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The population of locusts in a certain swarm doubles every
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Updated on: 16 Feb 2010, 07:32
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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 OA is D. I got this correct by using population chart, ie. multiplying every 2 years. Is there a formula I could apply?
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Originally posted by joyseychow on 12 Feb 2010, 23:33.
Last edited by joyseychow on 16 Feb 2010, 07:32, edited 1 time in total.




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Re: The population of locusts in a certain swarm doubles every
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31 Oct 2013, 00:51




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Re: Population Growthlocust
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12 Feb 2010, 23:43
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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Re: Population Growthlocust
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13 Feb 2010, 00:11
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 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 Growthlocust
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03 Jun 2012, 23:30
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
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05 Sep 2012, 11:43
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
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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^(n1) > 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
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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^(n1) > 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
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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^(n1)) or \(a_n=a_1R^{n1}\) and not \(a_1(R^n1)\).
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Re: The population of locusts in a certain swarm doubles every
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Updated on: 22 Oct 2012, 06:56
himanshuhpr wrote: If we use g.p in this question . And if we consider a=4000 ; r=2 then 4000*2^(n1) > 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 \(a_n=a_1R^{n1}\), which is the term on the \(n\)th place. Between the first term and the \(n\)th term, \(n1\) multiplications by the ratio \(R\) take place, and this is reflected in the exponent of \(n1\). 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 \(n1\) 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 \(n1\).
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Originally posted by EvaJager on 22 Oct 2012, 04:21.
Last edited by EvaJager on 22 Oct 2012, 06:56, edited 2 times in total.



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Re: The population of locusts in a certain swarm doubles every
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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^(n1) > 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 \(a_n=a_1R^{n1}\), which is the term on the \(n\)th place. Between the first term and the \(n\)th term, \(n1\) multiplications by the ratio \(R\) take place, and this is reflected in the exponent of \(n1\). 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 \(n1\) 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 \(n1\). Thank you !!! for the very helpful explanation ..



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Re: Population Growthlocust
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30 Oct 2013, 11:39
gurpreetsingh wrote: IMO D, OA pls
4 hours ago = 1000 2 hours ago = 2000 right now = 4000
now the given equation says
\(4000*2^n > 250000\)
=> \(2^n > 62.5\)since n must be an integer... thus take
\(2^n = 64 = 2^6\) thus it got doubled 6 times thus 12 hours
Thus D 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
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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? A. 6 B. 8 C. 10 D. 12 E. 14 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: acertainbacteriacolonydoublesinsizeeverydayfor144013.htmlittakes30daystofillalaboratorydishwithbacteria140269.htmlacertaincultureofbacteriaquadrupleseveryhourifa52258.htmlthenumberofwaterliliesonacertainlakedoublesevery142858.htmlthepopulationofgrasshoppersdoublesinaparticularfield160081.htmlHope this helps. Hi Bunuel, Can you help me solving this problem with this formula > [final population] = [initial population] (1 + r)^t



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The population of locusts in a certain swarm doubles every
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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!
 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



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Re: The population of locusts in a certain swarm doubles every
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15 Oct 2016, 12:38
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 164 = 12 hrs  in 12 hrs the population will exceed 250,000



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Re: The population of locusts in a certain swarm doubles every
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26 Mar 2017, 21:48
i am bit confused with this topic can anyone explain the theory behind this problem.
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Re: The population of locusts in a certain swarm doubles every
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28 Mar 2017, 03:08



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Re: The population of locusts in a certain swarm doubles every
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27 Mar 2018, 20:00
AkshayDavid wrote: 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 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: http://gmatclub.com/forum/acertainbac ... 44013.htmlhttp://gmatclub.com/forum/ittakes30d ... 40269.htmlhttp://gmatclub.com/forum/acertaincul ... 52258.htmlhttp://gmatclub.com/forum/thenumberof ... 42858.htmlhttp://gmatclub.com/forum/thepopulatio ... 60081.htmlHope this helps. Hi Bunuel, Can you help me solving this problem with this formula > [final population] = [initial population] (1 + r)^t Hi AkshayDavid The formulae you specified can be used if percentage is used. In this sum r = 100 The correct formulae that can be used here is final pop = initial pop ( 1 + rn/100)^(t/n) There rn is % change in n period n is the period when compounding happens Rn = 100 n = 2 T is. What is asked for
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