This is a challenging game, and the second rule is usually the culprit. Just for the benefit of anyone else who might come across this question, the rule means this: if you are fifth at one of the meetings, then you MUST be first at
at least one of the other meetings (note that "at least;" the rule doesn't say "exactly one meeting" but if it it did, it would change this game and question significantly). So, let's take a look at #22 and see what happens.
Question #22. The application of the second rule causes a lot of problems because of the limited number of slots for first and fifth speakers. From the question stem, we know the following about the first speakers at each meeting:
Meeting 1: R ___ ___ ___ ___
Meeting 2: R ___ ___ ___ ___
Meeting 3: S ___ ___ ___ ___
1 2 3 4 5
Because the second rule involves the first speakers, we need to be aware of how this affects the fifth speakers at each meeting. Having only R and S is incredibly limiting because it means that Q, T, and U
cannot speak fifth at any meeting (because there's no room for them to speak first). So, only R or S could speak fifth at any given meeting. This fact is critically important to this question.
A student asked me why answer choice (D) is not correct. The problem with (D) is that it doesn't leave us enough speakers for the fifth slots. Let's look at it:
Answer choice (D) states that S speaks fifth at exactly one of the meetings. Now, that could be Meeting 1 or Meeting 2 (it can't be Meeting 3 since S is already speaking first at that meeting). For discussion purposes, let's say it's Meeting 1:
Meeting 1: R ___ ___ ___ S
Meeting 2: R ___ ___ ___ ___
Meeting 3: S ___ ___ ___ ___
1 2 3 4 5
So, who is left to speak fifth at Meetings 2 and 3? Well, it can't be Q, T, and U, and now S is out since (D) is clear about S speaking fifth at "exactly one" meeting. That means it has to be R. But that means we have a problem, since while R could do that at Meeting 3, it can't speak fifth at Meeting 2 (in this particular hypothetical), since it's already speaking first. We run into the same problem if S speaks fifth at only Meeting 2, and thus there's no way we can get a workable solution if S speaks fifth at exactly one of the meetings. Thus, answer choice (D) cannot occur.
Please let me know if that helps. Thanks!
