Statements:\nI. There are monks among those who are felicitated for remarkable social service.\nII. Jitananda and Vidyananda are among those felicitated.\n\nConclusions:\nI. Jitananda and Vidyananda did remarkable social service.\nII. All monks do social service.\nIII. Jitananda and Vidyananda are not monks.\nIV. All monks are felicitated.

Introduction / Context:
We are given two set-membership statements: (1) the felicitation set consists of people recognized for remarkable social service and includes some monks; (2) two named individuals, Jitananda and Vidyananda, belong to the felicitation set. We must test which conclusions necessarily follow.

Given Data / Assumptions:

• Felicitation Set F = persons felicitated for remarkable social service.
• Some members of F are monks.
• Jitananda and Vidyananda ∈ F.

Concept / Approach:
From the definition of F, membership in F implies the property “did remarkable social service.” Therefore any named member of F inherits that property. However, the statements do not quantify monks beyond saying “there are monks among those felicitated.” That does not mean all monks do social service or that all monks are felicitated; nor does it tell us whether the two named individuals are monks or not.

Step-by-Step Solution:
1) Since J and V are in F, and all in F are felicitated for social service, J and V did remarkable social service → Conclusion I follows.2) Statement I does not state “all monks,” only that “there are monks” in F → Conclusion II does not follow.3) Nothing indicates J and V are not monks (or are monks) → Conclusion III does not follow.4) “There are monks in F” does not imply “all monks are in F” → Conclusion IV does not follow.

Verification / Alternative check:
Create a simple Venn diagram: F overlaps with the set of monks but is not identical to it. The named individuals lie inside F; their monk status is unspecified.

Why Other Options Are Wrong:
Options selecting II, III, or IV assume universality or negation not warranted by the premises.

Common Pitfalls:
Confusing “some” with “all” and assuming properties not specified for named members.

Final Answer:
Conclusion I.