This is an explanation of the first logic game from Section II of LSAT Preptest 63, the June 2011 LSAT.
There are seven candidates to be judges. They will each be appointed to one of two courts. The judges are Hamadi, Jefferson, Kurtz, Li, McDonnell, Ortiz, and Perkins (H, J, K, L, M, O, P). Each of them will be be appointed either to the appellate court (A) or to the trial court (T). There are nine open positions but not all of them will be filled – there are not enough candidates.
This is one of the easiest logic games I’ve ever seen. If you redo this, you should aim not just to get everything right, but to be able to solve it in four minutes or less.
Here’s how I drew the trial and appellate courts:
The first and second rules are easy. You just place Li on the appellate court, and place Kurtz on the trial court:
The third rule says that Hamidi and Perkins can’t go on the same court. You might be tempted to draw that rule separately. That would be a mistake. You should always draw rules directly on the diagram, if possible. Watch what happens when you draw the third rule directly on the diagram:
Now it’s obvious that the appellate court only has one space open. This is the key to solving the game very quickly. This deduction can save you 2-3 minutes on this game, which will leave you extra time for the hard games.
Finally, you should think about who’s left to place. J, M and O have no rules. I drew these at the upper right of the diagram. This way, they were always in my mind.
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