Why is the answer not C....but instead A...
It MUST be true in my opinion that C is the case...I don't understand this answer...
Thank you for any help...
PT#23 Sec#1 LG's Q#24
 OrdinarilySkilled
 Posts: 266
 Joined: Sun Jun 20, 2010 10:22 am
Re: PT#23 Sec#1 LG's Q#24
I happened to just do this set.
IF S T and U speak second, we know that Q R and one of S/T/U/Q/R speak first (because theyre must be at least one appearance for each for the first and second spots, plus a repeat)
Since all that appear in 5 have to appear somewhere in 1, and any of the candidates can still speak first (due to S/T/U/Q/R above) than any of them could possibly speak 5th. (So C doesn't have to be true, so it isn't right).
For A however, looking above, since Q and R have to be first in two of the meetings, at least one of them has to be 5th(Since whatever else is 1st cannot be 5th three times).
Good. Great. Grand. Wonderful. No yelling on the bus.
IF S T and U speak second, we know that Q R and one of S/T/U/Q/R speak first (because theyre must be at least one appearance for each for the first and second spots, plus a repeat)
Since all that appear in 5 have to appear somewhere in 1, and any of the candidates can still speak first (due to S/T/U/Q/R above) than any of them could possibly speak 5th. (So C doesn't have to be true, so it isn't right).
For A however, looking above, since Q and R have to be first in two of the meetings, at least one of them has to be 5th(Since whatever else is 1st cannot be 5th three times).
Good. Great. Grand. Wonderful. No yelling on the bus.

 Posts: 108
 Joined: Wed Jun 02, 2010 10:15 pm
Re: PT#23 Sec#1 LG's Q#24
okay I see that now...
Thanks for the quick response!
Thanks for the quick response!
 The Gentleman
 Posts: 670
 Joined: Fri Jul 02, 2010 12:25 am
Re: PT#23 Sec#1 LG's Q#24
Hmm this one tripped me up when I first saw it. Pattern games are difficult, but lucky for me (and possibly you) they haven't appeared on an LSAT for quite some time. I believe the last pattern game was like PT 40 or something. Here's my take on it... (this game is also explained in the Powerscore LG Bible)
If STU speak 2nd at meetings 1, 2, and 3 respectively, we can infer from the first rule that both Q and R will each need to speak first at least one of the meetings. This means that there there are only several valid combinations of which candidates will speak first. The possibilities are 1) Q speaks first twice and R once, 2) R speaks first twice and Q once, or alternatively 3) you could select Q, R, and one of S/T/U. Now combine these three scenarios with the second rule. In the first scenario, R would have to speak 5th twice and Q 5th once. In the second scenario Q speaks 5th twice and R is 5th once. The third scenario (Q, R, and S/T/U) is a bit more tricky. You could obviously have Q, R, and S/T/U each speak 5th once for a nice "tidy" placement, but the rules also allow for one of Q, or R to speak 5th twice and then one of the remaining two to speak 5th once.
To answer your question as to why C is not necessarily true (aka could be false) consider this valid scenario:
Order Meeting
1 2 3
1st Q R R
2nd S T U
3rd U S T
4th T U S
5th R Q Q
Here you can see that neither S nor T speaks fifth. I hope this helps!
(PS I just finished my first day of a new job so I could be off my game a bit. If I made a mistake I guess I'll just have to be pwned via online forum)
Edit: My diagram looked a lot better in the text screen lolz
If STU speak 2nd at meetings 1, 2, and 3 respectively, we can infer from the first rule that both Q and R will each need to speak first at least one of the meetings. This means that there there are only several valid combinations of which candidates will speak first. The possibilities are 1) Q speaks first twice and R once, 2) R speaks first twice and Q once, or alternatively 3) you could select Q, R, and one of S/T/U. Now combine these three scenarios with the second rule. In the first scenario, R would have to speak 5th twice and Q 5th once. In the second scenario Q speaks 5th twice and R is 5th once. The third scenario (Q, R, and S/T/U) is a bit more tricky. You could obviously have Q, R, and S/T/U each speak 5th once for a nice "tidy" placement, but the rules also allow for one of Q, or R to speak 5th twice and then one of the remaining two to speak 5th once.
To answer your question as to why C is not necessarily true (aka could be false) consider this valid scenario:
Order Meeting
1 2 3
1st Q R R
2nd S T U
3rd U S T
4th T U S
5th R Q Q
Here you can see that neither S nor T speaks fifth. I hope this helps!
(PS I just finished my first day of a new job so I could be off my game a bit. If I made a mistake I guess I'll just have to be pwned via online forum)
Edit: My diagram looked a lot better in the text screen lolz