Inequality DS

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Inequality DS

by bobdylan » Fri Jun 08, 2012 12:59 am
How do we attack this kind of problem????

Is | x^2 + y^2| > | x^2 - y^2|?
(1) x > y
(2) x > 0

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by aneesh.kg » Fri Jun 08, 2012 1:13 am
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
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.
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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by Anurag@Gurome » Fri Jun 08, 2012 1:15 am
bobdylan wrote:Is | x^2 + y^2| > | x^2 - y^2|?
(1) x > y
(2) x > 0
Picking Number Approach:

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|
Both above examples satisfy both the statements but in the first case the answer is YES but in the second it is NO. Hence, both statement together is not sufficient to answer the question.

The correct answer is E.
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