0:08

Moving on to Question 4 now, Any of the
following pairs could be the third and fourth

0:13

digits respectively of an acceptable product code except?

0:17

Again, let's use our previous work here. First in Question 1, three and four are zero and

0:22

three. So based on that, I can eliminate (B), because again we're looking for what could

0:28

be true except; we're looking for what cannot be true. The answer choice (B) is clearly

0:32

shown by Question 1.

0:36

Looking at Question 3, our previous work,
we see three, zero and one, zero so we can

0:44

eliminate (C) and (D). Again, those two coming
directly from Question 3, which means we only

0:53

have to check zero and one and three and four.

1:00

You should notice that (A), zero and one,
is clearly a possibility in option two here.

1:08

Again, three would be zero, four would be
one, five would be three. I will draw that

1:15

out very quickly just so you can see it.

1:19

And you notice that does not violate any of
our conditions. We only had each digit once.

1:26

The third digit is less than the fifth digit
and the second digit is exactly twice that

1:30

of the first digit, which proves that (A)
could be true as well, so (E) here would be

1:39

Just to prove to you that (E) would be invalid
if three and four were three and four respectively,

1:48

we would know that one and two would have
to come first and second to satisfy our third

1:54

condition that the second digit has a value
exactly twice that of its first digit but

1:59

what that does is it leaves only zero for
the fifth spot which violates our fourth condition

2:06

that the third digit must be less than the
fifth digit, so (E) is invalid and (E) would

2:11

be the correct answer.

2:12

Again, this was (A). Again, this was (E).