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Ordering — Who reached the station first among L, M, J, T, and R? (No ties) Statements: I. M reached only after J and T. II. L reached before R.

Difficulty: Medium

Correct Answer: Both Statements I and II together are not sufficient.

Explanation:


Introduction / Context:
The task is to infer the earliest (first) finisher from partial order constraints. DS problems require checking if the earliest can be uniquely identified.


Given Data / Assumptions:

  • I: M is after both J and T (so M cannot be first).
  • II: L is before R.
  • Participants: L, M, J, T, R; no ties.


Concept / Approach:
Construct compatible orders to test uniqueness of the first position.


Step-by-Step Solution:

From I alone: J and T are before M; L and R are unconstrained, so either J or T or even L could be first. Not sufficient.From II alone: L before R tells nothing about J, T, M; the first could be any of J/T/L (and even R if L is after someone else yet still before R). Not sufficient.Combining I & II: We only know (J,T) before M and L before R. Possible order 1: J < T < L < R < M → J is first. Possible order 2: T < J < L < R < M → T is first. Possible order 3: L < J < T < R < M → L is first. Multiple candidates remain.


Verification / Alternative check:
Try exhaustive reasoning for the first place; at least three valid winners exist under the constraints.


Why Other Options Are Wrong:

  • I alone: eliminates M but not others.
  • II alone: a single pairwise relation is too weak.
  • Either alone sufficient / Both together sufficient: contradicted by counterexamples.


Common Pitfalls:
Assuming unstated relations (e.g., transitivity to others) or inventing tie-breaking rules.


Final Answer:
Both Statements I and II together are not sufficient.

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