Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 22 May 2015, 11:31

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:
SVP
Joined: 28 Dec 2005
Posts: 1581
Followers: 2

Kudos [?]: 88 [0], given: 2

Is xy > x^2y^2 ? (1) 14x^2 = 3 (2) y^2 = 1 [#permalink]  17 Jul 2008, 05:41
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Is xy > $$x^2y^2$$?
(1) $$14x^2 = 3$$
(2) $$y^2 = 1$$
Director
Joined: 14 Aug 2007
Posts: 734
Followers: 8

Kudos [?]: 120 [0], given: 0

Re: DS: [#permalink]  17 Jul 2008, 05:58
pmenon wrote:
Is xy > $$x^2y^2$$?
(1) $$14x^2 = 3$$
(2) $$y^2 = 1$$

E for me.
we need to know if
$$xy - x^2 * y^2 > 0$$
= $$xy(1-xy) > 0$$

both 1 & 2 give us values of $$x^2$$ & $$y^2$$
x and y can be both +ve, both -ve, one positive other negative, fractions

thus E
SVP
Joined: 30 Apr 2008
Posts: 1891
Location: Oklahoma City
Schools: Hard Knocks
Followers: 34

Kudos [?]: 470 [0], given: 32

Re: DS: [#permalink]  17 Jul 2008, 06:01
I think it's C....Alpha, you're right. I forgot to think about negatives. I thought of fractions, but not negatives.

My first thought is that we need to know if x and y are fractions. If only one is a fraction, it's not enough to answer the question with certainty.

(1) Is Insufficient because we know that $$x = \sqrt{\frac{3}{14}}$$ but that doesn't tell us if Y is a fraction or not (or 1 as we'll see with statement 2)

(2) Insufficient because we know y = 1, we don't know the value of x.

x could be 10 so then we'd have (1)(10) > (10^2)(1^2) = 10 > 100...not true or x could be 1/2 so (1/2)(1) > (1/4)(1) = 1/2 > 1/4 true. Since we can answer yes or no, it is insufficient.

Together, if we know that $$x = \sqrt{\frac{3}{14}}$$ and y = 1, then $$xy > x^2y^2$$ always.
EDIT: Correct statement is $$x = +/-\sqrt{\frac{3}{14}}$$
pmenon wrote:
Is xy > $$x^2y^2$$?
(1) $$14x^2 = 3$$
(2) $$y^2 = 1$$

_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.

GMAT Club Premium Membership - big benefits and savings

Senior Manager
Joined: 07 Jan 2008
Posts: 298
Followers: 1

Kudos [?]: 26 [0], given: 0

Re: DS: [#permalink]  17 Jul 2008, 06:51
pmenon wrote:
Is xy > $$x^2y^2$$?
(1) $$14x^2 = 3$$
(2) $$y^2 = 1$$

xy > (xy)^2 can only happen if the product of x and y is a positive fraction.
1) tells me x is fractions but does not reveal anything about y.
2) tells me y is +1 or -1. Well +1 is workable but -1 aint - not enough

Combining y=1;x=fraction is good; y=-1;x=fraction is bad
E.
Senior Manager
Joined: 29 Aug 2005
Posts: 283
Followers: 1

Kudos [?]: 30 [0], given: 0

Re: DS: [#permalink]  17 Jul 2008, 07:00
pmenon wrote:
Is xy > $$x^2y^2$$?
(1) $$14x^2 = 3$$
(2) $$y^2 = 1$$

hi guys i have a different menthod of solving this ..just want u r comments on it

we are asked if xy > x^2y^2 or 1 >xy

statement 1 does not define what is value of y hence the statement is insufficenet
statement 2 does not define what is x so we still cant know what is the value of the expression

both statements put together we get
x = +(3/14) or -(3/14)
and y = -1 or 1
therefore a number of values are possible...hence both statements put together do not help

_________________

The world is continuous, but the mind is discrete

Current Student
Joined: 28 Dec 2004
Posts: 3390
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 182 [0], given: 2

Re: DS: [#permalink]  17 Jul 2008, 12:24
you cant do this.
you dont know if xy is positive or not..it might change the direction of the inequality

vdhawan1 wrote:
pmenon wrote:
Is xy > $$x^2y^2$$?
(1) $$14x^2 = 3$$
(2) $$y^2 = 1$$

hi guys i have a different menthod of solving this ..just want u r comments on it

we are asked if xy > x^2y^2 or 1 >xy

statement 1 does not define what is value of y hence the statement is insufficenet
statement 2 does not define what is x so we still cant know what is the value of the expression

both statements put together we get
x = +(3/14) or -(3/14)
and y = -1 or 1
therefore a number of values are possible...hence both statements put together do not help

Director
Joined: 27 May 2008
Posts: 550
Followers: 7

Kudos [?]: 215 [0], given: 0

Re: DS: [#permalink]  17 Jul 2008, 20:54
you can do it in this question, but not in general ....

actaully if you divide xy on both sides the inequality change into
1>xy (if xy is positive)
1<xy (if xy is negative) ... second case is not possible, as xy can not be both negative and greater than 1 at the same time.

so basically thequestion converts into 0<xy<1

fresinha12 wrote:
you cant do this.
you dont know if xy is positive or not..it might change the direction of the inequality

vdhawan1 wrote:
pmenon wrote:
Is xy > $$x^2y^2$$?
(1) $$14x^2 = 3$$
(2) $$y^2 = 1$$

hi guys i have a different menthod of solving this ..just want u r comments on it

we are asked if xy > x^2y^2 or 1 >xy

statement 1 does not define what is value of y hence the statement is insufficenet
statement 2 does not define what is x so we still cant know what is the value of the expression

both statements put together we get
x = +(3/14) or -(3/14)
and y = -1 or 1
therefore a number of values are possible...hence both statements put together do not help

Re: DS:   [#permalink] 17 Jul 2008, 20:54
Similar topics Replies Last post
Similar
Topics:
5 Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1 10 24 Nov 2009, 20:56
Is xy > x^2*y^2? (1) 14x^2 = 3 (2) y^2 = 1 4 19 Jan 2008, 05:15
Is xy > x^2y^2? (1) 14x^2 = 3 (2) y^2 = 1 A. Statement 4 19 Oct 2006, 02:22
Is xy > x^2y^2? (1) 14x^2 = 3 (2) y^2 = 1 A. Statement 7 23 Aug 2006, 19:39
Is xy>x^2y^2? (1)14x^2 = 3 (2)y^2 = 1 2 30 Nov 2005, 23:57
Display posts from previous: Sort by