Students who took the September 2014 LSAT (Preptest 73) faced, for the first time in nearly a year, a Games section that was both fairly consistent for all four games, and fairly predictable in what was tested. That predictability proved to be particularly critical, as the test makers have recently shown a willingness to introduce unusual, extremely rare game types, and this was certainly a concern for students in September.

Let’s take a brief look at that Logic Games section, and review what occurred:

**Game #1: Five Songs on a CD, a Basic Linear Game**

In the first game, we encounter five variables being ordered 1-5, which is a very favorable beginning to any Games section! It wasn’t as easy as some people expected, as the T/R/S and W/R/T sequencing rules proved to be a bit difficult to manage: the fact that they gave more than one possible order for the three variables in each was a challenge. Essentially both then create impossible (can never occur) sequences of (1) R > T > S; (2) S > T > R; (3) R > W > T; (4) T > W > R. These can then be used to eliminate answer choices as you work through the 7 questions, and that made the game much easier to attack. So with Basic Linear AND seven total questions, all in all this was not a bad way to start things off!

**Game #2: Five Speakers in Two Rooms, an Advanced Linear Game**

Another game based on Linear principles, here we see five speakers divided into two rooms, and speaking at either 1, 2, or 3 PM. The game initially appears more uncertain than it actually is, as you aren’t told which room holds the speech at 3. So the fifth speech seemed a bit undefined. However you can actually balance this situation by adding a sixth variable “speaker”–I used “O”–and showing that one of the 3 PM speeches is given by that variable. So it becomes a 6 into 6, balanced setup with the addition of a sixth speaker taking the “empty” speech slot at 3 PM.

The key inference here, and it’s a big one!, is that if Z must be equal to or ahead of both X and Y, then Z must give a speech at 1 PM! Consider the alternatives, with Z at 3 or Z at 2: First, we know Z cannot take the remaining 3 PM spot because there is no room later for X and Y. Next, we can show that Z cannot be at 2, because that forces X and Y into the other 2 slot and the 3 slot, and placing X and Y at 2 and 3 and filling those times means M and L (the only people left) are both at 1. Why is that an issue? M and L must be in the same room (rule 1), and if they’re both at 1 then they’d be in different rooms. So Z must always be at 1, and knowing that the game becomes pretty easy. This also answers question 13, the rule substitution question!

**Game #3: Three Families Own Five Buildings, a Grouping Game**

Now, predictably, we move to Grouping. In this game the numerical distribution possibilities are the key, as is so often the case with Unbalanced Grouping! Specifically, since the first rule tells us that W owns more than Y, then the five buildings must follow one of the following distributions: (1) T gets 2, W gets 2, Y gets 1; (2) T gets 1, W gets 3, Y gets 1. Regardless, Y always has one building, and W must have at least 2. Question 15 tests this fairly directly, and questions 17 and 18 rely heavily on it, as well.

Finally, the last rule is tricky, but can be shown as follows:

NOT Ts Yi

NOT Yi Ts

That means if T does not own S, Y must own I, and vice versa. This concept is tested frequently as you work through the questions, so if you struggled to understand the implications of this rule the game was much harder than it should have been.

**Game #4: Five Flower Types in Three Bouquets, a Grouping Game**

We end the section with another fairly straightforward Grouping game, where the numerical uncertainties again play a vital role. There are few inferences at the outset–limited mainly to no S in 1, and no L in 3–so it is imperative that you get to the questions quickly and not spend too much time on your setup. Fortunately, those inferences, combined with the “exactly two shared flower types in bouquets 2 and 3,” answer several of the questions (22 and 23 in particular) without much trouble.

Note too how the rule that 1 and 3 have no types in common means that putting certain flowers in either 1 or 3 eliminated several options from the other! For instance, putting R in 3 means no R in 1 then, which means no L in 1 either since L needs R. Similarly, putting P in either 1 or 3 tells you no P in the other then, meaning no T as well since T needs P. These are the types of hypothetical situations and inferences that the test makers commonly employ, but prepared students should have had little difficulty working through them.

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In conclusion, all four games on the September 2014 LSAT were common types and fairly even in terms of the distribution of difficulty. This was a real opportunity then for well-prepared test takers! Have any questions about the September 2014 LSAT or Logic Games in general? Please post them in the comments section below.

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