A C.I. PROBLEM

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plz..........solve
carbon decays at a constant rate in such a way tht it reduses to 50% in 5568 years find the age of an old wooden piece in which the carbon is only 12.5% of the original. Rolling Eyes

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plz....help guyz.im struck in it Sad

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hey pavan...wats d answer??is it 9744 yrs.??

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omg... thn i got it wrong......... tore the paper Very Happy

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even i got 9744yrs

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me too

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the ans is 16704 years
guyz.....i got the ans!!!!!!! after tearing paper!!!!!!!!!!!!
SOLUTION:
let it decay at a rate of R% p.a
(1-R/100)5568=50%=50/100=1/2
=> 1-R/100=(1/2)1/5568........................(1)
(1-R/100)n=12.5%=12.5/100=1/8
=> (1-R/100)n=(1/2)3
1-R/100=(1/2)3/n
from (1)
(1/2)1/5568=(1/2)3/n
3/n=1/5568
n=5568*3
n=16704

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oh...thanks

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how did u ppl get 9744? Suspect

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hey pavan i followed ur eg of tearin papers..............
sure to say it wrks!!!!! heheheheh:
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but i think we shud care abt de environment tooooooo!!!

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heee........ Laughing

so........try not my examlpe.
alll the best for all

Subject: Re: A C.I. PROBLEM Sat Mar 07, 2009 1:36 pm

that q is in evergreen
the q is wrong n it wont cum anyways

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the question is not wrong, the answer as solved by pavan is correct.

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P/2 = P ( 1- R/100) ^5568

=> ( 1- R /100) ^ 5568 = 1/2 ------- A

=> P/8 = P ( 1 - R /100)^ n

=>( 1 - R /100)^ n = 1/8 = (1/2 )^ 3

=> ( 1 - R /100)^ n = [ (1- R/100 )^5568]^3---- from A

=> n = 5568 * 3 = 16704 years

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i got thi in a state board PU buk

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yes that type of question is there in my text, but not about carbon...its about rate only...i forgot the q! but the solution is this method only

remem: when p and r are unknown, use this type of sol