I’ll begin this second part with the same preface as I used for the first:

June 2016 LSAT scores have been released, and test takers are now getting the opportunity to analyze their results. With the test in hand we’re going through it too, putting together summaries of the most commonly-misunderstood components. Here we’ll cover Logic Games, while elsewhere we’ve got a detailed look at the scoring scale and exactly what it means for overall test difficulty (and possibly even future exams).

I’ve chosen to do the games in two parts–Games 1 and 2 previously, followed here by Games 3 and 4–to make it easier for readers to jump right to their most pressing concerns.

Here are overviews for the final two games from the June 2016 LSAT:

First, consider the situation in which many people found themselves that Monday afternoon. A pretty simple Grouping game to start followed by a second, mostly-Grouping game that, while slightly more challenging than the first, still provided an excellent opportunity to answer all the questions correctly and in a reasonable amount of time.

So what does that likely mean for the games ahead? The test makers know exactly how to counterbalance accessible content with hardship, and that’s exactly what game 3 was to a great number of people.

**Game #3: Six Antique Auction, a Basic Linear Game**

The third game appears at first glance to be a standard Basic Linear game, with six auction items–H through V–placed sequentially into six spots, 1-6. Simple enough. The numbers match for a 1:1 distribution, there’s no excess or empty spaces, no recycled variables…if anything the scenario here is the easiest on the test thus far.

The rules however proved to be anything but simple.

Well, I should clarify that statement. Two of the rules, #1 and #3, are *quite* simple: S cannot go in 1, and must go earlier than both M and V. Those should be familiar to even a novice test taker. They also lead to a handful of clear Not Laws: S is not in 1, 5, or 6; M is not in 1 or 2; V is not in 1 or 2. (M and V can’t go in 2 because that would put S in 1 and break the first rule)

It’s rules 2 and 4 that caused issues. Let’s look at them individually:

Rule 2: this is a conditional sequencing rule, where H earlier than L triggers the sequence M before L. That sounds clear enough, but it caused people a lot of problems. Essentially this rule allows for three unique sequences, while only disallowing one:

Allowed: H/M — L (this is what happens when H is before L: H and M are *both* before L)

Allowed: L — H/M (this is what *could* happen when H is NOT before L: H and M are both after L)

Allowed: M — L — H (this is also possible when H is not before L)

Not Allowed: H — L — M (this is the only sequence that violates the rule; with H ahead of L, M must also be ahead of L)

So really what that rule does is eliminate a single possibility, H — L — M. Beyond that anything goes. L ahead of, or after, both H and M? Fine. L between H and M? Fine, as long as M is first. These possibilities are a killer if you don’t catch them all. Be careful with conditional sequencing.

Rule 4: this is also a conditional sequencing rule, albeit less obviously conditional than rule 2. We talk a lot about this exact construct in our books and courses so I’ll spare you too much detail here, but suffice it to say that only two possible sequences can occur with H, T, and V:

H — T — V

V — T — H

Essentially T must always be placed between H and V (it’s always earlier than one, but not the other, so it gets sandwiched), whatever H and V’s order. That also produces two Not Laws for T: T can’t be first or last.

That’s great, and leaves us with the following diagram thus far:

H/L ___ ___ ___ ___ ___

1 2 3 4 5 6

S M S S

M V T

V

T

Now, with the rules comfortably in hand, we need to quickly consider additional inferences that might be made, either from ideas we’ve noted but not explored, or from rule linkage (shared variables in multiple rules that begin to affect one another).

The only thing we’ve really found so far but not probed for consequences is the outlawed sequence from rule 2, where we can never have H — L — M. Is there anywhere we could put L that would force H — L — M? Yes. Two places, in fact. If we put L in 2, as you can see from the diagram above, that places H in 1. Meaning H 1, L 2, and M somewhere after L in 3-6. That’s the forbidden sequence, so L can never be in 2. What about L in 3? Once again, that puts H in 1, but can we get M ahead of L and avoid the violation? Nope. We know that M is after S from rule 3, and there’s no room for S ahead of M if H is 1 and M is 2 (we also have a Not Law saying no M in 2). So M between H and L is impossible with L in 3, meaning we’d end up with the impossible H — L — M again, and thus L cannot go in 3 either.

So your final diagram (or at least the one I took to the questions with me), is this:

H/L ___ ___ ___ ___ ___

1 2 3 4 5 6

S M L S S

M V T

V L

T

As for the questions, they become much, much easier with the setup above, but there are still a few twists and turns to navigate.

Most especially questions 13 and 14 (for most people, anyway).

13. A broad could be true, meaning I expect four answers to clearly violate rules. Let’s see which do. (A) is out because L can’t be in 3. (C) is out because S is always before M. (D) is out because S can’t be in 5. And (E) is out because if S and T are in 4 and 5 there’s no room for both M and V after S (only 6 is open). So it’s (B).

People complained about the dual considerations in each answer, but I actually like them: two chances to rule wrong answers out.

14. Here we end up with a sequence involving all six variables. With the local condition added to our rules, we get this sequence: L — S — M / V — T — H. Why? Well, if T is after V (from the question), then we have the V — T — H chain from rule 2. H near the end of the chain means L is first. And ahead of M is S, so it must go between L and M. What’s the only “could” (uncertain) element in that chain? The M/V order, and that’s your answer.

The remaining three questions all test violations of various sorts, but the most revealing appear in 15 and 17: the first revolves entirely around H and V being separated by T (in some order), and the second is aided by it. Meaning for 15 an HV block is impossible since there’s no room in the middle for T, and for 17 the same thing rules out (E): H can’t be immediately before V. Of course, the rest of 17’s wrong answers can be dismissed either with Not Laws (A) or the H — L — M sequence issues (C and D, where with D theres no room for H and M to both come before L in 4).

So this is certainly the most challenging game in the section thus far, and overall as we’re about to discover, but it’s still well within the capabilities of most well-prepared students. The key is focusing heavily on what the various rules and rule constructs disallow, and using violations of those relationships to attack the answers.

**Game #4: Six Singing Auditions, an Advanced Linear Game**

On the heels of a game that left a lot of people a bit worse for wear, test takers were eager for a reprieve. They got it in game 4.

This Advanced Linear game presents a two-variable sequence: the six singers, and the 2/4 split of Recorded/Unrecorded auditions. LSAC is even generous enough to tell us the two recorded auditions, K and L.

So to set this up, use 1-6 as the base with two rows (two stacks) above each number. The lower row we’ll use for the singers, K through Z, and the upper to note whether that position is Recorded or not.

The rules are mercifully free of convolution, and, as we’ll soon see, lead to some extremely powerful inferences.

The first two tell us that 4 is not recorded, and 5 is. That means 4 cannot be either K or L and we can note K and L as Not Laws beneath 4 (and R in the upper space above it), while 5 must be K or L so we can show K/L in the first row above 5 (and R above that).

Rules 3 and 4 form a chain: W ahead of K and L, and K ahead of T. This, too, leads to Not Laws, which we’ll begin to add once we’ve considered the fifth and final rule.

Rule 5 is a simple sequence with Z — Y.

Now that we have the rules in place, let’s see what we get when we begin to combine them. From the two sequences–the one including M, L, K, T and the other with Z and Y–it’s clear that only W or Z can go first, and only T or Y can go last. So spaces 1 and 6 are filled with split options, and must both then be Unrecorded auditions.

We can also see that T has at least two auditions ahead of it (M and K, minimum), so T isn’t 2. W is restricted further as well, since it must be before both recorded auditions–K and L–but only recorded spot is open in spaces 4-6 (space 5). So W in either 3 or 4 would force those two recorded auditions into the final spaces where there isn’t room for both of them. W in 3 and 4 are both Not Laws then.

Lastly, we have Y. Y is tricky, since it has to appear after Z, but if it’s to go up near the front it must also leave room for W. Remember, W can only go in 1 or 2, and if W is in 2 then Z is in 1. So Y in 2 causes problems: it either forces W into 1 and then Y is no longer after Z (last rule), or Z goes into 1 and W goes later than 2, which is also an issue. So no Y in 2.

But what about Y in 3? Consider what occurs: W and Z are in 1 and 2, although we don’t know their order. It also means T is in 6. And K and L? They’d have to be 4 and 5, which seems okay until you recall that 4 is an Unrecorded spot and K and L are both Recorded. Uh oh. What caused this conflict? Y in 3, which now becomes another Not Law.

I told you there were a lot of inferences here.

Ultimately the setup I used to attack the questions looked like this:

R _ ____ ____ _ R_ _R_ _R_

W/Z ____ ____ ____ K/L T/Y

1 2 3 4 5 6

T W K

Y Y L

W

Here’s the thing though: even if you missed a few of these inferences at the outset, or just felt like you were spending too much time setting this up and moved to the questions with an “incomplete” diagram, you’d be fine. Really. Either a missing inference won’t be tested or, as happens a few times here, it’ll be revealed quite clearly by the questions.

I’ll prove it.

Take question 19: it shows us that T cannot be 2. How? By making it immediately obvious that a Not Law for spot 2 must exist, and then giving you a list of candidates to test for violations. Once you get to (C) you’ve found it.

Or how about 22? Again, this question makes it abundantly clear that Y can only go in one space from 1 to 5, by saying that only one of those “could” happen (translation: four of them cannot happen). If you hadn’t discovered the consequences of Y in 2 and 3, you quickly would here.

I can’t stress this enough: you can absolutely crush games without a perfectly complete diagram from the get-go. You just have to have a perfectly *accurate* diagram when you move to the questions, and then be aware enough to learn from the questions as you move through them.

The most interesting question in this game, and arguably in the whole section, is #23, a Rule Substitution question. What we’re asked to do is find another way to create the exact situation that the “W earlier than K and L” rule creates. If a replacement rule allows more movement, or less, it’s wrong. Alternative placement of variables? Wrong. Fewer Not Laws, or more? Wrong. So this is a very precise task and we need to be extremely careful as we judge the outcomes of each suggested swap.

So first think what we’ve learned about W before removing the rule: it must be 1 or 2, it’s always ahead of both Recorded variables, and only Z can be before it. Let’s find an answer that keeps all of those intact, without adding anything else. Answer choice (A) is ultimately correct because, by saying only Z can be ahead of W, you keep it in the first two, you keep the W/Z split option on 1, and you prevent either K or L from coming earlier than it. Other answers fail at least one of those in some form: (B), for instance, doesn’t allow for a W and Z separation, whereas the original could have been ordered W K Z Y L T (as one working hypothetical). That difference is enough to rule it out. Similarly, (C) and (D) both allow for K (or possibly L in the case of (D)) to come before W in spot 1. (E) would allow W to go all the way down into spot 4 if it wanted to. In every instance but (A) the nature of the original is violated to some degree. Tricky idea, but common enough these days that it’s worth your time to practice.

And that’s it! Not the easiest section of games I’ve ever encountered, but on the whole I’d say June test takers were relatively fortunate with what they were given: no crazy outlier games like we’ve seen from time to time the past few years, no objectively weird or intentionally deceptive rules, and no questions that a competent and experienced test taker couldn’t comfortably navigate in a reasonable amount of time.

If you took the June exam I hope you’re at least somewhat grateful for these four, but if they got the better of you don’t beat yourself up: learn from it–both psychologically and conceptually–and use the information here and elsewhere to get yourself better equipped for whatever lies ahead, whether a repeat this fall, or the coming application season.

Whatever your next steps don’t hesitate to reach out with questions or concerns–we’re here to help!

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