|
Author |
Message |
|
TAGS:
|
|
|
Senior Manager
Joined: 20 Feb 2006
Posts: 376
Followers: 1
Kudos [?]:
0
[0], given: 0
|
A certain city with a population of 132,000 is to be divided [#permalink]
 02 Dec 2006, 11:38
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
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 percent reater 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
Anyone got a quick way to solve this one?
|
|
|
|
|
|
|
Director
Joined: 28 Dec 2005
Posts: 923
Followers: 1
Kudos [?]:
34
[0], given: 0
|
Use the formula for geometric progression and using approximations you can solve. I had to look it up also. Ans E?
I am surprised this was on GmatPrep. Are we expected to know these formulae? Maybe this is not the route expected by GMAT.
Sn = ( a1 - a1 r^n ) / ( 1 - r)
a1 = 1
r = 1/10
n = 11
If x is the least population: x(1+1/10+1/10^2...1/10^10) = 132000
x * (1-(1/10)^11)/(1-1/10) = 132000 I approximated 10^11 - 1 as 10^11.
|
|
|
|
|
|
Senior Manager
Joined: 08 Jun 2006
Posts: 347
Location: Washington DC
Followers: 1
Kudos [?]:
5
[0], given: 0
|
Getting E
Population of one district = A
A will be minimum only when all other districts’ populations are more than 10% of A
So A + (110A/100)*10 = 132,000
A = 11,000
|
|
|
|
|
|
Senior Manager
Joined: 20 Feb 2006
Posts: 376
Followers: 1
Kudos [?]:
0
[0], given: 0
|
This scared me a little when it came up - hope I dont get it on the GMAT
|
|
|
|
|
|
Senior Manager
Joined: 23 May 2005
Posts: 266
Location: Sing/ HK
Followers: 1
Kudos [?]:
3
[0], given: 0
|
Help pls... I don't get it...
"A will be minimum only when all other districts’ populations are more than 10% of A" --> why??
I also don't follow the arithmetic... A + (110A/100)^10 = 132,000A = 11,000
_________________
Impossible is nothing
|
|
|
|
|
|
Senior Manager
Joined: 20 Feb 2006
Posts: 376
Followers: 1
Kudos [?]:
0
[0], given: 0
|
I'm also missing the arithmetic - someone help!
|
|
|
|
|
|
Director
Joined: 28 Dec 2005
Posts: 923
Followers: 1
Kudos [?]:
34
[0], given: 0
|
Trying to explain:
Ok, according to this: Let's say district A has the minimum, then a district B cannot be more than 10% of A. ie: A+10/100A= 110/100A. Now it's given that "no district is to have a population that is more than 10 percent greater than the population of any other district", so for A being a minimum, this can only be satisfied when all of the other 10 districts are 10% more than A (if they were any higher, then they would be more than 10% of A) . Thus the equation that Anindya came up with.
|
|
|
|
|
|
Senior Manager
Joined: 20 Feb 2006
Posts: 376
Followers: 1
Kudos [?]:
0
[0], given: 0
|
thanks hsampath
Why when I try to do it the long way does the sum of the 11 districts not equal 132000?
1st district = 11,000 Multiplying each by 110/100 or 1.1
2nd = 12100
3rd = 13310
4th = 14641
5th = 16105.10 ...........
11th = 25686.8359
The sum of these does not equal 132000?
|
|
|
|
|
|
Director
Joined: 28 Dec 2005
Posts: 923
Followers: 1
Kudos [?]:
34
[0], given: 0
|
that's where I made a mistake also, the first time around. it's actually
1st district = 11,000 Multiplying each by 110/100 or 1.1
2nd = 11000* 1.1 = 12100
3rd = 11000 *1.1 = 12100 NOT (12100*1.1=13310)
4th = 11000 * 1.1 = 12100 NOT (13310*1.1)
etc
because think abt this, with your logic: 3rd district = 13310 which is greater than 10% of the 1st district. According to question, it cannot be more than 10% of any other district, including A. Thus, 3rd, or 4th or anything else can be a max of 110/100*A.
Is this clearer?
|
|
|
|
|
|
Senior Manager
Joined: 20 Feb 2006
Posts: 376
Followers: 1
Kudos [?]:
0
[0], given: 0
|
Dont worry - I've been misreading the question - I was looking at it as if it were a progression with a difference of 10% between districts ie. District 1 = x
District 2 = x+10%
District 3 = (x+10%)+10%
Thanks for your patience guys!
|
|
|
|
|
|
Senior Manager
Joined: 20 Feb 2006
Posts: 376
Followers: 1
Kudos [?]:
0
[0], given: 0
|
hsampath wrote: that's where I made a mistake also, the first time around. it's actually
1st district = 11,000 Multiplying each by 110/100 or 1.1
2nd = 11000* 1.1 = 12100
3rd = 11000 *1.1 = 12100 NOT (12100*1.1=13310)
4th = 11000 * 1.1 = 12100 NOT (13310*1.1)
etc
because think abt this, with your logic: 3rd district = 13310 which is greater than 10% of the 1st district. According to question, it cannot be more than 10% of any other district, including A. Thus, 3rd, or 4th or anything else can be a max of 110/100*A.
Is this clearer?
I see you just posted before me - thank hsampath - I was tearing my hair out trying to see what was wrong with my working when I'd just read the question wrong. Will add that to error log!!!
|
|
|
|
|
|
Senior Manager
Joined: 23 May 2005
Posts: 266
Location: Sing/ HK
Followers: 1
Kudos [?]:
3
[0], given: 0
|
Aaaaaaaaa..... Now I get it. Tricky tricky...
Thansk hsampath.
I still hope i don't get a question like that n my exam.... Ack!
_________________
Impossible is nothing
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
close
