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GMAT Instructor
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All scientists expect the population of penguins on the [#permalink]
 09 Jul 2006, 02:30
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All scientists expect the population of penguins on the island of Zeeberg is expected to rise by from 50,000 today to 800,000 18 years from now. Scientist A holds that the population will follow an arithmetic progression, that is that the absolute increase in population from year to year will be constant. Scientist B says that it will follow a geometric progression, which is to say that the percentage increase in population from year to year will be constant. If the two scientists are asked to predict the population of penguins in 9 years’ time, by how much will their predictions differ?
(A) 150,000 (B) 225,000 (C) 300,000 (D) 425,000 (E) none of these
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Manager
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For AP
Tn = a + (n-1)*d
T18 = 800,000 = 50,000 + (18-1) * d
=> d = 750,000/17
For GP
Tn = a * r ^ (n-1)
T18 = 800,000 = 50,000 * r ^ (18-1)
=> r^17 = 16 => r = 16 ^ 1/17
Now calculating the 9th term for the AP and GP
For AP
T9 = a*r^8 = 50,000 * 16 ^ (8/17) ~ 50,000 * 16^(1/2) = 200,000
For GP
T9 = a + 8d = 50,000+8*(750,000/17)~ 50,000 + (750,000/2) = 420,000
The difference between to T9s
= 420,000 - 200,000 = 220,000
Hence the closest answer is 225,000.
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Manager
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grepro wrote:
Now calculating the 9th term for the AP and GP
For GP
T9 = a*r^8 = 50,000 * 16 ^ (8/17) ~ 50,000 * 16^(1/2) = 200,000
For AP
T9 = a + 8d = 50,000+8*(750,000/17)~ 50,000 + (750,000/2) = 420,000
The difference between to T9s
= 420,000 - 200,000 = 220,000
Hence the closest answer is 225,000.
I follow your methodology until calculating for GP
How does 16 ^ (8/17) become 16 ^ (1/2)
It must be my weakness with exponent properties coming through here...
Your thoughts please...
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GMAT Instructor
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grepro wrote: For AP
Tn = a + (n-1)*d
T18 = 800,000 = 50,000 + (18-1) * d
=> d = 750,000/17
For GP
Tn = a * r ^ (n-1)
T18 = 800,000 = 50,000 * r ^ (18-1)
=> r^17 = 16 => r = 16 ^ 1/17
Now calculating the 9th term for the AP and GP
For AP
T9 = a*r^8 = 50,000 * 16 ^ (8/17) ~ 50,000 * 16^(1/2) = 200,000
For GP
T9 = a + 8d = 50,000+8*(750,000/17)~ 50,000 + (750,000/2) = 420,000
The difference between to T9s
= 420,000 - 200,000 = 220,000
Hence the closest answer is 225,000.
I assume you are saying that T1 is the value now, at the beginning of the first year. Then T 19=800,000
so r^18=16=4^2 and r^9=4
d=750,000/18 and 9d=375,000
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Manager
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I thought that Geometric Progression is out of scope of GMAt. Am I wrong? Should I put it back on my "cheat-sheet"?
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CEO
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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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The total terms are 19 not 18.
Using this I got 10th term (i.e after 9 years) of AP = 375,000
The 10th term for GP is = 200,000
Difference = 275,000
So answer should be E.
_________________
SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008
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