Similar questions to practice:
a-certain-bacteria-colony-doubles-in-size-every-day-for-144013.html
it-takes-30-days-to-fill-a-laboratory-dish-with-bacteria-140269.html
a-certain-culture-of-bacteria-quadruples-every-hour-if-a-52258.html
the-number-of-water-lilies-on-a-certain-lake-doubles-every-142858.html
the-population-of-grasshoppers-doubles-in-a-particular-field-160081.html

Hope this helps.
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IMO D, OA pls

4 hours ago = 1000
2 hours ago = 2000
right now = 4000

now the given equation says

\(4000*2^n > 250000\)

=> \(2^n > 62.5\)since n must be an integer... thus take

\(2^n = 64 = 2^6\) thus it got doubled 6 times thus 12 hours

Thus D
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n - number of 2 hours from now to cross 250000

so 2^(n+2)(1000) = 250000
now n must be integer as the possible answer are all integer
2^(n+2) = 256 (nearest 2 factor)
n+ 2 = 8
n = 6
total hours = 12 answer is D

1000*2^(4/2)=4000 (now)


4000*2^t/2=250000

2^t/2=62.5

since 2^6=64, t/2=6
t=12
ans is 12
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Is there a formula I could apply?

Yes there is formula you can apply but be careful while using it.

This question is a very simple application of Geometric progression wherein b/a = c/b = r i.e. the ratio of any two consecutive terms in a series is always constant.
Series in this question is 1 ,2 , 4,8,16,32........250
Nth term of GP is given by = (first term) ((Ratio^n)-1)

So the question is what is the Least value of n so that
250 < (first term) ((Ratio^n)-1)

I am leaving the remaining part so that others can apply this concept on their own.

In case, anyone didn't get the solution, let me know

If we use g.p in this question . And if we consider a=4000 ; r=2 then 4000*2^(n-1) > 250000 --------> 2^n > 125 <------> n=7 . Hence answer is 14 .... Pl. help me find anomaly to above mentioned description...

Hi,

You can refer to gurpreetsingh's post. He has used the same concept, though he has reduced the no of terms.

Nth term of GP is given by = (first term) ((Ratio^n)-1)

The correct formula is (first term)(Ratio^(n-1)) or \(a_n=a_1R^{n-1}\) and not \(a_1(R^n-1)\).
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The confusion comes from the interpretation of the formula:
\(a_1\) is the first term and then \(a_n=a_1R^{n-1}\), which is the term on the \(n\)th place. Between the first term and the \(n\)th term, \(n-1\) multiplications by the ratio \(R\) take place, and this is reflected in the exponent of \(n-1\).

Using the formula, you deduced that if \(a_1=4000\) is the first term, then the 7th term will be greater than 250,000. Between the first population and the 7th one, 6 cycles of 2 hours passed, a total of 12 hours, which is the correct answer.
According to the question, your answer should be \(n-1\) from your formula and not \(n\).
In other posts, \(2^n\) was considered, which means \(n\) represents the number of multiplications by \(n\), and obviously the \((n+1)\)th term is greater than 250,000. In both cases we talk about 6 and 7, the difference is whether you called 6 \(n\) or \(n-1\).
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I also did the same as fourteenstix, starting with adding a time for now and leaving it blank. The stem says that it doubles every 2 hours. We know that 4 hours ago their number was 1000. So, we add - 4 hours and the number 1000. In 2 hours, their number doubles, so we add - 2 hours and the number 2000. Using the same logic, their number is now 2000*2= 4000. We continue by adding hours to now, so now+2, now+4, now+6..... When we reach to + 12 their number is 256000, which is more than 250000, and we cn stop!

- 4 hours: 1,000
-2 hours: 2,000
Now: 4,000
+ 2 hours: 8,000
+ 4 hours: 16,000
+ 6 hours: 32,000
+ 8 hours: 64,000
+ 10 hours: 128,000
+ 12 hours: 256,000

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

In this question, 250,000 = 1000 * 2 ^(t/2) = > 250 = 2 ^ (t/2)
The questions asks in approximately how many hours will the swarm population exceed 250,000 locusts , 256 is 2^8

=> 2^8 = 2 ^(t/2) => 8 = t/2 => t =16...but the whole process started 4 hrs ago, which means that 16-4 = 12 hrs - in 12 hrs the population will exceed 250,000

i am bit confused with this topic can anyone explain the theory behind this problem.
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The question deals with geometric progressions: https://gmatclub.com/forum/math-sequenc ... 01891.html
Check similar topics here: https://gmatclub.com/forum/the-populati ... l#p1286223

Hope it helps.
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Since the population doubles every 2 hours, we get:
4 hours ago --> 1000
2 hours ago --> 2000
Now --> 4000

Formula for exponential change:
Final amount = (original amount) * (multiplier)^(number of changes)

Since the final amount must exceed 250,000, we get:
(original amount) * (multiplier)^(number of changes) > 250,000

Here:
original amount = current amount = 4000
multiplier = 2 (since the population keeps doubling)
x = number of changes

Plugging these values into the blue inequality above, we get::
\(4000 * 2^x > 250,000\)
\(2^x > 62.5\)

Since \(2^6 = 64\), the population must double 6 times.
Since the population doubles every 2 hours, the number of hours required = 2*6 = 12


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