Difficulty: Medium
Correct Answer: If only Conclusions I is true
Explanation:
Introduction / Context:
This problem tests deterministic inference from a single ordered chain that mixes strict (>, <) and non-strict (≤, =) relations. We must check which conclusions hold for all values that satisfy the chain simultaneously, not just for an illustrative example.
Given Data / Assumptions:
Concept / Approach:
Translate the chain into segment-by-segment facts and propagate them to the queried pairs. A single chain allows safe transitivity when directions align (e.g., if X ≤ Y and Y < Z, then X < Z). When checking a conclusion, attempt both “must be true” and “can fail” constructions.
Step-by-Step Solution:
From A = I and I ≤ 0, we get A ≤ 0 as well.
From 0 < E, the ordering around 0 is fixed: everything strictly greater than 0 is to the right of 0.
For Conclusion I (I ≤ E): we have I ≤ 0 and 0 < E. By transitivity, I ≤ 0 < E ⇒ I < E, hence certainly I ≤ E. So I is always true.
For Conclusion II (0 > U ≡ U < 0): we only know U > A and A = I ≤ 0. If I = 0 (hence A = 0), then U > 0 ⇒ 0 > U is false. If I = −10 (A = −10), U could be −9 (still > A), making U < 0 and rendering 0 > U true. Since both true and false realizations exist, II does not definitely follow.
Verification / Alternative check:
Construct two concrete assignments satisfying the chain: (i) I = 0, A = 0, U = 1, E = 2 → I ≤ E true, 0 > U false; (ii) I = −2, A = −2, U = −1, E = 5 → I ≤ E true, 0 > U true. Variability in (II) proves it doesn’t follow necessarily.
Why Other Options Are Wrong:
“Only II” is impossible because II can fail. “Either I or II” demands exactly one must hold in all cases, which is not guaranteed. “Neither” is false because I always holds.
Common Pitfalls:
Treating “U > (something ≤ 0)” as implying “U > 0”. That is invalid: U can be negative yet still greater than a more negative number.
Final Answer:
Only Conclusion I is true.
Discussion & Comments