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A certain bacteria colony doubles in size every day for 20

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gmatgeek
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A certain bacteria colony doubles in size every day for 20 [#permalink] New post   Updated on: 13 Dec 2012, 02:50
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A certain bacteria colony doubles in size every day for 20 days, at which point it reaches the limit of its habitat and can no longer grow. If two bacteria colonies start growing simultaneously, how many days will it take them to reach the habitat’s limit?

A. 6.33
B. 7.5
C. 10
D. 15
E. 19

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E

Originally posted by gmatgeek on 12 Dec 2012, 13:42.
Last edited by Bunuel on 13 Dec 2012, 02:50, edited 1 time in total.
Renamed the topic and edited the question.
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Re: A certain bacteria colony doubles in size every day for 20 [#permalink] New post   12 Dec 2012, 21:17
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gmatgeek wrote:
A certain bacteria colony doubles in size every day
for 20 days, at which point it reaches the limit of its
habitat and can no longer grow. If two bacteria
colonies start growing simultaneously, how many
days will it take them to reach the habitat’s limit?
 6.33
 7.5
 10
 15
 19


Say the colony has x bacteria in the beginning.
1st day - 2x bacteria
2nd day - \(2^2x\) bacteria
3rd day - \(2^3x\) bacteria
.
20th day - \(2^{20}x\) bacteria

If we start with 2 bacteria colonies, it means there are already 2x bacteria (equal to number of bacteria at the end of the first day in the first case). In how many more days will they reach \(2^{20}x\) ? Another 19 days!
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Re: A certain bacteria colony doubles in size every day for 20 [#permalink] New post   15 Dec 2012, 06:34
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I am not sure whether ny nethod works for everybody but this is how i did.

So let say our limit is 5 days (just for easiness of calculation) and we have one collony with 1 bacterium in the first day, so in 5 days we have 1+2+4+8+16=31 bacteria, but if we have two colonies, then in order to fill the max capacity of 31 bacteria we need overall 4 days (1+2+4+8=15 and 1+2+4+8=15).

The same principle applies for 20 days, then we need 19 days to fill the max capacity with two colonies.
E is the answer.

Hope thats helpful!
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Re: A certain bacteria colony doubles in size every day for 20 [#permalink] New post   12 Dec 2012, 13:55
samsonfred76 wrote:
simultaneous grow = same amount of time

In 20 days we should expect to have P(2)^20

So if we have two things doing the job of getting us there then;

2^x + 2^x = 20

2(2)^x=2
2^(x+1)=2^20

x+1 = 20

x = 19

We do not need additional information to answer this.
More often than not, if you start thinking the question is wrong chances are you are wrong!


If the colony doubles every day then on 20th day the size should be P 2^19 and not P 2^20..right ?
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Re: A certain bacteria colony doubles in size every day for 20 [#permalink] New post   12 Dec 2012, 15:09
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Let the size at the colony at the start of 20 days be = 1
At the end of the first day it would be 2...At the end of the second day it would be 2(2) and the day after that (2)(2)(2)...You can see then that following the same pattern at the end of 20 days you have (2)^20 since the population doubles...
Hope this helps...
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Re: A certain bacteria colony doubles in size every day for 20 [#permalink] New post   12 Dec 2012, 21:10
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Let a be the number of bacteria in the colony at start of day 1.

a * 2^(20) = 2a * 2^(x)
=> x = 20 -1 = 19

Option E
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Re: A certain bacteria colony doubles in size every day for 20 [#permalink] New post   12 Dec 2012, 21:24
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Minimal math answer.

If the bacteria doubles its size everyday and takes 20 days to fill up its habitat, then the previous day it filled up half the habitat.

So if it takes 1 bacteria 19 days to fill up half its habitat, it takes 2 bacteria 19 days to fill up the habitat.
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Re: A certain bacteria colony doubles in size every day for 20 [#permalink] New post   13 Dec 2012, 02:58
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gmatgeek wrote:
A certain bacteria colony doubles in size every day for 20 days, at which point it reaches the limit of its habitat and can no longer grow. If two bacteria colonies start growing simultaneously, how many days will it take them to reach the habitat’s limit?

A. 6.33
B. 7.5
C. 10
D. 15
E. 19


If there is one bacteria colony, then it will reach the limit of its habitat in 20 days.

If there are two bacteria colonies, then in order to reach the limit of habitat they would need to double one time less than in case with one colony. Thus colonies need to double 19 times.

Answer: E.

Similar questions to practice:
it-takes-30-days-to-fill-a-laboratory-dish-with-bacteria-140269.html
the-number-of-water-lilies-on-a-certain-lake-doubles-every-142858.html
a-certain-culture-of-bacteria-quadruples-every-hour-if-a-52258.html

Hope it helps.
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Re: A certain bacteria colony doubles in size every day for 20 [#permalink] New post   16 Dec 2012, 04:21
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If there is one bacteria colony, then it will reach the limit of its habitat in 20 days which is 2^19

If there are two bacteria colonies growing simultaneously, then they would multiply as 2, 2^2, 2^3 in order to reach the limit of habitat they would need one day less than normal which is 19 days. (E).
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Re: A certain bacteria colony doubles in size every day for 20 [#permalink] New post   16 Dec 2012, 18:24
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Two colonies double every day, the given limit value is 2^20 (the asked growth, x - number of days (doublings)):

2(2^x)=2^20

Manipulate the exponent:
2^(x+1)=2^20

Solve for x:
x=19

The answer is E.
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Re: A certain bacteria colony doubles in size every day for 20 [#permalink] New post   20 Jan 2014, 06:27
Let the final amount of bacteria or the MAX size of the habitat on day 20 be 2x. With single colony doubling every day, the quantity of bacteria on day 19 will be 2x/2 which is x. Now we have another colony growing at the same pace, its size on day 19 is also going to be x. Therefore on day 19 colony 1 size is x and colony 2 size is also x. So total size on day 19 will be 2x. Which is the MAX size of the habitat. Hence E is the answer.

PS: The question seems a little ambiguous because we have to assume each bacteria colony starts with same amount in the beginning.
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Re: A certain bacteria colony doubles in size every day for 20 [#permalink] New post   27 Feb 2015, 11:06
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20th day = 100%
19th day = 50% (coz it doubles every day)
so two colonies will take 19 days i.e 50% + 50% simultaneously. :)

Posted from my mobile device
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Re: A certain bacteria colony doubles in size every day for 20 [#permalink] New post   08 Apr 2016, 20:25
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Attached is a visual that should help. Technically the equation for "doubles in size x times" should be \(n(2^x)\) because we do not know that the colony is necessarily doubling in integer form, i.e, 1,2,4,8 etc. But it's easier to eliminate that part since "n" would be on both sides of the equation, and we are not given a specific starting number so we might as well use the integer powers of two.
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Screen Shot 2016-04-08 at 8.24.37 PM.png
Screen Shot 2016-04-08 at 8.24.37 PM.png [ 72.01 KiB | Viewed 23173 times ]

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Re: A certain bacteria colony doubles in size every day for 20 [#permalink] New post   15 Oct 2016, 12:56
I found this formula to be easy to apply.

Final population growth = S * P ^ (t/l)
S = starting population
P = progression (doubles = 2, triples = 3 etc.)
t/l = total amount of iterations
t = time
I = intervals

Let x be final population or habitat limit = > x = 1 bacteria * 2 ^(20 days/1 day) which is x = 2 ^20
if we have 2 bacterias, then each has to fill in half of the habitat => x/2 = 2^20 / 2 = > x/2 = 2 ^19 - so it will take 19 days for each bacteria to fill in half of the habitat, which mens that together they will reach habitat's limit in 19 days
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Re: A certain bacteria colony doubles in size every day for 20 [#permalink] New post   25 Mar 2018, 08:14
gmatgeek wrote:
A certain bacteria colony doubles in size every day for 20 days, at which point it reaches the limit of its habitat and can no longer grow. If two bacteria colonies start growing simultaneously, how many days will it take them to reach the habitat’s limit?

A. 6.33
B. 7.5
C. 10
D. 15
E. 19



I think the wording of the question is wrong here, ...first sentence talks about one bacteria colony, and second sentence talks about two bacteria colonies start growing simultaneously ...which really confuses me :) it looks like two colonies start growing in one space (room) lol so logically it should have been 10 days :)
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Re: A certain bacteria colony doubles in size every day for 20 [#permalink] New post   10 Aug 2019, 08:25
More than math it is simple logic... If a bacteria doubles daily for 20 days.. 2nd day there will be double the bacteria, so it completes a habitat in 20 days, it completes the habitat half in 19 days(As on the 20th day the amount of bacteria will double itself to complete the habitat) So if there happens to be two bacteria, they will cover the habit on 19th day itself.
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Re: A certain bacteria colony doubles in size every day for 20 [#permalink] New post   19 Apr 2020, 10:19
gmatgeek wrote:
A certain bacteria colony doubles in size every day for 20 days, at which point it reaches the limit of its habitat and can no longer grow. If two bacteria colonies start growing simultaneously, how many days will it take them to reach the habitat’s limit?

A. 6.33
B. 7.5
C. 10
D. 15
E. 19


**Solution**
We know 2^n is the growth rate of bacteria for one bacteria colony and once the growth reaches 2^20 threshold, done no more growth, does not matter how many bacteria colonies are.

There are two colonies (assuming equal initial population) (2^n)+ (2^n)=2^20 (threshold)
It implies 2*(2^n)=2^20
or 2^n=(2^20)/2
therefore 2^n=(2^19)
n=19 Answer
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Re: A certain bacteria colony doubles in size every day for 20 [#permalink] New post   09 Aug 2020, 06:23
BOTH METHODS FINE
A) Suppose the max limit is 65536000,32768000,16384000,8192000,4096000,2048000,1024000,512000,256000,12800,6400,3200,1600,800,400,200,100,50,25
which is achieved in 20 days with one bacteria colony. Now two colonies are simultaneously raised from 25 now at day 19 the size of each colony is 32768000 which sums up to 65536000.

B) Using G.P
T20th=a2^20-1====>a.2^19
T19th=a2^19-1====>a.2^18* 2( as two colonies are raised) which is equal to above a.2^19
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Re: A certain bacteria colony doubles in size every day for 20 [#permalink] New post   25 May 2021, 15:39
BPHASDEU wrote:
I found this formula to be easy to apply.

Final population growth = S * P ^ (t/l)
S = starting population
P = progression (doubles = 2, triples = 3 etc.)
t/l = total amount of iterations
t = time
I = intervals

Let x be final population or habitat limit = > x = 1 bacteria * 2 ^(20 days/1 day) which is x = 2 ^20
if we have 2 bacterias, then each has to fill in half of the habitat => x/2 = 2^20 / 2 = > x/2 = 2 ^19 - so it will take 19 days for each bacteria to fill in half of the habitat, which mens that together they will reach habitat's limit in 19 days


Are we assuming that the initial no. of bacteria is 1 and final is x?
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