M06-18

Question

The population of a bacteria colony doubles every day. If it was started 9 days ago with 2 bacteria and each bacteria lives for 12 days, how large is the colony today?

A. 512
B. 768
C. 1024
D. 2048
E. 4096

A. 512
B. 768
C. 1024
D. 2048
E. 4096

Bunuel · Accepted Answer

Official Solution:

The population of a bacteria colony doubles every day. If it was started 9 days ago with 2 bacteria and each bacteria lives for 12 days, how large is the colony today?

A. 512
B. 768
C. 1024
D. 2048
E. 4096

A. 512
B. 768
C. 1024
D. 2048
E. 4096

We know that the population doubles every day. If the population started ONE day ago (that is, yesterday), it would have grown to  2^2  today. If the population started  N  days ago, then today it would be  2^{(N+1)} . The provided information about the lifespan of each bacteria (12 days) indicates that on the 10th day, all bacteria are still alive. Therefore, given that the population started 9 days ago, today the population size is  2^{(9+1)} , which equals 1024.

Answer: C

notMD · Answer

The tricky part is "it was started 9 days ago" which means they are on day 10.

Thoughtosphere · Answer

This question was a sitter, still i got this wrong not because I didn't know the concepts well, but because I did a silly mistake. Since the information provided is of 9 days ago, I calculated the answer for 9 days itself. The calculation should be done for n+1 days.

It started with 2 bacteria, thus today, the number of bacteria will be  2^10   =1024

The other information about the lives of bacteria is redundant, this information would have been useful, had the question asked for more than 11 days.

Ans - 1024 
Choice C

samarthpawar · Answer

I got the correct answer, but I did it differently. I did not use the N+1 concept as I believe this could easily mean we are on the 9th day.

Let the population be : p
Rate of population expansion : 2^(N)

Therefore, the population after N days = p*2^(N).

In the question it is given p=2 and N =9. We plug it in and get 2*2^(9) = 2^(10) = 1024.

the logic of 2^(N+1) will not work if we started with a population of 100 and the population doubled everyday.

Let me know if I missed something here.

Sash143 · Answer

I solved the problem in this way- 
TIME ELAPSED | BACTERIA 
9 days ago ------ 2
8 days ago ------ 4
7 days ago ------ 8
6 days ago ------ 16
5 days ago ------ 32 
4 days ago ------ 64
3 days ago ------ 128
2 days ago ------ 256
1 day ago ------ 512 
NOW ------ 1024

Praveenksinha · Answer

got it wrong too, made same mistake 9 day ago means 10 days from now, good question
Thanks

san10789 · Answer

1 days ago means today is 2nd day
2 days ago means today is 3rd day
9 days ago means today is 10th day

2^10 = 1024

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

BottomJee · Answer

I think this is a  high-quality  question and I agree with explanation.

kop18 · Answer

Hi  Bunuel ,

I have a quick question - 
From this portion  "The population of a bacteria colony doubles every day. If the colony was started 9 days ago with 2 live bacteria"  why can't I infer that the day the colony started the no of bacteria doubled and hence the very first day we had 2*2= 4 bacteria and by the end of the 10th day we'd have 2048   where N is the no of times the population doubled i.e, N= 10?

Please can you or any other expert  KarishmaB  ,  AndrewN ,  chetan2u  help me out with the wording of this question?

KarishmaB · Answer

In questions involving "days ago," think in terms of an exact moment in time. In questions with "years ago" think of an exact day etc.

When I read "9 days ago," I say to myself, ok it all started at 6:00 AM on 1st August. At that time there were 2 live bacteria. 
So by 6:00 AM on 2nd August, there were 4 (= 2^2) live bacteria (1 day has passed)
So by 6:00 AM on 3rd August, there were 8 (= 2^3) live bacteria (2 days have passed)
...
For 9 days to have passed, it must now be 6:00 AM on 10th August and there must be 2^10 = 1024 live bacteria.

Answer (C)

kop18 · Answer

Thanks for response Karishma - just to add to that thought if it were explicitly stated that the experiment starts at 12.00am on Aug 1 with 2 live bacteria and the population doubles every day hence, by 12.00 am Aug 2 (24 hours later) we would have 2*2 live bacteria on the first day itself. 
Therefore, 2^11=2048 live bacteria at the end of the experiment is that correct?

WarasSingh · Answer

It can also be solved this way. Found it easy to visualize.

Bacteria doubles; thus Rate = 100%. Time period = 9 days.

So today value will be: 2(1+100%)^9
 = 2(2)^9 
 = 2^10.

Fr12milgap · Answer