A certain bacteria colony doubles in size every day for 20

Question

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

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

KarishmaB · Accepted Answer

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!

ziko · Answer

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!

gmatgeek · Answer

If the colony doubles every day then on 20th day the size should be P 2^19 and not P 2^20..right ?

samsonfred76 · Answer

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...

GyanOne · Answer

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

Abetron · Answer

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.

Bunuel · Answer

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.

priyamne · Answer

Ans:
 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).

HumptyDumpty · Answer

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.

gmatprav · Answer

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.

PS: The question seems a little ambiguous because we have to assume each bacteria colony starts with same amount in the beginning.

shubh09 · Answer

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

mcelroytutoring · Answer

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.

Attachments

Screen Shot 2016-04-08 at 8.24.37 PM.png [ 72.01 KiB | Viewed 28818 times ]

BPHASDEU · Answer

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

dave13 · Answer

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

abhisheksinha17 · Answer

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.

jkrishh7 · Answer

**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

varun325 · Answer

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

2021gre2021 · Answer

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

simplydecided720 · Answer

My approach was:
X (2^20) = X(2^a) + X(2^a)
X (2^20) = 2X(2^a)
2^20 = 2^a+1
20 = a+1
a=19

A certain bacteria colony doubles in size every day for 20

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A certain bacteria colony doubles in size every day for 20

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