Three-part chain; determine which conclusion(s) must follow: Statements: N < O ≥ R > T R < A B ≤ T Conclusions: I. N < A II. B < A

Introduction / Context: This inference task spans multiple chains that meet at R and T. We must decide which conclusions hold for every assignment consistent with all statements.

Given Data / Assumptions:

• N < O and O ≥ R and R > T
• R < A
• B ≤ T
• Conclusions: (I) N < A, (II) B < A

Concept / Approach: First pin A relative to R and T, then check B and N against A. Where there is no direct chain, attempt to build counterexamples.

Step-by-Step Solution: From R < A and R > T, we get T < R < A. Given B ≤ T and T < A, we have B ≤ T < A ⇒ B < A; hence (II) is guaranteed. For (I): N < O and O ≥ R, but O could be far greater than R. It does not force N below A. Counterexample: let R = 5, T = 1, A = 6, O = 100, N = 99, B = 1. All statements hold (N < O, O ≥ R, R > T, R < A, B ≤ T), yet N < A is false (99 ≮ 6). Therefore (I) does not necessarily follow.

Verification / Alternative check: Try edge case O = R: then N < R and R < A ⇒ N < A (true in that case), but since other valid cases make it false, it is not a must.

Why Other Options Are Wrong: “Only I”, “either”, and “neither” conflict with the proven necessity of (II).

Common Pitfalls: Assuming N < O and O ≥ R forces N < R — not necessarily; N can be just below a very large O, still exceeding A if A is only slightly above R.

Final Answer: Only conclusion II follows.