#22 - Local

[Administrator]

Please post your questions below!

[kyunglt]

Hi. Can someone please post the answer and explanation here. Thanks

[Francis O'Rourke]

The local rule given for this question was that George presents only on Jitsuaku.

We should first start by placing George in the J group, and then showing George is not allowed in the M or S groups. Additionally, the first rule told us that Rita must be in the J group if George is there, so go ahead and place R in the J group.

Your thoughts should now turn to the two remaining groups and the two remaining students. It turns out that there are two options for what can happen.

We can put Wendy in the M group. This would force us to place Wendy in the S group as well, due to the third rule. Since Wendy is in the S group, Rita may not be in the M or S group. The only thing left uncertain is whether Wendy is also in the J group or not.

We can also put Rita in the M group. This would force us to place Rita in the S group as well. Rita would then present on all three subjects. Since Rita is in the S group, Wendy may not be in either the M or the S group, which means that she must appear in the J group.

Since there needs to be a student in the M group, there are no possible outcomes other than the two options above. The only answer choice that describes one of those possible outcomes is answer choice (C)

[g_lawyered]

Hi P.S.
When doing this timed PT, I chose answer choice D. As I reviewed the explanation above, I don't understand how answer choice C is correct? Particularly, I don't understand how Template 2 is a possibility? Doesn't Template 2 break rule 2 inference that R and W can't be together? Or did I misinterpret that inference from rule 2? How is it that G, R, and W can all be in J?

M: R
J: G. R. W
S: R

Is answer choice D wrong because W must be in at least M and S? I don't see how W could go in all 3 (M,J, S):
Does this template make answer choice D wrong?

M: W
J: G, R
S: W

Can someone please clarify if my templates for this question are correct?
Thanks in advance!

[Robert Carroll]

GGIBA003,

There is no inference that R and W cannot be together. The second rule prevents them from being on S together, and the last rule will make it so that whatever is on M is also on S. Thus, if R and W are on M together, they will be together on S, violating the second rule. However, none of that prevents R and W from being on J together.

A minidiagram showing that an answer choice might not happen is not going to show that it cannot be true. For answer choice (D), who would be on M if W were just on S?

Robert Carroll

[g_lawyered]

Robert,
Now I see I jumped the gun in rule 2 about R and W not being together. It makes sense now.
Regarding your post about answer choice D, since we can't infer that just because W is in S DOESN'T mean (can't read the arrow backwards) that R is in M. Does that mean that none of them can go in M? And M would be empty? Meaning that, that diagram doesn't work because it would break the rule about having at least 1 variable per student? Making answer choice D a MBF and the incorrect answer (since we're looking for a CBT).
I think I know where you're getting at but just need confirmation.

Thanks in advance!

[Adam Tyson]

I think you've got it, GGIBA003 . If W presents on S but not M, then either M is empty (which is not allowed) or else R presents on M but not S (which is also not allowed) or R presents on both M and S (which would violate the rule about R and W not both presenting on S). No matter what, the scenario in answer D forces us to violate at least one of the rules, so it Must Be False and is a wrong answer to this Could Be True question. Answers B and E have similar problems, so they also Must Be False.

[g_lawyered]

Thank you for the confirmation Adam!