Minimum possible population of least populated district

This topic has expert replies
Senior | Next Rank: 100 Posts
Posts: 43
Joined: Sun Apr 06, 2014 10:50 am
Thanked: 1 times
Followed by:1 members
Hi Everyone,

Please help me in answering the below question from GMATPrep:

A certain city with a population of 132,000 is to be divided into 11 districts, and no district is to have a population that is more than 10 percent greater than the population of any other district. What is the minimum possible population that the least populated district could have?

1. 10,700
2. 10,800
3. 10,900
4. 11,000
5. 11,100

Please explain the solution also.

Many Thanks!!

User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Mon Jun 16, 2014 10:48 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
We can plug in the answer choices, which represent the minimum possible population of the least populated district.

To MINIMIZE the smallest population, we need to MAXIMIZE the other 10 populations.
Thus, each of the other 10 districts must have the maximum allowed population: 10% greater than the smallest population.
Since the total population of the city is 132,000 -- a multiple of 1,000 -- the correct answer choice is almost certainly a multiple of 1,000.

Answer choice D: Least populated district = 11,000.
Maximum value of each of the other 10 districts = 11,000 + .1(11,000) = 12,100.
Sum of the 11 districts = 11,000 + 10(12,100) = 132,000.
Success!

The correct answer is D.

Algebraically:
Let x = the population of the least populated district.
As noted above, to MINIMIZE the smallest population, we need to MAXIMIZE the other 10 populations.
Thus, each of the other 10 districts must have the maximum allowed population:
10% greater than the smallest population = 1.1x.
Thus, the sum of the populations in the other 10 districts = 10(1.1x) = 11x.
Since the sum of ALL the populations is equal to 132,000, we get:
x + 11x = 132,000
12x = 132,000
x = 11,000.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

Senior | Next Rank: 100 Posts
Posts: 43
Joined: Sun Apr 06, 2014 10:50 am
Thanked: 1 times
Followed by:1 members

by DevB » Mon Jun 16, 2014 11:00 am
Thank you Mitch for your quick replies and detailed explanations...!!

Answer is correct :)

Newbie | Next Rank: 10 Posts
Posts: 8
Joined: Sat Jun 07, 2014 3:12 am
Thanked: 2 times

by tass@ » Mon Jun 16, 2014 12:36 pm
let the least number is "x"
Now there are 10 numbers left which are (x+.10x)
So total sum of 10 numbers will be 10(1.1x)=11x
Now x+11x = 132000
x= 11000
So this is the answer