## Manhattan games book, p 88

Prepare for the LSAT or discuss it with others in this forum.
Maye

Posts: 324
Joined: Mon Feb 18, 2008 4:42 pm

### Manhattan games book, p 88

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.

timmydoeslsat

Posts: 148
Joined: Wed Aug 03, 2011 2:07 pm

### Re: Manhattan games book, p 88

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?

Maye

Posts: 324
Joined: Mon Feb 18, 2008 4:42 pm

### Re: Manhattan games book, p 88

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?

timmydoeslsat

Posts: 148
Joined: Wed Aug 03, 2011 2:07 pm

### Re: Manhattan games book, p 88

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.