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Official PYQ · 2026

GATE CS 2026 Set 2 Q24 — Number Representations (Fixed and Floating Point)

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The 3232-bit IEEE\text{IEEE} 754754 single precision representation of a number is 0xC27100000\text{xC}2710000. The number in decimal representation is ____\_\_\_\_. (rounded off to two decimal places)
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Answer: -60.25:-60.25

To find the decimal representation of the given 32-bit IEEE 754 single-precision floating-point number, we first convert the hexadecimal representation 0xC27100000\text{xC}2710000 into binary:

C271000016=1⏟Sign (S) 1000 0100⏟Biased Exponent (E) 111 0001 0000 0000 0000 0000⏟Mantissa (F)\text{C2710000}_{16} = \underbrace{1}_{\text{Sign (S)}} \ \underbrace{1000\ 0100}_{\text{Biased Exponent (E)}} \ \underbrace{111\ 0001\ 0000\ 0000\ 0000\ 0000}_{\text{Mantissa (F)}}

Now, we extract the three components of the IEEE 754 representation:

  1. Sign Bit (SS): S=1  ⟹  The number is negative.S = 1 \implies \text{The number is negative.}
  2. Biased Exponent (EE): E=100001002=128+4=13210E = 10000100_2 = 128 + 4 = 132_{10} The unbiased exponent (ee) is calculated as: e=E−127=132−127=5e = E - 127 = 132 - 127 = 5
  3. Mantissa (FF): F=0.11100012F = 0.1110001_2 The significand is (1+F)=1.11100012(1 + F) = 1.1110001_2.

The value of the floating-point number is given by:

Value=(−1)S×(1+F)×2e\text{Value} = (-1)^S \times (1 + F) \times 2^e Value=−1×(1.11100012)×25\text{Value} = -1 \times (1.1110001_2) \times 2^5

Multiplying by 252^5 shifts the binary point 55 places to the right:

Value=−111100.012\text{Value} = -111100.01_2

Converting this binary value to decimal:

  • Integer part: 1111002=32+16+8+4=6010111100_2 = 32 + 16 + 8 + 4 = 60_{10}
  • Fractional part: 0.012=0×2−1+1×2−2=0.25100.01_2 = 0 \times 2^{-1} + 1 \times 2^{-2} = 0.25_{10}

Thus, the decimal representation of the number is −60.25-60.25.

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