Difficulty: Easy
Correct Answer: None of these
Explanation:
Introduction / Context:
This is a straightforward chain of universal inclusions. Be careful not to reverse subset directions. From A ⊆ B ⊆ C ⊆ D, you can derive A ⊆ D and B ⊆ D, but not D ⊆ B or C ⊆ A.
Given Data / Assumptions:
Concept / Approach:
Transitivity of “All”: If X ⊆ Y and Y ⊆ Z, then X ⊆ Z. Do not invert inclusions unless explicitly given.
Step-by-Step Solution:
Verification / Alternative check:
Draw nested sets: P ⊆ B ⊆ S ⊆ H. It is visually clear that I and IV are true while II and III are false.
Why Other Options Are Wrong:
Common Pitfalls:
Accidentally reversing subset relations (assuming hills are birds, or skies are pencils).
Final Answer:
None of these (because the correct set is I and IV, a combination not listed).
Discussion & Comments