Difficulty: Easy
Correct Answer: If both conclusions I and II follow
Explanation:
Introduction / Context:
This question mixes strict and non-strict relations and asks you to compare B against E. Because E equals D and C is no greater than D, the strict part of the chain from B up to C ensures a strict comparison to E as well.
Given Data / Assumptions:
Concept / Approach:
From B < C and C ≤ E (since D = E), we have B < E directly. Once B < E is proved, the weaker statement B ≤ E follows automatically.
Step-by-Step Solution:
Since C ≤ D and D = E, we have C ≤ E.Given B < C and C ≤ E, transitivity yields B < E (II is true).Any strict inequality implies the corresponding non-strict one: from B < E we get B ≤ E (I true).
Verification / Alternative check:
Example values: let A=1, B=2, C=3, D=4, E=4. Then B < E (2 < 4) and B ≤ E (2 ≤ 4) both hold.
Why Other Options Are Wrong:
Options a/b assert only one of the two truths; d claims “either,” which undersells certainty; e contradicts the chain.
Common Pitfalls:
Mixing up ≤ and = when D = E; or thinking C ≤ E prevents a strict B vs E. The strict link is between B and C, which is sufficient to make B < E.
Final Answer:
If both conclusions I and II follow
Discussion & Comments