ISI 2004 ME - I Answer Key

[AJ]

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....

[archita]

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

[seema]

pls explain q 18

[Mr. Nobody]

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[Chinni18]

Yes. That's the very definition of linear relationship: the correlation is always perfect.

[archita]

plz expain q 28...

[aditi5000]

yeah i agree with Chinni

[AJ]

thankyou! :)

[sakshi]

plz explain q no.2
plzz rply fast

[Sinistral]

polynomial changes sign between 1 and 2.
hence one root will definitely lie between 1&2.
only option b lies between 1 & 2.

[neha 1]

ques 25
why part c
shouldnt it be a^2+b+c less than 0 becos f(1) is negative

[Bankelal]

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)

[neha 1]

thanks aditya
can you explain ques 18

[Bankelal]

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] <[hidden email]> wrote:

thanks aditya
can you explain ques 18

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NAML

[neha 1]

thanks:)

[neha 1]

hi aditya
pls explain ques 20 (2007 paper)

[Bankelal]

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] <[hidden email]> wrote:

hi aditya
pls explain ques 20 (2007 paper)

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[neha 1]

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

[Bankelal]

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] <[hidden email]> 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

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[neha 1]

thanks :)
pls explain ques 29 (2007)
why option d?
can you explain this with examples