GATE CS 2025 Set 2 Q39 — Number Representations (Fixed and Floating Point)

In the IEEE 754 single-precision floating-point representation, a 32-bit number is structured as:

• Sign bit ($S$): 1 bit
• Biased Exponent ($E$): 8 bits (with bias = 127)
• Fraction ($F$): 23 bits

The value of the number is given by: $(-1)^S \times (1.F)_2 \times 2^{E - 127}$.

Let us decode the three register values:

1. $\mathrm{R}_\mathrm{X} = \text{0xC1100000}$
   Binary: $\underbrace{1}_{S} \ \underbrace{1000\ 0010}_{E} \ \underbrace{0010\ 0000\ 0000\ 0000\ 0000\ 000}_{F}$
   $S = 1$ (negative)
   $E = 130_{10}$
   $F = 0.001_2 = 0.125_{10}$
   $X = -1 \times (1 + 0.125) \times 2^{130 - 127} = -1.125 \times 8 = -9$
2. $\mathrm{R}_\mathrm{Y} = \text{0x40C00000}$
   Binary: $\underbrace{0}_{S} \ \underbrace{1000\ 0001}_{E} \ \underbrace{1000\ 0000\ 0000\ 0000\ 0000\ 000}_{F}$
   $S = 0$ (positive)
   $E = 129_{10}$
   $F = 0.1_2 = 0.5_{10}$
   $Y = 1 \times (1 + 0.5) \times 2^{129 - 127} = 1.5 \times 4 = 6$
3. $\mathrm{R}_\mathrm{Z} = \text{0x41400000}$
   Binary: $\underbrace{0}_{S} \ \underbrace{1000\ 0010}_{E} \ \underbrace{1000\ 0000\ 0000\ 0000\ 0000\ 000}_{F}$
   $S = 0$ (positive)
   $E = 130_{10}$
   $F = 0.1_2 = 0.5_{10}$
   $Z = 1 \times (1 + 0.5) \times 2^{130 - 127} = 1.5 \times 8 = 12$

Now we evaluate the given options:

• Option A: $4(X+Y)+Z = 4(-9+6)+12 = 4(-3)+12 = 0$ (Correct)
• Option B: $2Y-Z = 2(6)-12 = 0$ (Correct)
• Option C: $4X+3Z = 4(-9)+3(12) = -36+36 = 0$ (Correct)
• Option D: $X+Y+Z = -9+6+12 = 9 \neq 0$ (Incorrect)