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# Is x^2 * y^2< xy ?

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Re: Is x^2 * y^2< xy ? [#permalink]
Is x2y2x2y2 < xy ?

x^2xY^2-XY<0
XY(XY-1)<0

So one of them should be negative one XY and XY-1. if XY is negative than obviously XY-1 is negative to. So XY-1 is Negative and XY is Positive. That means that XY>0 and XY-1<0, So Answer is C
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Re: Is x^2 * y^2< xy ? [#permalink]
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Is x2y2x2y2 < xy ?

x^2xY^2-XY<0
XY(XY-1)<0

So one of them should be negative one XY and XY-1. if XY is negative than obviously XY-1 is negative to. So XY-1 is Negative and XY is Positive. That means that XY>0 and XY-1<0, So Answer is C
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Re: Is x^2 * y^2< xy ? [#permalink]
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idontknowwhy94 wrote:
Is $$x^{2}y^{2}$$ < xy ?

1. 0 < xy

2. xy < 1

Kudos forms a nice way to say you liked the question ...

Square of any number is < number itself when the number x lies in range 0<x<1

so both statement necessary

Ans C
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Re: Is x^2 * y^2< xy ? [#permalink]
can anybody explain what is wrong with this approach!!

Q) Is x^2*y^2<xy? => (xy)^2<xy

1. 0 < xy => xy>0, hence positive. since xy is positive (xy)^2 will be positive. hence this statement is sufficient to answer the question. (we can eliminate choices B,C,E.

2. xy < 1
we can have three cases:
case1- where xy=0 in this case (xy)^2=xy
case2- xy is less than zero
xy is -ve thus (xy)^2 >xy
case3- 0<xy<1
here (xy)^2<xy

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Re: Is x^2 * y^2< xy ? [#permalink]
s1d1 wrote:
can anybody explain what is wrong with this approach!!

Q) Is x^2*y^2<xy? => (xy)^2<xy

1. 0 < xy => xy>0, hence positive. since xy is positive (xy)^2 will be positive. hence this statement is sufficient to answer the question. (we can eliminate choices B,C,E.

2. xy < 1
we can have three cases:
case1- where xy=0 in this case (xy)^2=xy
case2- xy is less than zero
xy is -ve thus (xy)^2 >xy
case3- 0<xy<1
here (xy)^2<xy

St1:

If xy =0.5, then (xy)^2=0.25

(xy)^2<xy

Yes

If xy =2, then (xy)^2=4

(xy)^2>xy

No

Insufficient

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Re: Is x^2 * y^2< xy ? [#permalink]
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Re: Is x^2 * y^2< xy ? [#permalink]
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