LSAT Preptest 78 Logic Games Explanations

Time on second attempt: 6 min

See “repeating games” at bottom of section

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This was an unusual game. I rarely drew any diagrams for the questions. Instead, I just drew the starting variables beside each question:

QRSTVWX

Note that because Xue was listed on the second line of the game setup, I forgot about them when I first did and first redid the game. This slowed me down until I noticed I was missing a variable!

The rules all involve elimination in some way, and so I would just eliminate workers by crossing them out on the list. I did this when workers couldn’t go in, or when they were placed. 

Rule 1, I didn’t draw much. I just wrote:

I’m remember that this means “If they’re in, they must be leader”. There’s no elegant way to draw that, so it’s simpler just to remember. Usually it’s fine to remember one rule if there’s no great way to draw it.

Rule 2 involves elimination, because if T is out, then S is out. That’s the contrapositive of the rule as given:

Rule 3 also involves elimination. W knocks out R and V:

(Note: I didn’t draw the contrapositives for rules 2 and 3 on my own page. But if you sometimes forget contrapositives, you must draw them.)

Now, I mentioned I used the list of all variables, and eliminated them. Here’s an example. Question 2 places T as project leader, and Wells as a member. So I drew this to start:

Why?

• T and W are on the diagram, so we can eliminate them from consideration.
• T is leader. That means Q and R can’t be in. (rule 1)
• W is in. That means R and V can’t be in. (rule 3)

So all of T, W, Q, R and V are out of consideration for the third slot. Only S and X are left to consider. Crossing out the variables that are already placed or eliminated makes it easier to see this.

I made zero up front deductions. They don’t tend to be common on modern logic games. 

Though it’s important to remember that you need exactly three workers. This helps on almost all the questions, as usually some workers are eliminated, leaving a limited selection to complete the set.

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Repeating Games

I’ve written elsewhere about the benefits of repeating games, to solidify your intuition for deductions. Note that the purpose of repeating games is to prove the answers right, so it doesn’t matter if you remember the right answer.

I repeated this game about three days after I first saw it, by which time I had forgotten the answers. I’ve written how long it took me on the second attempt. That time, or a couple minutes above it, is roughly the standard you should be aspiring to – a lot of people take 8-9 minutes on a repeat attempt, get everything right, and pat themselves on the back. But that’s too slow. The faster you go when repeating, the faster you’ll learn to go the first time you see a game.

(I say “a couple minutes above” my time because, after years of teaching the LSAT, I’m really, really fast. You should be almost as fast as me, but you don’t exactly need to match my pace to score -0.)

Time on second attempt: 6 min

The setup section explains how to build this diagram.

Main Diagram

For acceptable order questions, go through the rules and use them to eliminate answers one by one.

Note that I use the rules themselves. I don’t use my diagrams for these questions. Reading the rules again for this question will help you memorize them, and it’s also more efficient.

Rule 1 eliminates B. If Quinn is in, they need to be a leader.

Rule 2 eliminates D. Smith requires Taylor.

Rule 3 eliminates A and E. Wells can’t be with Ruiz or Verma.

C is CORRECT. It violates no rules.

This question places T as leader and W as a member. First, you should draw those:

Then draw T and W crossed off the worker list, because you no longer have to think about them:

Next, think about who else is eliminated:

• T is leader. That means Q and R are out, because they can’t be leader (rule 1)
• W is in. That means V is out (rule 3)

So you can add Q, R and V to the people not in consideration:

That leaves only S and X as possibilities. D is CORRECT.

This question places V as project leader. Let’s see what we can deduce:

• Q and R are out, because they can’t be leader (rule 1)
• W is out, because V is in (rule 3)

We can draw a diagram of all of the people out of consideration:

Only S, T and X are left as possibilities. And S needs T (rule 2). So we could pick ST or TX.

A is CORRECT. It’s possible to have Q and S out if we have TX. This places S out, and Q is already out.

All of the other answers are wrong because they don’t allow us to have ST or TX.

This question places T out. This also means S is out (rule 2). You can draw that on the list of all variables:

Next think about who is the most restricted variable. This is always important on logic games. W is the most restricted. 

If W is in, R and V are out. This only leaves QWX.

So that’s one possibility. What about if W is out? Then QRVX are left. We can only have one of Q and R (because they must be leader, rule 1), so we need to have Q/R + VX.

Either way, X must be in. E is CORRECT.

———-

Note: This is exactly how I solved this question in timed conditions. I first drew:

I figured out what happened if W was in by looking at the list (And visualizing R and V being out, since I had memorized rule 3). I wrote down the list of what’s left, QWX.

Having considered what happens if W is in, I then considered what happens if W is out by crossing it out.

I then figured out Q/R + VX were necessary, by seeing who was left, and remembering rule 1. 

This is the same process I described in the explanation above, but I wanted to show you the steps + diagrams I used in their real-time order.

This question asks how to completely determine who is in. You shouldn’t just try answers randomly – you should have a plan. 

The answers are all in terms of who is out. So the key is to ask which workers produce an effect if they are out. 

All of the answers place 2/7 workers out. So if a workers knocks out someone else, we are missing 3/7, with only 4/7 left to place in.

So, who knocks someone out, by being out? That’s pretty much only T. If T is out, S is out. 

Looking through the answers, only B has T: it knocks out both T and Q. So with B we get this setup:

So only RVWX are left to place. We need three of them. W can’t be in, because they would knock out both R and V. So RVW must be chosen. And R must be leader (rule 1).

So B is CORRECT. There’s no need to try other answers (we already checked to see if any others had T). 

This isn’t magic. I did this in timed conditions. I predicted that the right answer would have to knock out one other person, and tested B on that basis. The LSAT loves to leave shortcuts like this for you to find. 

(Of course, in timed conditions, I read all the answers. But that’s not the same thing as exhaustively disproving those. There’s no sense in that, when you’ve already made a prediction, found it, and proved the answer right.)

Time on second attempt: 4 min

See “repeating games” at bottom of section

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I found this game easier than average. I finished it in four minutes on my second attempt. The reason I found it easier is that it was possible to divide this game into three fairly restricted scenarios.

This division into scenarios used to be common on old logic games, but has become rare on new logic games. In fact, if you hadn’t done any old games, it’s possible you didn’t find this game particularly easy.

Games like this are rarer, but it is worth practicing the old games that allow up front deductions. Because when you do find one, you can go through it very quickly, allowing time for other games.

Now, how to divide the game in two? The key is that only L or T can be assigned to 1923. This means that there are only two possibilities for 1923. When there are only two possibilities, it makes sense to split a game.

Note that I’m going to number the years 1, 2, 3, 4, rather than 1921, 1920, etc. It’s just so much easier that way.

Let’s start with T in 3. First, draw that:

Rule 3 also mentions T. If T is in, then R must be in. And rule 4 says that if R is in, O must be directly in front of R. The only place to fit OR is in front of T:

So ORT fill the first three spaces. Rule 2 says that M can only go in slot 1 or 2. So, in this scenario M can’t be in. Only L or Y are possible. I drew by placing the remaining variables (M, L, Y) to the right, and crossing out M:

You could instead draw an “L/Y” in slot four, but somehow I found it clear to draw way clearer. I think it’s because it makes it clear that ORT are fixed and slot four is open.

So that’s the scenario with T in slot 3. Next, we have L in slot 3. We have O, R, M, T and Y left to place.

q

We saw in the prior scenario that placing R is very restrictive, because it forces OR to go in the first two slots. And that, in turn, forces M out. So let’s make a scenario with R in:

Again, I’ve drawn M, T, Y off to the side. You could instead draw T/Y in slot four, but I find this way is clearer for this game. This way, all six possible students are visually accounted for.

Finally, we the scenario with L in slot three, but R out. This will allow M in. Note that since R is out, T must be out as well. So that leaves M, O and Y to fill the remaining spaces. M must be in spots 1 or 2:

Placing M in the middle of 1/2 signifies that M goes in one of those two spaces. The O, Y above L signifies that one will go to the right of L, and one to the left. I like using physical placement of variables to clarify where they might go – I find it easy to visualize the variables drifting down into the open spots.

You might have thought “How the hell did he know to make the third scenario”? I made it because:

• T and R are very restrictive (they lead to the first two scenarios).
• The final scenario is just what things look like if R and T are both out.

Note that I drew no separate rules for this game. The scenarios cover all the rules. But you still must know the rules. I found them easiest to memorize, but if you wanted to draw something you could draw this:

These are combined rules. For instance, the second one says T ➞ OR because we know that R requires OR. Also, note that Rule 1 is entirely covered by the scenarios)

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Repeating Games

I’ve written elsewhere about the benefits of repeating games, to solidify your intuition for deductions. Note that the purpose of repeating games is to prove the answers right, so it doesn’t matter if you remember the right answer.

I repeated this game about three days after I first saw it, by which time I had forgotten the answers. I’ve written how long it took me on the second attempt. That time, or a couple minutes above it, is roughly the standard you should be aspiring to – a lot of people take 8-9 minutes on a repeat attempt, get everything right, and pat themselves on the back. But that’s too slow. The faster you go when repeating, the faster you’ll learn to go the first time you see a game.

(I say “a couple minutes above” my time because, after years of teaching the LSAT, I’m really, really fast. You should be almost as fast as me, but you don’t exactly need to match my pace to score -0.)

Time on second attempt: 4 min

The setup section explains how to build this diagram.

Main Diagram

Scenario 1

Scenario 2

Scenario 3

Note that I drew no separate rules for this game. The scenarios cover all the rules. But you still must know the rules. I found them easiest to memorize, but if you wanted to draw something you could draw this:

These are combined rules. For instance, the second one says T ➞ OR because we know that R requires OR.

Also, note that Rule 1 is entirely covered by the scenarios)

For acceptable order questions, go through the rules and use them to eliminate answers one by one.

Note that I use the rules themselves. I don’t use my diagrams for these questions. Reading the rules again for this question will help you memorize them, and it’s also more efficient.

Rule 1 eliminates A. Louis or Tiffany are the only ones who can go in 1923.

Rule 2 eliminates E. Mollie can only be assigned to 1921 or 1922.

Rule 3 eliminates B. Tiffany requires Ryan to be assigned with her.

Rule 4 eliminates D. If Ryan is assigned, Onyx must be assigned directly before Ryan.

C is CORRECT. It violates no rules.

This question asks how to force Mollie to be assigned to 1922. From our scenarios, only scenario 3 allows Mollie to be placed. Here’s that scenario:

Let’s think about how to force Molly to be second. We need to meet two conditions:

• It must be something that, in that diagram, would force Molly to 2nd
• It must be something that can’t occur in another diagram. (Because in those other scenarios, M isn’t second)

The second bit is less obvious. Don’t worry, I’ll explain below.

In the diagram above, either O or Y in 1st would force M 2nd. So it seems like either could be the right answer. 

But, the diagram above is only scenario 3. In scenarios 1 and 2, O is also first. And in those scenarios, M is not second. So, placing O first doesn’t force M to be second, in all scenarios.

Therefore, E is CORRECT. Placing Y 1st forces M to be second. (And this is the only scenario where Y can go first.)

This question assigns R and Y to the project. R is quite restrictive: rule 4 says that if R is in, then O must also be in, and that O must be placed directly before R.

Since one of L or T is always 3rd, that means OR can only go in spots 1 and 2. In the setup, I labelled this as scenarios 1 and 2. Here are those scenarios:

Scenario 1

Scenario 2

The scenarios are both almost full already. Since we have to place Y, that just means Y goes in slot four in both scenarios.

There’s no need to draw that. When doing this under timed conditions, I didn’t draw anything for this question. I just looked at these two scenarios as I had drawn them, in the setup area. Then I went through the answers.

A is CORRECT. In scenario 1, L is assigned to 1923. 

All of the other answers are wrong because they aren’t possible in either scenario 1 or 2.

This question asks how many people could be assigned to 1921. To solve, I just counted options based on the scenarios.

In scenarios 1 and 2, O is 1st. 

In scenario 3, it’s possible for M and Y to be first as well.

D is CORRECT. 

Here are the three scenarios, for reference:

Scenario 1

Scenario 2

Scenario 3

This question says Yoshio isn’t assigned to the project. This means we’re in scenarios 1 or 2, since in scenario 3 Y had to be assigned.

Scenario 1

Scenario 2

Note that I didn’t redraw anything for this in timed conditions. I just looked at my main setup, and noted who had to be in for each scenario if Y is out.

In the first scenario, normally L or Y can go fourth. But since this question eliminates Y, then L must go forth. The same logic means that T must go in 1924 in scenario two.

So that lets us predict the answer: L in 1924, or T in 1924.

E is CORRECT. L can be assigned to 1924 in scenario 1.

None of the other answers are possible. If Y is out, then both scenarios require L, R and T to be assigned. And O must be in 1921 in both scenarios.

This question asks who can’t be assigned to 1922. If you look at the three scenarios, the possibilities for 1922 are:

• R, in scenarios 1 and 2.
• M, O and Y, in scenario 3.

So that means L and T are the ones who can’t ever be assigned to 1922. A is CORRECT. 

Here are the three scenarios, for reference:

Scenario 1

Scenario 2

Scenario 3

Time on second attempt: 8:10

See “repeating games” at bottom of section

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This is a hard game, which is unusual for a sequencing game. After doing it three times, I came to the conclusion that the game is easier to solve by dividing the setup into three clear scenarios. With that, I easily solved all the questions.

But, it took me three tries to do that! So it is worth it for you to learn how to make the scenarios? 

Yes, I think so. Even if you don’t manage to get scenarios like this on all games, I think learning the process of generating them will help you create them on a larger percentage of games.

I’m going to first show the deductions I made during the setup when I first did the game timed. Then I’ll show the scenarios, and use those to solve the questions.

The Rules

This was a sequencing game, which makes it especially interesting that it was difficult. Normally, sequencing games are so easy to practice that none of them are challenging once you’ve mastered the basic form. Clearly, LSAC has put some thought into how to make these more difficult.

On sequencing games, I always start with the easiest rules to diagram. Here, that’s rules 1 and 3. 

Those rules say that S isn’t first, and S is earlier than M and V:

The blank space in front of S is a reminder that S can’t go first. This turned out to be the key to the game, for me.

Of course, you could just draw a “not rule” for S under slot 1, like this:

This is logically correct, but it’s not very helpful. It’s hard to combine with the other rules. Whereas drawing the blank space in front of S helped me make further deductions, and also helped me remember the rule in all contexts.

I approach rules from easiest to hardest. So let’s look at rule 4 next. Rule 2 can only be fully understood once we’ve drawn everything else.

Rule 4 says that T is earlier than one of H or V, but not the other. A lot of people draw this as “T – H or T – V”. Again, this is logically correct, but it’s not effective. Here’s how you should draw it:

Why? Because, if T can’t be before both H and V, but must be before one of them, then that means it must be between them. This is a common rule type on newer sequencing games, so you should learn to recognize the pattern.

Now we can turn to rule 2. This rule says that if H is before L, then M is also before L. Drawn alone, this isn’t that complicated:

The rules I’ve drawn above are all that I drew when I first tried this game. But, as I wrote above, we can combine the rules into three definite scenarios, starting with rule 2. Let’s see how.

We already have M and V in the other diagram. So, there must be a way to combine these two rules. We can make a version for two scenarios:

• Where H is before L (and the rule applies)
• Where L is not before H (and the rule does not apply)

Let’s try H before L first. We’ll call this scenario 1.

Scenario 1

Here’s the diagram we get. If H is before L, then we must draw both H and M before L. 

The diagram above already covers rules 1, 2 and 3. There are two open questions:

• How does rule 4 affect it? (H, T, V)
• Who can go first?

On this diagram, someone has to go in front of S, and only T or H are left to go in front of S. (H can go first: the only rule for H on this diagram is that they are before L. First is before L).

But look carefully at rule 4. T is always in the middle of V and H. So T can’t go first. 

So that leaves H. And since H is first, the order for rule 4 is H – T – V

So on the diagram, we can draw T before V and after H:

Note that the only thing that affects position are lines between two variables. So, for example. since there are no lines between V and L, either one of them could go last.

So that’s scenario 1, where H is before L. Now let’s look at what happens if we put L before H.

Scenarios 2 and 3

Here’s the diagram we’re starting from:

We already placed H before L. Now we’re placing L before H. But that doesn’t trigger any rules, so can we really make any deductions? 

Well, we saw that the first position is fairly restricted. Let’s see who can go first if L is before H.

• Not S, M, or V. The diagram shows they aren’t first.
• Not H, since we put it after L.
• Not T, because T is always in the middle of H and V (rule 4)

So if H isn’t first, then L is! So we can put L first, and also draw H after L, like this: 

The T is sort of floating up above to the right. Note that though V is visually after H, V could go before H in this diagram.

Now, are we done? Not quite. The diagram above covers rules 1-3. But we haven’t applied rule 4, the one that says T is between V and H, in either order. 

We can construct two more scenarios by placing them in both orders:

• H – T – V
• V – T – H

Here they are:

Scenario 2

Scenario 3

These diagrams are potentially confusing though. To read them, you must remember that there is only a relationship between variables if they are connected by lines. (e.g. S is before M because a line connects them)

But in, for example, scenario 2, S could be either before or after T. There are no lines connecting S and T.

I know sequencing diagrams well enough to avoid getting mixed up by this, but I would be careful drawing this type of diagram until you can avoid the error of thinking that left/right placement matters.

Here are all three scenarios, plus what triggers them.

Scenario 1 ( H before L)

Scenario 2 (L before H, H – T – V)

Scenario 3 (L before H, V – T – H)

You could also solve this game just by drawing the rules and making deductions. But I think you’ll find this method faster. If you prefer to try the rules based approach, here are the rules:

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Repeating Games

I’ve written elsewhere about the benefits of repeating games, to solidify your intuition for deductions. Note that the purpose of repeating games is to prove the answers right, so it doesn’t matter if you remember the right answer.

I repeated this game about three days after I first saw it, by which time I had forgotten the answers. I’ve written how long it took me on the second attempt. That time, or a couple minutes above it, is roughly the standard you should be aspiring to – a lot of people take 8-9 minutes on a repeat attempt, get everything right, and pat themselves on the back. But that’s too slow. The faster you go when repeating, the faster you’ll learn to go the first time you see a game.

(I say “a couple minutes above” my time because, after years of teaching the LSAT, I’m really, really fast. You should be almost as fast as me, but you don’t exactly need to match my pace to score -0.)

Time on second attempt: 8:10

The setup section explains how to build this diagram.

Main Diagram

In the setup section, I explained that you can solve the game by splitting the rules into three scenarios. However, they are rather complicated to construct. You might prefer to solve the game using just the rules written out. So, I’ve drawn the scenarios, and also the rules below for reference. I’ll be solving the questions by reference to the scenarios, though.

Scenario 1 ( H before L)

Scenario 2 (L before H, H – T – V)

Scenario 3 (L before H, V – T – H)

The Rules

For acceptable order questions, go through the rules and use them to eliminate answers one by one.

Note that I use the rules themselves. I don’t use my diagrams for these questions. Reading the rules again for this question will help you memorize them, and it’s also more efficient.

Rule 1 eliminates D. The sundial can’t be first.

Rule 2 eliminates A. If the harmonica is before the lamp, then the mirror also needs to be before the lamp.

Rule 3 eliminates E. The sundial is supposed to be before the vase.

Rule 4 eliminates B. The table has to be earlier than one of either the harmonica or the vase.

C is CORRECT. It violates no rules.

This is a general could be true question. It gives you no indication what to look for. This sort of question often takes longer to solve, as you have to try each choice. Fortunately, there are a couple of tricks you can try:

• Looking through the answers to determine which seems easiest to do within the rules (pick one that obeys a rule).
• Choose the hardest rule and use it to eliminate answers.

I’ll first use the most restrictive rule, and then explain what I mean by “obeys a rule”. The most restrictive rule was this diagram combining rules 1 and 3:

S must be before M and V. S can’t be first, and S needs at least two open spaces after it for M and V. 

This eliminates C, D and E!

• C: Places M before S
• D: Places V before S
• E: Places S fourth….and T fifth. There’s no space for M and V after S!

That’s convenient! This “most restricted factor” trick shows up in all types of logic games game and will eliminate many answers for you.

Next, what do I mean by obeys a rule? Let’s look at A and B:

• A: T before V….that isn’t mentioned in any rules.
• B: S before V. That helps comply with rule 3!

So B is making things easy for us by checking one item off the list of rules we have to obey. This doesn’t make B correct, of course….but it means it is a more promising place to start.

So we only really have to consider A and B, and B is the most likely answer. Now, if you’re using the scenarios (from the main diagram section), you’ll see that none of the scenarios allow for T to be directly before L. So, A is wrong and B must be CORRECT.

If you hadn’t found the scenarios, you could use this method to disprove A. It places T and L second and third:

Next, consider rule 4: we need to have V – T – H, or H – T – V.

We can’t put V in slot 1, because we need to put V after S. So we have to put H first. 

But that places H before L. And rule 2 says that if H is before L, then M must be before L. Which isn’t the case here (there’s no open space before L), so this diagram violates the rules.

So B must be CORRECT, by process of elimination.  Here’s scenario 3 from the setup, which proves that this works:

If you didn’t have the scenarios, you’d just have to sketch S second, V third, and fill in the rest of the spots while obeying the rules.

• M goes after S
• T – H need to go after V, since there’s no space to put them before it.
• That leaves L to go first

Note that when constructing scenarios like this you want to think in terms of rule blocks, not isolated variables.

This question really shows the value of the scenarios from the setup. It says that we must place T after both M and V. Only scenario 3 allows that:

To solve the question, just go through the answers and see which are consistent with the diagram.

• A is wrong. H is after T, not before.
• B is wrong. T is after L, not before.
• C is wrong, T is after S, not before.
• E is wrong, S is after L, not before.

D is CORRECT. No lines connect M and V, so they go in either order relative to each other.

This question asks who can’t go directly in front of the vase. To answer this, ask yourself what rules involve V.

That’s rules 3 and 4. But 4 is most useful. We saw V must go in one of these two orders:

We can see that H can’t be directly in front of V: T separates the two variables.

So A is CORRECT.

Like question 13, this is a general “could be true” question. That doesn’t give us an obvious starting point. But, if you drew the scenarios in the setup, it’s extremely easy to use them to eliminate answers. Here are the scenarios, for reference:

Scenario 1 ( H before L)

Scenario 2 (L before H, H – T – V)

Scenario 3 (L before H, V – T – H)

A, B, C and D aren’t possible in any scenario. 

• A: M always needs at least two others in front of it. 
• B: L can only be 1st (scenarios 2 and 3) or towards the end (scenarios 1). 
• C: V always needs at least three variables in front of it, so it can’t be second. 
• D: is wrong for the same reason as B.

E is CORRECT. In scenarios 2 or 3, M could be fifth. 

Note: As long as things aren’t directly connecting, there are no ordering rules between them. So, for example, in scenario 3 three you can place V and T before M.

Another general could be true question. Just like with question 16, the scenarios from the setup let you easily eliminate answers. Here are the scenarios, for reference: 

Scenario 1 ( H before L)

Scenario 2 (L before H, H – T – V)

Scenario 3 (L before H, V – T – H)

A is wrong because S needs M and V after it. So it can go fourth at latest. 

B is CORRECT. In scenario 2 you can place S fourth, with only M and V after it.

C is wrong because L can only go 5th in scenario 1. And in that scenario, V would have to be last, not M.

D is wrong because scenario 1 is the only scenario with L after T. And in that scenario L also needs H, S and M before it. So if L went after T, it would have four variables before it and go 5th at earliest. 

E is wrong because V can only go third in scenario 3, where L and S are before it.

Time on second attempt: 5:45

See “repeating games” at bottom of section

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This is a sequencing game with a small added factor: two singers are recorded. Overall, I found this game was much easier than game 3, which was also a sequencing game. 

There are a few upfront deductions you can make on this game. Note that I didn’t make all of these deductions the first time through the game. In fact, some of them I only figured out partway through my second time doing the game.

If that’s the case, why bother to learn about these deductions? Because on some games, you will spot all the deductions up front, and you’ll go faster. Or you might get some of them, which helps you go faster. You’re also more likely to make the deductions mid game if you’re aware of the patterns – mid-game deductions happen to me frequently. The deductions I’m going to describe are similar to those seen on other games.

I’ll point out which parts I figured out on my first attempt, and which I only figured out while doing the game a second time. Now, for the rules

First, you should draw a diagram and place some of the rules on it. Rules 1 and 2 say that five must be recorded, and that four can’t be:

Now, only two singers are recorded: Kammer and Lugo. That means that there are only two options for spot 5: K or L.

Whenever there are only two options, you should split that into two diagrams. Often this will lead to new deductions, and it will always makes the game easier to visualize. 

Next, let’s look at the other rules. Rule 3 says that W is before K and L (the recorded auditions). Rule 4 says that K is earlier than T. We can combine these two rules:

Rule 5 says that Z is before Y:

Right away, we can make one addition to our diagram. In scenario 1, T is in spot 6:

This was as far as I got with deductions on my first attempt. But I figured out more while doing the game a second time. Let’s see what else we can deduce.

“Not” rules are useful. That’s where we write the places that variables can’t go. K and L are particularly restrictive, so let’s add some not rules for them.

• They can’t go fourth, because the fourth session isn’t recorded.
• In scenario 2, K can’t go last, because K must be before T.

These are fairly obvious deductions. But drawing them on the diagram frees your mind to look for further deductions. 

Note that I’ve only drawn “not L” on the top diagram, rather than “not L and not K”. That’s because K is already placed, so it would just add clutter to put “not K” under slot 4. I did the same with the second diagram. You have to be careful not to add too much “not rule” clutter, or it becomes hard to read the diagram.

Now, what else can we deduce? Well, W has to be before both K and L. And since K/L can’t go fourth, that means W – K/L has to go 2 and 3rd at latest. You can draw it like this:

The vertical line is just a visual aid: it shows the division between spots 1-3 and 4-6, and that W – L/K must go in the first half.

It’s also possible to have made this deduction all in one step, and leave off the not rules. This is what I ended up drawing on my own page. I prefer it, because it’s simpler and conveys the same information:

So, with those diagrams, a fair amount of the game is determined. It turns out this game is quite restrictive. W can only go second or first. This upfront diagram actually solves question 23 entirely! Diagrams like this can pay off.

The only rules left to remember are:

• Z is before Y (both diagrams) 
• T is after K (scenario 2)

You can add those variables to the scenarios to make them even clearer. I’ll do each scenario separately, since they’re somewhat different.

Scenario 1

Only the Z before Y rules applies. Let’s think about where we can place them. W – L are already taking up two places in spots 1 – 3. That means there is no space to put both Z and Y in spots 1 – 3. So it turns out Y must go fourth in scenario 1!

I’ve placed Z with a comma alongside W – L to indicate that it can go anywhere amongst them, but is also in slots 1 – 3.

Scenario 2

I couldn’t find any further deductions for scenario 2. Instead, I just drew the remaining variables to the right of the diagram: Z – Y and T.

The only other thing to remember is that K is before T. This is the only thing that can’t be cleanly put on the diagram. So I just put it below the diagram and numbered it to visually indicate it’s a rule, and now part of the diagram.

—————

Repeating Games

I’ve written elsewhere about the benefits of repeating games, to solidify your intuition for deductions. Note that the purpose of repeating games is to prove the answers right, so it doesn’t matter if you remember the right answer.

I repeated this game about three days after I first saw it, by which time I had forgotten the answers. I’ve written how long it took me on the second attempt. That time, or a couple minutes above it, is roughly the standard you should be aspiring to – a lot of people take 8-9 minutes on a repeat attempt, get everything right, and pat themselves on the back. But that’s too slow. The faster you go when repeating, the faster you’ll learn to go the first time you see a game.

(I say “a couple minutes above” my time because, after years of teaching the LSAT, I’m really, really fast. You should be almost as fast as me, but you don’t exactly need to match my pace to score -0.)

Time on second attempt: 5:45

The setup section explains how to build this diagram.

Main Diagram

Scenario 1

Scenario 2

The setup explains how to make these scenarios.
The K – T below scenario 2 is just to indicate that that scenario has a rule that couldn’t be drawn on the diagram.

Rules

The scenarios above solve most questions. However, it is also helpful to have a visual representation of the combined rules, so I’ve drawn the ordering rules below. These combine rules 3, 4 and 5. (K and L are recorded, so rule 3 means W is before them)

For acceptable order questions, go through the rules and use them to eliminate answers one by one.

Note that I use the rules themselves. I don’t use my diagrams for these questions. Reading the rules again for this question will help you memorize them, and it’s also more efficient.

Rule 1 eliminates E. Lugo can’t go fourth, because the fourth session isn’t recorded.

Rule 2 eliminates C. Zinn can’t go fifth. Fifth is a recorded session, and only Kammer and Lugo are recorded.

Rule 3 eliminates A. Kammer is supposed to be after Waite.

Rule 4 eliminates no rules.

Rule 5 eliminates B. Yoshida is supposed to be after Zinn.

D is CORRECT. It violates no rules.

This question asks who can’t be placed second. You should look at the rules and see who has other auditions before theirs.

In the diagram above, you can see that T has both W and K before them. So T can’t be second.

C is CORRECT.

(The diagram is a combination of rules 3 and 4)

This question asks who can be placed sixth. Here are the diagrams from the setup:

In the first diagram, T is last. So, C is CORRECT.

This question adds a new rule. When that happens, you should combine it with the existing rules. Here were our original ordering rules:

This question places Y directly after K. So in the first diagram, change K to KY. And then place Z before KY:

I actually drew this diagram beside question 21 when doing this game timed. If you practice drawing this kind of diagram, it takes very little time. Perhaps 10 seconds. And it makes the question much faster to solve. 

Now you can use the diagram to eliminate answers. Answers A–D all contradict the diagram.

E is CORRECT. It’s possible for Z to go second if W is first.

Note: Just for fun, you could draw this diagram onto the sequence of six slots; the new rule has made things quite restrictive. From the setup, only L or K can go fifth, and now K can’t go fifth because Y and T are after K (making K fourth at latest). 

So we must place L fifth, and we get this diagram:

You definitely do not need to draw that to solve this question. Though it’s not an error to draw it. In some situations a diagram like that would speed you up. But it doesn’t add anything on this question.

(Though actually, drawing this diagram let me instantly solve question 22. It’s rarely useless to draw correct diagrams to help solve a question)

This question is asking where Y can go. If you drew the additional diagram I showed in question 21, you can solve question 22 instantly:
B is CORRECT, Y can go fourth.

If not, you can still solve this fairly quickly.

• A is wrong because only K or L can go fifth.
• E is wrong because Z needs to be in front of Y.
• The setup diagrams help us eliminate C and D. Here are the diagrams:

Even if you try to place Y as early as possible, Y can’t go second or third. W + one of K/L take up two out of the first three spots. Z would fill the third, because Z must go before Y.

Everyone hates rule substitution questions. But they’re not that hard if you consider that the only way to replace a rule is to use the other rules.

Looking at the full effect of all the rules helps you decide how to replace the rule. For instance, look at the setup diagrams for this game:

W is pretty restricted. They can only go first or second. Who can go in front of W?

• Not K and L. Rule 3 says they go after.
• Not T. It goes after K.
• Not Y, because Z needs to go before Y. 

(Don’t worry, I explain the third bullet below)

So only Z can go before W. A is CORRECT. 

The final bullet point is the hardest to understand, so I made a diagram to show what happens if we put Y before W:

Putting Z and Y before W pushes W third, and this pushes L fourth, which isn’t recorded. So that violates the rule from the setup that L must be recorded. You get the same result with K. 

(K and L can’t go sixth. See the setup for the full explanation. But basically, you either have KT in 5th and 6th, or L in 5th, and K somewhere earlier, because it goes before T)