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The number of antelope in a certain herd increases every

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Intern
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The number of antelope in a certain herd increases every  [#permalink]

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New post 09 Aug 2009, 09:50
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The number of antelope in a certain herd increases every year at a constant rate. If there are 500 antelope in the herd today, how many years will it take for the number of antelope to double?

(1) Ten years from now, there will be more than ten times the current number of antelope in the herd.

(2) If the herd were to grow in number at twice its current rate, there would be 980 antelope in the group in two years.

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Re: Good One!  [#permalink]

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New post 09 Aug 2009, 12:09
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IMO B

if r is the groth rate. after 1 years we will have 500(1+r) and after n years, we will have 500(1+r)^n

the question asks, for which n will we have 1000? (double 500)

to find n we need to know r

statement 1: for n=10, tne number will be graeter than 500x10, we will just have a min for r . INSUFF

Statement 2: if we considered a groth rate p=2r, we would have 980=500(1+P)². we can solve this to find p, then r=p/2 and plug this value in the first eq to find n. SUFF

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Re: Good One!  [#permalink]

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New post 09 Aug 2009, 18:19
madeinafrica wrote:
IMO B

if r is the groth rate. after 1 years we will have 500(1+r) and after n years, we will have 500(1+r)^n



can you explain how it is (1+r) after 1year ?
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Re: Good One!  [#permalink]

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New post 09 Aug 2009, 22:02
crejoc wrote:
madeinafrica wrote:
IMO B

if r is the groth rate. after 1 years we will have 500(1+r) and after n years, we will have 500(1+r)^n



can you explain how it is (1+r) after 1year ?


This is the same as a compound interest formula: A=P(1+R)^n
A-total amount
P-initial amount
R-rate
N-number of years

hope it helps
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Re: Good One!  [#permalink]

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New post 10 Aug 2009, 10:54
Answer is B and number of years after which the herd doubles = log2/log(6/5)

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Re: Good One! &nbs [#permalink] 10 Aug 2009, 10:54
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