Is 2x - 3y < x^2 ? 1) 2x - 3y = -2 2) x > 2 and y > : Quant Question Archive [LOCKED]
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# Is 2x - 3y < x^2 ? 1) 2x - 3y = -2 2) x > 2 and y >

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Is 2x - 3y < x^2 ? 1) 2x - 3y = -2 2) x > 2 and y > [#permalink]

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17 Oct 2006, 19:41
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Is 2x - 3y < x^2 ?

1) 2x - 3y = -2
2) x > 2 and y > 0
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17 Oct 2006, 19:53
1) SUFF
x^2 is always positive

2) SUFF
Is x^2 > 2x -3y
Is x^2 -2x + 3y > 0
Is x (x -2) + 3y >0

Here x Positive, (x-2) positive bcos x > 2 and y positive bcos y > 0
so the ineql is correct

Ans D
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18 Oct 2006, 07:18
Statement 2 is sufficient for sure.

Can someone explain in more detail how Statement 1 is sufficient?
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18 Oct 2006, 18:56
D.

for 1). u get x^2 > -2. obviously.
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19 Oct 2006, 10:51
I am not sure if it is possible. Pls. correct me if I am wrong..If we assume x^4=4 then can we take x^2 = +/- 2.In this case (A) is not sufficient and answer will be (B) ?
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19 Oct 2006, 12:19
I dont think so.....X^2 will always be positive so a negative number will always be less than it....so A is suff.
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19 Oct 2006, 12:51
is 2x - 3y < x^2 ?

1) 2x - 3y = -2
2) x > 2 and y > 0

from one

-2<x^2........suff

from two

x>2, y>0

id x= 2.1 , y = 0.1

4.2-3.3 = 1

if x =3 y = 1

6-3 = 3

if x=3 , y =2

6-6 = 0

suff

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20 Oct 2006, 07:48
Hey yezz,
Is it possible to assume x^4=4 => x^2 =2 or x^2 =-2. Is this possible ? Thanks.
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20 Oct 2006, 08:04
mukeshnathani wrote:
Hey yezz,
Is it possible to assume x^4=4 => x^2 =2 or x^2 =-2. Is this possible ? Thanks.

x^4=4 => x^2 =2 and not x^2 =2 or x^2 =-2, because x^2 is always positive

(we work with real numbers for GMAT)
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20 Oct 2006, 11:00
Thanks Fig, I have taken a note of that
20 Oct 2006, 11:00
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