Gate2018 cs Q58

  • Option : C
  • Explanation :
    Binary code: (b7b6b5b4b3 ⋅ b2b1b0)2
    (31.5)10 = (11111.1)2
    (0.875)10 = (0.111)2
    (12.100)10 = (01100.000110.....)2
              ↓
           Only 3 bits of fraction
           space available
           So can't be stored.
    (3.001)10 = (00011.000000..........)2
    It is also not accurate storage.
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