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Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1
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Updated on: 16 Feb 2012, 21:37
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Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1
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Originally posted by jade3 on 24 Nov 2009, 20:56.
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: Is xy > x2y2? (1) 14x2 = 3 (2) y2 = 1
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16 Feb 2012, 21:44




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Re: xy > x2y2?
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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?
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Updated on: 27 Nov 2009, 04:10
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]
Originally posted by shalva on 26 Nov 2009, 00:21.
Last edited by shalva on 27 Nov 2009, 04:10, edited 1 time in total.



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Re: xy > x2y2?
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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?
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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 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
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16 Feb 2012, 19:42



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Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1
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29 Apr 2015, 22:46
Bunuel wrote: 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 1xy > 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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Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1
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30 Apr 2015, 02:30
amnesia604 wrote: Bunuel wrote: 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 1xy > 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)\(xy>x^2*y^2\); \(xy>(xy)^2\); \(xy(1xy)>0\); \(0<xy<1\). Hope it helps.
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Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1
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Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1
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06 Jun 2016, 23:41
jade3 wrote: Is xy > x^2*y^2?
(1) 14*x^2 = 3 (2) y^2 = 1 xy > \(x^2\)*\(y^2\) xy  \(x^2\)*\(y^2\) > 0 xy (1xy) > 0 0<xy < 1 > This is our revised problem Statement 1 > clearly not sufficient, as no info about y. Statement 2 > clearly not sufficient, as no info about y. Combining (1) and (2) \(x^2\)*\(y^2\) = 3/14. One of the solution of xy lies between 0 and 1; Other solution does not lie between 0 and 1 . So not sufficient. (E) is the answer.



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Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1
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21 Jun 2017, 11:53
jade3 wrote: Is xy > x^2*y^2?
(1) 14*x^2 = 3 (2) y^2 = 1 This is definitely a trap question in DS. Given: xy > x^2y^2 (1) 14 * x^2 = 3 x^2 = 3/14 =====> x = +/\sqrt{3/14} ==== Clearly Not Sufficient as we do not know value of y.(2) y^2 = 1 y = +/ 1 =====> Clearly Not Sufficient as we do not know value of x.Combining both (1) & (2) We are still not sure of the value of x & y will be positive or negative. Hence, for the question Is xy > x^2*y^2, we get both the answers YES & NO. Hence, Answer is E
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Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1
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10 Feb 2019, 09:19
jade3 wrote: Is xy > x^2*y^2?
(1) 14*x^2 = 3 (2) y^2 = 1 KeyWord x and y can be +ive or ive from 1 x^2 = 3/14, y can take any value from 2 y^2 = 1, x can take any value Now both x and y can be +ive or ive different signs of x and y, can result in different answers for the given inequality E
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Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1
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