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# All scientists expect the population of penguins on the

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GMAT Instructor
Joined: 04 Jul 2006
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All scientists expect the population of penguins on the [#permalink]  09 Jul 2006, 01:30
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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 yearsâ€™ time, by how much will their predictions differ?

(A) 150,000 (B) 225,000 (C) 300,000 (D) 425,000 (E) none of these
Manager
Joined: 12 May 2006
Posts: 186
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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.
Manager
Joined: 08 Jul 2006
Posts: 90
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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.

How does 16 ^ (8/17) become 16 ^ (1/2)
It must be my weakness with exponent properties coming through here...

GMAT Instructor
Joined: 04 Jul 2006
Posts: 1268
Followers: 23

Kudos [?]: 148 [0], given: 0

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 T19=800,000

so r^18=16=4^2 and r^9=4
d=750,000/18 and 9d=375,000
Manager
Joined: 26 Jun 2006
Posts: 152
Followers: 1

Kudos [?]: 1 [0], given: 0

I thought that Geometric Progression is out of scope of GMAt. Am I wrong? Should I put it back on my "cheat-sheet"?
CEO
Joined: 20 Nov 2005
Posts: 2913
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 18

Kudos [?]: 110 [0], given: 0

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

_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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