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

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

GMAT Prep Q - Population

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 27
Joined: Sat Feb 27, 2010 12:05 pm
Location: England
by tito1545 » Sun Feb 20, 2011 10:23 am
A certain city with a population of 132,000 is to be divided into 11 voting districts, and no district is to have a population that is more than 10% grater than the population of any other district.What is the minimum possible population that the least populated district could have ?

A.10,700
B.10,800
C.10,900
D.11,000
E.11,100

ANS D
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 139
Joined: Fri Nov 26, 2010 4:45 pmFollowed by: 1 member
Location: Boston
Thanked: 20 times
GMAT Score:720

by stormier » Sun Feb 20, 2011 10:43 am
tito1545 wrote:A certain city with a population of 132,000 is to be divided into 11 voting districts, and no district is to have a population that is more than 10% grater than the population of any other district.What is the minimum possible population that the least populated district could have ?

A.10,700
B.10,800
C.10,900
D.11,000
E.11,100

ANS D
Each of the 10 districts, which have a greater population than the one with minimum population must have same population.

132000 = 10x + y

x <= 1.1y [ given that no district can have a population greater than 10% of other district]

=> 10x + y <= 11y + y

=> 132000 <= 12y

=> y >= 132000/12 = 11000

Thus y must be greater than or equal to 11000, which is its minimum value.
Each of the other 10 districts has a population of 12100.

If you are still thinking why the population of each of 10 districts has to be same, let us look at the following.

lets say 8 had a population of 12100, 1 had 12000 and 1 had 12200. => still a total of 121000 for the 10 districts. But now the difference between the town with 12200 population and the one with minimum (11000) is > 10 %.
Last edited by stormier on Sun Feb 20, 2011 10:51 am, edited 1 time in total.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1337
Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members
Thanked: 127 times

by Night reader » Sun Feb 20, 2011 10:46 am
again logic applied: one is the lowest x and the rest are different by as much as possible i.e. 10% --> (1.1x)*10+x=132,000 where x is the population of districts, solve for x and (1.1*10=11 :) ) 12x=132,000, approximately 10,900 answer D (11,000)
tito1545 wrote:A certain city with a population of 132,000 is to be divided into 11 voting districts, and no district is to have a population that is more than 10% grater than the population of any other district.What is the minimum possible population that the least populated district could have ?

A.10,700
B.10,800
C.10,900
D.11,000
E.11,100

ANS D
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members
Location: Milpitas, CA
Thanked: 1854 times
GMAT Score:770

by Anurag@Gurome » Sun Feb 20, 2011 11:50 pm
tito1545 wrote:A certain city with a population of 132,000 is to be divided into 11 voting districts, and no district is to have a population that is more than 10% grater than the population of any other district.What is the minimum possible population that the least populated district could have ?

A.10,700
B.10,800
C.10,900
D.11,000
E.11,100

ANS D
Solution:
Let the 11 voting districts have populations d1, d2, d3, d4,....d11 respectively.
Let d1 be the least population.
So, d1+d2+d3+...+d11 = 132,000.
Now, according to question d2 <= 1.1*d1, d3 <= 1.1*d1,....d11<=1.1*d1.
Adding all inequality, we get that (d2+d3+d4...d11) <= 10*1.1*d1.
Or (d2+d3+d4...d11) <= 11d1.
Or (d1+d2+...+d11) <= 12d1.
Or 132,000 <= 12d1.
Or d1 => 11,000.
So, the minimum possible population of least populated district is 11,000.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 27
Joined: Sat Feb 27, 2010 12:05 pm
Location: England

by tito1545 » Mon Feb 21, 2011 12:20 pm
Thanks Anurag , a good and simple approach
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1