Necessary-condition reasoning using “unless”: “Unless you study, you cannot crack CAT.” From the options below, identify the pair that can be true without violating this rule.

Introduction / Context:Statements with “unless” are common in logical connectives. “Unless you study, you cannot crack CAT” can be read as: if you do not study, then you will not crack CAT. Symbolically, not S -> not C. Equivalently (by contrapositive), C -> S. We must select the option pair that can both be true while respecting this rule.

Given Data / Assumptions:

• Rule: not study -> not crack; equivalently, crack -> study.
• (i) you cracked CAT (C)
• (ii) you could not crack CAT (not C)
• (iii) you did study (S)
• (iv) you didn't study (not S)

Concept / Approach:From not S -> not C, any case with not S and C together is impossible. From C -> S, if one cracked, studying must also be true; but studying does not guarantee cracking. So pairs must be checked for consistency with both implications.

Step-by-Step Solution:

Verification / Alternative check:Truth-table intuition: The allowed rows are (S, C), (S, not C), (not S, not C). The disallowed row is (not S, C). Pair (ii)&(iii) matches (S, not C), which is allowed.

Why Other Options Are Wrong:

• (i)&(iv) and (iv)&(i) force the forbidden combination (cracked without studying).
• None of these is false since (ii)&(iii) works.

Common Pitfalls:Reading “unless” as a biconditional. It creates a necessary condition (study is necessary for cracking) but not a sufficient one.

Final Answer:(ii) and (iii)