CareerCup

60 minutes only. As at 59 minutes it will be half filled , so the period of replicating is 1 minute (it can't replicate before that-- similar to 9 month human cycle) so there can;t be any time between 59 and 60. Jar will never be 75% . It will become 100% at 60 minute itself

Question: does that mean that one fly produces two of the same flies after a minute, or that one fly produces another fly after a minute? Does each fly that is produced reproduce?

Assuming that one fly produces another after a minute:
t[n]=total number of flies in the jar after n minutes
t[0]=1
t[1]=2
t[2]=4
..
t[60]=2^60 (when the Jar is 100% full)
t[n]=2^n

The Jar is 75% full when there are 3/4*2^60 flies. 2^n=3/4*2^60 => n ~ 59.584
But n is an integer. In the stage of reproducing, it is possible that a fly will increase in size gradually. Therefore the 75% capacity is achieved somewhere in between 59-60 minute.

at 45 minutes jar will be 75% filled. As there is only 1 fly who is replicating...

59min 35seconds?

what type of question is this?
when every minute the number of flies are getting doubled, How come you gonna have 75% filled jar....

It will be 60 min as rightly said by Tarun

correct answer is given by "anonymous on july 16th,2010"

From anonymous comment on July 16th,
The Jar is 75% full when there are 3/4*2^60 flies. 2^n=3/4*2^60
this is also equal to 3* 2^58 = (2+1) * 2 ^58 = 2^59 + 2^58.
It reaches to 2^59 in 59 mins.
now for it to double it takes 1 minute which means from 2^59 to 2^60 (which is 2^59 for 1 min).
If it takes 1 min for producing 2^59, how much time it will take to become 2^58?
2^58/2^59 = 1/2 min
So total time would be 59.5 mins...

correct me if I am wrong...

As every minute passes, the number of flies doubles. If 100% was filled in time "x", then at time "x-1" minutes, there would have only been 50% of the flies. Considering that the flies just pop out after replication at the end of a minute, I think 75% is never really observed

Yep. Anonymous on 16th July is correct. It follows exponential growth - x(t) = a.b^t/T
a = initial condn. = 1 fly
b = rate of growth = 2 (since it doubles)
T = time it takes from a to b = 1 min.
t = time to reach x(t)
First find 100% for t = 60mins and then solve for 75% with t unknown.

60 minutes.......... at 59th minute, its 50% full, at 60th minute, its 100% full....

Let n be the number of minutes the fly replicates.

In 60 minutes, the jar is full and the total number of flies will be 2 ^ 0 + 2 ^ 1 + .. + 2 ^ n = 2 ^ (n + 1) - 1 where n = 60.

When the jar is 3/4 full after m minutes, the number of of flies is 3/4 * [2 ^ (n + 1) - 1].

We need to find m such that 2 ^ (m + 1) - 1 = 3/4 * [2 ^ (n + 1) - 1] where n = 60

Take a logarithm of base 2 of both sides and simplifies

m + 3 = log3 + (n + 1) => m = 60 + 1 - 3 + log3 = 58 + 1.58 = 59.58 minutes, approximately 59 minutes and 35 seconds

Since fly get double every minute that means if Jar get full in 60 mins it was half full at 59 mins i.e. 50% full, now for the last minute remaining 50% of jar will get full. So it'll take 59.5 mins to fill 25% more of it as 1 min fill it completely.

It will be 60 minutes. In 59 minutes the jar is 50% filled . since it takes 1 minute to replicate it will be (both 75% and 100%) filled in 60th minute

2^t = 75% × 2^60
t = 58 + log3