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The population of locusts in a certain swarm doubles every

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The population of locusts in a certain swarm doubles every [#permalink] New post 12 Feb 2010, 22:33
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The population of locusts in a certain swarm doubles every two hours. If 4 hours ago there were 1,000 locusts in the swarm, in approximately how many hours will the swarm population exceed 250,000 locusts?

A. 6
B. 8
C. 10
D. 12
E. 14

[Reveal] Spoiler:
OA is D. I got this correct by using population chart, ie. multiplying every 2 years. Is there a formula I could apply?
[Reveal] Spoiler: OA

Last edited by joyseychow on 16 Feb 2010, 06:32, edited 1 time in total.
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Re: Population Growth-locust [#permalink] New post 12 Feb 2010, 22:43
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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 > 62.5since 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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Re: Population Growth-locust [#permalink] New post 12 Feb 2010, 23:11
joyseychow wrote:
The population of locusts in a certain swarm doubles every two hours. If 4 hours ago there were 1,000 locusts in the swarm, in approximately how many hours will the swarm population exceed 250,000 locusts?
A) 6
B) 8
C) 10
D) 12
E) 14

[Reveal] Spoiler:
OA is E. I got this correct by using population chart, ie. multiplying every 2 years. Is there a formula I could apply?


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
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Re: Population Growth-locust [#permalink] New post 03 Jun 2012, 22:30
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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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Re: The population of locusts in a certain swarm doubles every [#permalink] New post 05 Sep 2012, 10:43
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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
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Re: The population of locusts in a certain swarm doubles every [#permalink] New post 21 Oct 2012, 13:08
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...
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Re: The population of locusts in a certain swarm doubles every [#permalink] New post 21 Oct 2012, 21:52
himanshuhpr wrote:
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.
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Re: The population of locusts in a certain swarm doubles every [#permalink] New post 22 Oct 2012, 01:26
fameatop wrote:
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


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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Re: The population of locusts in a certain swarm doubles every [#permalink] New post 22 Oct 2012, 03:21
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himanshuhpr wrote:
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...


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 nth place. Between the first term and the nth 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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Last edited by EvaJager on 22 Oct 2012, 05:56, edited 2 times in total.
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Re: The population of locusts in a certain swarm doubles every [#permalink] New post 22 Oct 2012, 05:36
EvaJager wrote:
himanshuhpr wrote:
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...


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 nth place. Between the first term and the nth 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 case we talk about 6 and 7, the difference is whether you called 6 n or n-1.



Thank you !!! for the very helpful explanation ..
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Re: Population Growth-locust [#permalink] New post 30 Oct 2013, 10:39
gurpreetsingh wrote:
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 > 62.5since n must be an integer... thus take

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

Thus D



4 hours ago: 1,000
2 hours ago: 2,000
Now: 4,000
In 2 hours: 8,000
in 4 hours: 16,000
in 6 hours: 32,000
in 8 hours: 64,000
in 10 hours: 128,000
in 12 hours: 256,000
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Re: The population of locusts in a certain swarm doubles every [#permalink] New post 30 Oct 2013, 23:51
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joyseychow wrote:
The population of locusts in a certain swarm doubles every two hours. If 4 hours ago there were 1,000 locusts in the swarm, in approximately how many hours will the swarm population exceed 250,000 locusts?

A. 6
B. 8
C. 10
D. 12
E. 14

[Reveal] Spoiler:
OA is D. I got this correct by using population chart, ie. multiplying every 2 years. Is there a formula I could apply?


Similar questions to practice:
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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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Re: The population of locusts in a certain swarm doubles every [#permalink] New post 12 Sep 2014, 05:27
Bunuel wrote:
joyseychow wrote:
The population of locusts in a certain swarm doubles every two hours. If 4 hours ago there were 1,000 locusts in the swarm, in approximately how many hours will the swarm population exceed 250,000 locusts?

A. 6
B. 8
C. 10
D. 12
E. 14

[Reveal] Spoiler:
OA is D. I got this correct by using population chart, ie. multiplying every 2 years. Is there a formula I could apply?


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.



Hi Bunuel,

Can you help me solving this problem with this formula --> [final population] = [initial population] (1 + r)^t
Re: The population of locusts in a certain swarm doubles every   [#permalink] 12 Sep 2014, 05:27
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