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Is x>y? 1). x^2<y^2 2). y<0 [#permalink]
 27 May 2006, 23:36
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
Is x>y?
1). x^2<y^2
2). y<0
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Director
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I would go for C...
A is not sufficient becuase if x and y can both be negative or positive.
B alone is not sufficient for sure.
if A and B are true. We know that y is -ve so if x is positive then x > y.
If x is also negative and from A we know that if x is negative then x > y.
So C is the answer.
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Re: DS - Inequality [#permalink]
28 May 2006, 12:11
gmat_crack wrote: Is x>y?
1). x^2<y^2
2). y<0
if x = 2, and y = -5, yes.
if x = -2 and y = - 5, yes.
in any case x>y. so C.
updated.....
Last edited by Professor on 28 May 2006, 13:23, edited 1 time in total.
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Director
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Professor could you please explain...
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Re: DS - Inequality [#permalink]
28 May 2006, 18:28
Professor wrote: gmat_crack wrote: Is x>y?
1). x^2<y^2
2). y<0 if x = 2, and y = -5, yes. if x = -2 and y = - 5, yes. in any case x>y. so C. updated..... Professor I don't agree
here is my two cents
x^2 - y^2 < 0
--> (x+y) (x -y) < 0
For above condition to satisfy
x + y > 0 and x - y > 0
--> x > y ---- (A)
or
x + y < 0 and x -y < 0
----> x < y ------ (B)
So we can't say any thing.
St2. y < 0
even if y is (-ve) we can't say any thing about above two conditions.
Please let me know if I'm wrong..
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Re: DS - Inequality [#permalink]
28 May 2006, 20:44
gmat_crack wrote: Professor wrote: gmat_crack wrote: Is x>y?
1). x^2<y^2
2). y<0 if x = 2, and y = -5, yes. if x = -2 and y = - 5, yes. in any case x>y. so C. updated..... x^2 - y^2 < 0 --> (x+y) (x -y) < 0 For above condition to satisfy: x + y > 0 and x - y > 0x > y ---- (A) how is the red part correct?
x^2<y^2
x^2 - y^2<0
(x+y)(x-y) <0
either (x+y) <0 or (x-y) <0
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Senior Manager
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Re: DS - Inequality [#permalink]
28 May 2006, 21:34
Professor wrote: gmat_crack wrote: Professor wrote: gmat_crack wrote: Is x>y?
1). x^2<y^2
2). y<0 if x = 2, and y = -5, yes. if x = -2 and y = - 5, yes. in any case x>y. so C. updated..... x^2 - y^2 < 0 --> (x+y) (x -y) < 0 For above condition to satisfy: x + y > 0 and x - y > 0x > y ---- (A) how is the red part correct? x^2<y^2 x^2 - y^2<0 (x+y)(x-y) <0 either (x+y) <0 or (x-y) <0 OOps seems like I was sleeping  ..... Thanks for correction.
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GMAT Club Legend
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St1:
If x = 2, y = 5, x^2 < y^2 and x < y
If x = 2, y = -5, x^2 < y^2 but x > y
Insufficient.
St2:
Insufficient. X can be any value.
Using St1 and St2:
y must be a negative value.
y must be less than x if x^2 is to be less than y^2.
Ans C
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Re: DS - Inequality [#permalink]
30 May 2006, 17:01
gmat_crack wrote: Is x>y?
1). x^2<y^2
2). y<0
Combined:
(x+y)(x-y)>0
x could be positive or negative
if x>=0, x>y since y<0
if x<0, then x+y<0. But (x+y)(x-y)<0, that means (x-y)>0, ie x>y.
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Manager
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One more for C
The two statements together:
If x^2 < y^2 and Y is negative, then x is constrained to these values:
Y (negative) < X < -Y (positive)
From this, we know that X is always greater than Y.
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C it is.
St1: x^2 < y^2
example:
16 < 25, then x = 4,-4 and y = 5,-5 INSUFF
St2: y<0 clearly INSUFF
Combined: x = 4,-4 and y = -5. In both cases x>y so SUFF
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