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The population of locusts in a certain swarm doubles every [#permalink]

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12 Feb 2010, 23:33

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52% (02:02) correct
48% (01:01) wrong based on 474 sessions

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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?

Re: The population of locusts in a certain swarm doubles every [#permalink]

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05 Sep 2012, 11:43

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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 _________________

If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth-Game Theory

If you have any question regarding my post, kindly pm me or else I won't be able to reply

Re: The population of locusts in a certain swarm doubles every [#permalink]

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22 Oct 2012, 04:21

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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 \(n\)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 \(n-1\). _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Last edited by EvaJager on 22 Oct 2012, 06:56, edited 2 times in total.

Re: The population of locusts in a certain swarm doubles every [#permalink]

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31 Oct 2013, 00:51

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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?

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

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

Re: The population of locusts in a certain swarm doubles every [#permalink]

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21 Oct 2012, 14:08

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...

Re: The population of locusts in a certain swarm doubles every [#permalink]

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21 Oct 2012, 22:52

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. _________________

If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth-Game Theory

If you have any question regarding my post, kindly pm me or else I won't be able to reply

Re: The population of locusts in a certain swarm doubles every [#permalink]

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22 Oct 2012, 02:26

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 \(a_1(R^n-1)\). _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: The population of locusts in a certain swarm doubles every [#permalink]

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22 Oct 2012, 06:36

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 \(n\)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 case we talk about 6 and 7, the difference is whether you called 6 \(n\) or \(n-1\).

Re: The population of locusts in a certain swarm doubles every [#permalink]

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12 Sep 2014, 06:27

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?

The population of locusts in a certain swarm doubles every [#permalink]

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14 Feb 2015, 04:35

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!

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