Difficulty: Easy
Correct Answer: Only Conclusion II follows
Explanation:
Introduction / Context:
This is a straightforward application of subset and disjointness: all mangoes fall inside the set of golden things, and golden things are excluded from “cheap.”
Given Data / Assumptions:
Concept / Approach:
If all Mangoes are Golden, and Golden things are not Cheap, then Mangoes cannot be Cheap. A universal negative about Mangoes and Cheap follows; the offered conclusion II is a natural phrasing of that fact. Conclusion I contradicts the premises.
Step-by-Step Solution:
1) From Mango ⊆ Golden and Golden ∩ Cheap = ∅, deduce Mango ∩ Cheap = ∅.2) Therefore, “Golden-coloured mangoes are not cheap” is necessarily true (indeed, all mangoes are not cheap).3) “All mangoes are cheap” directly contradicts step 1 and is false.
Verification / Alternative check:
Diagram: Mangoes inside Golden; Cheap as a disjoint region from Golden. There is no overlap between Mangoes and Cheap.
Why Other Options Are Wrong:
Any option that includes I is inconsistent with the premises; “both” or “either” cannot hold.
Common Pitfalls:
Overlooking that a property of a superset (Golden not being Cheap) transfers to all of its subsets (Mangoes).
Final Answer:
Only Conclusion II follows.
Discussion & Comments