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by harsh.champ » Sun Feb 07, 2010 7:42 am
On Madagascar Island, there are x dodos in a particular year. 30 dodos in a thousand of the original population die every year and 25 dodos in a thousand of the original population are born every year. In how many years will the population of dodos halve itself ?

(A)200
(B)125
(C)100
(D)50
(E)None of the above
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by shashank.ism » Sun Feb 07, 2010 7:51 am
harsh.champ wrote:On Madagascar Island, there are x dodos in a particular year. 30 dodos in a thousand of the original population die every year and 25 dodos in a thousand of the original population are born every year. In how many years will the population of dodos halve itself ?

(A)200
(B)125
(C)100
(D)50
(E)None of the above
Let after Y years the population of dodos will be halved.
now death rate = 30/1000=0.03 total death in a year = 0.03X (30 dodos in a thousand of the original population die every year )

birth rate = 25/1000 = 0.025 total birth in a year = 0.025X (25 dodos in a thousand of the original population are born every year )

So X-0.03XY+0.025XY = X/2 --> X/2 = (0.03-0.025)XY --> Y = 0.5/0.005 = 100

The ans is C.
Last edited by shashank.ism on Sun Feb 07, 2010 8:53 am, edited 1 time in total.
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by ajith » Sun Feb 07, 2010 8:00 am
harsh.champ wrote:On Madagascar Island, there are x dodos in a particular year. 30 dodos in a thousand of the original population die every year and 25 dodos in a thousand of the original population are born every year. In how many years will the population of dodos halve itself ?

(A)200
(B)125
(C)100
(D)50
(E)None of the above
Rate of death = 3%
Rate of birth = 2.5%

The net rate = -0.5%

At the end of first year the no of dodos =x*.995
at the end of nth year no of dodos = x*(0.995)^n

x*(0.995)^n = x/2
(0.995)^n = 1/2
One obvious way to solve this is to apply log on both sides and write
n*log(0.995) = log (1/2)
n = log (1/2)/log(0.995)
n =138 and some days
So, I would say E , definitely not a gmat question
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by shashank.ism » Sun Feb 07, 2010 8:57 am
ajith wrote:
Rate of death = 3%
Rate of birth = 2.5%

The net rate = -0.5%

At the end of first year the no of dodos =x*.995
at the end of nth year no of dodos = x*(0.995)^n

x*(0.995)^n = x/2
(0.995)^n = 1/2
One obvious way to solve this is to apply log on both sides and write
n*log(0.995) = log (1/2)
n = log (1/2)/log(0.995)
n =138 and some days
So, I would say E , definitely not a gmat question
Ajith according to the question the death and birth is in terms of original population and not the population of the current year,
if u go through the question again (30 dodos in a thousand of the original population die every year and 25 dodos in a thousand of the original population are born every year.)


Well u have applied a very good approach for the problem, but it could be applied when the birth and death is in terms of th current year population, and certainly the question would be a bit tougher an would not likely to be a GMAT question.
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by ajith » Sun Feb 07, 2010 9:00 am
shashank.ism wrote:
Well u have applied a very good approach for the problem, but it could be applied when the birth and death is in terms of th current year population, and certainly the question would be a bit tougher an would not likely to be a GMAT question.
Oops, I overcooked it.... Thanks Shashank
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by shashank.ism » Sun Feb 07, 2010 11:35 am
ajith wrote:
shashank.ism wrote:
Well u have applied a very good approach for the problem, but it could be applied when the birth and death is in terms of th current year population, and certainly the question would be a bit tougher an would not likely to be a GMAT question.
Oops, I overcooked it.... Thanks Shashank
Anytime.. Ajith.. I rechecked many time as you put a big black strip. I was really scared.. :)
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