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

It is currently 22 Oct 2014, 20:11

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

algebra II qn8

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Manager
Manager
avatar
Joined: 24 Aug 2009
Posts: 151
Followers: 3

Kudos [?]: 33 [0], given: 46

algebra II qn8 [#permalink] New post 23 Oct 2009, 17:11
what is y?

1) cube(y)+2y = y +2*sqr(y)
2) sqr(y) = y


Taking statement 1, dividing both sides by y, we get sqr(y) - 2y +1 = 0 => (y-1)^2 = 0=> y-1 = 0 and hence y =1

Taking statement 2, dividing both sides by y, we straight away get y =1, but the OA is given as niether statement is sufficients...could some one please explain
Manager
Manager
User avatar
Joined: 05 Jul 2008
Posts: 139
GMAT 1: Q V
GMAT 2: 740 Q51 V38
Followers: 2

Kudos [?]: 68 [0], given: 40

Re: algebra II qn8 [#permalink] New post 23 Oct 2009, 18:17
ISBtarget wrote:
what is y?

1) cube(y)+2y = y +2*sqr(y)
2) sqr(y) = y


Taking statement 1, dividing both sides by y, we get sqr(y) - 2y +1 = 0 => (y-1)^2 = 0=> y-1 = 0 and hence y =1

Taking statement 2, dividing both sides by y, we straight away get y =1, but the OA is given as niether statement is sufficients...could some one please explain

Hey, never divide both sides for any thing that you are not sure whether or not it could be Zero. In both statements, y=1 or 0
Manager
Manager
avatar
Joined: 24 Aug 2009
Posts: 151
Followers: 3

Kudos [?]: 33 [0], given: 46

Re: algebra II qn8 [#permalink] New post 24 Oct 2009, 11:25
DavidArchuleta wrote:
ISBtarget wrote:
what is y?

1) cube(y)+2y = y +2*sqr(y)
2) sqr(y) = y


Taking statement 1, dividing both sides by y, we get sqr(y) - 2y +1 = 0 => (y-1)^2 = 0=> y-1 = 0 and hence y =1

Taking statement 2, dividing both sides by y, we straight away get y =1, but the OA is given as niether statement is sufficients...could some one please explain

Hey, never divide both sides for any thing that you are not sure whether or not it could be Zero. In both statements, y=1 or 0


thanks for the info

Even otherwise, take statement 1

take y out of left and right handside y ( y^2+2) = y ( 2y+1)

cancel out y on both sides , we get y^2 -2y+1 which is (y-1)^2 = 0 , which is y =1.

could you please explain
Intern
Intern
avatar
Joined: 16 Oct 2009
Posts: 11
Followers: 0

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

Re: algebra II qn8 [#permalink] New post 24 Oct 2009, 12:44
Here it goes..
2) You Cannot divided both sides by y and say y>1 so,
y2 = y
y2-y = 0
y(y-1) = 0
Solution is either y= 0 or y=1

Similarly for condition 1)
y3 + 2y = y +2(y2)
y3 -2y2 + y = 0 ... taking y as common equation becomes...
y(y2-2y+1) = 0
y = 0 or
(Y-1)2 = 0

Solution is either y = 0 or y = 1

Both the condition together are not able to provide single value of Y....
Hope this clears your confusion now...
Re: algebra II qn8   [#permalink] 24 Oct 2009, 12:44
    Similar topics Author Replies Last post
Similar
Topics:
Experts publish their posts in the topic Algebra II vinnik 1 12 Feb 2012, 10:35
Algebra milind1979 2 27 Apr 2009, 18:27
Algebra milind1979 3 25 Mar 2009, 22:04
Algebra ! chiragr 10 17 Jan 2006, 02:57
Display posts from previous: Sort by

algebra II qn8

  Question banks Downloads My Bookmarks Reviews Important topics  

Moderators: WoundedTiger, Bunuel



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

Powered by phpBB © phpBB Group and phpBB SEO

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®.