LSAT Preptest 84 Logic Games Explanations

Time on second attempt: 3:31

See “repeating games” at bottom of section

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This is a sequencing game, and a fairly standard one. Sequencing games are one of the most learnable types of logic game, so if you found this hard, you should redo it until it’s second nature. You can flip this from being a hard game type to routine.

The key to standard sequencing games is that all the rules can fit together. So, you should start by drawing the first rule and then adding the rest. S is before O:

Then, R, V and Z are all after S:

Then, Z is before L, and T is before L and M:

The most common point of confusion with these diagrams is that people try to read them left to right and ignore the lines. That’s a mistake – the lines are what make the diagram.

So T, for example, has no line to its left. That means that T could be last. This is true, even though O appears to the left of T on the diagram. This doesn’t matter, as there are no lines connecting T and O.

So, on this diagram, S or T could be first. R, V, L or M could be last. And M, for example, could be as early as second, or as late as last.

If this is confusing, play around with the rules and the diagram a bit. Make some examples, and follow the lines with your finger.

That’s pretty much all there is to the setup. The bulk of this game type is simply learning to read that diagram quickly and correctly.

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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.)

The setup section explains how to build this diagram.

Main Diagram

This question asks who can’t go in the first four spots. To solve this, look at the diagram and see who has many lines before it. Here’s the main diagram, who has the most diagrams to the left of it?

It’s L! A is CORRECT.

This is a skill you should practice doing until it is second nature. Here’s a skeleton diagram which only includes the elements to the left of L with a line. This is what I see in my head when I look at the main diagram and focus on L:

Note that if you picked M, your error was looking at a variable that was merely on the right of the diagram. It’s the lines that matter.

As for R, V and Z, they’re all equivalent: only S and O come before. Equivalent answers can’t be correct.

This question places T 6th. You should look at the diagram and see what other variables are affected. T has two people after it, L and M. Since T is 6th, those must fill spaces 7 and 8, in either order:

(The comma indicates that L and M are reversible – this is a quick way of drawing it)

The only people left to draw on the diagram are SO, RVZ. This section of the diagram:

You should practice looking at the main diagram in slices. Since T, L and M are on the diagram, they melt away from the main diagram and what’s above is what’s left. Now you can focus on placing it on the diagram. Since there is nothing before S – O, those go first:

R, V and Z go next, in either order. At this point, you should see if this solves the question. It does, A is CORRECT.

Note that it’s perfectly possible to do this question in your head – I did. But, the diagrams are there as a backup if you find it hard to visualize using the main diagram alone.

This question places O fourth. There are only four spaces left after O, and R, V and Z – L have to come after O. So the final spaces are taken:

That leaves S and T – M to go before O:

The only real restrictions are:

• T can’t go 3rd (M is after)
• M can’t go 1st (T is before)
• L can’t go 5th (Z is before)
• Z can’t go 8th (L is after)

You don’t need to write those out beforehand of course – I’m just pointing out all the limits. But in practice I just do those in my head when looking at each answer.

C is CORRECT. Everything else is possible in the diagram above.

This question asks who can be shipped second. To answer that, you should look at which variables don’t have any lines to the left of them. Then trace forward one line. Those are the variables that can go second. Like this:

On the diagram, if something has zero people in front of it, I marked a 0, and if it has one person in front, I marked a 1. Either group can go second.

You should learn to see sequencing diagrams this way. how many things are in front of a variable: 0, 1 or more? It determines where they can go.

(Something like S can easily go 2nd, as well, you just put T in front of S).

So, B is CORRECT.

If Z is 7th, then L is 8th:

This doesn’t tell us much on its own. It’s helpful to consider who else is left to place. Without Z and L, we’re left with these two diagrams:

You can mix these any way you like. For example, M could go in between O and V. Only the order rules represented by the lines matter (So, for example, M must be after T. Apart from that it is flexible.)

E is CORRECT. T can’t go 6th, because M has to go after it (and 6th is the final open slot).

This question places Z third. You should look at the main diagram and see who has to go before Z:

It’s O and V. So, you can draw that on a diagram:

Now, all that’s left is to consider who else is left to be placed from the main diagram above. There’s T, which is before L and M. And there’s R and V, which are now free to go anywhere – their only rule was that they’re after O:

So, on this diagram, once again, you follow the lines. T must be before L and M. But, you could have TRLMV for example.

Time on second attempt: 7:46

See “repeating games” at bottom of section

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This is a linear game, where you also have to match people with a project. For this, it’s easiest to draw one horizontal row. I placed the architect on top, and then draw their project below:

I’ve added the final rule as well, that H is last.

To explain what I meant about architects/projects, here’s are a couple of sample scenarios with architects on top and projects below, so you get a feel for it:

Note that there are other ways to do this. For instance, you could:

• Draw numbers beneath each slot
• Draw two rows

Both of those add more information, but increase clutter. Whether you need them depends on personal preference. In my case, I don’t need that info, so including it slows me down.

Next, there is one ordering rule. F is directly in front of G. I would normally draw this in a box, but for some reason here I preferred to draw it as two slots – maybe to distinguish from projects:

So we’ve covered rules 4 and 5. I left rules 1, 2 and 3 for last, because they’re all in/out type rules. It’s best handle similar rules at the same time.

Along with the fact that every rule is in/out, you should also notice that every rule has F. This means you can probably connect them together. It’s vital to look for these connections.

We can draw the first rule first, but we’ll place F in the necessary condition, in order to connect it. Either F or L has to have Z. So, if L doesn’t have Z, F does:

Now, think about the other rules. If F has Z, what does that mean? Well, F certainly doesn’t have W. How does this apply:

• Rule 2: G must have W (since F doesn’t)
• Rule 3: H must have Y (since F doesn’t have W)

We can add both of these deductions to the main diagram!

If this is confusing, remember:

• The L/F section is rule 1
• The F/G section is rule 2
• The F/H section is rule 3

So all the in/out rules can be connected! This really makes the game straightforward.

A note on rule two. If you have a rule in the format of “Gw —> Fw” then the contrapositive is “Fw —> Gw”. What this actually means is “one of either G or F has W”

The final thing you should do is take the contrapositive of the diagram above. Just reverse everything, and negate. Remember also to turn the “and” to “or”:

The final thing you can consider doing it mapping out the ordering rules. I actually didn’t do this, but there are only two possibilities, so it can be worth noticing upfront. The FG block makes this very restrictive:

You could probably even make some scenarios about which architects have which projects, but in my view that would be excessive.

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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.)

The setup section explains how to build this diagram.

Main Diagram

(Projects go on the bottom of these four slots: w, x, y, z)

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 against for this question will help you memorize them and it’s also more efficient.

Rule 1 eliminates A. Z must be assigned to either Fredericks or Lee.

Rule 2 eliminates E. Fredericks must be assigned W since Guerrero has not been assigned W.

Rule 3 eliminates C. Fredericks has not been assigned W, so Horowitz must be assigned Y.

Rule 4 eliminates B. Fredericks’ project must be completed immediately before Guerrero’s.

D is CORRECT. It violates no rules.

This question looked intimidating at first. It’s a general “could be true” question, with no guidance. But, it turns out that every answer in A–D has the same mistake. FG need to be together, and none of the first four answers allow that.

• A puts G first. (No space for F before)
• B puts F third (No space for G after, H is there)
• C puts L before G (Instead of F before G)
• D puts L after F (Instead of G after F)

It’s that simple. If you know the “FG” rule, these answers should be very easy.

E is CORRECT. L can go first, F can go second, G can go third.

This question places x last. We already know H is the architect that goes last, so that means H and x are together:

Next, you should consider the in/out rules. Since H has x, that means H doesn’t have y. I’ve underlined which parts of the diagram that triggers:

Since H doesn’t have y, F has w, and therefore L has z. (Since F doesn’t have z).

We can draw these two new deductions by the diagram:

In the setup, I showed how there were only two possible orderings: FGL, or LFG. That’s because FG have to be a block. Like this:

You can add in the deductions from this question: Hx, Fw, and Lz. G and y are the only ones left, so they must go together too:

This lets us solve the question. B is CORRECT. W can’t go third, because F can’t go third. Every other answer is possible.

This question asks which architect can have x. You can look at past questions to narrow down answers. For instance, question 9 placed x with H (since H is last). So the right answer has to have H.

Likewise, the correct answer to question 7 was D, where F had x. So, the right answer has to have F, too. That leaves only C and E.

E has both G and L. So, if you make any scenario where either G or L have x, then E is the right answer. To make an easy scenario, it’s useful to start by fulfilling rules. So, making H has y is easy, for example, as is making F have w:

Notice that I made sure a rule was fulfilled, and left x to go with either L or G. (The rules this fulfills are rule 3, Hy, and rule 2, Fw)

Since F doesn’t have z, L must. Which leaves G to have x:

And this proves E is CORRECT. But, you might be thinking “don’t we have to prove L can have x?”. Not really. We know the right answer has to have F, G and H, and only E has that combo.

But, if you want further proof, then this proves L can have x:

L: x

F: z

G: w

H: y

It’s the scenario from above, except L, F and G moves their letters one to the right. (From G to L, L to F, F to G, etc.)

This should be an easy question. We know that H is last, and that w can only go with F or G. So, therefore, W can’t go last. A is CORRECT.

For the other answers, this scenario from question 10 proves B and E are possible:

This diagram from question 9 proves that C is possible:

And, if you swap the y and x, then that proves D is possible as well. There’s no rule broken by this swap:

Time on second attempt: 6:16

See “repeating games” at bottom of section

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This is a grouping game. I found it fairly standard. It becomes fairly easy if you make dual scenarios. I’ll show you how to do that.

The third rule is what lets you make split scenarios. Either F is the nominee for treasurer, or H is. You should draw both scenarios:

I’ve added the first rule to the left hand scenario: if F is in, G must be out.

I’ve also added the fourth rule to the diagram. K can’t be councillor. Now, you could just leave it at that. But if you look closely, there’s another deduction. In either scenario F or T fill up the spot of treasurer. Which means that K can’t go there either. You should draw that:

So, if K is in, they must be mayor! All that’s left is to draw the remaining rules:

FG can’t go together (rule 1). If H is councillor, then J must be mayor (rule 2). And finally, L is random, which I’ve indicated with a circle.

I also drew K there, mostly as a visual reminder. Since F, G, H, J and L were already in the list, I preferred to draw the 6th character so I could count them. But, note that K can only be mayor, so they’re not totally unrestricted like L. (You could equally leave K out of this list)

That’s all there is to the setup!

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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.)

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 against for this question will help you memorize them and it’s also more efficient.

Rule 1 eliminates A. If Frost is nominated, Grant cannot be nominated.

Rule 2 eliminates D. Hu is a nominee for councillor, so Jensen must be the nominee for mayor.

Rule 3 eliminates C. The nominee for treasurer must be either Frost or Hu.

Rule 5 eliminates B. Kuno cannot be a nominee for councillor.

E is CORRECT. It violates no rules.

This question makes G a nominee for councillor. That means that F is out (rule 2). And that means H must be treasurer, since either F or H is treasurer (rule 3):

That solves the question! E is CORRECT.

This question asks who can be councillors together. The easiest way to solve this is to look for rule violations.

• A is wrong because F and G can’t both be in. (Rule 1)
• C is wrong because if H is a councillor, then F must be treasurer (rule 3). But, if F is in, then G can’t be in (rule 1).
• A is wrong because F and G can’t both be in. (Rule 1)
• D is wrong because of rule 2. If H is a councillor, then J has to be mayor, not councillor.
• E is wrong because of rule 4. K can’t be councillor.

B is CORRECT. If F and J are councillors, then H must be treasurer (rule 3). That leaves K or L to be mayor. So, this scenario works.

To solve a “is completely determined” question, you should ask yourself who is particularly restrictive and important. I would say F and H are both important. If one is out, then the other must be in, and be treasurer.

Of the two, F is more restrictive to have in. If F it in, then G is out. So, putting H out would be a fairly restrictive scenario. Let’s draw that.

H is out, so F is in, so G is out. There are only six people, so if two are out, then all the others have to be in. That’s K, J and L.

K can’t be in T or C, so they must go in M. That leaves J and L to go in C:

Everything is determined, so C is CORRECT.

A is wrong because it’s less restrictive than C: since F is out, F could be in. And K, L and J are all easy to place out: there are no rules forcing them to do anything (as long as H isn’t a councillor).

The key to solving these questions is to look for a chain reaction. H causes that chain reaction by forcing F in, and G out. J, K and L force nothing at all.

In the setup, we saw that K can’t be treasurer (rule 3, F or H must be) or councillor (rule 4). So, since K is in, they’re mayor:

Next, you should ask yourself which rules involve Mayor. Rule 2 does: if H is councillor, then J is mayor. But, in this scenario, J can’t be mayor. Therefore, we can say that H can’t be councillor.

C is CORRECT.

The deduction above is the contrapositive of rule 2:

Everyone hates rule substitution questions, but I actually think they can be solved quite easily. The trick is to look at all the factors which affect the variable in question. That lets you find the full effects of all the rule affecting that variable. This description of the full effects of the rule lets you replace the rule.

This question talks about K. We saw in the setup that K can’t be treasurer or councillor:

• Only F or H can be treasurer (rule 2)
• Rule 4 directly restricts K from being councillor

So, if K is in, they must be mayor. That’s the full effect of all rules affecting K. And, it’s in the answers: E is CORRECT!

That’s really all there is to rule substitution! They can be prephrased.

If you were stuck on eliminating other answers, there are shortcuts for that too. Since we’re just trying to replace the setup, that means the alternate rule must:

• Allow everything normally allowed
• Forbid everything normally forbidden (including mimicking the rule being replaced)

Anything that contradicts the normal setup is wrong.

• A is wrong because L can normally be mayor.
• B is wrong because it still allows K to be councillor! You could have J and K, for example.
• C contradicts the setup. Normally K can be in if F is out. Further, it also lets K be councillor if F is in!
• D is true in the ordinary setup, but it doesn’t get the effect we want. K could be councillor with this rule, if H wasn’t!Game 4 – Corporation Bonds

Time on second attempt:

See “repeating games” at bottom of section

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This is the unusual game of the section. Though, it’s not especially unusual. There are some other games that have similarities. For example, the car features game from preptest 35, section 3, game 2.

The best way to set up this kind of game is with a vertical list of corporations, since the corporations are the ones that hold the bonds. Oddly, the first question uses the bonds as the base, but I find this less efficient.

Here’s what I drew. I added a few rules on it directly:

Rule 1 says that H and L can’t have the same type of bonds. So I drew the arc with a line through it to represent that. I also drew two diagrams: there are only two ways for H and L to have different bonds.

In the first diagram, R has 10. That’s thanks to rule 3: if L has 10, R has 10. (note that R could still have 10 in the second diagram. It just doesn’t have to have it).

The final point are the vertical lines to the right of the bonds for H and L. Those show that the groups are closed. No more bonds can go there (thanks to rule 1). Whereas all the other corporations could have one or two bonds (for now). We’ll be seeing a lot of these vertical lines in this game.

There’s only one rule not not on the diagram: rule 2. If V has 5 year bonds, then S has both 5 and 10 year bonds. I draw any rules I can’t put on the diagram in a separate list. (In this case, just the one rule):

However, that’s not all there is to this setup!

Counting Games

This game is an example of what I call “counting games”. It’s not exactly a type, rather it’s a skill that cuts across game types.

In this game, there are 8 bonds: 5, 5, 5, 5, 10, 10, 10, 10

So, on a counting game, you need to….count these bonds! There will be some situations where only one 5 year bond has been placed, and only three spots are open. What does this mean? It means you need to put a 5 year bond in every open space.

It’s really very simple, except no rule directly tells you to do that. It’s instead just a feature of the numbers. So….count! And look for other games where there’s a high number of things, and you need to count.

For example, game 3 of this section had six variables, and four spaces. So, if two were eliminated, then the other four had to be in (6 – 2 = 4). The math is very simple, but because no rule mentions it, a lot of students overlook it. I’ve noticed counting games are more common on recent LSATs.

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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.)

Analysis

Count all the variables! This many bonds must be placed: 5, 5, 5, 5, 10, 10, 10, 10

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 against for this question will help you memorize them and it’s also more efficient.

Rule 1 eliminates C. HCN and Lorilou cannot offer the same type of bond.

Rule 2 eliminates E. If VELSOR offers 5-year bonds, then SamsonGonzales must offer both types of bonds.

Rule 3 eliminates A. If Lorilou offers 10-year-bonds, then RST must also offer them.

The introduction of the game eliminates B. Only two of the corporations can offer both types of bonds.

D is CORRECT. It violates no rules.

In this question, L and V can’t have 5 year bonds. Since every corporation needs at least one bond, that means they both have 10 year bonds. You should draw that, and combine it with the existing rules:

There are a few things to note here:

• H has 5, because H and L can’t be the same (rule 1)
• R has 10, because of rule 3 (L10 —> R10)
• H, L and V all have vertical lines to their right indicating that they are closed and can take no more bonds

In the setup, I said this was a counting game. So, let’s count what we’ve placed: 5, 10, 10, 10. That means the bonds left are 5, 5, 5, 10.

There are three 5’s, and only three open groups: G, R and S. That means they all must have a 5:

Either G or S could have the final ten. But since this is a must be true, we should look for something definite. The right answer could be any of: G must have 5, S must have 5, R must have 5 and 10.

B is CORRECT. R has to have both. A, C and D could be true, but don’t have to be. E has to be false.

This question took me a while when I first did it. Looking at it now though, it’s easy. I realized that I missed a key dynamic in the game the first time I did it. I’ll explain; this dynamic eliminates B and D, and also shows the right answer is correct.

But first, the easy answers. A and C are wrong because of rule 1. H and L have to have different bonds, and therefore they can each only have one bond.

Now, the dynamic that I missed is this:

• There are six corporations. Two of them have two bonds, the rest have one.
• If V has two bonds, then S also has two (rule 2, V5 —> S 10). That means that all the other corps have one bond.

The real key insight is the first point. There can only be two with two. But, knowing that insight lets you figure out the deduction about V. And V is present in three answers, so it’s worth thinking about what happens if it has both.

So that’s why C and D are wrong. If V has two, then S does, so G and R can’t. Only two corporations can have two.

E is CORRECT. As I showed above, if V has both then S has both (rule 3). So E works with that rule. The others are all fine with one bond. The only other rules are that H and L have different bonds, and that R has 10 if L does. That’s easy to fulfill.

Counting was the real constrain for question 20. You need to see how all eight of the bonds can be placed: if two corporations have two, then the others each have just one.

This question mentions S and V. You should think about what rules affect them. Rule 3 does: if V has 5, then S has both.

This question says that S and V do not have both (because they have different types of bonds). So, that means that V does not have 5. It therefore has 10, and S has 5:

Since H, L, S and V all have only one bond, that means both G and R will have both. (H and L only have 1 because of rule 1: H and L have different bonds).

This question asks which corporations can have only 10 years bonds. We know the answer must include V. Since only A has V, A is CORRECT.

Simple as that. You don’t need to fill in the rest of the diagram. That would be focussing on what could be true, which is usually a mistake. Instead, you should figure out what must be true, and then look for the answer.

This group asks who can be the only corporations to offer 5 year bonds. Unlike question 21, this offers no additional rule, so it seems very open ended. To reduce the complexity, you should try to eliminate answers. Think which rules involve 5 year bonds:

• Rule 3 says that if V has 5, S has both 10 and 5 —> This eliminates E
• Rule 1 says that H and L have different kinds of bonds. So, if one has 5, the other has 10 —> This eliminates C
• Further, H/L have at least one L between them. So the right answer has to have one of H/L — This eliminates B (and also E, again)

Only A and D are left. How to choose? Well, G has no rules, so that makes it more plausible for a “could be true”. Since R has a rule, that makes it more likely that D doesn’t work.

Actually, we saw it in the setup. If H has 5, L has 10 (rule 1). Then, R has 10 (rule 3). So, H and R can’t both have 5:

Therefore, A is CORRECT.

The elimination of wrong answers above proves it definitively, but let’s prove is a second way with a diagram. First, place 5 in G and H:

Then apply rules 1 and 3. H and L are different, and if L has 10, then so does R. That gives G, H, L and R one bond each. You can fill up S and V with both bonds, and the scenario is complete:

This question gives G and H different types of bonds. That could be either 5/10 or 10/5. You should draw both scenarios.

In both scenarios, you have to give L a different bond from H (rule 1). In the left-hand scenario, that also causes R to have 10 (rule 3).

In the setup, I said this was a counting game. You must place four 5’s, and four 10’s. Notice something funny about the scenarios:

• In scenario 1, there are three 5’s left to place, and three open groups.
• In scenario 2, there are three 10’s left to place, and three open groups.

LSAC leaves little tricks like this all over the place! In both scenarios, you can fill in something definite:

I also added the 2nd rule in scenario 1: if V has 5, S has 5 and 10.

This is a must be true question, so you should look for something both scenarios have in common. The only common points are that R has 10 in both scenarios, and S has 10 in both scenarios.

C is CORRECT. The wrong answers could be true in either scenario, but don’t have to be true.