Gate2020 cs Q41

  • Option : A
  • Explanation :
    Gate2020 cs
    ⇒ # of instruction possible = 24 = 16
    ∴ No. of free opcodes = 16 – x
    Let x is number of R-type instruction exist
    ∴ No. of I-type instruction possible = (16 – x) * 22
    8 = 64 – 4x
    4x = 64 – 8 = 56
    x = 14
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