Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 04 Dec 2008
Posts: 99

The population of locusts in a certain swarm doubles every [#permalink]
Show Tags
Updated on: 16 Feb 2010, 07:32
4
This post received KUDOS
21
This post was BOOKMARKED
Question Stats:
57% (00:53) correct 43% (00:55) wrong based on 854 sessions
HideShow timer Statistics
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?
Official Answer and Stats are available only to registered users. Register/ Login.
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.



CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2673
Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35

Re: Population Growthlocust [#permalink]
Show Tags
12 Feb 2010, 23:43
12
This post received KUDOS
4
This post was BOOKMARKED
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
_________________
Fight for your dreams :For all those who fear from Verbal lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
Support GMAT Club by putting a GMAT Club badge on your blog/Facebook
GMAT Club Premium Membership  big benefits and savings
Gmat test review : http://gmatclub.com/forum/670to710alongjourneywithoutdestinationstillhappy141642.html



Manager
Joined: 01 Feb 2010
Posts: 244

Re: Population Growthlocust [#permalink]
Show Tags
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



Senior Manager
Joined: 23 Oct 2010
Posts: 364
Location: Azerbaijan
Concentration: Finance

Re: Population Growthlocust [#permalink]
Show Tags
03 Jun 2012, 23:30
3
This post received KUDOS
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
_________________
Happy are those who dream dreams and are ready to pay the price to make them come true
I am still on all gmat forums. msg me if you want to ask me smth



Senior Manager
Joined: 24 Aug 2009
Posts: 485
Schools: Harvard, Columbia, Stern, Booth, LSB,

Re: The population of locusts in a certain swarm doubles every [#permalink]
Show Tags
05 Sep 2012, 11:43
1
This post received KUDOS
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



Intern
Joined: 29 Aug 2012
Posts: 26
GMAT Date: 02282013

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



Senior Manager
Joined: 24 Aug 2009
Posts: 485
Schools: Harvard, Columbia, Stern, Booth, LSB,

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



Director
Joined: 22 Mar 2011
Posts: 607
WE: Science (Education)

Re: The population of locusts in a certain swarm doubles every [#permalink]
Show Tags
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 \(a_1(R^n1)\).
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Director
Joined: 22 Mar 2011
Posts: 607
WE: Science (Education)

Re: The population of locusts in a certain swarm doubles every [#permalink]
Show Tags
Updated on: 22 Oct 2012, 06:56
1
This post received KUDOS
1
This post was BOOKMARKED
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 \(n1\).
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.
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.



Intern
Joined: 29 Aug 2012
Posts: 26
GMAT Date: 02282013

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



Intern
Joined: 26 Jun 2013
Posts: 2

Re: Population Growthlocust [#permalink]
Show Tags
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



Math Expert
Joined: 02 Sep 2009
Posts: 44573

Re: The population of locusts in a certain swarm doubles every [#permalink]
Show Tags
31 Oct 2013, 00:51
1
This post received KUDOS
Expert's post
5
This post was BOOKMARKED



Intern
Joined: 23 Jul 2013
Posts: 9

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



Senior Manager
Status: Math is psychological
Joined: 07 Apr 2014
Posts: 425
Location: Netherlands
GMAT Date: 02112015
WE: Psychology and Counseling (Other)

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



Manager
Joined: 08 Sep 2015
Posts: 71

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



Manager
Joined: 18 Aug 2014
Posts: 53
Location: India
GPA: 3.11

Re: The population of locusts in a certain swarm doubles every [#permalink]
Show Tags
26 Mar 2017, 21:48
i am bit confused with this topic can anyone explain the theory behind this problem.
_________________
Today's actions are tomorrow's results



Math Expert
Joined: 02 Sep 2009
Posts: 44573

Re: The population of locusts in a certain swarm doubles every [#permalink]
Show Tags
28 Mar 2017, 03:08



Intern
Joined: 20 Dec 2014
Posts: 36

Re: The population of locusts in a certain swarm doubles every [#permalink]
Show Tags
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
_________________
Kindly press "+1 Kudos" to appreciate




Re: The population of locusts in a certain swarm doubles every
[#permalink]
27 Mar 2018, 20:00






