ISI 2014 MEI

[aman]

Guys, have u solved isi 2014 sample mei? Lets discuss.

[SINGHAM]

Yes! it will be very helpful!

[neha 1]

pls explain ques 6 ,7 and 8

[The Villain]

Quest 12 ...anyone?

[Dreyfus]

M getting 100/101 for q12...

[kangkan]

can you provide a link to the paper?

[aman]

http://economicsentrance.weebly.com/uploads/1/1/0/5/1105777/syll-sample-msqe-pea-2014.pdf               link for isi 2014  mei

[SINGHAM]

Neha,

6. Determinant of matrix A is -ab. Now, note that p is prime, so it does not divide any integer less than p itself. But, we have to take p from 0 to p-1 only. So, a & b and for that -ab is divisible by p only if -ab is 0 in itself. So, ab can be zero if either one of a or b is zero. Suppose a is zero, so we have p-1 ways to select b from 1,...,p-1 . Similarly, if b is zero, so we have p-1 ways to select a from 1,...,p-1 . Thus, p-1+p-1 ways. But, if a and b both are 0, then also, -ab=0. So, the final answer is (p-1)+(p-1)+1=2p-1.


7. The first letter can be either A, B, or C. For each of the subsequent ones the only choice is whether it is to be the same or different from the previous one (because in each case there's only one "different" letter allowed).

There's 3 ways to make the first choice and 2 ways to make each of the 6 remaining choices.

Thus, 3*(2^6)=192


8. I am not sure, but I worked with (x-1)^3+(x-2)^3=0, which has only one real root 3/2, so in general, as no other option is there, 1 is the answer.

Please help me with systematic answer of 8.

[SINGHAM]

I also get the same

[The Villain]

Vaibhav..my answer is not matching...pls show me the steps

[SINGHAM]

Ron, First integrate: You get 1/(k*(k+1))

Now you have to sum it from k=1 to 100.
So, 1/(1*2)+1/(2*3)+...+1/(99*100)+1/(100*101)
Note that 1/1*2=1-(1/2), 1/2*3=(1/2)-(1/3),...,1/(100*101)=1/100-1/101
Now plug in those values! You get 1-(1/2)+(1/2)-(1/3)+...+(1/100)-(1/101)
Cancel terms and you end up with 1-(1/101)=100/101

[SINGHAM]

Will you please explain 16?
I get S1=0, S2=1/1!,S3=1/2! etc.
So, S1+S2+......=e-1, but there is no such option!
Where is my mistake?

[The Villain]

thnx singham

[neha 1]

thanks singham
ques 16     Sk = K/K!
So, S1 =1/1! = 1
answer is option (a) e

[neha 1]

pls explain ques 19

[SINGHAM]

Neha, ok. I also got the same Sk= K/K!, but if we simply write Sk,
Sk=(k-1)/k!+(k-1)/k!k+(k-1)/k!k^2+......
And if we plug in k=0, then we get (1-1)/1!+(1-1)/1!1+(1-1)/1!1^2+.....=0+0+0+.....=0, am i right?
So, S0=0
And we have e-1 (i think so)


19. I plugged in n=3 and got option a.

[neha 1]

Sum of infinite G.P. = a/1-r
first term of G.P. =(k-1)/k!
r = 1/k
SK = (k-1)/k! / 1/k = k/k!
S1 = 1/1! =1
S2 = 2/2! = 1
S3 = 3/3! = 1/2!
S4 = 4/4! =1/3!
S1 + s2 +S3 +S4 +............ =  1+ 1+ 1/2! + 1/3! +......
e^x = 1+x+x^2/2! +x^3/3!+........
therefore, e = 1+1+1/2! +1/3! +......
so, S1 + S2 +S3 +S4+.......= e

[neha 1]

ques 25??

[SINGHAM]

Thank you Neha!

25. I think it should be 20 as 1 can't be the answer. Because each one shakes hands with someone, and if there was one person was there who shook hands with odd number of persons, there must be other person who also shook hands with odd number of persons only. 20 can be answer as if each person shakes hand with only one, 20 are "oddies". But, by using the option 'all of the above', we can also eliminate 19. So answer is C

[Granpa Simpson]

The answer to Q6 I guess will be (p-1)^2,
Since for all a, b in the set {0,1,….p-1}, total matrices formed = p^2 and for the same the number of zero matrices = 2p-1.
So, number of p-special matrices = (p^2 -2p+1) = (p-1)^2 , option “a”