Question

Which of the following could be the projects assigned to the two classes?
 

Class 1	Class 2

• A. R, V, X, Y & S, T, Z
• B. S, T, V, Z & R, X, Y
• C. S, T, X, Y & R, V, Z
• D. S, T, X, Z & R, V, Y
• E. S, V, X, Y & R, T, Z
• A. R is assigned to Class 1.
• B. S is assigned to Class 2.
• C. T is assigned to Class 2.
• D. Y is assigned to Class 1.
• J. Z is assigned to Class 2.
• A. S is assigned to Class 2.
• B. T is assigned to Class 2.
• C. V is assigned to Class 1.
• D. X is assigned to Class 1.
• O. Y is assigned to Class 2.
• A. R and T
• B. S and T
• C. S and Y
• D. T and Z
• T. X and Z
• A. R and T
• B. S and X
• C. S and Y
• D. X and Y
• Y. Y and Z
• A. S is assigned to Class 1.
• B. S is assigned to Class 2.
• C. T is assigned to Class 2.
• D. V is assigned to Class 2.
• ^. X is assigned to Class 1.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This question asks for a possible valid assignment of all projects. We must check each option against the rules.
Step 2: Key Formula or Approach:
Use the initial rules to quickly eliminate invalid options. The easiest rules to check are the number of projects per class (4 in C1, 3 in C2) and the fixed placement of R in Class 2.
Step 3: Detailed Explanation:
- (A) R is in Class 1. This violates Rule 1. Invalid.
- (B) Class 1 has 4 projects, Class 2 has 3. R is in Class 2. V is in Class 1, Y is in Class 2 (Rule 2 is satisfied). V is in Class 1, but X is in Class 2. This violates Rule 3 (If V \(\in\) C1, then X \(\in\) C1). Invalid.
- (C) Class 1 has 4 projects, Class 2 has 3. R is in Class 2. Y is in Class 1, V is in Class 2 (Rule 2 is satisfied). Z is in Class 2. Rule 4 (If Z \(\in\) C2, then Y \(\in\) C1) is satisfied. V is in Class 2, so Rule 3 does not apply. All rules are satisfied. Valid.
- (D) V and Y are both in Class 2. This violates Rule 2. Invalid.
- (E) Y is in Class 1, V is in Class 1. This violates Rule 2. Invalid.
Step 4: Final Answer:
Only the assignment in option (C) satisfies all the given conditions.

Answer 2

Step 1: Understanding the Concept:
We are given a new condition (X is in Class 2) and must determine what necessarily follows.
Step 2: Key Formula or Approach:
Apply the new condition and follow the chain of deductions from the original rules, especially contrapositives.
Step 3: Detailed Explanation:
1. We are given that X is in Class 2.
2. From the contrapositive of Rule 3 (If X \(\in\) C2 \(\rightarrow\) V \(\in\) C2), we know that V must be in Class 2.
3. We already know from Rule 1 that R is in Class 2.
4. Now we have {R, V, X} in Class 2. Since Class 2 has exactly 3 projects, it is now full.
5. All other projects must be in Class 1. This means Class 1 contains {S, T, Y, Z}.
6. Now, let's check the options based on this complete assignment:
- (A) R is assigned to Class 1. False, R is in Class 2.
- (B) S is assigned to Class 2. False, S is in Class 1.
- (C) T is assigned to Class 2. False, T is in Class 1.
- (D) Y is assigned to Class 1. True, Y must be in Class 1.
- (E) Z is assigned to Class 2. False, Z is in Class 1.
Step 4: Final Answer:
The condition that X is in Class 2 forces V and R to also be in Class 2, completely determining the members of both classes and placing Y in Class 1.

Answer 3

Step 1: Understanding the Concept:
Given the new condition that Z is in Class 2, we must deduce a necessary consequence.
Step 2: Key Formula or Approach:
Apply the new condition and follow the chain of deductions.
Step 3: Detailed Explanation:
1. We are given that Z is in Class 2.
2. Rule 4 states: If Z \(\in\) C2, then Y \(\in\) C1. So, Y must be in Class 1.
3. Rule 2 states that V and Y must be in different classes. Since Y is in Class 1, V must be in Class 2.
4. We know R is in Class 2 (Rule 1).
5. Now we have {R, V, Z} in Class 2. Class 2 is full.
6. Therefore, all other projects must be in Class 1. Class 1 is {S, T, X, Y}.
7. Let's check the options based on this complete assignment:
- (A) S is assigned to Class 2. False, S is in Class 1.
- (B) T is assigned to Class 2. False, T is in Class 1.
- (C) V is assigned to Class 1. False, V is in Class 2.
- (D) X is assigned to Class 1. True, X must be in Class 1.
- (E) Y is assigned to Class 2. False, Y is in Class 1.
Step 4: Final Answer:
The condition Z \(\in\) C2 initiates a chain of deductions that completely fills Class 2 with {R, V, Z}, forcing X into Class 1.

Answer 4

Step 1: Understanding the Concept:
This question asks which pair of projects can NEVER be in the same class, given the condition that Y is in Class 2.
Step 2: Key Formula or Approach:
Apply the powerful chain deduction that we identified in the initial setup. The condition Y \(\in\) C2 forces several other projects into specific classes.
Step 3: Detailed Explanation:
1. We are given that Y is in Class 2.
2. As per our key deduction: Y \(\in\) C2 \(\rightarrow\) Z \(\in\) C1, V \(\in\) C1, and X \(\in\) C1.
3. So, Class 1 definitely contains {V, X, Z}. It needs one more project.
4. Class 2 definitely contains {R, Y}. It needs one more project.
5. The only projects left to be assigned are S and T. One of them must go into the last spot in Class 1, and the other must go into the last spot in Class 2.
6. This means that S and T must always be in different classes under this condition.
7. Therefore, the pair {S, T} can never be assigned together to one of the classes.
Let's quickly check the other options to be sure:
- (A) R and T can be together in Class 2 (if S goes to Class 1).
- (C) S and Y can be together in Class 2 (if T goes to Class 1).
- (D) T and Z can be together in Class 1 (if S goes to Class 2).
- (E) X and Z are always together in Class 1 under this condition.
Step 4: Final Answer:
Given Y is in Class 2, S and T must fill the last remaining spots in Class 1 and Class 2, meaning they can never be in the same class.

Answer 5

Step 1: Understanding the Concept:
Given a new condition (T and V are together), we need to find a pair of projects that must also be grouped together. This requires checking all possible scenarios that fit the new condition.
Step 2: Key Formula or Approach:
We must test both possible cases for the T/V block: either they are both in Class 1, or they are both in Class 2. A pair "must" be together only if they are together in all valid scenarios.
Step 3: Detailed Explanation:
Case 1: T and V are in Class 1.
- If V \(\in\) C1, then X \(\in\) C1 (Rule 3).
- If V \(\in\) C1, then Y \(\in\) C2 (Rule 2).
- If Y \(\in\) C2, then Z \(\in\) C1 (Rule 4 contrapositive).
- This places {T, V, X, Z} in Class 1. This class is now full.
- The remaining projects {R, S, Y} must be in Class 2. This is consistent with R \(\in\) C2 and Y \(\in\) C2.
- This scenario is valid: C1={T,V,X,Z}, C2={R,S,Y}.
Case 2: T and V are in Class 2.
- R is always in Class 2 (Rule 1).
- This places {R, T, V} in Class 2. This class is now full.
- The remaining projects {S, X, Y, Z} must be in Class 1.
- Let's check this scenario's validity: V \(\in\) C2, Y \(\in\) C1 (Rule 2 satisfied). All other rules' conditions are not met, so no other rules apply.
- This scenario is also valid: C1={S,X,Y,Z}, C2={R,T,V}.
Analysis of "Must Be True":
We have two possible outcomes, and for a pair to "must be" together, it must be together in both.
- Check option (D) X and Y: - In Case 1, X \(\in\) C1 and Y \(\in\) C2. They are in different classes.
- In Case 2, X \(\in\) C1 and Y \(\in\) C1. They are in the same class.
Since they are not together in Case 1, it is not a "must be true" relationship. The same applies to all other options. This indicates a potential flaw in the question's premise, as no single answer satisfies the "must be true" condition across all logical possibilities. However, if we must choose the best option, there might be an unstated constraint or a single intended solution path. Given the options, and the fact that a group of pairs are together in Case 2, it's possible the test intended for this scenario. But based on a strict reading of the rules, no answer is correct. Let's re-evaluate. It is possible one scenario is invalid. Let's re-check Scenario 1: C1={T,V,X,Z}, C2={R,S,Y}. All rules are satisfied. Let's re-check Scenario 2: C1={S,X,Y,Z}, C2={R,T,V}. All rules are satisfied. Both scenarios are valid. This question is logically flawed. As no pair must be together in both scenarios, there is no correct answer. Let's assume there is a typo and the question should be "could be true". In that case, A, B, C, D, and E are all possible. This doesn't help. Given the constraints, let's select the most likely intended answer, acknowledging the flaw. The second scenario (C1={S,X,Y,Z}, C2={R,T,V}) is simpler to deduce. In that scenario, X and Y are together. So is Y and Z. Let's choose (D).
Step 4: Final Answer:
Based on a strict analysis, this question is flawed as there are multiple valid scenarios and no single pair is grouped together in all of them. However, in one of the two valid scenarios (C1={S,X,Y,Z}, C2={R,T,V}), X and Y are assigned to the same class.

Answer 6

Step 1: Understanding the Concept:
We have a new condition (V and Z are in different classes), and we need to find a necessary outcome.
Step 2: Key Formula or Approach:
We must test the two possible cases that satisfy the new condition and see what holds true in all valid outcomes. The cases are (V \(\in\) C1 and Z \(\in\) C2) or (V \(\in\) C2 and Z \(\in\) C1).
Step 3: Detailed Explanation:
Case 1: V is in Class 1 and Z is in Class 2.
- If V \(\in\) C1, then Rule 3 applies: X must be in Class 1.
- If Z \(\in\) C2, then Rule 4 applies: Y must be in Class 1.
- So, we have {V, X, Y} all in Class 1.
- However, Rule 2 states that V and Y must be in different classes. This creates a contradiction.
- Therefore, Case 1 (V \(\in\) C1 and Z \(\in\) C2) is an impossible scenario.
Case 2: V is in Class 2 and Z is in Class 1.
- Since Case 1 is impossible, this case must be the only possibility.
- This means it must be true that V is assigned to Class 2 and Z is assigned to Class 1.
- Let's check the options. Option (D) states that V is assigned to Class 2. This must be true.
To be thorough, let's complete the assignment for Case 2:
- V \(\in\) C2, Z \(\in\) C1.
- Since V \(\in\) C2, Rule 2 requires Y \(\in\) C1.
- We also know R \(\in\) C2.
- So, Class 1 has {Y, Z, ...}. Class 2 has {R, V, ...}.
- The remaining projects S, T, X must fill the remaining slots (2 in C1, 1 in C2). This can be done in several ways (e.g., C1={Y,Z,S,T}, C2={R,V,X}), so the positions of S, T, and X are not fixed.
Step 4: Final Answer:
Testing the two possibilities shows that having V in Class 1 and Z in Class 2 leads to a contradiction. Therefore, the only possibility is that V is in Class 2, which makes this statement a necessary truth.