Explanation

(d) The given number is first converted from binary to decimal equivalence
$(100010)_{2} = 2^{5} \times 1 + 2^{4} \times 0 + 2^{3} \times 0$+$2^{2} \times 0 + 2^{1} \times 1 + 2^{0} \times 0$
= 32 + 0 + 0 + 2 + 0
= $(34)_{10}$
and $(11011)_{2} = 2^{4} \times 1 + 2^{3} \times 1 + 2^{2} \times 0 + 2^{1} \times 1 + 2^{0} \times 1$
= 16 + 8 + 0 + 2 + 1
= $(27)_{10}$
$\therefore$ Sum $(100010)_{2} + \ (11011)_{2} = \ (34)_{10} + (27)_{10}$
= $(61)_{10}$
Now,
[table] 2 | 61 ; 2 | 30-1 ; 2 | 15-0 ; 2 | 7-1 ; 2 | 3-1 ; 2 | 1-1 ; | 0-1 [/table]
$\therefore\ $Required sum (in binary system)
$(100010)_{2} + \ (11011)_{2} = (111101)_{2}$