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UGC NET Computer Science Nov 2021 (Paper II)

The Octal equivalent of hexadecimal $(D.C)_{16}$ is:

Solution

Convert hexadecimal to binary:

$D=(1101)_2$

$C=(1100)_2$

So,

$(D.C)_{16}=(1101.1100)_2$

Now convert binary to octal by making groups of $3$ bits.

Integer part:

$1101=(001\ 101)_2=(15)_8$

Fractional part:

$.1100=(.110\ 000)_2=(.6)_8$

Therefore,

$(D.C)_{16}=(15.6)_8$

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