GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 18 Jan 2020, 18:57

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# The population of locusts in a certain swarm doubles every

Author Message
TAGS:

### Hide Tags

Manager
Joined: 04 Dec 2008
Posts: 67
The population of locusts in a certain swarm doubles every  [#permalink]

### Show Tags

Updated on: 16 Feb 2010, 07:32
5
33
00:00

Difficulty:

65% (hard)

Question Stats:

58% (01:42) correct 42% (01:51) wrong based on 586 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?

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.
Math Expert
Joined: 02 Sep 2009
Posts: 60480
Re: The population of locusts in a certain swarm doubles every  [#permalink]

### Show Tags

31 Oct 2013, 00:51
2
7
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:
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
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.
_________________
SVP
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2469
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35

### Show Tags

12 Feb 2010, 23:43
18
1
5
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

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html
##### General Discussion
Manager
Joined: 01 Feb 2010
Posts: 171

### Show Tags

13 Feb 2010, 00:11
1
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: 317
Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38

### Show Tags

03 Jun 2012, 23:30
5
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: 440
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
2
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
Intern
Joined: 29 Aug 2012
Posts: 25
Schools: Babson '14
GMAT Date: 02-28-2013
Re: The population of locusts in a certain swarm doubles every  [#permalink]

### Show Tags

21 Oct 2012, 14:08
1
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...
Senior Manager
Joined: 24 Aug 2009
Posts: 440
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
1
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.
Director
Joined: 22 Mar 2011
Posts: 583
WE: Science (Education)
Re: The population of locusts in a certain swarm doubles every  [#permalink]

### Show Tags

22 Oct 2012, 02:26
2
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.
Director
Joined: 22 Mar 2011
Posts: 583
WE: Science (Education)
Re: The population of locusts in a certain swarm doubles every  [#permalink]

### Show Tags

Updated on: 22 Oct 2012, 06:56
2
2
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.

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: 25
Schools: Babson '14
GMAT Date: 02-28-2013
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^(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 $$n-1$$.

Thank you !!! for the very helpful explanation ..
Intern
Joined: 26 Jun 2013
Posts: 2

### 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
Intern
Joined: 23 Jul 2013
Posts: 5
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:
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
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
Senior Manager
Status: Math is psycho-logical
Joined: 07 Apr 2014
Posts: 401
Location: Netherlands
GMAT Date: 02-11-2015
WE: Psychology and Counseling (Other)
The population of locusts in a certain swarm doubles every  [#permalink]

### Show Tags

14 Feb 2015, 04:35
1
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: 62
Re: The population of locusts in a certain swarm doubles every  [#permalink]

### Show Tags

15 Oct 2016, 12:38
1
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
Intern
Joined: 18 Aug 2014
Posts: 46
Location: India
GMAT 1: 740 Q49 V40
GPA: 3.11
Re: The population of locusts in a certain swarm doubles every  [#permalink]

### Show Tags

26 Mar 2017, 21:48
1
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: 60480
Re: The population of locusts in a certain swarm doubles every  [#permalink]

### Show Tags

28 Mar 2017, 03:08
AARONRAMSEY wrote:
i am bit confused with this topic can anyone explain the theory behind this problem.

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.
_________________
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/a-certain-bac ... 44013.html
http://gmatclub.com/forum/it-takes-30-d ... 40269.html
http://gmatclub.com/forum/a-certain-cul ... 52258.html
http://gmatclub.com/forum/the-number-of ... 42858.html
http://gmatclub.com/forum/the-populatio ... 60081.html

Hope 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
Non-Human User
Joined: 09 Sep 2013
Posts: 13978
Re: The population of locusts in a certain swarm doubles every  [#permalink]

### Show Tags

13 Jul 2019, 14:55
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: The population of locusts in a certain swarm doubles every   [#permalink] 13 Jul 2019, 14:55
Display posts from previous: Sort by