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 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
Live EA class + 6 months of EA OnDemand
  • Expert-led weekly online sessions
  • EA Masterclass access between classes
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

130-point 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.

Ps- It was time consuming

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
User avatar
Master | Next Rank: 500 Posts
Posts: 319
Joined: Wed Feb 04, 2009 10:32 amFollowed by: 9 members
Location: Delhi
Thanked: 84 times

by sureshbala » Tue Feb 24, 2009 3:21 am
Let the population of the least district be x.

Now the maximum population of any other district is 110%(x).

Now x will be minimum if the population of the rest of the districts is maximum.

Hence for x to be minimum the population of each of the other 10 districts must be 110%(x).

Given that x + 10[110%(X)]= 132,000
i.e 12x = 132,000.
x=11,000
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 258
Joined: Thu Aug 07, 2008 5:32 am
Thanked: 16 times

by x2suresh » Tue Feb 24, 2009 3:45 am
sureshbala wrote:Let the population of the least district be x.

Now the maximum population of any other district is 110%(x).

Now x will be minimum if the population of the rest of the districts is maximum.

Hence for x to be minimum the population of each of the other 10 districts must be 110%(x).

Given that x + 10[110%(X)]= 132,000
i.e 12x = 132,000.
x=11,000
Agree with you.

Nice work
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 29
Joined: Sun Feb 08, 2009 11:29 am
Thanked: 3 times

by ven4gmat » Tue Feb 24, 2009 5:38 am
sureshbala wrote:Let the population of the least district be x.

Now the maximum population of any other district is 110%(x).

Now x will be minimum if the population of the rest of the districts is maximum.

Hence for x to be minimum the population of each of the other 10 districts must be 110%(x).

Given that x + 10[110%(X)]= 132,000
i.e 12x = 132,000.
x=11,000
ohh...you made it seem simple to me man...initially i was also a bit confused to tackle this question..
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1