Answer
\begin{equation}
a. 100111
\end{equation}
\begin{equation}
b. 1011.110
\end{equation}
\begin{equation}
c. 100000
\end{equation}
\begin{equation}
d. 1000.00
\end{equation}
also, see the following image:
Work Step by Step
some notes for additions in binary notation:
1- If the most significant bit is 1, then it represents the negative decimal number.
2- The most possible largest positive integer can be represented by using 5 bits pattern is 15.
3- The most possible least negative integer can be represented by using 5 bits pattern is, -16.
4- When two values are added, if there is an extra bit generated after the most significant bit, that is the 6’th bit. Then the 6’th bit is considered as carrying bit and needs to be truncated.
$\ \ \ $• When two positive numbers are added and the result obtained is negative.
$\ \ \ $• When two negative numbers are added and the result obtained is positive.
6- When you add and subtract binary numbers you will need to be careful when 'carrying' or 'borrowing' as these will take place more often.
Key Addition Results for Binary Numbers
1 + 0 = 1
1 + 1 = 10
1 + 1 + 1 = 11
_____________________________________________________
The solution of $(a)$
\begin{equation}
\begin{array}{r}{1 \ \ \ \ \ \ \ \ } \\ {11011} \\ {+\quad 1100} \\ \hline 100111\end{array}
\end{equation}
_____________________________________________________
The solution of $(b)$
\begin{equation}
\begin{array}{r}{1 \ \ } \\ {\quad 1010.001} \\ {+\quad 1.101} \\ \hline 1011.110\end{array}
\end{equation}
_____________________________________________________
The solution of $(c)$
\begin{equation}
\begin{array}{r}{11111 \ \ } \\ {\quad 11111} \\ {+\quad 0001} \\ \hline 100000\end{array}
\end{equation}
_____________________________________________________
The solution of $(d)$
\begin{equation}
\begin{array}{r}{1111 \ 1 \ \ } \\ {111.11} \\ {+\quad 00.01} \\ \hline 1000.00\end{array}
\end{equation}
