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# For each of the last 5 years, the population of a colony of beetles in

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Math Expert
Joined: 02 Sep 2009
Posts: 59572
For each of the last 5 years, the population of a colony of beetles in  [#permalink]

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22 Nov 2019, 03:33
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Difficulty:

25% (medium)

Question Stats:

79% (01:07) correct 21% (01:30) wrong based on 24 sessions

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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. $$(1.08)^{-5}P^5$$
Intern
Joined: 24 Sep 2019
Posts: 25
Re: For each of the last 5 years, the population of a colony of beetles in  [#permalink]

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22 Nov 2019, 09:32
first to second year: 1.08P
second to third year: 1.08x1.08xP = (1.08)^2P
....... AND SO ON.

SO TO GO BACK 5 Years, exponent P will be negative.... Answer (D)
Director
Joined: 05 Mar 2015
Posts: 985
Re: For each of the last 5 years, the population of a colony of beetles in  [#permalink]

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22 Nov 2019, 13:14
Bunuel 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. $$(1.08)^{-5}P^5$$

let x be population 5 years back ..then
x(1+8/100)^5 = P
x(1.08)^5 = P
x = P/(1.08)^5
x = P*(1.08)^-5

Ans D
Re: For each of the last 5 years, the population of a colony of beetles in   [#permalink] 22 Nov 2019, 13:14
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# For each of the last 5 years, the population of a colony of beetles in

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