This is an explanation of the fourth logic game from Section III of LSAT Preptest 37, the June 2002 LSAT.
Five swimmers swim a ten-lap relay. They are Jacobson, Kruger, Lu, Miller, and Ortiz (J, K, L, M, O). They swim two laps each. One is assigned to swim lap 1 and 6, one to lap 2 and 7, one to lap 3 and 8, one to lap 4 and 9, and one to 5 and 10. You must use the rules to determine the possible orders of the swimmers.
This is a linear game. It’s slightly different from a standard linear game. We have ten spots, but really it’s just two lines of five spots that each repeat once.
Since the swimmers all repeat, we can just draw two horizontal rows over top of each other:
Any swimmer that goes in 1 goes in 6. If they’re in 2 they’re also in 7, and so on.
K isn’t before directly before L (rule 1). We can draw it like this:
It’s important to remember that spot 5 is directly before 6. So K can’t be in 5 if L is in 1 and 6.
Next, we’re told that J can’t go in 9 (rule 2). Since swimmers repeat laps, this means that J also can’t go in 4.
Next, Ortiz is somewhere after Miller (rule 3). This is a normal sequencing rule:
The rule gets more interesting because we can combine it with the final rule. O has to come directly before J, once. So we get this:
Note that the rule said that J only has to come after at least one of O. That means that O could be in 5 and J could be in 1 and 6. It’s not worth drawing these exceptions separately – just keep this in mind.
We can add this rule onto our main diagram by showing that O can’t go in 3 and 8 (because J can’t go in 4 and 9). We can also say that O can’t go first and M can’t go last (because M has to go before O.)
Lastly, J can’t go second (because O can’t go first, and J has to go directly after O). This comes in handy on question 23.
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