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+27 votes

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In $32$ bit (single precision) IEEE-754 format, binary number is represented as

$S$($1$ bit) $E$($8$ bit) $M$($23$ bits) with implicit normalization and exponent is represented with Excess-127 code.

Here, Sign $bit =$ $0$ $\Rightarrow$ Number is positive.

Exponent bits $= 01111100 = 124$ $\Rightarrow$ $E = 124 - 127 = -3$ (Excess-127)

Mantissa bits $= 11011010000000000000000$ $\Rightarrow$ Number = $1.1101101$ (Implicit normalization).

$\therefore$ Number $= 1.1101101 * 2$$^{-3}$ $= 0.0011101101 = 0.227 = 2.27$ x $10$$^{-1}$

$\therefore$ Answer should be **C.**

+6 votes

in this qestion they are asking about IEEE 754

so total 32 bit representation in which 1 bit is for sign, 8 bit for biased exponent, 23 bit for mantissa.

biased exponent part:01111100 =124 in decimal

mantissa part:11011010000000000000000=$\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+\frac{1}{32}+\frac{1}{128}$

bias=$2^{n-1}-1$ where n=8 so biase=127

value=1.M*$2^{124-127}$

put all values you get closest ans C

so total 32 bit representation in which 1 bit is for sign, 8 bit for biased exponent, 23 bit for mantissa.

biased exponent part:01111100 =124 in decimal

mantissa part:11011010000000000000000=$\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+\frac{1}{32}+\frac{1}{128}$

bias=$2^{n-1}-1$ where n=8 so biase=127

value=1.M*$2^{124-127}$

put all values you get closest ans C

+4 votes

`Important`: The mistake to avoid here is to avoid the premature conversion of binary numbers to decimal digits.

Use: $-1^{S}*1.M*2^{e-127}$

You'll get $1.1101101 * 2^{-3}$

Now, don't convert 1.1101101 to decimal just yet. Get the whole value, then do the base conversion.

=> $0.0011101101$

=> $\frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{128} + \frac{1}{256} + \frac{1}{1024}$

I'll now ignore the last two terms, since they're really small.

=>$0.125+0.0625+0.03125+0.0078125$

=> $0.2265625$

=> $2.26 * 10^{-1}$

Option **C**

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