Math Problem from mba.com - Can anyone help me solve it?

[mya070]

Hi all! I got a math problem I could not solve when sitting one of tha mba.com CATs. It would be great if somebody could help me out with it! Here is the problem:

City X has a population of 132 000. If the population is to be devided in 11 voting districts, and no district is to have a population that is greater than 10% of the population of any other city district, what is the minimum possible popuation that the least population district could have?

Anwer: 11 000

But how do we get the answer?

[davegerstel]

Hi there,

An easy way to solve this question is by dividing 132,000 by the 11 districts to find the average district size. 11 goes into 132,000 12,000 times and so the average district size is 12,000. This means that for every number higher than 12,000 there must be a corresponding number lower than 12,000, to balance the scale. Think of an average as the middle of a scale with equal weights on either side of it. Now, one of the answer choices is 10,900. Let`s try this answer first. If we have a number that is 1100 less than the average, then there has to be 1100 higher than the average to keep the balance. Now, we can add this 1100 to one district of 12,000, creating a district of 13,100. But we can keep the districts much smaller by spreading the 1100 across the remaining ten numbers and say that they are all 12,110. But 12,110 is still more than ten percent more than 10,900 (which is 11,990. 10,900 plus 1090).

The next smallest district is 11,000. Lets try this one. 11,000 is 1000 less than the average, so there must be 1000 higher than the average. Lets spread it out across the remaining ten numbers and we have 10 districts of 12,100. 11,000 plus ten percent equals 12,110. So no district is MORE than the smallest district of 11,000.

The key in this question is to realize...

a) an average is like the middle of a scale with equal weights on either side

b) that we can spread the extra weight over as many districts as possible to keep them smaller

c) not to fall into the MORE than 10 percent trap.

[mya070]

Great! Thank you very much!!

[venkiteswaran.ramasamy]

A doubt in the steps leading to the deduction.

One of the key points that has been mentioned is

"that we can spread the extra weight over as many districts as
possible to keep them smaller"

Should we not try to add the extra weight entirely to one district, so as to make sure that even the district with maximum possible votes would still not be 10% more than the one with the minimum number (which is the one which gets stripped and reduced below the actual average). The question seems to ask for a solution where the district with minimum votes is still not 10% lower than ANY of the other districts. Kindly clarify.

[davegerstel]

I`m not entirely sure what you`re asking, but I`ll try to respond as best as I can.

The question states two requirements
A) Find the smallest possible district
B) No district can be MORE THAN 10% greater than any other district

Now, let`s see what happens if we add all the extra weight to one district. For example, let`s add 1000 to one district and create a district with 13,000. Now in order to satisfy the second requirement of the question, that no district be more than 10% smaller than another, the smallest possible district can be 11,818 (13,000/1.1).

However, if we spread the 1000 over ten districts and create ten 12,100 districts, then the greatest district is 12,100 and the greatest possible number that 12,100 is not more than ten percent greater than it, is 11,000 (12,100/1.1). So by keeping the greatest district smaller, (approx) 90% of it is also smaller, and the question asks for the smallest POSSIBLE district.

In this question we are walking a tightrope because the more weight we add to the bigger districts, the smaller we can make the smaller districts, but at the same time we can`t add too much weight or else the district will become more than ten percent apart. So we have to find the happy medium where we add the most weight possible to the big side while still keeping the two sides within ten percent of each other. This is accomplished by spreading out the extra weight as much as possible, thereby keeping the larger districts as small as possible, while taking as much weight as possible from the smaller side, thus creating the smallest possible district.

[tim]

Looks like we’ve got some good discussion here. Let me know if you need any further clarification from an instructor..

[venkiteswaran.ramasamy]

Firstly thanks for the reply.

The only problem I have with the solution is - the question asks for a solution in which the district with maximum votes is not greater than 10% compared to the district with the minimum votes. So should not this comparison involve the extreme possible case which will occur when x number of votes are removed from one district (thereby making it the minimum) and comparing it with the maximum single district that can be made? Adding the extra weight evenly to the remaining district seems to overlook one extreme possibility of adding the extra weight to one single district thereby making it the district with maximum possible votes. And the 10% test should still stand when we compare the minimum and the maximum thus built.

Please clarify or let me know if I am missing something here.

Thanks!
Venky

[davegerstel]

Hey Venky,
The question is not concerned with the largest possible district; the question only cares about the smallest possible district. So yes the more weight we take away from a district the smaller it will be, but the smaller we can keep the larger number, the smaller the small number can go. Ten percent less than a small number is less than ten percent less than a large number. So we want to take away the maximum amount from one number and keep the big numbers as small as possible. Remember, the question only cares about the smallest district.
I hope this answers your question,
David

[venkiteswaran.ramasamy]

Thanks David!

That really drives homes the point for me. The fact that the question did not have anything to do with the max size district was a little too subtle for me and I missed that. Thanks once again for explaining patiently.

Regards!
Venky

[tim]

:)

[lena]

The population in one district will be minimized by assigning the max possible value to all other districts. So if the district with min population is p then all other 10 districts should be 10% more than p, i.e. 1.1xp

p + 1.1xpx10=132,000

from here p= 11,000

[tim]

great solution lenam!

[krishnan.anju1987]

Lovely Solution Lena and Venky!!! :)

[tim]

:)