Difficulty: Easy
Correct Answer: Both Conclusions I and II follow.
Explanation:
Introduction / Context:We combine a universal inclusion with a universal exclusion and test both a universal negative and an existential conclusion.
Given Data / Assumptions:
Concept / Approach:If Erasers are disjoint from Papers and Pens are within Papers, then Erasers and Pens are disjoint (I). Also, if any pen exists, it is a paper, so at least some papers are pens (II).
Step-by-Step Solution:
I) Since Pens ⊆ Papers and Erasers ∩ Papers = ∅, it follows Erasers ∩ Pens = ∅ — no eraser is a pen.II) Pick any pen; it belongs to Papers, so 'Some papers are pens' holds.Verification / Alternative check:Venn diagrams depict Pens inside Papers and Erasers disjoint from Papers; both conclusions are immediate.
Why Other Options Are Wrong:Options denying either I or II contradict the straightforward set relations.
Common Pitfalls:Overlooking that an existential like 'some papers are pens' is justified by non-emptiness of Pens in typical exam logic.
Final Answer:Both Conclusions I and II follow.
Discussion & Comments