Logic Game 5

This is a Grouping game. You will select from the committee members and place them in three groups, the subcommittees. Because you can use each member more than once it is a Grouping Selection game, and our groups must be the subcommittees. If you try to make the members the groups and assign subcommittees to them you end up with a much messier diagram. Remember that in Grouping games the set that allows you to use more than one of each type (Members) must be the entities and the other set are the groups.
For more on Grouping Games see the Grouping Game Tutorial.

This game is taken from LSAT #46 (June 2005), it is the 4th game in the Analytical Reasoning section and includes questions 17 to 22.
To obtain a copy of this Analytical Reasoning Section or the full June 2005 LSAT please Contact Us or LSAC.

The six members are:

F G H I M P

Our groups (Subcommittees) are:

1 2 3
__ __ __
__ __ __
__ __ __

The rules are:

Rule one: One member on all three committees. This can be expressed by making one spot equal across all three committees with an equal = sign.

1 2 3
__ = __ = __
__ __ __
__ __ __

Rule two: F does not serve on any subcommittee with G.

No FG

Rule three: H does not serve on any subcommittee with Iriving.

No HI

Notice that since FG cannot be together and HI cannot be together that you could not use F, G, H, or I to satisfy Rule one. If F, G, H or I were in all three committees then you could not put each entity in a committee because of Rules two and three.
This means that only M or P can occur in each committee to satisfy Rule one.
So we have either:
MMM or PPPP

One other thing to remember is that there are no rules about specific committees. No entity is limited to committee 1, 2, or 3 and no entity cannot be placed in a specific committee. This just means that the committees themselves are interchangeable. So having FHM in committee 1 is the same as having FHM in committee 3. The number of the committee doesn’t matter, the only thing that matters is that you have three committees.

One more thing you can add to your rules and diagram if you combine all your information is that one entity occurs three times, one occurs twice, and all the remaining four occur once each. You cannot have any different than exactly one entity occurring three times, and exactly one occurring twice.
How do you get this new information?

When there a limited number of spots to fill and a limited number of entities to fill those spots, try to think of how those spots will be filled. Start with a rule that fills the most number of spots. We know that one entity serves on all three subcommittees. This means there are only 6 spots left and we have 5 entities to fill them. Each entity must occur at least once. And in order to fill the last spot one of those entities will have to occur twice.

This new information is not essential, and you can complete the game perfectly without it, but if you keep your eye out for it and understand how you find it some games will become much easier.

Answer and Wrong Answers:

  1. If French does not serve on any subcommittee with Magnus, which one of the following must be true?

If F is not with M then M cannot be on all committees. Therefore P must be on all committees.

1 2 3
P = P = P
__ __ __
__ __ __

Also we must have F and M in separate committees. Because the numbers of the committees don’t matter we can place F and M in any two committees as long as they are not together.

1 2 3
P = P = P
F M __
__ __ __

We also know that G cannot be with F so G is either in 2 or 3, or in both 2 and 3, remember that in a Grouping Selection game you can place multiple copies of each entity. You can represent G as being in either spot by putting G in brackets under 2 and 3.

1 2 3
P = P = P
F M __
__ __ __
(G) (G)

Or you can draw two diagrams and fill in both options.

Option 1:
1 2 3
P = P = P
F M __
__ G __

Option 2:
1 2 3
P = P = P
F M G
__ __ __

In this case creating two diagrams with two options does not give us any new information so I would probably just use the (G) in brackets above.

Now we move on to our questions. Remember that we still have to apply Rule three: No HI

Which one of the following must be true?

In a Must be True question you want to try to disprove each answer choice and the one that you can not disprove is the correct answer choice.

  1. F with H

 

We can place H in subcommittee 2. I in 3 another G in 3, and another M in 1, remember that in a Grouping Selection game you can place multiple copies of each entity.

1 2 3
P = P = P
F M I
M H G

We have satisfied all of our rules and F is not with H. (A) does not have to be true and so (A) is wrong.

  1. F with I

As we see in the explanation for (A) F need not be with I and so (B) is wrong.

  1. I with P

Since we have already determined that P needs to be in each subcommittee as it cannot be M or any other entity then P must be with every entity, including I. Therefore (C) must be true and (C) is correct.

You can solve this question quickly if you spot this right away. P must be in all subcommittees and so P must be with all entities.

  1. M with G

As we see in the explanation for (A) M need not be with G and so (D) is wrong.

  1. M with I

As we see in the explanation for (A) M need not be with I and so (E) is wrong.

  1. If P serves on every subcommittee on which F serves and every subcommittee on which G serves, then which one of the following could be true?

This essentially adds two new rules that are essentially if then statements as follows:
If F then P
If G then P

This means that If you have F or G you must also have P. It does NOT mean that if you have P you need F or G. If then statements only work in one direction.

Now take your Additional Information and plug it into your diagram. We know that we must have one subcommittee with FP and one with GP. They can not be the same committee due to Rule two: FG

1 2 3
__ = __ = __
F G __
P P __

The entity that serves on all subcommittees could be M or P. Try using M

1 2 3
M = M = M
F G __
P P __

This leaves only Subcommittee 3 for HI and we know from Rule three that HI cannot be together. Therefore P must serve on all three subcommittees.

1 2 3
__ __ __
F G __
P P P

Now we can move on to our question.

Which one of the following could be true?

You can immediately cross off (D) and (E) because they require an entity to serve on every subcommittee that P does. P is on every subcommittee and so this would leave no room for the other entities.

  1. M with every F and M with every G

This puts M on 1 and 2, and so leaves no room for HI to be on separate subcommittees.

1 2 3
M M __
F G __
P P P

(A) is wrong as HI cannot go together.

  1. M with every H and M with every I

 

1 2 3
__ __ __
F G __
P P P

So we need to place HM and IM and they cannot be on the same subcommittee as IHM because of Rule three. HI cannot be together.

Where can we place HM? We can put HM in the only subcommittee with two spots left, 3.
1 2 3
__ __ H
F G M
P P P

But this leaves us with no room to place IM. Therefore (B) cannot be true and (B) is wrong.

  1. H with every F and H with every G

We can place H with F on 1 and H with G on 2. Then we have two spots left on 3.

1 2 3
H H __
F G __
P P P

We still need to place M and I. MI can go together on 3.

1 2 3
H H M
F G I
P P P

(C) can be true and so (C) is correct.

  1. F with every P

If we put F with every P then F will have to serve on every subcommittee. This leaves no room for the other entities to be placed.

1 2 3
F F F
__ G __
P P P

We still have H, I and M to place and only two spots left. Remember that each entity must serve on at least one subcommittee. We have to use each entity at least once.
(D) cannot be true and so (D) is wrong.

  1. H with every P

If we put H with every P then H will have to serve on every subcommittee. This leaves no room for the other entities to be placed.

1 2 3
H H H
F G __
P P P

We still have I and M to place and only one spot left. Remember that each entity must serve on at least one subcommittee. We have to use each entity at least once.
(E) cannot be true and so (E) is wrong.

  1. If I serves on every subcommittee on which M serves, which one of the following could be true?

The Additional Information in this question adds a new rule.

If M then I

There is not a lot that this adds to our diagram as we don’t know where any Ms are in our diagram yet and so we must move on to our question without much new in our diagram.

Which one of the following could be true?

  1. MMM Or (M on all subcommittees)

This means there must be three Ms and from our Additional Information there must also be three Is.

1 2 3
M M M
I I I
__ __ __

Now we have only three spots left and we still have four entities to place. We do not have room for them all and so (A) cannot be true. (A) is wrong.

  1. I > 1 Or (I on more than 1 subcommittee)

Try to fit I into more than one subcommittee. This means that I will have to be on two committees for this answer choice. More than one and less than three. We know that only M or P can be placed in all subcommittees. But we know from (A) that M cannot serve on all three so it must be P on all three.

1 2 3
P P P
I I __
__ __ __

We know from Rule three that HI are not together so H must be on 3. We then have three spots left, one in each subcommittee and F, G and M left to place. They can fit in these three remaining spots. Check to see if this violates any rules. It does not and so (B) is correct. It can be true that I serves on more than one subcommittee.

1 2 3
P P P
I I M
F G H

  1. I with every P

This is the same as “If P then I”
à P à I

We know that P must be on all three committees as M cannot (see the explanation for (A). When we have P we also have I so we then have PI on each committee.

1 2 3
P P P
I I I
__ __ __

This leaves only three spots left and four entities left to place, F, G, H, and M. (C) cannot be true and so (C) is wrong.

  1. F serves on a subcommittee with M

We know from the explanation for (A) that P must occur three times.

1 2 3
P P P
__ __ __
__ __ __

We also know that we must have MF somewhere. From our answer choice (D)

1 2 3
P P P
M __ __
F __ __

But we know from our Additional Information in the Question Stem, that whenever we have M we must also have I. This means we need M, F, and I together but we also need P to occur in each subcommittee. There is no room in a subcommittee for M, F, I and P.
(D) cannot be true and so (D) is wrong.

  1. G serves on a subcommittee with M

(E) is wrong for the same reason that (D) is wrong. We have to have three Ps. We must have I for every M and we have to have MG somewhere. This means one subcommittee will have to have all of P, M, G, and I. A subcommittee can only hold three entities and so (E) is wrong.

We know from the explanation for (A) that P must occur three times.

1 2 3
P P P
__ __ __
__ __ __

We also know that we must have MG somewhere. From our answer choice (E)

1 2 3
P P P
M __ __
G __ __

But we know from our Additional Information in the Question Stem, that whenever we have M we must also have I. This means we need M, G, and I together but we also need P to occur in each subcommittee. There is no room in a subcommittee for M, G, I and P.
(E) cannot be true and so (E) is wrong.

  1. Which one of the following could be true?

This is a Could be True question without any Additional Information. Questions with no Additional Information can be more time consuming and you may consider skipping them if you are already struggling with the game or if you are running out of time. You can also leave them for last in this game.

  1. FFF

1 2 3
__ = __ = __
__ __ __
__ __ __

As we saw in the game set up, only M or P can be the entity that occurs three times, once in each subcommittee. If you try to put F in each subcommittee then there is nowhere for G to go. G cannot go with F. Therefore you cannot have FFF and so (A) is wrong.

  1. HHH

1 2 3
__ = __ = __
__ __ __
__ __ __

As we saw in the game set up, only M or P can be the entity that occurs three times, once in each subcommittee. If you try to put H in each subcommittee then there is nowhere for I to go. I cannot go with H. Therefore you cannot have HHH and so (A) is wrong.

  1. G with every M and G with every P

This gives us
If M then G
and
If G then M

We know that either M or P must occur three times. If M or P occurs three times then G will also have to occur three times. This will not leave room for the other entities. We have three spots left and four entities, H, I, F, and P or M left to place.

1 2 3
M M M
G G G
__ __ __

(C) cannot be true and so (C) is wrong.

  1. P with every I and P with every M

This is the same as:

If I then P
and
If M then P

We know that M or P must occur three times and if every time we have M we also have P we cannot have M occur three times or there will be no room for the remaining entities. Therefore we must have P three times.

1 2 3
P = P = P
__ __ __
__ __ __

Now we need to insert our other entities being careful not to break our rules. Remember FG and HI cannot go together. Our two rules added for this answer choice have no effect because we already have P in every subcommittee. Whenever I or M occur P will already be there.

1 2 3
P = P = P
I H F
G M M

This satisfies all of our rules and so (D) could be true and (D) is correct.

  1. M for every P and P for every M

This is the same as:

if M then P
and
if P then M

Or

MP are always together. Remember that we determined in the game setup that only M or P can be the entity that occurs three times, once in each subcommittee. If MP are always together then we must have MP in every subcommittee. This does not leave room for the other entities. We now have three spots left and four entities, F, G, H and I left to place.

1 2 3
P = P = P
M M M
__ __ __

So (E) cannot be true and so (E) is wrong.

  1. Which one of the following must be true?

In a Must be True question you want to try to disprove each answer choice and the one that you cannot disprove is the correct answer choice. We have no Additional Information and so we proceed to attempting to disprove each answer choice.

Questions with no Additional Information can be more time consuming because you need to test each answer choice, and without Additional Information there are far more possibilities. Often you will have to re-draw the diagram for each answer choice.

To save time, take a quick look at each answer choice remembering what you have learned from the previous questions and from your game set up. As you scan over the answer choices if nothing immediately jumps out at you just move on, however, here you may notice that (E) must be true. Why? Because we know that M or P have to occur three times, once in each subcommittee. This means that M and P will have to be together in at least one subcommittee.

Another way to speed up questions like this is to refer to question you have already done to see if the answer choice has already been tested. For example in many of the previous questions we say that G did not have to serve on at least two subcommittees and so we can cross off (A) without any new work. Often you can cross off 2 or more answer choices without doing any new work with this method. Just quickly look back at the diagrams you have already done.

Of course sometimes you have to do it the hard way and test the answer choice to see if it can be disproved.

  1. G on at least two subcommittees

We are trying to disprove that G must be on at least two subcommittees. All we need to do is put all the entities into the diagram without having G occur more than once and still satisfying all of our rules.

We can put P in all three subcommittees to satisfy Rule one.

1 2 3
P P P
__ __ __
__ __ __

Then apply each rule and try to limit G to only one subcommittee. If we can limit G to only one subcommittee then (A) need not be true and so we have disproved it and can cross (A) off and move on.

FG and HI cannot be together from Rules two and three.

1 2 3
P P P
F G H
I M M

This satisfies all of our rules and G only occurs in one subcommittee. We have disproved (A), it need not be true and so (A) is wrong.

  1. I on only one subcommittee

We are trying to disprove that I must be on only one subcommittees. All we need to do is put all the entities into the diagram and have I occur in more than one subcommittee while still satisfying all of our rules.

We can put P in all three subcommittees to satisfy Rule one.

1 2 3
P P P
__ __ __
__ __ __

Then apply each rule and try to put I in more than one subcommittee. If we can put I in more than one subcommittee then (B) need not be true and so we have disproved (B) and can cross (B) off and move on.

FG and HI cannot be together from Rules two and three.

1 2 3
P P P
F G H
I I M

This satisfies all of our rules and I occurs in more than one subcommittee. We have disproved (B), it need not be true and so (B) is wrong.

  1. FH

We are trying to disprove that F and H need to be together on one subcommittee. All we need to do is put all the entities into the diagram and have F and H not be together in any subcommittee while still satisfying all of our rules.

We can put P in all three subcommittees to satisfy Rule one.

1 2 3
P P P
__ __ __
__ __ __

Then apply each rule and try to put F and H in separate subcommittees. If we can keep F and H in separate subcommittees then (C) need not be true and so we have disproved (C) and can cross (C) off and move on.

FG and HI cannot be together from Rules two and three.

1 2 3
P P P
F G H
I I M

This satisfies all of our rules F and H are not in the same subcommittee. We have disproved (C), it need not be true and so (C) is wrong.

  1. GI

We are trying to disprove that G and I need to be together on one subcommittee. All we need to do is put all the entities into the diagram and have G and I not be together in any subcommittee while still satisfying all of our rules.

We can put P in all three subcommittees to satisfy Rule one.

1 2 3
P P P
__ __ __
__ __ __

Then apply each rule and try to put G and I in separate subcommittees. If we can keep G and I in separate subcommittees then (D) need not be true and so we have disproved (D) and can cross (D) off and move on.

FG and HI cannot be together from Rules two and three.

1 2 3
P P P
F G H
I H M

This satisfies all of our rules G and I are not in the same subcommittee. We have disproved (D), it need not be true and so (D) is wrong.

  1. MP

We know that M or P have to occur three times, once in each subcommittee. This means that M and P will have to be together in at least one subcommittee. (E) is correct.

If you try to put any other entity in each subcommittee then you end up violating either Rule two or Rule three.

not FG

not HI

If F, G, H or I occurs in each subcommittee, or in other words occurs three times, then one of these two rules will be broken when you try to put the corresponding entity. Example:

If F occurs three times then there is no place for G without breaking Rule two.

1 2 3
F F F
__ __ __
__ __ __

  1. Which one of the following must be true?

In a Must be True question you want to try to disprove each answer choice and the one that you cannot disprove is the correct answer choice. We have no Additional Information and so we proceed to attempting to disprove each answer choice.

Questions with no Additional Information can be more time consuming because you need to test each answer choice, and without Additional Information there are far more possibilities. Often you will have to re-draw the diagram for each answer choice.

To save time, take a quick look at each answer choice remembering what you have learned from the previous questions and from your game set up. As you scan over the answer choices if nothing immediately jumps out at you just move on, and answer attack the answer choices individually.

Another way to speed up questions like this is to refer to question you have already done to see if the answer choice has already been tested. Often you can cross off two or more answer choices without doing any new work with this method. Just quickly look back at the diagrams you have already done.

Of course sometimes you have to do it the hard way and test the answer choice to see if it can be disproved.

For most of these answer choices we can refer back to diagrams drawn in previous questions and cross off answer choices that need not be true. For example for (A) look back to 21. (A) we had the diagram:

1 2 3
P P P
F G H
I M M

Where F or G were not on every subcommittee allowing us to cross of (A), and also H or I are not on every subcommittee allowing us to cross off (B). There is also no subcommittee with FMP allowing us to cross off (C). Finally this diagram does not have either M on only one subcommittee or P on only one subcommittee so we can cross off (E). This leaves only (D) which is the correct answer. So by looking at our past diagrams we can solve the question very quickly with no need to re-diagram each answer choice.

However, if we missed the fact that we had already drawn a diagram to solve this question we can draw a new diagram to solve question 22.

We know that either M or P must occur three times. Let’s start this question with M occurring three times.

1 2 3
M M M
__ __ __
__ __ __

Then for each answer choice try to disprove it by filling in the diagram without what is requested by the answer choice. For example (A) asks if it must be true that F or G are on every subcommittee so we try to fill in the diagram without F or G on every subcommittee. If we can fill in the diagram with at least one subcommittee not having F or G in it and still satisfy all of our rule then we can cross off (A)

  1. F or G on every subcommittee

Start with our basic diagram and choose either M or P to fill in on each subcommittee.

1 2 3
M M M
__ __ __
__ __ __

Then put F and G into the diagram.

1 2 3
M M M
F G __
__ __ __

Then H and I which we know cannot go in the same subcommittee.

1 2 3
M M M
F G __
__ H I

And finally fill in the last two spots with P.

1 2 3
M M M
F G P
P H I

This completes our diagram satisfying all of our rules and F and G are not in every subcommittee so we have disproved (A) therefore (A) is wrong.

  1. H or I on every subcommittee

Start with our basic diagram and choose either M or P to fill in on each subcommittee.

1 2 3
M M M
__ __ __
__ __ __

Then H and I which we know cannot go in the same subcommittee.

1 2 3
M M M
H I __
__ __ __

Then put F and G into the diagram which we know cannot go in the same subcommittee.

1 2 3
M M M
H I __
__ F G

And finally fill in the last two spots with P.

1 2 3
M M M
H I P
P F G

This completes our diagram satisfying all of our rules and H or I are not in every subcommittee so we have disproved (A) therefore (A) is wrong.

  1. No subcommittee with F, M and P

Many of the diagrams we have already drawn do not have FMP in one subcommittee. See the solution for (B) above. It does not have a subcommittee with FMP and satisfies all the rules. Therefore (C) is wrong.

1 2 3
M M M
H I P
P F G

  1. Some committee member serves on exactly two subcommittees.

(D) is correct. The reason for this is that we have:
Nine Spots
Six members
One member that serves on all three
Each member must serve at least once.

This means that of the Nine spots, three are taken by MMM or PPP, and Five are taken by the five other members F, G, H, I, & P or M that must serve at least once. This leaves only one spot. Whichever entity fills this spot will serve on exactly two subcommittees.

Also if we look at every diagram we have draw so far we always have MMM or PPP and then one other entity that serves on exactly two subcommittees.

  1. Either M or P serves on only one subcommittee.

As we saw in the explanation for (B) above we can have M serve on three and P serve on two subcommittees, therefore neither serve on only one subcommittee. (E) is wrong.

1 2 3
M M M
H I P
P F G

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