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Two months from now, the population of a colony of insects in a remote

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Two months from now, the population of a colony of insects in a remote  [#permalink]

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New post 19 Oct 2016, 06:36
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Two months from now, the population of a colony of insects in a remote area will reach 3.2 * 10^4. If the population of the colony doubles every two months, what was the population eight months ago?

a) 3.6 * 10^2
b) 1.0 * 10^3
c) 2.0 * 10^3
d) 1.6 * 10^4
e) 2.6 * 10^4


Please assist with above problem.
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Re: Two months from now, the population of a colony of insects in a remote  [#permalink]

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New post 19 Oct 2016, 07:01
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We have to find population 10 months ago.

The population doubled 5 times in 10 months.

=32*10^3/2^5

=10^3

B


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Re: Two months from now, the population of a colony of insects in a remote  [#permalink]

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New post 19 Oct 2016, 07:15
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alanforde800Maximus wrote:
Two months from now, the population of a colony of insects in a remote area will reach 3.2 * 10^4. If the population of the colony doubles every two months, what was the population eight months ago?

a) 3.6 * 10^2
b) 1.0 * 10^3
c) 2.0 * 10^3
d) 1.6 * 10^4
e) 2.6 * 10^4


Please assist with above problem.


Let's use a growth table to work backwards

If the population DOUBLES every 2 months we travel into the future, we can also say that the population is HALVED every 2 months we travel into the past

2 months from now: population = (3.2)(10^4)
Now: population = (1/2)(3.2)(10^4) = (1.6)(10^4)
2 months ago: population = (1/2)(1.6)(10^4) = (0.8)(10^4)
4 months ago: population = (1/2)(0.8)(10^4) = (0.4)(10^4)
6 months ago: population = (1/2)(0.4)(10^4) = (0.2)(10^4)
8 months ago: population = (1/2)(0.2)(10^4) = (0.1)(10^4)

(0.1)(10^4) isn't among the answer choices, so let's REWRITE it.

(0.1)(10^4) = (1)(10^-1)(10^4)
= (1)(10^3)

Answer:

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Re: Two months from now, the population of a colony of insects in a remote  [#permalink]

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New post 19 Oct 2016, 07:31
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alanforde800Maximus wrote:
Two months from now, the population of a colony of insects in a remote area will reach 3.2 * 10^4. If the population of the colony doubles every two months, what was the population eight months ago?

a) 3.6 * 10^2
b) 1.0 * 10^3
c) 2.0 * 10^3
d) 1.6 * 10^4
e) 2.6 * 10^4


Please assist with above problem.


Draw a table a 10 sec answer
Attachment:
Capture.PNG
Capture.PNG [ 3.68 KiB | Viewed 2011 times ]

So, \(32x = 3.2*10^4\)

Or, \(x = 1.0 * 10^3\)

Hence correct answer will be (B)

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Two months from now, the population of a colony of insects in a remote  [#permalink]

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New post 01 Jul 2017, 15:24
2 months from now \(32 * 10^{3}\)
Now \(16 * 10^{3}\)
2 months ago \(8 * 10^{3}\)
4 months ago \(4 * 10^{3}\)
6 months ago \(2 * 10^{3}\)
8 months ago \(1 * 10^{3}\)

Answer is B
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Re: Two months from now, the population of a colony of insects in a remote  [#permalink]

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New post 24 Jan 2018, 21:12
Hi All,

Since the prompt involves Scientific Notation, you can either adjust each individual calculation (in terms of the Scientific Notation that would result) OR you could think in terms of what all of those 'doubles' would do to the total and do one adjustment right at the end:

We're told that a population doubles EVERY 2 MONTHS. If we go from 8 months ago to 2 months in the future, we would have:

8 months ago: X insects
6 months ago: 2X insects
4 months ago: 4X insects
2 months ago: 8X insects
Now: X insects: 16X
2 months in the future: 32X insects

So, in that time frame, the number of insects becomes 32 times what it started at. We can now divide (3.2 x 10^4) by 32 to figure out what the population was 8 months ago...

(3.2)(10^4)/(32) =
(.1)(10^4) =
(1)(10^3)

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Re: Two months from now, the population of a colony of insects in a remote &nbs [#permalink] 24 Jan 2018, 21:12
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