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For each of the last 5 years, the population of a colony of

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For each of the last 5 years, the population of a colony of [#permalink] New post 07 Nov 2019, 10:40
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For each of the last 5 years, the population of a colony of beetles increased by 8 percent of the preceding year's population. If P represents the current population of the colony, which of the following best represents the population 5 years ago, in terms of P?


A. \((5)(1.08 P^{-1})\)

B. \((1.08)^{-5}P^{-1}\)

C. \((1.08P)^{-5}\)

D. \((1.08)^{-5}P\)

E. \((1.08)^{-5}P^5\)
[Reveal] Spoiler: OA

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Re: For each of the last 5 years, the population of a colony of [#permalink] New post 08 Nov 2019, 06:44
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Carcass wrote:
For each of the last 5 years, the population of a colony of beetles increased by 8 percent of the preceding year's population. If P represents the current population of the colony, which of the following best represents the population 5 years ago, in terms of P?


A. \((5)(1.08 P^{-1})\)

B. \((1.08)^{-5}P^{-1}\)

C. \((1.08P)^{-5}\)

D. \((1.08)^{-5}P\)

E. \((1.08)^{-5}P^5\)


Here,

Let the population of beetles \(5\) yrs ago = \(x\)

Each year there is an increase of \(8\)%

that means in \(5\) yrs, total population = \(x * (1.08)^5\) = Current age = \(P\)

Therefore \(x = (1.08)^{-5}P\)
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Re: For each of the last 5 years, the population of a colony of [#permalink] New post 14 Nov 2019, 02:35
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HI,
I used the compound interest formula to do this question.
Let the, Population 5years ago be the principle- x
and Population now be- P, while time given is 5 years

SINCE,C.I = P(1+r%)^t

we will come down to
x(1+8%)^5=P
isolating x we get,
x=P(1.08)^-5

Please shoot a kudos if you like the solution.
Thank you in advance!
Re: For each of the last 5 years, the population of a colony of   [#permalink] 14 Nov 2019, 02:35
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