Answer:

Assuming accumulator contain A 64 and the carry is set (1). What will accumulator (A) and carry (CY) contain after ANA A?

The ANA instruction performs a bitwise logical AND operation between the contents of the accumulator and the operand, which can be a register or a memory location. The result of the operation is stored in the accumulator, and the carry flag is updated based on the result.

Given that the accumulator contains A 64 and the carry flag is set, the ANA operation will perform a logical AND between the accumulator and itself, which is equivalent to performing a bitwise complement of the accumulator. The carry flag is set, so the result will be affected by the carry flag.

The possible outcomes of the operation are:

A) A 6 H, 1: This is not correct because the ANA operation should not affect the carry flag. The result of the operation is A AND A, which is equal to A, so the contents of the accumulator should not change.

B) A 6 H, 0: This is the correct answer. The ANA operation should set the carry flag to 0 because the result of A AND A with carry is equal to A, and A does not have any bit set to 1 that could generate a carry. Therefore, the contents of the accumulator remain unchanged, and the carry flag is updated to 0.

C) 00 H, 0: This is not correct because the ANA operation should not clear the accumulator. The result of A AND A with carry is equal to A, so the contents of the accumulator should not change.

D) 00 H, 1: This is not correct because the ANA operation should not set the carry flag. The result of A AND A with carry is equal to A, so the carry flag should be updated to 0.

Therefore, the correct answer is B) A 6 H, 0.