Consider three registers R1, R2, and R3 that store numbers in IEEE-754 single precision floating point format. Assume that R1 and R2 contain the values (in hexadecimal notation) 0x42200000 and 0xC1200000, respectively. If $R3 = \frac{R1}{R2}$, what is the value stored in R3?

To find the value stored in $R3 = \frac{R1}{R2}$, let us decode the IEEE-754 single-precision (32-bit) representations of $R1$ and $R2$.

In IEEE-754 single precision:

• Bit 31: Sign bit ($S$)

• Bits 30–23 (8 bits): Biased exponent ($E$)

• Bits 22–0 (23 bits): Fractional mantissa ($M$)

• Value $= (-1)^S \times 1.M \times 2^{E - 127}$

### 1. Decoding $R1 = \text{0x42200000}$:

• $S = 0$ (Positive)

• $E = 10000100_2 = 132_{10} \implies \text{Actual Exponent} = 132 - 127 = 5$

• $M = 0.010_2 = 0.25$

• $R1 = +1.25 \times 2^5 = 1.25 \times 32 = +40.0$

### 2. Decoding $R2 = \text{0xC1200000}$:

• $S = 1$ (Negative)

• $E = 10000010_2 = 130_{10} \implies \text{Actual Exponent} = 130 - 127 = 3$

• $M = 0.010_2 = 0.25$

• $R2 = -1.25 \times 2^3 = -1.25 \times 8 = -10.0$

### 3. Computing $R3$:

### 4. Encoding $-4.0$ into IEEE-754 Single Precision:

• $S = 1$

• $-4.0 = -1.0 \times 2^2$

• Actual Exponent $= 2 \implies \text{Biased Exponent } E = 2 + 127 = 129_{10} = 10000001_2$

• Mantissa fraction $M = 0.000\dots$

Binary representation:

Grouped into 4-bit nibbles:

Thus, the correct option is B.