GATE CS 2022 Q31 — Number Representations (Fixed and Floating Point)

To find the values of $A$, $B$, and $C$, we decode their 32-bit IEEE-754 single-precision representation: $\text{Value} = (-1)^S \times (1 + M) \times 2^{E - 127}$.

1. For $R_A = \text{0xC1400000}$:
   • Binary: $\underbrace{1}_{S} \ \underbrace{1000\,0010}_{E} \ \underbrace{100\,0000\,0000\,0000\,0000\,0000}_{M}$
   • $S = 1$ (negative), $E = 130_{10}$, and $M = 0.5_{10}$
   • $A = -1 \times (1 + 0.5) \times 2^{130 - 127} = -1.5 \times 8 = -12$
2. For $R_B = \text{0x42100000}$:
   • Binary: $\underbrace{0}_{S} \ \underbrace{1000\,0100}_{E} \ \underbrace{001\,0000\,0000\,0000\,0000\,0000}_{M}$
   • $S = 0$ (positive), $E = 132_{10}$, and $M = 2^{-3} = 0.125_{10}$
   • $B = +1 \times (1 + 0.125) \times 2^{132 - 127} = 1.125 \times 32 = 36$
3. For $R_C = \text{0x41400000}$:
   • Binary: $\underbrace{0}_{S} \ \underbrace{1000\,0010}_{E} \ \underbrace{100\,0000\,0000\,0000\,0000\,0000}_{M}$
   • $S = 0$ (positive), $E = 130_{10}$, and $M = 0.5_{10}$
   • $C = +1 \times (1 + 0.5) \times 2^{130 - 127} = 1.5 \times 8 = 12$

Now, let's evaluate the options:

• (A) $A + C = -12 + 12 = 0$ (True)
• (B) $C = A + B \implies 12 = -12 + 36 = 24$ (False)
• (C) $B = 3C \implies 36 = 3 \times 12$ (True)
• (D) $(B - C) > 0 \implies 36 - 12 = 24 > 0$ (True)