It is currently 18 Oct 2017, 08:33

Live Now:

GMAT Verbal Live on YouTube: Join Now!


Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

No. Properties

  post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Senior Manager
Senior Manager
avatar
Joined: 05 Oct 2008
Posts: 270

Kudos [?]: 538 [0], given: 22

No. Properties [#permalink]

Show Tags

New post 27 Jan 2009, 09:48
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

is xy > 0?

x - y > -2
x - 2y < -6

Kudos [?]: 538 [0], given: 22

Manager
Manager
avatar
Joined: 04 Jan 2009
Posts: 237

Kudos [?]: 13 [0], given: 0

Re: No. Properties [#permalink]

Show Tags

New post 27 Jan 2009, 09:50
study wrote:
is xy > 0?

x - y > -2
x - 2y < -6

both are required. let us check if they are suff:
x-y>-2
-x+2y>6
y>4 and x>-2+y>-2+4>2
both together sufficient.
_________________

-----------------------
tusharvk

Kudos [?]: 13 [0], given: 0

SVP
SVP
avatar
Joined: 17 Jun 2008
Posts: 1534

Kudos [?]: 279 [0], given: 0

Re: No. Properties [#permalink]

Show Tags

New post 27 Jan 2009, 11:31
In the simplest form,

For stmt1: assume that y = 0, then x > -2 and xy = 0.

Similarly, for stmt2: if y = 0 then xy = 0.

Combining both,
-2 < x-y < y-6
or, y > 4
and hence x > 2 and accordingly, xy > 0.

Hence, C.

Kudos [?]: 279 [0], given: 0

Current Student
avatar
Joined: 28 Dec 2004
Posts: 3351

Kudos [?]: 319 [0], given: 2

Location: New York City
Schools: Wharton'11 HBS'12
Re: No. Properties [#permalink]

Show Tags

New post 27 Jan 2009, 15:11
I get E for some reason...

i am drawing this as a line

1)y<X+2 which is linear line and y is the region where below the line..in that case you can have where x, y>0 or x, y<0 or x<0 y>0 in this case xy<0

2) same thing y>x/2 + 3 y>0 as long as -x/2<3 however if -x/2<3 then y<0 and thus xy>0 but there is a region where y>0 and x<0 and thus xy<0

together too the posibility exists that x<0 and y>0

Kudos [?]: 319 [0], given: 2

Manager
Manager
avatar
Joined: 04 Jan 2009
Posts: 237

Kudos [?]: 13 [0], given: 0

Re: No. Properties [#permalink]

Show Tags

New post 27 Jan 2009, 15:30
fresinha12 wrote:
I get E for some reason...

i am drawing this as a line

1)y<X+2 which is linear line and y is the region where below the line..in that case you can have where x, y>0 or x, y<0 or x<0 y>0 in this case xy<0

2) same thing y>x/2 + 3 y>0 as long as -x/2<3 however if -x/2<3 then y<0 and thus xy>0 but there is a region where y>0 and x<0 and thus xy<0

together too the posibility exists that x<0 and y>0

Even if you draw these two lines you get the same result. I drew the lines y<x+2 (slope 1 and y intercept=2) y is below that line.
Then y>x/2+3. slope=1/2 and y intercept = 3. Y lies above this line. Thus x and y are to the right of the intersection point defined by the two lines viz. (2.4).
Hence, xy>0
_________________

-----------------------
tusharvk

Kudos [?]: 13 [0], given: 0

Re: No. Properties   [#permalink] 27 Jan 2009, 15:30
Display posts from previous: Sort by

No. Properties

  post reply Question banks Downloads My Bookmarks Reviews Important topics  


cron

GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.