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 →

Is M^3>=1

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Data Sufficiency |
Senior | Next Rank: 100 Posts
Posts: 76
Joined: Fri Dec 26, 2008 8:18 am
Thanked: 3 times

by srivas » Tue Sep 01, 2009 7:46 am
statement I p is even

p=2 m =-P+(-2)^P -2+(-2)^2 = -2+4 = 2
so m^3 = 8 >= 1

statement II

P^3 <= -1 suppose P^3 = -8 then P = -2
m = -P+(-2)^P
-(-2)+(-2)^-2 = 2- 1/4 = 7/4
so m^3 >= 1

it is D
Gmat710,, Hyd
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 18
Joined: Fri Jul 24, 2009 2:58 am
Thanked: 2 times

by Matmasi » Tue Sep 01, 2009 9:06 am
Unfortunately I have to disagree with B, that it seems, to me, not sufficient.
Infact, P^3 <= -1 and you suppose P^3 = -8 then P = -2.

But in this way you forget that P can be also equal to -1; let's try:
-(-1) + (-2^-1) = 1 - 1/2 = 1/2
m= 1/2
M^3 = 1/8 <1
So B is not sufficient.
Other opinions would be greatly appreciate
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 157
Joined: Tue Oct 07, 2008 5:47 am
Thanked: 3 times

by PussInBoots » Tue Sep 01, 2009 10:24 am
The asnwer is C

srivas you didn't test 0, -1/1. Those numbers usually screw everything up.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 128
Joined: Thu Jul 30, 2009 1:46 pm
Thanked: 1 times

by fruti_yum » Tue Sep 01, 2009 10:56 am
PussInBoots wrote:The asnwer is C

srivas you didn't test 0, -1/1. Those numbers usually screw everything up.
PussInBoots how do u know you have to test those specific numbers?.. I ended up picking D as well before coz i didn't check those numbers!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 75
Joined: Thu Aug 20, 2009 11:55 am
Location: vancouver, B.C
GMAT Score:640

by hpgmat » Thu Sep 03, 2009 11:03 pm
srivas wrote:statement I p is even

p=2 m =-P+(-2)^P -2+(-2)^2 = -2+4 = 2
so m^3 = 8 >= 1

statement II

P^3 <= -1 suppose P^3 = -8 then P = -2
m = -P+(-2)^P
-(-2)+(-2)^-2 = 2- 1/4 = 7/4
so m^3 >= 1

it is D
Sirivas, your calculation for statement 2 is wrong . Assuming P is -2, M would be 9/4. M cube still greater than 1.
just a clarification.
Will Win
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 60
Joined: Sat May 30, 2009 11:22 am
Thanked: 2 times

Re: Is M^3>=1

by qwe12 » Fri Sep 04, 2009 1:18 am
PAB2706 wrote:P is integer, m = -P+(-2)^P, Is M^3 is greater than or equal to 1.
1) P is even.
2) P^3 is less than or equal to -1.
please post the offical answer.

(1)
consider p=0, m=1 ... m^3>=1? YES
consider p=2, m=2 ... m^3>=1? YES
consider p=4, m=12 ... m^3>=1? YES

SUFFICIENT

(2)
consider p=-1, m=1/2 ... m^3>=1? NO
consider p=-2, m=9/4 ... m^3>=1? YES

NOT SUFFICIENT

Answer (A)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1