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For each of the last 5 years, the population of a colony of [#permalink]
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Expert's post 00:00

Question Stats: 100% (01:14) correct 0% (00:00) wrong based on 13 sessions
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]
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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]
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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. Re: For each of the last 5 years, the population of a colony of   [#permalink] 14 Nov 2019, 02:35
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