Absolute value inequality

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Absolute value inequality

by ollapodrida » Fri Jan 31, 2014 1:52 pm
Is |x^2+y^2| > |x^2-y^2|?
(1) x > y
(2) x > 0

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by [email protected] » Fri Jan 31, 2014 2:12 pm
Hi ollapodrida,

This DS question is perfect for TESTing Values.

We're asked if |x^2 + y^2| > |x^2 - y^2|? This is a YES/NO question. We are told nothing about x and y.

Fact 1: x > y

Let's TEST Values and track the results:
x = 1
y = 0
Is |1| > |1|? The answer to the question is NO.

x = 2
y = 1
Is |5| > |3|? The answer to the question is YES.
Fact 1 is INSUFFICIENT

Fact 2: x > 0

The same two examples from Fact 1 can be applied here.
We have a NO and a YES answer.
Fact 2 is INSUFFICIENT

Combined, we have the same TEST Cases for both Facts.
We have a NO and a YES answer.
Combined, INSUFFICIENT.

Final Answer: E

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by ceilidh.erickson » Fri Jan 31, 2014 2:19 pm
Both x^2 and y^2 must be either positive or 0. The absolute value of |pos + pos| will always be greater than |pos - pos|. The only time |x^2 + y^2| will not be greater than |x^2 - y^2| is if either x or y is 0, in which case the quantities will be equal.

Target question: is either x or y equal to 0?

Statement 1: x > y

This tells us nothing about whether either quantity is 0. Insufficient.

Statement 2: x > 0

This tells us that x is not equal to 0, but it doesn't tell us whether y is. Insufficient.

Together: The statements tell us nothing about whether either quantity equals 0. Insufficient.

The answer is E.
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by GMATGuruNY » Sat Feb 01, 2014 3:54 am
ollapodrida wrote:Is |x^2+y^2| > |x^2-y^2|?
(1) x > y
(2) x > 0
Alternate approach:

Let a=x² and b=y².
Substituting a=x² and b=y² into the question stem, we get:
|a+b| > |a-b|?

Since each side has absolute value -- implying that each side is NONNEGATIVE -- we can square both sides:
a² + 2ab + b² > a² - 2ab + b²
4ab > 0
ab > 0.

Substituting a=x² and b=y² into the resulting expression, we can rephrase the question stem as follows:
Is x²y² > 0?

When the statements are combined:
It's possible that x=1 and y=0, in which case x²y² = 0.
It's possible that x=2 and y=1, in which case x²y² > 0.
INSUFFICIENT.

The correct answer is E.
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