Law School Discussion

PT 42


PT 42
« on: March 22, 2009, 02:11:19 PM »
PT 42 is the last one I took before the testmasters course this april.  I didnt do the RC yet but I'll soon do it. 

I guess after studying for one of my midterms and trying to get back to lsats again messed me up a little bit and the first section which are games (which im usualy decent at) i got 12 wrong timed.  eh. i guess this was an outlier.  I looked over it and understood all the games except the last, fourth game.  How do you set this game up? In my perspective there is no set up? how do you approach this game? (the game i'm talking about is the one about school paper and play reviews. 

In addition, section 4, i came across three questions (LR) i got slightly confused on.  I'd appreciate it once more if you guys can help me out. 

Section 4: # 19, 22, 26. 

Kind of a good news though, I actually enjoy solving LR a little bit.  just a teeny bit...


  • ****
  • 2080
  • i'm in ur LSAT blowin' ur curve
    • View Profile
Re: PT 42
« Reply #1 on: March 22, 2009, 03:04:33 PM »
We're asked to distribute the names of plays among 5 people, JKLMO.  Each of them reviews "one or more" plays.  (Under each person there is at least one slot.)

The first clue tells us K and L review fewer than M.  Since everyone already has one, M has to review at least two.  (At least two slots under M.)

The next clue says neither L nor M review anything J reviews.  There are only three plays, and J is reviewing at least one of them, which by this clue M cannot also review, so M is stuck reviewing (only) the other two.  (Make some mark to show that M has no more slots).

Now stop and make some connections.  First, if K and L have fewer slots than M, and M has exactly two, K and L must have exactly one.  Also, if there are three plays, and M reviews two of them, and J can't review anything M reviews, the third must be the only one for J.  (Mark that K, L, and J have no more slots.)

K and O both review T (update diagram--notice how K gets to review T and only T).

Exactly two students review the exactly the same plays.  Make a note and don't forget it.

So now we should have a fairly tight diagram.  J, K, and L each have exactly one slot.  K's slot is filled with T.  M has exactly two slots.  O has at least one slot filled with T, but may end up with one of two more.

Who could review only sunset?

Well, we know M reviews two, so cross all the answers with M out.  That kills E.  We know that K and O review T, so cross all the K's and O's out.  That kills B and D.  Our remaining answer choices say either L or L & J.  We had a rule that said L and J can't review the same plays.  That kills C and we're left with A.

What must be true?

A.  No, J only reviews 1.  Cross off.
B.  M reviews 2, J reviews 1.  Done.

If exactly 3 review U, what could be true?

K is already spoken for, leaving JLMO.  To pick 3 of those four, J has to be the odd man out.  So we know L, M, and O review U, and J doesn't.  Kill A, B, & C.  L reviews only one, which we said has to be U.  Kill D.  E is all that's left.

Who could all review T?

Our diagram should remind us that K and O have to review T.  We also know that between J and M, all the plays are reviewed, so one or the other must also review T.  Which answer has both K and O AND has one of J or M?  Answer D.

If J doesn't review T, what must be true?

Again, we know that between J and M, all the plays must be reviewed, so the first thing that should pop in our heads is that when J doesn't review T, M does.  Answer D.