Is x^2 * y^2

Question

Is x^{2}y^{2} < xy ? 1. 0 < xy 2. xy < 1 Kudos forms a nice way to say you liked the question ... :-D

acegmat123 · Answer

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. 00 and xy<1 Sufficient C

Prakagmat · Answer

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

St1 and St2 together sufficient.
C

comedian1992 · Answer

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

comedian1992 · Answer

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

rohit8865 · Answer

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

so both statement necessary

Ans C

s1d1 · Answer

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

acegmat123 · 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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Is x^2 * y^2< xy ?

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