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A city with population of 132,000 is divided into 11

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A city with population of 132,000 is divided into 11 [#permalink] New post 26 May 2006, 09:30
A city with population of 132,000 is divided into 11 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 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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 [#permalink] New post 26 May 2006, 10:05
The average population is 12000, so min * 110% should be more than the average. Only D satisfy.
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 [#permalink] New post 26 May 2006, 10:49
deowl could you please explain this further...
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 [#permalink] New post 26 May 2006, 11:18
mahesh004 wrote:
deowl could you please explain this further...


Obviously the average of a district population is 12000 ( 132000 / 11 )
So for any district that has population less than the average exists another one that balances it ( there could be more than one that balance it exactly against the average , but certainly at least one exists ). That's it for any number less than the average exists another one that is more than the average.

Since for A, B and C certainly exists another district with population more than 10% , they are out.

D is less than E, so D is my answer.
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Re: Population [#permalink] New post 26 May 2006, 12:49
tl372 wrote:
A city with population of 132,000 is divided into 11 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 the least populated district could have?

a.) 10,700
b.) 10,800
c.) 10,900
d.) 11,000
e.) 11,100


x + 10(1.1)x = 132000
x + 11x = 132000
x = 11,000
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Re: Population [#permalink] New post 26 May 2006, 12:57
Professor wrote:
x + 10(1.1)x = 132000
x + 11x = 132000
x = 11,000


I don't get the reasoning behind your equation, please elaborate.
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 [#permalink] New post 26 May 2006, 13:50
deowl wrote:
mahesh004 wrote:
deowl could you please explain this further...


Obviously the average of a district population is 12000 ( 132000 / 11 )
So for any district that has population less than the average exists another one that balances it ( there could be more than one that balance it exactly against the average , but certainly at least one exists ). That's it for any number less than the average exists another one that is more than the average.

Since for A, B and C certainly exists another district with population more than 10% , they are out.

D is less than E, so D is my answer.


I follow your reasoning up until you rule out A, B, and C. Can you show this mathematically? I'm having a hard time visualizing why you rule out A, B, C but not D.

Thanks.
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 [#permalink] New post 26 May 2006, 14:07
tl372 wrote:
deowl wrote:
mahesh004 wrote:
deowl could you please explain this further...


Obviously the average of a district population is 12000 ( 132000 / 11 )
So for any district that has population less than the average exists another one that balances it ( there could be more than one that balance it exactly against the average , but certainly at least one exists ). That's it for any number less than the average exists another one that is more than the average.

Since for A, B and C certainly exists another district with population more than 10% , they are out.

D is less than E, so D is my answer.


I follow your reasoning up until you rule out A, B, and C. Can you show this mathematically? I'm having a hard time visualizing why you rule out A, B, C but not D.

Thanks.



Assume the answer is C , so one of the districts (C)has population of 10,900. So the biggest possible population of some other district can be 10,900 * 1.1 = 11,990. So how can we have the average of 12000 if the
the district with biggest population has less than that ?
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Re: Population [#permalink] New post 27 May 2006, 05:07
deowl wrote:
Professor wrote:
x + 10(1.1)x = 132000
x + 11x = 132000
x = 11,000


I don't get the reasoning behind your equation, please elaborate.


to get the least, put the maximums (110% of the minimum) at one side and the minimum at the other. then solve it.

the question syas that the maximum cannot be more than 10% of the leaset.

the can be many equations but to get the leaset only the following eq works: x + 10(1.1)x = 132000.

other eq could be:

2x + 9(1.1)x = 132000
3x + 8(1.1)x = 132000
.
.
.
.
10x + 1.1x = 132000

but only "x + 10(1.1)x = 132000" gives the least value for a city.
Re: Population   [#permalink] 27 May 2006, 05:07
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