Question 1

Implication and contrapositive reasoning: “If Rajesh studies, he will pass his exam.” Rajesh fails the exam. Based only on this information, determine the logically valid conclusion about whether Rajesh studied or not.

Accepted Answer

Introduction / Context: Many verbal reasoning questions test the truth-preserving rules of implication. The statement “If Rajesh studies, he will pass” is a conditional of the form P -> Q. We are also told Rajesh failed (not Q). The task is to decide what follows about whether he studied (P). Given Data / Assumptions: Rule: If study then pass (study -> pass). Observed outcome: Rajesh fails (not pass). No extra information beyond these two statements. Concept / Approach: Use modus tollens, the contrapositive rule: from (P -> Q) and (not Q) we infer (not P). The contrapositive of “study -> pass” is “not pass -> not study.” This is logically equivalent to the original conditional and always valid. Step-by-Step Solution: 1) Formalize: P = “Rajesh studies”; Q = “Rajesh passes”.2) Given: P -> Q and not Q.3) Apply contrapositive: not Q -> not P.4) Since not Q is true (he failed), conclude not P (he did not study). Verification / Alternative check: Try truth-table intuition: The only way a true conditional (P -> Q) can coexist with an observed failure of Q is when P is false. If P were true while Q is false, the conditional would be violated. Hence P must be false. Why Other Options Are Wrong: “Rajesh studied…” would make P true while Q is false, contradicting the conditional. “May or may not have studied” ignores the contrapositive; it is too weak. “None of these” is incorrect because a definite conclusion exists (not P). Common Pitfalls: A frequent error is treating conditionals as biconditionals (“pass only if and if and only if study”). We are not told that passing implies studying; we are told that studying implies passing. However, failing does imply not studying via the contrapositive. Final Answer: Rajesh did not study for the exam.

Question 2

Understanding what follows from “Whenever I go in the sun, I get a headache.” Decide which statement is logically entailed by this conditional, without assuming that I actually went in the sun today.

Accepted Answer

Introduction / Context: We analyze a everyday-language conditional: “If I go in the sun, then I get a headache.” Symbolically, Sun -> Headache. The question asks what we can conclude from this rule alone, without extra facts about today. Given Data / Assumptions: Rule: Sun -> Headache (whenever antecedent holds, consequent follows). No assertion that I actually went in the sun. No assertion that I currently have or do not have a headache. Concept / Approach: From a conditional P -> Q, we may infer Q if P happens (modus ponens). But unless we are told P is true, we cannot assert Q. Nor can we infer P from Q (that would be affirming the consequent). We also cannot infer not Q from not P (denying the antecedent). Step-by-Step Solution: 1) Translate: P = “went in sun”; Q = “have headache”.2) The statement gives only P -> Q.3) To conclude Q now, we would need evidence that P is true now; none is given.4) To conclude not Q from not P is invalid.5) To conclude P from Q is also invalid. Verification / Alternative check: Construct scenarios: If I did not go in the sun but have a headache (for other reasons), the rule still holds. If I went in the sun and got a headache, it holds. The rule does not fix what happens when I do not go in the sun. Why Other Options Are Wrong: “I went in sun so I have headache now.” Not supported; we are not told I went out. “I didn’t go in sun so I don’t have headache now.” Denying the antecedent—invalid. “I’m feeling headache so I went in sun.” Affirming the consequent—invalid. Common Pitfalls: Confusing sufficient with necessary conditions. “Sun” is sufficient for “headache”, not necessary. Final Answer: None of these.

Question 3

Exclusive choice reasoning: “Ram will either buy a car or a flat.” Identify the conclusion that must follow under an exclusive-or interpretation (exactly one of the two is purchased).

Accepted Answer

Introduction / Context: In many test items, “either … or …” is taken as an exclusive-or (XOR): exactly one of the two alternatives holds, not both. We are told: “Ram will either buy a car or a flat.” We must find which conclusion necessarily follows. Given Data / Assumptions: Interpret “either … or …” as exclusive: exactly one of {Car, Flat} is true. No other purchases beyond this pair are relevant. Concept / Approach: Under XOR, the two statements are mutually exclusive and collectively exhaustive. Therefore, if we learn one is false, the other must be true. Conversely, if we learn one is true, we may conclude the other is false. The only option that gives a must-follow conclusion independent of which item he actually chose is the conditional negation version. Step-by-Step Solution: 1) Encode: C = “bought car”; F = “bought flat”.2) Given: exactly one of C, F is true.3) From not C, we must infer F (because one must hold).4) Therefore, “Ram did not buy car so he must have bought flat” is necessarily true. Verification / Alternative check: Truth cases: (C, not F) or (not C, F). In the second case, not C implies F immediately; in the first, the statement’s antecedent is false, so no contradiction arises. Why Other Options Are Wrong: “Ram bought a flat so he will not buy a car.” This can be concluded if we already know he bought a flat, but it is not a universal conclusion derived solely from the original statement without that extra fact. “Ram bought a car so he will not buy a flat.” Same issue as above—requires extra fact. “None of these.” Incorrect because one option (the “not car → flat” conditional) is a necessary consequence of XOR. Common Pitfalls: Confusing inclusive-or with exclusive-or. Standard test convention favors exclusive unless context suggests otherwise. Final Answer: Ram did not bought Car so he must have bought Flat.

Question 4

Only-if reasoning with necessary condition: “The teacher gives a break only if students are exhausted.” From this, identify which pair of statements can both be true without violating the rule.

Accepted Answer

Introduction / Context: “Only if” signals a necessary condition. “The teacher gives a break, only if students are exhausted” means: If Break, then Exhausted. Symbolically, B -> E. We must choose the pair that is logically compatible with this rule. Given Data / Assumptions: Rule: B -> E (break implies exhausted). Statements: (i) The teacher gives break (B). (ii) The teacher did not give a break (not B). (iii) Students are not exhausted (not E). (iv) Students are exhausted (E). Concept / Approach: With B -> E: If B is true, E must be true. If E is true, nothing is guaranteed about B (E may occur without B). If E is false, B must be false (contrapositive). We check option pairs against these constraints. Step-by-Step Solution: Check (i)&(iii): B and not E contradict B -> E. Invalid.Check (i)&(iv): B and E are fully consistent with B -> E. Valid.Check (ii)&(iv): not B with E is also consistent (exhaustion without break is allowed). However, typical single-answer tests select the pair that most directly reflects the rule’s necessity linkage (B forces E). Among the printed options, (i)&(iv) cleanly exhibits the implication. Verification / Alternative check: Contrapositive: not E -> not B. This invalidates any pair containing B with not E (that is (i)&(iii)). Why Other Options Are Wrong: (i) and (iii) violates the conditional. (ii) and (iv) is logically possible but does not demonstrate the implication; many exams intend the pair that must hold together when B occurs, i.e., (i) with (iv). With single-correct-answer format, (i)&(iv) is the accepted key. “None of these” is false because at least one pair is undeniably consistent. Common Pitfalls: Misreading “only if” as “if and only if.” We are not told E -> B; we are told B -> E. Final Answer: (i) and (iv)

Question 5

Disjunction analysis: “Either CAT is tough or IIT JEE is easy.” Identify the pair of statements that can simultaneously be true without contradicting the given disjunction.

Accepted Answer

Introduction / Context: We assess truth conditions of a disjunction: “Either CAT is tough or IIT JEE is easy.” In standard logic, “either … or …” is often read exclusively in puzzles, but the safe consistency check works for inclusive-or as well: the disjunction is true if at least one disjunct is true. Given Data / Assumptions: Statement S: (CAT tough) OR (IIT JEE easy). Options reference: (ii) CAT is easy; (iii) IIT JEE is easy; (iv) IIT JEE is tough. Concept / Approach: We must find a pair that does not contradict S. If we pick CAT easy (ii) and IIT JEE easy (iii), S remains satisfied because “IIT JEE easy” makes the disjunction true. Pairs that force both “CAT not tough” and “IIT JEE not easy” would contradict S. Step-by-Step Solution: 1) Let C_t = “CAT tough”; J_e = “IIT JEE easy”.2) S asserts C_t OR J_e is true.3) Pair (ii)&(iii): “CAT easy” (which says nothing about toughness) and “IIT JEE easy” (J_e true). S is satisfied.4) Pair (ii)&(iv): “CAT easy” with “IIT tough” makes J_e false and does not guarantee C_t true; this can contradict S.5) Pair (iii)&(iv): asserts J_e true and J_e false simultaneously, a contradiction. Verification / Alternative check: If we treat “either … or …” as exclusive, (ii)&(iii) still keeps S true (since the clause only requires at least one to make the original sentence consistent; exclusivity is not enforced by the option pair itself). Why Other Options Are Wrong: (ii) and (iv) can force both disjuncts false relative to S (no guarantee CAT is tough). (iii) and (iv) is internally inconsistent. “None of these” is wrong because (ii)&(iii) is consistent with S. Common Pitfalls: Assuming “CAT easy” implies “CAT not tough” with certainty. The prompt does not define “tough vs easy” as strict negations; even if treated as opposites, (iii) alone already satisfies S. Final Answer: (ii) and (iii)

Question 6

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.

Accepted Answer

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: Check (i)&(iv): C and not S contradict contrapositive (C -> S). Invalid.Check (ii)&(iii): not C and S are compatible. Studying is not sufficient for success; failing despite studying does not violate the rule.Check (iv)&(i): same contradiction as first pair.Hence (ii)&(iii) is the consistent pair. 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)

Logical Connectives Questions