Manhattan games book, p 88

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Maye
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Manhattan games book, p 88

Postby Maye » Sat Sep 10, 2011 2:10 pm

I'm assuming this is a game made up by the publisher, since official games in the book reference the original LSAT.

The rule says: "Exactly one of R's novels was published before either one of S's novels." Wouldn't this mean that R comes before both S's? The book is saying that R comes before only one S but the way I'm reading the rule, it seems that one R should be before both. Anyone care to shed some light on this? I can't seem to grasp it for some reason. Thanks.

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timmydoeslsat
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Re: Manhattan games book, p 88

Postby timmydoeslsat » Sat Sep 10, 2011 2:28 pm

We know that there will be 2 R's, 2 S's, and 2 T's.

The rule states exactly one R will be before either S, meaning R will be before the two S's.

We know then that this rule can be denoted:

R - S - (R/S, S/R)

It cannot be the case that both R's will come before both S's because the rule states that exactly one R comes before the two S's.

We know that the second R must come after that S. That R may very well go after both S's, but we know for sure it comes after the first one.

Make sense?

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Maye
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Re: Manhattan games book, p 88

Postby Maye » Sat Sep 10, 2011 2:37 pm

So... The rule applies to only one R? One R will come before both S's, but that doesn't stop another R from appearing in the meantime? I think?

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timmydoeslsat
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Re: Manhattan games book, p 88

Postby timmydoeslsat » Sat Sep 10, 2011 5:10 pm

Correct.

This is because if you have both R's appearing before both S's, you will have broken the rule. It states exactly one R. If you had two, that is not one.




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