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Help Needed..

Expert replies

by Brian@VeritasPrep » Mon Sep 13, 2010 12:07 pm
Hey Aliva,

Nice catch - yes, that's exactly right...I fixed those numbers in my original post. Thank you for pointing that out (I have a nasty habit on these posts of doing the problem on paper then typing in the calculations from memory while I'm writing up the explanation...I usually do okay but here there may just have been too many 10s, 11s, and 12s to keep straight without looking back!).

Thanks!
Brian

alivapriyada wrote:
Brian@VeritasPrep wrote:Hey Sanalnnair

Good questions - you may want to even break these into individual posts so that people can devour each one thoroughly without confusion (I read your initial question, saw "pi" in the second post of this thread, and was completely lost).

Let me talk about that first one because I think it brings up a pretty good strategic point. When you're asked for the minimum of one city in this case, strategically you'll want to maximize everything else so that you can keep the district in question to a minimum.

With that in mind, let's see how we can minimize one district and maximize the others. We'd ideally want each of the other 10 districts to have the maximum allowable population which is 10% greater than (or 11/10 of) that of the smallest. That would mean that we have 10 districts of the maximum, and if we call the minimum population 'x', we could say that:

x (the smallest) + 10 districts * 11/10 * x = 132,000

x + 10x = 132,000
11x = 132,000

x = 11,000
wont that be X+11X=1,32,000
12X=1,32,000
X=11,000

M sorry if I'm mistaken.
Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep

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