ISI 2015 PEA Answer Key

[Amit Goyal]

 1. (b)
 2. (d)
 3. (c)
 4. (b)
 5. (d)
 6. (c)
 7. (a)
 8. (c)
 9. (a)
10. (c)
11. (d)
12. (c)
13. (b)
14. (d)
15. (c)
16. (c)
17. (b)
18. (a)
19. (a)
20. (d)
21. (d)
22. (b)
23. (b)
24. (b)
25. (a)
26. (d)
27. (a)
28. (b)
29. (b)
30. (d)

Ref: MSQE 2015

[L]

Onionknight, you were correct of about the square matrix of matrix (Q.5). Rest of the answers are same.

Amit sir can you please help in Q.2 of ISI PEB 2015. Should we take bribe as the price in demand equation or divide consumers' surplus by the quantity to get bribe/connection.

[deepti]

sir plz explain question no. 1 nd 5 from 2012 peb paper

[onionknight]

Q1.) In the short run, K is fixed as K* so the only decision the firm needs to make is how much labour to employ. Whatever labour the firm employs, the production function is Y= L^0.25 * K^0.25 and since capital is fixed as K*, L^0.25 = Y/K^0.25 which means L= Y^4/K. Now cost in the short run is wage rate times labour which is 10*Y^4/K.

In the long run, to maximize profits, set marginal derivative of profit with respect to labour=0, which gives
0.25*p*L^-0.25*K^0.25 = 10
Similarly, set marginal derivative of profit with respect to capital =0
0.25*p*K^-0.25*L^0.25 = 10
Divide the two equations to get K=L
Now K^0.25 * L^0.25 = Y, L and K are equal so L=K= Y^2
cost= w*L+r*K = 20*Y^2

[onionknight]

This image shows the budget set of the consumers and also their allotments and their final chosen sets : http://s28.postimg.org/cnt5m6kfx/20150503_203428.jpg. Yes, they can have the same preferences since there is nothing that suggests otherwise. For instance, both of them could have their bliss point for consumption of y at y=15 and spend all the money left on x.

[deepti]

Thanks it ws really helpful.. plz tell me the ans of question no. 6 from 2012 paper

[hs]

why is and to q20  d and not b



also in q19
if m =.2 and p=.1 x*=2
or m=2 and p=1   x*=0

so shouldnt ans be option C instead of A

[Jim_Moriarity]

Keep this post for 2015 discussions only.
please

Any body who have solved question 7 , 21 and 27. Please explain those.

[Rajat]

For q.7 put x+2y = a and x-2y = b
solve these simultaneous equations to get x and y in terms of a and b. This will give you f(a,b)

Q.21
P(A intersection B) may be zero or non-zero
=> P(A|B) may be zero or non-zero. If it is zero, it is definitely < P(A) because P(A) is non-zero ALWAYS.
Similar argument can be made for P(A|B)<P(B) (only those times when P(A|B) = 0)

so a,c can be ruled out.

now, P(A) = P(A|B) + P(A|B')
=> P(A|B) = P(A) - P(A|B') which obviously is =<P(A)
P(A|B) will be equal to P(A) if P(A|B') = 0. This can happen when A and B are mutually exclusive exhaustive events.
So, we cannot be sure that P(A|B) will only be 'LESS' than P(A) always. They can also be equal

Hope this helps !

[deepti]

So d should be the final ans.?? Of question 21

On May 6, 2015 7:40 AM, "Rajat [via Discussion forum]" <[hidden email]> wrote:

For q.7 put x+2y = a and x-2y = b
solve these simultaneous equations to get x and y in terms of a and b. This will give you f(a,b)

Q.21
P(A intersection B) may be zero or non-zero
=> P(A|B) may be zero or non-zero. If it is zero, it is definitely < P(A) because P(A) is non-zero ALWAYS.
Similar argument can be made for P(A|B)<P(B) (only those times when P(A|B) = 0)

so a,c can be ruled out.

now, P(A) = P(A|B) + P(A|B')
=> P(A|B) = P(A) - P(A|B') which obviously is =<P(A)
P(A|B) will be equal to P(A) if P(A|B') = 0. This can happen when A and B are mutually exclusive exhaustive events.
So, we cannot be sure that P(A|B) will only be 'LESS' than P(A) always. They can also be equal

Hope this helps !

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NAML

Thanks a lot @rajat. N please explain the 27th ques as well.

For w in the 1st draw , shudnt the prob be w/(w+b) * b/(w+b-1) and the same in the second draw as well

[dhruv]

A matrix is said to be zero matrix or null matrix if all its elements are zero.

Hence, answer of 5 should be a.

Copied and pasted above line from NCERT.

hi dhruv, the correct answer is d only. The ques is asking about the square of this matrix not the given matrix.

[dhruv]

Yeah, you are right.

Also, have you tried ISI PEB 2015, compared to previous sample papers, I found it really hard.

[onionknight]

It was but then again, solving 2-3 questions *correctly* was good enough to get you an interview call last year, from what I've heard (I didn't give the test myself).

Arey I filed an RTI regarding the cut offs. The cut off for 2014 PEB was 31 and PEA 83. As rightly said by onion , if the paper is tough cut off will go down for sure.

[deepti]

Can anyone hlp me wid the solow model questions from peb 2015

On May 7, 2015 7:24 PM, "[hidden email] [via Discussion forum]" <[hidden email]> wrote:

Arey I filed an RTI regarding the cut offs. The cut off for 2014 PEB was 31 and PEA 83. As rightly said by onion , if the paper is tough cut off will go down for sure.

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NAML

[deepti]

Question no. 7th n 8th

On May 7, 2015 8:18 PM, "deepti sharma" <[hidden email]> wrote:

Can anyone hlp me wid the solow model questions from peb 2015

On May 7, 2015 7:24 PM, "[hidden email] [via Discussion forum]" <[hidden email]> wrote:

Arey I filed an RTI regarding the cut offs. The cut off for 2014 PEB was 31 and PEA 83. As rightly said by onion , if the paper is tough cut off will go down for sure.

---

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NAML

[onionknight]

What is the marking scheme like for PEA?