If N and X are both scheduled in the same slot, which statement BEST describes the relationship between M, V and A?
Answer and solution
Correct answer: A
Solution
Answer: A — At least one of M and V is guided by A.
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 guides either U in Slot 2 or O in Slot 4, because X and Z are the two single-student guides assigned to the remaining Operations and Strategy students.
If X guides U, then N is in Slot 2. The Marketing student in Slot 2 is guided by A, so N is one of A's students. A's other student is in Slot 1. The remaining two Marketing students, M and V, occupy Slots 1 and 4, so one of them is guided by A.
If X guides O, then N is in Slot 4. The Marketing student in Slot 4 is guided by Y. In that case, the two remaining Marketing students M and V must both be guided by A.
In either case, at least one of M and V is guided by A. Option A is correct.
Verification: In either case, at least one of M and V is guided by A. Option A 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 guides either U in Slot 2 or O in Slot 4, because X and Z are the two single-student guides assigned to the remaining Operations and Strategy students.
If X guides U, then N is in Slot 2. The Marketing student in Slot 2 is guided by A, so N is one of A's students. A's other student is in Slot 1. The remaining two Marketing students, M and V, occupy Slots 1 and 4, so one of them is guided by A.
If X guides O, then N is in Slot 4. The Marketing student in Slot 4 is guided by Y. In that case, the two remaining Marketing students M and V must both be guided by A.
In either case, at least one of M and V is guided by A. Option A 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: In either case, at least one of M and V is guided by A. Option A 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: In either case, at least one of M and V is guided by A. Option A is correct.