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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
x+y/n < x+z/n and x^2 - z + y < 0
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Updated on: 05 Aug 2019, 06:52
Updated on: 05 Aug 2019, 06:52
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\(\frac{x+y}{n}<\frac{x+z}{n}\) 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
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05 Aug 2019, 08:49
05 Aug 2019, 08:49
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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
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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
