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The population of Jonasburgh will grow by r% per year for 30 Jul 2006, 01:43
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The population of Jonasburgh will grow by r% per year for the next 20 years. If the population of Jonasburgh is 50,000 today and is expected to be 6,500,000 in 14 years' time, which of the following best approximates what Jonasburgh's population will be five years from now?

(A) 280,000 (B) 500,000 (C) 560,000 (D) 2,000,000 (E) 2,400,000



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The population of Jonasburgh will grow by r% per year for 30 Jul 2006, 02:01
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A

130 = (1+r/100)^14
we can approx

140 = (1+r/100)^15

(140)^(1/3) = (1+r/100)^5

So Pop after 5 years = 50000 * (140)^(1/3) = 50000 * 5.3 approx
= 265,000

280,000 is closest. Also, for sure it can not be less than 250,000.
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The population of Jonasburgh will grow by r% per year for 30 Jul 2006, 19:26
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kevincan
The population of Jonasburgh will grow by r% per year for the next 20 years. If the population of Jonasburgh is 50,000 today and is expected to be 6,500,000 in 14 years' time, which of the following best approximates what Jonasburgh's population will be five years from now?

(A) 280,000 (B) 500,000 (C) 560,000 (D) 2,000,000 (E) 2,400,000
Show more


Using linear approximation..
slope = dp/dt = (650-50)/(14-0) = 42.8 (dp/dt is rate of change in population)
thus, 5years from now will be 50+43*5 = 50+215 = 265,000
Should be A..
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The population of Jonasburgh will grow by r% per year for 30 Jul 2006, 23:41
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ps_dahiya
A

130 = (1+r/100)^14
we can approx140 = (1+r/100)^15

(140)^(1/3) = (1+r/100)^5

So Pop after 5 years = 50000 * (140)^(1/3) = 50000 * 5.3 approx
= 265,000

280,000 is closest. Also, for sure it can not be less than 250,000.
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Nice approximation ;)
If i stop at the bold part, i use other approximation method:
130~ 128= 2^7 ----> (1+r/100) ^14= 2^7 ---> A=(1+r/100)=sqrt2
Population after 5 years = 50,000* A^5= 50,000* (sqrt2)^5= 200,000*sqrt2 . As we know sqrt2 ~ 1.4 ---> this approximate 280,000
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The population of Jonasburgh will grow by r% per year for 31 Jul 2006, 02:12
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kevincan
freetheking
kevincan
The population of Jonasburgh will grow by r% per year for the next 20 years. If the population of Jonasburgh is 50,000 today and is expected to be 6,500,000 in 14 years' time, which of the following best approximates what Jonasburgh's population will be five years from now?

(A) 280,000 (B) 500,000 (C) 560,000 (D) 2,000,000 (E) 2,400,000

Using linear approximation..
slope = dp/dt = (650???-50)/(14-0) = 42.8 (dp/dt is rate of change in population)
thus, 5years from now will be 50+43*5 = 50+215 = 265,000
Should be A..
Show more


He just didnot use 000 ... and added these zeros at the end:)
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The population of Jonasburgh will grow by r% per year for 31 Jul 2006, 02:58
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650000 = 50000(1+r/100) ^14

130 = (1+r/100)^14

(130)^1/14 = 1+r/100

Fives years from now

P = 50000(1+r/100)^5
P = 50000(130)^5/14
~ 50000(130)^1/3

We know 5^3 = 125 Hence 130^1/3 is > 5

P ~ 50000 * 5 = 250000
Closest choice is A
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The population of Jonasburgh will grow by r% per year for 31 Jul 2006, 21:56
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I basically made mistake while finding easiest way to solve this..
I assumed that r is not too big..

I guess linear approximation can't be used to those high power approximation.
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The population of Jonasburgh will grow by r% per year for 31 Jul 2006, 22:35
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great work by laxi. i would do using jaynayak's approach.

laxieqv
Nice approximation ;)
If i stop at the bold part, i use other approximation method:
130~ 128= 2^7 ----> (1+r/100) ^14= 2^7 ---> A=(1+r/100)=sqrt2
Population after 5 years = 50,000* A^5= 50,000* (sqrt2)^5= 200,000*sqrt2 . As we know sqrt2 ~ 1.4 ---> this approximate 280,000
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The population of Jonasburgh will grow by r% per year for 31 Jul 2006, 22:52
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Professor buddy, welcome you back!!!!! :-D . You've been shining in our forum!!! ^.^
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The population of Jonasburgh will grow by r% per year for 05 Aug 2006, 21:09
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since it is a linear increase, we can take
y-y1 = m(x-x1)
650-50=m(14-0)
m=600/14

Plugging into the same equation, y-50=600/14(5)
y=265
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The population of Jonasburgh will grow by r% per year for 05 Aug 2006, 22:33
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kevincan
The population of Jonasburgh will grow by r% per year for the next 20 years. If the population of Jonasburgh is 50,000 today and is expected to be 6,500,000 in 14 years' time, which of the following best approximates what Jonasburgh's population will be five years from now?

(A) 280,000 (B) 500,000 (C) 560,000 (D) 2,000,000 (E) 2,400,000
Show more


(50,000^14)(1+r/100)^14 =6,500,000
(1+r/100) = (6500,000)^1/14/(50,000)

stumped :roll:

Heman



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