ISI 2007 ME-I - Doubts

[Ridhika]

Question paper - http://economicsentrance.weebly.com/uploads/1/1/0/5/1105777/isi2007.pdf

Q8) if P= log(base x)xy and Q = log(base y)xy then P+Q =
A)PQ
B) P/Q
C) Q/P
D) (PQ)/2

Q12) - I got (A) exactly but Amit sirs solution says B -- someone please explain how? http://economicsentrance.weebly.com/uploads/1/1/0/5/1105777/isi2007.pdf

Q14) What does notation f(x;1) and f(x;0) mean?

Q21) Again I got (A) = 1 .. please explain how the answer is B?

23) How to approach this question?!

25) Again I got 240 possibilities.. please explain how the answer is 168

27) M=n/2 I get that.. but how to calculate variance?

28) ?? They all seem correct to me!! What am I missing?

Please respond!! Thank you!

[Homer Simpson]

For 12, take log on both sides and then differentiate wrt x. put x = 0 , you will get (B)

[Ridhika]

Thanks a lot Tsuki..

For 12 I differentiated directly and got (A).. why is it wrong to differentiate directly?

[Granpa Simpson]

Q14) Here, f(x;θ)=θ*f(x;1)+(1-θ)*f(x;0), where both f(x;1) and f(x;0) are pdf’s,
Now, ∫ f(x;θ)dx= θ*∫ f(x;1)dx+ (1-θ)*∫ f(x;0)dx.
Since ∫ f(x;1)dx=∫ f(x;0)dx=1.
∫ f(x;θ)dx=θ+(1-θ)=1.
Hence option d, f(x;θ) is a pdf for every θ

Q21) Integrate the function over (0,1), then (1,sqrt(2)), then (sqrt(2),3/2)) and you will get b as answer.

[Granpa Simpson]

[Granpa Simpson]

Q23) Just use normal algebraic equation and then compare the options, dats the easiest way to get the answer, u will get a relationship in the form of this identity (p1+2*p2) = 750*p1*p2, compare the options and u will get that both p1=p2 and p2=(3/4)p1 satisfies the identity.

[Granpa Simpson]

[Ridhika]

Thanks a lot Subhayu.. got these!!

Also I think you have posted the ans to 29 instead of 28..

[Granpa Simpson]

for Q28..all of the options seems to go with the problem given..

[rongmon]

Question 21, could someone share the workings for integral of fractional function.

Thanks

[Granpa Simpson]

The working has been shared..u just check the previous posts..!!

[rongmon]

Just noticed above workings.

thanks.

[rongmon]

Subhayu, could you briefly elaborate on how you derived the second step in the solution

[Granpa Simpson]

[rongmon]

Right, how does that lead to the next equation exactly?

Sorry for the bother!

[Granpa Simpson]

In the interval [0,1], [x] =[x^2]=0, in the interval [1, sqrt(2)], [x]=[x^2]=1 and in the interval [sqrt(2), 3/2], [x]=1 and [x^2]=2.

[Granpa Simpson]

Its not at all bothering brother..absolutely ok..!

[rongmon]

Grateful!

[Ridhika]

I got 25

number 3 can only be in 3rd , 4th or 5th positions as it must be preceeded and suceeded by 2 numbers.
Now if it is in 3rd place -> it can be preceeded in 2! ways and succeded in 4! ways. So total = 48. Exactly the reverse argument holds when 3 is in 5th place. So again there are 48 cases.

When 3 is in 4th place .. there are 3 places before and 3 after. in the 3 places before 1 and 2 must be there.. so we need to do 3P2 = 6. similarly, in the 3 places after 3 we can do 3P2 for finding correct places of 6 and 7.. so we have 6*6=36 ways so far and in addition the 2 empty places can be filled by 4 or 5.. so 36*2=72

now 48 + 48 + 72=168

I hope this helps someone :D

[sarawagisruti]

Hi Subhayu, can you please explain this question again?