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
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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
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If \(\frac{z-y}{n}\) is POSITIVE and z-y is POSITIVE, then we can be certain that
n is positiveWe get:
Quantity A: Some positive number
Quantity B: 0
Answer: A
Cheers,
Brent
_________________
Brent Hanneson - founder of Greenlight Test Prep