This can be solved by Max and Min concept.
Since , no two cities can have population greater than 10% of any city.
We can assume min population for one city be : x
Then , if we maximize all remaining 10 cities population, it will be : 1.1x for each of them

Thus , we get x+ 10*1.1x = 132000
x+11x= 132000
12x=132000
x=11000

Is OA D?

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

We can plug in the answer choices, which represent the minimum possible population of the least populated district.

To MINIMIZE the smallest population, we need to MAXIMIZE the other 10 populations.
Thus, each of the other 10 districts must have the maximum allowed population: 10% greater than the smallest population.
Since the total population of the city is 132,000 -- a multiple of 1,000 -- the correct answer choice is almost certainly a multiple of 1,000.

Answer choice D: Least populated district = 11,000.
Maximum value of each of the other 10 districts = 11,000 + .1(11,000) = 12,100.
Sum of the 11 districts = 11,000 + 10(12,100) = 132,000.
Success!

The correct answer is D.

Algebraically:
Let x = the population of the least populated district.
As noted above, to MINIMIZE the smallest population, we need to MAXIMIZE the other 10 populations.
Thus, each of the other 10 districts must have the maximum allowed population:
10% greater than the smallest population = 1.1x.
Thus, the sum of the populations in the other 10 districts = 10(1.1x) = 11x.
Since the sum of ALL the populations is equal to 132,000, we get:
x + 11x = 132,000
12x = 132,000
x = 11,000.