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PS Population (m01q09)

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Re: PS Population (m01q09) [#permalink]

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New post 09 Jan 2011, 09:51
This was my method:

The increase from 1990 to 1993 was 1200. Therefore when expressed as a percentage it comes to about 33.33% which is really 33.33/100.

If this were converted to per thousand, it would be 333 per thousand. From there on, [4800 + 4800(333.3/1000)] to find the increase.


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Re: PS Population [#permalink]

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New post 28 Nov 2011, 22:35
x2suresh wrote:
good discussion guys.

You view the question in differt way to..

You can rephrase the original problem "% increase is constant over the periods"

% increase inpopulation from 1990-1993 = % increase in population from 1993-96

(4800-3600)/3600 = 33.3%

1996 population = 4800 *1.333 = 6400

Your response makes sense. I'm not sure how close this is to a GMAT question.
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Re: PS Population (m01q09) [#permalink]

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New post 02 Dec 2011, 00:19
This is really good question, and very good discussion regarding wording of the Q

-------Analyze why option A in SC wrong-------

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Re: PS Population (m01q09) [#permalink]

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New post 30 Nov 2012, 12:17
Like you, I have a problem with this question, but because of wording. It is fine for the question to ask constant growth rate, but it cannot be ambiguous about base years, which is what it did. For instance, what was the population in 1991 or 1992? it isn't 11% (33%/3) year over year, because that ignores the compounding error that will result. By neglecting to specify that the base interval was 3 years, it falsely implies that the rate of pop. growth is YoY, when an answer like 6400 shows clearly that it is not. To compute YoY and go from 3600 to 4800 over 3 years: 3600(1+growthrate)^3=4800. Then dividing both sides by 3600 yields (1+growthrate)^3=(4/3). We don't have the calculating tools on the GMAT to compute logs and thus it is an unfair question.
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Re: PS Population (m01q09) [#permalink]

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New post 03 Dec 2013, 12:35
My approach:

Rate from 1,990 to 1,993: (4800-3600)/3600= 1/3
Population in 1,996= 4800* (1+1/3) = 4800*4/3=6,400
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Re: PS Population (m01q09) [#permalink]

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New post 06 Dec 2013, 20:45
4800/3600 = 4/3
So, population in 1996 would be 4800 * 4/3 = 6400.
So, B.

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Re: PS Population (m01q09) [#permalink]

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New post 02 Apr 2014, 12:45
\(4800=3600(1 + \frac{R}{100})^3\)

\((1 + \frac{R}{100})^3=48/36\)

\(Population-in-1996=4800(1+ \frac{R}{100})^3\)



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Re: PS Population (m01q09)   [#permalink] 02 Apr 2014, 12:45

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