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Home » Placement Papers » Mascot Placement Papers » Mascot Previous Year Placement Paper

 

Mascot Placement Paper -: Computer Knowledge Questions And Answers


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Mascot Previous Years Solved Sample Placement Papers

  1. A typical modern computer uses:

    • a) magnetic cores for secondary storage
    • b) LSI chips (Ans)
    • c) magnetic tape for primary memory
    • d) more than 10,000 vacuum tubes
  2. A collection of 8 bits is called:

    • a) byte (Ans)
    • b) record
    • c) word
    • d) nibble
  3. General-purpose computers are those that can be adopted to countless uses simply by changing their:

    • a) output device
    • b) input device
    • c) processor
    • d) program (Ans)
  4. The current generation of computers:

    • a) second
    • b) fifth
    • c) fourth (Ans)
    • d) third
  5. Non-volatility is an important advantage of:

    • a) CCDs
    • b) magnetic tapes and disks
    • c) magnetic bubbles
    • d) both b and c (Ans)
  6. What is the corresponding postfix expression for the given infix expression? a*(b+c)/d

    A) ab*+cd/

    B) ab+*cd/

    C) abc*+/d

    D) abc+*d/ (Ans)

  7. What is the corresponding postfix expression for the given infix expression? a+(b*c(d/e^f)*g)*h)

    A) ab*cdef/^*g-h+

    B) abcdef^/*g*h*+ (Ans)

    C) abcd*^ed/g*-h*+

    D) abc*de^fg/*-*h+

  8. What is the correct postfix expression for the following expression? a+b*(c^d-e)^(f+g*h)-i

    A) abc^de-fg+*^*+i

    B) abcde^-fg*+*^h*+i

    C) abcd^e-fgh*+^*+i (Ans)

    D) ab^-dc*+ef^gh*+i

  9. Why do we need a binary tree which is height balanced?

    • a) to avoid formation of skew trees (Ans)
    • b) to save memory
    • c) to attain faster memory access
    • d) to simplify storing

    Explanation: In real-world scenarios, dealing with random values is often not possible. The probability of encountering non-random values (like sequential ones) often leads to skew trees, causing worst-case scenarios. Thus, height balance is maintained by rotations.

  10. What is the maximum height of an AVL tree with p nodes?

    • a) p
    • b) log(p) (Ans)
    • c) log(p)/2
    • d) p^2

    Explanation: The number of nodes in terms of height follows the recurrence relation: N(he) = N(he-1) + 1 + N(he-2). Solving this relation gives N(he) = O(logp) as the worst-case height.

  11. To restore the AVL property after inserting an element, do we start at the insertion point and move towards the root of the tree?

    • a) true (Ans)
    • b) false

    Explanation: After insertion, only the path from the insertion point to the root or specific subtrees may become imbalanced in terms of height.

  12. Given an empty AVL tree, how would you construct an AVL tree when a set of numbers is given without performing any rotations?

    • a) just build the tree with the given input
    • b) find the median of the set of elements given, make it the root, and construct the tree (Ans)
    • c) use trial and error
    • d) use dynamic programming to build the tree

    Explanation: Constructing the tree with the median as the root ensures balance without requiring rotations.

  13. The memory address of element A[i][j] in row-major order is:

    A: loc(A[i][j]) = base(A) + W((i-LB) + n(j-LB))

    B: loc(A[i][j]) = base(A) + W(n(j-LB) + (i-LB))

    C: loc(A[j][i]) = base(A) + W(n(I-LB) + n(j-LB))

    D: loc(A[i][j]) = base(A) + W((I-LB) + n(j-LB))

    Ans: B
  14. The memory address of elements A[i][j] in column-major order is:

    A: loc(A[i][j]) = base(A) + W((I-LB) + m(j-LB))

    B: loc(A[i][j]) = base(A) + W(m(j-LB) + (i-LB))

    C: loc(A[j][i]) = base(A) + W(n(I-LB) + m(j-LB))

    D: loc(A[i][j]) = base(A) + W((I-LB) + m(j-LB))

    Ans: B
  15. The base address of an array is the address of:

    A: A[1]

    B: A[n1]

    C: A[0]

    D: Both A and B

    Ans: C
  16. Which data structure is used to implement queues, trees, graphs?

    A: Stack

    B: List

    C: Array

    D: None

    Ans: C
  17. In which data structure do addition and deletion of elements take place from the same position?

    A: Stack

    B: List

    C: Array

    D: Queue

    Ans: A




 


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