If M is scheduled in the slot immediately before X's, which of the following is NOT necessarily true?
Answer and solution
Correct answer: D
Solution
Answer: D — N is guided by Y.
1. There are exactly four Marketing students and four slots requiring one each, so each slot has exactly one Marketing student. J alone in Slot 3 is therefore Marketing. O is Strategy.
2. K, U are Operations. W guides K and L in the same department, so L is the third Operations student. The remaining J, M, N, V are Marketing.
3. W’s students K, L occupy Slots 1 and 2. U shares K’s slot and is consecutive to J in Slot 3, so K, U must be in Slot 2 and L in Slot 1.
4. Slot 2 also contains an A-guided Marketing student. A’s other student must be consecutive and cannot use Slot 3 (J is guided by Y), so is in Slot 1.
5. Y guides J in Slot 3 and the last Marketing student in Slot 4: Slot 2’s sole Marketing place is already assigned to A. Slots therefore have counts 2,3,1,2.
6. W guides L/K in Slots 1/2; A guides Marketing students in Slots 1/2; Y guides J and a Marketing student in Slots 3/4. X, Z guide U and O in either order. M, N, V can permute across Marketing places in Slots 1,2,4; 6×2=12 complete assignments remain.
Question-specific deduction:
X can guide either U in Slot 2 or O in Slot 4. If X guides O in Slot 4, the slot immediately before X's slot is Slot 3. But J is the only student in Slot 3, so M cannot be there. Therefore, X must guide U in Slot 2, and M must be in Slot 1.
Since A's two students are in Slots 1 and 2, M is guided by A. The other single-student guide, Z, must guide O. Thus, the first three statements are necessarily true.
The two Marketing students guided by Y occupy Slots 3 and 4. N could be the Marketing student in Slot 4, or N could be A's Marketing student in Slot 2, depending on the remaining assignment of N and V. Therefore, N is not necessarily guided by Y. Option D is correct.
Verification: The two Marketing students guided by Y occupy Slots 3 and 4. N could be the Marketing student in Slot 4, or N could be A's Marketing student in Slot 2, depending on the remaining assignment of N and V. Therefore, N is not necessarily guided by Y. Option D is correct.
1. There are exactly four Marketing students and four slots requiring one each, so each slot has exactly one Marketing student. J alone in Slot 3 is therefore Marketing. O is Strategy.
2. K, U are Operations. W guides K and L in the same department, so L is the third Operations student. The remaining J, M, N, V are Marketing.
3. W’s students K, L occupy Slots 1 and 2. U shares K’s slot and is consecutive to J in Slot 3, so K, U must be in Slot 2 and L in Slot 1.
4. Slot 2 also contains an A-guided Marketing student. A’s other student must be consecutive and cannot use Slot 3 (J is guided by Y), so is in Slot 1.
5. Y guides J in Slot 3 and the last Marketing student in Slot 4: Slot 2’s sole Marketing place is already assigned to A. Slots therefore have counts 2,3,1,2.
6. W guides L/K in Slots 1/2; A guides Marketing students in Slots 1/2; Y guides J and a Marketing student in Slots 3/4. X, Z guide U and O in either order. M, N, V can permute across Marketing places in Slots 1,2,4; 6×2=12 complete assignments remain.
Question-specific deduction:
X can guide either U in Slot 2 or O in Slot 4. If X guides O in Slot 4, the slot immediately before X's slot is Slot 3. But J is the only student in Slot 3, so M cannot be there. Therefore, X must guide U in Slot 2, and M must be in Slot 1.
Since A's two students are in Slots 1 and 2, M is guided by A. The other single-student guide, Z, must guide O. Thus, the first three statements are necessarily true.
The two Marketing students guided by Y occupy Slots 3 and 4. N could be the Marketing student in Slot 4, or N could be A's Marketing student in Slot 2, depending on the remaining assignment of N and V. Therefore, N is not necessarily guided by Y. Option D is correct.
G Strategy shortcut
Method: Use four Marketing students across four slots: exactly one per slot. W puts K in Slot 2 and L in Slot 1; A then fills Marketing Slots 1/2, Y fills 3/4. Only the M/N/V permutation and X/Z swap remain. For the extra condition, split only X=U(Slot 2) versus X=O(Slot 4).Verification: The two Marketing students guided by Y occupy Slots 3 and 4. N could be the Marketing student in Slot 4, or N could be A's Marketing student in Slot 2, depending on the remaining assignment of N and V. Therefore, N is not necessarily guided by Y. Option D is correct.
G Strategy: Method: Use four Marketing students across four slots: exactly one per slot. W puts K in Slot 2 and L in Slot 1; A then fills Marketing Slots 1/2, Y fills 3/4. Only the M/N/V permutation and X/Z swap remain. For the extra condition, split only X=U(Slot 2) versus X=O(Slot 4). Verification: The two Marketing students guided by Y occupy Slots 3 and 4. N could be the Marketing student in Slot 4, or N could be A's Marketing student in Slot 2, depending on the remaining assignment of N and V. Therefore, N is not necessarily guided by Y. Option D is correct.