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Senior Manager
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Is x > y^2 ? 1. x > y + 5 2. x^2 - y^2 = 0 [#permalink]
 08 Oct 2006, 19:18
Question Stats:
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Is x > y^2 ?
1. x > y + 5
2. x^2 - y^2 = 0
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Director
Joined: 23 Jun 2005
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GMAT 1: 740 Q48 V42
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B
Statement 1:
x>y+5
If y = 2, x>y^2
If y = 6, x<y^2
INSUFF
Statement 2:
x = +/- y
So, x<y^2 SUFF.
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GMAT Instructor
Joined: 04 Jul 2006
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anandsebastin wrote: B
Statement 1:
x>y+5
If y = 2, x>y^2
If y = 6, x<y^2
INSUFF
Statement 2:
x = +/- y
So, x<y^2 SUFF.
What did you wrongly assume here?
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Senior Manager
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OA C
C is the correct answer.
The best solution is to start from statement 2.
This gives us that |x| = |y| and x and y can be positive/negative or both equal to zero. It is not sufficient.
Statement 1 gives us nothing by itself, just some relationship.
However, when we combine the two statements - we see that absolute values of x and y are equal and that x is greater than y, meaning that x is positive and y is negative and neither are equal to zero. Therefore, x > y^2 is true.
I am not sure about this OA/OE , can some Math experts validate/invalidate this OE with a better example.
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Director
Joined: 23 Jun 2005
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kevincan wrote: What did you wrongly assume here?
 x=y=0?
Basics, basics, basics!
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Manager
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ivymba,
i follow the OE until x > y^2
to me, combining both (1) and (2) gives me x < y^2
even if X is greater than Y, is lXl = lYl, y^2 must be greater than X
what am i doing wrong here?
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Senior Manager
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keeeeeekse wrote: even if X is greater than Y, is lXl = lYl, y^2 must be greater than X
The question here is "Is x > y^2 ?" we can answer that using 1 and 1 . I agree the last part o the OE looks incorrect " Therefore, x > y^2 is true.
". But the OA still remains C.
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SVP
Joined: 05 Jul 2006
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Is x > y^2 ?
1. x > y + 5
2. x^2 - y^2 = 0
for x to be > y^2
from one
x-y> 5 ........insuff depends on values of x,y
from two
x^2 = y^2
we can only know that /y/ = /x/........" THEY MIGHT EVEN BE = ZERO OR 1 " insuff
BOTH
FROM TWO
if this is true then for x>y^2 to be true, they both has to be fractions where x is a positive fraction and y is a negative or positive
equal fraction ...........insuff
FROM ONE
BOTH can never be a fraction IN THE SAME TIME because
x-y>5
so x can never be > y ^2
mY answer is C
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Senior Manager
Joined: 28 Aug 2006
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It must be C.
If |x|=|y| we cannot conclude whether x<y^2 or not?
For exmaple let's take x=-y
let y=-3 clearly x= 3
x<y^2.
Let y=-1/2 clearly x=1/2
x>y^2
So 2 is not sufficient.
Combining both x must be positive and y must be -ve and also since x>y+5 and x=y or x=-y, they cannot take fractions............
So C
_________________
Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)
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VP
Joined: 25 Jun 2006
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C.
x must be -y and x > 0 and y < 0
and the distance between them must be over 5.
so x > 5/2. y^2 is equal to x^2. since x > 1, then x^2>x. So the answer is always No.
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Senior Manager
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agree C
I) INS x > y + 5, x - y > 5
II)INS x^2-y^2 = 0
using both
(x+y)*>5 = 0 therefore
(x+y) = 0
x = -y
x < y^2 therefore x cannot be > y^2
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