BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

gmatprep

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Legendary Member
Posts: 876
Joined: Thu Apr 10, 2008 8:14 am
Thanked: 13 times
by ketkoag » Sun Sep 06, 2009 9:07 am
please explain how to solve this one?
my method..
take 1. since 1/(k-1) > 0, k-1 should be positive.
therefore, k-1 < 1
hence, k < 2 insuff.

take 2. since 1/(k+2) > 0 , k+1 should be positive.
therefore, k+1 < 1
hence k<0 suff.

hence B..
please lemme know what i did wrong??
Attachments
2.jpg
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Data Sufficiency |
Master | Next Rank: 500 Posts
Posts: 153
Joined: Tue Jun 02, 2009 2:48 pm
Location: Ohio
Thanked: 16 times

Re: gmatprep

by JeffB » Sun Sep 06, 2009 1:59 pm
ketkoag wrote:please explain how to solve this one?
my method..
take 1. since 1/(k-1) > 0, k-1 should be positive.
therefore, k-1 < 1
hence, k < 2 insuff.

take 2. since 1/(k+2) > 0 , k+1 should be positive.
therefore, k+1 < 1
hence k<0 suff.

hence B..
please lemme know what i did wrong??
Were looking for:

k > 0

For Statement #1
1/(k-1) > 0 for all values of k > 1

A definitive Yes

For Statement #2

1/K+1 > 0

Works for K > -1

InSuff
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Mon Aug 10, 2009 7:12 am

by krumpy » Sun Sep 06, 2009 2:06 pm
Couldn't k be a fraction. If k = -1/2 than adding 1 will make 1/(k+1) > 0 even though 1/k < 0. Looks like you changed the direction of the inequality. I think it should read:

1/(k+1) > 0
k+1 should be positive.
therefore, k+1 > 0
hence k>-1

Same issue with 1 as well.

Otherwise I think your setup is fine.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1