Dear Bunuel Had the question asked that no two districts can have the same population, how the question could have been solved?

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

Answer: D

Cheers,
Brent
_________________

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,

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.
Probus
_________________

I took .90x + 10x = 132000 , I am getting different answer.

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

this is a very nice example for minimizing problem. Bunuel solution is very elegant.