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 →

Inequities

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Data Sufficiency |
User avatar
Legendary Member
Posts: 871
Joined: Wed Aug 13, 2008 7:48 am
Thanked: 48 times

by stop@800 » Sun Sep 28, 2008 7:45 am
A:
b4 will always be +ve
so a is positive

b2 will also be positive [Thank God, comples numbers does not exist in gmat land]

a can be greater and less
the problem will come when b is between 0 and 1

Ex:
let b = 1/2
so b^4 will be 1/16
and b^2 will be 1/4

for all numbers (=a) between 1/4 and 1/16
A will be greater than b^4 and less than b^2

b = 1/2 = .5
b^2 = 1/4 = .25
b^4 = 1/16 = .0625

a = 1/8 = .125

so a > b^4 but less than b^2

and for
a = 2 and b =2
a>b^2 and a>b^4

so Insuff



a > sqrt(b)
again
a +ve
b also +ve

again the similar problem

a = 2
b = 1

a > sqrt(b) and a > b^2


Next
a = 3
b = 4

sqrt(b) = 2

a > sqrt(b) and a < b^2
Insuff



Combining both
you can get the ans

so c in IMO
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 31
Joined: Mon Sep 15, 2008 6:16 pm
Thanked: 3 times

Re: Inequities

by dally_gmat » Sun Sep 28, 2008 8:02 am
Is question complete..?
Atleast I cannot break it.....
alsergi wrote:Hope you could help me with this:

Thank you all!!!


Is a > b^2 ?

1. a>b^4
c. a >sqrt (b)

The right answer is C. ¿?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 31
Joined: Mon Sep 15, 2008 6:16 pm
Thanked: 3 times

by dally_gmat » Sun Sep 28, 2008 8:07 am
@STOP

How you can deduce answer C from both Statements 1 and 2..I was not able to combine both and get final answer.

Thanks in advance

stop@800 wrote:A:
b4 will always be +ve
so a is positive

b2 will also be positive [Thank God, comples numbers does not exist in gmat land]

a can be greater and less
the problem will come when b is between 0 and 1

Ex:
let b = 1/2
so b^4 will be 1/16
and b^2 will be 1/4

for all numbers (=a) between 1/4 and 1/16
A will be greater than b^4 and less than b^2

b = 1/2 = .5
b^2 = 1/4 = .25
b^4 = 1/16 = .0625

a = 1/8 = .125

so a > b^4 but less than b^2

and for
a = 2 and b =2
a>b^2 and a>b^4

so Insuff



a > sqrt(b)
again
a +ve
b also +ve

again the similar problem

a = 2
b = 1

a > sqrt(b) and a > b^2


Next
a = 3
b = 4

sqrt(b) = 2

a > sqrt(b) and a < b^2
Insuff



Combining both
you can get the ans

so c in IMO
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 15
Joined: Mon Sep 22, 2008 1:47 am

by Gmatterz » Sun Sep 28, 2008 9:52 pm
In the above scenario ,how can one combine both the question and get the awnser as choice (C).Can anybody put in on this..?

Thanks
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 16
Joined: Sun Sep 21, 2008 9:12 am
Thanked: 1 times

Re: Inequities

by afterhours » Mon Sep 29, 2008 3:43 pm
alsergi wrote:Hope you could help me with this:

Thank you all!!!


Is a > b^2 ?

1. a>b^4
c. a >sqrt (b)

The right answer is C. ¿?
to make this answer into words

statement one is true only if b is greater than or equal to 1, below one and b^2 is more than b^4

in statement 1

so a>(1/2)^4=A>1/16


a>(1/2)^2=A>1/4

so A in this case could be between 1/4 and 1/16 and would make the original statement false

Statement two is similarbut the oppisite, if B is greater than 1(no negative #'s because it is a sqrt and this is the gmat) there is a similar jump. So lets take an easy number for b, say B=4

In statement 2
a>sqrt(4)= A>2

a>(4)^2= a>16

So if A were between 2-16 in this case, they aren't both true.

And because these are compliments of each other, using both would always assure getting some A>b^2


So answer=C
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1