Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 23 May 2017, 18:45

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# A city with population of 132,000 is divided into 11

Author Message
Manager
Joined: 10 May 2006
Posts: 186
Location: USA
Followers: 1

Kudos [?]: 5 [0], given: 0

A city with population of 132,000 is divided into 11 [#permalink]

### Show Tags

26 May 2006, 10:30
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A city with population of 132,000 is divided into 11 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 the least populated district could have?

a.) 10,700
b.) 10,800
c.) 10,900
d.) 11,000
e.) 11,100
Senior Manager
Joined: 09 Mar 2006
Posts: 445
Followers: 1

Kudos [?]: 8 [0], given: 0

### Show Tags

26 May 2006, 11:05
The average population is 12000, so min * 110% should be more than the average. Only D satisfy.
Director
Joined: 10 Oct 2005
Posts: 526
Location: US
Followers: 1

Kudos [?]: 63 [0], given: 0

### Show Tags

26 May 2006, 11:49
deowl could you please explain this further...
Senior Manager
Joined: 09 Mar 2006
Posts: 445
Followers: 1

Kudos [?]: 8 [0], given: 0

### Show Tags

26 May 2006, 12:18
mahesh004 wrote:
deowl could you please explain this further...

Obviously the average of a district population is 12000 ( 132000 / 11 )
So for any district that has population less than the average exists another one that balances it ( there could be more than one that balance it exactly against the average , but certainly at least one exists ). That's it for any number less than the average exists another one that is more than the average.

Since for A, B and C certainly exists another district with population more than 10% , they are out.

D is less than E, so D is my answer.
VP
Joined: 29 Dec 2005
Posts: 1343
Followers: 10

Kudos [?]: 62 [0], given: 0

### Show Tags

26 May 2006, 13:49
tl372 wrote:
A city with population of 132,000 is divided into 11 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 the least populated district could have?

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

x + 10(1.1)x = 132000
x + 11x = 132000
x = 11,000
Senior Manager
Joined: 09 Mar 2006
Posts: 445
Followers: 1

Kudos [?]: 8 [0], given: 0

### Show Tags

26 May 2006, 13:57
Professor wrote:
x + 10(1.1)x = 132000
x + 11x = 132000
x = 11,000

Manager
Joined: 10 May 2006
Posts: 186
Location: USA
Followers: 1

Kudos [?]: 5 [0], given: 0

### Show Tags

26 May 2006, 14:50
deowl wrote:
mahesh004 wrote:
deowl could you please explain this further...

Obviously the average of a district population is 12000 ( 132000 / 11 )
So for any district that has population less than the average exists another one that balances it ( there could be more than one that balance it exactly against the average , but certainly at least one exists ). That's it for any number less than the average exists another one that is more than the average.

Since for A, B and C certainly exists another district with population more than 10% , they are out.

D is less than E, so D is my answer.

I follow your reasoning up until you rule out A, B, and C. Can you show this mathematically? I'm having a hard time visualizing why you rule out A, B, C but not D.

Thanks.
Senior Manager
Joined: 09 Mar 2006
Posts: 445
Followers: 1

Kudos [?]: 8 [0], given: 0

### Show Tags

26 May 2006, 15:07
tl372 wrote:
deowl wrote:
mahesh004 wrote:
deowl could you please explain this further...

Obviously the average of a district population is 12000 ( 132000 / 11 )
So for any district that has population less than the average exists another one that balances it ( there could be more than one that balance it exactly against the average , but certainly at least one exists ). That's it for any number less than the average exists another one that is more than the average.

Since for A, B and C certainly exists another district with population more than 10% , they are out.

D is less than E, so D is my answer.

I follow your reasoning up until you rule out A, B, and C. Can you show this mathematically? I'm having a hard time visualizing why you rule out A, B, C but not D.

Thanks.

Assume the answer is C , so one of the districts (C)has population of 10,900. So the biggest possible population of some other district can be 10,900 * 1.1 = 11,990. So how can we have the average of 12000 if the
the district with biggest population has less than that ?
VP
Joined: 29 Dec 2005
Posts: 1343
Followers: 10

Kudos [?]: 62 [0], given: 0

### Show Tags

27 May 2006, 06:07
deowl wrote:
Professor wrote:
x + 10(1.1)x = 132000
x + 11x = 132000
x = 11,000

to get the least, put the maximums (110% of the minimum) at one side and the minimum at the other. then solve it.

the question syas that the maximum cannot be more than 10% of the leaset.

the can be many equations but to get the leaset only the following eq works: x + 10(1.1)x = 132000.

other eq could be:

2x + 9(1.1)x = 132000
3x + 8(1.1)x = 132000
.
.
.
.
10x + 1.1x = 132000

but only "x + 10(1.1)x = 132000" gives the least value for a city.
Re: Population   [#permalink] 27 May 2006, 06:07
Display posts from previous: Sort by