Question

Which of the following can be the musicians scheduled to teach the master classes, in the order in which they will teach, from first to sixth?

• A. F, J, G, T, Z, S
• B. F, W, H, T, G, Z
• C. G, F, H, T, S, Z
• D. S, G, W, H, R, J
• E. T, G, W, H, R, S
• A. F
• B. G
• C. J
• D. T
• J. W
• A. J is not scheduled to teach if R is scheduled to teach.
• B. J is not scheduled to teach if T is scheduled to teach.
• C. J is not scheduled to teach if W is scheduled to teach.
• D. W is not scheduled to teach if F is scheduled to teach.
• O. Z is not scheduled to teach if W is scheduled to teach.
• A. F is scheduled to teach the first class.
• B. G is scheduled to teach the first class.
• C. H is scheduled to teach an earlier class than the class Z is scheduled to teach.
• D. R is scheduled to teach an earlier class than the class T is scheduled to teach.
• T. S is scheduled to teach an earlier class than the class T is scheduled to teach.
• A. If F is scheduled to teach a class, then H is also scheduled to teach a class.
• B. If J is scheduled to teach a class, then R is also scheduled to teach a class.
• C. If J is scheduled to teach a class, then S is also scheduled to teach a class.
• D. If T is scheduled to teach a class, then R is also scheduled to teach a class.
• Y. If W is scheduled to teach a class, then Z is also scheduled to teach a class.
• A. F is scheduled to teach the fourth class.
• B. G is scheduled to teach the first class.
• C. H is scheduled to teach the third class.
• D. R is scheduled to teach the fifth class.
• ^. W is scheduled to teach the second class.
• A. F is scheduled to teach the second class.
• B. H is scheduled to teach the sixth class.
• C. R is scheduled to teach the fourth class.
• D. T is scheduled to teach the second class.
• c. W is scheduled to teach the third class.
• A. F is scheduled to teach the third class.
• B. G is scheduled to teach the first class.
• C. T is scheduled to teach the sixth class.
  (D) W is scheduled to teach the sixth class.
• D. Z is scheduled to teach the fifth class.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This question asks for a valid schedule. We must check each option against the set of rules provided. An option is correct only if it violates none of the rules. The musicians are Violinists (V) = \{F, G, H, J\} and Pianists (P) = \{R, S, T, W, Z\}. The schedule must have 3 V and 3 P.
Step 2: Detailed Explanation:
Let's evaluate each option:


(A) F, J, G, T, Z, S:

Musicians: 3 Violinists (F, J, G) and 3 Pianists (T, Z, S). This is valid.

J Rule: "If J teaches a class, it will be the sixth." Here, J is scheduled for the second class. VIOLATION.


(B) F, W, H, T, G, Z:

Musicians: 3 Violinists (F, H, G) and 3 Pianists (W, T, Z). This is valid.

F Rule: "F will not teach unless the first three classes are taught by violinists." Here, F is teaching, but the second class is taught by W, a pianist. VIOLATION.


(C) G, F, H, T, S, Z:

Musicians: 3 Violinists (G, F, H) and 3 Pianists (T, S, Z). This is valid.

F Rule: F is teaching. The first three classes are G (V), F (V), and H (V). All are violinists. This rule is satisfied.

J Rule: J is not teaching. This rule is not violated.

R Rule: R is not teaching. This rule is not violated.

W Rule: W is not teaching. This rule is not violated.

Since all rules are satisfied, this is a valid schedule.


(D) S, G, W, H, R, J:

Musicians: 3 Violinists (G, H, J) and 3 Pianists (S, W, R). This is valid.

R Rule: "R will teach only if T teaches the first class." Here, R is teaching (fifth class), but T is not teaching at all. VIOLATION.


(E) T, G, W, H, R, S:

Musicians: This schedule has 2 Violinists (G, H) and 4 Pianists (T, W, R, S). The schedule must have exactly 3 of each. VIOLATION.


Step 3: Final Answer:
The only option that satisfies all the given constraints is (C).

Answer 2

Step 1: Understanding the Concept:
We are given a new condition: R teaches the second class. We must first deduce all the consequences of this condition and then determine which musician could possibly teach the third class.
Step 2: Detailed Explanation:
1. Initial Condition: R is in class 2. The schedule starts as: __ , R, __, __, __, __.
2. Apply the R Rule: The rule states, "R will teach only if T teaches the first class." Since R is teaching, T must teach class 1. The schedule becomes: T, R, __, __, __, __.
3. Identify Musician Types: T and R are both pianists (P). So the first two classes are taught by pianists: P, P, __, __, __, __.
4. Apply the F Rule: The rule states, "F will not teach unless the first three classes are taught by violinists." Since the first two classes are taught by pianists, the condition for F to teach is not met. Therefore, F cannot teach in this schedule.
5. Apply the W Rule: The rule states, "No pianist will teach on a day that immediately precedes or immediately follows a day on which W teaches." R is a pianist in class 2. If W were to teach class 3, W would be immediately following R. Therefore, W cannot teach in class 3.
6. Evaluate the options for class 3:

(A) F: We deduced that F cannot teach at all in this schedule. Incorrect.

(B) G: G is a violinist (V). Placing G in class 3 results in the schedule: T(P), R(P), G(V), __, __, __. This does not create any immediate rule violations. It is a possibility.

(C) J: The J rule states that if J teaches, it must be the sixth class. J cannot teach the third class. Incorrect.

(D) T: T is already scheduled for the first class, and no musician can teach more than one class. Incorrect.

(E) W: We deduced that W cannot teach in class 3 due to its proximity to R (a pianist). Incorrect.

Step 3: Final Answer:
Based on the deductions, only G could be scheduled to teach the third class.

Answer 3

Step 1: Understanding the Concept:
This question asks for a statement that is a necessary consequence of the initial rules. To test each option, we assume the "if" part is true and see if the "then" part is forced to be true. If we can find even one valid counterexample (a schedule where the "if" is true but the "then" is false), the option is incorrect.
Step 2: Detailed Explanation:


(A) J is not scheduled to teach if R is scheduled to teach.
Assume R teaches. This means T must teach class 1. Can J also teach? J must be in class 6. Let's try to build a counterexample: T(P), G(V), W(P), H(V), R(P), J(V). This schedule has 3P/3V, T is 1st, J is 6th, and W is surrounded by violinists. All rules are met. Since we found a schedule where both R and J teach, this statement is not necessarily true.

(B) J is not scheduled to teach if T is scheduled to teach.
Assume T teaches. Can J teach? We can use a similar counterexample: T(P), G(V), S(P), H(V), W(P), J(V). This schedule is valid and contains both T and J. So, this statement is not necessarily true.

(C) J is not scheduled to teach if W is scheduled to teach.
Assume W teaches. Can J teach? The schedule T(P), G(V), W(P), H(V), S(P), J(V) is valid and contains both W and J. So, this statement is not necessarily true.

(D) W is not scheduled to teach if F is scheduled to teach.
Assume F teaches. According to the F rule, classes 1, 2, and 3 must be taught by violinists (V V V). Consequently, classes 4, 5, and 6 must be taught by pianists (P P P). Now, let's see if W can teach. W is a pianist, so W would have to be in slot 4, 5, or 6.

If W is in slot 4, it is immediately followed by a pianist in slot 5. VIOLATION.
If W is in slot 5, it is between two pianists in slots 4 and 6. VIOLATION.
If W is in slot 6, it is immediately preceded by a pianist in slot 5. VIOLATION.
There is no valid position for W if F teaches. Therefore, if F teaches, W cannot teach. This statement must be true.

(E) Z is not scheduled to teach if W is scheduled to teach.
Assume W teaches. Can Z teach? Let's try to build a counterexample with both W and Z. Schedule: G(V), W(P), H(V), Z(P), S(P), J(V). No, this is 2V/4P. How about: G(V), W(P), H(V), S(P), F(V), Z(P)? No, F rule violation. How about: G(V), W(P), H(V), T(P), Z(P), J(V)? No, 2V/4P. How about: T(P), G(V), W(P), H(V), Z(P), J(V)? Again, 2V/4P.
Let's try: Z(P), G(V), W(P), H(V), S(P), J(V). This is a valid schedule with 3V (G,H,J) and 3P (Z,W,S). J is 6th, W is surrounded by V. This is a valid schedule where both Z and W teach. Thus, the statement is not necessarily true.

Step 3: Final Answer:
The only statement that holds true under all conditions is (D).

Answer 4

Step 1: Understanding the Concept:
We are given a structural condition: the class types are fixed as Violinist, Violinist, Violinist, Pianist, Pianist, Pianist (V V V P P P). We must deduce which musicians can and cannot teach, and where they must be, and then evaluate the options.
Step 2: Detailed Explanation:
1. Initial Condition: The schedule structure is V, V, V, P, P, P.
2. Deduce the Violinists: The first three classes are taught by violinists. This satisfies the condition for F to teach. However, the J rule says if J teaches, it must be sixth. Since class 6 is taught by a pianist, J cannot teach. The three violinists must be the only ones remaining: F, G, and H. These three will occupy classes 1, 2, and 3 in some order.
3. Deduce the Pianists: Classes 4, 5, and 6 are taught by pianists.

R Rule: R teaches only if T is first. T is a pianist and cannot be in class 1 (a violinist slot). Therefore, the condition for R to teach is not met, so R cannot teach.
W Rule: W cannot be next to another pianist. In this schedule, all pianists are in a block (P P P). Any position for W (4, 5, or 6) would place it next to at least one other pianist. Therefore, W cannot teach.
The pianists who can teach are the ones remaining from \{R, S, T, W, Z\} after removing R and W. The three pianists must be S, T, and Z. These three will occupy classes 4, 5, and 6 in some order.
4. Summary of Deductions:

Classes 1, 2, 3: Taught by F, G, H (in any order).
Classes 4, 5, 6: Taught by S, T, Z (in any order).
5. Evaluate the options:

(A) F is scheduled to teach the first class. F must teach in slot 1, 2, or 3, but not necessarily slot 1. This is not a "must be true" statement.

(B) G is scheduled to teach the first class. G must teach in slot 1, 2, or 3, but not necessarily slot 1. Not a "must be true" statement.

(C) H is scheduled to teach an earlier class than the class Z is scheduled to teach. H must teach in one of the first three classes. Z must teach in one of the last three classes. Any class from \{1, 2, 3\} is earlier than any class from \{4, 5, 6\}. Therefore, H must teach before Z. This statement must be true.

(D) R is scheduled to teach an earlier class than the class T is scheduled to teach. We deduced that R cannot teach at all. This statement is false.

(E) S is scheduled to teach an earlier class than the class T is scheduled to teach. S and T can be in any order within slots 4, 5, and 6. For example, the pianists could be ordered T, S, Z. This statement is not a "must be true" statement.

Step 3: Final Answer:
The only statement that is a necessary consequence of the given condition is (C).

Answer 5

Step 1: Understanding the Concept:
This question asks for a statement that is a necessary consequence of the initial rules. We must evaluate each conditional statement ("If P, then Q"). The statement is true if whenever the condition P is met, the outcome Q is unavoidable.
Step 2: Detailed Explanation:
Let's analyze each option:


(A) If F is scheduled to teach a class, then H is also scheduled to teach a class.
Assume F is scheduled to teach. According to the F rule, "F will not teach unless the first three classes are taught by violinists." This means the schedule must begin with three violinists (V, V, V). Since there are only three violinist slots in total, the last three classes (4, 5, and 6) must be taught by pianists (P, P, P).
Now, let's consider the other violinists: G, H, and J.
The rule for J states, "If J teaches a class, it will be the sixth." However, the sixth class must be taught by a pianist in this scenario. Therefore, J cannot teach.
Since we need three violinists for the first three classes, and J cannot be one of them, the three violinists must be the only ones available from the pool {F, G, H, J}. These must be F, G, and H. So, if F teaches, H (and G) must also teach. This statement must be true.

(B) If J is scheduled to teach a class, then R is also scheduled to teach a class.
We can disprove this by finding a counterexample. Let's create a valid schedule where J teaches but R does not. For J to teach, it must be in slot 6. For R not to teach, T must not be in slot 1. Consider the schedule: G(V), H(V), T(P), S(P), W(P), J(V). This schedule has 3V and 3P, J is 6th, T is not 1st. However, the pianists T, S, W are in a block, which violates the W rule. Let's try another: T(P), G(V), S(P), H(V), W(P), J(V). Here W is between two violinists (H and J), so the W rule is satisfied. But this schedule has J and no R. Thus, the statement is not necessarily true.

(C) If J is scheduled to teach a class, then S is also scheduled to teach a class.
Let's try to build a valid schedule with J but without S. J must be in slot 6. Let's use pianists {R, T, W}. To use R, T must be in slot 1. Consider: T(P), G(V), W(P), H(V), R(P), J(V). This schedule is valid (W is surrounded by violinists) and includes J but not S. So, this statement is not necessarily true.

(D) If T is scheduled to teach a class, then R is also scheduled to teach a class.
The rule is "R will teach ONLY IF T teaches the first class." This means if R teaches, then T must be first (R \(\rightarrow\) T\(_{1}\)). This does not mean that if T teaches, R must teach. T could teach in a slot other than the first. The valid schedule G, F, H, T, S, Z (from question 9) shows T teaching while R does not. So, this statement is false.

(E) If W is scheduled to teach a class, then Z is also scheduled to teach a class.
We can find a counterexample. The schedule T(P), G(V), W(P), H(V), R(P), J(V) from option (C) is a valid schedule where W teaches, but Z does not. So, this statement is not necessarily true.

Step 3: Final Answer:
The only statement that is a necessary consequence of the rules is (A).

Answer 6

Step 1: Understanding the Concept:
The condition is that the schedule must be alternating between pianists (P) and violinists (V). This gives two possible patterns: PVPVPV or VPVPVP. We need to analyze which pattern is possible and then see which of the given options can occur in that valid pattern.
Step 2: Detailed Explanation:
1. Analyze the VPVPVP pattern:

The first three classes are V, P, V. The F rule requires the first three classes to be V, V, V for F to teach. The condition is not met, so F cannot teach.
The sixth class is a P slot. The rule for J states, "If J teaches, it will be the sixth." Since J is a violinist, J cannot teach in the sixth slot, which is reserved for a pianist. So, J cannot teach.
With both F and J unable to teach, the only available violinists are G and H. Since we need three violinists for the schedule, it's impossible to fill the three V slots. Therefore, the VPVPVP pattern is not possible.
2. Analyze the PVPVPV pattern:

This must be the only possible alternating pattern. Pianist (P) slots are 1, 3, 5. Violinist (V) slots are 2, 4, 6.
Violinists: The first class is a P slot, so the condition for the F rule is not met, and F cannot teach. The sixth slot is a V slot, so J can teach and must be in slot 6. With F out, the three violinists must be G, H, and J. J takes slot 6, and G and H will take slots 2 and 4 in some order.
Pianists: They occupy slots 1, 3, and 5.
3. Evaluate the options based on the PVPVPV pattern:

(A) F is scheduled to teach the fourth class. False. F cannot teach in this pattern.
(B) G is scheduled to teach the first class. False. Slot 1 must be a pianist.
(C) H is scheduled to teach the third class. False. Slot 3 must be a pianist.
(D) R is scheduled to teach the fifth class. This is possible. For R to teach, T must teach the first class. Let's build the schedule:

Slot 1: T (Pianist)
Slot 5: R (Pianist)
Slot 6: J (Violinist)
Slots 2 and 4 are G and H in any order.
Slot 3 is the remaining pianist (from S, W, Z).
Example: T, G, S, H, R, J. This schedule is valid. All rules are met. Therefore, this option can be true.

(E) W is scheduled to teach the second class. False. Slot 2 must be a violinist.
Step 3: Final Answer:
The only possible alternating pattern is PVPVPV. Within this pattern, it is possible for R to teach the fifth class.

Answer 7

Step 1: Understanding the Concept:
We are given a new condition: the schedule is V, __, __, __, __, V. This means there is one more violinist and three pianists to be placed in slots 2 through 5. We must deduce the consequences and check which option is possible. 
Step 2: Detailed Explanation: 
1. Deduce the Violinists: 

The rule for J states, ""If J teaches, it will be the sixth."" Since slot 6 is assigned to a violinist, J must be the violinist teaching the sixth class. 
The rule for F states, ""F will not teach unless the first three classes are taught by violinists."" Since we need to place three pianists in slots 2-5, it's impossible for slots 1, 2, and 3 to all be taught by violinists. Therefore, F cannot teach. 
With J assigned to slot 6 and F unable to teach, the remaining two violinists must be G and H. One teaches slot 1, and the other teaches one of the middle slots (2, 3, 4, or 5). 
2. Deduce the Pianists: 

The rule for R states, ""R will teach only if T teaches the first class."" Slot 1 is taken by a violinist (G or H). Thus, T cannot be first, which means R cannot teach. 
With R unable to teach, the three pianists must be chosen from the remaining pool: {S, T, W, Z}. 
3. Evaluate the options by trying to construct a valid schedule: 

(A) F is scheduled to teach the second class. False. As deduced, F cannot teach. 
(B) H is scheduled to teach the sixth class. False. As deduced, J must teach the sixth class. 
(C) R is scheduled to teach the fourth class. False. As deduced, R cannot teach. 
(D) T is scheduled to teach the second class. Let's try to build this schedule. 

Slot 1 is a violinist (say, G). Slot 2 is T (pianist). Slot 6 is J (violinist). 
We still need to place H (violinist) and two pianists from {S, W, Z} in slots 3, 4, and 5. 
To avoid violating the W rule (no P next to W), let's place H between the remaining two pianists. Let H teach slot 4. 
The schedule is now: G(V), T(P), \(\underline{Pianist}\), H(V), \(\underline{Pianist}\), J(V). 
We can place S in slot 3 and W in slot 5. The full schedule is: G, T, S, H, W, J. 
Let's check this schedule: It has 3V/3P. J is 6th. F and R are not teaching. W in slot 5 is surrounded by H(V) and J(V), which is valid. This is a possible schedule. 

(E) W is scheduled to teach the third class. Let's try. The schedule would be V, P, W(P), V, P, V. W would be in slot 3. For the W rule to be satisfied, slot 2 and slot 4 must be violinists. But we only have one violinist (G or H) left to place in the middle. So W would be next to a pianist in slot 2. This is impossible. 
Step 3: Final Answer: 
We successfully constructed a valid schedule where T teaches the second class. Therefore, option (D) can be true. 

Answer 8

Step 1: Understanding the Concept:
This question asks for a scenario that is impossible under any valid interpretation of the rules. We test each option. If we can create at least one valid schedule where the statement is true, then the option is possible. The correct answer will be the one for which no valid schedule can be created.
Step 2: Detailed Explanation:


(A) F is scheduled to teach the third class. This is possible. If F teaches, classes 1, 2, and 3 must be violinists. F can be third in an arrangement like G, H, F. The remaining classes 4, 5, 6 would be pianists (e.g., T, S, Z). The schedule G, H, F, T, S, Z is valid. So, this can be true.

(B) G is scheduled to teach the first class. This is possible. The schedule G, H, F, T, S, Z from option (A) shows G teaching the first class. So, this can be true.

(C) T is scheduled to teach the sixth class. Let's see if this is possible. If T (a pianist) teaches class 6, then J (a violinist) cannot teach. Therefore, the three violinists must be F, G, and H. For F to teach, classes 1, 2, and 3 must be taught by violinists. This works perfectly: F, G, H teach classes 1, 2, 3. The three pianists teach 4, 5, 6. We are told T is sixth. Since the pianists are in a block (4, 5, 6), W cannot teach. And R cannot teach because T is not first. So the pianists must be T, S, and Z. A valid schedule is F, G, H, S, Z, T. So, this can be true.

(D) W is scheduled to teach the sixth class. Let's analyze this.

If W (a pianist) teaches class 6, then J (a violinist) cannot teach.
The W rule states that no pianist can be immediately adjacent to W. This means the musician in class 5 must be a violinist.
Since J cannot teach, the three violinists must be selected from the pool {F, G, H}. But we have a problem.
Scenario 1: F teaches. If F teaches, classes 1, 2, and 3 must be violinists. The three violinists would be F, G, H. The three pianists would teach classes 4, 5, and 6. This would make the schedule V, V, V, P, P, P. If W is in class 6, it is preceded by a pianist in class 5. This violates the W rule. So, F cannot teach.
Scenario 2: F does not teach. If F does not teach, and J cannot teach (because W is in slot 6), then the only available violinists are G and H. It is impossible to form a schedule with only two violinists, as three are required.
Both scenarios lead to a contradiction. Therefore, it is impossible for W to teach the sixth class. This CANNOT be true.

(E) Z is scheduled to teach the fifth class. This is possible. The valid schedule G, F, H, T, S, Z from question 9(C) has Z in the sixth position. We can rearrange the pianists to get G, F, H, T, Z, S, which is also valid and has Z teaching fifth. So, this can be true.

Step 3: Final Answer:
It is impossible to create a valid schedule where W teaches the sixth class.