A five-day audit schedule: solving it, testing redundancy, and testing feasibility
A five-member internal audit team - Arun, Bela, Chetan, Divya and Esha - must each visit a different branch, one branch per working day from Monday to Friday. The branches are coded A, B, C, D and E, and the auditor assigned to a code is the one who visits that branch; the codes are not the auditors' initials and carry no other meaning. The planning note left by the previous head contains only five conditions: (i) the C visit falls on Wednesday; (ii) the auditor visiting A goes on the day immediately after the auditor visiting E; (iii) the B visit cannot be on Monday or on Friday; (iv) the D visit must not fall on the day immediately before or immediately after the C visit; (v) the D visit is not on Monday. The new head needs a defensible schedule, but she also wants to know which conditions are doing real work - she suspects one condition is redundant and that certain pairs of conditions cannot pin the schedule down on their own.
1. Build the schedule and show the deduction step by step.
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Fix positions 1–5 as Monday–Friday.
- (i) C = 3 (Wednesday).
- (iv) rules out D = 2 and D = 4; (v) rules out D = 1. Since position 3 is taken by C, D = 5 (Friday).
- Remaining positions for E, A, B are 1, 2, 4. Condition (ii) needs E and A on consecutive days; within {1, 2, 4} the only consecutive pair is 1–2, with E immediately before A. So E = 1 (Monday), A = 2 (Tuesday).
- B = 4 (Thursday), which satisfies (iii).
Schedule: Monday E, Tuesday A, Wednesday C, Thursday B, Friday D.
Verification of every clue: C on Wednesday ✓; A immediately after E ✓; B not Monday or Friday ✓; D (Friday) is not adjacent to C (Wednesday) ✓; D not on Monday ✓. The solution is unique - no other arrangement survives all five conditions simultaneously.
2. The new head claims one of the five conditions is redundant. Identify it and prove the point, or show that no condition is redundant.
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Condition (iii) - 'B is not on Monday or Friday' - is redundant. It is never needed to reach the unique schedule.
Proof by removing it and re-deriving: from (i), C = 3. From (iv), D ∉ {2, 4}; from (v), D ∉ {1}; position 3 is occupied by C, so D must be 5.
The free positions are then 1, 2, 4 for E, A, B. Condition (ii) requires E and A to occupy consecutive days; the only consecutive pair available is 1–2 (position 4 has no available neighbour, since 3 and 5 are taken). Hence E = 1, A = 2, and B = 4 - which happens to satisfy (iii) anyway.
So (iii) is implied by the other four conditions; it is a genuine clue but carries no extra information. By contrast, dropping (v) does break uniqueness: E1, A2, C3, B4, D5 and D1, B2, C3, E4, A5 both satisfy (i)–(iv), so (v) is doing real work. A systematic case check of this kind, rather than intuition, is what the question is testing.
3. The C branch manager is away on Wednesday, so the C visit must move to Tuesday; conditions (ii)–(v) are unchanged. Is a schedule still possible? Justify your answer.
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Now C = 2. Free positions are 1, 3, 4, 5 for A, B, D, E.
From (iv), D cannot be 1 or 3 (the neighbours of 2); combined with (v), D cannot be 1 either, so D ∈ {4, 5}.
Case D = 4: remaining positions 1, 3, 5 for E, A, B. No two of these are consecutive, so (ii) - E and A on consecutive days - cannot be satisfied. Rejected.
Case D = 5: remaining positions 1, 3, 4 for E, A, B. The only consecutive pair is 3–4, forcing E = 3 and A = 4, leaving B = 1 (Monday). That violates (iii). Rejected.
Both cases fail, so no schedule satisfies all the conditions. The correct professional answer is not to force a schedule but to report an inconsistency and name the cheapest relaxation.
Notice the link to the previous task: dropping (iii) makes C = 2 workable, giving E3, A4, B1, D5, C2 (check: D at 5 is not adjacent to C at 2; A immediately follows E). A clue that was redundant under the old date becomes essential under the new one - redundancy is relative to the whole set of conditions, not an absolute property.