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PREVIOUS YEAR SOLVED PAPERS - GATE 2017 Shift 2

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41. P and Q are considering to apply for a job.

  • Option : A
  • Explanation :
    Let A, B be the events denote that P, Q respectively applies for a job
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42. If the characteristic polynomial of a 3 × 3 matrix M over R (the set of real numbers) is λ3 – 4λ2 + aλ + 30, where a ∈ R, and one eigen value of M is 2, then the largest among the absolute values of the eigenvalues of M is _______

  • Option : D
  • Explanation :
    Since one eigen value of M is 2 thus it satisfies the characteristics equation so
    ∴ 23 - 4(2)2 + a(2) + 30 = 0
    ⇒ a = -11
    ∴ Charcteristic polynomial is
    λ3 - 4λ2 - 11λ + 30 = 0
    (λ - 2)(λ - 5)(λ + 3) = 0
    ∴ λ=2,5,-3
    Largest absolute value of 'λ' is 5
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43. Consider the following expression grammar G.
E -> E - T | T
T -> T + F | F
F -> (E) | id
Which of the following grammars are not left recursive, but equivalent to G.

  • Option : C
  • Explanation :
    The rule for removal of left recursion is
    A →Aα |β will be
    A →β A’
    A’ → αA’|ϵ
    The given grammar is:
    E → E – T | T; in this α is “– T” and β is T
    T →T + F | F, In this α is “+ F” and β is F
    F → (E) | id
    Hence after removal of the left recursion:
    E → TX
    X → - TX|ϵ
    T → FY
    Y → +FY|ϵ
    F → (E) | id
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44. In a two-level cache system, the access times of L1 and L2 1 and 8 clock cycles, respectively. The miss penalty from the L2 cache to main memory is 18 clock cycles. The miss rate of L1 cache is twice that of L2. The average memory access time(AMAT) of this cache system is 2 cycles. The miss rates of L1 and L2 respectively are:

  • Option : A
  • Explanation :
    --Access time of L1 = 1
    --Access Time of L2=8
    --miss penalty L1 cache (2*L2) = 18*2 = 2*a
    --miss penalty L2 cache say a = 18
    --AMAT (average memory access time) =2
    AMAT = Access time of L1 + (MissRate L1 * miss penalty L1) where miss penalty L1 = Access time of L2 + (MissRate L2 * miss penalty L2)
    2 = 1+ 2*a *(8 + a* 18)
    Solving the equation,
    a=0.111
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45. Consider two hosts X and Y, connected by a single direct link of rate 106 bits/sec. The distance between the two hosts is 10,000 km and the propagation speed along the link is 2 x 108 m/s. Hosts X send a file of 50,000 bytes as one large message to hosts Y continuously. Let the transmission and propagation delays be p milliseconds and q milliseconds, respectively. Then the vales of p and q are:

  • Option : D
  • Explanation :

    B = 106bits/sec
    d = 10,000km = 104 x 10 3m
    V = 2 x 108 m/s
    L = 50,000 Bytes
    ∴ Transmission time (p) =
    = 400 ms
    ∴ Propagation time (q) =
    = 50 ms
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Related Quiz.
GATE 2017 Shift 2