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# A certain city with population of 132,000 is to be divided into 11

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Re: A certain city with population of 132,000 is to be divided into 11  [#permalink]

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09 Feb 2016, 18:51
Dear Bunuel Had the question asked that no two districts can have the same population, how the question could have been solved?
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Re: A certain city with population of 132,000 is to be divided into 11  [#permalink]

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09 Feb 2016, 20:06
HI ankushbagwale,

The reason why this question 'works' the way it does (and can be approached in the way described) is because it was 'designed' that way. Every prompt on the GMAT comes with specific information/facts/restrictions that you're supposed to use to answer the question. If the populations were all different, then the question would have to be completely rewritten (with different facts and restrictions in place).

You could certainly see a question that tells you the average population of 12 districts, then compares the 'largest' to the 'smallest' (in some mathematical way), tells you that all 12 numbers are different and then asks you for either the largest or smallest. That is NOT this question though.

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Re: A certain city with population of 132,000 is to be divided into 11  [#permalink]

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19 Apr 2016, 18:12
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Flexxice 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 % greater than the population of any other district. What is the minimum possible population that the least populated district could have?

A) 10700

B) 10800

C) 10900

D) 11000

E) 11100

Let x = the population of the district with the LOWEST population.
To MINIMIZE the population in the smallest district, we must MAXIMIZE the population of the other 10 districts.

IMPORTANT: No other district can exceed x by more than 10%.
So 1.1x = the MAXIMUM population of each of the other 10 districts.

The TOTAL population is 132,000, so we can write:
(population of smallest district) + (population of other 10 districts) = 132,000
Rewrite as: x + [(10)(1.1x)] = 132,000
Simplify: 12x = 132,000
x = 11,000

Cheers,
Brent
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Re: A certain city with population of 132,000 is to be divided into 11  [#permalink]

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06 Jun 2017, 16:46
ugimba 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

Anytime we are presented with a “minimum value” problem, we must “maximize” all components except for one of them, thus leaving the last component as the “minimized” component of our set.

Let’s use an easy example to test this idea. For instance, we can say that Bob and Frank have a total of 100 apples between them. What is the minimum number of apples that Frank can have? We must “maximize” the number of apples that Bob has; this number is 99. Thus, the minimum number of apples that Frank can have is 1 apple.

Similarly, in this problem we are given 11 voting districts and we must minimize the population of one of those districts. This means that we want to maximize the population of the 10 other districts. We are also given that no district is to have a population that is more than 10% greater than the population of any other district.

Thus, if we label the population of the least populous district as x, we can then say that the maximum population in any other district must be: x + 0.1x = 1.1x. This satisfies the condition that no district has a population that is more than 10% greater than that of any other district.

Because we need to maximize the population of 10 of the 11 districts, all of these 10 districts must have populations of the maximum allowed number, which is 1.1x, and thus, the total population of these 10 districts is (1.1x)(10) = 11x.

We know that the total population of all the districts is 132,000, so we can say:

10 most populous districts + 1 least populous district = 132,000

11x + x = 132,000

12x = 132,000

x = 11,000

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Re: A certain city with population of 132,000 is to be divided into 11  [#permalink]

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20 Nov 2018, 09:04
I still don't get it. I think the answer is C.
Lets say that indeed the population in the first distinct is 10900 then the other distincts have a population of 12110.
10900+10*12110=132000. So in this scenario, the minimum population is 10900 and not 11000. If I did something wrong, please explain.
Thanks!!
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Re: A certain city with population of 132,000 is to be divided into 11  [#permalink]

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20 Nov 2018, 21:14
Giorgos92 wrote:
I still don't get it. I think the answer is C.
Lets say that indeed the population in the first distinct is 10900 then the other distincts have a population of 12110.
10900+10*12110=132000. So in this scenario, the minimum population is 10900 and not 11000. If I did something wrong, please explain.
Thanks!!

Hi Giorgos92,

The prompt states that no district has a population that is more than 10% greater than any other district.

IF.... the smallest district has 10,900, then the largest any of the other districts could be would be 10,900 + (10% of 10,900)....

10,900 + 1,090 = 11,990

In your calculation, the other districts each have a population of 12,110, which is not permissible given what we are told.

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A certain city with population of 132,000 is to be divided into 11  [#permalink]

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04 Feb 2019, 15:14
Hi,

Here are my two cents . Though not quick as Bunuel mentioned method but let me know if this makes sense.

We are told the the total population is 132000 which is to be be divided into 11 groups. So average population per group is 12000.

Now if we reduced population of any group Say Group 1 then the difference from average i.e Avergage - Grp 1 is at least added equally among remaining other 10 groups or to few groups.

here is what i mean

Say group 1 to group 11 each have population of 12000, average population is 12000 and total population is 132000.

Here no group has population more than 10% of other group.

Now if i reduced the population of group 1 by 2000 and make it 10000 then to maintain average we will have to add this population to some group of 10 .
Lets say we add equal members to each group then we will have add $$\frac{2000}{10}$$= 200 members to each of the remaing 10 groups .

Which makes the population of Group 2 to Group 10 as 12200 members each.

Now Population of Group 1 is 10,000 and any of the group can't have members more than 11000. Now this scenario is not possible since we have to divide a total 132000 and many will be left out. Atleast one of the group will have population more than or equal to 12200.

Similarly

if
(A)10700, then other groups must be atleast 12130 but other groups can't be more than 11770.
(B)10800, then other groups must be atleast 12120 but other groups can't be more than 11880.
(C)10900, then other groups must be atleast 12110 but other groups can't be more than 11990.
(D)11000, then other groups must be atleast 12100 Here other groups can't be more than 12100.
(E)10100, then other groups must be atleast 12090 but other groups must be atleast 12210

now from answer choices A through C if you reduce any of the subsequent groups members to match upto 10% more of Group 1 then the difference between them will have to add up to some group to make the total 132000,

Say Group 1 is 10700 and group 2 is 11770 then one of the groups will have members which will more than 10 % of group 1.

But in Option D we have if i reduced my group 1 members to 11000 than i can have other groups take that hit and balance out which will turn out that difference between any two groups will not be more than 10 %

In option E we have a similar scenario. But since we are asked for what is the minimum possible population clearly D is smaller than E

I am not sure if i was able to frame my understanding clearly.

Looking for hearing back from community.
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Re: A certain city with population of 132,000 is to be divided into 11  [#permalink]

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25 Aug 2019, 01:45
ugimba 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

Given: 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.

Asked: What is the minimum possible population that the least populated district could have?

Let the minimum possible population be s

Maximum possible population = 1.1s

To minimize s, populations of all other 10 districts have to be maximised (1.1s)

s + 10 * 1.1s = 132,000
12s = 132,000
s = 11,000
The minimum possible population that the least populated district could have = 11,000

IMO D
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Re: A certain city with population of 132,000 is to be divided into 11  [#permalink]

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18 Sep 2019, 06:10
I took .90x + 10x = 132000 , I am getting different answer.

Can someone help me to understand what is wrong in this ?
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Re: A certain city with population of 132,000 is to be divided into 11  [#permalink]

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18 Sep 2019, 06:15
prags1989 wrote:
I took .90x + 10x = 132000 , I am getting different answer.

Can someone help me to understand what is wrong in this ?

The difference between 0.9 and 1 is more that 10%: 0.9 + 0.1(0.9) = 0.99.
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Re: A certain city with population of 132,000 is to be divided into 11  [#permalink]

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21 Sep 2019, 02:06
this is a very nice example for minimizing problem. Bunuel solution is very elegant.
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Re: Funky Algebra Question from GmatPrep, need help!  [#permalink]

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07 Oct 2019, 02:30
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