Binary system

Solution

Option (c) is correct.

Explanation:

Perform binary addition: \[ \begin{aligned} & \ \ \ \ \ \ 1101101 \\ + & \ \ \ \ \ \ 1011011 \\ \hline & \ \ \ 11011000 \\ \end{aligned} \] Convert \( (11011000)_2 \) to decimal: \[ = 1 \times 2^7 + 1 \times 2^6 + 0 \times 2^5 + 1 \times 2^4 + 1 \times 2^3 + 0 \times 2^2 + 0 \times 2^1 + 0 \times 2^0 \] \[ = 128 + 64 + 0 + 16 + 8 + 0 + 0 + 0 \] \[ = 216 \] Thus, \[ 216_{10} \] Convert directly \(1101101_2 + 1011011_2\) gives \(11011000_2 = 200_{10}\).