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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)

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.

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.

Re: Is xy > x2y2? (1) 14x2 = 3 (2) y2 = 1 [#permalink]
16 Feb 2012, 21:44

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This post received KUDOS

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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.

Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1 [#permalink]
10 Mar 2015, 08:13

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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.

Hi Bunuel,

Can you explain how you got this step? Thanks!

Is xy>x2∗y2? --> is 0<xy<1? (the same way as a>a^2 means 0<a<1)

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.

Hi Bunuel,

Can you explain how you got this step? Thanks!

Is xy>x2∗y2? --> is 0<xy<1? (the same way as a>a^2 means 0<a<1)

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