How do we attack this kind of problem????
Is | x^2 + y^2| > | x^2 - y^2|?
(1) x > y
(2) x > 0
Inequality DS
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- aneesh.kg
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The inequality | x^2 + y^2| > | x^2 - y^2| does not hold true when either one or both of x and y are equal to 0.bobdylan wrote:How do we attack this kind of problem????
Is | x^2 + y^2| > | x^2 - y^2|?
(1) x > y
(2) x > 0
When x = 0,
|0 + y^2| = |0 - y^2|
When y = 0,
|x^2 + 0| = |x^2 - 0|
When both x and y are non-zero,
|x^2 + y^2| > |x^2 - y^2|
Therefore, the answer to the question is YES when both x and y are non-zero. If one of them is 0, the answer is NO.
Statement(1):
Does not tell if x or y are 0 or not.
INSUFFICIENT
Statement(2):
x is NOT equal to 0, but y can be 0.
INSUFFICIENT
Let's combine,
x > 0 AND x > y
y can still be 0 or not.
INSUFFICIENT
[spoiler](E)[/spoiler] is correct.
Last edited by aneesh.kg on Fri Jun 08, 2012 1:18 am, edited 1 time in total.
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Picking Number Approach:bobdylan wrote:Is | x^2 + y^2| > | x^2 - y^2|?
(1) x > y
(2) x > 0
Consider the following two examples,
- 1. x = 2 and y = 1 --> |x² + y²| = 5 > 3 = |x² - y²|
2. x = 2 and y = 0 --> |x² + y²| = 4 = |x² - y²|
The correct answer is E.
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