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Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1 Updated on: 16 Feb 2012, 22:37
Originally posted by jade3 on 24 Nov 2009, 21:56.
Last edited by Bunuel on 16 Feb 2012, 22:37, edited 1 time in total.
Edited the question and added the OA
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

55% (hard)

Question Stats:

61% (01:51) correct 39% (01:52) wrong based on 433 sessions

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Is xy > x^2*y^2?

(1) 14*x^2 = 3
(2) y^2 = 1
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E
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Bunuel
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Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1 16 Feb 2012, 22:44
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Baten80
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?
Show more

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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Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1 24 Nov 2009, 22:33
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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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Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1 Updated on: 27 Nov 2009, 05:10
Originally posted by shalva on 26 Nov 2009, 01:21.
Last edited by shalva on 27 Nov 2009, 05:10, edited 1 time in total.
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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]
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Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1 26 Nov 2009, 20:43
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shalva
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)
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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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Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1 27 Nov 2009, 05:11
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fall2009

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.
Show more


You're absolutely right :oops: In one case \(xy<x^2y^2\), in other - \(xy>x^2y^2\)
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Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1 16 Feb 2012, 20:42
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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?
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Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1 29 Apr 2015, 23:46
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Bunuel
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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.
Show more

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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Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1 30 Apr 2015, 03:30
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amnesia604
Bunuel
Baten80
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)
Show more


\(xy>x^2*y^2\);
\(xy>(xy)^2\);
\(xy(1-xy)>0\);
\(0<xy<1\).

Theory on Inequalities:
Solving Quadratic Inequalities - Graphic Approach: solving-quadratic-inequalities-graphic-approach-170528.html
Inequality tips: tips-and-hints-for-specific-quant-topics-with-examples-172096.html#p1379270

inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html

All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184
All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189

700+ Inequalities problems: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope it helps.
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Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1 13 Mar 2016, 08:34
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Here nothing is specified about the values of x adn y being >0 or <0
so => not sufficient.
hence E
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Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1 07 Jun 2016, 00:41
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jade3
Is xy > x^2*y^2?

(1) 14*x^2 = 3
(2) y^2 = 1
Show more

xy > \(x^2\)*\(y^2\)

xy - \(x^2\)*\(y^2\) > 0

xy (1-xy) > 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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Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1 21 Jun 2017, 12:53
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jade3
Is xy > x^2*y^2?

(1) 14*x^2 = 3
(2) y^2 = 1
Show more

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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Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1 10 Feb 2019, 10:19
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jade3
Is xy > x^2*y^2?

(1) 14*x^2 = 3
(2) y^2 = 1
Show more

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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Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1 25 May 2020, 16:49
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Hi Bunuel,

call me crazy but how do we get from

xy(1−xy)>0;

to

0<xy<10<xy<1.


this part is really confusing me?

thank you.


Bunuel
amnesia604
Bunuel
Baten80
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)


\(xy>x^2*y^2\);
\(xy>(xy)^2\);
\(xy(1-xy)>0\);
\(0<xy<1\).


Hope it helps.
Show more
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Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1 25 May 2020, 20:16
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mrcentauri
Hi Bunuel,

call me crazy but how do we get from

xy(1−xy)>0;

to

0<xy<10<xy<1.


this part is really confusing me?

thank you.


Bunuel
amnesia604
Bunuel
Baten80
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)


\(xy>x^2*y^2\);
\(xy>(xy)^2\);
\(xy(1-xy)>0\);
\(0<xy<1\).


Hope it helps.
Show more

Say xy = a, then xy(1- xy) > 0 becomes a(1- a) > 0, which is true when 0 < a < 1.
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Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1 31 Oct 2020, 07:19
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Is \(xy > x^2*y^2\)?

\(xy - x^2y^2 > 0?\)
\(xy - xy(xy) > 0\)?
\(xy (1-xy) > 0\)?
\(0 < xy < 1\)?

(1) \(14*x^2 = 3\)

No information about y. INSUFFICIENT.

(2) \(y^2 = 1\)

No information about x. INSUFFICIENT.

(1&2) Combining both, we can have a negative or positive answer. INSUFFICIENT.

Answer is E.
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Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1 10 Apr 2022, 03:38
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brunel, could you clarify here what do you mean by If x and y have the same sign then yes but no if not? Do you refer to square root sign here? Thanks
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Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1 22 Sep 2024, 02:54
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