Computer Arithmetic Questions and Answers

Answer: A

The flags are S = 1, Z = 0, AC = 0, P = 0, Cy = 0.

\[ \begin{aligned} 47\text{H} &= 0100\,0111 \\ 51\text{H} &= 0101\,0001 \\ \text{sum} &= 1001\,1000 = 98\text{H} \end{aligned} \]

• S = 1: bit 7 of the result is 1.
• Z = 0: the result is not zero.
• AC = 0: the low nibbles 7 + 1 = 8 produce no carry out of bit 3.
• P = 0: 98H contains three 1s, an odd number, and the 8085 sets P only for even parity.
• Cy = 0: 98H fits in 8 bits, so there is no carry out of bit 7.

Options B and C set Z = 1, which is impossible for a non-zero result, so neither they nor 'None of the above' fit.

Answer: D

The result is \(-\ \text{579}_{H}\).

In 15's complement subtraction, add the 15's complement of the subtrahend. The 15's complement of C14 is found by subtracting each digit from F: F - C = 3, F - 1 = E, F - 4 = B, giving 3EB.

\[ \begin{aligned} \text{69B} + \text{3EB} &= \text{A86} \\ \text{end carry} &= \text{none} \end{aligned} \]

With no end-around carry, the result is negative and in complement form. Its magnitude is the 15's complement of A86: F - A = 5, F - 8 = 7, F - 6 = 9, so the answer is -579 hexadecimal. Check: C14 - 69B = 579.

3EB is only the complement of C14, A86 is the uncorrected sum, and 579 without the sign is the magnitude, so all three miss the sign or the final step.

Answer: C

CB2H - 972H = 340H.

In 16's complement subtraction, add the 16's complement of the subtrahend and discard the end carry. The 15's complement of 972 is FFF - 972 = 68D; adding 1 gives the 16's complement 68E.

\[ \begin{aligned} \text{CB2} + \text{68E} &= \text{1340} \\ \text{discard carry} &\Rightarrow \text{340H} \end{aligned} \]

Check in decimal: CB2H = 3250, 972H = 2418, and 3250 - 2418 = 832 = 340H. 68DH is only the 15's complement of the subtrahend, and 1340H is the sum before the end carry is dropped.