LSAT Preptest 82 Logic Games Explanations

Time on first attempt: 7:52 (see note at bottom)

This game is a mix of grouping, linear and sequencing. A company has six presentations to schedule in two theatres (east and west). The presentations are Goals, Management, Organization, Personnel, Sales, and Utilization. The presentations are scheduled on an afternoon, one in each theatre at 1:00, 2:00 and 3:00.

For a mixed game, this isn’t that hard, as long as you create a framework to keep track of everything. Without that it will be difficult however. (I personally made a small slip-up and lost two minutes I shouldn’t have as a result.)

So first, you should draw the three spaces for the east and west theatres. I added rules 1 and 2: Sales can’t be in the east theatre, and Utilization can’t be in the west.

Note that I don’t number my diagrams for speed of reproduction. You quickly get used to seeing them as 1, 2, 3 without numbers.

Next, I drew the ordering rules separately, noting that S is random (apart from not being in the east):

Lastly, you should draw in the ordering rules as “not” rules on the diagram. Since O, P and G have variables before them, they can’t be first. And likewise, M and U can’t be last because they have things after them. These not rules are fairly restricting in this game. For instance, S, O, P and G all can’t be in the first presentation in the east theatre:

And that’s it! The setup is not overly complicated. The only real trouble is getting an intuition for making this kind of setup in the first place. If the above setup isn’t yet second nature to you, I recommend repeating the game until it is. This game is an archetype of many common logic games scenarios.

Being able to do the easier games fast is key to having enough time for the harder games. And this is a very common diagram type.

The setup section explains how to build this diagram.

Main Diagram

This question gives a new rule: P must be 2nd, in the west theatre.

Check the rules to see where P is mentioned. Rule 3 says M is before P. So, M must be first. Could the question be that easy? Yes! B is CORRECT.

Don’t second guess yourself: sometimes LSAT questions are indeed straightforward.

Whenever you make a major deduction you should always check if that’s the answer before trying to make more deductions. Especially if no further deductions are obvious.

This question tests your understanding of the rules

We know from rule 3 that Management must be presented earlier than both Organization and Personnel, so if Personnel is presented at 2:00 (regardless of whether it’s in the east or west theatre) Management must be presented at 1:00. So, B is the correct answer.

This is a “cannot be true” question. So we should look for something hard. Something that might violate one of the ordering rules from the setup. Here are those rules:

The main thing to notice is that M and U have to be early, and O, P, G have to be late (and S has no ordering rule).

So, it is difficult to put M or U late, or to put O, P or G early. Which answer does that?

Only A, which is CORRECT. Look what happens if we put G before M. You can combine the two ordering diagrams above:

This doesn’t work. This lists four things in a row: U, G, M, O/P. But there are only three presentations.

It is very important to think in terms of what is easy, and what is hard. That lets you predict which answer is likeliest to correct. This method would have worked just as quickly if A been E instead. It still would have been the clear choice to start with. All of the other choices do things which are easy according to the ordering rules.

This question places Goals in at 3:00 in the east theatre. So you should draw that to start:

As we saw in question 2, on a must be false question you can use an “easy/hard” framework to evaluate which answers are easy to do according to the rules and which aren’t. It makes sense to start with the answers which seem hard to do.

From the ordering rules, M has to be early, and OP have to be late. So any answer placing M late, or OP early is hard. These answers are likely correct and the best place to start.

Let’s look at the answers:

• A: This looks hard. It places M late.
• B: This looks easy. It places O late. (letting O go after M)
• C, D: These look easy to do, as S has no ordering rules. So any placement of S is likely possible. Also, the scenario I drew below shows that D specifically is possible.
• E: U is normally restricted, but since this scenario places G third, U can go anywhere before that. U’s only rule is that it must go before G.

So, only A looks hard. Let’s consider it. We know that O and P have to be after M. That means, if M is at 2:00, both O and P have to be at 3:00.

But that doesn’t work. This question already places G in one of the two places at 3:00. So if M is at 2:00, there isn’t enough space to put O and P after M. A is CORRECT.

The only way to legally place M in this scenario would be at 1:00. That leaves enough space for one of P/O to go at 2:00:

This question asks how many people can go at 3:00. In our setup, we saw that M and U can’t go at 3:00, because they have to go earlier than O/P and G, respectively. So, that leaves 4 as being the biggest possible answer: O, P, G, and S

Let’s see if we can put all of those at 3 pm. First, we can use diagrams from prior questions. This was from question 3:

It shows G, O and P all at or able to be at 3 pm: O and P are interchangeable. So, this proves that G, O and P can all go at 3:00.

What about S? Well, S has no ordering rules, so we can easily swap S and one of O/P. Here’s an example that obeys all the rules:

So S, O, P and G can all go at 3:00. C is CORRECT.

What I’d like you to take away from this is how to efficiently construct a diagram that lets you prove multiple things at once. Make things easy: use prior diagrams, see if you can prove that 3-4 variables work at once. Once you prove that 3 of the 4 contenders can work, focus directly on the final one.

This question places U at 2:00. And rule 2 says U must be in the east theatre. So U must be in the east theatre at 2 pm. We can also place G in 3:00, as rule 4 says U is before G:

I placed G in the east, but they could go west too. Nothing turns on whether G is placed East or West. So don’t get paralyzed: place them somewhere, and see what deductions flow from that.

Next, think about who has to fill in the remaining spaces. The most restrictive rule is that M must be before both O and P:

That’s three variables, and there are only two open spaces at 2:00 and 3:00. So, M must be at 1:00, and O/P in 2 and 3:

This diagram proves B is CORRECT. Management has to be at 1:00, not 2:00.

A and E have to be true. C and D both could be true, as OP are interchangeable.

(For E, S has to be at 1:00 because that’s the only open space left.)

This question places management at 2:00. Rule 3 says O and P have to be after M, so O and P must go third:

(It doesn’t matter which of O and P goes east and west; they’re reversible)

You should think about the next most restrictive rule. Rule 4 says U must be before G. Only spaces at 1 and 2 pm are left open. So G must go West at 2 pm, and U goes East at 1pm (U always goes East, rule 1):

B is CORRECT. 2 pm is full, so S can’t go at 2:00. It has to go at 1:00 instead, like this:

All of the other answers are possible (D, E) on this diagram, or even must be true (A, C)

This question asks us to substitute for the rule that sales must be in the west theatre. When thinking about how to replace a rule, consider which rules interact with the original rule, and consider rules similar to the original rule. In this case, we know that normally, S is in the west, and U is in the east. Because of rules 1 and 2, like this:

So the solution here is simple. U is still in the east (rule 2). If we also say “S can’t be with U”, then that has the effect that S is in the west. C is CORRECT.

The ordering rule answers are wrong, as there’s no relationship between ordering and theatre relationship in this game. And D is wrong because G can be in either theatre.

Time on first attempt: 5:03

In this game, Cafe Cosmopolitano is offering the following weekly specials on Monday to Saturday: gazpacho, linguini, nachos, pizza and quesadillas. You have to figure out which days the specials can go on.

This is a linear game. It’s fairly straightforward. But, even if you got most of the questions on this game right, you should redo it until you can do it in 6-7 minutes or less. Ideally less than 5. Going quickly on easier games like this is what saves you time for harder games like game 3.

You should first place the most definite rules. That’s rule 5. Pizza is on Thursday, but not Saturday:

Next, place the other rules that affect ordering on the diagram. Q is not Friday (rule 6), and specials can’t be beside each other (rule 2). Since pizza is on Thursday, that means it can’t go on Wednesday or Friday:

Now let’s consider rules 3 and 4. Q is before G, and G only goes once. So there is exactly one QG block (To be clear, Q could go a second time, without G).

This QG block needs two spaces, but it can’t go on Friday-Saturday, as Q can’t go Friday. So QG must go either on Monday-Tuesday or Tuesday-Wednesday.

You should draw both scenarios:

And remember, Q can go a second time, but G can’t. (rule 3).

Next, you should make a list of the remaining rules. One special is twice, but on separated days. You can draw this with two X’s and a box around them, and a + in the middle to show they’re separated by at least one:

In the list I also added all the variables. G and N are offered once (rule 3). L, Q and P could be offered twice, and must also be offered at least once (rule 1).

Also, I circled L and N as they have no ordering rules attached to them.

This is one of those old style games where an upfront deduction solves most questions. In this case, the deduction is the two scenarios I drew above, and most specifically that QG can only go Monday-Tuesday or Tuesday-Wednesday. These scenarios solve almost all questions, and quickly.

If you had any trouble making the scenarios or understanding them, repeat this game until they’re second nature.

The setup section explains how to build this diagram.

Main Diagram

Scenarios 1 and 2:

On this game, we can split the setup into two scenarios, with QG 2nd and 3rd or 1st and 2nd:

This question places L directly before P. That can only happen in the second scenario, with L on Wednesday. It looks like this:

(No other places work because P can’t go Friday or Saturday. And if P goes Monday, there’s no space for L to go in front)

Right away, this diagram lets us eliminate some answers.

• A and B are wrong because Monday is quesadillas, not pizza or gazpacho.
• C is wrong because quesadillas goes Monday, not Tuesday.
• D is wrong because Wednesday is linguini, not nachos.

E is CORRECT. Linguini can go more than once, so linguini could go Friday. We could then place anything else Saturday, such as Quesadillas.

This question asks where linguini can’t go. It’s a bit of a trick question. It’s really asking “which spot is already always filled”. The game is expecting you to have made a setup deduction to solve this question.

In the setup, we created these two scenarios:

You can see that Tuesday is filled in either scenario, by Q or G. So, L can’t go there. D is CORRECT.

This is a general could be true question. It’s easiest to use process of elimination. Here are the two scenarios from the setup:

• A is wrong because there’s only one G, and it has to have Q before it. If G is on Monday, there’s no space to put Q before it.
• B is wrong because in both scenarios Tuesday is full: either Q or G must go there. So N can’t go Tuesday.
• C is wrong because rule 2 says two specials can’t go in a row. P is already on Thursday, so it can’t go on Wednesday too.
• D is wrong because G needs Q before it (rule 4). Pizza is the special the day before Friday, so G can’t go on Friday.

That leaves E. This diagram proves it is possible. So, E is CORRECT.

To make a diagram like this, start by putting Q Saturday. Then see who is left to place (QG, L, N), and put them into valid positions. I started with QG first. Once that was set, I knew there were no ordering restrictions on L or N, so they could go anywhere.

In our main setup, we saw there were two main templates:

In both templates Tuesday is full, so Pizza can’t go there. Further, Pizza can’t go Wednesday or Friday (rule 2), or Saturday (rule 5).

So, that leaves only Monday:

A is CORRECT.

To solve this question, look at the main setup diagrams:

G can only go Tuesday or Wednesday. So, B is CORRECT.

Time on first attempt: 11:40

This game is a mixture of linear and matching. You have to match prosecutors to witnesses, and place them in order. The witnesses are Farrell, Greer, Hong and Ikaba. They are interviewed over four days by three investigators: Qin, Rivera and Shaw.

Each witness is interviewed once, and each investigator will interview at least one witness. (One investigator will do two interviews)

For me, this was the hardest game of the section by far. I took 11:40 min to do it. I had time to do this because I had gone very fast on the easy games and had a buffer of time. If you ran out of time due to  this game, part of your training should involve redoing the easy games to get better and faster at those so you have that buffer too.

With matching games, I like to place the second variables under the diagram. In this case, investigators go below. Since rule 4 says R does the third interview, we can draw this on the diagram:

Next, you should think about what is the most restrictive element in the remaining three rules. (I failed to do this, and it’s what made the game hard for me.)

The restrictive element is this: the interaction of rules 1 and 3. Note these three things:

1. Rule 1: G and H must be interviewed by the same investigator.
2. Rule 3: Qin must interview F
3. Setup: There are only three investigators

We can conclude that one of the three investigators interviews twice, and the other investigators interview once. e.g. QRRS could be the four investigator.

The interaction I am talking about is this:

• Exactly one of the investigators interviews twice, and they interview G and H
• Therefore, it is not Qin who interviews twice, as Qin must interview F.
• Therefore, it is Rivera or Shaw who must interview twice, and they interview G and H.

So Q can only be placed once. And since R is in third, that means there are three places that Q can go. So we can make three scenarios for this game: Q 1st, Q 2nd, and Q 4th.

I normally don’t make more than two upfront scenarios. But, when there is a strictly defined set of scenarios like this, it can be very efficient to make three. Scenarios tend to be much faster than simply using the rules. However, unusually for a scenario based game, you also have to remember the rules very well when examining the scenarios. That’s part of what makes this game difficult.

Before making the scenarios, let’s get clear on our rules. G and H need the same investigator: we can use an X to mark this. And, G and Ikaba can’t be consecutive:

I’ve also written the deduction that Q can only go once, as it will be important not to draw Q twice in any scenario. We could draw “QF”, but we’ll have that in the scenarios, so it’s not necessary.

Now, let’s make the scenarios, starting with placing QF first. Remember, when we’re making the scenarios and have placed Q, we only have two rules left. G and Ikaba are separate, and GH have the same investigator. So on any question all you have to do is draw the appropriate scenario then run through those two rules.

How to Build These Scenarios

It’s very possible that these scenarios will seem confusing if you just read them. So, I’d encourage you to build them yourself. Here’s how. Do the following for each placement of Q:

1. Place Q. In scenario 1, place Q 1st. In scenario 2, place them 2nd. In scenario 3, place Q 4th.
2. Place G and Ikaba separate from each other (rule 2). But draw a line to show they’re reversible.
3. Place the investigators. If either G or H is 3rd, then R interviews both G and H (rule 1 + rule 4). If instead Ikaba is third, then R interviews Ikaba and S interviews G and H.

Fiddle around with making these scenarios while reading the explanations and hopefully they’ll make sense. Make concrete examples placing G and Ikaba in definite spots to help it click. This kind of scenario construction under uncertainty but within the constraints of the rules is key to advanced logic games.

Scenario 1: Q first

If Q/F is first, then H has to go third in order to separate GI (rule 2):

The curved line shows G and I are reversible. So if a could be true question asked “could G go fourth”, this diagram would prove it could.

The only rule left is that GH have the same investigator. We can draw that with a subscript. H already has R, so G has to be interviewed by R too (rule 1), and Ikaba by S (because every investigator has to go at least once):

Scenario 2: Q second

For the next scenario, place QF 2nd. This leaves us with an open spot in 3-4. We can’t place GI there: they must be separate (rule 2). So one of GI goes there, along with H. The other GI goes first:

The comma between H, G/I tells us that H and G/I can go in either order. And I used x subscripts to remind us that GH have the same investigator, no matter where they are placed.

So, for example, if I is 1st, and HG in 3rd and 4th, then I is interviewed by S, and HG are interviewed by R (because R is third, and because both H and G need the same interviewer).

So, this scenario is a bit more open ended than scenario 1. But once you place Ikaba, it becomes quite clear how to place the rest and/or who investigates them.

Scenario 3: Q is fourth

Next, scenario 3. Q and F are fourth. This is like scenario 1: it forces H in the middle of G and I to keep them apart (rule 2):

G and I are reversible. The x’s indicate that H and G will have the same investigator. So, for example, if I goes third, then GH will have S. If I goes first, then GH will have R.

That’s it for this game. Here are the three scenarios all together:

These solve basically all the questions. Remember the outstanding rules:

1. GI don’t go together.
2. GH have the same investigator.
3. F only goes once. S needs to be placed.

I should also note that all three scenarios are valid and can obey the rules. But in some games, there are scenarios that seem possible at first, but don’t produce a valid scenario. So, it’s worth testing the scenarios a little to make sure they obey the rules in case you’re not 100% clear on how a valid scenario can be formed from the above drawings. (By valid scenario, I mean one where all the witnesses and investigators have been placed into a definite position on the board)

What if these scenarios don’t make sense

Rereading this explanation, I found it confusing to explain in writing. But, I have confidence these scenarios will make sense if you try them, and they aren’t overcomplicated to actually use. So, I recommend you try these scenarios on some questions and see how they work with examples, and then redo this game and use them.

Key Relationship: who interviews G and H?

If G or H is placed third, then both will be interviewed by R. Whereas if I is placed third, then GH will both be interviewed by S. This rule applies in all scenarios.

The setup section explains how to build this diagram.

Main Diagram

The three main scenarios:

The remaining rules to remember along with 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 C. Greer and Hong cannot be interviewed by different investigators.

Rule 2 eliminates B. Greer and Ikaba must not be interviewed on consecutive days.

Rule 3 eliminates A. Qin must interview Farrell.

Rule 4 eliminates D. Rivera needs to conduct the third interview.

E is CORRECT. It violates no rules.

This question says Rivers interviews Ikaba. To solve it, you should refer to the scenarios from the setup:

There are a couple things to note:

• The first scenario is out, as S interviews Ikaba there
• The same investigator interviews GH, so that investigator goes twice. The one that interviews Ikaba goes once.

Thus, if R interviews Ikaba, then that must be the only time R interviews anyone. So, we can say that Ikaba must be third.

Look at scenarios 2 and 3. What happens if Ikaba is third? We know IG can’t go together. So, in both scenarios, G must go first to get away from Ikaba.

B is CORRECT.

Here’s a full drawing of how it looks in scenarios 2 and 3:

You can see none of the other answers have to be true. In fact only even C and D could be true.

This question is a general could be true. There’s no upfront work, so you should refer to the three setup scenarios and look to see if the answers could be true in any of them.

Note, it’s helpful to be able to visually move the variables down onto the diagram correctly in your minds eye. When the variables are drawn so close to the diagram, this is a learnable skill. It saves a lot of drawing. For example, when looking at scenario 1, I can visualize G in 2nd and Ikaba in 4th, or vice versa. So you can imagine either variation without drawing them fully.

A and B are wrong because Q can only go once. That’s because the same investigator must interview G and H. Q can’t, because Q is already interviewing F. An interviewer can interview at most two people.

C doesn’t work in any scenario. Rule 4 says R must conduct the third interview. And no investigator can conduct three interviews. So R can’t do the 2nd, 3rd and 4th interviews!

D is CORRECT. In scenario 3, S could do the first and second interviews if Ikaba is third. Shaw would interview G and H, fulfilling rule 1.

E is wrong because it contradicts the scenarios above. In scenarios 2 and 3, Q is 2nd and 4th, respectively. And in scenario 1, Shaw only interviews Ikaba. It is Rivera who conducts two interviews (since Rivera interviews H, they must also interview G. See rule 1.)

This question asks what could be true if Shaw does the 4th interview. And it’s super easy if you made these upfront scenarios:

A is CORRECT. S can be fourth in the first scenario if Ikaba goes fourth. And in that scenario, F is first.

All of the other answers contradict the scenarios. Only read the rest of this explanation. if you were stuck on an answer: the analysis above is the main thing to know for solving this explanation.

The big challenge on the other answers is placing G and Ikaba. Rule 2 says they must be separate. So, for example, B and C place G 3rd and 4th, respectively. That means that Ikaba must be 1st in B, or 1st or 2nd in C.

And this leaves things very restrictive. To place Shaw 4th, you must have them interview Ikaba, G or H. They can’t interview F, because F is interviewed by Q.

And this just doesn’t work, because of rule 1: G and H must be interviewed by the same person. In B , that’s R, since G is interviewed by R. So H can’t be interviewed by Shaw.

In C, the issue is who goes third with Rivera. It can’t be Q, because Q has to interview F. So only H can go third. And therefore H and G are both interviewed by Rivera:

What about D? Here, H is 2nd. We also have to place G and Ikaba apart (rule 2). So one of G and Ikaba goes 1st, and the other goes in one of 3/4.

We could maybe make one of them S if they were 4th. The problem is we still have to place F. And they have to be investigated by Q, which means we can’t place Q 3rd. So Q/F are 4th, and once again S can’t be 4th:

Finally, E. We place Ikaba 1st. Then, there are only two ways to arrange things:

If G and H are 3rd and 4th, then they must both be interviewed by R (rules 3 and 4).

If G and Q are 3rd and 4th, then F interviews Q 4th (rule 3)

So, either way S can’t be 4th.

This question asks you to force Farrell to be interviewed fourth. If you made scenarios from the setup, you’ll see there’s only one scenario where F is fourth: the third one. Here are the setup scenarios:

So we somehow have to force the 3rd scenario. The notable feature of the 3rd  scenario is that H is 2nd, so B looks like the best answer to test. (In fact, if you were pressed for time, you could pick B with fairly high confidence based on the above reasoning)

Here’s what happens if you place H second:

Nothing obvious at first. But, we can’t place Q and F third, because R is already there. So we can only place QF 1st, or 4th. What happens if we place them third?

Hmm, GI are beside each other. That doesn’t work. So, if H goes second, Q must go fourth instead. B is CORRECT.

A and C could seem tempting. Why isn’t they right? First, Greer and Ikaba are semi interchangeable in a lot of scenarios. Interchangeable answers can’t both be right.

Second, look at scenario 2. Greer and Ikaba can both be first in that scenario, and yet Q and F are 2nd. This doesn’t work, we need Q and F to be 4th.

To solve this question, you should reference the setup scenarios (You may be noticing a pattern in these explanations.). Since this is a could be true exception question, you can eliminate an answer if it’s possible in any of the scenarios:

All of the wrong answers are possible in one of the scenarios:

• A is possible in scenario 1
• B is possible in scenario 2. Hong could be third in that scenario, with I and G 1st and 4th in either order.
• C is possible in scenarios 1 and 3
• E is possible in scenarios 1 and 3

D is CORRECT. It isn’t possible in any scenario. I:n scenario 1, H and F are in 1st and 3rd. In scenario 3, H and F are in 2nd and 4th. In scenario 2, H comes after F.

So, there’s no way to put H before F in any scenario. The real problem is this:

• If you put H before F in 1st and 2nd or 3rd and 4th, you get G and Ikaba together (in 3/4 or 1/2, respectively). This violates rule 2)
• If you try to put H and F in 2nd and 3rd, you violate rules and 4: F has to be interviewed by Qin, but Rivera must be 3rd. So, Q can’t be 3rd.

Time on first attempt: 6:05

In this game a teacher is assigning presentations to students. The three students are George, Rita and Wendy. The presentations are on Machiavellianism, jitsuaku, and Shakespeare’s villains.

This is a grouping game with three groups. It also has some in-out grouping features. So, this game could be difficult if you are not so familiar with in-out grouping. However, if you are, the game can be quite easy. If you had any trouble with this game, I recommend repeating it until the rules are second nature.

I should also note that it is possible to create long chain in-out grouping diagrams for this game, akin to the birds in the forest game from LSAT 33. However, those diagrams don’t end up being useful. It’s possible LSAC left this possibility as a trap for experienced students.

(I put the long chain diagrams at the bottom of this setup section if you’re curious to see them)

Instead of making long chains, the game is more easily solved by drawing rule 3 on the diagram itself, and then referencing rules 1 and 2 as situations call for it.

You can arrange the three groups like this:

Two things to note:

• Each subject has a single presenter, at least. So I drew one slot. If the rules hadn’t specified a minimum, I wouldn’t have put the slots.
• Rule 3 says everything in M is also in S, so I drew this directly on the diagram. That way it’s much harder to forget.

Apart from that diagram, there are only two rules in this game. Rule one is quite simple, it’s rule two that is tested by most of the questions. But let’s get rule one out of the way. It says that if G presents on J, then so does R:

If you’re razor sharp on contrapositives, that’s all you need to draw. But, if you can’t see contrapositives in your head yet, you should also draw the contrapositive, like so:

Rules 2 and 3 together are restrictive

Summary: You can’t have R and W together in S. And, if you place either of R and W in M, they’ll be in S, too. So, it is very easy to accidentally place R and W together in S by placing one of them in M.

So, it is very important to be careful when placing R and W in either S or M.

————

Rule two is quite simple, however it is more restrictive than rule 1. The rule says that if R presents on S, then W does not. Effectively, this means that W and R can’t be together in S. So, if W is in S, then R can’t be. We can draw both:

This is different from rule 1. Rule 1 just says “if one thing happens, do another”. But rule 2 actually presents a rule violation. If you put one of W or R in S, you can’t put the other.

This gets even more restrictive when you consider rule 3: anything in M is in S. So, suppose we have W in S, and R in M:

(The “not R” in S is a reminder of rule 2: W is already in S, so R can’t be)

On its face, the above diagram doesn’t violate rule 2. However, once we apply rule 3, we have to put R in S as well. Everything in M goes in S, too:

So this diagram violates rule 2. If W is in S, we can’t put R in S or in M. And vice versa. If R is in S, we can’t put W in S or in M.

This is the central point of the game. If you understand the paragraph above, you basically solve every question easily. This game is unusual in the LSAC is just repeatedly testing a single idea.

Also, note that in the case of W in S, we have to put R somewhere, so R goes in J. And if R is in S, we have to put W somewhere, so W goes in J. This comes up on a few questions too. If you didn’t follow that, review the bolded paragraph above, or the explanation under the conditional diagrams below.

Useless long chain diagrams

I wrote above that it’s possible (but useless) to make long conditional chain diagrams on this game. I made them, until I realized I never actually used them on any of the questions. Here are the diagrams I made:

Basically, if R is in M, then it’s in S (rule 2). So, W can’t be in S (rule 2), nor in M (contrapositive of rule 3). W has to go somewhere (setup paragraph), so it goes in J.

The same logic applies with W in M.

The setup section explains how to build this diagram.

Main Diagram

I kept the contrapositive for rule 2 as it’s vital to remember. I didn’t draw them for rules 1 and 3 to avoid clutter, as those rules are simpler: place a thing, another thing happens. But, these are the contrapositives below if you had any trouble identifying 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 D. If George presents on jitsuaku, then so does Rita.

Rule 2 eliminates B. If Rita presents on Shakespeare’s villains, then Wendy does not.

Rule 3 eliminates C. Wendy presents on Machiavellianism, so she must also present on Shakespeare’s villains.

Rule 3 also eliminates E. Rita presents on Machiavellianism, so she must also present on Shakespeare’s villains.

A is CORRECT. It violates no rules.

This game only has three rules. And only one of them mentions something impossible: R and W can’t go together in S. So, the right answer has to involve R and W, and somehow place them in S.

This question places students in Machiavellianism. But, rule 3 says anyone in Machiavellianism is also in S.

So, E is CORRECT. If R and W are in M, then they are also both in S (rule 3). This violates rule 2: R and W can’t be together in S.

This diagram illustrates it:

This question asks what happens if W presents on two subjects. Now, you could sit here staring at the question and “think” about what can happen.

This tends to be unproductive though. On logic games, thinking happens by drawing, not in your head. For this game, there are only three presentations. So, if W does two presentations, that means she is in two out of three groups.

There are only two ways to place W twice and two scenarios are easy to work with. So, draw them and see what happens:

(If W is in M, they’re also in S. That’s why there’s no third scenario with W only in J and M)

I drew that R can’t go in M or S. Why? Because rule 3 says everything in M must go in S. And rule 2 says R and W can’t go together in S. So, if we did place R in M, we’d also have to place R in S, with W.

So, R can’t go in M or S. But, we have to place R somewhere. So, J is the only place they can go:

When you make a major deduction like that, you should check if it solves the question. It does in this case: D is CORRECT. R has to present on jitsuaku.

This question restricts George to one presentation: jitsuaku. So, we must place G there, and also restrict them from going in the other two places.

I also placed R in J: rule 1 says that if G is in J, R is too.

This question asks what could be true. To figure that out, you should make a couple of scenarios to figure out how to place the other variables.

Basically, someone has to fill M, and only R and W are left. Anyone going in M also has to go in S, and R and W can’t be in S together. So, only one of R/W can go in M and S. We can make two scenarios based on that:

(You might wonder: why can’t G be in M? It’s because the local rule on this question said G is only in J)

Whenever you make major progress in your diagrams, you should check if that has already answered the question. No sense doing more work than necessary, this is just a could be true question. And in fact we have answered the question. C is CORRECT. In the second scenario, R can present on all three subjects.

Though, you might fairly wonder if the scenario works, because we didn’t complete it. So, what else do we need to add? Only W, and we can legally add them on J:

As for why the other answers are wrong:

• A is wrong because G always presents on J on this question. Rule 1 says that if G presents on J, so does R.
• B, D and E are all wrong for the same reason: they leave nobody to go on Machiavellianism. All three answers place either R or W on S, but not M. But, somebody has to go on M. And, it won’t be G, because this question says G only presents on J. So, we’re only left with the other of R or W to go on M. And, anyone in M goes in S (rule 3).

  So all three answers have the effect of placing both R and W in S, violating rule 2. For example, in B, R is in S. And since R isn’t in M, then W must be. This is because G can’t go there on this question, as G is only in J. So W is also on S (rule 3) and therefore R and W are together on S, violating rule 2.

  Every one of B, D and E has this same basic setup. Draw them out and see. You always end up with R and W together on S.

This shows why making scenarios is valuable. You can really waste a lot of time on B, C and E if you don’t have a framework that lets you predict the answer up front.

This question places W on J and S, but not M. So, we get the following diagram:

R can’t be in M or S. Why? Because anything in M is also in S (rule 3), and R and W can’t be in S together (rule 2).

This means R must go in J, as it’s the only space open to them. We have to place G, and we have to fill M. So G will have to go in M, since they’re the only one left:

This is a could be true question. B is CORRECT. G could go in jitsuaku of course, but he doesn’t have to, so this answer works.

All of the other answers contradict the diagram. A and C say that G isn’t presenting on M, but we need someone to present there. G is the only one who can.

D directly violates rule 3! If R is on M, they must be on S (rule 3). This forces R and W together, violating rule 2. Answers are rarely this flagrantly wrong.