# ISI 2014 PEA Answer Key

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## ISI 2014 PEA Answer Key

 In reply to this post by Rajat How to do question number 27? I'm getting the answer as B. not D. Please help me out with it.
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 In reply to this post by Amit Goyal Question 20. Please. Thanks
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 This is how i solved it: kth term of AP: Tk=a+(k-1)d Infinite sum of GP : a/(1-d) a+(k-1)d = a/(1-d) a-ad+(k-1)d-(k-1)d^2=a d[(k-1)-a-(k-1)d]=0 (k-1)(1-d)=a For d= 1/3    or    1/5      or    1/9 1-d  = 2/3    or    4/5      or    8/9 respectively a  =   2/3(k-1) ; 4/5(k-1) ; 8/9(k-1) ------> all three give composite numbers no matter what the value of "k" Only for 1-d = 1/2 => d=1/2 we can have "a" as prime
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## Re: ISI 2014 PEA Answer Key

 In reply to this post by Amit Goyal In question 14 won't the answer would be a since it is asking us head after even number of tosses so sample space should be H,TTH,TTTTH,.......... according to which prob will be 2/3
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## Re: ISI 2014 PEA Answer Key

 Even number of tosses include the last toss as well. So favourable outcomes are TH,TTTH,TTTTTH and so on.
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 In reply to this post by Varun Please provide the solution for ques 15
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 This is a binomial distribution with p(success)= 1/2 and n=10 Mean score = mean of this binomial distribution= np = 5 so P(SCORE =5)=p (NO. OF SUCCESS=5)= C(10,5)* (1/2)^5 * (1/2)^5  = 63/256
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 M not able to do question 13 plz help
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## Re: ISI 2014 PEA Answer Key

 eqn 1, x2>= x1-3  x2=x1-3+ h1 .......(1)eqn 2, x3>= x2+ 2  x3=x2 + 2 + h2 = x1+h1+h2-1 eqn 3, x4>= x3-10  x4=x1+h1+h2+h3-11   .......(2)where h1,h2,h3 are non negative values (2)- (1) gives   x4-x2 = h2 +h3 -8 since h2 and h3 are nonnegative so x4-x2 >= -8 now from eqn 4 x4-x2<=alpha so -8<=x4-x2<=alpha so alpha should be greater than -8 but all options are less than -8 hence option d none of these is the answer
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## Re: ISI 2014 PEA Answer Key

 Thanks a lot
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## Re: ISI 2014 PEA Answer Key

 Guys q 27 plz ?
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## Re: ISI 2014 PEA Answer Key

 This post was updated on . For q27 answer is (d). If the function is of the form f(z)=z, then from the given equation (let's call it it eq1): f(x+y)*f(x-y)= 􀀀(f(x) + f(y))^2 - 4*x^2*f(y), we have (x+y)*(x-y) = (x+y)^2 - 4*x^2*y Solving we get a quadratic equation 2*x^2-x-y=0. Solving for x, we get,  x=(1+(1+8y)^.5) or x=(1-(1+8y)^.5)..........(2) So for a given value of y, eq1 will not hold for any x not satisfying (2). But as per the question eq1 should hold for all x,y that are real. Contradiction! So a function of the form f(z)=z cannot hold. Or, f(2)=2 cannot hold.
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 In reply to this post by Ankita Hi, can you post the solution to Q15? I can't find out the value of the sum