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How do we approach this..???!!?

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Intern
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How do we approach this..???!!? [#permalink] New post 20 Apr 2007, 18:46
How do we approach this..???!!?? :roll:
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 [#permalink] New post 20 Apr 2007, 20:53
(B)

10 A + 0.9 A = 132,000
--> A = 1,211
---> 0.9A = 10,900 (this is the population of the least populated district)
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Re: GMATPrep Question [#permalink] New post 20 Apr 2007, 21:13
x + 1.1 x(10) = 132000
x = 11,000
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Re: GMATPrep Question [#permalink] New post 21 Apr 2007, 08:20
[quote="Himalayan"]x + 1.1 x(10) = 132000
x = 11,000[/quote]

Hi Himalayan,

Can you please explain your reasoning to get to the value of 11,000? Your numbers seem to be much nicer than mines :) so I wonder why my reasoning is incorrect?
Thanks!
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Re: GMATPrep Question [#permalink] New post 21 Apr 2007, 09:26
querio wrote:
Quote:
x + 1.1 x(10) = 132000
x = 11,000


Hi Himalayan,

Can you please explain your reasoning to get to the value of 11,000? Your numbers seem to be much nicer than mines :) so I wonder why my reasoning is incorrect?
Thanks!


if you tale x as the larger/largest and 0.9 as the smaler/smallest, the the difference could be grater than 10%. in your case, 10% of 1211 is correct. (calculate difference as % of both 10900 and 1211).

But if you take the difference as the % of 10900, then it is greater than 10%, which is not correct.

In my case, the difference is 10% or less than x (11000).
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 [#permalink] New post 21 Apr 2007, 09:56
Its always good to see how these types of problems can be simplified. I made this much more complicated on myself as I was plugging in each solution and narrowing down that way.

I think the formula is really:

x + (10) (1.1x) >= 132000

Really you just want to make sure that the 10 districts which have a population above the min population district can absorb the remaining population.

So if the min population is 11000, can the remaining districts absorb the remaining 121000 population without any one district going over 12100 (11000(1.1)).
  [#permalink] 21 Apr 2007, 09:56
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