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41) MOD DS

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Source: — Data Sufficiency |
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by liferocks » Tue May 11, 2010 10:20 pm
we can re write the question as
IS (x+y)(x-y)>0

From 1 x>|Y|

hence X is positive ..so (x+y) and (x-y) will be both positive...(x+y)(x-y)>0 ..sufficient

From 2
|x|>y
Image

So (x+y)(x-y)>0 ..sufficient
Ans option D
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by ballubalraj » Wed May 12, 2010 1:06 am
Another way to solve it:

Square of any positive or negative number is always positive. So irrespective of the sign of x and y, as long as x > y (leaving the sign), x^2 - y^2 will always be > 0.

As both the statements say that x > y (leaving the sign part), both statements alone are sufficient.

Answer: D
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by November Rain » Wed May 12, 2010 10:05 am
Hi,

I agree that statment 1 is sufficient, but i am not sure about 2.

take X = -3 and Y = -4

This means that |x|>y, but, x^2 - y^2 < 0
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by student22 » Wed May 12, 2010 10:10 am
Can't you just simply square both sides of each statement? That way we don't have to worry about sign changes. Or am I oversimplifying it?
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by ballubalraj » Fri May 14, 2010 2:13 am
I guess November Rain is correct. GMAT always tricks us with simple looking but tricky questions - ineuqalities and modes are probably the best examples.

So the answer should be A.

Any thoughts?
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by nakul_anand » Fri May 14, 2010 2:57 am
Yep...I think that the answer should be A

November Rain's values - -3 and -4 prove that a^2-b^2 can be less than 0.
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by ern5231 » Sun May 16, 2010 12:45 am
Also does not work out when x=2 and y=-1
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