A certain city

This topic has expert replies
Master | Next Rank: 500 Posts
Posts: 217
Joined: Sun Jan 10, 2010 1:39 pm
Thanked: 7 times
Followed by:1 members

A certain city

by zaarathelab » Mon Oct 17, 2011 10:09 am
A certain city with a population of 1, 32,000, is divided into 11 voting 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 that the least populated district could have?
A) 10700
B) 10800
C) 10900
D) 11000
E) 11100

What is the quickest way to solve this?

[spoiler]I figured out that this is a Geometric Progression

Sum of N terms = a(r^n-1)/r-1
132000 = a(1.10^11 - 1) / 1.10-1 we need to find a (first term also the lowest term)

But solving 1.10^11 is complicated[/spoiler]
Success = Max(Hardwork) + Min(Luck)
Source: — Problem Solving |

Legendary Member
Posts: 966
Joined: Sat Jan 02, 2010 8:06 am
Thanked: 230 times
Followed by:21 members

by shankar.ashwin » Mon Oct 17, 2011 10:14 am
1.1(10x) + x = 132000
12x = 132000
x = 11000.

Assume 10 towns have a population which is 10% greater than the 11th which is the least populated.

User avatar
Legendary Member
Posts: 504
Joined: Tue Apr 19, 2011 1:40 pm
Thanked: 114 times
Followed by:11 members

by knight247 » Mon Oct 17, 2011 10:17 am
The way I find quickest is as follows:

Since we have to minimise the population of 1 district we would need to maximise the population of the others.

If the population of the district with the least population is x then the population of the other 10 districts would be x+(10x/100)=1.1x. Multiplied by 10 would give the population of all the 10 districts i.e. 11x

We have x+11x=132000
12x=132000
x=11000 Hence D