LSAT Preptest 60 Logic Games Explanations

This game combines linear and grouping features. It’s a medium difficulty game, I’d say.

There are three days of the week, and slots for morning and afternoon. I find it most effective to set things up like this:

The first question of the game doesn’t do this; it uses a vertical setup, which I find strange. We are very used to seeing the week left-to-right, on calendars. Why depart from that convention? The diagram I drew also matches the normal way to show linear games on the LSAT.

The game gives three rules, and there aren’t really any way to combine them:

So this will be a rules based game. But, you can still think about restrictions. The third rule is quite restrictive. It’s not just “Q before K and N”. It’s “Q is on an earlier day than K and N”. So we can add some not rules to the diagram:

This leads to a few additional deductions:

• If J is last, it is with K, not Q.
• If R is first, it is with S, not N.

I think these deductions above are too complicated to draw, but it’s helpful to think about what’s restricted.

While reviewing these explanations, I also figured out an additional deduction: Q can’t go on Thursday morning.

Why? If they do, then K and N go after it, on Friday, in either order. We get this diagram:

The problem is that J needs to go in the morning, with Q and N. That can’t happen here.

I’m writing this to give you an example of deduction that’s possible, but that isn’t necessarily important to get upfront. I have solved this game multiple times, but never noticed this before.

However, on the questions, I was easily able to figure this out. So, basically, it’s ok to miss some upfront deductions if you can figure them out on the questions. But, if you’re unable to do that on the questions, then you should spend more time looking for deduction on the setup.

How would you find this one? You should look for the most restricted variable, and see what its limits are. In this case, Q is the most restricted: it is mentioned in two rules. Further, it forces things to happen for both K and N, whereas placing K only forces things to happen for one variable: Q.

So, you could try placing Q, and see what the limits are. It turns out you can only place Q on Wednesday (any time), or on Thursday afternoon.

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 E. J needs to be in the morning.

Rule 2 eliminates D. R has to be in the afternoon.

Rule 3 eliminates A and C. Q needs to be on an earlier day than K and N. The “day” part of this explains why C is wrong, in particular.

B is CORRECT. It violates no rules.

This question asks who can’t go on Thursday morning. To solve it, you should think about who is the most restricted. It’s Q: they force K and N to go after them. And Q is also in J’s rule (rule 1: J is in the morning, with K or Q in the afternoon).

Putting Q in the afternoon helps us fulfill J’s rule, by letting us put J in the morning. Conversely, putting Q in the morning makes things harder: It forces us to put J with K. So, Q in the morning seems a likely candidate for something impossible.

And then there’s the Q – K, N rule. If we put Q in Thursday morning, we have to put K and N Friday (rule 3). 

That leaves no place to fulfill rule 1 (J in the morning with Q or K in the afternoon.)

This question asks who can’t go on Wednesday morning and afternoon. If you look at the main diagram, this one is startlingly easy. Here’s the main diagram:

K and N can’t go on Wednesday. Yet A places K on Wednesday. A is CORRECT.

This question places K on Friday morning. When a question gives you a new rule, you should always see how the new rule affects other rules. 

In this case, we know that J must go with K or Q, and J must be in the morning (rule 1). So, J can’t go with K here: it must go with Q.

So JQ will either have to go on Wednesday or Thursday, with Q in the afternoon. Therefore I looked for answers that said “J W/T morning” or “Q W/T afternoon”.

Based on that, A looked possible. I drew a quick diagram to check:

This works, it obeys the NR rule and the rule that Q is before K and N. A is CORRECT.

C is wrong because Q must go in the afternoon.

B and E are wrong because one of either N or S needs to go with R, and be in the morning. And this pair must be on Wednesday or Thursday (So that K can go Friday).

JQ fills one of the Wednesday/Thursday slots. So, we need S/N and R, in the other slot, with S/N in the morning. Therefore, there’s no space to put S or N on Wednesday or Thursday afternoon.

D is wrong because R needs to go with N or S, not K.

These are the explanations for the wrong answers. But the real way to go quickly is to prephrase answers the way I did above. I was specifically looking for J on Wednesday or Thursday morning, or Q on Wednesday or Thursday afternoon. This process lets you answer much more quickly and confidently.

This question places Q in the morning. When a question gives you a new rule, you should always see how the new rule affects other rules. In this case, the new rule forces J to go with K. (Rule 1: J must be in the morning, with either Q or K).

That means Q must go on Wednesday. If Q went Thursday, we’d end up with this scenario:

If we place Q Wednesday, we’ll have to place both JK later in the week. We also have to place R with either N or S. It will have to be N. Why? Because we need S to go with Q:

No one else can go there. 

• Not JK: they must go together on this question.
• Not N: it has to go after Q (rule 3)
• Not R: it has to go with S or N (rule 2)

So only S is left to go on Wednesday. And therefore, E is CORRECT.

This question asks who can go on Wednesday morning. To solve this, I found it helpful to draw the variables by the question:

Next, let’s look at the rules:

• K and N can’t go Wednesday, because they have to be a day before Q (rule 3)
• R can’t go on Wednesday morning, because R must be in the afternoon. (rule 2)

So we get this list:

So C is CORRECT. J, Q and S can all go on Wednesday morning.

If pressed for time, I would just pick that. But you can also prove it quickly by using scenarios from past answers.

The correct answer for question one proves that J can go on Wednesday morning.

This diagram from question 5 proves that Q can go Wednesday morning 

This diagram from question 4 proves that S can go on Wednesday morning:

This is a fairly straightforward linear game. Though as per usual, the LSAC has added a few new twists. Linear games had been getting too easy as people prepped more, so LSAC needed to increase the difficulty somewhat.

The first step on linear games should be to draw the rules and combine them. Here’s the first rule, L and M before H:

The second rule also contains L, so you can add that directly (L and P before J):

I next drew the fourth rule. This doesn’t connect directly, but I nonetheless placed it close by so I could see all six variables:

The space is a reminder that someone comes after G.

Finally, the third rule adds a conditional rule. These are becoming common on linear games: 

I added the rule about G not being last here, as well. It’s easy to forget, so it’s best to make it automatic by drawing the extra space.

The only other thing you should ask in advance is: who can be last? Who can be first?

• Last is quite restricted: Only H and J. And if Rule 3 occurs (M before P), then only J can be last.
• First is not restricted. P, L, M and G can all go first.

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 4 eliminates D. G can’t be last. I did this rule first because it is the easiest rule to scan for quickly.

Rule 1 eliminates C. L and M must both be before H.

Rule 2 eliminates A. L and P must both be before J.

Rule 3 eliminates B. If M is before P, then H is before G.

E is CORRECT. It violates no rules.

This is a general “cannot be true” question. There’s no way to predict the answer in advance. Instead, I just skimmed the answers to see if any were obviously wrong once I checked the diagram. Here’s the diagram:

C is CORRECT. L has two people after her, so she can be fourth at latest.

I quickly checked each answer against the diagram. As for the others:

• A: H before G is what happens in rule 3.
• B: There’s no one after H in the diagram, so H can go last.
• D: There’s no line connecting M and P, so either one could go before the other.
• E: You can easily put P second. Just put L or M first, then put P next.

I think sometimes students get intimidated by LG answers. “Can I realllllly do that?”, you ask. Yes! The diagrams aren’t crazy traps. On a linear game, if there’s no direct rule against something, it’s allowed.

This question asks who can go last. Only J and H can. D is CORRECT. Neither J nor H have lines to the right of them, so either one could go last. Everyone else has a line to the right, so they can’t. Here’s the diagram for reference:

I redrew my diagram for this question to place J before M:

Normally, only J and H can be last. Since J isn’t last, then H must be, and G goes before H – rule 4 says that G has someone after them.

Scanning through the answers, you can see that only A places H last. A is CORRECT.

Making this kind of diagram on a question saves time by letting you spot the obvious.

This question adds a new rule. As with question 10, I drew a new diagram using the additional rule:

Well, that didn’t seem to do much. But are any other rules affected? Rule 3 is the only rule not on this diagram:

On this question, G is not after H. So that means that M can’t be before P. Thus, P is before M. We can redraw the diagram above to show that:

Honestly, it’s an awkward diagram. On my own page I had arced lines, and it was a bit clearer. It’s also potentially confusing: Visually, H is to the right of J. But, there is no line connecting them! So, J could be after H.

There might be a better way to draw that, but I can’t think of one. Do enough linear games and you’ll learn to read this – only connecting lines matter, not position.

I should also note that I didn’t draw this diagram in two steps, as I’ve done in this explanation. I just drew the diagram all at once. If you can think of the deduction with rule 3 before drawing, that’s the best way. Otherwise, you should go step by step.

Now go through the answers and use the diagram to eliminate them:

• A could be false if you put both P and M before LG.
• B doesn’t have to be true: J could go last. There’s no line connecting H and J.
• C is CORRECT. Only P and M can go before L, which would make L third at latest.
• D and E are wrong for the same reason: either LG or P – M could go first. So neither answer has to be true.

This question places M first. That triggers the third rule: If M is before P, then H is before G.

Further, since only H and G can be last, then if H is before G, J must be last. C is CORRECT.

Here’s the diagram for reference:

You could also redraw the deductions above into a new diagram, but I found it would have been too much work on this question.

This is a unique game. I enjoyed it quite a bit, personally. On this game, I made no upfront deduction. But, while doing the first question, I figured out the constraints.

I’m going to discuss the constraints here, in the setup section, but remember: I figured these out while doing the game, not beforehand. For some games, that’s the best approach. You can always revisit the setup if you try the questions and can’t figure things out while doing. Sitting and thinking about the setup without producing anything is a waste of time.

Here’s what I drew before starting, by the way:

The key to this game is that there are limited transfers. And, sticking Mulch right in the middle leads to two transfers by default:

We can only have three transfers, so we are quite restricted. We’re going to have to do one of two things:

• Cluster the M’s together, or
• Put an M on the side

(I hadn’t figured these out before starting the game, I had only drawn the M in 5)

Clustering M’s together

In this case, there are only two transfers. Note that the bloc of M’s could shift one to the right or to the left. i.e. The M’s could be in 345, 456, or 567. 

Placing an M on the side

The other possibility is keeping two M’s together near 5, and placing the other M on one of the ends. Here’s an example:

This only results in three transfers. Again, the M could be in 6 or 4.

Note that we could also can have two M’s on the side, like this:

The point here is not to map out all the scenarios (there are a bunch). It’s to recognize the general structure of the scenarios:

1. The “Three M’s together scenario”
2. The “Two M’s together, one M on an edge scenario”.

The setup section explains how to build this diagram.

Main Diagram

Below are examples of the two main scenarios.

Three M’s together scenario

Two M’s together, one M on an edge scenario

This is not an acceptable order question, which is unusual for the first question of a game. This type of question being first is normally a sign you should have made a bunch of upfront deductions.

However, in this case I actually used this question to figure out the game. I figured out the two main scenarios faster than I could have if I had just stared at the setup. I found having to work through answers  on this question let me understand the structure of the game better.

Through trial and error, I realized the mulches had to be close together, since having stone beside the mulch in spot 5 would lead to two transfers.

So, placing mulch in fourth and sixth makes things easy. E is CORRECT. It leads to a total of two transfers.

Note that there are no restrictions on what can be stone, as long as you keep transfers to a minimum.

This game is an unusual game. I find that on unusual games, I often figure out the setup better while doing the questions. On more standard games, you can do the setup upfront, but that’s only because you’re familiar with the forms.

In the setup, I described how the mulches have to be close together. Either you place mulches together around the mulch in spot 5, or you place a mulch at one of the ends.

What you can’t do is place all the mulches separately. You’d end up with too many transfers:

So, D is CORRECT. At least two mulches must be together.

When a question gives you a new rule, you should always see how the new rule affects other rules. In this case, you should draw the transfers that occur when you place M third:

If we don’t place mulches beside these two M’s, we have four transfers. That’s too many transfers, so we need to place our third M in the middle, to eliminate two of the transfers:

We can only place three M’s. So, that means all other loads must be stone. E is CORRECT: the first load is stone.

This question allows us only two transfers. That means we’re in the scenario where all three mulches are together. It looks like this:

This diagram shows the M’s in 456. We could also have them in 345, since that scenario also has only two transfers. Every other arrangement leads to more than two transfers. (Or, if the M’s are in 567, one transfer)

So, that means that 1, 2 and 7 are always stone.

A is CORRECT. 

If only two loads are consecutive, that means we’re in one of the “Two mulch together, with mulch on the end” scenarios. We also must space out the stones so that there aren’t two in a row. This is tricky. Here are our parameters:

• We must put two mulch together
• We must put at least one mulch on one of the ends
• We can’t have more than two stone together (or mulch, of course.)

We already start with a mulch in five. There are two spaces to the right of it. We could make those stone without consequence:

We can then put our end-mulch in spot 1:

The reverse of this (mulch 7, stone 1) wouldn’t have worked. Try it: you would have ended up with three stones in a row at the front, because you’d have to put your other mulch in spot 4 in order to avoid having too many transfers.

Next, we need to put two mulches together. There are two ways to do this:

Both scenarios only have three transfers, and no more than two loads of the same material in a row. I believe these are the only two possible scenarios, though it’s not impossible that I missed one.

These scenarios prove that B is CORRECT. The fourth load could be stone.

This is an interesting game, because it does something rare: it completely invalidates a potential scenario. To explain what I mean by that, I need to do some setup first.

Here’s the best way to draw the diagram:

You can clearly and quickly see the three countries and the two types of interns.

I started this game with the final two rules: J is in Tuscany, and K is not in Spain. J could be either a photographer or a writer, so I drew both:

Normally, when the LSAT gives you something that can only be drawn two ways, you should split the game into two scenarios. I did that on this game. But, the interesting thing here is that the second scenario doesn’t work!

To see why, let’s consider the rest of the rules, and count. Rule three says H will be a photographer’s assistant. I drew that under the diagram, to indicate that H could go anywhere (in the p column):

Rule two tells us that F and K are in different fields. That takes up one spot on each side of the diagram.

Finally, we have to place G and L. They go together, vertically, and take up two spaces. That means they can’t go with J. Because whichever side J is on already has two people: J, and one of F/K.

So in both diagrams GL go opposite J. Let’s see how that works in both diagrams:

In the second diagram, there isn’t enough space! P requires four people: H, GL, and one of F/K.

So, J must be a photographer’s assistant, in Tuscany. That’s a big deduction! GL will go in the other group.

And there’s no need to stop there. I said splitting things into two scenarios is usually helpful. Now, we can make two scenarios based on where F and K go:

Next, see what else is restricted. The main restriction is on K: they can’t go in Spain. In the first diagram, this means H must be in Spain, and K must be in Romania:

The diagram to the right is a fair bit more open ended. In fact, the writer’s group in the left diagram is also open ended. It’s important to realize that in these diagrams, the only restriction is you can’t put K in Spain. Everything else is allowed, because we covered all the other rules. That’s the power of scenarios!

The setup section explains how to build this diagram.

Main Diagram

See the setup section for how to build these two 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.

Remember, these answers are lists of the photographer’ss assistants. That means the unseen interns are writer’s assistants – you have to consider them too.

Rule 1 eliminates B. Gombarick and Lha must be in the same field.

Rule 2 eliminates C. F and K must be in different fields. In this answer, since neither F nor K appears, that means they are both writer’s assistants. Which isn’t allowed.

Rule 3 eliminates D. Hall needs to be a photographers assistant.

Rule 4 eliminates no answers.

Rule 5 eliminates E. K can’t be assigned to Spain.

A is CORRECT. It violates no rules.

This question confused me. I couldn’t see how assigning Farber to Romania determined anything. But once you look at both scenarios, you can spot a must be true. Here are the two scenarios:

Now put F in Romania:

It turns out that H is in Spain in both scenarios. I hadn’t expected that, but that’s the power of drawing things out. B is CORRECT.

(The reason this was confusing to me is that F didn’t cause H to be in Spain in the first scenario. It was already there. It’s possible if I had figured out less in advance I would have found this question easier. But of course you shouldn’t let confusion slow you on logic games: just draw what the question tells you to, and the rules will usually lead you to the answer. It’s sitting still and “thinking” that will slow you down and produce nothing.)

This question assigns F and H to the same story. Meaning they have different roles. H is always a photographer’s assistant, so that means F will have to be a writer’s assistant. F was a writer’s assistant in scenario 2 from our setup. Here’s that scenario:

Next, you should draw this scenario and do what the question says: place F with H.

G and L could go in either order, in Romania or Tuscany. Since this is a “could be true”, they’re likely going to be the answer.

B is CORRECT. G can be in Romania.

Farber is a writer’s assistant in scenario 1:

This question asks who can be assigned to Romania. It could be K plus one of G, L or F. D is CORRECT.

This question places G and K together on the same story. That means they must be in different roles. In the setup, G and K were in different roles in scenario 1:

In that scenario, K must be in Romania. Since this question places G and K together, G must be in Romania too.

That leaves L and F free to go in either Spain or Tuscany. E is CORRECT.

This question asks who can’t be in Tuscany. We saw in the setup that J is assigned to Tuscany in both scenarios:

That means we should look to the P column to see who can’t be assigned to Tuscany. Rule 3 says H must be in the P column. So, H can never be in Tuscany, since J is already there in the P column.
C is CORRECT.