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Circular Game

Six people—Alice, Bob, Carol, Dave, Emily, Frank—are seated evenly spaced around a circular table according to the following conditions:
Alice does not sit next to Carol.
Bob sits next to Carol or Dave.
Frank sits next to Dave.
If Emily sits next to Frank, then she does not sit next to Carol.


As usual we construct a diagram to aid in answering the questions. First, translate the given conditions into symbols—abbreviating each name with its first letter.
The most concrete condition is “Frank sits next to Dave”; it is naturally symbolized as FD.
Next, we symbolize the second most concrete condition, "Bob sits next to Carol or Dave" as
BC or BD. The “or” in this symbol is inclusive. That is, it includes the case in which B sits next to both C and D—in other words, immediately between them. (Unless otherwise stated, the meaning of “or” is inclusive on the LSAT.)
Next, “Alice does not sit next to Carol” is symbolized as ∼(ΑΧ).
Finally, we come to the last and most complicated condition, “If Emily sits next to Frank, then she does not sit next to Carol.” An initial symbol for this sentence might be (EF)—>~(EC), where the arrow stands for “if..., then....” However, we can derive a more descriptive symbol as follows: If Emily were to sit next to both Frank and Carol, then she would be seated immediately between them. This is not allowed. Thus the more concise symbol ~(FEC) is equivalent to our original symbol (EF)—>~(EC).
We now have the following schematic for our conditions:
BC or BD
We need, however, a diagram to fill out our scheme. To this end, draw the following circle with spokes inside:

Next, following the strategies developed earlier, we scan the initial conditions for a base axis. There is none. So we look for a base group. The only condition that fixes the relative position of two of the elements is FD; it forms our base group.

Since circle diagrams are not fixed, we may initially place FD anywhere on the circle. Following convention, however, we put F at the top of the circle, creating the following two possible diagrams:
In this problem, there is no mention of the circle’s orientation (left or right). We, therefore, need to consider only the first diagram, the other being the mirror image of it.

No other conditions fix the relative positions of the other elements, so our schematic is complete with diagram as follows:
BC or BD
This schematic is self-contained. There is no need to refer to the original problem, which should be avoided whenever possible. Now we’ll use this schematic to answer the questions.
Of the following, which one is a possible seating arrangement of the six people?
  1. Alice, Frank, Dave, Carol, Emily, Bob
  2. Alice, Bob, Carol, Frank, Dave, Emily
  3. Alice, Bob, Carol, Emily, Frank, Dave
  4. Alice, Emily, Frank, Dave, Bob, Carol
  5. Alice, Dave, Bob, Emily, Carol, Frank
This is a straightforward elimination question. We merely take the initial conditions in succession and test them against each answer-choice, eliminating any answer-choices that do not satisfy the conditions. The last remaining answer-choice will be the answer.

Let’s start the elimination process with the condition FD. All the answer-choices have F next to D except choice (E). This eliminates (E).
Next, we use the condition BC or BD. Choices (B), (C), and (D) all satisfy this condition; (A) does not. This eliminates (A).
Next, using the condition ~(AC), we eliminate choice (D), which has A next to C.
Note: Since this is a circular ordering, the list A, E, F, D, B, C does not end at C (recall that there is no first or last on a circle). Instead, the sequence returns to A and repeats the cycle. This is shown more clearly by the following “flow chart”:

Finally, choice (C) contradicts the condition ~(FEC). This eliminates (C).
Hence, by process of elimination, the answer is (B)—the only answer-choice remaining.
If Bob is seated next to Frank, then in which one of the following pairs must the people be seated next to each other?
  1. Alice and Emily
  2. Bob and Dave
  3. Bob and Emily
  4. Carol and Dave
  5. Carol and Frank
The new condition, “Bob is seated next to Frank,” is naturally symbolized as ΒΦ. Adding this condition to our original diagram gives

Next, from the condition BC or BD, we see that B must be next to C, as it is not next to D in the diagram. Our diagram, therefore, is as follows:

Finally, the condition ~(AC) forces A next to D (otherwise it would be next to C), which in turn forces E between A and C. Thus our uniquely determined diagram is

From this diagram, we see that A must sit next to E and therefore the answer is (A).
If Dave and Carol sit next to each other, then Alice could sit immediately between
  1. Bob and Carol
  2. Bob and Frank
  3. Dave and Emily
  4. Dave and Frank
  5. Frank and Emily
Remember that the questions in a game problem are independent of one another. So the condition BF, in Question 2, does not apply to this question.
Begin by adding the new condition “Dave sits next to Carol”DC—to the original diagram:

As in Question 2, the second condition, BC or BD, forces B next to C, and our diagram becomes

This diagram also satisfies the remaining initial conditions—~(AC) and ~(FEC). [Why?]
Therefore the placement of A and E is arbitrary, and the following two diagrams are possible:
The second diagram satisfies choice (E). The answer, therefore, is (E).
If Bob sits next to Carol, then which one of the following is a complete and accurate list of people who could also sit next to Bob?
  1. Alice
  2. Alice, Dave
  3. Dave, Frank
  4. Alice, Emily, Frank
  5. Alice, Dave, Emily, Frank
This question illustrates that during the test you should not erase previously derived diagrams, for we can use the diagrams derived in solving Questions 2 and 3 to help solve this question.
(Note by “a complete and accurate list” the writers of the LSAT mean a list of all possible people, and only those people.)

Referring to the final diagram in Question 2, which has B seated next to C, we see that F can be next to B. This eliminates both (A) and (B)—they don’t contain F.
Next, referring to the final two diagrams in Question 3, we see that both A and E can sit next to B. This eliminates (C).
Finally, we need to decide between choices (D) and (E).
Choice (E) differs from choice (D) only in that it contains D.
So we place D next to BC in our original diagram and then check whether this leads to a contradiction of the conditions:

Now if we place A next to F, and E next to C, then all the initial conditions are satisfied by the following diagram:

Hence, it is possible for D to be next to B, and ADEF is therefore the complete and accurate list of people who can sit next to B.
The answer is (E).
Which of the following must be false if Bob sits next to Dave?
I.   Emily sits next to Frank.
II.  Carol sits directly opposite Bob.
III. Carol sits immediately between Emily and Bob.
  1. I only
  2. III only
  3. I and II only
  4. II and III only
  5. I, II, and III
The most efficient way to solve triple-multiple-choice questions is to eliminate answer-choices as you check each sub-statement. Additionally, this method often gives a bonus: you may not need to check the final statement, which typically is the hardest. Even if you’re not able to solve the problem, elimination allows you to make an educated guess. (Remember there is no guessing penalty on the LSAT.)

The logic of this question is convoluted because the correct answer will always make a false statement! This question would be much easier if it were worded, “Which of the following is possible?” (See Obfuscation.)

Let’s begin our solution by adding the new condition “Bob sits next to Dave”BD—to the original diagram:

Next, test the first statement “Emily sits next to Frank”—EF. To this end, place it on the diagram as follows

This diagram forces A and C next to each other, which violates the condition ~(AC). Hence, Statement I is always false and therefore it is a correct choice.* This eliminates choices (B) and (D); they don’t contain I.
Next, test the second statement “Carol sits directly opposite Bob,” C<—>B, which forms a base axis. Place it on the original diagram as follows:

Then placing A next to B—otherwise it would be next to C, which violates the condition ~(AC)—and placing E next to C gives

This diagram satisfies all the initial conditions, so it is true. Hence, the statement C<—>B is not always false.
Hence, Statement II is incorrect (Whew!).
This eliminates both (C) and (E), since they both contain II. By elimination, therefore, the answer is (A), and there is no need to check the third statement.
If Alice sits next to Emily, then Bob CANNOT sit immediately between
  1. Alice and Carol
  2. Alice and Dave
  3. Carol and Dave
  4. Carol and Frank
  5. Dave and Emily
This question is hard (or at least long) because there are many places where Alice and Emily may sit.
However, the answers and diagrams we derived for previous questions will help here. The final diagram in Question 4 has B immediately between C and D. This eliminates choice (C).
Furthermore, the final diagrams in Questions 2 and 3 have B immediately between C and F, and A and C, respectively.
This eliminates choices (D) and (A).
Now the question is not so daunting: we need only to decide between choices (B) and (E).

Let’s test choice (B), first. If Bob sits immediately between Alice and Dave, i.e., ABD, then combining this condition with “Alice sits next to Emily,” AE, generates the base group EABD. Adding this to the original diagram gives

(Note the circle around C indicates that it was forced into that position by the other conditions.)
This diagram satisfies all four of the initial conditions, which eliminates choice (B).
Therefore, by the elimination method, we have learned that the answer is (E).
It is, however, instructive to verify that B cannot sit immediately between D and E.
To this end, form the symbol ΔΒΕΑ and place it on the diagram as follows:

This diagram clearly violates the condition ~(AC).

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