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# GMATPrep Question

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Manager
Joined: 14 May 2006
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GMATPrep Question [#permalink]  14 May 2006, 17:48
Hi, I am new here. Great site by the way. Got the below question on GMATPrep today and couldn't figure out:

Q: 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 districts. What is the minimum possible population that the least populated district could have?

10,700
10,800
10,900
11,000
11,100

Cheers,
SJ
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Senior Manager
Joined: 08 Jun 2004
Posts: 499
Location: Europe
Followers: 1

Kudos [?]: 20 [0], given: 0

11,000 it is.

X - the least populated district could have.
1.1 X - no more than 10% + to the population of any other districts, so

X+10(1.1X) = 132,000
12X=132000
X=11,000
Intern
Joined: 23 Feb 2006
Posts: 40
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Kudos [?]: 0 [0], given: 0

I think it's 10900. Solution:
x/1.1+10*x=132000 => x=12100 => x/1.1=10909
The possible mistake could happen if one take
Intern
Joined: 23 Feb 2006
Posts: 40
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Kudos [?]: 0 [0], given: 0

I'm sorry - made a rounding error...
12*x=145200 => x=145200/12=12.1 and then cheapest 12.1/1.1=11
Manager
Joined: 17 Jan 2006
Posts: 92
Followers: 1

Kudos [?]: 3 [0], given: 0

i got 10,800

132,000/11 * 0.9 = 10,800

what is the OA?
Director
Joined: 24 Oct 2005
Posts: 660
Location: London
Followers: 1

Kudos [?]: 7 [0], given: 0

Got 11000.

1 district minimum population = x
Remaining 10 districts - Max population =10% greater than x

Same method as M8.
Manager
Joined: 14 May 2006
Posts: 202
Followers: 1

Kudos [?]: 10 [0], given: 0

OA is 11000. Thanks everyone.
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