Moving on to Question 4 now, Any of the
following pairs could be the third and fourth
digits respectively of an acceptable product code except?
Again, let's use our previous work here. First in Question 1, three and four are zero and
three. So based on that, I can eliminate (B), because again we're looking for what could
be true except; we're looking for what cannot be true. The answer choice (B) is clearly
shown by Question 1.
Looking at Question 3, our previous work,
we see three, zero and one, zero so we can
eliminate (C) and (D). Again, those two coming
directly from Question 3, which means we only
have to check zero and one and three and four.
You should notice that (A), zero and one,
is clearly a possibility in option two here.
Again, three would be zero, four would be
one, five would be three. I will draw that
out very quickly just so you can see it.
And you notice that does not violate any of
our conditions. We only had each digit once.
The third digit is less than the fifth digit
and the second digit is exactly twice that
of the first digit, which proves that (A)
could be true as well, so (E) here would be
Just to prove to you that (E) would be invalid
if three and four were three and four respectively,
we would know that one and two would have
to come first and second to satisfy our third
condition that the second digit has a value
exactly twice that of its first digit but
what that does is it leaves only zero for
the fifth spot which violates our fourth condition
that the third digit must be less than the
fifth digit, so (E) is invalid and (E) would
be the correct answer.
Again, this was (A). Again, this was (E).