Gate CS 2024 (1)| Question - 13 | Digital Logic

$$\begin{gathered} As we know the range of 2\text{'}s complement number is \\ \mathbf{-}{\mathbf{2}}^{\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{ }}\mathbf{ }\mathit{t}\mathit{o}\mathbf{ }\mathbf{(}{\mathbf{2}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{ }\mathbf{-}\mathbf{ }\mathbf{1}\mathbf{)}\mathbf{ }because the given data is five-bit long so its \\ range is -16 to +15. \\ A = 01010 = +10 as the MSB bit is zero so the number is positive \\ B = 11010 = -2+4-8=-6 as MSB bit is one so the number is negative. \\ \mathit{O}\mathit{p}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ }\mathit{A}\mathbf{:} A+B is 4 which is 00100, there is no overflow \\ \mathit{O}\mathit{p}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ }\mathit{B}\mathbf{:} A-B is 16 which is 010000, see that we used 6 bits \\ hence it is an overflow. \\ \mathit{O}\mathit{p}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ }\mathit{C}\mathbf{:} B-A is -16 which is 10000, there is no overflow. \\ \mathit{O}\mathit{p}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ }\mathit{D}\mathbf{:} 2*B is -12 which is 10100, there is no overflow. \\ Hence, the correct option will be \mathit{O}\mathit{p}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ }\mathit{B}. \end{gathered}$$