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Population Growth-locust

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Population Growth-locust

by joyseychow » Fri Feb 12, 2010 10:33 pm
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

[spoiler]OA is D. I got this correct by using population chart, ie. multiplying every 2 years. Is there a formula I could apply?[/spoiler]
Last edited by joyseychow on Tue Feb 16, 2010 6:33 am, edited 1 time in total.
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by fibbonnaci » Fri Feb 12, 2010 10:55 pm
are u sure of OA? What is the source?

I am getting the answer as D.
here is my approach:

4 hours ago = 1000
right now = 4000

4000 * 2^n = 250,000
2^n = 62.5, sunce n has to be an integer- we round off the number to the nearest multiple of 2 ie 64.

2^n= 64 => n=6. ie after 6 spaces down the chart the population exceeds 250,000. Note that each space is at a 2 hours difference. Therefore the total time is 6*2 =>12 hours.
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by thephoenix » Sat Feb 13, 2010 12:00 am
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

[spoiler]OA is E. I got this correct by using population chart, ie. multiplying every 2 years. Is there a formula I could apply?[/spoiler]
there is a st. formula for such problem

P=s*n^(t/I)

where as
P=population at time=t
s=population at time=0
n=multplication rate=2,3,4or doble triple or........
t=time
I=the amnt of time given for the Qunatity to double

here
when time=-4 ( 4 hrs ago) P=1000
let S=x
n=2(doubles)
I=2(every two hrs)
so 1000=x*2^(-4/2)-------->x=4000

now when p=250000 we need to find t

250000=4000*2(t/2)

---->2^(t/2)=250---->approximately.......t/2=7 to 8--->t=14 to 16
as at t=16 P=256000
so we have to take the lower value
which is 14
HTH
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by fibbonnaci » Sat Feb 13, 2010 12:53 am
2^(t/2)=250---->approximately.......t/2=7 to 8
u missed dividing 250 by 4.

2^(t/2)=250/4
ie. 2^(t/2) = 62.5
round it off to nearest power of 2, it is 64.
2^6 = 64

t/2= 6
therefore t=12.

How did u get 14/ 16??
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by joyseychow » Tue Feb 16, 2010 6:27 am
fibbonnaci wrote:are u sure of OA? What is the source?

I am getting the answer as D.
here is my approach:

4 hours ago = 1000
right now = 4000

4000 * 2^n = 250,000
2^n = 62.5, sunce n has to be an integer- we round off the number to the nearest multiple of 2 ie 64.

2^n= 64 => n=6. ie after 6 spaces down the chart the population exceeds 250,000. Note that each space is at a 2 hours difference. Therefore the total time is 6*2 =>12 hours.
Sorry, made an error....the OA is D. Source: Mgmat.
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by komal » Tue Feb 16, 2010 8:22 am
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

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