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

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If x=3y, is x^2>y^2? 1). y+x<y*x 2). x^2=9y^2 [#permalink]  08 Oct 2006, 22:14
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If x=3y, is x^2>y^2?

1). y+x<y*x

2). x^2=9y^2
Senior Manager
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[#permalink]  09 Oct 2006, 00:16
x=3y

Stat1 : x+y<x*y

4y<3y^2

4<3y

4<x

x>4

therfore x is Positive and >4

Therefore x^2>y^2

Stat2 : is neway given in the question itself

Statement 1 is enough in itself to arrive at the anwser.

choice A

Not sure if this is the rite way to do it. It's just an attempt
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[#permalink]  09 Oct 2006, 01:26
(D) for me.

No need statments

x = 3*y
=> x^2 = 9*y^2 Thus, x^2 > y^2
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[#permalink]  09 Oct 2006, 06:34
iced_tea wrote:
Fig wrote:
(D) for me.

No need statments

x = 3*y
=> x^2 = 9*y^2 Thus, x^2 > y^2

what if y = 0 ?

Yes :D ... U are right .... always this case of y=0 or x=0, the one to keep in mind

Thus, it turns to be (A)
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Re: DS - x, y [#permalink]  09 Oct 2006, 07:22
if x^2>y^2
implies 9y^2>y^2
implies 8y^2>0 whic is always true for all values of y except zero.
so whichever option can let me know that either x or y are not zero.i'm through.

1. x+y<y*x
so 4y<3y^2
so y(y-4/3)<0
solving which we get either y>4/3 or <zero.
hence y cant be zero.hence x cant be zero.
so this choice helpsme ensure that y ,x cant assume zero value.

2. in this x=y=0 is a posiblity.

hence choice A should be the answer.

iced_tea wrote:
If x=3y, is x^2>y^2?

1). y+x<y*x

2). x^2=9y^2
Re: DS - x, y   [#permalink] 09 Oct 2006, 07:22
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# If x=3y, is x^2>y^2? 1). y+x<y*x 2). x^2=9y^2

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