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by szapiszapo » Tue Jul 01, 2008 9:14 am
total population = 132,000
11 voting districts named D1, D2, D3, ..., D11

we know that D1 + D2 + D3 + ... + D11 = 132,000 (a)
and that the maximum gap between all districts is 110% in terms of population count, i.e. Di <= 1.1 Dj (b)

the minimum possible population for a given district is the one for which all the others districts have the same population (thus maximizing the gap)

hence if D1 is the least populated district, D2 = D3 = ... = D11
(a) can then be written as D1 + 10 D2 = 132,000
and when the gap is maximum, (b) translates into D2 = 1.1 D1

combining the two gives us D1 + 11 D1 = 132,000
and D1 = 11,000

You can verify that equations (a) and (b) above are respected
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by andes1 » Sun Jul 06, 2008 1:33 pm
well why 110%
LEARNING ENGLIS H
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by swdatta » Sun Jul 06, 2008 2:30 pm
I got B as the answer for this question.

If all the 11 district have the same population then each district will have a population of 12000. 10% of 12000 is 1200. Then the district with greatest population is 13200 and the district with the smallest population is 10800.

Am I missing something obvious in this question?
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by Stuart@KaplanGMAT » Sun Jul 06, 2008 6:29 pm
Great question for backsolving!

To get the smallest possible population in 1 district, we want the biggest possible populations in the other 10. We know that the total number of voters is 132000.

Let's look at the choices:

a) 10,700
b) 10,800
c) 10,900
d) 11,000
e) 11,100

When we backsolve, we usually start with either (b) or (d). (d) is a much easier number to work with, so let's start there.

Assuming (d) is correct, the smallest district has 11000 voters. The maximum size of every other district is 10% more than that, or 11000 + 1100 = 12,100.

So, we have 10 districts * 12,100 = 121,000 + 1 district of 11000 = 132000 people! 132000 is exactly what we wanted to get, so (d) is correct.

If we ended up with fewer than 132000, we'd know we needed a bigger number and that (e) was correct. If we ended up with more than 132000, we'd know that we neededa smaller number and we'd try (b). Using the same reasoning (i.e. too small, just right or too big), after our 2nd try we'd definitely know which answer is correct.
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