GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 13 Oct 2019, 23:03

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 19 Nov 2007
Posts: 137
Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1  [#permalink]

### Show Tags

Updated on: 16 Feb 2012, 22:37
9
00:00

Difficulty:

55% (hard)

Question Stats:

60% (01:55) correct 40% (01:54) wrong based on 182 sessions

### HideShow timer Statistics

Is xy > x^2*y^2?

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

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
Math Expert
Joined: 02 Sep 2009
Posts: 58335
Re: Is xy > x2y2? (1) 14x2 = 3 (2) y2 = 1  [#permalink]

### Show Tags

16 Feb 2012, 22:44
3
4
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.

_________________
##### General Discussion
Manager
Joined: 05 Jun 2009
Posts: 66

### Show Tags

24 Nov 2009, 22:33
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)
Tuck School Moderator
Joined: 20 Aug 2009
Posts: 247
Location: Tbilisi, Georgia
Schools: Stanford (in), Tuck (WL), Wharton (ding), Cornell (in)

### Show Tags

Updated on: 27 Nov 2009, 05:10
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, 01:21.
Last edited by shalva on 27 Nov 2009, 05:10, edited 1 time in total.
Manager
Joined: 24 Sep 2009
Posts: 58

### Show Tags

26 Nov 2009, 20:43
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.
_________________
http://www.online-stopwatch.com/
http://gmatsentencecorrection.blogspot.com/
Tuck School Moderator
Joined: 20 Aug 2009
Posts: 247
Location: Tbilisi, Georgia
Schools: Stanford (in), Tuck (WL), Wharton (ding), Cornell (in)

### Show Tags

27 Nov 2009, 05:11
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 - $$xy>x^2y^2$$
Senior Manager
Status: No dream is too large, no dreamer is too small
Joined: 14 Jul 2010
Posts: 428
Re: Is xy > x2y2? (1) 14x2 = 3 (2) y2 = 1  [#permalink]

### Show Tags

16 Feb 2012, 20:42
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?
_________________
Intern
Joined: 11 May 2013
Posts: 7
GMAT 1: 770 Q51 V44
GPA: 2.57
WE: Consulting (Consulting)
Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1  [#permalink]

### Show Tags

29 Apr 2015, 23:46
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 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.

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)
Math Expert
Joined: 02 Sep 2009
Posts: 58335
Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1  [#permalink]

### Show Tags

30 Apr 2015, 03:30
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 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.

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.
_________________
Current Student
Joined: 12 Aug 2015
Posts: 2574
Schools: Boston U '20 (M)
GRE 1: Q169 V154
Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1  [#permalink]

### Show Tags

13 Mar 2016, 08:34
Here nothing is specified about the values of x adn y being >0 or <0
so => not sufficient.
hence E
_________________
Manager
Joined: 18 Jan 2010
Posts: 246
Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1  [#permalink]

### Show Tags

07 Jun 2016, 00:41
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 (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.

Retired Moderator
Joined: 19 Mar 2014
Posts: 923
Location: India
Concentration: Finance, Entrepreneurship
GPA: 3.5
Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1  [#permalink]

### Show Tags

21 Jun 2017, 12:53
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.

_________________
"Nothing in this world can take the place of persistence. Talent will not: nothing is more common than unsuccessful men with talent. Genius will not; unrewarded genius is almost a proverb. Education will not: the world is full of educated derelicts. Persistence and determination alone are omnipotent."

Best AWA Template: https://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html#p470475
Director
Joined: 09 Mar 2018
Posts: 997
Location: India
Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1  [#permalink]

### Show Tags

10 Feb 2019, 10:19
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
_________________
If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.
Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1   [#permalink] 10 Feb 2019, 10:19
Display posts from previous: Sort by