GATE 2006 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 2006 CS Solution Keys

 1.    A
2.    D
3.    C
4.    A
5.    C
6.    B
7.    C
8.    C
9.    C
10.   A
11.   B
12.   A
13.   D
14.   C
15.   C
16.   B
17.   B
18.   D
19.   D
20.   B
21.   A
22.   C
23.   D
24.   D
25.   B
26.   D
27.   D
28.   D
29.   B
30.   B
31.   A
32.   B
33.   B
34.   D
35.   A
36.   A
37.   C
38.   A
39.   B
40.   C
41.   D
42.   C
43.   A
44.   B
45.   C
46.   C
47.   D
48.   D
49.   A
50.   D
51.   B
52.   B
53.   C
54.   C
55.   C
56.   A
57.   B
58.   A
59.   D
60.   D
61.   A
62.   C
63.   C
64.   A
65.   B
66.   B
67.   C
68.   A
69.   C
70.   C
71.   C
72.   A
73.   B
74.   A
75.   D
76.   D
77.   A
78.   B
79.   B
80.   C
81.   B
82.   A
83.   A
84.   D
85.   A

Comments 15

  1. Ankur sir, can you please explain ans to qno12 because i am confused with the fact that why cant we implement it with a heap and how do we get the vertices that will give the minimum hop count to reach destination from source??

    Reply
    1. Post
      Author

  2. Here #vertices we will take care of to reach from source to destination as the graph is unweighted. So, how will we get the vertices that will be the shortest path as there can be some vertices during level order traversing that doesn’t fall within the shortest path. eg the entry into the queue ca be ABCDEF when the actual path can be ABDF?? how to get this actual path from ABCDEF??
    If possible, can you send me a snap with the solution to my mail??

    Reply
    1. Post
      Author

      Entering ABCDEF does not mean that there is a path from A to F. First of all you will insert the source node in the queue. Then you will remove the source node and insert all the adjacent nodes to it. The source node will become the parent of those nodes inserted in the queue. Then You will remove one of the nodes in queue and insert its adjacent nodes and make that node the parent of inserted nodes. Please read the BFS again. You are talking about something else. Try to make the picture in your mind. It seems you are applying DFS algorithm.

      Reply

  3. Suppose A is adjacent to B and C, D is adjacent to B & also E is adjacent to B. F is adjacent to both D & E. Now tell me to find the shortest path from A to D how to proceed and what to do using BFS(queue)??

    Reply
    1. Post
      Author

      It will be like this
      A
      /
      B C
      /
      D E
      /
      F
      First of all you will insert A in queue. Then you will remove A and insert B and C in queue. A will become parent of B and C. No you will remove B from queue and insert D and E in the queue. Now B will become parent of D and E. So you can traverse the shortest path from A to D by traversing the parent nodes from D to A which will give you DBA. I hope it clears your doubt. If it does not clear, then I again suggest you to go through the book and try to make diagram and practice to apply the algorithm.

      Reply

  4. A
    / \
    B–C
    /\
    D E
    \ /
    F
    Sorry i forgot to mention B is also adjacent to C. Now there are three paths to F( one is ABDF, other is ACBEF & another is ACBDF). Among them shortest is ABDF, now how we can manage to get that path when C will get inserted before DF and will get removed also before them. Then how do we will be getting this shortest path??

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    1. Post
      Author

  9. I need your help to solve 59 question how do you get answer such type of question regarding evaluation of expression. I have not seen such type of problems in my uimaan book but getting confused when i see solution in differenr link because all link have different wrong logics and different answer question.

    Reply

  11. Consider two cache organizations: The first one is 32 KB 2-way set associative with 32-byte block size. The second one is of the same size but direct mapped. The size of an address is 32 bits in both cases. A 2-to-1 multiplexer has a latency of 0.6 ns while a kbit comparator has a latency of k/10 ns. The hit latency of the set associative
    organization is h1 while that of the direct mapped one is h2.

    The value of h2 is:
    (A) 2.4 ns
    (B) 2.3
    (C) 1.8
    (D) 1.7

    Answer is 1.7 ns. Why are we not adding mux latency in direct mapped cache?

    Reply

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