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A certain city with a population of 132,000 is to be divided

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A certain city with a population of 132,000 is to be divided [#permalink]

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New post 13 Aug 2005, 06:19
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A certain city with a population of 132,000 is to be divided into 11 voting districts, and no district is to have 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
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New post 13 Aug 2005, 07:10
I think it's best to choose one district and designate it to be the smallest one. We make all the big districts of identical size, with enough capacity to accommodate exactly 110% of the population of the smallest one.

x + 10*1.1x=132k
x = 132k / 12 = 11k

Is (D) the right answer?

I can't really explain the solution clearly, but I'll try. We are looking for the case where the average of the 10 large districts is exactly 110% of the population of the smallest district. But if the population of 1 of the 10 is greater than their average, this would break condition of "not greater than 10%", so the 10 big ones must be of equal size.

I am sure someone can do better with the explanation...
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New post 13 Aug 2005, 07:25
yes it should be D

to make x min; we should consider max population for 10 districts i.e 1.1X

x+10*1.1x=13200

x= 11000
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New post 13 Aug 2005, 16:41
D

the strategy is to have one (let's say x) smallest and the rest 10 equal (let's say y)

x + 10y = 132000
y=1.1x

x = 11000
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New post 14 Aug 2005, 20:38
minimum = x
maximum = 1.1x

total = x+10(1.1x)=132000
so x = 11,000
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New post 15 Aug 2005, 05:18
i will vote for 10 800 which is 10% less than the average-12000
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A certain city with a population of 132,000 is to be divided

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