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

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A certain city with population of 132,000 is to be divided [#permalink] New post 23 Oct 2005, 17:21
A certain city with 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
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here goes [#permalink] New post 23 Oct 2005, 17:31
My guess is 11,000
don't know for sure but here's what I did. If no population can be more than 10% of any other population, and you have to have 11 populations- then the biggest one can be is 1.1X and if 10 of those are 1.1x then the smallest can be just X.
So 10*1.1X + X = 132,000
that means 11X+X = 132,000
12X=132,000
X = 11,000

final answer
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 [#permalink] New post 24 Oct 2005, 18:56
woo hoo! i finally got one of these right the first time!
Maybe I've finally turned the corner in my gmat prep. thanks!
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 [#permalink] New post 24 Oct 2005, 21:14
I got D.

Consider all of the rest of the voting districts (10) are in fact 10% more populated than the smallest district (say x population)
10(1.1x) + x = 132K
12x = 132K
x=11K
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 [#permalink] New post 25 Oct 2005, 04:20
This is the first time I absolutely disagree with the OA.

I got C
12000-x>9/10(12000+x/10) where x should be integer.

Solved and got
120000>109X where x should be the biggest possible value. It is 1100.
12000-1100=10900.

IT can be easily proved: 10900 + 10(12000+110)=132000, where no term ecxeeds the other more than by 10%

Could someone explain, where is the flaw?
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 [#permalink] New post 25 Oct 2005, 04:39
Dilshod,

Here is the long explanation for this answer I posted in an earlier thread:

Quote:
The correct answer is D. I used picking numbers.

Before starting, realize that aside from the lowest populated district, the other 10 disctricts will all have the same size. This will maximize the difference between the least and 2nd least populated districts.

Starting with C:
-If you choose 10,900 as the lowest possible population, then the remaining ten districts will have an average population of 12,100 [(132,000-10,900)/10].
-12,100 is more than 10% higher than 10,900
Be careful on this option. 10,900 is 10% less than 12,100. However, the question says that "no district is to have a population that is more than 10% greater than the population of any other district."

So... looking at answer D:
-If the lowest district has 11,000 people, then the average of the remaining 10 districts is 12,100 [(132,000-11,000)/10]
-12,100 is 10% greater than 11,000.


Better yet, here is the short, algebraic solution:

Quote:
1.1x = (132,000-x)/10
11x = 132,000-x
12x = 132,000
x = 11,000
  [#permalink] 25 Oct 2005, 04:39
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