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m01#9

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m01#9 [#permalink] New post 21 Jul 2011, 03:48
The population of Linterhast was 3,600 people in 1990 and 4,800 people in 1993. If the population growth rate per thousand is constant, then what will be the population in 1996?

6,000
6,400
7,200
8,000
9,600

Official solution:
This is a set rate problem. If the population grew by 1,200 people in the past three years, then it grew by 33 percent:
\frac{4800-3600}{3600} = \frac{1200}{3600} = \frac{1}{3} \approx 33%

Therefore in the next three years the population will grow at the same rate of 33% because the growth rate has been constant.

4,800 + \frac{1}{3}*4,800 = 4,800 + 1,600 = 6,400

Another approach is to backsolve by comparing the ratio of each answer to 4,800. For example, the ratio of 7,200 to 4,800 is not the same as the ratio of 4,800 to 3,600.
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Re: m01#9 [#permalink] New post 27 Aug 2011, 00:14
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a much faster way would be:
growth rate = 4800/3600 = 4/3

pop in '96 = 4800*4/3 = 6400
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Re: m01#9 [#permalink] New post 21 Jul 2011, 03:51
Problem with the solution is that it is assuming that rate is constant per 3 years... i.e. 33 % over a period o where mentioned in the question....
If assumed that increase rate is constant over the period of 1 year then problem is very diff.
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Re: m01#9 [#permalink] New post 20 Oct 2011, 00:50
Why can't we add the average yearly increase (400) to following three years (1994, 95 and 96) to get 6000?
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Re: m01#9 [#permalink] New post 21 Oct 2011, 12:52
saeedt wrote:
Why can't we add the average yearly increase (400) to following three years (1994, 95 and 96) to get 6000?


Because the question says "If the population growth rate per thousand is constant, then what will be the population in 1996?"

If it is mentioned anything like "if the population increased each year at constant number" then your method is correct
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Re: m01#9   [#permalink] 21 Oct 2011, 12:52
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