Sign reasoning around zero: Given S > C ≥ 0 and P < C, determine which conclusions definitely follow. Conclusions: I) 0 < P; II) S > P.

Introduction / Context:
This item tests careful handling of comparisons with zero. Knowing C is non-negative and S is above C, while P is below C, tells us something definitive about S vs P, but not necessarily about P vs 0.

Given Data / Assumptions:

• S > C
• C ≥ 0
• P < C
• Conclusions: I) 0 < P; II) S > P.

Concept / Approach:
If P < C and C ≥ 0, P could still be negative (e.g., C=0, P=-1) or positive (e.g., C=5, P=4). Thus I is not forced. However, S > C and P < C imply S > P strictly by “squeezing” around C.

Step-by-Step Solution:
I) Not forced: take C=0; then P can be -1 (P < 0), so 0 < P fails.II) Since S > C and P < C, we have S > C > P or at least S > P in all compliant assignments. Hence II is guaranteed.

Verification / Alternative check:
Example values: C=3, S=5, P=1 ⇒ S > P. Or C=0, S=2, P=-3 ⇒ S > P. In every case consistent with the premises, S > P holds.

Why Other Options Are Wrong:
a claims I is true—counterexample shown; c claims “either,” which understates certainty; d contradicts the valid deduction in II.

Common Pitfalls:
Assuming “P < C and C ≥ 0” forces P ≥ 0—false when C=0 or small positive numbers.

Final Answer:
If only Conclusion II is true