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# Is x>y? 1). x^2<y^2 2). y<0

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Senior Manager
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Is x>y? 1). x^2<y^2 2). y<0 [#permalink]

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27 May 2006, 23:36
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Is x>y?
1). x^2<y^2
2). y<0
Director
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27 May 2006, 23:45
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.

VP
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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.
Director
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28 May 2006, 12:48
Senior Manager
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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..
VP
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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 > 0
x > 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
Senior Manager
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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 > 0
x > 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.
GMAT Club Legend
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28 May 2006, 21:46
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
SVP
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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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31 May 2006, 09:47
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.
CEO
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31 May 2006, 10:43
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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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

31 May 2006, 10:43
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