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

Question

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

Bunuel · Accepted Answer

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

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

So, we want to minimize x. To minimize x, we assign the minimum possible population, x, to  only one  district and the maximum possible population, 1.1x, to the  remaining 10  districts.

x+10*1.1x=132 ; 
 
 12x=132 ; 
 
 x=11 .

Answer: D.

Hope it's clear.

AbhayPrasanna · Answer

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!

EMPOWERgmatRichC · Answer

Hi All,

Certain Quant questions have built in shortcuts IF the answer choices are numbers and if certain other conditions are met. Here, we can TEST THE ANSWERS....

Logically, when a GMAT question asks you to figure out the LEAST or GREATEST value of something, then there are going to be restrictions on how the values will relate to one another. Here we have 11 cities; to make one as SMALL as possible, I'd think to make all of the others as LARGE as possible.

The answers: 
10,700 
10,800 
10,900 
11,000 
11,100

Statistically, it's best to TEST either B or D first. D seems like an easier number to manipulate, so I'll TEST that one first.

IF the least city = 11,000 
Then 10% greater would be 12,100 
IF the other 10 cities are 12,100 each, then they would sum to 121,000 
Add in the least city: 121,000 + 11,000 = 132,000

Since all the math that I've done matches the data that I was given....

Final Answer:  D

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

Bunuel · Answer

The populations of 11 districts should be in the range x and 1.1x, but as we want to minimize x then all other district must have max number of population possible so 1.1x (rule: to maximize one quantity, minimize the others; to minimize one quantity, maximize the others).

ezhilkumarank · Answer

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.

marijose · Answer

why is it that the remaining 10 popullations have all to be equal? cant you have more then two different populations?

I ask because I am having trouble getting to the equation
 p+10*1.1p =132

here you are stating that the 10 remaining popullations have the same popul. I am confused because I dont get why the HAVE to be the same...

omerrauf · Answer

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

ENAFEX · Answer

Bunuel,

Can you tell me how you arrived at the range? x and 1.1x.

Bunuel · Answer

Sure. We are told that "no district has a population that is more than  10%  greater than the population of  ANY  other district."

Now, if the least population is x then the greatest population cannot be more than 1.1x. So, all 11 districts must have population between x and 1.1x (between the least and the greatest).

Hope it's clear.

avrgmat · Answer

I got this problem on my recent GMATPrep mock. This was the 8th question and I ended up getting this one wrong. This was my first mistake on the test in the quant section. I ended up getting 49Q with a total of 10 incorrect.

Coming to the problem, at the test time, I started with the below logic --:

let city 1 =x

Now city2 can have a max of 10% greater of city1 i.e., city2 = 1.1x

At this point, I guess I missed the trick and I am still trying to digest the solutions provided above.

For city3 = 10% of city2 i.e. 1.1(city2) = 1.1*1.1x = 1.21x

I lost the problem at this stage and could not proceed further. I guessed the answer choice as 10,900.

Going over the mentioned solutions, I think it makes sense to say that the sum of the other 10 would have been approximately 10*1.1x only since the value is going to increase only on the decimal side.

Thus, the equation would have indeed been x + 10(1.1x) = 132

Sometimes it is the easily worded problems which cause the most difficulty in understanding.

maibhihun · Answer

Like any other verbose weighted avergae problem idea here is to get some idea about average and then how other data are associated wit hit. 132000 for 11 districts means an average of 12000. Now you have to find a number in 10% of this range so on lower side if N is the number than 1.1N >= 12000 or N>= 12000/1.1 i.e >=10900 so 11000 is closest to it.

mojorising666 · Answer

If we take the max polulation to be x

then

10x + 0.9x=132000
10.9x=132000
x= 12110.09

0.9 x = 121110* .9
=10899.08

Whats the error. Pls tell

up4gmat · Answer

Hi Bunuel,
could u please explain the maximize rule you mentioned with a small e.g..???
also,other populations cant exceed 10 % so why are we taking the max permissible value of the population for 10 of them? pls explain

Bunuel · Answer

Sure. Consider the following problem: If x and y are positive integers and x+y=10, what is the maximum possible value of y? We want to maximize y, thus we need to minimize x. Since x is a positive integer then the least value of x is 1. In this case 1+y=10 --> y=9 --> the maximum possible value of y is 9. DS min/max questions: search.php?search_id=tag&tag_id=42 PS min/max questions: search.php?search_id=tag&tag_id=63 Next, the part saying "no district is to have a population that is more than 10 percent greater than the population of any other district" is about the maximum difference in population of districts. For example, if the least population of a district is 10 then no district can have a population more than 10+0.1*10=11. Hope it's clear.

russ9 · Answer

Hi Bunuel,

Two questions:

1)One area that stumped me was why did we want to minimize ONLY 1 and maximize 10, versus minimizing 10 and maximizing 1? In algebraic form -- why would it not be (10x + 1.1x = 132?)

2) I interpreted the 10% increase as 90% of something, meaning, I wrote, .9Y + 10Y = 132,000 and didn't get the right answer. Why is that wrong?

Thanks,
R

Bunuel · Answer

1. If you minimize 10 and maximize 1, you'd get the maximum possible population that the most populated district could have.

General rule for such kind of problems, when the total is fixed: 
to maximize one quantity, minimize the others; 
to minimize one quantity, maximize the others.

2. a is 10% greater than b (a=1.1b --> a=11/10*b), is not the same as b is 10% less than a (b=0.9a --> a=10/9*b).

Hope this helps.

BrushMyQuant · Answer

"no district is to have a population that is more than 10 percent greater than the population of any other district" and "they are asking for the least populated district" So, we can take one district's population as x and remaining 10 district's population as 10% greater than x

Now, we are satisfying all the conditions in the question
x + 10*(x+(0.10x)) = 132,000
x + 11x = 132,000
12x = 132,000
=> x = 11,000

So, Answer will be D
Hope it helps!

honchos · Answer

Do you think that this question has no verbiage issue. How come this is possible that all districts exist that no district has population grater than 10% of any other district?

EMPOWERgmatRichC · Answer

Hi honchos,

If you read the other posts in this thread, you'll learn the EXACT situation described by the prompt: If 10 of the districts have the SAME population, then the 11th could be the one with the 10% difference in population.

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

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

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