Stacey (and Guest),

Thanks for your earlier responses, I am truly appreciative. Here's another one similar to the DS question I posted earlier. I don't have a systematic approach to solving these types of problems. Any thoughts would be appreciated.

Thanks.

Ladan,

Following information is given in the question: -

1) 11 districts - 132000 population
2) No district to have a population > 10% of any other

We need to consider (2) while calculating the minimum possible population of the least populates district.

If you divide the population equally among 11 districts the population in district = 132000/11 = 12000

Now, the key is to decrease the population of one district and compensate(or increase) that amount equally into rest.

If you look into the answer choices.

1) 10,700 - the amount decreased 1300, if we compensate this amount equally into others, the population of rest 10 districts will be 12000 + 130 = 12300, Now to satisfy (2) - calculate 10% of 10,700 = 1070 if we add it to 10700

10,700 + 1070 = 11770, and 11770 < 12300. Not possible.

the minimum possible population that will satisfy (2) wil be 11,000. So (D) is the answer.

Unable to crack this one. Totally stumped.

Took about 4 or more minutes just to crack the idea behind this! Would've probably had to guess this in the pressure of the exam.

Let us assume the minimum town has a population x. The maximum population town cannot have a population more than 1.1*x. Now comes the difficult part. For x to be he lowest value it can be, every other town has to be equal to the maximum population value. (Hope that makes sense, think about a set of numbers where you know the total and the maximum value. The minimum value will be at its least possible when every other number in the set is equal to the max value.).

So that settled we know 10 towns have the value 1.1 x. So 10*1.1*x+x = 132,000. So 12x=132,000 So x is 11,000.

I also thought 132,000 was an odd choice with the 11 districts. (132 = 11*12)

nicely done

One last possible approach (although nov1907's approach is excellent and leaves very little to be desired):

You could realize that the answer choices are for the LEAST populated district, and that, therefore, the other districts have at most 10% more people than the number in the answer choice.

Take the answer choices one at a time, starting with the middle choice (to minimize the necessary # of guesses), and find out whether that number is big enough to produce an overall population of 132 000.

Try (C): Minimum population is 10 900.
All other districts are 10% more than 10 900 = 11 990.
Total population = 10(11 990) + 10 900 = 130 800, which isn't big enough. Therefore, choice (C) is too small.

Since C is too small, so are A and B (that's the wisdom of trying C first: If you try A and, surprise of surprises, it's too small, then you've eliminated NOTHING else).

Try D. It works, producing a population of exactly 132 000.

to minimize the number of ppl in the least populated district, we should push the population as much as possible to other 10 districts. what is the maximum pushing that can be done? it is 10%. remember this 10% deviation applies to all the districts and even to the least populated district. let the least populated district be p. other 10 districts can have maximum 10% deviation from p.

so p + (0.1*p)*10 = 13200
11p = 13200
p = 11,000 :)

Just a quick question - When answering this questions, I had done the following math and got the wrong answer:

.9Y + 10Y = 132,000

Why doesn't that work?

this won't work because, if X is 10% greater than Y, then Y is not 10% less than X.

this is a universal truth for all nonzero percentages, of which you should be acutely aware on test day: an increase of p% and a decrease of p% DO NOT cancel other out.
this is in fact a corollary of a much more general principle stating that percent changes are never additive. for instance, if a quantity is increased by 20% and then the resulting quantity is increased by 30%, then the original quantity did not increase by 50%.
this is a common trap on percent problems. if you know it, then you'll be able to eliminate the "sucker answer" right away on many such problems.

you can also figure this out for yourself by thinking about things that are intuitively easier than 10% changes. for instance, to undo cutting something in half (50% decrease), you'd have to double it (100% increase).

--

in your solution, you're assuming that the least populated district should be 10% less populated than the most populated district. as noted above, this is not the same as assuming that the most populated district should be 10% more populated than the least populated district.

The math ain't right:
p + 0.1p*10 = 2p=13200 and what follows is incorrect. Nobody else noticed this error?
The author probably meant p + 0.1p*10 +10p=132000

I agree that that is what the author probably meant. Your equation is correct.

_________________
Jamie Nelson
ManhattanGMAT Instructor

Actually, my guess is that he or she meant p + (1.1p)*10 = 132000.... equivalent but a bit different.

Okay, same equation, just looks slightly different.

_________________
Jamie Nelson
ManhattanGMAT Instructor

Excellent explanation...

Just a tip about choosing the right one to guess. Following the same logic, it's more useful to try B or/then D.
In this particular case, you would not have to pick another one, even if the answer were E.

In summary, for both, the max # of attempts is two, but you have more chances to solve in 1 guess, if you try B or D.

Regards!!