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For questions with a fixed sum (here the total population of 132,000), if you need to minimize one term, then you maximize all the others. Likewise, if asked to maximize one term, you minimize all the others.
To see that chetan2u is doing just that, I find it useful to create placeholders for the terms (here, the districts) on paper, filling in with numbers or variables:
(In order, smallest to largest) x / 1.1x / 1.1x / 1.1x / 1.1x / 1.1x /...../ 1.1x / 1.1x = sum of 132,000
As many know, there are many ways to the correct answer in PS. I'm very good at algebra but if I, for example, use logical reasoning and I get to the correct answer more quickly, then that's better.
...Or, if I see that using the answer choices may be quicker than algebra, then I may do that too. On any multiple choice test, the answer choices are your best friend.
Coming to the question, as esledge points out, in any minimum question the first thing you should think of is what you have to maximize (and vice-versa). We need to minimize the population of the smallest voting district.
So, we'll have to maximize the populations of the other 10. Each of these can be at most 10% larger than the population of the smallest.
Look at the answer choices...which number is it easiest to take 10% of?
Clearly, choice D (since it ends in a "0"). 10% of 11, 000 is obviously 1, 100.
So, the biggest we can make the other each of the other ten populations is 11, 000 + 1, 100 = 12, 100.
The total sum, then, is 10*12,100 + 11,000 = 132,000. Success--happy ending to the story--all initial conditions satisfied--correct answer must be choice D!
TAKEAWAY: often algebraic approaches will, for a large chunk of the test-taking population, be less efficient than alternative approaches based on logical reasoning, picking numbers or, as here, backsolving. During review, return to questions you've answered correctly, and ask yourself whether you could have answered them more quickly had you made use of non-traditional approaches.
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