GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
x+y/n < x+z/n and x^2 - z + y < 0
[#permalink]
Updated on: 05 Aug 2019, 06:52
Updated on: 05 Aug 2019, 06:52
2
13
Bookmarks
00:00
Question Stats:
52% (01:55) correct
47% (01:26) wrong based on 141 sessions
Hide Show timer Statistics
\(\frac{x+y}{n}<\frac{x+z}{n}\) and \(x^2-z+y<0\)
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.
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.
Originally posted by GreenlightTestPrep on 05 Aug 2019, 06:46.
Last edited by Carcass on 05 Aug 2019, 06:52, edited 1 time in total.
Last edited by Carcass on 05 Aug 2019, 06:52, edited 1 time in total.
Edited by Carcass
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: x+y/n < x+z/n and x^2 - z + y < 0
[#permalink]
05 Aug 2019, 08:49
05 Aug 2019, 08:49
2
3
Bookmarks
GreenlightTestPrep wrote:
\(\frac{x+y}{n}<\frac{x+z}{n}\) and \(x^2-z+y<0\)
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.
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
General Discussion
Re: x+y/n < x+z/n and x^2 - z + y < 0
[#permalink]
14 Aug 2020, 20:39
14 Aug 2020, 20:39
1
1
Bookmarks
GreenlightTestPrep wrote:
\(\frac{x+y}{n}<\frac{x+z}{n}\) and \(x^2-z+y<0\)
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.
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.
We have two cases:
\(n > 0\) and \(n < 0\). \(n\) cannot equal 0 or the fractions would be undefined.
Case 1: \(n < 0\)
\(\frac{x+y}{n}<\frac{x+z}{n}\)
Multiply both sides by \(n\) to get:
\(x + y > x + z\)
Since \(n\) is negative, the sign switches. Continuing:
\(y > z\)
Looking at the second equation:
\(x^2-z+y<0\)
\(x^2 < z - y\)
Since \(y > z\):
\(0 > z - y\), so \(0 > x^2\).
\(0 > x^2\) is a contradiction and is false. Therefore, n cannot b less than 0.
Case 2: \(n < 0\)
\(\frac{x+y}{n}<\frac{x+z}{n}\)
Multiply both sides by \(n\) to get:
\(x + y < x + z\)
\(y < z\)
Looking at the second equation:
\(x^2-z+y<0\)
\(x^2 < z - y\)
Since \(y < z\):
\(0 < z - y\), so \(0 ≤ x^2 < z - y\)
In other words, \(x^2\) is less than a positive number \((z-y)\), which is possible, and so the statement is true.
Example:
\((2^2 < 10 - 4)\)
It follows that \(n > 0\) for all \(n\), so A is the answer
