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

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

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New post 22 Nov 2019, 03:33
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A
B
C
D
E

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  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. \((1.08)^{-5}P^5\)
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Re: For each of the last 5 years, the population of a colony of beetles in  [#permalink]

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New post 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)
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Re: For each of the last 5 years, the population of a colony of beetles in  [#permalink]

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New post 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
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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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