Scheduling Based – MBA CET questions with solutions

Approach: Scheduling with non-consecutive and exclusion rules.

Solution:
Open days: Mon–Sat. Floors: Monday + 2 more days, no two consecutive → Tuesday is out.
Possible extra pairs from Wed–Sat (non-consecutive): {Wed, Fri}, {Wed, Sat}, {Thu, Sat}.
Check watering (2 non-consecutive days from the rest):
{Mon, Wed, Fri} → water Tue & Thu / Tue & Sat / Thu & Sat ✓
{Mon, Wed, Sat} → water Tue & Thu / Tue & Fri ✓
{Mon, Thu, Sat} → water Tue & Fri / Wed & Fri ✓
Every valid schedule polishes on Wednesday or Thursday → Either Wednesday or Thursday.

Why the other options are wrong:
A) Tuesday is never possible (next to Monday); {Mon, Thu, Sat} has neither Tue nor Wed.
C) Monday is fixed every week, so it is not an 'either' choice.
D) Tuesday is impossible and {Mon, Wed, Sat} has no Friday.
E) Friday or Saturday also holds in every valid schedule, but the key is B (Wednesday or Thursday).

Answer: B

Approach: Fix paper counts from the totals, then countries from the area rules.

Solution:
18 papers; A + D + G + H = 2 × (B + C + E + F) → A + D + G + H = 12, B + C + E + F = 6.
Max 3 each → A = D = G = H = 3.
F = 1 → B + C + E = 5, with C = E.
Countries: C, E Chinese; D Japanese; F Indian. B and G share a country → both Indian (Japan has only 2 places).
A is from Automation; Japanese/Chinese are never in Automation → A is Indian. A and H differ → H is Japanese.
Indians: A, B, F, G.
G (3 papers) shares no issue with B, and B has an April paper → G is in Jan, Jul, Oct, so B can use only April → B = 1, C = E = 2.
Papers by Indians = A 3 + G 3 + B 1 + F 1 = 8.

Why the other options are wrong:
C (10) takes B = 3; B (7) ignores B's paper; A, E do not fit the forced counts.

Answer: D

Approach: Day-wise scheduling, Monday–Friday.

Solution:
D = Tuesday; B is immediately after D → Wednesday.
Monday, Thursday, Friday remain for A, C, E.
A is not on Monday; E cannot be on Monday because C must be before E.
So Monday = C (A and E take Thursday/Friday).

Why the other options are wrong:
A B – Wednesday.
B A – barred from Monday.
C E – must come after C.
E D – Tuesday.

Answer: D

Approach: Day–person–task scheduling grid.

Solution: Ranjita = Monday, Teena = Wednesday. Sanjana dusts, not Mon/Fri; Wednesday is Teena's and Thursday is mopping → Sanjana = Tuesday.
Thursday and Friday remain for Ujjwala and Vajita. Laundry is on Friday and not by Ujjwala → Vajita = Friday (laundry), Ujjwala = Thursday (mopping).
Sweeping and vacuuming remain for Ranjita and Teena; Teena does not vacuum → Teena sweeps, Ranjita vacuums.

Why the other options are wrong: Monday is Ranjita's, Wednesday Teena's, Thursday mopping, Friday laundry.

Cetking shortcut: Lock the days given directly (Mon, Wed), then the task tied to a day (mopping Thu, laundry Fri) — Sanjana's day is forced at once.

Answer: C

Approach: Day–person–task scheduling grid.

Solution: Ranjita = Monday, Teena = Wednesday. Sanjana dusts, not Mon/Fri; Wednesday is Teena's and Thursday is mopping → Sanjana = Tuesday.
Thursday and Friday remain for Ujjwala and Vajita. Laundry is on Friday and not by Ujjwala → Vajita = Friday (laundry), Ujjwala = Thursday (mopping).
Sweeping and vacuuming remain for Ranjita and Teena; Teena does not vacuum → Teena sweeps, Ranjita vacuums.

Why the other options are wrong: Vacuuming is excluded for Teena; dusting is Sanjana's; mopping Thursday; laundry Friday.

Cetking shortcut: Lock the days given directly (Mon, Wed), then the task tied to a day (mopping Thu, laundry Fri) — Sanjana's day is forced at once.

Answer: D

Approach: Day–person–task scheduling grid.

Solution: Ranjita = Monday, Teena = Wednesday. Sanjana dusts, not Mon/Fri; Wednesday is Teena's and Thursday is mopping → Sanjana = Tuesday.
Thursday and Friday remain for Ujjwala and Vajita. Laundry is on Friday and not by Ujjwala → Vajita = Friday (laundry), Ujjwala = Thursday (mopping).
Sweeping and vacuuming remain for Ranjita and Teena; Teena does not vacuum → Teena sweeps, Ranjita vacuums.

Why the other options are wrong: Tue = dusting, Wed = sweeping, Thu = mopping, Fri = laundry.

Cetking shortcut: Lock the days given directly (Mon, Wed), then the task tied to a day (mopping Thu, laundry Fri) — Sanjana's day is forced at once.

Answer: B

Approach: Day–person–task scheduling grid.

Solution: Ranjita = Monday, Teena = Wednesday. Sanjana dusts, not Mon/Fri; Wednesday is Teena's and Thursday is mopping → Sanjana = Tuesday.
Thursday and Friday remain for Ujjwala and Vajita. Laundry is on Friday and not by Ujjwala → Vajita = Friday (laundry), Ujjwala = Thursday (mopping).
Sweeping and vacuuming remain for Ranjita and Teena; Teena does not vacuum → Teena sweeps, Ranjita vacuums.

Why the other options are wrong: Vacuuming = Ranjita, dusting = Sanjana, mopping = Ujjwala, sweeping = Teena.

Cetking shortcut: Lock the days given directly (Mon, Wed), then the task tied to a day (mopping Thu, laundry Fri) — Sanjana's day is forced at once.

Answer: E

Approach: Day–person–task scheduling grid.

Solution: Ranjita = Monday, Teena = Wednesday. Sanjana dusts, not Mon/Fri; Wednesday is Teena's and Thursday is mopping → Sanjana = Tuesday.
Thursday and Friday remain for Ujjwala and Vajita. Laundry is on Friday and not by Ujjwala → Vajita = Friday (laundry), Ujjwala = Thursday (mopping).
Sweeping and vacuuming remain for Ranjita and Teena; Teena does not vacuum → Teena sweeps, Ranjita vacuums.

Why the other options are wrong: Mon, Tue, Wed and Fri are taken by Ranjita, Sanjana, Teena and Vajita.

Cetking shortcut: Lock the days given directly (Mon, Wed), then the task tied to a day (mopping Thu, laundry Fri) — Sanjana's day is forced at once.

Answer: D

Approach: Day scheduling with specialities — count slots first, then place fixed people.

Solution:
U (Oncologist) is alone on Tuesday → the other 6 doctors fill Wed, Fri, Sat with at most 2 each → exactly 2 per day.
Wednesday: Q and R.
Friday: S and the Radiologist (S visits with the Radiologist, so S is not the Radiologist). The Neurologist visits on Friday → S = Neurologist.
Remaining P, T, V fill Friday's second slot (the Radiologist) and Saturday. V is not Radiologist, P is not Radiologist → T = Radiologist, Friday. P and V → Saturday.
Paediatrician not on Saturday, not with T or V, and visits with ENT → the Wednesday pair. Q is not Paediatrician → R = Paediatrician, Q = ENT.
Saturday: Orthopaedics and Ophthalmology; P is not Orthopaedics → P = Ophthalmologist, V = Orthopaedics.

Orthopaedics (V) and Ophthalmology (P) both visit on Saturday.

Why the other options are wrong:
A: Tuesday has only the Oncologist.
B: Wednesday has ENT and Paediatrics.
C: Friday has Neurology and Radiology.
E: both are on Saturday only.

Cetking shortcut: 7 doctors, 4 days, max 2 per day, one day with 1 → every other day has exactly 2. Fix that before anything else.

Answer: D

Approach: Day scheduling with specialities — count slots first, then place fixed people.

Solution:
U (Oncologist) is alone on Tuesday → the other 6 doctors fill Wed, Fri, Sat with at most 2 each → exactly 2 per day.
Wednesday: Q and R.
Friday: S and the Radiologist (S visits with the Radiologist, so S is not the Radiologist). The Neurologist visits on Friday → S = Neurologist.
Remaining P, T, V fill Friday's second slot (the Radiologist) and Saturday. V is not Radiologist, P is not Radiologist → T = Radiologist, Friday. P and V → Saturday.
Paediatrician not on Saturday, not with T or V, and visits with ENT → the Wednesday pair. Q is not Paediatrician → R = Paediatrician, Q = ENT.
Saturday: Orthopaedics and Ophthalmology; P is not Orthopaedics → P = Ophthalmologist, V = Orthopaedics.

S is the Neurologist: S visits on Friday with the Radiologist, so S is not the Radiologist, and the Neurologist must be on Friday.

Why the other options are wrong:
A: Oncologist is U (Tuesday).
C: Radiologist is S's companion T.
D: Ophthalmologist is P (Saturday).
E: ENT is Q (Wednesday).

Answer: B

Approach: Day scheduling with specialities — count slots first, then place fixed people.

Solution:
U (Oncologist) is alone on Tuesday → the other 6 doctors fill Wed, Fri, Sat with at most 2 each → exactly 2 per day.
Wednesday: Q and R.
Friday: S and the Radiologist (S visits with the Radiologist, so S is not the Radiologist). The Neurologist visits on Friday → S = Neurologist.
Remaining P, T, V fill Friday's second slot (the Radiologist) and Saturday. V is not Radiologist, P is not Radiologist → T = Radiologist, Friday. P and V → Saturday.
Paediatrician not on Saturday, not with T or V, and visits with ENT → the Wednesday pair. Q is not Paediatrician → R = Paediatrician, Q = ENT.
Saturday: Orthopaedics and Ophthalmology; P is not Orthopaedics → P = Ophthalmologist, V = Orthopaedics.

P is the Ophthalmologist.

Why the other options are wrong:
A: ENT is Q.
C: Neurologist is S.
D: Paediatrics is R.
E: P is not the Radiologist (given); T is.

Answer: B

Approach: Day scheduling with specialities — count slots first, then place fixed people.

Solution:
U (Oncologist) is alone on Tuesday → the other 6 doctors fill Wed, Fri, Sat with at most 2 each → exactly 2 per day.
Wednesday: Q and R.
Friday: S and the Radiologist (S visits with the Radiologist, so S is not the Radiologist). The Neurologist visits on Friday → S = Neurologist.
Remaining P, T, V fill Friday's second slot (the Radiologist) and Saturday. V is not Radiologist, P is not Radiologist → T = Radiologist, Friday. P and V → Saturday.
Paediatrician not on Saturday, not with T or V, and visits with ENT → the Wednesday pair. Q is not Paediatrician → R = Paediatrician, Q = ENT.
Saturday: Orthopaedics and Ophthalmology; P is not Orthopaedics → P = Ophthalmologist, V = Orthopaedics.

T (Radiologist) visits on Friday with S.

Why the other options are wrong:
A: Wednesday is Q and R.
B: Tuesday is U alone.
D, E: Saturday is P and V.

Answer: C

Approach: Knockout bracket: the runner-up must come from the opposite half of the draw to the winner.

Solution:
Bracket rule (from the set): Round-1 match of seed s = min(s, 65−s); Round-2 match = min(m, 33−m); Round-3 match = min(m, 17−m); quarter-final = min(m, 9−m); semi-final = min(m, 5−m). Two seeds meet in the first round in which their numbers become equal.
Seed 43: 22 → 11 → 6 → QF 3 → SF 2 (lower half).
Seed 46: 19 → 14 → 3 → QF 3 → SF 2. Same Round-3 match as 43, so 46 would meet 43 in Round 3 and can never reach the final against him.
Seed 44: 21 → 12 → 5 → QF 4 → SF 1 (upper half) – can be runner-up. Seed 45: 20 → 13 → 4 → QF 4 → SF 1 – can. Seed 40: 25 → 8 → 8 → QF 1 → SF 1 – can.
So seed 46 cannot be the runner-up.

Why the other options are wrong:
B (44), D (45), E (40): all are in the other half (SF 1), so each can meet 43 in the final.
A (42): 23 → 10 → 7 → QF 2 → SF 2 — it is also in 43's half (they would meet in the semi-final), so strictly 42 also cannot be runner-up; the key takes 46, which sits in 43's own quarter.

Cetking shortcut: Fold the seed number: s → 65−s if s>32, then 33−, 17−, 9−, 5−. Same final number = same half.

Answer: C

Approach: Knockout bracket: the final opponent must come from the half opposite to seed 12.

Solution:
Bracket rule (from the set): Round-1 match of seed s = min(s, 65−s); Round-2 match = min(m, 33−m); Round-3 match = min(m, 17−m); quarter-final = min(m, 9−m); semi-final = min(m, 5−m). Two seeds meet in the first round in which their numbers become equal.
Seed 12: 12 → 12 → 5 → QF 4 → SF 1 (upper half). Opponent must be in SF 2.
Try the lowest seeds first: seed 64: 1 → 1 → 1 → 1 → SF 1 (same half as 12) – not possible.
Seed 63: 2 → 2 → 2 → QF 2 → SF 2 (opposite half) – possible.
So the lowest-seeded player who can face seed 12 in the final is seed 63.

Why the other options are wrong:
A (57), C (62), D (59): all are higher seeds than 63, so not the lowest possible.
E (65): no such seed in a 64-player draw.

Cetking shortcut: Start from 64 and move upward; the first seed that lands in the other half is the answer.

Answer: B

Approach: Knockout bracket: if 23 met 46, find the round of that meeting and who was eliminated on the way.

Solution:
Bracket rule (from the set): Round-1 match of seed s = min(s, 65−s); Round-2 match = min(m, 33−m); Round-3 match = min(m, 17−m); quarter-final = min(m, 9−m); semi-final = min(m, 5−m). Two seeds meet in the first round in which their numbers become equal.
Seed 23: 23 → 10 → 7 → QF 2 → SF 2. Seed 46: 19 → 14 → 3 → QF 3 → SF 2.
Different quarters, same half ⇒ 23 vs 46 is the semi-final. So every other player of quarters 2 and 3 was knocked out inside his own quarter and never played anyone outside it.
9 and 13: 9 → 9 → 8 → QF 1; 13 → 13 → 4 → QF 4 — both in the other half; QF 1 vs QF 4 is the other semi-final, so 9 vs 13 is possible.
6 (QF 3) & 18 (QF 2): could only meet in the semi-final, which is 23 vs 46 – impossible.
5 & 51, 20 & 35, 17 & 15: each pair could meet only in the final, but 51, 35 (QF 3) and 15 (QF 2) were already eliminated – impossible.

Cetking shortcut: Players from the quarters of 23 and 46 are "frozen" inside their own quarters; reject any pair that needs one of them to go beyond it.

Answer: A

Approach: Turn the clues into equations and list all cases (scores are integers 0–4).

Solution:
Let both score a in test 1; Puneet scores p (≠ 0) in each of tests 2–8; Urvisha scores U2, U3, U4 and then p in tests 5–8.
Puneet's first three = Urvisha's first two: a + 2p = a + U2 → U2 = 2p ≤ 4 → p = 1 or 2.
GP: a, a + 2p, total (a + 6p + U3 + U4) → (a + 2p)² = a(a + 6p + U3 + U4) → a(2p + U3 + U4) = 4p².
Valid cases:

Cetking shortcut: U2 = 2p ≤ 4 immediately caps all of Puneet's tests 2–8 at 2 – that kills A, B, D, E.

Answer: C

Approach: Use the case table from the set with a = 4.

Solution:
Let both score a in test 1; Puneet scores p (≠ 0) in each of tests 2–8; Urvisha scores U2, U3, U4 and then p in tests 5–8.
Puneet's first three = Urvisha's first two: a + 2p = a + U2 → U2 = 2p ≤ 4 → p = 1 or 2.
GP: a, a + 2p, total (a + 6p + U3 + U4) → (a + 2p)² = a(a + 6p + U3 + U4) → a(2p + U3 + U4) = 4p².
Puneet's 1st test = a = 4 → a(2p + U3 + U4) = 4p² → 2p + U3 + U4 = p².
p = 1: 2 + U3 + U4 = 1 ✗; p = 2: 4 + U3 + U4 = 4 → U3 = U4 = 0.
Urvisha's 3rd test = 0.

Answer: A

Approach: Slot counting with minimums (7 experts, one slot each; W at lunch).

Solution:
Before ≥ 2, After ≥ 3, W is at lunch → Lunch ≤ 7 − 2 − 3 = 2.
So After (≥ 3) can never equal Lunch (≤ 2) → statement D is impossible.
Check the rest: A — Before 2, After 4, Lunch 1 (W) is possible.
B — J is only after school, so J is never at lunch (true).
C — Before 2, Lunch 2, After 3 is possible (e.g. N, G before; W, M lunch; J, K, L after).
E — Before 3, After 3, Lunch 1 is possible (N, G, M before; J, K, L after).

Cetking shortcut: Lunch has at most 2 people and After has at least 3 — any option equating them is out.

Answer: D

Approach: Option testing against two conditions.

Solution:
Condition 1: C cannot come immediately after B.
Condition 2: once A is inspected, E cannot come later.
A ABCDE — B then C ✗.
B DCABE — A before E ✗.
C DCBEA — B is followed by E (not C) ✓; A is last, so nothing comes after it ✓.
D DCBAE — A before E ✗.
So DCBEA is a valid order.

Note: CDBEA (option E) also satisfies both conditions as printed; the key marks DCBEA.

Answer: C