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

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

Answer: D

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.

GMAT assassins aren't born, they're made,
Rich

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

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

Bunuel - what would be the equation if we want to maximize one quantity and minimize the others in this problem? I read your general rule for maximizing and minimizing but could not formulate the equation.

Hi anindhya25,

Assuming that all of the other details for this questions were the same (re: 11 total districts, a total populations of 132,000 and no district has a population that's more than 10% of any other district), the equation that would maximize the possible number of people in any one district would be fairly similar to the one that calculates the minimum population for any one district:

To find the minimum: 1(X) + 10(1.1X) = 132,000 and we solve for the value of X

To find the maximum: 10(X) + 1(1.1X) = 132,000 and we solve for the value of 1.1X

Keep in mind that when all of the parameters in this question are maintained, the maximum possible population for a city is NOT a nice, "round number."

GMAT assassins aren't born, they're made,
Rich

Tried a different approach using Answer Choices. This seems to be impractical to solve in the exam timing. But would be timesaver, if someone can solve this logically in mind

D. \(11000 \)is easy to solve.

Suppose that each district has equal numbers => \(\frac{132,000}{11} = 12,000\) population .

Now, If least population is 11,000 , then maximum acceptable limit is \(\frac{10*11,000}{100} + 11,000= 12,100 \)

\(12,000 - 11,000 = 1,000\). If one district population is \(11,000\). The remaining \(1,000\) population must be shared with other district.

but \(12,100 - 12,000 = 100\).

Also, \(10 * (12,100 - 12,000) = 1,000\)

So, \(10 \) district of \(12,100 \) population and \(1\) district of \(11,000\) population satisfy the given condition.

D is the right Answer.

Answer: Option D

Video solution by GMATinsight with Concept of Mixima-Minima

Let's say, the minimum population in 11 districts be X.
its given that, no district is to have a population that is more than 10 percent greater than the population of any other district

So the maximum population in any other districts will be 1.1X.

Total population in 11 districts = 132,000.

In order to find the least population X, we will assume that all other 10 districts will have max population possible i.e. 1.1X

X + 1.1 X* 10 = 132000
12X = 132000
X= 11,000.

Option D is the answer.

Thanks,
Clifin J Francis,
GMAT SME

Approach 1: You can try checking the options - for all options less than 11000, you will find that even if you take the rest of the districts to be of the max population i.e. 1.1 of the lowest, it will not add up to 132000.

Approach 2: Given the sum is constant, the least population would be the one that will be left with the residual number of people after we have distributed the maximum to the rest of the districts.

In such questions, minimizing one thing means maximizing all others. Now if minimum is x, the question is what could be maximum possible value. The questions states that no value can be more than 10% greater than any other value. So 1.1x is the maximum for any other value.

See the video solution:

https://www.youtube.com/watch?v=vse3FlF ... GL&index=3

 

Classic optimization. To minimize one district, maximize the others: