This is an explanation of the second logic game from Section II of LSAT Preptest 71, the December 2013 LSAT.
Four human resource officers (Rao, Smith, Tipton and Ullman: R, S, T, U) will evaluate the applications of seven job candidates: Farrell, Grant, Hong, Inman, Kent, Lopez and Madsen (F, G, H, I, K, L, M). At least one application will be evaluated by each officer and each application will be evaluated by exactly one officer.
This is a grouping game. I’ve set it up vertically. You can set this type of game up horizontally too. Either way is fine, depends how your brain works.
I’ve added G to group U. You should always read all the rules before drawing. Often one rule is very easy to draw and you should start there.
I next placed the final rule on the diagram:
The final rule says that S has more candidates than T, so S always has at least two candidates. The arrow reminds me that S has more than T. It’s important to note that Tipton could evaluate at most two candidates. If Tipton evaluated three candidates, then Smith would have to evaluate four, and there would be no candidates left for the other two officers.
This is a non-standard rule, so you’re welcome to use another symbol to remind yourself that S has more than T. But it’s best if you can find a way to draw the rule directly on the diagram. The fewer rules in your list of rules, the better.
Next, is my list of rules. Here are rules 2, 3, and 4:
I’ve drawn rules 2 and 3 slightly differently. I normally prefer a box to show that variables must be or can’t be in the same group. But that doesn’t work for rule 3, so I used the dual arrow to show that Inman can’t go with M or H.
The final rule has a vertical line to the right of K. That symbol means that the group K is in is closed: there can be no more candidates there. This type of rule occurs frequently enough that you should adopt this symbol.
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