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

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New post 08 Oct 2006, 19:18
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Is x > y^2 ?

1. x > y + 5
2. x^2 - y^2 = 0
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New post 08 Oct 2006, 19:31
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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New post 09 Oct 2006, 15:13
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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New post 09 Oct 2006, 16:03
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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New post 10 Oct 2006, 04:27
kevincan wrote:
What did you wrongly assume here?

:oops: x=y=0?

Basics, basics, basics!
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New post 10 Oct 2006, 07:29
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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New post 10 Oct 2006, 14:52
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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New post 10 Oct 2006, 15:28
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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New post 10 Oct 2006, 23:06
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
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New post 12 Oct 2006, 19:26
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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New post 13 Oct 2006, 13:31
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
  [#permalink] 13 Oct 2006, 13:31
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