Given that Numbers are in IEEE-754 single precision.
Representation :: $\text{1 sign bit, 8 exponent bits and 23 Mantissa Bits}$
$\text{Decimal value} = (-1)^{\text{Sign bit}}\times(1.Mantissa)\times(2)^{exponent-127}$
$R_A = 0xC1400000 = 1100 \;0001\;0100 \; 0000\; 0000\; 0000\; 0000\; 0000$
$R_A = 0xC1400000 = \underbrace{1}\underbrace{100 \;0001\;0}\underbrace{100 \; 0000\; 0000\; 0000\; 0000\; 0000}$
$\text{Decimal value of R}_A = (-1)^{\text{1}}\times(1.1000000000..)\times(2)^{130-127}=-(1100)_2 = -12$
$R_B = 0x42100000 = 0100 \;0010\;0001 \; 0000\; 0000\; 0000\; 0000\; 0000$
$R_B = 0x42100000 = \underbrace{0}\underbrace{100 \;0010\;0}\underbrace{001 \; 0000\; 0000\; 0000\; 0000\; 0000}$
$\text{Decimal value of R}_B = (-1)^{\text{0}}\times(1.001000000000..)\times(2)^{132-127}=-(100100)_2 = 36$
$R_C = 0x41400000 = 0100 \;0001\;0100 \; 0000\; 0000\; 0000\; 0000\; 0000$
$R_C = 0x41400000 = \underbrace{0}\underbrace{100 \;0001\;0}\underbrace{100 \; 0000\; 0000\; 0000\; 0000\; 0000}$
$\text{Decimal value of R}_C= (-1)^{\text{0}}\times(1.1000000000..)\times(2)^{130-127}=(1100)_2 = 12$
$\therefore C=A+B\;\; is\; false.$
Alternate :
$R_A = 0xC1400000, R_B = 0x42100000 \;\;and\;\; R_C = 0x41400000$
- By observing $R_A$ and $R_C$ only MSB bit is different ===> $R_C$ = -$R_A$ ===> $R_C$ + $R_A$ = 0
- By observing $R_B$ and $R_C$, $R_B$ exponent part is higher than $R_C$ exponent part ===> $R_B$ > $R_C$ ===> $R_B$ - $R_C$ > 0
Now Verify other two options.