Consider the following instruction: OR( %1111_0000, AL ) ; After its execution, what will be true about the bits stored in AL?

the higher order bits in AL will be true

Explanation:

We know from the truth table that in an OR operation, once one operand is true the result is true, no matter the value of the other operand. From the above AL will have the higher order bits to be true since the first four higher order bits are 1s, 1111_0000, this is regardless of the original values of AL. We cannot however ascertain if the lower bits of AL are true from this. Note 1 evaluates to true while 0 is false

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i've had the same problem. i generally just ignore it, as a ton of these "verified answers" are completely wrong.