# ISI 2004 ME - I Answer Key

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## Re: ISI 2004 ME - I Answer Key

 Sum is S... multiply the whole series by 1/4 this wud be S/4, say.... Now, find... S-S/4 .... this will give a G.P. .. sum it.. we will get.. 3S/4 equals something... then solve for S....
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## Re: ISI 2004 ME - I Answer Key

 In reply to this post by AJ in question 8 lets sn=1+3/2^2 +7/2^4 +15/2^6 +31/2^8 +.........                      sn/4=    1       +3/2^4 +7/2^6 +15/2^8 +31/2^10 +.........        sn-sn/4=3/4sn=1+2/2^2 +4/2^4 +8/2^6 +16/2^8+.........                           =1+1/2 +1/2^2 +1/2^3 +......                           =2                        sn=8/3
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## Re: ISI 2004 ME - I Answer Key

 pls explain q 18
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## Re: ISI 2004 ME - I Answer Key

 CONTENTS DELETED The author has deleted this message.
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## Re: ISI 2004 ME - I Answer Key

 In reply to this post by AJ Yes. That's the very definition of linear relationship: the correlation is always perfect.
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## Re: ISI 2004 ME - I Answer Key

 plz expain q 28...
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## Re: ISI 2004 ME - I Answer Key

 In reply to this post by Chinni18 yeah i agree with Chinni
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## Re: ISI 2004 ME - I Answer Key

 thankyou! :)
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## Re: ISI 2004 ME - I Answer Key

 In reply to this post by AJ plz explain q no.2 plzz rply fast
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## Re: ISI 2004 ME - I Answer Key

 polynomial changes sign between 1 and 2. hence one root will definitely lie between 1&2. only option b lies between 1 & 2. ---  "You don't have to believe in God, but you should believe in The Book." -Paul Erdős
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## Re: ISI 2004 ME - I Answer Key

 In reply to this post by Amit Goyal ques 25 why part c shouldnt it be a^2+b+c less than 0 becos f(1) is negative
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## Re: ISI 2004 ME - I Answer Key

 Hi n1, Say x1 and x2 are the roots of eqn with X1>1 and X2<1. Then (a^2)*((X2)^2) + 2b(X2) + c = 0 As we know that X2<1, so (a^2)*((X2)^2)> 0 => 2b(X2) + c<0 or c/(-2b) < X2 and X2<1(given) thus c/(-2b)< 1 hence (c) "Woh mara papad wale ko!"
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## Re: ISI 2004 ME - I Answer Key

 thanks aditya can you explain ques 18
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## Re: ISI 2004 ME - I Answer Key

 Yea sure,if the series is arranged like this y= 1! + 2 ! + 3! + 4!  + 5! + 36( ... )it can be observed that terms from 6! are divisible by  36, so remainder is 0 for all those terms. so, remainder now depends on terms upto 5!.=> remainder is :  1! + 2 ! + 3! + 4!  + 5!  =153 % 36 = 9.Regards,Aditya On Thu, Mar 27, 2014 at 5:18 PM, n1 [via Discussion forum] wrote: thanks aditya can you explain ques 18 If you reply to this email, your message will be added to the discussion below: http://discussion-forum.2150183.n2.nabble.com/ISI-2004-ME-I-Answer-Key-tp7515784p7585591.html To start a new topic under General Discussions, email [hidden email] To unsubscribe from General Discussions, click here. NAML "Woh mara papad wale ko!"
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## Re: ISI 2004 ME - I Answer Key

 thanks:)
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## Re: ISI 2004 ME - I Answer Key

 In reply to this post by Bankelal hi aditya pls explain ques 20 (2007 paper)
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## Re: ISI 2004 ME - I Answer Key

 Restaurant can't accommodate if 52 or 51 turn up  P(can't accomodate) =  P(52) + P(51)  = (52C52) *((4/5)^52) *((1/5)^0) + (52C51) *((4/5)^51) *(1/5) = (4/5)^52 + 52 *(1/5)*(4/5)^51  = 1 - 14* (4/5)^52 P(can accomodate) = 1 - P(can't accommodate) = 1 - 14* (4/5)^52 On Fri, Mar 28, 2014 at 10:38 AM, n1 [via Discussion forum] wrote: hi aditya pls explain ques 20 (2007 paper) If you reply to this email, your message will be added to the discussion below: http://discussion-forum.2150183.n2.nabble.com/ISI-2004-ME-I-Answer-Key-tp7515784p7585626.html To start a new topic under General Discussions, email [hidden email] To unsubscribe from General Discussions, click here. NAML "Woh mara papad wale ko!"
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## Re: ISI 2004 ME - I Answer Key

 thanks :) i have a doubt in ques 25 (2007) _ _ 3_ _ _ _ _ = 2!* 4! =48 _ _ _ 3_ _ _ _ =3!*3! =36 _ _ _ _ 3_ _ _ =4!* 2! =48     48+36+48=132     why cant we solve it like ths ...pls explain
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## Re: ISI 2004 ME - I Answer Key

 Hey n1,Yes we can solve it like this but we just need to be careful about 2nd case for which there exist 2 scenarios:_ _ _ 3_ _ _ {1,2,4} 3 { 5,6,7} =3!*3! =36  {1,2,5} 3 { 4,6,7} =3!*3! =36 so 132 + 36 = 168.  On Fri, Mar 28, 2014 at 10:01 PM, n1 [via Discussion forum] wrote: thanks :) i have a doubt in ques 25 (2007) _ _ 3_ _ _ _ _ = 2!* 4! =48 _ _ _ 3_ _ _ _ =3!*3! =36 _ _ _ _ 3_ _ _ =4!* 2! =48     48+36+48=132     why cant we solve it like ths ...pls explain If you reply to this email, your message will be added to the discussion below: http://discussion-forum.2150183.n2.nabble.com/ISI-2004-ME-I-Answer-Key-tp7515784p7585651.html To start a new topic under General Discussions, email [hidden email] To unsubscribe from General Discussions, click here. NAML "Woh mara papad wale ko!"