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GMAT Prep Q - Population

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tito1545: 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

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Source: Beat The GMAT — Problem Solving | Sun Feb 20, 2011 10:23 am

stormier: 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.

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Night reader: 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

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Anurag@Gurome: 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)

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tito1545: 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

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