CAT 2026 · Slot 3 · Expected paper

CAT 2026 › Slot 3 paper › DILR

CAT 2026 Slot 3 DILR: Set 5: DILR – Doctoral Students and Seminar Scheduling with Solutions

5 questions: Q18, Q19, Q20, Q21, Q22. Solve each one here: tap the answer, press Check, then open the solution.

Logical Reasoning

The set

A college is organizing eight seminars, one for each of its doctoral students — J, K, L, M, N, O, U and V. Four students belong to Marketing, three to Operations and one to Strategy. Each student is guided by one of W, X, Y, Z and A. W, Y and A guide two students each, while X and Z guide one student each. Every student and their guide belong to the same department.

The seminars are scheduled in four consecutive 30-minute slots: 9:00 am, 9:30 am, 10:00 am and 10:30 am. At most three seminars can take place in a slot, and a guide cannot supervise two seminars in the same slot. Students guided by the same guide must be scheduled in consecutive slots.

The following information is also known:
1. At least one Marketing student has a seminar in each of the four slots.
2. J is the only student scheduled at 10:00 am. J is guided by Y.
3. O is the Strategy student and is scheduled at 10:30 am.
4. At least one Operations student is scheduled at 9:00 am.
5. K and U are Operations students and are scheduled in the same slot. A Marketing student guided by A is also scheduled in that slot.
6. W guides K and L, and W's two students are scheduled in the first two slots.
7. J and U are scheduled in consecutive slots.

Question 18Scheduling

MCQ · +3 / −1 · medium

Which one of the following statements is true?

Answer and solution

Correct answer: D

Solution

Answer: D — Two seminars are scheduled in the first slot.

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:
Start with the fixed positions. J is alone in Slot 3, and O is in Slot 4. Since J and U are in consecutive slots, while K and U share a slot, K and U must both be in Slot 2. W guides K and L, and W's students must occupy the first two slots, so L is in Slot 1.

The Marketing student guided by A who shares K and U's slot is in Slot 2. Since A guides two students in consecutive slots, A's other student is in Slot 1. Therefore, Slot 1 has L and one Marketing student: exactly two seminars. Slot 2 has K, U and one Marketing student: three seminars. Slot 3 has only J. Slot 4 has O and the remaining Marketing student.

Hence, exactly two seminars are scheduled in the first slot. Option D is correct.

G Strategy shortcut

Question 19Scheduling

MCQ · +3 / −1 · medium

Who among the following are NOT guiding any Marketing students?

Answer and solution

Correct answer: C

Solution

Answer: C — W, X and Z

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:
There are four Marketing students: J and the remaining three students M, N and V. J is guided by Y. The Marketing student in Slot 2 and the Marketing student in Slot 1 are guided by A, because A's two students must be in consecutive slots. The Marketing student in Slot 4 is guided by Y, since J is Y's student in Slot 3 and Y's two students must be in consecutive slots.

Thus, A and Y guide all four Marketing students. W guides K and L, both Operations students. The remaining guides X and Z guide U and O, who belong to Operations and Strategy respectively. Therefore, W, X and Z do not guide any Marketing student. Option C 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.
Verification: Thus, A and Y guide all four Marketing students. W guides K and L, both Operations students. The remaining guides X and Z guide U and O, who belong to Operations and Strategy respectively. Therefore, W, X and Z do not guide any Marketing student. Option C 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. Verification: Thus, A and Y guide all four Marketing students. W guides K and L, both Operations students. The remaining guides X and Z guide U and O, who belong to Operations and Strategy respectively. Therefore, W, X and Z do not guide any Marketing student. Option C is correct.

Question 20Scheduling

MCQ · +3 / −1 · medium

Which of the following statements is necessarily true?

Answer and solution

Correct answer: B

Solution

Answer: B — V is a Marketing student.

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:
K and U are Operations students, and L is also an Operations student because W guides both K and L and guides belong to the same department as their students. O is the only Strategy student. That accounts for all three Operations students and the one Strategy student.

The four remaining students — J, M, N and V — must therefore be the four Marketing students. Hence, V is definitely a Marketing student. X and Z can exchange the individual assignments to U and O, so neither of those guide assignments is individually certain. K is in Slot 2, not Slot 1. Option B 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.
Verification: The four remaining students — J, M, N and V — must therefore be the four Marketing students. Hence, V is definitely a Marketing student. X and Z can exchange the individual assignments to U and O, so neither of those guide assignments is individually certain. K is in Slot 2, not Slot 1. Option B 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. Verification: The four remaining students — J, M, N and V — must therefore be the four Marketing students. Hence, V is definitely a Marketing student. X and Z can exchange the individual assignments to U and O, so neither of those guide assignments is individually certain. K is in Slot 2, not Slot 1. Option B is correct.

Question 21Scheduling

MCQ · +3 / −1 · hard

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.

G Strategy shortcut

Question 22Scheduling

MCQ · +3 / −1 · hard

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.

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.

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