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Is x^2 * y^2< xy ?

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idontknowwhy94
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Is x^2 * y^2< xy ? [#permalink] New post   06 Oct 2016, 13:25
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62% (01:20) correct 38% (01:21) wrong based on 213 sessions
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Is \(x^{2}y^{2}\) < xy ?

1. 0 < xy

2. xy < 1

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C
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acegmat123
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Re: Is x^2 * y^2< xy ? [#permalink] New post   06 Oct 2016, 19:18
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 ... :-D :-D


x^2*y^2 < xy ?

Divide both sides by x^2*y^2

1/xy > 1 ?

The above condition will be satisfied only if xy is positive and less than 1.

i.e. 0<xy<1?

Statement 1:

Insufficient

Statement 2:

Insufficient


St1 & 2:

xy>0 and xy<1

Sufficient

C
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Re: Is x^2 * y^2< xy ? [#permalink] New post   06 Oct 2016, 22:35
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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 ... :-D :-D


(x^2)(y^2) - xy < 0
xy(xy-1) < 0
xy<0 or xy-1<0

St1 and St2 together sufficient.
C
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Re: Is x^2 * y^2< xy ? [#permalink] New post   07 Oct 2016, 06:40
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] New post   07 Oct 2016, 06:41
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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] New post   07 Oct 2016, 10:55
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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 ... :-D :-D

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] New post   07 Oct 2016, 12:32
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

thus A is the answer
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Re: Is x^2 * y^2< xy ? [#permalink] New post   07 Oct 2016, 12:46
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

thus A is the answer


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] New post   27 Mar 2023, 00:22
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Re: Is x^2 * y^2< xy ? [#permalink]
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