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

Question

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)

Kudos for a correct solution.

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

Skywalker18 · Accepted Answer

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

BrentGMATPrepNow · Answer

This is a great candidate for creating a   growth table

ELAPSED TIME (in months)  | NUMBER OF ANIMALS  
 0  |  3  
 1  | ( 3 ) (2)  animals   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

lipsi18 · Answer

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

v12345 · Answer

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

BPHASDEU · Answer

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

EMPOWERgmatRichC · Answer

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:  D

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

ScottTargetTestPrep · Answer

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

Shiv2016 · Answer

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

SIDDHARTH1901 · Answer

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

GMATinsight · Answer

Answer: Option D

Video solution by  GMATinsight

https://www.youtube.com/watch?v=eGq9twKdPzA

avigutman · Answer

Video solution from Quant Reasoning: Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=-dWmh4gmEIM

NiksM · Answer

I understand the general way of solution. However as it's the arithmetic progression with common ratio r=2, how to apply the AP sum formula : (First Term)x (r^n-1 / r-1) ? Because with this formula, and is 3(2^9). What am I missing ?

Posted from my mobile device

EMPOWERgmatRichC · Answer

Hi NiksM, The prompt tells us that there are 3 animals at the START of the experiment - and that the population doubles at the end of each month for 10 months. This means that you have to double the population 10 times. Your calculation assumes that "3" is the value of the 'first term' in the sequence, but 3 is the 'zero term' - so the 'first term' would actually be 3(2) = 6. GMAT assassins aren't born, they're made, Rich Contact Rich at: Rich.C@empowergmat.com

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards