Difficulty: Easy
Correct Answer: Only II, III and IV follow
Explanation:
Introduction / Context:This is a classic chain of subset relations. The key skill is keeping track of inclusions and identifying which existential statements are warranted by the universals given the usual non emptiness convention in such exam problems.
Given Data / Assumptions:
Concept / Approach:Transitivity of subset works for universal affirmatives: if F ⊆ Bk and Bk ⊆ B, then F ⊆ B. From universal plus non emptiness, we may infer some statements like some buses are books (pick any book, it is a bus). However, without an explicit link between fruits and jungles, we cannot infer their overlap.
Step-by-Step Solution:
Conclusion I (Some fruits are jungles): There is no premise connecting fruits to jungles. They could be disjoint while all premises remain true. So I does not follow.Conclusion II (Some buses are books): Since all books are buses and books are assumed to exist (because fruits are books), at least one bus is a book. II follows.Conclusion III (Some buses are jungles): Since all jungles are buses and jungles are assumed to exist, at least one bus is a jungle. III follows.Conclusion IV (All fruits are buses): From F ⊆ Bk and Bk ⊆ B, transitivity gives F ⊆ B. IV follows.Verification / Alternative check:Set diagram: place F inside Bk inside B. Place J also inside B but not intersecting F to see that I need not hold, while II, III, IV remain valid.
Why Other Options Are Wrong:
Common Pitfalls:Assuming overlap between unrelated subclasses (fruits and jungles); forgetting to apply subset transitivity; overlooking that universal statements with existing classes permit some conclusions.
Final Answer:Only II, III and IV follow
Discussion & Comments