

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_________________

Brent Hanneson – Creator of greenlighttestprep.com

Intern Joined: 29 Mar 2017
Posts: 3
Location: United States
WE: Engineering (Consulting)
Followers: 0

Kudos [?]: 5 , given: 2

One can also investigate intuitively without doing any calculation that fewer x's being in the mean average when x is greater will ALWAYS be greater when there are fewer comparative numbers.
GRE Instructor Joined: 10 Apr 2015
Posts: 2569
Followers: 91

Kudos [?]: 2731 , given: 40

Expert's post
ryanmcgrawse wrote:
One can also investigate intuitively without doing any calculation that fewer x's being in the mean average when x is greater will ALWAYS be greater when there are fewer comparative numbers.

Sounds reasonable - I think Can you provide an example or two?

Cheers,
Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com

Intern Joined: 29 Mar 2017
Posts: 3
Location: United States
WE: Engineering (Consulting)
Followers: 0

Kudos [?]: 5 , given: 2

Greenlighttestprep - I suppose I'm not too helpful on the explanation. I just used reason, which Sandy clearly illustrated. Let's assume you have x > y and there is the mean of x,y for A and the mean of x,x,y for B, you'd get a similar answer. Essentially, more power is given to the higher number when the mean has a more substantial ratio between the higher and lower number. I don't have a mathematical proof, but I can put it together logically.
GRE Instructor Joined: 10 Apr 2015
Posts: 2569
Followers: 91

Kudos [?]: 2731 , given: 40

Expert's post
ryanmcgrawse wrote:
Greenlighttestprep - I suppose I'm not too helpful on the explanation. I just used reason, which Sandy clearly illustrated. Let's assume you have x > y and there is the mean of x,y for A and the mean of x,x,y for B, you'd get a similar answer. Essentially, more power is given to the higher number when the mean has a more substantial ratio between the higher and lower number. I don't have a mathematical proof, but I can put it together logically.

Sounds perfect to me! Thanks for the clarification!

Cheers,
Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com Re: if X>Y   [#permalink] 10 Apr 2017, 10:10
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