If x = 3y, is x^2 > y^2? 1. y+x < x-y 2. x^2 = 9 y^2

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If x = 3y, is x^2 > y^2? 1. y+x < x-y 2. x^2 = 9 y^2 [#permalink]

14 Sep 2006, 06:41

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If x = 3y, is x^2 > y^2?

1. y+x < x-y
2. x^2 = 9y^2

Edited: Thanks yezz...again!! Too sleepy early in the morning

Last edited by haas_mba07 on 14 Sep 2006, 08:56, edited 2 times in total.

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14 Sep 2006, 06:47

If x = 3y, is x^2 > y^2?

1. y+x < y-x
2. x^2 = 3y^2

i think something is wrong in the stem

if x = 3y thus x^2 =9y^2

the the question asks whether

9y^2>y^2.....ie y^2(8)>0 and y^2 has to be positive thus we dont even need the given
:roll:

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Re: DS: Inequalities [#permalink]

14 Sep 2006, 15:00

[quote="haas_mba07"]If x = 3y, is x^2 > y^2?

1. y+x < x-y
2. x^2 = 9y^2

[quote]

hmmm..this question is different

Essentially it is asking if 9y^2 > y^2 . This is true for all numbers except 0

1. x+y < x-y . For this to happen y should be negative or both x and y should be negative. so x is not 0 which implies y is not zero so SUFF

2. states 9y^2 = 9y^2 so nothing new here. It is true for all values of y except 0 so INSUFF

I would go with A

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15 Sep 2006, 01:01

I think i was absent yesterday ... and as a part of the human beings psychological defense mechanism .... i am denying that the one who sent my post is not me :lol:

I can never be that carless

thanks every one

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15 Sep 2006, 02:15

If x = 3y, is x^2 > y^2?

1. y+x < x-y
2. x^2 = 9y^2

if x = 3y thus x^2 =9y^2

the the question asks whether

9y^2>y^2.....ie y^2(8)>0 and y^2 larger than zero ,so we are left to find out whether y >0 or not

from one

y+x+y-x<0

ie : y<0 ....suff

from two

x^2 = 9y^2 and from the given x = 3y

rephrasing

9y^2 = 9Y^2 still y could be anything {0,1 , 2....} ...insuff

answer thus is A

[#permalink] 15 Sep 2006, 02:15

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If x = 3y, is x^2 > y^2? 1. y+x < x-y 2. x^2 = 9 y^2

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If x = 3y, is x^2 > y^2? 1. y+x < x-y 2. x^2 = 9 y^2

If x = 3y, is x^2 > y^2? 1. y+x < x-y 2. x^2 = 9 y^2

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