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A certain city with population of 132,000 is to be divided [#permalink]
02 Mar 2009, 19:52

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Question Stats:

53% (01:58) correct
47% (01:17) wrong based on 559 sessions

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?

Re: PS (gmatprep1) -- city population [#permalink]
12 Dec 2010, 08:32

103

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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

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

So we want to minimize x. To minimize x we should make only one district to have that # of population (minimum possible) and the rest 10 districts to have 1.1x # of population (maximum possible).

Re: PS (gmatprep1) -- city population [#permalink]
30 Dec 2010, 08:37

Bunuel wrote:

metallicafan wrote:

So we want to minimize x. To minimize x we should make only one district to have that # of population (minimum possible) and the rest 10 districts to have 1.1x # of population (maximum possible).

i appreciate this fact to understand the fomulas _________________

Consider giving me kudos if you find my explanations helpful so i can learn how to express ideas to people more understandable.

Re: PS (gmatprep1) -- city population [#permalink]
30 Dec 2010, 14:32

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marijose wrote:

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...

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). _________________

Re: A certain city with population of 132,000 is to be divided [#permalink]
17 Apr 2012, 04:51

Bunuel,

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

Quote:

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

Re: A certain city with population of 132,000 is to be divided [#permalink]
17 Apr 2012, 05:03

3

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Expert's post

ENAFEX wrote:

Bunuel,

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

Quote:

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

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).

Re: A certain city with population of 132,000 is to be divided [#permalink]
21 Sep 2012, 00:35

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. _________________

My attempt to capture my B-School Journey in a Blog : tranquilnomadgmat.blogspot.com

A certain city with population of 132,000 is to [#permalink]
31 Dec 2012, 11:19

2

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joyce wrote:

OA: D

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.

Re: PS (gmatprep1) -- city population [#permalink]
18 May 2013, 13:11

Bunuel wrote:

marijose wrote:

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...

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).

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

Re: PS (gmatprep1) -- city population [#permalink]
19 May 2013, 02:58

1

This post received KUDOS

Expert's post

up4gmat wrote:

Bunuel wrote:

marijose wrote:

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...

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).

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

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.

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.

Re: A certain city with population of 132,000 is to be divided [#permalink]
22 Dec 2013, 05:39

[quote="ugimba"]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

how do you solve this kind of problems .. any general formula(e)? I just split and try ... so usually it takes long time. Some times endup having wrong answer after somuch time spent..[/q

i approached the answer like this, 132000/11= 12000, hence the max population that each city can have is 12000. Let min. population be x. Therefore, x+10x/100=12000, which gives x as 12000/1.1 and x=10909.9. is this method right plz help

Re: A certain city with population of 132,000 is to be divided [#permalink]
22 Dec 2013, 06:06

Expert's post

ddp123 wrote:

i approached the answer like this, 132000/11= 12000, hence the max population that each city can have is 12000. Let min. population be x. Therefore, x+10x/100=12000, which gives x as 12000/1.1 and x=10909.9. is this method right plz help

Re: A certain city with population of 132,000 is to be divided [#permalink]
17 Jun 2014, 15:58

Bunuel wrote:

ENAFEX wrote:

Bunuel,

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

Quote:

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

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.

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?

Re: A certain city with population of 132,000 is to be divided [#permalink]
18 Jun 2014, 04:35

Expert's post

russ9 wrote:

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

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).

A certain city with population of 132,000 is to be divided [#permalink]
01 Oct 2014, 02:41

Bunuel wrote:

russ9 wrote:

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

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.

If the question had been "What is the minimum possible population that the most populated district could have?" Would that just be the average--12000?

Makes sense logically and you really cannot do 10x + 1.1x = 132 as that will yield a fraction of a person. A fetus?!

gmatclubot

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