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# Is x > y^2 ? 1. x > y + 5 2. x^2 - y^2 = 0

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Senior Manager
Joined: 11 Jul 2006
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Is x > y^2 ? 1. x > y + 5 2. x^2 - y^2 = 0 [#permalink]  08 Oct 2006, 18:18
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Is x > y^2 ?

1. x > y + 5
2. x^2 - y^2 = 0
Director
Joined: 23 Jun 2005
Posts: 847
GMAT 1: 740 Q48 V42
Followers: 5

Kudos [?]: 36 [0], given: 1

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.
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1266
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Kudos [?]: 167 [0], given: 0

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?
Senior Manager
Joined: 11 Jul 2006
Posts: 383
Location: TX
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Kudos [?]: 6 [0], given: 0

OA C

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.
Director
Joined: 23 Jun 2005
Posts: 847
GMAT 1: 740 Q48 V42
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Kudos [?]: 36 [0], given: 1

kevincan wrote:
What did you wrongly assume here?

x=y=0?

Basics, basics, basics!
Manager
Joined: 12 Sep 2006
Posts: 92
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Kudos [?]: 2 [0], given: 0

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?
Senior Manager
Joined: 11 Jul 2006
Posts: 383
Location: TX
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Kudos [?]: 6 [0], given: 0

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.
SVP
Joined: 05 Jul 2006
Posts: 1516
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Kudos [?]: 124 [0], given: 39

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

Senior Manager
Joined: 28 Aug 2006
Posts: 302
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Kudos [?]: 99 [0], given: 0

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
_________________
VP
Joined: 25 Jun 2006
Posts: 1173
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Kudos [?]: 79 [0], given: 0

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.
Senior Manager
Joined: 30 Aug 2006
Posts: 374
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Kudos [?]: 24 [0], given: 0

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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