There are five people P1, P2, P3, P4 and P5. Each has a different profession…
2021
There are five people P1, P2, P3, P4 and P5. Each has a different profession among D1, D2, D3, D4 and D5 (not necessarily in the same order). Each uses a different transport to go to the office among T1, T2, T3, T4 and T5 (not necessarily in the same order).
The person having profession D3 does not use T1 or T4.
P2’s profession is D1.
P4 uses T2.
P3’s profession is not D4 or D5.
P5 does not use T3 or T5.
The persons having professions D4 and D2 use transports T5 and T3 respectively.
Which of the following combination of person – profession – transport is definitely correct (true in every valid arrangement)?
- A.
P5 – D3 – T4
- B.
P2 – D1 – T4
- C.
P4 – D3 – T2
- D.
P5 – D5 – T1
Attempted by 4 students.
Show answer & explanation
Correct answer: C
Concept
An attribute-matching puzzle is solved by chaining the fixed links first, then forcing the open slots by elimination. A combination is the answer only when every other assignment for that person is ruled out by a constraint — never by a guess. Two fixed links here are: profession D4 ↔ transport T5, and profession D2 ↔ transport T3. Two negative limits decide the rest: D3 may not take T1 or T4, and P5 may not take T3 or T5.
Applying it step by step
P4 uses T2 and P5 avoids T3 and T5, so neither P4 nor P5 can hold D4 (needs T5) or D2 (needs T3). P2 is fixed as D1. So D4 and D2 must sit with P1 and P3 only.
P3 cannot be D4 (and not D5), so P3 takes D2 and therefore rides T3; that forces P1 to be D4 and ride T5.
Only D3 and D5 are left, for P4 and P5. The free transports left for P2 and P5 are T1 and T4 (since P4 already holds T2).
P5 is limited to T1 or T4, but D3 is barred from both T1 and T4 — so P5 cannot be D3. Hence P5 = D5 and P4 = D3.
P4 holds D3 on transport T2, and T2 is neither T1 nor T4, so the D3 limit is respected. This gives the forced triple: person P4, profession D3, transport T2.
Cross-check
Read back every clue against P1–D4–T5, P2–D1, P3–D2–T3, P4–D3–T2, P5–D5: D4 takes T5 and D2 takes T3 (links hold), the D3 holder uses T2 so it avoids T1 and T4, P5 avoids T3 and T5, and P3 is neither D4 nor D5. The only freedom left is that P2 and P5 share the leftover pair T1 and T4 in either order — so any triple that depends on P2’s or P5’s exact transport is only sometimes true. P4 – D3 – T2 is the single combination forced in every valid arrangement, which is why it is the definitely-correct answer.