Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
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.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 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
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
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
GMATOnDemandStart 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

Population

Expert replies
by eitijan » Thu May 12, 2016 2:28 am
OA D
You do not have the required permissions to view the files attached to this post.
Join the discussion
Source: — Problem Solving |

by Brent@GMATPrepNow » Thu May 12, 2016 9:35 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% greater 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
Let x = the population of the district with the LOWEST population.
To MINIMIZE the population in the smallest district, we must MAXIMIZE the population of the other 10 districts.

IMPORTANT: No other district can exceed x by more than 10%.
So 1.1x = the MAXIMUM population of each of the other 10 districts.

The TOTAL population is 132,000, so we can write:
(population of smallest district) + (population of other 10 districts) = 132,000
Rewrite as: x + [(10)(1.1x)] = 132,000
Simplify: 12x = 132,000
x = 11,000

Answer: D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion

by [email protected] » Thu May 12, 2016 9:37 am
Hi eitijan,

Certain Quant questions have built in shortcuts IF the answer choices are numbers and if certain other conditions are met. Here, we can TEST THE ANSWERS....

Logically, when a GMAT question asks you to figure out the LEAST or GREATEST value of something, then there are going to be restrictions on how the values will relate to one another. Here we have 11 cities; to make one as SMALL as possible, I'd think to make all of the others as LARGE as possible.

The answers:
10,700
10,800
10,900
11,000
11,100

Statistically, it's best to TEST either B or D first. D seems like an easier number to manipulate, so I'll TEST that one first.

IF the least city = 11,000
Then 10% greater would be 12,100
IF the other 10 cities are 12,100 each, then they would sum to 121,000
Add in the least city: 121,000 + 11,000 = 132,000

Since all the math that I've done matches the data that I was given....

Final Answer: D

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
Image
Join the discussion

by Matt@VeritasPrep » Fri May 13, 2016 2:54 pm
Let's say that the least populous town has population p. To get everyone else as far as possible from this town, we'll make every OTHER town as big as it can be: 10% greater than the least populous town.

That means that we've got one town with p and ten towns with 1.1p, for a total of

p + 10*1.1p = 132,000

12p = 132,000

p = 11,000

and we're set!
Join the discussion

by Brent@GMATPrepNow » Wed Jan 10, 2018 12:09 pm
Let x = the population of the district with the LOWEST population.
To MINIMIZE the population in the smallest district, we must MAXIMIZE the population of the other 10 districts.

IMPORTANT: No other district can exceed x by more than 10%.
So 1.1x = the MAXIMUM population of each of the other 10 districts.

The TOTAL population is 132,000, so we can write:
(population of smallest district) + (population of other 10 districts) = 132,000
Rewrite as: x + [(10)(1.1x)] = 132,000
Simplify: 12x = 132,000
x = 11,000

Answer: D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion