Re: ISI 2014 MEI

Posted by Granpa Simpson on Apr 06, 2014; 7:44am
URL: http://discussion-forum.276.s1.nabble.com/ISI-2014-MEI-tp7585912p7586175.html

The answer to Q21 will be b...this can be done using sum of a G.P series for the minimum time required and using normal algebraic equation for the maximum case:

For minimum T occurs when all in a given period gives 2 chicks each, thus it can be mathematically expressed as; {3+3*2+3*2*2+........+3*2*(T-1)times} = 31,
 or, 3{(2^T)-1/2-1}= 31,
 or, 3{1*(2^(T)-1)/(2-1)}= 31,...................................(Using sum of a gp series with common ratio 2).
 Solving for T we get T= least Int {3.ab}= 4.

similarly for maximum case, it cannot occur that neither of the chicks do not lay, so at most one has to carry on the process, thus mathematically;
 {3+2+2+....+(T-1) times 2}=31,
or, 3+2*(T-1)= 31
Solving for T we get, T=15.
hence option b.