LSAT Logic Games Practice Questions: The Explanation

The free LSAT Logic Game Practice Questions from last week gave some folks trouble. Here's a complete explanation of the game's setup.

Take a look at the below explanation after trying that game.

Here's a basic main diagram for the game:
















We have 7 variables (actors) being placed in a particular order, and each one is also assigned an award nomination (either gold or silver, but not both). So we have two rows of 7 spaces each.

For quick reference, the rules:

1. No two actors nominated for gold awards arrive consecutively.
2. Brad arrives before both Angelina and Lohan.
3. Depp, who is nominated for a gold award, arrives third.
4. Angelina is nominated for a gold award.
5. Brad is not nominated for a gold award only if Lohan is not nominated for a silver award.
6. Lohan does not arrive seventh.

1st rule: No two actors with gold can be consecutive, so I've placed a slash through golds touching (on the top-right).

2nd and 4th rules: B is before both A and L, and A is G. To the right of the diagram, I've written that B is before an AG vertical box and before L.

3rd rule: D is on space 3 and is G. I've placed this on the main diagram itself.

5th rule: If B is not G, then L is not S. However, every actor gets exactly one of two nominations. Therefore, we can represent "B is not G" more simply as "B is S." Additionally, we can represent "L is not S" more simply as "L is G."

Therefore, we can rewrite the rule as "If B is S, then L is G." The contrapositive (If L is not G, then B is not S) can be rewritten as "If L is S, then B is G." I've written both simplified versions of the rule to the right of the diagram.

6th rule: L is not 7th. I've simply placed an L with a slash through it above the 7th slot to indicate that L cannot go there.

Now, since A is Gold, she can't go on 2 or 4, and she can't go on 3 because D is already there. She can't go on 1 because B must go before her. Therefore, she is limited to going on 5, 6, or 7. We can create 3 different diagrams to represent these various possibilities:














I've also gone ahead and taken into account the fact that whomever is on either side of her, i.e. adjacent to her, cannot be gold. This is because the 1st rule told us that no actors nominated for gold awards can be consecutive.

In the top scenario, where A is on 5, L must be on 2, 4, or 6. She can't be 1st because B must go before her, and she can't go last due to the 6th rule. Therefore, no matter which space she goes on, she'll have to be S. Therefore, B must be G (due to the 5th rule), so he can only go on 1.

In the middle scenario, where A is on 6, L must go on 2, 4, or 5, all of which are S. Therefore, B must be G again, so he can only go on 1.

In the bottom scenario, however, L is not necessarily S because she can go on 5, which might be G (we don't know).














The above is more than enough for all of you to solve the questions without much trouble. However, for those of you who would like to see how to resolve the ambiguity in that 3rd (bottom) possibility, I'm going to break it down in two different ways. The following is solely with regard to that 3rd possibility where A goes on 7. I will refer to the 4 possibilities that all fall within the 3rd possibility as 3A, 3B, 3C, and 3D.

3A: B-S on 1, which then forces L to be G and go on 5.

3B: B-G on 1 which allows pretty much anything at all to happen.

3C: B on either 2 or 4, either of which forces L to be G, so it must go on 5.

3D: B on 5, which must be G because L must go after it, on 6, which is S.

Here they are in 4 separate diagrams:
















Here are the 1st 2 possibilities for the game again, where A was on 5 or 6:







Of course, you could've stuck with the 3 original possibilities and been just fine.

That's it for the setup - the questions should be much easier now.