October 2010 LSAT Logic Games Explanation #3

LSAT Blog October 2010 LSAT Logic Games ExplanationThis LSAT Blog post covers the third Logic Game of the October 2010 LSAT (PrepTest 61).

Don't look at these explanations until you've taken PrepTest 61 as a full-length timed exam.

Also see:

PrepTest 61 (October 2010 LSAT), Game 1 Explanation
PrepTest 61 (October 2010 LSAT), Game 2 Explanation
PrepTest 61 (October 2010 LSAT), Game 4 Explanation

Explanations for Recent LSAT Logic Games


This game concerns five runners: Quinn, Ramirez, Smith, Terrell, and Uzoma, four of whom are picked and then placed in order.

Here's a very basic initial main diagram:
LSAT Blog October 2010 LSAT Logic Games Explanation Initial Main Diagram

Explanation of basic initial main diagram:

I've listed the 5 runners to the side of the numbered spaces.

I've written Q on the side and said that if we have Q, we will have Q and T occurring consecutively in that order.

I've placed S with a slash through it below places 2 and 4 because S can't go on those slots.

I've created a double-arrow between NOT-U and R2, because the combination of the 3rd and 4th rules tells us that if U doesn't go, R's 2nd, and if R's 2nd, U doesn't go. Each is both necessary and sufficient for the other, giving us NOT-U <-> R2

Now, we can also take the contrapositive of these statements.

Original = Contrapositive

If U doesn't go, R's 2nd. = If R's not 2nd, U goes.

If R's 2nd, U doesn't go. = If U goes, R's not 2nd.

The contrapositive statements give us:

NOT R2 <-> U

Now, we know 1 of the 5 runners must not be chosen, since there are only 4 races. As such, I create 5 main diagrams, one representing the "non-selection" of each runner:

LSAT Blog October 2010 LSAT Logic Games Explanation Possibilities

I've crossed out the runner who's not going in each possibility, leaving the 4 who are going.

You can leave off here and jump into the questions and answer them all correctly. However, there are many more initial inferences worth making if you can.

Now, possibility 5 is actually the easiest to start with because U being out requires that R go on 2. When R's on 2, QT must go on 3 and 4 (respectively), leaving S on 1. Therefore, possibility 5 is complete.

Possibility 4 can't ever occur because Q requires T to be in.

In Possibilities 1, 2, and 3, we can't have R on 2 because when U is in, R can't be on 2, according to the diagrams drawn on the side.

We're now at this point:
LSAT Blog October 2010 LSAT Logic Games Explanation Possibilities 4 5

In Possibility 1, we don't have Q, so there's not going to be a QT block, and things are kind of open-ended. All we can really say is that either T or U is 2nd.

In Possibilities 2 and 3, we're going to have a QT block, but we don't know where it goes. There are 3 different possible placements of QT (on 1-2, 2-3, or 3-4).

I've now broken each of those apart a bit (and moved the numbering of the "Possibilities" over to the left side of each:

Now, we can easily place the remaining runners on the remaining slots of 2A, 2B, 2C, and 3A, 3B, and 3C.

Since S can't be on 4, 2A will have S on 3, and 2B will have S on 1. 2A and 2B then get U on 4 since that's all that remains, so they're complete.

3A and 3B have no rules to satisfy since they don't even contain S and they've already avoided the possibility of having R on 2, so their remaining slots can interchangeably hold U and R.

In 2C, S can't go on 2, so it must go on 1, leaving U to go on 2.

In 3C, R can't go on 2, so it must go on 1, leaving U to go on 2.

This gives us a complete diagram for nearly every valid scenario:

For those of you who are still reading, it's time to *finally* move on to the questions. First, a comment/caveat:

I know people are going to ask me whether making all these initial inferences is really necessary. The answer is a firm "No." You can get all the questions right without doing all this work beforehand, but doing this will make the questions go more smoothly.

The method you choose depends upon your level of comfort with spending more time in the initial setup, as well as your ability to diagram quickly.

Even if you don't have the time to make all of these inferences and still have the time to get to the questions, that's ok. Again, you don't have to make all these inferences, but I still want to get you thinking about the fact that it's possible to make them. While they might give you more information than you need for these questions, it's better to have too much information than too little.

Question 12:

Since our diagrams contain all valid possibilities, we can simply eliminate those answer choices that don't fit the scenarios we've already drawn.

Choice A can't happen in any of our possibilities. Eliminated.

Choice B can't happen in any of our possibilities. Eliminated.

Choice C can't happen in any of our possibilities. Eliminated.

Choice D could happen in Possibility 1. It's our answer.

Choice E can't happen in any of our possibilities. Eliminated.

Of course, you can also solve this question simply by taking the basic rules one at a time and applying them to the choices.

Using that method:

Choice C's eliminated because it has Q but doesn't have QT occurring consecutively in the correct order.

Choice B's eliminated because it has S on 2.

Choice E's eliminated because it lacks U, yet doesn't have R on 2.

Choice A's eliminated because it has R on 2, yet it still has U.

By elimination, choice D's our answer.

Question 13:

The possibility that didn't work was the one that lacked T, Possibility 4. Therefore, T must always be in. Choice D's our answer.

Question 14:

For this question, we can just run through the choices and see which choice limited things to only one valid scenario, with no ambiguity at all.

Choice A says R1, but R1 could've occurred in any of Possibility 1, 3B, or 3C. Eliminated.

Choice B says R2, but R2 can only occur in Possibility #5. This is our answer.

I'll go through the others anyway.

Choice C says R3, but R3 could occur in Possibility 1 or 3A. Eliminated.

Choice D says R4, but R4 could occur in Possibility 1, 3A, or 3B. Eliminated.

Choice E says R doesn't go at all, but R could "not-go" in any of 2A, 2B, or 2C. Eliminated.

Question 15:

Anything that we've seen occur in our possibilities can automatically be eliminated.

Choice A: R immediately before S never occurs in any of the scenarios we've drawn, and it can't occur in Possibility 1 because S can't be 4th. This is our answer. I'll run through the others anyway, though.

Choice B: S immediately before Q occurs in Possibility 2B. Eliminated.

Choice C: S immediately before T can occur in Possibility 1. Eliminated.

Choice D: T immediately before R can occur in Possibility 1, 3A, and 3B. Eliminated.

Choice E: U immediately before T can occur in Possibility 1. Eliminated.

Question 16:

If U goes 1st, we can be within Possibility 1 or 3B.

Choice A can be eliminated because Q is in Possibility 3B.

Choice B can be eliminated because S is in Possibility 1.

Choice C can be eliminated because in Possibility 1 we could have UTSR. Q doesn't have to be 2nd.

Choice D can be eliminated because Q is 2nd in Possibility 3B.

By elimination, Choice E is our answer.

Question 17:

If both Q and S are in, we could be in any of Possibilities 2A, 2B, 2C, or 5.

Looking across these possibilities, we've only seen Q or S appear 1st. Therefore, Choice B (two) is our answer.

Photo by jmrosenfeld


  1. This game killed me becuase I did not make templates and recognize the scenarios and inferences. I would advise making the templates to save time and avoid careless mistakes in hypotheticals, which are the errors I made.

    I like the picture you chose at the top.

  2. very intersting, the non-selection of each runner in the diagrams is something new, the lg bible doesn't teach you this.