Is x^2-y^2> 0?
1) x>|y|
2) |x|>y
41) MOD DS
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- ballubalraj
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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
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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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
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
- ballubalraj
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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?
So the answer should be A.
Any thoughts?
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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.
November Rain's values - -3 and -4 prove that a^2-b^2 can be less than 0.