What is the result of the following addition problem (using two's compliment notation)? $\begin{array} { r }
00001111 \
+ 10101010 \
\hline
\end{array}$
A) 011000101
B) 10111001
C) 01010101
D) 10110101

Question

What is the result of the following addition problem (using two's compliment notation)? $\begin{array} { r } 
00001111 \
+ 10101010 \
\hline
\end{array}$
A) 011000101
B) 10111001
C) 01010101
D) 10110101

Quizplus · Accepted Answer

First, we need to convert the numbers to binary. 
11ea32c0 = 0001 0001 1110 1010 0011 0010 1100 0000 
TB1806_00 = 1011 1000 0001 1000 0000 0110 0000 0000 
Next, we add the two numbers together (carry in overflow is ignored):

0001 0001 1110 1010 0011 0010 1100 0000
+  1011 1000 0001 1000 0000 0110 0000 0000
----------------------------------------
  1 0100 1001 1111 0010 0011 1000 1100 0000
  
Finally, we convert the result back to hexadecimal, and since it is a negative number in two's complement, we take the complement and add 1:

Two's complement of 0100 1001 1111 0010 0011 1000 1100 0000 is 1011 0110 0000 1101 1100 0111 0100 0000

Adding 1, we get: 1011 0110 0000 1101 1100 0111 0100 0001

This is equal to B6 0D C7 41 in hexadecimal, which is negative.

Thus, our answer is:

B) 10111001

What Is the Result of the Following Addition Problem (Using 00001111+10101010\begin{array} { r } 00001111 \\+ 10101010 \\\hline\end{array}00001111+10101010​​

What Is the Result of the Following Addition Problem (Using $\begin{array} { r } 00001111 \\+ 10101010 \\\hline\end{array}$