Question

Perform the arithmetic addition of the two decimal numbers given in List-I using the signed-complement system. Match the corresponding output of List-I with binary number representation given in List-II. 
\[\begin{array}{|c|c|}\hline \text{List-I} & \text{List-II} \\ \hline \text{(A) +6, +13} & \text{(I) 00000111} \\ \hline \text{(B) -6, +13} & \text{(II) 00010011} \\ \hline \text{(C) +6, -13} & \text{(III) 11101011} \\ \hline \text{(D) +6, -13} & \text{(IV) 11111101} \\ \hline \end{array}\]

• (A) - (III), (B) - (I), (C) - (III), (D) - (IV)
• (A) - (II), (B) - (I), (C) - (II), (D) - (III)
• (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
• (A) - (II), (B) - (IV), (C) - (I), (D) - (III)

Hint:

In signed-complement arithmetic, converting negative numbers to two's complement is necessary before performing addition.

Answer 1

The Correct Option is C

Step 1: Converting Decimal Numbers to Signed Binary.
- For **+6** and **+13**, we convert them to signed 8-bit binary: - \(+6 = 00000110_2\) - \(+13 = 00001101_2\) - For **-6** and **+13**, we convert them using two's complement: - \(-6 = 11111010_2\) - \(+13 = 00001101_2\) - For **+6** and **-13**, we convert **-13** using two's complement: - \(+6 = 00000110_2\) - \(-13 = 11110011_2\) - For **-6** and **-13**, we convert both using two's complement: - \(-6 = 11111010_2\) - \(-13 = 11110011_2\)

Step 2: Conclusion.
The correct mapping is: - (A) +6, +13 → **(I)** 00000111 - (B) -6, +13 → **(II)** 00001011 - (C) +6, -13 → **(III)** 11110101 - (D) -6, -13 → **(IV)** 11111100