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# Is xy > x^2y^2 ? (1) 14x^2 = 3 (2) y^2 = 1

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Is xy > x^2y^2 ? (1) 14x^2 = 3 (2) y^2 = 1 [#permalink]

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17 Jul 2008, 06:41
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Is xy > $$x^2y^2$$?
(1) $$14x^2 = 3$$
(2) $$y^2 = 1$$
Director
Joined: 14 Aug 2007
Posts: 733
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17 Jul 2008, 06:58
pmenon wrote:
Is xy > $$x^2y^2$$?
(1) $$14x^2 = 3$$
(2) $$y^2 = 1$$

E for me.
we need to know if
$$xy - x^2 * y^2 > 0$$
= $$xy(1-xy) > 0$$

both 1 & 2 give us values of $$x^2$$ & $$y^2$$
x and y can be both +ve, both -ve, one positive other negative, fractions

thus E
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17 Jul 2008, 07:01
I think it's C....Alpha, you're right. I forgot to think about negatives. I thought of fractions, but not negatives.

My first thought is that we need to know if x and y are fractions. If only one is a fraction, it's not enough to answer the question with certainty.

(1) Is Insufficient because we know that $$x = \sqrt{\frac{3}{14}}$$ but that doesn't tell us if Y is a fraction or not (or 1 as we'll see with statement 2)

(2) Insufficient because we know y = 1, we don't know the value of x.

x could be 10 so then we'd have (1)(10) > (10^2)(1^2) = 10 > 100...not true or x could be 1/2 so (1/2)(1) > (1/4)(1) = 1/2 > 1/4 true. Since we can answer yes or no, it is insufficient.

Together, if we know that $$x = \sqrt{\frac{3}{14}}$$ and y = 1, then $$xy > x^2y^2$$ always.
EDIT: Correct statement is $$x = +/-\sqrt{\frac{3}{14}}$$
pmenon wrote:
Is xy > $$x^2y^2$$?
(1) $$14x^2 = 3$$
(2) $$y^2 = 1$$

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17 Jul 2008, 07:51
pmenon wrote:
Is xy > $$x^2y^2$$?
(1) $$14x^2 = 3$$
(2) $$y^2 = 1$$

xy > (xy)^2 can only happen if the product of x and y is a positive fraction.
1) tells me x is fractions but does not reveal anything about y.
2) tells me y is +1 or -1. Well +1 is workable but -1 aint - not enough

Combining y=1;x=fraction is good; y=-1;x=fraction is bad
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17 Jul 2008, 08:00
pmenon wrote:
Is xy > $$x^2y^2$$?
(1) $$14x^2 = 3$$
(2) $$y^2 = 1$$

hi guys i have a different menthod of solving this ..just want u r comments on it

we are asked if xy > x^2y^2 or 1 >xy

statement 1 does not define what is value of y hence the statement is insufficenet
statement 2 does not define what is x so we still cant know what is the value of the expression

both statements put together we get
x = +(3/14) or -(3/14)
and y = -1 or 1
therefore a number of values are possible...hence both statements put together do not help

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17 Jul 2008, 13:24
you cant do this.
you dont know if xy is positive or not..it might change the direction of the inequality

vdhawan1 wrote:
pmenon wrote:
Is xy > $$x^2y^2$$?
(1) $$14x^2 = 3$$
(2) $$y^2 = 1$$

hi guys i have a different menthod of solving this ..just want u r comments on it

we are asked if xy > x^2y^2 or 1 >xy

statement 1 does not define what is value of y hence the statement is insufficenet
statement 2 does not define what is x so we still cant know what is the value of the expression

both statements put together we get
x = +(3/14) or -(3/14)
and y = -1 or 1
therefore a number of values are possible...hence both statements put together do not help

Director
Joined: 27 May 2008
Posts: 549
Followers: 8

Kudos [?]: 267 [0], given: 0

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17 Jul 2008, 21:54
you can do it in this question, but not in general ....

actaully if you divide xy on both sides the inequality change into
1>xy (if xy is positive)
1<xy (if xy is negative) ... second case is not possible, as xy can not be both negative and greater than 1 at the same time.

so basically thequestion converts into 0<xy<1

fresinha12 wrote:
you cant do this.
you dont know if xy is positive or not..it might change the direction of the inequality

vdhawan1 wrote:
pmenon wrote:
Is xy > $$x^2y^2$$?
(1) $$14x^2 = 3$$
(2) $$y^2 = 1$$

hi guys i have a different menthod of solving this ..just want u r comments on it

we are asked if xy > x^2y^2 or 1 >xy

statement 1 does not define what is value of y hence the statement is insufficenet
statement 2 does not define what is x so we still cant know what is the value of the expression

both statements put together we get
x = +(3/14) or -(3/14)
and y = -1 or 1
therefore a number of values are possible...hence both statements put together do not help

Re: DS:   [#permalink] 17 Jul 2008, 21:54
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