LSAT Preptest 81 Logic Games Explanations

Time on first attempt: 6:30

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This is a linear game. And, unusually for new logic games, this one allows you to figure out a lot in advance.

You should draw the game like this, with five slots:

I’ve added rules 1 and 4 to the diagram above: essay 1 is rural, and essay 3 is G.

Note that I don’t number my own diagrams. It’s faster to draw, and with practice you can quickly see which number a slot is at a glance. If you practice not numbering the slots, your brain will develop the ability of knowing where “4” is.

For explanations, I do normally number my diagrams for clarity, but on this type of two attribute game I’m not going to. That’s because I place the second attribute (rural/urban) just under the slot, and adding a number would add distraction.

Next, we can draw rules 2, 3 and 5:

I’ve combined rules 2 and 3. K is right before F, and they can’t have the same theme. That’s what the crossed line underneath them represents.

And, J has urban as a theme.

Now, we could just start the game, and probably do fine. But, you should always look for deductions. There are a few ways to look for them. Here are a couple we can use on this game:

• Look for the most restricted element
• If something can only be one of two ways, draw both

KF are the most restricted element. They need two spaces, and G has cut the game in two. So, we can make two diagrams: one with KF before G, and one after.

We know that KF have one r and one u. In the first diagram, the order is ru, because rule 1 says that r must be 1st. In the second diagram, they could be either order.

Next, you should think about numerical distribution. There are three r’s and two u’s. KF take up one of the u’s (rule 3), and J takes the other (rule 5). This means that both G and H must be r!

We can add this to our diagram:

Having H and J floating above the diagram separated by a comma signifies that they can go in either order. I’ve also put the r and u underneath them for reference.

So, nearly everything is determined in this game. Once you have these two diagrams, the questions fly by. It’s definitely worth taking a bit of extra time to make deductions where they are available, as you get your time spent back threefold.

(Some wonder: how to know where there are deductions. As a rule, deductions are about what must be true in general or in a set of very limited scenarios. If you’re just generally drawing what “could be true”, you’re not making deductions – you’re wasting time. That’s the distinction.)

The setup section explains how to build this diagram.

Main Diagram

These two scenarios cover both possibilities. As G is third, the KF block must be either 1st/2nd or 4th/5th, and they have different themes.

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 first essay must have a rural theme, and Jordt’s essay must have an urban theme.

Rule 2 eliminates A and C. Kim’s essay must be in an issue immediately preceding the issue in which Fetter’s essay is featured.

Rule 4 eliminates B. Gonzalez’s essay must be in the third issue.

E is CORRECT. It violates no rules.

Use your diagram to eliminate answers which aren’t possible. Here’s the diagram. See the setup for how to create these two scenarios. One with KF in front of G, one with KF after G:

A is wrong. F is either in front of G, or 5th. (Rule 2 + rule 4)

C is wrong. H, J and KF are always on opposite sides of G. (Rule 2 + rule 4)

D is wrong. K is either first, or directly after G. (Rule 2 + rule 4)

E is wrong. K is always right before F, not G. (Rule 2)

B is CORRECT. In the first scenario above, H can go 4th or 5th, so it can go directly after G.

note: this explanation is wrong. It’s on the list of corrections to make

To answer this question efficiently, you should use the setup diagram. Here’s that for reference. See the setup for how to create these two scenarios. One with KF in front of G, one with KF after G:

We can see that there are only two possibilities for the fourth issue being urban:

• scenario 1: J is fourth, H is fifth. (The commas between H, J indicate they can go in either order in the two open slots below)
• scenario 2: KH are 4th and 5th, and K is urban.

C is CORRECT. H can’t be fourth, because H is always rural.

A and B are possible in scenarios 1 and 2, respectively. D and E are possible in scenario 1, where J is urban in 4th and H goes 5th.

We saw in the setup that both G and H have to be rural. D is CORRECT.

This is because there are only two urbans. One of KF takes one of the two urbans (rule 3), and J takes the other (rule 5). Thus, one of KF + G and H have to be rural.

A must be false. B, C and E could be true, but don’t have to be.

Here’s the setup diagram. See the setup section for how to create these two scenarios. One with KF in front of G, one with KF after G:

The only people who can’t be fourth are G and F. 

So, A is CORRECT. G is third, and F always has to be after K, so F is either 2nd or 5th. 

(Either H or J could go third as they are reversible)

To solve rule substitutions, you must consider all consequences of the setup diagram, including deductions and other rules. The right answer will tend to use some of those consequences to replicate the rule.

In the setup, we made the deduction that both G and H must have rural themes. This is because there are only two essays with urban themes. Since KF have different themes, they take one urban. And rule 5 says J takes the other.

That means everyone else has to be rural, so G and H must be.

So to replicate the full effects of the rule, we can do things in reverse. There are only three rurals. If G and H are rural, that leaves only two urbans and one rural. Add in J as urban (rule 5), and we’re left with one rural, and one urban theme for KF.

So, B is CORRECT.

A only gets us part of the way there. C, D and E add new restrictions that weren’t in the main game. The correct answer has to keep the constraints exactly the same, not change them.

Game Setup 

Time on first attempt: 4:38

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LSAT 81 Game 2 is a linear game, and a fairly straightforward one, I think. That said, it’s worth trying to get fast at these. I did this in 4:38 on my first try. This left me with a massive time surplus at the end, which was fortunate. I made an error on the final game, and took 14 minutes to do it. 

I could not have done that without the time surplus from games 1-3. In fact, I often have a game on LG where I take 12-13 minutes. But then the others are really fast. Aiming for even times is a mistake. You need to go really fast on the easy ones to create a time buffer for the hard ones.

Split the game using the 5th rule

The 5th rule splits this game in two. Parker performs either first or seventh.

Whenever a rule allows you to split a game, you should do it. And you should skim the rules to look for such a rule first. Drawing the rules in order slows you down. You should draw them in the order of most definite to least definite.

Rule 4 forces M into position

Why? Well, this lets you add subsequent rules directly on the diagrams. This lets you avoid wasted drawing, and make deductions quickly. For example, it’s so obvious how to draw rule 4 now that we have P on the diagram. One space between M and P:

Drawing the rules out of order expands on what the rules directly give you, and lets you see a deduction immediately (M is 3rd or 5th).

How not to draw LSAT 81 Game 2

Below are how people tend to draw games. Would you have been better off drawing things in order, like this, and then looking for deductions and scenarios? I think not:

These two diagrams above are logically accurate, but add nothing beyond what the rules literally said. Such diagrams would just clutter your page and slow your down. And the more cluttered the page is, the harder it is to see deductions. Deductions are just combinations of elements, after all. If you have clutter, you can’t separate the wheat from the chaff.

You should also approach the rules in terms of which ones can be combined what what’s already been drawn. So, for example, rule 4 was good, because P was already on the diagram.

Rule 2 places T

Now, since M is on the diagram, we should next draw rule 2, since it mentions M is before T. We get this:

In the first scenario, we just know T is anywhere after M (represented as a floating variable above the diagram, to signify it goes anywhere, after that point.)

But in the second scenario, we see that T can only go sixth. Excellent. Definite diagrams are very good.

Put the remaining rules in a list

LSAT 81 game 2 has a few rules we can’t easily put on the diagram:

L is before N, L and O are one space apart in either order, and S is random.

Note that using these rules, you could make scenarios out of the bottom diagram (since it’s more limited). I count four scenarios in total. But, it’s better to stick to two main scenarios, and draw the others when a question requires it. Having five total scenarios would take time and be confusing. And it would offer little gain beyond what we’ve already got.

“Not” rules help you remember where N and L can’t go

I normally don’t add that many “not” rules to my diagram, but they may be helpful here. You could optionally add that L can’t go last and N can’t go 1st, as a reminder. Specifically, first and last in the open spots:

A few questions test this explicitly, so it’s probably worth noting these down if you find you sometimes don’t notice that about L/N. (We can deduce these “not” rules because L must be before N. If N is 2nd in the top diagram, then L can’t be before N, and so on)

The setup section explains how to build this diagram.

Main Diagram

The fact that this is a general “could be true” suggests that you should have made some upfront deductions/scenarios. If you ever find one of these as the first question and haven’t made deductions, you should revisit the setup.

Here’s the main diagram and rules, for reference:

None of the answers in the list below work:

• A. Miller can only be 3rd or 5th. (Rules 4 + 5)
• B. N can’t be first, because they must be after L (rule 1)
• C. O could only be 5th in scenario 1. We’d have to place L last (rule 3). But that wouldn’t work, since L has to go before N (rule 1).
• E. T has to be fourth or later in either scenario.

D is CORRECT. Sen can perform seventh. This scenario shows one way to do it:

Note that when I construct a scenario like that, I try to make things easy for myself:

• I first thing I place is S seventh, since that’s what I’m trying to prove
• I next place L_O, since they’re the most restricted.
• I also make sure to place L as early as I can in the diagram, since they have to be before N. So I will always place L_O, and not O_L. (If I’m trying to make a working scenario)

This makes it easy to obey the rules (if possible). Then only T and N are left to place. And since their rules have already been met, I can safely place them in either 5th or 6th.

This question places O earlier than M, and then asks who can be fifth. To answer this, we’ll have to build on our setup scenarios. Here are the scenarios for reference:

This question is asking who can be fifth. In scenario 2, we can see that:

• O must be before M
• M is fifth

So, M can be the fifth performer and we can safely eliminate B. We’ll also have to concentrate on scenario 1 for the other answers.

In scenario 1, M is third. So, to place O earlier than M, we have to place O second. And, thanks to rule 3, that means we also must place L fourth: L and O must be one space away from each other:

Only T, N and S are left to place. They can go in any order, since we’ve already obeyed all their rules. 

A is CORRECT. L can’t be fifth in either scenario. As we saw just above, in scenario 1, L has to be fourth, in order to be two spaces away from O. (And in scenario 2, M if 5th, so L isn’t.).

This question asks who can’t be third. You should look at the setup diagrams for reference:

In scenario 1, M is third. So, we can eliminate answer B, and concentrate on the other scenario. In scenario 2, there are only four open spaces, and we have to place L_O, S and N. When I drew this, I switched to a four space diagram, since MTP are fixed in 5-6-7. So the four space diagram shows spaces 1-2-3-4:

This diagram shows that both L and O can be third (since they’re reversible). Next, place L_O 2nd and 4th. Place L early, in order to allow N to go after:

This proves that N can go third. And it also shows that S can’t. If we swapped S and N, then N would be before L, violating rule 1. E is CORRECT.

With these diagrams, I hope to show not just why the right answer is right, but also how to get to it quickly. You should think in terms of which placement is easy, and which is hard. Placing L early is easy, because of N and rule 1. Place L later is hard or impossible. So always place L early when trying to make a working diagram.

You should also think in terms of what can be reversed. Often, one diagram disproves two answers.

This question places ST together. Here are the scenarios, for reference. As ST are together, only scenario 1 works:

Now that ST are together, you should think about the most restricted places. Space 2 is suddenly quite restricted. Normally, in the first diagram, L, O or S could go in space 2. (T couldn’t, since they’re after M. N couldn’t, since they must be after L)

But now, only L or O can go in those positions. L and O are reversible:

This leaves only N and ST left to place. They can go in either order. You can use this to eliminate answers. 

A and B are both possible when we reverse L_O. C can be true if we make the order STN. E can be true if we make the order NST.

D is CORRECT. We have to place T after S, so T can only go 6th or 7th.

We have two scenarios in this game. Here they are for reference: 

Since this is a “order completely determined” question, we should look for an answer which forces us into only one scenario.

A is not a good candidate since it could happen in either scenario.

B is not a good answer, since it puts us in scenario 2, but doesn’t make anything else happen.

C, D and E take more consideration. In timed conditions, you should go with the one that feels more restrictive. But, that kind of intuition only comes with practice, and an understanding of the constraints. So, I’ll try to explain what makes each answer effective or ineffective. 

(Though, mainly I solved this by recognizing that all of C, D and E were in scenario 2, and that I could use my diagrams from question 9 to solve things.)

Each of these answers takes place in scenario two. D and E because 1st and 3rd are full in scenario 1. C, because if you place N 4th in scenario 1, there’s no place to put L earlier and also fit O.

So this is like question 9. If you drew some scenarios there, you might recognize that placing N 4th ends up in the following reversible scenario:

That eliminates C: L and O are reversible, so placing N fourth does not completely determine the order. This also helps eliminate D. Not everyone can visualize variables in their head. But you may be able to look at the diagram above, imagine we’re placing O third, and then imagine S and N swapping places. This still obeys all the rules, and there’s no need to draw another diagram. So, placing O third does not completely determine the order (since S and N could swap).

(If you can’t visualize, you can just quickly sketch LSON and LNOS)

That leaves us with E. Normally, placing the random variable doesn’t do much. But here, it forces us into scenario 2. Since MTP are already fixed, I won’t draw them. Here are the first 4 spaces, and who must be placed:

It should be pretty clear that you can’t place L fourth, since N must come after. So, we have to place L 2nd, N 3rd, and O 4th:

(L and O are 2nd and 4th because they must be a space apart from each other.)

Time on first attempt: 6:30

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This is another linear game. Unlike the last one, there’s almost nothing you can figure out in advance. Instead, the questions will give you new constraints which reduce the options and let you pick answers. That makes this a “rules based game”.

That said, I did draw split scenarios for the first rule. It led to no deductions, but this kind of setup still helps me visualize how the variables can be placed in each scenario:

The next two rules are quite standard. The wall must be right ahead of the zipline, and the rope bridge can’t be beside the vaulting apparatus. Here’s how you draw them:

I’ve also draw T as random. You can use a circle to symbolize this.

The only real restriction is that you can’t end up with a situation where there are two sets of two open spaces, and you have RV and WZ left to place. 

That’s because WZ must be together, and RV can’t go together. If you fill one of the spaces with WZ, the only place for RV is togheter, which is illegal.

I’ve noticed this type of situation as a constraint on at least two recent linear games. I’ll mark this pattern with X’s whenever it shows up, like this:

E.g. in either diagram, we need to place WZ (together) and RV apart. There’s no way to do that.

MainDiagram

The setup section explains how to build this 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 C. The spinning platform must be the third or fourth obstacle.

Rule 2 eliminates A. The wall must be placed immediately before the zipline. 

Rule 3 eliminates B and E. The rope bridge cannot immediately precede or immediately follow the vaulting apparatus.

D is CORRECT. It violates no rules.

If T is first, we can make two diagrams. One with S third, and one with S 4th (rule 1):

The second diagram doesn’t work. In the setup I mentioned that having two blocs of two open spaces will force WZ to go in one, and RV in the other. This doesn’t work, because RV can’t be beside each other. So, we’re in scenario 1.

RV and WZ are left to place. Since RV must be separate, we can put one of RV second, and the other after S with WZ:

Though this is more work than we had to do. Just knowing S is 3rd is enough: C is CORRECT. Whenever you make a major deduction such as “S can’t go fourth”, it’s a good idea to stop and check if that answers the question. (S can only go 3rd or 4th in this game, so “S not 4th” = “S 3rd”)

The other answers are all things that could be true in the scenario above, but don’t have to be. 

This question asks where T can be placed. A good shortcut on this type of question is to see which spots are in all the answers. There’s no need to test first and second, because all answers have them.

This leaves two options. Solving the question now, or doing the rest of the game in order to get more scenarios from those questions and use those to eliminate answers.

Either strategy is fine. Personally, I just answered this question right away, since it didn’t feel very hard. If it had felt harder, I would have gone on to get more scenarios.

I saw right away that placing T third and fourth wouldn’t work, since S always has to be third or fourth. Placing S and T in the middle leaves two blocs of two spaces each:

I mentioned in the setup this doesn’t work. The people left to place are WZ (together), and RV (apart). This kind of setup forces RV together, e.g .WZ–ST–RV. That violates rule 3: RV can’t be together.

So we’re between A and B. Sixth is the only difference between those two, so we should test that:

The scenario above proves it’s possible to place T sixth. So, B is CORRECT.

To answer this, you should place R second, and make two scenarios: one with S 3rd, one with S 4th (rule 1).

I’ve drawn the not rules for V beside R (rule 3). Next, we should place WZ, since it’s the hardest to fit:

This eliminates the first scenario. In that scenario, we’re forced to put V beside R. In the second scenario, we can see that WZ and V have a couple options, but T has to go first. So, B is CORRECT.

This question places R and V before T. This new restriction makes T quite difficult to place. Consider the other variables in the game:

• R_V (reversible)
• S
• WZ

And consider how many variables have to go before T:

• R_V (reversible): because of this new rule
• S: because rule 1 says S must be third or fourth. 

Let’s try to place R_V and T as early as we can. R_V as a block take up three spaces, so S is fourth. That means T is fifth at the earliest. That’s quite restricted! 

Now, looking back a the list of variables, who can go after T, if T is 5th? …..nobody. Only WZ are left, but they need two spaces.

So actually, T must go 6th. We can add this to our setup diagram:

The first option doesn’t work, because we have the two blocs of two. That forces WZ in one bloc, and RV in another. That violates rule 3.

So, A is CORRECT. S must be fourth.

In case you were curious, these are the ways it can be this question can be arranged. R/V are reversible in both cases:

Time on first attempt: 14 minutes

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Boy, was I slow on this one. This happens to me fairly often on newer LG sections; I go extremely fast on 2-3 games, then quite slowly on another game. I usually get the whole section right, but I need the extra time to get through the harder one. This is why it’s very important to finish easy games fast. You’ll appreciate the extra time on hard ones.

(I was slow because I missed a key deduction I should have made.)

This is a grouping game. And here’s the secret I missed: 

• Each group needs two people. And there are only three options: G, I and one of F/H
• Deduction: Either G or I must be in a group. If one is out, the other is in.

They test you on this deduction in the final question. It’s phrased as a rule substitution question, but the answer is also a “must be true” deduction that really makes the game easier if you get it earlier.

On to the rules. Here’s how you should set up the game:

I’ve added rule 4: G can’t visit Sydney. I’ve also added the constraint from the setup that each city must have exactly two visitors. This becomes very important. (And I missed drawing it in my own setup, which was part of the reason I was slow)

Next, we can draw the other rules. Then we’ll think about deductions:

When looking for deductions, it helps to think of the most restricted point. Here, that’s one of the three cities: S. Normally, there are four people who can go in a city: G, I, F and H

But, S only has three: I, F and H. And, thanks to rule 2, it can only be one of F and H. So actually, the people who can go in S are: I, F/H

This is important. I has to go in S:

This is the major deduction of the game. I actually didn’t get it until partway through, so don’t feel bad if you didn’t spot it.

The other major deduction is that either G or I has to be in each group. That’s because, again, only G, I, or F/H can go. Since we need two, that means you need at least one of G and I (since only one of F/H can be in)

You can draw this as an additional rule as a reminder:

Note that we could have both G and I in a city. The rule above only means we need at least one. It does not mean we can’t have both. (“G —> not I” is the “not both” format of a rule. This is “not I —> G”)

Now let’s consider the other rules. We know that Ibanez must go in exactly two places. Since they have to go in Sydney, that means we can split the game into two scenarios: one with Ibanez in Manila, and one with Ibanez in Tokyo:

I’ve added two other things:

• G must go in the other city, since that’s the only way to have two people in each group without I (G + F/H).
• In the second scenario, rule 3 applies. Since G is in Manila, then H is in Tokyo.

That’s pretty much it for the scenarios. Remember that you have to place both F and H at least once in each scenario, as well. (But not together)

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 A. Ibanez must visit exactly two cities. 

Rule 2 eliminates E. Fan and Haley cannot visit the same city as each other.

Rule 3 eliminates D. Manila is visited by Gleeson, so Tokyo must be visited by Haley.

Rule 4 eliminates B. Gleeson cannot visit Sydney. 

C is CORRECT. It violates no rules.

To answer a “completely determines” question, you should look at the scenarios + the rules and think about what makes something else happen. That “something else” is the key, since determining everything requires a chain reaction that forces 3-5 things to occur.

Here’s are the scenarios and rules for reference:

We can more or less ignore A and C for now. F and H aren’t totally interchangeable, but they practically are, so these are very nearly the same answer. Interchangeable answers can’t be right. (While rule 3 does place H in Tokyo, H in Tokyo isn’t a sufficient condition for anything else.)

D and E have the same issue. They’re the same, except one is F and one is H. Interchangeable answers can’t both be right. (And, since G or H in Tokyo isn’t a sufficient condition for anything, these answers don’t force anything else to happen.)

So let’s try B. Placing G in two cities forces G into Manila and Tokyo, since G can’t go in Sydney (rule 4)

This in turn triggers rule 3 (G Manila ➞ H Tokyo). So, this is a promising answer, because it forces another thing to happen. Let’s draw it:

Next, we know Ibanez has to go in two spots. I placed one Ibanez in Sydney. There are only two cities with space open: Manila and Sydney. So Ibanez must go in both:

We have to place each person at least once, and we haven’t placed F. So, we have to put F in Sydney:

Everything is determined. B is CORRECT.

We saw in the setup that Ibanez must visit Sydney. D is CORRECT.

Let’s recap why this is true:

1. Every city needs two people
2. Only G, I, F and H are available. 
3. But, F and H can’t go together, so only G, I and one of F/H can go
4. G can’t go in Sydney
5. That leaves only I + one of F/H to go there

Before doing a general could be true, it’s good to skim all the answers and see which are plausible and which aren’t. I find narrowing it down to a couple of candidates in order to brute force only those helps clear the mind.

D and E are clearly wrong. F and H can’t go together, and every product manager must go at least once. So, neither F nor H can go everywhere (because then the other would have nowhere to go)

If you got question 19 right, you’ll recognize C is wrong. Ibanez goes exactly twice (rule 1), and as we saw in the setup, they must go in Sydney (because G can’t go there, so I + F/H must). Therefore Ibanez can’t also go in Manila and Tokyo: that would be three cities.

That leaves A and B. Now that there are only two contenders, we can try drawing both.

A places F and I in Manila:

We know that G must go in Tokyo, since it can’t go in Sydney. Then we can put either F or H in the other spots. Here’s one example:

So, this works. A is CORRECT.

Let’s try B, just to be sure. It places G and I in Tokyo:

Manila is the big space left to fill. We have already placed both I’s, so the only managers who can go in Manila are G + F/H. And here’s the problem: if we place G in Manila, then we’d have to put H in Tokyo (rule 3). But, Tokyo is full. So, B doesn’t work.

This question places G and H together. G can only go in Manila or Tokyo (rule 4), so we can try both scenarios:

The first scenario triggered rule 3: G in Manila ➞ H in Tokyo

It looks like H is in Tokyo in either scenario. So, D is CORRECT.

There’s no need to do any more diagramming. We would merely be drawing out “could be true” scenarios, which is a great waste of time on “must be true” questions you have already solved. You should focus on certainties.

This question places Ibanez in Tokyo. You should then consider the next most restricted factor: G. G must go either in Manila or Tokyo:

In the left hand scenario, Ibanez and G fill up Tokyo. This means that G can’t go in Manila, since that would trigger rule 3 (H in Tokyo). This can’t happen, since Tokyo is full.

Ibanez also can’t go in Manila, since they have already gone twice. That leaves just H and F, but they can’t go together (rule 2). 

So the left hand scenario actually doesn’t work!

That leaves us in the right hand scenario. G in Manila has triggered rule 3, so H must go in Tokyo. Since we have placed our two Ibanez and can’t place G in Sydney, this means both open spaces will be filled with F or H (and at least one must be F, because F hasn’t gone yet.)

A is CORRECT, because it’s possible. B through D contradict the diagram. E is a trap answer. It seems like it would work, but if you filled both open spots with H then there would be no room for F. Every product manager must go at least once.

On a rule substitution question, the correct answer must:

• Forbid everything normally forbidden
• Allow everything normally allowed

A lot of people miss this second quality. It’s very useful. It means that if one of the answers contradicts a scenario that would normally be allowed, then that answer is wrong.

Likewise, if an answer makes something happen that didn’t happen under the normal rules, then it is wrong.

A is wrong because normally G and Ibanez can go together. Here’s an example; it’s a scenario I made for question 20. It also disproves D:

This scenario proves B is too restrictive (I made this diagram just for this question):

C is a trickier answer. It’s true according to the rules, but it’s not restrictive enough. It allows F and H to be together elsewhere. This scenario is possible with C:

So we can eliminate four answers by recycling one scenario from question 20, and then making two quick custom scenarios. Not bad!

That leaves us with E, which is CORRECT. I covered why in the setup, but I’ll repeat it here. 

• Each city must have two managers. F/H can’t go together.
• The three options for each city are G, I and one of F/H
• So, if you are missing one of G or I, you must have the other in order to fill the two spots.

If you are missing both G and I, you’d end up with F + H, which violates rule 2.

Ok, but why does this substitute for the rule? Because it ensures every city is filled with at least one of G or I. There are then no open spaces for F + H to go together. For example:

As per E, every space has either I or G. Do you see any place to put F and H together? I don’t. So E indirectly replaces the rule.

It’s best to prove the right answer. But, if you were short on time, you could also have merely eliminated the first four answers as I did with scenarios above.