GATE 1996 CS Solution Keys

I have posted these solution keys here because all the books, available in the market, have lots of wrong solutions. Most of the students, like me, find themselves in dilemma and waste a lot of time in finding the correct keys. These solution keys are those which I had corrected while preparing for GATE Computer Science 2012 Exam. You will find most of these solution keys to be correct, except few for which I could not find the correct solution. Still I suggest you that please keep your mind open while using these keys.

GATE 1996 CS Solution Keys

1.1.   A
1.2.   B
1.3.   A
1.4.   C
1.5.   D
1.6.   D
1.7.   C
1.8.   C
1.9.   D
1.10.   D
1.11.   B
1.12.   A
1.13.   C
1.14.   B
1.15.   C
1.16.   B
1.17.   B
1.18.   B
1.19.   C
1.20.   A
1.21.   D
1.22.   C
1.23.   D
1.24.   A
1.25.   D
2.1.   C
2.2.   B
2.3.   D
2.4.   C
2.5.   C
 2.6.   A
2.7.   C
2.8.   B
2.9.   C
2.10.   D
2.11.   C
2.12.   A
2.13.   D
2.14.   B
2.15.   C
2.16.   A
2.17.   B
2.18.   D
2.19.   C
2.20.   D
2.21.   D
2.22.   D
2.23.   B
2.24.   C
2.25.   C

Comments 5

  1. answer for Q. 23 is A, As Booths algorithm gives worst performance when the multiplier consists of alternating 1’s and 0’s.

  2. answer for 17 is D,
    Activation Record – Subroutine Call
    Location counter -Assembler(Assembler uses 2 counters- location and line counter)
    Reference count – garbage collector
    address relocation – linking loader(or relocating loader)

  3. hi amit,
    solution of 1.3 should be B , since it has been given that there is exactly 97 function from X->Y :
    by If we consider solution is A |x| = 1 and |y| = 97
    meaning element(only one in our case) of x is related to all the 97 elements y (which cant be a function ).
    as per function definition each element of domain can correspond exactly one value of co-domain. that is the only possible in over case
    in most extereme case when |x| = 97
    and |y| = 1 So I think Answer should be B. (What’s Your Opinion).

    1. Answer of 2.2 should be D.
      Let R is Symmetric Over (A,B)
      then if there is xRy then there should be yRx this implies that y is in set A also that means A intersection B is not phi (but in question it has been given that A intersection B = phi ).
      hence R can not be Symmetric.
      and it cannot be Reflexive also because reflexive means aRa holds .that means a is in Set A and a is Also in Set B (wich is not possible if A intersection B is = phi)
      (Am I Right or wrong ???)

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