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

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## Re: ISI 2014 PEA Answer Key

 In reply to this post by Amit Goyal Question 20. Please. Thanks
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## Re: ISI 2014 PEA Answer Key

 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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## Re: ISI 2014 PEA Answer Key

 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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## Re: ISI 2014 PEA Answer Key

 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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## Re: ISI 2014 PEA Answer Key

 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