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A cockroach population doubles every 3 days. In 30 days, by

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A cockroach population doubles every 3 days. In 30 days, by [#permalink] New post 02 Jun 2018, 03:39
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A cockroach population doubles every 3 days. In 30 days, by what percent would a cockroach population increase?

(A) 900%
(B) 1,000%
(C) 9,999%
(D) 102,300%
(E) 102,400%
[Reveal] Spoiler: OA

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Re: A cockroach population doubles every 3 days. In 30 days, by [#permalink] New post 10 Jun 2018, 07:04
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sandy wrote:
A cockroach population doubles every 3 days. In 30 days, by what percent would a cockroach population increase?

(A) 900%
(B) 1,000%
(C) 9,999%
(D) 102,300%
(E) 102,400%


Given every 3 days population doubles and this continues for 30 days.

This double increase overall happens for 10 times. (30/3)

Now let's take the initial population is 10.

Now for every 3 days population doubles and total 10 times.

10 - 20 - 40 - 80 - 160 - 320 - 640 - 1280 - 2560 - 5120 - 10240.

Exactly after 10th time i.e. 30 days , population is 10240.

Now take the percentage increase = 10240 - 10 / 10 = 10230 / 10 = 1023 = 102300%.

Option D.
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Re: A cockroach population doubles every 3 days. In 30 days, by [#permalink] New post 11 Jul 2018, 13:36
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Explanation

The percent increase is the difference between the amounts divided by the original, converted to a percent. If the population doubles, mathematically the increase can be written as a
power of 2. In the 30-day interval, if the original population is 1, it will double to 2 after three days —so, 21 represents the population after the first increase, the second increase would then be 22 and so on. Since there are 10 increases, the final population would be 210 or 1,024. Therefore, the difference, 1,024 – 1, is 1,023. Use the percent change formula to calculate percent increase:

Percent Change=\((\frac{Difference}{original} \times 100)\)
Percent Change=\((\frac{1023}{1} \times 100)=102300 \%\)

Note that the new number is 102,400% of the original, but that was not the question asked—the percent increase is 102,300%.
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Re: A cockroach population doubles every 3 days. In 30 days, by [#permalink] New post 17 Sep 2019, 04:55
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sandy wrote:
Explanation

The percent increase is the difference between the amounts divided by the original, converted to a percent. If the population doubles, mathematically the increase can be written as a
power of 2. In the 30-day interval, if the original population is 1, it will double to 2 after three days —so, 21 represents the population after the first increase, the second increase would then be 22 and so on. Since there are 10 increases, the final population would be 210 or 1,024. Therefore, the difference, 1,024 – 1, is 1,023. Use the percent change formula to calculate percent increase:

Percent Change=\((\frac{Difference}{original} \times 100)\)
Percent Change=\((\frac{1023}{1} \times 100)=102300 \%\)

Note that the new number is 102,400% of the original, but that was not the question asked—the percent increase is 102,300%.



FROM 2ND LINE 21 represents the population after the first increase, the second increase would then be 22
This will be: \(2^1\) represents the population after the first increase, the second increase would then be \(2^2\)
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Re: A cockroach population doubles every 3 days. In 30 days, by [#permalink] New post 30 Aug 2020, 09:02
What I am confused by is the fact that if we use the equation of a geometric series, we get the value of the tenth term to be \(512*(\text{the original population})\). This is because we know that \(a_n=a_1*r^{n-1}\). Using this formula, given that r=2, \(a_n=a_1*2^{10-1}=512*a_1\).

How is it that we cannot use the geometric series to find the 10th term in this setting?
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Re: A cockroach population doubles every 3 days. In 30 days, by [#permalink] New post 30 Aug 2020, 09:16
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dazedandconfused wrote:
What I am confused by is the fact that if we use the equation of a geometric series, we get the value of the tenth term to be \(512*(\text{the original population})\). This is because we know that \(a_n=a_1*r^{n-1}\). Using this formula, given that r=2, \(a_n=a_1*2^{10-1}=512*a_1\).

How is it that we cannot use the geometric series to find the 10th term in this setting?




If \(a_1\) represents the INITIAL population, then \(a_2\) represents the population after 1 "doubling"
So, \(a_2=a_1*2^{1}=a_1*(2)\)

Likewise, \(a_3\) represents the population after 2 "doublings"
So, \(a_3=a_1*2^{2}=a_1*(4)\)

Likewise, \(a_4\) represents the population after 3 "doublings"
So, \(a_4=a_1*2^{3}=a_1*(8)\)

In general, \(a_n\) represents the population after n-1 "doublings"

So, \(a_n=a_1*r^{n-1}\)

30/3 = 10
So, the population gets doubled 10 times.
Since \(a_n\) represents the population after n-1 "doublings," we need to find the value of \(a_{11}\)

We get: \(a_{11}=a_1*2^{10}=a_1*1024\)

Cheers,
Brent
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Re: A cockroach population doubles every 3 days. In 30 days, by [#permalink] New post 30 Aug 2020, 09:27
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Hi Brent,

Just re-read your response. This was very very helpful. Thanks so much!
Re: A cockroach population doubles every 3 days. In 30 days, by   [#permalink] 30 Aug 2020, 09:27
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