A=0x2C, B=0x01, OV=0 (correct multiplication with no overflow)

A=0x2C, B=0x01, OV=1 (3CH×05H=12CH: low byte 2CH→A, high byte 01H→B, result>255 so OV=1)

A=0x01, B=0x2C, OV=1 (bytes reversed with overflow indication)

A=0x012C stored as 16-bit value, B=0x05 unchanged, OV=0

A=0x00, C=0 (both addition and rotation clear all values)

A=0x02, C=0 (RLC shifts the result left by one bit position)

A=0x00, C=1 (addition produces zero with carry, RLC doesn't change the value)

A=0x01, C=0 (FFH+01H=00H with C=1, then RLC shifts: 0000 0000 with C=1 → 0000 0001, C=0)

0x05, because SWAP has no effect after the masking operation

0x50, because ANL extracts lower nibble (55H AND 0FH = 05H), then SWAP moves it to upper nibble (05H → 50H)

0xA5, by inverting the bits during the swap operation

0x50, because SWAP multiplies the value by 16 in hex notation

A=0xF0, C=1 (10H - 20H = -10H in signed, wraps to F0H in unsigned with borrow indicated by C=1)

A=0xF0, C=0 (correct subtraction result with no borrow needed)

A=0x10, C=0 (subtraction cannot proceed because minuend < subtrahend)

A=0xE0, C=1 (two's complement result with borrow indication)

0x0084 with C=0 (simple shift with no overflow)

0x8042 with C=0 (unchanged because RLC preserves the value)

0x0084 with C=1 (shifted left once: 1000 0000 0100 0010 → 0000 0000 1000 0100, bit 15 went to carry)

0x1084 with C=0 (complete 17-bit result stored in 16 bits)

A=0x61, C=0 (arithmetic left shift preserving sign bit)

A=0x87, C=unchanged (RL rotates left without involving carry: 1100 0011 → 1000 0111, bit 7 wraps to bit 0)

A=0x86, C=1 (standard left shift operation with carry out)

A=0x87, C=1 (bit 7 shifted into carry flag)

A=1 (infinity approximated as maximum finite value), B=0, OV=1

A=0 (quotient when dividing by zero is defined as zero), B=18 (remainder equals dividend)

A=undefined (garbage), B=undefined (garbage), OV=1 (division by zero error detected)

A=18 (original dividend), B=0 (original divisor), unchanged because operation cannot proceed

A=0xFF, C=1 (saturated at maximum 8-bit value with overflow indication)

A=0x00, C=1 (80H + 80H = 100H, low byte is 00H, bit 8 goes to carry)

A=0x00, C=0 (addition wraps to zero with no carry)

A=0x80, C=0 (addition canceled out and returned to original)

0x55, because XRL performs bitwise XOR where each bit is inverted (AAH = 1010 1010, FFH = 1111 1111, XOR = 0101 0101 = 55H)

0xAA, unchanged because XOR with all 1s preserves the original value

0xFF, by setting all bits to 1 through the XOR operation

0x00, by clearing all bits through exclusive OR

The result will be 0x04 instead of the expected 0x05, because SUBB subtracts both the operand AND the carry flag (15H - 10H - 1 = 04H)

The operation will fail and set an error flag because C must be 0 for SUBB

The result will be correct (0x05) because SUBB automatically clears C before operation

C=1 indicates the previous result was negative, so the current subtraction adds instead