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x+y/n < x+z/n and x^2 - z + y < 0

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GRE Instructor
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x+y/n < x+z/n and x^2 - z + y < 0 [#permalink] New post 05 Aug 2019, 06:46
Expert's post
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Question Stats:

58% (01:19) correct 41% (00:26) wrong based on 12 sessions
\(\frac{x+y}{n}<\frac{x+z}{n}\) and \(x^2-z+y<0\)

Quantity A
Quantity B
n
0


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

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Brent Hanneson – Creator of greenlighttestprep.com
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Last edited by Carcass on 05 Aug 2019, 06:52, edited 1 time in total.
Edited by Carcass
GRE Instructor
User avatar
Joined: 10 Apr 2015
Posts: 2311
Followers: 73

Kudos [?]: 2208 [0], given: 27

CAT Tests
Re: x+y/n < x+z/n and x^2 - z + y < 0 [#permalink] New post 05 Aug 2019, 08:49
Expert's post
GreenlightTestPrep wrote:
\(\frac{x+y}{n}<\frac{x+z}{n}\) and \(x^2-z+y<0\)

Quantity A
Quantity B
n
0


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.


Given: \(\frac{x+y}{n}<\frac{x+z}{n}\)

Rewrite as: \(\frac{x}{n}+\frac{y}{n}<\frac{x}{n}+\frac{z}{n}\)

Subtract \(\frac{x}{n}\) from both sides to get: \(\frac{y}{n}<\frac{z}{n}\)

Subtract \(\frac{y}{n}\) from both sides to get: \(0<\frac{z}{n}-\frac{y}{n}\)

Simplify to get: \(0<\frac{z-y}{n}\)

In other words, \(\frac{z-y}{n}\) is POSITIVE
------------------------------------------------------

Also given: \(x^2-z+y<0\)
Add z to both sides to get: \(x^2+y<z\)
Subtract y from both sides to get: \(x^2<z-y\)
Since 0 ≤ x², we can write 0 ≤ x² < z - y
This tells us that: 0 < z - y
In other words, z - y is POSITIVE
-------------------------------------------------------

If \(\frac{z-y}{n}\) is POSITIVE and z-y is POSITIVE, then we can be certain that n is positive

We get:
Quantity A: Some positive number
Quantity B: 0

Answer: A

Cheers,
Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com
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Re: x+y/n < x+z/n and x^2 - z + y < 0   [#permalink] 05 Aug 2019, 08:49
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