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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.

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.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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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)

Please don't begin new threads for math questions that are already discussed here on GMAT Club. This particular question is already posted here: is-xy-x-2-y-2-1-14-x-2-3-2-y-87269.html I will ask Bunuel to merge the two posts.

You can find a detailed discussion at that link, and if you still have a question, that would be the appropriate place to ask it, once these posts are merged.

Please don't begin new threads for math questions that are already discussed here on GMAT Club. This particular question is already posted here: is-xy-x-2-y-2-1-14-x-2-3-2-y-87269.html I will ask Bunuel to merge the two posts.

You can find a detailed discussion at that link, and if you still have a question, that would be the appropriate place to ask it, once these posts are merged.

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