# ISI 2014 PEA Answer Key Classic List Threaded 55 messages 123
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## Re: ISI 2014 PEA Answer Key

 Pr{Min(X,Y)<=Z} = Pr(X<=Z,Y<=Z)=Pr(X<=Z)*Pr(Y<=Z) since the are independent, so Pr{Min(X,Y)<=Z}=Z*Z=Z^2, Proceeding in similar manner I am getting Pr{Max(X,Y)<=(1-Z)}=(1-Z)^2, Now (1-Z)^2=Z^2, using this I am getting Z=1/2 as the answer...Amit sir plz let me know where am I am doing the mistake..!!!!  "I don't ride side-saddle. I'm as straight as a submarine"
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## Re: ISI 2014 PEA Answer Key

 In reply to this post by Arushi For q 25..it is one of d standard prob based on "'Pigeon hole principal' ." soluiton in detail :max n min no of handshakes(HSs) a person can do = 19 & 1 respectively (u can't handshake urself.) Assume for a moment, a random person ,say , A shook hand wid B n leaves d party implying A being one of d person to have shaken hand in odd number i e 1 in dis case. now, max HS B can do  is 19(18 +1) while for odrs would be 18 (since A is gone home,nly 19 members r left) For d sake of argument,to make all HS even,  lets say dat B does HS wid wid 17 ppl in party leaving C, dat way his total handshakes are 17 + 1 = 18 now. Now out of18 ppl( excluding A B ) , odr than C, everyone HS wid 17 making der total HS to 18 .(still even). but for B, d max HS can only b 17 ( A is gone n B refused) making his HS always odd.in total A n C have shaken hand wid odd ppl. Dis way,  der cant be '1' person to have shaken hand wid odd no off ppl,wtever way u try. Since 1 cant b answer, option D is not d answer implyiing option C as answer.( u dun need to check for 19, sicne der is only 1 correct ans)... For more details, u can refer to 'Pigeon hole principal' .
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## Re: ISI 2014 PEA Answer Key

 Questions 6, 8 and 12 anyone? Thank you.
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## Re: ISI 2014 PEA Answer Key

 This post was updated on . Q6) The matrix is called p-special if detA is not divisible by p..DetA=-(a*b)....now any number can be broken down in to product of prime numbers. This number will be divisible by p only if p is one of the factors of -(a*b) or (a*b)=0. So in order to form a p-special matrix we will have to select a and b in such a way that none of them has p as a factor or their product not equal to 0. Now if you look at the set then it will be clear that since both a and b are less than p except 0 you can take any of the (p-1) elements for a and similarly (p-1) elements for b.(taking 0 there are total of p elements). Thus (a*b) can be taken in (p-1)*(p-1)=(p-1)^2 ways. Q8) If you look at the equation then both are of odd degree, their sum can be zero only if they are equal and opposite in sign. This can occur only if x lies in [a,b] s.t it is equidistant from both a and b, thus x=(a+b)/2 is the only solution that you can get. Thats the way I have solved, however you may use theorems related to polynomials.Alternative Method: Differentiate the function wrt you will get, (2n+1)*[(x-a)^(2n)+(x-b)^(2n)], now since (2n+1) not equal to 0 and 2n is even for all n so f'(x) cannot be zero for any x, so there is no real root for f'(x) so f(x) can have either 1 or no real root, however since the polynomial is of odd degree it has atleast one real root, so it has exactly one root.  "I don't ride side-saddle. I'm as straight as a submarine"
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## Re: ISI 2014 PEA Answer Key

 In reply to this post by rongmon "I don't ride side-saddle. I'm as straight as a submarine"
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## Re: ISI 2014 PEA Answer Key

 In reply to this post by Amit Goyal can someone help me with q 23 thanks
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## Re: ISI 2014 PEA Answer Key

 In reply to this post by Granpa Simpson pls explain ques 3
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## Re: ISI 2014 PEA Answer Key

 In reply to this post by Granpa Simpson Please can anyone help me with Q13? Also if you can mention from where to read for such questions.
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## Re: ISI 2014 PEA Answer Key

 In reply to this post by neha 1 "I don't ride side-saddle. I'm as straight as a submarine"
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## Re: ISI 2014 PEA Answer Key

 In reply to this post by Spiti Q13) Arrange the equation in matrix form of the form AX=b, form the augmented matrix by incorporating the constant terms. Now compare the coefficient matrix and the augmented matrix, by comparing you will find out that the Rank of Coefficient Matrix =4 and the Rank of Augmented Matrix=5 whatever be the value of alpha. The necessary and sufficient condition for a solution to exist is Rank of Coefficient Matrix=Rank of Augmented Matrix. Here this condition does not hold for any value of alpha...hence option d is the answer.  "I don't ride side-saddle. I'm as straight as a submarine"
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## Re: ISI 2014 PEA Answer Key

 In reply to this post by shietal Q23) The equation to have two unique roots there can be two cases Case 1: the equation is of the form f(x)=(x-1)*(x-a)^2, for this to hold alpha = (49/4), which is not given in the options. Case 2: f(x)=(x-1)*(x-1)*(x-a), in this case one of the roots is 1 which occurs twice and the other is a, for this to hold alpha=6. thus answer is option a.  "I don't ride side-saddle. I'm as straight as a submarine"
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## Re: ISI 2014 PEA Answer Key

 In reply to this post by Granpa Simpson Hey Subhayu.. So this condition [condition for a solution to exist is Rank of Coefficient Matrix=Rank of Augmented Matrix] I know hold for a system of linear equations.. But does it always hold for a system of inequalities as well? Also Im not being able to solve 11, 20 and 27.. Please help?
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## Re: ISI 2014 PEA Answer Key

 Vaibhav has posted the right solution to Q13 is this link...https://www.facebook.com/groups/659129117445776/, thanx Vaibhav..check it..!!!!  "I don't ride side-saddle. I'm as straight as a submarine"
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## Re: ISI 2014 PEA Answer Key

 My pleasure bro....:-)
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## Re: ISI 2014 PEA Answer Key

 In reply to this post by Ridhika Ridhika....in q11...fx is PDF...therefore total area under fx should be sum to 1 So when u integratee it...ie x from 0 to a and 1/2 from a to 2 their sum should be equal to 1 You'll get a=0,1 now a cannot be 0 as prob with x<=a is positive and rv is b/w 0 and 2 only..therefore a is 1 Now Prob(X>=1) = integration ½ from 1 to 2...
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## Re: ISI 2014 PEA Answer Key

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## Re: ISI 2014 PEA Answer Key

 In reply to this post by Granpa Simpson Can you please post the solution in this forum, for the benefit of those who are not in the facebook group
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## Re: ISI 2014 PEA Answer Key

 In reply to this post by Dreyfus Thanks a lot vaibhav ! :) and good luck for the exam tomorrow! ----- Reply message ----- From: "Vaibhav Garg [via Discussion forum]" To: "Ridhika" <[hidden email]> Subject: ISI 2014 PEA Answer Key Date: Fri, May 9, 2014 7:30 PM Ridhika....in q11...fx is PDF...therefore total area under fx should be sum to 1 So when u integratee it...ie x from 0 to a and 1/2 from a to 2 their sum should be equal to 1 You'll get a=0,1 now a cannot be 0 as prob with x<=a is positive and rv is b/w 0 and 2 only..therefore a is 1 Now Prob(X>=1) = integration ½ from 1 to 2... If you reply to this email, your message will be added to the discussion below: http://discussion-forum.2150183.n2.nabble.com/ISI-2014-PEA-Answer-Key-tp7587867p7589167.html To unsubscribe from ISI 2014 PEA Answer Key, click here. NAML
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## Re: ISI 2014 PEA Answer Key

 In reply to this post by Spiti Spiti...for q13...add equation 2 and equation 4 u'll get x4-x3<=a-2 And we are given 5th eq as x4-x3<=-4 Since, this system of equation has solution that means these two inequalities needs to be consistent na for this a-2=-4 a=-2 thus option d nonoe of the abovr