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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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27 Apr 2010, 09:59
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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 percent 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

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-certain-city-with-population-of-132-000-is-to-be-divided-76217.html
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Re: GMAT PREP (PS) [#permalink]

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27 Apr 2010, 12:00
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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 percent 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

As "no district is to have a population that is more than 10 percent greater than the population of any other district", then the populations of 11 districts should be in the range: $$x$$ and $$1.1x$$.

So we want to minimize $$x$$. To minimize $$x$$ we should make only one district to have that # of population (minimum possible) and the rest 10 districts to have $$1.1x$$ # of population (maximum possible).

$$x+10*1.1x=132$$ --> $$12x=132$$ --> $$x=11$$.

Answer: D.
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Re: GMAT PREP (PS) [#permalink]

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28 Apr 2010, 03:23
Bunuel wrote:
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 percent 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

As "no district is to have a population that is more than 10 percent greater than the population of any other district", then the populations of 11 districts should be in the range: $$x$$ and $$1.1x$$.

So we want to minimize $$x$$. To minimize $$x$$ we should make only one district to have that # of population (minimum possible) and the rest 10 districts to have $$1.1x$$ # of population (maximum possible).

$$x+10*1.1x=132$$ --> $$12x=132$$ --> $$x=11$$.

Answer: D.

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15 Jun 2010, 06:00
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First thing to realize:

You want to minimize the lowest possible quantity. For this we need to maximize all the others, however, we are limited by the question which tells us the greatest qty can only be 10% greater than the minimum one. So if the minimum value is x, the maximum value is 1.1x

Now, here's the trick, and this may not occur to many people - if you want to minimize this lowest qty, the only way to do this is to set ALL the other 10 to the maximum value. This is because those 10 cannot take a value more than that, and any value below that would lead to a higher value of the minimum one.

So: x + 10*1.1x = 132000

=> 12x = 132000

=> x = 11000

Hope this helps!
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Re: GMAT PREP PS [#permalink]

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23 Aug 2010, 19:33
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uzzy12 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 percent greater than the population of any other district. What is the minimum possible population taht the least populated district could have?

A. 10,700
B. 10,800
C. 10,900
D. 11,000
E. 11,100

My attempt:

132,000 divided by 11 districts would give us 12,000 people on average. Also we need to keep a population of a district minimum within the condition that the population of the no district is greater than 10% of the population of the least populated district.

Hence forming equation we get

D1 + D2 + D3 + .....D11 = 132,000
Let us assume D1 is the least populated district. If we have to reduce the number of people in the district D1 to a minimum and bound to the condition, we should equally distribute the difference of the minimized population of D1 and the average population of D1 (12000) equally to the rest of the 10 districts.

Hence D2 will be equal to D3 = D4 = D5 = ....= D11. Let D2 be x and b be the population of D1
hence 10 * x + b = 132000

also we know that x $$<=$$ 1.1 b. Let us take the boundary case - x = 1.1b

Hence the equation becomes 10 * 1.1 b + b = 132000
11b + b = 132000

12b = 132000 => b = 11000.

x = 12100. Hence the answer is 11,000 (D).

I hope my reasoning is sound and in understandable format.
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Re: GMAT PREP (PS) [#permalink]

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28 Jan 2011, 11:56
Great explanations.

Thanks.
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Re: percentages [#permalink]

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08 Jan 2012, 06:09
kotela wrote:
Can anyone please help me in solving this problem.......

lets assume all districts have same population = 132000/11 = 12000

Consider option C - 10900 as correct (POE method)

then 10900 is 1100 less than avg of 12000. Even if equally divided the population of others will be 12100 - which is 1200 more than 10900 and more than 10% of 10900. So we need a higher value

Eliminate A, B and C

D - 11000 - this is 1000 less than assumed avg 12000 and if equally divided population will be 12100 which is exactly 10% greater than 11000.
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Re: percentages [#permalink]

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11 Jan 2012, 05:34
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Weighted averages... That would be my solution to the question.

Voting districts = 11

Total = 132,000

Now how do we get "the least" value.... We make one min and the rest of the district equal to each other. It is absolutely imperative that you understand this statement. If all the rest are maximum possible value only then will you get the least possible value for one voting district.

From here on in, it is just about setting up the equation:

Let minimum = a

So max value possible = (10/100)a + a

Now write the equation...

1 of minimum and 10 of maximum value so.....
1*a + 10*(a + 0.1a) = 132,000

Solve for a = 11,000
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Re: A certain city with a population of 132,000 [#permalink]

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16 Jan 2012, 00:08
Thanks a lot..great explanation...
Re: A certain city with a population of 132,000   [#permalink] 16 Jan 2012, 00:08
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