ISI 2014 MEI

[Granpa Simpson]

Σk*C(n,k)= 1*(n!/1!*(n-1)!)+2*(n!/2!*(n-2)!)+3*(n!/3!*(n-3)!)+……n*(n!/n!*0!)
                = {n*(n-1)!/0!*(n-1-0)!}+{n*(n-1)!/1!*(n-1-1)!}+{n*(n-1)!/2!*(n-2-1)!+……
                =n*{C(n-1,0)+C(n-1,1)+C(n-1,2)+…..+C(n-1,n-1)}
                =n*2^(n-1) (option a)

[kangkan]

hi is the ans to 21 b...i did it by manual couting..is there a formal method?

[Granpa Simpson]

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.

[kangkan]

thanks

[SINGHAM]

Sorry, I found the number that is not p-special.

[SoniaKapoor]

How to do question 19?

[kangkan]

Hi sonia..take the series (1+x)^n and then differentiate it..then put x=1..you will get the ans

[manisha]

can somebody explain Q-15 and Q-20

[Granpa Simpson]

for question 20, (1/2) will be the answer...this is because if we put d=1/3 then a=2*(k-1)/3 which can be either an even integer or any other fraction..it cannot be a prime number..put the other values of d and u will find the same..only for d=1/2, a=(k-1)/2 which can be a prime number...so option a...!!!!!

[Dreyfus]

Is it d for both q27 and q28 ..?

[Rajat]

Hi,

I will be taking ISI next year. Can someone please tell me the pattern of the examination - duration, markin scheme, number of questions etc.