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

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Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1 [#permalink] New post 24 Nov 2009, 20:56
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Is xy > x^2*y^2?

(1) 14*x^2 = 3
(2) y^2 = 1
[Reveal] Spoiler: OA

Last edited by Bunuel on 16 Feb 2012, 21:37, edited 1 time in total.
Edited the question and added the OA
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Re: xy > x2y2? [#permalink] New post 24 Nov 2009, 21:33
E

both statements dont reference both X and Y

and statement 2 leaves Y with both positive and negative values thus it can be greater or less. (also the square of X is less than X but that doesnt matter if the products can be positive or negative)
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Re: xy > x2y2? [#permalink] New post 26 Nov 2009, 00:21
Let's consider both statements together

xy could be positive or negative;

if xy<0, clearly xy \leq x^2y^2, no calculations are needed. Inequality in question stem is false

If xy>0, than y=1; x=sqrt{3/14} or y=-1; x=-sqrt{3/14}. once more no need to calculate this, just note that xy<1. In this case (xy)^2<xy<1

[strike]So, in both cases xy<x^2y^2[/strike]

Last edited by shalva on 27 Nov 2009, 04:10, edited 1 time in total.
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Re: xy > x2y2? [#permalink] New post 26 Nov 2009, 19:43
shalva wrote:
Let's consider both statements together

xy could be positive or negative;

if xy<0, clearly xy \leq x^2y^2, no calculations are needed. Inequality in question stem is false

If xy>0, than y=1; x=sqrt{3/14} or y=-1; x=-sqrt{3/14}. once more no need to calculate this, just note that xy<1. In this case (xy)^2<xy<1

So, in both cases xy<x^2y^2

(C)

I don't get this, in the blue part, you mentioned that (xy)^2<xy, then your conclusion (the red part) was in both cases xy<x^2y^2.

Anyway, IMO, E is the answer because (xy)^2 = 3/14.
If xy<0, clearly that xy<(xy)^2.
If xy>0, because (xy)^2 =3/14 <1 => xy>(xy)^2.

Hence E.
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Re: xy > x2y2? [#permalink] New post 27 Nov 2009, 04:11
fall2009 wrote:
I don't get this, in the blue part, you mentioned that (xy)^2<xy, then your conclusion (the red part) was in both cases xy<x^2y^2.

Anyway, IMO, E is the answer because (xy)^2 = 3/14.
If xy<0, clearly that xy<(xy)^2.
If xy>0, because (xy)^2 =3/14 <1 => xy>(xy)^2.

Hence E.



You're absolutely right :oops: In one case xy<x^2y^2, in other - xy>x^2y^2
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Re: Is xy > x2y2? (1) 14x2 = 3 (2) y2 = 1 [#permalink] New post 16 Feb 2012, 19:42
if as follows:
xy - x^2*y^2 > 0
xy(1- xy) > 0
so, xy>0
or, x>0 or, y>0
and 1-xy > 0
xy<1
Thus no statements are sufficient.
Ans. E

What are wrongs with this approach?
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Re: Is xy > x2y2? (1) 14x2 = 3 (2) y2 = 1 [#permalink] New post 16 Feb 2012, 21:44
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Baten80 wrote:
if as follows:
xy - x^2*y^2 > 0
xy(1- xy) > 0
so, xy>0
or, x>0 or, y>0
and 1-xy > 0
xy<1
Thus no statements are sufficient.
Ans. E

What are wrongs with this approach?


Is xy > x^2*y^2?

Is xy>x^2*y^2? --> is 0<xy<1? (the same way as a>a^2 means 0<a<1)

(1) 14*x^2 = 3. Clearly insufficient, since no info about y.
(2) y^2 = 1. Clearly insufficient, since no info about x.

(1)+(2) If x and y have the same sign then xy=\sqrt{\frac{3}{14}} and the answer will be YES but if x and y have the opposite signs then xy=-\sqrt{\frac{3}{14}} and the answer will be NO. Not sufficient.

Answer: E.
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Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1 [#permalink] New post 23 Feb 2014, 05:02
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Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1   [#permalink] 23 Feb 2014, 05:02
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