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# Is x > y^2? (1) x > y + 5 (2) x^2 - y^2 = 0 This is a

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Is x > y^2? (1) x > y + 5 (2) x^2 - y^2 = 0 This is a [#permalink]

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20 Nov 2007, 14:28
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Is x > y^2?

(1) x > y + 5
(2) x^2 - y^2 = 0

This is a GMAT Club Challenge Question. The OA is (c). I am getting (E). Please explain.
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20 Nov 2007, 14:32
(Edit. incorrect)
I think E

will thy to draw.....

Last edited by walker on 20 Nov 2007, 14:37, edited 1 time in total.
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20 Nov 2007, 14:35
sujitmgmat wrote:
Is x > y^2?

(1) x > y + 5
(2) x^2 - y^2 = 0

This is a GMAT Club Challenge Question. The OA is (c). I am getting (E). Please explain.

from i, x > y + 5

from ii, x^2 = y^2
x = + or - y

togather, x is +ve and is equal to -y. so enough.
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20 Nov 2007, 14:45
GMAT TIGER, so which answer will you opt for? Since x is +ve and y is -ve and both have the same numerical value, x can be > or < than y^2, depending on whether the numerical value is <or> than 1. That's why I think the answer should be (E).
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20 Nov 2007, 15:07
explanation for C. in drawing
green: x=y+5
red: x^2=y^2
blue: 1&2
Attachments

GC55915.gif [ 13.87 KiB | Viewed 1002 times ]

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20 Nov 2007, 16:57
sujitmgmat wrote:
GMAT TIGER, so which answer will you opt for? Since x is +ve and y is -ve and both have the same numerical value, x can be > or < than y^2, depending on whether the numerical value is <or> than 1. That's why I think the answer should be (E).

Quote:
Is x > y^2?

(1) x > y + 5
(2) x^2 - y^2 = 0

x^2 - y^2 = 0 is possible only when either x = y or x = -y but since x > y+5, x must be +ve and therefore y is -ve.

x > y+5
x - y > 5
- y - y > 5
y < -2.5

suppose, y = - 3, x = 6, which is grater than -3+5 = 2.
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20 Nov 2007, 17:23
Im a little lost here - why isn't it B?

if X^2 = Y^2, then x = +/- y

In that case is it possible that y^2 is ever less than x? Even if y is negative...

For instance - x = 2.5 y=-2.5

x is not greater than -2.5^2 (which is 6.25)...

I can't think of a number that would fit x>y^2...
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20 Nov 2007, 18:58
sujitmgmat wrote:
Is x > y^2?

(1) x > y + 5
(2) x^2 - y^2 = 0

This is a GMAT Club Challenge Question. The OA is (c). I am getting (E). Please explain.

Ya def C.

1) Just test values. X=10 and Y=2. X=10 and Y=-6 Insuff.

2) x^2>y^2 try X=10 y=2 and X=-10 and y=2 Insuff.

Just use the same values, ul see that x>y^2
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20 Nov 2007, 23:19
alrussell wrote:
Im a little lost here - why isn't it B?

if X^2 = Y^2, then x = +/- y

In that case is it possible that y^2 is ever less than x? Even if y is negative...

For instance - x = 2.5 y=-2.5

x is not greater than -2.5^2 (which is 6.25)...

I can't think of a number that would fit x>y^2...

it cannot be because x^2 = y^2, in whixh x = y or -y.
supoose of x = 5, and y = -5, then x cannot be > y^2.
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21 Nov 2007, 08:41
Thanks folks.

I get it now. Since (1) and (2) imply that x=+/-y, x and y have different signs, and y<-2.5, we can conclude that x<y^2. So answer is (C).
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21 Nov 2007, 17:03
alrussell wrote:
Im a little lost here - why isn't it B?

if X^2 = Y^2, then x = +/- y

In that case is it possible that y^2 is ever less than x? Even if y is negative...

For instance - x = 2.5 y=-2.5

x is not greater than -2.5^2 (which is 6.25)...

I can't think of a number that would fit x>y^2...

bump...I too chose (B). Anyone?

For any value of x such that x = +/- y, x > y^2 is always false.
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21 Nov 2007, 17:13
GMATBLACKBELT wrote:
sujitmgmat wrote:
Is x > y^2?

(1) x > y + 5
(2) x^2 - y^2 = 0

This is a GMAT Club Challenge Question. The OA is (c). I am getting (E). Please explain.

Ya def C.

1) Just test values. X=10 and Y=2. X=10 and Y=-6 Insuff.

2) x^2>y^2 try X=10 y=2 and X=-10 and y=2 Insuff.

Just use the same values, ul see that x>y^2

GBB, for case (2), x^2=y^2, which means x = +/-y. How can you use x = -10 and y = 2?
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21 Nov 2007, 20:41
Another one for B.
When I looked to OA I became frustrated ...

II. x = y or x = -y
in this case x is not more than y^2 (or (-y)^2), so the answer is clear NO
Actually, x can be only less than y^2 or equal to y^2
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22 Nov 2007, 11:06
1) insuff.

2) insuff. this statement means (x+y) (x-y)=0
therefore:

x=y or
x=-y

statement 2 alone would be true if we know at least that the variables are integers, however, we weren't told that they are integers. Therefore they could be fractions, which can mess up the whole logic and give different results.

Both statement together

worst case scenario:
scenario a: if y=1/2, then x is at least 6
6>1/4 true

scenario b: if y=-1/2, then x is at least 5
5> 1/4 true

therefore, the answer should be C
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22 Nov 2007, 12:57
Tino,
Can you please look into this challenges question and, if needed, change the OA? Or, if it is right, please explain the OA.
Thanks.
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22 Nov 2007, 13:39
Is x > y^2?

(1) x > y + 5
(2) x^2 - y^2 = 0

from 1

insuff

from 2

x = y or x = -y insuff

both

they contradict ( subst by x = y or -y)

#
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23 Nov 2007, 00:04
I am for B too.

X = +/- Y

so Y > Y^2 - NO
is -Y > ^ (-Y)^2 - NO

Isnt the answer NO in both cases!!
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23 Nov 2007, 06:14
Is x > y^2?

(1) x > y + 5
(2) x^2 - y^2 = 0

Firstly, looking at the statement: y^2 will always be positive, unless y = 0. Therefore, for x > y^2, one of the conditions is that x must be > 0.

Looking at the statements:

1) INSUFFICIENT: we cannot determine from this equation whether x is > 0. For example: -5 > -11 + 5 AND 5 > -11 + 5

2) x^2 - y^2 = 0, therefore (x-y)(x+y) = 0, therefore x = -y OR + y

INSUFFICIENT: we cannot determine from this equation whether x is > 0. All we know is that x = y OR x is equal to y * -1

1) & 2) Together:

Subsitute x = y into the equation in Statement 1: y > y + 5 This cannot be true. Therefore, x is NOT equal to y! Therefore x = -y
Substitute x = -y into the equation in Statement 1:

-y > y + 5
-2y > 5
-y > 5/2.... x > 5/2
y < -5/2 Because y is <1 and x = -y, as x gets bigger, y gets smaller. Therefore y^2 will always be greater than x. The answer to the question "Is x > y^2" is NO. Therefore it's C.
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03 Dec 2007, 19:04
OA is C.

OE
when we combine the two statements, we see that the absolute values of x and y are equal and that x > y , implying that x is positive and y is negative and neither is equal to zero. Therefore, x > y2 is true.
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03 Dec 2007, 20:46
(1) or (2) alone is not sufficient
Together we have lxl = lyl and x>y+5
so x= -y and y<-2.5
thus x>2.5 and x<x2 = y2
The answer is with (1)(2) together we have x<y2
03 Dec 2007, 20:46

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