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if x is an integer, is x even? 1) x^2-y^2=0 2) x^2+y^2=18

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VP
VP
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Joined: 25 Nov 2004
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if x is an integer, is x even? 1) x^2-y^2=0 2) x^2+y^2=18 [#permalink] New post 26 Feb 2005, 22:41
00:00
A
B
C
D
E

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0% (00:00) correct 0% (00:00) wrong based on 3 sessions
if x is an integer, is x even?
1) x^2-y^2=0
2) x^2+y^2=18
Director
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Joined: 07 Jun 2004
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Location: PA
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 [#permalink] New post 27 Feb 2005, 04:19
from 1 x = y or x <> Y

from 2 x can be 4 or 3 and y can be any ting as its not an integer

from 1 and 2 x has to be 3 or -3 so its C
Current Student
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Joined: 28 Dec 2004
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Schools: Wharton'11 HBS'12
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 [#permalink] New post 28 Feb 2005, 09:46
c it is...

statement 1

X^2-y^2=0, pick 0 (which is even) and 3 which is odd.

we dont know what Y is... so insuff

statement 2

X^2+y^2=18,

x can be 3 or 0 if y is 3 or some value like 4.4 <--we dont know if y is integer or not.

taking together we now know that x=y therfore 3 is the only possible soution.

C it is
SVP
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 [#permalink] New post 28 Feb 2005, 10:52
It the stem is changed to "X and Y are integers", will your answer change?
Senior Manager
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 [#permalink] New post 28 Feb 2005, 11:56
Why does it become B
how is that enough to ditermine if X is an even integer or not?
VP
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 [#permalink] New post 28 Feb 2005, 12:02
Rupstar wrote:
Why does it become B
how is that enough to ditermine if X is an even integer or not?


If both r ints...then state I satisfies when X = 1 and Y = 1 and X = 2 and Y = 2....insuff.....as Xcan be even or odd

In state 2.....if X = 0,1,2,4...Y can't be an integer and vice versa, also either of X or Y can't be >= 5.....so only for X = Y = #3 satisfies....so X is always odd...so ANS is NO.....suff
  [#permalink] 28 Feb 2005, 12:02
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if x is an integer, is x even? 1) x^2-y^2=0 2) x^2+y^2=18

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