At the start of an experiment, a certain population consisted of 3

At time when the experiment started , Population N = 3
At end of the first month , Population N1= 3x2 = 6
At end of the second month , Population N2= 6x2 = 12 =\(3 * 2^{2}\)

At the end of nth month , Population Nn = \(3 * 2^{n}\)

Answer D when n =10
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This is a great candidate for creating a growth table

ELAPSED TIME (in months) | NUMBER OF ANIMALS
0 | 3
1 | (3)(2) animals [since the population doubles each month]
2 | (3)(2)(2) animals
3 | (3)(2)(2)(2) animals
4 | (3)(2)(2)(2)(2) animals
.
.
.
We can see the pattern and recognize that, at the end of month 10, we'll have the product of one 3 and ten 2's.
.
.
10 | (3)(2)^10

Answer: D

Cheers,
Brent

3*2^n where n is the #of month --> 3*2^10 Answer (D)

My Solution:

At the end of every month the population size doubles, so,

After 10 months the population becomes 3072 (3,6,12,24,48,96,192,384,768,1536,3072)

Then find the prime factors of 3072 i.e 3 (2^10) Option D
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At the beginning of the month the number is 3;
At the end of 1st month number is 3*2
At the end of 2nd month number is 3*2^2
hence at the end of 10th month =3*2^10
Hence answer is D

population at starting of experiment = 3
at the end of 1st month population = 3 * 2
at the end of 10th month population = 3 * 2^10

Answer option D

Common ratio of increase = 2 and doubling started from end of first month, therefore nth term will be provided by \(initialvalue*r^n\), i.e. in this case 3*2^10.
Answer D.

A very simple formula that always helps me with these types of questions:

Final population = S * P ^ (t/i)
S -> starting population
P -> progression (2 - doubles, 3- triples etc.)
t -> time
i -> interval

In our case it is a simple solution, which is : Final population = 3 * 2 ^ (10/1)
so it is 3 * 2^10 -> D

Hi All,

This is a great 'concept' question - if you recognize the concept involved, then you don't have to do any actual math to solve this problem. If you're not sure how the math 'works' however, you can still use a bit of 'brute force' to define the pattern involved and still get the correct answer.

We're told that a population starts with 3 animals and at the end of each month, the population DOUBLES. We're asked for the size of the population at the end of 10 months.

Based on the 'growth' that's defined, we can list out the first few months worth of data for reference....

Start = 3 animals
Mo. 1 = 3(2) = 6 animals
Mo. 2 = 6(2) = 12 animals
Mo. 3 = 12(2) = 24 animals
Etc.

Notice how each month we multiply the prior month by TWO. That is the pattern involved. We start with 3, then multiply by 2 over-and-over. Since there are 10 months, there will be 10 "multiply by 2s", so the correct answer must be....

Final Answer:

GMAT assassins aren't born, they're made,
Rich

\(Final\quad Population\quad =\quad { S\ast P }^{ \left( \frac { t }{ i } \right) }\)

Very useful - I did not recognize the pattern and ended up doubling the values until the 10th month. Then I prime factorized 3072 and when I saw 1024, I knew there was an easier way to solve it (Actually, I had known it from the very beginning but just couldn't figure it out).
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We are given that a certain population consisted of 3 animals and that at the end of each month after the start of the experiment, the population size was double what it had been at the beginning of that month. Thus:

End of Month 1 = 3 x 2 = 3 x 2^1

End of Month 2 = 3 x 2 x 2 = 3 x 2^2

End of Month 3 = 3 x 2 x 2 x 2 = 3 x 2^3

We see that after each month the value increases by a factor of two. Thus:

End of Month 10 = 3 x 2^10

Answer: D
_________________

I think its a GP:
3*2, 3*2*2, 3*2*2,.....................

First term= 6
Common ratio= 2

10th term= first term*common ratio^(n-1)
= 6*2^9
=2*3*2^9
=3*2^10

each month doubled means for each month multiply by 2
so 3*(2*2*2*10times)
3*(2^10)

End of 1st Month population =3*2

End of 2nd Month population =3*2*2

Therefore at the end of 10 th Month it will be
3*(2^10)

Hence D


Sent from my iPhone using GMAT Club Forum

At the start of an experiment, a certain population consisted of 3 animals. At the end of each month after the start of the experiment, the population size was double its size at the beginning of that month. Which of the following represents the population size at the end of 10 months?

(A) 2^3
(B) 3^2
(C) 2(3^10)
(D) 3(2^10)
(E) 3(10^2)

Start of the month: 3
At the end of each month after the start of the experiment, the population size was double its size at the beginning : 6
1st month end = 6
2nd month end = 12
3rd month end = 24

So a= 6 , r= 2 (6*2=12, 12*2=24) so r=2; lets put the formula.

nth term in GP = ar^n-1= 6*2^10-1= (3*2) * 2^9 = 3*2^10 (D)

End of Month 1 = 3 x 2 = 3 x(2^1)

End of Month 2 = 3 x 2 x 2 = 3x(2^2)

End of Month 3 = 3 x 2 x 2 x 2 = 3x(2^3)

We see that after each month the value increases by a factor of two. Thus:

End of Month 10 = 3 x (2^10)

Answer: D
_________________

Posted from my mobile device

Answer: Option D

Video solution by GMATinsight



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month --- start--- end
1-----3-------6
2---6-----12
3---12---24
4---24--48
5---48---96
6---96---192
7--192---384
8---384---768
9----768---1536
10----1536---3072

3072 = 3*1024 ; 3*2^10
option D

Video solution from Quant Reasoning:
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