[warrior]

well guys time to gear up for PGDCM

to begin please answer some of the questions (sample PGDCM paper sent by IIMC)

Please try question no 6,14,15,18,20,23 from the attached .pdf file

Also key for the paper is also attached

[siddhesh_j]

Originally Posted by warrior

well guys time to gear up for PGDCM

to begin please answer some of the questions (sample PGDCM paper sent by IIMC)

Please try question no 6,14,15,18,20,23 from the attached .pdf file

Also key for the paper is also attached

14 .

There is a radar, a computer and a gyroscope on board an airplane. The probability that the radar fails is 0.2. If the radar fails, the gyroscope will also fail, and the probability that the computer fails is 0.3. If the radar functions correctly, then the computer will also function correctly, and the probability that the gyroscope fails is 0.2. 14. The probability that the computer or the gyroscope functions correctly while the other does not is 1. 0.7 2. 0.14 3. 0 4. None of these


qn is on Bayes theorem :

Acc to me one term is missing over here ...

they havennt considered the case whr Radar works , computer works , but gyroscope doesnt ... tht shud have given them an extra 0.16 ... and my answer wud then have been 0.30 (option 4 none of these )

Option 2. 0.14 only tks into account the case when radar fails, gyroscope fails and comp is running

warrior , plz chk if analysis is ok ...

[subodh_iit]

Prob that comp works correctly while others do not = 0.7*0.2 = 0.14

Prob that gs works correctly while others do not = 0.2*0 = 0

Answer = 0.14

Its just a brain twister I guess..

PS: - The question says that other two should fail..Hence the case whr Radar works , computer works , but gyroscope doesnt does not come into pic.

Sorry, I might have misinterpreted it..

If it is a combi of (1) C but not G and (2) G but not C.

Then its is .7*.2 + .8*.2 = .14+.16

[warrior]

Please confirm the answers to following questions also please solve the unaswered question
Also attached is the question paper for reference

1...4
2...3
3...4
4...1
5...2
6...1
7...4
8...1
9...2
10...2
11...3
12...1
13...4
14...2
15...3
16...2
17...2
18...3
19...4
20...2
21...2
22...4
23...1
24...3
25...4
26...3
27...2
28---not clear what qn is asking
29...4
30...2
31...2
32...4
33...3
34...3
35...3
36...4
37...3
38...2
39...1
40...3

Junta esp vinnet, amitabh, amar and sid do check them

[amar_kashyap]

Originally Posted by warrior

well guys time to gear up for PGDCM

to begin please answer some of the questions (sample PGDCM paper sent by IIMC)

Please try question no 6,14,15,18,20,23 from the attached .pdf file

Also key for the paper is also attached

18.sin^3x+cos^3x= (sinx+cosx)^3- 3sinx.cosx(sinx+cosx)

we have sin2x= 2sinx.cosx= -3/4


Also use 2sinx.cosx= (sinx+cosx)^2 -1
sinx+cosx= (1+2sinx.cosx)^1/2

now do it..!!

20. t^3-3t^2+3t+1=0
=> (t-1)^3+2=0
That is of the form x^3+p^3=0 The equation will have two imaginary roots and one real root= -p.
So the sum of real roots is -p..


23. Ist and last digit of 2^43
Last is prett simple.

For first, i dont think there is a proper method.
But we can try with..8. (2^10)^4= 8. (1024)^4= 8.( 1000^4 +4 . 1000^3. 24 + 6.1000^2. 24^2 + 4. 1000. 24^3 + 24^4)= The fiest digit will be 8.......

[siddhesh_j]

My doubts from PGDCM 2002 : qn no 21,22,32,38

38 was the toughest ... btw TIME has posted slns to th papers but some solns as usual are pathetic

[evergreen_cool]

i have few doubts in the sample paper sent by IIMC guys...
ques number 6..nt able to visualize the figure itself..
Four spheres lie on a plane: two spheres each of radius a and two spheres of each of
radius x (with x < a) lie so that each sphere touches the other three and the plane, the
radius x will be
1. a(2-3) 2. a3/2 3. a/2 4. None of these

[vineet.nitd]

Originally Posted by evergreen_cool

i have few doubts in the sample paper sent by IIMC guys...
ques number 6..nt able to visualize the figure itself..
Four spheres lie on a plane: two spheres each of radius a and two spheres of each of
radius x (with x < a) lie so that each sphere touches the other three and the plane, the
radius x will be
1. a(2-3) 2. a3/2 3. a/2 4. None of these

I guess they have mistke in this question.

THe ways in which u can pack spheres on a single plane is hexagonally or in a squared structure .We cant make all of the four touch each other in a single plane like that.

For that we have to place them as the corner of the tetrahedron or octahedron..

Any other thoughts ?

[ashish abraham]

is there no separate thread for ATM yet????

[ashish abraham]

Originally Posted by ashish abraham

is there no separate thread for ATM yet????

anyways.. i attempted only 19 questions.. dint go very well..

there were only 11 students in Chennai centre.. ppl do tel abt the number of students in other centres..