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 →

Inequality DS

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Data Sufficiency |
Master | Next Rank: 500 Posts
Posts: 189
Joined: Thu Apr 03, 2008 2:03 pm
Location: USA
Thanked: 21 times

by rohan_vus » Fri Jan 08, 2010 8:21 pm
papgust wrote:Is x between -1 and 1?

(1) x^3 is less than x.
(2) x^3 is less than x^2.

Why it isn't A? OA is E
Consider x = -2 ..x^3<x...So answer to original question that if x is between -1 and 1 is 'No'.
Consider x = .5 then x^3<x..So answer to oroginal question that if x is between -1 and 1 is 'Yes'.
Thus A is ruled out

Algebrically A is reduced to x^3-X<0 --> x(x^2 - 1) < 0..Now situation varies depending on x is _ve or +ve..
If x is +ve then x^2 - 1 < 0 ..this means |x| < 1..which satisfies the original questionas 'yes'.
But of x = -ve then x^2 - 1 >0 ...which means |x| > 1.. so x could be any integer greater than 1 or any integer less than -1..This condition comes as 'No' to the original question asked in the question stem
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 527
Joined: Mon Jun 02, 2008 9:14 am
Location: Atlanta
Thanked: 17 times

by pandeyvineet24 » Fri Jan 08, 2010 8:23 pm
for stmt 1,
put x = -5, x^ 3 = -125, x = -5 therefore x^3 < x and x not between -1 and 1
put x = 1/2, x^3 =1/8, x =1/2 , therefore x^3 < x and x between -1 and 1
clearly insuff.

for stmt 2
put x = 1/2, x^2 =1/4, x^3 = 1/8, --> x^3 < x ^2 and x between -1 and 1
put x = -5 , x^3 = -125, x^2 = 25 --> x^3 < x ^2 and x not between -1 and 1
insuff.

combine 2, use same 2 values as above. together insuff
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Community Manager
Posts: 1537
Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members
Thanked: 653 times

by papgust » Sat Jan 09, 2010 3:20 am
rohan_vus wrote: Algebrically A is reduced to x^3-X<0 --> x(x^2 - 1) < 0..Now situation varies depending on x is _ve or +ve..
If x is +ve then x^2 - 1 < 0 ..this means |x| < 1..which satisfies the original questionas 'yes'.
But of x = -ve then x^2 - 1 >0 ...which means |x| > 1.. so x could be any integer greater than 1 or any integer less than -1..This condition comes as 'No' to the original question asked in the question stem
Hey Rohan,

Thanks for replying. I have a doubt here.

In (1), simplify the eqn to x(x^2-1) < 0 -> from which you get x^2 < 1 and x < 0. As x < 0, we only get a single scenario and we don't have to consider x as +ve.

What is wrong in this approach?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 189
Joined: Thu Apr 03, 2008 2:03 pm
Location: USA
Thanked: 21 times

by rohan_vus » Sat Jan 09, 2010 4:33 am
Hey papgust

You are right !.. Its errata on my part as i wrote it my midnight time just before sleeping:) ..

First inequlaity we agree its reduced to x(x^2-1) < 0... So this equantion is satisfied in following ways
a) x < 0 and (x^2-1) > 0
b) x > 0 and ( x^2-1) < 0

We need to consider 2 possibilities , if x would have been considered to be INTEGER then possibility (b) we wont get and then in that case only possibility (a) exists and which is what agrees to your solution

But since x could be non integer so we must consider both the possibilities

Now if x < 0 then |x| > 1..which means x < -1 as x is -ve..So this value ranges DONT fit into the ranges ( -1<x<1)---- So its 'NO' in this case

Now when x is > 0 , then x^2-1 < 0 ..==>|x| < 1..Here the ranges for 0<x<1 ( as is +ve so ignoring the -ve possibilities)..These values ranges fit into the ranges which is asked ( -1<x<1)---- -- So 'Yes' in this case ..

But still situation is same , i.e, A is not sufficient to give a definite answer
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Community Manager
Posts: 1537
Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members
Thanked: 653 times

by papgust » Sat Jan 09, 2010 7:54 am
Now it makes sense. I failed to consider 2 possibilities here with opposite signs. Thank you very much!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1